Introduction
A classical theorem of Schur [43 ] states that the exponential Taylor polynomial
E n ( X ) = ∑ j = 0 n X j j ! E_n(X) = \sum_{j=0}^{n} \frac{X^j}{j!} E n ( X ) = j = 0 ∑ n j ! X j is irreducible over Q \mathbb{Q} Q n ≥ 1 n \ge1 n ≥ 1 A n A_n A n 4 ∣ n 4 \mid n 4 ∣ n S n S_n S n [44 ] (see also [10 ] ).
In fact, Schur family occurs among the generalized Laguerre polynomials. More precisely, we use the monic normalization
H n , s ( X ) = ∑ k = 0 n ( n k ) ( s ) k X n − k , ( s ) 0 = 1 , ( s ) k = s ( s + 1 ) ⋯ ( s + k − 1 ) . (1) H_{n,s}(X) = \sum_{k=0}^{n} \binom{n}{k} (s)_k X^{n-k}, \qquad(s)_0 = 1, \qquad(s)_k = s(s+1)\cdots(s+k-1). \tag*{(1)} H n , s ( X ) = k = 0 ∑ n ( k n ) ( s ) k X n − k , ( s ) 0 = 1 , ( s ) k = s ( s + 1 ) ⋯ ( s + k − 1 ) . (1) Here n n n s s s H 0 , s = 1 H_{0,s} = 1 H 0 , s = 1 n , s n,s n , s H n , s ∈ Z [ X ] H_{n,s} \in\mathbb{Z}[X] H n , s ∈ Z [ X ] [26 ] ):
y n ( X ) = ∑ j = 0 n ( n + j ) ! ( n − j ) ! j ! ( X 2 ) j . y_n(X) = \sum_{j=0}^{n} \frac{(n+j)!}{(n-j)!\,j!}\left(\frac{X}{2}\right)^j. y n ( X ) = j = 0 ∑ n ( n − j )! j ! ( n + j )! ( 2 X ) j . Filaseta proved y n y_n y n Q \mathbb{Q} Q n n n [21 ] . After that, together with Trifonov, they proved the result for every n n n [24 ] .
Note that the two classical families are the specializations
H n , 1 ( X ) = n ! E n ( X ) , H n , n + 1 ( X ) = X n y n ( 2 / X ) . (2) H_{n,1}(X) = n!E_n(X), \qquad H_{n,n+1}(X) = X^n y_n(2/X). \tag*{(2)} H n , 1 ( X ) = n ! E n ( X ) , H n , n + 1 ( X ) = X n y n ( 2/ X ) . (2) The second identity uses the usual normalization of the Bessel polynomial [40 (18.34.1)] . Multiplication by a nonzero rational constant and the reciprocal substitution in (2) preserve irreducibility and splitting fields.
Hajir proposed extending these arithmetic properties to every positive integral pair ( n , s ) (n,s) ( n , s ) [29 Conjectures 1.1–1.2] . His first conjecture asserts that every H n , s H_{n,s} H n , s Q \mathbb{Q} Q H n , s H_{n,s} H n , s A n A_n A n s s s
Uniform arithmetic
For positive integers n , s n,s n , s G n , s = Gal ( H n , s / Q ) G_{n,s}=\operatorname{Gal}(H_{n,s}/\mathbb{Q}) G n , s = Gal ( H n , s / Q ) H n , s H_{n,s} H n , s Q \mathbb{Q} Q
Theorem 1.1. There is an effectively computable absolute integer N N N
n ≥ N , s ∈ Z ≥ 1 ⟹ H n , s is irreducible over Q and A n ≤ G n , s . n \ge N,\qquad s \in\mathbb{Z}_{\ge1}\quad\Longrightarrow\quad H_{n,s}\text{ is irreducible over }\mathbb{Q}\quad\text{and}\quad A_n\le G_{n,s}. n ≥ N , s ∈ Z ≥ 1 ⟹ H n , s is irreducible over Q and A n ≤ G n , s . The bound is uniform in the parameter: s s s n n n s = s ( n ) s=s(n) s = s ( n ) s s s
Corollary 1.2. The set of positive integral pairs ( n , s ) (n,s) ( n , s ) H n , s H_{n,s} H n , s A n ≰ G n , s A_n\not\le G_{n,s} A n ≤ G n , s
For reducibility, the fixed-degree bound is explicit:
H n , s reducible , n ≥ 2 ⟹ s < ⌊ n 2 4 ⌋ n . (3) H_{n,s}\text{ reducible},\qquad n\ge2\quad\Longrightarrow\quad s<\left\lfloor\frac{n^2}{4}\right\rfloor^n. \tag*{(3)} H n , s reducible , n ≥ 2 ⟹ s < ⌊ 4 n 2 ⌋ n . (3) For irreducible specializations with smaller Galois group, a decomposition-group criterion at infinity permits an effective application of Runge’s theorem. Combining these fixed-degree results with Theorem 1.1 proves the corollary in Section 5. The exceptional set has not been enumerated, and we do not prove that it is empty. Thus the result gives uniform eventual forms of Hajir’s conjectures and an effective finite reduction of their all-degree forms.
Earlier work controls complementary parts of the parameter plane. Hajir proved both conclusions for each fixed s s s n n n [29 Theorem 1.3] . Jindal–Laishram–Sarma gave the explicit sufficient condition
r = s − 1 ≥ 2 , n > exp ( r ( 1 + 1.2762 log r ) ) . r=s-1\ge2,\qquad n>\exp\left(r\left(1+\frac{1.2762}{\log r}\right)\right). r = s − 1 ≥ 2 , n > exp ( r ( 1 + log r 1.2762 ) ) . and proved both conclusions in every degree for 1 ≤ s ≤ 61 1\le s\le61 1 ≤ s ≤ 61 [32 Theorems 1.1–1.2] . The published theorem of Jindal–Nair–Shorey extends that interval to 1 ≤ s ≤ 93 1\le s\le93 1 ≤ s ≤ 93 94 ≤ s ≤ 101 94\le s\le101 94 ≤ s ≤ 101 [33 ] . Jindal–Laishram studied the neighboring Bessel lines n ≤ s ≤ n + 4 n\le s\le n+4 n ≤ s ≤ n + 4 [31 ] . In the other direction, Hajir–Wong proved that for each fixed n ≥ 5 n\ge5 n ≥ 5 [30 Theorem 1(a)] ; the finiteness there is ineffective [29 Corollary 6.2 and the following remark] . Finiteness on each fixed row and column does not exclude infinitely many exceptional pairs with both coordinates unbounded. Theorem 1.1 excludes such pairs.
The fixed-parameter results for L n ( α ) L_n^{(\alpha)} L n ( α )
H n , s ( X ) = ( − 1 ) n n ! L n ( − n − s ) ( X ) . (4) H_{n,s}(X)=(-1)^n n!L_n^{(-n-s)}(X). \tag*{(4)} H n , s ( X ) = ( − 1 ) n n ! L n ( − n − s ) ( X ) . (4) with α ∈ Q ∖ Z < 0 \alpha\in\mathbb{Q}\setminus\mathbb{Z}_{<0} α ∈ Q ∖ Z < 0 [22 Theorem 1] [28 Theorem 1.1] . Here α = − n − s \alpha=-n-s α = − n − s S n S_n S n [24 ] [27 Chapter 12, Theorem 1] .
Root geometry and the proof
A root theorem valid for every n ≥ 2 n \ge2 n ≥ 2 n ≥ 2 n \ge2 n ≥ 2 H n , s H_{n,s} H n , s s > 0 s > 0 s > 0 q ( s ) q(s) q ( s ) [6 ] . The next theorem also describes their motion.
Theorem 1.3. For every such branch there is a unique finite, absolutely continuous, nonnegative measure μ \mu μ [ 0 , n − 1 ] [0,n-1] [ 0 , n − 1 ]
q ( s ) = s + ∫ 0 n − 1 s s + t d μ ( t ) ( s > 0 ) . q(s)=s+\int_{0}^{n-1}\frac{s}{s+t}\,d\mu(t) \qquad(s>0). q ( s ) = s + ∫ 0 n − 1 s + t s d μ ( t ) ( s > 0 ) . If ξ ≥ 0 \xi\ge0 ξ ≥ 0 α ( s ) = − s + i ξ s + O ( 1 ) \alpha(s)=-s+i\xi\sqrt{s}+O(1) α ( s ) = − s + i ξ s + O ( 1 ) s → + ∞ s \to+\infty s → + ∞ ξ = 0 \xi=0 ξ = 0
μ ( [ 0 , n − 1 ] ) = n − 1 3 + ξ 2 6 . \mu([0,n-1])=\frac{n-1}{3}+\frac{\xi^2}{6}. μ ([ 0 , n − 1 ]) = 3 n − 1 + 6 ξ 2 . For every integer r ≥ 1 r \ge1 r ≥ 1 s > 0 s>0 s > 0
( − 1 ) r − 1 d r d s r ( q ( s ) − s ) > 0. (-1)^{r-1}\frac{d^r}{ds^r}(q(s)-s)>0. ( − 1 ) r − 1 d s r d r ( q ( s ) − s ) > 0. Here the Hermite polynomials have the probabilists’ normalization, specified in (23) . Theorem 2.5 , Corollary 2.6, and Proposition A.2 prove the statement and identify the exact support of each measure. In particular, q − s q-s q − s [9 ] ; the present representation concerns complex Laguerre roots in the range α < − n \alpha<-n α < − n
Outline of the proof. The proof separates three possibilities: a rational factor, an imprimitive action on the roots, and a primitive Galois group not containing A n A_n A n
First, a tridiagonal realization and Fan’s variational principle [20 ] give the strict trace estimate
∣ Re ∑ i = 1 d ( α i + s ) ∣ < d ( n − d ) ( 0 < d < n ) . (5) \left|\operatorname{Re}\sum_{i=1}^{d}(\alpha_i+s)\right|<d(n-d)\qquad(0<d<n). \tag*{(5)} Re i = 1 ∑ d ( α i + s ) < d ( n − d ) ( 0 < d < n ) . (5) The centered trace of a rational factor is therefore an integer in a bounded interval independent of s s s s = 0 , − 1 , − 2 , − 3 s=0,-1,-2,-3 s = 0 , − 1 , − 2 , − 3 n ≥ 2 n \ge2 n ≥ 2
rad ( ( s + i ) ( s + j ) ) < n 6 for some 0 ≤ i < j ≤ 3. \operatorname{rad}((s+i)(s+j))<n^6\qquad\text{for some }0\le i<j\le3. rad (( s + i ) ( s + j )) < n 6 for some 0 ≤ i < j ≤ 3. Matveev’s theorem [38 ] then gives log s < n ε \log s<n^\varepsilon log s < n ε ε > 0 \varepsilon>0 ε > 0 n n n Theorem 2.12 isolates this argument for general monic families; Theorem 2.15 gives the all-degree bound, including rational parameters.
Second, we attach to a factor or an invariant partition a Galois-invariant positive semidefinite matrix Q Q Q Q 1 = 0 Q\mathbf{1}=0 Q 1 = 0 s s s
For the Galois group, local permutations first exclude primitive groups not containing A n A_n A n [7 ] ; this is the only input to the main theorem that uses the classification of finite simple groups.
Consequences and related families
The exponential Padé table. The same question has a natural formulation in rational approximation. For a , b ≥ 0 a,b \ge0 a , b ≥ 0 A a , b / B a , b A_{a,b}/B_{a,b} A a , b / B a , b [ a / b ] [a/b] [ a / b ] e X e^X e X A a , b ( 0 ) = B a , b ( 0 ) = 1 A_{a,b}(0)=B_{a,b}(0)=1 A a , b ( 0 ) = B a , b ( 0 ) = 1
deg A a , b = a , deg B a , b = b , B a , b ( X ) e X − A a , b ( X ) = O ( X a + b + 1 ) . \deg A_{a,b}=a,\qquad\deg B_{a,b}=b,\qquad B_{a,b}(X)e^X-A_{a,b}(X)=O(X^{a+b+1}). deg A a , b = a , deg B a , b = b , B a , b ( X ) e X − A a , b ( X ) = O ( X a + b + 1 ) . Its numerator and denominator are
A a , b ( X ) = b ! ( a + b ) ! H a , b + 1 ( X ) , B a , b ( X ) = a ! ( a + b ) ! H b , a + 1 ( − X ) . (6) A_{a,b}(X)=\frac{b!}{(a+b)!}H_{a,b+1}(X),\qquad B_{a,b}(X)=\frac{a!}{(a+b)!}H_{b,a+1}(-X). \tag*{(6)} A a , b ( X ) = ( a + b )! b ! H a , b + 1 ( X ) , B a , b ( X ) = ( a + b )! a ! H b , a + 1 ( − X ) . (6) Cullinan–Scheel used these identities to study diagonal approximants and special cases of irreducibility [11 ] . The boundary b = 0 b=0 b = 0 a = b a=b a = b
Corollary 1.4. For the absolute constant N N N Theorem 1.1 , A a , b A_{a,b} A a , b Q \mathbb{Q} Q A a A_a A a a ≥ N a\ge N a ≥ N b ≥ 0 b\ge0 b ≥ 0 B a , b B_{a,b} B a , b A b A_b A b b ≥ N b\ge N b ≥ N a ≥ 0 a\ge0 a ≥ 0
Proof. Apply Theorem 1.1 to (6) . Multiplication by a nonzero rational constant and X ↦ − X X\mapsto-X X ↦ − X
This is an equivalent form of the main theorem: the required bound depends only on the degree of the numerator or denominator being studied. It also applies to the exponential family in [41 Question 1.3 and Theorem 1.4] .
The symmetric and alternating cases. Schur’s discriminant formula [45 p. 54, (7)] , in the present normalization, is
Δ n , s = ( − 1 ) n ( n − 1 ) / 2 ∏ j = 2 n j j ( s + n − j ) j − 1 . (7) \Delta_{n,s}=(-1)^{n(n-1)/2}\prod_{j=2}^{n}j^{j}(s+n-j)^{j-1}. \tag*{(7)} Δ n , s = ( − 1 ) n ( n − 1 ) /2 j = 2 ∏ n j j ( s + n − j ) j − 1 . (7) Once alternating containment is known, the square class of Δ n , s \Delta_{n,s} Δ n , s G n , s = A n G_{n,s}=A_n G n , s = A n S n S_n S n G n , s = S n G_{n,s}=S_n G n , s = S n n ≡ 2 , 3 ( m o d 4 ) n\equiv2,3\pmod{4} n ≡ 2 , 3 ( mod 4 ) n ≥ N n\ge N n ≥ N [2 ] gives an exact description in sufficiently large degree: the alternating cases lie on the ten parameter lines
s ∈ { 1 , 2 , 3 , 4 , 5 , 6 , n − 4 , n , n + 2 , n + 4 } , s\in\{1,2,3,4,5,6,n-4,n,n+2,n+4\}, s ∈ { 1 , 2 , 3 , 4 , 5 , 6 , n − 4 , n , n + 2 , n + 4 } , with the conditions in Table 1 . The classification is due to the cited authors; the uniform theorem supplies alternating containment throughout the parameter range.
Table 1.
Other families and geometric applications. The methods also apply beyond this family. The finite-fiber criterion proves large-parameter irreducibility for the Appell family generated by e X z ( 1 − z ) − s ( 1 − 2 z ) − s e^{Xz(1-z)^{-s}(1-2z)^{-s}} e X z ( 1 − z ) − s ( 1 − 2 z ) − s [3 ] .
Finally, the uniform theorem permits an arbitrary positive integral parameter in each degree in the root-set criterion of Bary-Soroker–Kadets [4 ] and in the hyperelliptic-Jacobian theorems of Zarhin [47 , 48 ] . Together with Lombardo’s product criterion [35 ] , this gives the geometric consequences in Section 6.3. The general geometric criteria are due to these authors; our contribution is their uniform application to the present family.
Organization. Section 2 develops the root geometry and finite-fiber arithmetic. Sections 3 and 4 prove uniform irreducibility and alternating containment, respectively. Section 5 proves the effective finite reduction. Section 6 contains the applications and the remaining questions. Appendix A gives the analytic-continuation argument for the root-modulus representation; Appendix B provides the exact certificate for the Appell family.
Notation. All logarithms are natural. We normalize v p ( p ) = 1 v_p(p)=1 v p ( p ) = 1 Q ‾ p \overline{\mathbb{Q}}_p Q p m ∈ Z > 0 m \in\mathbb{Z}_{>0} m ∈ Z > 0 rad ( m ) = ∏ p ∣ m p \operatorname{rad}(m)=\prod_{p\mid m}p rad ( m ) = ∏ p ∣ m p ω ( m ) = # { p : p ∣ m } \omega(m)=\#\{p:p\mid m\} ω ( m ) = # { p : p ∣ m } P ( m ) P(m) P ( m ) m = 1 m=1 m = 1 1 , 0 , 1 1,0,1 1 , 0 , 1 y y y y y y a / b a/b a / b b > 0 b>0 b > 0 H ( a / b ) = max { ∣ a ∣ , b } H(a/b)=\max\{|a|,b\} H ( a / b ) = max { ∣ a ∣ , b } h ( a / b ) = log H ( a / b ) h(a/b)=\log H(a/b) h ( a / b ) = log H ( a / b ) g g g Ω \Omega Ω supp ( g ) \operatorname{supp}(g) supp ( g ) fpr ( g ) = ∣ Fix ( g ) ∣ / ∣ Ω ∣ \operatorname{fpr}(g)=|\operatorname{Fix}(g)|/|\Omega| fpr ( g ) = ∣ Fix ( g ) ∣/∣Ω∣ μ ( G ) \mu(G) μ ( G ) G G G
Geometry of the roots and integral traces
Root geometry and a preliminary parameter bound
Two estimates on the roots will be used in the arithmetic proof. After translation by s s s s s s s s s n n n ∣ α − s + 1 ∣ |\alpha-s+1| ∣ α − s + 1∣ s ≥ 1 s\ge1 s ≥ 1
C n , s ( Y ) = H n , s ( Y − s ) . C_{n,s}(Y)=H_{n,s}(Y-s). C n , s ( Y ) = H n , s ( Y − s ) . Identities, root counts, and an annulus.
The exponential generating functions are
∑ n ≥ 0 H n , s ( X ) t n n ! = e X t ( 1 − t ) − s . (8) \sum_{n\ge0}H_{n,s}(X)\frac{t^n}{n!}=e^{Xt}(1-t)^{-s}. \tag*{(8)} n ≥ 0 ∑ H n , s ( X ) n ! t n = e X t ( 1 − t ) − s . (8) ∑ n ≥ 0 C n , s ( Y ) t n n ! = exp ( Y t + s ∑ j ≥ 2 t j j ) . (9) \sum_{n\ge0}C_{n,s}(Y)\frac{t^n}{n!}=\exp\left(Yt+s\sum_{j\ge2}\frac{t^j}{j}\right). \tag*{(9)} n ≥ 0 ∑ C n , s ( Y ) n ! t n = exp ( Y t + s j ≥ 2 ∑ j t j ) . (9) These are identities of formal power series. Coefficient comparison gives
H n , s ′ ( X ) = n H n − 1 , s ( X ) , (10) H_{n,s}^{\prime}(X)=nH_{n-1,s}(X), \tag*{(10)} H n , s ′ ( X ) = n H n − 1 , s ( X ) , (10) H n + 1 , s ( X ) = ( X + s + n ) H n , s ( X ) − n X H n − 1 , s ( X ) , (11) H_{n+1,s}(X)=(X+s+n)H_{n,s}(X)-nXH_{n-1,s}(X), \tag*{(11)} H n + 1 , s ( X ) = ( X + s + n ) H n , s ( X ) − n X H n − 1 , s ( X ) , (11) X H n , s ′ ′ ( X ) − ( X + n + s − 1 ) H n , s ′ ( X ) + n H n , s ( X ) = 0. (12) XH_{n,s}^{\prime\prime}(X)-(X+n+s-1)H_{n,s}^{\prime}(X)+nH_{n,s}(X)=0. \tag*{(12)} X H n , s ′′ ( X ) − ( X + n + s − 1 ) H n , s ′ ( X ) + n H n , s ( X ) = 0. (12) For real s > 0 s>0 s > 0
H n , s ( X ) = 1 Γ ( s ) ∫ 0 ∞ ( X + t ) n t s − 1 e − t d t . (13) H_{n,s}(X)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}(X+t)^n t^{s-1}e^{-t}\,\mathrm{d}t. \tag*{(13)} H n , s ( X ) = Γ ( s ) 1 ∫ 0 ∞ ( X + t ) n t s − 1 e − t d t . (13) Thus H n , s H_{n,s} H n , s R \mathbb{R} R (10) , so it has exactly one real root, which is negative because H n , s ( 0 ) = ( s ) n > 0 H_{n,s}(0)=(s)_n>0 H n , s ( 0 ) = ( s ) n > 0
For any complex parameter, a multiple nonzero root is impossible. Indeed, if a ≠ 0 a\ne0 a = 0 r ≥ 2 r\ge2 r ≥ 2 X H n , s ′ ′ XH_{n,s}^{\prime\prime} X H n , s ′′ (12) would have a nonzero coefficient of ( X − a ) r − 2 (X-a)^{r-2} ( X − a ) r − 2 s = 0 , − 1 , … , 1 − n s=0,-1,\ldots,1-n s = 0 , − 1 , … , 1 − n s > 0 s>0 s > 0
The identity H n , s ( X ) = X n y n ( 2 / X ; s − n + 1 ) H_{n,s}(X)=X^n y_n(2/X;s-n+1) H n , s ( X ) = X n y n ( 2/ X ; s − n + 1 ) [40 (18.34.1)] , relates these roots to those of generalized Bessel polynomials. Classical annular estimates for the latter were given by de Bruin, Saff, and Varga [6 Theorem 5.1] . The following matrix argument gives the bound needed here.
Lemma 2.1. For n ≥ 2 n\ge2 n ≥ 2 s > 0 s>0 s > 0 α \alpha α H n , s H_{n,s} H n , s
s < ∣ α ∣ < s + n − 1. (14) s<|\alpha|<s+n-1. \tag*{(14)} s < ∣ α ∣ < s + n − 1. (14) Proof. Let T T T 1 , … , n − 1 \sqrt{1},\ldots,\sqrt{n-1} 1 , … , n − 1 D 0 = diag ( s , s + 1 , … , s + n − 1 ) D_0=\operatorname{diag}(s,s+1,\ldots,s+n-1) D 0 = diag ( s , s + 1 , … , s + n − 1 ) (11) give
H n , s ( z 2 ) = det ( z 2 I − z T + D 0 ) . H_{n,s}(z^2)=\det(z^2I-zT+D_0). H n , s ( z 2 ) = det ( z 2 I − z T + D 0 ) . Choose z 2 = α z^2=\alpha z 2 = α v v v v ∗ v^* v ∗
z 2 − ( v ∗ T v ) z + v ∗ D 0 v = 0. z^2-(v^*Tv)z+v^*D_0v=0. z 2 − ( v ∗ T v ) z + v ∗ D 0 v = 0. Both coefficients are real. Since α \alpha α z z z z ‾ \overline{z} z
∣ α ∣ = ∣ z ∣ 2 = v ∗ D 0 v ∈ [ s , s + n − 1 ] . |\alpha|=|z|^2=v^*D_0v\in[s,s+n-1]. ∣ α ∣ = ∣ z ∣ 2 = v ∗ D 0 v ∈ [ s , s + n − 1 ] . Equality at either endpoint would force v v v T T T ( z 2 I − z T + D 0 ) v = 0 (z^2I-zT+D_0)v=0 ( z 2 I − z T + D 0 ) v = 0
Centered roots and integral factor traces
After centering at − s -s − s s s s [20 Theorem 1] ; we include the matrix proof to obtain the strict inequality.
Proposition 2.2. Let n ≥ 2 n\ge2 n ≥ 2 s > 0 s>0 s > 0 d d d α 1 , … , α d \alpha_1,\ldots,\alpha_d α 1 , … , α d H n , s H_{n,s} H n , s 0 < d < n 0<d<n 0 < d < n
∣ Re ∑ i = 1 d ( α i + s ) ∣ < d ( n − d ) . \left|\operatorname{Re}\sum_{i=1}^{d}(\alpha_i+s)\right|<d(n-d). Re i = 1 ∑ d ( α i + s ) < d ( n − d ) . In particular, if a a a C n , s C_{n,s} C n , s d d d
∣ a ∣ < d ( n − d ) . (15) |a|<d(n-d). \tag*{(15)} ∣ a ∣ < d ( n − d ) . (15) For positive integral s s s a ∈ Z a\in\mathbb{Z} a ∈ Z α \alpha α H n , s H_{n,s} H n , s
∣ α + s ∣ ≤ n − 1 + 2 ( n − 1 ) ( s + n − 1 ) . (16) |\alpha+s|\le n-1+2\sqrt{(n-1)(s+n-1)}. \tag*{(16)} ∣ α + s ∣ ≤ n − 1 + 2 ( n − 1 ) ( s + n − 1 ) . (16) Proof. The centered polynomial is the characteristic polynomial of the tridiagonal matrix A A A
A j j = 2 j − n − 1 ( 1 ≤ j ≤ n ) , A_{jj}=2j-n-1 \qquad(1\le j\le n), A j j = 2 j − n − 1 ( 1 ≤ j ≤ n ) , A j , j + 1 = j , A j + 1 , j = − ( n + s − j ) ( 1 ≤ j < n ) . A_{j,j+1}=j,\qquad A_{j+1,j}=-(n+s-j)\qquad(1\le j<n). A j , j + 1 = j , A j + 1 , j = − ( n + s − j ) ( 1 ≤ j < n ) . To verify this, let P k ( X ) P_k(X) P k ( X ) k k k k k k A − s I A-sI A − s I
P k = ( X + n + s + 1 − 2 k ) P k − 1 + ( k − 1 ) ( n + s − k + 1 ) P k − 2 , P_k=(X+n+s+1-2k)P_{k-1}+(k-1)(n+s-k+1)P_{k-2}, P k = ( X + n + s + 1 − 2 k ) P k − 1 + ( k − 1 ) ( n + s − k + 1 ) P k − 2 , with P 0 = 1 P_0=1 P 0 = 1 P 1 = X + n + s − 1 P_1=X+n+s-1 P 1 = X + n + s − 1
P k ( X ) = ∑ j = 0 k k ! j ! ( n + s − 1 − j k − j ) X j . P_k(X)=\sum_{j=0}^{k}\frac{k!}{j!}\binom{n+s-1-j}{k-j}X^j. P k ( X ) = j = 0 ∑ k j ! k ! ( k − j n + s − 1 − j ) X j . At k = n k=n k = n X j X^j X j ( n j ) ( s ) n − j \binom{n}{j}(s)_{n-j} ( j n ) ( s ) n − j
A positive diagonal similarity transforms A A A D + K D+K D + K
D = diag ( 1 − n , 3 − n , … , n − 1 ) , K ∗ = − K , D=\operatorname{diag}(1-n,3-n,\ldots,n-1),\qquad K^*=-K, D = diag ( 1 − n , 3 − n , … , n − 1 ) , K ∗ = − K , and
K j , j + 1 = j ( n + s − j ) , K j + 1 , j = − K j , j + 1 . K_{j,j+1}=\sqrt{j(n+s-j)},\qquad K_{j+1,j}=-K_{j,j+1}. K j , j + 1 = j ( n + s − j ) , K j + 1 , j = − K j , j + 1 . Eigenvectors belonging to the selected roots span a d d d V V V U U U V V V U ∗ ( D + K ) U U^*(D+K)U U ∗ ( D + K ) U ∑ i ( α i + s ) \sum_i(\alpha_i+s) ∑ i ( α i + s ) K K K
Re ∑ i = 1 d ( α i + s ) = tr ( U ∗ D U ) = ∑ j = 1 n ( 2 j − n − 1 ) w j , 0 ≤ w j ≤ 1 , ∑ j = 1 n w j = d , \operatorname{Re}\sum_{i=1}^{d}(\alpha_i+s)=\operatorname{tr}(U^*DU)=\sum_{j=1}^{n}(2j-n-1)w_j,\qquad0\leq w_j\leq1,\qquad\sum_{j=1}^{n}w_j=d, Re i = 1 ∑ d ( α i + s ) = tr ( U ∗ D U ) = j = 1 ∑ n ( 2 j − n − 1 ) w j , 0 ≤ w j ≤ 1 , j = 1 ∑ n w j = d , where w j w_j w j V V V d d d D D D − d ( n − d ) -d(n-d) − d ( n − d ) d ( n − d ) d(n-d) d ( n − d ) V V V D + K D+K D + K (15) .
Finally, the row and column sum bounds for the operator norm give
∥ D ∥ = n − 1 , ∥ K ∥ ≤ 2 max 1 ≤ j < n j ( n + s − j ) ≤ 2 ( n − 1 ) ( s + n − 1 ) . \lVert D\rVert=n-1,\qquad\lVert K\rVert\leq2\max_{1\leq j<n}\sqrt{j(n+s-j)}\leq2\sqrt{(n-1)(s+n-1)}. ∥ D ∥ = n − 1 , ∥ K ∥ ≤ 2 1 ≤ j < n max j ( n + s − j ) ≤ 2 ( n − 1 ) ( s + n − 1 ) . Each α + s \alpha+s α + s D + K D+K D + K (16) . If s s s
To turn the trace bound into a divisibility condition, we first consider a polynomial that reduces to a power of X X X
Lemma 2.3. Let P ∈ Z [ X ] P\in\mathbb{Z}[X] P ∈ Z [ X ] N ≥ 2 N\geq2 N ≥ 2
P ( X ) ≡ X N − 1 ( X − c ) ( m o d M ) P(X)\equiv X^{N-1}(X-c)\pmod{M} P ( X ) ≡ X N − 1 ( X − c ) ( mod M ) for integers M ≥ 1 M\geq1 M ≥ 1 c c c P P P τ \tau τ τ ≡ A ( m o d M ) \tau\equiv A\pmod{M} τ ≡ A ( mod M ) A ( A − c ) ≠ 0 A(A-c)\ne0 A ( A − c ) = 0
M ∣ gcd ( ∣ A ∣ , ∣ A − c ∣ ) lcm ( ∣ A ∣ , ∣ A − c ∣ ) N − 1 . (17) M\mid\gcd(|A|,|A-c|)\operatorname{lcm}(|A|,|A-c|)^{N-1}. \tag*{(17)} M ∣ g cd( ∣ A ∣ , ∣ A − c ∣ ) lcm ( ∣ A ∣ , ∣ A − c ∣ ) N − 1 . (17) Proof. Fix p e ∥ M p^e\mathbin{\Vert}M p e ∥ M
t = min { v p ( A ) , v p ( A − c ) } , b = max { v p ( A ) , v p ( A − c ) } , u = v p ( c ) ≥ t , t=\min\{v_p(A),v_p(A-c)\},\qquad b=\max\{v_p(A),v_p(A-c)\},\qquad u=v_p(c)\geq t, t = min { v p ( A ) , v p ( A − c )} , b = max { v p ( A ) , v p ( A − c )} , u = v p ( c ) ≥ t , where v p ( 0 ) = + ∞ v_p(0)=+\infty v p ( 0 ) = + ∞ e > t + ( N − 1 ) b e>t+(N-1)b e > t + ( N − 1 ) b e > N t e>Nt e > N t p − N t P ( p t Z ) p^{-Nt}P(p^tZ) p − N t P ( p t Z ) p p p u > t u>t u > t Z N Z^N Z N P P P t t t v p ( A ) = v p ( A − c ) = t v_p(A)=v_p(A-c)=t v p ( A ) = v p ( A − c ) = t τ ≡ A ( m o d p e ) \tau\equiv A\pmod{p^e} τ ≡ A ( mod p e ) u = t u=t u = t Z N − 1 ( Z − γ ) Z^{N-1}(Z-\gamma) Z N − 1 ( Z − γ ) γ ≠ 0 \gamma\ne0 γ = 0 Z p \mathbb{Z}_p Z p Z N − 1 Z^{N-1} Z N − 1 β \beta β P P P t t t t t t α \alpha α
v p ( α N − 1 ( α − c ) ) ≥ e . v_p\left(\alpha^{N-1}(\alpha-c)\right)\ge e. v p ( α N − 1 ( α − c ) ) ≥ e . For α ≠ β \alpha\ne\beta α = β v p ( α − c ) = t v_p(\alpha-c)=t v p ( α − c ) = t
v p ( α ) ≥ e − t N − 1 > b . v_p(\alpha)\ge\frac{e-t}{N-1}>b. v p ( α ) ≥ N − 1 e − t > b . For β \beta β
v p ( β − c ) ≥ e − ( N − 1 ) t > b . v_p(\beta-c)\ge e-(N-1)t>b. v p ( β − c ) ≥ e − ( N − 1 ) t > b . Thus, according as the selection excludes or includes β \beta β v p ( τ ) > b v_p(\tau)>b v p ( τ ) > b v p ( τ − c ) > b v_p(\tau-c)>b v p ( τ − c ) > b e > b e>b e > b τ ≡ A ( m o d p e ) \tau\equiv A\pmod{p^e} τ ≡ A ( mod p e ) e ≤ t + ( N − 1 ) b e\le t+(N-1)b e ≤ t + ( N − 1 ) b p p p (17) . □
At s = 0 s=0 s = 0 s = − 1 s=-1 s = − 1 H n , s H_{n,s} H n , s
Proposition 2.4. Let n ≥ 2 n\ge2 n ≥ 2 s ≥ 1 s\ge1 s ≥ 1 C n , s C_{n,s} C n , s d d d a a a
s ∣ a n . (18) s\mid a^n. \tag*{(18)} s ∣ a n . (18) If a = 0 a=0 a = 0
s + 1 ∣ gcd ( d , n − d ) lcm ( d , n − d ) n − 1 . (19) s+1\mid\gcd(d,n-d)\operatorname{lcm}(d,n-d)^{n-1}. \tag*{(19)} s + 1 ∣ g cd( d , n − d ) lcm ( d , n − d ) n − 1 . (19) In particular, writing B d = d ( n − d ) B_d=d(n-d) B d = d ( n − d ) g = gcd ( d , n − d ) g=\gcd(d,n-d) g = g cd( d , n − d )
{ s ≤ ( B d − 1 ) n , a ≠ 0 , s ≤ B d n − 1 g 2 − n − 1 , a = 0. (20) \begin{cases}
s\le(B_d-1)^n, & a\ne0,\\
s\le B_d^{n-1}g^{2-n}-1, & a=0. \tag*{(20)}
\end{cases} { s ≤ ( B d − 1 ) n , s ≤ B d n − 1 g 2 − n − 1 , a = 0 , a = 0. (20) Consequently, reducibility of H n , s H_{n,s} H n , s
s < ⌊ n 2 4 ⌋ n , log s < 2 n log n . (21) s<\left\lfloor\frac{n^2}{4}\right\rfloor^n,\qquad\log s<2n\log n. \tag*{(21)} s < ⌊ 4 n 2 ⌋ n , log s < 2 n log n . (21) Proof. The coefficient formula gives C n , s ( Y ) ≡ Y n ( m o d s ) C_{n,s}(Y)\equiv Y^n\pmod{s} C n , s ( Y ) ≡ Y n ( mod s ) p e ∥ s p^e\mathrel{\|}s p e ∥ s e / n e/n e / n v p ( a ) ≥ e / n v_p(a)\ge e/n v p ( a ) ≥ e / n (18) .
