Introduction
Let K K K G K → G L 2 ( F p ) G_K \to\mathrm{GL}_2(\mathbb{F}_p) G K → GL 2 ( F p ) p p p
Over Q \mathbb{Q} Q [58 Section 3, Problem 1] , [60 ] . The lifting theorems of Wiles, Taylor–Wiles and Skinner–Wiles, together with potential modularity, laid the foundations for its proof [74 , 67 , 62 , 66 ] .
An essential step in this development was the construction of compatible systems containing a given representation. Dieulefait proved such an existence theorem over Q \mathbb{Q} Q [18 Theorem 1.1] . This permits a change of residual characteristic while retaining a single characteristic-zero compatible system. The method was used to establish cases of the Fontaine–Mazur conjecture and small-level cases of Serre’s conjecture [19 , 39 ] . Khare and Wintenberger subsequently proved Serre’s conjecture in full generality [40 , 41 ] , with Kisin’s two-adic modularity lifting theorem providing the final dyadic input [43 ] . Later lifting theorems permit shorter versions of this proof, as in Dieulefait–Pacetti [24 ] . Over totally real fields, the corresponding strategy of connecting residual representations through compatible systems is formulated in the Rayuela conjecture [23 ] .
Over a real quadratic field, discriminant bounds need not exclude all small-characteristic representations of small conductor. For example, the elliptic curve
y 2 + 6 x y + ( 1 + 6 ) y = x 3 + ( 1 − 6 ) x 2 − ( 1 + 3 6 ) x − 1 − 2 6 y^2 + \sqrt{6}xy + (1+\sqrt{6})y = x^3 + (1-\sqrt{6})x^2 - (1+3\sqrt{6})x - 1 - 2\sqrt{6} y 2 + 6 x y + ( 1 + 6 ) y = x 3 + ( 1 − 6 ) x 2 − ( 1 + 3 6 ) x − 1 − 2 6 has everywhere good reduction over Q ( 6 ) \mathbb{Q}(\sqrt{6}) Q ( 6 ) [45 ] . Thus the base cases must establish modularity of surviving representations, rather than only nonexistence. In addition, residual reducibility and ordinarity of a characteristic-zero source are different conditions. A congruence is useful only when modularity can be transferred back to that actual source.
Our starting point is the characteristic-three method of Dieulefait–Pacetti over Q ( 13 ) \mathbb{Q}(\sqrt{13}) Q ( 13 ) [25 ] . Successive discriminant comparisons restrict the residual images, and a conductor reduction relates arbitrary representations to the remaining base case. We retain this distinction between the residual base and the conductor argument, remove the restriction at two, and supply new source theorems where residual modularity alone is insufficient. Fontaine’s local bounds, refined local ramification calculations, and global discriminant estimates provide the arithmetic input [31 , 46 , 47 ] ; finite-flat groups and integral trace orders control the scalar and dihedral cases.
The results below use the lifting and automorphic inputs specified in Sections 2 and 5. In particular, statements involving the split-dyadic scalar argument are relative to the dyadic structural and propagation results listed there, including the cited preprints. These inputs are distinct from the analytic hypothesis GRH. Write d d d
Theorem 1.1. Let
d ∈ { 2 , 3 , 5 , 6 , 7 , 10 , 11 , 13 , 17 , 21 , 33 , 37 , 41 , 57 , 65 , 73 , 89 } , K = Q ( d ) . d \in\{2,3,5,6,7,10,11,13,17,21,33,37,41,57,65,73,89\}, \qquad K = \mathbb{Q}(\sqrt{d}). d ∈ { 2 , 3 , 5 , 6 , 7 , 10 , 11 , 13 , 17 , 21 , 33 , 37 , 41 , 57 , 65 , 73 , 89 } , K = Q ( d ) . Assume GRH for Dedekind zeta functions. For every odd prime p p p
ρ ‾ : G K ⟶ G L 2 ( F ‾ p ) \overline{\rho}: G_K \longrightarrow\mathrm{GL}_2(\overline{\mathbb{F}}_p) ρ : G K ⟶ GL 2 ( F p ) is modular.
For d = 17 d = 17 d = 17 d = 33 d = 33 d = 33 § 9.10 . Further nonordinary source arguments and individual residual-image exclusions hold without GRH; these refinements are recorded in the corresponding proofs.
Structural theorems
The field list is an application of three mechanisms: conductor reduction to a specified terminal characteristic, scalar modularity from a quadratic Selmer condition, and coefficient-field rigidity from simultaneous integral trace orders. We state these mechanisms separately, since their hypotheses have different roles.
Conductor reduction. A strongly compatible system, and the conventions for weight two, are specified in § 2.1 . The following theorem concerns a system already constructed in characteristic zero; it does not supply an initial lift of an arbitrary residual representation.
Theorem 1.2 (Conductor reduction at a split terminal prime). Let K K K K K K ℓ \ell ℓ K K K
(i) every absolutely irreducible totally odd representation G K → GL 2 ( F ‾ ℓ ) G_K \to\operatorname{GL}_2(\overline{\mathbb{F}}_\ell) G K → GL 2 ( F ℓ ) χ ˉ ℓ \bar{\chi}_\ell χ ˉ ℓ ℓ \ell ℓ
(ii) for every reducible representation with that determinant and ramification support, its constituent quotient is totally odd and extends to G Q G_{\mathbb{Q}} G Q
Then every nondihedral potentially modular weight-two strongly compatible system over K K K
The proof is given in § 2.5 . The quotient in hypothesis (ii) is automatically totally odd, since its two constituents have product χ ˉ ℓ \bar{\chi}_\ell χ ˉ ℓ
A scalar Selmer condition. Suppose two splits in K K K w 0 , w 1 w_0,w_1 w 0 , w 1
V w 0 + ( K ) = { [ a ] ∈ K × / K × 2 : a is totally positive , v q ( a ) ≡ 0 ( m o d 2 ) ( q ∤ 2 ) , a ∈ K w 1 × 2 } . (1) V^+_{w_0}(K) = \left\{[a] \in K^\times/K^{\times2} : \begin{array}{l} a\text{ is totally positive},\\ v_{\mathfrak q}(a) \equiv0 \pmod{2}\quad(\mathfrak q \nmid2),\\ a \in K_{w_1}^{\times2}\end{array} \right\}. \tag*{(1)} V w 0 + ( K ) = ⎩ ⎨ ⎧ [ a ] ∈ K × / K × 2 : a is totally positive , v q ( a ) ≡ 0 ( mod 2 ) ( q ∤ 2 ) , a ∈ K w 1 × 2 ⎭ ⎬ ⎫ . (1) This finite squareclass space classifies totally real quadratic extensions unramified outside w 0 w_0 w 0 w 1 w_1 w 1
V w 0 + ( K ) = 0. V^+_{w_0}(K) = 0. V w 0 + ( K ) = 0. (SV)
It is not an odd-narrow-class-number condition.
Theorem 1.3 (Scalar dyadic modularity). Let K K K § 2.4 , let E / Q 2 E/\mathbb{Q}_2 E / Q 2 r : G K → GL 2 ( E ) r : G_K \to\operatorname{GL}_2(E) r : G K → GL 2 ( E ) det r = ε 2 \det r = \varepsilon_2 det r = ε 2 r s s = 1 ⊕ 1 r^{\mathrm{ss}} = \mathbf{1} \oplus\mathbf{1} r ss = 1 ⊕ 1 r r r r r r
Sections 7–8 construct the auxiliary fields and prove the theorem. Together with a residual base at two and a two-power ordinary class number, it gives the split-dyadic Serre criterion of Theorem 10.1. The seven split-dyadic fields in the main theorem satisfy these arithmetic conditions. The Selmer vanishing is also valid when the narrow class number is even.
Integral trace orders and symmetry. Let R \mathcal{R} R K K K D D D E 0 E_0 E 0 § 3.1 . Its rank is four times the sum of the trace-field degrees. The integral order, rather than a single residual two-dimensional realization, is essential here.
Theorem 1.4 (A trace-order bound and quadratic symmetry). Let K / Q K/\mathbb{Q} K / Q σ \sigma σ R \mathcal{R} R D D D h , η h,\eta h , η G Q G_{\mathbb{Q}} G Q
H ( R ) = R ⊗ h , G ( R ) = R σ ⊗ η . H(\mathcal{R})=\mathcal{R}\otimes h,\qquad G(\mathcal{R})=\mathcal{R}^{\sigma}\otimes\eta. H ( R ) = R ⊗ h , G ( R ) = R σ ⊗ η . Suppose R ≄ R ⊗ h \mathcal{R}\not\simeq\mathcal{R}\otimes h R ≃ R ⊗ h dim F 2 ( E 0 / 2 E 0 ) < 32 \dim_{\mathbb{F}_2}(E_0/2E_0)<32 dim F 2 ( E 0 /2 E 0 ) < 32
R σ ≃ R ⊗ η or R σ ≃ R ⊗ η h , \mathcal{R}^{\sigma}\simeq\mathcal{R}\otimes\eta\quad\text{or}\quad\mathcal{R}^{\sigma}\simeq\mathcal{R}\otimes\eta h, R σ ≃ R ⊗ η or R σ ≃ R ⊗ η h , or [ D : Q ] ≤ 2 [D:\mathbb{Q}]\leq2 [ D : Q ] ≤ 2 D / Q D/\mathbb{Q} D / Q
The arithmetic work in applying Theorem 1.4 lies in bounding the actual order. The finite-flat ramification theorem 11.7 and the regular-order construction in § 3.1 supply different parts of this argument. The bound of twenty-six over Q ( 11 ) \mathbb{Q}(\sqrt{11}) Q ( 11 ) Q ( 3 ) \mathbb{Q}(\sqrt{3}) Q ( 3 )
Two further source theorems have a separate role. Proposition 20.3 constructs the initial lift in the exceptional characteristic-five case over Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 ) Q ( 21 ) \mathbb{Q}(\sqrt{21}) Q ( 21 )
Field applications
The terminal characteristics of the outer conductor reductions are listed below. A source proof may use other coefficient characteristics internally. At every backward step the lifting theorem is applied to the actual old source, with its weight and local ordering retained.
Part I develops the lifting constructions, conductor criteria, scalar theorem and integral-order methods. Part II proves the required source theorems. Part III verifies the remaining residual bases and assembles the field results. The appendices specify the longer finite calculations; their programs and outputs are provided in [49 ] .
Buzzard–Diamond–Jarvis formulated the weight conjecture at unramified residual primes [10 ] ; see also [34 ] . Our modularity results yield the weight part of Serre’s conjecture under the image hypotheses of Gee–Liu–Savitt [36 Theorem 6.2.1] . Namely, suppose the residual representation remains absolutely irreducible after adjoining the roots of unity of order the residual characteristic and, in characteristic five, that the projective image of this restriction is not the alternating group on five letters. Then the occurring Serre weights are exactly those prescribed by the local crystalline-lift recipes, including at ramified residual primes and for all predicted companion weights. These consequences inherit the assumptions of the corresponding modularity theorem; the weight theorem introduces no further analytic hypothesis.
General criteria and local methods
Compatible systems and congruences
We use d d d D D D q q q [49 ] ; the three longer certificate descriptions appear in the appendices.
Throughout, G F = Gal ( F ‾ / F ) G_F = \operatorname{Gal}(\overline{F}/F) G F = Gal ( F / F ) χ p \chi_p χ p − 1 -1 − 1
A system over a totally real field is a pure, regular, totally odd, absolutely irreducible, strongly compatible rank-two system. We require compatibility of the Frobenius-semisimplified Weil–Deligne representations at every finite place, including a place above the coefficient characteristic, and compatibility of labelled weights. Its conductor support consists of the places where the common Weil–Deligne representation is ramified or has nonzero monodromy. Outside this support its p p p p p p { 0 , 1 } \{0,1\} { 0 , 1 }
On the closure of a special local component the Frobenius eigenvalues have ratio N ( v ) N(v) N ( v )
We write R ⇝ R ′ \mathcal{R} \rightsquigarrow\mathcal{R}' R ⇝ R ′ R ′ \mathcal{R}' R ′ R \mathcal{R} R
Residual witnesses and lifting
Lemma 2.1. Let F F F
Proof. Transfer the witness to the quaternion algebra over F F F U = U p U p U = U^pU_p U = U p U p V V V k k k U p U_p U p G G G U p ′ U'_p U p ′ V → Map ( G , V ) V \to\operatorname{Map}(G,V) V → Map ( G , V ) v ↦ ( g ↦ g v ) v \mapsto(g \mapsto gv) v ↦ ( g ↦ g v ) ( h f ) ( g ) = f ( g h ) (hf)(g)=f(gh) ( h f ) ( g ) = f ( g h )
::: Pointwise application and induction in stages give the same away-from-p p p U p U p ′ U^pU'_p U p U p ′ O X \mathcal{O}^X O X X X X [17 Lemme 6.11] lifts the eigencharacter to characteristic zero. A norm character would have reducible residue; inverse Jacquet–Langlands therefore gives a parallel-weight-two cuspidal form.
Retaining both operators T v T_v T v S v S_v S v X 2 − T v X + N ( v ) S v X^2-T_vX+N(v)S_v X 2 − T v X + N ( v ) S v
Totally odd solvable residual representations have solvable Artin witnesses, followed by regular weight raising, in the scopes of automorphic induction and Langlands–Tunnell; see [3 Theorem 1.2(1), Lemma 2.1 and proof of Theorem 2.5] . Lemma 2.1 gives weight two whenever it is needed. We use Dickson’s classification and the rank-two adequacy criterion [6 Proposition A.2.1] throughout.
Lemma 2.2. Let K K K p > 2 p > 2 p > 2 r : G K → GL 2 ( Q ‾ p ) r : G_K \to\operatorname{GL}_2(\overline{\mathbb{Q}}_p) r : G K → GL 2 ( Q p ) G K ( ζ p ) G_{K(\zeta_p)} G K ( ζ p ) r r r
(i) p p p K K K r r r p p p
(ii) r r r { 0 , 1 } \{0,1\} { 0 , 1 } p p p
Proof. Apply respectively [70 Theorem 4.7] and [3 Theorem 2.3] , dualizing to their inverse-cyclotomic normalization when required. In (ii), the parallel-weight-two witness gives the required weight-zero automorphic convention.
Lemma 2.3. Let F F F r : G F → GL 2 ( Q ‾ 2 ) r : G_F \to\operatorname{GL}_2(\overline{\mathbb{Q}}_2) r : G F → GL 2 ( Q 2 ) χ 2 ψ \chi_2\psi χ 2 ψ ψ \psi ψ r r r
(i) r r r
(ii) r r r
Proof. The first case is [2 Theorem 5.2.1] . If the line and quotient have weights b τ > a τ b_\tau> a_\tau b τ > a τ w τ = a τ w_\tau= a_\tau w τ = a τ k τ = b τ − a τ + 1 k_\tau= b_\tau- a_\tau+ 1 k τ = b τ − a τ + 1 k τ + 2 w τ = 2 k_\tau+ 2w_\tau= 2 k τ + 2 w τ = 2
In the second case choose finite Galois extensions of the dyadic completions over which the source is Barsotti–Tate. The solvable local realization theorem [5 Chapter X, Section 2] , in the form [14 Theorem 3.1 and Example 3.3] , realizes them in a solvable totally real Galois extension. No degree is prescribed. The nonsolvable residual projective image is PSL 2 ( F 2 a ) \operatorname{PSL}_2(\mathbb{F}_{2^a}) PSL 2 ( F 2 a ) a ≥ 2 a \ge2 a ≥ 2 [43 Theorem 3.3.5] . Its condition that a dyadic completion equal Q 2 \mathbb{Q}_2 Q 2
For descent take a cyclic series. Strong multiplicity one makes the automorphic representation invariant at each step because the Galois representation extends downstairs. The descended Galois representation differs from the given one by a character of the cyclic quotient. Twisting and iterating proves modularity over F F F [7 Lemma 2.2.2] .
Globalization and compatibility
Lemma 2.4. Let K K K S S S r ˉ \bar{r} r ˉ S S S [28 Definition 4.1.1] , these components admit a global lift unramified outside S S S
(i) odd coefficient characteristic and parallel weight two, with irreducible cyclotomic residual restriction;
(ii) odd coefficient characteristic, adequate cyclotomic residue, and regular potentially diagonalizable crystalline local components;
(iii) coefficient characteristic two, determinant χ 2 \chi_2 χ 2
Proof. Existence is [28 Theorem 4.2.1 and Remark 1.1.2] . Away from the coefficient prime, inertial conditions and conductor bounds specify unions of components. At that prime, take components through the supplied points; potential diagonalizability persists after a fixed sufficiently large local extension.
For weight two first use the avoiding weight-two residual witness of [64 Theorem 5.1.1] , followed by the actual-source theorem [3 Theorem 2.3] . The potential-modularity field is chosen Galois and disjoint from the residual kernel and cyclotomic fields. Thus the actual source becomes modular after this extension. This argument starts with the weight-two point already constructed; it does not assume the existence of an exceptional initial characteristic-five point.
For regular potentially diagonalizable sources with p ≥ 7 p \ge7 p ≥ 7 [7 Theorem C, Theorem 4.5.1 and Corollary 4.5.2] , with avoidance imposed before the CM quadratic base change. For the smaller odd characteristics, Snowden’s avoiding residual theorem gives an ordinary weight-two witness over a totally real field, disjoint from the residual and cyclotomic fields. After a further disjoint CM quadratic extension, its local representations are potentially diagonalizable by [7 Lemma 1.4.3(1)] . Apply the potentially diagonalizable automorphic alternative of [22 Theorem 1] to the actual source, with polarization multiplier its determinant; adequacy and exclusion of S p \mathfrak{S}_p S p
At two, first make the finite local preparation which trivializes the residual dyadic restrictions, inside a totally real extension avoiding the global residual kernel. The ordinary weight-two residual witness of [69 Proposition 6.7] can then be chosen over a further avoiding field, with trivial finite central character. Its residual ordinary ordering agrees with that of the source because the prepared residual characters are trivial. Apply Lemma 2.3 to the actual source. This proves potential modularity before a compatible system is invoked.
The construction of compatible systems by potential modularity and descent originates in Dieulefait’s work over Q \mathbb{Q} Q [18 Theorem 1.1] . For the totally real fields and local conditions here, use the Brauer-induction construction of [64 Proposition 9.2.1 and Remark 9.2.2] , retaining the finite characters in the Brauer identity. Mackey’s formula and Frobenius reciprocity express the self-pairing of each virtual member in terms of pairings of restricted modular representations with finite character twists. Compatibility and Chebotarev make these pairings independent of the coefficient prime. At the original prime the self-pairing is one, so each virtual member is an irreducible or its negative. Its dimension two excludes the negative. Local–global compatibility for the modular systems [61 Theorem 1 and equation (1)] includes the coefficient prime and arbitrary cohomological weights. Restriction–induction identities give the common Weil representations; purity determines monodromy [7 Lemma 1.3.6(2)–(3)] . Labelled weights are compatible by the same restriction argument.
For the quadratic fields here other than Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 ) S 5 S_5 S 5 K ( ζ 5 ) K(\zeta_5) K ( ζ 5 ) S 5 a b = C 2 S_5^{\mathrm{ab}} = C_2 S 5 ab = C 2 Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 )
An auxiliary prime for congruences
In this section K K K S S S
Proposition 3.1. Let R R R χ ψ \chi\psi χ ψ ψ \psi ψ S S S 2 , 3 2, 3 2 , 3 K K K N > max ( S ∪ { 5 } ) N > \max(S \cup\{5\}) N > max ( S ∪ { 5 }) λ \lambda λ K K K l > N l > N l > N R ′ R' R ′ R ⇝ R ′ R \rightsquigarrow R' R ⇝ R ′ N N N χ ψ \chi\psi χ ψ λ \lambda λ λ \lambda λ N N N
While this type is retained and conductor is supported at places over S ∪ { l } S \cup\{l\} S ∪ { l } S S S
Proof. Let K S / K K_S/K K S / K S S S 2 a 2^a 2 a ψ \psi ψ [23 Proposition 4.6] , choose N > max S N > \max S N > max S N > 5 N > 5 N > 5 K K K N ≡ 1 ( m o d 2 max { 2 , a + 1 } ) N \equiv1 \pmod{2^{\max\{2,a+1\}}} N ≡ 1 ( mod 2 m a x { 2 , a + 1 } ) S L 2 ( F N ) \mathrm{SL}_2(\mathbb{F}_N) SL 2 ( F N ) N N N G L 2 ( F N ) \mathrm{GL}_2(\mathbb{F}_N) GL 2 ( F N )
Let R / K R/K R / K K S K_S K S K ( ζ N ) K(\zeta_N) K ( ζ N ) R R R K ( ζ N ) / K K(\zeta_N)/K K ( ζ N ) / K N N N K S / K K_S/K K S / K P G L 2 ( F N ) \mathrm{PGL}_2(\mathbb{F}_N) PGL 2 ( F N ) ψ ‾ \overline{\psi} ψ χ ‾ N ( N − 1 ) / 2 \overline{\chi}_N^{(N-1)/2} χ N ( N − 1 ) /2 K ( ζ N ) K(\zeta_N) K ( ζ N ) P S L 2 ( F N ) \mathrm{PSL}_2(\mathbb{F}_N) PSL 2 ( F N ) − 1 -1 − 1 N N N R ∩ K S K ( ζ N ) R \cap K_S K(\zeta_N) R ∩ K S K ( ζ N )
Chebotarev gives a positive-density set of such prime ideals. Those of residue degree greater than one have zero prime-ideal density. We may consequently choose λ \lambda λ l l l l l l K K K l ≡ − 1 ( m o d N ) l \equiv-1 \pmod N l ≡ − 1 ( mod N ) a , − a a,-a a , − a
Put δ = χ N ψ \delta= \chi_N\psi δ = χ N ψ α \alpha α a a a α 2 = − δ ( Frob λ ) \alpha^2 = -\delta(\operatorname{Frob}_\lambda) α 2 = − δ ( Frob λ )
T = ( ζ N 0 0 ζ N − 1 ) , Φ = ( 0 α α 0 ) . T = \begin{pmatrix} \zeta_N & 0 \\ 0 & \zeta_N^{-1} \end{pmatrix}, \qquad\Phi= \begin{pmatrix} 0 & \alpha\\ \alpha& 0 \end{pmatrix}. T = ( ζ N 0 0 ζ N − 1 ) , Φ = ( 0 α α 0 ) . They satisfy Φ T Φ − 1 = T l \Phi T \Phi^{-1} = T^l Φ T Φ − 1 = T l δ \delta δ N N N
For a later coefficient characteristic t ∈ S t \in S t ∈ S N N N λ \lambda λ N > 5 N > 5 N > 5 A 4 A_4 A 4 S 4 S_4 S 4 A 5 A_5 A 5 S S S λ \lambda λ λ \lambda λ λ \lambda λ K S K_S K S N N N t > 5 t > 5 t > 5
Local lifts and persistent inertial types
Lemma 4.1. Let F / Q p F/\mathbb{Q}_p F / Q p p > 2 p > 2 p > 2 r ˉ : G F → GL 2 ( F ‾ p ) \bar{r} : G_F \to\operatorname{GL}_2(\overline{\mathbb{F}}_p) r ˉ : G F → GL 2 ( F p ) χ ‾ p \overline{\chi}_p χ p r ˉ \bar{r} r ˉ χ p \chi_p χ p
Proof. Apply [9 Proposition 2.7] with prescribed determinant χ p \chi_p χ p [7 Lemma 1.4.3(1)] applies. The unramified scalar correction in the fixed-determinant construction preserves both properties.
Lemma 4.2. Let F / Q p F/\mathbb{Q}_p F / Q p r : G F → GL 2 ( Q ‾ p ) r : G_F \to\operatorname{GL}_2(\overline{\mathbb{Q}}_p) r : G F → GL 2 ( Q p ) { 0 , 1 } \{0,1\} { 0 , 1 } χ p \chi_p χ p r r r
If F = Q p F = \mathbb{Q}_p F = Q p r r r p p p χ p \chi_p χ p a p = s ∈ { 1 , − 1 } a_p = s \in\{1,-1\} a p = s ∈ { 1 , − 1 } s s s
Proof. Pass to a finite extension over which r r r r ∨ r^\vee r ∨ f f f N ≠ 0 N \ne0 N = 0 N φ = p φ N N\varphi= p\varphi N N φ = pφN α , p f α \alpha, p^f\alpha α , p f α φ \varphi φ N N N v p ( α ) = 0 v_p(\alpha) = 0 v p ( α ) = 0
Every labelled Hodge number on this line is zero or one. Weak admissibility bounds their average by the Newton slope zero, so all are zero. The line is weakly admissible and corresponds to a crystalline weight-zero, hence unramified, subrepresentation of r ∨ r^\vee r ∨ r r r N = 0 N = 0 N = 0
In the last assertion, local–global compatibility identifies the Frobenius eigenvalues on the dual module as s , p s s, ps s , p s s s s s − 1 = s s^{-1} = s s − 1 = s
Lemma 4.3. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 r ˉ : G F → GL 2 ( F ‾ 2 ) \bar{r} : G_F \to\operatorname{GL}_2(\overline{\mathbb{F}}_2) r ˉ : G F → GL 2 ( F 2 )
If r ˉ \bar{r} r ˉ χ 2 \chi_2 χ 2 F / Q 2 F/\mathbb{Q}_2 F / Q 2 { 1 − n , n } \{1-n,n\} { 1 − n , n }
(ii) If r ˉ \bar{r} r ˉ χ 2 \chi_2 χ 2 { 0 , 1 } \{0,1\} { 0 , 1 }
Proof. In the reducible case use the proof of [9 Proposition 2.7] , starting with the given residual line and quotient. Choose crystalline character lifts with product χ 2 \chi_2 χ 2 χ 2 \chi_2 χ 2 χ ˉ 2 = 1 \bar{\chi}_2=1 χ ˉ 2 = 1 a τ < b τ a_\tau<b_\tau a τ < b τ a τ + b τ = 1 a_\tau+b_\tau=1 a τ + b τ = 1 b τ b_\tau b τ
For fixed regular weights, the ordinary crystalline deformation ring is a reduced quotient of the crystalline ring, with the same pure dimension, and both generic fibres are smooth [7 Section 1.4] . The ordinary locus is therefore a union of irreducible components of the crystalline generic fibre. Smoothness makes this union open and closed; its intersection with the fixed-determinant fibre is again a union of components. We choose a component through the constructed point.
For a ramified quadratic F / Q 2 F/\mathbb{Q}_2 F / Q 2 O F × \mathcal{O}_F^\times O F × χ 2 n \chi_2^n χ 2 n χ 2 1 − n \chi_2^{1-n} χ 2 1 − n n n n
Suppose now that r ˉ \bar{r} r ˉ F / Q 2 F/\mathbb{Q}_2 F / Q 2 θ \theta θ θ 2 \theta^2 θ 2 θ 3 = 1 \theta^3=1 θ 3 = 1 θ \theta θ G L 2 ( F 2 ) ≃ S 3 \mathrm{GL}_2(\mathbb{F}_2)\simeq S_3 GL 2 ( F 2 ) ≃ S 3
The elliptic curve
A / Q : y 2 + y = x 3 , A/\mathbb{Q}: \quad y^2+y=x^3, A / Q : y 2 + y = x 3 , with Cremona label 27a3 (LMFDB label 27.a4) [45 ] , realizes it over Q 2 \mathbb{Q}_2 Q 2 − 27 -27 − 27 y = − 1 / 2 y=-1/2 y = − 1/2 4 x 3 + 1 = 0 4x^3+1=0 4 x 3 + 1 = 0
Q 2 ( A [ 2 ] ) = Q 2 ( ζ 3 , 2 3 ) . \mathbb{Q}_2(A[2])=\mathbb{Q}_2(\zeta_3,\sqrt[3]{2}). Q 2 ( A [ 2 ]) = Q 2 ( ζ 3 , 3 2 ) . This is an S 3 S_3 S 3 X 3 − 2 X^3-2 X 3 − 2 Q 2 ( ζ 3 ) / Q 2 \mathbb{Q}_2(\zeta_3)/\mathbb{Q}_2 Q 2 ( ζ 3 ) / Q 2 F F F A [ 2 ] ∣ G F A[2]|_{G_F} A [ 2 ] ∣ G F V 2 ( A ) ∣ G F V_2(A)|_{G_F} V 2 ( A ) ∣ G F χ 2 \chi_2 χ 2 [30 Theorem 6.2] .
Suppose instead that F / Q 2 F/\mathbb{Q}_2 F / Q 2 E / F E/F E / F r ˉ = Ind G E G F θ ˉ \bar{r}=\operatorname{Ind}_{G_E}^{G_F}\bar{\theta} r ˉ = Ind G E G F θ ˉ θ ˉ ∣ I = ω 4 e \bar{\theta}|_{I}=\omega_4^e θ ˉ ∣ I = ω 4 e ω 4 \omega_4 ω 4 F 16 × \mathbb{F}_{16}^\times F 16 × F F F 5 e ≡ 0 ( m o d 15 ) 5e\equiv0\pmod{15} 5 e ≡ 0 ( mod 15 ) e = 0 e=0 e = 0 e ∈ { 3 , 6 , 9 , 12 } e\in\{3,6,9,12\} e ∈ { 3 , 6 , 9 , 12 }
Index the four embeddings of E E E i i i i + 2 i+2 i + 2 F F F ω 4 \omega_4 ω 4 k i k_i k i ∑ i 2 i k i \sum_i 2^i k_i ∑ i 2 i k i
( k 0 , k 1 , k 2 , k 3 ) = ( a , b , 1 − a , 1 − b ) , a , b ∈ { 0 , 1 } , (k_0,k_1,k_2,k_3)=(a,b,1-a,1-b),\qquad a,b\in\{0,1\}, ( k 0 , k 1 , k 2 , k 3 ) = ( a , b , 1 − a , 1 − b ) , a , b ∈ { 0 , 1 } , have paired sums one and exponents 12 − 3 a − 6 b ∈ { 3 , 6 , 9 , 12 } 12-3a-6b\in\{3,6,9,12\} 12 − 3 a − 6 b ∈ { 3 , 6 , 9 , 12 } θ ˉ \bar{\theta} θ ˉ { 0 , 1 } \{0,1\} { 0 , 1 } F F F χ 2 \chi_2 χ 2 δ \delta δ ν \nu ν ν 2 = δ − 1 \nu^2=\delta^{-1} ν 2 = δ − 1 ν \nu ν
Finally, a potentially ordinary lift of these weights has, over a finite Galois extension, a line of constant labelled weight one. This line is unique, since two such lines would give weight one twice at every embedding. Its Galois conjugates are the same line, so it descends to G F G_F G F r ˉ \bar r r ˉ
Lemma 4.4. For a finite extension F / Q l F/\mathbb{Q}_l F / Q l l ≠ 2 l \ne2 l = 2 r ˉ : G F → S L 2 ( F 2 ) \bar r:G_F\to\mathrm{SL}_2(\mathbb{F}_2) r ˉ : G F → SL 2 ( F 2 ) χ 2 \chi_2 χ 2 r ˉ \bar r r ˉ
Proof. If r ˉ \bar r r ˉ α , α − 1 \alpha,\alpha^{-1} α , α − 1 α ≠ 1 \alpha\ne1 α = 1 H 1 ( G F , α 2 ) = 0 H^1(G_F,\alpha^2)=0 H 1 ( G F , α 2 ) = 0 H 0 ( G F , α 2 ) H^0(G_F,\alpha^2) H 0 ( G F , α 2 ) H 0 ( G F , α − 2 χ ˉ 2 ) H^0(G_F,\alpha^{-2}\bar\chi_2) H 0 ( G F , α − 2 χ ˉ 2 ) χ ˉ 2 = 1 \bar\chi_2=1 χ ˉ 2 = 1 r ˉ \bar r r ˉ α ~ ⊕ χ 2 α ~ − 1 \widetilde\alpha\oplus\chi_2\widetilde\alpha^{-1} α ⊕ χ 2 α − 1
If α = 1 \alpha=1 α = 1 r ˉ = ( 1 c 0 1 ) \bar r=\begin{pmatrix}1&c\\0&1\end{pmatrix} r ˉ = ( 1 0 c 1 ) c c c τ \tau τ Φ \Phi Φ Q Q Q Φ τ Φ − 1 = τ Q \Phi\tau\Phi^{-1}=\tau^Q Φ τ Φ − 1 = τ Q n , x n,x n , x c ( τ ) , c ( Φ ) c(\tau),c(\Phi) c ( τ ) , c ( Φ )
τ ⟼ ( 1 n 0 1 ) , Φ ⟼ ( Q x 0 1 ) . \tau\longmapsto\begin{pmatrix}1&n\\0&1\end{pmatrix},\qquad
\Phi\longmapsto\begin{pmatrix}Q&x\\0&1\end{pmatrix}. τ ⟼ ( 1 0 n 1 ) , Φ ⟼ ( Q 0 x 1 ) . This satisfies the relation and has determinant χ 2 \chi_2 χ 2 n = 0 n=0 n = 0 r ˉ \bar r r ˉ
If r ˉ \bar r r ˉ r ˉ = Ind G E G F θ ˉ \bar r=\operatorname{Ind}_{G_E}^{G_F}\bar\theta r ˉ = Ind G E G F θ ˉ E / F E/F E / F θ ˉ \bar\theta θ ˉ θ ˉ ∣ F × = 1 \bar\theta|_{F^\times}=1 θ ˉ ∣ F × = 1 θ 0 \theta_0 θ 0 ϵ E / F \epsilon_{E/F} ϵ E / F F × F^\times F × ϕ \phi ϕ E × E^\times E × χ 2 ϵ E / F \chi_2\epsilon_{E/F} χ 2 ϵ E / F F × F^\times F ×
For unramified E / F E/F E / F ϕ \phi ϕ E / F E/F E / F O E × \mathcal{O}_E^\times O E × ϵ E / F \epsilon_{E/F} ϵ E / F O F × \mathcal{O}_F^\times O F × π F = u π E 2 \pi_F=u\pi_E^2 π F = u π E 2 ϕ ( π E ) \phi(\pi_E) ϕ ( π E ) ( χ 2 ϵ E / F ) ( π F ) / ϕ ( u ) (\chi_2\epsilon_{E/F})(\pi_F)/\phi(u) ( χ 2 ϵ E / F ) ( π F ) / ϕ ( u ) Ind G E G F ( θ 0 ϕ ) \operatorname{Ind}_{G_E}^{G_F}(\theta_0\phi) Ind G E G F ( θ 0 ϕ ) ϵ E / F ( θ 0 ϕ ) ∣ F × = χ 2 \epsilon_{E/F}(\theta_0\phi)|_{F^\times}=\chi_2 ϵ E / F ( θ 0 ϕ ) ∣ F × = χ 2
A type containing every relevant Serre weight
Let F / Q p F/\mathbb{Q}_p F / Q p p > 2 p>2 p > 2 k k k q q q χ p \chi_p χ p K 0 = G L 2 ( O F ) K_0=\mathrm{GL}_2(\mathcal{O}_F) K 0 = GL 2 ( O F ) π \pi π K 1 = 1 + π M 2 ( O F ) K_1=1+\pi M_2(\mathcal{O}_F) K 1 = 1 + π M 2 ( O F )
I + = { ( 1 + π a b π c 1 + π d ) : a , b , c , d ∈ O F } , J = O F × I + . I^+=\left\{\begin{pmatrix}1+\pi a&b\\ \pi c&1+\pi d\end{pmatrix}:a,b,c,d\in\mathcal{O}_F\right\},\qquad J=\mathcal{O}_F^\times I^+. I + = { ( 1 + π a π c b 1 + π d ) : a , b , c , d ∈ O F } , J = O F × I + . Fix a nontrivial additive character ψ : k → Q ‾ p × \psi:k\to\overline{\mathbb{Q}}_p^\times ψ : k → Q p × J J J
θ ( z ( 1 + π a b π c 1 + π d ) ) = ψ ( b ˉ + c ˉ ) , z ∈ O F × . \theta\left(z\begin{pmatrix}1+\pi a&b\\ \pi c&1+\pi d\end{pmatrix}\right)=\psi(\bar b+\bar c),\qquad z\in\mathcal{O}_F^\times. θ ( z ( 1 + π a π c b 1 + π d ) ) = ψ ( b ˉ + c ˉ ) , z ∈ O F × . The ambiguity in z z z Z Z Z U U U G L 2 ( k ) \mathrm{GL}_2(k) GL 2 ( k )
Lemma 4.5. Every irreducible F ‾ p \overline{\mathbb{F}}_p F p G L 2 ( k ) \mathrm{GL}_2(k) GL 2 ( k ) Ind Z U G L 2 ( k ) 1 \operatorname{Ind}_{ZU}^{\mathrm{GL}_2(k)} 1 Ind Z U GL 2 ( k ) 1
Proof. Let V V V p p p p p p V U ≠ 0 V^U \ne0 V U = 0 Z Z Z V Z U ≠ 0 V^{ZU} \ne0 V Z U = 0 Ind Z U G L 2 ( k ) 1 → V \operatorname{Ind}_{ZU}^{\mathrm{GL}_2(k)} 1 \to V Ind Z U GL 2 ( k ) 1 → V
Lemma 4.6. The representation σ = Ind J K 0 θ \sigma= \operatorname{Ind}_{J}^{K_0} \theta σ = Ind J K 0 θ q 2 − 1 q^2-1 q 2 − 1 τ \tau τ Ind Z U G L 2 ( k ) 1 \operatorname{Ind}_{ZU}^{\mathrm{GL}_2(k)} 1 Ind Z U GL 2 ( k ) 1
Proof. On K 1 = 1 + π M 2 ( O F ) K_1 = 1 + \pi M_2(\mathcal{O}_F) K 1 = 1 + π M 2 ( O F ) 1 + π A ↦ ψ ( A ‾ 21 ) 1+\pi A \mapsto\psi(\overline{A}_{21}) 1 + π A ↦ ψ ( A 21 ) G L 2 ( k ) \mathrm{GL}_2(k) GL 2 ( k ) Z U ZU Z U M 2 ( k ) M_2(k) M 2 ( k ) k k k Z U ZU Z U σ \sigma σ K 1 K_1 K 1 q 2 − 1 q^2-1 q 2 − 1 K 0 K_0 K 0
The simple-supercuspidal construction identifies σ \sigma σ θ \theta θ ⟨ g ⟩ F × I + \langle g\rangle F^\times I^+ ⟨ g ⟩ F × I + g = ( 0 1 π 0 ) g=\begin{pmatrix}0&1\\ \pi&0\end{pmatrix} g = ( 0 π 1 0 ) [1 Section 3.1] . To check the occurrence condition, let an irreducible smooth representation contain θ \theta θ J J J θ \theta θ F × I + F^\times I^+ F × I + [1 Section 3.1, p. 6] . Changing the extension on g g g σ \sigma σ σ \sigma σ σ ( τ ) \sigma(\tau) σ ( τ ) [35 Theorem 1.1.3] . The Langlands parameter is the induced parameter of [1 Section 3.4 and Theorem 3.18] . The central character on O F × \mathcal{O}_F^\times O F × det τ = 1 \det\tau= 1 det τ = 1
All values of θ \theta θ p p p K 0 / J K_0/J K 0 / J K 1 ⊂ J K_1 \subset J K 1 ⊂ J J / K 1 = Z U J/K_1 = ZU J / K 1 = Z U
Theorem 4.7 (A universal wild inertial type). Let r ‾ : G F → G L 2 ( F ‾ p ) \overline{r}:G_F \to\mathrm{GL}_2(\overline{\mathbb{F}}_p) r : G F → GL 2 ( F p ) χ ‾ p \overline{\chi}_p χ p r ‾ \overline{r} r { 0 , 1 } \{0,1\} { 0 , 1 } χ p \chi_p χ p τ \tau τ F = Q p F=\mathbb{Q}_p F = Q p τ p , w \tau_{p,w} τ p , w p p p
Proof. We use the potentially Barsotti–Tate Breuil–Mézard theorem of Gee–Kisin, [35 Corollaries 3.5.3 and 3.5.6] . To match their conventions, apply their theorem to the dual residual representation, and dualize the resulting lift at the end. Their cyclotomic character has Hodge–Tate weight − 1 -1 − 1
At least one of the nonnegative multiplicities μ V \mu_V μ V [35 Theorem A.2] : the irreducible case is contained in [64 Proposition 7.8.1] , and the reducible case follows from the crystalline-extension construction cited there. The nonzero potentially Barsotti–Tate ring for that lift has positive Hilbert–Samuel multiplicity; the Breuil–Mézard formula makes at least one μ V \mu_V μ V V V V τ \tau τ p p p r r r
Finally det r / χ p \det r/\chi_p det r / χ p p p p r r r
For F = Q p F = \mathbb{Q}_p F = Q p p p p [1 Section 3.4 and Theorem 3.18] . In coefficient characteristic greater than p p p 2 2 2 p p p
Remark 4.8. For p = 3 p = 3 p = 3
Lemma 4.9. Let ℓ ≠ 3 \ell\ne3 ℓ = 3 ℓ ≡ − 1 ( m o d 3 ) \ell\equiv-1 \pmod{3} ℓ ≡ − 1 ( mod 3 ) s ˉ \bar{s} s ˉ χ 3 \chi_3 χ 3 G Q ℓ G_{\mathbb{Q}_\ell} G Q ℓ s ˉ \bar{s} s ˉ χ 3 \chi_3 χ 3
Proof. Remove the scalar quadratic character. Wild inertia is trivial; tame inertia is trivial or unipotent of order three. Frobenius has characteristic polynomial X 2 − 1 X^2 - 1 X 2 − 1 ℓ α , α \ell\alpha, \alpha ℓ α , α α = ± 1 \alpha= \pm1 α = ± 1 ℓ \ell ℓ
For a cuspidal lift take
T = ( 0 − 1 1 − 1 ) , S = ( 0 1 1 0 ) , Φ = b S , b 2 = − ℓ , b ≡ 1 ( m o d 3 ) . T = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}, \qquad S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad\Phi= bS,\quad b^2 = -\ell,\quad b \equiv1 \pmod{3}. T = ( 0 1 − 1 − 1 ) , S = ( 0 1 1 0 ) , Φ = b S , b 2 = − ℓ , b ≡ 1 ( mod 3 ) . Hensel’s lemma supplies b b b T 3 = 1 T^3 = 1 T 3 = 1 S T S = T − 1 STS = T^{-1} S T S = T − 1 det Φ = ℓ \det\Phi= \ell det Φ = ℓ − 1 -1 − 1 Φ \Phi Φ − Φ -\Phi − Φ T = diag ( ζ 3 , ζ 3 − 1 ) T = \operatorname{diag}(\zeta_3,\zeta_3^{-1}) T = diag ( ζ 3 , ζ 3 − 1 ) Q 3 ( ζ 3 ) \mathbb{Q}_3(\zeta_3) Q 3 ( ζ 3 )
For a special lift use
Φ = diag ( ℓ , 1 ) , U ( t ) = ( 1 t 0 1 ) , Φ U ( t ) Φ − 1 = U ( ℓ t ) . \Phi= \operatorname{diag}(\ell,1), \qquad U(t) = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}, \qquad\Phi U(t)\Phi^{-1} = U(\ell t). Φ = diag ( ℓ , 1 ) , U ( t ) = ( 1 0 t 1 ) , Φ U ( t ) Φ − 1 = U ( ℓ t ) . Choose its three-primary tame parameter to be a unit multiple when residual inertia is nontrivial, and three times a unit multiple otherwise. Both choices have nonzero monodromy. An unramified quadratic sign selects either orientation. Restoring the removed quadratic character preserves the determinant and gives the required inertial order.
Lemma 4.10. Let p ∈ { 2 , 3 } p \in\{2,3\} p ∈ { 2 , 3 } F / Q p F/\mathbb{Q}_p F / Q p p − 1 p-1 p − 1 r : G F → GL 2 ( E ) r:G_F \to\operatorname{GL}_2(E) r : G F → GL 2 ( E ) { 0 , 1 } \{0,1\} { 0 , 1 } χ p \chi_p χ p r r r
Proof. After a finite unramified extension the residual image is a p p p F p \mathbb{F}_p F p G G G G G G r ∨ ( 1 ) ≃ r r^\vee(1) \simeq r r ∨ ( 1 ) ≃ r
The schematic closure of the fixed F p \mathbb{F}_p F p p p p p − 1 p-1 p − 1 p = 3 p=3 p = 3 F F F
Dyadic automorphy and pro-modularity inputs
The local weight-two lifts and their support-preserving global construction are proved in Section 9, using the fixed-determinant Breuil–Mézard theorem [71 Corollary 8.0.2] and the prescribed-lift theorem [28 Theorem 4.2.1] . Potential modularity is established separately there. For backward modularity we use Kisin’s and Tung’s theorems with their local-source and residual-image hypotheses [43 Theorem 3.3.5] , [71 Theorem 8.0.3] .
The scalar argument uses the representation spaces, relative presentations and local geometry of [48 Section 3] , and the completed-forms and local block comparisons of [48 Propositions 4.1, 4.2 and 4.4, Lemma 4.6] . The localized propagation theorem [48 Theorem 5.1] applies over totally real solvable fields of even degree completely split at two, to a potentially pro-modular characteristic-two curve whose generic representation is not virtually solvable and whose allowed odd local images are finite. In these conventions the completed-form central character is ( det r ) ε 2 (\det r)\varepsilon_2 ( det r ) ε 2 r ⊗ ε 2 r \otimes\varepsilon_2 r ⊗ ε 2 [12 Theorems 1.1 and 1.3] .
For irreducible solvable residue we use [21 Proposition 10.3 and Theorem 10.6] . The application is over an abelian totally real field completely split at two, with every dyadic restriction of the actual source absolutely irreducible, so there are no locally reducible places. The auxiliary fields and scalar-source deduction needed here are proved in Sections 7 and 8.
Conductor reduction at a split odd prime
We prove Theorem 1.2. The local constructions above preserve the actual residual extension; the lifting implications in Section 2 concern the old characteristic-zero source at each congruence.
Proof of Theorem 1.2. Twist by the inverse square root of the odd-order determinant factor, leaving determinant χ ψ \chi\psi χ ψ ψ \psi ψ S S S 2 , 3 , 5 , ℓ 2,3,5,\ell 2 , 3 , 5 , ℓ K K K q \mathfrak{q} q q > N > max S q>N>\max S q > N > max S N N N S S S
In characteristic two, use Lemma 4.3 at the unique quadratic dyadic completion to choose a crystalline target of exact determinant χ 2 \chi_2 χ 2 χ 2 \chi^2 χ 2 q q q
Reset in characteristic ℓ \ell ℓ ℓ \ell ℓ q q q
For each remaining odd conductor characteristic t ≠ ℓ t \ne\ell t = ℓ ℓ \ell ℓ t t t ℓ \ell ℓ t t t N N N t t t S ∪ { q } S \cup\{q\} S ∪ { q }
Finally work in coefficient characteristic q > max { 5 , ℓ } q > \max\{5,\ell\} q > max { 5 , ℓ } ℓ \ell ℓ q q q K ( ζ q ) / K K(\zeta_q)/K K ( ζ q ) / K q q q q q q q q q q q q ℓ \ell ℓ
Its actual ℓ \ell ℓ ℓ \ell ℓ [76 Theorem 6.1.1] give modularity. If it is reducible, apply (ii) and [76 Theorem 5.1.1] . Here K K K ℓ \ell ℓ ℓ \ell ℓ ω ℓ a \omega_\ell^a ω ℓ a ω ℓ 1 − 2 a \omega_\ell^{1-2a} ω ℓ 1 − 2 a ℓ − 1 \ell-1 ℓ − 1
Auxiliary fields with scalar Selmer vanishing
Let K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) d > 1 d > 1 d > 1 d ≡ 1 ( m o d 8 ) d \equiv1 \pmod{8} d ≡ 1 ( mod 8 ) V w 0 + ( K ) V_{w_0}^+(K) V w 0 + ( K )
Lemma 7.1 (Cyclic towers with bounded odd splitting). Let K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) Σ \Sigma Σ n ≥ 2 n \ge2 n ≥ 2 r ∉ Σ r \notin\Sigma r ∈ / Σ L / Q L/\mathbb{Q} L / Q 2 n 2^n 2 n
L L L r r r
r r r K K K
2 2 2 L L L
every ℓ ∈ Σ \ell\in\Sigma ℓ ∈ Σ K L KL K L
The prime r r r
Proof. Let P P P Σ \Sigma Σ d d d
B n = Q ( ζ 2 n + 1 , 2 1 / 2 n ) , C n = B n ( t 1 / 2 n : t ∈ P ) . B_n = \mathbb{Q}(\zeta_{2n+1}, 2^{1/2^n}), \qquad C_n = B_n(t^{1/2^n} : t \in P). B n = Q ( ζ 2 n + 1 , 2 1/ 2 n ) , C n = B n ( t 1/ 2 n : t ∈ P ) . The field B n B_n B n Q \mathbb{Q} Q B n B_n B n t t t t t t t ∈ P t \in P t ∈ P Z / 2 n Z \mathbb{Z}/2^n\mathbb{Z} Z / 2 n Z B n × / ( B n × ) 2 n B_n^\times/(B_n^\times)^{2^n} B n × / ( B n × ) 2 n ∏ t ∈ P t b t \prod_{t\in P}t^{b_t} ∏ t ∈ P t b t 2 n 2^n 2 n 2 n ∣ b t 2^n \mid b_t 2 n ∣ b t t t t
Gal ( C n / B n ) ≃ ( Z / 2 n Z ) P , \operatorname{Gal}(C_n/B_n) \simeq(\mathbb{Z}/2^n\mathbb{Z})^P, Gal ( C n / B n ) ≃ ( Z / 2 n Z ) P , where a vector ( a t ) (a_t) ( a t ) t 1 / 2 n ↦ ζ 2 n a t t 1 / 2 n t^{1/2^n} \mapsto\zeta_{2^n}^{a_t}t^{1/2^n} t 1/ 2 n ↦ ζ 2 n a t t 1/ 2 n
Let j j j d d d j j j a t = 1 a_t=1 a t = 1 t ∈ P t\in P t ∈ P j j j a t = 2 a_t=2 a t = 2 d d d a t = 1 a_t=1 a t = 1 t ∈ P t\in P t ∈ P ∑ t ∣ d a t \sum_{t\mid d}a_t ∑ t ∣ d a t a t a_t a t C n C_n C n Q \mathbb{Q} Q h ∈ Gal ( C n / B n ) h\in\operatorname{Gal}(C_n/B_n) h ∈ Gal ( C n / B n ) r r r B n B_n B n h h h h ( d ) = − d h(\sqrt{d})=-\sqrt{d} h ( d ) = − d r r r K K K r r r B n B_n B n
r ≡ 1 ( m o d 2 n + 1 ) , 2 ∈ ( F r × ) 2 n . r\equiv1\pmod{2^{n+1}},\qquad2\in(\mathbb{F}_r^\times)^{2^n}. r ≡ 1 ( mod 2 n + 1 ) , 2 ∈ ( F r × ) 2 n . Take L L L Q ( ζ r ) \mathbb{Q}(\zeta_r) Q ( ζ r ) 2 n 2^n 2 n L L L t ∈ P t\in P t ∈ P Gal ( L / Q ) \operatorname{Gal}(L/\mathbb{Q}) Gal ( L / Q ) 2 n 2^n 2 n a t a_t a t 2 n − 1 2^{n-1} 2 n − 1 a t = 2 a_t=2 a t = 2 2 n 2^n 2 n r r r ℓ ∤ d \ell\nmid d ℓ ∤ d a ℓ = 1 a_\ell=1 a ℓ = 1 L L L ℓ \ell ℓ K L / L KL/L K L / L ℓ ∣ d \ell\mid d ℓ ∣ d K K K L L L L L L K K K ℓ \ell ℓ K L KL K L L L L
Lemma 7.2 (Propagation of one-dyadic scalar Selmer vanishing). Let K K K w 1 , w 2 w_1,w_2 w 1 , w 2 K K K w 1 w_1 w 1 w 2 w_2 w 2 F / K F/K F / K 2 n 2^n 2 n w 1 , w 2 w_1,w_2 w 1 , w 2 r ∤ 2 r\nmid2 r ∤ 2 F F F w 1 w_1 w 1 w 2 w_2 w 2
Proof. Let V V V H 1 ( G F , F 2 ) H^1(G_F,\mathbb{F}_2) H 1 ( G F , F 2 ) Γ = Gal ( F / K ) \Gamma=\operatorname{Gal}(F/K) Γ = Gal ( F / K ) V ≠ 0 V\ne0 V = 0 c ∈ V Γ c\in V^\Gamma c ∈ V Γ H / F H/F H / F c c c H / K H/K H / K G G G Γ \Gamma Γ C 2 C_2 C 2 Γ \Gamma Γ G G G
Because H / F H/F H / F r r r F / K F/K F / K I ⊂ G I\subset G I ⊂ G Γ \Gamma Γ Gal ( H / F ) \operatorname{Gal}(H/F) Gal ( H / F ) G = I × Gal ( H / F ) G=I\times\operatorname{Gal}(H/F) G = I × Gal ( H / F ) E = H I E=H^I E = H I K K K r r r w 1 w_1 w 1 r r r H / F H/F H / F F / K F/K F / K E / K E/K E / K w 2 w_2 w 2 E / K E/K E / K K K K
Theorem 7.3 (Uniform scalar field preparation). Let K K K Σ \Sigma Σ n > a ≥ 1 n>a\ge1 n > a ≥ 1 K ⊂ A ⊂ F K\subset A\subset F K ⊂ A ⊂ F v 0 v_0 v 0 A A A
D = [ F : Q ] , M = { v ∣ v 0 in F } , m = ∣ M ∣ , D=[F:\mathbb{Q}],\qquad M=\{v\mid v_0\text{ in }F\},\qquad m=|M|, D = [ F : Q ] , M = { v ∣ v 0 in F } , m = ∣ M ∣ , and let S A , S F S_A, S_F S A , S F Σ \Sigma Σ
D = 2 n + 1 D = 2^{n+1} D = 2 n + 1 [ F : A ] = 2 n − a [F : A] = 2^{n-a} [ F : A ] = 2 n − a m / D = 2 − a − 1 m/D = 2^{-a-1} m / D = 2 − a − 1 u A , u F u_A, u_F u A , u F 2 ∣ Σ ∣ 2|\Sigma| 2∣Σ∣
For every finite field k k k
{ c ∈ H 1 ( G F , S F , k ) : loc v c = 0 for all v ∈ S F ∖ M } = 0. \{c \in H^1(G_{F,S_F}, k) : \operatorname{loc}_v c = 0 \text{ for all } v \in S_F \setminus M\} = 0. { c ∈ H 1 ( G F , S F , k ) : loc v c = 0 for all v ∈ S F ∖ M } = 0. Here and below k k k
Restriction H 1 ( G A , S A , k ) ⟶ H 1 ( G F , S F , k ) H^1(G_{A,S_A}, k) \longrightarrow H^1(G_{F,S_F}, k) H 1 ( G A , S A , k ) ⟶ H 1 ( G F , S F , k )
Let H s t r 1 ( F , k ) H^1_{\mathrm{str}}(F, k) H str 1 ( F , k ) M M M S F ∖ M S_F \setminus M S F ∖ M S F S_F S F
dim k H s t r 1 ( F , k ) ≤ 2 D − 3 m + u F . \dim_k H^1_{\mathrm{str}}(F, k) \le2D - 3m + u_F. dim k H str 1 ( F , k ) ≤ 2 D − 3 m + u F . The degrees [ A : Q ] [A : \mathbb{Q}] [ A : Q ] [ F : A ] [F : A] [ F : A ]
Proof. Apply Lemma 7.1, take the fields A = K L a A = KL_a A = K L a F = K L F = KL F = K L Lemma 7.2 , using (SV) as its base vanishing hypothesis. The first three assertions follow from the two preceding lemmas and complete splitting at two. For injectivity, a class killed under restriction factors through the cyclic group Gal ( F / A ) \operatorname{Gal}(F/A) Gal ( F / A ) r r r S A S_A S A H 1 ( G A , S A , k ) H^1(G_{A,S_A}, k) H 1 ( G A , S A , k )
For the last assertion use the Selmer dimension formula from local and global Tate duality, with the real local conditions included [51 Theorem 4.2] . The scalar module k k k M M M S F ∖ M S_F \setminus M S F ∖ M dim H 1 ( G Q 2 , k ) − 1 = 2 \dim H^1(G_{\mathbb{Q}_2}, k) - 1 = 2 dim H 1 ( G Q 2 , k ) − 1 = 2 − 1 -1 − 1 1 − 1 = 0 1 - 1 = 0 1 − 1 = 0 S F S_F S F
2 ( D − m ) − m + u F = 2 D − 3 m + u F , 2(D - m) - m + u_F = 2D - 3m + u_F, 2 ( D − m ) − m + u F = 2 D − 3 m + u F , as claimed. Taking a a a n − a n - a n − a
When Σ = ∅ \Sigma= \varnothing Σ = ∅ n = a + 2 n = a + 2 n = a + 2 Theorem 7.3 gives [ F : A ] = 4 [F : A] = 4 [ F : A ] = 4 [ A : Q ] [A : \mathbb{Q}] [ A : Q ] dim k H s t r 1 ( F , k ) = 2 [ F : Q ] − 12 \dim_k H^1_{\mathrm{str}}(F, k) = 2[F : \mathbb{Q}] - 12 dim k H str 1 ( F , k ) = 2 [ F : Q ] − 12
Remark 7.4. Condition (SV) is a genuine restriction. For K = Q ( 113 ) K = \mathbb{Q}(\sqrt{113}) K = Q ( 113 ) α = ( 11 + 113 ) / 2 \alpha= (11 + \sqrt{113})/2 α = ( 11 + 113 ) /2 u u u u + 2 / u = 11 u + 2/u = 11 u + 2/ u = 11 u ≡ 1 ( m o d 8 ) u \equiv1 \pmod{8} u ≡ 1 ( mod 8 ) Q 2 \mathbb{Q}_2 Q 2 [ α ] [\alpha] [ α ] V 0 + ( K ) V^+_0(K) V 0 + ( K )
The scalar dyadic endpoint
We prove Theorem 1.3 using the field preparation of § 2.6 and the dyadic inputs listed in § 2.4 . The argument treats arbitrary finite ramification and regular weights.
Proof of Theorem 1.3. First suppose r r r K K K K K K r r r r r r
The seed and the auxiliary fields. A cuspidal modular seed with scalar semisimple reduction is available over Q \mathbb{Q} Q Δ \Delta Δ ε 2 − 5 \varepsilon_2^{-5} ε 2 − 5 ε 2 \varepsilon_2 ε 2 2 2 2 1 ⊕ 1 1 \oplus1 1 ⊕ 1 Δ ≡ ∑ j ≥ 0 q ( 2 j + 1 ) 2 ( m o d 2 ) \Delta\equiv\sum_{j\ge0} q^{(2j+1)^2} \pmod{2} Δ ≡ ∑ j ≥ 0 q ( 2 j + 1 ) 2 ( mod 2 ) K K K
Choose a finite set Σ \Sigma Σ r r r d d d Theorem 7.3 supplies
K ⊂ A ⊂ F , D A = [ A : Q ] , D = [ F : Q ] , m = [ F : A ] , K \subset A \subset F,\qquad D_A=[A:\mathbb{Q}],\qquad D=[F:\mathbb{Q}],\qquad m=[F:A], K ⊂ A ⊂ F , D A = [ A : Q ] , D = [ F : Q ] , m = [ F : A ] , with A / Q A/\mathbb{Q} A / Q F / Q F/\mathbb{Q} F / Q 2 2 2 u A , u F u_A,u_F u A , u F M M M m m m F F F A A A D A D_A D A m m m D = 1 / D A D=1/D_A D = 1/ D A A ⊂ F A\subset F A ⊂ F
{ c ∈ H 1 ( G F , S F ; k ) : loc v c = 0 for all v ∈ S F ∖ M } = 0. (2) \{c\in H^1(G_F,S_F;k):\operatorname{loc}_v c=0\ \text{for all }v\in S_F\setminus M\}=0. \tag*{(2)} { c ∈ H 1 ( G F , S F ; k ) : loc v c = 0 for all v ∈ S F ∖ M } = 0. (2) Here S F S_F S F k k k
For the strict Selmer group, Theorem 7.3 (iv) gives
dim k H s t r 1 ( F , k ) ≤ 2 D − 3 m + u F . (3) \dim_k H^1_{\mathrm{str}}(F,k)\le2D-3m+u_F. \tag*{(3)} dim k H str 1 ( F , k ) ≤ 2 D − 3 m + u F . (3) Let c c c G Q 2 G_{\mathbb{Q}_2} G Q 2 1 ⊕ 1 1\oplus1 1 ⊕ 1 A A A Q 2 \mathbb{Q}_2 Q 2
D A > max { 4 , u A + c + 3 } , m > u F + 4. (4) D_A>\max\{4,u_A+c+3\},\qquad m>u_F+4. \tag*{(4)} D A > max { 4 , u A + c + 3 } , m > u F + 4. (4) The uniform bound on u A , u F u_A,u_F u A , u F D A , m D_A,m D A , m D = m D A D=mD_A D = m D A D > u F + 4 D>u_F+4 D > u F + 4
( 2 D − 2 m − 4 − u F ) − D = m ( D A − 2 ) − ( u F + 4 ) > 0. (2D-2m-4-u_F)-D=m(D_A-2)-(u_F+4)>0. ( 2 D − 2 m − 4 − u F ) − D = m ( D A − 2 ) − ( u F + 4 ) > 0. They also give the dimension bound needed for entry from A A A
Connected transverse charts. Over the characteristic-two fixed-determinant pseudodeformation base for G F G_F G F
At a transverse point, the product of the real and marked dyadic local line problems has dimension 2 D + 5 m 2D+5m 2 D + 5 m [48 Lemma 3.12] . The relative global presentation of [48 Proposition 3.6] has number of generators minus relations at least − 6 m -6m − 6 m 2 D − m 2D-m 2 D − m 2 D − m − 1 2D-m-1 2 D − m − 1 [65 Lemma 52.14.1, Tag 0ECR] .
The globally reducible locus has dimension at most
2 D − 2 m + u F + 3. 2D-2m+u_F+3. 2 D − 2 m + u F + 3. Indeed, fixed-product diagonal characters contribute at most one parameter by Leopoldt for the abelian field F F F 2 D − 3 m + u F 2D-3m+u_F 2 D − 3 m + u F m m m m m m 2 D − m − 1 2D-m-1 2 D − m − 1
Therefore a maximal-dimensional intersection in a component chain is generically globally irreducible. Forgetting the global frame and marked lines loses at most 3 + m 3+m 3 + m 2 D − 2 m − 4 2D-2m-4 2 D − 2 m − 4 u F u_F u F D D D [48 Lemma 3.8] and its curve-avoidance argument supply a non-virtually-solvable characteristic-two curve. The finite local-image conclusion is exactly [48 Lemma 3.9] , applied after its fixed preliminary local extensions; the determinant on the characteristic-two curve is one. The prescribed residual Frobenius traces thus make all allowed odd local images finite. Theorem 5.1 of [48 ] then propagates potential pro-modularity across this intersection. Thus all components through a transverse residual point have the same potential pro-modularity status.
The transverse residual fibre is geometrically irreducible. Fixing the global line, its extensions form the nonzero vectors of the strict scalar cohomology space. This space contains the strict space with no odd ramification, of dimension 2 D − 3 m > 0 2D-3m>0 2 D − 3 m > 0 A m \mathbb{A}^m A m G L 2 \mathrm{GL}_2 GL 2 Y 0 Y_0 Y 0 Y 0 Y_0 Y 0
Entering the transverse charts from the seed and target. We include this step because connecting deformation components for one fixed residual extension would not suffice. Work first over A A A 1 + 2 D A 1+2D_A 1 + 2 D A [48 Proposition 4.4] . An unrestricted fixed-determinant framed component through r ∣ G A r|_{G_A} r ∣ G A 2 D A − 2 2D_A-2 2 D A − 2
In each component pass to the special fibre, prescribe the residual Frobenius traces at the allowed odd places, and prescribe the constant scalar local pseudorepresentation at the selected dyadic place of A A A c c c 1 + u A + c 1+u_A+c 1 + u A + c
2 D A − 2 − ( 1 + u A + c ) = 2 D A − u A − c − 3 > D A 2D_A-2-(1+u_A+c)=2D_A-u_A-c-3>D_A 2 D A − 2 − ( 1 + u A + c ) = 2 D A − u A − c − 3 > D A by (S3). The bad-locus bound and curve avoidance therefore supply a non-virtually-solvable characteristic-two curve in each component. Lemma 3.9 of [48 ] makes its odd local images finite, as above.
At the selected dyadic place the curve representation has an invariant line after a finite coefficient extension. Choose a nonzero vector on that line and take its full G A G_A G A A ⊂ F A\subset F A ⊂ F F F F F / A F/A F / A F F F
The seed curve is potentially pro-modular by solvable base change. Propagation through the transverse charts, through the common component Y 0 Y_0 Y 0 [48 Theorem 5.1] back over A A A r r r K K K 2 2 2
Classicality and descent. At every dyadic place of the resulting field the local representation is still the original absolutely irreducible regular de Rham representation over Q 2 \mathbb{Q}_2 Q 2 2 2 2 [48 ] , together with the all-prime local correspondence [12 Theorems 1.1 and 1.3] , gives a nonzero locally algebraic vector in its Hecke eigenmodule, including when the residual local block is scalar. Proposition 4.1 identifies this vector with a regular algebraic definite-quaternionic eigensystem. Global irreducibility excludes reduced-norm characters, so Jacquet–Langlands produces a Hilbert cusp form. Local-global compatibility and Chebotarev identify its Galois representation with the restriction of r r r
Finally descend through a tower of prime-degree cyclic extensions to K K K r r r [4 Chapter III, Theorems 4.2(d) and 5.1] yields a cuspidal representation downstairs. Its Galois representation differs from the given one by a character of the cyclic quotient, because their restrictions are absolutely irreducible and isomorphic upstairs; twisting corrects that character. The archimedean types remain regular algebraic because every real place splits in a totally real extension. Iterating gives modularity of r r r K K K [48 Section 8.8] ; none of its ordinary dense-point arguments over Q \mathbb{Q} Q
Dyadic lifts and compatible systems
We construct two local types which suffice at every dyadic place and then globalize them without enlarging the prescribed ramification support. Steinberg lifts enter only in the local multiplicity calculation; the global lifts will all be potentially Barsotti–Tate and potentially nonordinary.
Let k k k s ˉ : G Q 2 → G L 2 ( k ) \bar{s}: G_{\mathbb{Q}_2} \to\mathrm{GL}_2(k) s ˉ : G Q 2 → GL 2 ( k ) ψ s ˉ = ε 2 [ det s ˉ ] \psi_{\bar{s}} = \varepsilon_2[\det\bar{s}] ψ s ˉ = ε 2 [ det s ˉ ] E / Q 2 E/\mathbb{Q}_2 E / Q 2 k k k O \mathcal{O} O ϖ \varpi ϖ τ 0 = ω ~ 2 ⊕ ω ~ 2 2 \tau_0 = \widetilde{\omega}_2 \oplus\widetilde{\omega}_2^2 τ 0 = ω 2 ⊕ ω 2 2 ω 2 \omega_2 ω 2 τ ∗ \tau_\ast τ ∗
y 2 = x 3 − 2 x , y^2 = x^3 - 2x, y 2 = x 3 − 2 x , with Cremona label 256b1 (LMFDB label 256.b1) [45 ] .
Proposition 9.1. Every s ˉ \bar{s} s ˉ ψ s ˉ \psi_{\bar{s}} ψ s ˉ { 0 , 1 } \{0,1\} { 0 , 1 } τ 0 \tau_0 τ 0 τ ∗ \tau_\ast τ ∗ τ ∗ \tau_\ast τ ∗ Q 8 Q_8 Q 8
Proof. Every character G Q 2 → k × G_{\mathbb{Q}_2} \to k^\times G Q 2 → k × Z 2 × \mathbb{Z}_2^\times Z 2 × k × k^\times k × ψ s ˉ \psi_{\bar{s}} ψ s ˉ δ \delta δ ν \nu ν ν 2 = δ − 1 \nu^2 = \delta^{-1} ν 2 = δ − 1 ν \nu ν
We first identify the residual representations for which no such crystalline lift exists. An irreducible s ˉ \bar{s} s ˉ G Q 2 G_{\mathbb{Q}_2} G Q 2 A [ 2 ] A[2] A [ 2 ] A : y 2 + y = x 3 A:y^2+y=x^3 A : y 2 + y = x 3 α ˉ ⊕ β ˉ \bar{\alpha}\oplus\bar{\beta} α ˉ ⊕ β ˉ ε 2 [ α ˉ ] ⊕ [ β ˉ ] \varepsilon_2[\bar{\alpha}]\oplus[\bar{\beta}] ε 2 [ α ˉ ] ⊕ [ β ˉ ] η ˉ \bar{\eta} η ˉ η ˉ ≠ 1 \bar{\eta}\ne1 η ˉ = 1 [8 Corollary 3.8.4] give
dim E H 1 ( G Q 2 , E ( ε 2 [ η ˉ ] ) ) = dim E H f 1 ( G Q 2 , E ( ε 2 [ η ˉ ] ) ) = 1. \dim_E H^1(G_{\mathbb{Q}_2}, E(\varepsilon_2[\bar{\eta}])) = \dim_E H^1_f(G_{\mathbb{Q}_2}, E(\varepsilon_2[\bar{\eta}])) = 1. dim E H 1 ( G Q 2 , E ( ε 2 [ η ˉ ])) = dim E H f 1 ( G Q 2 , E ( ε 2 [ η ˉ ])) = 1. Moreover H 2 ( G Q 2 , O ( ε 2 [ η ˉ ] ) ) = 0 H^2(G_{\mathbb{Q}_2}, \mathcal{O}(\varepsilon_2[\bar{\eta}]))=0 H 2 ( G Q 2 , O ( ε 2 [ η ˉ ])) = 0 [ η ˉ ] − 1 − 1 [\bar{\eta}]^{-1}-1 [ η ˉ ] − 1 − 1 H 1 H^1 H 1
s ˉ = α ˉ ⊗ ( 1 c 0 1 ) , 0 ≠ c ∈ H 1 ( G Q 2 , k ) , \bar{s}=\bar{\alpha}\otimes
\begin{pmatrix}
1 & c\\
0 & 1
\end{pmatrix},
\qquad0\ne c\in H^1(G_{\mathbb{Q}_2},k), s ˉ = α ˉ ⊗ ( 1 0 c 1 ) , 0 = c ∈ H 1 ( G Q 2 , k ) , with α ˉ \bar{\alpha} α ˉ
For this last representation, Kummer theory identifies H 1 ( G Q 2 , O ( ε 2 ) ) H^1(G_{\mathbb{Q}_2},\mathcal{O}(\varepsilon_2)) H 1 ( G Q 2 , O ( ε 2 )) Q 2 × ^ ⊗ Z 2 O \widehat{\mathbb{Q}_2^\times}\otimes_{\mathbb{Z}_2}\mathcal{O} Q 2 × ⊗ Z 2 O H 1 ( G Q 2 , k ) H^1(G_{\mathbb{Q}_2},k) H 1 ( G Q 2 , k ) κ \kappa κ c c c ϖ O \varpi\mathcal{O} ϖ O 1 1 1 ε 2 \varepsilon_2 ε 2 κ \kappa κ [ α ˉ ] [\bar{\alpha}] [ α ˉ ] ε 2 [ α ˉ ] 2 = ψ s ˉ \varepsilon_2[\bar{\alpha}]^2=\psi_{\bar{s}} ε 2 [ α ˉ ] 2 = ψ s ˉ
We now apply the full fixed-determinant Breuil–Mézard formula [71 Conjectures 5.1.2–5.1.3 and Corollary 8.0.2] . The two Serre weights are W 0 = 1 W_0=1 W 0 = 1 W 1 = Sym 1 W_1=\operatorname{Sym}^1 W 1 = Sym 1 μ 0 ( s ˉ ) , μ 1 ( s ˉ ) \mu_0(\bar{s}),\mu_1(\bar{s}) μ 0 ( s ˉ ) , μ 1 ( s ˉ ) σ \sigma σ W i W_i W i a i a_i a i ϖ \varpi ϖ
a 0 μ 0 ( s ˉ ) + a 1 μ 1 ( s ˉ ) . a_0\mu_0(\bar{s})+a_1\mu_1(\bar{s}). a 0 μ 0 ( s ˉ ) + a 1 μ 1 ( s ˉ ) . These deformation rings are taken with their reduced, O \mathcal{O} O W 0 W_0 W 0 μ 0 > 0 \mu_0>0 μ 0 > 0 τ 0 \tau_0 τ 0 GL 2 ( F 2 ) ≃ S 3 \operatorname{GL}_2(\mathbb{F}_2)\simeq S_3 GL 2 ( F 2 ) ≃ S 3 W 0 W_0 W 0 τ 0 \tau_0 τ 0 μ 0 = 0 \mu_0=0 μ 0 = 0 μ 1 > 0 \mu_1>0 μ 1 > 0 W 1 W_1 W 1 det τ 0 = 1 \det\tau_0=1 det τ 0 = 1 [ det s ˉ ] [\det\bar{s}] [ det s ˉ ]
For τ ∗ \tau_\ast τ ∗ [11 Definition A.3 and Lemma A.4] ; in particular it has no crystalline weight-two lift. Its three-division polynomial is 3 X 4 − 12 X 2 − 4 3X^4-12X^2-4 3 X 4 − 12 X 2 − 4 u 2 = − 3 u^2=-3 u 2 = − 3 i 2 = − 1 i^2=-1 i 2 = − 1 s = i u s=iu s = i u ± α , ± β \pm\alpha,\pm\beta ± α , ± β
α 2 = 2 + 4 / s , β 2 = 2 − 4 / s , α β = 2 u / 3. \alpha^2=2+4/s,\qquad\beta^2=2-4/s,\qquad\alpha\beta=2u/3. α 2 = 2 + 4/ s , β 2 = 2 − 4/ s , α β = 2 u /3. They recover s , u s,u s , u L = Q 2 ( u , i , α ) L=\mathbb{Q}_2(u,i,\alpha) L = Q 2 ( u , i , α ) U = Q 2 ( u ) U=\mathbb{Q}_2(u) U = Q 2 ( u ) T = U ( i ) T=U(i) T = U ( i ) π = 1 − i \pi=1-i π = 1 − i α 2 / π 2 = i + 2 / u \alpha^2/\pi^2=i+2/u α 2 / π 2 = i + 2/ u v π ( i + 2 / u − 1 ) = 1 v_\pi(i+2/u-1)=1 v π ( i + 2/ u − 1 ) = 1 T T T π 2 \pi^2 π 2 L / T L/T L / T e ( L / Q 2 ) = 4 e(L/\mathbb{Q}_2)=4 e ( L / Q 2 ) = 4 f ( L / Q 2 ) = 2 f(L/\mathbb{Q}_2)=2 f ( L / Q 2 ) = 2 V 4 ⊂ A 4 V_4\subset A_4 V 4 ⊂ A 4 SL 2 ( F 3 ) \operatorname{SL}_2(\mathbb{F}_3) SL 2 ( F 3 ) Q 8 Q_8 Q 8 V 4 V_4 V 4 j = 1728 j=1728 j = 1728 T 3 T_3 T 3 Q 8 Q_8 Q 8 τ ∗ \tau_\ast τ ∗
The Tate module of this curve gives a nonempty type-τ ∗ \tau_\ast τ ∗ μ 0 = 0 \mu_0=0 μ 0 = 0 a 1 ( τ ∗ ) > 0 a_1(\tau_\ast)>0 a 1 ( τ ∗ ) > 0 s ˉ \bar{s} s ˉ μ 1 ( s ˉ ) > 0 \mu_1(\bar{s})>0 μ 1 ( s ˉ ) > 0 τ ∗ \tau_\ast τ ∗ α ˉ \bar{\alpha} α ˉ
The two characters of τ 0 \tau_0 τ 0 τ ∗ \tau_\ast τ ∗
Lemma 9.2. Let F F F r : G F → GL 2 ( Q ‾ ) r:G_F\to\operatorname{GL}_2(\overline{\mathbb{Q}}) r : G F → GL 2 ( Q ) det r = ε 2 η \det r=\varepsilon^2\eta det r = ε 2 η η \eta η r r r
Proof. A finite character twist makes the residual determinant one. Apply [69 Proposition 6.7] , avoiding its kernel field, to make the residue modular without shrinking its image. A further solvable totally real extension makes every dyadic source Barsotti–Tate. Nonsolvability and residual modularity persist. Kisin’s theorem [43 Theorem 3.3.5] applies: its condition that a completion equal Q 2 \mathbb{Q}_2 Q 2
Corollary 9.3 (Prescribed dyadic lifts in compatible systems). Let K K K r ˉ : G K → S L 2 ( k ) \bar{r}: G_K \to\mathrm{SL}_2(k) r ˉ : G K → SL 2 ( k ) S S S S S S ε 2 \varepsilon_2 ε 2 r ˉ \bar{r} r ˉ S S S Proposition 9.1 .