At the next integral fiber,
H n , s ( X ) ≡ X n − 1 ( X − n ) ( m o d s + 1 ) . H_{n,s}(X)\equiv X^{n-1}(X-n)\pmod{s+1}. H n , s ( X ) ≡ X n − 1 ( X − n ) ( mod s + 1 ) . The uncentered root sum is τ = a − d s ≡ a + d ( m o d s + 1 ) \tau=a-ds\equiv a+d\pmod{s+1} τ = a − d s ≡ a + d ( mod s + 1 ) a = 0 a=0 a = 0 A = d A=d A = d c = n c=n c = n 0 < d < n 0<d<n 0 < d < n (19) .
Since a ∈ Z a\in\mathbb{Z} a ∈ Z (15) gives ∣ a ∣ ≤ B d − 1 |a|\le B_d-1 ∣ a ∣ ≤ B d − 1 gcd ( d , n − d ) lcm ( d , n − d ) n − 1 = B d n − 1 g 2 − n \gcd(d,n-d)\operatorname{lcm}(d,n-d)^{n-1}=B_d^{n-1}g^{2-n} g cd( d , n − d ) lcm ( d , n − d ) n − 1 = B d n − 1 g 2 − n (20) . Now B d ≤ B = ⌊ n 2 / 4 ⌋ B_d\le B=\lfloor n^2/4\rfloor B d ≤ B = ⌊ n 2 /4 ⌋ a ≠ 0 a\ne0 a = 0 s ≤ ( B − 1 ) n < B n s\le(B-1)^n<B^n s ≤ ( B − 1 ) n < B n a = 0 a=0 a = 0 s + 1 ≤ B n − 1 ≤ B n s+1\le B^{n-1}\le B^n s + 1 ≤ B n − 1 ≤ B n B = 1 B=1 B = 1 (21) follows in every case. □
The motion of the root moduli
The annulus bounds the excess ∣ α ∣ − s |\alpha|-s ∣ α ∣ − s s s s s = 1 s=1 s = 1
Theorem 2.5. Fix n ≥ 2 n \ge2 n ≥ 2 H n , s H_{n,s} H n , s s > 0 s>0 s > 0 q ( s ) q(s) q ( s ) n n n q ( s ) q(s) q ( s ) μ \mu μ [ 0 , n − 1 ] [0,n-1] [ 0 , n − 1 ]
q ( s ) = s + ∫ 0 n − 1 s s + t d μ ( t ) ( s > 0 ) . (22) q(s)=s+\int_{0}^{n-1}\frac{s}{s+t}\,d\mu(t) \qquad(s>0). \tag*{(22)} q ( s ) = s + ∫ 0 n − 1 s + t s d μ ( t ) ( s > 0 ) . (22) The measure is absolutely continuous. In particular, q ( s ) − s q(s)-s q ( s ) − s
Appendix A proves the representation by continuation across the singular fibers and Cauchy’s formula. Proposition A.2 identifies the exact support of the measure for each branch. In the standard terminology, q q q q − s q-s q − s [9 Theorem 1.4] .
The expansion at infinity identifies the mass of the measure. We fix the Hermite normalization used throughout the paper:
h m ( T ) = m ! [ u m ] e T u + u 2 / 2 = i − m He m ( i T ) , (23) h_m(T)=m![u^m]e^{Tu+u^2/2}=i^{-m}\operatorname{He}_m(iT), \tag*{(23)} h m ( T ) = m ! [ u m ] e T u + u 2 /2 = i − m He m ( i T ) , (23) where He m \operatorname{He}_m He m
F n ( T , v ) : = v n C n , v − 2 ( T / v ) = n ! [ u n ] exp ( T u + ∑ m ≥ 2 v m − 2 u m m ) . (24) F_n(T,v):=v^n C_{n,v^{-2}}(T/v)=n![u^n]\exp\left(Tu+\sum_{m\ge2}\frac{v^{m-2}u^m}{m}\right). \tag*{(24)} F n ( T , v ) := v n C n , v − 2 ( T / v ) = n ! [ u n ] exp ( T u + m ≥ 2 ∑ m v m − 2 u m ) . (24) The right side extends polynomially to v = 0 v=0 v = 0 F n ( T , 0 ) = h n ( T ) F_n(T,0)=h_n(T) F n ( T , 0 ) = h n ( T ) h n h_n h n He n ( x ) = ( − 1 ) n e x 2 / 2 ( d / d x ) n e − x 2 / 2 \operatorname{He}_n(x)=(-1)^n e^{x^2/2}(d/dx)^n e^{-x^2/2} He n ( x ) = ( − 1 ) n e x 2 /2 ( d / d x ) n e − x 2 /2 He n \operatorname{He}_n He n n n n
Corollary 2.6. Let q q q μ \mu μ ξ ≥ 0 \xi\ge0 ξ ≥ 0 He n \operatorname{He}_n He n
α ( s ) = − s + i ξ s + O ( 1 ) ( s ⟶ + ∞ ) . \alpha(s)=-s+i\xi\sqrt{s}+O(1) \qquad(s\longrightarrow+\infty). α ( s ) = − s + i ξ s + O ( 1 ) ( s ⟶ + ∞ ) . For the real root in odd degree, take ξ = 0 \xi=0 ξ = 0
μ ( [ 0 , n − 1 ] ) = lim s → ∞ ( q ( s ) − s ) = n − 1 3 + ξ 2 6 . (25) \mu([0,n-1])=\lim_{s\to\infty}(q(s)-s)=\frac{n-1}{3}+\frac{\xi^2}{6}. \tag*{(25)} μ ([ 0 , n − 1 ]) = s → ∞ lim ( q ( s ) − s ) = 3 n − 1 + 6 ξ 2 . (25) In particular, for every integer r ≥ 1 r\ge1 r ≥ 1
( − 1 ) r − 1 d r d s r ( q ( s ) − s ) = r ! ∫ 0 n − 1 t ( s + t ) r + 1 d μ ( t ) > 0. (26) (-1)^{r-1}\frac{d^r}{ds^r}(q(s)-s)=r!\int_{0}^{n-1}\frac{t}{(s+t)^{r+1}}\,d\mu(t)>0. \tag*{(26)} ( − 1 ) r − 1 d s r d r ( q ( s ) − s ) = r ! ∫ 0 n − 1 ( s + t ) r + 1 t d μ ( t ) > 0. (26) Thus q ′ ( s ) > 1 q'(s)>1 q ′ ( s ) > 1 q ′ ′ ( s ) < 0 q''(s)<0 q ′′ ( s ) < 0
Proof. The polynomials in (2.16) satisfy h j ′ = j h j − 1 h'_j=jh_{j-1} h j ′ = j h j − 1 h j = T h j − 1 + ( j − 1 ) h j − 2 h_j=Th_{j-1}+(j-1)h_{j-2} h j = T h j − 1 + ( j − 1 ) h j − 2 n ≥ 3 n\ge3 n ≥ 3
F n ( w , z ) = h n ( w ) + n ( n − 1 ) ( n − 2 ) 3 z 2 h n − 3 ( w ) + O ( z 2 ) . F_n(w,z)=h_n(w)+\frac{n(n-1)(n-2)}{3}z^2h_{n-3}(w)+O(z^2). F n ( w , z ) = h n ( w ) + 3 n ( n − 1 ) ( n − 2 ) z 2 h n − 3 ( w ) + O ( z 2 ) . Let w 0 = i ξ w_0=i\xi w 0 = i ξ w ( z ) = w 0 + b z + O ( z 2 ) w(z)=w_0+bz+O(z^2) w ( z ) = w 0 + b z + O ( z 2 ) h n ( w 0 ) = 0 h_n(w_0)=0 h n ( w 0 ) = 0 h n ′ ( w 0 ) ≠ 0 h'_n(w_0)\ne0 h n ′ ( w 0 ) = 0
b = − ( n − 1 ) ( n − 2 ) 3 h n − 3 ( w 0 ) h n − 1 ( w 0 ) . b=-\frac{(n-1)(n-2)}{3}\frac{h_{n-3}(w_0)}{h_{n-1}(w_0)}. b = − 3 ( n − 1 ) ( n − 2 ) h n − 1 ( w 0 ) h n − 3 ( w 0 ) . The recurrence, applied twice, yields
( n − 1 ) ( n − 2 ) h n − 3 ( w 0 ) = ( n − 1 + w 0 2 ) h n − 1 ( w 0 ) . (n-1)(n-2)h_{n-3}(w_0)=(n-1+w_0^2)h_{n-1}(w_0). ( n − 1 ) ( n − 2 ) h n − 3 ( w 0 ) = ( n − 1 + w 0 2 ) h n − 1 ( w 0 ) . Consequently
b = − n − 1 + w 0 2 3 = − ξ 2 − n + 1 3 . b=-\frac{n-1+w_0^2}{3}=-\frac{\xi^2-n+1}{3}. b = − 3 n − 1 + w 0 2 = − 3 ξ 2 − n + 1 . For n = 2 n=2 n = 2 F n ( w , z ) = w 2 + 1 F_n(w,z)=w^2+1 F n ( w , z ) = w 2 + 1 ξ 2 = 1 \xi^2=1 ξ 2 = 1 b = 0 b=0 b = 0 z = s − 1 / 2 z=s^{-1/2} z = s − 1/2
α ( s ) = − s + i ξ s + ξ 2 − n + 1 3 + O ( s − 1 / 2 ) . \alpha(s)=-s+i\xi\sqrt{s}+\frac{\xi^2-n+1}{3}+O(s^{-1/2}). α ( s ) = − s + i ξ s + 3 ξ 2 − n + 1 + O ( s − 1/2 ) . Taking absolute values shows that
q ( s ) − s ⟶ − ξ 2 − n + 1 3 + ξ 2 2 = n − 1 3 + ξ 2 6 . q(s)-s\longrightarrow-\frac{\xi^2-n+1}{3}+\frac{\xi^2}{2}=\frac{n-1}{3}+\frac{\xi^2}{6}. q ( s ) − s ⟶ − 3 ξ 2 − n + 1 + 2 ξ 2 = 3 n − 1 + 6 ξ 2 . This also applies to ξ = 0 \xi=0 ξ = 0 q = − α q=-\alpha q = − α (22) identifies this limit with the mass of μ \mu μ (25) .
Differentiation under the integral is valid on every compact subinterval of ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) (26) . Its right-hand side is strictly positive: the mass in (25) is positive, and absolute continuity excludes an atom at 0. □
The quadratic example. For n = 2 n=2 n = 2 − s ± i s -s\pm i\sqrt{s} − s ± i s q ( s ) = s ( s + 1 ) q(s)=\sqrt{s(s+1)} q ( s ) = s ( s + 1 )
d μ ( t ) = 1 π 1 − t t 1 ( 0 , 1 ) ( t ) d t . d\mu(t)=\frac{1}{\pi}\sqrt{\frac{1-t}{t}}\,\mathbf{1}_{(0,1)}(t)\,dt. d μ ( t ) = π 1 t 1 − t 1 ( 0 , 1 ) ( t ) d t . Its mass is 1 / 2 1/2 1/2 (25) . This example exhibits both the full support and the strict derivative signs directly.
Proposition 2.7. Let n ≥ 2 n\ge2 n ≥ 2 s > 0 s>0 s > 0 H n , s H_{n,s} H n , s
α j ( s ) ∼ ( ( n − 1 ) ! s ) 1 / n e ( 2 j + 1 ) π i / n , 0 ≤ j ≤ ⌊ n − 1 2 ⌋ , s ↓ 0. \alpha_j(s)\sim\bigl((n-1)!s\bigr)^{1/n}e^{(2j+1)\pi i/n},\qquad0\le j\le\left\lfloor\frac{n-1}{2}\right\rfloor,\quad s\downarrow0. α j ( s ) ∼ ( ( n − 1 )! s ) 1/ n e ( 2 j + 1 ) π i / n , 0 ≤ j ≤ ⌊ 2 n − 1 ⌋ , s ↓ 0. Then the corresponding Hermite zeros in α j ( s ) = − s + i ξ j s + O ( 1 ) \alpha_j(s)=-s+i\xi_j\sqrt{s}+O(1) α j ( s ) = − s + i ξ j s + O ( 1 )
ξ j > ξ j + 1 ( 0 ≤ j < ⌊ n − 1 2 ⌋ ) , ξ ⌊ ( n − 1 ) / 2 ⌋ ≥ 0. \xi_j>\xi_{j+1}\quad\left(0\le j<\left\lfloor\frac{n-1}{2}\right\rfloor\right),\qquad\xi_{\lfloor(n-1)/2\rfloor}\ge0. ξ j > ξ j + 1 ( 0 ≤ j < ⌊ 2 n − 1 ⌋ ) , ξ ⌊( n − 1 ) /2 ⌋ ≥ 0. Thus the exact support endpoint n − 1 − j n-1-j n − 1 − j Proposition A.2 corresponds to the j j j j = 0 j=0 j = 0
Proof. Put A = n + s − 1 A=n+s-1 A = n + s − 1 κ = n − A / 2 \kappa=n-A/2 κ = n − A /2 X = ρ e i θ X=\rho e^{i\theta} X = ρ e i θ H ( X ) = e i θ ρ / 2 ρ A / 2 u ( ρ ) H(X)=e^{i\theta\rho/2}\rho^{A/2}u(\rho) H ( X ) = e i θ ρ /2 ρ A /2 u ( ρ ) (12) gives
u ′ ′ + ( − e 2 i θ 4 + κ e i θ ρ − A ( A + 2 ) 4 ρ 2 ) u = 0. (27) u''+\left(-\frac{e^{2i\theta}}{4}+\frac{\kappa e^{i\theta}}{\rho}-\frac{A(A+2)}{4\rho^2}\right)u=0. \tag*{(27)} u ′′ + ( − 4 e 2 i θ + ρ κ e i θ − 4 ρ 2 A ( A + 2 ) ) u = 0. (27) Suppose two roots lie at radii ρ 1 < ρ 2 \rho_1<\rho_2 ρ 1 < ρ 2 Proposition 2.2 gives ρ cos θ < n − 1 − s = 2 κ − 2 \rho\cos\theta<n-1-s=2\kappa-2 ρ cos θ < n − 1 − s = 2 κ − 2 u ‾ \overline{u} u
0 = sin θ ∫ ρ 1 ρ 2 ( κ ρ − cos θ 2 ) ∣ u ∣ 2 d ρ , (28) 0=\sin\theta\int_{\rho_1}^{\rho_2}\left(\frac{\kappa}{\rho}-\frac{\cos\theta}{2}\right)|u|^2\,d\rho, \tag*{(28)} 0 = sin θ ∫ ρ 1 ρ 2 ( ρ κ − 2 cos θ ) ∣ u ∣ 2 d ρ , (28) although the parenthesized factor is greater than 1 / ρ 1/\rho 1/ ρ s > 0 s>0 s > 0 arg α j ( s ) = π − ξ j / s + O ( s − 1 ) \arg\alpha_j(s)=\pi-\xi_j/\sqrt{s}+O(s^{-1}) arg α j ( s ) = π − ξ j / s + O ( s − 1 ) (24) , proving the strict inequalities. □
The arithmetic applications require a lower bound uniform in s s s s = 1 s=1 s = 1
Corollary 2.8. Let ρ 0 > 0 \rho_0>0 ρ 0 > 0 log ρ 0 + ρ 0 = − 1 \log\rho_0+\rho_0=-1 log ρ 0 + ρ 0 = − 1 n ≥ 2 n\ge2 n ≥ 2 s ≥ 1 s\ge1 s ≥ 1 α \alpha α H n , s H_{n,s} H n , s
∣ α ∣ − s + 1 > ρ 0 ( n + 1 ) > n + 1 4 . (29) |\alpha|-s+1>\rho_0(n+1)>\frac{n+1}{4}. \tag*{(29)} ∣ α ∣ − s + 1 > ρ 0 ( n + 1 ) > 4 n + 1 . (29) Proof. At s = 1 s=1 s = 1 H n , 1 ( X ) = n ! ∑ j = 0 n X j / j ! H_{n,1}(X)=n!\sum_{j=0}^{n}X^j/j! H n , 1 ( X ) = n ! ∑ j = 0 n X j / j ! α \alpha α
e α = α n + 1 n ! ∫ 0 1 e t α ( 1 − t ) n d t . e^\alpha=\frac{\alpha^{n+1}}{n!}\int_0^1 e^{t\alpha}(1-t)^n\,\mathrm{d}t. e α = n ! α n + 1 ∫ 0 1 e t α ( 1 − t ) n d t . Dividing by e α e^\alpha e α R = ∣ α ∣ R=|\alpha| R = ∣ α ∣ N = n + 1 N=n+1 N = n + 1
1 ≤ R n + 1 n ! ∫ 0 1 e − ( 1 − t ) ℜ α ( 1 − t ) n d t ≤ R N e R N ! . 1\le\frac{R^{n+1}}{n!}\int_0^1 e^{-(1-t)\Re\alpha}(1-t)^n\,\mathrm{d}t\le\frac{R^N e^R}{N!}. 1 ≤ n ! R n + 1 ∫ 0 1 e − ( 1 − t ) ℜ α ( 1 − t ) n d t ≤ N ! R N e R . Since log ( N ! ) > N log N − N \log(N!)>N\log N-N log ( N !) > N log N − N log ( R / N ) + R / N > − 1 \log(R/N)+R/N>-1 log ( R / N ) + R / N > − 1 u ↦ log u + u u\mapsto\log u+u u ↦ log u + u − ∞ -\infty − ∞ + ∞ +\infty + ∞ R > ρ 0 N R>\rho_0N R > ρ 0 N log 4 > 5 / 4 \log4>5/4 log 4 > 5/4 e < 3 e<3 e < 3 3 5 < 4 4 3^5<4^4 3 5 < 4 4 ρ 0 > 1 / 4 \rho_0>1/4 ρ 0 > 1/4
Continue the root or its conjugate pair from parameter 1 1 1 s s s Theorem 2.5 gives q ( s ) − s + 1 ≥ q ( 1 ) > ρ 0 ( n + 1 ) q(s)-s+1\ge q(1)>\rho_0(n+1) q ( s ) − s + 1 ≥ q ( 1 ) > ρ 0 ( n + 1 ) 1 / 4 1/4 1/4
Finite fibers and large-parameter irreducibility
A monic rational factor of a monic integral polynomial has an integral root sum. At a fiber where all but a bounded number of roots vanish, reduction modulo a prime restricts this sum to a subset sum of the residual roots. If the corresponding trace relation is nonzero, its size bounds the prime divisors of the parameter difference. Two such relations give a height bound by linear forms in logarithms. We formulate this argument first, then apply it to Laguerre polynomials. The Appell application is given in § 6.1 .
The arithmetic input comes from linear forms in logarithms. We record the exact estimate before deriving the bound used here.
Linear forms in logarithms
We use the following rational specialization of Matveev’s theorem [38 Corollary 2.3, p. 1219] . The number of logarithms is allowed to vary; its dependence in the estimate will be needed when the prime support grows with the degree.
Lemma 2.9. Let a 1 , … , a t a_1,\ldots,a_t a 1 , … , a t 1 1 1 b 1 , … , b t b_1,\ldots,b_t b 1 , … , b t
Λ = b 1 log a 1 + ⋯ + b t log a t ≠ 0. \Lambda=b_1\log a_1+\cdots+b_t\log a_t\ne0. Λ = b 1 log a 1 + ⋯ + b t log a t = 0. Put B = max { 1 , ∣ b 1 ∣ , … , ∣ b t ∣ } B=\max\{1,|b_1|,\ldots,|b_t|\} B = max { 1 , ∣ b 1 ∣ , … , ∣ b t ∣ }
A j ≥ max { h ( a j ) , ∣ log a j ∣ , 0.16 } ( 1 ≤ j ≤ t ) . A_j\ge\max\{h(a_j),|\log a_j|,0.16\}\qquad(1\le j\le t). A j ≥ max { h ( a j ) , ∣ log a j ∣ , 0.16 } ( 1 ≤ j ≤ t ) . Here h ( M / N ) = log max { M , N } h(M/N)=\log\max\{M,N\} h ( M / N ) = log max { M , N } M , N M,N M , N
log ∣ Λ ∣ > − C t A 1 ⋯ A t log ( e B ) , C t = min { e 2 30 t + 3 t 9 / 2 , 2 6 t + 20 } . (30) \log|\Lambda|>-C_tA_1\cdots A_t\log(eB),\qquad C_t=\min\left\{\frac{e}{2}30^{t+3}t^{9/2},2^{6t+20}\right\}. \tag*{(30)} log ∣Λ∣ > − C t A 1 ⋯ A t log ( e B ) , C t = min { 2 e 3 0 t + 3 t 9/2 , 2 6 t + 20 } . (30) Proof. For t ≥ 2 t \ge2 t ≥ 2 A t = max j A j A_t = \max_j A_j A t = max j A j [38 (1.3)] then satisfies
B M = max { 1 , max j ∣ b j ∣ A j / A t } ≤ B . B_M = \max\{1,\max_j |b_j|A_j/A_t\} \le B. B M = max { 1 , j max ∣ b j ∣ A j / A t } ≤ B . Corollary 2.3 of [38 p. 1219] , with D = κ = 1 D=\kappa=1 D = κ = 1 (30) . The corollary allows dependent logarithms. For t = 1 t=1 t = 1 a 1 = M / N a_1=M/N a 1 = M / N M ≠ N M\ne N M = N 1 / x 1/x 1/ x ∣ log a 1 ∣ > e − h ( a 1 ) |\log a_1|>e^{-h(a_1)} ∣ log a 1 ∣ > e − h ( a 1 )
Nearby integers with bounded prime support.
Lemma 2.10. Let M , h M,h M , h
R ≥ e 4 , log rad ( M ( M + h ) ) ≤ R . R \ge e^4,\qquad\log\operatorname{rad}(M(M+h))\le R. R ≥ e 4 , log rad ( M ( M + h )) ≤ R . Put
T = 2 R log R , C R = e 2 30 3 T 9 / 2 ( 15 log R ) T . T=\frac{2R}{\log R},\qquad C_R=\frac{e}{2}30^{3}T^{9/2}(15\log R)^T. T = log R 2 R , C R = 2 e 3 0 3 T 9/2 ( 15 log R ) T . Then
log M < max { 2 log ( 2 h ) , 4 C R log ( 12 e C R ) } . (31) \log M<\max\{2\log(2h),\,4C_R\log(12eC_R)\}. \tag*{(31)} log M < max { 2 log ( 2 h ) , 4 C R log ( 12 e C R )} . (31) Proof. The assertion is immediate if M = 1 M=1 M = 1 M ≤ h M\le h M ≤ h L = log M L=\log M L = log M H = log h H=\log h H = log h ( M + h ) / M (M+h)/M ( M + h ) / M
0 < Λ = log M + h M = ∑ ℓ = 1 t b ℓ log p ℓ < h M , ∑ ℓ = 1 t log p ℓ ≤ R , 0<\Lambda=\log\frac{M+h}{M}=\sum_{\ell=1}^{t}b_\ell\log p_\ell<\frac{h}{M},\qquad\sum_{\ell=1}^{t}\log p_\ell\le R, 0 < Λ = log M M + h = ℓ = 1 ∑ t b ℓ log p ℓ < M h , ℓ = 1 ∑ t log p ℓ ≤ R , where the p ℓ p_\ell p ℓ b ℓ ≠ 0 b_\ell\ne0 b ℓ = 0 ℓ \ell ℓ ℓ + 1 \ell+1 ℓ + 1
R ≥ log ( ( t + 1 ) ! ) ≥ t log t − t . R\ge\log((t+1)!)\ge t\log t-t. R ≥ log (( t + 1 )!) ≥ t log t − t . We claim that t ≤ T t\le T t ≤ T w = log R ≥ 4 w=\log R\ge4 w = log R ≥ 4
T ( log T − 1 ) − R = R w ( w − 2 log w + 2 log 2 − 2 ) > 0. T(\log T-1)-R=\frac{R}{w}(w-2\log w+2\log2-2)>0. T ( log T − 1 ) − R = w R ( w − 2 log w + 2 log 2 − 2 ) > 0. The expression in parentheses is positive at w = 4 w=4 w = 4 z ( log z − 1 ) z(\log z-1) z ( log z − 1 ) z > 1 z>1 z > 1
Use the first constant in Lemma 2.9, the rational case of [38 Corollary 2.3, p. 1219] :
log ∣ Λ ∣ > − e 2 30 t + 3 t 9 / 2 ( ∏ ℓ = 1 t log p ℓ ) log ( e B ) , B = max ℓ ∣ b ℓ ∣ . \log|\Lambda|>-\frac{e}{2}30^{t+3}t^{9/2}\left(\prod_{\ell=1}^{t}\log p_\ell\right)\log(eB),\qquad B=\max_\ell|b_\ell|. log ∣Λ∣ > − 2 e 3 0 t + 3 t 9/2 ( ℓ = 1 ∏ t log p ℓ ) log ( e B ) , B = ℓ max ∣ b ℓ ∣. By the arithmetic–geometric mean inequality and t ≤ T t\le T t ≤ T
e 2 30 t + 3 t 9 / 2 ∏ ℓ = 1 t log p ℓ ≤ e 2 30 3 t 9 / 2 ( 30 R / t ) t ≤ C R . \frac{e}{2}30^{t+3}t^{9/2}\prod_{\ell=1}^{t}\log p_\ell\le\frac{e}{2}30^{3}t^{9/2}(30R/t)^t\le C_R. 2 e 3 0 t + 3 t 9/2 ℓ = 1 ∏ t log p ℓ ≤ 2 e 3 0 3 t 9/2 ( 30 R / t ) t ≤ C R . The last inequality uses the increase of z log ( 30 R / z ) z\log(30R/z) z log ( 30 R / z ) 0 < z ≤ T 0<z\le T 0 < z ≤ T T ≤ R / 2 T\le R/2 T ≤ R /2 M ≥ 2 M\ge2 M ≥ 2 h < M h<M h < M
B ≤ log ( M + h ) log 2 < 3 L . B\le\frac{\log(M+h)}{\log2}<3L. B ≤ log 2 log ( M + h ) < 3 L . Consequently
L − H < C R log ( 3 e L ) . L-H<C_R\log(3eL). L − H < C R log ( 3 e L ) . If L ≤ 2 H L\le2H L ≤ 2 H L < 2 C R log ( 3 e L ) L<2C_R\log(3eL) L < 2 C R log ( 3 e L ) a , b ≥ 1 a,b\ge1 a , b ≥ 1 y > 0 y>0 y > 0 y < a log ( b y ) y<a\log(by) y < a log ( b y ) y < 2 a log ( 2 a b ) y<2a\log(2ab) y < 2 a log ( 2 ab ) 2 a b ≥ e 2ab\ge e 2 ab ≥ e y 0 = 2 a log ( 2 a b ) y_0=2a\log(2ab) y 0 = 2 a log ( 2 ab ) a log ( b y 0 ) ≤ y 0 a\log(by_0)\le y_0 a log ( b y 0 ) ≤ y 0 y − a log ( b y ) y-a\log(by) y − a log ( b y ) y ≥ y 0 y\ge y_0 y ≥ y 0 a = 2 C R a=2C_R a = 2 C R b = 3 e b=3e b = 3 e L < 4 C R log ( 12 e C R ) L<4C_R\log(12eC_R) L < 4 C R log ( 12 e C R )
The finite-fiber criterion
If R ∈ Z [ X ] R \in\mathbb{Z}[X] R ∈ Z [ X ] r r r ρ 1 , … , ρ r \rho_1,\ldots,\rho_r ρ 1 , … , ρ r subset-trace polynomial by
T R ( Z ) = ∏ J ⊆ { 1 , … , r } ( Z − ∑ j ∈ J ρ j ) . T_R(Z)=\prod_{J\subseteq\{1,\ldots,r\}}\left(Z-\sum_{j\in J}\rho_j\right). T R ( Z ) = J ⊆ { 1 , … , r } ∏ Z − j ∈ J ∑ ρ j . It is monic of degree 2 r 2^r 2 r Z [ Z ] \mathbb{Z}[Z] Z [ Z ] Z Z Z
Proposition 2.11. Let P ( T , X ) ∈ Z [ T , X ] P(T,X)\in\mathbb{Z}[T,X] P ( T , X ) ∈ Z [ T , X ] n n n X X X c ( T ) ∈ Z [ T ] c(T)\in\mathbb{Z}[T] c ( T ) ∈ Z [ T ] t 1 , … , t q t_1,\ldots,t_q t 1 , … , t q
P ( t i , X ) = X n − r i R i ( X ) , 0 ≤ r i ≤ n , R i ∈ Z [ X ] monic of degree r i . P(t_i,X)=X^{n-r_i}R_i(X),\qquad0\leq r_i\leq n,\qquad R_i\in\mathbb{Z}[X]\text{ monic of degree }r_i. P ( t i , X ) = X n − r i R i ( X ) , 0 ≤ r i ≤ n , R i ∈ Z [ X ] monic of degree r i . Let t ∈ Z ∖ { t 1 , … , t q } t\in\mathbb{Z}\setminus\{t_1,\ldots,t_q\} t ∈ Z ∖ { t 1 , … , t q } P ( t , X ) P(t,X) P ( t , X ) d d d τ \tau τ A = τ − d c ( t ) A=\tau-dc(t) A = τ − d c ( t )
rad ( ∣ t − t i ∣ ) ∣ T R i ( A + d c ( t i ) ) . (32) \operatorname{rad}(|t-t_i|)\mid T_{R_i}\left(A+dc(t_i)\right). \tag*{(32)} rad ( ∣ t − t i ∣ ) ∣ T R i ( A + d c ( t i ) ) . (32) Whenever two of the integers on the right are nonzero, the radical of the product of the corresponding two parameter differences divides the absolute value of their product.
Proof. For a prime p ∣ t − t i p\mid t-t_i p ∣ t − t i p p p X n − r i R i ( X ) X^{n-r_i}R_i(X) X n − r i R i ( X ) R i R_i R i T R i T_{R_i} T R i p p p A + d c ( t i ) A+dc(t_i) A + d c ( t i ) T R i T_{R_i} T R i p p p
The next theorem reduces large-parameter irreducibility to a strip bound and a condition on the subset traces of finitely many fibers.
Theorem 2.12. Fix an integer n 0 ≥ 2 n_0\geq2 n 0 ≥ 2 n ≥ n 0 n\geq n_0 n ≥ n 0 P n ( T , X ) ∈ Z [ T , X ] P_n(T,X)\in\mathbb{Z}[T,X] P n ( T , X ) ∈ Z [ T , X ] n n n X X X c n ( T ) ∈ Z [ T ] c_n(T)\in\mathbb{Z}[T] c n ( T ) ∈ Z [ T ] C ≥ 1 C\geq1 C ≥ 1 κ ≥ 1 \kappa\geq1 κ ≥ 1 q ≥ 3 q\geq3 q ≥ 3 r 0 ≥ 0 r_0\geq0 r 0 ≥ 0
For every positive integer t t t α \alpha α P n ( t , X ) P_n(t,X) P n ( t , X )
∣ Re ( α − c n ( t ) ) ∣ ≤ C n κ . \left|\operatorname{Re}(\alpha-c_n(t))\right|\leq Cn^\kappa. ∣ Re ( α − c n ( t )) ∣ ≤ C n κ . There are distinct integers t n , 1 , … , t n , q t_{n,1},\ldots,t_{n,q} t n , 1 , … , t n , q
∣ t n , i ∣ + ∣ c n ( t n , i ) ∣ ≤ C n κ |t_{n,i}|+|c_n(t_{n,i})|\leq Cn^\kappa ∣ t n , i ∣ + ∣ c n ( t n , i ) ∣ ≤ C n κ and specializations
P n ( t n , i , X ) = X n − r n , i R n , i ( X ) , 0 ≤ r n , i ≤ min { n , r 0 } , P_n(t_{n,i},X)=X^{n-r_{n,i}}R_{n,i}(X),\qquad0\leq r_{n,i}\leq\min\{n,r_0\}, P n ( t n , i , X ) = X n − r n , i R n , i ( X ) , 0 ≤ r n , i ≤ min { n , r 0 } , where R n , i R_{n,i} R n , i r n , i r_{n,i} r n , i C n κ Cn^\kappa C n κ
Put U n , i = T R n , i U_{n,i}=T_{R_{n,i}} U n , i = T R n , i 1 ≤ d ≤ n / 2 1\leq d\leq n/2 1 ≤ d ≤ n /2 A ∈ Z A\in\mathbb{Z} A ∈ Z
U n , i ( A + d c n ( t n , i ) ) , 1 ≤ i ≤ q , U_{n,i}\left(A+dc_n(t_{n,i})\right),\qquad1\leq i\leq q, U n , i ( A + d c n ( t n , i ) ) , 1 ≤ i ≤ q , are nonzero.
Then, for every ε > 0 \varepsilon>0 ε > 0 N = N ( C , κ , q , r 0 , ε , n 0 ) N=N(C,\kappa,q,r_0,\varepsilon,n_0) N = N ( C , κ , q , r 0 , ε , n 0 ) P n ( t , X ) P_n(t,X) P n ( t , X ) Q \mathbb{Q} Q
n ≥ N , t ∈ Z > 0 , log t ≥ n ε . n\geq N,\qquad t\in\mathbb{Z}_{>0},\qquad\log t\geq n^\varepsilon. n ≥ N , t ∈ Z > 0 , log t ≥ n ε . The last condition says equivalently that, after omitting any one index j j j
gcd Q [ Z ] { U n , i ( Z + d c n ( t n , i ) ) : i ≠ j } = 1 for every d , j as above . (33) \underset{\mathbb{Q}[Z]}{\gcd}\{U_{n,i}(Z+dc_n(t_{n,i})):i\ne j\}=1\qquad\text{for every }d,j\text{ as above}. \tag*{(33)} Q [ Z ] g cd { U n , i ( Z + d c n ( t n , i )) : i = j } = 1 for every d , j as above . (33) The distinction is useful: the theorem only needs exclusion of integer roots, whereas a resultant can verify the stronger condition.
Proof. Suppose a monic factor has degree 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2 τ \tau τ τ ∈ Z \tau\in\mathbb{Z} τ ∈ Z A = τ − d c n ( t ) A=\tau-dc_n(t) A = τ − d c n ( t )
∣ A ∣ ≤ d C n κ ≤ C n κ + 1 . |A|\le dCn^\kappa\le Cn^{\kappa+1}. ∣ A ∣ ≤ d C n κ ≤ C n κ + 1 . By the third hypothesis, at least two of the integers
U n , i ( A + d c n ( t n , i ) ) U_{n,i}(A+dc_n(t_{n,i})) U n , i ( A + d c n ( t n , i )) are nonzero.
There are constants C 1 , K 1 C_1,K_1 C 1 , K 1 C , κ , r 0 C,\kappa,r_0 C , κ , r 0 C 1 n K 1 C_1n^{K_1} C 1 n K 1 R n , i R_{n,i} R n , i 1 + C n κ 1+Cn^\kappa 1 + C n κ r 0 ( 1 + C n κ ) r_0(1+Cn^\kappa) r 0 ( 1 + C n κ ) 2 r 0 2^{r_0} 2 r 0 U n , i U_{n,i} U n , i
∣ A + d c n ( t n , i ) ∣ ≤ 2 C n κ + 1 . |A+dc_n(t_{n,i})|\le2Cn^{\kappa+1}. ∣ A + d c n ( t n , i ) ∣ ≤ 2 C n κ + 1 . These estimates give, for example, a uniform bound
∣ U n , i ( A + d c n ( t n , i ) ) ∣ ≤ ( 2 C + r 0 ( 1 + C ) ) 2 r 0 n ( κ + 1 ) 2 r 0 . |U_{n,i}(A+dc_n(t_{n,i}))|\le(2C+r_0(1+C))^{2^{r_0}}n^{(\kappa+1)2^{r_0}}. ∣ U n , i ( A + d c n ( t n , i )) ∣ ≤ ( 2 C + r 0 ( 1 + C ) ) 2 r 0 n ( κ + 1 ) 2 r 0 . Choose indices i ≠ j i\ne j i = j n n n log t ≥ n ε \log t\ge n^\varepsilon log t ≥ n ε t > 2 C n κ t>2Cn^\kappa t > 2 C n κ t − t n , i t-t_{n,i} t − t n , i t − t n , j t-t_{n,j} t − t n , j Proposition 2.11 gives
log rad ( ( t − t n , i ) ( t − t n , j ) ) ≤ 2 log C 1 + 2 K 1 log n . \log\operatorname{rad}\bigl((t-t_{n,i})(t-t_{n,j})\bigr)\le2\log C_1+2K_1\log n. log rad ( ( t − t n , i ) ( t − t n , j ) ) ≤ 2 log C 1 + 2 K 1 log n . Their difference h h h 1 1 1 2 C n κ 2Cn^\kappa 2 C n κ M M M M ≥ t / 2 M\ge t/2 M ≥ t /2 Lemma 2.10 with R = max { e 4 , 2 log C 1 + 2 K 1 log n } R=\max\{e^4,2\log C_1+2K_1\log n\} R = max { e 4 , 2 log C 1 + 2 K 1 log n }
log M ≤ exp ( O ( log n log log log n log log n ) ) , \log M\le\exp\left(O\left(\frac{\log n\log\log\log n}{\log\log n}\right)\right), log M ≤ exp ( O ( log log n log n log log log n ) ) , for all sufficiently large n n n n o ( 1 ) n^{o(1)} n o ( 1 ) log M < n ε / 2 \log M<n^{\varepsilon/2} log M < n ε /2 n n n M ≥ t / 2 M\ge t/2 M ≥ t /2 log M ≥ n ε − log 2 > n ε / 2 \log M\ge n^\varepsilon-\log2>n^{\varepsilon/2} log M ≥ n ε − log 2 > n ε /2
Corollary 2.13. Under the hypotheses of Theorem 2.12 , suppose that, for some fixed δ > 0 \delta>0 δ > 0 P n ( t , X ) P_n(t,X) P n ( t , X ) 1 ≤ t ≤ exp ( n δ ) 1\le t\le\exp(n^\delta) 1 ≤ t ≤ exp ( n δ ) n n n P n ( t , X ) P_n(t,X) P n ( t , X ) t t t n n n
Proof. Apply the theorem with ε = δ \varepsilon=\delta ε = δ
Scope. The geometric hypothesis is stronger than necessary: a uniform bound for the centered trace of every rational factor suffices. Likewise, one may replace U n , i U_{n,i} U n , i δ \delta δ
The same proof also permits fibers of the form
P n ( t n , i , X ) = ( X − b n , i ) n − r n , i R n , i ( X − b n , i ) , b n , i ∈ Z . P_n(t_{n,i},X)=(X-b_{n,i})^{n-r_{n,i}}R_{n,i}(X-b_{n,i}),\qquad b_{n,i}\in\mathbb{Z}. P n ( t n , i , X ) = ( X − b n , i ) n − r n , i R n , i ( X − b n , i ) , b n , i ∈ Z . One adds the bound ∣ b n , i ∣ ≤ C n κ |b_{n,i}| \le Cn^{\kappa} ∣ b n , i ∣ ≤ C n κ c n ( t n , i ) c_n(t_{n,i}) c n ( t n , i ) c n ( t n , i ) − b n , i c_n(t_{n,i})-b_{n,i} c n ( t n , i ) − b n , i d d d d b n , i db_{n,i} d b n , i
The trace condition excludes persistent factors that the other hypotheses do not detect. For example, the Appell family n ! [ z n ] e X z ( 1 − z 2 ) − t n![z^n]e^{Xz}(1-z^2)^{-t} n ! [ z n ] e X z ( 1 − z 2 ) − t X X X n n n H n , t ( X ) H_{n,t}(X) H n , t ( X ) ( − 1 ) n H n , t ( − X ) (-1)^nH_{n,t}(-X) ( − 1 ) n H n , t ( − X ) ∣ Re X ∣ ≤ 2 ( n − 1 ) |\operatorname{Re} X| \le2(n-1) ∣ Re X ∣ ≤ 2 ( n − 1 ) t t t X X X
Corollary 2.14. Assume the hypotheses of Theorem 2.12 , with its strip bound valid for every real parameter t > 0 t > 0 t > 0 c n ( T ) c_n(T) c n ( T )
deg T ( [ X n − k ] P n ( T , X ) ) ≤ k ( 0 ≤ k ≤ n ) . \deg_T([X^{n-k}]P_n(T,X)) \le k \qquad(0 \le k \le n). deg T ([ X n − k ] P n ( T , X )) ≤ k ( 0 ≤ k ≤ n ) . For every fixed B ≥ 0 B \ge0 B ≥ 0 ε > 0 \varepsilon> 0 ε > 0 N = N ( C , κ , q , r 0 , n 0 , B , ε ) N = N(C,\kappa,q,r_0,n_0,B,\varepsilon) N = N ( C , κ , q , r 0 , n 0 , B , ε ) P n ( a / b , X ) P_n(a/b,X) P n ( a / b , X ) Q \mathbb{Q} Q
n ≥ N , a , b ∈ Z > 0 , ( a , b ) = 1 , b ≤ n B , log a ≥ n ε . n \ge N,\qquad a,b \in\mathbb{Z}_{>0},\qquad(a,b)=1,\qquad b \le n^B,\qquad\log a \ge n^\varepsilon. n ≥ N , a , b ∈ Z > 0 , ( a , b ) = 1 , b ≤ n B , log a ≥ n ε . Proof. The coefficient-degree assumption makes
F ( X ) = b n P n ( a / b , X / b ) F(X) = b^nP_n(a/b,X/b) F ( X ) = b n P n ( a / b , X / b ) monic and integral. If it has a monic integral factor of degree 1 ≤ d ≤ n / 2 1 \le d \le n/2 1 ≤ d ≤ n /2 τ \tau τ A = τ − d b c n ( a / b ) A = \tau- dbc_n(a/b) A = τ − d b c n ( a / b ) A ∈ Z A \in\mathbb{Z} A ∈ Z ∣ A / b ∣ ≤ C n κ + 1 |A/b| \le Cn^{\kappa+1} ∣ A / b ∣ ≤ C n κ + 1 e i = deg U n , i ≤ 2 r 0 e_i = \deg U_{n,i} \le2^{r_0} e i = deg U n , i ≤ 2 r 0 p ∣ a − b t n , i p \mid a-bt_{n,i} p ∣ a − b t n , i p ∤ b p \nmid b p ∤ b t n , i t_{n,i} t n , i
p ∣ b e i U n , i ( A / b + d c n ( t n , i ) ) . p \mid b^{e_i}U_{n,i}(A/b+dc_n(t_{n,i})). p ∣ b e i U n , i ( A / b + d c n ( t n , i )) . These evaluations are integers. At least two are nonzero: otherwise A / b A/b A / b
For all sufficiently large n n n a > 2 C b n κ a > 2Cb n^\kappa a > 2 C b n κ a − b t n , i a-bt_{n,i} a − b t n , i a / 2 a/2 a /2 Theorem 2.12 bound each nonzero evaluation by b 2 r 0 C 1 n K 1 b^{2r_0}C_1n^{K_1} b 2 r 0 C 1 n K 1 C 1 , K 1 C_1,K_1 C 1 , K 1 a − b t n , i a-bt_{n,i} a − b t n , i a − b t n , j a-bt_{n,j} a − b t n , j
log rad ( ( a − b t n , i ) ( a − b t n , j ) ) ≤ 2 log C 1 + 2 K 1 log n + 2 r 0 + 1 log b . \log\operatorname{rad}((a-bt_{n,i})(a-bt_{n,j})) \le2\log C_1+2K_1\log n+2^{r_0+1}\log b. log rad (( a − b t n , i ) ( a − b t n , j )) ≤ 2 log C 1 + 2 K 1 log n + 2 r 0 + 1 log b . Their positive difference is at most 2 C b n κ 2Cb n^\kappa 2 C b n κ b ≤ n B b \le n^B b ≤ n B O B ( log n ) O_B(\log n) O B ( log n ) O B ( log n ) O_B(\log n) O B ( log n ) Lemma 2.10 gives log a = n o B ( 1 ) \log a = n^{o_B(1)} log a = n o B ( 1 ) b b b log a ≥ n ε \log a \ge n^\varepsilon log a ≥ n ε F F F
Four degenerate Laguerre fibers
For relatively prime positive integers a , b a,b a , b
F n , a , b ( X ) = b n H n , a / b ( X / b ) = ∑ k = 0 n ( n k ) ( ∏ j = 0 k − 1 ( a + j b ) ) X n − k . F_{n,a,b}(X)=b^nH_{n,a/b}(X/b)=\sum_{k=0}^{n}\binom{n}{k}\left(\prod_{j=0}^{k-1}(a+jb)\right)X^{n-k}. F n , a , b ( X ) = b n H n , a / b ( X / b ) = k = 0 ∑ n ( k n ) ( j = 0 ∏ k − 1 ( a + j b ) ) X n − k . This polynomial is monic and integral. Its irreducibility over Q \mathbb{Q} Q H n , a / b H_{n,a/b} H n , a / b
Theorem 2.15. Let n ≥ 2 n \ge2 n ≥ 2 a , b a,b a , b F n , a , b F_{n,a,b} F n , a , b Q \mathbb{Q} Q 0 ≤ i < j ≤ 3 0 \le i < j \le3 0 ≤ i < j ≤ 3
rad ( ( a + i b ) ( a + j b ) ) < b 3 n 6 . \operatorname{rad}((a+ib)(a+jb)) < b^3n^6. rad (( a + ib ) ( a + j b )) < b 3 n 6 . Proof. For n = 2 n=2 n = 2 H 2 , a / b H_{2,a/b} H 2 , a / b − 4 a / b < 0 -4a/b<0 − 4 a / b < 0 n ≥ 3 n\ge3 n ≥ 3 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2 β 1 , … , β d \beta_1,\ldots,\beta_d β 1 , … , β d
A = ∑ ν = 1 d ( β ν + a ) A=\sum_{\nu=1}^{d}(\beta_\nu+a) A = ν = 1 ∑ d ( β ν + a ) is an integer. Since β ν / b \beta_\nu/b β ν / b H n , a / b H_{n,a/b} H n , a / b Proposition 2.2 gives
∣ A ∣ < b d ( n − d ) . (34) |A|<bd(n-d). \tag*{(34)} ∣ A ∣ < b d ( n − d ) . (34) At parameters s = 0 , − 1 , − 2 , − 3 s=0,-1,-2,-3 s = 0 , − 1 , − 2 , − 3 H n , s H_{n,s} H n , s
1 , X − n , X 2 − 2 n X + n ( n − 1 ) , R ( X ) , 1,\qquad X-n,\qquad X^2-2nX+n(n-1),\qquad R(X), 1 , X − n , X 2 − 2 n X + n ( n − 1 ) , R ( X ) , where
R ( X ) = ( X − n ) 3 − 3 n ( X − n ) − 2 n . R(X)=(X-n)^3-3n(X-n)-2n. R ( X ) = ( X − n ) 3 − 3 n ( X − n ) − 2 n . Polynomials annihilating all subset sums of the nonzero roots are
T 0 ( Z ) = Z , T_0(Z)=Z, T 0 ( Z ) = Z , T 1 ( Z ) = Z ( Z − n ) , T_1(Z)=Z(Z-n), T 1 ( Z ) = Z ( Z − n ) , T 2 ( Z ) = Z ( Z − 2 n ) ( ( Z − n ) 2 − n ) , T_2(Z)=Z(Z-2n)\bigl((Z-n)^2-n\bigr), T 2 ( Z ) = Z ( Z − 2 n ) ( ( Z − n ) 2 − n ) , T 3 ( Z ) = − Z ( Z − 3 n ) R ( Z ) R ( 3 n − Z ) . T_3(Z)=-Z(Z-3n)R(Z)R(3n-Z). T 3 ( Z ) = − Z ( Z − 3 n ) R ( Z ) R ( 3 n − Z ) . For the last identity, the subset sums are 0 0 0 3 n 3n 3 n R R R 3 n 3n 3 n
If p ∣ a + i b p\mid a+ib p ∣ a + ib p ∤ b p\nmid b p ∤ b gcd ( a , b ) = 1 \gcd(a,b)=1 g cd( a , b ) = 1 a = − i b a=-ib a = − ib A + i b d A+ibd A + ib d p p p
p ∣ b deg T i T i ( A / b + i d ) . (35) p\mid b^{\deg T_i}T_i(A/b+id). \tag*{(35)} p ∣ b d e g T i T i ( A / b + i d ) . (35) The quantity on the right is an integer. The assertion remains valid in every characteristic, since roots are counted with multiplicity. Whenever A / b A/b A / b b b b
First suppose A ∉ { 0 , − b d , b ( n − d ) } A\notin\{0,-bd,b(n-d)\} A ∈ / { 0 , − b d , b ( n − d )} ( i , j ) = ( 0 , 1 ) (i,j)=(0,1) ( i , j ) = ( 0 , 1 ) (35) shows that the required radical divides the nonzero integer
N = A ( A + b d ) ( A + b d − b n ) . N=A(A+bd)(A+bd-bn). N = A ( A + b d ) ( A + b d − bn ) . By (34) ,
∣ A ∣ < b d n , ∣ A + b d ∣ < b d n , ∣ A + b d − b n ∣ < b ( d + 1 ) n , |A|<bdn,\qquad|A+bd|<bdn,\qquad|A+bd-bn|<b(d+1)n, ∣ A ∣ < b d n , ∣ A + b d ∣ < b d n , ∣ A + b d − bn ∣ < b ( d + 1 ) n , and hence ∣ N ∣ < b 3 n 3 d 2 ( d + 1 ) < b 3 n 6 |N|<b^3n^3d^2(d+1)<b^3n^6 ∣ N ∣ < b 3 n 3 d 2 ( d + 1 ) < b 3 n 6
For the three exceptional trace values, the following integers contain the required prime supports whenever they are nonzero:
Table 2.28.
This follows by substitution in T 0 , T 1 , T 2 T_0,T_1,T_2 T 0 , T 1 , T 2 n 4 n^4 n 4 ( n , d ) = ( 4 , 2 ) (n,d)=(4,2) ( n , d ) = ( 4 , 2 ) n = m 2 n=m^2 n = m 2 n − d = m n-d=m n − d = m m 2 − m ≤ m 2 / 2 m^2-m\le m^2/2 m 2 − m ≤ m 2 /2 m ≤ 2 m\le2 m ≤ 2 A = − 2 b A=-2b A = − 2 b 0 0 0 3 3 3 T 0 ( − 2 ) = − 2 T_0(-2)=-2 T 0 ( − 2 ) = − 2 T 3 ( 4 ) = − 2048 T_3(4)=-2048 T 3 ( 4 ) = − 2048 2 2 2
The other two entries leave only the following cases:
n = m 2 , ( A , d ) = ( 0 , m 2 − m 2 ) , or n = m 2 , ( A , d ) = ( b ( m 2 − m ) , m ) . n=m^2,\quad(A,d)=\left(0,\frac{m^2-m}{2}\right),\quad\text{or}\quad n=m^2,\quad(A,d)=\left(b(m^2-m),m\right). n = m 2 , ( A , d ) = ( 0 , 2 m 2 − m ) , or n = m 2 , ( A , d ) = ( b ( m 2 − m ) , m ) . Here m ≥ 2 m\ge2 m ≥ 2
In the first case use fibers 1 and 3. Writing d = m ( m − 1 ) / 2 d=m(m-1)/2 d = m ( m − 1 ) /2
T 1 ( d ) = − m 2 ( m 2 − 1 ) 4 T_1(d)=-\frac{m^2(m^2-1)}{4} T 1 ( d ) = − 4 m 2 ( m 2 − 1 ) and
T 3 ( 3 d ) = 9 m 6 ( m 2 − 1 ) 3 ( m 2 − 9 m + 16 ) ( m 2 + 9 m + 16 ) 256 . T_3(3d)=\frac{9m^6(m^2-1)^3(m^2-9m+16)(m^2+9m+16)}{256}. T 3 ( 3 d ) = 256 9 m 6 ( m 2 − 1 ) 3 ( m 2 − 9 m + 16 ) ( m 2 + 9 m + 16 ) . Every prime divisor of either integer divides
N = 3 m ( m 2 − 1 ) ( m 2 − 9 m + 16 ) ( m 2 + 9 m + 16 ) . N=3m(m^2-1)(m^2-9m+16)(m^2+9m+16). N = 3 m ( m 2 − 1 ) ( m 2 − 9 m + 16 ) ( m 2 + 9 m + 16 ) . Neither quadratic factor vanishes at an integer: the first has discriminant 17, and the second is positive. For m ≥ 3 m\ge3 m ≥ 3
∣ m 4 − 49 m 2 + 256 ∣ ≤ m 4 + 49 m 2 + 256 < 10 m 4 , \left|m^4-49m^2+256\right|\le m^4+49m^2+256<10m^4, m 4 − 49 m 2 + 256 ≤ m 4 + 49 m 2 + 256 < 10 m 4 , so 0 < ∣ N ∣ < 30 m 7 < m 12 = n 6 0<|N|<30m^7<m^{12}=n^6 0 < ∣ N ∣ < 30 m 7 < m 12 = n 6 m = 2 m=2 m = 2 1368 < 4096 = m 12 1368<4096=m^{12} 1368 < 4096 = m 12
In the second case use fibers 0 and 3. Put
J ( m ) = m ( m − 1 ) ( m − 4 ) + 2. J(m)=m(m-1)(m-4)+2. J ( m ) = m ( m − 1 ) ( m − 4 ) + 2. Then
T 3 ( n + 2 m ) = 4 m 6 ( m − 1 ) 3 ( m + 2 ) J ( m ) . T_3(n+2m)=4m^6(m-1)^3(m+2)J(m). T 3 ( n + 2 m ) = 4 m 6 ( m − 1 ) 3 ( m + 2 ) J ( m ) . The prime supports of n − m = m ( m − 1 ) n-m=m(m-1) n − m = m ( m − 1 )
N = 2 m ( m − 1 ) ( m + 2 ) J ( m ) . N=2m(m-1)(m+2)J(m). N = 2 m ( m − 1 ) ( m + 2 ) J ( m ) . Here J ( 2 ) = − 2 J(2)=-2 J ( 2 ) = − 2 J ( 3 ) = − 4 J(3)=-4 J ( 3 ) = − 4 0 < J ( m ) < m 3 0<J(m)<m^3 0 < J ( m ) < m 3 m ≥ 4 m\ge4 m ≥ 4 0 < ∣ J ( m ) ∣ < m 3 0<|J(m)|<m^3 0 < ∣ J ( m ) ∣ < m 3 m ≥ 2 m\ge2 m ≥ 2 0 < ∣ N ∣ < 4 m 6 < m 12 = n 6 0<|N|<4m^6<m^{12}=n^6 0 < ∣ N ∣ < 4 m 6 < m 12 = n 6 b b b
Corollary 2.16. For every fixed B ≥ 0 B\ge0 B ≥ 0 ε > 0 \varepsilon>0 ε > 0 N ( B , ε ) N(B,\varepsilon) N ( B , ε ) F n , a , b F_{n,a,b} F n , a , b Q \mathbb{Q} Q
n ≥ N ( B , ε ) , gcd ( a , b ) = 1 , 1 ≤ b ≤ n B , log a ≥ n ε . n\ge N(B,\varepsilon),\qquad\gcd(a,b)=1,\qquad1\le b\le n^B,\qquad\log a\ge n^\varepsilon. n ≥ N ( B , ε ) , g cd( a , b ) = 1 , 1 ≤ b ≤ n B , log a ≥ n ε . More explicitly, a reducible member with n ≥ 2 n\ge2 n ≥ 2
log a < max { 2 log ( 6 b ) , 4 C R log ( 12 e C R ) } , R = max { e 4 , 3 log b + 6 log n } , \log a<\max\{2\log(6b),\,4C_R\log(12eC_R)\},\qquad R=\max\{e^4,\,3\log b+6\log n\}, log a < max { 2 log ( 6 b ) , 4 C R log ( 12 e C R )} , R = max { e 4 , 3 log b + 6 log n } , where C R C_R C R
Proof. Theorem 2.15 supplies M = a + i b ≥ a M=a+ib\ge a M = a + ib ≥ a h = ( j − i ) b ≤ 3 b h=(j-i)b\le3b h = ( j − i ) b ≤ 3 b b ≤ n B b\le n^B b ≤ n B u = log n u=\log n u = log n R = O B ( u ) R=O_B(u) R = O B ( u )
log C R = log ( ( e / 2 ) 30 3 ) + 9 2 log 2 R log R + 2 R log ( 15 log R ) log R = o B ( u ) . \log C_R=\log((e/2)^{30^3})+\frac{9}{2}\log\frac{2R}{\log R}+\frac{2R\log(15\log R)}{\log R}=o_B(u). log C R = log (( e /2 ) 3 0 3 ) + 2 9 log log R 2 R + log R 2 R log ( 15 log R ) = o B ( u ) . It follows that 4 C R log ( 12 e C R ) = exp ( o B ( u ) ) = n o B ( 1 ) 4C_R\log(12eC_R)=\exp(o_B(u))=n^{o_B(1)} 4 C R log ( 12 e C R ) = exp ( o B ( u )) = n o B ( 1 ) 2 log ( 6 b ) = O B ( u ) 2\log(6b)=O_B(u) 2 log ( 6 b ) = O B ( u ) n ε n^\varepsilon n ε
Positive rational parameters can give reducible polynomials. For example,
H 3 , 1 / 4 ( X ) = ( X + 3 4 ) ( X 2 + 15 16 ) , F 3 , 1 , 4 ( X ) = ( X + 3 ) ( X 2 + 15 ) . H_{3,1/4}(X)=\left(X+\frac{3}{4}\right)\left(X^2+\frac{15}{16}\right),\qquad F_{3,1,4}(X)=(X+3)(X^2+15). H 3 , 1/4 ( X ) = ( X + 4 3 ) ( X 2 + 16 15 ) , F 3 , 1 , 4 ( X ) = ( X + 3 ) ( X 2 + 15 ) . Thus the numerator and denominator conditions in the rational theorem have genuine content; positivity alone is not sufficient.
Uniform irreducibility
Local arithmetic and energy
A rational factor selects a Galois-invariant set of roots; a block system gives an invariant partition. Both restrict how roots can occupy a local cluster. We express these restrictions by positive semidefinite matrices and compare the resulting local contributions with a bound from complex root separation. The same inequality will therefore exclude factors and block systems.
We first determine the tame clusters of H n , s H_{n,s} H n , s
Tame root clusters and inertia
The local calculation rests on the following form of Hensel’s lemma. No irreducibility assumption is needed.