Proof. Lemma 4.4 makes the odd local lifting rings nonempty and permits an unramified choice wherever the residue is unramified. At the dyadic places use Proposition 9.1 . Choose a finite extension of each completion killing its type. In the corresponding potentially semistable generic fibre of weights { 0 , 1 } \{0,1\} { 0 , 1 } Proposition 9.1 gives potential nonordinarity.
The global determinant ε 2 \varepsilon_2 ε 2 K ( ζ 2 ) = K K(\zeta_2)=K K ( ζ 2 ) = K [28 Theorem 4.2.1] therefore supplies a characteristic-zero point unramified outside S S S
A Serre criterion when two splits
Theorem 10.1 (The split-dyadic criterion). Let K K K § 2.1 and § 2.4 , and suppose:
(i) V ω + ( K ) = 0 V_{\omega}^{+}(K)=0 V ω + ( K ) = 0
(ii) the ordinary ideal class group C l ( K ) \mathrm{Cl}(K) Cl ( K )
(iii) every continuous absolutely irreducible representation G K → G L 2 ( F ‾ 2 ) G_K \to\mathrm{GL}_2(\overline{\mathbb{F}}_2) G K → GL 2 ( F 2 )
Then every continuous absolutely irreducible totally odd representation G K → G L 2 ( F ‾ p ) G_K \to\mathrm{GL}_2(\overline{\mathbb{F}}_p) G K → GL 2 ( F p ) p p p
Condition (iii) is a residual base case, separate from the scalar source theorem. The proof below has no analytic assumption beyond whatever is needed to establish that condition. Strong compatibility includes the coefficient prime, so a crystalline replacement clears that conductor from every companion.
Initial lifts and the protective place
Proof. Use the global construction and compatible systems of Lemma 2.4, the modular witnesses of Lemma 2.1, and the backward lifting statements in Lemma 2.2. For arbitrary regular dyadic sources at the split completions Q 2 \mathbb{Q}_2 Q 2 [71 Theorem 8.0.3] . The initial characteristic-five exception is absent for this K K K K ≠ Q ( 5 ) K \ne\mathbb{Q}(\sqrt{5}) K = Q ( 5 ) [ K ( ζ 5 ) : K ] = 4 [K(\zeta_5):K]=4 [ K ( ζ 5 ) : K ] = 4 P G L 2 ( F 5 ) \mathrm{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) K ( ζ 5 ) K(\zeta_5) K ( ζ 5 ) [28 Definition 4.1.1 and Proposition 4.1.2] . The subsequent protective images exclude all small exceptional images as well.
The protective place in Proposition 3.1 can be chosen with the additional property that the order of two modulo its rational norm is odd. We give the simultaneous field conditions, since this property is needed when its own inertia ceases to protect the last congruence.
Write K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) d > 1 d > 1 d > 1 det r t = ε t ψ 2 \det r_t = \varepsilon_t\psi_2 det r t = ε t ψ 2 ψ 2 \psi_2 ψ 2 s > 5 s > 5 s > 5 K K K d d d S S S
Choose N N N Proposition 3.1 , with N N N N N N K K K PGL 2 ( F N ) \operatorname{PGL}_2(\mathbb{F}_N) PGL 2 ( F N ) ψ 2 \psi_2 ψ 2 N N N
Let K S / K K_S/K K S / K S S S λ \lambda λ ℓ \ell ℓ
ℓ ≡ − 1 ( m o d N ) , λ split in K S , ℓ ≡ 9 ( m o d 16 ) , 2 ∈ ( F ℓ × ) 8 . (5) \ell\equiv-1 \pmod{N}, \qquad\lambda\text{ split in } K_S, \qquad\ell\equiv9 \pmod{16}, \qquad2 \in(\mathbb{F}_{\ell}^{\times})^8. \tag*{(5)} ℓ ≡ − 1 ( mod N ) , λ split in K S , ℓ ≡ 9 ( mod 16 ) , 2 ∈ ( F ℓ × ) 8 . (5) The last two conditions make ord ℓ ( 2 ) \operatorname{ord}_{\ell}(2) ord ℓ ( 2 ) B = Q ( ζ 8 , 2 1 / 8 ) B = \mathbb{Q}(\zeta_8, 2^{1/8}) B = Q ( ζ 8 , 2 1/8 ) ( a , b ) (a,b) ( a , b ) 2 1 / 8 2^{1/8} 2 1/8 ζ 8 \zeta_8 ζ 8 ( a , b ) ( c , e ) = ( a + b c , b e ) (a,b)(c,e) = (a+bc,be) ( a , b ) ( c , e ) = ( a + b c , b e ) ( 1 , 3 ) , ( 0 , 7 ) (1,3),(0,7) ( 1 , 3 ) , ( 0 , 7 ) ( 2 , 1 ) (2,1) ( 2 , 1 ) Gal ( B / Q ( ζ 8 ) ) \operatorname{Gal}(B/\mathbb{Q}(\zeta_8)) Gal ( B / Q ( ζ 8 )) Q ( ζ 8 ) \mathbb{Q}(\zeta_8) Q ( ζ 8 ) B B B K K K K ∩ B = Q K \cap B = \mathbb{Q} K ∩ B = Q Q ( ζ 16 ) \mathbb{Q}(\zeta_{16}) Q ( ζ 16 ) B B B K S / K K_S/K K S / K K S K_S K S K ( ζ 16 ) ⊄ B K S K(\zeta_{16}) \not\subset B K_S K ( ζ 16 ) ⊂ B K S B K S B K_S B K S K ( ζ 8 ) K(\zeta_8) K ( ζ 8 ) B K S B K_S B K S ζ 16 \zeta_{16} ζ 16 N N N N N N K ( ζ N ) K(\zeta_N) K ( ζ N ) ζ N \zeta_N ζ N N ≡ 1 ( m o d 4 ) N \equiv1 \pmod{4} N ≡ 1 ( mod 4 ) ℓ \ell ℓ K K K
Install the order-N N N λ \lambda λ N N N Proposition 3.1 . Its image argument makes every residue in S S S
Dyadic preparation and odd conductor
Reduce the protected system modulo two. Its determinant is one, since ψ 2 \psi_2 ψ 2 Lemma 4.4 to choose a component of determinant ε 2 \varepsilon_2 ε 2 λ \lambda λ N N N diag ( ζ N , ζ N − 1 ) \operatorname{diag}(\zeta_N,\zeta_N^{-1}) diag ( ζ N , ζ N − 1 ) − ℓ -\ell − ℓ ℓ \ell ℓ τ 0 \tau_0 τ 0 τ ∗ \tau_* τ ∗ K K K Q 2 \mathbb{Q}_2 Q 2
Retain the dyadic types τ 0 , τ ∗ \tau_0,\tau_* τ 0 , τ ∗ s , ℓ s,\ell s , ℓ t t t s s s s s s [64 Proposition 7.8.1] , followed by prescribed lifting. At s s s s s s [64 Section 7.9] . Unramified local components preserve cleared places, and the old components preserve retained types. Since s s s
In characteristic t t t t t t t t t [3 Theorem 2.3] , with the witness from Lemma 2.1, gives the backward implication. For split t t t K K K t t t s s s s s s
The last prime and the terminal source
Remove the protective place last, in characteristic ℓ \ell ℓ N N N τ ∗ \tau_* τ ∗ Q 8 Q_8 Q 8 ℓ \ell ℓ − 1 -1 − 1 τ 0 \tau_0 τ 0 ℓ > 3 \ell>3 ℓ > 3 (5) , the cyclotomic residue degree is odd, so that interchange survives restriction to G K ( ζ ℓ ) G_{K(\zeta_\ell)} G K ( ζ ℓ )
If the residue is solvable, its Artin lift and regular weight raising give a modular witness; Tung at the split prime ℓ \ell ℓ ℓ \ell ℓ ℓ \ell ℓ λ \lambda λ
The terminal system has cyclotomic determinant, conductor supported above two, and both dyadic supercuspidal types. Its actual 2-adic member r 2 r_2 r 2 [21 Theorem 10.6] : K / Q K/\mathbb{Q} K / Q Σ n \Sigma_n Σ n Σ o = ∅ \Sigma_o=\varnothing Σ o = ∅ Q 2 \mathbb{Q}_2 Q 2 Theorem 1.3 apply. Virtually solvable actual sources are alternatively covered by the CM-induction argument in its proof.
In every case the actual terminal source is modular. Compatibility identifies its entire system with the automorphic system, and all backward implications return to the initial lift. Reducing proves Theorem 10.1.
Integral orders and finite-flat groups
For the quadratic applications write F d = Q ( d ) F_d=\mathbb{Q}(\sqrt{d}) F d = Q ( d )
Trace orders
We use simultaneous realizations at every dyadic coefficient place. This is essential when a local calculation bounds a global trace-field degree. Let R 1 , … , R m \mathcal{R}_1,\ldots,\mathcal{R}_m R 1 , … , R m D i D_i D i d i = [ D i : Q ] d_i = [D_i : \mathbb{Q}] d i = [ D i : Q ] R 0 R_0 R 0 Z 2 \mathbb{Z}_2 Z 2 E 0 E_0 E 0
Lemma 11.1. The rings R 0 R_0 R 0 E 0 E_0 E 0 Z 2 \mathbb{Z}_2 Z 2
R 0 [ 1 / 2 ] = ∏ i ( D i ⊗ Q Q 2 ) , rank Z 2 R 0 = ∑ i d i , rank Z 2 E 0 = 4 ∑ i d i . R_0[1/2] = \prod_i (D_i \otimes_{\mathbb{Q}} \mathbb{Q}_2), \qquad\operatorname{rank}_{\mathbb{Z}_2} R_0 = \sum_i d_i, \qquad\operatorname{rank}_{\mathbb{Z}_2} E_0 = 4\sum_i d_i. R 0 [ 1/2 ] = i ∏ ( D i ⊗ Q Q 2 ) , rank Z 2 R 0 = i ∑ d i , rank Z 2 E 0 = 4 i ∑ d i . The order E 0 E_0 E 0 Z 2 \mathbb{Z}_2 Z 2 R 0 R_0 R 0 E 0 / 2 E 0 E_0/2E_0 E 0 /2 E 0 F 2 \mathbb{F}_2 F 2
Proof. All the chosen matrices preserve integral lattices, so their span is contained in a finite free Z 2 \mathbb{Z}_2 Z 2
T ( g ) 1 = g + det ( g ) g − 1 . T(g)1 = g + \det(g)g^{-1}. T ( g ) 1 = g + det ( g ) g − 1 . Since det ( g ) ∈ Z 2 × \det(g) \in\mathbb{Z}_2^\times det ( g ) ∈ Z 2 × R 0 R_0 R 0
det ( x + y ) = det ( x ) + det ( y ) + T ( x ) T ( y ) − T ( x y ) , \det(x+y) = \det(x) + \det(y) + T(x)T(y) - T(xy), det ( x + y ) = det ( x ) + det ( y ) + T ( x ) T ( y ) − T ( x y ) , first for linear combinations of group elements and then by continuity.
Good Frobenius traces generate each D i D_i D i ∏ i D i \prod_i D_i ∏ i D i M 2 M_2 M 2
In characteristic two a scalar residual semisimplification with cyclotomic determinant is trivial. Each actual residual image is unipotent, and the principal congruence kernels are pro-two. The simultaneous image is therefore pro-two. Its finite image in E 0 / 2 E 0 E_0/2E_0 E 0 /2 E 0 F 2 \mathbb{F}_2 F 2 F 2 \mathbb{F}_2 F 2 □ \square □
The next finite-flat assertion is restricted to a category in which the order-two model is unique. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 F F F
Lemma 11.2. Among biconnected finite flat groups killed by two whose generic Jordan–Hölder factors over F 2 \mathbb{F}_2 F 2 u u u
Proof. Schematic closure and finite-flat quotient preserve biconnectedness. Indeed a closed subgroup of a connected finite group over the perfect residue field has local coordinate algebra and is connected; connectedness of quotients follows from surjectivity. Cartier duality gives the other two assertions.
For a group of order two, the height-one exponent is j j j 0 ≤ j ≤ 2 0 \le j \le2 0 ≤ j ≤ 2 j = 0 , 2 j=0,2 j = 0 , 2 j = 1 j=1 j = 1 [42 Theorem 4.1, Proposition 4.2 and Corollaries 4.3–4.4] .
A generic isomorphism between two objects extends as an isomorphism by induction on their common order. Close its graph in the product. For each projection, choose a generic subgroup of order two and close it in source and target. The restriction of the projection is an isomorphism by uniqueness of the middle model. Its induced map on the finite-flat quotients is an isomorphism by induction, hence the projection is an isomorphism. For an arbitrary generic morphism, apply this to the first projection of its graph closure. The second projection then extends the morphism. Flatness gives uniqueness. Taking schematic closures and finite-flat quotients gives the asserted exact sequences.
Let k 0 k_0 k 0 [42 Corollary 4.3 and Proposition 5.7.3] represents a self-extension of the simple model by
Φ = ( u h ( u ) 0 u ) , h ∈ k 0 [ [ u ] ] . \Phi= \begin{pmatrix} u & h(u) \\ 0 & u \end{pmatrix}, \qquad h \in k_0[[u]]. Φ = ( u 0 h ( u ) u ) , h ∈ k 0 [[ u ]] . Changing the upper lift replaces h h h h + u ( φ ( f ) − f ) h+u(\varphi(f)-f) h + u ( φ ( f ) − f ) φ − 1 \varphi-1 φ − 1 u k 0 [ [ u ] ] uk_0[[u]] u k 0 [[ u ]] h = h 0 + h 1 u h=h_0+h_1u h = h 0 + h 1 u u − 1 u^{-1} u − 1 φ \varphi φ h 0 / u + h 1 h_0/u+h_1 h 0 / u + h 1
To compare ramification, choose a uniformizer π \pi π F F F F n = F ( π 1 / 2 n ) F_n=F(\pi^{1/2^n}) F n = F ( π 1/ 2 n ) F n / F n − 1 F_n/F_{n-1} F n / F n − 1 2 e ( F n − 1 ) + 1 2e(F_{n-1})+1 2 e ( F n − 1 ) + 1 4 ⋅ 2 n − 1 4\cdot2^{n-1} 4 ⋅ 2 n − 1 [31 Theorem A] ; his numbering is shifted by one from classical upper numbering. Herbrand’s theorem therefore preserves this character’s break at each transition. Wintenberger’s comparison [75 Proposition 3.3.2 and Corollary 3.3.6] identifies it with the field-of-norms break. Its conductor is consequently at most two. The argument is unchanged after finite unramified extension. □
Lemma 11.3. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 V V V { 0 , 1 } \{0,1\} { 0 , 1 } χ 2 \chi_2 χ 2 Z 2 \mathbb{Z}_2 Z 2 V V V E 0 E_0 E 0
Proof. The crystalline lattice classification gives a Barsotti–Tate group for each constituent. Every stable lattice in their sum is realized by an isogenous group, by schematic closure of the finite generic isogeny kernel. Connectedness and dual connectedness are invariant under isogeny.
The connected–étale decomposition is stable under coefficient actions. A nonzero rational étale factor in a rank-two constituent has coefficient rank one and gives an unramified quotient; the determinant and weights then give the ordinary filtration. Nonordinarity excludes this. Applying the same argument to r ∨ ( 1 ) ≃ r r^\vee(1)\simeq r r ∨ ( 1 ) ≃ r E 0 E_0 E 0
Here the trace and determinant are those of the original integral order. A subsequent nonfaithful matrix quotient need not preserve the information they carry. In particular a vanishing matrix in such a quotient does not imply vanishing of its original trace.
Lemma 11.4. Suppose every dyadic coefficient constituent is crystalline of labelled weights { 0 , 1 } \{0,1\} { 0 , 1 } χ 2 \chi_2 χ 2 m = 2 m=2 m = 2 6 6 6 F m F_m F m E 0 E_0 E 0
Its Breuil–Kisin module is free of rank one as a right E 0 [ [ u ] ] E_0[[u]] E 0 [[ u ]] Φ ( u ) \Phi(u) Φ ( u )
D ( Φ ( u ) ) ∈ ( u 2 − m ) R 0 [ [ u ] ] , T ( Φ ( 0 ) ) = a 2 ∈ R 0 . \mathcal{D}(\Phi(u)) \in(u^2-m)R_0[[u]], \qquad T(\Phi(0))=a_2\in R_0. D ( Φ ( u )) ∈ ( u 2 − m ) R 0 [[ u ]] , T ( Φ ( 0 )) = a 2 ∈ R 0 . The point field of E 0 / 2 E 0 E_0/2E_0 E 0 /2 E 0 ( F m ) 2 (F_m)_2 ( F m ) 2
Proof. The regular lattice is Barsotti–Tate and biconnected by Lemma 11.3. Lemma 11.2 gives exactness and full faithfulness for its torsion subquotients with trivial simple generic factors. We use the covariant Breuil–Kisin functor M c o v ( G ) = M c o n t r a ( G D ) \mathcal{M}_{\mathrm{cov}}(G)=\mathcal{M}_{\mathrm{contra}}(G^D) M cov ( G ) = M contra ( G D ) [42 Corollary 4.3 and Proposition 5.7.3] , so it preserves the right action and the quotient used below.
The commuting right action of E 0 E_0 E 0 J 0 = rad ( E 0 ) J_0=\operatorname{rad}(E_0) J 0 = rad ( E 0 ) M \mathcal{M} M J 0 / 2 E 0 J_0/2E_0 J 0 /2 E 0 E 0 / J 0 = F 2 E_0/J_0=F_2 E 0 / J 0 = F 2 M / M J 0 = F 2 [ [ u ] ] \mathcal{M}/\mathcal{M}J_0=F_2[[u]] M / M J 0 = F 2 [[ u ]] u u u E 0 [ [ u ] ] → M E_0[[u]]\to\mathcal{M} E 0 [[ u ]] → M Z 2 [ [ u ] ] \mathbb{Z}_2[[u]] Z 2 [[ u ]]
Height one supplies Ψ \Psi Ψ Φ Ψ = Ψ Φ = u 2 − m \Phi\Psi=\Psi\Phi=u^2-m ΦΨ = ΨΦ = u 2 − m [43 Lemmas 2.3.2 and 2.3.4] gives divisibility of its determinant by u 2 − m u^2-m u 2 − m R 0 [ [ u ] ] R_0[[u]] R 0 [[ u ]] u = 0 u=0 u = 0 V ( − 1 ) = V ∨ V(-1)=V^\vee V ( − 1 ) = V ∨ V V V 2 2 2 a 2 a_2 a 2 V ∨ ≃ V ( − 1 ) V^\vee\simeq V(-1) V ∨ ≃ V ( − 1 ) 2 e ( ( F m ) 2 ) − 1 = 3 2e((F_m)_2)-1=3 2 e (( F m ) 2 ) − 1 = 3
Lemma 11.5. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 M / F M/F M / F v 2 ( 2 ) = 1 v_2(2)=1 v 2 ( 2 ) = 1 { 0 , 1 } \{0,1\} { 0 , 1 }
Proof. Fontaine’s bound [31 Theorem A] , converted to classical upper numbering, is 2 e ( F ) − 1 = 3 2e(F)-1=3 2 e ( F ) − 1 = 3 e ( F / Q 2 ) = 2 e(F/\mathbb{Q}_2)=2 e ( F / Q 2 ) = 2
Rank-one selfextensions at the endpoint exponents
Lemma 11.6. Let F / Q p F/\mathbb{Q}_p F / Q p e e e f f f k k k i ∈ Z / f Z i\in\mathbb{Z}/f\mathbb{Z} i ∈ Z / f Z b i u r i b_i u^{r_i} b i u r i b i ∈ k × b_i\in k^\times b i ∈ k × r i ∈ { 0 , e } r_i\in\{0,e\} r i ∈ { 0 , e } k k k u u u u p u^p u p k [ [ u ] ] k[[u]] k [[ u ]]
A i = b i u r i ( 1 a i 0 1 ) , a i ∈ k , A_i=b_i u^{r_i}
\begin{pmatrix}
1 & a_i\\
0 & 1
\end{pmatrix},
\qquad a_i\in k, A i = b i u r i ( 1 0 a i 1 ) , a i ∈ k , after changing the lifts of the quotient generators. The extension classes form a one-dimensional k k k
Proof. Before changing lifts, write A i = ( b i u r i h i 0 b i u r i ) A_i=\begin{pmatrix}b_i u^{r_i}&h_i\\0&b_i u^{r_i}\end{pmatrix} A i = ( b i u r i 0 h i b i u r i ) u e A i − 1 u^eA_i^{-1} u e A i − 1 u e ∣ h i u^e\mid h_i u e ∣ h i r i = e r_i=e r i = e r i = 0 r_i=0 r i = 0 h i = b i u r i f i ( u ) h_i=b_i u^{r_i}f_i(u) h i = b i u r i f i ( u ) f i ∈ k [ [ u ] ] f_i\in k[[u]] f i ∈ k [[ u ]] μ i \mu_i μ i f i f_i f i φ ( μ i − 1 ) − μ i \varphi(\mu_{i-1})-\mu_i φ ( μ i − 1 ) − μ i
On u k [ [ u ] ] uk[[u]] u k [[ u ]] 1 − φ f 1-\varphi^f 1 − φ f ∑ n ≥ 0 φ f n \sum_{n\geq0}\varphi^{fn} ∑ n ≥ 0 φ f n u u u ( μ i − 1 − μ i ) i (\mu_{i-1}-\mu_i)_i ( μ i − 1 − μ i ) i k k k ∑ i a i \sum_i a_i ∑ i a i
The restriction r i ∈ { 0 , e } r_i\in\{0,e\} r i ∈ { 0 , e } h i h_i h i
Ramification of nonordinary torsion
We first prove a local statement independent of K K K
Theorem 11.7. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 r F : G F → G L 2 ( E ) r_F:G_F\to\mathrm{GL}_2(E) r F : G F → GL 2 ( E ) { 0 , 1 } \{0,1\} { 0 , 1 } χ 2 \chi_2 χ 2 v 2 ( 2 ) = 1 v_2(2)=1 v 2 ( 2 ) = 1 S 3 S_3 S 3
Proof. As in Lemma 11.3, the lattice is realized by a biconnected Barsotti–Tate group G G G G [ ϖ E ] G[\varpi_E] G [ ϖ E ]
After enlarging the residue coefficient field, suppose first that the reduction is reducible. Its diagonal characters are unramified, since O F × \mathcal{O}_F^\times O F × 1 + c ( g ) E 12 1+c(g)E_{12} 1 + c ( g ) E 12 c c c k k k F 2 \mathbb{F}_2 F 2 λ : k → F 2 \lambda:k\to\mathbb{F}_2 λ : k → F 2 F 2 \mathbb{F}_2 F 2 ker λ \ker\lambda ker λ λ ∘ c \lambda\circ c λ ∘ c e e e 2 ( e − 1 ) 2(e-1) 2 ( e − 1 ) 2 e 2e 2 e
If the reduction is locally absolutely irreducible, the local classification in Lemma 4.3(ii) gives full image S 3 S_3 S 3 1 / 3 1/3 1/3
Inert odd coefficient primes
Lemma 11.8. Let ( K , p ) (K,p) ( K , p ) ( F 2 , 5 ) (\mathbb{F}_2,5) ( F 2 , 5 ) ( F 3 , 5 ) (\mathbb{F}_3,5) ( F 3 , 5 ) ( F 3 , 7 ) (\mathbb{F}_3,7) ( F 3 , 7 ) r r r p p p ψ , ψ ψ − 1 \psi,\psi\psi^{-1} ψ , ψ ψ − 1 ψ \psi ψ p p p r r r
Proof. The completion K p K_p K p p p p a a a ψ \psi ψ p 2 − 1 p^2-1 p 2 − 1 ψ 2 \psi^2 ψ 2
Table 4.
For the first order, ( 1 + 2 ) 3 = 2 (1+\sqrt{2})^3=2 ( 1 + 2 ) 3 = 2 ( 1 + 2 ) 6 = − 1 (1+\sqrt{2})^6=-1 ( 1 + 2 ) 6 = − 1 ϵ 3 2 + ϵ 3 + 1 = 0 \epsilon_3^2+\epsilon_3+1=0 ϵ 3 2 + ϵ 3 + 1 = 0 ϵ 3 4 = − 1 \epsilon_3^4=-1 ϵ 3 4 = − 1 ϵ 3 8 = 1 \epsilon_3^8=1 ϵ 3 8 = 1 a = 0 , p + 1 a=0,p+1 a = 0 , p + 1 [53 ] .
This implies actual ordinarity. Otherwise the Barsotti–Tate group of a stable coefficient lattice and its dual would be connected: a coefficient-stable etale factor would give the ordinary coefficient-rank-one filtration. Schematic closure and finite-flat quotient, after killing unramified character factors, produce a rank-one finite-flat subquotient with one of the two endpoint generic characters. At absolute ramification index one, finite-flat uniqueness [53 Theorem 3.3.3 and Corollary 3.3.6] identifies its model with the etale or multiplicative endpoint, contradicting biconnectedness. The unique ordinary filtration descends through the unramified extension. Apply [26 Corollary 1.2.2] to r ∨ r^\vee r ∨ K K K p p p r r r
Lemma 11.9. Let p ≥ 5 p \ge5 p ≥ 5 K K K [ K ( ζ p ) : K ] = p − 1 [K(\zeta_p):K]=p-1 [ K ( ζ p ) : K ] = p − 1 p p p K ( ζ p ) K(\zeta_p) K ( ζ p )
Proof. Otherwise Clifford theory induces it from the unique quadratic subfield of K ( ζ p ) / K K(\zeta_p)/K K ( ζ p ) / K p p p p p p 0 , 1 , p , p + 1 0,1,p,p+1 0 , 1 , p , p + 1 p 2 − 1 p^2-1 p 2 − 1 ( 0 , p + 1 ) (0,p+1) ( 0 , p + 1 ) ( 1 , p ) (1,p) ( 1 , p ) p − 1 p-1 p − 1 p + 1 p+1 p + 1
p + 1 , p 2 + p , p 3 + 1 , p 3 + p 2 ( m o d p 4 − 1 ) . p+1,\quad p^2+p,\quad p^3+1,\quad p^3+p^2 \pmod{p^4-1}. p + 1 , p 2 + p , p 3 + 1 , p 3 + p 2 ( mod p 4 − 1 ) . They are paired by multiplication by p 2 p^2 p 2 ( p 2 + 1 ) / 2 (p^2+1)/2 ( p 2 + 1 ) /2 gcd ( p 2 + 1 , p ± 1 ) = 2 \gcd(p^2+1,p\pm1)=2 g cd( p 2 + 1 , p ± 1 ) = 2 p ≥ 5 p\ge5 p ≥ 5 [53 ] , so the induced case is impossible.
For either field at five, the cyclotomic degree is four; for F 3 F_3 F 3 [3 Theorem 2.5] , or the projective witness specified below, can therefore be lifted to the actual source by [3 Theorem 2.3] using Lemma 11.9. The prime three is treated separately, since its order p − 1 = 2 p-1=2 p − 1 = 2
A common cubic coefficient-field screen
Lemma 11.10. Let D D D d ≤ 3 d\le3 d ≤ 3 d = 3 d=3 d = 3 D = Q ( α ) D=\mathbb{Q}(\alpha) D = Q ( α ) α \alpha α α \alpha α [ − 8 , 8 ] [-\sqrt{8},\sqrt{8}] [ − 8 , 8 ] X 3 X^3 X 3 D D D F 3 F_3 F 3 F 9 F_9 F 9 F 5 F_5 F 5
Proof. For d ≤ 2 d \le2 d ≤ 2 f ( z ) = z 3 + A z 2 + B z + C f(z) = z^3 + Az^2 + Bz + C f ( z ) = z 3 + A z 2 + B z + C ∣ A ∣ ≤ 8 |A| \le8 ∣ A ∣ ≤ 8 − 8 ≤ B ≤ 24 -8 \le B \le24 − 8 ≤ B ≤ 24 ∣ C ∣ ≤ 22 |C| \le22 ∣ C ∣ ≤ 22 B B B [ − 8 , 8 ] 3 [-\sqrt{8}, \sqrt{8}]^3 [ − 8 , 8 ] 3 − f ( − z ) -f(-z) − f ( − z )
Table 1 (PDF p. 31).
The enumeration discards C = 0 C=0 C = 0 Q ( 8 ) \mathbf{Q}(\sqrt{8}) Q ( 8 ) F 3 \mathbf{F}_3 F 3 − f ( − z ) -f(-z) − f ( − z ) [49 common] .
Symmetry and quadratic descent
The integral-order bounds can force either a small coefficient field or an isomorphism with a quadratic conjugate. We record the two rigidity mechanisms used below. The prime-to-characteristic cotangent argument is distinct from the dyadic order argument.
Finite-order actions on complete local rings
Lemma 12.1 (Cotangent rigidity). Let O \mathcal{O} O A A A O \mathcal{O} O Γ \Gamma Γ ∣ Γ ∣ ∈ O × |\Gamma| \in\mathcal{O}^{\times} ∣Γ∣ ∈ O × O \mathcal{O} O A A A m A / ( m A 2 + m O A ) \mathfrak{m}_A/(\mathfrak{m}_A^2+\mathfrak{m}_{\mathcal{O}}A) m A / ( m A 2 + m O A ) A A A
Proof. Average lifts x 1 , … , x n x_1,\ldots,x_n x 1 , … , x n Γ \Gamma Γ O [ [ X 1 , … , X n ] ] \mathcal{O}[[X_1,\ldots,X_n]] O [[ X 1 , … , X n ]] A A A
The invertibility hypothesis is essential: T ↦ − T T \mapsto-T T ↦ − T Z 2 [ [ T ] ] \mathbf{Z}_2[[T]] Z 2 [[ T ]]
Lemma 12.2. Let k k k V = k s ⊕ k d V=ks\oplus kd V = k s ⊕ k d σ ( s ) = s \sigma(s)=s σ ( s ) = s σ ( d ) = − d \sigma(d)=-d σ ( d ) = − d V ∨ ⊗ k V ∨ V^{\vee}\otimes_k V^{\vee} V ∨ ⊗ k V ∨
ι ( a ⊗ b ) = σ ∨ ( b ) ⊗ σ ∨ ( a ) , ℓ ± = s ∨ ± d ∨ . \iota(a\otimes b)=\sigma^{\vee}(b)\otimes\sigma^{\vee}(a),\qquad\ell_{\pm}=s^{\vee}\pm d^{\vee}. ι ( a ⊗ b ) = σ ∨ ( b ) ⊗ σ ∨ ( a ) , ℓ ± = s ∨ ± d ∨ . If an ι \iota ι ℓ + ⊗ ℓ + \ell_+\otimes\ell_+ ℓ + ⊗ ℓ + ℓ − ⊗ ℓ − \ell_-\otimes\ell_- ℓ − ⊗ ℓ − ι \iota ι
Proof. The ( − 1 ) (-1) ( − 1 ) ι \iota ι s ∨ ⊗ d ∨ + d ∨ ⊗ s ∨ s^{\vee}\otimes d^{\vee}+d^{\vee}\otimes s^{\vee} s ∨ ⊗ d ∨ + d ∨ ⊗ s ∨
ℓ + ⊗ ℓ + − ℓ − ⊗ ℓ − = 2 ( s ∨ ⊗ d ∨ + d ∨ ⊗ s ∨ ) . \ell_+\otimes\ell_+-\ell_-\otimes\ell_-=2(s^{\vee}\otimes d^{\vee}+d^{\vee}\otimes s^{\vee}). ℓ + ⊗ ℓ + − ℓ − ⊗ ℓ − = 2 ( s ∨ ⊗ d ∨ + d ∨ ⊗ s ∨ ) . Since two is invertible, the two eigenspaces split and the quotient has only the eigenvalue one.
Descent from a twisted conjugacy
Lemma 12.3 (Quadratic descent). Let K / Q K/\mathbb{Q} K / Q σ \sigma σ R = ( r λ ) \mathcal{R}=(r_\lambda) R = ( r λ ) K K K
r λ σ ≃ r λ ⊗ η , r_\lambda^\sigma\simeq r_\lambda\otimes\eta, r λ σ ≃ r λ ⊗ η , where η \eta η G Q G_{\mathbb{Q}} G Q R \mathcal{R} R
Proof. The identity of good Frobenius traces transfers the displayed isomorphism to a companion r r r p ≥ 5 p \ge5 p ≥ 5 H = G K H=G_K H = G K t ∈ G Q ∖ H t\in G_{\mathbb{Q}}\setminus H t ∈ G Q ∖ H A A A
A r ( h ) A − 1 = η ( h ) r ( t h t − 1 ) ( h ∈ H ) . Ar(h)A^{-1}=\eta(h)r(tht^{-1}) \qquad(h\in H). A r ( h ) A − 1 = η ( h ) r ( t h t − 1 ) ( h ∈ H ) . Since η σ η = 1 \eta^\sigma\eta=1 η σ η = 1 A 2 r ( t 2 ) − 1 A^2r(t^2)^{-1} A 2 r ( t 2 ) − 1 [ r ] [r] [ r ] [ A ] [A] [ A ] G Q G_{\mathbb{Q}} G Q
After a finite coefficient extension, scale A A A det A = χ p ( t ) \det A=\chi_p(t) det A = χ p ( t ) L ( h ) = r ( h ) L(h)=r(h) L ( h ) = r ( h ) L ( h t ) = r ( h ) A L(ht)=r(h)A L ( h t ) = r ( h ) A h ∈ H h\in H h ∈ H det L ( g ) = χ p ( g ) \det L(g)=\chi_p(g) det L ( g ) = χ p ( g ) g g g
c ( g 1 , g 2 ) = L ( g 1 ) L ( g 2 ) L ( g 1 g 2 ) − 1 c(g_1,g_2)=L(g_1)L(g_2)L(g_1g_2)^{-1} c ( g 1 , g 2 ) = L ( g 1 ) L ( g 2 ) L ( g 1 g 2 ) − 1 takes values in { ± 1 } \{\pm1\} { ± 1 } H × H H\times H H × H [54 Theorem 6.3] , for the discrete trivial G Q G_{\mathbb{Q}} G Q Q ‾ × \overline{\mathbb{Q}}^{\times} Q × b : G Q → Q ‾ × b:G_{\mathbb{Q}}\to\overline{\mathbb{Q}}^{\times} b : G Q → Q × c = δ b c=\delta b c = δ b b b b Q ‾ ↪ Q ‾ p \overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}}_p Q ↪ Q p r ~ = b − 1 L \widetilde r=b^{-1}L r = b − 1 L G Q G_{\mathbb{Q}} G Q c ∣ H × H = 1 c|_{H\times H}=1 c ∣ H × H = 1 b ∣ H b|_H b ∣ H c 2 = 1 c^2=1 c 2 = 1 b 2 b^2 b 2 G Q G_{\mathbb{Q}} G Q
The representation r ~ \widetilde r r H H H r r r b b b K / Q K/\mathbb{Q} K / Q r r r H H H r ~ \widetilde r r r ~ \widetilde r r p p p H H H b ∣ H b|_H b ∣ H r ~ \widetilde r r [50 Theorem 1.0.4] . Cyclic base change to K K K b ∣ H b|_H b ∣ H r r r □ \square □
Coefficient orbits and the integral-order bound
Proof of Theorem 1.4. The two operations commute and are involutions on isomorphism classes. Let V = C 2 V=\mathbb{C}^2 V = C 2 m ∈ { 1 , 2 , 4 } m\in\{1,2,4\} m ∈ { 1 , 2 , 4 } e = [ D : Q ] e=[D:\mathbb{Q}] e = [ D : Q ] 4 m e = dim F 2 ( E 0 / 2 E 0 ) < 32 4me=\dim_{\mathbb{F}_2}(E_0/2E_0)<32 4 m e = dim F 2 ( E 0 /2 E 0 ) < 32
The stabilizer V 0 V_0 V 0 R \mathcal{R} R 4 / m 4/m 4/ m D D D D D D V 0 → Aut ( D ) V_0\to\operatorname{Aut}(D) V 0 → Aut ( D ) H H H G G G G H GH G H V 0 V_0 V 0 Aut ( D ) \operatorname{Aut}(D) Aut ( D ) 4 / m ∣ e 4/m \mid e 4/ m ∣ e 4 ∣ m e 4 \mid me 4 ∣ m e 0 < m e < 8 0 < me < 8 0 < m e < 8 m e = 4 me = 4 m e = 4
( m , e ) = ( 4 , 1 ) , ( 2 , 2 ) , ( 1 , 4 ) . (m,e)=(4,1),(2,2),(1,4). ( m , e ) = ( 4 , 1 ) , ( 2 , 2 ) , ( 1 , 4 ) . In the last case D / Q D/\mathbb{Q} D / Q C 2 2 C_2^2 C 2 2 Lemma 12.3 . □ \square □
Corollary 12.4. (The two-orbit alternative). Let R \mathcal{R} R D D D h h h h h h R \mathcal{R} R R ⊗ h \mathcal{R} \otimes h R ⊗ h D = Q D = \mathbb{Q} D = Q D D D h h h
Proof. Write m ∈ { 1 , 2 } m \in\{1,2\} m ∈ { 1 , 2 } e = [ D : Q ] e = [D : \mathbb{Q}] e = [ D : Q ] m e < 4 me < 4 m e < 4 m = 2 m = 2 m = 2 e = 1 e = 1 e = 1 m = 1 m = 1 m = 1 C 2 C_2 C 2 D D D 2 ∣ e 2 \mid e 2 ∣ e e < 4 e < 4 e < 4 e = 2 e = 2 e = 2 □ \square □
The bounds concern the actual simultaneous order, not one residual realization. The small-residue-degree alternative must still be combined with residual modularity and a lifting theorem for the actual source.
II Terminal source theorems
Five ramified-dyadic reductions
Write F d = Q ( d ) F_d = \mathbb{Q}(\sqrt{d}) F d = Q ( d )
Table 5.
In every case ℓ \ell ℓ F d F_d F d F 2 \mathbb{F}_2 F 2 8 / 2 < 2 \sqrt{8}/2 < 2 8 /2 < 2 12 / 2 < 2 \sqrt{12}/2 < 2 12 /2 < 2 F 2 , F 3 F_2,F_3 F 2 , F 3
Theorem 13.1. Assume GRH for Dedekind zeta functions. For d ∈ { 2 , 3 , 6 , 10 , 11 } d \in\{2,3,6,10,11\} d ∈ { 2 , 3 , 6 , 10 , 11 } G F d → GL 2 ( F p ) G_{F_d} \to\operatorname{GL}_2(\mathbb{F}_p) G F d → GL 2 ( F p ) p p p
The field-dependent arguments below prove the supported residual bases. We first give the reduction which is common to all five fields.
The required local choices are Lemma 4.1 at odd removal characteristics, Lemma 4.3 at the ramified dyadic completion, and Theorem 4.7 at the split terminal prime. Each retains the actual residual representation and has exact cyclotomic determinant.
Reduction to the supported residual base
Proposition 13.2. Let ( K , ℓ ) (K,\ell) ( K , ℓ )
every absolutely irreducible totally odd residual representation of determinant χ ‾ ℓ \overline{\chi}_{\ell} χ ℓ ℓ \ell ℓ
for every totally odd reducible representation of determinant χ ‾ ℓ \overline{\chi}_{\ell} χ ℓ ℓ \ell ℓ G Q G_{\mathbb{Q}} G Q
Then every continuous absolutely irreducible totally odd representation G K → GL 2 ( F ‾ p ) G_K \to\operatorname{GL}_2(\overline{\mathbb{F}}_p) G K → GL 2 ( F p ) p p p
Proof. The solvable-image case has an odd solvable Artin witness and is modular after weight raising. Otherwise the cyclotomic restriction is irreducible. The initial weight-two lift of [3 Theorem 2.2] and Lemma 2.4 supply a potentially modular compatible system, with cyclotomic determinant times a finite totally even character. Its nonsolvable residue makes it nondhedral. Apply Theorem 1.2 and reduce at the original coefficient place. □
Lemma 13.3. Let ( K , ℓ ) = ( F 2 , 7 ) (K,\ell)=(F_2,7) ( K , ℓ ) = ( F 2 , 7 ) ( F 3 , 13 ) (F_3,13) ( F 3 , 13 ) r ‾ : G K → GL 2 ( F ‾ ℓ ) \overline{r}:G_K\to\operatorname{GL}_2(\overline{\mathbb{F}}_{\ell}) r : G K → GL 2 ( F ℓ ) ω ℓ \omega_{\ell} ω ℓ ℓ \ell ℓ ω ℓ e \omega_{\ell}^{e} ω ℓ e e e e G Q G_{\mathbb{Q}} G Q ℓ \ell ℓ
Proof. Write the constituents as ψ , ω ℓ ψ − 1 \psi,\omega_{\ell}\psi^{-1} ψ , ω ℓ ψ − 1 ψ \psi ψ ℓ \ell ℓ ℓ \ell ℓ a , b a,b a , b ℓ − 1 \ell-1 ℓ − 1 u , v ∈ Z / 2 u,v\in\mathbb{Z}/2 u , v ∈ Z /2
For K = Q ( 2 ) K=\mathbb{Q}(\sqrt{2}) K = Q ( 2 ) 1 + 2 1+\sqrt{2} 1 + 2 ( + , − ) (+,-) ( + , − ) 4 = 3 4 4=3^4 4 = 3 4 5 = 3 5 5=3^5 5 = 3 5
a + b + u + v ≡ 0 ( m o d 2 ) , 4 a + 5 b + 3 v ≡ 0 ( m o d 6 ) . a+b+u+v\equiv0\pmod{2},\qquad4a+5b+3v\equiv0\pmod{6}. a + b + u + v ≡ 0 ( mod 2 ) , 4 a + 5 b + 3 v ≡ 0 ( mod 6 ) . The second relation modulo three gives a ≡ b ( m o d 3 ) a\equiv b\pmod{3} a ≡ b ( mod 3 ) 2 a ≡ 2 b ( m o d 6 ) 2a\equiv2b\pmod{6} 2 a ≡ 2 b ( mod 6 ) K = F 3 K=F_3 K = F 3 2 + 3 2+\sqrt{3} 2 + 3 6 = 2 5 6=2^5 6 = 2 5 11 = 2 7 11=2^7 11 = 2 7 5 a + 7 b ≡ 0 ( m o d 12 ) 5a+7b\equiv0\pmod{12} 5 a + 7 b ≡ 0 ( mod 12 ) a ≡ b ( m o d 12 ) a\equiv b\pmod{12} a ≡ b ( mod 12 ) ψ 2 ω ℓ − 2 a \psi^2\omega_{\ell}^{-2a} ψ 2 ω ℓ − 2 a ω ℓ 1 − 2 a \omega_{\ell}^{1-2a} ω ℓ 1 − 2 a
Nonsolvable dyadic sources
The following ray auxiliary is used only over F 3 F_3 F 3 F 2 F_2 F 2
Proposition 14.1. Let K = Q ( 3 ) K=\mathbb{Q}(\sqrt{3}) K = Q ( 3 ) A / K A/K A / K [ A : Q ] = 6144 [A:\mathbb{Q}]=6144 [ A : Q ] = 6144
A / K A/K A / K
its dyadic classical upper ramification groups are trivial above three;
at both primes over thirteen its completion is the full tame biquadratic extension of Q 13 \mathbb{Q}_{13} Q 13 ( e , f ) = ( 2 , 2 ) (e,f)=(2,2) ( e , f ) = ( 2 , 2 )
In particular, if M / K M/K M / K L = A M L = AM L = A M
rd ( L ) < 8 39 , 2 N 169 ( L ) [ L : Q ] = 1 2 . \operatorname{rd}(L) < 8\sqrt{39}, \qquad\frac{2N_{169}(L)}{[L:\mathbb{Q}]} = \frac{1}{2}. rd ( L ) < 8 39 , [ L : Q ] 2 N 169 ( L ) = 2 1 . Here N 169 ( L ) N_{169}(L) N 169 ( L )
Proof. Put
ϵ = 2 + 3 , π = 4 + 3 , π ′ = 4 − 3 , \epsilon= 2 + \sqrt{3}, \qquad\pi= 4 + \sqrt{3}, \qquad\pi' = 4 - \sqrt{3}, ϵ = 2 + 3 , π = 4 + 3 , π ′ = 4 − 3 , and let
B = K ( − 1 , ϵ , π , π ′ ) . B = K(\sqrt{-1},\sqrt{\epsilon},\sqrt{\pi},\sqrt{\pi'}). B = K ( − 1 , ϵ , π , π ′ ) . The four squareclasses are independent: the valuations at the two thirteen-primes separate π , π ′ \pi,\pi' π , π ′ − 1 -1 − 1 ϵ \epsilon ϵ [ B : K ] = 16 [B:K] = 16 [ B : K ] = 16 B / K B/K B / K ϵ = ( 1 + 3 ) 2 / 2 \epsilon= (1+\sqrt{3})^2/2 ϵ = ( 1 + 3 ) 2 /2 π π ′ = 13 \pi\pi' = 13 π π ′ = 13
There are two primes of B B B ( e , f ) = ( 8 , 2 ) (e,f) = (8,2) ( e , f ) = ( 8 , 2 ) ( e , f ) = ( 2 , 2 ) (e,f) = (2,2) ( e , f ) = ( 2 , 2 ) Q 13 \mathbb{Q}_{13} Q 13 ϵ \epsilon ϵ π , π ′ \pi,\pi' π , π ′
We record the dyadic filtration explicitly. In the sixteen-element squareclass group generated above, the counts of relative quadratic conductor exponents 0 , 2 , 4 0,2,4 0 , 2 , 4 K 2 K_2 K 2 4 , 4 , 8 4,4,8 4 , 4 , 8 2 , 2 , 4 2,2,4 2 , 2 , 4 0 < u ≤ 1 0 < u \le1 0 < u ≤ 1 1 < u ≤ 3 1 < u \le3 1 < u ≤ 3
ψ B w / K 2 ( 3 ) = 1 + 2 ( 3 − 1 ) = 5. \psi_{B_w/K_2}(3) = 1 + 2(3 - 1) = 5. ψ B w / K 2 ( 3 ) = 1 + 2 ( 3 − 1 ) = 5. These conductor counts follow directly by taking the dyadic valuation of the relative discriminant of X 2 − a X^2-a X 2 − a a a a [49 ramified/d2] .
Let P 1 , P 2 \mathfrak{P}_1,\mathfrak{P}_2 P 1 , P 2 B B B m = P 1 6 P 2 6 \mathfrak{m} = \mathfrak{P}_1^6\mathfrak{P}_2^6 m = P 1 6 P 2 6 B B B m \mathfrak{m} m A A A B B B B B B Gal ( B / K ) \operatorname{Gal}(B/K) Gal ( B / K ) A / K A/K A / K B B B
The ray exponent six implies that the classical upper breaks of A / B A/B A / B A / K A/K A / K
Here are the exact arithmetic data defining the extension. An absolute equation for B B B i i i 2 \sqrt{2} 2 13 \sqrt{13} 13 X 4 − 8 X 2 + 13 X^4-8X^2+13 X 4 − 8 X 2 + 13
2 72 3 16 13 16 , 2^{72}3^{16}13^{16}, 2 72 3 16 1 3 16 , and class group C 12 C_{12} C 12 m \mathfrak{m} m
C 24 × C 8 × C 2 4 . C_{24}\times C_8\times C_2^4. C 24 × C 8 × C 2 4 . Quotienting by the classes of all eight thirteen-primes gives
Gal ( A / B ) ≃ C 12 × C 2 4 . \operatorname{Gal}(A/B) \simeq C_{12} \times C_{2}^{4}. Gal ( A / B ) ≃ C 12 × C 2 4 . Its order is 192, so [ A : Q ] = 32 ⋅ 192 = 6144 [A:\mathbb{Q}] = 32 \cdot192 = 6144 [ A : Q ] = 32 ⋅ 192 = 6144 verify_d3_aux6144.gp constructs the degree-32 polynomial and computes these ray-class data under GRH. The exact calculation is supplied in [49 ramified/d3/aux32] .
For the last assertion, upper-break bounds are preserved under compositum. At two, the relative different contribution over K 2 K_2 K 2 v 2 ( 2 ) = 1 v_2(2) = 1 v 2 ( 2 ) = 1 ( 1 + 3 ) / 2 = 2 (1 + 3)/2 = 2 ( 1 + 3 ) /2 = 2 K K K 1 / 2 1/2 1/2 1 / 2 1/2 1/2 rd ( L ) < 2 3 3 13 = 8 39 \operatorname{rd}(L) < 2^{3}\sqrt{3}\sqrt{13} = 8\sqrt{39} rd ( L ) < 2 3 3 13 = 8 39 Q 13 \mathbb{Q}_{13} Q 13 N 169 ( L ) = [ L : Q ] / 4 N_{169}(L) = [L:\mathbb{Q}]/4 N 169 ( L ) = [ L : Q ] /4
Proposition 14.2. Assume GRH. Let ( K , ℓ ) (K,\ell) ( K , ℓ ) ( F 2 , 7 ) (F_2,7) ( F 2 , 7 ) ( F 3 , 13 ) (F_3,13) ( F 3 , 13 ) R \mathcal{R} R ℓ \ell ℓ F 2 F_2 F 2 F 3 F_3 F 3
Proof. Let M / K M/K M / K G = SL 2 ( F q ) G = \operatorname{SL}_2(\mathbb{F}_q) G = SL 2 ( F q ) q = 2 f ≥ 4 q = 2^f \ge4 q = 2 f ≥ 4 K 2 K_2 K 2 e ( K 2 ) = 2 e(K_2) = 2 e ( K 2 ) = 2 3 / 2 3/2 3/2 F 2 F_2 F 2 1 1 1 F 3 F_3 F 3 7 / 2 7/2 7/2 3 3 3
Table 14.
For F 2 F_2 F 2 disc ( K ( i ) ) = 8 ( − 4 ) ( − 8 ) = 256 \operatorname{disc}(K(i)) = 8(-4)(-8) = 256 disc ( K ( i )) = 8 ( − 4 ) ( − 8 ) = 256 256 / 8 2 = 4 256/8^2 = 4 256/ 8 2 = 4
For F 3 F_3 F 3 ± 14 \pm14 ± 14 C 2 2 C_2^2 C 2 2 Q 13 \mathbb{Q}_{13} Q 13 Proposition 14.1 contains precisely that local field, has dyadic upper bound three, and has absolute degree 6144. That proposition proves the displayed discriminant bound and the norm-169 density. Over F 2 F_2 F 2
The group G G G M M M K K K B b , 100 B_{b,100} B b , 100 F b F_b F b
Table 15.
These exceed respectively 8 14 < 29.93325910 8\sqrt{14}<29.93325910 8 14 < 29.93325910 8 39 < 49.95998399 8\sqrt{39}<49.95998399 8 39 < 49.95998399 b b b [49 ramified/d2] .
For K = F 3 K=\mathbb{F}_3 K = F 3 q = 8 q=8 q = 8 σ \sigma σ K / Q K/\mathbb{Q} K / Q M ≠ M σ M\ne M^\sigma M = M σ K K K G × G G\times G G × G A M M σ AMM^\sigma A M M σ 6144 ⋅ 5042 = 1560674304 6144\cdot5042=1560674304 6144 ⋅ 5042 = 1560674304 b = 20 b=20 b = 20 51.79576050 51.79576050 51.79576050 M = M σ M=M^\sigma M = M σ S L 2 ( F 8 ) \mathrm{SL}_2(\mathbb{F}_8) SL 2 ( F 8 ) Gal ( K / Q ) \operatorname{Gal}(K/\mathbb{Q}) Gal ( K / Q ) G G G Gal ( M / Q ) → G \operatorname{Gal}(M/\mathbb{Q})\to G Gal ( M / Q ) → G Gal ( M / K ) \operatorname{Gal}(M/K) Gal ( M / K ) K K K q = 4 q=4 q = 4 [3 Theorem 2.5] supplies the witness.
In every remaining case some residual real conjugation is nontrivial. Otherwise M M M 120 120 120 n = 120 n=120 n = 120 b = 4.8 b=4.8 b = 4.8 J b ( 1 / 2 ) − J b ( 3 / 2 ) + J b ( 5 / 2 ) − J b ( 7 / 2 ) J_b(1/2)-J_b(3/2)+J_b(5/2)-J_b(7/2) J b ( 1/2 ) − J b ( 3/2 ) + J b ( 5/2 ) − J b ( 7/2 ) rd ( M ) > 56.06306031 \operatorname{rd}(M)>56.06306031 rd ( M ) > 56.06306031 [69 Theorem 6.1] to the dual actual crystalline two-adic source and dualize back. The nonsolvable modular residue and checked real-place condition meet its hypotheses. It imposes no Q 2 \mathbb{Q}_2 Q 2 R \mathcal{R} R
Sources over Q(2 \sqrt{2} 2
In this section K = F 2 K=\mathbb{F}_2 K = F 2 s = 2 s=\sqrt{2} s = 2 ℓ = 7 \ell=7 ℓ = 7 ϵ = 1 + s \epsilon=1+s ϵ = 1 + s
Lemma 15.1. Put α ± = 5 ± 4 s \alpha_\pm=5\pm4s α ± = 5 ± 4 s K ( α ± ) / K K(\sqrt{\alpha_\pm})/K K ( α ± ) / K K ( − 7 ) / K K(\sqrt{-7})/K K ( − 7 ) / K
Proof. We have α + = ϵ ( 3 + s ) \alpha_+=\epsilon(3+s) α + = ϵ ( 3 + s ) α − = α + σ \alpha_-=\alpha_+^\sigma α − = α + σ α + α − = − 7 \alpha_+\alpha_-=-7 α + α − = − 7 ( 1 + α ± ) / 2 (1+\sqrt{\alpha_\pm})/2 ( 1 + α ± ) /2 X 2 − X − ( 1 ± s ) X^2-X-(1\pm s) X 2 − X − ( 1 ± s ) X 2 + X + 1 X^2+X+1 X 2 + X + 1 − 7 -7 − 7 Q 2 \mathbb{Q}_2 Q 2
Nonscalar dyadic residues
Lemma 15.2. Let K = Q ( 2 ) K=\mathbb{Q}(\sqrt{2}) K = Q ( 2 ) r ˉ : G K → S L 2 ( F ‾ 2 ) \bar{r}:G_K\to\mathrm{SL}_2(\overline{\mathbb{F}}_2) r ˉ : G K → SL 2 ( F 2 ) r ˉ s s = 1 ⊕ 1 \bar{r}^{\mathrm{ss}}=1\oplus1 r ˉ ss = 1 ⊕ 1
Proof. If the representation is reducible, its diagonal characters have odd order and are unramified at seven. Their inertia at two is trivial because the dyadic residue field is F 2 \mathbb{F}_2 F 2
Otherwise its image is an odd dihedral group, and it is induced from an odd-order character θ \theta θ L / K L/K L / K K K K ⟨ − 1 , ϵ ⟩ \langle-1,\epsilon\rangle ⟨ − 1 , ϵ ⟩ − 1 -1 − 1 ϵ \epsilon ϵ s s s 3 + s 3+s 3 + s 3 − s 3-s 3 − s s 2 = 2 s^2=2 s 2 = 2 K 2 K_2 K 2 2 e ( K 2 ) + 1 = 5 2e(K_2)+1=5 2 e ( K 2 ) + 1 = 5 s s s
− 1 , ϵ , 3 + s , 3 − s . -1,\quad\epsilon,\quad3+s,\quad3-s. − 1 , ϵ , 3 + s , 3 − s . The character θ \theta θ L / K L/K L / K L / K L/K L / K θ \theta θ L L L
That quotient is trivial in all fifteen cases. Here is a complete finite certificate: use the bits of an integer 1 ≤ m ≤ 15 1 \le m \le15 1 ≤ m ≤ 15 A ( y ) A(y) A ( y )
P m ( X ) = Res y ( ( X − y ) 2 − A ( y ) , y 2 − 2 ) . P_m(X)=\operatorname{Res}_y\left((X-y)^2-A(y),y^2-2\right). P m ( X ) = Res y ( ( X − y ) 2 − A ( y ) , y 2 − 2 ) . The accompanying program verify_d2_solvable_two.gp computes the maximal order and certified class group of each quartic field, and its ray group modulo two. It applies bnfcertify in every row and asserts that every ray invariant is a power of two. The possible absolute discriminants and ray invariants are
Table 6.
The modulus ( 2 ) (2) ( 2 ) θ \theta θ [49 ramified/d2] .
Ordinary scalar sources
Proposition 15.3 (Ordinary scalar sources). Let R \mathcal{R} R K = Q ( 2 ) K=\mathbb{Q(\sqrt{2})} K = Q ( 2 ) { 0 , 1 } \{0,1\} { 0 , 1 }
Proof. Write s = 2 s=\sqrt{2} s = 2 ϵ = 1 + s \epsilon=1+s ϵ = 1 + s π = 3 + s \pi=3+s π = 3 + s π ′ = 3 − s \pi'=3-s π ′ = 3 − s δ = − 7 \delta=-7 δ = − 7
V = ⟨ [ − 1 ] , [ ϵ ] , [ π ] , [ π ′ ] ⟩ . V=\langle[-1],[\epsilon],[\pi],[\pi']\rangle. V = ⟨[ − 1 ] , [ ϵ ] , [ π ] , [ π ′ ]⟩ . Localization at K 2 K_2 K 2 ⟨ [ δ ] ⟩ \langle[\delta]\rangle ⟨[ δ ]⟩ 8 O K 8\mathcal{O}_K 8 O K ( 1 , 0 ) , ( 1 , 4 ) , ( 3 , 2 ) , ( 3 , 6 ) (1,0),(1,4),(3,2),(3,6) ( 1 , 0 ) , ( 1 , 4 ) , ( 3 , 2 ) , ( 3 , 6 ) 1 , s 1,s 1 , s 1 , δ 1,\delta 1 , δ V V V π \pi π π ′ \pi' π ′ − 1 -1 − 1 ϵ \epsilon ϵ ( − , − ) (-,-) ( − , − ) ( − , + ) (-,+) ( − , + )
Let r r r E / Q 2 E/\mathbb{Q}_2 E / Q 2 ϖ \varpi ϖ k k k v v v 1 + χ δ E 12 1+\chi_\delta E_{12} 1 + χ δ E 12 δ \delta δ c c c ( ( r ( c ) − 1 ) v , v ) ((r(c)-1)v,v) (( r ( c ) − 1 ) v , v )
r ( c ) = C : = ( − 1 1 0 1 ) , ordinary line e 2 , r ‾ = 1 + χ δ E 12 . r(c)=C:=\begin{pmatrix}-1&1\\0&1\end{pmatrix},\qquad\text{ordinary line }e_2,\qquad\overline{r}=1+\chi_\delta E_{12}. r ( c ) = C := ( − 1 0 1 1 ) , ordinary line e 2 , r = 1 + χ δ E 12 . The normalized residue is defined over F 2 \mathbb{F}_2 F 2
We compare two actual sources with these normalizations and the same exact unramified ordinary quotient ν \nu ν r 1 ≡ r 2 ( m o d ϖ n ) r_1\equiv r_2\pmod{\varpi^n} r 1 ≡ r 2 ( mod ϖ n ) n ≥ 1 n\geq1 n ≥ 1
r 2 ( g ) = ( 1 + ϖ n X ( g ) ) r 1 ( g ) ( m o d ϖ n + 1 ) , X ( g ) = ( x ( g ) y ( g ) z ( g ) x ( g ) ) . r_2(g)=(1+\varpi^nX(g))r_1(g)\pmod{\varpi^{n+1}},\qquad X(g)=\begin{pmatrix}x(g)&y(g)\\z(g)&x(g)\end{pmatrix}. r 2 ( g ) = ( 1 + ϖ n X ( g )) r 1 ( g ) ( mod ϖ n + 1 ) , X ( g ) = ( x ( g ) z ( g ) y ( g ) x ( g ) ) . The determinant gives the trace-zero form. The adjoint action of 1 + χ δ E 12 1+\chi_\delta E_{12} 1 + χ δ E 12 χ δ ( g ) z ( h ) ( 1 + E 12 ) \chi_\delta(g)z(h)(1+E_{12}) χ δ ( g ) z ( h ) ( 1 + E 12 ) z z z x ∣ G K 2 = y ∣ G K 2 = 0 x|_{G_{K_2}}=y|_{G_{K_2}}=0 x ∣ G K 2 = y ∣ G K 2 = 0 z z z ν , χ 2 ν − 1 \nu,\chi_2\nu^{-1} ν , χ 2 ν − 1 ϖ n \varpi^n ϖ n E E E
At a seven-place residual inertia has a unit upper entry, so its special line is uniquely determined. Congruent representations have congruent special lines: take the kernel of a rank-one inertia difference with a unit entry. If the exact special characters agree modulo ϖ n + 1 \varpi^{n+1} ϖ n + 1 t v e 2 t_v e_2 t v e 2
z ∣ G v = 0 , x ∣ G v = t v χ δ ∣ G v . z|_{G_v}=0,\qquad x|_{G_v}=t_v\chi_\delta|_{G_v}. z ∣ G v = 0 , x ∣ G v = t v χ δ ∣ G v . If they differ by an unramified quadratic sign, put e = v ϖ ( 2 ) e=v_\varpi(2) e = v ϖ ( 2 ) n < e n<e n < e n = e n=e n = e
z ∣ G v = 0 , x ∣ G v = t v χ δ ∣ G v + u U v , u = 2 ϖ e ≠ 0. z|_{G_v}=0,\qquad x|_{G_v}=t_v\chi_\delta|_{G_v}+uU_v,\qquad u=\frac{2}{\varpi^e}\ne0. z ∣ G v = 0 , x ∣ G v = t v χ δ ∣ G v + u U v , u = ϖ e 2 = 0. where U v U_v U v
In either case z z z V V V z = 0 z=0 z = 0 x = A χ δ x=A\chi_\delta x = A χ δ y = B χ δ y=B\chi_\delta y = B χ δ X ( c ) = 0 X(c)=0 X ( c ) = 0 A = B = 0 A=B=0 A = B = 0 χ δ \chi_\delta χ δ U v U_v U v
Choose a lift σ \sigma σ Gal ( K / Q ) \operatorname{Gal}(K/\mathbb{Q}) Gal ( K / Q ) r σ r^\sigma r σ ν \nu ν K 2 / Q 2 K_2/\mathbb{Q}_2 K 2 / Q 2 1 + χ δ E 12 1+\chi_\delta E_{12} 1 + χ δ E 12 δ \delta δ r r r n < e n<e n < e ϖ e \varpi^e ϖ e n = e n=e n = e n ≥ 1 n\geq1 n ≥ 1 r σ ≃ r r^\sigma\simeq r r σ ≃ r
Lemma 12.3 , with η = 1 \eta=1 η = 1
Corollary 15.4. There is no ordinary scalar-residual source as above which is good at one seven-prime and good or special at the other.
Proof. The cyclic ordinary lattice supplies a nonzero actual unipotent residue trivial at two. Its supported quadratic class must be χ δ \chi_\delta χ δ δ = − 7 \delta=-7 δ = − 7
Nonordinary scalar sources
Proposition 15.5. Let R \mathcal{R} R K = Q ( 2 ) K=\mathbb{Q}(\sqrt{2}) K = Q ( 2 ) { 0 , 1 } \{0,1\} { 0 , 1 } q \mathfrak{q} q Q \mathbb{Q} Q
Proof. After conjugating K / Q K/\mathbb{Q} K / Q β = 5 + 4 2 \beta=5+4\sqrt{2} β = 5 + 4 2 q \mathfrak{q} q K ( β ) / K K(\sqrt{\beta})/K K ( β ) / K q \mathfrak{q} q q \mathfrak{q} q
Apply the common simultaneous regular trace-order construction to all dyadic coefficient factors; they are all nonordinary. Write its orders as R 0 , E 0 R_0,E_0 R 0 , E 0 R = R 0 / 2 R 0 R=R_0/2R_0 R = R 0 /2 R 0 E = E 0 / 2 E 0 E=E_0/2E_0 E = E 0 /2 E 0 J = rad E J=\operatorname{rad} E J = rad E ( J / J 2 ) ∨ (J/J^2)^\vee ( J / J 2 ) ∨ q \mathfrak{q} q
α = [ − 1 ] , β = [ 5 + 4 2 ] . \alpha=[-1],\qquad\beta=[5+4\sqrt{2}]. α = [ − 1 ] , β = [ 5 + 4 2 ] . Choose the real conjugation at 2 > 0 \sqrt{2}>0 2 > 0 q \mathfrak{q} q A , C A,C A , C ( α , β ) (\alpha,\beta) ( α , β ) ( 1 , 0 ) , ( 0 , 1 ) (1,0),(0,1) ( 1 , 0 ) , ( 0 , 1 ) A , C A,C A , C E E E
A 2 = C 2 = 0 , A C + C A = X = T ( A C ) . A^2=C^2=0,\qquad AC+CA=X=T(AC). A 2 = C 2 = 0 , A C + C A = X = T ( A C ) . All words lie in the R R R 1 , A , C , A C 1,A,C,AC 1 , A , C , A C q \mathfrak{q} q − 1 -1 − 1 β \beta β B B B A A A B − A ∈ J 2 B-A\in J^2 B − A ∈ J 2 B C = 0 BC=0 B C = 0 A C ∈ J 3 AC\in J^3 A C ∈ J 3
Work over R ′ = R / m R 2 R'=R/\mathfrak{m}_R^2 R ′ = R / m R 2 X X X 1 , A , C 1,A,C 1 , A , C T ( J 3 ) = 0 T(J^3)=0 T ( J 3 ) = 0 R ′ R' R ′ A C A = X A ACA=XA A C A = X A C A C = X C CAC=XC C A C = X C T ( A C A C ) = X 2 T(ACAC)=X^2 T ( A C A C ) = X 2 R ′ R' R ′ m R E \mathfrak{m}_R E m R E m R \mathfrak{m}_R m R X = T ( A C ) = 0 X=T(AC)=0 X = T ( A C ) = 0 R ′ R' R ′ R = F 2 R=\mathbb{F}_2 R = F 2 R 0 R_0 R 0 Q \mathbb{Q} Q
Lemma 15.7 now proves modularity.