Lemma 3.1. Let F ∈ Z p [ X ] F \in\mathbb{Z}_p[X] F ∈ Z p [ X ] F ‾ = X N R ‾ \overline{F}=X^N\overline{R} F = X N R 1 ≤ N ≤ deg F 1 \le N \le\deg F 1 ≤ N ≤ deg F R ‾ \overline{R} R R ‾ ( 0 ) ≠ 0 \overline{R}(0) \ne0 R ( 0 ) = 0 A A A N N N X N X^N X N p ∤ N p \nmid N p ∤ N e > 0 e>0 e > 0 A A A e e e v p ( A ( 0 ) ) = e v_p(A(0))=e v p ( A ( 0 )) = e g = gcd ( N , e ) g=\gcd(N,e) g = g cd( N , e ) q = N / g q=N/g q = N / g A A A e / N e/N e / N A A A Q p \mathbb{Q}_p Q p q q q q g 1 deg F − N q^g1^{\deg F-N} q g 1 d e g F − N
Proof. Write N = g q N=gq N = g q e = g a e=ga e = g a gcd ( a , q ) = 1 \gcd(a,q)=1 g cd( a , q ) = 1 K 0 / Q p K_0/\mathbb{Q}_p K 0 / Q p R ‾ \overline{R} R Y N + c Y^N+c Y N + c c c c p − e A ( 0 ) p^{-e}A(0) p − e A ( 0 ) q q q p ∤ q p\nmid q p ∤ q π q = p \pi^q=p π q = p p − e A ( π a Y ) p^{-e}A(\pi^aY) p − e A ( π a Y ) Y N + c Y^N+c Y N + c p ∤ N p\nmid N p ∤ N K 0 ( π ) K_0(\pi) K 0 ( π ) A A A K 0 K_0 K 0
The splitting field is therefore contained in the tame extension K 0 ( π ) K_0(\pi) K 0 ( π ) q q q a / q a/q a / q A A A q q q q q q g g g q q q □ \square □
For the Laguerre family, a prime above the degree can divide only one of the consecutive parameter nodes s , … , s + n − 1 s,\ldots,s+n-1 s , … , s + n − 1
Proposition 3.2. Let n ≥ 2 n \ge2 n ≥ 2 s ≥ 1 s \ge1 s ≥ 1 p > n p>n p > n p ∣ s + j p\mid s+j p ∣ s + j 0 ≤ j < n 0 \le j<n 0 ≤ j < n
e = v p ( s + j ) , N = n − j , g = gcd ( N , e ) , q = N / g . e=v_p(s+j),\qquad N=n-j,\qquad g=\gcd(N,e),\qquad q=N/g. e = v p ( s + j ) , N = n − j , g = g cd( N , e ) , q = N / g . In the splitting field of H n , s H_{n,s} H n , s p p p q q q q g 1 j q^g1^j q g 1 j N ∣ e N\mid e N ∣ e
Proof. There is at most one node divisible by p p p n n n p p p
H n , s ( X ) ≡ X N R n , j ( X ) ( m o d p ) , R n , j ( X ) = ∑ k = 0 j ( − 1 ) k ( n k ) j ! ( j − k ) ! X j − k . H_{n,s}(X) \equiv X^N R_{n,j}(X) \pmod p,\qquad R_{n,j}(X)=\sum_{k=0}^{j}(-1)^k\binom{n}{k}\frac{j!}{(j-k)!}X^{j-k}. H n , s ( X ) ≡ X N R n , j ( X ) ( mod p ) , R n , j ( X ) = k = 0 ∑ j ( − 1 ) k ( k n ) ( j − k )! j ! X j − k . The constant coefficient of R n , j R_{n,j} R n , j R n , j = ( − 1 ) j j ! L j ( N ) R_{n,j}=(-1)^j j!L_j^{(N)} R n , j = ( − 1 ) j j ! L j ( N )
X R n , j ′ ′ + ( N + 1 − X ) R n , j ′ + j R n , j = 0. (36) X R_{n,j}''+(N+1-X)R_{n,j}'+jR_{n,j}=0. \tag*{(36)} X R n , j ′′ + ( N + 1 − X ) R n , j ′ + j R n , j = 0. (36) The residual polynomial is separable. Indeed, a repeated nonzero root, followed through successive derivatives of (3.1), would make all derivatives through order j j j R n , j ( j ) = j ! ≠ 0 R_{n,j}^{(j)}=j!\ne0 R n , j ( j ) = j ! = 0 p p p j = 0 j=0 j = 0
Hensel factorization gives H n , s = F U H_{n,s}=FU H n , s = F U Z p \mathbb{Z}_p Z p F , U F,U F , U deg F = N \deg F=N deg F = N F ‾ = X N \overline{F}=X^N F = X N U ‾ = R n , j \overline{U}=R_{n,j} U = R n , j H n , s H_{n,s} H n , s N N N e e e U ( 0 ) U(0) U ( 0 ) H n , s = F U H_{n,s}=FU H n , s = F U
v p ( F ( 0 ) ) = e , v p ( [ X h ] F ) ≥ e ( 0 < h < N ) . v_p(F(0))=e,\qquad v_p([X^h]F)\ge e\quad(0<h<N). v p ( F ( 0 )) = e , v p ([ X h ] F ) ≥ e ( 0 < h < N ) . Lemma 3.1 now applies and gives the stated inertia group and cycle type. Its order is one exactly when N ∣ e N\mid e N ∣ e
Root separation
A valuation packet is the full set of roots with a specified valuation at a fixed prime. For the Laguerre family put
ϵ p = { 1 / ( p − 1 ) , p ≤ n , 0 , p > n . \epsilon_p=
\begin{cases}
1/(p-1),&p\le n,\\
0,&p>n.
\end{cases} ϵ p = { 1/ ( p − 1 ) , 0 , p ≤ n , p > n . The differential equation bounds the distance between distinct roots at every place. The p p p
Lemma 3.3. Let n ≥ 2 n\ge2 n ≥ 2 s ≥ 1 s\ge1 s ≥ 1 H n , s H_{n,s} H n , s
v p ( α − β ) ≤ v p ( α ) + ϵ p . (37) v_p(\alpha-\beta)\le v_p(\alpha)+\epsilon_p. \tag*{(37)} v p ( α − β ) ≤ v p ( α ) + ϵ p . (37) If s ≥ n s\ge n s ≥ n
∣ α − β ∣ > 1 12 s n . (38) |\alpha-\beta|>\frac{1}{12}\sqrt{\frac{s}{n}}. \tag*{(38)} ∣ α − β ∣ > 12 1 n s . (38) Proof. The roots are simple by § 2.1 . Fix a root α \alpha α
D k = H n , s ( k ) ( α ) H n , s ′ ( α ) , D 0 = 0 , D 1 = 1. D_k=\frac{H_{n,s}^{(k)}(\alpha)}{H_{n,s}'(\alpha)},\qquad D_0=0,\qquad D_1=1. D k = H n , s ′ ( α ) H n , s ( k ) ( α ) , D 0 = 0 , D 1 = 1. Differentiating (2.5) k k k 0 ≤ k ≤ n − 2 0\le k\le n-2 0 ≤ k ≤ n − 2
α D k + 2 − ( α + n + s − 1 − k ) D k + 1 + ( n − k ) D k = 0. (39) \alpha D_{k+2}-(\alpha+n+s-1-k)D_{k+1}+(n-k)D_k=0. \tag*{(39)} α D k + 2 − ( α + n + s − 1 − k ) D k + 1 + ( n − k ) D k = 0. (39) The roots are algebraic integers. If a = v p ( α ) ≥ 0 a=v_p(\alpha)\ge0 a = v p ( α ) ≥ 0 v p ( D k ) ≥ − ( k − 1 ) a v_p(D_k)\ge-(k-1)a v p ( D k ) ≥ − ( k − 1 ) a k ≥ 1 k\ge1 k ≥ 1 v p ( z ) > a + ϵ p v_p(z)>a+\epsilon_p v p ( z ) > a + ϵ p
H n , s ( α + z ) H n , s ′ ( α ) z = 1 + ∑ k = 2 n D k k ! z k − 1 \frac{H_{n,s}(\alpha+z)}{H_{n,s}'(\alpha)z}=1+\sum_{k=2}^{n}\frac{D_k}{k!}z^{k-1} H n , s ′ ( α ) z H n , s ( α + z ) = 1 + k = 2 ∑ n k ! D k z k − 1 has positive valuation. Here we use v p ( k ! ) ≤ ( k − 1 ) / ( p − 1 ) v_p(k!)\le(k-1)/(p-1) v p ( k !) ≤ ( k − 1 ) / ( p − 1 ) p ≤ n p\le n p ≤ n v p ( k ! ) = 0 v_p(k!)=0 v p ( k !) = 0 p > n p>n p > n
For the complex estimate put t = n / s ≤ 1 t=n/s\le1 t = n / s ≤ 1 (14) and centered-radius estimate (16) imply, for 0 ≤ k ≤ n − 2 0\le k\le n-2 0 ≤ k ≤ n − 2
∣ α + n + s − 1 − k α ∣ ≤ 2 t + 2 2 t < 5 t , ∣ n − k α ∣ ≤ t . \left|\frac{\alpha+n+s-1-k}{\alpha}\right|\le2t+2\sqrt{2t}<5\sqrt{t},\qquad\left|\frac{n-k}{\alpha}\right|\le t. α α + n + s − 1 − k ≤ 2 t + 2 2 t < 5 t , α n − k ≤ t . With M = 6 t M=6\sqrt{t} M = 6 t 5 t M + t < M 2 5\sqrt{t}M+t<M^2 5 t M + t < M 2 ∣ D k ∣ ≤ M k − 1 |D_k|\le M^{k-1} ∣ D k ∣ ≤ M k − 1 0 < ∣ z ∣ ≤ 1 / ( 2 M ) 0<|z|\le1/(2M) 0 < ∣ z ∣ ≤ 1/ ( 2 M )
∑ k ≥ 2 2 1 − k k ! = 2 e 1 / 2 − 3 < 1. \sum_{k\ge2}\frac{2^{1-k}}{k!}=2e^{1/2}-3<1. k ≥ 2 ∑ k ! 2 1 − k = 2 e 1/2 − 3 < 1. This proves (38) . □
An adelic energy inequality
The local positivity and the complex estimate below are elementary logarithmic-energy arguments. Galois invariance allows their comparison by the product formula, even when the matrix entries are arbitrary real numbers. The condition Q 1 = 0 Q\mathbf{1}=0 Q 1 = 0 Q Q Q [12 Sections 4–5] .
Theorem 3.4. Let F ∈ Z [ X ] F\in\mathbb{Z}[X] F ∈ Z [ X ] n ≥ 2 n\ge2 n ≥ 2 α 1 , … , α n \alpha_1,\ldots,\alpha_n α 1 , … , α n p p p v p v_p v p Q ‾ \overline{\mathbb{Q}} Q v p ( p ) = 1 v_p(p)=1 v p ( p ) = 1 ϵ p ≥ 0 \epsilon_p\ge0 ϵ p ≥ 0 p p p
v p ( α i − α j ) ≤ v p ( α i ) + ϵ p ( i ≠ j ) . (40) v_p(\alpha_i-\alpha_j)\le v_p(\alpha_i)+\epsilon_p \qquad(i\ne j). \tag*{(40)} v p ( α i − α j ) ≤ v p ( α i ) + ϵ p ( i = j ) . (40) Let Q = ( q i j ) Q=(q_{ij}) Q = ( q ij ) Q 1 = 0 Q\mathbf{1}=0 Q 1 = 0 τ = tr Q \tau=\operatorname{tr}Q τ = tr Q
E p ( Q ) = ∑ i ≠ j q i j v p ( α i − α j ) + ∑ i q i i ( v p ( α i ) + ϵ p ) . E_p(Q)=\sum_{i\ne j}q_{ij}v_p(\alpha_i-\alpha_j)+\sum_i q_{ii}\bigl(v_p(\alpha_i)+\epsilon_p\bigr). E p ( Q ) = i = j ∑ q ij v p ( α i − α j ) + i ∑ q ii ( v p ( α i ) + ϵ p ) . Then only finitely many E p ( Q ) E_p(Q) E p ( Q )
E p ( Q ) ≥ ∫ 0 ∞ 1 C p ( t ) T Q 1 C p ( t ) d t ≥ 0 , C p ( t ) = { i : v p ( α i ) ≥ t } . (41) E_p(Q)\ge\int_0^\infty\mathbf{1}_{C_p(t)}^{\mathsf T}Q\mathbf{1}_{C_p(t)}\,\mathrm{d}t\ge0,\qquad C_p(t)=\{i:v_p(\alpha_i)\ge t\}. \tag*{(41)} E p ( Q ) ≥ ∫ 0 ∞ 1 C p ( t ) T Q 1 C p ( t ) d t ≥ 0 , C p ( t ) = { i : v p ( α i ) ≥ t } . (41) Furthermore,
∑ p E p ( Q ) log p = ∑ i ≠ j q i j log ∣ α i − α j ∣ + ∑ i q i i log ∣ α i ∣ + τ ∑ p ϵ p log p . (42) \sum_p E_p(Q)\log p=\sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|+\sum_iq_{ii}\log|\alpha_i|+\tau\sum_p\epsilon_p\log p. \tag*{(42)} p ∑ E p ( Q ) log p = i = j ∑ q ij log ∣ α i − α j ∣ + i ∑ q ii log ∣ α i ∣ + τ p ∑ ϵ p log p . (42) If the closed disks of radii ρ i > 0 \rho_i>0 ρ i > 0
0 ≤ ∑ p E p ( Q ) log p ≤ ∑ i q i i log ∣ α i ∣ ρ i + τ ∑ p ϵ p log p . (43) 0\le\sum_p E_p(Q)\log p\le\sum_iq_{ii}\log\frac{|\alpha_i|}{\rho_i}+\tau\sum_p\epsilon_p\log p. \tag*{(43)} 0 ≤ p ∑ E p ( Q ) log p ≤ i ∑ q ii log ρ i ∣ α i ∣ + τ p ∑ ϵ p log p . (43) In particular, if ∣ α i ∣ ≤ R |\alpha_i|\le R ∣ α i ∣ ≤ R ρ i = ρ \rho_i=\rho ρ i = ρ τ ( log ( R / ρ ) + ∑ p ϵ p log p ) \tau\bigl(\log(R/\rho)+\sum_p\epsilon_p\log p\bigr) τ ( log ( R / ρ ) + ∑ p ϵ p log p )
If, at some prime with ϵ p = 0 \epsilon_p=0 ϵ p = 0 C C C N ≥ 1 N\ge1 N ≥ 1 e / N e/N e / N e > 0 e>0 e > 0 e / N e/N e / N C C C p p p
E p ( Q ) = e N 1 C T Q 1 C . (44) E_p(Q)=\frac{e}{N}\mathbf{1}_C^{\mathsf T}Q\mathbf{1}_C. \tag*{(44)} E p ( Q ) = N e 1 C T Q 1 C . (44) The theorem can also be applied after replacing each α i \alpha_i α i α i − a \alpha_i-a α i − a a a a F ( a ) ≠ 0 F(a)\ne0 F ( a ) = 0 (40) is verified for the shifted roots; the correction terms may change.
Proof. Write h i = v p ( α i ) + ϵ p h_i=v_p(\alpha_i)+\epsilon_p h i = v p ( α i ) + ϵ p t > 0 t>0 t > 0 h i ≥ t h_i\ge t h i ≥ t v p ( α i − α j ) ≥ t v_p(\alpha_i-\alpha_j)\ge t v p ( α i − α j ) ≥ t t t t
E p ( Q ) = ∫ 0 ∞ ∑ C 1 C T Q 1 C d t . E_p(Q)=\int_0^\infty\sum_C \mathbf{1}_C^{\mathsf T}Q\mathbf{1}_C\,dt. E p ( Q ) = ∫ 0 ∞ C ∑ 1 C T Q 1 C d t . Every term is nonnegative. The indices with v p ( α i ) ≥ t v_p(\alpha_i)\ge t v p ( α i ) ≥ t t t t ϵ p \epsilon_p ϵ p F ( 0 ) F(0) F ( 0 ) Disc F \operatorname{Disc}F Disc F
To apply the product formula with real matrix entries, first group the terms by Galois orbits. The entries q i i q_{ii} q ii I I I q i j q_{ij} q ij O \mathcal{O} O
∏ i ∈ I α i and ∏ { i , j } ∈ O ( α i − α j ) 2 \prod_{i\in I}\alpha_i\qquad\text{and}\qquad\prod_{\{i,j\}\in\mathcal{O}}(\alpha_i-\alpha_j)^2 i ∈ I ∏ α i and { i , j } ∈ O ∏ ( α i − α j ) 2 are nonzero rational numbers. Apply the product formula to each one, weight the resulting identity by the corresponding real matrix entry, and sum over the orbits. The squares in the pair products account for the two ordered terms q i j q_{ij} q ij q j i q_{ji} q j i τ ∑ p ϵ p log p \tau\sum_p\epsilon_p\log p τ ∑ p ϵ p log p
Decompose Q = ∑ ν c ν c ν T Q=\sum_\nu c_\nu c_\nu^{\mathsf T} Q = ∑ ν c ν c ν T c ν c_\nu c ν 1 \mathbf{1} 1 c c c σ i \sigma_i σ i ρ i \rho_i ρ i α i \alpha_i α i η = ∑ i c i σ i \eta=\sum_i c_i\sigma_i η = ∑ i c i σ i
U ( z ) = ∑ i c i log max { ρ i , ∣ z − α i ∣ } . U(z)=\sum_i c_i\log\max\{\rho_i,|z-\alpha_i|\}. U ( z ) = i ∑ c i log max { ρ i , ∣ z − α i ∣ } . Because ∑ i c i = 0 \sum_i c_i=0 ∑ i c i = 0 U ( z ) = O ( ∣ z ∣ − 1 ) U(z)=O(|z|^{-1}) U ( z ) = O ( ∣ z ∣ − 1 ) ∇ U ( z ) = O ( ∣ z ∣ − 2 ) \nabla U(z)=O(|z|^{-2}) ∇ U ( z ) = O ( ∣ z ∣ − 2 ) Δ U = 2 π η \Delta U=2\pi\eta Δ U = 2 π η
∬ log ∣ z − w ∣ d η ( z ) d η ( w ) = − 1 2 π ∫ C ∣ ∇ U ∣ 2 d A ≤ 0. \iint\log|z-w|\,d\eta(z)\,d\eta(w)=-\frac{1}{2\pi}\int_{\mathbb{C}}|\nabla U|^2\,dA\le0. ∬ log ∣ z − w ∣ d η ( z ) d η ( w ) = − 2 π 1 ∫ C ∣∇ U ∣ 2 d A ≤ 0. The radial derivative jumps by c i / ρ i c_i/\rho_i c i / ρ i i i i log ∣ α i − α j ∣ \log|\alpha_i-\alpha_j| log ∣ α i − α j ∣ i i i log ρ i \log\rho_i log ρ i ν \nu ν
∑ i ≠ j q i j log ∣ α i − α j ∣ + ∑ i q i i log ρ i ≤ 0. \sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|+\sum_iq_{ii}\log\rho_i\le0. i = j ∑ q ij log ∣ α i − α j ∣ + i ∑ q ii log ρ i ≤ 0. Together with (3.7), this proves (3.8). Translating by an integer preserves monicity, integrality, the differences, and the Galois action. □
Lemma 3.5. Let n ≥ 2 n\ge2 n ≥ 2 s ≥ 1 s\ge1 s ≥ 1 Q Q Q H n , s H_{n,s} H n , s
ϵ p = { 1 / ( p − 1 ) , p ≤ n , 0 , p > n . \epsilon_p=
\begin{cases}
1/(p-1), & p\le n,\\
0, & p>n.
\end{cases} ϵ p = { 1/ ( p − 1 ) , 0 , p ≤ n , p > n . Then (3.6)–(3.8) hold. If p > n p>n p > n p e ∥ s + j p^e\Vert s+j p e ∥ s + j 0 ≤ j < n 0\le j<n 0 ≤ j < n N = n − j N=n-j N = n − j C C C p p p ∣ C ∣ = N |C|=N ∣ C ∣ = N
E p ( Q ) = e N 1 C T Q 1 C . (45) E_p(Q)=\frac{e}{N}\mathbf{1}_C^{\mathsf T}Q\mathbf{1}_C. \tag*{(45)} E p ( Q ) = N e 1 C T Q 1 C . (45) If s ≥ n s \ge n s ≥ n log n ≥ 10000 \log n \ge10000 log n ≥ 10000 Q ≠ 0 Q \ne0 Q = 0
∑ p E p ( Q ) log p < tr ( Q ) ( 1 2 log s + 3 2 log n + log 48 ) . \sum_p E_p(Q)\log p < \operatorname{tr}(Q)\left(\frac{1}{2}\log s+\frac{3}{2}\log n+\log48\right). p ∑ E p ( Q ) log p < tr ( Q ) ( 2 1 log s + 2 3 log n + log 48 ) . Proof. Lemma 3.3 supplies the local-separation hypothesis. The proof of Proposition 3.2 identifies the N N N ∣ α i ∣ < s + n ≤ 2 s |\alpha_i|<s+n\le2s ∣ α i ∣ < s + n ≤ 2 s ρ = s / n / 24 \rho=\sqrt{s/n}/24 ρ = s / n /24 ∑ p ≤ n log p / ( p − 1 ) < log n \sum_{p\le n}\log p/(p-1)<\log n ∑ p ≤ n log p / ( p − 1 ) < log n
Proposition 3.6. In the notation of Theorem 3.4, a Galois-invariant subset A A A k k k 0 < k < n 0<k<n 0 < k < n Q = c c T Q=cc^{\mathsf T} Q = c c T
c i = { 1 − k / n , i ∈ A , − k / n , i ∉ A , tr Q = k ( n − k ) n . c_i=
\begin{cases}
1-k/n, & i\in A,\\
-k/n, & i\notin A,
\end{cases}
\qquad
\operatorname{tr} Q=\frac{k(n-k)}{n}. c i = { 1 − k / n , − k / n , i ∈ A , i ∈ / A , tr Q = n k ( n − k ) . For an exact packet C C C N N N m = ∣ A ∩ C ∣ m=|A\cap C| m = ∣ A ∩ C ∣
E p ( Q ) = e N ( m − k N n ) 2 . E_p(Q)=\frac{e}{N}\left(m-\frac{kN}{n}\right)^2. E p ( Q ) = N e ( m − n k N ) 2 . A Galois-invariant partition into a ≥ 2 a\ge2 a ≥ 2 B 1 , … , B a B_1,\ldots,B_a B 1 , … , B a b ≥ 2 b\ge2 b ≥ 2 n = a b n=ab n = ab
q i j = 1 { i , j in the same block } b − 1 n , tr Q = a − 1. q_{ij}=\frac{\mathbf{1}_{\{i,j\text{ in the same block}\}}}{b}-\frac{1}{n},
\qquad
\operatorname{tr} Q=a-1. q ij = b 1 { i , j in the same block } − n 1 , tr Q = a − 1. For an exact packet and k ℓ = ∣ B ℓ ∩ C ∣ k_\ell=|B_\ell\cap C| k ℓ = ∣ B ℓ ∩ C ∣
E p ( Q ) = e N ( 1 b ∑ ℓ = 1 a k ℓ 2 − N 2 n ) . (46) E_p(Q)=\frac{e}{N}\left(\frac{1}{b}\sum_{\ell=1}^{a}k_\ell^2-\frac{N^2}{n}\right). \tag*{(46)} E p ( Q ) = N e ( b 1 ℓ = 1 ∑ a k ℓ 2 − n N 2 ) . (46) If an inertia element of prime order P > a P>a P > a C C C N = P ℓ N=P\ell N = P ℓ ℓ = a u + r \ell=au+r ℓ = a u + r 0 ≤ r < a 0\le r<a 0 ≤ r < a
E p ( Q ) ≥ e P 2 r ( a − r ) n N . (47) E_p(Q)\ge\frac{eP^2r(a-r)}{nN}. \tag*{(47)} E p ( Q ) ≥ n N e P 2 r ( a − r ) . (47) When the uniform disk and modulus bounds ρ , R \rho,R ρ , R log p \log p log p ( a − 1 ) ( log ( R / ρ ) + ∑ p ϵ p log p ) (a-1)(\log(R/\rho)+\sum_p\epsilon_p\log p) ( a − 1 ) ( log ( R / ρ ) + ∑ p ϵ p log p ) P > b P>b P > b C C C b P bP b P ∣ C ∣ |C| ∣ C ∣
Proof. The rank-one formula is immediate. The block matrix is the orthogonal projector onto the vectors constant on each block and orthogonal to 1 \mathbf{1} 1 P > a P>a P > a a a a P P P k ℓ k_\ell k ℓ P P P ℓ \ell ℓ a − r a-r a − r u u u r r r u + 1 u+1 u + 1
∑ ℓ k ℓ 2 ≥ P 2 ( ℓ 2 a + r ( a − r ) a ) . \sum_{\ell}k_\ell^2\ge P^2\left(\frac{\ell^2}{a}+\frac{r(a-r)}{a}\right). ℓ ∑ k ℓ 2 ≥ P 2 ( a ℓ 2 + a r ( a − r ) ) . Substitution proves (3.12); summing it and using nonnegativity at all other primes proves the energy criterion. For the final divisibility assertion, any fixed block has fewer than P P P P P P b b b
Corollary 3.7. Under the hypotheses of Theorem 3.4 , suppose ∣ α i ∣ ≤ R |\alpha_i| \le R ∣ α i ∣ ≤ R ρ > 0 \rho> 0 ρ > 0
B = log ( R / ρ ) + ∑ p ϵ p log p . \mathcal{B}=\log(R/\rho)+\sum_p\epsilon_p\log p. B = log ( R / ρ ) + p ∑ ϵ p log p . Let P \mathcal{P} P ϵ p = 0 \epsilon_p=0 ϵ p = 0 N p N_p N p e p / N p e_p/N_p e p / N p q p q_p q p 1 ≤ k ≤ ⌊ n / 2 ⌋ 1\le k\le\lfloor n/2\rfloor 1 ≤ k ≤ ⌊ n /2 ⌋
M p , k = { m ∈ Z : max ( 0 , k − n + N p ) ≤ m ≤ min ( k , N p ) , q p ∣ m } . \mathcal{M}_{p,k}=\{m\in\mathbb{Z}:\max(0,k-n+N_p)\le m\le\min(k,N_p),\quad q_p\mid m\}. M p , k = { m ∈ Z : max ( 0 , k − n + N p ) ≤ m ≤ min ( k , N p ) , q p ∣ m } . If, for every such k k k M p , k \mathcal{M}_{p,k} M p , k
∑ p ∈ P e p log p n N p k ( n − k ) min m ∈ M p , k ( n m − k N p ) 2 > B , (48) \sum_{p\in\mathcal{P}}\frac{e_p\log p}{nN_pk(n-k)}\min_{m\in\mathcal{M}_{p,k}}(nm-kN_p)^2>\mathcal{B}, \tag*{(48)} p ∈ P ∑ n N p k ( n − k ) e p log p m ∈ M p , k min ( nm − k N p ) 2 > B , (48) then F F F Q \mathbb{Q} Q Lemma 3.1 , one may take q p = N p / gcd ( N p , e p ) q_p=N_p/\gcd(N_p,e_p) q p = N p / g cd( N p , e p )
Proof. A proper rational factor has degree k ≤ n / 2 k\le n/2 k ≤ n /2 M p , k \mathcal{M}_{p,k} M p , k Proposition 3.6 . Dividing its local energy by tr Q = k ( n − k ) / n \operatorname{tr}Q=k(n-k)/n tr Q = k ( n − k ) / n B \mathcal{B} B
Differential equations as a source of energy bounds
The proof of Lemma 3.3 applies more generally to a second-order differential equation. Its leading coefficient replaces the root in the diagonal term of the energy. Since that coefficient need only be nonzero at the roots, the polynomial itself may have a zero root.
Proposition 3.8. Let P ∈ Z [ X ] P\in\mathbb{Z}[X] P ∈ Z [ X ] n ≥ 2 n\ge2 n ≥ 2
A P ′ ′ + B P ′ + C P = 0 , A , B ∈ Z [ X ] , C ∈ Z , deg A ≤ 2 , deg B ≤ 1 , gcd ( A , P ) = 1. AP''+BP'+CP=0,\qquad A,B\in\mathbb{Z}[X],\quad C\in\mathbb{Z},\quad\deg A\le2,\quad\deg B\le1,\quad\gcd(A,P)=1. A P ′′ + B P ′ + C P = 0 , A , B ∈ Z [ X ] , C ∈ Z , deg A ≤ 2 , deg B ≤ 1 , g cd( A , P ) = 1. Write its roots as α i \alpha_i α i ϵ p = 1 / ( p − 1 ) \epsilon_p=1/(p-1) ϵ p = 1/ ( p − 1 ) p ≤ n p\le n p ≤ n ϵ p = 0 \epsilon_p=0 ϵ p = 0
v p ( α i − α j ) ≤ min { v p ( A ( α i ) ) , v p ( A ( α j ) ) } + ϵ p ( i ≠ j ) . v_p(\alpha_i-\alpha_j)\le\min\{v_p(A(\alpha_i)),v_p(A(\alpha_j))\}+\epsilon_p\qquad(i\ne j). v p ( α i − α j ) ≤ min { v p ( A ( α i )) , v p ( A ( α j ))} + ϵ p ( i = j ) . For any real symmetric Galois-invariant positive semidefinite matrix Q Q Q Q 1 = 0 Q\mathbf{1}=0 Q 1 = 0
E p A ( Q ) = ∑ i ≠ j q i j v p ( α i − α j ) + ∑ i q i i ( v p ( A ( α i ) ) + ϵ p ) . E_p^A(Q)=\sum_{i\ne j}q_{ij}v_p(\alpha_i-\alpha_j)+\sum_iq_{ii}\bigl(v_p(A(\alpha_i))+\epsilon_p\bigr). E p A ( Q ) = i = j ∑ q ij v p ( α i − α j ) + i ∑ q ii ( v p ( A ( α i )) + ϵ p ) . These energies are nonnegative, and with τ = tr Q \tau=\operatorname{tr}Q τ = tr Q
∑ p E p A ( Q ) log p = ∑ i ≠ j q i j log ∣ α i − α j ∣ + ∑ i q i i log ∣ A ( α i ) ∣ + τ S ( n ) \sum_pE_p^A(Q)\log p=\sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|+\sum_iq_{ii}\log|A(\alpha_i)|+\tau S(n) p ∑ E p A ( Q ) log p = i = j ∑ q ij log ∣ α i − α j ∣ + i ∑ q ii log ∣ A ( α i ) ∣ + τ S ( n ) ≤ τ ( log ( 4 M R ) + S ( n ) ) , S ( n ) = ∑ p ≤ n log p p − 1 , \le\tau\bigl(\log(4MR)+S(n)\bigr),\qquad S(n)=\sum_{p\le n}\frac{\log p}{p-1}, ≤ τ ( log ( 4 M R ) + S ( n ) ) , S ( n ) = p ≤ n ∑ p − 1 log p , where R = max { 1 , max i ∣ A ( α i ) ∣ } R=\max\{1,\max_i|A(\alpha_i)|\} R = max { 1 , max i ∣ A ( α i ) ∣ } M = max { 1 , U + V } M=\max\{1,U+\sqrt{V}\} M = max { 1 , U + V }
U = max i , 0 ≤ k ≤ n − 2 ∣ k A ′ ( α i ) + B ( α i ) A ( α i ) ∣ , U=\max_{\substack{i,\,0\le k\le n-2}}\left|\frac{kA'(\alpha_i)+B(\alpha_i)}{A(\alpha_i)}\right|, U = i , 0 ≤ k ≤ n − 2 max A ( α i ) k A ′ ( α i ) + B ( α i ) , V = max i , 0 ≤ k ≤ n − 2 ∣ ( k 2 ) A ′ ′ ( α i ) + k B ′ ( α i ) + C A ( α i ) ∣ . V=\max_{\substack{i,\,0\le k\le n-2}}\left|\frac{\binom{k}{2}A''(\alpha_i)+kB'(\alpha_i)+C}{A(\alpha_i)}\right|. V = i , 0 ≤ k ≤ n − 2 max A ( α i ) ( 2 k ) A ′′ ( α i ) + k B ′ ( α i ) + C . Proof. At a root α \alpha α D k = P ( k ) ( α ) / P ′ ( α ) D_k=P^{(k)}(\alpha)/P'(\alpha) D k = P ( k ) ( α ) / P ′ ( α ) k k k
A ( α ) D k + 2 + ( k A ′ ( α ) + B ( α ) ) D k + 1 + ( ( k 2 ) A ′ ′ ( α ) + k B ′ ( α ) + C ) D k = 0 , A(\alpha)D_{k+2}+\left(kA'(\alpha)+B(\alpha)\right)D_{k+1}+\left(\binom{k}{2}A''(\alpha)+kB'(\alpha)+C\right)D_k=0, A ( α ) D k + 2 + ( k A ′ ( α ) + B ( α ) ) D k + 1 + ( ( 2 k ) A ′′ ( α ) + k B ′ ( α ) + C ) D k = 0 , with D 0 = 0 D_0=0 D 0 = 0 D 1 = 1 D_1=1 D 1 = 1 v p ( D k ) ≥ − ( k − 1 ) v p ( A ( α ) ) v_p(D_k)\geq-(k-1)v_p(A(\alpha)) v p ( D k ) ≥ − ( k − 1 ) v p ( A ( α )) v p ( k ! ) ≤ ( k − 1 ) ϵ p v_p(k!)\leq(k-1)\epsilon_p v p ( k !) ≤ ( k − 1 ) ϵ p 2 ≤ k ≤ n 2\leq k\leq n 2 ≤ k ≤ n
P ( α + z ) z P ′ ( α ) = 1 + ∑ k = 2 n D k k ! z k − 1 \frac{P(\alpha+z)}{zP'(\alpha)}=1+\sum_{k=2}^{n}\frac{D_k}{k!}z^{k-1} z P ′ ( α ) P ( α + z ) = 1 + k = 2 ∑ n k ! D k z k − 1 cannot vanish if v p ( z ) > v p ( A ( α ) ) + ϵ p v_p(z)>v_p(A(\alpha))+\epsilon_p v p ( z ) > v p ( A ( α )) + ϵ p
Use the cluster-integral proof of Theorem 3.4 , now with diagonal heights h i = v p ( A ( α i ) ) + ϵ p h_i=v_p(A(\alpha_i))+\epsilon_p h i = v p ( A ( α i )) + ϵ p E p A ( Q ) ≥ 0 E_p^A(Q)\geq0 E p A ( Q ) ≥ 0 A ( α i ) A(\alpha_i) A ( α i ) Theorem 3.4 therefore gives the displayed global identity. Only primes dividing Disc ( P ) Res ( P , A ) \operatorname{Disc}(P)\operatorname{Res}(P,A) Disc ( P ) Res ( P , A ) n n n
For complex separation, M 2 ≥ U M + V M^2\geq UM+V M 2 ≥ U M + V ∣ D k ∣ ≤ M k − 1 |D_k|\leq M^{k-1} ∣ D k ∣ ≤ M k − 1 0 < ∣ z ∣ ≤ 1 / ( 2 M ) 0<|z|\leq1/(2M) 0 < ∣ z ∣ ≤ 1/ ( 2 M ) ∑ k ≥ 2 2 1 − k / k ! = 2 e 1 / 2 − 3 < 1 \sum_{k\geq2}2^{1-k}/k!=2e^{1/2}-3<1 ∑ k ≥ 2 2 1 − k / k ! = 2 e 1/2 − 3 < 1 1 / ( 4 M ) 1/(4M) 1/ ( 4 M ) Theorem 3.4 gives ∑ i ≠ j q i j log ∣ α i − α j ∣ ≤ τ log ( 4 M ) \sum_{i\ne j}q_{ij}\log|\alpha_i-\alpha_j|\leq\tau\log(4M) ∑ i = j q ij log ∣ α i − α j ∣ ≤ τ log ( 4 M )
For Laguerre polynomials, A ( X ) = X A(X)=X A ( X ) = X P ′ ′ − X P ′ + n P = 0 P''-XP'+nP=0 P ′′ − X P ′ + n P = 0 A = 1 A=1 A = 1 P P P X X X
Padé norms and the prime-square obstruction
The root-modulus gap also has an arithmetic consequence. We use the exponential Padé pair to construct a nonzero integral resultant. Its local valuations and its complex absolute value give opposite bounds, which constrain the prime divisors of a factor’s constant term. A Newton-polygon obstruction then excludes factors throughout a polynomial range of parameters. The Padé connection for this family also appears in Cullinan and Scheel [11 ] . For positive integers K K K d d d
T d ( K ) = ∑ p ∣ K min { v p ( K ) , d p − 1 } log p . (49) T_d(K)=\sum_{p\mid K}\min\left\{v_p(K),\frac{d}{p-1}\right\}\log p. \tag*{(49)} T d ( K ) = p ∣ K ∑ min { v p ( K ) , p − 1 d } log p . (49) The exponential Padé pair
Proposition 3.9. Let n ≥ 3 n\geq3 n ≥ 3 s ≥ 1 s\geq1 s ≥ 1 G G G H n , s H_{n,s} H n , s d ≥ 1 d\geq1 d ≥ 1 K = ∣ G ( 0 ) ∣ K=|G(0)| K = ∣ G ( 0 ) ∣ ρ 0 \rho_0 ρ 0
log ( ρ 0 ( n + 1 ) ) < T d ( K ) d + ω ( K ) log s n + 1 . (50) \log(\rho_0(n+1))<\frac{T_d(K)}{d}+\frac{\omega(K)\log s}{n+1}. \tag*{(50)} log ( ρ 0 ( n + 1 )) < d T d ( K ) + n + 1 ω ( K ) log s . (50) If s ≥ n + 2 s\geq n+2 s ≥ n + 2
log s + log n 2 − 2 < T d ( K ) d + ω ( K ) log s n + 1 . (51) \frac{\log s+\log n}{2}-2<\frac{T_d(K)}{d}+\frac{\omega(K)\log s}{n+1}. \tag*{(51)} 2 log s + log n − 2 < d T d ( K ) + n + 1 ω ( K ) log s . (51) In the latter inequality, the constant 2 2 2 log 2 + 1 / 2 \log2+1/2 log 2 + 1/2
Proof. Gauss’s lemma gives G ∈ Z [ X ] G\in\mathbb{Z}[X] G ∈ Z [ X ] r = s − 1 r=s-1 r = s − 1 P ( Z ) = H n , s ( − Z ) P(Z)=H_{n,s}(-Z) P ( Z ) = H n , s ( − Z ) Q ( Z ) = H r , n + 1 ( Z ) Q(Z)=H_{r,n+1}(Z) Q ( Z ) = H r , n + 1 ( Z ) H 0 , n + 1 = 1 H_{0,n+1}=1 H 0 , n + 1 = 1 ( 1 − t ) j − r − 1 (1-t)^{j-r-1} ( 1 − t ) j − r − 1
( n + r − j n ) = ∑ a = 0 n ( − 1 ) a ( n + r − a n − a ) ( j a ) . (52) \binom{n+r-j}{n}=\sum_{a=0}^{n}(-1)^a\binom{n+r-a}{n-a}\binom{j}{a}. \tag*{(52)} ( n n + r − j ) = a = 0 ∑ n ( − 1 ) a ( n − a n + r − a ) ( a j ) . (52) Consequently, as formal power series,
e Z P ( Z ) n ! = ∑ j ≥ 0 ( n + r − j n ) Z j j ! = Q ( Z ) r ! + O ( Z n + r + 1 ) . (53) e^{Z}\frac{P(Z)}{n!}=\sum_{j\geq0}\binom{n+r-j}{n}\frac{Z^{j}}{j!}=\frac{Q(Z)}{r!}+O(Z^{n+r+1}). \tag*{(53)} e Z n ! P ( Z ) = j ≥ 0 ∑ ( n n + r − j ) j ! Z j = r ! Q ( Z ) + O ( Z n + r + 1 ) . (53) The generalized binomial coefficients in the infinite sum are integers. Those with r + 1 ≤ j ≤ n + r r+1\leq j\leq n+r r + 1 ≤ j ≤ n + r E = e Z P / n ! − Q / r ! E=e^{Z}P/n!-Q/r! E = e Z P / n ! − Q / r !
P E ′ − ( P ′ + P ) E = − P Q ′ − P ′ Q − P Q 1 = O ( Z n + r ) . PE'-(P'+P)E=-\frac{PQ'-P'Q-PQ}{\phantom{1}}=O(Z^{n+r}). P E ′ − ( P ′ + P ) E = − 1 P Q ′ − P ′ Q − P Q = O ( Z n + r ) . The numerator is a polynomial of degree at most n + r n+r n + r ( − 1 ) n + 1 (-1)^{n+1} ( − 1 ) n + 1
P Q ′ − P ′ Q − P Q = ( − 1 ) n + 1 Z n + r . (54) PQ'-P'Q-PQ=(-1)^{n+1}Z^{n+r}. \tag*{(54)} P Q ′ − P ′ Q − P Q = ( − 1 ) n + 1 Z n + r . (54) Since P ( 0 ) ≠ 0 P(0)\neq0 P ( 0 ) = 0 P P P Q Q Q − M 1 , … , − M d -M_1,\ldots,-M_d − M 1 , … , − M d G G G
Fix a prime p p p e = v p ( M i ) > 0 e=v_p(M_i)>0 e = v p ( M i ) > 0 e e e s p ( r ) s_p(r) s p ( r ) p p p r r r
v p Q ( M i ) ≥ ( n + s ) e − ( n + 1 ) min { e , ( p − 1 ) − 1 } − s p ( r ) p − 1 . (55) v_pQ(M_i)\geq(n+s)e-(n+1)\min\{e,(p-1)^{-1}\}-\frac{s_p(r)}{p-1}. \tag*{(55)} v p Q ( M i ) ≥ ( n + s ) e − ( n + 1 ) min { e , ( p − 1 ) − 1 } − p − 1 s p ( r ) . (55) If e > 1 / ( p − 1 ) e>1/(p-1) e > 1/ ( p − 1 ) M i M_i M i P ( M i ) = 0 P(M_i)=0 P ( M i ) = 0 Q ( M i ) / r ! Q(M_i)/r! Q ( M i ) / r ! j = n + s j=n+s j = n + s j ≥ 1 j\geq1 j ≥ 1
j e − v p ( j ! ) ≥ j ( e − 1 p − 1 ) + 1 p − 1 . je-v_p(j!)\geq j\left(e-\frac{1}{p-1}\right)+\frac{1}{p-1}. j e − v p ( j !) ≥ j ( e − p − 1 1 ) + p − 1 1 . It follows that
v p Q ( M i ) ≥ ( n + s ) e − n + s p ( r ) p − 1 , v_pQ(M_i)\geq(n+s)e-\frac{n+s_p(r)}{p-1}, v p Q ( M i ) ≥ ( n + s ) e − p − 1 n + s p ( r ) , which is stronger than (55) . If 0 < e ≤ 1 / ( p − 1 ) 0<e\leq1/(p-1) 0 < e ≤ 1/ ( p − 1 )
Q ( M i ) = r ! ∑ j = 0 r ( n + r − j n ) M i j j ! . Q(M_i)=r!\sum_{j=0}^{r}\binom{n+r-j}{n}\frac{M_i^j}{j!}. Q ( M i ) = r ! j = 0 ∑ r ( n n + r − j ) j ! M i j . For 0 ≤ j ≤ r 0\leq j\leq r 0 ≤ j ≤ r j e − v p ( j ! ) ≥ r ( e − 1 / ( p − 1 ) ) je-v_p(j!)\geq r(e-1/(p-1)) j e − v p ( j !) ≥ r ( e − 1/ ( p − 1 )) v p Q ( M i ) ≥ r e − s p ( r ) / ( p − 1 ) v_pQ(M_i)\geq re-s_p(r)/(p-1) v p Q ( M i ) ≥ r e − s p ( r ) / ( p − 1 ) r = 0 r=0 r = 0
We shall use the elementary digit estimate
s p ( r ) p − 1 ≤ r + 1 = s . (56) \frac{s_p(r)}{p-1}\leq r+1=s. \tag*{(56)} p − 1 s p ( r ) ≤ r + 1 = s . (56) Indeed, concavity of log ( a + 1 ) \log(a+1) log ( a + 1 ) 0 ≤ a ≤ p − 1 0\leq a\leq p-1 0 ≤ a ≤ p − 1 p a / ( p − 1 ) ≤ a + 1 p^{a/(p-1)}\leq a+1 p a / ( p − 1 ) ≤ a + 1 a i a_i a i r r r ∏ i ( a i + 1 ) ≤ r + 1 \prod_i(a_i+1)\leq r+1 ∏ i ( a i + 1 ) ≤ r + 1 a i a_i a i [ 0 , r ] [0,r] [ 0 , r ]
The resultant
R = Res ( ( − 1 ) d G ( − Z ) , Q ( Z ) ) = ∏ i = 1 d Q ( M i ) \mathcal{R}=\operatorname{Res}((-1)^dG(-Z),Q(Z))=\prod_{i=1}^{d}Q(M_i) R = Res (( − 1 ) d G ( − Z ) , Q ( Z )) = i = 1 ∏ d Q ( M i ) is a nonzero integer. Both M i M_i M i Q ( M i ) Q(M_i) Q ( M i ) p p p ∑ i v p ( M i ) = v p ( K ) \sum_i v_p(M_i)=v_p(K) ∑ i v p ( M i ) = v p ( K )
∑ i : v p ( M i ) > 0 min { v p ( M i ) , ( p − 1 ) − 1 } ≤ min { v p ( K ) , d / ( p − 1 ) } . \sum_{i:v_p(M_i)>0}\min\{v_p(M_i),(p-1)^{-1}\}\leq\min\{v_p(K),d/(p-1)\}. i : v p ( M i ) > 0 ∑ min { v p ( M i ) , ( p − 1 ) − 1 } ≤ min { v p ( K ) , d / ( p − 1 )} . For roots with zero valuation, use only v p Q ( M i ) ≥ 0 v_pQ(M_i)\geq0 v p Q ( M i ) ≥ 0 (55) over the remaining roots, then over p ∣ K p\mid K p ∣ K (56) , yields
log ∣ R ∣ ≥ ( n + s ) log K − ( n + 1 ) T d ( K ) − d ω ( K ) log s . (57) \log|\mathcal{R}|\geq(n+s)\log K-(n+1)T_d(K)-d\omega(K)\log s. \tag*{(57)} log ∣ R ∣ ≥ ( n + s ) log K − ( n + 1 ) T d ( K ) − d ω ( K ) log s . (57) Primes not dividing K K K
For R = ∣ M i ∣ > s R = |M_i| > s R = ∣ M i ∣ > s
∣ Q ( M i ) ∣ ≤ R r n ! ∫ 0 ∞ t n e − t ( 1 + t R ) r d t ≤ R n + s ( R − r ) n + 1 . |Q(M_i)| \le\frac{R^r}{n!}\int_0^\infty t^n e^{-t}\left(1+\frac{t}{R}\right)^r\,dt
\le\frac{R^{n+s}}{(R-r)^{n+1}}. ∣ Q ( M i ) ∣ ≤ n ! R r ∫ 0 ∞ t n e − t ( 1 + R t ) r d t ≤ ( R − r ) n + 1 R n + s . The root gap (2.21) gives R − r > ρ 0 ( n + 1 ) R-r>\rho_0(n+1) R − r > ρ 0 ( n + 1 )
log ∣ R ∣ < ( n + s ) log K − d ( n + 1 ) log ( ρ 0 ( n + 1 ) ) . \log|\mathcal{R}| < (n+s)\log K-d(n+1)\log\bigl(\rho_0(n+1)\bigr). log ∣ R ∣ < ( n + s ) log K − d ( n + 1 ) log ( ρ 0 ( n + 1 ) ) . This is strict also when r = 0 r=0 r = 0 (57) proves (50) .
For the second inequality, the tangent line to log ( 1 + x ) \log(1+x) log ( 1 + x ) v > 0 v>0 v > 0
log ( 1 + x ) ≤ ϕ ( v ) + x 1 + v , ϕ ( v ) = log ( 1 + v ) − v 1 + v ≤ v 2 2 ( x ≥ 0 ) . \log(1+x)\le\phi(v)+\frac{x}{1+v},\qquad\phi(v)=\log(1+v)-\frac{v}{1+v}\le\frac{v^2}{2}\quad(x\ge0). log ( 1 + x ) ≤ ϕ ( v ) + 1 + v x , ϕ ( v ) = log ( 1 + v ) − 1 + v v ≤ 2 v 2 ( x ≥ 0 ) . The last bound follows by integrating ϕ ′ ( v ) = v / ( 1 + v ) 2 ≤ v \phi'(v)=v/(1+v)^2\le v ϕ ′ ( v ) = v / ( 1 + v ) 2 ≤ v R > s > r R>s>r R > s > r
∣ Q ( M i ) ∣ < R n + s L v n + 1 , L v = e − r ϕ ( v ) / ( n + 1 ) ( R − r 1 + v ) . (58) |Q(M_i)|<\frac{R^{n+s}}{L_v^{n+1}},\qquad L_v=e^{-r\phi(v)/(n+1)}\left(R-\frac{r}{1+v}\right). \tag*{(58)} ∣ Q ( M i ) ∣ < L v n + 1 R n + s , L v = e − r ϕ ( v ) / ( n + 1 ) ( R − 1 + v r ) . (58) The inequality is strict because r > 0 r>0 r > 0 v = ( n + 1 ) / r ≤ 1 v=\sqrt{(n+1)/r}\le1 v = ( n + 1 ) / r ≤ 1 s ≥ n + 2 s\ge n+2 s ≥ n + 2
L v > e − 1 / 2 ( 1 + r v 1 + v ) > r ( n + 1 ) 2 e ≥ s n 2 e . L_v>e^{-1/2}\left(1+\frac{rv}{1+v}\right)>\frac{\sqrt{r(n+1)}}{2\sqrt{e}}\ge\frac{\sqrt{sn}}{2\sqrt{e}}. L v > e − 1/2 ( 1 + 1 + v r v ) > 2 e r ( n + 1 ) ≥ 2 e s n . The last step uses ( s − 1 ) ( n + 1 ) ≥ s n (s-1)(n+1)\ge sn ( s − 1 ) ( n + 1 ) ≥ s n (58) over the roots and comparing with (57) proves the assertion with constant log 2 + 1 / 2 < 2 \log2+1/2<2 log 2 + 1/2 < 2
In the estimates below we use the weaker bound ρ 0 > 1 / 4 \rho_0>1/4 ρ 0 > 1/4
Prime estimates
Write
π ( x ) = # { p ≤ x : p prime } , ϑ ( x ) = ∑ p ≤ x log p , S ( x ) = ∑ p ≤ x log p p − 1 . \pi(x)=\#\{p\le x:p\text{ prime}\},\qquad\vartheta(x)=\sum_{p\le x}\log p,\qquad S(x)=\sum_{p\le x}\frac{\log p}{p-1}. π ( x ) = # { p ≤ x : p prime } , ϑ ( x ) = p ≤ x ∑ log p , S ( x ) = p ≤ x ∑ p − 1 log p . We use the following explicit estimates:
∣ ϑ ( x ) − x ∣ ≤ x 100 ( log x ) 2 ( x > 7 713 133 853 ) , (59) |\vartheta(x)-x|\le\frac{x}{100(\log x)^2}\qquad(x>7\,713\,133\,853), \tag*{(59)} ∣ ϑ ( x ) − x ∣ ≤ 100 ( log x ) 2 x ( x > 7 713 133 853 ) , (59) π ( x ) ≤ x log x − 1.1 ( x > 60 184 ) , (60) \pi(x)\le\frac{x}{\log x-1.1}\qquad(x>60\,184), \tag*{(60)} π ( x ) ≤ log x − 1.1 x ( x > 60 184 ) , (60) S ( x ) < log x ( x ≥ 319 ) . (61) S(x)<\log x\qquad(x\ge319). \tag*{(61)} S ( x ) < log x ( x ≥ 319 ) . (61) The first two estimates are [15 Theorem 5.2, p. 4, and Theorem 6.9, (6.6)] . For the third, Rosser and Schoenfeld [42 Theorem 6, (3.22)] prove
∑ p ≤ x log p p < log x + E + 1 2 log x ( x ≥ 319 ) , \sum_{p\le x}\frac{\log p}{p}<\log x+E+\frac{1}{2\log x}\qquad(x\ge319), p ≤ x ∑ p log p < log x + E + 2 log x 1 ( x ≥ 319 ) , where [42 (2.8)] defines
E = − γ − ∑ p log p p ( p − 1 ) . E=-\gamma-\sum_p\frac{\log p}{p(p-1)}. E = − γ − p ∑ p ( p − 1 ) log p . Adding the sum over p ≤ x p \le x p ≤ x log p / ( p ( p − 1 ) ) \log p/(p(p-1)) log p / ( p ( p − 1 ))
S ( x ) < log x − γ + 1 2 log x < log x . S(x) < \log x-\gamma+\frac{1}{2\log x}<\log x. S ( x ) < log x − γ + 2 log x 1 < log x . The last inequality follows, for example, from γ > 1 − log 2 > 3 / 10 \gamma>1-\log2>3/10 γ > 1 − log 2 > 3/10 log x > 2 \log x>2 log x > 2 γ > 1 − log 2 \gamma>1-\log2 γ > 1 − log 2 γ \gamma γ ∑ k = 1 N ( 1 / k − log ( 1 + 1 / k ) ) \sum_{k=1}^{N}(1/k-\log(1+1/k)) ∑ k = 1 N ( 1/ k − log ( 1 + 1/ k ))
We shall repeatedly use the elementary comparisons
log u u < 1 1000 , e u > 10 30 u 5 , e u / 2 > 10 25 u 4 ( u ≥ 10000 ) . (62) \frac{\log u}{u}<\frac{1}{1000},\qquad e^u>10^{30}u^5,\qquad e^{u/2}>10^{25}u^4\qquad(u\ge10000). \tag*{(62)} u log u < 1000 1 , e u > 1 0 30 u 5 , e u /2 > 1 0 25 u 4 ( u ≥ 10000 ) . (62) At u = 10000 u=10000 u = 10000 log 10000 < 10 \log10000<10 log 10000 < 10 log 10 < 7 / 3 \log10<7/3 log 10 < 7/3 ( log u ) / u (\log u)/u ( log u ) / u 1 − 5 / u > 0 1-5/u>0 1 − 5/ u > 0 1 / 2 − 4 / u > 0 1/2-4/u>0 1/2 − 4/ u > 0 2 / 3 < log 2 < 7 / 10 2/3<\log2<7/10 2/3 < log 2 < 7/10 log 13 < 3 \log13<3 log 13 < 3 log 52 < 4 \log52<4 log 52 < 4
log 2 = 2 ∑ j ≥ 0 1 ( 2 j + 1 ) 3 2 j + 1 . \log2=2\sum_{j\ge0}\frac{1}{(2j+1)3^{2j+1}}. log 2 = 2 j ≥ 0 ∑ ( 2 j + 1 ) 3 2 j + 1 1 . The Newton-polygon arguments require lower bounds for the number of primes in intervals proportional to n n n n n n
Lemma 3.10. If u = log n ≥ 10000 u=\log n\ge10000 u = log n ≥ 10000
π ( n ) < 1003 n 1000 u , π ( n / ℓ ) − π ( n / ( 4 ℓ ) ) > 749 n 1000 ℓ u ( 1 ≤ ℓ ≤ 13 ) . (63) \pi(n)<\frac{1003n}{1000u},\qquad\pi(n/\ell)-\pi(n/(4\ell))>\frac{749n}{1000\ell u}\qquad(1\le\ell\le13). \tag*{(63)} π ( n ) < 1000 u 1003 n , π ( n / ℓ ) − π ( n / ( 4 ℓ )) > 1000 ℓ u 749 n ( 1 ≤ ℓ ≤ 13 ) . (63) In particular, π ( n ) < n / 100 \pi(n)<n/100 π ( n ) < n /100
Proof. The estimate (60) gives directly
π ( n ) ≤ n u − 11 / 10 < 1003 n 1000 u ( u ≥ 10000 ) . \pi(n)\le\frac{n}{u-11/10}<\frac{1003n}{1000u}\qquad(u\ge10000). π ( n ) ≤ u − 11/10 n < 1000 u 1003 n ( u ≥ 10000 ) . For t = n / ℓ t=n/\ell t = n / ℓ (59) , by (62) , and log ( t / 4 ) > u − 4 \log(t/4)>u-4 log ( t /4 ) > u − 4
ϑ ( t ) − ϑ ( t / 4 ) ≥ 3 t 4 − 5 t 400 ( u − 4 ) 2 > 749 t 1000 . \vartheta(t)-\vartheta(t/4)\ge\frac{3t}{4}-\frac{5t}{400(u-4)^2}>\frac{749t}{1000}. ϑ ( t ) − ϑ ( t /4 ) ≥ 4 3 t − 400 ( u − 4 ) 2 5 t > 1000 749 t . Division by log t ≤ u \log t\le u log t ≤ u
The Padé inequalities involve ω ( K ) \omega(K) ω ( K ) T d ( K ) T_d(K) T d ( K ) log K \log K log K
Lemma 3.11. Let K , d K,d K , d log K < X \log K<X log K < X X ≥ 10000 X\ge10000 X ≥ 10000
ω ( K ) < 2 X log X , T d ( K ) d < log X + 1.001. (64) \omega(K)<\frac{2X}{\log X},\qquad\frac{T_d(K)}{d}<\log X+1.001. \tag*{(64)} ω ( K ) < log X 2 X , d T d ( K ) < log X + 1.001. (64) Proof. Put w = ω ( K ) w=\omega(K) w = ω ( K ) i i i i + 1 i+1 i + 1 K ≥ ( w + 1 ) ! K\ge(w+1)! K ≥ ( w + 1 )! w ≥ 1 w\ge1 w ≥ 1 log t \log t log t log K > w log w − w \log K>w\log w-w log K > w log w − w w = 2 X / log X w=2X/\log X w = 2 X / log X X X X y = log X ≥ 4 y=\log X\ge4 y = log X ≥ 4 y / 2 − log y + log 2 − 1 > 0 y/2-\log y+\log2-1>0 y /2 − log y + log 2 − 1 > 0 y = 4 y=4 y = 4 w log w − w w\log w-w w log w − w w > 1 w>1 w > 1 w = 0 w=0 w = 0
For the second bound, use T d ( K ) / d ≤ ∑ p ∣ K log p / ( p − 1 ) T_d(K)/d\le\sum_{p\mid K}\log p/(p-1) T d ( K ) / d ≤ ∑ p ∣ K log p / ( p − 1 ) X X X S ( X ) < log X S(X)<\log X S ( X ) < log X (61) . The remaining contribution is at most log K / ( X − 1 ) < X / ( X − 1 ) < 1.001 \log K/(X-1)<X/(X-1)<1.001 log K / ( X − 1 ) < X / ( X − 1 ) < 1.001
When s s s n n n
Lemma 3.12. If log n ≥ 10000 \log n \ge10000 log n ≥ 10000 1 ≤ s ≤ n 3 1 \le s \le n^{3} 1 ≤ s ≤ n 3 H n , s H_{n,s} H n , s d d d
d > n 100 log n . (65) d > \frac{n}{100\log n}. \tag*{(65)} d > 100 log n n . (65) Proof. Let G G G d d d u = log n u=\log n u = log n t = log ( s + n ) t=\log(s+n) t = log ( s + n ) X = d t X=dt X = d t K = ∣ G ( 0 ) ∣ K=\lvert G(0)\rvert K = ∣ G ( 0 )∣ (14) gives log K < X \log K<X log K < X u < t < 3.01 u u<t<3.01u u < t < 3.01 u d ≤ n / ( 100 u ) d\le n/(100u) d ≤ n / ( 100 u ) 10000 ≤ u < X < n / 32 10000\le u<X<n/32 10000 ≤ u < X < n /32
T d ( K ) d < u − 5 log 2 + 1.001 < u − 2.3. \frac{T_{d}(K)}{d}<u-5\log2+1.001<u-2.3. d T d ( K ) < u − 5 log 2 + 1.001 < u − 2.3. As X / log X X/\log X X / log X
ω ( K ) log s n + 1 < 3 u 16 ( u − log 32 ) < 1 5 . \frac{\omega(K)\log s}{n+1}<\frac{3u}{16(u-\log32)}<\frac{1}{5}. n + 1 ω ( K ) log s < 16 ( u − log 32 ) 3 u < 5 1 . The right side of (3.15) is less than u − 2.1 u-2.1 u − 2.1 u − log 4 > u − 2 u-\log4>u-2 u − log 4 > u − 2
The prime-square obstruction
For primes p p p n < p ≤ n \sqrt{n}<p\le n n < p ≤ n p 2 p^{2} p 2 − 1 / p -1/p − 1/ p
Proposition 3.13. Suppose H n , s H_{n,s} H n , s d d d p p p p 2 > n p^{2}>n p 2 > n p ∣ ( n d ) p\mid\binom{n}{d} p ∣ ( d n ) [ s , s + n − 1 ] [s,s+n-1] [ s , s + n − 1 ] p 2 p^{2} p 2 n ≥ 5 n\ge5 n ≥ 5 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2
π ( n ) − π ( n − d ) ≤ ⌊ s + n − 1 ( n − d + 1 ) 2 ⌋ − ⌊ s n 2 ⌋ + 1. (66) \pi(n)-\pi(n-d)\le\left\lfloor\frac{s+n-1}{(n-d+1)^{2}}\right\rfloor-\left\lfloor\frac{s}{n^{2}}\right\rfloor+1. \tag*{(66)} π ( n ) − π ( n − d ) ≤ ⌊ ( n − d + 1 ) 2 s + n − 1 ⌋ − ⌊ n 2 s ⌋ + 1. (66) Proof. Suppose that the interval contains no multiple of p 2 p^{2} p 2 ( n d ) \binom{n}{d} ( d n ) p ≤ n p\le n p ≤ n n = a p + b n=ap+b n = a p + b s − 1 = c p + t s-1=cp+t s − 1 = c p + t 0 ≤ b , t < p 0\le b,t<p 0 ≤ b , t < p a < p a<p a < p j = ℓ p + v j=\ell p+v j = ℓ p + v 0 ≤ v < p 0\le v<p 0 ≤ v < p X j X^{j} X j
a − ℓ + ⌊ t + b − v p ⌋ + 1 v > b . (67) a-\ell+\left\lfloor\frac{t+b-v}{p}\right\rfloor+1_{v>b}. \tag*{(67)} a − ℓ + ⌊ p t + b − v ⌋ + 1 v > b . (67) Indeed, the rising factorial ( s ) n − j (s)_{n-j} ( s ) n − j p p p n < p 2 n<p^{2} n < p 2 v p ( n j ) = 1 v > b v_{p}\binom{n}{j}=1_{v>b} v p ( j n ) = 1 v > b
If t + b < p t+b<p t + b < p
( 0 , a ) , ( a p , 0 ) , ( n , 0 ) . (0,a),\qquad(ap,0),\qquad(n,0). ( 0 , a ) , ( a p , 0 ) , ( n , 0 ) . If t + b ≥ p t+b\ge p t + b ≥ p v 0 = t + b − p < b v_{0}=t+b-p<b v 0 = t + b − p < b
( 0 , a + 1 ) , ( v 0 + 1 , a ) , ( a p + v 0 + 1 , 0 ) , ( n , 0 ) . (0,a+1),\qquad(v_{0}+1,a),\qquad(ap+v_{0}+1,0),\qquad(n,0). ( 0 , a + 1 ) , ( v 0 + 1 , a ) , ( a p + v 0 + 1 , 0 ) , ( n , 0 ) . These assertions follow directly from (3.31); coincident vertices are omitted. In the second case the initial slope is − 1 / ( v 0 + 1 ) -1/(v_{0}+1) − 1/ ( v 0 + 1 ) 1 ≤ v 0 + 1 ≤ b < p 1\le v_{0}+1\le b<p 1 ≤ v 0 + 1 ≤ b < p − 1 / p -1/p − 1/ p a p ap a p 1 / p 1/p 1/ p b b b
The roots selected by the rational factor form a set invariant under the local Galois group. Intersecting it with the set of roots of valuation 1 / p 1/p 1/ p Q × \mathbb{Q}^{\times} Q × 1 / p 1/p 1/ p p p p d = k p + e d=kp+e d = k p + e 0 ≤ k ≤ a 0\le k\le a 0 ≤ k ≤ a 0 ≤ e ≤ b 0\le e\le b 0 ≤ e ≤ b p p p v p ( n d ) = 0 v_{p}\binom{n}{d}=0 v p ( d n ) = 0
For the second assertion, every prime p ∈ ( n − d , n ] p\in(n-d,n] p ∈ ( n − d , n ] p > n / 2 > n p>n/2>\sqrt{n} p > n /2 > n ( n d ) \binom{n}{d} ( d n ) k p ≥ 1 k_{p}\ge1 k p ≥ 1 s ≤ k p p 2 ≤ s + n − 1 s\le k_{p}p^{2}\le s+n-1 s ≤ k p p 2 ≤ s + n − 1 p < q p<q p < q k k k k ( q 2 − p 2 ) ≥ 2 p + 1 > n − 1 k(q^{2}-p^{2})\ge2p+1>n-1 k ( q 2 − p 2 ) ≥ 2 p + 1 > n − 1
of length n − 1 n-1 n − 1 ⌈ s / n 2 ⌉ \lceil s/n^2\rceil ⌈ s / n 2 ⌉ ⌊ ( s + n − 1 ) / ( n − d + 1 ) 2 ⌋ \lfloor(s+n-1)/(n-d+1)^2\rfloor ⌊( s + n − 1 ) / ( n − d + 1 ) 2 ⌋ (66) .