Put K = Q ( 2 ) K = \mathbb{Q}(\sqrt{2}) K = Q ( 2 ) s = 2 s = \sqrt{2} s = 2 ϵ = 1 + s \epsilon= 1+s ϵ = 1 + s π = 3 + s \pi= 3+s π = 3 + s π ′ = 3 − s \pi' = 3-s π ′ = 3 − s
Proposition 15.6. Let R R R K K K { 0 , 1 } \{0,1\} { 0 , 1 } D = Q ( a 2 ) D = \mathbb{Q}(a_2) D = Q ( a 2 ) d ≤ 3 d \le3 d ≤ 3 R 0 = Z 2 [ a 2 ] R_0 = \mathbb{Z}_2[a_2] R 0 = Z 2 [ a 2 ] R 0 / 2 R 0 ≅ F 2 [ t ] / ( t d ) R_0/2R_0 \cong\mathbb{F}_2[t]/(t^d) R 0 /2 R 0 ≅ F 2 [ t ] / ( t d ) t = a ‾ 2 t = \overline{a}_2 t = a 2
Proof. Let D D D d = [ D : Q ] d = [D:\mathbb{Q}] d = [ D : Q ] R 0 R_0 R 0 E 0 E_0 E 0 T T T D \mathcal{D} D rank Z 2 R 0 = d \operatorname{rank}_{\mathbb{Z}_2} R_0 = d rank Z 2 R 0 = d E = E 0 / 2 E 0 E = E_0/2E_0 E = E 0 /2 E 0 R = R 0 / 2 R 0 R = R_0/2R_0 R = R 0 /2 R 0 J = rad E J = \operatorname{rad} E J = rad E ( J / J 2 ) ∨ (J/J^2)^\vee ( J / J 2 ) ∨
The elementary quadratic quotient of the narrow ray group of modulus p 2 2 π π ′ ∞ 1 ∞ 2 \mathfrak{p}_2^2\pi\pi'\infty_1\infty_2 p 2 2 π π ′ ∞ 1 ∞ 2
α = [ − 1 ] , β = [ 5 + 4 s ] = [ ϵ π ] , δ = [ − 7 ] . \alpha=[-1],\qquad\beta=[5+4s]=[\epsilon\pi],\qquad\delta=[-7]. α = [ − 1 ] , β = [ 5 + 4 s ] = [ ϵ π ] , δ = [ − 7 ] . Indeed the full narrow ray group has invariants ( 6 , 2 , 2 ) (6,2,2) ( 6 , 2 , 2 ) 4 , 7 , 49 4,7,49 4 , 7 , 49 α \alpha α β \beta β δ \delta δ − 7 ≡ 1 ( m o d 8 ) -7 \equiv1 \pmod{8} − 7 ≡ 1 ( mod 8 ) Q 2 \mathbb{Q}_2 Q 2
Choose the seven-place where s ≡ 3 ( m o d 7 ) s \equiv3 \pmod{7} s ≡ 3 ( mod 7 ) s > 0 s>0 s > 0 [ 7 ] [7] [ 7 ] ( α , β , δ ) (\alpha,\beta,\delta) ( α , β , δ )
( 1 , 0 , 1 ) , ( 1 , 1 , 1 ) , ( 0 , 0 , 1 ) . (1,0,1),\qquad(1,1,1),\qquad(0,0,1). ( 1 , 0 , 1 ) , ( 1 , 1 , 1 ) , ( 0 , 0 , 1 ) . Let A , B , C A,B,C A , B , C E E E A , B , C A,B,C A , B , C J / J 2 J/J^2 J / J 2 E E E L ( g ) L(g) L ( g ) L ( g ) L ( h ) = ( χ 2 ( g ) − 1 ) L ( h ) L(g)L(h)=(\chi_2(g)-1)L(h) L ( g ) L ( h ) = ( χ 2 ( g ) − 1 ) L ( h ) E E E
A 2 = B 2 = C 2 = B C = C B = 0 , A B + B A = X = T ( A B ) , A C + C A = Y = T ( A C ) . A^2=B^2=C^2=BC=CB=0,\qquad AB+BA=X=T(AB),\qquad AC+CA=Y=T(AC). A 2 = B 2 = C 2 = B C = C B = 0 , A B + B A = X = T ( A B ) , A C + C A = Y = T ( A C ) . Every word is in the R R R 1 , A , B , C , A B , A C 1,A,B,C,AB,AC 1 , A , B , C , A B , A C
We calculate in R ′ = R / m R 2 R'=R/\mathfrak{m}_R^2 R ′ = R / m R 2 X , Y X,Y X , Y E 0 [ [ u ] ] E_0[[u]] E 0 [[ u ]] Φ ( u ) \Phi(u) Φ ( u ) u 2 − 2 u^2-2 u 2 − 2 u = 0 u=0 u = 0 a 2 a_2 a 2 ( 1 , 0 , 0 ) (1,0,0) ( 1 , 0 , 0 ) ( α , β , δ ) (\alpha,\beta,\delta) ( α , β , δ )
Φ = N + u ( 1 + M ) + O ( u 2 ) , N = A + C + Z , M = m A A + m B B + m C C + Z ′ , Z , Z ′ ∈ J 2 . \Phi=N+u(1+M)+O(u^2),\qquad N=A+C+Z,\qquad M=m_AA+m_BB+m_CC+Z',\qquad Z,Z'\in J^2. Φ = N + u ( 1 + M ) + O ( u 2 ) , N = A + C + Z , M = m A A + m B B + m C C + Z ′ , Z , Z ′ ∈ J 2 . The unramified functional takes value one on β \beta β m B = 1 m_B=1 m B = 1 Z Z Z A + C A+C A + C R ′ R' R ′ 0 = D ( N ) = D ( A + C ) = Y 0=\mathcal{D}(N)=\mathcal{D}(A+C)=Y 0 = D ( N ) = D ( A + C ) = Y
a 2 = T ( N ) = T ( N M ) = m B X + ( m A + m C ) Y = X in R ′ . a_2=T(N)=T(NM)=m_BX+(m_A+m_C)Y=X\quad\text{in }R'. a 2 = T ( N ) = T ( N M ) = m B X + ( m A + m C ) Y = X in R ′ . Nakayama proves
R = F 2 [ t ] / ( t d ) , t = a ‾ 2 , D = Q ( a 2 ) . R = \mathbf{F}_2[t]/(t^d), \qquad t = \overline{a}_2, \qquad D = \mathbf{Q}(a_2). R = F 2 [ t ] / ( t d ) , t = a 2 , D = Q ( a 2 ) . If d ≥ 2 d \ge2 d ≥ 2 X X X Y = v X Y = vX Y = v X v ∈ m R v \in\mathfrak{m}_R v ∈ m R 1 , a 2 , … , a 2 d − 1 1,a_2,\ldots,a_2^{d-1} 1 , a 2 , … , a 2 d − 1 R 0 R_0 R 0
Suppose d ≥ 4 d \ge4 d ≥ 4 Z 0 = C + v B Z_0 = C + vB Z 0 = C + v B Z 0 2 = B Z 0 = X Z 0 = 0 Z_0^2 = BZ_0 = XZ_0 = 0 Z 0 2 = B Z 0 = X Z 0 = 0 T ( Z 0 ) = T ( A Z 0 ) = 0 T(Z_0) = T(AZ_0) = 0 T ( Z 0 ) = T ( A Z 0 ) = 0 S = F 2 [ X ] / ( X 3 ) S = \mathbf{F}_2[X]/(X^3) S = F 2 [ X ] / ( X 3 )
Q = S ⟨ a , b ⟩ / ( a 2 , b 2 , b a + a b − X ) . Q = S\langle a,b\rangle/(a^2,b^2,ba+ab-X). Q = S ⟨ a , b ⟩ / ( a 2 , b 2 , ba + ab − X ) . There is a surjection E → Q E \to Q E → Q A , B , Z 0 A,B,Z_0 A , B , Z 0 a , b , 0 a,b,0 a , b , 0 1 , A , B , A B , Z 0 , A Z 0 1,A,B,AB,Z_0,AZ_0 1 , A , B , A B , Z 0 , A Z 0 1 , A , B , A B 1,A,B,AB 1 , A , B , A B f i f_i f i X f 3 , X f 2 , X f 1 , X f 0 + X 2 f 3 Xf_3,Xf_2,Xf_1,Xf_0+X^2f_3 X f 3 , X f 2 , X f 1 , X f 0 + X 2 f 3 f i f_i f i ( X d − 1 ) (X^{d-1}) ( X d − 1 ) S S S
Compose this quotient with a ↦ E 12 a \mapsto E_{12} a ↦ E 12 b ↦ X E 21 b \mapsto XE_{21} b ↦ X E 21 θ : G K → G L 2 ( S ) \theta:G_K \to\mathrm{GL}_2(S) θ : G K → GL 2 ( S ) S S S α + β \alpha+\beta α + β β \beta β A , B , C A,B,C A , B , C ( 1 , 0 , 0 ) (1,0,0) ( 1 , 0 , 0 ) ( 0 , 1 , 0 ) (0,1,0) ( 0 , 1 , 0 ) v v v
F = K ( i , 5 + 4 s ) . F = K(i,\sqrt{5+4s}). F = K ( i , 5 + 4 s ) . On G F G_F G F I + X M + X 2 N I+XM+X^2N I + X M + X 2 N M M M M M M θ ( G F ) \theta(G_F) θ ( G F ) r r r tr θ ( r ) = X \operatorname{tr}\theta(r)=X tr θ ( r ) = X
θ ( r ) 4 = I + X 2 I ≠ I . \theta(r)^4 = I+X^2I \ne I. θ ( r ) 4 = I + X 2 I = I . Since r 2 ∈ G F r^2 \in G_F r 2 ∈ G F F F F
This contradicts a small ray calculation. The field F F F 3211264 3211264 3211264 P \mathfrak{P} P ( e , f ) = ( 4 , 2 ) (e,f)=(4,2) ( e , f ) = ( 4 , 2 ) ( e , f ) = ( 2 , 2 ) , ( 1 , 2 ) , ( 1 , 2 ) (e,f)=(2,2),(1,2),(1,2) ( e , f ) = ( 2 , 2 ) , ( 1 , 2 ) , ( 1 , 2 ) F / K F/K F / K K 2 K_2 K 2 F F F E E E θ ∣ G F \theta|_{G_F} θ ∣ G F H H H P 6 ∏ w ∣ 7 w \mathfrak{P}^6\prod_{w\mid7}w P 6 ∏ w ∣ 7 w 4 H 4H 4 H 2 I 7 2I_7 2 I 7 I 7 I_7 I 7
A defining polynomial, and the exact ray quotient, are
z 8 − 4 z 7 + 8 z 6 − 8 z 5 + 7 z 4 − 8 z 3 + 8 z 2 − 4 z + 1 , z^8-4z^7+8z^6-8z^5+7z^4-8z^3+8z^2-4z+1, z 8 − 4 z 7 + 8 z 6 − 8 z 5 + 7 z 4 − 8 z 3 + 8 z 2 − 4 z + 1 , H ≃ C 48 × C 8 × C 2 3 , H / ( 4 H + 2 I 7 ) ≃ C 2 5 . H \simeq C_{48}\times C_8\times C_2^3,\qquad H/(4H+2I_7)\simeq C_2^5. H ≃ C 48 × C 8 × C 2 3 , H / ( 4 H + 2 I 7 ) ≃ C 2 5 . Equivalently, let H 0 H_0 H 0 P 6 \mathfrak{P}^6 P 6 H → H 0 H \to H_0 H → H 0 I 7 I_7 I 7 H 0 ≃ C 2 2 H_0 \simeq C_2^2 H 0 ≃ C 2 2 4 H + 2 ker ( H → H 0 ) 4H+2\ker(H\to H_0) 4 H + 2 ker ( H → H 0 ) verify_sp7_ray_quotient_independent.gp. The quotient has exponent two, contradicting the order-four image just constructed. Hence d ≤ 3 d \le3 d ≤ 3 [49 ramified/d2] . □ \square □
Lemma 15.7. A good/special-seven system as above is modular if its trace field has degree at most three and is generated by a 2 a_2 a 2 Z 2 [ a 2 ] / 2 = F 2 [ t ] / ( t d ) \mathbf{Z}_2[a_2]/2=\mathbf{F}_2[t]/(t^d) Z 2 [ a 2 ] /2 = F 2 [ t ] / ( t d ) d = 3 d=3 d = 3
Proof. Lemma 11.10 gives a coefficient prime over three with residue field F 3 \mathbf{F}_3 F 3 F 9 \mathbf{F}_9 F 9 F 5 \mathbf{F}_5 F 5 1 1 1 ω 3 \omega_3 ω 3 [3 Theorem 2.5] , followed by [3 Theorem 2.3] , modularizes the actual weight-two source.
At the selected prime over five the residue field is F 5 \mathbb{F}_5 F 5
Theorem 15.8. Assume GRH. Every pure totally odd absolutely irreducible weight-two cyclotomic compatible system over K K K
Proof. First remove any ramified quadratic special scalar factors using Lemma 15.1; the corrections are unramified at two and three and at the other seven-prime. Hence all actual special scalar characters may be taken unramified. If some dyadic coefficient residue is nonsolvable, apply Proposition 14.2, which covers both ordinary and nonordinary sources under the present GRH hypothesis. The irreducible solvable alternative is excluded by Lemma 15.2; hence we may assume every dyadic coefficient residue has scalar semisimplification. If some dyadic member is ordinary, the ordinary proposition or its one-good-place corollary proves the assertion. Otherwise all are nonordinary. With exactly one special place, Proposition 15.5 applies. With two special places, Proposition 15.6 and Lemma 15.7 apply.
If both seven-places are good, the first-layer quadratic space of the regular nonordinary trace algebra is just ⟨ [ − 1 ] ⟩ \langle[-1]\rangle ⟨[ − 1 ]⟩ p 2 2 ∞ 1 ∞ 2 \mathfrak{p}_2^2\infty_1\infty_2 p 2 2 ∞ 1 ∞ 2 A = c − 1 A=c-1 A = c − 1 E = E 0 / 2 E 0 E=E_0/2E_0 E = E 0 /2 E 0 A 2 = 0 A^2=0 A 2 = 0 1 1 1 A A A R = R 0 / 2 R 0 R=R_0/2R_0 R = R 0 /2 R 0 F 2 \mathbb{F}_2 F 2 E E E 1 , A 1,A 1 , A dim F 2 E = 4 [ D : Q ] ≥ 4 \dim_{\mathbb{F}_2}E=4[D:\mathbb{Q}]\ge4 dim F 2 E = 4 [ D : Q ] ≥ 4
An unconditional observation. The separate nonordinary exclusion remains available without GRH. If a nonordinary dyadic member in the preceding theorem had nonsolvable residue, its kernel field M M M rd ( M ) < 4 14 < 14.967 \operatorname{rd}(M)<4\sqrt{14}<14.967 rd ( M ) < 4 14 < 14.967 b = 7 b=7 b = 7 rd ( M ) > 16.661 \operatorname{rd}(M)>16.661 rd ( M ) > 16.661
The supported-seven problem over Q ( 2 ) \mathbb{Q}(\sqrt{2}) Q ( 2 )
Put K = Q ( s ) K=\mathbb{Q}(s) K = Q ( s ) s 2 = 2 s^2=2 s 2 = 2 ϵ = 1 + s \epsilon=1+s ϵ = 1 + s l ± = ( 3 ± s ) \mathfrak{l}_{\pm}=(3\pm s) l ± = ( 3 ± s ) K 3 / Q 3 K_3/\mathbb{Q}_3 K 3 / Q 3
Lemma 15.9. Let R \mathcal{R} R G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 )
Proof. Twist away the scalar quadratic inertia using Lemma 15.1. Modulo three the inertia at both seven-primes is unipotent. A reducible global residue therefore has characters unramified outside three. The prime-to-three narrow ray group modulo three is C 2 C_2 C 2 ϵ \epsilon ϵ F 9 × \mathbb{F}_9^\times F 9 × ( + , − ) (+,-) ( + , − ) − 1 -1 − 1 1 , ω 3 1,\omega_3 1 , ω 3 1 , ω 3 1,\omega_3 1 , ω 3 e ( K 3 ) = 1 < 3 − 1 e(K_3)=1<3-1 e ( K 3 ) = 1 < 3 − 1 [62 §4.5, Theorem A] applies: the ratio is locally nontrivial, totally odd, and its splitting field K ( ζ 3 ) K(\zeta_3) K ( ζ 3 ) Q \mathbb{Q} Q χ 3 \chi_3 χ 3
For the second assertion, a cyclotomic-reducible irreducible residue would be induced from a prime-to-three character θ \theta θ H = K ( ζ 3 ) H=K(\zeta_3) H = K ( ζ 3 ) H H H 8 ⋅ ( − 3 ) ⋅ ( − 24 ) = 576 8\cdot(-3)\cdot(-24)=576 8 ⋅ ( − 3 ) ⋅ ( − 24 ) = 576 36 / π 2 < 4 36/\pi^2<4 36/ π 2 < 4 H H H F 9 \mathbb{F}_9 F 9 ϵ \epsilon ϵ θ \theta θ
Lemma 15.10. Every local determinant-ω 7 \omega_7 ω 7 Q 7 \mathbb{Q}_7 Q 7 χ 7 \chi_7 χ 7
Proof. The central-trivial Serre weights are Sym r det b \operatorname{Sym}^r\det^b Sym r det b r = 0 , 2 , 4 , 6 r=0,2,4,6 r = 0 , 2 , 4 , 6 r + 2 b = 0 ( m o d 6 ) r+2b=0\pmod6 r + 2 b = 0 ( mod 6 ) ω ~ 7 t , ω ~ 7 − t \widetilde{\omega}_7^t,\widetilde{\omega}_7^{-t} ω 7 t , ω 7 − t
Table 7.
Together with the two scalar crystalline and two scalar semistable weights, these cover all eight weights. The fixed-determinant formula [70 Theorem 1.2] therefore gives a nonempty weight-two local ring in this menu, for the actual residual extension. A scalar quadratic twist interchanges the principal-series rows and changes order six to order three. The independent global characters in Lemma 15.1 implement these choices at the two split seven-primes without disturbing good reduction at two or three.
:::
Lemma 15.11. Let r ˉ \bar{r} r ˉ
Proof. If inertia is trivial, lift the entire unramified Frobenius matrix, with fixed determinant seven; unequal Frobenius eigenvalues are allowed. Otherwise write tame inertia as 1 + t E 12 1+tE_{12} 1 + t E 12 t ≠ 0 t\ne0 t = 0 7 = 1 ( m o d 3 ) 7=1\pmod3 7 = 1 ( mod 3 ) a I + b E 12 aI+bE_{12} a I + b E 12 a = ± 1 a=\pm1 a = ± 1
Φ = ( 7 a b ~ 0 a ) , U ( x ) = 1 + t ~ x E 12 \Phi=\begin{pmatrix}7a & \tilde{b}\\ 0 & a\end{pmatrix},\qquad U(x)=1+\tilde{t}xE_{12} Φ = ( 7 a 0 b ~ a ) , U ( x ) = 1 + t ~ x E 12 has determinant seven, satisfies Φ U ( x ) Φ − 1 = U ( 7 x ) \Phi U(x)\Phi^{-1}=U(7x) Φ U ( x ) Φ − 1 = U ( 7 x )
Proposition 15.12. Assume GRH. Every irreducible determinant-ω 7 \omega_7 ω 7
Proof. A solvable image has a modular Artin witness. For a nonsolvable image the cyclotomic restriction is irreducible. Use Lemma 15.10, prescribed globalization, actual avoiding potential modularity and Brauer descent to construct a weight-two compatible system with the stated bounded menu, good outside seven. Normalize the independent principal-series signs as above.
Reduce at three. A reducible residue already modularizes the actual source by Lemma 15.9. Otherwise that lemma gives cyclotomic irreducibility. There is no exceptional-five exclusion in characteristic three. Prescribe the actual good/special lifts of Lemma 15.11, retaining the old Barsotti–Tate component at three, and globalize by EGPZ. Actual potential modularity and compatibility give the new system. Purity ensures nonzero special monodromy wherever the special component was selected, since its two Frobenius eigenvalues have ratio seven. By Theorem 15.8 this new system is modular. AKT applies backwards to the old weight-two three-adic source. Undoing the quadratic twists and reducing at seven proves the assertion.
Integral-order rigidity over Q ( 3 ) \mathbb{Q}(\sqrt{3}) Q ( 3 )
In this section K = F 3 K = \mathbb{F}_3 K = F 3 ℓ = 13 \ell= 13 ℓ = 13 ϵ = 2 + 3 \epsilon= 2 + \sqrt{3} ϵ = 2 + 3
Nonscalar dyadic sources
Proposition 16.1. Let a pure cyclotomic weight-two system over K K K
Proof. An irreducible solvable residue is necessarily odd dihedral. Write it as the induction of an odd-order character θ \theta θ L / K L/K L / K K K K
⟨ − 1 , ϵ , a , b ⟩ , ϵ = 2 + 3 , a = 4 + 3 , b = 4 − 3 , a b = 13. \langle-1,\epsilon,a,b\rangle,\qquad\epsilon= 2 + \sqrt{3},\quad a = 4 + \sqrt{3},\quad b = 4 - \sqrt{3},\quad ab = 13. ⟨ − 1 , ϵ , a , b ⟩ , ϵ = 2 + 3 , a = 4 + 3 , b = 4 − 3 , ab = 13. If the source is nonordinary, Theorem 11.7 makes the inducing dyadic conductor at most two. The seven nontrivial possibilities are the masks 1, 6, 7, 10, 11, 12, 13 in this displayed basis. Their ordinary ray groups modulo ( 2 ) (2) ( 2 )
1 , 1 , C 8 , 1 , C 8 , C 2 , C 4 × C 2 . 1,\ 1,\ C_8,\ 1,\ C_8,\ C_2,\ C_4 \times C_2. 1 , 1 , C 8 , 1 , C 8 , C 2 , C 4 × C 2 . The odd rotation character θ \theta θ ( 2 ) (2) ( 2 ) ( 2 ) (2) ( 2 ) θ \theta θ
For an ordinary source use the conductor-at-most-four list, which includes all fifteen nontrivial classes above. Their ray groups modulo ( 2 ) (2) ( 2 )
L = K ( − 13 ϵ ) = K ( − 26 ) , Cl ( 2 ) ( L ) = C 12 × C 2 × C 2 . L = K(\sqrt{-13\epsilon}) = K(\sqrt{-26}),\qquad\operatorname{Cl}_{(2)}(L) = C_{12} \times C_2 \times C_2. L = K ( − 13 ϵ ) = K ( − 26 ) , Cl ( 2 ) ( L ) = C 12 × C 2 × C 2 . Here the equality of fields follows from ϵ \epsilon ϵ K K K
The program d3_quads13_all.gp computes these fifteen ray groups, certifying each class group with bnfcertify. The exact calculation is supplied in [49 ramified/d3] .
The scalar envelope
Lemma 16.2. Put K = Q ( 3 ) K = \mathbb{Q}(\sqrt{3}) K = Q ( 3 ) ϵ = 2 + 3 \epsilon= 2 + \sqrt{3} ϵ = 2 + 3 δ 1 = ϵ ( 4 + 3 ) \delta_1 = \epsilon(4 + \sqrt{3}) δ 1 = ϵ ( 4 + 3 ) δ 2 = ϵ ( 4 − 3 ) \delta_2 = \epsilon(4 - \sqrt{3}) δ 2 = ϵ ( 4 − 3 )
F 0 = K ( i , δ 1 , δ 2 ) . F_0 = K(i,\sqrt{\delta_1},\sqrt{\delta_2}). F 0 = K ( i , δ 1 , δ 2 ) . This is a Galois extension of K K K 2 24 3 8 13 8 2^{24}3^{8}13^{8} 2 24 3 8 1 3 8 C 2 C_2 C 2 F 1 F_1 F 1 F 1 / K F_1/K F 1 / K Γ = Gal ( F 1 / K ) \Gamma= \operatorname{Gal}(F_1/K) Γ = Gal ( F 1 / K ) a , b , c , t a,b,c,t a , b , c , t
a 2 = b 2 = c 2 = t 2 = 1 , t central , a^2 = b^2 = c^2 = t^2 = 1,\qquad t\ \text{central}, a 2 = b 2 = c 2 = t 2 = 1 , t central , [ a , b ] = [ a , c ] = t , [ b , c ] = 1. [a,b] = [a,c] = t,\qquad[b,c] = 1. [ a , b ] = [ a , c ] = t , [ b , c ] = 1. Here b b b c c c ⟨ b , t ⟩ \langle b,t\rangle ⟨ b , t ⟩ ⟨ c , t ⟩ \langle c,t\rangle ⟨ c , t ⟩ D 8 × C 2 D_8 \times C_2 D 8 × C 2
Let P 1 , P 2 \mathfrak{P}_1,\mathfrak{P}_2 P 1 , P 2 F 0 F_0 F 0
P 1 6 P 2 6 \mathfrak{P}_1^6\mathfrak{P}_2^6 P 1 6 P 2 6 is C 4 C_4 C 4 C 2 C_2 C 2
Proof. We have δ 1 δ 2 = 13 ϵ 2 \delta_1\delta_2 = 13\epsilon^2 δ 1 δ 2 = 13 ϵ 2 δ j \delta_j δ j F 0 = K ( i , δ 1 , 13 ) F_0 = K(i,\sqrt{\delta_1},\sqrt{13}) F 0 = K ( i , δ 1 , 13 ) [ F 0 : K ] = 8 [F_0:K] = 8 [ F 0 : K ] = 8 ( e , f ) = ( 4 , 2 ) (e,f) = (4,2) ( e , f ) = ( 4 , 2 ) ( e , f ) = ( 2 , 1 ) (e,f) = (2,1) ( e , f ) = ( 2 , 1 ) K 2 / Q 2 K_2/\mathbb{Q}_2 K 2 / Q 2 ψ ( 3 ) = 1 + 2 ( 3 − 1 ) = 5 \psi(3) = 1 + 2(3 - 1) = 5 ψ ( 3 ) = 1 + 2 ( 3 − 1 ) = 5
The exact arithmetic computation starts from the displayed radical description of F 0 F_0 F 0 2 24 3 8 13 8 2^{24}3^{8}13^{8} 2 24 3 8 1 3 8 C 2 C_2 C 2 bnfcertify=1. Computing the ray group at the stated dyadic modulus gives C 4 C_4 C 4 C 2 C_2 C 2
The program verify_d3_classfield_source.gp constructs the Hilbert quadratic extension from the certified class field of F 0 F_0 F 0 2 48 3 16 13 16 2^{48}3^{16}13^{16} 2 48 3 16 1 3 16 X 2 − 3 X^2 - 3 X 2 − 3 ⟨ t ⟩ \langle t\rangle ⟨ t ⟩ b , c b,c b , c a a a ⟨ b , c , t ⟩ \langle b,c,t\rangle ⟨ b , c , t ⟩ t t t ⟨ b , t ⟩ \langle b,t\rangle ⟨ b , t ⟩ ⟨ c , t ⟩ \langle c,t\rangle ⟨ c , t ⟩ F 1 F_1 F 1 [49 ramified/d3] .
The group relation follows; putting z = b c z = bc z = b c t t t ⟨ a , b ⟩ ≃ D 8 \langle a,b\rangle\simeq D_8 ⟨ a , b ⟩ ≃ D 8 Γ = ⟨ a , b ⟩ × ⟨ z ⟩ \Gamma= \langle a,b\rangle\times\langle z\rangle Γ = ⟨ a , b ⟩ × ⟨ z ⟩ □ \square □
Lemma 16.3. Let Γ \Gamma Γ
( 1 + t ) ( 1 + b ) , ( 1 + t ) ( 1 + c ) (1+t)(1+b),\qquad(1+t)(1+c) ( 1 + t ) ( 1 + b ) , ( 1 + t ) ( 1 + c ) has dimension ten over F 2 \mathbb{F}_2 F 2
Proof. Put z = b c z = bc z = b c ⟨ t , z ⟩ ≃ C 2 2 \langle t,z\rangle\simeq C_2^2 ⟨ t , z ⟩ ≃ C 2 2 F 2 [ Γ ] \mathbb{F}_2[\Gamma] F 2 [ Γ ] 1 , a , b , a b 1,a,b,ab 1 , a , b , ab D = 1 + t D = 1 + t D = 1 + t Z = 1 + z Z = 1 + z Z = 1 + z D ( 1 + b ) D(1+b) D ( 1 + b ) D Z DZ D Z c = b z c=bz c = b z D D D D Z DZ D Z D D D D a Da D a D b Db D b D a b Dab D ab a a a D b = D Db=D D b = D D a b = D a Dab=Da D ab = D a D D D 8 + 2 = 10 8+2=10 8 + 2 = 10
The multiplication-table computation in verify_d3_group_algebra_independent.py gives the same quotient directly. The exact calculation is supplied in [49 ramified] .
The characteristic-five inner-twist endpoint
Lemma 16.4 (Finite-field projective descent). Let k = F 25 k=\mathbb{F}_{25} k = F 25 ϕ : x ↦ x 5 \phi:x\mapsto x^5 ϕ : x ↦ x 5 r ˉ : G → GL 2 ( k ) \bar r:G\to\operatorname{GL}_2(k) r ˉ : G → GL 2 ( k ) η : G → { ± 1 } \eta:G\to\{\pm1\} η : G → { ± 1 }
r ˉ ϕ ≃ r ˉ ⊗ η . \bar r^{\phi}\simeq\bar r\otimes\eta. r ˉ ϕ ≃ r ˉ ⊗ η . Then the projectivization of r ˉ \bar r r ˉ PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) G = G F G=G_F G = G F r ˉ \bar r r ˉ η \eta η
Proof. Choose A ∈ GL 2 ( k ) A\in\operatorname{GL}_2(k) A ∈ GL 2 ( k ) r ˉ ϕ = A r ˉ A − 1 η \bar r^{\phi}=A\bar rA^{-1}\eta r ˉ ϕ = A r ˉ A − 1 η ϕ \phi ϕ A ϕ A A^{\phi}A A ϕ A r ˉ \bar r r ˉ A ϕ A = c I A^{\phi}A=cI A ϕ A = c I ϕ \phi ϕ A A A c ϕ = c c^{\phi}=c c ϕ = c c ∈ F 5 × c\in\mathbb{F}_5^{\times} c ∈ F 5 × k × → F 5 × k^{\times}\to\mathbb{F}_5^{\times} k × → F 5 × A A A A ϕ A = I A^{\phi}A=I A ϕ A = I B ∈ GL 2 ( k ) B\in\operatorname{GL}_2(k) B ∈ GL 2 ( k ) A = B ϕ B − 1 A=B^{\phi}B^{-1} A = B ϕ B − 1
r ˉ ′ = B − 1 r ˉ B , ( r ˉ ′ ) ϕ = η r ˉ ′ . \bar r'=B^{-1}\bar rB,\qquad(\bar r')^{\phi}=\eta\bar r'. r ˉ ′ = B − 1 r ˉ B , ( r ˉ ′ ) ϕ = η r ˉ ′ . Every matrix class [ r ˉ ′ ( g ) ] [\bar r'(g)] [ r ˉ ′ ( g )] ϕ \phi ϕ PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) η ( g ) = − 1 \eta(g)=-1 η ( g ) = − 1 r ˉ ′ ( g ) \bar r'(g) r ˉ ′ ( g ) u ∈ k u\in k u ∈ k u ϕ = − u u^{\phi}=-u u ϕ = − u F 5 \mathbb{F}_5 F 5
For a real conjugation c v c_v c v C = r ˉ ′ ( c v ) C=\bar r'(c_v) C = r ˉ ′ ( c v ) c v 2 = 1 c_v^2=1 c v 2 = 1 C 2 = I C^2=I C 2 = I det C = − 1 \det C=-1 det C = − 1 η ( c v ) = − 1 \eta(c_v)=-1 η ( c v ) = − 1 C ϕ = − C C^{\phi}=-C C ϕ = − C F 5 \mathbb{F}_5 F 5 u C uC u C − u 2 -u^2 − u 2 u 2 ∈ F 5 × u^2\in\mathbb{F}_5^{\times} u 2 ∈ F 5 × u 2 = a 2 u^2=a^2 u 2 = a 2 a ∈ F 5 a\in\mathbb{F}_5 a ∈ F 5 u = ± a ∈ F 5 u=\pm a\in\mathbb{F}_5 u = ± a ∈ F 5 − 1 -1 − 1 F 5 \mathbb{F}_5 F 5 det ( u C ) \det(uC) det ( u C )
Proposition 16.5 (The quadratic inner-twist source at five). Put K = Q ( 3 ) K=\mathbb{Q}(\sqrt{3}) K = Q ( 3 ) h = χ − 1 h=\chi_{-1} h = χ − 1 K ( i ) / K K(i)/K K ( i ) / K R \mathcal{R} R K K K D D D D = Q D=\mathbb{Q} D = Q D D D τ \tau τ
R τ ≃ R ⊗ h . \mathcal{R}^{\tau}\simeq\mathcal{R}\otimes h. R τ ≃ R ⊗ h . Then R \mathcal{R} R
Proof. Choose a coefficient place above five and write r r r r ˉ \bar r r ˉ { 0 , 1 } \{0,1\} { 0 , 1 } F 5 \mathbb{F}_5 F 5 F 25 \mathbb{F}_{25} F 25 r ˉ \bar r r ˉ 1 , − 1 1,-1 1 , − 1 r ˉ \bar r r ˉ
Now suppose r ˉ \bar r r ˉ D = Q D=\mathbb{Q} D = Q D D D F 5 \mathbb{F}_5 F 5 [3 Theorem 2.5] supplies a modular witness. In the remaining case five is inert in D D D τ \tau τ F 25 \mathbb{F}_{25} F 25 ( ∗ ) (*) ( ∗ )
r ˉ ϕ ≃ r ˉ ⊗ h . \bar r^\phi\simeq\bar r \otimes h. r ˉ ϕ ≃ r ˉ ⊗ h . Lemma 16.4 descends its projectivization to PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) [ K ( ζ 5 ) : K ] = 4 [K(\zeta_5):K]=4 [ K ( ζ 5 ) : K ] = 4 Q ( ζ 5 ) \mathbb{Q}(\zeta_5) Q ( ζ 5 ) Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 ) [3 Theorem 1.2(3)] applies. Its Lemma 2.1(2) returns a modular witness for the original linear representation r ˉ \bar r r ˉ
Lemma 11.9 proves the required absolute irreducibility over K ( ζ 5 ) K(\zeta_5) K ( ζ 5 ) [3 Theorem 2.3] now applies to r r r { 0 , 1 } \{0,1\} { 0 , 1 } R \mathcal{R} R
All scalar dyadic residues: a ten-dimensional group algebra
Put K = Q ( 3 ) K=\mathbb{Q}(\sqrt{3}) K = Q ( 3 ) h h h K ( i ) / K K(i)/K K ( i ) / K K ( i ) K(i) K ( i ) K K K − 1 -1 − 1 Q 13 \mathbb{Q}_{13} Q 13
Proposition 16.6. Let R \mathcal{R} R K K K { 0 , 1 } \{0,1\} { 0 , 1 }
Proof. We first give a bound for a finite packet of such systems whose exact quadratic special characters agree at each thirteen-place. Apply Lemmas 11.1 and 11.3 to the whole simultaneous coefficient tuple. Write R 0 R_0 R 0 E 0 E_0 E 0 s s s
rank Z 2 R 0 = s , rank Z 2 E 0 = 4 s . \operatorname{rank}_{\mathbb{Z}_2} R_0=s,\qquad\operatorname{rank}_{\mathbb{Z}_2} E_0=4s. rank Z 2 R 0 = s , rank Z 2 E 0 = 4 s . Put E = E 0 / 2 E 0 E=E_0/2E_0 E = E 0 /2 E 0 G G G E E E L / K L/K L / K K 2 K_2 K 2 L L L
At a thirteen-place the integral special matrices have the form
ξ ( τ ) ( 1 v 0 1 ) , ξ ( F ) ( 13 w 0 1 ) , ξ ( τ ) , ξ ( F ) ∈ { ± 1 } . \xi(\tau)\begin{pmatrix}1&v\\0&1\end{pmatrix},\qquad
\xi(F)\begin{pmatrix}13&w\\0&1\end{pmatrix},\qquad
\xi(\tau),\xi(F)\in\{\pm1\}. ξ ( τ ) ( 1 0 v 1 ) , ξ ( F ) ( 13 0 w 1 ) , ξ ( τ ) , ξ ( F ) ∈ { ± 1 } . Normalize by the common rational signs. The inertia difference is in the actual order and squares to zero. The normalized Frobenius has polynomial X 2 − 14 X + 13 X^2-14X+13 X 2 − 14 X + 13
( F − 1 ) ( τ − 1 ) = ( τ − 1 ) ( F − 1 ) = 0 in E . (6) (F-1)(\tau-1)=(\tau-1)(F-1)=0\quad\text{in }E. \tag*{(6)} ( F − 1 ) ( τ − 1 ) = ( τ − 1 ) ( F − 1 ) = 0 in E . (6) The integral normalized products are zero in one order and twelve times the inertia difference in the other. Exact common signs justify these normalizations in the actual joint order.
Put ϵ = 2 + 3 \epsilon=2+\sqrt{3} ϵ = 2 + 3 δ 1 = ϵ ( 4 + 3 ) \delta_1=\epsilon(4+\sqrt{3}) δ 1 = ϵ ( 4 + 3 ) δ 2 = ϵ ( 4 − 3 ) \delta_2=\epsilon(4-\sqrt{3}) δ 2 = ϵ ( 4 − 3 ) F 0 = K ( i , δ 1 , δ 2 ) F_0=K(i,\sqrt{\delta_1},\sqrt{\delta_2}) F 0 = K ( i , δ 1 , δ 2 ) Lemma 16.2 gives [ F 0 : K ] = 8 [F_0:K]=8 [ F 0 : K ] = 8 C 4 C_4 C 4 C 2 C_2 C 2 H = Gal ( L F 0 / F 0 ) H=\operatorname{Gal}(LF_0/F_0) H = Gal ( L F 0 / F 0 ) K 2 K_2 K 2 F 0 / K F_0/K F 0 / K ψ F 0 / K ( 3 ) = 1 + 2 ( 3 − 1 ) = 5 \psi_{F_0/K}(3)=1+2(3-1)=5 ψ F 0 / K ( 3 ) = 1 + 2 ( 3 − 1 ) = 5 H a b H^{\mathrm{ab}} H ab
At thirteen the local field of L L L Q 13 \mathbb{Q}_{13} Q 13 F 0 F_0 F 0 L F 0 / F 0 LF_0/F_0 L F 0 / F 0 H a b H^{\mathrm{ab}} H ab C 2 C_2 C 2 H H H F 1 / F 0 F_1/F_0 F 1 / F 0 Cl ( F 0 ) = C 2 \operatorname{Cl}(F_0)=C_2 Cl ( F 0 ) = C 2 L ⊂ F 1 L\subset F_1 L ⊂ F 1 F 0 ⊂ L F_0\subset L F 0 ⊂ L F 0 F_0 F 0 G G G
Lemma 16.2 identifies Γ = Gal ( F 1 / K ) \Gamma=\operatorname{Gal}(F_1/K) Γ = Gal ( F 1 / K ) b , c b,c b , c ⟨ b , t ⟩ , ⟨ c , t ⟩ \langle b,t\rangle,\langle c,t\rangle ⟨ b , t ⟩ , ⟨ c , t ⟩ t t t t t t (6) makes E E E
F 2 [ T ] / ( ( 1 + t ) ( 1 + b ) , ( 1 + t ) ( 1 + c ) ) . \mathbb{F}_2[T]/\bigl((1+t)(1+b),(1+t)(1+c)\bigr). F 2 [ T ] / ( ( 1 + t ) ( 1 + b ) , ( 1 + t ) ( 1 + c ) ) . By Lemma 16.3 this algebra has dimension ten. We have proved the packet bound
4 s ≤ 10. (7) 4s\le10. \tag*{(7)} 4 s ≤ 10. (7) It bounds the actual matrix algebra and assumes no faithfulness of the scalar action R 0 / 2 ⟶ E 0 / 2 R_0/2\longrightarrow E_0/2 R 0 /2 ⟶ E 0 /2
Let D D D { R , R ⊗ h } \{\mathcal{R},\mathcal{R}\otimes h\} { R , R ⊗ h } h h h K ( i ) K(i) K ( i ) D D D ι \iota ι
R ι ≃ R ⊗ h . (8) \mathcal{R}^{\iota}\simeq\mathcal{R}\otimes h. \tag*{(8)} R ι ≃ R ⊗ h . (8) The hypotheses of Proposition 16.5 are now satisfied: either D = Q D=\mathbb{Q} D = Q D D D
Theorem 16.7. Assume GRH. Every pure, totally odd, absolutely irreducible, cyclotomic-determinant weight-two strongly compatible system over K K K
Proof. If a dyadic residue is nonsolvable, Proposition 14.2 applies. If it is irreducible and solvable, Proposition 16.1 applies. Otherwise every dyadic coefficient residue is reducible. Its characters have odd order and are unramified at thirteen because inertia has order at most two. At two wild inertia is pro-two and the residue unit group is F 2 × = 1 \mathbb{F}_2^\times=1 F 2 × = 1 Proposition 16.6 proves modularity. □
The supported residual base at thirteen
Lemma 16.8. Every local residual representation of G Q 13 G_{\mathbb{Q}_{13}} G Q 13
Proof. Write W ( r , b ) = Sym r F 13 2 ⊗ det b W(r,b) = \operatorname{Sym}^{r}\mathbb{F}_{13}^{2} \otimes\det^{b} W ( r , b ) = Sym r F 13 2 ⊗ det b r + 2 b = 0 r + 2b = 0 r + 2 b = 0 r ∈ { 0 , 2 , 4 , 6 , 8 , 10 , 12 } r \in\{0, 2, 4, 6, 8, 10, 12\} r ∈ { 0 , 2 , 4 , 6 , 8 , 10 , 12 } b b b t = 1 , … , 6 t = 1,\ldots,6 t = 1 , … , 6 12 t 12t 12 t
W ( 2 t − 2 , 1 − t ) and W ( 12 − 2 t , t ) . W(2t-2,1-t) \qquad\text{and} \qquad W(12-2t,t). W ( 2 t − 2 , 1 − t ) and W ( 12 − 2 t , t ) . These twelve weights are distinct and exhaust those with r < 12 r < 12 r < 12 [70 Theorem 1.2] , with its nonnegative multiplicities therefore gives the required nonempty local lifting ring for the actual residual extension. The inertial orders are fourteen for odd t t t t t t □ \square □
Lemma 16.9. Let R R R K = Q ( 3 ) K = \mathbb{Q}(\sqrt{3}) K = Q ( 3 )
Proof. Suppose first that the constituents are ψ \psi ψ ω 7 ψ − 1 \omega_{7}\psi^{-1} ω 7 ψ − 1 ψ 2 \psi^{2} ψ 2 ( K , p ) = ( F 3 , 7 ) (K,p) = (\mathbb{F}_{3}, 7) ( K , p ) = ( F 3 , 7 ) □ \square □
Lemma 16.10. At either thirteen-place, the actual reduction in characteristic seven of any member of the cuspidal/special menu admits a genuine quadratic-special lift of cyclotomic determinant. This includes the case of trivial residual inertia.
Proof. Retain the exact quadratic scalar inertial character and first remove it locally. Cuspidal Frobenius has trace zero and determinant thirteen, so its residual eigenvalues are 1 , − 1 1,-1 1 , − 1 13 = − 1 13 = -1 13 = − 1 u = ± 1 u = \pm1 u = ± 1
T ( x ) = 1 + t x E 12 , Φ = diag ( 13 u , u ) , T(x) = 1 + txE_{12}, \qquad\Phi= \operatorname{diag}(13u,u), T ( x ) = 1 + t x E 12 , Φ = diag ( 13 u , u ) , interchanging the basis vectors if needed. Choose t ≠ 0 t \ne0 t = 0 t t t Φ T ( x ) Φ − 1 = T ( 13 x ) \Phi T(x)\Phi^{-1} = T(13x) Φ T ( x ) Φ − 1 = T ( 13 x ) □ \square □
Proposition 16.11. Assume GRH. Every irreducible cyclotomic-determinant residual representation unramified outside thirteen is modular.
Proof. Solvable residual images have modular Artin witnesses. Otherwise the cyclotomic residual restriction in characteristic thirteen is irreducible. Use Lemma 16.8 and prescribed global lifting to construct a weight-two source, then establish its actual avoiding potential modularity and compatible system. In characteristic seven, a reducible residue gives modularity by Lemma 16.9 . For an irreducible residue that lemma verifies the cyclotomic hypothesis. Prescribe the genuine special lifts of Lemma 16.10 at both thirteen-primes and retain the old Barsotti–Tate component above seven. Global lift existence in characteristic seven and actual potential modularity give the replacement system. Purity makes its special monodromy nonzero. Theorem 16.7 modularizes this actual target. Allen–Khare–Thorne then applies backwards to the old actual weight-two seven-adic source. Compatibility and reduction at thirteen return the required original residual witness.
The bounded source conditions for d = 6 , 11 d=6,11 d = 6 , 11
For K = F 6 K=F_6 K = F 6 F 11 F_{11} F 11 { 0 , 1 } \{0,1\} { 0 , 1 } α χ p \alpha\chi_p α χ p α \alpha α α \alpha α
Trace relations over Q ( 6 ) \mathbb{Q}(\sqrt{6}) Q ( 6 )
Put K = Q ( s ) K=\mathbb{Q}(s) K = Q ( s ) s 2 = 6 s^2=6 s 2 = 6
Theorem 17.1. For a bounded source with types SC/SC or SP/SP, a solvable residual image at a coefficient prime above three implies modularity.
Theorem 17.2. Assume GRH. A bounded source with types SP/SP and nonsolvable residual image at a coefficient prime above two is modular. GRH is unnecessary if that dyadic member is nonordinary.
Characters and cubic classes
Write
ε = 5 + 2 s , a = 1 + s , b = 1 − s , π = 3 + s , γ = 3 + 2 s , H = K ( ζ 3 ) . \varepsilon=5+2s,\qquad a=1+s,\qquad b=1-s,\qquad\pi=3+s,\qquad\gamma=3+2s,\qquad H=K(\zeta_3). ε = 5 + 2 s , a = 1 + s , b = 1 − s , π = 3 + s , γ = 3 + 2 s , H = K ( ζ 3 ) . Then N ( ε ) = 1 N(\varepsilon)=1 N ( ε ) = 1 a b = − 5 ab=-5 ab = − 5 N ( π ) = 3 N(\pi)=3 N ( π ) = 3 N ( γ ) = − 15 N(\gamma)=-15 N ( γ ) = − 15 σ \sigma σ K / Q K/\mathbb{Q} K / Q ω = χ ˉ 3 \omega=\bar{\chi}_3 ω = χ ˉ 3
Lemma 17.3. The class number of K K K
[ − 3 ] = [ − ε ] , [ 5 ] , [ γ ] . [-3]=[-\varepsilon],\qquad[5],\qquad[\gamma]. [ − 3 ] = [ − ε ] , [ 5 ] , [ γ ] . For a reducible r ˉ 3 \bar r_3 r ˉ 3 ω \omega ω K ( 5 ) / K K(\sqrt{5})/K K ( 5 ) / K { 1 , ω } \{1,\omega\} { 1 , ω } { ψ γ , ψ γ ω } \{\psi_\gamma,\psi_\gamma\omega\} { ψ γ , ψ γ ω }
Proof. The Minkowski bound is 24 / 2 < 2.45 \sqrt{24}/2<2.45 24 /2 < 2.45 2 + s 2+s 2 + s − 2 -2 − 2 ε \varepsilon ε ( 5 , 2 ) (5,2) ( 5 , 2 ) x 2 − 6 y 2 = 1 x^2 - 6y^2 = 1 x 2 − 6 y 2 = 1 − 1 -1 − 1 [ − 1 ] [-1] [ − 1 ] [ ϵ ] [\epsilon] [ ϵ ] [ π ] [\pi] [ π ] [ a ] [a] [ a ] [ b ] [b] [ b ]
All five generators are dyadic units. Such a unit u u u u u u 4 O K 2 4\mathcal{O}_{K_2} 4 O K 2 A + B s A + Bs A + B s ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 3 , 2 ) (3,2) ( 3 , 2 ) B B B 3 / ϵ = ( 3 − s ) 2 3/\epsilon= (3 - s)^2 3/ ϵ = ( 3 − s ) 2 [ − ϵ ] = [ − 3 ] [-\epsilon] = [-3] [ − ϵ ] = [ − 3 ]
Let a residual constituent be ψ \psi ψ F 3 \mathbb{F}_3 F 3 ω \omega ω ψ 2 = 1 \psi^2 = 1 ψ 2 = 1 K 3 = Q 3 ( ζ 3 ) K_3 = \mathbb{Q}_3(\zeta_3) K 3 = Q 3 ( ζ 3 ) − 2 -2 − 2 Q 3 \mathbb{Q}_3 Q 3
Let F = K 3 F = K_3 F = K 3 v F ( s ) = 1 v_F(s) = 1 v F ( s ) = 1 U i = 1 + p F i U_i = 1 + \mathfrak{p}_F^i U i = 1 + p F i
Lemma 17.4. The global S S S S = { v : v ∣ 15 } S = \{v : v \mid15\} S = { v : v ∣ 15 } [ ϵ ] [\epsilon] [ ϵ ] [ π ] [\pi] [ π ] [ a ] [a] [ a ] [ b ] [b] [ b ] F × / F × 3 F^\times/F^{\times3} F × / F × 3 U 2 / U 4 U_2/U_4 U 2 / U 4
V = ⟨ q , d ⟩ , q = [ 5 ] , d = [ 5 ] + [ ϵ b ] . V = \langle q,d\rangle,\qquad q = [5],\qquad d = [5] + [\epsilon b]. V = ⟨ q , d ⟩ , q = [ 5 ] , d = [ 5 ] + [ ϵ b ] . We have σ ( q ) = q \sigma(q) = q σ ( q ) = q σ ( d ) = − d \sigma(d) = -d σ ( d ) = − d ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 1 , − 1 ) (1,-1) ( 1 , − 1 )
Proof. Since μ 3 ⊂ F \mu_3 \subset F μ 3 ⊂ F U 1 = μ 3 U 2 U_1 = \mu_3 U_2 U 1 = μ 3 U 2 U 2 → U 4 U_2 \to U_4 U 2 → U 4 1 1 1 − 1 -1 − 1 9 O F 9\mathcal{O}_F 9 O F
The π \pi π ϵ i a j b k \epsilon^i a^j b^k ϵ i a j b k 0 ≤ i , j , k < 3 0 \le i,j,k < 3 0 ≤ i , j , k < 3 ± 1 \pm1 ± 1 ( Z / 9 Z ) [ s ] / ( s 2 − 6 ) (\mathbb{Z}/9\mathbb{Z})[s]/(s^2 - 6) ( Z /9 Z ) [ s ] / ( s 2 − 6 ) ϵ = − ( 1 + s ) \epsilon= -(1 + s) ϵ = − ( 1 + s ) a = 1 + s a = 1 + s a = 1 + s b = 1 − s b = 1 - s b = 1 − s s 2 = 0 s^2 = 0 s 2 = 0 { ± 1 } U 2 \{\pm1\}U_2 { ± 1 } U 2 i + j − k = 0 i + j - k = 0 i + j − k = 0 [ 5 ] = [ a b ] [5] = [ab] [ 5 ] = [ ab ] w = [ ϵ b ] w = [\epsilon b] w = [ ϵ b ] N ( ϵ b ) = − 5 N(\epsilon b) = -5 N ( ϵ b ) = − 5 σ ( w ) = q − w \sigma(w) = q - w σ ( w ) = q − w d = q + w d = q + w d = q + w
Lemma 17.5. The three-part of the ray class group modulo p 3 3 \mathfrak{p}_3^3 p 3 3 C 3 C_3 C 3 σ \sigma σ
Proof. The unit group of O K / p 3 3 \mathcal{O}_K/\mathfrak{p}_3^3 O K / p 3 3 − 1 -1 − 1 ϵ \epsilon ϵ C 3 C_3 C 3 p 3 3 = ( 9 , 3 s ) \mathfrak{p}_3^3 = (9,3s) p 3 3 = ( 9 , 3 s ) a a a b b b ± 1 , ± ( 5 + 2 s ) , ± ( 4 + 2 s ) \pm1,\pm(5 + 2s),\pm(4 + 2s) ± 1 , ± ( 5 + 2 s ) , ± ( 4 + 2 s ) a = ϵ 2 b = 72 + 30 s ∈ ( 9 , 3 s ) a = \epsilon^2 b = 72 + 30s \in(9,3s) a = ϵ 2 b = 72 + 30 s ∈ ( 9 , 3 s ) σ \sigma σ
Lemma 17.6. No absolutely irreducible characteristic-three residue of a bounded source becomes reducible over H H H
Proof. The field H = Q ( − 2 , − 3 ) H = \mathbb{Q}(\sqrt{-2},\sqrt{-3}) H = Q ( − 2 , − 3 ) 576 576 576 36 / π 2 < 4 36/\pi^2 < 4 36/ π 2 < 4 A 2 = − 2 A^2 = -2 A 2 = − 2 u 2 − u + 1 = 0 , u^2 - u + 1 = 0, u 2 − u + 1 = 0 , 1 + A u 1 + Au 1 + A u − 1 + A -1 + A − 1 + A Q ( A ) \mathbb{Q}(A) Q ( A )
A cyclotomic-reducible irreducible residue would be Ind H K ξ \operatorname{Ind}_{H}^{K}\xi Ind H K ξ ξ c = ξ − 1 \xi^{c}=\xi^{-1} ξ c = ξ − 1 ξ 2 \xi^{2} ξ 2 H H H F 3 \mathbb{F}_{3} F 3 ξ 2 \xi^{2} ξ 2 ξ = ξ c \xi=\xi^{c} ξ = ξ c
Ordinary sources and an unramified quotient
We prove the ordinary part of Theorem 17.1 without a local-distinctness assumption at three. Such an assumption would fail here because ω ∣ G F = 1 \omega|_{G_F}=1 ω ∣ G F = 1
Proposition 17.7. Suppose that the residual constituents are unramified at three. Then the actual system is modular. After the quadratic normalization of Lemma 17.3, it is invariant under σ \sigma σ
Proof. By Lemma 17.3, a twist by ψ 5 \psi_{5} ψ 5 1 + ω 1+\omega 1 + ω σ \sigma σ
Lemma 4.10 makes the crystalline source at F F F ν \nu ν χ 3 ν − 1 \chi_{3}\nu^{-1} χ 3 ν − 1
Put A = Z 3 [ [ U ] ] A=\mathbb{Z}_{3}[[U]] A = Z 3 [[ U ]] ψ U \psi_{U} ψ U G F G_{F} G F 1 + U 1+U 1 + U 1 + ω 1+\omega 1 + ω [72 Theorems 2.5.3 and 2.5.5] , and extend its coefficients to A A A g , h ∈ G F g,h\in G_{F} g , h ∈ G F
( g − χ 3 ( g ) ψ U ( g ) − 1 ) ( h − ψ U ( h ) ) = 0. (9) \left(g-\chi_{3}(g)\psi_{U}(g)^{-1}\right)\left(h-\psi_{U}(h)\right)=0. \tag*{(9)} ( g − χ 3 ( g ) ψ U ( g ) − 1 ) ( h − ψ U ( h ) ) = 0. (9) No local type is imposed at five. To construct the Cayley–Hamilton quotient, take the closed two-sided ideal I I I D ( 1 − x t ) D(1-xt) D ( 1 − x t ) x ∈ I x\in I x ∈ I [72 Lemma 2.4.5] . Thus no coefficient-A A A R R R m \mathfrak{m} m
( R B C R ) . \begin{pmatrix} R & B \\ C & R \end{pmatrix}. ( R C B R ) . The actual source factors through it with U = ν ( Φ 3 ) − 1 U=\nu(\Phi_{3})-1 U = ν ( Φ 3 ) − 1
The offdiagonal cotangent duals embed into the finite-flat global extension spaces by [72 Theorem 4.3.5] ; scalar extension and further quotients retain these inclusions at 3 = U = 0 3=U=0 3 = U = 0 X 3 − u X^{3}-u X 3 − u v F ( u ) = 1 v_{F}(u)=1 v F ( u ) = 1
V 3 = ⟨ [ ε ] , [ a ] , [ b ] ⟩ . V_{3}=\langle[\varepsilon],[a],[b]\rangle. V 3 = ⟨[ ε ] , [ a ] , [ b ]⟩ . Their restriction to G F G_{F} G F F 3 \mathbb{F}_{3} F 3
Let J J J
Q = m / ( 3 R + U R + m 2 ) , Q r e d = m / ( 3 R + U R + m 2 + J ) . Q=\mathfrak{m}/(3R+UR+\mathfrak{m}^{2}),\qquad Q_{\mathrm{red}}=\mathfrak{m}/(3R+UR+\mathfrak{m}^{2}+J). Q = m / ( 3 R + U R + m 2 ) , Q red = m / ( 3 R + U R + m 2 + J ) . For g , h ∈ G F g,h\in G_{F} g , h ∈ G F Q Q Q b g c h = 0 b_{g}c_{h}=0 b g c h = 0 m \mathfrak{m} m G F G_{F} G F ( B / m B ) ⊗ ( C / m C ) (B/\mathfrak{m}B) \otimes(C/\mathfrak{m}C) ( B / m B ) ⊗ ( C / m C ) Q Q Q J J J Q ≃ Q r e d Q \simeq Q_{\mathrm{red}} Q ≃ Q red
A diagonal tangent relative to A A A 1 + η f 1+\eta f 1 + η f ω ( 1 − η f ) \omega(1-\eta f) ω ( 1 − η f ) η 2 = 0 \eta^2=0 η 2 = 0 p 3 3 \mathfrak{p}_3^3 p 3 3 f f f Q r e d Q_{\mathrm{red}} Q red Hom ( Cl 3 p ( K ) , F 3 ) \operatorname{Hom}(\operatorname{Cl}_3^p(K),\mathbb{F}_3) Hom ( Cl 3 p ( K ) , F 3 ) σ \sigma σ Q Q Q
The involution σ \sigma σ U U U G F G_F G F F / Q 3 F/\mathbb{Q}_3 F / Q 3 R R R A A A Q Q Q A A A σ \sigma σ R R R r 3 σ ≃ r 3 r_3^\sigma\simeq r_3 r 3 σ ≃ r 3
Lemma 12.3, with η = 1 \eta=1 η = 1
Finally, σ \sigma σ
Nonordinary reducible sources
The other pair in Lemma 17.3 requires a different finite-flat extension space.
Lemma 17.8. For either ramified quadratic character θ \theta θ G F G_F G F θ \theta θ U 2 / U 4 U_2/U_4 U 2 / U 4
Proof. The rank-one model has Breuil–Kisin exponent one. An extension has Frobenius matrix
( c u h ( u ) 0 c u ) \begin{pmatrix}
cu & h(u) \\
0 & cu
\end{pmatrix} ( c u 0 h ( u ) c u ) The height-one condition imposes no further restriction. A change of upper lift changes h h h c u ( f ( u 3 ) − f ( u ) ) cu(f(u^3)-f(u)) c u ( f ( u 3 ) − f ( u )) f f f 1 , u 1,u 1 , u [44 Theorem 1.1.3, Proposition 1.1.11 and Lemmas 1.2.4–1.2.5] . Coordinate pullback and quotient give the same bound for arbitrary coefficients.
The unramified selfextension is finite flat and gives U 3 / U 4 U_3/U_4 U 3 / U 4 1 + 6 s 1+6s 1 + 6 s α = 1 + s z \alpha=1+sz α = 1 + sz α 3 = 1 + 6 s \alpha^3=1+6s α 3 = 1 + 6 s
z 3 + ( s / 2 ) z 2 + ( 1 / 2 ) z − 1 = 0 , z^3+(s/2)z^2+(1/2)z-1=0, z 3 + ( s /2 ) z 2 + ( 1/2 ) z − 1 = 0 , whose reduction z 3 − z − 1 z^3-z-1 z 3 − z − 1 F 3 \mathbb{F}_3 F 3
For an independent class take
E : y 2 = x 3 − s 8 x 2 + x . E:y^2=x^3-\frac{s}{8}x^2+x. E : y 2 = x 3 − 8 s x 2 + x . Its discriminant is − 125 / 2 -125/2 − 125/2
ψ 3 ( x ) = ( x − 1 / s ) ( 3 x 3 + 6 x + s ) . \psi_3(x)=(x-1/s)(3x^3+6x+s). ψ 3 ( x ) = ( x − 1/ s ) ( 3 x 3 + 6 x + s ) . At x = 1 / s x=1/s x = 1/ s s ( 5 / 12 ) 2 s(5/12)^2 s ( 5/12 ) 2 θ s \theta_s θ s 2 / ( 3 s ) 2/(3s) 2/ ( 3 s ) s 3 s^3 s 3 [2 ] . It lies in U 2 ∖ U 3 U_2 \setminus U_3 U 2 ∖ U 3 − 1 -1 − 1 γ / ( 2 s ) = 1 + s / 4 \gamma/(2s) = 1 + s/4 γ / ( 2 s ) = 1 + s /4 U 2 / U 4 U_2/U_4 U 2 / U 4
Proposition 17.9. A pure bounded source with common ramified quadratic residual constituents at three is modular.
Proof. Normalize its residual pair to ψ γ + ψ γ ω \psi_\gamma+ \psi_{\gamma\omega} ψ γ + ψ γ ω Lemmas 17.4 and 17.8 , the two global offdiagonal extension spaces are contained in V = ⟨ q , d ⟩ V = \langle q,d\rangle V = ⟨ q , d ⟩ ℓ 1 = ( 1 , 1 ) \ell_1 = (1,1) ℓ 1 = ( 1 , 1 ) ℓ 2 = ( 1 , − 1 ) \ell_2 = (1,-1) ℓ 2 = ( 1 , − 1 ) Proposition 17.7 , now over Z 3 \mathbb{Z}_3 Z 3 T 2 + T + 1 = 0 T^2 + T + 1 = 0 T 2 + T + 1 = 0 Φ 2 + 5 = 0 \Phi^2 + 5 = 0 Φ 2 + 5 = 0 α v \alpha_v α v ( g − χ 3 ( g ) α v ( g ) ) ( h − α v ( h ) ) = 0 (g - \chi_3(g)\alpha_v(g))(h - \alpha_v(h)) = 0 ( g − χ 3 ( g ) α v ( g )) ( h − α v ( h )) = 0 g , h g,h g , h Lemma 17.5 . Thus products of the two offdiagonal cotangents surject onto m / ( 3 R + m 2 ) \mathfrak{m}/(3R + \mathfrak{m}^2) m / ( 3 R + m 2 )
Since N ( γ ) = − 15 N(\gamma) = -15 N ( γ ) = − 15 σ \sigma σ ψ 5 \psi_5 ψ 5 σ ( q ) = q \sigma(q) = q σ ( q ) = q σ ( d ) = − d \sigma(d) = -d σ ( d ) = − d ( T − 1 ) 2 = − 3 T (T - 1)^2 = -3T ( T − 1 ) 2 = − 3 T ℓ v ⊗ ℓ v \ell_v \otimes\ell_v ℓ v ⊗ ℓ v Lemma 12.2 , with s = q s = q s = q
For two special conditions, choose the place with s ≡ 1 ( m o d 5 ) s \equiv1 \pmod{5} s ≡ 1 ( mod 5 ) ψ γ \psi_\gamma ψ γ R U RU R U R R R Q 5 ( 5 ) \mathbb{Q}_5(\sqrt{5}) Q 5 ( 5 ) U U U R U e RU^e R U e U f U^f U f e = 0 e = 0 e = 0 f = 1 f = 1 f = 1 e ≠ f e \ne f e = f ℓ 1 , ℓ 2 \ell_1,\ell_2 ℓ 1 , ℓ 2 e = f e = f e = f R R R σ \sigma σ ψ 5 \psi_5 ψ 5
Lemma 12.1 makes the combined involution trivial on the complete ring. We obtain r 3 σ ≅ r 3 ⊗ ψ 5 r_3^\sigma\cong r_3 \otimes\psi_5 r 3 σ ≅ r 3 ⊗ ψ 5 ψ 5 \psi_5 ψ 5 G Q G_{\mathbb{Q}} G Q Lemma 12.3 proves modularity. □
Proof of Theorem 17.1. If the residue is reducible, Propositions 17.7 and 17.9 apply. If it is irreducible, Lemma 17.6 gives irreducibility on G H G_H G H Lemma 2.1 gives one in the weight-zero convention. Then [3 Theorem 2.3] applies to the actual weight-two three-adic source. Strong compatibility gives modularity of the system. □
The mixed reducible case
Proposition 17.10. A mixed bounded source cannot have a globally reducible three-adic residue.
Proof. The unramified constituent pair is excluded by Proposition 17.7 , since the involution σ \sigma σ q − = ( 1 − s ) \mathfrak{q}_- = (1 - s) q − = ( 1 − s ) q + = ( 1 + s ) \mathfrak{q}_+ = (1 + s) q + = ( 1 + s ) ψ γ + ψ γ ω \psi_\gamma+ \psi_{\gamma\omega} ψ γ + ψ γ ω V = ⟨ q , d ⟩ V = \langle q,d\rangle V = ⟨ q , d ⟩ ℓ + = ( 1 , 1 ) , ℓ − = ( 1 , − 1 ) \ell_+ = (1,1),\ell_- = (1,-1) ℓ + = ( 1 , 1 ) , ℓ − = ( 1 , − 1 )
We need the relation q ′ ⊗ q ′ ′ = 0 q' \otimes q'' = 0 q ′ ⊗ q ′′ = 0
by three whose generic factors are the same ramified quadratic θ \theta θ θ \theta θ θ \theta θ p − 1 p-1 p − 1
Let R R R ( R B ; C R ) (R\ B; C\ R) ( R B ; C R ) R ′ = R / ( 3 R + m 2 ) R' = R/(3R + \mathfrak{m}^{2}) R ′ = R / ( 3 R + m 2 ) B ′ = B / m B B' = B/\mathfrak{m}B B ′ = B / m B C ′ = C / m C C' = C/\mathfrak{m}C C ′ = C / m C E ′ E' E ′ B ′ B' B ′ C ′ C' C ′ m R ′ \mathfrak{m}_{R'} m R ′ I I I I 3 = 0 I^{3} = 0 I 3 = 0 E ′ E' E ′ [72 Definition 2.5.2, Theorem 2.5.5] . Every finite left E ′ E' E ′ θ \theta θ
For such a module V V V M ( V ) \mathcal{M}(V) M ( V ) [42 Corollary 4.3 and Proposition 5.7.3(2)] therefore gives an exact sequence of the underlying Kisin modules. These are finite free F 3 [ [ u ] ] \mathbb{F}_{3}[[u]] F 3 [[ u ]] u u u M ( V ) / u = P 0 ⊗ E ′ V \mathcal{M}(V)/u = P_{0} \otimes_{E'} V M ( V ) / u = P 0 ⊗ E ′ V E ′ E' E ′ P 0 P_{0} P 0 P 0 P_{0} P 0 P 0 ≃ E ′ P_{0} \simeq E' P 0 ≃ E ′ M ( E ′ ) ≃ E ′ [ [ u ] ] \mathcal{M}(E') \simeq E'[[u]] M ( E ′ ) ≃ E ′ [[ u ]]
Normalize the simple Frobenius maps to the same c u cu c u c u + X ( u ) cu + X(u) c u + X ( u ) X ( u ) ∈ I [ [ u ] ] X(u) \in I[[u]] X ( u ) ∈ I [[ u ]]
u 2 ( c u + X ( u ) ) − 1 = u / c − X ( u ) / c 2 + X ( u ) 2 / ( c 3 u ) . u^{2}(cu + X(u))^{-1} = u/c - X(u)/c^{2} + X(u)^{2}/(c^{3}u). u 2 ( c u + X ( u ) ) − 1 = u / c − X ( u ) / c 2 + X ( u ) 2 / ( c 3 u ) . Integrality forces X ( 0 ) 2 = 0 X(0)^{2} = 0 X ( 0 ) 2 = 0 R ′ R' R ′ h ( 0 ) h(0) h ( 0 ) Lemma 17.8 . The constant functional vanishes on the unramified line d d d 5 ϵ ( 1 − s ) ≡ 1 + 3 s ( m o d 9 ) 5\epsilon(1-s) \equiv1 + 3s \pmod{9} 5 ϵ ( 1 − s ) ≡ 1 + 3 s ( mod 9 ) q q q q ′ ⊗ q ′ q' \otimes q' q ′ ⊗ q ′
In the mixed problem the special relation kills one extension orientation at q − q_{-} q − ℓ + ⊗ ℓ + \ell_{+} \otimes\ell_{+} ℓ + ⊗ ℓ + C 3 C_{3} C 3 Lemma 17.5 , killed by the exact cuspidal Frobenius trace zero. Thus the relative pseudocotangent is zero, and every characteristic-zero point factors through Z 3 \mathbb{Z}_{3} Z 3
Compare such a point with the CM curve
E : y 2 = x 3 + γ 3 ( 7 + 3 s ) 2 / 4. E : y^{2} = x^{3} + \gamma^{3}(7 + 3s)^{2}/4. E : y 2 = x 3 + γ 3 ( 7 + 3 s ) 2 /4. It is the γ \gamma γ y 2 + ( 7 + 3 s ) y = x 3 y^{2} + (7 + 3s)y = x^{3} y 2 + ( 7 + 3 s ) y = x 3 q + 2 q^{2}_{+} q + 2 q − q_{-} q − q + q_{+} q + x = 0 x = 0 x = 0 ψ γ \psi_{\gamma} ψ γ q − q_{-} q − Q ( − 3 ) \mathbb{Q}(\sqrt{-3}) Q ( − 3 )
The unknown source has rational three-adic traces by the vanishing cotangent. Its trace algebra is split: nonzero special monodromy provides a nonzero nilpotent, impossible in a division algebra. Realize it over Q 3 \mathbb{Q}_{3} Q 3 E E E q − q_{-} q − 1 + x , 1 + y 1+x,1+y 1 + x , 1 + y x , y ∈ 3 Z 3 x,y \in3\mathbb{Z}_3 x , y ∈ 3 Z 3 − 1 -1 − 1 x y − 3 xy-3 x y − 3 ker ℓ − \ker\ell_- ker ℓ − β \beta β β ( q ∨ ) \beta(q^\vee) β ( q ∨ ) β ( ℓ + ) \beta(\ell_+) β ( ℓ + )
Let f , g f,g f , g q + q_+ q + q − q_- q − δ = ( f − g ) / 3 \delta=(f-g)/3 δ = ( f − g ) /3 c f , c g c_f,c_g c f , c g δ ( B C ) = β ( B ) ( c f − c g ) ( C ) \delta(BC)=\beta(B)(c_f-c_g)(C) δ ( B C ) = β ( B ) ( c f − c g ) ( C ) q ∨ q^\vee q ∨
( c f − c g ) ( q ∨ ) = ( c f − c g ) ( ℓ + ) = 0. (c_f-c_g)(q^\vee)=(c_f-c_g)(\ell_+)=0. ( c f − c g ) ( q ∨ ) = ( c f − c g ) ( ℓ + ) = 0. These functionals are independent on V V V q − q_- q − ± 6 = 0 \pm6=0 ± 6 = 0
Nonsolvable characteristic-two sources
We give the complete ramification and explicit-formula argument for Theorem 17.2. The quadratic extension H / K H/K H / K
Lemma 17.11. Let a finite-flat representation G K 2 → S L 2 ( F q ) G_{K_2}\to\mathrm{SL}_2(\mathbb{F}_q) G K 2 → SL 2 ( F q ) L L L 2 − 2 / q 2-2/q 2 − 2/ q
Proof. Fontaine’s classical upper-break bound is 2 ( 1 + 1 / ( 2 − 1 ) ) − 1 = 3 2(1+1/(2-1))-1=3 2 ( 1 + 1/ ( 2 − 1 )) − 1 = 3 G K 2 G_{K_2} G K 2 F 2 × \mathbb{F}_2^\times F 2 × q q q
d 2 ≤ 1 2 { 1 − 1 / q + 3 ( 1 − 1 / q ) } = 2 − 2 / q . d_2\le\frac{1}{2}\{1-1/q+3(1-1/q)\}=2-2/q. d 2 ≤ 2 1 { 1 − 1/ q + 3 ( 1 − 1/ q )} = 2 − 2/ q . If wild inertia is trivial, a nontrivial irreducible tame inertia has order three, since Frobenius squares and exchanges its determinant-one eigencharacters. Its contribution is 1 / 3 1/3 1/3
At a special five-place the full residual decomposition image is contained in C 2 2 C_2^2 C 2 2 H = K ( ζ 3 ) H=K(\zeta_3) H = K ( ζ 3 ) M / K M/K M / K S L 2 ( F q ) \mathrm{SL}_2(\mathbb{F}_q) SL 2 ( F q ) M ∩ H = K M\cap H=K M ∩ H = K
[ M H : Q ] = 4 q ( q 2 − 1 ) , rd ( M H ) ≤ 24 2 2 − 2 / q 5 , 2 N 25 / [ M H : Q ] ≥ 1 2 . [MH:\mathbb{Q}]=4q(q^2-1),\qquad\operatorname{rd}(MH)\le\sqrt{24}\,2^{2-2/q}\sqrt{5},\qquad2N_{25}/[MH:\mathbb{Q}]\ge\frac{1}{2}. [ M H : Q ] = 4 q ( q 2 − 1 ) , rd ( M H ) ≤ 24 2 2 − 2/ q 5 , 2 N 25 / [ M H : Q ] ≥ 2 1 . If the actual dyadic member is nondyadic, Theorem 11.7 improves the upper bound to rd ( M H ) < 2 120 < 21.910 \operatorname{rd}(MH)<2\sqrt{120}<21.910 rd ( M H ) < 2 120 < 21.910 q ≥ 8 q\ge8 q ≥ 8 b = 10 b=10 b = 10 n ≥ 2016 n\ge2016 n ≥ 2016 rd ( M H ) > 23.29 \operatorname{rd}(MH)>23.29 rd ( M H ) > 23.29 rd ( M ) > 41.69 \operatorname{rd}(M)>41.69 rd ( M ) > 41.69 b = 7 b=7 b = 7
For an arbitrary dyadic member, assume GRH and apply the explicit formula of Section 22, with B b , 100 B_{b,100} B b , 100 A b , 4 A_{b,4} A b , 4
Table 8 (PDF p. 58).
Each lower entry is strict and each upper entry is a strict outward upper bound. At fixed b b b n n n C b > 0 C_b > 0 C b > 0 q ≥ 8 q \ge8 q ≥ 8 [49 ramified/d6_d10_d11/shared] .
For q = 4 q = 4 q = 4 [3 Theorem 2.5] supplies a regular cuspidal residual witness. If all real conjugations acted trivially, the kernel field would be totally real of degree at least 120, contradicting the preceding unconditional comparison for a nonordinary member, or the last table row under GRH. Thus some residual real conjugation is nontrivial. All hypotheses of [69 Theorem 6.1] now hold for the actual crystalline two-adic source, including nonsolvability and this real-place condition. Its modularity proves Theorem 17.2 .
Two special conditions at two
Proposition 17.12. For a source of types SP/SP, the semisimplified residue r ˉ 2 \bar r_2 r ˉ 2 1 ⊕ 1 1 \oplus1 1 ⊕ 1 r ˉ 2 \bar r_2 r ˉ 2
Proof. An absolutely irreducible solvable subgroup of GL 2 ( F ‾ 2 ) \operatorname{GL}_2(\overline{\mathbb{F}}_2) GL 2 ( F 2 ) SL 2 ( F 2 r ) \operatorname{SL}_2(\mathbb{F}_{2^r}) SL 2 ( F 2 r ) r ≥ 2 r \ge2 r ≥ 2 Ind G L G K ξ \operatorname{Ind}_{G_L}^{G_K} \xi Ind G L G K ξ L / K L/K L / K ξ \xi ξ
Special inertia at five reduces to the identity or a group of order two: its scalar quadratic factor becomes trivial in characteristic two. Its intersection with the odd rotation subgroup is trivial. Consequently ξ \xi ξ L / K L/K L / K ( 2 ) (2) ( 2 ) L L L ( 2 ) (2) ( 2 )
Fontaine’s upper-break bound three gives dyadic conductor at most four for L / K L/K L / K 2 e + 1 = 5 2e + 1 = 5 2 e + 1 = 5 e ( K 2 / Q 2 ) = 2 e(K_2/\mathbb{Q}_2) = 2 e ( K 2 / Q 2 ) = 2 ⟨ − 1 , ϵ , a , b ⟩ \langle-1, \epsilon, a, b \rangle ⟨ − 1 , ϵ , a , b ⟩ ( 2 ) (2) ( 2 )
C 4 , 1 , C 2 , C 4 , C 4 , C 4 , C 4 , C 4 , C 4 , C 4 , C 8 × C 4 , C 2 , C 4 × C 2 , C 2 . C_4,1,C_2,C_4,C_4,C_4,C_4,C_4,C_4,C_4,C_8 \times C_4,C_2,C_4 \times C_2,C_2. C 4 , 1 , C 2 , C 4 , C 4 , C 4 , C 4 , C 4 , C 4 , C 4 , C 8 × C 4 , C 2 , C 4 × C 2 , C 2 . These are exact degree-four ray computations; certification of each class group returns one. All groups are two-primary, so none admits the odd-order character ξ \xi ξ [49 ramified] .
If the residue is reducible, both constituent characters have odd order. They are unramified at five by the special calculation, and at two since the residue units are F 2 × \mathbb{F}_2^\times F 2 × Theorem 17.2 . □
Ordinary scalar sources
Proposition 17.13. A bounded source with two special places and scalar residual semisimplification at two is modular if some two-adic member is ordinary.
Proof. Let r r r H 0 = χ − 3 ∣ G K H_0=\chi_{-3}|_{G_K} H 0 = χ − 3 ∣ G K ψ 5 = χ 5 ∣ G K \psi_5=\chi_5|_{G_K} ψ 5 = χ 5 ∣ G K θ 1 , θ 2 \theta_1,\theta_2 θ 1 , θ 2 η ∈ { 1 , H 0 , ψ 5 , H 0 ψ 5 } \eta\in\{1,H_0,\psi_5,H_0\psi_5\} η ∈ { 1 , H 0 , ψ 5 , H 0 ψ 5 } θ 1 θ 2 \theta_1\theta_2 θ 1 θ 2 r r r r σ ⊗ η r^\sigma\otimes\eta r σ ⊗ η
Choose a vector v v v [ − 15 ] [-15] [ − 15 ] − 1 , ϵ , 1 + s , 1 − s -1,\epsilon,1+s,1-s − 1 , ϵ , 1 + s , 1 − s 8 O K 2 8\mathcal{O}_{K_2} 8 O K 2 ( 1 , 0 ) , ( 1 , 4 ) , ( 7 , 2 ) , ( 7 , 6 ) (1,0),(1,4),(7,2),(7,6) ( 1 , 0 ) , ( 1 , 4 ) , ( 7 , 2 ) , ( 7 , 6 ) [ − 15 ] [-15] [ − 15 ] 1 + χ − 15 E 12 1+\chi_{-15}E_{12} 1 + χ − 15 E 12 c c c ( ( r ( c ) − 1 ) v , v ) ((r(c)-1)v,v) (( r ( c ) − 1 ) v , v )
r ( c ) = ( − 1 1 0 1 ) , r(c)=\begin{pmatrix}-1&1\\0&1\end{pmatrix}, r ( c ) = ( − 1 0 1 1 ) , ordinary mark e 2 e_2 e 2
It puts both aligned representations in the same marked problem.
Kisin’s marked ordinary functor, Proposition 2.4.4 and Corollary 2.4.5 of [43 ] , is represented at arbitrary ramification, including two. Pull it back to the global matrix functor with the displayed real matrix and the ordinary section e 2 e_2 e 2
We compute the homogeneous tangent in this gauge. Write its cocycle as X ( g ) = ( x ( g ) y ( g ) z ( g ) x ( g ) ) X(g)=\begin{pmatrix}x(g)&y(g)\\z(g)&x(g)\end{pmatrix} X ( g ) = ( x ( g ) z ( g ) y ( g ) x ( g ) ) 1 + χ − 15 ( g ) E 12 1+\chi_{-15}(g)E_{12} 1 + χ − 15 ( g ) E 12 χ − 15 ( g ) z ( g ) ( 1 + E 12 ) \chi_{-15}(g)z(g)(1+E_{12}) χ − 15 ( g ) z ( g ) ( 1 + E 12 ) z z z y ∣ G K 2 = 0 y|_{G_{K_2}}=0 y ∣ G K 2 = 0 x ∣ G K 2 x|_{G_{K_2}} x ∣ G K 2 z ∣ G K 2 z|_{G_{K_2}} z ∣ G K 2 e 1 + t v e 2 e_1+tv e_2 e 1 + t v e 2 z ∣ G v = 0 z|_{G_v}=0 z ∣ G v = 0 x ∣ G v = t v χ − 15 ∣ G v x|_{G_v}=t_v\chi_{-15}|_{G_v} x ∣ G v = t v χ − 15 ∣ G v z ∈ k χ − 1 z\in k\chi_{-1} z ∈ k χ − 1 z ( c ) = 0 z(c)=0 z ( c ) = 0 z = 0 z=0 z = 0 x , y x,y x , y y = B χ − 15 y=B\chi_{-15} y = B χ − 15 x = A χ − 15 + D ψ 5 x=A\chi_{-15}+D\psi_5 x = A χ − 15 + D ψ 5 χ − 15 = R + U \chi_{-15}=R+U χ − 15 = R + U ψ 5 = R \psi_5=R ψ 5 = R R , U R,U R , U D = 0 D=0 D = 0 A = B = 0 A=B=0 A = B = 0 W ( k ) W(k) W ( k ) W ( k ) W(k) W ( k )
We have r σ = r ⊗ η r^\sigma=r\otimes\eta r σ = r ⊗ η η \eta η G Q G_{\mathbb{Q}} G Q
The scalar two-adic source with two special places
Let K = Q ( 6 ) K=\mathbb{Q}(\sqrt{6}) K = Q ( 6 ) a v a_v a v
Theorem 17.14. Let R R R K K K { 0 , 1 } \{0,1\} { 0 , 1 }
The integral trace order and its regular group
Let D D D d = [ D : Q ] d=[D:\mathbb{Q}] d = [ D : Q ] R 0 , E 0 , T , D R_0,E_0,T,D R 0 , E 0 , T , D R 0 R_0 R 0 Z 2 \mathbb{Z}_2 Z 2 d d d E 0 / 2 E 0 E_0/2E_0 E 0 /2 E 0 F 2 \mathbb{F}_2 F 2
The two special generators and the dyadic trace
Put E = E 0 / 2 E 0 E=E_0/2E_0 E = E 0 /2 E 0 R = R 0 / 2 R 0 R=R_0/2R_0 R = R 0 /2 R 0 J J J E E E J / J 2 J/J^2 J / J 2 Theorem 11.7 , its Kisin matrix is ( u h 0 u ) \begin{pmatrix} u & h \\ 0 & u \end{pmatrix} ( u 0 h u ) n / u + m n/u+m n / u + m n , m ∈ F 2 n,m\in\mathbb{F}_2 n , m ∈ F 2
At five these quadratic characters are tame. They belong to the quadratic dual of the narrow ray group modulo the square of the dyadic prime and both five-primes. Its invariants are 4 , 2 , 2 4,2,2 4 , 2 , 2
α = [ − 1 ] , β = [ ϵ ] = [ 2 ] , γ = [ − 15 ] , ϵ = 5 + 2 6 . \alpha=[-1],\qquad\beta=[\epsilon]=[2],\qquad\gamma=[-15],\qquad\epsilon=5+2\sqrt{6}. α = [ − 1 ] , β = [ ϵ ] = [ 2 ] , γ = [ − 15 ] , ϵ = 5 + 2 6 . Their relative discriminant norms are 4 , 4 , 25 4,4,25 4 , 4 , 25 i , 2 , 5 i,\sqrt{2},\sqrt{5} i , 2 , 5 K K K − 1 -1 − 1 2 2 2 − 15 -15 − 15 γ \gamma γ A , B , C A,B,C A , B , C ( α , β , γ ) (\alpha,\beta,\gamma) ( α , β , γ ) ( 1 , 0 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) (1,0,1),(0,1,0),(0,0,1) ( 1 , 0 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) E E E J / J 2 J/J^2 J / J 2 J J J E E E
At any one five-place normalize by its fixed rational quadratic special sign. In a basis of the special line, for local g g g L ( g ) L(g) L ( g ) ( χ 2 ( g ) − 1 , ∗ ) (\chi_2(g)-1,*) ( χ 2 ( g ) − 1 , ∗ ) L ( g ) L ( h ) = ( χ 2 ( g ) − 1 ) L ( h ) L(g)L(h)=(\chi_2(g)-1)L(h) L ( g ) L ( h ) = ( χ 2 ( g ) − 1 ) L ( h ) ( 1 + A ) 2 = 1 (1+A)^2=1 ( 1 + A ) 2 = 1 T ( A ) = 0 T(A)=0 T ( A ) = 0 E E E
A 2 = B 2 = C 2 = B C = C B = 0 , A B + B A = X = T ( A B ) , A C + C A = Y = T ( A C ) . (10) A^2=B^2=C^2=BC=CB=0,
\qquad
AB+BA=X=T(AB),\qquad AC+CA=Y=T(AC). \tag*{(10)} A 2 = B 2 = C 2 = B C = C B = 0 , A B + B A = X = T ( A B ) , A C + C A = Y = T ( A C ) . (10) In particular all words reduce to the R R R 1 , A , B , C , A B , A C 1,A,B,C,AB,AC 1 , A , B , C , A B , A C
Lemma 17.15. The ring R R R F 2 [ t ] / ( t d ) \mathbb{F}_2[t]/(t^d) F 2 [ t ] / ( t d ) t t t a 2 a_2 a 2 d ≥ 2 d\geq2 d ≥ 2 X X X Y / X Y/X Y / X D = Q ( a 2 ) D=\mathbb{Q}(a_2) D = Q ( a 2 )
Proof. First work in the relative cotangent quotient R ′ = R / m R 2 R'=R/\mathfrak{m}_R^2 R ′ = R / m R 2 X , Y X,Y X , Y R ′ R' R ′ m R ′ \mathfrak{m}_{R'} m R ′
The ramified functional n n n ( α , β , γ ) (\alpha,\beta,\gamma) ( α , β , γ ) ( 1 , 1 , 0 ) (1,1,0) ( 1 , 1 , 0 ) α \alpha α γ \gamma γ α + β + γ = [ 5 ] \alpha+\beta+\gamma=[5] α + β + γ = [ 5 ] J / J 2 J/J^2 J / J 2 Φ \Phi Φ A + B + C A+B+C A + B + C u u u Φ \Phi Φ [ 5 ] [5] [ 5 ] u ( φ − 1 ) z u(\varphi-1)z u ( φ − 1 ) z
Φ = N + u ( 1 + M ) + O ( u 2 ) , N = A + B + C + Z , \Phi=N+u(1+M)+O(u^2),\qquad N=A+B+C+Z, Φ = N + u ( 1 + M ) + O ( u 2 ) , N = A + B + C + Z , M = m A A + m B B + m C C + Z ′ , Z , Z ′ ∈ J 2 . M=m_AA+m_BB+m_CC+Z',\qquad Z,Z'\in J^2. M = m A A + m B B + m C C + Z ′ , Z , Z ′ ∈ J 2 . Evaluation of [ 5 ] [5] [ 5 ] A , B , C A,B,C A , B , C 0 , 1 , 1 0,1,1 0 , 1 , 1 m B + m C = 1 m_B+m_C=1 m B + m C = 1 − 1 , 0 -1,0 − 1 , 0 M M M
We give the determinant calculation in R ′ R' R ′ J 2 ⊂ m R ′ E + R ′ A B + R ′ A C J^2 \subset\mathfrak{m}_{R'}E+R'AB+R'AC J 2 ⊂ m R ′ E + R ′ A B + R ′ A C m R ′ E \mathfrak{m}_{R'}E m R ′ E m R ′ 2 = 0 \mathfrak{m}_{R'}^2=0 m R ′ 2 = 0 m R ′ \mathfrak{m}_{R'} m R ′ A B AB A B A C AC A C X Y − T ( A B A C ) = 0 XY-T(ABAC)=0 X Y − T ( A B A C ) = 0 A + B + C A+B+C A + B + C (10) . Thus D ( N ) = D ( A + B + C ) = X + Y D(N)=D(A+B+C)=X+Y D ( N ) = D ( A + B + C ) = X + Y Lemma 11.4 gives X = Y X=Y X = Y
The same normal form gives T ( J 3 ) = 0 T(J^3)=0 T ( J 3 ) = 0 X , Y X,Y X , Y u u u T ( N ) T ( 1 + M ) − T ( N ( 1 + M ) ) T(N)T(1+M)-T(N(1+M)) T ( N ) T ( 1 + M ) − T ( N ( 1 + M )) R ′ R' R ′
T ( N ) = T ( N M ) = ( m B + m C ) X = X in R ′ . T(N)=T(NM)=(m_B+m_C)X=X \quad\text{in } R'. T ( N ) = T ( N M ) = ( m B + m C ) X = X in R ′ . Lemma 11.4 identifies T ( N ) T(N) T ( N ) a 2 a_2 a 2 R R R F 2 [ t ] \mathbb{F}_2[t] F 2 [ t ] t t t d d d R 0 R_0 R 0 t d = 0 t^d=0 t d = 0 X X X Y Y Y d ≥ 2 d\geq2 d ≥ 2 1 , a 2 , … , a 2 d − 1 1,a_2,\ldots,a_2^{d-1} 1 , a 2 , … , a 2 d − 1 R 0 R_0 R 0 Z 2 \mathbb{Z}_2 Z 2 a 2 ∈ D a_2\in D a 2 ∈ D d d d D = Q ( a 2 ) D=\mathbb{Q}(a_2) D = Q ( a 2 )
An abelian restriction of the regular image
Suppose d ≥ 4 d\geq4 d ≥ 4 X X X R R R v ∈ R × v\in R^\times v ∈ R × Y = v X Y=vX Y = v X Z = C + v B Z=C+vB Z = C + v B (10) , Z Z Z Z 2 = B Z = X Z = 0 Z^2=BZ=XZ=0 Z 2 = B Z = X Z = 0 T ( Z ) = T ( A Z ) = 0 T(Z)=T(AZ)=0 T ( Z ) = T ( A Z ) = 0 X C = ( A B + B A ) C = B A C = B ( Y + C A ) = Y B XC=(AB+BA)C=BAC=B(Y+CA)=YB X C = ( A B + B A ) C = B A C = B ( Y + C A ) = Y B E E E R R R 1 , A , B , A B , Z , A Z 1,A,B,AB,Z,AZ 1 , A , B , A B , Z , A Z
Let S = F 2 [ X ] / ( X 3 ) S=\mathbb{F}_2[X]/(X^3) S = F 2 [ X ] / ( X 3 ) Q = S ⟨ a , b ⟩ / ( a 2 , b 2 , b a + a b − X ) Q=S\langle a,b\rangle/(a^2,b^2,ba+ab-X) Q = S ⟨ a , b ⟩ / ( a 2 , b 2 , ba + ab − X ) Proposition 15.6 . The same trace-pairing argument gives a surjection E → Q E\to Q E → Q A , B , Z A,B,Z A , B , Z a , b , 0 a,b,0 a , b , 0 1 , A , B , A B , Z , A Z 1,A,B,AB,Z,AZ 1 , A , B , A B , Z , A Z 1 , A , B , A B 1,A,B,AB 1 , A , B , A B ( X d − 1 ) (X^{d-1}) ( X d − 1 ) S S S 1 , a , b , a b 1,a,b,ab 1 , a , b , ab E E E
Compose with a ↦ E 12 a\mapsto E_{12} a ↦ E 12 b ↦ X E 21 b\mapsto XE_{21} b ↦ X E 21 θ : G K → GL 2 ( S ) \theta:G_K\to\operatorname{GL}_2(S) θ : G K → GL 2 ( S )
( f 0 + X f 3 f 1 X f 2 f 0 ) \begin{pmatrix}
f_0+Xf_3 & f_1\\
Xf_2 & f_0
\end{pmatrix} ( f 0 + X f 3 X f 2 f 1 f 0 ) Its trace is the reduction of T T T det θ ( g ) = 1 \det\theta(g)=1 det θ ( g ) = 1 f 1 ( 0 ) , f 2 ( 0 ) f_1(0),f_2(0) f 1 ( 0 ) , f 2 ( 0 ) a a a b b b F 2 [ z ] / ( z 2 ) \mathbb{F}_2[z]/(z^2) F 2 [ z ] / ( z 2 ) X X X A , B , C A,B,C A , B , C ( 1 , 0 , 0 ) , ( 0 , 1 , 1 ) (1,0,0),(0,1,1) ( 1 , 0 , 0 ) , ( 0 , 1 , 1 ) f 1 ( 0 ) = α f_1(0)=\alpha f 1 ( 0 ) = α f 2 ( 0 ) = [ 5 ] f_2(0)=[5] f 2 ( 0 ) = [ 5 ]
Put F = K ( i , 5 ) F=K(i,\sqrt{5}) F = K ( i , 5 ) G F G_F G F I + X M + X 2 N I+XM+X^2N I + X M + X 2 N M M M f 2 ( 0 ) = 0 f_2(0)=0 f 2 ( 0 ) = 0 M M M I , E 12 I,E_{12} I , E 12 X 3 X^3 X 3 θ ( G F ) \theta(G_F) θ ( G F ) r r r X X X
θ ( r ) 2 = I + X θ ( r ) , θ ( r ) 4 = I + X 2 I . \theta(r)^2=I+X\theta(r),\qquad\theta(r)^4=I+X^2I. θ ( r ) 2 = I + X θ ( r ) , θ ( r ) 4 = I + X 2 I . The element r 2 r^2 r 2 G F G_F G F F / K F/K F / K
The four Frobenius classes above five
Fix a five-prime w w w F F F h h h F w F_w F w k k k K K K [ 5 ] ( k ) = 1 [5](k)=1 [ 5 ] ( k ) = 1
T ( h − 1 ) = T ( k − 1 ) = 0 , ( h − 1 ) ( k − 1 ) = ( k − 1 ) 2 = 0 in E . T(h-1)=T(k-1)=0,\qquad(h-1)(k-1)=(k-1)^2=0\quad\text{in }E. T ( h − 1 ) = T ( k − 1 ) = 0 , ( h − 1 ) ( k − 1 ) = ( k − 1 ) 2 = 0 in E . where the trace equalities are in R R R a b ab ab Q Q Q X f 3 = 0 Xf_3=0 X f 3 = 0 R R R h − 1 = P + U a + V b h-1=P+Ua+Vb h − 1 = P + U a + V b k − 1 = P ′ + U ′ a + V ′ b k-1=P'+U'a+V'b k − 1 = P ′ + U ′ a + V ′ b Q Q Q b b b a b ab ab
P V ′ + V P ′ = U V ′ + V U ′ = 0. PV'+VP'=UV'+VU'=0. P V ′ + V P ′ = U V ′ + V U ′ = 0. Now V ′ V' V ′ V ∈ X S V\in XS V ∈ X S [5 ] . Thus h − 1 = ( V / V ′ ) ( k − 1 ) h-1=(V/V')(k-1) h − 1 = ( V / V ′ ) ( k − 1 ) abstract algebra Q Q Q k k k I I I X X X − 1 -1 − 1 θ ( k ) − I = X M 0 + O ( X 2 ) \theta(k)-I=XM_0+O(X^2) θ ( k ) − I = X M 0 + O ( X 2 ) M 0 2 = 0 M_0^2=0 M 0 2 = 0
θ ( h ) = I + X 2 N w , N w 2 = 0. (11) \theta(h)=I+X^2N_w,\qquad N_w^2=0. \tag*{(11)} θ ( h ) = I + X 2 N w , N w 2 = 0. (11) There is one dyadic prime P \mathfrak{P} P F F F ( e , f ) = ( 4 , 2 ) (e,f)=(4,2) ( e , f ) = ( 4 , 2 ) K K K K ( i ) K(i) K ( i ) 24 24 24 − 4 -4 − 4 − 24 -24 − 24 24 2 24^2 2 4 2 K ( 5 ) K(\sqrt{5}) K ( 5 ) F / K F/K F / K F F F 1 + 2 ( 3 − 1 ) = 5 1+2(3-1)=5 1 + 2 ( 3 − 1 ) = 5 K K K K ( 5 ) K(\sqrt{5}) K ( 5 ) F F F
Write H H H P 6 \mathfrak{P}^6 P 6 F F F θ ∣ G F \theta|_{G_F} θ ∣ G F
H / ( 4 H + ⟨ 2 [ w ] : w ∣ 5 ⟩ ) . H/(4H+\langle2[w]:w\mid5\rangle). H / ( 4 H + ⟨ 2 [ w ] : w ∣ 5 ⟩) . The following coordinates are from the ordinary ray group, with the ordered primes as in the certificate. They concern abelian extensions of the degree-eight field F F F
Table 1 (PDF p. 62).
For reproducibility one defining polynomial for F F F
y 8 − 4 y 7 + 2 y 6 + 8 y 5 + 13 y 4 − 44 y 3 + 164 y 2 − 140 y + 145. y^8-4y^7+2y^6+8y^5+13y^4-44y^3+164y^2-140y+145. y 8 − 4 y 7 + 2 y 6 + 8 y 5 + 13 y 4 − 44 y 3 + 164 y 2 − 140 y + 145. The certificate computes a certified class group and units, obtains the ray cyclic generators and the ideal logarithms L L L diag ( 8 , 4 , 2 , 2 ) \operatorname{diag}(8,4,2,2) diag ( 8 , 4 , 2 , 2 ) 4 I 4I 4 I 2 L 2L 2 L [ w 1 ] − [ w 4 ] [w_1]-[w_4] [ w 1 ] − [ w 4 ] I + X 2 I I+X^2I I + X 2 I θ ( r 2 ) \theta(r^2) θ ( r 2 ) (11) this gives N w 1 + N w 4 = I N_{w_1}+N_{w_4}=I N w 1 + N w 4 = I ( I + N w 1 ) 2 = I (I+N_{w_1})^2=I ( I + N w 1 ) 2 = I d ≤ 3 d\leq3 d ≤ 3 [49 ramified/d6_d10_d11/d6] .
Passage to an odd coefficient prime
The trace-field bound and Lemma 11.10 give a coefficient prime above three with residue field F 3 \mathbb{F}_3 F 3 F 9 \mathbb{F}_9 F 9 F 5 \mathbb{F}_5 F 5
Choose the indicated coefficient place. For an absolutely irreducible residue, the traces in its finite residue field give a realization over that field by finite-field descent (the Brauer group of a finite field is zero). Allen–Khare–Thorne, Modularity of PGL 2 ( F p ) \operatorname{PGL}_2(\mathbb{F}_p) PGL 2 ( F p ) , Theorems 1.1 and 2.3, give residual automorphy and lift the actual weight-two member when its restriction to the cyclotomic subgroup is absolutely irreducible. In the nonsolvable case this restriction is automatic: reducibility on a normal cyclic quotient would force an induced dihedral representation, with solvable image. At three all solvable cases, including globally reducible residues, are covered by the pure special characteristic-three source theorem. At five a solvable irreducible residue either remains irreducible on the cyclotomic subgroup or is induced from K ( 5 ) K(\sqrt{5}) K ( 5 )
When the five-adic residual semisimplification is reducible, the dual source has determinant χ 5 − 1 \chi_5^{-1} χ 5 − 1 χ 5 − 1 \chi_5^{-1} χ 5 − 1 ( 0 , 0 ) (0,0) ( 0 , 0 ) [26 §1.6] . Corollary 1.2.2 there applies since K K K K K K
Local lifts and a switch in characteristic three
Lemma 17.16 (Bounded five-adic lifts). Every continuous representation r ˉ : G Q 5 → GL 2 ( F 5 ) \bar r:G_{\mathbb{Q}_5}\to\operatorname{GL}_2(\mathbb{F}_5) r ˉ : G Q 5 → GL 2 ( F 5 ) of determinant χ ˉ 5 \bar\chi_5 χ ˉ 5 has a lift of determinant χ 5 \chi_5 χ 5 labelled weights { 0 , 1 } \{0,1\} { 0 , 1 } and type SC or SP.
Proof. The weights of GL 2 ( F 5 ) \operatorname{GL}_2(\mathbb{F}_5) GL 2 ( F 5 )
1 , det , Sym 2 det , Sym 2 det 3 , Sym 4 , Sym 4 det 2 . 1,\quad\det,\quad\operatorname{Sym}^2\det,\quad\operatorname{Sym}^2\det^3,\quad\operatorname{Sym}^4,\quad\operatorname{Sym}^4\det^2. 1 , det , Sym 2 det , Sym 2 det 3 , Sym 4 , Sym 4 det 2 . The two cuspidal compact types reduce with constituents
σ ˉ ( τ 6 ) s s = 1 ⊕ Sym 2 det , σ ˉ ( τ 3 ) s s = det ⊕ Sym 2 det 3 . \bar\sigma(\tau_6)^{\mathrm{ss}}=1\oplus\operatorname{Sym}^2\det,\qquad\bar\sigma(\tau_3)^{\mathrm{ss}}=\det\oplus\operatorname{Sym}^2\det^3. σ ˉ ( τ 6 ) ss = 1 ⊕ Sym 2 det , σ ˉ ( τ 3 ) ss = det ⊕ Sym 2 det 3 . Here τ 3 = ω ~ 2 8 ⊕ ω ~ 2 16 \tau_3=\widetilde{\omega}_2^8\oplus\widetilde{\omega}_2^{16} τ 3 = ω 2 8 ⊕ ω 2 16 τ 6 = ω ~ 2 4 ⊕ ω ~ 2 20 \tau_6=\widetilde{\omega}_2^4\oplus\widetilde{\omega}_2^{20} τ 6 = ω 2 4 ⊕ ω 2 20 ω 2 \omega_2 ω 2 [70 Theorem 1.2] has nonnegative multiplicities; some multiplicity is positive because a potentially Barsotti–Tate lift exists. The determinant confines it to the six displayed weights. The first four therefore give the cuspidal alternatives.
The last two give the semistable compact type for a scalar quadratic inertial type. If the resulting point has nonzero monodromy, it is SP. If all such points were crystalline, the corresponding crystalline compact type, 1 1 1 det 2 \det^2 det 2
Lemma 17.17 (The actual-residual switch). Let s ˉ \bar s s ˉ be the reduction of a determinant- χ 3 \chi_3 χ 3 representation of G Q 5 G_{\mathbb{Q}_5} G Q 5 of type SC or SP, with any stable lattice. After finite coefficient extension,
s ˉ \bar{s} s ˉ χ 3 \chi_3 χ 3 τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
Proof. Apply Lemma 4.9 with ℓ = 5 \ell= 5 ℓ = 5 τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
The supported residual base
Proposition 17.18. Assume GRH. Every bounded system over K = Q ( 6 ) K = \mathbb{Q}(\sqrt{6}) K = Q ( 6 ) ω 5 \omega_5 ω 5
Proof. Consider first a bounded system of two special types. The quartic-ray calculation excludes an irreducible solvable two-residue. A nonsolvable two-residue gives modularity by the bounded special source theorem under GRH. In the remaining case its semisimplification is 1 ⊕ 1 1 \oplus1 1 ⊕ 1
Now consider any bounded system at three. A reducible residue is handled by Theorem 17.1 for a pure menu and is excluded by Proposition 17.10 for a mixed menu. An irreducible solvable residue has a regular cuspidal Artin witness; Lemma 17.6 and [3 Theorem 2.3] then modularize the actual weight-two source.
In the nonsolvable case the residue stays absolutely irreducible over K ( ζ 3 ) K(\zeta_3) K ( ζ 3 )
Let r ˉ \bar{r} r ˉ χ ˉ 5 \bar{\chi}_5 χ ˉ 5 [ K ( ζ 5 ) : K ] = 4 [K(\zeta_5):K] = 4 [ K ( ζ 5 ) : K ] = 4 PGL 2 ( F 5 ) ≅ S 5 \operatorname{PGL}_2(\mathbb{F}_5) \cong S_5 PGL 2 ( F 5 ) ≅ S 5 K w = Q 5 K_w = \mathbb{Q}_5 K w = Q 5
Dihedral sources over Q ( 10 ) \mathbb{Q}(\sqrt{10}) Q ( 10 )
Put K = Q ( s ) K = \mathbb{Q}(s) K = Q ( s ) s 2 = 10 s^2 = 10 s 2 = 10
Theorem 18.1. Assume GRH. Every continuous irreducible representation G K → GL 2 ( F ‾ 3 ) G_K \to\operatorname{GL}_2(\overline{\mathbb{F}}_3) G K → GL 2 ( F 3 ) χ ˉ 3 \bar{\chi}_3 χ ˉ 3
The two places v + , v − v_+,v_- v + , v − s = 1 , − 1 s = 1,-1 s = 1 , − 1 Q 3 \mathbb{Q}_3 Q 3 F = K 2 F = K_2 F = K 2 F 2 \mathbb{F}_2 F 2 v + v_+ v + v − v_- v − A 5 A_5 A 5 Theorem 13.1 then follows by a conductor-removal argument with protected residual images.
The local conditions at three
Lemma 18.2. Let r ˉ : G Q 3 → G L 2 ( F 3 ) \bar{r}:G_{\mathbb{Q}_3}\to\mathrm{GL}_2(\mathbb{F}_3) r ˉ : G Q 3 → GL 2 ( F 3 ) χ ˉ 3 \bar{\chi}_3 χ ˉ 3 { 0 , 1 } \{0,1\} { 0 , 1 } χ 3 \chi_3 χ 3
τ 4 = ω ~ 2 2 ⊕ ω ~ 2 6 , \tau_4=\widetilde{\omega}_2^{\,2}\oplus\widetilde{\omega}_2^{\,6}, τ 4 = ω 2 2 ⊕ ω 2 6 , where ω 2 \omega_2 ω 2
E ′ = Q 3 ( 3 , − 1 ) , e ( E ′ / Q 3 ) = 2 , f ( E ′ / Q 3 ) = 2 , d ( E ′ / Q 3 ) = 1 2 . (12) E'=\mathbb{Q}_3(\sqrt{3},\sqrt{-1}),\qquad e(E'/\mathbb{Q}_3)=2,\qquad f(E'/\mathbb{Q}_3)=2,\qquad d(E'/\mathbb{Q}_3)=\frac{1}{2}. \tag*{(12)} E ′ = Q 3 ( 3 , − 1 ) , e ( E ′ / Q 3 ) = 2 , f ( E ′ / Q 3 ) = 2 , d ( E ′ / Q 3 ) = 2 1 . (12) Proof. Apply Lemma 23.5; its two local choices have the same determinant, weights and characteristic-two kernel envelope.
At v + v_+ v + Lemma 23.3 , with its notation K 0 K_0 K 0 J J J ϑ \vartheta ϑ σ + \sigma_+ σ + τ + \tau_+ τ +
Proposition 18.3. The representation σ + = Ind K 0 G ϑ \sigma_+=\operatorname{Ind}_{K_0}^{G}\vartheta σ + = Ind K 0 G ϑ τ + \tau_+ τ + r ˉ : G Q 3 → G L 2 ( F 3 ) \bar{r}:G_{\mathbb{Q}_3}\to\mathrm{GL}_2(\mathbb{F}_3) r ˉ : G Q 3 → GL 2 ( F 3 ) χ ˉ 3 \bar{\chi}_3 χ ˉ 3 { 0 , 1 } \{0,1\} { 0 , 1 } χ 3 \chi_3 χ 3 τ + \tau_+ τ +
Its inertia remains irreducible after reduction in every characteristic greater than three. For a determinant-cyclotomic compatible system of this type, its residual characteristic-two kernel field at this place is the unique totally ramified S 3 S_3 S 3 E + / Q 3 E_+/\mathbb{Q}_3 E + / Q 3 Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 ) 7 / 6 7/6 7/6
Proof. The construction is that of Lemma 23.3 , and Proposition 23.4 gives its lifts with the specified determinant and weights. Lemma 23.3 also gives the irreducibility of inertia in characteristic greater than three and the unique characteristic-two kernel field E + E_+ E +
Proposition 18.4. Let ρ ˉ \bar{\rho} ρ ˉ R = ( r λ ) \mathcal{R}=(r_\lambda) R = ( r λ ) χ \chi χ { 0 , 1 } \{0,1\} { 0 , 1 } τ + \tau_+ τ + v + v_+ v + Lemma 18.2 at v − v_- v −
Proof. Proposition 18.3 and Lemma 18.2 supply the local lifts at v + v_+ v + v − v_- v − § 2.1 retains nonzero monodromy. Nonsolvability implies that the restriction to G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 )
Apply Lemma 2.4 to these nonempty weight-two components, with determinant χ 3 \chi_3 χ 3
strongly compatible system, including local compatibility at the coefficient places. In particular its two-adic member is crystalline of weights { 0 , 1 } \{0,1\} { 0 , 1 }
Choose a place above two in the coefficient field and write r 2 r_2 r 2 r ‾ 2 \overline{r}_2 r 2 r ‾ 2 \overline{r}_2 r 2 M M M v + v_+ v + E + E_+ E + v − v_- v − E ′ E' E ′
The controlled characteristic-two image
Lemma 18.5. The splitting field A / K A/K A / K X 3 − 3 X + 1 + s X^3-3X+1+s X 3 − 3 X + 1 + s S 3 S_3 S 3 K 2 K_2 K 2 v + v_+ v + v − v_- v − E + E_+ E + Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 )
Proof. The cubic reduces to X 3 + X + 1 X^3+X+1 X 3 + X + 1 F 2 \mathbb{F}_2 F 2
− 27 ( 7 + 2 s ) = − 6 ( 3 + 3 2 s ) 2 , N K / Q ( − 27 ( 7 + 2 s ) ) = 3 8 . -27(7+2s)=-6\left(3+\frac{3}{2}s\right)^2,\qquad N_{K/\mathbb{Q}}\bigl(-27(7+2s)\bigr)=3^8. − 27 ( 7 + 2 s ) = − 6 ( 3 + 2 3 s ) 2 , N K / Q ( − 27 ( 7 + 2 s ) ) = 3 8 . It follows that Gal ( A / K ) = S 3 \operatorname{Gal}(A/K)=S_3 Gal ( A / K ) = S 3 K ( − 6 ) K(\sqrt{-6}) K ( − 6 )
At v + v_+ v + d = s − 1 d=s-1 d = s − 1 d ( s + 1 ) = 9 d(s+1)=9 d ( s + 1 ) = 9 v 3 ( d ) = 2 v_3(d)=2 v 3 ( d ) = 2 X = 1 + y X=1+y X = 1 + y y 3 + 3 y 2 + d = 0 y^3+3y^2+d=0 y 3 + 3 y 2 + d = 0 2 / 3 2/3 2/3 z = y 2 / 3 z=y^2/3 z = y 2 /3
z 3 − 3 z 2 − 2 d 3 z − d 2 27 = 0. z^3-3z^2-\frac{2d}{3}z-\frac{d^2}{27}=0. z 3 − 3 z 2 − 3 2 d z − 27 d 2 = 0. This polynomial is Eisenstein. In the resulting cubic extension the three derivative terms have valuations 5 , 4 , 3 5,4,3 5 , 4 , 3 6 , 3 , 1 6,3,1 6 , 3 , 1 ( 5 + 2 ) / 6 = 7 / 6 (5+2)/6=7/6 ( 5 + 2 ) /6 = 7/6 Q 3 ( − 6 ) = Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{-6})=\mathbb{Q}_3(\sqrt{3}) Q 3 ( − 6 ) = Q 3 ( 3 ) E + E_+ E +
At v − v_- v − 1 + s 1+s 1 + s − 1 -1 − 1 − 1 / 2 -1/2 − 1/2 Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 )
Proposition 18.6. There is a solvable Galois extension N / K N/K N / K v + v_+ v + E + E_+ E + v − v_- v − E ′ E' E ′
Proof. We use an abelian ray extension of the field A A A A A A
P A ( X ) = X 12 − 2 X 11 + 5 X 10 + 19 X 8 + 2 X 7 + 31 X 6 + 2 X 5 + 19 X 4 + 5 X 2 − 2 X + 1. P_A(X)=X^{12}-2X^{11}+5X^{10}+19X^8+2X^7+31X^6
+2X^5+19X^4+5X^2-2X+1. P A ( X ) = X 12 − 2 X 11 + 5 X 10 + 19 X 8 + 2 X 7 + 31 X 6 + 2 X 5 + 19 X 4 + 5 X 2 − 2 X + 1. It is obtained by taking the compositum of X 6 − 6 X 4 + 2 X 3 + 9 X 2 − 6 X − 9 X^6-6X^4+2X^3+9X^2-6X-9 X 6 − 6 X 4 + 2 X 3 + 9 X 2 − 6 X − 9 X 2 + 6 X^2+6 X 2 + 6 u u u K K K s = − u 3 + 3 u − 1 s=-u^3+3u-1 s = − u 3 + 3 u − 1 P 1 , P 2 \mathfrak{P}_1,\mathfrak{P}_2 P 1 , P 2 v + v_+ v +
residue degree one, lying over v − v_{-} v − ( 0 , 6 ) (0, 6) ( 0 , 6 )
Let R R R P 1 4 P 2 4 \mathfrak{P}_{1}^{4}\mathfrak{P}_{2}^{4} P 1 4 P 2 4 R R R v + v_{+} v + v − v_{-} v −
R ≃ C 4 2 × C 2 5 , R / ⟨ [ q + ] , 2 [ q − ] ⟩ ≃ C 4 × C 2 5 . R \simeq C_{4}^{2} \times C_{2}^{5}, \qquad R/\langle[\mathfrak{q}_{+}],2[\mathfrak{q}_{-}]\rangle\simeq C_{4} \times C_{2}^{5}. R ≃ C 4 2 × C 2 5 , R / ⟨[ q + ] , 2 [ q − ]⟩ ≃ C 4 × C 2 5 . Here the bracket in the denominator includes all three primes over v − v_{-} v − bnfcertify = 1 for the class-group calculation. The exact conductor of the quotient has exponent four at both dyadic primes. The exact calculation is supplied in [49 ramified/d6_d10_d11/d10] .
Let N / A N/A N / A Gal ( A / K ) \operatorname{Gal}(A/K) Gal ( A / K ) N / K N/K N / K S 3 S_{3} S 3 [ N : K ] = 128 ⋅ 6 = 768 [N:K] = 128 \cdot6 = 768 [ N : K ] = 128 ⋅ 6 = 768 v + v_{+} v + N / A N/A N / A E + E_{+} E + v − v_{-} v − E ′ E' E ′ A A A A / K A/K A / K K 2 K_{2} K 2
Theorem 18.7. Assume GRH for Dedekind zeta functions. There is no finite Galois extension M / K M/K M / K
Gal ( M / K ) ≃ SL 2 ( F q ) \operatorname{Gal}(M/K) \simeq\operatorname{SL}_{2}(\mathbb{F}_{q}) Gal ( M / K ) ≃ SL 2 ( F q ) q = 2 r ≥ 8 q = 2^{r} \ge8 q = 2 r ≥ 8
its dyadic completions have no upper break greater than three;
its completion at v + v_{+} v + E + E_{+} E + v − v_{-} v − E ′ E' E ′
Proof. The group SL 2 ( F q ) \operatorname{SL}_{2}(\mathbb{F}_{q}) SL 2 ( F q ) M ∩ N = K M \cap N = K M ∩ N = K Proposition 18.6 . For L = M N L = MN L = M N
[ L : Q ] = 2 ⋅ 768 q ( q 2 − 1 ) ≥ 774144. [L:\mathbb{Q}] = 2 \cdot768q(q^{2}-1) \ge774144. [ L : Q ] = 2 ⋅ 768 q ( q 2 − 1 ) ≥ 774144. Upper numbering is compatible with quotients, so the compositum still has no dyadic upper break greater than three. Since K 2 K_{2} K 2
rd ( L ) < 4 40 3 5 / 6 = 63.1962098567 … \operatorname{rd}(L) < 4\sqrt{40}\,3^{5/6} = 63.1962098567\ldots rd ( L ) < 4 40 3 5/6 = 63.1962098567 … Apply the GRH explicit formula of § 9.1 with the retained coefficients c 3 = 1 / 6 c_{3} = 1/6 c 3 = 1/6 c 9 = 1 / 4 c_{9} = 1/4 c 9 = 1/4 v + v_{+} v +
At b = 13.3 b = 13.3 b = 13.3 rd ( L ) > 64.5683408714 \operatorname{rd}(L) > 64.5683408714 rd ( L ) > 64.5683408714 [49 ramified/d6_d10_d11/shared] .
Corollary 18.8. Assume GRH. Let r : G K ⟶ GL 2 ( Q ‾ 2 ) r:G_{K} \longrightarrow\operatorname{GL}_{2}(\overline{\mathbb{Q}}_{2}) r : G K ⟶ GL 2 ( Q 2 ) χ 2 \chi_{2} χ 2 Theorem 18.7 . Then r r r
Proof. Dickson’s classification and Theorem 18.7 leave only residual image SL 2 ( F 4 ) ≃ A 5 \operatorname{SL}_2(\mathbb{F}_4) \simeq A_5 SL 2 ( F 4 ) ≃ A 5 A 5 A_5 A 5 [3 Theorem 2.5] supplies residual modularity. If its kernel field M M M rd ( M ) > 66.10839 \operatorname{rd}(M) > 66.10839 rd ( M ) > 66.10839 b = 5 b = 5 b = 5 [69 Theorem 6.1] are satisfied, so r r r
Lemma 18.9. Every finite characteristic-three character ψ \psi ψ G K G_K G K ν \nu ν K ( 2 ) / K K(\sqrt{2})/K K ( 2 ) / K χ ˉ 3 ψ − 2 \bar{\chi}_3\psi^{-2} χ ˉ 3 ψ − 2 G Q G_{\mathbb{Q}} G Q
Proof. The Minkowski bound is 40 / 2 = 10 < 3.17 \sqrt{40}/2 = \sqrt{10} < 3.17 40 /2 = 10 < 3.17 p 2 \mathfrak{p}_2 p 2 ± 2 \pm2 ± 2 x 2 ≡ ± 2 ( m o d 5 ) x^2 \equiv\pm2 \pmod5 x 2 ≡ ± 2 ( mod 5 ) ( 2 ) (2) ( 2 ) 2 + s 2+s 2 + s − 6 -6 − 6 p 2 \mathfrak{p}_2 p 2 h K = 2 h_K = 2 h K = 2 3 + s 3+s 3 + s − 1 -1 − 1
The principal units at three are pro-three, and each residue field is F 3 \mathbb{F}_3 F 3 ψ 2 \psi^2 ψ 2 ν \nu ν K ( 2 ) = Q ( 2 , 5 ) K(\sqrt{2}) = \mathbb{Q}(\sqrt{2},\sqrt{5}) K ( 2 ) = Q ( 2 , 5 ) 8 ⋅ 5 ⋅ 40 = 40 2 8 \cdot5 \cdot40 = 40^2 8 ⋅ 5 ⋅ 40 = 4 0 2 K K K Q ( 2 ) / Q \mathbb{Q}(\sqrt{2})/\mathbb{Q} Q ( 2 ) / Q χ ˉ 3 \bar{\chi}_3 χ ˉ 3
The residual field at two
Let K = Q ( s ) K = \mathbb{Q}(s) K = Q ( s ) s 2 = 10 s^2 = 10 s 2 = 10 v + v_+ v + v − v_- v − s ≡ 1 s \equiv1 s ≡ 1 s ≡ − 1 s \equiv-1 s ≡ − 1 v + v_+ v + v − v_- v − { 0 , 1 } \{0, 1\} { 0 , 1 } v + v_+ v + S 3 S_3 S 3
The possible inducing quadratics
For a nonsolvable residual image, Proposition 18.6 and Theorem 18.7 leave only SL 2 ( F 4 ) \operatorname{SL}_2(\mathbb{F}_4) SL 2 ( F 4 )
In the solvable case Dickson’s classification gives induction from an odd-order character θ \theta θ H / K H/K H / K H H H Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 ) v + v_+ v +
− 1 , ϵ = 3 + s , 2 , 3 , 2 + s . -1,\qquad\epsilon= 3+s,\qquad2,\qquad3,\qquad2+s. − 1 , ϵ = 3 + s , 2 , 3 , 2 + s . Lemma 18.9 gives the class group of order two, generated by the dyadic class, and identifies both triadic classes with it. The fundamental unit is ϵ = 3 + s \epsilon= 3+s ϵ = 3 + s 2 2 2 3 3 3 2 + s 2+s 2 + s
For a representative a a a 3 a 3a 3 a v + v_+ v + X 2 − a X^2-a X 2 − a
Only L = K ( − 6 ) L=K(\sqrt{-6}) L = K ( − 6 ) [2 Lemma 5.1.2 and Theorem 5.2.1] applies for each of the three non-CM choices, with its ordinary witness construction and without a dyadic nonsplitting hypothesis for these choices. If the source is nonordinary, its residual dyadic point field has upper breaks at most one, by Theorem 11.7 . Both conductor-four quadratics are therefore impossible.
The odd character and the cubic
It remains to consider H = K ( d ) H=K(\sqrt{d}) H = K ( d ) d = 3 d=3 d = 3 − 6 -6 − 6 F 2 \mathbb{F}_{2} F 2 θ \theta θ v + v_{+} v + 3 = 1 + a H ( θ ) 3=1+a_{H}(\theta) 3 = 1 + a H ( θ ) v − v_{-} v − θ \theta θ
Write P + , P − P_{+},P_{-} P + , P − H H H v + , v − v_{+},v_{-} v + , v − P + P_{+} P + P − P_{-} P − [ P − ] [P_{-}] [ P − ] 1 + σ 1+\sigma 1 + σ σ \sigma σ Gal ( H / K ) \operatorname{Gal}(H/K) Gal ( H / K )
Table 1 (PDF p. 69).
Here 1 + σ 1+\sigma 1 + σ z = s + d z=s+\sqrt{d} z = s + d z 4 − 2 ( 10 + d ) z 2 + ( 10 − d ) 2 z^{4}-2(10+d)z^{2}+(10-d)^{2} z 4 − 2 ( 10 + d ) z 2 + ( 10 − d ) 2 s = ( z + ( 10 − d ) / z ) / 2 s=(z+(10-d)/z)/2 s = ( z + ( 10 − d ) / z ) /2 z ↦ 2 s − z z\mapsto2s-z z ↦ 2 s − z σ \sigma σ bnrinit and bnrgaloismatrix, adjoin the column bnrisprincipal(P_-) and the columns of 1 + σ 1+\sigma 1 + σ H H H C 2 C_{2} C 2 C 4 C_{4} C 4 bnfcertify returns one in both fields. For L L L Cl ( L ) = C 4 \operatorname{Cl}(L)=C_{4} Cl ( L ) = C 4 [49 ramified/d6_d10_d11/d10] .
There is consequently no such residual representation induced from K ( 3 ) K(\sqrt{3}) K ( 3 ) L L L A A A X 3 − 3 X + 1 + s X^{3}-3X+1+s X 3 − 3 X + 1 + s Lemma 18.5 , so it realizes this cubic quotient. In particular, once a conjugate with one of the modular images has been disposed of, every remaining two-adic residual representation has this same kernel and is the standard representation over F 2 \mathbb{F}_{2} F 2
The two normal ray modules
Proposition 18.10. Let V n V_{n} V n A A A n n n S 3 S_{3} S 3 W 0 W_{0} W 0 Hom S 3 ( V 2 , W 0 ) = 0 \operatorname{Hom}_{S_{3}}(V_{2},W_{0})=0 Hom S 3 ( V 2 , W 0 ) = 0 C 3 C_{3} C 3 F 4 \mathbb{F}_{4} F 4
Proof. Write W 0 W_{0} W 0 F 2 \mathbb{F}_{2} F 2 Gal ( A / K ) = S 3 \operatorname{Gal}(A/K)=S_{3} Gal ( A / K ) = S 3 P 1 , P 2 \mathfrak{P}_{1},\mathfrak{P}_{2} P 1 , P 2 q + \mathfrak{q}_{+} q + v + v_{+} v +
D n = Cl P 1 n P 2 n ( A ) , V n = D n / ( 2 D n + [ q + ] ) . D_{n}=\operatorname{Cl}_{\mathfrak{P}_{1}^{n}\mathfrak{P}_{2}^{n}}(A),\qquad V_{n}=D_{n}/(2D_{n}+[q_{+}]). D n = Cl P 1 n P 2 n ( A ) , V n = D n / ( 2 D n + [ q + ]) . There is no infinite modulus. The normal first-error module in the basis of the wild order-three element is W 0 W_0 W 0 q + q_+ q + n n n Hom S 3 ( V n , W 0 ) \operatorname{Hom}_{S_3}(V_n,W_0) Hom S 3 ( V n , W 0 ) G A G_A G A W 0 W_0 W 0 S 3 S_3 S 3
Use the maximal order and certified class group of A A A C 3 C_3 C 3
Table 1 (PDF p. 70).
In particular Hom S 3 ( V 2 , W 0 ) = 0 \operatorname{Hom}_{S_3}(V_2,W_0)=0 Hom S 3 ( V 2 , W 0 ) = 0
The exact matrices and local unit columns in Appendix A give a two-dimensional space of equivariant maps. For α 2 + α + 1 = 0 \alpha^2+\alpha+1=0 α 2 + α + 1 = 0 C 3 C_3 C 3 1 + α 1+\alpha 1 + α α \alpha α □ \square □
Lemma 18.11. Let R R R Q 2 \mathbb{Q}_2 Q 2 ζ 3 \zeta_3 ζ 3 r : G K → GL 2 ( R ) r:G_K\to\operatorname{GL}_2(R) r : G K → GL 2 ( R ) A A A J J J R R R v − v_- v − G A G_A G A J J J r r r J J J η = χ − 6 \eta=\chi^{-6} η = χ − 6 v + v_+ v +
Proof. The order-three eigenprojectors are integral. In their basis the normal space to the diagonal torus is the standard S 3 S_3 S 3 W 0 W_0 W 0 u u u H i ( S 3 , W 0 ) = 0 H^i(S_3,W_0)=0 H i ( S 3 , W 0 ) = 0 i > 0 i>0 i > 0 m \mathfrak{m} m R / J R/J R / J m n \mathfrak{m}^n m n m n + 1 \mathfrak{m}^{n+1} m n + 1 I = m n / m n + 1 I=\mathfrak{m}^n/\mathfrak{m}^{n+1} I = m n / m n + 1 η ( g ) \eta(g) η ( g ) κ \kappa κ W 0 ⊗ I W_0\otimes I W 0 ⊗ I κ ( u ) = 0 \kappa(u)=0 κ ( u ) = 0
At v + v_+ v + u u u v − v_- v − κ ∣ G A \kappa|_{G_A} κ ∣ G A
Here the conductor bounds apply to actual finite-flat subquotients. At the dyadic place L / K L/K L / K S = ( R / J ) / m n + 1 S=(R/J)/\mathfrak{m}^{n+1} S = ( R / J ) / m n + 1 S 2 S^2 S 2 2 2 2 Z 2 \mathbb{Z}_2 Z 2 R 2 R^2 R 2 S e 2 + I e 1 Se_2+Ie_1 S e 2 + I e 1 m S e 2 \mathfrak{m}Se_2 m S e 2 I I I A A A Theorem 11.7 . The depth-two vanishing in Proposition 18.10 therefore kills the global class.
For an ordinary tuple, the two dyadic residual Frobenius eigenvalues are distinct. The common ordinary line is consequently the same Frobenius projector in every constituent, and extends over R R R G A G_A G A Proposition 18.10 again kills the global class. The unramified local class is zero because Frobenius minus one is invertible on the nontrivial unramified cubic character.
Thus κ \kappa κ u u u W 0 u ⊗ I = 0 W_0^u \otimes I = 0 W 0 u ⊗ I = 0
The quartic mild type
Proposition 18.12. Suppose that the controlled two-adic source has the residual field A A A v − v_- v − G L G_L G L L = K ( − 6 ) L = K(\sqrt{-6}) L = K ( − 6 )
Proof. Choose a coefficient ring containing ζ 3 \zeta_3 ζ 3 v + v_+ v + v − v_- v − η = χ − 6 \eta= \chi_{-6} η = χ − 6 G A G_A G A J = 0 J = 0 J = 0
The component character is η \eta η r = Ind G L G K θ r = \operatorname{Ind}_{G_L}^{G_K}\theta r = Ind G L G K θ θ \theta θ r r r [57 Chapter III] . Its conjugate is distinct, by the absolute irreducibility of r r r L L L r r r [4 Chapter 3] .
The coefficient field of the special source
Let R \mathcal{R} R v + v_+ v + Proposition 18.3 and its type at v − v_- v − B B B Gal ( A / K ) \operatorname{Gal}(A/K) Gal ( A / K ) k = F 4 k = \mathbb{F}_4 k = F 4 W = W ( k ) W = W(k) W = W ( k ) ζ ∈ W \zeta\in W ζ ∈ W
Proposition 18.13. Let d 0 d_0 d 0 B B B Q ‾ 2 \overline{\mathbb{Q}}_2 Q 2 d 0 ≤ 2 d_0 \le2 d 0 ≤ 2 Q 2 \mathbb{Q}_2 Q 2 [ B : Q ] ≤ 4 [B : \mathbb{Q}] \le4 [ B : Q ] ≤ 4
We first describe the integral representation to which we shall apply the finite-flat conditions. Partition the embeddings of B B B ζ \zeta ζ ζ − 1 \zeta^{-1} ζ − 1 W [ 1 / 2 ] W[1/2] W [ 1/2 ] B ⊗ Q W [ 1 / 2 ] B \otimes_{\mathbb{Q}} W[1/2] B ⊗ Q W [ 1/2 ] R R R W \mathcal{W} W B B B Q \mathbb{Q} Q R R R W \mathcal{W} W k k k
The joint representation has a model on R 2 R^{2} R 2 k k k M 2 ( k ) M_{2}(k) M 2 ( k ) R R R M 2 ( k ) M_{2}(k) M 2 ( k ) R R R M 2 ( R ) M_{2}(R) M 2 ( R ) r r r Z 2 \mathbb{Z}_{2} Z 2 R 2 R^{2} R 2 K 2 K_{2} K 2 [42 Theorem 4.1] . All its generic subquotients killed by two acquire finite-flat models by schematic closure and quotient. For the nonordinary set their special fibres and their Cartier duals are connected, by the isogeny argument in the proof of Theorem 11.7 .
Induction modulo rational two
Write T = R / 2 R T=R/2R T = R /2 R n \mathfrak{n} n u u u v + v_{+} v + r ( u ) = diag ( ζ , ζ − 1 ) r(u)=\operatorname{diag}(\zeta,\zeta^{-1}) r ( u ) = diag ( ζ , ζ − 1 ) η = χ − 6 \eta=\chi_{-6} η = χ − 6 χ a \chi_{a} χ a K ( a ) / K K(\sqrt{a})/K K ( a ) / K W ⊗ F 2 k W\otimes_{\mathbb{F}_{2}} k W ⊗ F 2 k Proposition 18.10 . In particular W u = 0 W^{u}=0 W u = 0 H i ( S 3 , W ) = 0 H^{i}(S_{3},W)=0 H i ( S 3 , W ) = 0 i > 0 i>0 i > 0 W W W
Lemma 18.14. In the fixed u u u r r r T T T η \eta η
Proof. A tame generator U U U v − v_{-} v − ϵ , ϵ \epsilon,\epsilon ϵ , ϵ ϵ = ± 1 \epsilon=\pm1 ϵ = ± 1 U 2 = 2 ϵ U − 1 U^{2}=2\epsilon U-1 U 2 = 2 ϵ U − 1 U 2 = 1 U^{2}=1 U 2 = 1 G A G_{A} G A J = 2 R J=2R J = 2 R □ \square □
The residual scalar character
Lemma 18.15. For the nonordinary set dim k n / n 2 ≤ 1 \dim_{k}\mathfrak{n}/\mathfrak{n}^{2}\leq1 dim k n / n 2 ≤ 1 χ − 2 \chi_{-2} χ − 2 k η k\eta k η n = 0 \mathfrak{n}=0 n = 0
Proof. A k k k n / n 2 \mathfrak{n}/\mathfrak{n}^{2} n / n 2 k [ δ ] / ( δ 2 ) k[\delta]/(\delta^{2}) k [ δ ] / ( δ 2 ) Lemma 18.14 makes the specialized representation monomial. Compare it with the constant residual matrices. Their determinant agrees, since the cyclotomic determinant is one in characteristic two. The diagonal infinitesimal error therefore has equal entries, even on an antidiagonal coset after multiplying by the inverse residual matrix. The error is δ c ( g ) \delta c(g) δ c ( g ) c : G K → k c:G_{K}\to k c : G K → k η \eta η c c c k η k\eta k η η \eta η
At v + v_{+} v + t t t t 2 = − 1 t^{2}=-1 t 2 = − 1 t t t u u u t t t − u i t -u^{i}t − u i t i i i u u u − 1 -1 − 1 λ , − λ \lambda,-\lambda λ , − λ λ 2 = − 3 \lambda^{2}=-3 λ 2 = − 3 u u u λ ( ζ − ζ − 1 ) = ± 3 \lambda(\zeta-\zeta^{-1})=\pm3 λ ( ζ − ζ − 1 ) = ± 3 v + v_{+} v + λ \lambda λ ± ( ζ − ζ − 1 ) \pm(\zeta-\zeta^{-1}) ± ( ζ − ζ − 1 ) W W W u u u c ∣ D v + c|_{D_{v_{+}}} c ∣ D v + η \eta η c c c
Use the supported squareclass basis ⟨ − 1 , 3 + s , 2 , 3 , 2 + s ⟩ \langle-1,3+s,2,3,2+s\rangle ⟨ − 1 , 3 + s , 2 , 3 , 2 + s ⟩ § 6.4.3 . The condition at v + v_{+} v +
Table 9.
For an explicit verification form the 32 products a a a a a a − 6 a -6a − 6 a v + v_{+} v + x 2 − a x^{2}-a x 2 − a scalar_characters.gp. Scalar extension from F 2 \mathbb{F}_{2} F 2 k k k η \eta η χ − 2 \chi_{-2} χ − 2 n = 0 \mathfrak{n}=0 n = 0 [49 ramified] . □
The local height condition
Proof of Proposition 18.13. Lemma 18.15 shows that each nonempty ordinary set has cardinality dim k T = 1 \dim_k T=1 dim k T = 1 T = k [ z ] / ( z h ) T=k[z]/(z^{h}) T = k [ z ] / ( z h ) h = dim k T h=\dim_k T h = dim k T χ − 2 \chi_{-2} χ − 2 k η k\eta k η z z z F / K 2 F/K_2 F / K 2 L / K L/K L / K Lemma 18.14 splits the local representation as two rank-one T T T η \eta η F F F χ − 2 \chi_{-2} χ − 2
Take one of the rank-one summands as a finite-flat subquotient of the regular lattice modulo two. It and its dual have connected special fibres. By Lemma 11.2 , the generic coefficient action, the images of powers of z z z T [ [ u ] ] T[[u]] T [[ u ]] F F F z z z z z z
The Frobenius multiplier on each factor reduces modulo z z z u u u u − λ i u-\lambda_i u − λ i λ i ∈ ( z ) \lambda_i\in(z) λ i ∈ ( z ) u 2 ∈ ( u − λ i ) u^2\in(u-\lambda_i) u 2 ∈ ( u − λ i ) u 2 u^2 u 2 λ i \lambda_i λ i λ i 2 = 0 \lambda_i^2=0 λ i 2 = 0 λ i \lambda_i λ i z 2 z^2 z 2 u u u Theorem 11.7 , including its comparison of breaks with the field of norms. It contradicts the ramification just proved. Thus one λ i \lambda_i λ i z z z z 2 = 0 z^2 = 0 z 2 = 0 h ≤ 2 h \le2 h ≤ 2
Finally suppose that the completion of B B B Q 2 \mathbb{Q}_2 Q 2 Lemma 11.2 splits its finite-flat model as the direct sum of middle models. Therefore Frobenius at u = 0 u = 0 u = 0 F 2 \mathbb{F}_2 F 2 2 O E 2\mathcal{O}_E 2 O E E E E [42 Proposition 4.2] . This contradiction proves the ramification assertion. Summing the three bounds proves the proposition.
The small coefficient prime
Let B B B v − v_- v − B ↪ Q 2 B \hookrightarrow\mathbb{Q}_2 B ↪ Q 2 S 3 S_3 S 3 A / K A/K A / K
Proposition 18.16. If all the dyadic embeddings are ordinary, or all are nonordinary, then [ B : Q ] ≤ 2 [B : \mathbb{Q}] \le2 [ B : Q ] ≤ 2 B B B a 2 a_2 a 2 X 2 ( X 2 + X + 1 ) X^2(X^2 + X + 1) X 2 ( X 2 + X + 1 )
Proof. The trace order in each set bounds the number of embeddings in that set. Residue Frobenius interchanges the ordinary sets, so if they occur each has exactly one embedding. A nonordinary completion must ramify and its block has at most two embeddings. In the mixed case it has exactly two. At an ordinary embedding one crystalline Frobenius root is the unramified quotient and the other is divisible by the maximal ideal; a 2 a_2 a 2 a 2 ∈ B a_2 \in B a 2 ∈ B [ B : Q ( a 2 ) ] [B : \mathbb{Q}(a_2)] [ B : Q ( a 2 )]
Lemma 18.17. Let B B B Proposition 18.16 . Then at least one of the following holds:
(i) Every prime of B B B
(ii) There is a degree-one prime of B B B
(iii) Up to replacing a 2 a_2 a 2 − a 2 -a_2 − a 2
X 4 − X 3 − 7 X 2 + 8 X + 2 or X 4 − X 3 − 5 X 2 + 2 X + 2. X^4 - X^3 - 7X^2 + 8X + 2 \quad\text{or} \quad X^4 - X^3 - 5X^2 + 2X + 2. X 4 − X 3 − 7 X 2 + 8 X + 2 or X 4 − X 3 − 5 X 2 + 2 X + 2. Proof. Every conjugate of the algebraic integer a 2 a_2 a 2 [ − 8 , 8 ] [-\sqrt{8}, \sqrt{8}] [ − 8 , 8 ] P = X 4 + a X 3 + b X 2 + c X + d P = X^4 + aX^3 + bX^2 + cX + d P = X 4 + a X 3 + b X 2 + c X + d Proposition 18.16 , a , b a,b a , b c , d c,d c , d
The elementary symmetric functions and Cauchy’s inequality give
∣ a ∣ ≤ 11 , ∣ b ∣ ≤ 48 , ∣ c ∣ ≤ 90 , 0 < ∣ d ∣ ≤ 64 , 0 ≤ a 2 − 2 b ≤ 32 , 8 b ≤ 3 a 2 . |a| \le11,\quad|b| \le48,\quad|c| \le90,\quad0 < |d| \le64,\quad0 \le a^2 - 2b \le32,\quad8b \le3a^2. ∣ a ∣ ≤ 11 , ∣ b ∣ ≤ 48 , ∣ c ∣ ≤ 90 , 0 < ∣ d ∣ ≤ 64 , 0 ≤ a 2 − 2 b ≤ 32 , 8 b ≤ 3 a 2 . Replacing a 2 a_2 a 2 − a 2 -a_2 − a 2 a > 0 a > 0 a > 0 [ − 283 / 100 , 283 / 100 ] [-283/100,283/100] [ − 283/100 , 283/100 ] ( p , r , e ) (p,r,e) ( p , r , e ) P ( r ) ≡ 0 P(r) \equiv0 P ( r ) ≡ 0 P ′ ( r ) ≡ e ≠ 0 P'(r) \equiv e \ne0 P ′ ( r ) ≡ e = 0 p p p
Table 1 (PDF p. 75).
The accompanying certificate uses only integer and rational arithmetic: it tests irreducibility by Gauss’s lemma and counts real roots by Sturm’s theorem. Thus this step requires no maximal-order computation.
A simple root modulo five or seven gives an embedding B ↪ Q p B \hookrightarrow\mathbb{Q}_p B ↪ Q p Q = X 2 − X − 1 Q = X^2 - X - 1 Q = X 2 − X − 1
P = Q 2 + 3 ( X 3 − 2 X 2 − 4 X − 1 ) . P = Q^2 + 3(X^3 - 2X^2 - 4X - 1). P = Q 2 + 3 ( X 3 − 2 X 2 − 4 X − 1 ) . The polynomial Q Q Q Q Q Q − 4 X − 2 -4X - 2 − 4 X − 2 [ O B : Z [ a 2 ] ] [\mathcal{O}_B : \mathbb{Z}[a_2]] [ O B : Z [ a 2 ]] B B B
For the penultimate row, put α = a 2 \alpha= a_2 α = a 2 β = ( α 3 + α 2 + α ) / 2 \beta= (\alpha^3 + \alpha^2 + \alpha)/2 β = ( α 3 + α 2 + α ) /2 P P P
β 4 − 29 β 2 + 10 β + 19 = 0. \beta^4 - 29\beta^2 + 10\beta+ 19 = 0. β 4 − 29 β 2 + 10 β + 19 = 0. Thus β \beta β disc P = 8900 \operatorname{disc} P = 8900 disc P = 8900 1 , α , α 2 , β 1,\alpha,\alpha^2,\beta 1 , α , α 2 , β Proposition 18.16 . The two remaining rows are exactly the polynomials in (iii), after replacing X X X − X -X − X
Lemma 18.18. If an irreducible residual representation of determinant χ ˉ 3 \bar{\chi}_3 χ ˉ 3 G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 )
Proof. Otherwise the representation is induced from a character θ \theta θ H = K ( ζ 3 ) H = K(\zeta_3) H = K ( ζ 3 ) θ σ = θ − 1 \theta^\sigma= \theta^{-1} θ σ = θ − 1 H H H X 4 + 10 X 2 + 100 X^4 + 10X^2 + 100 X 4 + 10 X 2 + 100 bnfcertify; its ray group for this modulus is also C 4 C_4 C 4 S S S K K K 1 + S 1+S 1 + S 1 + σ 1+\sigma 1 + σ C 2 C_2 C 2 θ 2 = 1 \theta^2 = 1 θ 2 = 1 θ = θ σ \theta= \theta^\sigma θ = θ σ cyclotomic_three_ray.gp.
Proof of Theorem 18.1. Soluble residues have solvable odd Artin lifts, which give modularity by Hecke induction or Langlands–Tunnell and regular weight raising. Assume therefore that the prescribed three-residue is nonsolvable. The bounded lift, the discriminant estimate and the soluble-image classification reduce the proof to the special source with the fixed S 3 S_3 S 3 [49 ramified] .
If the degree of B B B F 3 \mathbb{F}_3 F 3 F 9 \mathbb{F}_9 F 9
At a degree-one coefficient prime of characteristic p = 5 p = 5 p = 5 7 7 7 v + v_+ v + p p p v + v_+ v + G K ( ζ p ) G_{K(\zeta_p)} G K ( ζ p ) [3 ] give respectively modularity of its residue and modularity of the actual de Rham member with weights { 0 , 1 } \{0,1\} { 0 , 1 }
The last two polynomials in Lemma 18.17 have a coefficient place above three with residue field F 3 \mathbb{F}_3 F 3 ψ \psi ψ χ ‾ 3 ψ − 1 \overline{\chi}_3\psi^{-1} χ 3 ψ − 1 ψ 2 = 1 \psi^2 = 1 ψ 2 = 1 ν \nu ν G K v − G_{K_{v_-}} G K v − α ⊕ α χ 3 \alpha\oplus\alpha\chi_3 α ⊕ α χ 3 α \alpha α ψ 2 \psi^2 ψ 2 ψ 2 = 1 \psi^2 = 1 ψ 2 = 1 ψ ( Frob 2 ) ( 1 + 2 ) = 0 \psi(\operatorname{Frob}_2)(1 + 2) = 0 ψ ( Frob 2 ) ( 1 + 2 ) = 0 λ ∣ 3 \lambda\mid3 λ ∣ 3 a 2 ∈ λ a_2 \in\lambda a 2 ∈ λ 3 ∣ N B / Q ( a 2 ) 3 \mid N_{B/\mathbb{Q}}(a_2) 3 ∣ N B / Q ( a 2 ) N B / Q ( a 2 ) = 2 N_{B/\mathbb{Q}}(a_2) = 2 N B / Q ( a 2 ) = 2 [3 Theorems 2.5 and 2.3] make the actual member modular. This proves the residual theorem.
Quadratic symmetry over Q ( 11 ) \mathbb{Q}(\sqrt{11}) Q ( 11 )
Put K = Q ( s ) K = \mathbb{Q}(s) K = Q ( s ) s 2 = 11 s^2 = 11 s 2 = 11
Arithmetic and modular witnesses
We will use the elements
ε = 10 + 3 s , π = 3 + s , α = 4 + s , α ′ = 4 − s . \varepsilon= 10 + 3s,\qquad\pi= 3 + s,\qquad\alpha= 4 + s,\qquad\alpha' = 4 - s. ε = 10 + 3 s , π = 3 + s , α = 4 + s , α ′ = 4 − s . They satisfy N ( ε ) = 1 N(\varepsilon) = 1 N ( ε ) = 1 N ( π ) = − 2 N(\pi) = -2 N ( π ) = − 2 N ( α ) = N ( α ′ ) = 5 N(\alpha) = N(\alpha') = 5 N ( α ) = N ( α ′ ) = 5
Lemma 19.1. The class number of K K K − 1 -1 − 1 ε \varepsilon ε C 8 × C 2 C_8 \times C_2 C 8 × C 2 C 4 × C 2 C_4 \times C_2 C 4 × C 2
Proof. Minkowski’s bound is 44 / 2 < 3.32 \sqrt{44}/2 < 3.32 44 /2 < 3.32 π \pi π x 2 − 11 y 2 = 1 x^2 - 11y^2 = 1 x 2 − 11 y 2 = 1 ( 10 , 3 ) (10,3) ( 10 , 3 ) y = 1 , 2 y = 1,2 y = 1 , 2 ε \varepsilon ε − 1 -1 − 1
Modulo three the residue group is F 3 × \mathbb{F}_3^\times F 3 × ε = 1 \varepsilon= 1 ε = 1 { ± 1 } \{\pm1\} { ± 1 }
( C 8 × C 2 2 ) / ( ( − 1 , − 1 , − 1 ) ) ≅ C 8 × C 2 . (C_8 \times C_2^2)/((-1,-1,-1)) \cong C_8 \times C_2. ( C 8 × C 2 2 ) / (( − 1 , − 1 , − 1 )) ≅ C 8 × C 2 . At the two five-places, ordered by s = 1 , − 1 s=1,-1 s = 1 , − 1 5 5 5 ϵ \epsilon ϵ ( 3 , 2 ) (3,2) ( 3 , 2 ) ( − 1 , − 1 ) (-1,-1) ( − 1 , − 1 ) − 1 -1 − 1 ( u , v ) ↦ u v (u,v)\mapsto uv ( u , v ) ↦ uv F 5 × \mathbb{F}_5^\times F 5 × C 4 × C 2 C_4 \times C_2 C 4 × C 2
Characteristic-three sources
Lemma 19.2. Let F / Q 3 F/\mathbb{Q}_3 F / Q 3 r r r { 0 , 1 } \{0,1\} { 0 , 1 } χ 3 \chi_3 χ 3 1 , ω 3 1,\omega_3 1 , ω 3 r r r
Proof. Since e ( F ) = 1 < 3 − 1 e(F)=1<3-1 e ( F ) = 1 < 3 − 1 r ∨ ( 1 ) ≃ r r^\vee(1)\simeq r r ∨ ( 1 ) ≃ r
Proposition 19.3. If r ˉ 3 s s = ψ ⊕ ω 3 ψ − 1 \bar r_3^{\mathrm{ss}}=\psi\oplus\omega_3\psi^{-1} r ˉ 3 ss = ψ ⊕ ω 3 ψ − 1 R \mathcal{R} R ψ 2 = 1 \psi^2=1 ψ 2 = 1 r 3 r_3 r 3
Proof. At a five-place the semisimplified residual inertia of our two types is scalar quadratic: the cubic part of the cuspidal inertia disappears modulo three, as does the unipotent part of the special condition. Hence ψ 2 \psi^2 ψ 2 ψ 2 \psi^2 ψ 2 ( 3 ) (3) ( 3 )
Fontaine–Laffaille theory [32 ] over the unramified quadratic K 3 K_3 K 3 0 , 1 , 3 , 4 0,1,3,4 0 , 1 , 3 , 4 8 8 8 { 0 , 4 } \{0,4\} { 0 , 4 } { 1 , 3 } \{1,3\} { 1 , 3 } ψ 2 \psi^2 ψ 2 α , α ′ \alpha,\alpha' α , α ′ 1 + s , 1 − s 1+s,1-s 1 + s , 1 − s F 9 \mathbb{F}_9 F 9
( 1 + s ) 2 = − s , ( 1 + s ) 4 = − 1 , 1 − s = ( 1 + s ) 3 . (1+s)^2=-s,\qquad(1+s)^4=-1,\qquad1-s=(1+s)^3. ( 1 + s ) 2 = − s , ( 1 + s ) 4 = − 1 , 1 − s = ( 1 + s ) 3 . It follows that ψ 2 ( Frob v ) \psi^2(\operatorname{Frob}_v) ψ 2 ( Frob v )
The quotient of the two residual local characters would therefore have Frobenius value of order four. For both of our five-adic types that quotient is instead − 1 -1 − 1 { 1 , 3 } \{1,3\} { 1 , 3 }
For the pair { 0 , 4 } \{0,4\} { 0 , 4 } ψ 2 \psi^2 ψ 2 ψ 2 = 1 \psi^2=1 ψ 2 = 1 ω 3 \omega_3 ω 3 G Q G_{\mathbb{Q}} G Q K 3 K_3 K 3 K ( ζ 3 ) K(\zeta_3) K ( ζ 3 ) Q \mathbb{Q} Q [62 Theorem A] are satisfied, so r 3 r_3 r 3
Proposition 19.4. Let r : G K → G L 2 ( Q ‾ 3 ) r:G_K\to\mathrm{GL}_2(\overline{\mathbb{Q}}_3) r : G K → GL 2 ( Q 3 ) χ 3 \chi_3 χ 3 r ˉ \bar r r ˉ G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 ) r r r
Proof. Put H = K ( ζ 3 ) H=K(\zeta_3) H = K ( ζ 3 ) r ˉ = Ind G H G K ξ \bar r=\operatorname{Ind}_{G_H}^{G_K}\xi r ˉ = Ind G H G K ξ ξ \xi ξ H 3 / K 3 H_3/K_3 H 3 / K 3 t t t G H 3 G_{H_3} G H 3
det r ˉ ( t ) = − ξ ( t 2 ) = ω 3 ( t ) = − 1. \det\bar{r}(t)=-\xi(t^2)=\omega_3(t)=-1. det r ˉ ( t ) = − ξ ( t 2 ) = ω 3 ( t ) = − 1. Thus r ˉ ( t ) 2 = 1 \bar{r}(t)^2=1 r ˉ ( t ) 2 = 1 G H 3 G_{H_3} G H 3 r ˉ ( t ) \bar{r}(t) r ˉ ( t ) ± 1 \pm1 ± 1 β , β ω 3 \beta,\beta\omega_3 β , β ω 3 β \beta β
Induce the Teichmuller lift of ξ \xi ξ ± 1 \pm1 ± 1 [73 Theorem 1.4.1] gives a regular ordinary specialization with the same residual quotient. This application is also explicit in [27 pp. 1133–1134] .
We have obtained the quotient-compatible modular lift required by [63 Theorem 5.1] . The actual source is ordinary and distinguished at three, its determinant is cyclotomic, and its remaining geometric and parity hypotheses are the stated ones. That theorem permits reducibility on the cyclotomic subgroup and proves modularity of r r r
Nonsolvable and dihedral dyadic residues
In this section the system is special at both five-primes. Let M M M Q 5 \mathbb{Q}_5 Q 5
Lemma 19.5. Put F = K ( i , 5 ) F=K(i,\sqrt{5}) F = K ( i , 5 ) H H H [ H : Q ] = 16 [H:\mathbb{Q}]=16 [ H : Q ] = 16 ( e , f ) = ( 2 , 4 ) (e,f)=(2,4) ( e , f ) = ( 2 , 4 ) ( e , f ) = ( 2 , 2 ) (e,f)=(2,2) ( e , f ) = ( 2 , 2 ) C 4 2 × C 2 4 C_4^2 \times C_2^4 C 4 2 × C 2 4
Proof. The defining polynomials for F , H F,H F , H
y 8 − 4 y 7 + 24 y 6 − 58 y 5 + 141 y 4 − 190 y 3 + 186 y 2 − 100 y + 20 , y^8-4y^7+24y^6-58y^5+141y^4-190y^3+186y^2-100y+20, y 8 − 4 y 7 + 24 y 6 − 58 y 5 + 141 y 4 − 190 y 3 + 186 y 2 − 100 y + 20 , y 16 − 9 y 14 + 39 y 12 − 111 y 10 + 214 y 8 − 189 y 6 + 69 y 4 − 6 y 2 + 1. y^{16}-9y^{14}+39y^{12}-111y^{10}+214y^8-189y^6+69y^4-6y^2+1. y 16 − 9 y 14 + 39 y 12 − 111 y 10 + 214 y 8 − 189 y 6 + 69 y 4 − 6 y 2 + 1. The class groups are C 2 , 1 C_2,1 C 2 , 1 H H H 12 , 4 , 4 , 2 , 2 , 2 , 2 , 2 12,4,4,2,2,2,2,2 12 , 4 , 4 , 2 , 2 , 2 , 2 , 2 sp_auxiliary.gp checks the polynomial identification, both class groups with bnfcertify, the prime decompositions and the Smith factors. The exact calculation is supplied in [49 ramified/d6_d10_d11/d11] .