The degree bound and the prime estimates ensure that ( n − d , n ] (n-d,n] ( n − d , n ] s s s
Proposition 3.14. If log n ≥ 10000 \log n \ge10000 log n ≥ 10000
1 ≤ s ≤ n 3 10 log n , (68) 1 \le s \le\frac{n^3}{10\log n}, \tag*{(68)} 1 ≤ s ≤ 10 log n n 3 , (68) then H n , s H_{n,s} H n , s Q \mathbb{Q} Q
Proof. If H n , s H_{n,s} H n , s 1 ≤ d ≤ n / 2 1 \le d \le n/2 1 ≤ d ≤ n /2 u = log n u=\log n u = log n d > n / ( 100 u ) d>n/(100u) d > n / ( 100 u ) n n n n − d n-d n − d (59) , and log ( n − d ) ≥ u − log 2 \log(n-d)\ge u-\log2 log ( n − d ) ≥ u − log 2
ϑ ( n ) − ϑ ( n − d ) > d − 3 n 100 u 2 > 4 d 5 , π ( n ) − π ( n − d ) > 4 d 5 u . \vartheta(n)-\vartheta(n-d)>d-\frac{3n}{100u^2}>\frac{4d}{5}, \qquad\pi(n)-\pi(n-d)>\frac{4d}{5u}. ϑ ( n ) − ϑ ( n − d ) > d − 100 u 2 3 n > 5 4 d , π ( n ) − π ( n − d ) > 5 u 4 d . On the other hand, the right side of (66) is less than
6 s d n 3 + 1 + 4 n ≤ 3 d 5 u + 1 + 4 n . \frac{6sd}{n^3}+1+\frac{4}{n}\le\frac{3d}{5u}+1+\frac{4}{n}. n 3 6 s d + 1 + n 4 ≤ 5 u 3 d + 1 + n 4 . Here we used n − d ≥ n / 2 n-d\ge n/2 n − d ≥ n /2
1 ( n − d ) 2 − 1 n 2 = d ( 2 n − d ) n 2 ( n − d ) 2 ≤ 6 d n 3 . \frac{1}{(n-d)^2}-\frac{1}{n^2}=\frac{d(2n-d)}{n^2(n-d)^2}\le\frac{6d}{n^3}. ( n − d ) 2 1 − n 2 1 = n 2 ( n − d ) 2 d ( 2 n − d ) ≤ n 3 6 d . But d / ( 5 u ) > n / ( 500 u 2 ) > 1 + 4 / n d/(5u)>n/(500u^2)>1+4/n d / ( 5 u ) > n / ( 500 u 2 ) > 1 + 4/ n (62) . This is a contradiction.
For larger parameters, the tangent estimate in Proposition 3.9 gives a stronger restriction. The resulting degree bound depends on s s s log ( s + n ) \log(s+n) log ( s + n )
Lemma 3.15. Suppose log n ≥ 10000 \log n\ge10000 log n ≥ 10000 s ≥ n 5 / 2 s\ge n^{5/2} s ≥ n 5/2 H n , s H_{n,s} H n , s d ≥ 1 d\ge1 d ≥ 1 A = log ( s + n ) / log n A=\log(s+n)/\log n A = log ( s + n ) / log n
d > n 16 A . (69) d>\frac{n}{16A}. \tag*{(69)} d > 16 A n . (69) Proof. Write u = log n u=\log n u = log n d ≤ n / ( 16 A ) d\le n/(16A) d ≤ n / ( 16 A ) X = d A u X=dAu X = d A u K = ∣ G ( 0 ) ∣ K=|G(0)| K = ∣ G ( 0 ) ∣ G G G d d d log K < X \log K<X log K < X 10000 ≤ A u ≤ X ≤ n u / 16 10000\le Au\le X\le nu/16 10000 ≤ A u ≤ X ≤ n u /16
T d ( K ) d < u + log u , ω ( K ) < n 4 . \frac{T_d(K)}{d}<u+\log u,\qquad\omega(K)<\frac{n}{4}. d T d ( K ) < u + log u , ω ( K ) < 4 n . The estimate for T d ( K ) T_d(K) T d ( K ) log 16 > 1.001 \log16>1.001 log 16 > 1.001 ω ( K ) \omega(K) ω ( K ) X / log X X/\log X X / log X log ( n u / 16 ) > u / 2 \log(nu/16)>u/2 log ( n u /16 ) > u /2 A u − 1 < log s < A u Au-1<\log s<Au A u − 1 < log s < A u
A u + u 2 − 3 < u + log u + A u 4 . \frac{Au+u}{2}-3<u+\log u+\frac{Au}{4}. 2 A u + u − 3 < u + log u + 4 A u . This is impossible: A ≥ 5 / 2 A\ge5/2 A ≥ 5/2
u ( A 4 − 1 2 ) − log u − 3 ≥ u 8 − log u − 3 > 0 u\left(\frac{A}{4}-\frac{1}{2}\right)-\log u-3\ge\frac{u}{8}-\log u-3>0 u ( 4 A − 2 1 ) − log u − 3 ≥ 8 u − log u − 3 > 0 by (62) .
Root energy and uniform eventual irreducibility
We apply the matrix-energy inequality to a rational factor. Its proportion of roots in a local cluster need not equal its proportion in the full root set; the square of this discrepancy contributes to the energy. The resulting degree restrictions will combine with the trace criterion to cover all parameters. Throughout this section, n n n s s s u = log n u=\log n u = log n Q \mathbb{Q} Q
The energy of a rational factor
Suppose H n , s = G J H_{n,s}=GJ H n , s = G J G G G J J J 1 ≤ d = deg G ≤ n / 2 1\le d=\deg G\le n/2 1 ≤ d = deg G ≤ n /2 x = d / n x=d/n x = d / n Q = c c T Q=cc^{\mathsf{T}} Q = c c T c i = 1 − x c_i=1-x c i = 1 − x G G G c i = − x c_i=-x c i = − x J J J
∑ i c i = 0 , ∑ i c i 2 = n x ( 1 − x ) . (70) \sum_i c_i=0,\qquad\sum_i c_i^2=nx(1-x). \tag*{(70)} i ∑ c i = 0 , i ∑ c i 2 = n x ( 1 − x ) . (70) At a prime p p p
h p , i = v p ( α i ) + ϵ p , E p = ∑ i ≠ j c i c j v p ( α i − α j ) + ∑ i c i 2 h p , i . (71) h_{p,i}=v_p(\alpha_i)+\epsilon_p,\qquad E_p=\sum_{i\ne j}c_i c_j v_p(\alpha_i-\alpha_j)+\sum_i c_i^2h_{p,i}. \tag*{(71)} h p , i = v p ( α i ) + ϵ p , E p = i = j ∑ c i c j v p ( α i − α j ) + i ∑ c i 2 h p , i . (71) There are only finitely many nonzero contributions: outside the primes at most n n n
Lemma 3.16. Every E p E_p E p N p ( t ) N_p(t) N p ( t ) t t t K p ( t ) K_p(t) K p ( t ) t t t
E p ≥ ∫ 0 ∞ ( K p ( t ) − x N p ( t ) ) 2 d t . (72) E_p\ge\int_0^\infty\left(K_p(t)-xN_p(t)\right)^2\,dt. \tag*{(72)} E p ≥ ∫ 0 ∞ ( K p ( t ) − x N p ( t ) ) 2 d t . (72) Write
K = ∣ G ( 0 ) ∣ , P = ( s ) n , S ( n ) = ∑ p ≤ n log p p − 1 , K=\lvert G(0)\rvert,\qquad P=(s)_n,\qquad S(n)=\sum_{p\le n}\frac{\log p}{p-1}, K = ∣ G ( 0 )∣ , P = ( s ) n , S ( n ) = p ≤ n ∑ p − 1 log p , C ∞ = ∑ i ≠ j c i c j log ∣ α i − α j ∣ . C_\infty=\sum_{i\ne j}c_i c_j\log\lvert\alpha_i-\alpha_j\rvert. C ∞ = i = j ∑ c i c j log ∣ α i − α j ∣ . Then the global energy E = ∑ p E p log p \mathcal{E}=\sum_p E_p\log p E = ∑ p E p log p
E = C ∞ + ( 1 − x ) 2 log K + x 2 log ( P / K ) + n x ( 1 − x ) S ( n ) . (73) \mathcal{E}=C_\infty+(1-x)^2\log K+x^2\log(P/K)+nx(1-x)S(n). \tag*{(73)} E = C ∞ + ( 1 − x ) 2 log K + x 2 log ( P / K ) + n x ( 1 − x ) S ( n ) . (73) If the closed complex disks of radius ρ \rho ρ
C ∞ ≤ − n x ( 1 − x ) log ρ . (74) C_\infty\le-nx(1-x)\log\rho. \tag*{(74)} C ∞ ≤ − n x ( 1 − x ) log ρ . (74) Proof. The root sets of the rational factors are Galois invariant, so Q = c c T Q=cc^{\mathsf{T}} Q = c c T n x ( 1 − x ) nx(1-x) n x ( 1 − x ) Lemma 3.5 therefore applies. On the roots of valuation at least t t t K p ( t ) − x N p ( t ) K_p(t)-xN_p(t) K p ( t ) − x N p ( t ) (72) . The diagonal contribution to the product formula is ( 1 − x ) 2 log K + x 2 log ( P / K ) (1-x)^2\log K+x^2\log(P/K) ( 1 − x ) 2 log K + x 2 log ( P / K ) (73) . The complex bound follows from the same lemma.
The annulus and separation estimates now give an upper bound.
Lemma 3.17. Suppose u = log n ≥ 10000 u=\log n\ge10000 u = log n ≥ 10000 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2 s ≥ n 5 / 2 s\ge n^{5/2} s ≥ n 5/2
E < 3 n x ( 1 − x ) log s . (75) \mathcal{E}<3nx(1-x)\log s. \tag*{(75)} E < 3 n x ( 1 − x ) log s . (75) Proof. For s ≥ n s\ge n s ≥ n Lemma 3.5 to Q = c c T Q=cc^{\mathsf{T}} Q = c c T n x ( 1 − x ) nx(1-x) n x ( 1 − x )
E < n x ( 1 − x ) ( 1 2 log s + 3 2 u + log 48 ) . (76) \mathcal{E}<nx(1-x)\left(\frac{1}{2}\log s+\frac{3}{2}u+\log48\right). \tag*{(76)} E < n x ( 1 − x ) ( 2 1 log s + 2 3 u + log 48 ) . (76) For s ≥ n 5 / 2 s\ge n^{5/2} s ≥ n 5/2 (75) .
Packets and deletion of small-prime contributions
For p > n p>n p > n Proposition 3.2 concerns a single node s + j s+j s + j
Lemma 3.18. Suppose p > n p>n p > n p e ∥ s + j p^e\mathrel{\|}s+j p e ∥ s + j 0 ≤ j < n 0\le j<n 0 ≤ j < n N = n − j N=n-j N = n − j N N N e / N e/N e / N j j j k k k
N ∣ k e , E p log p = N ( k N − x ) 2 log ( p e ) . (77) N\mid ke,\qquad E_p\log p=N\left(\frac{k}{N}-x\right)^2\log(p^e). \tag*{(77)} N ∣ k e , E p log p = N ( N k − x ) 2 log ( p e ) . (77) Proof. The tame-packet construction in Proposition 3.2 and Lemma 3.1 gives the stated valuations and inertia orbit length q = N / gcd ( N , e ) q=N/\gcd(N,e) q = N / g cd( N , e ) q ∣ k q\mid k q ∣ k N ∣ k e N\mid ke N ∣ k e (45) , applied to Q = c c T Q=c c^{\mathsf T} Q = c c T
E p = e N ( k − x N ) 2 , E_p=\frac{e}{N}(k-xN)^2, E p = N e ( k − x N ) 2 , which proves the energy identity, including when N = 1 N=1 N = 1
Before summing over large primes, we bound the parts of the nodes supported on small primes. For each small prime, remove its full contribution from one node of maximal valuation. The remaining product divides a factorial. Deleting those nodes altogether gives the form used below. This is the allocation argument of Erdős and Selfridge [17 Lemma 2] .
Lemma 3.19. Let I ⊆ { 0 , … , n − 1 } \mathcal{I}\subseteq\{0,\ldots,n-1\} I ⊆ { 0 , … , n − 1 } Y ≥ 2 Y\ge2 Y ≥ 2 P \mathcal{P} P p ≤ Y p\le Y p ≤ Y s + j s+j s + j j ∈ I j\in\mathcal{I} j ∈ I p ∈ P p\in\mathcal{P} p ∈ P j p ∈ I j_p\in\mathcal{I} j p ∈ I p p p p p p p p p
R j = ∏ p ∈ P j ≠ j p p v p ( s + j ) . R_j=\prod_{\substack{p\in\mathcal{P}\\j\ne j_p}}p^{v_p(s+j)}. R j = p ∈ P j = j p ∏ p v p ( s + j ) . Then ∏ j ∈ I R j ∣ ( n − 1 ) ! \prod_{j\in\mathcal{I}}R_j\mid(n-1)! ∏ j ∈ I R j ∣ ( n − 1 )! π ( Y ) \pi(Y) π ( Y ) J \mathcal{J} J S Y ( m ) S_Y(m) S Y ( m ) m m m Y Y Y
∏ j ∈ J S Y ( s + j ) ∣ ( n − 1 ) ! . (78) \prod_{j\in\mathcal{J}}S_Y(s+j)\mid(n-1)!. \tag*{(78)} j ∈ J ∏ S Y ( s + j ) ∣ ( n − 1 )! . (78) Proof. For j ≠ j p j\ne j_p j = j p v p ( s + j ) ≤ v p ( j − j p ) v_p(s+j)\le v_p(j-j_p) v p ( s + j ) ≤ v p ( j − j p ) j p j_p j p j p ! ( n − 1 − j p ) ! j_p!(n-1-j_p)! j p ! ( n − 1 − j p )! ( n − 1 ) ! (n-1)! ( n − 1 )! Y Y Y
From now on take Y = n Y=n Y = n S j = S n ( s + j ) S_j=S_n(s+j) S j = S n ( s + j ) B j = ( s + j ) / S j B_j=(s+j)/S_j B j = ( s + j ) / S j B j B_j B j n n n n n n p e ∥ B j p^e\mathrel{\|}B_j p e ∥ B j
e < A : = log ( s + n ) log n . (79) e<A:=\frac{\log(s+n)}{\log n}. \tag*{(79)} e < A := log n log ( s + n ) . (79) The reduced denominator of a selected fraction q = k / ( n − j ) q=k/(n-j) q = k / ( n − j ) n − j n-j n − j e e e (77) .
A restriction on the factor degree
Since e < A e<A e < A A A A A − 2 A^{-2} A − 2 x = d / n x=d/n x = d / n
Lemma 3.20. Suppose u = log n ≥ 10000 u=\log n\ge10000 u = log n ≥ 10000 s ≥ n 5 / 2 s\ge n^{5/2} s ≥ n 5/2 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2
d < 4800 or A > ( n / 4800 ) 1 / 4 . (80) d<4800\qquad\text{or}\qquad A>(n/4800)^{1/4}. \tag*{(80)} d < 4800 or A > ( n /4800 ) 1/4 . (80) Proof. Put x = d / n x=d/n x = d / n η = min { x / 2 , 1 / ( 4 A 2 ) } \eta=\min\{x/2,1/(4A^2)\} η = min { x /2 , 1/ ( 4 A 2 )} A A A A − 2 A^{-2} A − 2 ( x − η , x + η ) (x-\eta,x+\eta) ( x − η , x + η ) b b b 0 < x − η < x + η < 1 0<x-\eta<x+\eta<1 0 < x − η < x + η < 1 ⌈ n / 8 ⌉ ≤ N ≤ n \lceil n/8\rceil\le N\le n ⌈ n /8 ⌉ ≤ N ≤ n b b b 7 n / 16 − 2 7n/16-2 7 n /16 − 2 A A A N N N η \eta η x x x
Apply Lemma 3.19 to the corresponding indices j = n − N j=n-N j = n − N π ( n ) < n / 100 \pi(n)<n/100 π ( n ) < n /100 (63) leaves more than 0.42 n 0.42n 0.42 n
∑ j ∈ J log B j > 0.42 n log s − log ( ( n − 1 ) ! ) > n 50 log s \sum_{j\in\mathcal{J}}\log B_j>0.42n\log s-\log((n-1)!)>\frac{n}{50}\log s j ∈ J ∑ log B j > 0.42 n log s − log (( n − 1 )!) > 50 n log s since log s ≥ ( 5 / 2 ) u \log s\ge(5/2)u log s ≥ ( 5/2 ) u log ( ( n − 1 ) ! ) < n u \log((n-1)!)<nu log (( n − 1 )!) < n u B j B_j B j (79) and the preceding denominator exclusion give ∣ k / N − x ∣ ≥ η |k/N-x|\ge\eta ∣ k / N − x ∣ ≥ η N ≥ n / 8 N\ge n/8 N ≥ n /8
ε > n 2 η 2 400 log s . \varepsilon>\frac{n^2\eta^2}{400}\log s. ε > 400 n 2 η 2 log s . Comparison with (75) gives n η 2 < 1200 x ( 1 − x ) n\eta^2<1200x(1-x) n η 2 < 1200 x ( 1 − x ) η = x / 2 \eta=x/2 η = x /2 d < 4800 d<4800 d < 4800 n / ( 16 A 4 ) < 1200 x ( 1 − x ) ≤ 300 n/(16A^4)<1200x(1-x)\le300 n / ( 16 A 4 ) < 1200 x ( 1 − x ) ≤ 300
Uniform eventual irreducibility
For a reducible member, the four-fiber bound makes log s \log s log s n n n n n n
Proposition 3.21. There is an effective absolute integer N 0 N_0 N 0 H n , s H_{n,s} H n , s n ≥ N 0 n\ge N_0 n ≥ N 0 s s s
Proof. Suppose a factor of degree 1 ≤ d ≤ n / 2 1\le d\le n/2 1 ≤ d ≤ n /2 B = 0 B=0 B = 0 ε = 1 / 8 \varepsilon=1/8 ε = 1/8 log s < n 1 / 8 \log s<n^{1/8} log s < n 1/8 n n n n n n n 5 / 2 ≤ n 3 / ( 10 log n ) n^{5/2}\le n^3/(10\log n) n 5/2 ≤ n 3 / ( 10 log n ) s ≤ n 5 / 2 s\le n^{5/2} s ≤ n 5/2 u = log n u=\log n u = log n A = log ( s + n ) / u A=\log(s+n)/u A = log ( s + n ) / u n n n
A < 2 n 1 / 8 u < ( n / 4800 ) 1 / 4 , d > n 16 A > 4800 , A<\frac{2n^{1/8}}{u}<(n/4800)^{1/4},\qquad d>\frac{n}{16A}>4800, A < u 2 n 1/8 < ( n /4800 ) 1/4 , d > 16 A n > 4800 , where the middle degree bound is Lemma 3.15. These inequalities contradict Lemma 3.20. Every threshold used here is effective, so their maximum gives the asserted N 0 N_0 N 0
Galois groups
Assuming irreducibility, we prove that the Galois group contains A n A_n A n n n n A n A_n A n s s s
Prime valuations in short intervals
We begin with an arithmetic criterion for alternating containment. A short interval cannot contain many integers whose large-prime valuations are all divisible by exponents comparable to its length. Newton polygons turn this statement into a criterion for alternating containment. We apply it first to Laguerre polynomials; the trimmed-exponential application is given in Section 6.2. Sections 4.2–4.5 will place every remaining Laguerre obstruction in its parameter range.
Independent nodes and simultaneous logarithms.
For a rational number a / b a/b a / b H ( a / b ) = max { ∣ a ∣ , ∣ b ∣ } H(a/b)=\max\{|a|,|b|\} H ( a / b ) = max { ∣ a ∣ , ∣ b ∣ } h ( a / b ) = log H ( a / b ) h(a/b)=\log H(a/b) h ( a / b ) = log H ( a / b ) [36 Theorem 4] , in the form stated by Stewart [46 Lemma 1] . Let r ≥ 2 r\ge2 r ≥ 2 1 ≤ m ≤ r 1\le m\le r 1 ≤ m ≤ r β 1 , … , β r \beta_1,\ldots,\beta_r β 1 , … , β r ( b i j ) (b_{ij}) ( b ij ) m m m
Λ i = ∑ j = 1 r b i j log β j ( 1 ≤ i ≤ m ) . \Lambda_i=\sum_{j=1}^{r}b_{ij}\log\beta_j \qquad(1\le i\le m). Λ i = j = 1 ∑ r b ij log β j ( 1 ≤ i ≤ m ) . For A j , B ≥ 4 A_j,B\ge4 A j , B ≥ 4 H ( β j ) ≤ A j H(\beta_j)\le A_j H ( β j ) ≤ A j H ( b i j ) ≤ B H(b_{ij})\le B H ( b ij ) ≤ B Ω = ∏ j = 1 r log A j \Omega=\prod_{j=1}^{r}\log A_j Ω = ∏ j = 1 r log A j
max i ∣ Λ i ∣ > exp ( − ( 16 r ) 200 r ( Ω log Ω ) 1 / m log ( B Ω ) ) . (81) \max_i|\Lambda_i|>\exp\left(-(16r)^{200r}(\Omega\log\Omega)^{1/m}\log(B\Omega)\right). \tag*{(81)} i max ∣ Λ i ∣ > exp ( − ( 16 r ) 200 r ( Ω log Ω ) 1/ m log ( B Ω ) ) . (81) The forms must retain full rank after dependent bases are removed. We ensure this by first choosing multiplicatively independent integers in the parameter interval. The following elementary version of Stewart’s argument is convenient [46 Theorem 1 and Corollary 3] .
Lemma 4.1. Fix an integer t ≥ 2 t\ge2 t ≥ 2
K t = t ( ( 4 ( t − 1 ) ) t − 1 + 1 ) . K_t=t\bigl((4(t-1))^{t-1}+1\bigr). K t = t ( ( 4 ( t − 1 ) ) t − 1 + 1 ) . Let M , s M,s M , s s > 2 M 2 s>2M^2 s > 2 M 2 K t K_t K t [ s , s + M ] [s,s+M] [ s , s + M ] t t t
Proof. Suppose every t t t k k k ( t − 1 ) (t-1) ( t − 1 )
⌊ ( k t ) ( k t − 1 ) ⌋ = ⌊ k − t + 1 t ⌋ \left\lfloor\frac{\binom{k}{t}}{\binom{k}{t-1}}\right\rfloor=\left\lfloor\frac{k-t+1}{t}\right\rfloor ⌊ ( t − 1 k ) ( t k ) ⌋ = ⌊ t k − t + 1 ⌋ subsets. Write its elements as a 1 , … , a t − 1 a_1,\ldots,a_{t-1} a 1 , … , a t − 1 z z z
z ℓ = ∏ i = 1 t − 1 a i c i , ℓ > 0 , ∣ c i ∣ ≤ ℓ . z^\ell=\prod_{i=1}^{t-1}a_i^{c_i},\qquad\ell>0,\qquad|c_i|\le\ell. z ℓ = i = 1 ∏ t − 1 a i c i , ℓ > 0 , ∣ c i ∣ ≤ ℓ . Divide each coordinate interval of [ − 1 , 1 ] t − 1 [-1,1]^{t-1} [ − 1 , 1 ] t − 1 4 ( t − 1 ) 4(t-1) 4 ( t − 1 ) k ≥ K t k\ge K_t k ≥ K t z , z ′ z,z' z , z ′
∣ c i ℓ − c i ′ ℓ ′ ∣ ≤ 1 2 ( t − 1 ) ( 1 ≤ i < t ) . \left|\frac{c_i}{\ell}-\frac{c_i'}{\ell'}\right|\le\frac{1}{2(t-1)}\qquad(1\le i<t). ℓ c i − ℓ ′ c i ′ ≤ 2 ( t − 1 ) 1 ( 1 ≤ i < t ) . Raise the quotient of the two relations to clear denominators. The identities h ( x j ) = ∣ j ∣ h ( x ) h(x^j)=|j|h(x) h ( x j ) = ∣ j ∣ h ( x ) h ( x y ) ≤ h ( x ) + h ( y ) h(xy)\le h(x)+h(y) h ( x y ) ≤ h ( x ) + h ( y )
h ( z / z ′ ) ≤ ∑ i = 1 t − 1 ∣ c i ℓ − c i ′ ℓ ′ ∣ h ( a i ) ≤ 1 2 log ( 2 s ) . h(z/z')\le\sum_{i=1}^{t-1}\left|\frac{c_i}{\ell}-\frac{c_i'}{\ell'}\right|h(a_i)\le\frac{1}{2}\log(2s). h ( z / z ′ ) ≤ i = 1 ∑ t − 1 ℓ c i − ℓ ′ c i ′ h ( a i ) ≤ 2 1 log ( 2 s ) . Here s + M < 2 s s+M<2s s + M < 2 s gcd ( z , z ′ ) ∣ z − z ′ \gcd(z,z')\mid z-z' g cd( z , z ′ ) ∣ z − z ′ 0 < ∣ z − z ′ ∣ ≤ M 0<|z-z'|\le M 0 < ∣ z − z ′ ∣ ≤ M
h ( z / z ′ ) ≥ log ( s / M ) > 1 2 log ( 2 s ) , h(z/z')\ge\log(s/M)>\frac{1}{2}\log(2s), h ( z / z ′ ) ≥ log ( s / M ) > 2 1 log ( 2 s ) , a contradiction. □
The next bound allows a fixed number of small primes to have large exponents. It will be used after assigning each small prime to one node where its valuation is maximal.
Lemma 4.2. Fix C ≥ 1 C \ge1 C ≥ 1 m ≥ 2 m \ge2 m ≥ 2 R ≥ 0 R \ge0 R ≥ 0 C m , R , C > 0 C_{m,R,C} > 0 C m , R , C > 0 n ≥ 3 n \ge3 n ≥ 3 s s s L = log s L = \log s L = log s
L ≥ 2 log ( C n ) , 2 ≤ H ≤ L . L \ge2\log(Cn), \qquad2 \le H \le L. L ≥ 2 log ( C n ) , 2 ≤ H ≤ L . Let z 0 , … , z m z_0,\ldots,z_m z 0 , … , z m [ s , s + C n ] [s,s+Cn] [ s , s + C n ] 1 ≤ i ≤ m 1 \le i \le m 1 ≤ i ≤ m
z i = d i x i k i ∏ ℓ = 1 R i p i ℓ e i ℓ , z_i=d_i x_i^{k_i}\prod_{\ell=1}^{R_i}p_{i\ell}^{e_{i\ell}}, z i = d i x i k i ℓ = 1 ∏ R i p i ℓ e i ℓ , where d i , x i , k i d_i,x_i,k_i d i , x i , k i p i ℓ p_{i\ell} p i ℓ
log d i ≤ H , n / 2 ≤ k i ≤ n , R i ≤ R , p i ℓ ≤ C n , 0 ≤ e i ℓ ≤ n . \log d_i \le H,\qquad n/2 \le k_i \le n,\qquad R_i \le R,\qquad p_{i\ell} \le Cn,\qquad0 \le e_{i\ell} \le n. log d i ≤ H , n /2 ≤ k i ≤ n , R i ≤ R , p i ℓ ≤ C n , 0 ≤ e i ℓ ≤ n . Then
L 1 − 1 / m ≤ C m , R , C H ( log n ) R max { 1 , L / n } ( log ( n L ) ) 1 + 1 / m . (82) L^{1-1/m} \le C_{m,R,C} H(\log n)^R\max\{1,L/n\}(\log(nL))^{1+1/m}. \tag*{(82)} L 1 − 1/ m ≤ C m , R , C H ( log n ) R max { 1 , L / n } ( log ( n L ) ) 1 + 1/ m . (82) Proof. Use the m m m
Λ i = k i log x i + log d i + ∑ ℓ = 1 R i e i ℓ log p i ℓ − log z 0 = log ( z i / z 0 ) . \Lambda_i=k_i\log x_i+\log d_i+\sum_{\ell=1}^{R_i}e_{i\ell}\log p_{i\ell}-\log z_0=\log(z_i/z_0). Λ i = k i log x i + log d i + ℓ = 1 ∑ R i e i ℓ log p i ℓ − log z 0 = log ( z i / z 0 ) . Their values are linearly independent over Q \mathbb{Q} Q z 0 , … , z m z_0,\ldots,z_m z 0 , … , z m
0 < max i ∣ Λ i ∣ ≤ C n / s . (83) 0<\max_i|\Lambda_i|\le Cn/s. \tag*{(83)} 0 < i max ∣ Λ i ∣ ≤ C n / s . (83) List the positive integers occurring as bases, omitting 1 1 1 β 1 , … , β r \beta_1,\ldots,\beta_r β 1 , … , β r r 0 = m ( R + 2 ) + 1 r_0=m(R+2)+1 r 0 = m ( R + 2 ) + 1 m ≤ r ≤ r 0 m\le r\le r_0 m ≤ r ≤ r 0 2 L 2L 2 L s ≥ ( C n ) 2 s\ge(Cn)^2 s ≥ ( C n ) 2
log x i ≤ log ( s + C n ) k i ≤ 4 L n . \log x_i\le\frac{\log(s+Cn)}{k_i}\le\frac{4L}{n}. log x i ≤ k i log ( s + C n ) ≤ n 4 L . Choose r r r β j \beta_j β j 4 L 4L 4 L det V \det V det V
D 0 = r ! ( 4 L ) r . D_0=r!(4L)^r. D 0 = r ! ( 4 L ) r . These are relations at every prime: the selected valuation vectors form a basis of the span of all original valuation vectors, and restriction to the chosen r r r Λ i \Lambda_i Λ i m m m B 0 = ( R + 3 ) n D 0 B_0=(R+3)nD_0 B 0 = ( R + 3 ) n D 0 R + 3 R+3 R + 3 n n n
log max { 4 , B 0 } ≪ m , R log ( n L ) . (84) \log\max\{4,B_0\}\ll_{m,R}\log(nL). \tag*{(84)} log max { 4 , B 0 } ≪ m , R log ( n L ) . (84) Put U = max { 1 , L / n } U=\max\{1,L/n\} U = max { 1 , L / n } H H H 4 U 4U 4 U 2 log ( C n ) 2\log(Cn) 2 log ( C n ) 2 L 2L 2 L d i d_i d i x i x_i x i p i ℓ p_{i\ell} p i ℓ z 0 z_0 z 0 log 4 \log4 log 4 1 1 1
Ω ≤ 2 L ( 4 H U ( 2 log ( C n ) ) R ) m , log Ω ≪ m , R , C log ( n L ) . (85) \Omega\le2L\left(4HU(2\log(Cn))^R\right)^m,\qquad\log\Omega\ll_{m,R,C}\log(nL). \tag*{(85)} Ω ≤ 2 L ( 4 H U ( 2 log ( C n ) ) R ) m , log Ω ≪ m , R , C log ( n L ) . (85) Repeated bases can be assigned any one of their corresponding bounds; removing repetitions or dependent bases decreases the upper bound for the product. Applying (4.1) and using (4.4)–(4.5), we obtain
− log max i ∣ Λ i ∣ ≪ m , R , C L 1 / m H U ( log n ) R ( log ( n L ) ) 1 + 1 / m . -\log\max_i|\Lambda_i|\ll_{m,R,C}L^{1/m}HU(\log n)^R(\log(nL))^{1+1/m}. − log i max ∣ Λ i ∣ ≪ m , R , C L 1/ m H U ( log n ) R ( log ( n L ) ) 1 + 1/ m . By (83) , the left side is at least L − log ( C n ) ≥ L / 2 L-\log(Cn)\ge L/2 L − log ( C n ) ≥ L /2 (82) . □
Large-prime valuations at nearby integers
Theorem 4.3. Fix real numbers C ≥ 1 C\ge1 C ≥ 1 c > 0 c>0 c > 0 D > 0 D>0 D > 0 κ > ⌊ C / c ⌋ + 3 \kappa>\lfloor C/c\rfloor+3 κ > ⌊ C / c ⌋ + 3
There is an effectively computable integer N = N ( C , c , D , κ ) N=N(C,c,D,\kappa) N = N ( C , c , D , κ ) n ≥ N n\ge N n ≥ N s ∈ Z > 0 s\in\mathbb{Z}_{>0} s ∈ Z > 0
( log n ) κ ≤ log s ≤ n D (\log n)^\kappa\le\log s\le n^D ( log n ) κ ≤ log s ≤ n D If
Z ⊆ [ s , s + ⌊ C n ⌋ ] ∩ Z , ∣ Z ∣ ≥ c n / log n , \mathcal{Z}\subseteq[s,s+\lfloor Cn\rfloor]\cap\mathbb{Z},\qquad|\mathcal{Z}|\ge cn/\log n, Z ⊆ [ s , s + ⌊ C n ⌋] ∩ Z , ∣ Z ∣ ≥ c n / log n , then, for every assignment of integers k z k_z k z n / 2 ≤ k z ≤ n n/2\le k_z\le n n /2 ≤ k z ≤ n z ∈ Z z\in\mathcal{Z} z ∈ Z p > C n p>Cn p > C n
k z ∤ v p ( z ) . k_z\nmid v_p(z). k z ∤ v p ( z ) . Proof. Put M = ⌊ C n ⌋ M=\lfloor Cn\rfloor M = ⌊ C n ⌋ L = log s L=\log s L = log s k z ∣ v p ( z ) k_z\mid v_p(z) k z ∣ v p ( z ) z ∈ Z z\in\mathcal{Z} z ∈ Z p > C n p>Cn p > C n p ≤ M p\le M p ≤ M p p p z p z_p z p v p ( z p ) v_p(z_p) v p ( z p ) z z z k z k_z k z d z d_z d z
z = d z x z k z ∏ p a s s i g n e d t o z p r z , p , 0 ≤ r z , p < k z . z=d_zx_z^{k_z}\prod_{p\ \mathrm{assigned\ to}\ z}p^{r_{z,p}},\qquad0\le r_{z,p}<k_z. z = d z x z k z p assigned to z ∏ p r z , p , 0 ≤ r z , p < k z . There are no remainders at primes exceeding M M M M M M C n Cn C n
The unassigned parts satisfy
∏ z ∈ Z d z ∣ M ! . (86) \prod_{z\in\mathcal{Z}}d_z\mid M!. \tag*{(86)} z ∈ Z ∏ d z ∣ M ! . (86) To check this at a prime p ≤ M p\le M p ≤ M h ≥ 1 h\ge1 h ≥ 1 ⌊ M / p h ⌋ + 1 \lfloor M/p^h\rfloor+1 ⌊ M / p h ⌋ + 1 M M M p h p^h p h z p z_p z p ∑ h ≥ 1 ⌊ M / p h ⌋ = v p ( M ! ) \sum_{h\ge1}\lfloor M/p^h\rfloor=v_p(M!) ∑ h ≥ 1 ⌊ M / p h ⌋ = v p ( M !) k z k_z k z (86) .