Proposition 19.6. Assume GRH. A two-special bounded system with a nonsolvable dyadic residue is modular.
Proof. Its image is SL 2 ( q ) \operatorname{SL}_2(q) SL 2 ( q ) q = 2 f ≥ 4 q=2^f\geq4 q = 2 f ≥ 4 B B B Lemma 19.5 , of absolute degree 4096 4096 4096 K K K B M / K BM/K B M / K H / K H/K H / K v 2 ( 2 ) = 1 v_2(2)=1 v 2 ( 2 ) = 1 4 / 2 = 2 4/2=2 4/2 = 2 H H H Q 5 ( 5 ) \mathbb{Q}_5(\sqrt{5}) Q 5 ( 5 ) L = B M L = BM L = B M L = B M M σ L = BMM^{\sigma} L = B M M σ
rd ( L ) ≤ 4 44 ⋅ 5 < 59.329588 , 2 N 25 ( L ) [ L : Q ] = 1 / 2. \operatorname{rd}(L) \le4\sqrt{44\cdot5} < 59.329588,\qquad\frac{2N_{25}(L)}{[L:\mathbb{Q}]} = 1/2. rd ( L ) ≤ 4 44 ⋅ 5 < 59.329588 , [ L : Q ] 2 N 25 ( L ) = 1/2. Here N Q N_Q N Q Q Q Q
Apply Section 22 with B b , 100 B_{b,100} B b , 100 A b , 4 A_{b,4} A b , 4
Table 19.7.
Simplicity gives disjointness from the solvable auxiliary. Distinct simple conjugate fields are disjoint over K K K [49 ramified/d6_d10_d11/shared] .
When q = 8 q=8 q = 8 M = M σ M=M^{\sigma} M = M σ Gal ( M / Q ) \operatorname{Gal}(M/\mathbb{Q}) Gal ( M / Q ) SL 2 ( 8 ) \operatorname{SL}_2(8) SL 2 ( 8 ) G Q G_{\mathbb{Q}} G Q [40 , 41 ] and quadratic base change give a regular cuspidal witness over K K K q = 4 q=4 q = 4 [3 Theorem 2.5] . A real conjugation is nontrivial in either case: otherwise M M M F 2 \mathbb{F}_2 F 2 q q q 2 − 2 / q 2-2/q 2 − 2/ q rd ( M ) ≤ 44 ⋅ 5 7 / 4 < 49.890038 \operatorname{rd}(M)\le\sqrt{44\cdot5^{7/4}}<49.890038 rd ( M ) ≤ 44 ⋅ 5 7/4 < 49.890038 [69 Theorem 6.1] , applied to the dual in the opposite weight convention, give modularity of the actual dyadic source.
Proposition 19.7. A two-special system with irreducible solvable two-residue is modular. Every reducible two-residue has trivial semisimplification.
Proof. An irreducible solvable subgroup in determinant one is odd dihedral. Let L / K L/K L / K L L L L L L − 1 , ϵ , π , α , α ′ -1,\epsilon,\pi,\alpha,\alpha' − 1 , ϵ , π , α , α ′ ⟨ − 1 , ϵ , α , α ′ ⟩ \langle-1,\epsilon,\alpha,\alpha'\rangle ⟨ − 1 , ϵ , α , α ′ ⟩ − 5 ϵ -5\epsilon − 5 ϵ C 3 C_3 C 3 sp_dihedral_rays.gp verifies these quartic ray quotients. Since π 2 = 2 ϵ \pi^2=2\epsilon π 2 = 2 ϵ K ( − 10 ) K(\sqrt{-10}) K ( − 10 ) [2 Theorem 5.2.1] applies. For a reducible residue the odd-order constituents are unramified at two and five and everywhere else, so class number one makes both trivial. The exact calculation is supplied in [49 ramified/d6_d10_d11/d11] .
Ordinary scalar systems
Proposition 19.8. A two-special system with a scalar semisimple two-residue and an ordinary member above two is modular.
Proof. In that member choose a vector v v v v v v v v v χ = χ − 5 \chi=\chi_{-5} χ = χ − 5 ⟨ − 1 , ϵ , α , α ′ ⟩ \langle-1,\epsilon,\alpha,\alpha'\rangle ⟨ − 1 , ϵ , α , α ′ ⟩ 1 1 1 5 5 5 − 1 -1 − 1 − 5 -5 − 5 1 1 1 − 5 -5 − 5 ordinary_squareclasses.gp. Every coordinate of the additive residual extension is a multiple of χ \chi χ c c c ( ( r ( c ) − 1 ) v , v ) ((r(c)-1)v,v) (( r ( c ) − 1 ) v , v )
1 + χ E 12 , ( − 1 1 0 1 ) , e 2 . 1+\chi E_{12},\qquad
\begin{pmatrix}
-1 & 1\\
0 & 1
\end{pmatrix},\qquad e_2. 1 + χ E 12 , ( − 1 0 1 1 ) , e 2 . Use the marked ordinary crystalline functor of [43 Proposition 2.4.4, Corollary 2.4.5] , pulled back to the matrix problem with this fixed real matrix, support and determinant. At five fix the quadratic characters on inertia, and allow the unramified special signs to vary. The special line is represented without an extra free parameter: for a tame generator normalized by the given sign, its difference from one has unit upper-right entry, so the coordinate of a fixed line in the e 1 e_1 e 1 [49 ramified] .
Write a homogeneous tangent as ( 1 + t X ( g ) ) r ˉ ( g ) (1+tX(g))\bar r(g) ( 1 + tX ( g )) r ˉ ( g ) X ( g ) = ( x ( g ) y ( g ) z ( g ) x ( g ) ) X(g)=\begin{pmatrix}x(g)&y(g)\\z(g)&x(g)\end{pmatrix} X ( g ) = ( x ( g ) z ( g ) y ( g ) x ( g ) ) χ z ( 1 + E 12 ) \chi z(1+E_{12}) χz ( 1 + E 12 ) z z z y = 0 y=0 y = 0 x x x z z z e 1 + t t w e 2 e_1+tt_we_2 e 1 + t t w e 2 z = 0 z=0 z = 0 α , α ′ \alpha,\alpha' α , α ′ ϵ \epsilon ϵ z z z χ − 1 \chi_{-1} χ − 1 x , y x,y x , y y ∈ k χ − 5 y\in k\chi_{-5} y ∈ k χ − 5 y ( c ) = 0 y(c)=0 y ( c ) = 0 x ( c ) = 0 x(c)=0 x ( c ) = 0 x ∈ k χ 5 x\in k\chi_5 x ∈ k χ 5 ν \nu ν χ 5 \chi_5 χ 5 W ( k ) [ [ ν ( Frob 2 ) − 1 ] ] W(k)[[\nu(\operatorname{Frob}_2)-1]] W ( k ) [[ ν ( Frob 2 ) − 1 ]]
Choose η = 1 \eta=1 η = 1 χ − 5 \chi_{-5} χ − 5 r r r r σ ⊗ η r^\sigma\otimes\eta r σ ⊗ η
The integral order of nonordinary scalar systems
Take the tuple at all dyadic coefficient places of the bounded systems under consideration. Every member is nonordinary with trivial residual semisimplification, and the systems have common quadratic inertial signs at each special place. Use the orders R 0 , E 0 R_0,E_0 R 0 , E 0 m m m d d d
rank Z 2 R 0 = m d , rank Z 2 E 0 = 4 m d . \operatorname{rank}_{\mathbb Z_2} R_0=md,\qquad
\operatorname{rank}_{\mathbb Z_2} E_0=4md. rank Z 2 R 0 = m d , rank Z 2 E 0 = 4 m d . Write E = E 0 / 2 E 0 E=E_0/2E_0 E = E 0 /2 E 0 F 2 \mathbb F_2 F 2
Lemma 19.9. The left regular lattice of E 0 E_0 E 0 J J J E E E A , B , C A,B,C A , B , C E E E F 2 \mathbb F_2 F 2
Proof. Apply Lemmas 11.3 and 11.2 to the regular lattice and its trivial-simple torsion subquotients. These lemmas also hold after the unramified base changes used below.
A radical functional gives a middle/middle extension by quotienting the regular model. At the dyadic place its quadratic conductor is at most two by the height-one Artin–Schreier calculation in the proof of Theorem 11.7 . These characters belong to the narrow ray modulo π 2 ( 5 ) \pi^{2}(5) π 2 ( 5 ) 4 , 2 , 2 4,2,2 4 , 2 , 2
[ − 1 ] , [ ϵ α ] , [ ϵ α ′ ] . [-1],\qquad[\epsilon\alpha],\qquad[\epsilon\alpha']. [ − 1 ] , [ ϵ α ] , [ ϵ α ′ ] . Their dyadic discriminant exponents are respectively 0 , 2 , 2 0,2,2 0 , 2 , 2 A , B , C A,B,C A , B , C J / J 2 J/J^{2} J / J 2 scalar_regular_rays.gp. The exact calculation is supplied in [49 ramified] .
A quaternion obstruction over the Hilbert class field
The next result is used both for the original reduction and a central quotient of it. Suppose E ′ E' E ′ M ′ M' M ′ P = Gal ( M ′ H / H ) P=\operatorname{Gal}(M'H/H) P = Gal ( M ′ H / H ) M ′ H / H M'H/H M ′ H / H H H H
Lemma 19.10. Assume GRH. The group P P P E ′ E' E ′
Proof. Denote the dyadic primes of H H H v 1 , v 2 v_{1},v_{2} v 1 , v 2 ( 4 , 0 ) (4,0) ( 4 , 0 ) ( 0 , 4 ) (0,4) ( 0 , 4 ) ( 2 , 2 ) (2,2) ( 2 , 2 ) C 2 2 C_{2}^{2} C 2 2 P P P P P P
Let C 0 C_{0} C 0 P P P E ′ E' E ′ J 0 J_{0} J 0 C 0 ⊂ E ′ C_{0}\subset E' C 0 ⊂ E ′ J 0 / J 0 2 J_{0}/J_{0}^{2} J 0 / J 0 2 Theorem 11.7 . The ( 2 , 2 ) (2,2) ( 2 , 2 ) dim J 0 / J 0 2 ≤ 2 \dim J_{0}/J_{0}^{2}\leq2 dim J 0 / J 0 2 ≤ 2 F 2 \mathbf{F}_{2} F 2 P P P
The right action of C 0 C_{0} C 0 ( k ⊗ F 2 C 0 ) [ [ u ] ] (k\otimes_{\mathbf{F}_{2}} C_{0})[[u]] ( k ⊗ F 2 C 0 ) [[ u ]] k = F 16 k=\mathbf{F}_{16} k = F 16 u + V ( u ) u+V(u) u + V ( u ) V ( u ) ∈ ( k ⊗ J 0 ) [ [ u ] ] V(u)\in(k\otimes J_{0})[[u]] V ( u ) ∈ ( k ⊗ J 0 ) [[ u ]] u 2 ( u + V ) − 1 u^{2}(u+V)^{-1} u 2 ( u + V ) − 1 J 0 3 J_{0}^{3} J 0 3 u − V + V 2 / u u-V+V^{2}/u u − V + V 2 / u V ( 0 ) 2 = 0 V(0)^{2}=0 V ( 0 ) 2 = 0
With one radical generator its nonzero ramified coefficient gives J 0 2 = J 0 3 J_{0}^{2}=J_{0}^{3} J 0 2 = J 0 3 J 0 2 = 0 J_{0}^{2}=0 J 0 2 = 0 V ( 0 ) = α X + β Y ( m o d J 0 2 ) V(0)=\alpha X+\beta Y\pmod{J_{0}^{2}} V ( 0 ) = α X + β Y ( mod J 0 2 ) α , β ∈ F 16 \alpha,\beta\in\mathbf{F}_{16} α , β ∈ F 16 F 2 \mathbf{F}_{2} F 2 β / α ∉ F 4 \beta/\alpha\notin\mathbf{F}_{4} β / α ∈ / F 4 1 , β / α , ( β / α ) 2 1,\beta/\alpha,(\beta/\alpha)^{2} 1 , β / α , ( β / α ) 2 1 + t + t 2 = 0 1+t+t^{2}=0 1 + t + t 2 = 0 X 2 , X Y + Y X , Y 2 X^{2},XY+YX,Y^{2} X 2 , X Y + Y X , Y 2 J 0 2 / J 0 3 J_{0}^{2}/J_{0}^{3} J 0 2 / J 0 3 z z z z = 0 z=0 z = 0 F 2 [ X , Y ] / ( X 2 , Y 2 ) \mathbf{F}_{2}[X,Y]/(X^{2},Y^{2}) F 2 [ X , Y ] / ( X 2 , Y 2 ) J 0 3 = 0 J_{0}^{3}=0 J 0 3 = 0 C 0 C_{0} C 0
If z ≠ 0 z\neq0 z = 0 J 0 3 J_{0}^{3} J 0 3 1 + z 1+z 1 + z ⟨ 1 + z ⟩ \langle1+z\rangle ⟨ 1 + z ⟩ P P P H H H H ′ / H H'/H H ′ / H scalar_quaternion.gp gives dyadic break one at both places and the split-five ray group C 4 × C 2 C_4 \times C_2 C 4 × C 2 H ′ / H H'/H H ′ / H C 2 C_2 C 2 S + I S+I S + I 1 + 2 ( 3 − 1 ) = 5 1+2(3-1)=5 1 + 2 ( 3 − 1 ) = 5 [49 ramified/d6_d10_d11/d11] .
We have proved that C 0 C_0 C 0 F 2 [ X , Y ] / ( X 2 , Y 2 ) \mathbb{F}_2[X,Y]/(X^2,Y^2) F 2 [ X , Y ] / ( X 2 , Y 2 ) P P P ∣ P ∣ ≤ 8 |P|\leq8 ∣ P ∣ ≤ 8 C 0 C_0 C 0 P P P
u ( 1 + f X + g Y + h X Y ) , f , g , h ∈ u − 1 k [ [ u ] ] . u(1+fX+gY+hXY), \qquad f,g,h\in u^{-1}k[[u]]. u ( 1 + f X + g Y + h X Y ) , f , g , h ∈ u − 1 k [[ u ]] . After removing the simple factor u u u 1 + x X + y Y + w X Y 1+xX+yY+wXY 1 + x X + y Y + w X Y
x 2 − x = f , y 2 − y = g , ( w + x y ) 2 − ( w + x y ) = h + f g . x^2-x=f,\qquad y^2-y=g,\qquad(w+xy)^2-(w+xy)=h+fg. x 2 − x = f , y 2 − y = g , ( w + x y ) 2 − ( w + x y ) = h + f g . Multiplication by 1 + a X + b Y + c X Y ∈ P 1+aX+bY+cXY\in P 1 + a X + bY + c X Y ∈ P ( x , y , w + x y ) (x,y,w+xy) ( x , y , w + x y ) ( a , b , c + a b ) (a,b,c+ab) ( a , b , c + ab ) P P P k k k P P P F 2 \mathbb{F}_2 F 2 Theorem 11.7 identifies them with the field-of-norms breaks. Every character consequently has conductor at most two over each H v i H_{v_i} H v i P P P C 2 2 C_2^2 C 2 2
Quadratic conjugation and the scalar theorem
Lemma 19.11. In the setting of Lemma 19.10, if E ′ E' E ′ dim F 2 E ′ ≤ 24 \dim_{\mathbb{F}_2}E'\leq24 dim F 2 E ′ ≤ 24
Proof. The group Gal ( H / K ) \operatorname{Gal}(H/K) Gal ( H / K ) Gal ( F / K ) \operatorname{Gal}(F/K) Gal ( F / K )
Set Γ = Gal ( M ′ H / K ) \Gamma=\operatorname{Gal}(M'H/K) Γ = Gal ( M ′ H / K ) a , b , c a,b,c a , b , c b , c b,c b , c Gal ( H / K ) \operatorname{Gal}(H/K) Gal ( H / K ) r = a b r=ab r = ab c = j b c=jb c = j b j ∈ P j\in P j ∈ P R , B R,B R , B r , b r,b r , b P P P b r b = r − 1 b r b=r^{-1} b r b = r − 1 B j = j B j=j B j = j Γ \Gamma Γ Aut ( P ) \operatorname{Aut}(P) Aut ( P ) R 2 = 1 R^2=1 R 2 = 1
At the five-place belonging to b b b f = p r 2 f=pr^2 f = p r 2 p ∈ P p\in P p ∈ P M ′ H M'H M ′ H f 2 = 1 f^2=1 f 2 = 1 R 2 = 1 R^2=1 R 2 = 1 P P P f 2 = r 4 f^2=r^4 f 2 = r 4 r 4 = 1 r^4=1 r 4 = 1 ⟨ r , b ⟩ \langle r,b\rangle ⟨ r , b ⟩ Γ 0 = ⟨ a , b , c ⟩ \Gamma_0=\langle a,b,c\rangle Γ 0 = ⟨ a , b , c ⟩ P 0 × ⟨ r , b ⟩ P_0\times\langle r,b\rangle P 0 × ⟨ r , b ⟩ P 0 = ⟨ j , R j ⟩ P_0=\langle j,Rj\rangle P 0 = ⟨ j , R j ⟩ B B B j j j R R R E ′ E' E ′ Γ 0 \Gamma_0 Γ 0 Γ \Gamma Γ dim P 0 ≤ 1 \dim P_0 \le1 dim P 0 ≤ 1 dim E ′ ≤ ∣ Γ 0 ∣ ≤ 16 \dim E' \le|\Gamma_0| \le16 dim E ′ ≤ ∣ Γ 0 ∣ ≤ 16 P 0 = P P_0=P P 0 = P R = U R=U R = U j j j B ∈ { 1 , U } B\in\{1,U\} B ∈ { 1 , U } j j j B = 1 B=1 B = 1
Now Γ = Γ 0 = P ⋊ D 8 \Gamma=\Gamma_0=P\rtimes D_8 Γ = Γ 0 = P ⋊ D 8 q = R ( j ) j q=R(j)j q = R ( j ) j q q q r 2 r^2 r 2 r + r − 1 r+r^{-1} r + r − 1 E ′ E' E ′ j j j
( 1 + q ) ( 1 + r 2 ) = 0 in E ′ . (1+q)(1+r^2)=0 \qquad\text{in }E'. ( 1 + q ) ( 1 + r 2 ) = 0 in E ′ . Indeed r j = q j r rj=qjr r j = q j r r − 1 j = q j r − 1 r^{-1}j=qjr^{-1} r − 1 j = q j r − 1 F 2 [ Γ ] \mathbf F_2[\Gamma] F 2 [ Γ ] F 2 [ ⟨ q , r 2 ⟩ ] \mathbf F_2[\langle q,r^2\rangle] F 2 [⟨ q , r 2 ⟩] ( 1 + q ) ( 1 + r 2 ) (1+q)(1+r^2) ( 1 + q ) ( 1 + r 2 ) F 2 [ Γ ] \mathbf F_2[\Gamma] F 2 [ Γ ] dim E ′ ≤ 32 − 8 = 24 \dim E'\le32-8=24 dim E ′ ≤ 32 − 8 = 24
Theorem 19.12. Assume GRH. A two-special bounded system all of whose dyadic members are nonordinary and residually scalar is modular.
Proof. Let h = χ − 1 h=\chi^{-1} h = χ − 1 η = 1 \eta=1 η = 1 χ − 5 \chi^{-5} χ − 5 r r r r σ ⊗ η r^\sigma\otimes\eta r σ ⊗ η σ \sigma σ h h h η \eta η r ↦ r ⊗ h r\mapsto r\otimes h r ↦ r ⊗ h g : r ↦ r σ ⊗ η g:r\mapsto r^\sigma\otimes\eta g : r ↦ r σ ⊗ η m m m d d d E E E σ \sigma σ r , r h r,rh r , r h g ( r ) , g ( r ) h g(r),g(r)h g ( r ) , g ( r ) h σ \sigma σ h h h η \eta η σ \sigma σ
Write S S S + 1 +1 + 1 − 1 -1 − 1 T ( F ) = ± 6 S T(F)=\pm6S T ( F ) = ± 6 S t = 2 S ∈ R 0 t=2S\in R_0 t = 2 S ∈ R 0 B , C B,C B , C B F = ± S B BF=\pm SB B F = ± S B C F = ± S C CF=\pm SC C F = ± S C E E E t B = t C = 0 tB=tC=0 tB = tC = 0 t 2 = 0 t^2=0 t 2 = 0 F 2 − T ( F ) F + 5 = 0 F^2-T(F)F+5=0 F 2 − T ( F ) F + 5 = 0 E ′ = E / t E E'=E/tE E ′ = E / tE E ′ = E E'=E E ′ = E H H H E E E A , B , C A,B,C A , B , C E ′ E' E ′ R = R 0 / 2 R 0 R=R_0/2R_0 R = R 0 /2 R 0 m m m A , B , C A,B,C A , B , C X = T ( A B ) X=T(AB) X = T ( A B ) Y = T ( A C ) Y=T(AC) Y = T ( A C ) Z = T ( B C ) Z=T(BC) Z = T ( B C ) X , Y , Z X,Y,Z X , Y , Z W = T ( A B C ) W=T(ABC) W = T ( A B C ) m m m A , B , C A,B,C A , B , C F 2 \mathbf F_2 F 2 R R R t A B , t A C , t B C , t A B C tAB,tAC,tBC,tABC t A B , t A C , tB C , t A B C t X = t Y = t Z = t W = 0 tX=tY=tZ=tW=0 tX = t Y = tZ = t W = 0 R R R t m = 0 tm=0 t m = 0 t t t B , C B,C B , C t E tE tE F 2 \mathbf F_2 F 2 t , t A t,tA t , t A
4 m d = dim E ≤ 24 + 2 = 26. 4md=\dim E\le24+2=26. 4 m d = dim E ≤ 24 + 2 = 26. An actual selftwist by h h h K ( i ) K(i) K ( i ) dim E ≤ 26 < 32 \dim E\le26<32 dim E ≤ 26 < 32 σ \sigma σ Q \mathbf Q Q
In the small-residue-field cases use the triadic source results above for a solvable residue. A nonsolvable residue stays absolutely irreducible over K ( ζ 3 ) K(\zeta_{3}) K ( ζ 3 ) F 3 \mathbb{F}_{3} F 3 F 9 \mathbb{F}_{9} F 9 [3 ] , with Lemma 2.1, give modularity of the actual weight-two member and hence of the system.
The supported residual base at five
Proposition 19.13. Assume GRH. Every bounded system is modular. Every absolutely irreducible r ˉ : G K → GL 2 ( F 5 ) \bar{r}: G_K \to\operatorname{GL}_{2}(\mathbb{F}_{5}) r ˉ : G K → GL 2 ( F 5 ) ω 5 \omega_{5} ω 5
Proof. For the two-special menu, Propositions 19.6, 19.7, 19.8 and Theorem 19.12 exhaust the dyadic reductions. Consider now any bounded system in characteristic three. The preceding triadic results handle reducible and cyclotomic-dihedral residues. For an irreducible solvable residue not of the latter form, solvable Artin modularity and weight raising supply a weight-zero witness, and [3 Theorem 2.3] applies. Otherwise the residue remains absolutely irreducible on K ( ζ 3 ) K(\zeta_{3}) K ( ζ 3 ) [28 Theorem 4.2.1] and the weight-two construction in Lemma 2.4 give a pure compatible replacement. A point on the special component has Frobenius eigenvalue ratio five; with zero monodromy this would contradict purity. Thus the replacement is special at both places and modular. A common modular residual witness lifts the actual old weight-two triadic member by [3 Theorem 2.3] .
Finally, an irreducible solvable supported five-residue has a regular cuspidal witness by solvable Artin modularity. In the nonsolvable case its restriction to K ( ζ 5 ) K(\zeta_{5}) K ( ζ 5 ) [ K ( ζ 5 ) : K ] = 4 [K(\zeta_{5}):K]=4 [ K ( ζ 5 ) : K ] = 4 PGL 2 ( F 5 ) \operatorname{PGL}_{2}(\mathbb{F}_{5}) PGL 2 ( F 5 )
Exceptional lifts and sources over Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 )
Throughout this section put K = Q ( 5 ) K=\mathbb{Q}(\sqrt{5}) K = Q ( 5 )
Arithmetic and terminal characters
Write
ϕ = ( 1 + 5 ) / 2 , π = 3 + ϕ , π ′ = 4 − ϕ , p = ( π ) , p ′ = ( π ′ ) . \phi=(1+\sqrt{5})/2,\qquad\pi=3+\phi,\qquad\pi'=4-\phi,\qquad\mathfrak{p}=(\pi),\qquad\mathfrak{p}'=(\pi'). ϕ = ( 1 + 5 ) /2 , π = 3 + ϕ , π ′ = 4 − ϕ , p = ( π ) , p ′ = ( π ′ ) . Then π π ′ = 11 \pi\pi'=11 π π ′ = 11 N ( ϕ ) = − 1 N(\phi)=-1 N ( ϕ ) = − 1 N ( π ) = 11 N(\pi)=11 N ( π ) = 11 5 / 2 < 2 \sqrt{5}/2<2 5 /2 < 2 { ± ϕ n : n ∈ Z } \{\pm\phi^{n}:n\in\mathbb{Z}\} { ± ϕ n : n ∈ Z } K 2 K_{2} K 2 Q 2 \mathbb{Q}_{2} Q 2
Proposition 20.1. Let r ˉ : G K → GL 2 ( F 11 ) \bar{r}:G_K\to\operatorname{GL}_{2}(\mathbb{F}_{11}) r ˉ : G K → GL 2 ( F 11 ) χ ˉ 11 \bar{\chi}_{11} χ ˉ 11
χ ˉ 11 e , e ∈ { 1 , 3 , 5 , 7 , 9 } , \bar{\chi}_{11}^{e},\qquad e\in\{1,3,5,7,9\}, χ ˉ 11 e , e ∈ { 1 , 3 , 5 , 7 , 9 } , up to inversion. In particular it extends to G Q G_{\mathbb{Q}} G Q
Proof. Write r ˉ s s = ψ ⊕ χ ˉ 11 ψ − 1 \bar{r}^{\mathrm{ss}}=\psi\oplus\bar{\chi}_{11}\psi^{-1} r ˉ ss = ψ ⊕ χ ˉ 11 ψ − 1 ϕ \phi ϕ 8 = 2 3 8=2^3 8 = 2 3 4 = 2 2 4=2^2 4 = 2 2 F 11 × \mathbb{F}_{11}^{\times} F 11 × 2 2 2 ϕ \phi ϕ ( + , − ) (+,-) ( + , − )
Write a , b ∈ Z / 10 Z a,b\in\mathbb{Z}/10\mathbb{Z} a , b ∈ Z /10 Z u , v ∈ Z / 2 Z u,v\in\mathbb{Z}/2\mathbb{Z} u , v ∈ Z /2 Z − 1 -1 − 1 ϕ \phi ϕ
a + b + u + v ≡ 0 ( m o d 2 ) , 3 a + 2 b + 5 v ≡ 0 ( m o d 10 ) . a+b+u+v\equiv0\pmod{2},\qquad3a+2b+5v\equiv0\pmod{10}. a + b + u + v ≡ 0 ( mod 2 ) , 3 a + 2 b + 5 v ≡ 0 ( mod 10 ) . They give a ≡ b ( m o d 5 ) a\equiv b\pmod{5} a ≡ b ( mod 5 ) v ≡ a ( m o d 2 ) v\equiv a\pmod{2} v ≡ a ( mod 2 ) u ≡ b ( m o d 2 ) u\equiv b\pmod{2} u ≡ b ( mod 2 ) χ ˉ 11 2 \bar{\chi}_{11}^{2} χ ˉ 11 2 ψ 2 \psi^2 ψ 2 χ ˉ 11 \bar{\chi}_{11} χ ˉ 11 χ ˉ 11 ψ − 2 \bar{\chi}_{11}\psi^{-2} χ ˉ 11 ψ − 2
Put
α = ϕ π = 1 + 4 ϕ , α ′ = 5 − 4 ϕ , α α ′ = − 11. \alpha=\phi\pi=1+4\phi,\qquad\alpha'=5-4\phi,\qquad\alpha\alpha'=-11. α = ϕ π = 1 + 4 ϕ , α ′ = 5 − 4 ϕ , α α ′ = − 11. The quadratic extensions K ( α ) K(\sqrt{\alpha}) K ( α ) K ( α ′ ) K(\sqrt{\alpha'}) K ( α ′ ) ( 1 + α ) / 2 (1+\sqrt{\alpha})/2 ( 1 + α ) /2 X 2 − X − ϕ = 0 X^2-X-\phi=0 X 2 − X − ϕ = 0 F 4 \mathbb{F}_4 F 4 Tr F 4 / F 2 ( ϕ ) = 1 \operatorname{Tr}_{\mathbb{F}_4/\mathbb{F}_2}(\phi)=1 Tr F 4 / F 2 ( ϕ ) = 1 α ′ \alpha' α ′
H = K ( − 11 ) , H=K(\sqrt{-11}), H = K ( − 11 ) , which is split at two: K 2 K_2 K 2 5 \sqrt{5} 5 − 11 / 5 -11/5 − 11/5 Q 2 \mathbb{Q}_2 Q 2 1 1 1
We use Lemma 2.4 for prescribed lifts and compatible systems; the exceptional initial characteristic-five point is constructed separately in § 7.2.2 .
The notation F b F_b F b J b J_b J b B b B_b B b C b C_b C b § 9.1 . We use the explicit formula (43) , including its finite-tail bound and retained prime densities. The unconditional comparison below uses the positive kernel F b ( x ) / cosh ( x / 2 ) F_b(x)/\cosh(x/2) F b ( x ) / cosh ( x /2 )
Starting an exceptional characteristic-five representation
The ring-theoretic theorem of Khare–Thorne gives the finiteness needed below before a characteristic-zero global point has been found. We use its published nilpotent-kernel statement, rather than its finite-length approximation corollary.
Corollary 20.2 (A classical quotient of the prepared KT ring). In the exact setup of [38 §7] , assume the fixed set V 2 V_2 V 2 D S t ( u n i ) D^{\mathrm{St}(\mathrm{uni})} D St ( uni ) R S c l R^{\mathrm{cl}}_{\mathcal{S}} R S cl Λ \Lambda Λ O \mathcal{O} O
Proof. In the notation of [38 §7] , let T \mathbb{T} T T \mathbb{T} T Λ \Lambda Λ [38 ] gives a surjection
R S ⟶ T R_{\mathcal{S}}\longrightarrow\mathbb{T} R S ⟶ T with nilpotent kernel I I I V 2 ≠ ∅ V_2\ne\varnothing V 2 = ∅ R S R_{\mathcal{S}} R S I j / I j + 1 I^j/I^{j+1} I j / I j + 1 R S / I = T R_S/I=\mathbb{T} R S / I = T Λ \Lambda Λ I I I R S R_S R S Λ \Lambda Λ Λ → O \Lambda\to\mathcal{O} Λ → O R S R_S R S
Proposition 20.3 (An initial exceptional lift). Let K = Q ( 5 ) K=\mathbb{Q}(\sqrt{5}) K = Q ( 5 ) r ˉ : G K ⟶ GL 2 ( F 5 ) \bar{r}:G_K\longrightarrow\operatorname{GL}_2(\mathbb{F}_5) r ˉ : G K ⟶ GL 2 ( F 5 ) PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) K ( ζ 5 ) K(\zeta_5) K ( ζ 5 )
ν = [ det r ˉ χ ˉ 5 − 1 ] . \nu=[\det\bar{r}\bar{\chi}_5^{-1}]. ν = [ det r ˉ χ ˉ 5 − 1 ] . where brackets denote the Teichmuller lift. There is a continuous lift r : G K ⟶ GL 2 ( Q 5 ) r:G_K\longrightarrow\operatorname{GL}_2(\mathbb{Q}_5) r : G K ⟶ GL 2 ( Q 5 ) ν χ 5 \nu\chi_5 ν χ 5
Proof. Fixed local rings over K K K S S S 5 5 5 r ˉ \bar{r} r ˉ v ∈ S v\in S v ∈ S [64 Propositions 7.7.1 and 7.8.1] , weight two when v ∣ 5 v\mid5 v ∣ 5 [64 §7.9] twists by the square root of a finite character reducing to 1 1 1 ν χ 5 \nu\chi_5 ν χ 5 O \mathcal{O} O R v † R_v^\dagger R v † [64 Propositions 7.3.1 and 7.4.1] . Its framed dimension is 4 4 4 v ∤ 5 v\nmid5 v ∤ 5 4 + [ K v : Q 5 ] 4+[K_v:\mathbb{Q}_5] 4 + [ K v : Q 5 ] v ∣ 5 v\mid5 v ∣ 5 5 5 5 A A A B B B C C C C C C
Fix finite Galois local extensions killing these fixed inertial types, the local residual representations, and their finite determinant factors. After these extensions, the local characteristic-zero points are unramified or in the Steinberg closure away from 5 5 5 5 5 5 A A A C C C B B B K ν K_\nu K ν ν \nu ν ν \nu ν
Preparation and the modular residual witness. Real approximation and Krasner’s lemma, followed by taking a totally real Galois closure, supply a totally real finite extension containing K ν K_\nu K ν [64 ] over this field. This theorem is for any odd residual representation and has no exceptional-image exclusion. It gives a weight-two modular residual witness after a further totally real extension.
Let R R R 5 5 5 σ \sigma σ 5 5 5 A A A C C C [38 Lemma 9.2] to the known modular witness, this R R R σ \sigma σ π \pi π σ \sigma σ 5 5 5 R R R 5 5 5 R R R
Finally make a totally real quadratic extension split at the current places above 5 5 5 R R R π \pi π F F F [ F : K ] [F:K] [ F : K ] # R F \#R_F # R F ψ = 1 \psi=1 ψ = 1 5 5 5 1 1 1 F F F 5 5 5 5 5 5 R F R_F R F
A fresh fixed unipotent prime. Choose the auxiliary prime only after F F F M / K M/K M / K F g / K F^{g}/K F g / K F / K F/K F / K F g F^{g} F g M ∩ F g = K M \cap F^{g} = K M ∩ F g = K S 5 S_{5} S 5 M / K M/K M / K K ( ζ 5 ) K(\zeta_{5}) K ( ζ 5 ) M M M L / K L/K L / K G = Gal ( L / K ) G = \operatorname{Gal}(L/K) G = Gal ( L / K ) Z Z Z N = Gal ( L / ( L ∩ F g ) ) N = \operatorname{Gal}(L/(L \cap F^{g})) N = Gal ( L / ( L ∩ F g )) Z N = G ZN = G Z N = G G / N G/N G / N Z Z Z G G G N N N K ( ζ 5 ) K(\zeta_{5}) K ( ζ 5 )
Chebotarev therefore gives a prime q ∉ S q \notin S q ∈ / S F g F^{g} F g N q ≡ 1 ( m o d 5 ) Nq \equiv1 \pmod{5} N q ≡ 1 ( mod 5 ) ν ( q ) = 1 \nu(q) = 1 ν ( q ) = 1 D S t ( u n i ) D^{\mathrm{St(uni)}} D St ( uni ) q q q K K K q q q F F F [38 Proposition 6.6] . Write Q F Q_{F} Q F q q q [ F : K ] [F : K] [ F : K ] V 1 = ∅ V_{1} = \varnothing V 1 = ∅ V 2 = Q F V_{2} = Q_{F} V 2 = Q F [38 §7] . No congruence modulo 25 is needed for this fixed local problem.
Choose the good-subgroup auxiliary place a a a [38 Lemma 7.1] , away from the places above 5, R F R_{F} R F Q F Q_{F} Q F [38 Lemma 5.2 and the proof of Theorem 9.1] to π \pi π Q F Q_{F} Q F [68 Lemma 4.13] preserves the local conditions at 5 and R F R_{F} R F Q F Q_{F} Q F a a a [38 §7] , with V 1 = ∅ V_{1} = \varnothing V 1 = ∅ V 2 = Q F V_{2} = Q_{F} V 2 = Q F
Restriction at the level of local rings. Let R K , q † R^{\dagger}_{K,q} R K , q † K K K R v † R^{\dagger}_{v} R v † D S t ( u n i ) D^{\mathrm{St(uni)}} D St ( uni ) q q q S ∪ { q } S \cup\{q\} S ∪ { q } R F c l R^{\mathrm{cl}}_{F} R F cl
R F c l ⟶ R K , q † . R^{\mathrm{cl}}_{F} \longrightarrow R^{\dagger}_{K,q}. R F cl ⟶ R K , q † . Here it is essential to verify the type conditions before passing to the global ring. On each original reduced, O \mathcal{O} O R v † R^{\dagger}_{v} R v † [64 Lemma 6.4.1] .
The ordinary local ring is the reduced, O \mathcal{O} O [68 §5.4.1 and Proposition 5.13] . Its classical characteristic-zero points require an ordinary line with trivial inertial character, without marking a particular line. After preparation and dualization each type A A A C C C B B B q q q q q q F F F D S t ( u n i ) D^{\mathrm{St(uni)}} D St ( uni )
The projective image over F F F S 5 S_{5} S 5 [64 Lemma 6.3.1] . Passing to the local quotients preserves finiteness, since their defining ideals have already been shown to vanish. Corollary 20.2 therefore makes R K , q † R^{\dagger}_{K,q} R K , q † O \mathcal{O} O
A characteristic-zero point. Put n = # ( S ∪ { q } ) n = \#(S \cup\{q\}) n = # ( S ∪ { q }) 1 + [ K : Q ] + 3 n 1 + [K : \mathbb{Q}] + 3n 1 + [ K : Q ] + 3 n [64 Proposition 6.2.1] uses r + n − 1 r + n - 1 r + n − 1 r + [ K : Q ] r + [K : \mathbb{Q}] r + [ K : Q ] 4 n 4n 4 n 4 n − 1 4n - 1 4 n − 1 dim R K , q □ ≥ 1 \dim R^{\square}_{K,q} \ge1 dim R K , q □ ≥ 1 K ( ζ 5 ) K(\zeta_5) K ( ζ 5 ) O \mathcal{O} O
Actual potential modularity and the compatible system. For any prescribed finite extension B / K B/K B / K [64 Theorem 5.1.1] to r ˉ \bar r r ˉ K ′ / K K'/K K ′ / K K ′ K' K ′ r ∣ G K ′ r|_{G_{K'}} r ∣ G K ′ G K ′ ( ζ 5 ) G_{K'(\zeta_5)} G K ′ ( ζ 5 ) [38 Theorem 1.1] , after dualizing, to obtain the automorphy of this actual lift.
The proof of [64 Proposition 9.2.1 and Remark 9.2.2] , using solvable descent and Brauer induction, now constructs the strongly compatible system. In this use the proposition’s old exceptional exclusion is replaced in its proof by the actual avoiding potential modularity just proved; the remaining argument uses only that input and absolute irreducibility, not an adequacy hypothesis. In particular this is an application of that proof with its modularity input supplied by KT, not an invocation of the original proposition outside its stated hypotheses.
Remark 20.4 (Scope). This construction permits the extra prime q q q K K K
A one-prime source obstruction
Proposition 20.5. Let r ˉ : G K → S L 2 ( k ) \bar r : G_K \to\mathrm{SL}_2(k) r ˉ : G K → SL 2 ( k ) 2 p 2\mathfrak{p} 2 p p \mathfrak{p} p k k k r s s = 1 ⊕ 1 r^{\mathrm{ss}} = 1 \oplus1 r ss = 1 ⊕ 1
Proof. Fontaine’s finite-flat bound over the unramified dyadic base gives classical upper breaks at most one and normalized different d 2 < 2 d_2 < 2 d 2 < 2 [31 §2.1, Theorem 1, and §1.2(i)] . The latter reference converts his upper numbering to the classical one. At eleven the ramification is tame of index at most two at just one of the two places. A nonsolvable image is S L 2 ( F q ) \mathrm{SL}_2(\mathbb{F}_q) SL 2 ( F q ) q ≥ 4 q \ge4 q ≥ 4
rd ( M ) < 4 5 11 1 / 4 < 16.28896. \operatorname{rd}(M) < 4\sqrt{5}11^{1/4} < 16.28896. rd ( M ) < 4 5 1 1 1/4 < 16.28896. With b = 7 b = 7 b = 7 [ M : Q ] ≥ 120 [M : \mathbb{Q}] \ge120 [ M : Q ] ≥ 120 rd ( M ) > 16.661 \operatorname{rd}(M) > 16.661 rd ( M ) > 16.661 [49 d5/new] .
An irreducible solvable image is odd dihedral, induced from an odd-order character θ \theta θ L / K L/K L / K L L L ⟨ − 1 , ϕ , π ⟩ \langle-1,\phi,\pi\rangle ⟨ − 1 , ϕ , π ⟩
The bound in the fifth row is strict. Integral bases are 1 , ϕ , t , ϕ t 1,\phi,t,\phi t 1 , ϕ , t , ϕt t 2 = a t^2=a t 2 = a t 2 − t = ϕ t^2-t=\phi t 2 − t = ϕ π \pi π ( 2 ) (2) ( 2 ) N ( t − 1 ) = 5 N(t-1)=5 N ( t − 1 ) = 5 − π -\pi − π N ( t − ϕ ) = 20 N(t-\phi)=20 N ( t − ϕ ) = 20
If L / K L/K L / K α = ϕ π \alpha=\phi\pi α = ϕ π L = K ( α ) L=K(\sqrt{\alpha}) L = K ( α ) t = ( 1 + α ) / 2 t=(1+\sqrt{\alpha})/2 t = ( 1 + α ) /2 X 4 + X + 1 X^4+X+1 X 4 + X + 1 F 16 \mathbb{F}_{16} F 16 θ \theta θ
Finally, in the reducible case the odd-order constituent characters are unramified at eleven. At two their inertia factors through F 4 × \mathbb{F}_4^\times F 4 × ϕ \phi ϕ
Put R = O K [ 1 / π ] R=\mathcal{O}_K[1/\pi] R = O K [ 1/ π ] F = K 2 F=K_2 F = K 2
Lemma 20.6. We have Ext R 1 ( μ 2 , Z / 2 Z ) = 0 \operatorname{Ext}^1_R(\mu_2,\mathbb{Z}/2\mathbb{Z})=0 Ext R 1 ( μ 2 , Z /2 Z ) = 0
Proof. Over the completion at two, an extension of μ 2 \mu_2 μ 2 Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z H 1 ( K , F 2 ) = K × / K × 2 H^1(K,\mathbb{F}_2)=K^\times/K^{\times2} H 1 ( K , F 2 ) = K × / K × 2 F F F
For R R R − 1 , ϕ , 2 , π -1,\phi,2,\pi − 1 , ϕ , 2 , π F × / F × 2 F^\times/F^{\times2} F × / F × 2 2 2 2 Q 2 \mathbb{Q}_2 Q 2 ϕ \phi ϕ π \pi π − 1 , 11 , − 11 -1,11,-11 − 1 , 11 , − 11 Q 2 \mathbb{Q}_2 Q 2 − 1 -1 − 1 F F F
A generically split extension is split over R R R [55 Proposition 2.3 and Corollary 2.4] .
:::
Lemma 20.7. Let G G G 2 2 2 R R R G [ 2 n ] G[2^n] G [ 2 n ] G G G 2 2 2
Proof. A simple generic factor has order two. Schematic closure realizes a generic composition series as a finite flat composition series. At the unique dyadic prime, the prime-order classification over an unramified ring gives either the etale or the multiplicative model. Away from two the model is etale and its generic action is trivial. Since the generic group of order two has no nontrivial automorphism, gluing gives exactly Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z μ 2 \mu_2 μ 2 R R R
By Lemma 20.6, adjacent factors in the order Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z μ 2 \mu_2 μ 2 G [ 2 n ] G[2^n] G [ 2 n ]
This subgroup is unique. Its localization at two is the connected part, since its quotient is etale; its global subgroup is the flat closure of that generic subgroup. Consequently it is functorial in G [ 2 n ] G[2^n] G [ 2 n ] [55 , 56 ] .
Theorem 20.8. Let E 0 / Q 2 E_0/\mathbb{Q}_2 E 0 / Q 2 r : G K → GL 2 ( E 0 ) r:G_K\to\operatorname{GL}_2(E_0) r : G K → GL 2 ( E 0 ) det r = χ 2 \det r=\chi_2 det r = χ 2 r r r 2 p 2\mathfrak{p} 2 p p \mathfrak{p} p r r r
0 ⟶ E 0 ( 1 ) ⟶ r ⟶ E 0 ⟶ 0. 0 \longrightarrow E_0(1) \longrightarrow r \longrightarrow E_0 \longrightarrow0. 0 ⟶ E 0 ( 1 ) ⟶ r ⟶ E 0 ⟶ 0. In particular there is no absolutely irreducible such representation. The same assertion holds with p ′ \mathfrak{p}' p ′ p \mathfrak{p} p
Proof of Theorem 20.8. Choose a global stable O E 0 \mathcal{O}_{E_0} O E 0 [55 Proposition 2.3] gives a 2-divisible group over R R R O E 0 \mathcal{O}_{E_0} O E 0
The reduction modulo the maximal ideal of O E 0 \mathcal{O}_{E_0} O E 0 F 2 \mathbb{F}_2 F 2 G [ 2 n ] G[2^n] G [ 2 n ]
The height of G G G 2 [ E 0 : Q 2 ] 2[E_0:\mathbb{Q}_2] 2 [ E 0 : Q 2 ] [ E 0 : Q 2 ] [E_0:\mathbb{Q}_2] [ E 0 : Q 2 ] { 0 , 1 } \{0,1\} { 0 , 1 } [ E 0 : Q 2 ] [E_0:\mathbb{Q}_2] [ E 0 : Q 2 ] E 0 E_0 E 0 E 0 E_0 E 0 r r r χ 2 \chi_2 χ 2
Two cuspidal places above eleven
We use one finite-flat ordinarity observation repeatedly. A nonordinary rank-two Barsotti–Tate source of cyclotomic determinant is biconnected up to isogeny. Its coefficient-uniformizer torsion and finite-flat subquotients are then biconnected. Over an unramified dyadic ring, a prime-order model with trivial generic action is Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z μ 2 \mu_2 μ 2 p p p e < p − 1 e<p-1 e < p − 1 1 1 1 ω p \omega_p ω p [53 Theorem 3.3.3 and Corollary 3.3.6] . Neither model is biconnected. Their occurrence as simple factors therefore forces the actual source to be ordinary.
Proposition 20.9. Let K = Q ( 5 ) K = \mathbb{Q}(\sqrt{5}) K = Q ( 5 ) R \mathcal{R} R
Proof. Modulo two the scalar quadratic factors disappear. At either eleven-adic place tame inertia has eigencharacters of order three, and Frobenius interchanges them because 11 ≡ − 1 ( m o d 3 ) 11 \equiv-1 \pmod{3} 11 ≡ − 1 ( mod 3 ) S 3 S_3 S 3 GL 2 ( F 2 ) \operatorname{GL}_2(\mathbb{F}_2) GL 2 ( F 2 ) L / K L/K L / K
The quadratic quotient is unramified at eleven: its local inertia is the odd rotation subgroup. It is unramified outside two, and Fontaine’s finite-flat upper bound one over the unramified dyadic completion gives dyadic conductor at most two. Since K K K ϕ = ( 1 + 5 ) / 2 \phi= (1 + \sqrt{5})/2 ϕ = ( 1 + 5 ) /2 − 1 -1 − 1 ϕ \phi ϕ 2 2 2 − 1 -1 − 1 ϕ \phi ϕ − ϕ -\phi − ϕ
At the two eleven-primes, the values of ϕ \phi ϕ 8 8 8 4 4 4 F 11 \mathbb{F}_{11} F 11 − ϕ -\phi − ϕ 3 3 3 7 7 7 − 1 -1 − 1 L = K ( i ) L = K(i) L = K ( i )
At two this quadratic extension is ramified. Its nontrivial wild inertia image is a reflection of the odd dihedral group. The normalizer of a reflection subgroup is that subgroup itself. Since wild inertia is normal in the decomposition group, the whole dyadic decomposition image is C 2 C_2 C 2
Filtering the coefficient-uniformizer torsion over F 2 \mathbb{F}_2 F 2
The main theorem of [2 Theorem 5.2.1] applies: the source is ordinary and totally odd, the residue is absolutely irreducible and solvable, and its CM inducing quadratic extension does not split at the dyadic place. Equivalently combine his Lemma 5.1.2 with Theorem 5.2.1. This proves modularity of the source and hence of the system.
Lemma 20.10. Let R \mathcal{R} R K K K G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 )
Proof. The quadratic characters ramified at just one eleven-prime, constructed from α = 1 + 4 ϕ \alpha= 1 + 4\phi α = 1 + 4 ϕ
If the residue is reducible, its characters are unramified at eleven. They factor through the narrow ray modulo three, whose order is two: the local/sign group has order 8 ⋅ 4 = 32 8 \cdot4 = 32 8 ⋅ 4 = 32 ϕ \phi ϕ − 1 -1 − 1 ϕ \phi ϕ ( + , − ) (+,-) ( + , − ) − 1 -1 − 1 ω 3 \omega_3 ω 3 1 , ω 3 1,\omega_3 1 , ω 3
At three the absolute ramification index is 1 < 3 − 1 1 < 3 - 1 1 < 3 − 1 [62 §4.5, Theorem A] now applies. The quotient splitting field K ( ζ 3 ) K(\zeta_{3}) K ( ζ 3 ) Q \mathbb{Q} Q χ 3 \chi_{3} χ 3
If the cyclotomic restriction of an irreducible residue were reducible, the residue would be induced from a prime-to-three character θ \theta θ H = K ( ζ 3 ) H = K(\zeta_{3}) H = K ( ζ 3 ) G H G_{H} G H θ \theta θ H H H 5 ( − 3 ) ( − 15 ) = 225 5(-3)(-15) = 225 5 ( − 3 ) ( − 15 ) = 225 22.5 / π 2 < 2.28 22.5/\pi^{2} < 2.28 22.5/ π 2 < 2.28 F 9 \mathbb{F}_{9} F 9 ϕ \phi ϕ θ \theta θ
Lemma 20.11. Let r : G K → G L 2 ( E ) r:G_{K} \to\mathrm{GL}_{2}(E) r : G K → GL 2 ( E ) χ 5 \chi_{5} χ 5 r r r
Proof. The narrow ray modulo the prime above five is C 2 C_{2} C 2 1 , ω 5 1,\omega_{5} 1 , ω 5 2 < 5 − 1 2 < 5 - 1 2 < 5 − 1 [62 §4.5, Theorem A] . Its quotient field K ( ω 5 ) = Q ( ζ 5 ) K(\omega_{5}) = \mathbb{Q}(\zeta_{5}) K ( ω 5 ) = Q ( ζ 5 ) Q \mathbb{Q} Q k = 2 k = 2 k = 2 K K K
Theorem 20.12. Assume GRH for Dedekind zeta functions. Let K = Q ( 5 ) K = \mathbb{Q}(\sqrt{5}) K = Q ( 5 ) R R R { 0 , 1 } \{0,1\} { 0 , 1 } R R R
Proof. Choose a coefficient place above two and let M / K M/K M / K 11 ≡ − 1 ( m o d 3 ) 11 \equiv-1 \pmod{3} 11 ≡ − 1 ( mod 3 ) S 3 S_{3} S 3
If its image is solvable, Proposition 20.9 proves modularity. We may therefore suppose the image is G = S L 2 ( F q ) G = \mathrm{SL}_{2}(\mathbb{F}_{q}) G = SL 2 ( F q ) q = 2 f ≥ 4 q = 2^{f} \ge4 q = 2 f ≥ 4
We construct a solvable auxiliary extension. Put H = K ( i ) H = K(i) H = K ( i ) 5 ( − 4 ) ( − 20 ) = 400 5(-4)(-20) = 400 5 ( − 4 ) ( − 20 ) = 400 30 / π 2 < 3.04 30/\pi^{2} < 3.04 30/ π 2 < 3.04 i i i ϕ = ( 1 + 5 ) / 2 \phi= (1+\sqrt{5})/2 ϕ = ( 1 + 5 ) /2 K K K ϕ 2 m \phi^{2m} ϕ 2 m ϕ m \phi^{m} ϕ m H H H μ 4 \mu_{4} μ 4
The two eleven-primes of K K K H / K H/K H / K F 121 \mathbb{F}_{121} F 121 H H H C 3 2 C_{3}^{2} C 3 2 120 120 120 C 3 C_{3} C 3 ϕ \phi ϕ F 11 × \mathbb{F}_{11}^{\times} F 11 × i i i
Let B / H B/H B / H K K K C 3 2 C_{3}^{2} C 3 2 B / K B/K B / K C 3 2 × C 2 C_3^2 \times C_2 C 3 2 × C 2 [ B : Q ] = 36 [B:\mathbb{Q}]=36 [ B : Q ] = 36 B / H B/H B / H H / K H/K H / K B / K B/K B / K C 2 C_2 C 2 C 3 2 × C 2 C_3^2 \times C_2 C 3 2 × C 2 C 2 C_2 C 2 K 2 ( i ) K_2(i) K 2 ( i )
At either eleven-prime the ray inertia is C 3 C_3 C 3 Gal ( H / K ) \operatorname{Gal}(H/K) Gal ( H / K ) C 3 2 C_3^2 C 3 2 S 3 S_3 S 3 Q 11 \mathbb{Q}_{11} Q 11 ( e , f ) = ( 3 , 2 ) (e,f)=(3,2) ( e , f ) = ( 3 , 2 ) μ 3 \mu_3 μ 3 3 ∤ 10 3\nmid10 3 ∤ 10
Since G G G M ∩ B = K M\cap B=K M ∩ B = K L = B M L=BM L = B M 36 q ( q 2 − 1 ) 36q(q^2-1) 36 q ( q 2 − 1 ) S 3 S_3 S 3
2 N 121 ( L ) [ L : Q ] = 1 3 , d 11 = 2 3 . \frac{2N_{121}(L)}{[L:\mathbb{Q}]}=\frac{1}{3},\qquad d_{11}=\frac{2}{3}. [ L : Q ] 2 N 121 ( L ) = 3 1 , d 11 = 3 2 . At two the residual Barsotti–Tate point field has no upper break exceeding one by Fontaine’s bound over the unramified base. The same is true of B B B d 2 < 2 d_2<2 d 2 < 2
rd ( L ) < 4 5 11 2 / 3 . \operatorname{rd}(L)<4\sqrt{5}11^{2/3}. rd ( L ) < 4 5 1 1 2/3 . For q = 16 q=16 q = 16 M M M G K 2 G_{K_2} G K 2 F 4 × \mathbb{F}_4^\times F 4 × q q q B B B w w w e e e w ≤ 2 q w\le2q w ≤ 2 q e ≤ 3 w e\le3w e ≤ 3 w
d 2 ≤ ( 1 − 1 / e ) + ( 1 − 1 / w ) ≤ 2 − 2 3 q . d_2\le(1-1/e)+(1-1/w)\le2-\frac{2}{3q}. d 2 ≤ ( 1 − 1/ e ) + ( 1 − 1/ w ) ≤ 2 − 3 q 2 . If wild inertia in M M M d 2 < 3 / 2 d_2<3/2 d 2 < 3/2 K 2 K_2 K 2
rd ( B M ) ≤ 5 2 2 − 2 / ( 3 q ) 11 2 / 3 . \operatorname{rd}(BM)\le\sqrt{5}2^{2-2/(3q)}11^{2/3}. rd ( B M ) ≤ 5 2 2 − 2/ ( 3 q ) 1 1 2/3 . Apply (39) with B b , 100 B_{b,100} B b , 100 1 / 3 1/3 1/3
Table 1 (PDF p. 93).
Here σ \sigma σ K / Q K/\mathbb{Q} K / Q q ≥ 16 q\ge16 q ≥ 16 b b b [49 d5] .
For the third line use B M M σ BMM^\sigma B M M σ M ≠ M σ M\ne M^\sigma M = M σ K K K B B B 36 ⋅ 504 2 36\cdot504^2 36 ⋅ 50 4 2 S 3 S_3 S 3
If q = 8 q=8 q = 8 M = M σ M=M^\sigma M = M σ M / Q M/\mathbb{Q} M / Q Γ = Gal ( M / Q ) \Gamma=\operatorname{Gal}(M/\mathbb{Q}) Γ = Gal ( M / Q ) G = Gal ( M / K ) = SL 2 ( 8 ) G=\operatorname{Gal}(M/K)=\operatorname{SL}_2(8) G = Gal ( M / K ) = SL 2 ( 8 ) Γ \Gamma Γ Aut ( G ) \operatorname{Aut}(G) Aut ( G ) Γ / G → Out ( G ) \Gamma/G\to\operatorname{Out}(G) Γ/ G → Out ( G ) G G G Γ → Inn ( G ) = G \Gamma\to\operatorname{Inn}(G) = G Γ → Inn ( G ) = G G ↪ SL 2 ( F 8 ) G \hookrightarrow\operatorname{SL}_{2}(\mathbb{F}_{8}) G ↪ SL 2 ( F 8 ) G Q G_{\mathbb{Q}} G Q [40 , 41 ] and quadratic base change give a cuspidal modular witness over K K K q = 4 q = 4 q = 4 [3 Theorem 2.5] .
Finally some real complex conjugation acts nontrivially in the residue. Otherwise M M M 4 5 11 2 / 3 4\sqrt{5}11^{2/3} 4 5 1 1 2/3 n = 120 n = 120 n = 120 b = 4.8 b = 4.8 b = 4.8 J b ( 1 / 2 ) − J b ( 3 / 2 ) + J b ( 5 / 2 ) − J b ( 7 / 2 ) J_b(1/2) - J_b(3/2) + J_b(5/2) - J_b(7/2) J b ( 1/2 ) − J b ( 3/2 ) + J b ( 5/2 ) − J b ( 7/2 ) rd ( M ) > 56.06306031 \operatorname{rd}(M) > 56.06306031 rd ( M ) > 56.06306031 [69 Theorem 6.1] to the dual of the actual crystalline two-adic source, in his Hodge–Tate convention: its residual image is nonsolvable, its residue is modular, and the required real-place condition holds. This proves modularity of the source and hence of R \mathcal{R} R
The supported residual base at eleven
Fixed-determinant positivity. We use the following observation for both eleven-adic menus. Let r ˉ : G Q 11 → GL 2 ( F 11 ) \bar{r}: G_{\mathbb{Q}_{11}} \to\operatorname{GL}_{2}(\mathbb{F}_{11}) r ˉ : G Q 11 → GL 2 ( F 11 ) ω 11 \omega_{11} ω 11 [9 Propositions 2.6–2.7] it has a regular crystalline lift of determinant χ 11 \chi_{11} χ 11 h 1 < h 2 h_1 < h_2 h 1 < h 2 h 1 + h 2 = 1 h_1 + h_2 = 1 h 1 + h 2 = 1
Sym h 2 − h 1 − 1 ⊗ det h 1 . \operatorname{Sym}^{h_2-h_1-1} \otimes\det^{h_1}. Sym h 2 − h 1 − 1 ⊗ det h 1 . Its scalar central character has exponent h 1 + h 2 − 1 = 0 h_1 + h_2 - 1 = 0 h 1 + h 2 − 1 = 0 [70 Theorem 1.2] therefore give a nonzero effective cycle for at least one Serre weight with trivial scalar central character. These cycles are independent of the Hodge type and inertial type for this fixed determinant. Consequently, in any collection of weight-two compact types containing all such Serre weights, at least one type has a nonzero lifting ring. That ring is eleven-torsion-free and has a characteristic-zero point. This argument applies to the actual residual extension.
Lemma 20.13 (The bounded eleven-adic menu). Every continuous r ˉ : G Q 11 → GL 2 ( F 11 ) \bar{r}: G_{\mathbb{Q}_{11}} \to\operatorname{GL}_{2}(\mathbb{F}_{11}) r ˉ : G Q 11 → GL 2 ( F 11 ) ω 11 \omega_{11} ω 11 χ 11 \chi_{11} χ 11
Proof. Write W ( r , b ) = Sym r ( F 11 2 ) ⊗ det b W(r,b) = \operatorname{Sym}^{r}(\mathbb{F}_{11}^{2}) \otimes\det^{b} W ( r , b ) = Sym r ( F 11 2 ) ⊗ det b 0 ≤ r ≤ 10 0 \le r \le10 0 ≤ r ≤ 10 0 ≤ b < 10 0 \le b < 10 0 ≤ b < 10 r + 2 b ≡ 0 ( m o d 10 ) r + 2b \equiv0 \pmod{10} r + 2 b ≡ 0 ( mod 10 ) ω t , ω − t \omega^{t}, \omega^{-t} ω t , ω − t
Table 10.
The trivial crystalline type and its quadratic twist contribute W ( 0 , 0 ) , W ( 0 , 5 ) W(0,0), W(0,5) W ( 0 , 0 ) , W ( 0 , 5 ) W ( 10 , 0 ) , W ( 10 , 5 ) W(10,0), W(10,5) W ( 10 , 0 ) , W ( 10 , 5 )
A common quadratic twist adds five to b b b t = 2 , 4 t=2,4 t = 2 , 4 □ \square □
The global quadratic characters cut out by K ( ϕ ( 3 + ϕ ) ) K(\sqrt{\phi(3+\phi)}) K ( ϕ ( 3 + ϕ ) ) Gal ( K / Q ) \operatorname{Gal}(K/\mathbb{Q}) Gal ( K / Q ) Lemma 20.13 simultaneously without destroying good reduction at two. We use them only as finite twists, which preserve the determinant and can be reversed after proving modularity.
Lemma 20.14 (The actual residual switch at eleven). Let s ˉ : G Q 11 → GL 2 ( F ‾ 5 ) \bar{s}:G_{\mathbb{Q}_{11}}\to\operatorname{GL}_{2}(\overline{\mathbb{F}}_{5}) s ˉ : G Q 11 → GL 2 ( F 5 ) ω 5 \omega_{5} ω 5 χ 5 \chi_{5} χ 5
Proof. If inertia is trivial, lift the full residual Frobenius matrix unramifiedly and correct its determinant to eleven by a scalar square root reducing to one. This includes unequal unramified eigencharacters; no special lift is asserted in that case.
Otherwise choose a tame generator with matrix T = 1 + t E 12 T=1+tE_{12} T = 1 + t E 12 t ≠ 0 t\ne0 t = 0 Φ \Phi Φ T T T Φ T Φ − 1 = T 11 = T \Phi T\Phi^{-1}=T^{11}=T Φ T Φ − 1 = T 11 = T Φ = a I + b E 12 \Phi=aI+bE_{12} Φ = a I + b E 12 a = ± 1 a=\pm1 a = ± 1 t ~ , b ~ \tilde{t},\tilde{b} t ~ , b ~
Φ ~ = ( 11 a b ~ 0 a ) , U ( x ) = 1 + t ~ x E 12 . \widetilde{\Phi}=
\begin{pmatrix}
11a & \widetilde{b}\\
0 & a
\end{pmatrix},
\qquad
U(x)=1+\widetilde{t}xE_{12}. Φ = ( 11 a 0 b a ) , U ( x ) = 1 + t x E 12 . On the tame Z 5 \mathbb{Z}_{5} Z 5 Φ ~ U ( x ) Φ ~ − 1 = U ( 11 x ) \widetilde{\Phi}U(x)\widetilde{\Phi}^{-1}=U(11x) Φ U ( x ) Φ − 1 = U ( 11 x ) □ \square □
Lemma 20.15. Let r ˉ : G K → GL 2 ( F ‾ 5 ) \bar{r}:G_{K}\to\operatorname{GL}_{2}(\overline{\mathbb{F}}_{5}) r ˉ : G K → GL 2 ( F 5 ) r ˉ ∣ G K ( ζ 5 ) \left.\bar{r}\right|_{G_{K(\zeta_{5})}} r ˉ ∣ G K ( ζ 5 )
Proof. Otherwise Clifford theory gives induction from a prime-to-five character θ \theta θ H = K ( ζ 5 ) = Q ( ζ 5 ) H=K(\zeta_{5})=\mathbb{Q}(\zeta_{5}) H = K ( ζ 5 ) = Q ( ζ 5 ) P \mathfrak{P} P H H H 125 125 125 3 125 / ( 2 π 2 ) < 2 3\sqrt{125}/(2\pi^{2})<2 3 125 / ( 2 π 2 ) < 2 ϕ \phi ϕ P \mathfrak{P} P 3 3 3 F 5 × \mathbb{F}_{5}^{\times} F 5 × P \mathfrak{P} P θ \theta θ □ \square □
Proposition 20.16. Assume GRH. Let r ˉ : G K → GL 2 ( F ‾ 5 ) \bar{r}:G_{K}\to\operatorname{GL}_{2}(\overline{\mathbb{F}}_{5}) r ˉ : G K → GL 2 ( F 5 ) χ ˉ 5 \bar{\chi}_{5} χ ˉ 5 S 5 S_{5} S 5 S 5 / A 5 S_{5}/A_{5} S 5 / A 5 K ( ζ 5 ) / K K(\zeta_{5})/K K ( ζ 5 ) / K
Proof. At the five-adic place, the absolute ramification index of K K K { 0 , 1 , 2 } \{0,1,2\} { 0 , 1 , 2 } a + 5 b a+5b a + 5 b a + b = 2 a+b=2 a + b = 2 0 ≤ a , b ≤ 2 0\leq a,b\leq2 0 ≤ a , b ≤ 2
If projective wild inertia is nontrivial, it is a C 5 C_{5} C 5 C 5 ⋊ C 4 C_{5}\rtimes C_{4} C 5 ⋊ C 4 S 5 S_{5} S 5 C 5 ⋊ C 2 C_{5}\rtimes C_{2} C 5 ⋊ C 2 A 5 A_5 A 5 K ( ζ 5 ) / K K(\zeta_5)/K K ( ζ 5 ) / K
Thus wild inertia is trivial. Since the quadratic quotient is ramified, projective inertia has order two. The absolute ramification index at five in the projective kernel field N / Q N/\mathbb{Q} N / Q 3 / 4 3/4 3/4 4 / 5 4/5 4/5 [ N : Q ] = 240 [N:\mathbb{Q}]=240 [ N : Q ] = 240
rd ( N ) ≤ 5 3 / 4 11 4 / 5 < 22.76889. \operatorname{rd}(N) \le5^{3/4}11^{4/5} < 22.76889. rd ( N ) ≤ 5 3/4 1 1 4/5 < 22.76889. The GRH lower bound (43) at n = 240 n=240 n = 240 b = 5 b=5 b = 5 23.24040 23.24040 23.24040
Proposition 20.17. Assume GRH for Dedekind zeta functions. Every pure, totally odd, absolutely irreducible strongly compatible weight-two system over K K K
Proof. Take a coefficient place above three. If its residue is reducible, Lemma 20.10 proves modularity of the actual source. Otherwise the same lemma gives absolute irreducibility after restriction to K ( ζ 3 ) K(\zeta_3) K ( ζ 3 )
At each eleven-prime apply Lemma 4.9 to the actual residual representation of the old special source. It gives a tame determinant-one cuspidal lift of order three, with determinant χ 3 \chi_3 χ 3
Prescribe these tame order-three cuspidal types at both eleven-primes, retain the old crystalline weight-two component at three, and retain unramified conditions elsewhere. EGPZ [28 ] , Theorem 4.2.1 gives a global point. Lemma 2.4 gives actual potential modularity and the pure strongly compatible replacement. It is modular by Theorem 20.12 . AKT [3 ] , Theorem 2.3, applied to the old weight-two three-adic member, gives the backward implication.
Theorem 20.18. Assume GRH for Dedekind zeta functions. Every continuous absolutely irreducible totally odd representation r ˉ : G K → G L 2 ( F 11 ) \bar{r}:G_K\to\mathrm{GL}_2(\mathbb{F}_{11}) r ˉ : G K → GL 2 ( F 11 ) ω 11 \omega_{11} ω 11
Proof. A solvable image has an odd solvable Artin witness and hence is modular after regular weight raising. Otherwise the cyclotomic restriction is absolutely irreducible. Use Lemma 20.13 at both split eleven-places, and unramified conditions elsewhere. EGPZ gives a weight-two global lift of exact cyclotomic determinant; there is no exceptional case in characteristic eleven. Lemma 2.4 constructs its pure strongly compatible system.
Make the independent global quadratic corrections described above. The resulting system is good outside eleven and its two local conditions are good, special, or principal series of odd inertial order five. All twists are reversed at the end. Its five-adic member is Barsotti–Tate above five. At eleven its actual residual inertia is trivial or unipotent of order five: the fifth-order eigencharacters coalesce, while a special representation has procyclic tame pro-five unipotent inertia.
If the five-residue is globally reducible, Lemma 20.11 gives modularity of the actual member and hence of the entire system. Otherwise Lemma 20.15 proves cyclotomic irreducibility. The controlled exceptional exclusion Proposition 20.16 applies: the representation is finite flat at five, supported only at five and eleven, and has the stated eleven-inertia. Thus the residual five-representation is nonexceptional in the precise sense required by EGPZ.
At each eleven-prime choose the good or special local lift of Lemma 20.14 . Retain a component through the old Barsotti–Tate point at five and unramified conditions elsewhere. EGPZ gives the weight-two replacement, and Lemma 2.4 gives its pure strongly compatible system. By the purity observation in § 2.1 , the replacement is special or good at both eleven-places, and good away from eleven.
If either eleven-place is good, Theorem 20.8 excludes this pure absolutely irreducible system. If both are special, Proposition 20.17 proves its modularity. In the latter case the common modular residual witness lifts the actual old weight-two five-adic source by [3 Theorem 2.3] . Compatibility, reversal of the quadratic twist and reduction at the original eleven-place prove the theorem.
Removal of arbitrary conductor
Proof for K = Q ( 5 ) K = \mathbb{Q}(\sqrt{5}) K = Q ( 5 ) A solvable residual image has an odd solvable Artin witness and is modular after regular weight raising. Otherwise its cyclotomic restriction is absolutely irreducible. For an initial weight-two lift use [3 Theorem 2.2] , except in exceptional characteristic five, where Proposition 20.3 supplies the point with its allowed auxiliary ramification. Lemma 2.4 , or the potential-modularity assertion of that proposition, gives a potentially modular compatible system. Its nonsolvable residue makes it nondihedral, and its determinant is cyclotomic times a finite totally even character.
Apply Theorem 1.2 with ℓ = 11 \ell= 11 ℓ = 11 Theorem 20.18 ; Proposition 20.1 gives the constituent quotient χ ‾ 11 \overline{\chi}_{11} χ 11 e ∈ { 1 , 3 , 5 , 7 , 9 } e \in\{1,3,5,7,9\} e ∈ { 1 , 3 , 5 , 7 , 9 } G Q G_{\mathbb{Q}} G Q
Put
K = Q ( 21 ) , a = 1 + 21 2 , ϵ = a + 2 , γ = a + 1 , H = K ( ζ 3 ) . K = \mathbb{Q}(\sqrt{21}), \qquad a = \frac{1+\sqrt{21}}{2}, \qquad\epsilon= a+2, \qquad\gamma= a+1, \qquad H = K(\zeta_3). K = Q ( 21 ) , a = 2 1 + 21 , ϵ = a + 2 , γ = a + 1 , H = K ( ζ 3 ) . Thus a 2 − a − 5 = 0 a^2-a-5=0 a 2 − a − 5 = 0 γ 2 = 3 ϵ \gamma^2=3\epsilon γ 2 = 3 ϵ N ( γ ) = − 3 N(\gamma)=-3 N ( γ ) = − 3
The proof first handles the bounded local menu at the two places above five. Nonsolvable dyadic residues are controlled by auxiliary fields and the common discriminant inequality. The remaining mixed source is treated by finite-flat rigidity and a toric realization. The pure menus are then reduced to three, where actual ordinary, dihedral and residually reducible sources require different arguments.
We use the compatible-system conventions of Section 2. Potential modularity supplies a cuspidal realization after base change, so every companion used below is absolutely irreducible.
Let ω 2 \omega_2 ω 2 Q 5 \mathbb{Q}_5 Q 5
τ 3 = ω ~ 2 8 ⊕ ω ~ 2 16 , τ 6 = ω ~ 2 4 ⊕ ω ~ 2 20 . \tau_3=\widetilde{\omega}_2^8\oplus\widetilde{\omega}_2^{16}, \qquad\tau_6=\widetilde{\omega}_2^4\oplus\widetilde{\omega}_2^{20}. τ 3 = ω 2 8 ⊕ ω 2 16 , τ 6 = ω 2 4 ⊕ ω 2 20 . The notation SC means one of these types, and SP means a fixed quadratic twist of Steinberg with nonzero monodromy. The exact quadratic character, including its unramified sign, is retained whenever a special deformation condition is imposed.
Proposition 21.1 (Mixed source). Assume GRH. A potentially modular compatible system over K K K { 0 , 1 } \{0,1\} { 0 , 1 } K K K G L 2 \mathrm{GL}_2 GL 2
The proof of this proposition is given below, including both the nonsolvable and solvable dyadic branches. The special condition retains its actual nonzero monodromy.
For a pure special menu with reducible five-adic residue, Proposition 21.2 applies directly. For either pure menu, the three-adic member r 3 r_3 r 3
Table 11.
The last line is essential: good supersingular elliptic curves in the present conductor class realize it.