Set
R = ⌊ C / c ⌋ , γ = c − C R + 1 > 0 , A = max { 2 , 4 ( C + 1 ) γ } , H = A ( log n ) 2 . R=\lfloor C/c\rfloor,\qquad\gamma=c-\frac{C}{R+1}>0,\qquad A=\max\left\{2,\frac{4(C+1)}{\gamma}\right\},\qquad H=A(\log n)^2. R = ⌊ C / c ⌋ , γ = c − R + 1 C > 0 , A = max { 2 , γ 4 ( C + 1 ) } , H = A ( log n ) 2 . For all effectively sufficiently large n n n (60) and the elementary factorial bound give
π ( M ) ≤ ( C + γ ( R + 1 ) 4 ) n log n , ∑ z log d z ≤ log ( M ! ) ≤ ( C + 1 ) n log n . \pi(M)\le\left(C+\frac{\gamma(R+1)}{4}\right)\frac{n}{\log n},\qquad\sum_z\log d_z\le\log(M!)\le(C+1)n\log n. π ( M ) ≤ ( C + 4 γ ( R + 1 ) ) log n n , z ∑ log d z ≤ log ( M !) ≤ ( C + 1 ) n log n . The number of nodes with more than R R R
π ( M ) R + 1 ≤ ( c − 3 γ 4 ) n log n . \frac{\pi(M)}{R+1}\le\left(c-\frac{3\gamma}{4}\right)\frac{n}{\log n}. R + 1 π ( M ) ≤ ( c − 4 3 γ ) log n n . At most γ n / ( 4 log n ) \gamma n/(4\log n) γ n / ( 4 log n ) log d z > H \log d_z>H log d z > H γ n / ( 2 log n ) \gamma n/(2\log n) γ n / ( 2 log n ) R R R log d z ≤ H \log d_z\le H log d z ≤ H
Choose an integer
m > max { 2 , D , κ + 1 κ − R − 3 } . m>\max\left\{2,D,\frac{\kappa+1}{\kappa-R-3}\right\}. m > max { 2 , D , κ − R − 3 κ + 1 } . Then D / m < 1 D/m < 1 D / m < 1
κ ( 1 − 1 / m ) > R + 3 + 1 / m . \kappa(1 - 1/m) > R + 3 + 1/m. κ ( 1 − 1/ m ) > R + 3 + 1/ m . For sufficiently large n n n K m + 1 K_{m+1} K m + 1 s > 2 M 2 s > 2M^2 s > 2 M 2 L ≥ 2 log ( C n ) L \ge2\log(Cn) L ≥ 2 log ( C n ) 2 ≤ H ≤ L 2 \le H \le L 2 ≤ H ≤ L z 0 , … , z m z_0,\ldots,z_m z 0 , … , z m z 1 , … , z m z_1,\ldots,z_m z 1 , … , z m z 0 z_0 z 0
L 1 − 1 / m ≪ C , c , D , κ ( log n ) R + 3 + 1 / m max { 1 , L / n } , L^{1-1/m} \ll_{C,c,D,\kappa} (\log n)^{R+3+1/m}\max\{1,L/n\}, L 1 − 1/ m ≪ C , c , D , κ ( log n ) R + 3 + 1/ m max { 1 , L / n } , where we used L ≤ n D L \le n^D L ≤ n D log ( n L ) \log(nL) log ( n L ) L ≥ n L \ge n L ≥ n
n ≪ C , c , D , κ n D / m ( log n ) R + 3 + 1 / m , n \ll_{C,c,D,\kappa} n^{D/m}(\log n)^{R+3+1/m}, n ≪ C , c , D , κ n D / m ( log n ) R + 3 + 1/ m , which is impossible for large n n n D / m < 1 D/m < 1 D / m < 1 L < n L < n L < n
( log n ) κ ( 1 − 1 / m ) ≪ C , c , D , κ ( log n ) R + 3 + 1 / m , (\log n)^{\kappa(1-1/m)} \ll_{C,c,D,\kappa} (\log n)^{R+3+1/m}, ( log n ) κ ( 1 − 1/ m ) ≪ C , c , D , κ ( log n ) R + 3 + 1/ m , again a contradiction by the choice of m m m
For every fixed 0 < ε ≤ D 0 < \varepsilon\le D 0 < ε ≤ D n ε ≤ log s ≤ n D n^\varepsilon\le\log s \le n^D n ε ≤ log s ≤ n D C , c , ε , D C,c,\varepsilon,D C , c , ε , D κ = ⌊ C / c ⌋ + 4 \kappa= \lfloor C/c\rfloor+ 4 κ = ⌊ C / c ⌋ + 4 ( log n ) κ ≤ n ε (\log n)^\kappa\le n^\varepsilon ( log n ) κ ≤ n ε n n n
A Newton-polygon criterion
A single edge whose slope has a sufficiently large prime denominator forces alternating containment in a transitive group [28 Theorems 2.1 and 2.2] . The preceding theorem supplies such a denominator from a collection of nearby integers.
Theorem 4.4. Fix C ≥ 1 C \ge1 C ≥ 1 c > 0 c > 0 c > 0 D > 0 D > 0 D > 0 κ > ⌊ C / c ⌋ + 3 \kappa> \lfloor C/c\rfloor+ 3 κ > ⌊ C / c ⌋ + 3 n n n s ∈ Z > 0 s \in\mathbb{Z}_{>0} s ∈ Z > 0 ( log n ) κ ≤ log s ≤ n D (\log n)^\kappa\le\log s \le n^D ( log n ) κ ≤ log s ≤ n D F ∈ Z [ X ] F \in\mathbb{Z}[X] F ∈ Z [ X ] n n n F ( 0 ) ≠ 0 F(0) \ne0 F ( 0 ) = 0
P ⊆ ( n / 2 , n − 3 ] , ∣ P ∣ ≥ c n / log n , \mathcal{P} \subseteq(n/2,n-3], \qquad|\mathcal{P}| \ge cn/\log n, P ⊆ ( n /2 , n − 3 ] , ∣ P ∣ ≥ c n / log n , and distinct integers j P ∈ { 0 , … , ⌊ C n ⌋ } j_P \in\{0,\ldots,\lfloor Cn\rfloor\} j P ∈ { 0 , … , ⌊ C n ⌋} p > C n p > Cn p > C n e = v p ( s + j P ) > 0 e = v_p(s+j_P) > 0 e = v p ( s + j P ) > 0 P P P p p p F F F − e / P -e/P − e / P ( p , P ) (p,P) ( p , P ) F ( X + a p , P ) F(X+a_{p,P}) F ( X + a p , P ) a p , P a_{p,P} a p , P
Then Gal ( F / Q ) \operatorname{Gal}(F/\mathbb{Q}) Gal ( F / Q ) P P P P ∈ P P \in\mathcal{P} P ∈ P F F F Q \mathbb{Q} Q
A n ≤ Gal ( F / Q ) . A_n \le\operatorname{Gal}(F/\mathbb{Q}). A n ≤ Gal ( F / Q ) . The threshold depends only on C , c , D , κ C,c,D,\kappa C , c , D , κ
Proof. Apply Theorem 4.3 to the distinct integers z P = s + j P z_P = s+j_P z P = s + j P k z P = P k_{z_P}=P k z P = P p > C n p > Cn p > C n P ∈ P P \in\mathcal{P} P ∈ P P ∤ e = v p ( z P ) P \nmid e = v_p(z_P) P ∤ e = v p ( z P ) e > 0 e > 0 e > 0 p p p e / P e/P e / P F F F P P P P P P P P P P P P 2 P > n 2P > n 2 P > n
A group preserving a blocks of size b b b a , b > 2 a,b > 2 a , b > 2 a b = n ab = n ab = n S b ≀ S a S_b \wr S_a S b ≀ S a ( b ! ) a a ! (b!)^a a! ( b ! ) a a ! P > n / 2 ≥ max { a , b } P > n/2 \ge\max\{a,b\} P > n /2 ≥ max { a , b } F F F P P P P ≤ n − 3 P \le n-3 P ≤ n − 3 A n A_n A n
Corollary 4.5. Fix D > 0 D > 0 D > 0 κ > 5 \kappa> 5 κ > 5 n n n H n , s H_{n,s} H n , s
( log n ) κ ≤ log s ≤ n D . (\log n)^\kappa\le\log s \le n^D. ( log n ) κ ≤ log s ≤ n D . If H n , s H_{n,s} H n , s A n ≤ G n , s A_n \le G_{n,s} A n ≤ G n , s n ε ≤ log s ≤ n D n^\varepsilon\le\log s \le n^D n ε ≤ log s ≤ n D 0 < ε ≤ D 0 < \varepsilon\le D 0 < ε ≤ D ε , D \varepsilon,D ε , D
Proof. Use all primes n / 2 < P ≤ n − 3 n/2 < P \le n-3 n /2 < P ≤ n − 3 j P = n − P j_P=n-P j P = n − P 2 n / ( 5 log n ) 2n/(5\log n) 2 n / ( 5 log n ) n n n p > n p>n p > n e = v p ( s + j P ) > 0 e=v_p(s+j_P)>0 e = v p ( s + j P ) > 0 H n , s ( X ) = ∑ k = 0 n A k X k H_{n,s}(X)=\sum_{k=0}^n A_kX^k H n , s ( X ) = ∑ k = 0 n A k X k
A k = ( n k ) ( s ) n − k , v p ( A k ) = { e , 0 ≤ k < P , 0 , P ≤ k ≤ n . A_k=\binom{n}{k}(s)_{n-k},\qquad v_p(A_k)=
\begin{cases}
e, & 0\le k<P,\\
0, & P\le k\le n.
\end{cases} A k = ( k n ) ( s ) n − k , v p ( A k ) = { e , 0 , 0 ≤ k < P , P ≤ k ≤ n . Indeed s + j P s+j_P s + j P p p p s , … , s + n − 1 s,\ldots,s+n-1 s , … , s + n − 1 p ∤ ( n k ) p\nmid\binom{n}{k} p ∤ ( k n ) ( 0 , e ) (0,e) ( 0 , e ) ( P , 0 ) (P,0) ( P , 0 ) C = 1 C=1 C = 1 c = 2 / 5 c=2/5 c = 2/5 ⌊ C / c ⌋ + 3 = 5 \lfloor C/c\rfloor+3=5 ⌊ C / c ⌋ + 3 = 5 κ = 6 \kappa=6 κ = 6 ( log n ) 6 ≤ n ε (\log n)^6\le n^\varepsilon ( log n ) 6 ≤ n ε
Powers in dense subsets of a progression
The preceding criterion requires an upper bound on log s \log s log s k 2 + ε k^{2+\varepsilon} k 2 + ε [16 ] with the determinant count of Filaseta–Laishram–Saradha [23 Lemma 3.1] . For earlier work on products after terms have been omitted, see Erdős–Turk [18 Theorem 6.1] .
For ε > 0 \varepsilon>0 ε > 0
a ε = min { ε , 3 7 − 1 } , δ ε = a ε ( 1 − 7 2 a ε ) 6 ( 1 + a ε ) . a_\varepsilon=\min\left\{\varepsilon,\frac{3}{\sqrt{7}}-1\right\},\qquad
\delta_\varepsilon=\frac{a_\varepsilon\left(1-\frac{7}{2}a_\varepsilon\right)}{6(1+a_\varepsilon)}. a ε = min { ε , 7 3 − 1 } , δ ε = 6 ( 1 + a ε ) a ε ( 1 − 2 7 a ε ) . Theorem 4.6. Fix positive integers d , C d,C d , C C ≥ d C\ge d C ≥ d ε > 0 \varepsilon>0 ε > 0 0 < δ < δ ε 0<\delta<\delta_\varepsilon 0 < δ < δ ε k 0 ( d , C , ε , δ ) k_0(d,C,\varepsilon,\delta) k 0 ( d , C , ε , δ ) k , A k,A k , A
k ≥ k 0 , A ≥ k 2 + ε , gcd ( A , d ) = 1 , k\ge k_0,\qquad A\ge k^{2+\varepsilon},\qquad\gcd(A,d)=1, k ≥ k 0 , A ≥ k 2 + ε , g cd( A , d ) = 1 , the following assertions hold simultaneously for every integer r ≥ 2 r\ge2 r ≥ 2
(i) Fewer than ( 1 − δ ) k (1-\delta)k ( 1 − δ ) k 0 ≤ j < k 0\le j<k 0 ≤ j < k
A + j d = a j x j r , a j , x j ∈ Z > 0 , P ( a j ) ≤ C k . A+jd=a_jx_j^r,\qquad a_j,x_j\in\mathbb{Z}_{>0},\qquad P(a_j)\le Ck. A + j d = a j x j r , a j , x j ∈ Z > 0 , P ( a j ) ≤ C k . (ii) More than δ k \delta k δ k p > C k p>Ck p > C k
r ∤ v p ( ∏ j = 0 k − 1 ( A + j d ) ) . r\nmid v_p\left(\prod_{j=0}^{k-1}(A+jd)\right). r ∤ v p ( j = 0 ∏ k − 1 ( A + j d ) ) . (iii) If I ⊆ { 0 , … , k − 1 } I\subseteq\{0,\ldots,k-1\} I ⊆ { 0 , … , k − 1 } ( 1 − δ ) k (1-\delta)k ( 1 − δ ) k
∏ j ∈ I ( A + j d ) ≠ c y r for all c , y ∈ Z > 0 with P ( c ) ≤ C k . \prod_{j\in I}(A+jd)\ne cy^r\quad\text{for all }c,y\in\mathbb{Z}_{>0}\text{ with }P(c)\le Ck. j ∈ I ∏ ( A + j d ) = c y r for all c , y ∈ Z > 0 with P ( c ) ≤ C k . In particular, for ε ≥ 1 / 10 \varepsilon\ge1/10 ε ≥ 1/10 δ = 1 / 102 \delta=1/102 δ = 1/102 A A A r r r
Proof. Choose fixed real numbers a , b > 0 a,b>0 a , b > 0
a ≤ ε , 7 2 a + 6 b < 1 , δ < a b 1 + a . (87) a\le\varepsilon,\qquad\frac{7}{2}a+6b<1,\qquad\delta<\frac{ab}{1+a}. \tag*{(87)} a ≤ ε , 2 7 a + 6 b < 1 , δ < 1 + a ab . (87) Such a choice exists by the definition of δ ε \delta_\varepsilon δ ε a ( 1 − 7 2 a ) / ( 6 ( 1 + a ) ) a(1-\frac{7}{2}a)/(6(1+a)) a ( 1 − 2 7 a ) / ( 6 ( 1 + a )) a = a ε a=a_\varepsilon a = a ε I I I ( 1 − δ ) k (1-\delta)k ( 1 − δ ) k a j a_j a j r r r A + j d A+jd A + j d p ≤ C k p\le Ck p ≤ C k I I I v p ( A + j d ) v_p(A+jd) v p ( A + j d ) J J J m = ∣ J ∣ m=|J| m = ∣ J ∣
m ≥ ( 1 − δ ) k − π ( C k ) = ( 1 − δ − o ( 1 ) ) k , ∏ j ∈ J a j ∣ ( k − 1 ) ! . (88) m\ge(1-\delta)k-\pi(Ck)=(1-\delta-o(1))k,\qquad\prod_{j\in J}a_j\mid(k-1)!. \tag*{(88)} m ≥ ( 1 − δ ) k − π ( C k ) = ( 1 − δ − o ( 1 )) k , j ∈ J ∏ a j ∣ ( k − 1 )! . (88) Indeed, if j p j_p j p p ∤ d p\nmid d p ∤ d v p ( A + j d ) ≤ v p ( j − j p ) v_p(A+jd)\le v_p(j-j_p) v p ( A + j d ) ≤ v p ( j − j p ) j j j j p j_p j p j p ! ( k − 1 − j p ) ! j_p!(k-1-j_p)! j p ! ( k − 1 − j p )! ( k − 1 ) ! (k-1)! ( k − 1 )!
v p ( a j ) ≤ ⌊ log ( k − 1 ) log p ⌋ ( j ∈ J ) . v_p(a_j)\le\left\lfloor\frac{\log(k-1)}{\log p}\right\rfloor\qquad(j\in J). v p ( a j ) ≤ ⌊ log p log ( k − 1 ) ⌋ ( j ∈ J ) . Put α = k a \alpha=k^a α = k a β = k − b \beta=k^{-b} β = k − b U = { j ∈ J : a j ≤ α k } U=\{j\in J:a_j\le\alpha k\} U = { j ∈ J : a j ≤ α k }
O d ( α 7 / 2 β − 5 ) (89) O_d(\alpha^{7/2}\beta^{-5}) \tag*{(89)} O d ( α 7/2 β − 5 ) (89) primes in [ β k , k ) [\beta k,k) [ β k , k ) U U U p p p ( i , j ) (i,j) ( i , j ) ( α / β ) 2 (\alpha/\beta)^2 ( α / β ) 2 ( a i / p , a j / p ) (a_i/p,a_j/p) ( a i / p , a j / p ) ( h , h ′ ) (h,h') ( h , h ′ ) p p p ( x i , x j ) (x_i,x_j) ( x i , x j ) q q q ( x u , x v ) (x_u,x_v) ( x u , x v ) k k k
L = ( A / ( α k ) ) 1 / r ≤ x ≤ R = ( 2 A / ( β k ) ) 1 / r , L=(A/(\alpha k))^{1/r}\le x\le R=(2A/(\beta k))^{1/r}, L = ( A / ( α k ) ) 1/ r ≤ x ≤ R = ( 2 A / ( β k ) ) 1/ r , and
∣ h x i r − h ′ x j r ∣ ≤ d / β , ∣ h x u r − h ′ x v r ∣ ≤ d / β . |hx_i^r-h'x_j^r|\le d/\beta,\qquad|hx_u^r-h'x_v^r|\le d/\beta. ∣ h x i r − h ′ x j r ∣ ≤ d / β , ∣ h x u r − h ′ x v r ∣ ≤ d / β . Multiplying by x v r x_v^r x v r x j r x_j^r x j r
h ∣ ( x i x v ) r − ( x j x u ) r ∣ ≤ 4 d A β 2 k . h|(x_ix_v)^r-(x_jx_u)^r|\le\frac{4dA}{\beta^2k}. h ∣ ( x i x v ) r − ( x j x u ) r ∣ ≤ β 2 k 4 d A . Consequently
∣ x i x v − x j x u ∣ ≤ 4 d r α 2 − 2 / r β − 2 ( A / k ) − 1 + 2 / r ≤ 2 d α β − 2 . |x_ix_v-x_jx_u|\le\frac{4d}{r}\alpha^{2-2/r}\beta^{-2}(A/k)^{-1+2/r}\le2d\alpha\beta^{-2}. ∣ x i x v − x j x u ∣ ≤ r 4 d α 2 − 2/ r β − 2 ( A / k ) − 1 + 2/ r ≤ 2 d α β − 2 . The last inequality uses A / k ≥ α A/k\ge\alpha A / k ≥ α − 1 + 2 / r ≤ 0 -1+2/r\le0 − 1 + 2/ r ≤ 0 e = gcd ( x i , x j ) e=\gcd(x_i,x_j) e = g cd( x i , x j ) p e r pe^r p e r d d d p e r ≤ k − 1 pe^r\le k-1 p e r ≤ k − 1 e ≤ β − 1 / r e\le\beta^{-1/r} e ≤ β − 1/ r x i x v − x j x u x_ix_v-x_jx_u x i x v − x j x u ( x u , x v ) (x_u,x_v) ( x u , x v ) x i / e x_i/e x i / e x j / e x_j/e x j / e [ L , R ] 2 [L,R]^2 [ L , R ] 2
2 R e / L + 1 ≤ 2 ( 2 α ) 1 / r β − 2 / r + 1 ≤ 2 2 α β − 1 + 1. 2Re/L+1\le2(2\alpha)^{1/r}\beta^{-2/r}+1\le2\sqrt{2}\alpha\beta^{-1}+1. 2 R e / L + 1 ≤ 2 ( 2 α ) 1/ r β − 2/ r + 1 ≤ 2 2 α β − 1 + 1. Since gcd ( A , d ) = 1 \gcd(A,d)=1 g cd( A , d ) = 1 x j x_j x j d d d x j x_j x j j ∈ U j\in U j ∈ U x j r ∣ i − j x_j^r\mid i-j x j r ∣ i − j x j r ≥ A / ( α k ) ≥ k > ∣ i − j ∣ x_j^r\ge A/(\alpha k)\ge k>|i-j| x j r ≥ A / ( α k ) ≥ k > ∣ i − j ∣ ( x u , x v ) (x_u,x_v) ( x u , x v )
Since b < 1 / 6 b<1/6 b < 1/6 p ≥ β k p\ge\beta k p ≥ β k a j a_j a j 2 / β 2/\beta 2/ β [ β k , k ) [\beta k,k) [ β k , k ) U U U k k k
∑ j ∈ U log a j ≤ k ∑ p ≤ β k log p p − 1 + ∑ p < k log p + O d ( α 7 / 2 β − 6 log k ) ≤ ( 1 − b ) k log k + o ( k log k ) . \begin{aligned}
\sum_{j\in U}\log a_j\le k\sum_{p\le\beta k}\frac{\log p}{p-1}+\sum_{p<k}\log p+O_d(\alpha^{7/2}\beta^{-6}\log k) \\
&\le(1-b)k\log k+o(k\log k).
\end{aligned} j ∈ U ∑ log a j ≤ k p ≤ β k ∑ p − 1 log p + p < k ∑ log p + O d ( α 7/2 β − 6 log k ) ≤ ( 1 − b ) k log k + o ( k log k ) . Here (4.7) makes the last error o ( k log k ) o(k\log k) o ( k log k ) [42 Theorems 4 and 6] .
All cofactors a j a_j a j j ∈ J j \in J j ∈ J a i = a j = c a_i=a_j=c a i = a j = c
d ∣ i − j ∣ = c ∣ x i r − x j r ∣ ≥ r c 1 / r A 1 − 1 / r ≥ 2 A > d k , d|i-j|=c|x_i^r-x_j^r|\ge rc^{1/r}A^{1-1/r}\ge2\sqrt{A}>dk, d ∣ i − j ∣ = c ∣ x i r − x j r ∣ ≥ r c 1/ r A 1 − 1/ r ≥ 2 A > d k , a contradiction. The uniform Stirling bound
log ( u ! ) ≥ u log k − O ( k ) ( 0 ≤ u ≤ k ) \log(u!)\ge u\log k-O(k)\qquad(0\le u\le k) log ( u !) ≥ u log k − O ( k ) ( 0 ≤ u ≤ k ) therefore shows that ∣ U ∣ ≤ ( 1 − b + o ( 1 ) ) k |U|\le(1-b+o(1))k ∣ U ∣ ≤ ( 1 − b + o ( 1 )) k W = J ∖ U W=J\setminus U W = J ∖ U ∣ W ∣ ≥ ( b − δ − o ( 1 ) ) k |W|\ge(b-\delta-o(1))k ∣ W ∣ ≥ ( b − δ − o ( 1 )) k
∑ j ∈ J log a j ≥ log ( ∣ U ∣ ! ) + ( 1 + a ) ∣ W ∣ log k ≥ ( 1 + a b − ( 1 + a ) δ − o ( 1 ) ) k log k . \begin{aligned}
\sum_{j\in J}\log a_j &\ge\log(|U|!)+(1+a)|W|\log k\\
&\ge(1+ab-(1+a)\delta-o(1))k\log k.
\end{aligned} j ∈ J ∑ log a j ≥ log ( ∣ U ∣ !) + ( 1 + a ) ∣ W ∣ log k ≥ ( 1 + ab − ( 1 + a ) δ − o ( 1 )) k log k . This contradicts (4.8), since a b > ( 1 + a ) δ ab>(1+a)\delta ab > ( 1 + a ) δ
Every index excluded from (i) has a prime p > C k p>Ck p > C k r r r p > C k ≥ d k > d ∣ i − j ∣ p>Ck\ge dk>d|i-j| p > C k ≥ d k > d ∣ i − j ∣ ε ≥ 1 / 10 \varepsilon\ge1/10 ε ≥ 1/10 a = 1 / 10 a=1/10 a = 1/10 b = 27 / 250 b=27/250 b = 27/250 δ = 1 / 102 \delta=1/102 δ = 1/102 ( 7 / 2 ) a + 6 b = 499 / 500 < 1 (7/2)a+6b=499/500<1 ( 7/2 ) a + 6 b = 499/500 < 1 a b / ( 1 + a ) = 27 / 2750 > 1 / 102 ab/(1+a)=27/2750>1/102 ab / ( 1 + a ) = 27/2750 > 1/102
Theorem 4.6(iii) allows a fixed positive proportion of arbitrary omissions. It does not settle the corresponding problem for unions of widely separated intervals, or the problem of powerful products: a valuation not divisible by r r r
Local cycles and primitive groups
We first obtain a large prime cycle when the parameter is at most cubic in the degree. For larger parameters, the local packets give elements with small support. We combine these with a progression estimate and the primitive-group theorem of Burness and Guralnick. Irreducibility is assumed only in the statements that require transitivity.
A first parameter range
For a prime close to the degree, the absence of a square multiple in the parameter interval produces the required prime cycle.
Lemma 4.7. Let n / 2 < p < n n/2<p<n n /2 < p < n [ s , s + n − 1 ] [s,s+n-1] [ s , s + n − 1 ] p 2 p^2 p 2 p p p H n , s H_{n,s} H n , s p p p − 1 / p -1/p − 1/ p p p p ∣ G n , s ∣ |G_{n,s}| ∣ G n , s ∣
Proof. Write n = p + b n=p+b n = p + b 0 ≤ b < p 0\le b<p 0 ≤ b < p Proposition 3.13 , show that exactly p p p 1 / p 1/p 1/ p p p p [28 Theorems 2.1 and 2.2] . □
The obstruction in the lemma is a square multiple in a short interval. Such multiples cannot occur simultaneously for all primes close to n n n s s s
Proposition 4.8. There is an absolute effective integer n c u b n_{\mathrm{cub}} n cub n ≥ n c u b n\ge n_{\mathrm{cub}} n ≥ n cub H n , s H_{n,s} H n , s Q \mathbb{Q} Q
1 ≤ s ≤ n 3 10 log n , 1\le s\le\frac{n^3}{10\log n}, 1 ≤ s ≤ 10 log n n 3 , then G n , s G_{n,s} G n , s A n A_n A n
Proof. Put u = log n u=\log n u = log n
P = { p prime : 9 n / 10 < p < n − 2 } . \mathcal{P}=\{p\text{ prime}:9n/10<p<n-2\}. P = { p prime : 9 n /10 < p < n − 2 } . The effective prime number theorem gives
∣ P ∣ > n 20 u (90) |\mathcal{P}| > \frac{n}{20u} \tag*{(90)} ∣ P ∣ > 20 u n (90) for all sufficiently large n n n ϑ \vartheta ϑ [14 Theorem 4.2] , after removing the at most three primes in the upper endpoint interval.
Suppose every p ∈ P p \in\mathcal{P} p ∈ P k p 2 ∈ [ s , s + n − 1 ] kp^2 \in[s, s+n-1] k p 2 ∈ [ s , s + n − 1 ] p < q p<q p < q k k k
k ( q 2 − p 2 ) ≥ q 2 − p 2 > 2 p > n − 1 , k(q^2-p^2) \ge q^2-p^2 > 2p > n-1, k ( q 2 − p 2 ) ≥ q 2 − p 2 > 2 p > n − 1 , which is impossible for two integers in this interval. The multipliers all lie in
[ s n 2 , 100 ( s + n − 1 ) 81 n 2 ] . \left[\frac{s}{n^2},\frac{100(s+n-1)}{81n^2}\right]. [ n 2 s , 81 n 2 100 ( s + n − 1 ) ] . The number of integers in this real interval is at most its length plus one. Consequently
∣ P ∣ ≤ 19 s 81 n 2 + 100 81 n + 1 ≤ 19 n 810 u + 100 81 n + 1 < n 20 u |\mathcal{P}| \le\frac{19s}{81n^2}+\frac{100}{81n}+1 \le\frac{19n}{810u}+\frac{100}{81n}+1 < \frac{n}{20u} ∣ P ∣ ≤ 81 n 2 19 s + 81 n 100 + 1 ≤ 810 u 19 n + 81 n 100 + 1 < 20 u n for sufficiently large n n n
Choose p ∈ P p \in\mathcal{P} p ∈ P p ∣ ∣ G n , s ∣ p \mid|G_{n,s}| p ∣ ∣ G n , s ∣ p p p p p p 2 p > n 2p>n 2 p > n p > n / 2 p>n/2 p > n /2 p p p p p p 2 p 2p 2 p p p p p > n / 2 p>n/2 p > n /2 G n , s ⊇ A n G_{n,s}\supseteq A_n G n , s ⊇ A n p < n − 2 p<n-2 p < n − 2 [28 Theorem 2.2] . Every numerical threshold used above is effective. □
Small supports and Cartesian products
The progression estimate first gives small nonidentity permutations uniformly in the parameter. No rate of decay is needed for the subsequent argument.
Proposition 4.9. For every fixed δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 ) n δ n_\delta n δ
μ ( G n , s ) ≤ δ n \mu(G_{n,s}) \le\delta n μ ( G n , s ) ≤ δ n whenever n ≥ n δ n \ge n_\delta n ≥ n δ s ≥ 1 s \ge1 s ≥ 1 H n , s H_{n,s} H n , s
Proof. For s ≤ n 3 / ( 10 log n ) s \le n^3/(10\log n) s ≤ n 3 / ( 10 log n ) G n , s ⊇ A n G_{n,s}\supseteq A_n G n , s ⊇ A n 3 ≤ δ n 3 \le\delta n 3 ≤ δ n n n n
Suppose therefore that s > n 3 / ( 10 log n ) s>n^3/(10\log n) s > n 3 / ( 10 log n ) k = ⌊ δ n / 2 ⌋ k=\lfloor\delta n/2\rfloor k = ⌊ δ n /2 ⌋ μ ( G n , s ) > δ n \mu(G_{n,s})>\delta n μ ( G n , s ) > δ n N = 2 , 4 , … , 2 k N=2,4,\ldots,2k N = 2 , 4 , … , 2 k p > n p>n p > n s + n − N s+n-N s + n − N N ≤ δ n N\le\delta n N ≤ δ n N ∣ v p ( s + n − N ) N\mid v_p(s+n-N) N ∣ v p ( s + n − N )
The resulting nodes form a progression of length k k k h = gcd ( s + n , 2 ) h=\gcd(s+n,2) h = g cd( s + n , 2 )
A = s + n − 2 k h , d = 2 h ∈ { 1 , 2 } , gcd ( A , d ) = 1. A=\frac{s+n-2k}{h}, \qquad d=\frac{2}{h}\in\{1,2\}, \qquad\gcd(A,d)=1. A = h s + n − 2 k , d = h 2 ∈ { 1 , 2 } , g cd( A , d ) = 1. Every term is a square times an n n n C ≥ 4 / δ C\ge4/\delta C ≥ 4/ δ n ≤ C k n\le Ck n ≤ C k n n n A ≥ s / 2 > n 3 / ( 20 log n ) > k 5 / 2 A\ge s/2>n^3/(20\log n)>k^{5/2} A ≥ s /2 > n 3 / ( 20 log n ) > k 5/2 n n n ε = 1 / 2 \varepsilon=1/2 ε = 1/2 δ = 1 / 102 \delta=1/102 δ = 1/102 r = 2 r=2 r = 2 k k k
The next observation relates fixed-point counts to divisors. Its strength is that it applies to the full symmetric group on each coordinate, irrespective of the smaller group acting there.