Proposition 21.2. Let R \mathcal{R} R K K K R \mathcal{R} R
Proof. After removing the scalar quadratic character, the filtered-module argument of Lemma 4.2 gives an ordinary quotient at each five-place. Restoring that character preserves ordinarity. Apply [26 Corollary 1.2.2] to the dual of the actual five-adic source: K / Q K/\mathbb{Q} K / Q K K K χ 5 − 1 \chi_5^{-1} χ 5 − 1 R \mathcal{R} R
Local lifts and a switch in characteristic three
Lemma 21.3 (Bounded five-adic lifts). Every continuous representation r ˉ : G Q 5 → GL 2 ( F 5 ) \bar{r}: G_{\mathbb{Q}_5} \to\operatorname{GL}_2(\mathbb{F}_5) r ˉ : G Q 5 → GL 2 ( F 5 ) χ ˉ 5 \bar{\chi}_5 χ ˉ 5 χ 5 \chi_5 χ 5 { 0 , 1 } \{0,1\} { 0 , 1 }
Proof. The weights of GL 2 ( F 5 ) \operatorname{GL}_2(\mathbb{F}_5) GL 2 ( F 5 )
1 , det , Sym 2 det , Sym 2 det , Sym 4 , Sym 4 det 2 . 1,\quad\det,\quad\operatorname{Sym}^{2}\det,\quad\operatorname{Sym}^{2}\det,\quad\operatorname{Sym}^{4},\quad\operatorname{Sym}^{4}\det^{2}. 1 , det , Sym 2 det , Sym 2 det , Sym 4 , Sym 4 det 2 . The two cuspidal compact types reduce with constituents
σ ˉ ( τ 6 ) s s = 1 ⊕ Sym 2 det , σ ˉ ( τ 3 ) s s = det 2 ⊕ Sym 2 det 3 . \bar{\sigma}(\tau_6)^{\mathrm{ss}}=1\oplus\operatorname{Sym}^{2}\det,\qquad\bar{\sigma}(\tau_3)^{\mathrm{ss}}=\det^{2}\oplus\operatorname{Sym}^{2}\det^{3}. σ ˉ ( τ 6 ) ss = 1 ⊕ Sym 2 det , σ ˉ ( τ 3 ) ss = det 2 ⊕ Sym 2 det 3 . These identities follow by evaluating the cuspidal Brauer characters on split and nonsplit regular elements. The cuspidal character is zero on split regular elements, equals − θ ( z ) − θ ( z 5 ) -\theta(z)-\theta(z^5) − θ ( z ) − θ ( z 5 ) 4 θ ( a ) 4\theta(a) 4 θ ( a ) a a a
The last two give the semistable compact type for a scalar quadratic inertial type. If the resulting point has nonzero monodromy, it is SP. If all such points were crystalline, the corresponding crystalline compact type, 1 1 1 det 2 \det^2 det 2
Lemma 21.4 (The actual-residual switch). Let s ˉ \bar{s} s ˉ χ 3 \chi_3 χ 3 G Q 5 G_{\mathbb{Q}_5} G Q 5 s ˉ \bar{s} s ˉ χ 3 \chi_3 χ 3 τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
Proof. Apply Lemma 4.9 with ℓ = 5 \ell= 5 ℓ = 5
Controlled characteristic-two residues
Put F = K 2 F = K_2 F = K 2 Q 2 \mathbb{Q}_2 Q 2 S C 2 \mathrm{SC}_2 SC 2 S P 2 \mathrm{SP}_2 SP 2 S C , S P \mathrm{SC},\mathrm{SP} SC , SP
Definition 21.5. The two permitted conditions at a place v ∣ 5 v \mid5 v ∣ 5 K K K
(i) S C 2 \mathrm{SC}_2 SC 2 S 3 S_3 S 3 Q 5 \mathbb{Q}_5 Q 5
(ii) S P 2 \mathrm{SP}_2 SP 2 E 5 ′ = Q 5 ( 2 , 5 ) E'_5 = \mathbb{Q}_5(\sqrt{2},\sqrt{5}) E 5 ′ = Q 5 ( 2 , 5 )
The S C 2 \mathrm{SC}_2 SC 2 X 3 − 5 X^3 - 5 X 3 − 5 + 1 +1 + 1 − 1 -1 − 1
Theorem 21.6. Assume G R H GRH GR H
r ˉ : G K ⟶ S L 2 ( F ‾ 2 ) \bar{r}: G_K \longrightarrow\mathrm{SL}_2(\overline{\mathbb{F}}_2) r ˉ : G K ⟶ SL 2 ( F 2 ) be continuous and absolutely irreducible. Suppose that r ˉ \bar{r} r ˉ r ˉ \bar{r} r ˉ
The theorem covers the pairs ( S C 2 , S C 2 ) (\mathrm{SC}_2,\mathrm{SC}_2) ( SC 2 , SC 2 ) ( S C 2 , S P 2 ) (\mathrm{SC}_2,\mathrm{SP}_2) ( SC 2 , SP 2 ) ( S P 2 , S P 2 ) (\mathrm{SP}_2,\mathrm{SP}_2) ( SP 2 , SP 2 )
Corollary 21.7. Assume G R H GRH GR H R = ( r λ ) λ \mathcal{R} = (r_\lambda)_\lambda R = ( r λ ) λ K K K { 0 , 1 } \{0,1\} { 0 , 1 } v ∤ 5 v \nmid5 v ∤ 5 v v v R \mathcal{R} R
The distinction between these two assertions matters. Theorem 21.6 provides a modular witness for a residue. Corollary 21.7 uses an automorphy lifting theorem to identify the given characteristic-zero representation with a modular one.
Lemma 21.8. In a determinant-cyclotomic compatible system with all-place Weil–Deligne compatibility, reduction at a place above two of a local member of quadratic-special type satisfies S P 2 \mathrm{SP}_2 SP 2 S C 2 \mathrm{SC}_2 SC 2
Proof. In the special case the diagonal characters are, up to a quadratic twist, χ 2 ν , ν \chi_2\nu,\nu χ 2 ν , ν ν \nu ν E 5 ′ E'_5 E 5 ′ e ≤ 2 e \leq2 e ≤ 2 f ≤ 2 f \leq2 f ≤ 2 d ≤ 1 / 2 d \leq1/2 d ≤ 1/2
For either cuspidal type the reduced inertia has exact order three. A Frobenius lift swaps its two eigenlines. In characteristic zero its square is − 5 I -5I − 5 I S 3 S_3 S 3 e = 3 e = 3 e = 3 f = 2 f = 2 f = 2 d = 2 / 3 d = 2/3 d = 2/3
Finite-flat ramification at two
We use Fontaine’s ramification theorem [31 Theorem A and its corollary, pp. 515–516] . His numbering is G ( u ) = G u − 1 G^{(u)}=G^{u-1} G ( u ) = G u − 1 d < 2 d<2 d < 2 p p p d < 1 + 1 / ( p − 1 ) d<1+1/(p-1) d < 1 + 1/ ( p − 1 )
Lemma 21.9. Let q = 2 f ≥ 4 q=2^f\geq4 q = 2 f ≥ 4 ρ ˉ : G F → S L 2 ( F q ) \bar{\rho}:G_F\to\mathrm{SL}_2(\mathbb{F}_q) ρ ˉ : G F → SL 2 ( F q ) E E E
d ( E / F ) ≤ 2 − 2 q . d(E/F)\leq2-\frac{2}{q}. d ( E / F ) ≤ 2 − q 2 . If its tame diagonal characters are nontrivial and wild inertia is nontrivial, the stronger bound d ( E / F ) < 4 / 3 d(E/F)<4/3 d ( E / F ) < 4/3
Proof. First suppose wild inertia is trivial. If the representation is reducible over F 2 \mathbb{F}_2 F 2 # F 4 × = 3 \#\mathbb{F}_4^\times=3 # F 4 × = 3 2 / 3 2/3 2/3 θ , θ 4 \theta,\theta^4 θ , θ 4 θ 5 = 1 \theta^5=1 θ 5 = 1 4 / 5 4/5 4/5
If wild inertia is nontrivial, its unique fixed line is rational over F q \mathbb{F}_q F q η , η − 1 \eta,\eta^{-1} η , η − 1 T T T P ≤ q P\leq q P ≤ q
If η \eta η T / F T/F T / F E / T E/T E / T d ( E / F ) ≤ 2 ( 1 − 1 / P ) ≤ 2 − 2 / q d(E/F)\leq2(1-1/P)\leq2-2/q d ( E / F ) ≤ 2 ( 1 − 1/ P ) ≤ 2 − 2/ q
Suppose the tame diagonal order is three. Now e ( T / F ) = 3 e(T/F)=3 e ( T / F ) = 3 T T T [16 §1] . A conductor-six character has upper break five over T T T 5 / 3 5/3 5/3 F F F
d ( E / F ) ≤ 2 3 + 2 3 ( 1 − 1 / P ) < 4 3 . d(E/F)\leq\frac{2}{3}+\frac{2}{3}(1-1/P)<\frac{4}{3}. d ( E / F ) ≤ 3 2 + 3 2 ( 1 − 1/ P ) < 3 4 . The solvable auxiliary fields
Let
H = K ( − 3 ) = Q ( − 3 , − 7 ) . H=K(\sqrt{-3})=\mathbb{Q}(\sqrt{-3},\sqrt{-7}). H = K ( − 3 ) = Q ( − 3 , − 7 ) . It is quadratic over K K K ( − 3 ) ( − 7 ) ( 21 ) = 21 2 (-3)(-7)(21)=21^2 ( − 3 ) ( − 7 ) ( 21 ) = 2 1 2 S P 2 \mathrm{SP}_2 SP 2
Let S S S Q \mathbb{Q} Q
f ( X ) = X 3 − 3 X 2 − 2 X − 1 , f(X)=X^3-3X^2-2X-1, f ( X ) = X 3 − 3 X 2 − 2 X − 1 , and put A = K S A=KS A = K S − 175 -175 − 175 S / Q S/\mathbb{Q} S / Q S 3 S_3 S 3 Q ( − 7 ) \mathbb{Q}(\sqrt{-7}) Q ( − 7 ) A / K A/K A / K S 3 S_3 S 3 H H H τ \tau τ
Lemma 21.10. The extension A / K A/K A / K F F F S C 2 \mathrm{SC}_2 SC 2
Proof. The splitting field S S S f ( X + 1 ) = X 3 − 5 X − 5 f(X+1)=X^3-5X-5 f ( X + 1 ) = X 3 − 5 X − 5 − 7 -7 − 7 S 3 S_3 S 3 Q 7 ( − 7 ) = Q 7 ( 21 ) \mathbb{Q}_7(\sqrt{-7})=\mathbb{Q}_7(\sqrt{21}) Q 7 ( − 7 ) = Q 7 ( 21 ) − 3 -3 − 3 Q 7 \mathbb{Q}_7 Q 7 K K K f f f F F F
We now specify four unit squareclasses in A A A u u u f f f z = − 3 z=\sqrt{-3} z = − 3 A A A
v = 9 + 26 u − 8 u 2 + ( 2 u − 7 ) z 2 . v=\frac{9+26u-8u^2+(2u-7)z}{2}. v = 2 9 + 26 u − 8 u 2 + ( 2 u − 7 ) z . The choices of roots below are made compatibly with τ \tau τ
v 3 + 15 z − 1 2 v 2 + 15 z + 1 2 v − 1 = 0. v^3+\frac{15z-1}{2}v^2+\frac{15z+1}{2}v-1=0. v 3 + 2 15 z − 1 v 2 + 2 15 z + 1 v − 1 = 0. In particular u , v u,v u , v Q \mathbb{Q} Q
f ( X ) , X 6 − X 5 + 170 X 4 + 335 X 3 + 170 X 2 − X + 1. f(X),\qquad X^6-X^5+170X^4+335X^3+170X^2-X+1. f ( X ) , X 6 − X 5 + 170 X 4 + 335 X 3 + 170 X 2 − X + 1. Both have norm one for A / H A/H A / H
Lemma 21.11. In W = O A × / ( O A × ) 2 W=\mathcal{O}_A^\times/(\mathcal{O}_A^\times)^2 W = O A × / ( O A × ) 2
[ u ] , [ τ u ] , [ v ] , [ τ v ] [u],\ [\tau u],\ [v],\ [\tau v] [ u ] , [ τ u ] , [ v ] , [ τ v ] are independent and span U = ( τ − 1 ) W U=(\tau-1)W U = ( τ − 1 ) W A A A
Proof. Here A A A ( 0 , 6 ) (0,6) ( 0 , 6 ) dim F 2 W = 6 \dim_{\mathbb{F}_2}W=6 dim F 2 W = 6 A / H A/H A / H τ \tau τ W W W H H H ( τ − 1 ) W (\tau-1)W ( τ − 1 ) W
For completeness we give a finite exact certificate of independence and of the dyadic ranks. In the ordered list ( u , τ u , v , τ v ) (u,\tau u,v,\tau v) ( u , τ u , v , τ v ) m = ∑ j = 0 3 m j 2 j m=\sum_{j=0}^{3}m_j2^j m = ∑ j = 0 3 m j 2 j
Table 12.
Each kernel has order four and their intersection is zero. This proves local rank two and global rank four. The calculation can be performed by reduction modulo eight: over an unramified dyadic field a unit is a square precisely when its residue modulo eight has a square root. Appendix B specifies an exact number-field certificate for this table. The exact calculation is supplied in [49 d21] .
Finally, inertia at five contains τ \tau τ [ w / τ w ] [w/\tau w] [ w / τ w ] w w w U U U
Define
B = A ( U ) . B = A(\sqrt{U}). B = A ( U ) . Conjugation in Gal ( A / K ) \operatorname{Gal}(A/K) Gal ( A / K ) τ \tau τ τ − 1 \tau^{-1} τ − 1 U U U
[ B : K ] = 96 , B / K Galois and solvable . (13) [B:K] = 96,\qquad B/K\ \text{Galois and solvable}. \tag*{(13)} [ B : K ] = 96 , B / K Galois and solvable . (13) The field B / A B/A B / A C 2 2 C_2^2 C 2 2 C 3 C_3 C 3
Lemma 21.12. If E / F E/F E / F B B B d < 2 d < 2 d < 2
Proof. A quadratic extension defined by a unit is the point field of a finite flat extension of Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z μ 2 \mu_2 μ 2
The stable degree-four image at two
The following refinement is the only place where the local rank two in Lemma 21.11 is essential.
Lemma 21.13. Let ρ ˉ : G F → SL 2 ( F 16 ) \bar{\rho}: G_F \to\operatorname{SL}_2(\mathbb{F}_{16}) ρ ˉ : G F → SL 2 ( F 16 ) E E E G Q 2 G_{\mathbb{Q}_2} G Q 2 x ↦ x 4 x \mapsto x^4 x ↦ x 4 B B B
d ( E B / F ) ≤ 15 8 . d(EB/F) \le\frac{15}{8}. d ( E B / F ) ≤ 8 15 . Proof. Write T / F T/F T / F B B B T T T Gal ( T / F ) \operatorname{Gal}(T/F) Gal ( T / F )
If ρ ˉ \bar{\rho} ρ ˉ B B B
d ( E B / F ) ≤ 4 5 + 3 ⋅ 6 4 ⋅ 5 = 17 10 < 15 8 . d(EB/F) \le\frac{4}{5}+\frac{3\cdot6}{4\cdot5}=\frac{17}{10}<\frac{15}{8}. d ( E B / F ) ≤ 5 4 + 4 ⋅ 5 3 ⋅ 6 = 10 17 < 8 15 . If the tame diagonal order is three, let P ≤ 16 P \le16 P ≤ 16 B B B
d ( E B / F ) ≤ 2 3 + 3 P ⋅ 4 + ( P − 1 ) ⋅ 2 3 ⋅ 4 P = 11 6 + 1 6 P < 15 8 . d(EB/F) \le\frac{2}{3}+\frac{3P\cdot4+(P-1)\cdot2}{3\cdot4P}=\frac{11}{6}+\frac{1}{6P}<\frac{15}{8}. d ( E B / F ) ≤ 3 2 + 3 ⋅ 4 P 3 P ⋅ 4 + ( P − 1 ) ⋅ 2 = 6 11 + 6 P 1 < 8 15 . It remains to consider unramified diagonal characters. If wild inertia is zero, E / F E/F E / F B B B 3 / 2 3/2 3/2 s s s s 2 s^2 s 2 a a 4 ∈ F 4 × aa^4 \in\mathbb{F}_4^\times a a 4 ∈ F 4 × F F F
In the order-three case, over T T T B B B Gal ( T / F ) \operatorname{Gal}(T/F) Gal ( T / F ) T T T F F F 2 ( 1 − 1 / 16 ) = 15 / 8 2(1 - 1/16) = 15/8 2 ( 1 − 1/16 ) = 15/8
In the order-one case, the unit squareclasses from F F F B B B
The discriminant comparisons and residual modularity
Use the common test function and explicit formula of Section 22, with the finite-tail bound B b , N B_{b,N} B b , N A b , 4 A_{b,4} A b , 4
For every nonsolvable kernel field considered here, the group is G = SL 2 ( F q ) G = \operatorname{SL}_{2}(\mathbb{F}_{q}) G = SL 2 ( F q ) q = 2 f ≥ 4 q = 2^{f} \ge4 q = 2 f ≥ 4 K K K G G G S C 2 \mathrm{SC}_{2} SC 2 n n n n / 12 n/12 n /12 c 25 = 1 / 6 c_{25} = 1/6 c 25 = 1/6 S P 2 \mathrm{SP}_{2} SP 2 H H H c 25 ≥ 1 / 4 c_{25} \ge1/4 c 25 ≥ 1/4
Table 1 (PDF p. 103).
The common dyadic factor four comes from Fontaine’s strict d 2 < 2 d_{2} < 2 d 2 < 2 2 2 − 2 / q 2^{2-2/q} 2 2 − 2/ q q = 16 q = 16 q = 16 S C 2 \mathrm{SC}_{2} SC 2 2 15 / 8 2^{15/8} 2 15/8 B B B
Table 1 records rigorous evaluations of (39). In every row N = 100 N = 100 N = 100 m m m r 1 / n = 1 r_{1}/n = 1 r 1 / n = 1
Table 1 (PDF p. 103, second). Certified root-discriminant contradictions.
Proof of the residual theorem
Proof of Theorem 21.6. The solvable absolutely irreducible images in characteristic two are dihedral and are modular by the usual Hecke theta-series argument, as recalled in [3 proof of Theorem 2.5] . For nonsolvable images Dickson’s classification gives, after conjugation, G = SL 2 ( F q ) G = \operatorname{SL}_{2}(\mathbb{F}_{q}) G = SL 2 ( F q ) q = 2 f ≥ 4 q = 2^{f} \ge4 q = 2 f ≥ 4 q = 4 q = 4 q = 4 [3 Theorem 2.5] . We may assume q ≥ 8 q \ge8 q ≥ 8 M / K M/K M / K S P 2 \mathrm{SP}_{2} SP 2 H H H M M M
[ M H : Q ] = 4 q ( q 2 − 1 ) . [MH:\mathbb{Q}]=4q(q^{2}-1). [ M H : Q ] = 4 q ( q 2 − 1 ) . For a mixed pair, the three mixed rows of Table 1 exclude q = 8 q=8 q = 8 q = 16 q=16 q = 16 q ≥ 32 q\geq32 q ≥ 32 S P 2 \mathrm{SP}_{2} SP 2 q ≥ 8 q\geq8 q ≥ 8
Table 1.
We are left with two S C 2 \mathrm{SC}_{2} SC 2 B B B
[ M B : Q ] = 192 q ( q 2 − 1 ) , rd ( M B ) < 4 21 5 2 / 3 . [MB:\mathbb{Q}]=192q(q^{2}-1),\qquad\operatorname{rd}(MB)<4\sqrt{21}5^{2/3}. [ M B : Q ] = 192 q ( q 2 − 1 ) , rd ( M B ) < 4 21 5 2/3 . Its five-adic completions remain exactly S 3 S_{3} S 3 M M M A A A S C 2 \mathrm{SC}_{2} SC 2 B / A B/A B / A Table 1 excludes q ≥ 64 q\geq64 q ≥ 64
Let σ \sigma σ K / Q K/\mathbb{Q} K / Q M ≠ M σ M\ne M^{\sigma} M = M σ M ∩ M σ = K M\cap M^{\sigma}=K M ∩ M σ = K K K K G × G G\times G G × G B B B
[ M M σ B : Q ] = 192 ∣ G ∣ 2 ≥ 48 771 072. [MM^{\sigma}B:\mathbb{Q}]=192|G|^{2}\geq48\,771\,072. [ M M σ B : Q ] = 192∣ G ∣ 2 ≥ 48 771 072. The first row of the table excludes this possibility. Hence M = M σ M=M^{\sigma} M = M σ M / Q M/\mathbb{Q} M / Q
For q = 8 , 16 , 32 q=8,16,32 q = 8 , 16 , 32
C 2 ⟶ Out ( S L 2 ( F 2 f ) ) = C f . C_{2}\longrightarrow\operatorname{Out}(\mathrm{SL}_{2}(\mathbb{F}_{2^{f}}))=C_{f}. C 2 ⟶ Out ( SL 2 ( F 2 f )) = C f . If this action is trivial, the centerless extension Gal ( M / Q ) \operatorname{Gal}(M/\mathbb{Q}) Gal ( M / Q ) G × C 2 G\times C_{2} G × C 2 G G G G G G G Q G_{\mathbb{Q}} G Q Q \mathbb{Q} Q [40 , 41 ] and quadratic base change [4 ] prove modularity over K K K
For q = 8 q=8 q = 8 32 32 32 f f f q = 16 q=16 q = 16 x ↦ x 4 x\mapsto x^{4} x ↦ x 4 K K K
[ M B : Q ] = 783 360 , rd ( M B ) ≤ 21 2 15 / 8 5 2 / 3 . [MB:\mathbb{Q}]=783\,360,\qquad\operatorname{rd}(MB)\leq\sqrt{21}2^{15/8}5^{2/3}. [ M B : Q ] = 783 360 , rd ( M B ) ≤ 21 2 15/8 5 2/3 . The third row of the table contradicts this. All cases have therefore been excluded or proved modular. □ \square □
Lifting the given nonsolvable source
Lemma 21.14. Under the hypotheses of Theorem 21.6, if the image is nonsolvable, then r ˉ ( c v ) ≠ 1 \bar r(c_v)\ne1 r ˉ ( c v ) = 1 v v v K K K
Proof. If all complex conjugations were trivial, the kernel field M M M 2 ∣ S L 2 ( F 4 ) ∣ = 120 2|\mathrm{SL}_{2}(\mathbb{F}_{4})|=120 2∣ SL 2 ( F 4 ) ∣ = 120 4 21 5 2 / 3 4\sqrt{21}5^{2/3} 4 21 5 2/3 Table 1 contradicts this, even without retaining any Euler term. □ \square □
Proof of Corollary 21.7. Choose the two-adic member of R \mathcal{R} R { 0 , 1 } \{0,1\} { 0 , 1 } [42 Theorems 2.3(3), 4.1 and Corollary 4.3] . All-place compatibility and Lemma 21.8 give the actual local fields required by Theorem 21.6. That theorem provides a regular algebraic cuspidal modular witness for the residue. Lemma 21.14 supplies a nontrivial residual complex conjugation at some real place.
Apply [69 Theorem 6.1] . Its hypotheses are finite ramification, potential crystallinity with labelled weights { 0 , 1 } \{0,1\} { 0 , 1 } G L 2 \mathrm{GL}_2 GL 2 H T ( χ 2 ) = − 1 \mathrm{HT}(\chi_2)=-1 HT ( χ 2 ) = − 1
A separate finite-flat theorem in characteristic five
Let
H = K ( − 3 ) = Q ( − 3 , − 7 ) , η : G K ⟶ { 1 , − 1 } H = K(\sqrt{-3}) = \mathbb{Q}(\sqrt{-3},\sqrt{-7}), \qquad\eta:G_K \longrightarrow\{1,-1\} H = K ( − 3 ) = Q ( − 3 , − 7 ) , η : G K ⟶ { 1 , − 1 } be its quadratic character. Section 21.3 identifies a unique S 3 S_3 S 3 A 1 / K A_1/K A 1 / K H H H ( 10 ) (10) ( 10 ) S 3 S_3 S 3 e = 3 e=3 e = 3 f = 2 f=2 f = 2 C 3 C_3 C 3 r ˉ 1 : G K ⟶ S L 2 ( F 2 ) \bar{r}_1:G_K\longrightarrow\mathrm{SL}_2(\mathbb{F}_2) r ˉ 1 : G K ⟶ SL 2 ( F 2 )
Theorem 21.15. Let r : G K ⟶ G L 2 ( Q ‾ 2 ) r:G_K\longrightarrow\mathrm{GL}_2(\overline{\mathbb{Q}}_2) r : G K ⟶ GL 2 ( Q 2 ) r ˉ 1 \bar{r}_1 r ˉ 1 χ 2 \chi_2 χ 2 r r r { 0 , 1 } \{0,1\} { 0 , 1 } τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
r ≃ Ind G H G K ψ r \simeq\operatorname{Ind}_{G_H}^{G_K}\psi r ≃ Ind G H G K ψ for a geometric character ψ \psi ψ G H G_H G H r r r
The compatible-system assumption in Proposition 21.1 will be used to compare every coefficient conjugate and then to construct an abelian variety. Theorem 21.15 instead uses a quadratic selftwist of the actual two-adic source.
Solvable residual representations and their fields
Reduction in characteristic two of either cuspidal type has the exact tame S 3 S_3 S 3 e = 3 e=3 e = 3 f = 2 f=2 f = 2 Q 5 ( 2 , 5 ) \mathbb{Q}_5(\sqrt{2},\sqrt{5}) Q 5 ( 2 , 5 )
The cuspidal place makes every characteristic-two residue in Proposition 21.1 absolutely irreducible. Its nonsolvable case is already modular as an actual source, by Corollary 21.7. In that result the GRH comparisons exclude simple images S L 2 ( F q ) \mathrm{SL}_2(\mathbb{F}_q) SL 2 ( F q ) q ≥ 8 q\geq8 q ≥ 8 q = 4 q=4 q = 4 [69 Theorem 6.1] . We therefore concentrate on solvable residues.
Such an absolutely irreducible determinant-one image in characteristic two is dihedral D 2 n D_{2n} D 2 n n n n L / K L/K L / K
Lemma 21.16. Let r : G F → G L 2 ( E ) r:G_F\to\mathrm{GL}_2(E) r : G F → GL 2 ( E ) { 0 , 1 } \{0,1\} { 0 , 1 } χ 2 \chi_2 χ 2 E / Q 2 E/\mathbb{Q}_2 E / Q 2 r r r
Proof. Here F = K 2 F=K_2 F = K 2 Q 2 \mathbb{Q}_2 Q 2 p = 2 p=2 p = 2 e ( F / Q 2 ) = 1 e(F/\mathbb{Q}_2)=1 e ( F / Q 2 ) = 1
Proposition 21.17. For a residue occurring in the mixed theorem, either its given source is modular by the ordinary theorem of Allen, or its quadratic inducing field is H H H
Proof. If L / K L/K L / K n n n C 2 C_2 C 2 D 2 n D_{2n} D 2 n C 2 C_2 C 2 C 2 C_2 C 2 L F / F L_F/F L F / F [31 Theorem A] . Its Kummer class has even dyadic valuation.
The actual crystalline source is ordinary by Lemma 21.16. Allen’s ordinary modular witness lemma [2 Lemma 5.1.2] supplies a quotient-compatible nearly ordinary cuspidal lift of the globally irreducible dihedral residue. It has no CM nonsplitting hypothesis. The actual-source theorem [2 Theorem 5.2.1] then applies: if L L L L L L
It remains to identify the possible inducing fields that are split or unramified at two. Put α = ( 1 + 21 ) / 2 \alpha=(1+\sqrt{21})/2 α = ( 1 + 21 ) /2 ϵ = ( 5 + 21 ) / 2 \epsilon=(5+\sqrt{21})/2 ϵ = ( 5 + 21 ) /2 K K K 21 / 2 < 2.3 \sqrt{21}/2<2.3 21 /2 < 2.3 ϵ \epsilon ϵ
[ − 1 ] , [ ϵ ] , [ 2 ] , [ α ] , [ α 2 ] . [-1],\ [\epsilon],\ [2],\ [\alpha],\ [\alpha^2]. [ − 1 ] , [ ϵ ] , [ 2 ] , [ α ] , [ α 2 ] . Even dyadic valuation removes [ 2 ] [2] [ 2 ] [ − 1 ] [-1] [ − 1 ] [ ϵ ] [\epsilon] [ ϵ ] [ 5 ] [5] [ 5 ] [ α ] [\alpha] [ α ] F F F [ 5 ] = 1 [5]=1 [ 5 ] = 1 [ ϵ ] = [ − 1 ] [\epsilon]=[-1] [ ϵ ] = [ − 1 ]
( 3 + 21 2 ) 2 = 3 ϵ , \left(\frac{3+\sqrt{21}}{2}\right)^2=3\epsilon, ( 2 3 + 21 ) 2 = 3 ϵ , and − 3 -3 − 3 F F F w = α m o d 2 ∈ F 4 w=\alpha\bmod2\in\mathbb{F}_4 w = α mod 2 ∈ F 4 α 3 ≡ 1 + 2 α ( m o d 4 ) \alpha^3\equiv1+2\alpha\pmod4 α 3 ≡ 1 + 2 α ( mod 4 ) [ − 1 ] [-1] [ − 1 ] [ α ] [\alpha] [ α ] [ − α ] [-\alpha] [ − α ] U 1 / U 2 U_1/U_2 U 1 / U 2 1 1 1 w w w 1 + w 1+w 1 + w L L L
The kernel of the local squareclass map is generated by [ 5 ] [5] [ 5 ] [ − ϵ ] [-\epsilon] [ − ϵ ] 5 5 5 − ϵ -\epsilon − ϵ − 5 ϵ -5\epsilon − 5 ϵ L = K ( − ϵ ) = K ( − 3 ) = H L=K(\sqrt{-\epsilon})=K(\sqrt{-3})=H L = K ( − ϵ ) = K ( − 3 ) = H
We next enumerate all odd inducing characters for this last quadratic field. The quartic H H H Y 4 + 20 Y 2 + 16 Y^4+20Y^2+16 Y 4 + 20 Y 2 + 16 441 441 441 bnfcertify:
C l H ( 5 ) = C 24 , C l H ( 10 ) = C 24 × C 3 . \mathrm{Cl}_H(5)=C_{24},\qquad\mathrm{Cl}_H(10)=C_{24}\times C_3. Cl H ( 5 ) = C 24 , Cl H ( 10 ) = C 24 × C 3 . An odd inducing character is tame at two and five; its conductor divides ( 10 ) (10) ( 10 ) F 3 \mathbb{F}_3 F 3 H / K H/K H / K diag ( 5 , 2 ) \operatorname{diag}(5,2) diag ( 5 , 2 ) − I -I − I K K K
( 5 ) , ( 10 ) , ( 2 ) q 1 , ( 2 ) q 2 , (14) (5),\qquad(10),\qquad(2)\mathfrak{q}_1,\qquad(2)\mathfrak{q}_2, \tag*{(14)} ( 5 ) , ( 10 ) , ( 2 ) q 1 , ( 2 ) q 2 , (14) where q i q_i q i H H H K K K
For the mixed orientation, the first two quotients ramify cubically at the special place and are impossible. Exactly one of the last two ramifies at the selected cuspidal place. Denote its field over K K K A m A_m A m S 3 S_3 S 3 GL 2 ( F 2 ) \operatorname{GL}_2(\mathbb{F}_2) GL 2 ( F 2 )
For a pair of cuspidal places, the first two fields in (11) are the possibilities. We denote them A 0 A_0 A 0 A 1 A_1 A 1 C 3 C_3 C 3 A m A_m A m A 1 A_1 A 1 F 4 \mathbb{F}_4 F 4 ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , 1 ) (0,1) ( 0 , 1 ) A m A_m A m
Finite-flat tangent extensions at two
We use the exact Honda-system description at the boundary e ( F ) = 1 = p − 1 e(F) = 1 = p - 1 e ( F ) = 1 = p − 1 [13 pp. 4–6] . On biconnected objects the relevant categories are abelian, with the usual schematic kernels and quotients, and the generic-fibre functor is fully faithful. The integral classification and exactness at two can also be expressed using [42 Corollaries 4.3–4.4] .
Lemma 21.18. Let k = F 4 k = \mathbb{F}_4 k = F 4 R = k [ ε ] / ( ε 2 ) R = k[\varepsilon]/(\varepsilon^2) R = k [ ε ] / ( ε 2 ) θ : G F → k × \theta: G_F \to k^\times θ : G F → k × θ ( 1 + ε c ) \theta(1+\varepsilon c) θ ( 1 + ε c ) F 2 \mathbb{F}_2 F 2 c c c
Proof. Every finite-flat model of θ \theta θ θ \theta θ R R R
By the exact coefficient version of the classification [42 Corollaries 4.3–4.4 and Proposition 5.7.3] , its two coefficient factors have exponents r 0 , r 1 ∈ { 0 , 1 } r_0,r_1 \in\{0,1\} r 0 , r 1 ∈ { 0 , 1 } r 0 + 2 r 1 r_0 + 2r_1 r 0 + 2 r 1 ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , 1 ) (0,1) ( 0 , 1 ) p = 2 p = 2 p = 2 e = 1 e = 1 e = 1 f = 2 f = 2 f = 2 θ \theta θ
Lemma 21.19. For A = A m A = A_m A = A m A 1 A_1 A 1 A A A
Proof. Over k = F 4 k = \mathbb{F}_4 k = F 4 θ ⊕ θ − 1 \theta\oplus\theta^{-1} θ ⊕ θ − 1 k [ ε ] k[\varepsilon] k [ ε ] F F F
Apply Fontaine’s theorem there: classical upper breaks are at most one. Killing the two residual characters then makes a tame cubic base change, which multiplies positive upper breaks by three. Each resulting quadratic character has break at most three, hence conductor at most four.
The order of these operations supplies the bound for every dual-number tangent point. A finite-flat group defined only over the ramified cubic field can instead have conductor six. No division by a dual-number parameter, and no characteristic-zero lift of the tangent point, is used in Lemma 21.19.
The complete normal Selmer calculations
Write r ˉ = Ind H K χ \bar{r}=\operatorname{Ind}_{H}^{K}\chi r ˉ = Ind H K χ k = F 4 k=\mathbb{F}_4 k = F 4
ad 0 ( r ˉ ) = k I ⊕ V , V = Ind H K χ 2 . (15) \operatorname{ad}^{0}(\bar{r})=kI\oplus V,\qquad V=\operatorname{Ind}_{H}^{K}\chi^{2}. \tag*{(15)} ad 0 ( r ˉ ) = k I ⊕ V , V = Ind H K χ 2 . (15) The scalar summand consists of the equal diagonal entries; the two offdiagonal entries form V V V
Shapiro’s lemma and Kummer theory identify the global normal classes with the appropriate nontrivial C 3 C_{3} C 3 A A A C 3 = Gal ( A / H ) C_{3}=\operatorname{Gal}(A/H) C 3 = Gal ( A / H )
For A m A_m A m
P m ( X ) = X 12 − 3 X 11 + 3 X 10 − X 8 + 3 X 7 − 6 X 6 + 3 X 5 + X 4 + 6 X 3 − 3 X 2 − 9 X + 9. P_m(X)=X^{12}-3X^{11}+3X^{10}-X^{8}+3X^{7}-6X^{6}
+3X^{5}+X^{4}+6X^{3}-3X^{2}-9X+9. P m ( X ) = X 12 − 3 X 11 + 3 X 10 − X 8 + 3 X 7 − 6 X 6 + 3 X 5 + X 4 + 6 X 3 − 3 X 2 − 9 X + 9. Its class group is trivial, with certified full unit group. The unit-squareclass space has dimension six; there are four primes over five, giving an S 5 S_{5} S 5 C 3 C_{3} C 3 H H H C 3 C_{3} C 3 F 2 \mathbb{F}_{2} F 2 C 3 C_{3} C 3
Enumerate the 64 products in the ordered basis of the attached certificate, encoded by their six-bit masks. The conductor bound from Lemma 21.19 retains
0 , 17 , 33 , 48. 0,17,33,48. 0 , 17 , 33 , 48. At the special place the fixed exact line character will force homogeneous Frobenius trace zero in the tangent. Its necessary cut retains
0 , 5 , 8 , 13 , 16 , 21 , 24 , 29 , 34 , 39 , 42 , 47 , 50 , 55 , 58 , 63. 0,5,8,13,16,21,24,29,34,39,42,47,50,55,58,63. 0 , 5 , 8 , 13 , 16 , 21 , 24 , 29 , 34 , 39 , 42 , 47 , 50 , 55 , 58 , 63. The intersection consists only of zero. To identify the second cut, take residual Frobenius S = ( 0 1 1 0 ) S=\begin{pmatrix}0&1\\1&0\end{pmatrix} S = ( 0 1 1 0 ) z = ( c a a b ) z=\begin{pmatrix}c&a\\a&b\end{pmatrix} z = ( c a a b ) tr ( z S ) = b + c \operatorname{tr}(zS)=b+c tr ( z S ) = b + c F 25 \mathbb{F}_{25} F 25 ( u , 5 ) (u,5) ( u , 5 ) 25 ≡ 1 ( m o d 4 ) 25\equiv1\pmod4 25 ≡ 1 ( mod 4 ) [49 d21] .
For A 1 A_1 A 1
P 1 ( X ) = X 12 − 9 X 11 + 32 X 10 − 57 X 9 + 66 X 8 − 171 X 7 + 559 X 6 − 1020 X 5 + 1092 X 4 − 396 X 3 − 40 X 2 − 180 X + 148. P_1(X)=X^{12}-9X^{11}+32X^{10}-57X^{9}+66X^{8}-171X^{7}
+559X^{6}-1020X^{5}+1092X^{4}-396X^{3}-40X^{2}-180X+148. P 1 ( X ) = X 12 − 9 X 11 + 32 X 10 − 57 X 9 + 66 X 8 − 171 X 7 + 559 X 6 − 1020 X 5 + 1092 X 4 − 396 X 3 − 40 X 2 − 180 X + 148. Its field discriminant is 8576612100000000, its certified class group is C 3 C_{3} C 3 C 3 C_{3} C 3 C 3 C_{3} C 3
Table 13.
Only zero satisfies conductor at most four at both primes. At either five-adic place the normal character χ 2 \chi^2 χ 2 H 0 H^0 H 0 H 2 H^2 H 2 H 1 = 0 H^1 = 0 H 1 = 0
Proposition 21.20. For the mixed residue, a finite-flat normal tangent with homogeneous trace zero at the special place is a coboundary. For r ˉ 1 \bar{r}_1 r ˉ 1
Proof. The exhaustive squareclass computations above and Lemma 21.19 leave only zero in each of the corresponding global Kummer spaces.
These computations use certified class and unit data only in degrees four and twelve. The script names and exact outputs are listed in Appendix § 9.14 . The exact calculation is supplied in [49 d21] .
Closed deformation conditions and the full tangent
We now pass from the necessary normal cuts to the actual global deformation functor. Residual S 3 S_3 S 3 F 2 \mathbb{F}_2 F 2 Z 2 \mathbb{Z}_2 Z 2 χ 2 \chi_2 χ 2 [13 Definition 1.1.1 and the following discussion] . Here its residual model is biconnected, so models and coefficient actions are unique. Local framed conditions give closed quotients without a local-centralizer restriction.
At a cuspidal five-adic place, require the fixed finite inertia kernel to act trivially and a chosen generator to have the characteristic polynomial for the selected τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
For the mixed case we impose a stronger special condition retaining the exact quadratic character. Fix α : G K v S P → { 1 , − 1 } \alpha: G_{K_{\mathrm{vSP}}} \to\{1,-1\} α : G K vSP → { 1 , − 1 } Φ \Phi Φ a = α ( Φ ) ∈ { 1 , − 1 } a = \alpha(\Phi) \in\{1,-1\} a = α ( Φ ) ∈ { 1 , − 1 } M = r ( Φ ) − a I M = r(\Phi) - aI M = r ( Φ ) − a I S − I S-I S − I
tr r ( Φ ) = 6 a , r ( g ) v = χ 2 ( g ) α ( g ) v ( g ∈ G K v S P ) , (16) \operatorname{tr} r(\Phi) = 6a,\qquad r(g)v = \chi_2(g)\alpha(g)v\quad(g \in G_{K_{\mathrm{vSP}}}), \tag*{(16)} tr r ( Φ ) = 6 a , r ( g ) v = χ 2 ( g ) α ( g ) v ( g ∈ G K vSP ) , (16) where v v v M M M
det M = 5 − a ( 6 a ) + a 2 = 0. \det M = 5-a(6a)+a^2 = 0. det M = 5 − a ( 6 a ) + a 2 = 0. The unit pivot therefore makes im M \operatorname{im} M im M v v v ( r ( Φ ) − 5 a I ) M = 0 (r(\Phi)-5aI)M = 0 ( r ( Φ ) − 5 a I ) M = 0 χ 2 α \chi_2\alpha χ 2 α a a a α \alpha α
Let D m D_m D m D 1 D_1 D 1 r ˉ 1 \bar{r}_1 r ˉ 1
Proposition 21.21. Both Z 2 → D m \mathbb{Z}_2 \to D_m Z 2 → D m Z 2 → D 1 \mathbb{Z}_2 \to D_1 Z 2 → D 1 Z 2 \mathbb{Z}_2 Z 2
Proof. We calculate the true relative tangent, after scalar extension to k = F 4 k = \mathbb{F}_4 k = F 4 ad 0 ( r ˉ ) \operatorname{ad}^0(\bar{r}) ad 0 ( r ˉ ) End ( r ˉ ) \operatorname{End}(\bar{r}) End ( r ˉ )
Z 1 ( ad 0 r ˉ ) / d End ( r ˉ ) , (17) Z^1(\operatorname{ad}^0\bar{r})/d\operatorname{End}(\bar{r}), \tag*{(17)} Z 1 ( ad 0 r ˉ ) / d End ( r ˉ ) , (17) with the stated local conditions. In characteristic two this need not equal H 1 ( ad 0 r ˉ ) H^1(\operatorname{ad}^0\bar{r}) H 1 ( ad 0 r ˉ )
For the mixed problem, (16) gives homogeneous trace zero on tangent Frobenius, since 6 a 6a 6 a k [ ε ] k[\varepsilon] k [ ε ] D 1 D_1 D 1 V V V (15) . After a strict global change of basis the tangent is therefore scalar, c I cI c I c c c c c c c c c
The narrow class number of K K K k k k δ H \delta_H δ H H / K H/K H / K δ H I \delta_H I δ H I X = diag ( 1 , 0 ) X = \operatorname{diag}(1,0) X = diag ( 1 , 0 ) 1 + ε X = diag ( 1 + ε , 1 ) 1 + \varepsilon X = \operatorname{diag}(1 + \varepsilon,1) 1 + εX = diag ( 1 + ε , 1 ) 1 + ε 1 + \varepsilon 1 + ε δ H I \delta_H I δ H I X X X
The special marking adds no tangent of its own: with representation and eigencharacter fixed, an infinitesimal line motion w w w ( S − I ) w = 0 (S-I)w = 0 ( S − I ) w = 0
All the data are defined over Z 2 \mathbb{Z}_2 Z 2 F 4 \mathbb{F}_4 F 4 F 2 \mathbb{F}_2 F 2 Z 2 \mathbb{Z}_2 Z 2 F 2 \mathbb{F}_2 F 2 Z 2 \mathbb{Z}_2 Z 2 Z 2 \mathbb{Z}_2 Z 2
The cuspidal pair with tame dyadic inertia
Proof of Theorem 21.15. The actual source r r r D 1 D_1 D 1 D 1 = Z 2 D_1 = \mathbb{Z}_2 D 1 = Z 2
The quadratic character η \eta η F F F H / K H/K H / K r ⊗ η r \otimes\eta r ⊗ η r r r τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6
r ≃ r ⊗ η . r \simeq r \otimes\eta. r ≃ r ⊗ η . Absolute irreducibility and the nontrivial quadratic selftwist give r = Ind H K ψ r = \operatorname{Ind}_H^K \psi r = Ind H K ψ
The character is geometric as a constituent of r ∣ G H r|_{G_H} r ∣ G H [57 Chapter III] . Since H / K H/K H / K r r r [4 Chapter III] .
Rational traces and a toric place
We first isolate the rational endomorphism argument needed for the mixed source. It removes a possible Schur-index ambiguity in passing from rational traces to an elliptic curve.
Proposition 21.22. Let B / K ′ B/K' B / K ′ g g g ℓ \ell ℓ r r r Q ℓ \mathbb{Q}_{\ell} Q ℓ V ℓ ( B ) ≃ r ⊕ g V_{\ell}(B) \simeq r^{\oplus g} V ℓ ( B ) ≃ r ⊕ g ℓ \ell ℓ r r r B B B K ′ K' K ′ E g E^{g} E g E / K ′ E/K' E / K ′ V ℓ ( E ) ≃ r V_{\ell}(E) \simeq r V ℓ ( E ) ≃ r
Proof. Make the reduction semistable and its torus split by a finite local extension. Monodromy on V ℓ ( B ) V_{\ell}(B) V ℓ ( B ) g g g B B B T T T g g g C = End K ′ ( B ) ⊗ Q C = \operatorname{End}_{K'}(B) \otimes\mathbb{Q} C = End K ′ ( B ) ⊗ Q [29 ] gives
C ⊗ Q ℓ = End G K ′ ( V ℓ B ) = M g ( Q ℓ ) . C \otimes\mathbb{Q}_{\ell} = \operatorname{End}_{G_{K'}}(V_{\ell}B) = M_g(\mathbb{Q}_{\ell}). C ⊗ Q ℓ = End G K ′ ( V ℓ B ) = M g ( Q ℓ ) . so dim Q C = g 2 \dim_{\mathbb{Q}} C = g^2 dim Q C = g 2 X ∗ ( T ) ⊗ Q X_*(T) \otimes\mathbb{Q} X ∗ ( T ) ⊗ Q
Let I I I N N N V = V ℓ ( B ) V = V_{\ell}(B) V = V ℓ ( B )
W = im N = ker N = V ℓ ( T ) . W = \operatorname{im} N = \ker N = V_{\ell}(T). W = im N = ker N = V ℓ ( T ) . Every f ∈ I f \in I f ∈ I W W W f N = N f fN = Nf f N = N f ker N = W \ker N = W ker N = W I I I V V V I 2 = 0 I^2 = 0 I 2 = 0 C C C I = 0 I = 0 I = 0
Thus C C C End Q ( X ∗ ( T ) ⊗ Q ) = M g ( Q ) \operatorname{End}_{\mathbb{Q}}(X_*(T) \otimes\mathbb{Q}) = M_g(\mathbb{Q}) End Q ( X ∗ ( T ) ⊗ Q ) = M g ( Q ) r r r E / K ′ E/K' E / K ′ g g g K ′ K' K ′
Proof of Proposition 21.1. Assume that R \mathcal{R} R Q ‾ 2 \overline{\mathbb{Q}}_2 Q 2 Corollary 21.7 , and all dihedral cases other than H H H Proposition 21.17 . The ray calculation therefore puts every coefficient conjugate in the unique mixed residual class A m A_m A m
The cuspidal type is rational: coefficient conjugation exchanges its two inverse constituents, preserving τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6 α \alpha α { 1 , − 1 } \{1,-1\} { 1 , − 1 } Φ \Phi Φ
tr r 2 ( Φ ) = 6 α ( Φ ) ∈ { 6 , − 6 } . \operatorname{tr} r_2(\Phi) = 6\alpha(\Phi) \in\{6,-6\}. tr r 2 ( Φ ) = 6 α ( Φ ) ∈ { 6 , − 6 } . Every coefficient embedding fixes this number, its sign and the quadratic inertia character. Thus every conjugate belongs to the same closed problem D m D_m D m § 7.4.5 . The actual characteristic-zero point gives D m = Z 2 D_m = \mathbb{Z}_2 D m = Z 2 Q 2 \mathbb{Q}_2 Q 2 Q \mathbb{Q} Q Q 2 \mathbb{Q}_2 Q 2
Potential modularity and the special finite place give a G L 2 \mathrm{GL}_2 GL 2 B / K B/K B / K [64 Lemma 9.4.3] ; [64 Lemma 9.4.4] descends the realization to K K K
Let g = dim B g = \dim B g = dim B Q 2 \mathbb{Q}_2 Q 2 r 2 r_2 r 2 Q \mathbb{Q} Q
V 2 ( B ) ⊗ Q ‾ 2 ≃ ( r 2 ⊗ Q ‾ 2 ) ⊕ g V_2(B) \otimes\overline{\mathbb{Q}}_2 \simeq(r_2 \otimes\overline{\mathbb{Q}}_2)^{\oplus g} V 2 ( B ) ⊗ Q 2 ≃ ( r 2 ⊗ Q 2 ) ⊕ g The space Hom G K ( r 2 , V 2 ( B ) ) \operatorname{Hom}_{G_K}(r_2,V_2(B)) Hom G K ( r 2 , V 2 ( B )) Q 2 \mathbb{Q}_2 Q 2 g g g Q ‾ 2 \overline{\mathbb{Q}}_2 Q 2 V 2 ( B ) ≃ r 2 ⊕ g V_2(B) \simeq r_2^{\oplus g} V 2 ( B ) ≃ r 2 ⊕ g
At the special place monodromy on r 2 r_2 r 2 E / K E/K E / K V 2 ( E ) ≃ r 2 V_2(E) \simeq r_2 V 2 ( E ) ≃ r 2 [33 Theorem 1] . Thus r 2 r_2 r 2
Switching the pure menus to characteristic three
Proposition 21.23 (Switchable sources). Assume GRH. Let R R R r ˉ 3 ∣ G H \bar r_3|_{G_H} r ˉ 3 ∣ G H R R R
Proof. At one five-place apply Lemma 21.4 to obtain the opposite kind of lift; at the other retain its kind. At three retain a crystalline weight-two component containing the original member. Prescribe no ramification elsewhere and determinant χ 3 \chi_3 χ 3 [28 ] gives a global lift with precisely these component choices. Its hypotheses in characteristic three are total oddness and irreducibility on G H G_H G H [64 Theorem 5.1.2 and Section 9.2] give a potentially modular compatible system with support at five.
Choose the special component through the constructed point with nonzero monodromy. The purity observation in § 2.1 retains this condition in the global system. Proposition 21.1 makes the replacement modular. Apply [3 Theorem 2.3] to the old weight-two three-adic member, whose cyclotomic residual restriction is irreducible, to return modularity to the original system.
Reducible residual characters and the ordinary case
Lemma 21.24. If the three-adic residue of a bounded-menu system of determinant cyclotomic and support at five is reducible, its semisimplification is ψ ⊕ ψ ˉ χ ˉ 3 \psi\oplus\bar{\psi}\bar{\chi}_3 ψ ⊕ ψ ˉ χ ˉ 3 ψ \psi ψ
ψ = 1 , ψ 5 , ψ γ , ψ 5 ψ γ . \psi= 1,\quad\psi_5,\quad\psi_\gamma,\quad\psi_5\psi_\gamma. ψ = 1 , ψ 5 , ψ γ , ψ 5 ψ γ . Here ψ x \psi_x ψ x K ( x ) / K K(\sqrt{x})/K K ( x ) / K
Proof. Write the constituents initially as ψ \psi ψ χ ˉ 3 ψ − 1 \bar{\chi}_3\psi^{-1} χ ˉ 3 ψ − 1 ψ 2 \psi^2 ψ 2 K K K 21 / 2 < 2.3 \sqrt{21}/2 < 2.3 21 /2 < 2.3 { ± ϵ n } \{\pm\epsilon^n\} { ± ϵ n } η 0 = χ ‾ 3 \eta_0=\overline{\chi}_3 η 0 = χ 3 ψ ( c ) 2 = 1 \psi(c)^2=1 ψ ( c ) 2 = 1 η 0 \eta_0 η 0 ψ 2 \psi^2 ψ 2 η 0 \eta_0 η 0 ψ 2 = 1 \psi^2=1 ψ 2 = 1
The quadratic squareclasses supported at three and five have basis [ − 1 ] , [ ϵ ] , [ γ ] , [ 5 ] , [ a ] [-1],[\epsilon],[\gamma],[5],[a] [ − 1 ] , [ ϵ ] , [ γ ] , [ 5 ] , [ a ]
⟨ [ − ϵ ] , [ 5 ] , [ γ ] ⟩ . \langle[-\epsilon],[5],[\gamma]\rangle. ⟨[ − ϵ ] , [ 5 ] , [ γ ]⟩ . For an explicit check in the unramified quadratic dyadic ring, a unit class is unramified exactly when its cube is one modulo four. The identities
a 3 = 6 a + 5 , ϵ 3 = 24 a + 43 , γ 3 = 12 a + 21 a^3=6a+5,\qquad\epsilon^3=24a+43,\qquad\gamma^3=12a+21 a 3 = 6 a + 5 , ϵ 3 = 24 a + 43 , γ 3 = 12 a + 21 give this kernel: the ramified images of − 1 -1 − 1 a a a [ ϵ ] = [ − 1 ] [\epsilon]=[-1] [ ϵ ] = [ − 1 ] 5 5 5 − ϵ -\epsilon − ϵ γ \gamma γ ( γ 3 − 1 ) / 4 = 3 a + 5 (\gamma^3-1)/4=3a+5 ( γ 3 − 1 ) /4 = 3 a + 5
At three both 5 5 5 − ϵ -\epsilon − ϵ γ \gamma γ a ≡ 0 , 1 a\equiv0,1 a ≡ 0 , 1
Table 16.
where U U U R R R Q 5 ( 5 ) \mathbb{Q}_5(\sqrt{5}) Q 5 ( 5 ) ψ \psi ψ η 0 \eta_0 η 0
Proposition 21.25. A bounded-menu compatible system whose three-adic residue has constituents unramified at three is modular.
Proof. By Lemmas 4.10 and 21.24, the actual crystalline weight-two three-adic source is ordinary, with constituent quotient χ ‾ 3 \overline{\chi}_3 χ 3 [62 ] , applies: K K K K ( χ ‾ 3 ) = H K(\overline{\chi}_3)=H K ( χ 3 ) = H Q \mathbb{Q} Q G K 3 G_{K_3} G K 3 χ 3 \chi_3 χ 3
The cyclotomic-dihedral exception at three
Write ω = χ ‾ 3 \omega=\overline{\chi}_3 ω = χ 3 σ \sigma σ K / Q K/\mathbb{Q} K / Q v 1 , v 2 v_1,v_2 v 1 , v 2 Q 5 \mathbb{Q}_5 Q 5 χ 3 \chi_3 χ 3 τ 3 , τ 6 \tau_3,\tau_6 τ 3 , τ 6 − 5 I -5I − 5 I
The residual representation and the quartic fields
Lemma 21.26. Let r ‾ : G K ⟶ GL 2 ( F 3 ) \overline{r}:G_K\longrightarrow\operatorname{GL}_2(\mathbb{F}_3) r : G K ⟶ GL 2 ( F 3 ) ω \omega ω G H G_H G H
r ‾ ≃ Ind G H G K ξ , ξ c = ξ − 1 , ord ( ξ ) = 4 , \overline{r}\simeq\operatorname{Ind}_{G_H}^{G_K}\xi,\qquad\xi^c=\xi^{-1},\qquad\operatorname{ord}(\xi)=4, r ≃ Ind G H G K ξ , ξ c = ξ − 1 , ord ( ξ ) = 4 , where c c c ξ 2 \xi^2 ξ 2 H ( γ ) / H H(\sqrt{\gamma})/H H ( γ ) / H D 8 D_8 D 8 F 3 \mathbb{F}_3 F 3 η , η ′ \eta,\eta' η , η ′
L = K ( γ ) , L ′ = K ( − 3 γ ) , L = K(\sqrt{\gamma}), \qquad L' = K(\sqrt{-3\gamma}), L = K ( γ ) , L ′ = K ( − 3 γ ) , then
ad 0 r ˉ = ω ⊕ η ⊕ η ′ , η ′ = ω η . (18) \operatorname{ad}^0 \bar r = \omega\oplus\eta\oplus\eta', \qquad\eta' = \omega\eta. \tag*{(18)} ad 0 r ˉ = ω ⊕ η ⊕ η ′ , η ′ = ω η . (18) The summand normal to induction from H H H η ⊕ η ′ \eta\oplus\eta' η ⊕ η ′
Proof. Clifford theory gives the displayed induction. Restriction of the determinant to G H G_H G H ξ c = ξ − 1 \xi^c=\xi^{-1} ξ c = ξ − 1 ξ 2 \xi^2 ξ 2 F ‾ 3 × \overline{\mathbb{F}}_3^\times F 3 × ξ 2 \xi^2 ξ 2 H H H P 3 \mathfrak{P}_3 P 3
Here is an elementary calculation of this ray group. The field H = Q ( − 3 , − 7 ) H=\mathbb{Q}(\sqrt{-3},\sqrt{-7}) H = Q ( − 3 , − 7 ) 441 441 441 63 / ( 2 π 2 ) < 4 63/(2\pi^2)<4 63/ ( 2 π 2 ) < 4 h H = 1 h_H=1 h H = 1
u = − 3 + − 7 2 , u 4 + 5 u 2 + 1 = 0 , u=\frac{\sqrt{-3}+\sqrt{-7}}{2}, \qquad u^4+5u^2+1=0, u = 2 − 3 + − 7 , u 4 + 5 u 2 + 1 = 0 , is a unit. Its image at P 3 \mathfrak{P}_3 P 3 F 9 × \mathbb{F}_9^\times F 9 × − 1 -1 − 1 Cl H ( P 3 ) \operatorname{Cl}_H(\mathfrak{P}_3) Cl H ( P 3 )
The quadratic extension H ( γ ) / H H(\sqrt{\gamma})/H H ( γ ) / H P 3 \mathfrak{P}_3 P 3 γ \gamma γ γ 3 ≡ 1 ( m o d 4 ) \gamma^3\equiv1\pmod4 γ 3 ≡ 1 ( mod 4 )
Cl H ( P 3 ) ≃ C 2 . \operatorname{Cl}_H(\mathfrak{P}_3)\simeq C_2. Cl H ( P 3 ) ≃ C 2 . Thus ξ \xi ξ ξ c = ξ \xi^c=\xi ξ c = ξ
The relation γ γ σ = − 3 \gamma\gamma^\sigma=-3 γ γ σ = − 3 L ′ L' L ′ L L L F 3 \mathbb{F}_3 F 3
( 0 − 1 1 0 ) , ( 0 1 1 0 ) . \begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad
\begin{pmatrix}0&1\\1&0\end{pmatrix}. ( 0 1 − 1 0 ) , ( 0 1 1 0 ) . At each five-prime of H H H u u u F 25 × \mathbb{F}_{25}^\times F 25 × u 4 = − 1 u^4=-1 u 4 = − 1 − 1 -1 − 1 u u u − 1 -1 − 1 u u u
Finally the order-four inertia character at three has just two possibilities, inverse to one another. Once it and the two five-adic quadratic inertia characters are fixed, two possible global characters differ by an everywhere unramified character of H H H h H = 1 h_H=1 h H = 1
Remark 21.27. The scalar-inertia hypothesis in Lemma 21.26 is automatic for the cyclotomic-irreducible residue of either source considered below. An induction from a character with values in F ‾ 3 \overline{\mathbb{F}}_3 F 3
Lemma 21.28. Let ι \iota ι L / K L/K L / K
y = γ , z = a + y 2 = y 2 + y − 1 2 , g = 1 − y 2 − 5 y 2 = 2 a − 5 z . y=\sqrt{\gamma},\qquad z=\frac{a+y}{2}=\frac{y^{2}+y-1}{2},\qquad g=\frac{1-y^{2}-5y}{2}=2a-5z. y = γ , z = 2 a + y = 2 y 2 + y − 1 , g = 2 1 − y 2 − 5 y = 2 a − 5 z . Then h L = 1 h_L=1 h L = 1 disc ( L ) = − 1323 \operatorname{disc}(L)=-1323 disc ( L ) = − 1323 S 3 , 5 S_{3,5} S 3 , 5 L L L [ g ] , [ z ] [g],[z] [ g ] , [ z ] F 3 \mathbb{F}_3 F 3 q \mathfrak{q} q L q L_{\mathfrak{q}} L q
1 , [ g ] , [ g ] 2 . 1,\quad[g],\quad[g]^2. 1 , [ g ] , [ g ] 2 . Each of the two nontrivial classes has cubic conductor six at three after adjoining ζ 3 \zeta_3 ζ 3
Proof. The identities
N L / K ( z ) = 1 , z + z − 1 = a , N L / K ( g ) = − a 3 , g + g ι = − a (19) N_{L/K}(z)=1,\qquad z+z^{-1}=a,\qquad N_{L/K}(g)=-a^3,\qquad g+g^{\iota}=-a \tag*{(19)} N L / K ( z ) = 1 , z + z − 1 = a , N L / K ( g ) = − a 3 , g + g ι = − a (19) follow directly from y 2 = γ y^2=\gamma y 2 = γ a 2 = a + 5 a^2=a+5 a 2 = a + 5 z z z
X 4 − X 3 − 3 X 2 − X + 1. X^4-X^3-3X^2-X+1. X 4 − X 3 − 3 X 2 − X + 1. The relative discriminant of O K [ z ] \mathcal{O}_K[z] O K [ z ] ( a 2 − 4 ) = ( γ ) (a^2-4)=(\gamma) ( a 2 − 4 ) = ( γ ) a = z + z − 1 a=z+z^{-1} a = z + z − 1 O L = Z [ z ] \mathcal{O}_L=\mathbb{Z}[z] O L = Z [ z ] disc ( L ) = 2 1 2 N K / Q ( γ ) = − 1323 \operatorname{disc}(L)=2^{1^2}N_{K/\mathbb{Q}}(\gamma)=-1323 disc ( L ) = 2 1 2 N K / Q ( γ ) = − 1323 z z z Φ 5 ( X ) \Phi_5(X) Φ 5 ( X ) F 2 \mathbb{F}_2 F 2 L L L ( 2 , 1 ) (2,1) ( 2 , 1 ) y y y − 3 -3 − 3 h L = 1 h_L=1 h L = 1
The unit ranks of L L L K K K ι \iota ι K K K L L L γ = a + 1 \gamma=a+1 γ = a + 1 S 3 , 5 S_{3,5} S 3 , 5 S 3 , 5 S_{3,5} S 3 , 5
Both g g g z z z g g g g = 2 a − 5 z g=2a-5z g = 2 a − 5 z − a -a − a
At q \mathfrak{q} q a = 1 a=1 a = 1 y 2 = 2 y^2=2 y 2 = 2 F 25 \mathbb{F}_{25} F 25
g = 2 , z = ( 1 + y ) / 2. g=2,\qquad z=(1+y)/2. g = 2 , z = ( 1 + y ) /2. The element 2 ∈ F 5 × 2\in\mathbb{F}_5^\times 2 ∈ F 5 × F 25 × \mathbb{F}_{25}^\times F 25 × z z z − 1 -1 − 1 ± 1 \pm1 ± 1 [ z ] [z] [ z ] [ g ] [g] [ g ] q \mathfrak{q} q
At three, y y y L 3 L_3 L 3 M = L 3 ( ζ 3 ) / L 3 M=L_3(\zeta_3)/L_3 M = L 3 ( ζ 3 ) / L 3
g ≡ − 1 − y ( m o d y 2 ) , g\equiv-1-y\pmod{y^2}, g ≡ − 1 − y ( mod y 2 ) , the polynomial ( X + 1 ) 3 + g (X+1)^3+g ( X + 1 ) 3 + g M M M v E ( 3 ) = 12 v_E(3)=12 v E ( 3 ) = 12 X + 1 X+1 X + 1 ζ 3 \zeta_3 ζ 3 M M M 12 / ( 3 − 1 ) = 6 12/(3-1)=6 12/ ( 3 − 1 ) = 6 g 2 g^2 g 2 ( X + 1 ) 3 − g 2 (X+1)^3-g^2 ( X + 1 ) 3 − g 2
Corollary 21.29. A class in H 1 ( K , F 3 ( η ′ ) ) H^1(K,\mathbf{F}_3(\eta')) H 1 ( K , F 3 ( η ′ )) η ′ ( Frob 5 ) = 1 \eta'(\operatorname{Frob}_5)=1 η ′ ( Frob 5 ) = 1 η \eta η
Proof. Since η ′ = ω η \eta'=\omega\eta η ′ = ω η
H 1 ( K , F 3 ( η ′ ) ) ≃ ( L × / L × 3 ) − . H^1(K,\mathbf{F}_3(\eta'))\simeq(L^\times/L^{\times3})^{-}. H 1 ( K , F 3 ( η ′ )) ≃ ( L × / L × 3 ) − . The support condition and h L = 1 h_L=1 h L = 1 S 3 , 5 S_{3,5} S 3 , 5 η ( Frob 5 ) = − 1 \eta(\operatorname{Frob}_5)=-1 η ( Frob 5 ) = − 1 L / K L/K L / K q \mathfrak{q} q 1 1 1 g g g g 2 g^2 g 2 L L L L ′ L' L ′
The normal tangent space
Lemma 21.30. Let a class in H 1 ( K , η ⊕ η ′ ) H^1(K,\eta\oplus\eta') H 1 ( K , η ⊕ η ′ )
Proof. We verify the conditions of Corollary 21.29 for each normal summand.
At a five-place where a normal character ν \nu ν + 1 +1 + 1 H 1 ( K v , F 3 ( ν ) ) H^1(K_v,\mathbf{F}_3(\nu)) H 1 ( K v , F 3 ( ν )) ν ( Frob 5 ) = 5 = − 1 \nu(\operatorname{Frob}_5)=5=-1 ν ( Frob 5 ) = 5 = − 1 F 3 \mathbf{F}_3 F 3 Φ ˉ \bar{\Phi} Φ ˉ Φ ˉ 2 = I \bar{\Phi}^2=I Φ ˉ 2 = I b Φ ˉ b\bar{\Phi} b Φ ˉ
b tr ( Φ ˉ 2 ) = 2 b . b\operatorname{tr}(\bar{\Phi}^2)=2b. b tr ( Φ ˉ 2 ) = 2 b . The trace condition therefore kills this unramified class. At the other five-place, where ν ( Frob 5 ) = − 1 \nu(\operatorname{Frob}_5)=-1 ν ( Frob 5 ) = − 1
At three, work first over H 3 H_3 H 3 K 3 K_3 K 3 F 3 \mathbf{F}_3 F 3 F 9 \mathbf{F}_9 F 9 F 9 \mathbf{F}_9 F 9 H 3 H_3 H 3 F 9 \mathbf{F}_9 F 9
Fontaine’s ramification bound, applied over this original field of absolute ramification index two, bounds the classical upper breaks of each extracted point field by
2 ( 1 + 1 3 − 1 ) − 1 = 2. (20) 2\left(1+\frac{1}{3-1}\right)-1=2. \tag*{(20)} 2 ( 1 + 3 − 1 1 ) − 1 = 2. (20) Here the shift by one comes from the convention G ( u ) = G u − 1 G^{(u)}=G^{u-1} G ( u ) = G u − 1 [31 ] , Theorem A. Pass to the projective kernel field, which is a quotient of the point field and retains the bound. The diagonal ratio has order two, although the original diagonal characters have order four. Killing the ratio by its tame quadratic extension multiplies positive upper breaks by two. The resulting cubic characters have upper break at most four, hence conductor at most five.
If extracting a quadratic-adjoint summand requires combining the two extensions, take the compositum of their point fields first. The direct sum of their finite-flat groups satisfies the same bound, and every projected character field is a quotient. The relevant field over L 3 L_3 L 3 ζ 3 \zeta_3 ζ 3 F 3 \mathbf{F}_3 F 3
Remark 21.31. The projective quotient in this proof is essential. We apply Fontaine’s bound to actual finite-flat extensions over H 3 H_3 H 3
Two cuspidal places
Theorem 21.32. Let r : G K ⟶ GL 2 ( Q ‾ 3 ) r:G_K \longrightarrow\operatorname{GL}_2(\overline{\mathbb{Q}}_3) r : G K ⟶ GL 2 ( Q 3 ) χ 3 \chi_3 χ 3 { 0 , 1 } \{0,1\} { 0 , 1 } τ 3 \tau_3 τ 3 τ 6 \tau_6 τ 6 H H H r r r G H G_H G H r r r
Proof. By Lemma 21.26, the residue is r ˉ = Ind H K ξ \bar{r}=\operatorname{Ind}_{H}^{K}\xi r ˉ = Ind H K ξ F 9 \mathbf{F}_9 F 9 J = diag ( 1 , − 1 ) J=\operatorname{diag}(1,-1) J = diag ( 1 , − 1 ) J r ˉ ( g ) J − 1 = ω ( g ) r ˉ ( g ) J\bar{r}(g)J^{-1}=\omega(g)\bar{r}(g) J r ˉ ( g ) J − 1 = ω ( g ) r ˉ ( g ) J 2 = 1 J^2=1 J 2 = 1
Let D D D r ˉ \bar{r} r ˉ [72 Corollary 3.3.5 and Section 5.2] . The actual source gives a characteristic-zero point of D D D
The character ω \omega ω K K K
r A ⟼ J − 1 ( r A ⊗ ω ) J r_A \longmapsto J^{-1}(r_A\otimes\omega)J r A ⟼ J − 1 ( r A ⊗ ω ) J preserves these conditions, the determinant and the identified residue. Unramified twisting preserves finite-flatness by descent from the unramified extension killing the twist. At five, twisting preserves the prescribed inertia and the zero Frobenius trace. This operation defines an involution of D D D
On ad 0 r = ω ⊕ η ⊕ η ′ \operatorname{ad}^{0}r=\omega\oplus\eta\oplus\eta' ad 0 r = ω ⊕ η ⊕ η ′ + 1 +1 + 1 − 1 -1 − 1 m = 2 m=2 m = 2 D D D
It follows that the actual irreducible source satisfies r ≃ r ⊗ ω r\simeq r\otimes\omega r ≃ r ⊗ ω G H G_H G H r = Ind H K ψ r=\operatorname{Ind}_{H}^{K}\psi r = Ind H K ψ ψ \psi ψ [57 Chapter III] . Since H / K H/K H / K [4 Chapter III] .
A finite-flat selfextension calculation
Lemma 21.33. Let F = H 3 F = H_3 F = H 3 k k k F 9 F_9 F 9 θ : G F ⟶ k × \theta: G_F \longrightarrow k^\times θ : G F ⟶ k × Lemma 21.26 . A finite-flat selfextension of θ \theta θ k k k θ \theta θ
Proof. Take the schematic closure of the generic subline and the finite-flat quotient. The exact finite-flat/Kisin-module equivalence and its coefficient versions apply to this sequence [44 Theorem 1.1.3, Proposition 1.1.11 and Lemmas 1.2.4–1.2.5] . The two coefficient idempotents give rank-one endpoints with Frobenius exponents ( r 0 , r 1 ) (r_0,r_1) ( r 0 , r 1 ) e ( F ) = 2 e(F) = 2 e ( F ) = 2 0 ≤ r i ≤ 2 0 \le r_i \le2 0 ≤ r i ≤ 2 r 0 + 3 r 1 r_0 + 3r_1 r 0 + 3 r 1
( 2 , 0 ) , ( 0 , 2 ) . (2,0), \qquad(0,2). ( 2 , 0 ) , ( 0 , 2 ) . The rank-one models of the two endpoints are consequently isomorphic, including their unramified constants.
Lemma 11.6 applies with p = 3 p = 3 p = 3 e = 2 e = 2 e = 2 f = 2 f = 2 f = 2 θ \theta θ k [ [ u ] ] k[[u]] k [[ u ]] k [ t ] / ( t 2 ) [ [ u ] ] k[t]/(t^2)[[u]] k [ t ] / ( t 2 ) [[ u ]]
Two special places
Theorem 21.34. Let R R R K K K { 0 , 1 } \{0,1\} { 0 , 1 } v i ∣ 5 v_i \mid5 v i ∣ 5 α i : G Q 5 ⟶ { ± 1 } \alpha_i : G_{\mathbb{Q}_5} \longrightarrow\{\pm1\} α i : G Q 5 ⟶ { ± 1 } H H H R R R
Proof. Again Lemma 21.26 applies. Fix its model over F 3 F_3 F 3 χ 3 \chi_3 χ 3 v i v_i v i Φ i \Phi_i Φ i b i = α i ( Φ i ) ∈ { ± 1 } b_i = \alpha_i(\Phi_i) \in\{\pm1\} b i = α i ( Φ i ) ∈ { ± 1 }
tr r ( Φ i ) = 6 b i , r ( g ) ∣ ℓ i = χ 3 ( g ) α i ( g ) , (21) \operatorname{tr} r(\Phi_i) = 6b_i, \qquad r(g)|_{\ell_i} = \chi_3(g)\alpha_i(g), \tag*{(21)} tr r ( Φ i ) = 6 b i , r ( g ) ∣ ℓ i = χ 3 ( g ) α i ( g ) , (21) where ℓ i \ell_i ℓ i 5 b i 5b_i 5 b i r ( Φ i ) r(\Phi_i) r ( Φ i ) 5 b i 5b_i 5 b i b i b_i b i 4 b i 4b_i 4 b i α i \alpha_i α i
For a precise construction of this quotient, impose the stable condition on finite-cardinality Z 3 [ G K , { 3 , 5 } ] \mathbb{Z}_3[G_{K,\{3,5\}}] Z 3 [ G K , { 3 , 5 } ]
( g − χ 3 ( g ) α i ( g ) ) ( h − α i ( h ) ) = 0 ( g , h ∈ G K v i ) (g-\chi_3(g)\alpha_i(g))(h-\alpha_i(h)) = 0 \qquad(g,h \in G_{K_{v_i}}) ( g − χ 3 ( g ) α i ( g )) ( h − α i ( h )) = 0 ( g , h ∈ G K v i ) at each five-place. Finite-flatness is stable under subquotients and finite direct sums, as are these identities. Wake–Wang-Erickson’s stable-condition theorem gives the universal pseudodeformation quotient [72 Theorem 2.5.5 and Section 5.2] ; impose the fixed determinant χ 3 \chi_3 χ 3 M 2 ( D ) M_2(D) M 2 ( D ) D D D e = p − 1 e=p-1 e = p − 1 D D D
We show that its relative tangent over Z 3 \mathbb{Z}_3 Z 3 H 1 ( K , η ⊕ η ′ ) H^1(K,\eta\oplus\eta') H 1 ( K , η ⊕ η ′ ) (21) has zero first trace variation in characteristic three. The tangent construction of Lemma 21.30, including its finite-flat extraction, kills both normal summands. After a strict conjugation, the tangent lies in the remaining diagonal summand ω \omega ω
At a five-place, residual inertia is scalar and residual Frobenius is a reflection with distinct eigenlines. In an inducing basis, a traceless diagonal matrix takes either Frobenius eigenline to the other. The prescribed character on ℓ i \ell_i ℓ i
At three, compute the tangent over F 3 \mathbb{F}_3 F 3 F 9 \mathbb{F}_9 F 9 H 3 H_3 H 3 F 9 \mathbb{F}_9 F 9 H 3 H_3 H 3 ω \omega ω K 3 K_3 K 3
H u n r 1 ( K 3 , F 3 ( ω ) ) = F 3 / ( ω ( Frob 3 ) − 1 ) F 3 = 0. H^1_{\mathrm{unr}}(K_3,\mathbb{F}_3(\omega))=\mathbb{F}_3/(\omega(\operatorname{Frob}_3)-1)\mathbb{F}_3=0. H unr 1 ( K 3 , F 3 ( ω )) = F 3 / ( ω ( Frob 3 ) − 1 ) F 3 = 0. The extension H 3 / K 3 H_3/K_3 H 3 / K 3
Globally Kummer theory identifies H 1 ( K , F 3 ( ω ) ) = K × / K × 3 H^1(K,\mathbb{F}_3(\omega))=K^\times/K^{\times3} H 1 ( K , F 3 ( ω )) = K × / K × 3 [ ϵ ] [\epsilon] [ ϵ ] [ γ ] [\gamma] [ γ ] γ \gamma γ ϵ = 1 + γ \epsilon=1+\gamma ϵ = 1 + γ 1 1 1 γ 3 \gamma^3 γ 3 v γ ( 3 ) = 2 v_\gamma(3)=2 v γ ( 3 ) = 2 ϵ \epsilon ϵ ϵ 2 \epsilon^2 ϵ 2
All deformation data are defined over Z 3 \mathbb{Z}_3 Z 3 D D D Z 3 \mathbb{Z}_3 Z 3
D = Z 3 . (22) D=\mathbb{Z}_3. \tag*{(22)} D = Z 3 . (22) Thus the source is the unique lift with its ordered pair of exact special characters.
It remains to compare this unique lift with its quadratic conjugate; this also avoids any need to control all other three-adic coefficient places. Let η 0 \eta_0 η 0 H / K H/K H / K ψ 5 \psi_5 ψ 5 K ( 5 ) / K K(\sqrt{5})/K K ( 5 ) / K η 0 \eta_0 η 0 ψ 5 \psi_5 ψ 5 Q 5 ( 5 ) \mathbb{Q}_5(\sqrt{5}) Q 5 ( 5 ) β = α 1 α 2 \beta=\alpha_1\alpha_2 β = α 1 α 2
δ ∈ { 1 , η 0 , ψ 5 , η 0 ψ 5 } \delta\in\{1,\eta_0,\psi_5,\eta_0\psi_5\} δ ∈ { 1 , η 0 , ψ 5 , η 0 ψ 5 } whose restriction at both places is β \beta β β \beta β
δ = ψ 5 or δ = η 0 ψ 5 . \delta=\psi_5\quad\text{or}\quad\delta=\eta_0\psi_5. δ = ψ 5 or δ = η 0 ψ 5 . Both characters are unramified at three. The extension H / K H/K H / K F 3 \mathbb{F}_3 F 3
The representation r 3 σ ⊗ δ r_3^\sigma\otimes\delta r 3 σ ⊗ δ ( α 1 , α 2 ) (\alpha_1,\alpha_2) ( α 1 , α 2 ) D D D (22) gives
r 3 σ ≃ r 3 ⊗ δ . (23) r_3^\sigma\simeq r_3\otimes\delta. \tag*{(23)} r 3 σ ≃ r 3 ⊗ δ . (23) Both possible characters δ \delta δ G Q G_{\mathbb{Q}} G Q ψ 5 \psi_5 ψ 5 Q ( 5 ) \mathbb{Q}(\sqrt{5}) Q ( 5 ) η 0 ψ 5 \eta_0\psi_5 η 0 ψ 5 Q ( − 15 ) \mathbb{Q}(\sqrt{-15}) Q ( − 15 ) (23) to the whole system. Lemma 12.3, with η = δ \eta=\delta η = δ
Remark 21.35. The class, unit and conductor calculations above are elementary and use no GRH. The corresponding PARI/GP certificates have, in the basis [ g ] [g] [ g ] [ z ] [z] [ z ] { 0 } \{0\} { 0 }
The exact calculation is supplied in [49 d21] .
The nonordinary reducible characteristic-three source
This section proves a source modularity theorem for both pure local menus. Its input is an actual characteristic-zero compatible system. The proof uses finite-flat extensions at the ramified prime above 3, a universal pseudodeformation ring with explicit local conditions, and descent under quadratic conjugation. In particular, it does not replace source modularity by modularity of a semisimplified residue.
Let σ \sigma σ Gal ( K / Q ) \operatorname{Gal}(K/\mathbb{Q}) Gal ( K / Q ) ω = χ ‾ 3 ∣ G K \omega=\overline{\chi}_3|_{G_K} ω = χ 3 ∣ G K ω \omega ω K ( − 3 ) / K K(\sqrt{-3})/K K ( − 3 ) / K ψ γ \psi_\gamma ψ γ K ( γ ) / K K(\sqrt{\gamma})/K K ( γ ) / K
ψ γ σ = ψ γ ω . (24) \psi_\gamma^\sigma=\psi_\gamma\omega. \tag*{(24)} ψ γ σ = ψ γ ω . (24) Theorem 21.36 (Nonordinary reducible source modularity). *Let R R R K K K
Suppose the semisimplification of its characteristic-three residue is globally reducible, and both constituent characters are ramified quadratic at the prime over 3. Then R R R
More precisely, after a quadratic twist unramified outside 5, the system has Frobenius traces invariant under quadratic conjugation. Its 5-adic companion descends to an odd geometric representation of G Q G_{\mathbb{Q}} G Q Q \mathbb{Q} Q
Quadratic normalization
In the reducible characteristic-three case, each constituent has quadratic inertia at a prime over 5. The residues of the fundamental unit ϵ \epsilon ϵ ψ 5 \psi_5 ψ 5 K ( 5 ) / K K(\sqrt{5})/K K ( 5 ) / K
Lemma 21.24 now gives, up to interchanging labels,
χ 1 = ψ γ , χ 2 = ψ γ ω (25) \chi_1=\psi_\gamma,\qquad\chi_2=\psi_\gamma\omega \tag*{(25)} χ 1 = ψ γ , χ 2 = ψ γ ω (25) with distinct characters defined over F 3 \mathbb{F}_3 F 3 (25) .
The finite-flat extension line at three
Let F = K p 3 = Q 3 ( 21 ) F=K_{p^3}=\mathbb{Q}_3(\sqrt{21}) F = K p 3 = Q 3 ( 21 ) F 3 \mathbb{F}_3 F 3 ω ∣ G F \omega|_{G_F} ω ∣ G F F F F − 3 \sqrt{-3} − 3 − 7 -7 − 7 (25) have the same ramified quadratic inertia and opposite unramified factors.
Lemma 21.37 (The local finite-flat line). For either orientation of the pair (22), the finite-flat subspace of H 1 ( F , F 3 ( ω ) ) H^1(F,\mathbb{F}_3(\omega)) H 1 ( F , F 3 ( ω )) [ a ϵ ] [a\epsilon] [ a ϵ ] k / F 3 k/\mathbb{F}_3 k / F 3 k k k
Proof. First work over F 3 \mathbb{F}_3 F 3 r = 1 r=1 r = 1 0 ≤ r ≤ 2 0\leq r\leq2 0 ≤ r ≤ 2 − 1 -1 − 1
Any finite-flat extension gives an exact extension of these rank-one Kisin modules. Indeed, take the schematic closure of the generic rank-one subrepresentation and then the finite-flat quotient. The exact correspondence on objects killed by p p p [44 Theorem 1.1.3 and Proposition 1.1.11] ; the coefficient versions are [44 Lemmas 1.2.4–1.2.5] . In suitable bases the Frobenius matrix is, up to a common unramified factor,
Φ = ( u h ( u ) 0 − u ) , h ∈ F 3 [ [ u ] ] . (26) \Phi=\begin{pmatrix}u&h(u)\\0&-u\end{pmatrix},\qquad h\in\mathbb{F}_3[[u]]. \tag*{(26)} Φ = ( u 0 h ( u ) − u ) , h ∈ F 3 [[ u ]] . (26) Every such matrix has height at most one, since u 2 Φ − 1 u^2\Phi^{-1} u 2 Φ − 1 f ( u ) f(u) f ( u ) h h h
u ( f ( u ) + f ( u 3 ) ) . u\bigl(f(u)+f(u^3)\bigr). u ( f ( u ) + f ( u 3 ) ) . The operator f ↦ f + f ( u 3 ) f\mapsto f+f(u^3) f ↦ f + f ( u 3 ) F 3 [ [ u ] ] \mathbb{F}_3[[u]] F 3 [[ u ]] h h h G F ∞ G_{F_\infty} G F ∞
We identify a nonzero class using the curve with number field label 2.2.21.1–25.1–a4 in [45 ] . In generalized Weierstrass notation its model is
E = [ a , a , a + 1 , − a − 3 , − a − 2 ] . (27) E=[a,a,a+1,-a-3,-a-2]. \tag*{(27)} E = [ a , a , a + 1 , − a − 3 , − a − 2 ] . (27) Its conductor is (5) [45 ] , so it has good reduction at 3; counting its points gives a p 3 ( E ) = 0 a_{p_3}(E)=0 a p 3 ( E ) = 0
3 ( x − a + 1 3 ) ( x 3 + ( 2 a + 2 ) x 2 + ( 2 a + 3 ) x + 4 a + 7 3 ) . (28) 3\left(x-\frac{a+1}{3}\right)\left(x^3+(2a+2)x^2+(2a+3)x+\frac{4a+7}{3}\right). \tag*{(28)} 3 ( x − 3 a + 1 ) ( x 3 + ( 2 a + 2 ) x 2 + ( 2 a + 3 ) x + 3 4 a + 7 ) . (28) The point above the rational root has squared doubled generalized y y y ( 61 a + 109 ) / 9 (61a+109)/9 ( 61 a + 109 ) /9
61 a + 109 9 γ = ( 2 a + 3 3 ) 2 \frac{61a+109}{9\gamma}=\left(\frac{2a+3}{3}\right)^2 9 γ 61 a + 109 = ( 3 2 a + 3 ) 2 identifies its squareclass as γ \gamma γ χ 1 = ψ γ \chi_1=\psi_\gamma χ 1 = ψ γ χ 2 = ψ γ ω \chi_2=\psi_\gamma\omega χ 2 = ψ γ ω
Twisting E [ 3 ] E[3] E [ 3 ] χ 2 \chi_2 χ 2 ω \omega ω
z ⟼ ω ( g ) z + c ( g ) , [ c ] ∈ H 1 ( K , F 3 ( ω ) ) . z\longmapsto\omega(g)z+c(g),\qquad[c]\in H^1(K,\mathbb{F}_3(\omega)). z ⟼ ω ( g ) z + c ( g ) , [ c ] ∈ H 1 ( K , F 3 ( ω )) . The cubic factor of (28) parametrizes those three lines. Depressing it gives
P = − 2 a − 5 , Q = 34 a + 55 9 , Q 2 4 + P 3 27 = 50 a 9 + 1075 108 = ( 5 ( 4 a + 7 ) 18 ) 2 . P=-2a-5,\qquad Q=\frac{34a+55}{9},\qquad\frac{Q^2}{4}+\frac{P^3}{27}=\frac{50a}{9}+\frac{1075}{108}=\left(\frac{5(4a+7)}{18}\right)^2. P = − 2 a − 5 , Q = 9 34 a + 55 , 4 Q 2 + 27 P 3 = 9 50 a + 108 1075 = ( 18 5 ( 4 a + 7 ) ) 2 . One Cardano radicand is − 3 a − 5 = − a ϵ -3a-5=-a\epsilon − 3 a − 5 = − a ϵ K ( − a ϵ 3 , ζ 3 ) K(\sqrt[3]{-a\epsilon},\zeta_3) K ( 3 − a ϵ , ζ 3 ) ⟨ [ a ϵ ] ⟩ \langle[a\epsilon]\rangle ⟨[ a ϵ ]⟩ − 1 -1 − 1 [ a ϵ ] [a\epsilon] [ a ϵ ] 3 3 3 E [ 3 ] E[3] E [ 3 ] 3 3 3
Finally let k / F 3 k/\mathbb{F}_3 k / F 3 c c c k k k Z 3 \mathbb{Z}_3 Z 3 F 3 \mathbb{F}_3 F 3 F 3 \mathbb{F}_3 F 3 λ : k → F 3 \lambda:k\to\mathbb{F}_3 λ : k → F 3 ker ( λ ) χ 1 \ker(\lambda)\chi_1 ker ( λ ) χ 1 F 3 \mathbb{F}_3 F 3 λ ∘ c \lambda\circ c λ ∘ c c c c k k k
The global plane and local evaluations
Write classes in K × / K × 3 K^\times/K^{\times3} K × / K × 3
b = [ a ϵ ] , s = [ 5 ] , d = [ 5 a ϵ ] = s + b . b=[a\epsilon],\qquad s=[5],\qquad d=[5a\epsilon]=s+b. b = [ a ϵ ] , s = [ 5 ] , d = [ 5 a ϵ ] = s + b . Lemma 21.38 (The global finite-flat extension plane). The global classes unramified outside { 3 , 5 } \{3,5\} { 3 , 5 } 3 3 3
V = ⟨ b , s ⟩ = ⟨ s , d ⟩ . (29) V=\langle b,s\rangle=\langle s,d\rangle. \tag*{(29)} V = ⟨ b , s ⟩ = ⟨ s , d ⟩ . (29) Quadratic conjugation acts by σ ( s ) = s \sigma(s)=s σ ( s ) = s σ ( d ) = − d \sigma(d)=-d σ ( d ) = − d If q \mathfrak q q 5 5 5 a a a ( s , d ) (s,d) ( s , d )
ℓ q = ( 1 , − 1 ) , ℓ q ‾ = ( 1 , 1 ) . (30) \ell_{\mathfrak q}=(1,-1),\qquad\ell_{\overline{\mathfrak q}}=(1,1). \tag*{(30)} ℓ q = ( 1 , − 1 ) , ℓ q = ( 1 , 1 ) . (30) Moreover
χ 1 ( Frob q ) = 1 , χ 1 ( Frob q ‾ ) = − 1 , (31) \chi_1(\operatorname{Frob}_{\mathfrak q})=1,\qquad\chi_1(\operatorname{Frob}_{\overline{\mathfrak q}})=-1, \tag*{(31)} χ 1 ( Frob q ) = 1 , χ 1 ( Frob q ) = − 1 , (31) and the values of χ 2 \chi_2 χ 2
Proof. The class and unit calculation in Lemma 21.24 shows that the global cubic classes with the indicated support have basis
[ ϵ ] , [ γ ] , [ a ] , [ a ‾ ] , a ‾ = 1 − a , [\epsilon],\ [\gamma],\ [a],\ [\overline a],\qquad\overline a=1-a, [ ϵ ] , [ γ ] , [ a ] , [ a ] , a = 1 − a , and dimension four. We compute their local-cube kernel at 3 3 3 v F ( γ ) = 1 v_F(\gamma)=1 v F ( γ ) = 1 U i = 1 + p F i U_i=1+\mathfrak p_F^i U i = 1 + p F i
ϵ 3 − 5 a ϵ = 9 ϵ \epsilon^3-5a\epsilon=9\epsilon ϵ 3 − 5 a ϵ = 9 ϵ gives
5 a ϵ ϵ 3 = 1 − 9 ϵ 2 ∈ U 4 . \frac{5a\epsilon}{\epsilon^3}=1-\frac{9}{\epsilon^2}\in U_4. ϵ 3 5 a ϵ = 1 − ϵ 2 9 ∈ U 4 . The 3 3 3 U 2 → U 4 U_2\to U_4 U 2 → U 4 3 3 3 d d d
ϵ = 1 + γ ϵ ∈ U 1 ∖ U 2 , − a ϵ = 1 − 3 ϵ ∈ U 2 ∖ U 3 . (32) \epsilon=1+\gamma\epsilon\in U_1\setminus U_2,\qquad-a\epsilon=1-3\epsilon\in U_2\setminus U_3. \tag*{(32)} ϵ = 1 + γ ϵ ∈ U 1 ∖ U 2 , − a ϵ = 1 − 3 ϵ ∈ U 2 ∖ U 3 . (32) Every unit cube lies in { ± 1 } U 3 \{\pm1\}U_3 { ± 1 } U 3 [ ϵ ] [\epsilon] [ ϵ ] [ a ϵ ] [a\epsilon] [ a ϵ ] γ \gamma γ μ 3 ⊄ F \mu_3 \not\subset F μ 3 ⊂ F F × / F × 3 F^{\times}/F^{\times3} F × / F × 3 [ F : Q 3 ] + 1 = 3 [F:\mathbb{Q}_3]+1=3 [ F : Q 3 ] + 1 = 3 ⟨ d ⟩ \langle d\rangle ⟨ d ⟩
The norm identity N ( a ϵ ) = − 5 N(a\epsilon)=-5 N ( a ϵ ) = − 5 σ ( b ) = s − b \sigma(b)=s-b σ ( b ) = s − b σ ( d ) = − d \sigma(d)=-d σ ( d ) = − d σ ( b ) \sigma(b) σ ( b ) b b b d d d − 2 = 1 -2=1 − 2 = 1 F 3 \mathbb{F}_3 F 3 3 ∤ 5 − 1 3\nmid5-1 3 ∤ 5 − 1 s s s ( 1 , 1 ) (1,1) ( 1 , 1 ) d d d ( 2 , 1 ) (2,1) ( 2 , 1 ) γ \gamma γ 1 1 1 q q q 2 2 2 q ˉ \bar q q ˉ
A universal pseudoring with explicit local conditions
Fix a finite residue coefficient field k / F 3 k/\mathbb{F}_3 k / F 3 W = W ( k ) W=W(k) W = W ( k ) G = G K , { 3 , 5 } G=G_{K,\{3,5\}} G = G K , { 3 , 5 } C \mathcal{C} C Z 3 [ G ] \mathbb{Z}_3[G] Z 3 [ G ] [72 §5.2] ; an algebra annihilator condition is stable by definition.
The universal pseudodeformation ring with residual pseudorepresentation χ 1 ⊕ χ 2 \chi_1\oplus\chi_2 χ 1 ⊕ χ 2 C \mathcal{C} C [72 Theorem 2.5.5] . It is a complete noetherian local W W W χ 3 \chi_3 χ 3 R R R R R R
Because the residual pseudorepresentation is multiplicity-free, its universal Cayley–Hamilton algebra has generalized-matrix-algebra form
E = ( R B C R ) . \mathcal{E}=\begin{pmatrix}R & B\\ C & R\end{pmatrix}. E = ( R C B R ) . The reducibility ideal J J J B ⊗ R C → R B\otimes_R C\to R B ⊗ R C → R [72 Proposition 4.2.2] . Writing m \mathfrak{m} m R R R [72 Theorem 4.3.5] gives
( B / m B ) ∨ ⊆ Ext 1 ( χ 2 , χ 1 ) , ( C / m C ) ∨ ⊆ Ext 1 ( χ 1 , χ 2 ) . (33) (B/\mathfrak{m}B)^\vee\subseteq\operatorname{Ext}^{1}(\chi_2,\chi_1),\qquad(C/\mathfrak{m}C)^\vee\subseteq\operatorname{Ext}^{1}(\chi_1,\chi_2). \tag*{(33)} ( B / m B ) ∨ ⊆ Ext 1 ( χ 2 , χ 1 ) , ( C / m C ) ∨ ⊆ Ext 1 ( χ 1 , χ 2 ) . (33) Before the fixed-determinant quotient these are equalities; the inclusions suffice afterwards. The construction is functorial for quadratic conjugation whenever the local conditions are exchanged.
For a normalized SP type, let α v \alpha_v α v
( g − χ 3 ( g ) α v ( g ) ) ( h − α v ( h ) ) = 0 ( g , h ∈ G K v ) . (34) \bigl(g-\chi_3(g)\alpha_v(g)\bigr)\bigl(h-\alpha_v(h)\bigr)=0\qquad(g,h\in G_{K_v}). \tag*{(34)} ( g − χ 3 ( g ) α v ( g ) ) ( h − α v ( h ) ) = 0 ( g , h ∈ G K v ) . (34) These are relations in the completed group algebra. They hold on every extension of α v \alpha_v α v χ 3 α v \chi_3\alpha_v χ 3 α v χ 3 α v \chi_3\alpha_v χ 3 α v
For a normalized SC3 type, choose arithmetic Frobenius Φ \Phi Φ T T T Φ T Φ − 1 = T 5 \Phi T\Phi^{-1}=T^5 Φ T Φ − 1 = T 5
T 2 + T + 1 = 0 , Φ 2 + 5 = 0. (35) T^2+T+1=0,\qquad\Phi^2+5=0. \tag*{(35)} T 2 + T + 1 = 0 , Φ 2 + 5 = 0. (35) The actual normalized SC representation has inertial eigenvalues ζ 3 , ζ 3 − 1 \zeta_3,\zeta_3^{-1} ζ 3 , ζ 3 − 1 Φ 2 = − 5 \Phi^2=-5 Φ 2 = − 5
( T − 1 ) 2 = 0. (36) (T-1)^2=0. \tag*{(36)} ( T − 1 ) 2 = 0. (36) The diagonal pseudotangent
Lemma 21.39 (Vanishing of the reducible tangent). For either pure local menu, the ring R R R
m / ( 3 R + m 2 + J ) = 0. (37) \mathfrak{m}/(3R+\mathfrak{m}^{2}+J)=0. \tag*{(37)} m / ( 3 R + m 2 + J ) = 0. (37) Consequently multiplication induces a surjection
( B / m B ) ⊗ k ( C / m C ) ↠ m / ( 3 R + m 2 ) . (38) (B/\mathfrak{m}B)\otimes_k(C/\mathfrak{m}C)\twoheadrightarrow\mathfrak{m}/(3R+\mathfrak{m}^{2}). \tag*{(38)} ( B / m B ) ⊗ k ( C / m C ) ↠ m / ( 3 R + m 2 ) . (38) Proof. A tangent in the reducible quotient R / J R/J R / J
χ 1 ( 1 + ε f ) , χ 2 ( 1 − ε f ) , f ∈ H 1 ( G , k ) , \chi_1(1+\varepsilon f),\qquad\chi_2(1-\varepsilon f),\qquad f\in H^1(G,k), χ 1 ( 1 + ε f ) , χ 2 ( 1 − ε f ) , f ∈ H 1 ( G , k ) , where the opposite signs follow from the fixed determinant. The constituent character deformations are themselves finite-flat modules at 3, by the stability of C C C [72 Proposition 4.3.4] .
Fontaine’s bound for a finite-flat module killed by p n p^n p n G ( u ) G^{(u)} G ( u ) u > e ( n + 1 / ( p − 1 ) ) u>e(n+1/(p-1)) u > e ( n + 1/ ( p − 1 )) G ( u ) = G u − 1 G^{(u)}=G^{u-1} G ( u ) = G u − 1 [31 pp. 515–516, Theorem A] . For p = 3 p=3 p = 3 n = 1 n=1 n = 1 e = 2 e=2 e = 2 f f f p 3 3 \mathfrak{p}_3^3 p 3 3 3 ∤ 5 − 1 3\nmid5-1 3 ∤ 5 − 1
We need only the ray group of conductor p 3 3 \mathfrak{p}_3^3 p 3 3 ( O K / p 3 3 ) × (\mathcal{O}_K/\mathfrak{p}_3^3)^\times ( O K / p 3 3 ) × ϵ = 1 + γ \epsilon=1+\gamma ϵ = 1 + γ γ 3 \gamma^3 γ 3 ⟨ − 1 , ϵ ⟩ \langle-1,\epsilon\rangle ⟨ − 1 , ϵ ⟩ C 3 C_3 C 3 q = ( a ) \mathfrak{q}=(a) q = ( a ) − a = 1 − γ -a=1-\gamma − a = 1 − γ ϵ , ϵ 2 \epsilon,\epsilon^2 ϵ , ϵ 2 γ 3 \gamma^3 γ 3
In the SP case, (30) fixes the two Frobenius characters, hence forces f ( Φ ) = 0 f(\Phi)=0 f ( Φ ) = 0 Φ 2 = − 5 \Phi^2=-5 Φ 2 = − 5 ± 1 \pm1 ± 1 f = 0 f=0 f = 0 J J J B B B C C C m J ⊆ m 2 \mathfrak{m}J\subseteq\mathfrak{m}^2 m J ⊆ m 2
The SP/SP action of quadratic conjugation
Lemma 21.40 (The two SP signs agree). An irreducible actual source in the situation of Theorem 21.36 has equal normalized unramified Steinberg signs at the two primes over 5. For these signs, quadratic conjugation acts trivially on the relative cotangent space of R R R
Proof. Suppose first that both signs are − 1 -1 − 1 χ 3 α v \chi^3\alpha_v χ 3 α v − α v -\alpha_v − α v
ker ℓ q = ⟨ b ⟩ , ker ℓ q ′ = ⟨ σ ( b ) ⟩ , (39) \ker\ell_{\mathfrak{q}}=\langle b\rangle,\qquad\ker\ell_{\mathfrak{q}'}=\langle\sigma(b)\rangle, \tag*{(39)} ker ℓ q = ⟨ b ⟩ , ker ℓ q ′ = ⟨ σ ( b )⟩ , (39) respectively. If both signs are + 1 +1 + 1
ker ℓ q ∩ ker ℓ q ˉ = 0. \ker\ell_q \cap\ker\ell_{\bar{q}} = 0. ker ℓ q ∩ ker ℓ q ˉ = 0. Nakayama gives B = 0 B = 0 B = 0 C = 0 C = 0 C = 0 J = 0 J = 0 J = 0 R R R
Quadratic conjugation preserves the resulting local conditions, interchanges the two residual characters, and interchanges the two lines (35). Its action on their product is + 1 +1 + 1 b = d − s b = d - s b = d − s σ ( b ) = s − b = − ( d + s ) \sigma(b) = s - b = -(d + s) σ ( b ) = s − b = − ( d + s ) h ∈ G Q h \in G_{\mathbb{Q}} h ∈ G Q σ \sigma σ ζ 3 \zeta_3 ζ 3 K ∩ Q ( ζ 3 ) = Q K \cap\mathbb{Q}(\zeta_3) = \mathbb{Q} K ∩ Q ( ζ 3 ) = Q ω \omega ω χ 1 ( h 2 ) = χ 2 ( h 2 ) \chi_1(h^2) = \chi_2(h^2) χ 1 ( h 2 ) = χ 2 ( h 2 ) + 1 +1 + 1
The SC/SC action of quadratic conjugation
Lemma 21.41 (The SC anti-invariant product vanishes). For the normalized SC/SC menu, quadratic conjugation acts trivially on the relative cotangent space of R R R
Proof. Both extension spaces in (29) are subspaces of V ⊗ F 3 k V \otimes_{\mathbb{F}_3} k V ⊗ F 3 k V ∨ ⊗ V ∨ V^\vee\otimes V^\vee V ∨ ⊗ V ∨ s ↦ s s \mapsto s s ↦ s d ↦ − d d \mapsto-d d ↦ − d h h h ℓ v ⊗ ℓ v \ell_v \otimes\ell_v ℓ v ⊗ ℓ v
T − 1 = ( x b T c T y ) , x , y ∈ m . T - 1 =
\begin{pmatrix}
x & b_T \\
c_T & y
\end{pmatrix},
\qquad x,y \in\mathfrak{m}. T − 1 = ( x c T b T y ) , x , y ∈ m . Its upper-left squared entry is x 2 + b T c T x^2 + b_Tc_T x 2 + b T c T R R R ( T − 1 ) 2 = − 3 T (T - 1)^2 = -3T ( T − 1 ) 2 = − 3 T b T c T ∈ m 2 + 3 R b_Tc_T \in\mathfrak{m}^2 + 3R b T c T ∈ m 2 + 3 R b T c T b_Tc_T b T c T m / ( 3 R + m 2 ) \mathfrak{m}/(3R + \mathfrak{m}^2) m / ( 3 R + m 2 ) b T b_T b T c T c_T c T
The evaluations in (27) are s ∨ − d ∨ s^\vee- d^\vee s ∨ − d ∨ s ∨ + d ∨ s^\vee+ d^\vee s ∨ + d ∨ Lemma 12.2 makes the action on the relative cotangent trivial. The local calculation above concerns products of opposite offdiagonal extensions, not only traces of representation tangents.
From the cotangent to the entire pseudoring
Quadratic conjugation defines a W ( k ) W(k) W ( k ) R R R G K G_K G K Lemma 12.1 , show that this involution is the identity. Consequently every normalized actual source satisfies
tr r 3 ( g ) = tr r 3 ( h g h − 1 ) ( g ∈ G K ) . (40) \operatorname{tr} r_3(g) = \operatorname{tr} r_3(hgh^{-1}) \qquad(g \in G_K). \tag*{(40)} tr r 3 ( g ) = tr r 3 ( h g h − 1 ) ( g ∈ G K ) . (40) No assertion that R R R W ( k ) W(k) W ( k )
Whole-system descent and the rational theorem
Completion of the proof of Theorem 21.36. Strong compatibility transports (36), first at every good Frobenius, to all companions of the normalized system. Chebotarev identifies the semisimplifications of the system and its quadratic conjugate. In particular its absolutely irreducible 5-adic member is invariant under σ \sigma σ
Lemma 12.3 makes the normalized system modular. Undoing the finite quadratic normalization proves the claim.
Remark 21.42. The seed has a v = − 1 a_{\mathrm{v}}=-1 a v = − 1 a p 3 = 0 a_{\mathrm{p}3}=0 a p 3 = 0
The exact calculation is supplied in [49 d21] .
Completion of the bounded source theorem
Theorem 21.43. Assume GRH. Let R \mathcal{R} R K K K { 0 , 1 } \{0,1\} { 0 , 1 } R \mathcal{R} R
Proof. The mixed case is Proposition 21.1. A pure special source with reducible five-adic residue is covered directly by Proposition 21.2. For the other pure sources, consider the three-adic member, which is crystalline at three. If its residue remains irreducible on G H G_H G H H H H
Otherwise Lemma 21.24 gives the complete constituent list. The two unramified-at-three possibilities are modular by Proposition 21.25. In the remaining two, twist by ψ 5 \psi_5 ψ 5
Theorem 21.44 (Unrestricted residual base at five). Assume GRH. Every irreducible representation
r ˉ : G K ⟶ GL 2 ( F 5 ) , det r ˉ = χ ˉ 5 , \bar{r}: G_K \longrightarrow\operatorname{GL}_2(\mathbb{F}_5), \qquad\det\bar{r}=\bar{\chi}_5, r ˉ : G K ⟶ GL 2 ( F 5 ) , det r ˉ = χ ˉ 5 , unramified outside five is modular. No local finite-flatness, tameness, coefficient-field or conductor-exponent restriction is imposed.
Proof. Solvable projective image is handled by an odd Artin lift and solvable Artin automorphy, with weight raising to obtain a regular residual witness. In the nonsolvable case the restriction to G K ( ζ 5 ) G_{K(\zeta_5)} G K ( ζ 5 ) PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 ) [ K ( ζ 5 ) : K ] = 4 [K(\zeta_5):K]=4 [ K ( ζ 5 ) : K ] = 4 PGL 2 ( F 5 ) \operatorname{PGL}_2(\mathbb{F}_5) PGL 2 ( F 5 )
Lemma 21.3 provides actual weight-two local lifts at both five-places with determinant χ 5 \chi_5 χ 5 [28 ] gives a global lift. Snowden’s potential-modularity and compatible-system construction gives a potentially modular weight-two system with the selected menu. At a selected special component, the purity observation in Section 2 retains nonzero monodromy. The system is crystalline at two and three by all-place compatibility. Theorem 21.43 makes the actual system modular, and its five-adic reduction is the original r ˉ \bar{r} r ˉ
:::
The arbitrary-conductor transfer
We verify the terminal character condition and apply Theorem 1.2 with reset prime five. Its terminal source is allowed arbitrary regular weights.
Characters supported at five
Lemma 21.45. Every finite characteristic-five character of G K G_K G K G Q G_{\mathbb{Q}} G Q χ ˉ 5 \bar{\chi}_5 χ ˉ 5 χ ˉ 5 \bar{\chi}_5 χ ˉ 5 χ ˉ 5 − 1 \bar{\chi}_5^{-1} χ ˉ 5 − 1
Proof. Such a character kills principal units at five. Since h K = 1 h_K=1 h K = 1 m , n m,n m , n ϵ \epsilon ϵ 2 , 3 = 2 − 1 2,3=2^{-1} 2 , 3 = 2 − 1 m = n m=n m = n − 1 -1 − 1 F 5 × \mathbb{F}_5^{\times} F 5 ×
The equality of the tame exponents shows that ψ 2 \psi^2 ψ 2 χ ˉ 5 \bar{\chi}_5 χ ˉ 5 1 1 1 χ ˉ 5 2 \bar{\chi}_5^2 χ ˉ 5 2 h K = 1 h_K=1 h K = 1 ψ 2 ∈ { 1 , χ ˉ 5 2 } \psi^2\in\{1,\bar{\chi}_5^2\} ψ 2 ∈ { 1 , χ ˉ 5 2 } χ ˉ 5 ψ − 2 \bar{\chi}_5\psi^{-2} χ ˉ 5 ψ − 2 χ ˉ 5 ± 1 \bar{\chi}_5^{\pm1} χ ˉ 5 ± 1
Remark 21.46. The generic mixed-case theorem of [20 Theorem A] excludes the local ratio χ ˉ 5 ± 1 \bar{\chi}_5^{\pm1} χ ˉ 5 ± 1
Removal of arbitrary conductor over Q ( 21 ) \mathbb{Q}(\sqrt{21}) Q ( 21 ) A solvable residual image is modular by an odd solvable Artin lift and regular weight raising. Otherwise its cyclotomic restriction is absolutely irreducible. The initial weight-two lift and Lemma 2.4 give a potentially modular compatible system with cyclotomic determinant times a finite totally even character. In characteristic five the exception is excluded by [ K ( ζ 5 ) : K ] = 4 [K(\zeta_5):K]=4 [ K ( ζ 5 ) : K ] = 4 Theorem 21.44 . The nonsolvable residue makes the system nondihedral.
Apply Theorem 1.2 with ℓ = 5 \ell=5 ℓ = 5 Theorem 21.44 is the supported irreducible residual assertion, and Lemma 21.45 gives constituent quotient χ ˉ 5 ± 1 \bar{\chi}_5^{\pm1} χ ˉ 5 ± 1 G Q G_{\mathbb{Q}} G Q
Part III
Residual bases and applications
A common discriminant comparison
For b > 0 b>0 b > 0 a = π / ( 2 b ) a=\pi/(2b) a = π / ( 2 b )
F b ( x ) = { ( 1 − ∣ x ∣ 2 b ) cos ( a ∣ x ∣ ) + π − 1 sin ( a ∣ x ∣ ) , ∣ x ∣ ≤ 2 b , 0 , ∣ x ∣ > 2 b . F_b(x)=
\begin{cases}
\left(1-\frac{|x|}{2b}\right)\cos(a|x|)+\pi^{-1}\sin(a|x|), & |x|\le2b,\\
0, & |x|>2b.
\end{cases} F b ( x ) = { ( 1 − 2 b ∣ x ∣ ) cos ( a ∣ x ∣ ) + π − 1 sin ( a ∣ x ∣ ) , 0 , ∣ x ∣ ≤ 2 b , ∣ x ∣ > 2 b . This is the normalized autocorrelation of cos ( a t ) \cos(at) cos ( a t ) [ − b , b ] [-b,b] [ − b , b ] F b ( 0 ) = 1 F_b(0)=1 F b ( 0 ) = 1 0 ≤ F b ≤ 1 0\le F_b\le1 0 ≤ F b ≤ 1
J b ( s ) = ∫ 0 2 b F b ( x ) e − s x d x = s s 2 + a 2 + a 2 ( 1 + e − 2 b s ) b ( s 2 + a 2 ) 2 . (41) J_b(s)=\int_{0}^{2b}F_b(x)e^{-sx}\,dx=\frac{s}{s^2+a^2}+\frac{a^2(1+e^{-2bs})}{b(s^2+a^2)^2}. \tag*{(41)} J b ( s ) = ∫ 0 2 b F b ( x ) e − s x d x = s 2 + a 2 s + b ( s 2 + a 2 ) 2 a 2 ( 1 + e − 2 b s ) . (41) Put
B b = ∫ 0 ∞ 1 − F b ( x ) 2 sinh ( x / 2 ) d x , C b = 2 { J b ( − 1 / 2 ) + J b ( 1 / 2 ) } . B_b=\int_{0}^{\infty}\frac{1-F_b(x)}{2\sinh(x/2)}\,dx,\qquad C_b=2\{J_b(-1/2)+J_b(1/2)\}. B b = ∫ 0 ∞ 2 sinh ( x /2 ) 1 − F b ( x ) d x , C b = 2 { J b ( − 1/2 ) + J b ( 1/2 )} . Expansion of ( 2 sinh ( x / 2 ) ) − 1 (2\sinh(x/2))^{-1} ( 2 sinh ( x /2 ) ) − 1 B b = ∑ j ≥ 0 ( ( j + 1 / 2 ) − 1 − J b ( j + 1 / 2 ) ) B_b=\sum_{j\ge0}((j+1/2)^{-1}-J_b(j+1/2)) B b = ∑ j ≥ 0 (( j + 1/2 ) − 1 − J b ( j + 1/2 )) a 2 ( j + 1 / 2 ) − 3 a^2(j+1/2)^{-3} a 2 ( j + 1/2 ) − 3 N ≥ 1 N\ge1 N ≥ 1
B b ≤ B b , N : = ∑ j = 0 N − 1 ( ( j + 1 / 2 ) − 1 − J b ( j + 1 / 2 ) ) + a 2 2 ( N − 1 / 2 ) 2 . (42) B_b\le B_{b,N}:=\sum_{j=0}^{N-1}((j+1/2)^{-1}-J_b(j+1/2))+\frac{a^2}{2(N-1/2)^2}. \tag*{(42)} B b ≤ B b , N := j = 0 ∑ N − 1 (( j + 1/2 ) − 1 − J b ( j + 1/2 )) + 2 ( N − 1/2 ) 2 a 2 . (42) The midpoint inequality for x − 3 x^{-3} x − 3 B b ≤ 2 − J b ( 1 / 2 ) + a 2 / 2 B_b\le2-J_b(1/2)+a^2/2 B b ≤ 2 − J b ( 1/2 ) + a 2 /2
A b , 4 = J b ( 1 / 2 ) − J b ( 3 / 2 ) + J b ( 5 / 2 ) − J b ( 7 / 2 ) A_{b,4}=J_b(1/2)-J_b(3/2)+J_b(5/2)-J_b(7/2) A b , 4 = J b ( 1/2 ) − J b ( 3/2 ) + J b ( 5/2 ) − J b ( 7/2 ) is a lower bound for ∫ 0 ∞ F b ( x ) / ( 2 cosh ( x / 2 ) ) d x \int_0^\infty F_b(x)/(2\cosh(x/2))\,dx ∫ 0 ∞ F b ( x ) / ( 2 cosh ( x /2 )) d x
Proposition 22.1. Assume GRH for the Dedekind zeta function of L L L n = [ L : Q ] n=[L:\mathbb{Q}] n = [ L : Q ] r 1 r_1 r 1 N Q N_Q N Q Q Q Q B b ∗ B_b^* B b ∗ B b B_b B b
log rd ( L ) ≥ γ + log ( 8 π ) − B b ∗ − C b n + r 1 n A b , 4 + ∑ Q 2 N Q n log Q ∑ m ≥ 1 Q − m / 2 F b ( m log Q ) . (43) \log\operatorname{rd}(L)\ge\gamma+\log(8\pi)-B_b^*-\frac{C_b}{n}+\frac{r_1}{n}A_{b,4}+\sum_Q\frac{2N_Q}{n}\log Q\sum_{m\ge1}Q^{-m/2}F_b(m\log Q). \tag*{(43)} log rd ( L ) ≥ γ + log ( 8 π ) − B b ∗ − n C b + n r 1 A b , 4 + Q ∑ n 2 N Q log Q m ≥ 1 ∑ Q − m /2 F b ( m log Q ) . (43) Any subset of the prime terms may be retained. At fixed b b b n n n
Proof. Apply Poitou’s explicit formula [52 equation (7) and Proposition 4] with F b F_b F b B b B_b B b C b > 0 C_b>0 C b > 0 F b F_b F b
For comparisons without GRH, use G b ( x ) = F b ( x ) / cosh ( x / 2 ) G_b(x)=F_b(x)/\cosh(x/2) G b ( x ) = F b ( x ) / cosh ( x /2 ) B ( G b ) , C ( G b ) B(G_b),C(G_b) B ( G b ) , C ( G b ) F b F_b F b G b G_b G b [47 equation (2.3)] and the strip positivity calculation in § 9.10 . The bound 1 − F b ( x ) ≤ π 2 x 2 / ( 8 b 2 ) 1-F_b(x)\le\pi^2x^2/(8b^2) 1 − F b ( x ) ≤ π 2 x 2 / ( 8 b 2 )
B ( G b ) ≤ log 2 + 7 π 2 ζ ( 3 ) 16 b 2 , C ( G b ) = 32 b π 2 , ∫ 0 ∞ F b ( x ) 2 cosh 2 ( x / 2 ) d x ≥ 1 − π 4 24 b 2 . B(G_b)\le\log2+\frac{7\pi^2\zeta(3)}{16b^2},\qquad C(G_b)=\frac{32b}{\pi^2},\qquad\int_0^\infty\frac{F_b(x)}{2\cosh^2(x/2)}\,dx\ge1-\frac{\pi^4}{24b^2}. B ( G b ) ≤ log 2 + 16 b 2 7 π 2 ζ ( 3 ) , C ( G b ) = π 2 32 b , ∫ 0 ∞ 2 cosh 2 ( x /2 ) F b ( x ) d x ≥ 1 − 24 b 2 π 4 . Consequently, without any hypothesis on the zeros,
log rd ( L ) ≥ log ( 4 π ) + γ − 7 π 2 ζ ( 3 ) 16 b 2 − 32 b π 2 n + r 1 n ( 1 − π 4 24 b 2 ) + ∑ Q 2 N Q n log Q ∑ m ≥ 1 2 F b ( m log Q ) Q m + 1 . (44) \begin{aligned}
\log\operatorname{rd}(L)&\ge\log(4\pi)+\gamma-\frac{7\pi^2\zeta(3)}{16b^2}-\frac{32b}{\pi^2n}+\frac{r_1}{n}\left(1-\frac{\pi^4}{24b^2}\right)\\
&\quad+\sum_Q\frac{2N_Q}{n}\log Q\sum_{m\ge1}\frac{2F_b(m\log Q)}{Q^m+1}. \tag*{(44)}
\end{aligned} log rd ( L ) ≥ log ( 4 π ) + γ − 16 b 2 7 π 2 ζ ( 3 ) − π 2 n 32 b + n r 1 ( 1 − 24 b 2 π 4 ) + Q ∑ n 2 N Q log Q m ≥ 1 ∑ Q m + 1 2 F b ( m log Q ) . (44) As before, one may discard prime terms or the real-place term. The numerical uses below take ζ ( 3 ) < 1.203 \zeta(3)<1.203 ζ ( 3 ) < 1.203 F b F_b F b [49 common/verify_unconditional_refinements.py] .
The comparisons below specify the degree, signature, retained Euler densities and local discriminant upper bound. Their strict numerical inequalities include the tail in (42) ; no truncated infinite sum is used without a remainder bound.
Local constructions for a split terminal prime
We record the local characteristic-three calculations used in the three supported residual bases below. The universal existence statement is Theorem 4.7 ; here we also need the exact characteristic-two reduction of its type over Q 3 \mathbb{Q}_3 Q 3
Lemma 23.1. Let K K K G K → F ‾ 3 × G_K \to\overline{\mathbb{F}}_3^{\times} G K → F 3 ×
Proof. Such a character has order prime to three and kills the principal units at the two places above three. The ray-class sequence, with both real places included, makes its group a quotient of ( F 3 × ) 2 × { ± 1 } 2 (\mathbb{F}_3^{\times})^2 \times\{\pm1\}^2 ( F 3 × ) 2 × { ± 1 } 2
For a finite Galois extension L / Q 3 L/\mathbb{Q}_3 L / Q 3 d ( L ) = v 3 ( Δ ( L / Q 3 ) ) / [ L : Q 3 ] d(L)=v_3(\Delta(L/\mathbb{Q}_3))/[L:\mathbb{Q}_3] d ( L ) = v 3 ( Δ ( L / Q 3 )) / [ L : Q 3 ] v 3 ( 3 ) = 1 v_3(3)=1 v 3 ( 3 ) = 1
Lemma 23.2. Let r ˉ : G Q 3 → GL 2 ( F 3 ) \bar{r}:G_{\mathbb{Q}_3}\to\operatorname{GL}_2(\mathbb{F}_3) r ˉ : G Q 3 → GL 2 ( F 3 ) χ ˉ 3 \bar{\chi}_3 χ ˉ 3 L L L r ˉ \bar{r} r ˉ χ ˉ 3 α \bar{\chi}_3\alpha χ ˉ 3 α α \alpha α Q Q Q
(i) If α ≠ 1 \alpha\ne1 α = 1 d ( L ) ≤ 3 / 2 − 1 / Q < 3 / 2 d(L)\le3/2-1/Q<3/2 d ( L ) ≤ 3/2 − 1/ Q < 3/2
(ii) If α = 1 \alpha=1 α = 1 Q ∈ { 3 , 9 } Q\in\{3,9\} Q ∈ { 3 , 9 } d ( L ) ≤ 13 / 6 − 1 / Q d(L)\le13/6-1/Q d ( L ) ≤ 13/6 − 1/ Q 2 Q 2Q 2 Q
(iii) In the second case with Q = 9 Q=9 Q = 9
E 0 = Q 3 ( ζ 3 , 3 3 , 4 3 ) , [ E 0 : Q 3 ] = 18 , d ( E 0 ) = 37 / 18. (45) E_0=\mathbb{Q}_3(\zeta_3,\sqrt[3]{3},\sqrt[3]{4}),\qquad[E_0:\mathbb{Q}_3]=18,\qquad d(E_0)=37/18. \tag*{(45)} E 0 = Q 3 ( ζ 3 , 3 3 , 3 4 ) , [ E 0 : Q 3 ] = 18 , d ( E 0 ) = 37/18. (45) In the tame case d ( L ) < 1 d(L)<1 d ( L ) < 1 d ( L ) ≤ 37 / 18 d(L)\le37/18 d ( L ) ≤ 37/18
Proof. A nontrivial normal 3 3 3 Q 3 \mathbb{Q}_3 Q 3 1 , χ ˉ 3 1,\bar{\chi}_3 1 , χ ˉ 3 χ ˉ 3 α \bar{\chi}_3\alpha χ ˉ 3 α α \alpha α
The kernel P P P P P P Q Q Q
Put T = L P T=L^P T = L P μ 3 \mu_3 μ 3 T = U ( ζ 3 ) T=U(\zeta_3) T = U ( ζ 3 ) U / Q 3 U/\mathbb{Q}_3 U / Q 3 k k k ϖ = ζ 3 − 1 \varpi=\zeta_3-1 ϖ = ζ 3 − 1 T × / ( T × ) 3 → P T^\times/(T^\times)^3\to P T × / ( T × ) 3 → P U j = 1 + ϖ j O T U_j=1+\varpi^j\mathcal{O}_T U j = 1 + ϖ j O T U 2 3 = U 4 U_2^3=U_4 U 2 3 = U 4
( 1 + a ϖ ) 3 = 1 + ( a ˉ 3 − a ˉ ) ϖ 3 ( m o d ϖ 4 ) , (1+a\varpi)^3=1+(\bar{a}^3-\bar{a})\varpi^3\pmod{\varpi^4}, ( 1 + aϖ ) 3 = 1 + ( a ˉ 3 − a ˉ ) ϖ 3 ( mod ϖ 4 ) , since ϖ 2 = − 3 ( 1 + ϖ ) \varpi^2=-3(1+\varpi) ϖ 2 = − 3 ( 1 + ϖ ) 1 , 2 , 3 1,2,3 1 , 2 , 3 k / { a 3 − a : a ∈ k } k/\{a^3-a:a\in k\} k / { a 3 − a : a ∈ k } F 3 \mathbb{F}_3 F 3 j j j ( − 1 ) j (-1)^j ( − 1 ) j
and on the valuation quotient trivially. Only grades one and three can map to P P P j j j j + 1 j+1 j + 1 [59 , 46 ] .
A Frobenius fixing ζ 3 \zeta_{3} ζ 3 α ≠ 1 \alpha\ne1 α = 1 P P P F 3 \mathbb{F}_{3} F 3 P P P
d ( L ) = 1 2 + 1 2 Q ∑ ξ ≠ 1 a T ( ξ ) ≤ 3 2 − 1 Q . (46) d(L) = \frac{1}{2} + \frac{1}{2Q}\sum_{\xi\ne1} a_T(\xi) \le\frac{3}{2} - \frac{1}{Q}. \tag*{(46)} d ( L ) = 2 1 + 2 Q 1 ξ = 1 ∑ a T ( ξ ) ≤ 2 3 − Q 1 . (46) If α = 1 \alpha=1 α = 1
H 1 ( Q 3 , F ‾ 3 ( χ ‾ 3 ) ) = ( Q 3 × / ( Q 3 × ) 3 ) ⊗ F 3 F ‾ 3 . H^{1}(\mathbb{Q}_{3},\overline{\mathbb{F}}_{3}(\overline{\chi}_{3})) = \left(\mathbb{Q}_{3}^{\times}/(\mathbb{Q}_{3}^{\times})^{3}\right)\otimes_{\mathbb{F}_{3}}\overline{\mathbb{F}}_{3}. H 1 ( Q 3 , F 3 ( χ 3 )) = ( Q 3 × / ( Q 3 × ) 3 ) ⊗ F 3 F 3 . The rational Kummer space has basis the classes of 3 3 3 4 4 4 Q / 3 Q/3 Q /3
∑ ξ ≠ 1 a T ( ξ ) ≤ 2 ( Q / 3 − 1 ) + 4 ( 2 Q / 3 ) = 10 Q / 3 − 2. \sum_{\xi\ne1} a_T(\xi) \le2(Q/3-1)+4(2Q/3)=10Q/3-2. ξ = 1 ∑ a T ( ξ ) ≤ 2 ( Q /3 − 1 ) + 4 ( 2 Q /3 ) = 10 Q /3 − 2. Equation (46) gives d ( L ) ≤ 13 / 6 − 1 / Q ≤ 37 / 18 d(L)\le13/6-1/Q\le37/18 d ( L ) ≤ 13/6 − 1/ Q ≤ 37/18
In the projective representation, a Frobenius fixing ζ 3 \zeta_{3} ζ 3 2 Q 2Q 2 Q Q = 9 Q=9 Q = 9 E 0 E_{0} E 0 Q 3 ( ζ 3 ) \mathbb{Q}_{3}(\zeta_{3}) Q 3 ( ζ 3 )
d ( E 0 ) = 1 2 + 2 ⋅ 2 + 6 ⋅ 4 18 = 37 18 . d(E_{0})=\frac{1}{2}+\frac{2\cdot2+6\cdot4}{18}=\frac{37}{18}. d ( E 0 ) = 2 1 + 18 2 ⋅ 2 + 6 ⋅ 4 = 18 37 . The tame formula is 1 − 1 / e < 1 1-1/e<1 1 − 1/ e < 1
The bound Q ≤ 9 Q\le9 Q ≤ 9
d ( L ) ≤ 13 6 − 1 Q (47) d(L)\le\frac{13}{6}-\frac{1}{Q} \tag*{(47)} d ( L ) ≤ 6 13 − Q 1 (47) when the wild ramification index is Q > 1 Q>1 Q > 1 d ( L ) < 1 d(L)<1 d ( L ) < 1 [46 Theorem 3 and the following Remark (1)] .
A wild type over Q 3 \mathbb{Q}_{3} Q 3
Specialize the construction of Lemma 4.6 to F = Q 3 F=\mathbb{Q}_{3} F = Q 3 + 3 +3 + 3 ψ ( x ) = ζ 3 x \psi(x)=\zeta_{3}^{x} ψ ( x ) = ζ 3 x K 0 K_{0} K 0 K 1 K_{1} K 1 J J J ϑ \vartheta ϑ
Lemma 23.3. The representation σ + : = Ind K 0 J ϑ \sigma_{+}:=\operatorname{Ind}_{K_{0}}^{J}\vartheta σ + := Ind K 0 J ϑ τ + \tau_{+} τ +
σ ‾ + s s = 1 ⊕ det ⊕ Sym 2 F 3 2 ⊕ ( det ⊗ Sym 2 F 3 2 ) . (48) \overline{\sigma}_{+}^{\mathrm{ss}}=1\oplus\det\oplus\operatorname{Sym}^{2}\mathbb{F}_{3}^{2}\oplus(\det\otimes\operatorname{Sym}^{2}\mathbb{F}_{3}^{2}). \tag*{(48)} σ + ss = 1 ⊕ det ⊕ Sym 2 F 3 2 ⊕ ( det ⊗ Sym 2 F 3 2 ) . (48) The inertia representation τ + \tau_{+} τ + S 3 \mathrm{S}_{3} S 3 E + / Q 3 E_{+}/\mathbb{Q}_{3} E + / Q 3 Q 3 ( 3 ) \mathbb{Q}_{3}(\sqrt{3}) Q 3 ( 3 ) 7 / 6 7/6 7/6
Proof. The construction of Theorem 4.7 at ℓ = 3 \ell=3 ℓ = 3 σ + \sigma_{+} σ + τ + \tau_{+} τ +
In the induced integral lattice, ϑ \vartheta ϑ σ ˉ + = Ind Z U G L 2 ( F 3 ) 1 \bar{\sigma}_{+}=\operatorname{Ind}_{Z U}^{\mathrm{GL}_{2}(\mathbb{F}_{3})}1 σ ˉ + = Ind Z U GL 2 ( F 3 ) 1 B B B B / Z U B/ZU B / Z U 1 , det 1,\det 1 , det
σ ˉ + = Ind B G L 2 ( F 3 ) 1 ⊕ det ⊗ Ind B G L 2 ( F 3 ) 1. \bar{\sigma}_{+}=\operatorname{Ind}_{B}^{\mathrm{GL}_{2}(\mathbb{F}_{3})}1\oplus\det\otimes\operatorname{Ind}_{B}^{\mathrm{GL}_{2}(\mathbb{F}_{3})}1. σ ˉ + = Ind B GL 2 ( F 3 ) 1 ⊕ det ⊗ Ind B GL 2 ( F 3 ) 1. The first summand is the permutation module on P 1 ( F 3 ) \mathbb{P}^{1}(\mathbb{F}_{3}) P 1 ( F 3 ) Sym 2 F 3 2 \operatorname{Sym}^{2}\mathbb{F}_{3}^{2} Sym 2 F 3 2 (48) ; its four constituents are exactly the Serre weights with trivial scalar central character.
By [1 Section 3.4 and Theorem 3.18] , the parameter is induced from a character of a ramified quadratic extension of Q 3 \mathbb{Q}_{3} Q 3
u = ( ζ 3 0 0 ζ 3 − 1 ) , v = ( 0 1 − 1 0 ) . u=\begin{pmatrix}\zeta_{3}&0\\0&\zeta_{3}^{-1}\end{pmatrix},
\qquad
v=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. u = ( ζ 3 0 0 ζ 3 − 1 ) , v = ( 0 − 1 1 0 ) . The off-diagonal matrix has this form because inertia has determinant one. These matrices generate a group of order twelve. In characteristic ℓ > 3 \ell>3 ℓ > 3 u u u v v v ℓ \ell ℓ [1 Sections 3.1, 3.4 and Theorem 3.18] , the affine-generic character b + c b+c b + c + 3 +3 + 3 Q 3 ( 3 ) \mathbb{Q}_{3}(\sqrt{3}) Q 3 ( 3 ) S 3 \mathrm{S}_{3} S 3 S L 2 ( F 2 ) \mathrm{SL}_{2}(\mathbb{F}_{2}) SL 2 ( F 2 ) S 3 \mathrm{S}_{3} S 3 e = 6 e=6 e = 6 f = 1 f=1 f = 1
The wild character has conductor two over the ramified quadratic field. The two-dimensional Artin conductor is three and remains three after reduction modulo two: the wild inertia has odd order and the inertia invariants remain zero. The lower groups of the residual field have orders 6 , 3 , 1 6,3,1 6 , 3 , 1 ( 5 + 2 ) / 6 = 7 / 6 (5+2)/6=7/6 ( 5 + 2 ) /6 = 7/6 Q 3 ( 3 ) \mathbb{Q}_{3}(\sqrt{3}) Q 3 ( 3 ) F 3 \mathbb{F}_{3} F 3 S 3 \mathrm{S}_{3} S 3 x 3 − 3 x − 3 x^{3}-3x-3 x 3 − 3 x − 3
Proposition 23.4. Every continuous r ˉ : G Q 3 → G L 2 ( F ‾ 3 ) \bar{r}:G_{\mathbb{Q}_{3}}\to\mathrm{GL}_{2}(\overline{\mathbb{F}}_{3}) r ˉ : G Q 3 → GL 2 ( F 3 ) χ ‾ 3 \overline{\chi}_{3} χ 3 χ 3 \chi_{3} χ 3 { 0 , 1 } \{0,1\} { 0 , 1 } τ + \tau_{+} τ +
Proof. Apply Theorem 4.7 with F = Q 3 F=\mathbb{Q}_{3} F = Q 3
Lemma 23.5. Let r ˉ : G Q 3 → G L 2 ( F ‾ 3 ) \bar{r}:G_{\mathbb{Q}_{3}}\to\mathrm{GL}_{2}(\overline{\mathbb{F}}_{3}) r ˉ : G Q 3 → GL 2 ( F 3 ) χ ‾ 3 \overline{\chi}_{3} χ 3 { 0 , 1 } \{0,1\} { 0 , 1 } χ 3 \chi_{3} χ 3
τ 4 = ω ~ 2 2 ⊕ ω ~ 2 6 , \tau_{4}=\widetilde{\omega}_{2}^{2}\oplus\widetilde{\omega}_{2}^{6}, τ 4 = ω 2 2 ⊕ ω 2 6 , where ω 2 \omega_2 ω 2
E ′ = Q 3 ( 3 , − 1 ) , e ( E ′ / Q 3 ) = 2 , f ( E ′ / Q 3 ) = 2 , d ( E ′ / Q 3 ) = 1 2 . (49) E'=\mathbb{Q}_3(\sqrt{3},\sqrt{-1}),\qquad e(E'/\mathbb{Q}_3)=2,\qquad f(E'/\mathbb{Q}_3)=2,\qquad d(E'/\mathbb{Q}_3)=\frac{1}{2}. \tag*{(49)} E ′ = Q 3 ( 3 , − 1 ) , e ( E ′ / Q 3 ) = 2 , f ( E ′ / Q 3 ) = 2 , d ( E ′ / Q 3 ) = 2 1 . (49) Proof. We first obtain a quadratic twist of a semistable lift. If r ˉ \bar{r} r ˉ χ ˉ 3 λ \bar{\chi}_3\lambda χ ˉ 3 λ λ − 1 \lambda^{-1} λ − 1 λ \lambda λ η \eta η λ \lambda λ λ 2 ≠ 1 \lambda^2\ne1 λ 2 = 1 H 2 ( G Q 3 , O ( χ 3 η 2 ) ) = 0 H^2(G_{\mathbb{Q}_3},\mathcal{O}(\chi_3\eta^2))=0 H 2 ( G Q 3 , O ( χ 3 η 2 )) = 0 η − 1 \eta^{-1} η − 1 χ 3 η \chi_3\eta χ 3 η λ 2 = 1 \lambda^2=1 λ 2 = 1 η 2 = 1 \eta^2=1 η 2 = 1 Q 3 × / ( Q 3 × ) 3 \mathbb{Q}_3^\times/(\mathbb{Q}_3^\times)^3 Q 3 × / ( Q 3 × ) 3 r ˉ \bar{r} r ˉ { 1 , 3 } \{1,3\} { 1 , 3 } { 5 , 7 } \{5,7\} { 5 , 7 } ( 1 , 0 ) (1,0) ( 1 , 0 ) [64 proof of Proposition 7.8.1] .
Retain this lift if its monodromy is nonzero. Otherwise its inertial type is scalar quadratic, say κ ⊕ κ \kappa\oplus\kappa κ ⊕ κ κ ∘ det \kappa\circ\det κ ∘ det 1 1 1 det \det det [70 Theorem 1.2] , throughout with determinant χ 3 \chi_3 χ 3 μ 1 ( r ˉ ) > 0 \mu_1(\bar{r})>0 μ 1 ( r ˉ ) > 0 μ det ( r ˉ ) > 0 \mu_{\det}(\bar{r})>0 μ d e t ( r ˉ ) > 0
The cuspidal type for τ 4 \tau_4 τ 4 PGL 2 ( F 3 ) ≃ S 4 \operatorname{PGL}_2(\mathbb{F}_3)\simeq S_4 PGL 2 ( F 3 ) ≃ S 4 S 3 S_3 S 3 Z 3 3 \mathbb{Z}_3^3 Z 3 3 det \det det 1 ⊕ det 1\oplus\det 1 ⊕ det ( 1 , 0 ) (1,0) ( 1 , 0 ) τ 4 \tau_4 τ 4 μ 1 ( r ˉ ) + μ det ( r ˉ ) > 0 \mu_1(\bar{r})+\mu_{\det}(\bar{r})>0 μ 1 ( r ˉ ) + μ d e t ( r ˉ ) > 0 χ 3 \chi_3 χ 3
For the semistable choice, the characteristic-two local member is a quadratic twist of a special representation with diagonal characters χ 2 ν \chi_2\nu χ 2 ν ν \nu ν ν \nu ν T , Φ T,\Phi T , Φ
T 2 = − I , Φ T Φ − 1 = T − 1 , Φ 2 = − 3 I . T^2=-I,\qquad\Phi T\Phi^{-1}=T^{-1},\qquad\Phi^2=-3I. T 2 = − I , Φ T Φ − 1 = T − 1 , Φ 2 = − 3 I . The last equality follows from trace zero and determinant three. Modulo two these matrices become commuting involutions, again giving an elementary abelian two-group for every stable lattice. The field E ′ E' E ′ Q 3 \mathbb{Q}_3 Q 3 □ \square □
Over a real quadratic field with a single dyadic place, Lemma 2.4 applies to every finitely ramified, totally odd lift of determinant χ 2 \chi_2 χ 2
The characteristic-three base over Q ( 7 ) \mathbb{Q}(\sqrt{7}) Q ( 7 )
Throughout this section K = Q ( 7 ) K=\mathbb{Q}(\sqrt{7}) K = Q ( 7 ) C = K ( ζ 3 ) C=K(\zeta_3) C = K ( ζ 3 ) p 1 = ( 2 + 7 ) \mathfrak{p}_1=(2+\sqrt{7}) p 1 = ( 2 + 7 ) p 2 = ( 2 − 7 ) \mathfrak{p}_2=(2-\sqrt{7}) p 2 = ( 2 − 7 )
Theorem 24.1. Assume G R H GRH GR H ρ ˉ : G K → GL 2 ( F 3 ) \bar{\rho}:G_K\to\operatorname{GL}_2(\mathbb{F}_3) ρ ˉ : G K → GL 2 ( F 3 ) χ 3 \chi_3 χ 3 3 3 3
Lemma 24.2. The ordinary class number of K K K K ( i ) / K K(i)/K K ( i ) / K 3 3 3 G K → F 3 G_K \to\mathbb{F}_3 G K → F 3 3 3 3
Proof. The Minkowski bound is 7 < 3 \sqrt{7}<3 7 < 3 2 2 2 3 + 7 3+\sqrt{7} 3 + 7 x 2 − 7 y 2 = − 1 x^2-7y^2=-1 x 2 − 7 y 2 = − 1 x 2 − 7 y 2 = 1 x^2-7y^2=1 x 2 − 7 y 2 = 1 y > 0 y>0 y > 0 ( 8 , 3 ) (8,3) ( 8 , 3 ) ε = 8 + 3 7 \varepsilon=8+3\sqrt{7} ε = 8 + 3 7
The discriminant of the biquadratic field K ( i ) K(i) K ( i ) 28 ( − 4 ) ( − 7 ) = 28 2 28(-4)(-7)=28^2 28 ( − 4 ) ( − 7 ) = 2 8 2 K K K − 1 -1 − 1 F 3 \mathbb{F}_3 F 3
The character assertion follows from Lemma 23.1 .
Lemma 24.3. Assume GRH. Retain only a proved number N 3 N_3 N 3 c = 2 N 3 / [ L : Q ] c=2N_3/[L:\mathbb{Q}] c = 2 N 3 / [ L : Q ] c c c
In particular, every totally imaginary K K K E 0 E_0 E 0
[ L : Q ] < 20 000 000 000. (50) [L:\mathbb{Q}]<20\,000\,000\,000. \tag*{(50)} [ L : Q ] < 20 000 000 000. (50) Proof. Use the explicit formula of § 9.1 , discarding the real-place term and retaining only the specified primes of norm three. Its one-term bound for B b B_b B b
log rd ( L ) ≥ γ + log ( 8 π ) − B b ∗ − C b n + c log 3 ∑ m ≥ 1 3 − m / 2 F b ( m log 3 ) . (51) \log\operatorname{rd}(L)\geq\gamma+\log(8\pi)-B_b^*-\frac{C_b}{n}+c\log3\sum_{m\geq1}3^{-m/2}F_b(m\log3). \tag*{(51)} log rd ( L ) ≥ γ + log ( 8 π ) − B b ∗ − n C b + c log 3 m ≥ 1 ∑ 3 − m /2 F b ( m log 3 ) . (51) The prime sum is finite. Outward evaluation gives the four rows; the bound increases with n n n c c c
For the last assertion, the local degree is eighteen and residue degree one at both primes. Thus N 3 = n / 18 N_3=n/18 N 3 = n /18 c = 1 / 9 c=1/9 c = 1/9
rd ( L ) = 28 3 37 / 18 < 50.62072. (52) \operatorname{rd}(L)=\sqrt{28}\,3^{37/18}<50.62072. \tag*{(52)} rd ( L ) = 28 3 37/18 < 50.62072. (52) The last row contradicts this equality for degree at least twenty billion.
Lemma 24.4. Assume GRH. A representation as in Theorem 24.1 is modular unless its projective group is
PSL 2 ( F 3 2 e ) ( e ≥ 2 ) , or PGL 2 ( F 3 2 e + 1 ) ( e ≥ 1 ) . \operatorname{PSL}_2(\mathbb{F}_{3^{2e}})\quad(e\geq2),\qquad\text{or}\qquad\operatorname{PGL}_2(\mathbb{F}_{3^{2e+1}})\quad(e\geq1). PSL 2 ( F 3 2 e ) ( e ≥ 2 ) , or PGL 2 ( F 3 2 e + 1 ) ( e ≥ 1 ) . In every remaining case its projective completion at each place above three is exactly E 0 E_0 E 0 K K K C C C
Proof. Let L L L M M M G G G H H H K K K Z Z Z G G G 1 1 1 − 1 -1 − 1 A 5 A_5 A 5 PSL 2 ( F q ) \operatorname{PSL}_2(\mathbb{F}_q) PSL 2 ( F q ) PGL 2 ( F q ) \operatorname{PGL}_2(\mathbb{F}_q) PGL 2 ( F q ) q = 3 f ≥ 9 q=3^f\ge9 q = 3 f ≥ 9
We first consider q ≥ 27 q\ge27 q ≥ 27 H H H ∣ Z ∣ ≥ 4 |Z|\ge4 ∣ Z ∣ ≥ 4 ∣ Z ∣ ≤ 2 |Z|\le2 ∣ Z ∣ ≤ 2 Z Z Z H H H H = PGL 2 ( F q ) H=\operatorname{PGL}_2(\mathbb{F}_q) H = PGL 2 ( F q ) Z Z Z PSL 2 ( F q ) \operatorname{PSL}_2(\mathbb{F}_q) PSL 2 ( F q ) SL 2 ( F 3 ) \operatorname{SL}_2(\mathbb{F}_3) SL 2 ( F 3 ) [ L : Q ] ≥ 78624 [L:\mathbb{Q}]\ge78624 [ L : Q ] ≥ 78624
If one place above three is tame or has nontrivial unramified factor α \alpha α Lemma 23.2 gives
rd ( L ) < 28 3 ( 3 / 2 + 37 / 18 ) / 2 = 28 3 16 / 9 < 37.30737 , \operatorname{rd}(L)<\sqrt{28}\,3^{(3/2+37/18)/2}=\sqrt{28}\,3^{16/9}<37.30737, rd ( L ) < 28 3 ( 3/2 + 37/18 ) /2 = 28 3 16/9 < 37.30737 , contrary to the first row of Lemma 24.3 . Both diagonal quotients are therefore cyclotomic with nontrivial wild inertia. If one wild rank is one, the projective field satisfies
rd ( M ) ≤ 28 3 35 / 18 < 44.80380. \operatorname{rd}(M)\le\sqrt{28}\,3^{35/18}<44.80380. rd ( M ) ≤ 28 3 35/18 < 44.80380. Its degree is at least 2 ∣ PSL 2 ( F 27 ) ∣ = 19656 2|\operatorname{PSL}_2(\mathbb{F}_{27})|=19656 2∣ PSL 2 ( F 27 ) ∣ = 19656 c = 1 / e 1 + 1 / e 2 ≥ 2 / 9 c=1/e_1+1/e_2\ge2/9 c = 1/ e 1 + 1/ e 2 ≥ 2/9 E 0 E_0 E 0
For odd f f f − 1 -1 − 1 F q \mathbb{F}_q F q PSL 2 ( F q ) \operatorname{PSL}_2(\mathbb{F}_q) PSL 2 ( F q ) f f f PGL 2 ( F q ) \operatorname{PGL}_2(\mathbb{F}_q) PGL 2 ( F q ) − 1 -1 − 1 Lemma 24.2 . This gives the asserted parity restrictions.
In the remaining projective general linear case, the squareclass quotient agrees with χ 3 \chi_3 χ 3 C C C
It remains to consider full PGL 2 ( F 9 ) \operatorname{PGL}_2(\mathbb{F}_9) PGL 2 ( F 9 ) ± 1 \pm1 ± 1 F 9 \mathbb{F}_9 F 9
rd ( L ) < 28 3 55 / 36 < 28.34747. \operatorname{rd}(L)<\sqrt{28}\,3^{55/36}<28.34747. rd ( L ) < 28 3 55/36 < 28.34747. If the squareclass character is unramified everywhere finite, it is the character of K ( i ) K(i) K ( i ) rd ( L ) < 28 3 3 / 2 < 27.49546 \operatorname{rd}(L)<\sqrt{28}\,3^{3/2}<27.49546 rd ( L ) < 28 3 3/2 < 27.49546 [ L : Q ] ≥ 2880 [L:\mathbb{Q}]\ge2880 [ L : Q ] ≥ 2880 Lemma 24.3 .
The remaining small nonsolvable groups are A 5 A_5 A 5 PSL 2 ( F 9 ) \operatorname{PSL}_2(\mathbb{F}_9) PSL 2 ( F 9 ) PGL 2 ( F 9 ) \operatorname{PGL}_2(\mathbb{F}_9) PGL 2 ( F 9 ) [3 ] give modularity of the projective representation and of its given linear lift. □ \square □
A classical form of level eighty-four
Let η = χ 28 \eta=\chi_{28} η = χ 28 K / Q K/\mathbb{Q} K / Q
Lemma 24.5. There is a normalized newform f ∈ S 2 ( Γ 0 ( 84 ) , η ) f\in S_2(\Gamma_0(84),\eta) f ∈ S 2 ( Γ 0 ( 84 ) , η ) F = Q ( α ) F=\mathbb{Q}(\alpha) F = Q ( α )
α 4 + 14 α 2 + 32 = 0 , a 5 ( f ) = α , a 11 ( f ) = α 3 + 10 α 4 . (53) \alpha^4+14\alpha^2+32=0,\qquad a_5(f)=\alpha,\qquad a_{11}(f)=\frac{\alpha^3+10\alpha}{4}. \tag*{(53)} α 4 + 14 α 2 + 32 = 0 , a 5 ( f ) = α , a 11 ( f ) = 4 α 3 + 10 α . (53) Moreover, a 3 ( f ) = 1 a_3(f)=1 a 3 ( f ) = 1 F F F α \alpha α − α -\alpha − α Q ( 17 ) \mathbb{Q}(\sqrt{17}) Q ( 17 ) 17 = α 2 + 7 \sqrt{17}=\alpha^2+7 17 = α 2 + 7
Proof. Take the normalized newform with LMFDB label 84.2.b.b [45 ] . Its nebentypus is χ 84 ( 55 , ⋅ ) \chi_{84}(55,\cdot) χ 84 ( 55 , ⋅ ) η \eta η Q ( ν ) \mathbb{Q}(\nu) Q ( ν ) ν 4 − ν 3 − 2 ν + 4 = 0 \nu^{4}-\nu^{3}-2\nu+4=0 ν 4 − ν 3 − 2 ν + 4 = 0
a 3 ( f ) = 1 , a 5 ( f ) = ν 3 + ν − 2 , a 11 ( f ) = ν − ν 2 . a_{3}(f)=1,\qquad a_{5}(f)=\nu^{3}+\nu-2,\qquad a_{11}(f)=\nu-\nu^{2}. a 3 ( f ) = 1 , a 5 ( f ) = ν 3 + ν − 2 , a 11 ( f ) = ν − ν 2 . Put α = ν 3 + ν − 2 \alpha=\nu^{3}+\nu-2 α = ν 3 + ν − 2 ν \nu ν
α 4 + 14 α 2 + 32 = 0 , 4 ( ν − ν 2 ) = α 3 + 10 α , \alpha^{4}+14\alpha^{2}+32=0,\qquad4(\nu-\nu^{2})=\alpha^{3}+10\alpha, α 4 + 14 α 2 + 32 = 0 , 4 ( ν − ν 2 ) = α 3 + 10 α , which proves (49).
Modulo three, P ( X ) = X 4 + 14 X 2 + 32 P(X)=X^{4}+14X^{2}+32 P ( X ) = X 4 + 14 X 2 + 32 X 4 − X 2 − 1 X^{4}-X^{2}-1 X 4 − X 2 − 1 y 2 = y + 1 y^{2}=y+1 y 2 = y + 1 F 9 \mathbb{F}_{9} F 9 y 4 = − 1 y^{4}=-1 y 4 = − 1 y y y F 9 \mathbb{F}_{9} F 9 P P P Q \mathbb{Q} Q F = Q ( ν ) = Q ( α ) F=\mathbb{Q}(\nu)=\mathbb{Q}(\alpha) F = Q ( ν ) = Q ( α ) 2 13 17 2 2^{13}17^{2} 2 13 1 7 2 F F F α 2 = − 7 ± 17 \alpha^{2}=-7\pm\sqrt{17} α 2 = − 7 ± 17 F F F α \alpha α − α -\alpha − α Q ( 17 ) \mathbb{Q}(\sqrt{17}) Q ( 17 )
Proposition 24.6. There exists a finite Galois extension B / K B/K B / K
Gal ( B / K ) ≃ D : = { g ∈ GL 2 ( F 9 ) : det g ∈ { 1 , − 1 } } , [ B : K ] = 1440. \operatorname{Gal}(B/K)\simeq D:=\{g\in\operatorname{GL}_{2}(\mathbb{F}_{9}):\det g\in\{1,-1\}\},\qquad[B:K]=1440. Gal ( B / K ) ≃ D := { g ∈ GL 2 ( F 9 ) : det g ∈ { 1 , − 1 }} , [ B : K ] = 1440. It is unramified outside three, and its two local completions at three are contained in E 0 E_{0} E 0 C 2 C_{2} C 2 C C C
Proof. Let λ \lambda λ F F F f f f G K G_{K} G K r ‾ \overline{r} r λ \lambda λ χ 3 \chi_{3} χ 3 η χ 3 \eta\chi_{3} η χ 3 F 81 \mathbb{F}_{81} F 81
Write t = α ‾ t=\overline{\alpha} t = α x = t 2 x=t^{2} x = t 2 x 2 = x + 1 x^{2}=x+1 x 2 = x + 1 K K K G K G_{K} G K
s 5 = x + 1 , s 11 = 1 − x . s_{5}=x+1,\qquad s_{11}=1-x. s 5 = x + 1 , s 11 = 1 − x . Here a ‾ 11 2 = − x \overline{a}_{11}^{2}=-x a 11 2 = − x
s 5 2 = 2 , s 11 2 = − x − 1 , ( − x − 1 ) 2 = − 1. (54) s_{5}^{2}=2,\qquad s_{11}^{2}=-x-1,\qquad(-x-1)^{2}=-1. \tag*{(54)} s 5 2 = 2 , s 11 2 = − x − 1 , ( − x − 1 ) 2 = − 1. (54) If z z z u u u u + u − 1 = z − 2 u+u^{-1}=z-2 u + u − 1 = z − 2 z 2 + 1 = 0 z^{2}+1=0 z 2 + 1 = 0 u 4 + u 3 + u 2 + u + 1 = 0 u^{4}+u^{3}+u^{2}+u+1=0 u 4 + u 3 + u 2 + u + 1 = 0
All traces on G K G_{K} G K F 9 \mathbb{F}_{9} F 9 q q q a q ( f ) = η ( q ) a q ( f ) a_{q}(f)=\eta(q)a_{q}(f) a q ( f ) = η ( q ) a q ( f ) K K K a q ( f ) a_{q}(f) a q ( f ) a q ( f ) 2 − 2 η ( q ) q a_{q}(f)^{2}-2\eta(q)q a q ( f ) 2 − 2 η ( q ) q F + = Q ( 17 ) F^{+}=\mathbb{Q}(\sqrt{17}) F + = Q ( 17 ) F 9 \mathbb{F}_{9} F 9 G K G_{K} G K
This also proves that r ‾ \overline{r} r ψ 1 ⊕ ψ 2 \psi_{1}\oplus\psi_{2} ψ 1 ⊕ ψ 2
Absolute irreducibility, together with the traces and determinant in F 9 \mathbb{F}_{9} F 9 r ‾ \overline{r} r F 9 \mathbb{F}_{9} F 9
Both 1 and − 1 -1 − 1 F 9 \mathbb{F}_{9} F 9 PSL 2 ( F 9 ) \operatorname{PSL}_{2}(\mathbb{F}_{9}) PSL 2 ( F 9 ) A 4 A_4 A 4 S 4 S_4 S 4 A 5 A_5 A 5 PSL 2 ( F 9 ) ≃ A 6 \operatorname{PSL}_2(\mathbb{F}_9) \simeq A_6 PSL 2 ( F 9 ) ≃ A 6
The scalar subgroup of the linear image is contained in μ 4 \mu_4 μ 4 4 ⋅ 360 = 1440 4 \cdot360 = 1440 4 ⋅ 360 = 1440 D D D D = μ 4 SL 2 ( F 9 ) D = \mu_4 \operatorname{SL}_2(\mathbb{F}_9) D = μ 4 SL 2 ( F 9 ) SL 2 ( F 9 ) \operatorname{SL}_2(\mathbb{F}_9) SL 2 ( F 9 ) D ′ = SL 2 ( F 9 ) D' = \operatorname{SL}_2(\mathbb{F}_9) D ′ = SL 2 ( F 9 ) C C C
We check the local kernels, not merely the projective conductor. At 2 and 7, local–global compatibility for the newform gives conductor exponents respectively two and one [61 equation (1)] . These equal the exponents of its determinant character η \eta η a ( V , N ) = a ( det V ) a(V,N) = a(\det V) a ( V , N ) = a ( det V ) 1 ⊕ det V 1 \oplus\det V 1 ⊕ det V N = 0 N = 0 N = 0 V I = 0 V^I = 0 V I = 0
a ( V , N ) ≥ 2 + Swan ( V ) ≥ 2 + Swan ( det V ) > a ( det V ) . a(V,N) \ge2 + \operatorname{Swan}(V) \ge2 + \operatorname{Swan}(\det V) > a(\det V). a ( V , N ) ≥ 2 + Swan ( V ) ≥ 2 + Swan ( det V ) > a ( det V ) . Thus V I V^I V I 1 ⊕ η 1 \oplus\eta 1 ⊕ η K K K
At three, η ∣ G Q 3 = 1 \eta|_{G_{\mathbb{Q}_3}} = 1 η ∣ G Q 3 = 1 3 ∥ 84 3 \mathrel{\Vert} 84 3 ∥ 84 a 3 ( f ) = 1 a_3(f) = 1 a 3 ( f ) = 1 Lemma 4.2 give the trivial unramified ordinary quotient at both places above three. Its determinant is χ 3 \chi_3 χ 3 1 1 1 χ ‾ 3 \overline{\chi}_3 χ 3
Its extension class lies in H 1 ( Q 3 , F 3 ( χ ‾ 3 ) ) H^1(\mathbb{Q}_3, \mathbb{F}_3(\overline{\chi}_3)) H 1 ( Q 3 , F 3 ( χ 3 )) F 3 \mathbb{F}_3 F 3 E 0 E_0 E 0 B = K ker r ‾ B = K^{\ker\overline{r}} B = K k e r r
Auxiliary composita and the residual theorem
Proposition 24.7. There exists a totally imaginary K K K F F F K K K E 0 E_0 E 0 C C C
Proof. Put π = 2 + 7 \pi= 2 + \sqrt{7} π = 2 + 7 π ′ = 2 − 7 \pi' = 2 - \sqrt{7} π ′ = 2 − 7 ϵ = 8 + 3 7 \epsilon= 8 + 3\sqrt{7} ϵ = 8 + 3 7
E = K ( ζ 3 , π 3 , π ′ 3 , ϵ 3 ) . E = K\left(\zeta_3, \sqrt[3]{\pi}, \sqrt[3]{\pi'}, \sqrt[3]{\epsilon}\right). E = K ( ζ 3 , 3 π , 3 π ′ , 3 ϵ ) . The three cube classes are independent over K K K p 1 , p 2 \mathfrak{p}_1,\mathfrak{p}_2 p 1 , p 2 C C C K K K [ E : C ] = 27 [E:C] = 27 [ E : C ] = 27 [ E : K ] = 54 [E:K] = 54 [ E : K ] = 54 ( C 3 ) 3 ⋊ C 2 (C_3)^3 \rtimes C_2 ( C 3 ) 3 ⋊ C 2
The radicands are units outside three, so their Kummer extensions are unramified there. At the two three-adic embeddings, 7 \sqrt{7} 7 π , π ′ \pi,\pi' π , π ′ E 0 E_0 E 0
Let B B B C 2 C_2 C 2 E / K E/K E / K C C C B ∩ E = C B \cap E = C B ∩ E = C F = B E F = BE F = B E
[ F : K ] = 1440 ⋅ 54 2 = 38880 = 2 5 ⋅ 3 3 ⋅ 5. (55) [F:K] = \frac{1440 \cdot54}{2} = 38880 = 2^5 \cdot3^3 \cdot5. \tag*{(55)} [ F : K ] = 2 1440 ⋅ 54 = 38880 = 2 5 ⋅ 3 3 ⋅ 5. (55) Its local completions are still E 0 E_0 E 0 B B B E 0 E_0 E 0 E E E E 0 E_0 E 0 C ⊂ F C \subset\mathcal{F} C ⊂ F
Whenever two K K K E 0 E_0 E 0 E 0 E_0 E 0 Q ‾ 3 \overline{\mathbb{Q}}_3 Q 3 (45) is normal and is the full Kummer extension of the rational cube-class space. We shall repeatedly apply the degree bound (50) after taking such composita.
Proof of Theorem 24.1. By Lemma 24.4, it suffices to treat the two displayed families there, with projective field M M M E 0 E_0 E 0 H = Gal ( M / K ) H = \operatorname{Gal}(M/K) H = Gal ( M / K ) F \mathcal{F} F
Suppose H = PSL 2 ( F 3 2 e ) H = \operatorname{PSL}_2(\mathbb{F}_{3^{2e}}) H = PSL 2 ( F 3 2 e ) e ≥ 2 e \ge2 e ≥ 2 M ∩ F M \cap\mathcal{F} M ∩ F K K K M M M ∣ H ∣ ≥ ∣ PSL 2 ( F 81 ) ∣ = 265680 > 38880 |H| \ge|\operatorname{PSL}_2(\mathbb{F}_{81})| = 265680 > 38880 ∣ H ∣ ≥ ∣ PSL 2 ( F 81 ) ∣ = 265680 > 38880
[ M F : Q ] ≥ 2 ⋅ 265680 ⋅ 38880 = 20 659 276 800 , [ M\mathcal{F} : \mathbb{Q} ] \ge2 \cdot265680 \cdot38880 = 20\,659\,276\,800, [ M F : Q ] ≥ 2 ⋅ 265680 ⋅ 38880 = 20 659 276 800 , contradicting (50) .
Suppose H = PGL 2 ( F 3 2 e + 1 ) H = \operatorname{PGL}_2(\mathbb{F}_{3^{2e+1}}) H = PGL 2 ( F 3 2 e + 1 ) e ≥ 2 e \ge2 e ≥ 2 C 2 C_2 C 2 C C C M ∩ F = C M \cap\mathcal{F} = C M ∩ F = C
[ M F : Q ] ≥ ∣ PGL 2 ( F 243 ) ∣ ⋅ 38880 = 557 876 056 320 , [ M\mathcal{F} : \mathbb{Q} ] \ge|\operatorname{PGL}_2(\mathbb{F}_{243})| \cdot38880 = 557\,876\,056\,320, [ M F : Q ] ≥ ∣ PGL 2 ( F 243 ) ∣ ⋅ 38880 = 557 876 056 320 , again impossible.
It remains to consider H = PGL 2 ( F 27 ) H = \operatorname{PGL}_2(\mathbb{F}_{27}) H = PGL 2 ( F 27 ) h = 19656 h = 19656 h = 19656 σ \sigma σ K / Q K/\mathbb{Q} K / Q M ≠ M σ M \ne M^\sigma M = M σ C C C C C C P = M M σ P = MM^\sigma P = M M σ
Gal ( P / K ) ≃ H × C 2 H , [ P : K ] = h 2 / 2. \operatorname{Gal}(P/K) \simeq H \times_{C_2} H, \qquad[P:K] = h^2/2. Gal ( P / K ) ≃ H × C 2 H , [ P : K ] = h 2 /2. Its derived subgroup is PSL 2 ( F 27 ) 2 \operatorname{PSL}_2(\mathbb{F}_{27})^2 PSL 2 ( F 27 ) 2 Gal ( F / K ) \operatorname{Gal}(\mathcal{F}/K) Gal ( F / K ) P ∩ F = C P \cap\mathcal{F} = C P ∩ F = C
[ P F : Q ] = h 2 ⋅ 19440 = 7 510 806 051 840 , [P\mathcal{F}:\mathbb{Q}] = h^2 \cdot19440 = 7\,510\,806\,051\,840, [ P F : Q ] = h 2 ⋅ 19440 = 7 510 806 051 840 , contradicting (50) . Consequently M = M σ M = M^\sigma M = M σ M / Q M/\mathbb{Q} M / Q
We now prove modularity rather than nonexistence. Write Γ = Gal ( M / Q ) \Gamma= \operatorname{Gal}(M/\mathbb{Q}) Γ = Gal ( M / Q ) Γ → Aut ( H ) \Gamma\to\operatorname{Aut}(H) Γ → Aut ( H ) Γ / H = C 2 \Gamma/H = C_2 Γ/ H = C 2 Out ( PGL 2 ( F 27 ) ) ≃ C 3 \operatorname{Out}(\operatorname{PGL}_2(\mathbb{F}_{27})) \simeq C_3 Out ( PGL 2 ( F 27 )) ≃ C 3 [37 Section 13, Lemma 52] . The outer action is trivial. As H H H Inn ( H ) ≃ H \operatorname{Inn}(H) \simeq H Inn ( H ) ≃ H G Q → Γ G_{\mathbb{Q}} \to\Gamma G Q → Γ G Q G_{\mathbb{Q}} G Q K K K
Its complex conjugations are nontrivial projective involutions. Tate’s lifting theorem gives a finite linear lift in characteristic three, necessarily odd and irreducible [3 Section 2.1] . Serre modularity over Q \mathbb{Q} Q [40 , 41 ] and cyclic base change give residual modularity over K K K K K K
The residual base over Q ( 13 ) \mathbb{Q}(\sqrt{13}) Q ( 13 )
In this section K = Q ( 13 ) K = \mathbb{Q}(\sqrt{13}) K = Q ( 13 ) Q 3 \mathbb{Q}_3 Q 3
Lemma 25.1. The narrow class number of K K K G K → F ‾ 3 × G_K \to\overline{\mathbb{F}}_3^{\times} G K → F 3 × 3 3 3
Proof. The Minkowski bound 13 / 2 < 2 \sqrt{13}/2 < 2 13 /2 < 2 ε = ( 3 + 13 ) / 2 \varepsilon= (3+\sqrt{13})/2 ε = ( 3 + 13 ) /2 − 1 -1 − 1 − 1 -1 − 1
The character assertion follows from Lemma 23.1.
Lemma 25.2. Under GRH, a totally imaginary number field of degree n n n n ≥ 531360 n \ge531360 n ≥ 531360 n ≥ 78624 n \ge78624 n ≥ 78624 n ≥ 1440 n \ge1440 n ≥ 1440
Proof. Use the explicit formula of Section 22, with no Euler terms or real-place term retained, using B b ≤ B b ∗ B_b \le B_b^* B b ≤ B b ∗
b n rd ( L ) > 45 / 4 531360 39.01373 21 / 2 78624 37.63959 8 1440 28.02230 \begin{array}{c|c|c}
b & n & \operatorname{rd}(L) > \\
45/4 & 531360 & 39.01373 \\
21/2 & 78624 & 37.63959 \\
8 & 1440 & 28.02230
\end{array} b 45/4 21/2 8 n 531360 78624 1440 rd ( L ) > 39.01373 37.63959 28.02230 These are outward interval evaluations of the displayed finite expressions. Each bound increases with n n n C b > 0 C_b > 0 C b > 0 [49 classical] .
Proposition 25.3. Assume GRH. Every continuous irreducible totally odd representation ρ ˉ : G K → GL 2 ( F ‾ 3 ) \bar{\rho}: G_K \to\operatorname{GL}_2(\overline{\mathbb{F}}_3) ρ ˉ : G K → GL 2 ( F 3 ) 3 3 3
Proof. Let L / K L/K L / K G = Gal ( L / K ) G = \operatorname{Gal}(L/K) G = Gal ( L / K ) Z Z Z H = G / Z H = G/Z H = G / Z L L L K K K (47) give
rd ( L ) ≤ 13 3 13 / 6 < 38.971. \operatorname{rd}(L) \le\sqrt{13}\,3^{13/6} < 38.971. rd ( L ) ≤ 13 3 13/6 < 38.971. The solvable case is already modular. Otherwise Dickson’s classification gives H ≃ A 5 H \simeq A_5 H ≃ A 5 PSL 2 ( F 3 f ) \operatorname{PSL}_2(\mathbb{F}_{3^f}) PSL 2 ( F 3 f ) PGL 2 ( F 3 f ) \operatorname{PGL}_2(\mathbb{F}_{3^f}) PGL 2 ( F 3 f ) f ≥ 2 f \ge2 f ≥ 2 3 f ≥ 81 3^f \ge81 3 f ≥ 81 [ L : Q ] ≥ 2 ∣ PSL 2 ( F 81 ) ∣ = 531360 [L:\mathbb{Q}] \ge2|\operatorname{PSL}_2(\mathbb{F}_{81})| = 531360 [ L : Q ] ≥ 2∣ PSL 2 ( F 81 ) ∣ = 531360
Suppose 3 f = 27 3^f = 27 3 f = 27
rd ( L ) ≤ 13 3 13 / 6 − 1 / 27 < 37.417. \operatorname{rd}(L) \le\sqrt{13}\,3^{13/6-1/27} < 37.417. rd ( L ) ≤ 13 3 13/6 − 1/27 < 37.417. If H = PSL 2 ( F 27 ) H = \operatorname{PSL}_2(\mathbb{F}_{27}) H = PSL 2 ( F 27 ) ∣ Z ∣ ≥ 4 |Z| \ge4 ∣ Z ∣ ≥ 4 ∣ Z ∣ ≤ 2 |Z| \le2 ∣ Z ∣ ≤ 2 Z Z Z H H H H = PGL 2 ( F 27 ) H = \operatorname{PGL}_2(\mathbb{F}_{27}) H = PGL 2 ( F 27 ) ∣ Z ∣ ≥ 2 |Z| \ge2 ∣ Z ∣ ≥ 2 PSL 2 ( F 27 ) \operatorname{PSL}_2(\mathbb{F}_{27}) PSL 2 ( F 27 ) SL 2 ( F ‾ 3 ) \operatorname{SL}_2(\overline{\mathbb{F}}_3) SL 2 ( F 3 ) − I -I − I [ L : Q ] ≥ 78624 [L:\mathbb{Q}] \ge78624 [ L : Q ] ≥ 78624
Suppose now that H = PGL 2 ( F 9 ) H = \operatorname{PGL}_2(\mathbb{F}_9) H = PGL 2 ( F 9 ) p i \mathfrak{p}_i p i F 9 \mathbb{F}_9 F 9 a 3 = a a^3 = a a 3 = a a = ± 1 a = \pm1 a = ± 1
squares in F 9 × \mathbf{F}_9^\times F 9 ×
rd ( L ) < 13 3 ( 1 + 13 / 6 ) / 2 = 13 3 19 / 12 < 20.532. \operatorname{rd}(L)<\sqrt{13}\,3^{(1+13/6)/2}=\sqrt{13}\,3^{19/12}<20.532. rd ( L ) < 13 3 ( 1 + 13/6 ) /2 = 13 3 19/12 < 20.532. But [ L : Q ] ≥ 2 ∣ PGL 2 ( F 9 ) ∣ = 1440 [L:\mathbf{Q}]\ge2|\operatorname{PGL}_2(\mathbf{F}_9)|=1440 [ L : Q ] ≥ 2∣ PGL 2 ( F 9 ) ∣ = 1440 b = 16 b=16 b = 16 rd ( L ) > 21.15 \operatorname{rd}(L)>21.15 rd ( L ) > 21.15
The remaining nonsolvable groups are A 5 A_5 A 5 PSL 2 ( F 9 ) \operatorname{PSL}_2(\mathbf{F}_9) PSL 2 ( F 9 ) PGL 2 ( F 9 ) \operatorname{PGL}_2(\mathbf{F}_9) PGL 2 ( F 9 ) [3 Theorem 1.2(5)] , and its given linear lift is modular by [3 Lemma 2.1] . □
The residual base over Q(sqrt(37))
Put K = Q ( 37 ) K=\mathbf{Q}(\sqrt{37}) K = Q ( 37 )
Theorem 26.1. Assume GRH. Every continuous irreducible representation ρ ˉ : G K → GL 2 ( F ‾ 3 ) \bar{\rho}:G_K\to\operatorname{GL}_2(\overline{\mathbf{F}}_3) ρ ˉ : G K → GL 2 ( F 3 ) χ ‾ 3 \overline{\chi}_3 χ 3
Write
w = 1 + 37 2 , ϵ = 5 + 2 w , π = 3 + w , π ′ = 4 − w . w=\frac{1+\sqrt{37}}{2},\qquad\epsilon=5+2w,\qquad\pi=3+w,\qquad\pi'=4-w. w = 2 1 + 37 , ϵ = 5 + 2 w , π = 3 + w , π ′ = 4 − w . Thus w 2 − w − 9 = 0 w^2-w-9=0 w 2 − w − 9 = 0 N K / Q ( ϵ ) = − 1 N_{K/\mathbf{Q}}(\epsilon)=-1 N K / Q ( ϵ ) = − 1 π π ′ = 3 \pi\pi'=3 π π ′ = 3 p = ( π ) \mathfrak{p}=(\pi) p = ( π ) p ′ = ( π ′ ) \mathfrak{p}'=(\pi') p ′ = ( π ′ ) Q 3 \mathbf{Q}_3 Q 3 Q 2 \mathbf{Q}_2 Q 2
Lemma 26.2. The ordinary and narrow class numbers of K K K G K → F ‾ 3 × G_K\to\overline{\mathbf{F}}_3^\times G K → F 3 ×
Proof. The Minkowski bound is 37 / 2 < 3.05 \sqrt{37}/2<3.05 37 /2 < 3.05 ϵ \epsilon ϵ − 1 -1 − 1
Proposition 26.3. Let ρ ˉ \bar{\rho} ρ ˉ R = ( r λ ) R=(r_\lambda) R = ( r λ ) χ \chi χ { 0 , 1 } \{0,1\} { 0 , 1 } τ + \tau_+ τ + p \mathfrak{p} p p ′ \mathfrak{p}' p ′
Proof. Proposition 23.4 and Lemma 23.5 supply the local lifts at p \mathfrak{p} p p ′ \mathfrak{p}' p ′ G K ( ζ 3 ) G_{K(\zeta_3)} G K ( ζ 3 )
Lemma 2.4(i) now globalizes these weight-two components with determinant χ 3 \chi_3 χ 3 { 0 , 1 } \{0,1\} { 0 , 1 }
Choose a place above two in the coefficient field and write r 2 r_2 r 2 r ˉ 2 \bar{r}_2 r ˉ 2 r ˉ 2 \bar{r}_2 r ˉ 2 M M M p \mathfrak{p} p E + E_+ E + p ′ \mathfrak{p}' p ′ E ′ E' E ′
Ramification at two
For a finite Galois extension L / F L/F L / F d ( L / F ) d(L/F) d ( L / F ) v F ( 2 ) = 1 v_F(2)=1 v F ( 2 ) = 1 d ( L / F ) d(L/F) d ( L / F )
Lemma 26.4. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 J J J O F \mathcal{O}_F O F J ( F ) J(F) J ( F )
Proof. Fontaine’s Theorem A and its corollary [31 pp. 515–516] apply to finite flat group schemes of any rank. His notation is G ( u ) = G u − 1 G^{(u)}=G^{u-1} G ( u ) = G u − 1 p n p^n p n e e e
G ( u ) acts trivially if u > e ( n + 1 p − 1 ) . G^{(u)}\text{ acts trivially if }u>e\left(n+\frac{1}{p-1}\right). G ( u ) acts trivially if u > e ( n + p − 1 1 ) . Here e = n = 1 e=n=1 e = n = 1 p = 2 p=2 p = 2 G v G^v G v v > 1 v>1 v > 1
Corollary 26.5. A compositum of finite Galois extensions of F F F
Proof. Upper-numbering ramification is compatible with quotients. If all quotient fields have trivial groups above one, so does their compositum. The integral formula for the different then gives d < 2 d<2 d < 2 t t t Q Q Q d ≤ 2 − ( t + 1 ) / ( t Q ) d\le2-(t+1)/(tQ) d ≤ 2 − ( t + 1 ) / ( tQ )
We also need to recognize ordinary lifts in the dihedral case.
Lemma 26.6. Let F / Q 2 F/\mathbb{Q}_2 F / Q 2 r : G F → GL 2 ( E ) r:G_F\to\operatorname{GL}_2(E) r : G F → GL 2 ( E ) χ 2 \chi_2 χ 2 r r r
Proof. A two-group has only trivial irreducible representations in characteristic two, so both residual constituents are unramified. Apply Lemma 4.10 with p = 2 p=2 p = 2 e ( F / Q 2 ) = 1 e(F/\mathbb{Q}_2)=1 e ( F / Q 2 ) = 1
An explicit solvable auxiliary field
Put
Q ( X ) = X 4 − 2 X 3 + ( 3 − 2 w ) X 2 + ( 4 + 2 w ) X − w − 2 , C = K ( π , π ′ , − ϵ ) , Q(X)=X^4-2X^3+(3-2w)X^2+(4+2w)X-w-2,\qquad C=K\left(\sqrt{\pi},\sqrt{\pi'},\sqrt{-\epsilon}\right), Q ( X ) = X 4 − 2 X 3 + ( 3 − 2 w ) X 2 + ( 4 + 2 w ) X − w − 2 , C = K ( π , π ′ , − ϵ ) , and let H H H Q Q Q K K K N = C H N=CH N = C H
Proposition 26.7. The extension N / K N/K N / K p \mathfrak{p} p p ′ \mathfrak{p}' p ′ E + E_+ E + E ′ E' E ′
Proof. The cubic resolvent and discriminant of Q Q Q
R ( T ) = T 3 + ( 2 w − 3 ) T 2 + 4 − 4 w , disc ( Q ) = disc ( R ) = 144 δ , δ = 99 − 28 w = − ϵ − 3 π 3 . R(T)=T^3+(2w-3)T^2+4-4w,\qquad\operatorname{disc}(Q)=\operatorname{disc}(R)=144\delta,\qquad\delta=99-28w=-\epsilon^{-3}\pi^3. R ( T ) = T 3 + ( 2 w − 3 ) T 2 + 4 − 4 w , disc ( Q ) = disc ( R ) = 144 δ , δ = 99 − 28 w = − ϵ − 3 π 3 . The quartic is irreducible modulo the prime above eleven with w = 7 w=7 w = 7 X 4 − 2 X 3 + 7 X + 2 X^4-2X^3+7X+2 X 4 − 2 X 3 + 7 X + 2 X 121 − X X^{121}-X X 121 − X X 14641 − X X^{14641}-X X 14641 − X F 11 \mathbb{F}_{11} F 11 Q Q Q K K K
At p \mathfrak{p} p w ≡ 18 ( m o d 27 ) w \equiv18 \pmod{27} w ≡ 18 ( mod 27 )
Q ( Y − 1 ) = Y 4 − 6 Y 3 + ( 15 − 2 w ) Y 2 + ( 6 w − 12 ) Y − 5 w . Q(Y-1)=Y^4-6Y^3+(15-2w)Y^2+(6w-12)Y-5w. Q ( Y − 1 ) = Y 4 − 6 Y 3 + ( 15 − 2 w ) Y 2 + ( 6 w − 12 ) Y − 5 w . Its Newton polygon has segments of slopes − 1 -1 − 1 − 1 / 3 -1/3 − 1/3 Q 3 \mathbb{Q}_3 Q 3 1 / 3 1/3 1/3 v p ( disc ( Q ) ) = 5 v_{\mathfrak{p}}(\operatorname{disc}(Q))=5 v p ( disc ( Q )) = 5 − ϵ π -\epsilon\pi − ϵ π 3 3 3 S 3 S_3 S 3 7 / 6 7/6 7/6 Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 ) Lemma 23.3 it is E + E_+ E +
The global quartic group is transitive and contains this point-stabilizer S 3 S_3 S 3 S 4 S_4 S 4 D = K ( − ϵ π ) D=K(\sqrt{-\epsilon\pi}) D = K ( − ϵ π ) p \mathfrak{p} p p ′ \mathfrak{p}' p ′ − ϵ -\epsilon − ϵ [ C : K ] = 8 [C:K]=8 [ C : K ] = 8 H ∩ C H\cap C H ∩ C D D D S 4 S_4 S 4 [ N : K ] = 24 ⋅ 8 / 2 = 96 [N:K]=24\cdot8/2=96 [ N : K ] = 24 ⋅ 8/2 = 96
At p ′ \mathfrak{p}' p ′ Q Q Q X 2 ( X − 1 ) 2 X^2(X-1)^2 X 2 ( X − 1 ) 2 v p ′ ( disc ( Q ) ) = 2 v_{\mathfrak{p}'}(\operatorname{disc}(Q))=2 v p ′ ( disc ( Q )) = 2 E ′ E' E ′ π \pi π π ′ \pi' π ′ − ϵ -\epsilon − ϵ 3 , 1 , 1 3,1,1 3 , 1 , 1 p \mathfrak{p} p 1 , 3 , − 1 1,3,-1 1 , 3 , − 1 p ′ \mathfrak{p}' p ′ C C C E + E_+ E + E ′ E' E ′
At two put h = X 2 + X + 1 − w h=X^2+X+1-w h = X 2 + X + 1 − w
Q = h 2 + 2 g , g = − 2 X 3 + ( 2 w + 1 ) X − 6. Q=h^2+2g,\qquad g=-2X^3+(2w+1)X-6. Q = h 2 + 2 g , g = − 2 X 3 + ( 2 w + 1 ) X − 6. Over the residue field F 4 \mathbb{F}_4 F 4 h ˉ \bar{h} h ˉ Tr F 4 / F 2 ( 1 − w ˉ ) = 1 \operatorname{Tr}_{\mathbb{F}_4/\mathbb{F}_2}(1-\bar{w})=1 Tr F 4 / F 2 ( 1 − w ˉ ) = 1 g ˉ = X \bar{g}=X g ˉ = X h h h Q Q Q h h h
The reduction of R R R T 2 ( T + 1 ) T^2(T+1) T 2 ( T + 1 ) Q Q Q S 4 S_4 S 4 G 0 = G 1 G_0=G_1 G 0 = G 1 G 0 G_0 G 0 G 1 G_1 G 1 G 2 = 1 G_2=1 G 2 = 1
Finally, the radicands defining C C C Corollary 26.5 gives the same bound for N N N
The discriminant comparisons
Let L / K L/K L / K p \mathfrak{p} p E + E_+ E + p ′ \mathfrak{p}' p ′ E ′ E' E ′
rd ( L ) < 4 37 5 / 6 < 60.78017. (56) \operatorname{rd}(L)<4\sqrt{37}^{\,5/6}<60.78017. \tag*{(56)} rd ( L ) < 4 37 5/6 < 60.78017. (56) Each three-adic exponent is multiplied by one half, since [ K : Q ] = 2 [K:\mathbb{Q}]=2 [ K : Q ] = 2
Proposition 26.8. Assume GRH for the Dedekind zeta function of L L L n = [ L : Q ] n=[L:\mathbb{Q}] n = [ L : Q ]
(i) If its completions above p , p ′ \mathfrak{p},\mathfrak{p}' p , p ′ E + , E ′ E_{+},E' E + , E ′ n ≥ 96768 n\ge96768 n ≥ 96768 rd ( L ) > 60.80125 \operatorname{rd}(L)>60.80125 rd ( L ) > 60.80125
(ii) If L L L p \mathfrak{p} p E + E_{+} E + n = 120 n=120 n = 120 rd ( L ) > 66.10839 \operatorname{rd}(L)>66.10839 rd ( L ) > 66.10839
Proof. Use the explicit formula of Section 22 with its one-term bound B b ≤ B b ∗ B_b\le B_b^* B b ≤ B b ∗ n / 12 n/12 n /12 n / 8 n/8 n /8 c 3 = 1 / 6 c_3=1/6 c 3 = 1/6 c 9 = 1 / 4 c_9=1/4 c 9 = 1/4 c 3 = 1 / 6 c_3=1/6 c 3 = 1/6 r 1 / n = 1 r_1/n=1 r 1 / n = 1
Table 26.8.
The first expression increases with n n n C b > 0 C_b>0 C b > 0
Proposition 26.9. Assume GRH. The image of r ˉ 2 \bar{r}_2 r ˉ 2 SL 2 ( F 4 ) ≃ A 5 \operatorname{SL}_2(\mathbb{F}_4)\simeq A_5 SL 2 ( F 4 ) ≃ A 5
Proof. Dickson’s classification makes a nonsolvable finite subgroup of SL 2 ( F 2 ) \operatorname{SL}_2(\mathbb{F}_2) SL 2 ( F 2 ) SL 2 ( F q ) \operatorname{SL}_2(\mathbb{F}_q) SL 2 ( F q ) q = 2 r q=2^r q = 2 r r ≥ 2 r\ge2 r ≥ 2 q ≥ 8 q\ge8 q ≥ 8 M M M K K K N N N L = M N L=MN L = M N
[ L : Q ] = 2 ⋅ 96 ⋅ q ( q 2 − 1 ) ≥ 96768. [L:\mathbb{Q}]=2\cdot96\cdot q(q^2-1)\ge96768. [ L : Q ] = 2 ⋅ 96 ⋅ q ( q 2 − 1 ) ≥ 96768. By Lemma 26.4 , Proposition 26.7 and Corollary 26.5 , its dyadic different is less than two. Its completions at three are exactly E + E_{+} E + E ′ E' E ′ (56) contradicts Proposition 26.8 (i).
Modularity of the remaining images
The sign of the type τ + \tau_{+} τ +
Lemma 26.10. Let D / K D/K D / K p \mathfrak{p} p Q 3 ( − 3 ) \mathbb{Q}_3(\sqrt{-3}) Q 3 ( − 3 )
Proof. Class number one shows that D = K ( α ) D=K(\sqrt{\alpha}) D = K ( α ) α \alpha α − 1 , ϵ , π , π ′ -1,\epsilon,\pi,\pi' − 1 , ϵ , π , π ′ ( Z / 4 ) [ w ] / ( w 2 − w − 9 ) (\mathbb{Z}/4)[w]/(w^2-w-9) ( Z /4 ) [ w ] / ( w 2 − w − 9 )
1 , ϵ π , − π π ′ = − 3 , − ϵ π ′ . 1,\qquad\epsilon\pi,\qquad-\pi\pi'=-3,\qquad-\epsilon\pi'. 1 , ϵ π , − π π ′ = − 3 , − ϵ π ′ . At p \mathfrak{p} p w = 18 ( m o d 27 ) w=18\pmod{27} w = 18 ( mod 27 ) π / 3 = 1 ( m o d 3 ) \pi/3=1\pmod{3} π /3 = 1 ( mod 3 ) ϵ = − 1 ( m o d 3 ) \epsilon=-1\pmod{3} ϵ = − 1 ( mod 3 ) ϵ π \epsilon\pi ϵ π − 3 -3 − 3 − 3 -3 − 3
Proposition 26.11. Assume GRH. The characteristic-two member r 2 r_2 r 2
Proof. Suppose first that the residual image is solvable. Since r ˉ 2 \bar{r}_2 r ˉ 2 D 2 m D_{2m} D 2 m D / K D/K D / K p \mathfrak{p} p Q 3 ( 3 ) \mathbb{Q}_3(\sqrt{3}) Q 3 ( 3 ) Lemma 23.3 . Lemma 26.10 shows that D / K D/K D / K
The dyadic wild inertia image is a nontrivial normal two-subgroup of the local decomposition image in D 2 m D_{2m} D 2 m C 2 C_2 C 2 r 2 r_2 r 2 D D D
Suppose now that the residual image is nonsolvable. Proposition 26.9 makes it A 5 = SL 2 ( F 4 ) A_5 = \operatorname{SL}_2(\mathbb{F}_4) A 5 = SL 2 ( F 4 ) [3 Theorem 2.5] . We check the real-place hypothesis in Thorne’s characteristic-two lifting theorem. If the kernel field M M M [ M : Q ] = 120 [M : \mathbb{Q}] = 120 [ M : Q ] = 120
rd ( M ) < 4 37 35 / 6 < 60.78017. \operatorname{rd}(M) < 4\sqrt{37}^{35/6} < 60.78017. rd ( M ) < 4 37 35/6 < 60.78017. This contradicts Proposition 26.8(ii). Hence r ˉ 2 ( c v ) ≠ 1 \bar{r}_2(c_v) \ne1 r ˉ 2 ( c v ) = 1 v v v K K K r 2 r_2 r 2 [69 ] applies directly. It permits both ordinary and nonordinary local representations and imposes no condition K 2 = Q 2 K_2 = \mathbb{Q}_2 K 2 = Q 2
Proof of Theorem 26.1. Solvable projective images are modular by automorphic induction and Langlands–Tunnell, with the characteristic-three lifting and twisting argument used in [3 proof of Theorem 2.5] . We may therefore suppose that the image is nonsolvable. Proposition 26.3 constructs a compatible system lifting ρ ˉ \bar{\rho} ρ ˉ ρ ˉ \bar{\rho} ρ ˉ
The split-three field applications
For the rest of this section let K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) d ∈ { 7 , 13 , 37 } d \in\{7, 13, 37\} d ∈ { 7 , 13 , 37 } p 1 , p 2 \mathfrak{p}_1,\mathfrak{p}_2 p 1 , p 2
Theorem 27.1. Assume GRH for Dedekind zeta functions. For K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) d ∈ { 7 , 13 , 37 } d \in\{7, 13, 37\} d ∈ { 7 , 13 , 37 } G K → GL 2 ( F ℓ ) G_K \to\operatorname{GL}_2(\mathbb{F}_\ell) G K → GL 2 ( F ℓ ) ℓ > 2 \ell> 2 ℓ > 2
Proof of Theorem 27.1. For ℓ = 3 \ell= 3 ℓ = 3 d = 7 , 13 , 37 d = 7, 13, 37 d = 7 , 13 , 37 χ ˉ 3 \bar{\chi}_3 χ ˉ 3 χ ˉ 3 \bar{\chi}_3 χ ˉ 3 G Q G_{\mathbb{Q}} G Q
Solvable residual images are already modular. Otherwise the initial weight-two lift and Lemma 2.4 give a non dihedral potentially modular compatible system. Two does not split and three splits in all three fields, so Theorem 1.2 proves its modularity. Reduce at the original coefficient place. GRH is used only in the supported residual assertions.
The ramified-dyadic field applications
Proof of Theorem 13.1. For d = 2 , 3 d = 2, 3 d = 2 , 3 d = 6 , 10 , 11 d = 6, 10, 11 d = 6 , 10 , 11 d = 6 d = 6 d = 6 ψ \psi ψ ω 5 ψ − 1 \omega_5\psi^{-1} ω 5 ψ − 1 5 + 2 6 5 + 2\sqrt{6} 5 + 2 6 2 , 3 2, 3 2 , 3 ψ \psi ψ ψ 2 \psi^2 ψ 2 1 1 1 ω 5 \omega_5 ω 5 ψ 2 \psi^2 ψ 2 1 1 1 ω 5 2 \omega_5^2 ω 5 2 ω 5 \omega_5 ω 5 ω 5 − 1 \omega_5^{-1} ω 5 − 1
For d = 10 d = 10 d = 10 Lemma 18.9 gives the quotient ω 3 \omega_3 ω 3 ω 3 ν \omega_3\nu ω 3 ν ν \nu ν G Q G_{\mathbb{Q}} G Q
For d = 11 d = 11 d = 11 C 4 C_4 C 4 C 4 × C 2 C_4 \times C_2 C 4 × C 2 ψ 2 \psi^2 ψ 2 ω 5 2 \omega_5^2 ω 5 2 ω 5 ± 1 \omega_5^{\pm1} ω 5 ± 1 G Q G_{\mathbb{Q}} G Q
All hypotheses of Proposition 13.2 hold.
The split-dyadic theorem
Put D = { 17 , 33 , 41 , 57 , 65 , 73 , 89 } \mathcal{D} = \{17,33,41,57,65,73,89\} D = { 17 , 33 , 41 , 57 , 65 , 73 , 89 } d d d K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) § 2.4 .
Theorem 29.1. Assume GRH for Dedekind zeta functions. Let K = Q ( d ) K = \mathbb{Q}(\sqrt{d}) K = Q ( d ) d ∈ D d \in\mathcal{D} d ∈ D p p p r ˉ : G K → GL 2 ( F p ) \bar{r} : G_K \to\operatorname{GL}_2(\mathbb{F}_p) r ˉ : G K → GL 2 ( F p )
Corollary 29.2. Theorem 29.1 holds without GRH for d = 17 d = 17 d = 17 d = 33 d = 33 d = 33 Re s > 31 / 32 \operatorname{Re} s > 31/32 Re s > 31/32 [15 Theorem 1.1] .
The general criterion is Theorem 10.1. The next two sections verify its arithmetic and residual hypotheses. The source theorems used for the terminal representation are already proved in the general part.
Arithmetic of the split-dyadic fields
Write w 0 , w 1 w_0, w_1 w 0 , w 1 K K K
Lemma 30.1. Let S 2 S_2 S 2 K K K G K → F ‾ 2 G_K \to\overline{\mathbb{F}}_2 G K → F 2
H 1 ( G K , S 2 , F 2 ) = ⟨ [ − 1 ] , [ ϵ ] , [ 2 ] , [ α ] ⟩ F 2 H^1(G_{K,S_2},\mathbb{F}_2) = \langle[-1], [\epsilon], [2], [\alpha]\rangle_{\mathbb{F}_2} H 1 ( G K , S 2 , F 2 ) = ⟨[ − 1 ] , [ ϵ ] , [ 2 ] , [ α ] ⟩ F 2 has dimension four, with the choices in Table 2 .
Table 2. Units, dyadic generators, and ordinary and narrow class numbers.
Proof. Minkowski’s bound is d / 2 \sqrt{d}/2 d /2
N ( 15 + 2 57 ) = − 3 , N ( 17 + 2 73 ) = − 3. N(15 + 2\sqrt{57}) = -3,\qquad N(17 + 2\sqrt{73}) = -3. N ( 15 + 2 57 ) = − 3 , N ( 17 + 2 73 ) = − 3. For d ≠ 65 d \ne65 d = 65 α \alpha α ± 2 \pm2 ± 2 ± 2 \pm2 ± 2 x 2 − 65 y 2 = ± 8 x^2 - 65y^2 = \pm8 x 2 − 65 y 2 = ± 8 h 65 = 2 h_{65} = 2 h 65 = 2 S 2 S_2 S 2
The unit squareclass group has dimension two. If N ϵ = − 1 N\epsilon=-1 N ϵ = − 1 ϵ \epsilon ϵ [ − 1 ] [-1] [ − 1 ] d = 33 , 57 d=33,57 d = 33 , 57 [ − 1 ] [-1] [ − 1 ] [ ϵ ] [\epsilon] [ ϵ ]
Kummer theory identifies the group in the lemma with O K , S 2 × / ( O K , S 2 × ) 2 O_{K,S_2}^{\times}/(O_{K,S_2}^{\times})^2 O K , S 2 × / ( O K , S 2 × ) 2 d ≠ 65 d\ne65 d = 65 α \alpha α ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 2 , 0 ) (2,0) ( 2 , 0 ) α \alpha α − 1 -1 − 1
Finally, a characteristic-two finite character has odd-order image. At either dyadic completion Q 2 \mathbb{Q}_2 Q 2
For i = 0 , 1 i=0,1 i = 0 , 1 V i ( K ) = V w i + ( K ) V_i(K)=V^+_{w_i}(K) V i ( K ) = V w i + ( K ) (1) . We verify its vanishing using the full quadratic Selmer group, including the ideal-class contribution.
Lemma 30.2 (Base scalar Selmer vanishing). For every d ∈ D d\in\mathcal{D} d ∈ D V i ( K ) = 0 V_i(K)=0 V i ( K ) = 0
Proof. Let H + H^+ H + H 1 ( G K , S 2 , F 2 ) H^1(G_{K,S_2},\mathbb{F}_2) H 1 ( G K , S 2 , F 2 ) ϵ \epsilon ϵ α \alpha α d > 0 \sqrt{d}>0 d > 0 d = 17 , 41 , 65 , 73 , 89 d=17,41,65,73,89 d = 17 , 41 , 65 , 73 , 89 − 1 , + 1 -1,+1 − 1 , + 1 d = 33 , 57 d=33,57 d = 33 , 57 + 1 , − 1 +1,-1 + 1 , − 1
H + = ⟨ [ 2 ] , [ β ] ⟩ , β = { α , d = 17 , 41 , 65 , 73 , 89 , ϵ , d = 33 , 57. H^+=\langle[2],[\beta]\rangle,\qquad
\beta=
\begin{cases}
\alpha, & d=17,41,65,73,89,\\
\epsilon, & d=33,57.
\end{cases} H + = ⟨[ 2 ] , [ β ]⟩ , β = { α , ϵ , d = 17 , 41 , 65 , 73 , 89 , d = 33 , 57. Index the dyadic places by w = ( 1 + d ) / 2 ≡ 0 , 1 ( m o d 2 ) w=(1+\sqrt{d})/2\equiv0,1\pmod2 w = ( 1 + d ) /2 ≡ 0 , 1 ( mod 2 )
Table 3.
These entries follow by Hensel lifting w 2 − w + ( 1 − d ) / 4 w^2-w+(1-d)/4 w 2 − w + ( 1 − d ) /4 { 1 , [ 2 ] } ⊂ Q 2 × / ( Q 2 × ) 2 \{1,[2]\}\subset\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2 { 1 , [ 2 ]} ⊂ Q 2 × / ( Q 2 × ) 2 H + H^+ H + w 1 − i w_{1-i} w 1 − i V i ( K ) V_i(K) V i ( K )
For d = 17 , 41 , 73 , 89 d=17,41,73,89 d = 17 , 41 , 73 , 89 Table 2 gives narrow class number one. The totally positive element α \alpha α 5 , 7 , 9 , 217 5,7,9,217 5 , 7 , 9 , 217 u + 2 u − 1 = Tr ( α ) u+2u^{-1}=\operatorname{Tr}(\alpha) u + 2 u − 1 = Tr ( α ) 7 , 5 , 3 , 3 7,5,3,3 7 , 5 , 3 , 3 [21 Lemma 9.1] .
Residual representations unramified outside two
For a finite Galois extension E / Q 2 E/\mathbb{Q}_2 E / Q 2 d ( E ) d(E) d ( E ) v 2 ( 2 ) = 1 v_2(2)=1 v 2 ( 2 ) = 1 E 2 = Q 2 ( − 1 , 2 , 5 ) E_2=\mathbb{Q}_2(\sqrt{-1},\sqrt{2},\sqrt{5}) E 2 = Q 2 ( − 1 , 2 , 5 )
Lemma 31.1. Let r : G Q 2 → SL 2 ( F q ) r:G_{\mathbb{Q}_2}\to\operatorname{SL}_2(\mathbb{F}_q) r : G Q 2 → SL 2 ( F q ) q = 2 a q=2^a q = 2 a E E E d ( E ) ≤ 2 d(E)\le2 d ( E ) ≤ 2 E ⊂ E 2 E\subset E_2 E ⊂ E 2 Q ≤ q Q\le q Q ≤ q E E E E 2 E_2 E 2 d ( E ) ≤ 2 ( 1 − 1 / Q ) d(E)\le2(1-1/Q) d ( E ) ≤ 2 ( 1 − 1/ Q ) E ∩ Q 2 ( − 1 , 5 ) = Q 2 E\cap\mathbb{Q}_2(\sqrt{-1},\sqrt{5})=\mathbb{Q}_2 E ∩ Q 2 ( − 1 , 5 ) = Q 2 S 3 S_3 S 3
Proof. If wild inertia is trivial, its tame generator has eigencharacters θ , θ − 1 \theta,\theta^{-1} θ , θ − 1 S 3 S_3 S 3 2 / 3 2/3 2/3
Otherwise the wild subgroup fixes a unique line, stable under the whole decomposition group. The diagonal characters are θ , θ − 1 \theta,\theta^{-1} θ , θ − 1 Z 2 × \mathbb{Z}_2^\times Z 2 × θ = 1 \theta=1 θ = 1 E ⊂ E 2 E\subset E_2 E ⊂ E 2 E 2 E_2 E 2 0 , 2 , 2 , 3 , 3 , 3 , 3 0,2,2,3,3,3,3 0 , 2 , 2 , 3 , 3 , 3 , 3 d ( E 2 ) = 2 d(E_2)=2 d ( E 2 ) = 2 θ ≠ 1 \theta\ne1 θ = 1 P P P θ ( Frob 2 ) 2 \theta(\operatorname{Frob}_2)^2 θ ( Frob 2 ) 2 [ P , Frob 2 ] = P [P,\operatorname{Frob}_2]=P [ P , Frob 2 ] = P P P P E P / Q 2 E^P/\mathbb{Q}_2 E P / Q 2 Z / 2 Z \mathbb{Z}/2\mathbb{Z} Z /2 Z
Lemma 31.2. Put H = K ( − 1 , ϵ ) H=K(\sqrt{-1},\sqrt{\epsilon}) H = K ( − 1 , ϵ ) K K K Q \mathbb{Q} Q Q 2 ( − 1 , 5 ) \mathbb{Q}_2(\sqrt{-1},\sqrt{5}) Q 2 ( − 1 , 5 ) L / K L/K L / K SL 2 ( F q ) \operatorname{SL}_2(\mathbb{F}_q) SL 2 ( F q ) q ≥ 8 q\ge8 q ≥ 8 M = L H M=LH M = L H k k k E 2 E_2 E 2
[ M : Q ] = 8 q ( q 2 − 1 ) , rd ( M ) ≤ d 2 2 − 1 / q + k / ( 2 q ) . (57) [M:\mathbb{Q}]=8q(q^2-1),\qquad\operatorname{rd}(M)\le\sqrt{d}\,2^{2-1/q+k/(2q)}. \tag*{(57)} [ M : Q ] = 8 q ( q 2 − 1 ) , rd ( M ) ≤ d 2 2 − 1/ q + k / ( 2 q ) . (57) Each of these k k k [ M : Q ] / 16 [M:\mathbb{Q}]/16 [ M : Q ] /16
Proof. The classes [ − 1 ] , [ ϵ ] [-1],[\epsilon] [ − 1 ] , [ ϵ ] ϵ \sqrt{\epsilon} ϵ 1 / ϵ 1/\sqrt{\epsilon} 1/ ϵ − 1 / ϵ \sqrt{-1}/\sqrt{\epsilon} − 1 / ϵ H / Q H/\mathbb{Q} H / Q Q 2 \mathbb{Q}_2 Q 2 [ − 1 ] , [ 5 ] [-1],[5] [ − 1 ] , [ 5 ] L ∩ H = K L\cap H=K L ∩ H = K
At a dyadic place with E ⊂ E 2 E\subset E_2 E ⊂ E 2 E 2 E_2 E 2
wild case of Lemma 31.1 , adjoining the ramified quadratic part of H H H P P P P × C 2 P \times C_{2} P × C 2 2 ( 1 − 1 / ( 2 Q ) ) ≤ 2 − 1 / q 2(1 - 1/(2Q)) \le2 - 1/q 2 ( 1 − 1/ ( 2 Q )) ≤ 2 − 1/ q ( 1 − 1 / 6 ) + ( 1 − 1 / 2 ) = 4 / 3 < 2 − 1 / q (1 - 1/6) + (1 - 1/2) = 4/3 < 2 - 1/q ( 1 − 1/6 ) + ( 1 − 1/2 ) = 4/3 < 2 − 1/ q
At a counted place, residue degree two gives at least [ M : Q ] / 16 [M:\mathbb{Q}]/16 [ M : Q ] /16 [ M : Q ] / 8 [M:\mathbb{Q}]/8 [ M : Q ] /8
Use the functions F b F_b F b J b J_b J b B b B_b B b C b C_b C b § 9.1 . Put a = π / ( 2 b ) a = \pi/(2b) a = π / ( 2 b )
log L ( n , b , k ) = γ + log ( 8 π ) − ∑ j = 0 3 { 1 j + 1 / 2 − J b ( j + 1 / 2 ) } − 2 a 2 49 − C b n + k log 4 8 ∑ m ≥ 1 4 − m / 2 F b ( m log 4 ) . \log\mathcal{L}(n,b,k) = \gamma+ \log(8\pi) - \sum_{j=0}^{3} \left\{ \frac{1}{j+1/2} - J_b(j+1/2) \right\} - \frac{2a^2}{49} - \frac{C_b}{n} + \frac{k\log4}{8} \sum_{m\ge1} 4^{-m/2}F_b(m\log4). log L ( n , b , k ) = γ + log ( 8 π ) − j = 0 ∑ 3 { j + 1/2 1 − J b ( j + 1/2 ) } − 49 2 a 2 − n C b + 8 k log 4 m ≥ 1 ∑ 4 − m /2 F b ( m log 4 ) . The tail estimate in that theorem makes this a lower bound for the root discriminant, under GRH, when the indicated norm-four Euler contributions are present. It increases with n n n b , k b,k b , k B ( G ) , C ( G ) B(G), C(G) B ( G ) , C ( G ) G G G
Theorem 31.3. Assume G R H GRH GR H d ∈ D d \in\mathcal{D} d ∈ D G K G_K G K
Proof. Twist by the inverse square root of the odd-order determinant to make the image determinant one. Linear and projective kernels then coincide. For a solvable irreducible image use [21 Proposition 10.3] , since K / Q K/\mathbb{Q} K / Q SL 2 ( F 4 ) \operatorname{SL}_{2}(\mathbb{F}_{4}) SL 2 ( F 4 ) [3 Theorem 1.2(1)] and its projective-to-linear modularity statement. Otherwise Dickson’s classification gives SL 2 ( F q ) \operatorname{SL}_{2}(\mathbb{F}_{q}) SL 2 ( F q ) q ≥ 8 q \ge8 q ≥ 8
For M M M Lemma 31.2 , the following strict comparisons hold with n = 8 q ( q 2 − 1 ) n = 8q(q^{2} - 1) n = 8 q ( q 2 − 1 ) d ≤ 89 d \le89 d ≤ 89
Table 17.
The last row drops Euler terms and uses n ≥ 261888 n \ge261888 n ≥ 261888 rd ( M ) ≤ 4 89 \operatorname{rd}(M) \le4\sqrt{89} rd ( M ) ≤ 4 89 k k k log 4 8 ∑ m 4 − m / 2 F b ( m log 4 ) \frac{\log4}{8}\sum_m 4^{-m/2}F_b(m\log4) 8 l o g 4 ∑ m 4 − m /2 F b ( m log 4 ) log 2 / ( 2 q ) \log2/(2q) log 2/ ( 2 q ) F b ( log 4 ) > 1 / 2 F_b(\log4) > 1/2 F b ( log 4 ) > 1/2 q = 8 q = 8 q = 8 k = 0 k = 0 k = 0 [49 split_dyadic] .
Let σ \sigma σ K / Q K/\mathbb{Q} K / Q L σ ≠ L L^\sigma\ne L L σ = L [ L L σ : Q ] = 2 ⋅ 504 2 = 508032 [LL^\sigma:\mathbb{Q}] = 2 \cdot504^{2} = 508032 [ L L σ : Q ] = 2 ⋅ 50 4 2 = 508032 k = 0 k = 0 k = 0 rd ( L L σ ) ≤ 4 89 \operatorname{rd}(LL^\sigma) \le4\sqrt{89} rd ( L L σ ) ≤ 4 89
Hence L / Q L/\mathbb{Q} L / Q Out ( SL 2 ( F 8 ) ) = C 3 \operatorname{Out}(\operatorname{SL}_{2}(\mathbb{F}_{8})) = C_{3} Out ( SL 2 ( F 8 )) = C 3 σ \sigma σ Gal ( L / Q ) = SL 2 ( F 8 ) × C 2 \operatorname{Gal}(L/\mathbb{Q}) = \operatorname{SL}_{2}(\mathbb{F}_{8}) \times C_{2} Gal ( L / Q ) = SL 2 ( F 8 ) × C 2 G Q G_{\mathbb{Q}} G Q [40 , 41 , 43 ] makes that extension modular: in characteristic two oddness is the determinant condition − 1 = 1 -1 = 1 − 1 = 1 K K K
Comparisons without GRH
For an even test function G G G (43) , the zero terms associated to Φ G ( s ) = ∫ R G ( x ) e ( s − 1 / 2 ) x d x \Phi_G(s)=\int_{\mathbb{R}}G(x)e^{(s-1/2)x}\,dx Φ G ( s ) = ∫ R G ( x ) e ( s − 1/2 ) x d x [47 equation (2.3)] . If the zeros lie in ∣ Re s − 1 / 2 ∣ ≤ a |\operatorname{Re}s-1/2|\le a ∣ Re s − 1/2∣ ≤ a G = F b / cosh ( a x ) G=F_b/\cosh(ax) G = F b / cosh ( a x ) ∣ h ∣ < a |h|<a ∣ h ∣ < a cosh ( h x ) / cosh ( a x ) \cosh(hx)/\cosh(ax) cosh ( h x ) / cosh ( a x )
π a cos ( π h / ( 2 a ) ) cosh ( π t / ( 2 a ) ) sinh 2 ( π t / ( 2 a ) ) + cos 2 ( π h / ( 2 a ) ) ≥ 0. \frac{\pi}{a}\frac{\cos(\pi h/(2a))\cosh(\pi t/(2a))}{\sinh^2(\pi t/(2a))+\cos^2(\pi h/(2a))}\ge0. a π sinh 2 ( π t / ( 2 a )) + cos 2 ( π h / ( 2 a )) cos ( π h / ( 2 a )) cosh ( π t / ( 2 a )) ≥ 0. The endpoints follow by continuity, and multiplication by the positive definite function F b F_b F b B ( G ) , C ( G ) B(G),C(G) B ( G ) , C ( G ) (43) .
First, a totally real kernel field with group SL 2 ( F q ) \operatorname{SL}_2(\mathbb{F}_q) SL 2 ( F q ) q ≥ 8 q\ge8 q ≥ 8 (44) , with b = 24 b=24 b = 24 n ≥ 1008 n\ge1008 n ≥ 1008 rd ( L ) > 55.42 \operatorname{rd}(L)>55.42 rd ( L ) > 55.42 rd ( L ) ≤ 4 89 < 37.736 \operatorname{rd}(L)\le4\sqrt{89}<37.736 rd ( L ) ≤ 4 89 < 37.736
For d = 17 d=17 d = 17 1008 1008 1008 rd ( L ) ≤ 4 17 < 16.50 \operatorname{rd}(L)\le4\sqrt{17}<16.50 rd ( L ) ≤ 4 17 < 16.50 (44) with b = 10 b=10 b = 10 rd ( L ) > 20.57 \operatorname{rd}(L)>20.57 rd ( L ) > 20.57
For d = 33 d=33 d = 33 b = 24 b=24 b = 24 k ≥ 1 k\ge1 k ≥ 1 4032 4032 4032 (44) gives rd ( M ) > 23.31 \operatorname{rd}(M)>23.31 rd ( M ) > 23.31 (57) gives rd ( M ) ≤ 4 33 < 22.979 \operatorname{rd}(M)\le4\sqrt{33}<22.979 rd ( M ) ≤ 4 33 < 22.979 q = 8 q=8 q = 8 k = 0 k=0 k = 0 rd ( M ) > 21.75 > 21.07115 \operatorname{rd}(M)>21.75>21.07115 rd ( M ) > 21.75 > 21.07115 (57) . For k = 0 k=0 k = 0 L L L [ L : Q ] = 2 q ( q 2 − 1 ) [L:\mathbb{Q}]=2q(q^2-1) [ L : Q ] = 2 q ( q 2 − 1 ) rd ( L ) ≤ 33 2 2 − 2 / q \operatorname{rd}(L)\le\sqrt{33}\,2^{2-2/q} rd ( L ) ≤ 33 2 2 − 2/ q
Table 18 (PDF p. 148).
Only k = 0 k=0 k = 0 q ≥ 64 q\ge64 q ≥ 64
Assume now the additional zero-free statement in Corollary 29.2. The functional equation puts the zeros of every relevant zeta function in the strip with a = 15 / 32 a=15/32 a = 15/32 G = F b / cosh ( 15 x / 32 ) G=F_b/\cosh(15x/32) G = F b / cosh ( 15 x /32 )
B ( G ) < 13 20 + 3 π 2 4 b 2 , C ( G ) ≤ 4 J b ( − 1 / 32 ) . B(G)<\frac{13}{20}+\frac{3\pi^2}{4b^2},\qquad C(G)\le4J_b(-1/32). B ( G ) < 20 13 + 4 b 2 3 π 2 , C ( G ) ≤ 4 J b ( − 1/32 ) . Here the first bound follows from B ( G ) ≤ B ∞ ( a ) + π 2 Q ( a ) / ( 8 b 2 ) B(G)\le B_\infty(a)+\pi^2Q(a)/(8b^2) B ( G ) ≤ B ∞ ( a ) + π 2 Q ( a ) / ( 8 b 2 )
B ∞ ( a ) = ∫ 0 ∞ 1 − sech ( a x ) 2 sinh ( x / 2 ) d x < 13 / 20 , Q ( a ) = ∫ 0 ∞ x 2 2 sinh ( x / 2 ) cosh ( a x ) d x < 6. B_\infty(a)=\int_0^\infty\frac{1-\operatorname{sech}(ax)}{2\sinh(x/2)}\,dx<13/20,\qquad Q(a)=\int_0^\infty\frac{x^2}{2\sinh(x/2)\cosh(ax)}\,dx<6. B ∞ ( a ) = ∫ 0 ∞ 2 sinh ( x /2 ) 1 − sech ( a x ) d x < 13/20 , Q ( a ) = ∫ 0 ∞ 2 sinh ( x /2 ) cosh ( a x ) x 2 d x < 6. For completeness these constants have finite rational bounds: expand B ∞ ( a ) = ∑ j ≥ 0 ( − 1 ) j S j B_\infty(a)=\sum_{j\ge0}(-1)^jS_j B ∞ ( a ) = ∑ j ≥ 0 ( − 1 ) j S j S j = ∑ m ≥ 0 2 a 2 / [ s ( s + a ) ( s + 2 a ) ] S_j=\sum_{m\ge0}2a^2/[s(s+a)(s+2a)] S j = ∑ m ≥ 0 2 a 2 / [ s ( s + a ) ( s + 2 a )] s = m + 1 / 2 + 2 j a s=m+1/2+2ja s = m + 1/2 + 2 j a j = 80 , 81 j=80,81 j = 80 , 81 a 2 / ( 499.5 + 2 j a ) 2 a^2/(499.5+2ja)^2 a 2 / ( 499.5 + 2 j a ) 2 0.64911 < B ∞ ( 15 / 32 ) < 0.64922 0.64911<B_\infty(15/32)<0.64922 0.64911 < B ∞ ( 15/32 ) < 0.64922 Q ( 15 / 32 ) ≤ 4 ∑ j = 0 ∞ ( j + 31 / 32 ) − 3 + 2 / ( 3 + 31 / 32 ) 2 < 6 Q(15/32)\le4\sum_{j=0}^\infty(j+31/32)^{-3}+2/(3+31/32)^2<6 Q ( 15/32 ) ≤ 4 ∑ j = 0 ∞ ( j + 31/32 ) − 3 + 2/ ( 3 + 31/32 ) 2 < 6 q ≥ 64 q\ge64 q ≥ 64 b = 30 b=30 b = 30 n ≥ 32640 n\ge32640 n ≥ 32640
rd ( M ) ≥ 8 π e γ exp ( − 13 20 − 3 π 2 3600 − 4 J 30 ( − 1 / 32 ) 32640 ) > 23.06000 > 4 33 . \operatorname{rd}(M)\ge8\pi e^\gamma\exp\left(-\frac{13}{20}-\frac{3\pi^2}{3600}-\frac{4J_{30}(-1/32)}{32640}\right)>23.06000>4\sqrt{33}. rd ( M ) ≥ 8 π e γ exp ( − 20 13 − 3600 3 π 2 − 32640 4 J 30 ( − 1/32 ) ) > 23.06000 > 4 33 . This proves the residual assertion needed for the second part of Corollary 29.2. All numerical inequalities in this section follow from the displayed finite expressions and tail bounds by directed interval arithmetic. The exact calculation is supplied in [49 split_dyadic] .
Proof of Theorem 29.1 and Corollary 29.2. Lemma 30.1 gives the two-power ordinary class number, and Lemma 30.2 verifies (SV). Theorem 31.3 supplies the irreducible residual base. Apply Theorem 10.1. For d = 17 d = 17 d = 17 d = 33 d = 33 d = 33 □ \square □
Proof of the main theorem
Proof of Theorem 1.1 . Theorem 27.1 proves the cases d = 7 , 13 , 37 d = 7, 13, 37 d = 7 , 13 , 37 Theorem 13.1 proves d = 2 , 3 , 6 , 10 , 11 d = 2, 3, 6, 10, 11 d = 2 , 3 , 6 , 10 , 11 § 7.2 and 21 give d = 5 , 21 d = 5, 21 d = 5 , 21 Theorem 29.1 . Every congruence returns modularity to the original residual representation, without a restriction on its conductor. □ \square □
Depth-four normal ray calculation
We give the matrices used in Proposition 18.10 . The fields K K K A A A W 0 W_0 W 0 S = ( 0 1 1 0 ) S=\begin{pmatrix}0&1\\1&0\end{pmatrix} S = ( 0 1 1 0 ) U = ( 0 1 1 1 ) U=\begin{pmatrix}0&1\\1&1\end{pmatrix} U = ( 0 1 1 1 ) W 0 W_0 W 0 [ q + ] [q_+] [ q + ] v = ( 1 , 0 , 0 , 1 , 1 , 0 , 1 ) t v=(1,0,0,1,1,0,1)^{\mathsf{t}} v = ( 1 , 0 , 0 , 1 , 1 , 0 , 1 ) t S , U S,U S , U
C 2 = ( 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 ) , C 3 = ( 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 1 1 1 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 ) . C_2=\begin{pmatrix}
1&0&0&0&0&0&0\\
0&1&0&0&0&0&0\\
0&1&0&0&0&1&0\\
0&1&1&1&0&1&0\\
0&0&0&0&1&0&0\\
0&1&1&0&0&0&0\\
0&0&1&1&1&0&1
\end{pmatrix},
\qquad
C_3=\begin{pmatrix}
1&0&0&0&0&0&0\\
0&1&0&0&0&0&0\\
1&0&1&0&0&1&1\\
1&1&1&0&0&1&0\\
0&0&0&0&1&0&0\\
0&0&0&1&0&0&1\\
0&0&1&1&1&0&1
\end{pmatrix}. C 2 = 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 , C 3 = 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 1 0 0 1 1 . The solution space to X v = 0 Xv=0 X v = 0 X C 2 = S X XC_2=SX X C 2 = S X X C 3 = U X XC_3=UX X C 3 = U X
X 1 = ( 1 1 1 1 0 1 0 1 0 0 1 0 0 0 ) , X 2 = ( 1 0 1 1 1 1 1 1 1 1 0 0 0 1 ) . X_1=\begin{pmatrix}1&1&1&1&0&1&0\\1&0&0&1&0&0&0\end{pmatrix},
\qquad
X_2=\begin{pmatrix}1&0&1&1&1&1&1\\1&1&1&0&0&0&1\end{pmatrix}. X 1 = ( 1 1 1 0 1 0 1 1 0 0 1 0 0 0 ) , X 2 = ( 1 1 0 1 1 1 1 0 1 0 1 0 1 1 ) . Chinese remainder lifts of local units, chosen to be one at the other dyadic prime, have binary ray logarithm columns
g = ( 0 , 1 , 1 , 0 , 0 , 0 , 0 ) t , h = ( 0 , 0 , 0 , 1 , 0 , 0 , 0 ) t , j = ( 0 , 0 , 0 , 1 , 0 , 1 , 1 ) t . g=(0,1,1,0,0,0,0)^{\mathsf{t}},\qquad
h=(0,0,0,1,0,0,0)^{\mathsf{t}},\qquad
j=(0,0,0,1,0,1,1)^{\mathsf{t}}. g = ( 0 , 1 , 1 , 0 , 0 , 0 , 0 ) t , h = ( 0 , 0 , 0 , 1 , 0 , 0 , 0 ) t , j = ( 0 , 0 , 0 , 1 , 0 , 1 , 1 ) t . The first prime supplies g , h g,h g , h g , j g,j g , j α 2 + α + 1 = 0 \alpha^2+\alpha+1=0 α 2 + α + 1 = 0 U U U ( 1 , α ) (1,\alpha) ( 1 , α )
( 0 1 + α 1 1 ) , ( 0 α 1 1 + α ) . \begin{pmatrix}0&1+\alpha\\1&1\end{pmatrix},
\qquad
\begin{pmatrix}0&\alpha\\1&1+\alpha\end{pmatrix}. ( 0 1 1 + α 1 ) , ( 0 1 α 1 + α ) . Both determinants are nonzero, and the conjugate eigenrow has the same property. Hence vanishing on local units in either one of the two C 3 C_3 C 3 [49 ramified/d6_d10_d11/d10] .
Auxiliary fields over the field of discriminant twenty-one
The unit assertions involve only a number field of degree twelve. One defining polynomial for A A A
P ( X ) = X 12 − 3 X 11 + 8 X 10 − 9 X 9 + 9 X 8 + 6 X 7 + X 6 + 6 X 5 + 9 X 4 − 9 X 3 + 8 X 2 − 3 X + 1. P(X)=X^{12}-3X^{11}+8X^{10}-9X^9+9X^8+6X^7+X^6
+6X^5+9X^4-9X^3+8X^2-3X+1. P ( X ) = X 12 − 3 X 11 + 8 X 10 − 9 X 9 + 9 X 8 + 6 X 7 + X 6 + 6 X 5 + 9 X 4 − 9 X 3 + 8 X 2 − 3 X + 1. The field defined by P P P 216 5 8 216^5 8 21 6 5 8 ( 0 , 6 ) (0,6) ( 0 , 6 ) u u u z z z ± 1 \pm1 ± 1 nfinit, exact field arithmetic, Galois conjugates and local-power tests; no computation of a full unit group or class group is used in the proof. The exact calculation is supplied in [49 d21] .
The arithmetic and interval certificates are:
check_small_unit_formulas.gp: the two-orbit unit identities and local squareclass kernels in Lemma 21.11;
verify_split5_local.py: the two compact-type Brauer-character identities, calculated in Z [ ζ 24 ] \mathbb{Z}[\zeta_{24}] Z [ ζ 24 ]
certify_K21_SC_SC.py: real-ball evaluations of (39), with each comparison asserted only when its lower and upper balls are disjoint;
K21_SC_SC_arb_certificate.json: the full-precision interval enclosures for Table 1 .
The smallest gap, in the distinct-conjugate row, is greater than 0.00974417509. The analytic calculation certifies the displayed inequalities; the group and local hypotheses are proved in the preceding sections. The exact calculation is supplied in [49 d21/inherited_python, d21/split5] .
Normal Selmer calculations over Q(sqrt(21))
The following scripts in [49 d21] print certificates for the normal Selmer calculations. All class and unit calculations have absolute degree at most twelve. The accompanying code includes output records, the certificate s3_mixed_full_sunit_portable.gp, and the diagnostic-checking runner run_gp_checked.py. The certificate uses relative discriminants and Hilbert symbols to test the stated invariant subspace conditions.
Table 1 (PDF p. 151).
The displayed masks in Section 21.3.2 refer to the chosen unit and S S S C 3 C_3 C 3 S 5 S_5 S 5 F 2 \mathbb{F}_2 F 2 A 1 A_1 A 1 bnfcertify=1, the full normal rank and every retained mask; a zero process exit status alone does not certify success in PARI/GP. The exact calculation is supplied in [49 d21] .
The nonsolvable branch uses the discriminant comparisons above and the certify_K21_SC_SC.py and K21_SC_SC_arb_certificate.json. It is used only for the nonsolvable branch of the mixed theorem; Theorem 21.15 has no analytic hypothesis. The exact calculation is supplied in [49 d21/split5] .
AI disclosure
During the preparation of this article, the authors used ChatGPT 6-Astra and Sol 6.1 as an assistant in mathematical research and manuscript preparation. Its use included exploring proof strategies, checking mathematical arguments, investigating possible simplifications, assisting with computations and verification scripts, checking references, and drafting and revising portions of the exposition. The authors take full responsibility for the mathematical content, proofs, computations, and conclusions of the article.