Lemma 4.10. Let m ≥ 2 m \ge2 m ≥ 2 a ≥ 2 a \ge2 a ≥ 2 n = m a n=m^{a} n = m a S m ≀ S a S_m \wr S_a S m ≀ S a { 1 , … , m } a \{1,\ldots,m\}^{a} { 1 , … , m } a 15 n / 16 15n/16 15 n /16 f 1 ⋯ f a f_1\cdots f_a f 1 ⋯ f a 15 m / 16 ≤ f i ≤ m 15m/16 \le f_i \le m 15 m /16 ≤ f i ≤ m ( m / 2 , m ] (m/2,m] ( m /2 , m ]
Proof. A cycle of r ≥ 2 r \ge2 r ≥ 2 m m m m r m^r m r 1 / m ≤ 1 / 2 1/m \le1/2 1/ m ≤ 1/2 ∏ i ( f i / m ) \prod_i(f_i/m) ∏ i ( f i / m ) [ 0 , 1 ] [0,1] [ 0 , 1 ] 15 / 16 15/16 15/16 15 / 16 15/16 15/16
Ford’s divisor estimate makes the set of possible fixed-point counts in this lemma sparse when m m m
Proposition 4.11. For all sufficiently large n n n s ≥ 1 s \ge1 s ≥ 1 H n , s H_{n,s} H n , s G n , s G_{n,s} G n , s S m ≀ S a S_m \wr S_a S m ≀ S a m ≥ 2 m \ge2 m ≥ 2 a ≥ 2 a \ge2 a ≥ 2 n = m a n=m^a n = m a
Proof. In this action a nontrivial coordinate permutation moves at least ( 1 − 1 / m ) n (1-1/m)n ( 1 − 1/ m ) n 2 n / m 2n/m 2 n / m n min { 1 − 1 / m , 2 / m } n\min\{1-1/m,2/m\} n min { 1 − 1/ m , 2/ m } m → ∞ m \to\infty m → ∞ n → ∞ n \to\infty n → ∞
For large n n n 3 3 3 m ≥ 2 m \ge2 m ≥ 2 a ≥ 2 a \ge2 a ≥ 2 A n A_n A n s > n 3 / ( 10 log n ) s > n^3/(10\log n) s > n 3 / ( 10 log n ) k = ⌊ n / 32 ⌋ k=\lfloor n/32\rfloor k = ⌊ n /32 ⌋ N ∈ { 2 , 4 , … , 2 k } N \in\{2,4,\ldots,2k\} N ∈ { 2 , 4 , … , 2 k } j = n − N j=n-N j = n − N p > n p>n p > n s + j s+j s + j N ≤ n / 16 N \le n/16 N ≤ n /16 j j j ( m / 2 , m ] (m/2,m] ( m /2 , m ]
Let H ( x , y , z ) H(x,y,z) H ( x , y , z ) x x x ( y , z ] (y,z] ( y , z ]
H ( x , y , 2 y ) ≪ x ( log y ) κ ( log log y ) 3 / 2 , 3 ≤ y ≤ x , κ = 1 − 1 + log log 2 log 2 > 0 ; (91) H(x,y,2y)\ll\frac{x}{(\log y)^{\kappa}(\log\log y)^{3/2}},\qquad3\le y\le\sqrt{x},\qquad\kappa=1-\frac{1+\log\log2}{\log2}>0; \tag*{(91)} H ( x , y , 2 y ) ≪ ( log y ) κ ( log log y ) 3/2 x , 3 ≤ y ≤ x , κ = 1 − log 2 1 + log log 2 > 0 ; (91) see [25 Theorem 1(v) and Corollary 2] . Its upper bound is effective. Since a ≥ 2 a\ge2 a ≥ 2 x = n x=n x = n y = m / 2 y=m/2 y = m /2 m → ∞ m\to\infty m → ∞ j j j o ( n ) = o ( k ) o(n)=o(k) o ( n ) = o ( k )
At every other label all inertia at primes above n n n N N N o ( k ) o(k) o ( k )
s + n − 2 k , s + n − 2 k + 2 , … , s + n − 2 s+n-2k,\ s+n-2k+2,\ldots,s+n-2 s + n − 2 k , s + n − 2 k + 2 , … , s + n − 2 are squares times n n n h = gcd ( s + n , 2 ) h=\gcd(s+n,2) h = g cd( s + n , 2 ) d = 1 d=1 d = 1 2 2 2 A ≥ s / 2 A\ge s/2 A ≥ s /2 k 5 / 2 / 2 k^{5/2}/2 k 5/2 /2 k k k n n n n ≤ 33 k n\le33k n ≤ 33 k C = 33 C=33 C = 33 r = 2 r=2 r = 2 ϵ = 1 / 2 \epsilon=1/2 ϵ = 1/2 δ = 1 / 102 \delta=1/102 δ = 1/102 m 0 m_0 m 0 1 / 102 1/102 1/102 min 2 ≤ m < m 0 min { 1 − 1 / m , 2 / m } \min_{2\le m<m_0}\min\{1-1/m,2/m\} min 2 ≤ m < m 0 min { 1 − 1/ m , 2/ m } m m m
Primitive groups
After Cartesian products have been excluded, the primitive-group classification leaves actions of symmetric or alternating groups on subsets. Their homogeneous permutations have a simple restriction.
Lemma 4.12. Suppose 2 ≤ ℓ < m / 2 2\le\ell<m/2 2 ≤ ℓ < m /2 S m S_m S m ℓ \ell ℓ { 1 , … , m } \{1,\ldots,m\} { 1 , … , m } 15 / 16 15/16 15/16
Proof. Let σ ∈ S m \sigma\in S_m σ ∈ S m u u u I I I ℓ \ell ℓ
E ∣ I △ σ ( I ) ∣ = 2 u ℓ ( m − ℓ ) m ( m − 1 ) . \mathbb{E}|I \mathbin{\triangle} \sigma(I)| = \frac{2u\ell(m-\ell)}{m(m-1)}. E ∣ I △ σ ( I ) ∣ = m ( m − 1 ) 2 u ℓ ( m − ℓ ) . Since the symmetric difference is at most 2 ℓ 2\ell 2 ℓ
1 − fpr ( σ ) ≥ u ( m − ℓ ) m ( m − 1 ) > u 2 m . 1-\operatorname{fpr}(\sigma) \ge\frac{u(m-\ell)}{m(m-1)} > \frac{u}{2m}. 1 − fpr ( σ ) ≥ m ( m − 1 ) u ( m − ℓ ) > 2 m u . Thus u < m / 8 u<m/8 u < m /8 ℓ \ell ℓ r r r ℓ − 1 \ell-1 ℓ − 1 r r r q q q q q q q > 2 q>2 q > 2 q q q ℓ − 2 \ell-2 ℓ − 2 q / 2 > 1 q/2>1 q /2 > 1 □ \square □
The lemma converts odd valuations into forbidden permutations on a full progression of widths.
Proposition 4.13. For all sufficiently large n n n s ≥ 1 s \ge1 s ≥ 1 H n , s H_{n,s} H n , s S m S_m S m ℓ \ell ℓ 2 ≤ ℓ < m / 2 2 \le\ell< m/2 2 ≤ ℓ < m /2
Proof. The small-parameter case follows from Proposition 4.8 , since a proper subset action has order at most m ! < ∣ A n ∣ m!<|A_n| m ! < ∣ A n ∣ n = ( m ℓ ) n=\binom{m}{\ell} n = ( ℓ m ) s > n 3 / ( 10 log n ) s>n^3/(10\log n) s > n 3 / ( 10 log n ) k = ⌊ n / 64 ⌋ k=\lfloor n/64\rfloor k = ⌊ n /64 ⌋ N = 4 , 8 , … , 4 k N=4,8,\ldots,4k N = 4 , 8 , … , 4 k e e e p > n p>n p > n
g = gcd ( N , e ) odd , q = N / g ≡ 0 ( m o d 4 ) . g=\gcd(N,e)\text{ odd}, \qquad q=N/g\equiv0\pmod{4}. g = g cd( N , e ) odd , q = N / g ≡ 0 ( mod 4 ) . The inertia is nonidentity, homogeneous, and has fixed-point ratio at least 15 / 16 15/16 15/16 Lemma 4.12 excludes it. Therefore every valuation at a prime above n n n
s + n − 4 k , s + n − 4 k + 4 , … , s + n − 4. s+n-4k,\ s+n-4k+4,\ldots,s+n-4. s + n − 4 k , s + n − 4 k + 4 , … , s + n − 4. Divide by h = gcd ( s + n , 4 ) h=\gcd(s+n,4) h = g cd( s + n , 4 ) d = 4 / h ∈ { 1 , 2 , 4 } d=4/h\in\{1,2,4\} d = 4/ h ∈ { 1 , 2 , 4 } A = ( s + n − 4 k ) / h A=(s+n-4k)/h A = ( s + n − 4 k ) / h d d d n n n n n n n ≤ 65 k n\le65k n ≤ 65 k A ≥ s / 4 > k 5 / 2 A\ge s/4>k^{5/2} A ≥ s /4 > k 5/2 C = 65 C=65 C = 65 r = 2 r=2 r = 2 ε = 1 / 2 \varepsilon=1/2 ε = 1/2 δ = 1 / 102 \delta=1/102 δ = 1/102 □ \square □
We can now apply the primitive-group theorem. The inequality in that theorem is strict, so the numerical margin below also handles elements of order two.
Theorem 4.14. For all effectively sufficiently large n n n s s s G n , s G_{n,s} G n , s A n ≤ G n , s A_n\le G_{n,s} A n ≤ G n , s
Proof. Primitivity includes transitivity, so H n , s H_{n,s} H n , s Proposition 4.9 , a sufficiently large G n , s G_{n,s} G n , s n / 3 n/3 n /3 r r r 2 / 3 > 1 / r + 1 2/3>1/\sqrt{r+1} 2/3 > 1/ r + 1 [7 Corollary 2] imply that the primitive action has the form
( A m ) a ≤ G n , s ≤ S m ≀ S a , n = ( m ℓ ) a , 1 ≤ ℓ < m / 2 , (A_m)^a \le G_{n,s} \le S_m \wr S_a,\qquad n=\binom{m}{\ell}^{a},\qquad1\le\ell<m/2, ( A m ) a ≤ G n , s ≤ S m ≀ S a , n = ( ℓ m ) a , 1 ≤ ℓ < m /2 , where each coordinate is the action on ℓ \ell ℓ ℓ = m / 2 \ell=m/2 ℓ = m /2 a ≥ 2 a\ge2 a ≥ 2 Proposition 4.13 excludes ℓ ≥ 2 \ell\ge2 ℓ ≥ 2 a = ℓ = 1 a=\ell=1 a = ℓ = 1 G n , s ≥ A n G_{n,s}\ge A_n G n , s ≥ A n □ \square □
Invariant partitions
We exclude invariant partitions in three steps. Progressions of cubes rule out blocks of size two. Linear forms in logarithms bound the parameter when both the block size and the number of blocks are at least three. The partition energy then excludes all partitions in the complementary range. Only partitions into two large blocks remain for the next section.
Blocks of size two
For blocks of size two, an odd packet cannot support inertia whose nontrivial cycles have length divisible by three.
Lemma 4.15. There is an effective absolute integer n 0 n_0 n 0 n ≥ n 0 n \ge n_0 n ≥ n 0 s > n 3 / ( 10 log n ) s > n^3/(10\log n) s > n 3 / ( 10 log n ) G n , s G_{n,s} G n , s
Proof. Suppose such a partition exists. Put k = ⌊ n / 12 ⌋ k = \lfloor n/12\rfloor k = ⌊ n /12 ⌋ N t = 3 ( 1 + 2 t ) N_t = 3(1 + 2t) N t = 3 ( 1 + 2 t ) 0 ≤ t < k 0 \le t < k 0 ≤ t < k n / 2 n/2 n /2 p > n p > n p > n s + n − N t s + n - N_t s + n − N t e e e Proposition 3.2 gives an inertia permutation whose nontrivial cycles all have length q = N t / gcd ( N t , e ) q = N_t/\gcd(N_t,e) q = N t / g cd( N t , e ) 3 ∣ q 3 \mid q 3 ∣ q q > 2 q > 2 q > 2 N t N_t N t n n n
In reverse order the nodes form A , A + 6 , … , A + 6 ( k − 1 ) A, A + 6, \ldots, A + 6(k - 1) A , A + 6 , … , A + 6 ( k − 1 ) A > s A > s A > s g = gcd ( A , 6 ) g = \gcd(A,6) g = g cd( A , 6 ) A ′ = A / g A' = A/g A ′ = A / g d = 6 / g d = 6/g d = 6/ g gcd ( A ′ , d ) = 1 \gcd(A',d) = 1 g cd( A ′ , d ) = 1 A ′ + j d A' + jd A ′ + j d n n n n n n n ≤ 13 k n \le13k n ≤ 13 k
A ′ > s / 6 > n 3 60 log n > k 5 / 2 . A' > s/6 > \frac{n^3}{60\log n} > k^{5/2}. A ′ > s /6 > 60 log n n 3 > k 5/2 . Theorem 4.6(i), with C = 13 C = 13 C = 13 r = 3 r = 3 r = 3 ε = 1 / 2 \varepsilon= 1/2 ε = 1/2 δ = 1 / 102 \delta= 1/102 δ = 1/102 d ∈ { 1 , 2 , 3 , 6 } d \in\{1,2,3,6\} d ∈ { 1 , 2 , 3 , 6 } k k k □ \square □
A parameter bound for invariant partitions
It is not necessary to discard a whole node for each small prime. Instead, assign each small prime to one node where its valuation is maximal. A pair of nodes with a common large exponent then has only a bounded number of exceptional prime factors.
Proposition 4.16. There are effective absolute constants C 1 , n 3 > 0 C_1, n_3 > 0 C 1 , n 3 > 0 n ≥ n 3 n \ge n_3 n ≥ n 3 G n , s G_{n,s} G n , s
3 ≤ b ≤ n / 3 , 3 \le b \le n/3, 3 ≤ b ≤ n /3 , one has
log s ≤ C 1 ( log n ) 9 . \log s \le C_1(\log n)^9. log s ≤ C 1 ( log n ) 9 . Irreducibility is not required for this assertion.
Proof. Put a = n / b a = n/b a = n / b u = log n u = \log n u = log n L = log s L = \log s L = log s L > 2 u L > 2u L > 2 u C 1 C_1 C 1
n / 3 < P ≤ n / 2 n/3 < P \le n/2 n /3 < P ≤ n /2 use the two widths P P P 2 P 2P 2 P a , b ≥ 3 a,b \ge3 a , b ≥ 3 a b = n ab = n ab = n max { a , b } ≤ n / 3 < P \max\{a,b\} \le n/3 < P max { a , b } ≤ n /3 < P P P P ( b ! ) a a ! (b!)^a a! ( b ! ) a a ! p > n p > n p > n P P P Proposition 3.2 supplies an inertia element whose order is divisible by P P P
z P , ℓ : = s + n − ℓ P = a P , ℓ x P , ℓ P , P ( a P , ℓ ) ≤ n , ℓ ∈ { 1 , 2 } , z_{P,\ell} := s + n - \ell P = a_{P,\ell}x_{P,\ell}^{P}, \qquad P(a_{P,\ell}) \le n,\qquad\ell\in\{1,2\}, z P , ℓ := s + n − ℓ P = a P , ℓ x P , ℓ P , P ( a P , ℓ ) ≤ n , ℓ ∈ { 1 , 2 } , where each a P , ℓ a_{P,\ell} a P , ℓ P P P p ≤ n p \le n p ≤ n z P , ℓ z_{P,\ell} z P , ℓ v p ( z P , ℓ ) v_p(z_{P,\ell}) v p ( z P , ℓ ) p p p E P , ℓ \mathcal{E}_{P,\ell} E P , ℓ
r P , ℓ , p = v p ( a P , ℓ ) ( p ∈ E P , ℓ ) , d P , ℓ = a P , ℓ ∏ p ∈ E P , ℓ p − r P , ℓ , p . r_{P,\ell,p}=v_p(a_{P,\ell}) \quad(p \in\mathcal{E}_{P,\ell}), \qquad d_{P,\ell}=a_{P,\ell}\prod_{p\in\mathcal{E}_{P,\ell}}p^{-r_{P,\ell,p}}. r P , ℓ , p = v p ( a P , ℓ ) ( p ∈ E P , ℓ ) , d P , ℓ = a P , ℓ p ∈ E P , ℓ ∏ p − r P , ℓ , p . These are positive integers, and 0 ≤ r P , ℓ , p < P 0 \le r_{P,\ell,p} < P 0 ≤ r P , ℓ , p < P
∏ P , ℓ d P , ℓ ∣ ( n − 1 ) ! . (92) \prod_{P,\ell}d_{P,\ell}\mid(n-1)!. \tag*{(92)} P , ℓ ∏ d P , ℓ ∣ ( n − 1 )! . (92) This is the allocation assertion of Lemma 3.19 : the remaining cofactor at each node divides its unassigned small-prime part.
Let R R R
R = ( 1 6 + o ( 1 ) ) n u , π ( n ) = ( 1 + o ( 1 ) ) n u . R=\left(\frac{1}{6}+o(1)\right)\frac{n}{u}, \qquad\pi(n)=(1+o(1))\frac{n}{u}. R = ( 6 1 + o ( 1 ) ) u n , π ( n ) = ( 1 + o ( 1 )) u n . At most π ( n ) / 7 \pi(n)/7 π ( n ) /7 (92) , at most n / ( 1000 u ) n/(1000u) n / ( 1000 u )
log d P , 1 + log d P , 2 > 1000 u 2 . \log d_{P,1}+\log d_{P,2}>1000u^2. log d P , 1 + log d P , 2 > 1000 u 2 . Since
1 6 − 1 7 − 1 1000 > 0 , \frac{1}{6}-\frac{1}{7}-\frac{1}{1000}>0, 6 1 − 7 1 − 1000 1 > 0 , there is, for all sufficiently large n n n
log d P , 1 + log d P , 2 ≤ 1000 u 2 . \log d_{P,1}+\log d_{P,2}\le1000u^2. log d P , 1 + log d P , 2 ≤ 1000 u 2 . Write its assigned primes as p 1 , … , p h p_1,\ldots,p_h p 1 , … , p h h ≤ 6 h\le6 h ≤ 6 c j c_j c j ∣ c j ∣ < P |c_j|<P ∣ c j ∣ < P
z P , 1 z P , 2 = d P , 1 d P , 2 ( x P , 1 x P , 2 ) P ∏ j = 1 h p j c j . \frac{z_{P,1}}{z_{P,2}}=\frac{d_{P,1}}{d_{P,2}}\left(\frac{x_{P,1}}{x_{P,2}}\right)^P\prod_{j=1}^{h}p_j^{c_j}. z P , 2 z P , 1 = d P , 2 d P , 1 ( x P , 2 x P , 1 ) P j = 1 ∏ h p j c j . Taking real logarithms gives a nonzero linear form Λ = log ( z P , 1 / z P , 2 ) \Lambda=\log(z_{P,1}/z_{P,2}) Λ = log ( z P , 1 / z P , 2 )
0 < ∣ Λ ∣ ≤ n / s . 0<|\Lambda|\le n/s. 0 < ∣Λ∣ ≤ n / s . The rational bases have logarithmic heights at most
1000 u 2 , log ( s + n ) P ≤ 2 L P , u , … , u , 1000u^2,\qquad\frac{\log(s+n)}{P}\le\frac{2L}{P},\qquad u,\ldots,u, 1000 u 2 , P log ( s + n ) ≤ P 2 L , u , … , u , respectively. Omit any base equal to one and any term with zero coefficient. The number of logarithms is at most eight and all integer coefficients have modulus at most P P P
If only one logarithm remains, its elementary rational-height bound gives the inequality below directly. Otherwise Matveev’s theorem [38 ] , Corollary 2.3, in degree one with coefficient bound B ∗ ≤ P B^*\le P B ∗ ≤ P
L − u ≤ C 2 u 9 max { 1 , L P } ≤ C 3 u 9 max { 1 , L n } . L-u\le C_2u^9\max\left\{1,\frac{L}{P}\right\}\le C_3u^9\max\left\{1,\frac{L}{n}\right\}. L − u ≤ C 2 u 9 max { 1 , P L } ≤ C 3 u 9 max { 1 , n L } . Here the factors u 2 u^2 u 2 u u u log ( e P ) = O ( u ) \log(eP)=O(u) log ( e P ) = O ( u ) x P , 1 = x P , 2 x_{P,1}=x_{P,2} x P , 1 = x P , 2
Since L > 2 u L>2u L > 2 u
L ≤ C 4 u 9 max { 1 , L n } . L\le C_4u^9\max\left\{1,\frac{L}{n}\right\}. L ≤ C 4 u 9 max { 1 , n L } . For n > 2 C 4 u 9 n > 2C_4u^9 n > 2 C 4 u 9 L ≥ n L \ge n L ≥ n L < C 4 u 9 L < C_4u^9 L < C 4 u 9
In particular, for partitions whose block size and number are both at least three, it now suffices to treat a parameter range of fixed polynomial growth in log n \log n log n
The energy of a partition
The matrix energy identity applies even when individual blocks are not defined over Q \mathbb{Q} Q
Suppose now that the Galois group preserves a partition into a a a b b b n = a b n = ab n = ab a , b ≥ 2 a, b \ge2 a , b ≥ 2
Q i j = 1 { i , j in the same block } b − 1 n . Q_{ij} = \frac{\mathbf{1}_{\{i,j\ \text{in the same block}\}}}{b} - \frac{1}{n}. Q ij = b 1 { i , j in the same block } − n 1 . This is the orthogonal projection onto the block-constant vectors orthogonal to the constant vector. It has rank a − 1 a-1 a − 1 ( a − 1 ) / n (a-1)/n ( a − 1 ) / n
Lemma 4.17. Put u = log n u = \log n u = log n T = log s T = \log s T = log s
E p = ∑ i ≠ j Q i j v p ( α i − α j ) + ∑ i Q i i ( v p ( α i ) + ϵ p ) . E_p = \sum_{i\ne j} Q_{ij}v_p(\alpha_i-\alpha_j) + \sum_i Q_{ii}(v_p(\alpha_i)+\epsilon_p). E p = i = j ∑ Q ij v p ( α i − α j ) + i ∑ Q ii ( v p ( α i ) + ϵ p ) . Then E p ≥ 0 E_p \ge0 E p ≥ 0 s ≥ n s \ge n s ≥ n log n ≥ 10000 \log n \ge10000 log n ≥ 10000
E : = ∑ p E p log p < ( a − 1 ) ( T 2 + 3 u 2 + log 48 ) . (93) \mathcal{E} := \sum_p E_p\log p < (a-1)\left(\frac{T}{2}+\frac{3u}{2}+\log48\right). \tag*{(93)} E := p ∑ E p log p < ( a − 1 ) ( 2 T + 2 3 u + log 48 ) . (93) If p > n p > n p > n s + j s+j s + j N = n − j N=n-j N = n − j e = v p ( s + j ) e=v_p(s+j) e = v p ( s + j ) k i k_i k i i i i
E p = e N ( 1 b ∑ i = 1 a k i 2 − N 2 n ) . (94) E_p = \frac{e}{N}\left(\frac{1}{b}\sum_{i=1}^{a} k_i^2-\frac{N^2}{n}\right). \tag*{(94)} E p = N e ( b 1 i = 1 ∑ a k i 2 − n N 2 ) . (94) Proof. The displayed matrix Q Q Q Lemma 3.5 , with tr Q = a − 1 \operatorname{tr} Q=a-1 tr Q = a − 1 (93) . For a packet C C C N N N 1 C Q 1 C = b − 1 ∑ i k i 2 − N 2 / n {\mathbf{1}}_C Q {\mathbf{1}}_C=b^{-1}\sum_i k_i^2-N^2/n 1 C Q 1 C = b − 1 ∑ i k i 2 − N 2 / n (45) proves (4.14). □
An order-P P P P P P P P P
Lemma 4.18. Suppose p > n p>n p > n N = ℓ P N=\ell P N = ℓ P P P P P ∤ e = v p ( s + n − N ) P\nmid e=v_p(s+n-N) P ∤ e = v p ( s + n − N ) P > a P>a P > a a ∤ ℓ a\nmid\ell a ∤ ℓ
E p ≥ e P 2 ( a − 1 ) n N . (95) E_p \ge\frac{eP^2(a-1)}{nN}. \tag*{(95)} E p ≥ n N e P 2 ( a − 1 ) . (95) If P > b P>b P > b b ∤ N b\nmid N b ∤ N
Proof. The homogeneous inertia cycle length q = N / gcd ( N , e ) q=N/\gcd(N,e) q = N / g cd( N , e ) P P P q / P q/P q / P P P P P > a P>a P > a k i = P t i k_i=Pt_i k i = P t i ∑ i t i = ℓ \sum_i t_i=\ell ∑ i t i = ℓ ℓ = a h + r \ell=ah+r ℓ = ah + r 0 < r < a 0<r<a 0 < r < a
1 b ∑ i k i 2 − N 2 n ≥ P 2 r ( a − r ) n ≥ P 2 ( a − 1 ) n . \frac{1}{b}\sum_i k_i^2-\frac{N^2}{n}\ge\frac{P^2r(a-r)}{n}\ge\frac{P^2(a-1)}{n}. b 1 i ∑ k i 2 − n N 2 ≥ n P 2 r ( a − r ) ≥ n P 2 ( a − 1 ) . (94) proves (4.15). For the last assertion, q ≥ P > b q \ge P > b q ≥ P > b b ∣ N b \mid N b ∣ N
We next choose many widths to which this occupation bound applies. Fix θ = 101 / 200 \theta= 101/200 θ = 101/200 δ = log ( 1 / θ ) \delta= \log(1/\theta) δ = log ( 1/ θ ) δ / 2 > 1 / 3 \delta/2 > 1/3 δ /2 > 1/3 d ≥ 2 d \ge2 d ≥ 2
N = ℓ P ≤ n , P > n θ prime , d ∤ ℓ N = \ell P \le n,\qquad P > n^\theta\ \text{prime},\qquad d \nmid\ell N = ℓ P ≤ n , P > n θ prime , d ∤ ℓ is at least ( δ ( 1 − 1 / d ) − o ( 1 ) ) n (\delta(1 - 1/d) - o(1))n ( δ ( 1 − 1/ d ) − o ( 1 )) n d d d P > n P > \sqrt{n} P > n
∑ n θ < P ≤ n ( ⌊ n P ⌋ − ⌊ n d P ⌋ ) ≥ n ( 1 − 1 / d ) ∑ n θ < P ≤ n 1 P − π ( n ) . \sum_{n^\theta<P\le n}\left(\left\lfloor\frac{n}{P}\right\rfloor-\left\lfloor\frac{n}{dP}\right\rfloor\right)\ge n(1-1/d)\sum_{n^\theta<P\le n}\frac{1}{P}-\pi(n). n θ < P ≤ n ∑ ( ⌊ P n ⌋ − ⌊ d P n ⌋ ) ≥ n ( 1 − 1/ d ) n θ < P ≤ n ∑ P 1 − π ( n ) . The prime reciprocal sum tends effectively to δ \delta δ π ( n ) = o ( n ) \pi(n) = o(n) π ( n ) = o ( n ) d d d
Proposition 4.19. For all effectively sufficiently large n n n
s > n 3 10 log n , log ( s + n ) log n < n 101 / 200 , s > \frac{n^3}{10\log n},\qquad\frac{\log(s+n)}{\log n}<n^{101/200}, s > 10 log n n 3 , log n log ( s + n ) < n 101/200 , then G n , s G_{n,s} G n , s
Proof. Suppose there are a a a b b b a ≤ n θ a \le n^\theta a ≤ n θ d = a d = a d = a p > n p > n p > n e < n θ < P e < n^\theta< P e < n θ < P E p > e ( a − 1 ) n 2 θ − 2 E_p > e(a - 1)n^{2\theta-2} E p > e ( a − 1 ) n 2 θ − 2
Delete one selected node with maximal valuation for every prime at most n n n π ( n ) \pi(n) π ( n ) ( n − 1 ) ! (n - 1)! ( n − 1 )! W W W n n n
W ≥ ( δ / 2 − o ( 1 ) ) n T − n u ≥ c n T W \ge(\delta/2 - o(1))nT - nu \ge cnT W ≥ ( δ /2 − o ( 1 )) n T − n u ≥ c n T for an absolute c > 0 c > 0 c > 0 T > 3 u − log ( 10 u ) T > 3u - \log(10u) T > 3 u − log ( 10 u ) δ / 2 > 1 / 3 \delta/2 > 1/3 δ /2 > 1/3 E > c ( a − 1 ) n 2 θ − 1 T E > c(a - 1)n^{2\theta-1}T E > c ( a − 1 ) n 2 θ − 1 T 2 ( a − 1 ) T 2(a - 1)T 2 ( a − 1 ) T
If a > n θ a > n^\theta a > n θ b < n 1 − θ < n θ b < n^{1-\theta} < n^\theta b < n 1 − θ < n θ d = b d = b d = b P > b P > b P > b b ∤ ℓ b \nmid\ell b ∤ ℓ b ∤ ℓ P b \nmid\ell P b ∤ ℓ P n n n W ≥ c n T W \ge cnT W ≥ c n T
Laurent interpolation and quadratic block fields
Let E E E G B G_B G B K = E G B K = E^{G_B} K = E G B G B G_B G B K K K
The real quadratic case
The centered block trace is bounded at both real embeddings. Its norm allows us to combine this bound with the fibers at 0 0 0 − 1 -1 − 1
Lemma 4.20. Let n = 2 b ≥ 4 n = 2b \ge4 n = 2 b ≥ 4 H n , s H_{n,s} H n , s b b b
log s < 2 n log n . \log s < 2n\log n. log s < 2 n log n . Proof. Let K K K B B B t = ∑ α ∈ B ( α + s ) t=\sum_{\alpha\in B}(\alpha+s) t = ∑ α ∈ B ( α + s ) K K K − t -t − t ∣ t ∣ < b 2 |t|<b^2 ∣ t ∣ < b 2
If t ≠ 0 t\ne0 t = 0 A = N K / Q ( t ) = − t 2 A=N_{K/\mathbb{Q}}(t)=-t^2 A = N K / Q ( t ) = − t 2 ∣ A ∣ < b 4 |A|<b^4 ∣ A ∣ < b 4 p e ∥ s p^e\parallel s p e ∥ s H n , s ( Y − s ) ≡ Y n ( m o d s ) H_{n,s}(Y-s)\equiv Y^n\pmod{s} H n , s ( Y − s ) ≡ Y n ( mod s ) e / n e/n e / n v p v_p v p v p ( A ) ≥ 2 e / n = e / b v_p(A)\ge2e/n=e/b v p ( A ) ≥ 2 e / n = e / b s ∣ A b s\mid A^b s ∣ A b
log s ≤ b log ∣ A ∣ < 4 b log b < 2 n log n . \log s\le b\log|A|<4b\log b<2n\log n. log s ≤ b log ∣ A ∣ < 4 b log b < 2 n log n . If t = 0 t=0 t = 0 − b s -bs − b s H n , s ( X ) ≡ X n − 1 ( X − n ) ( m o d s + 1 ) H_{n,s}(X)\equiv X^{n-1}(X-n)\pmod{s+1} H n , s ( X ) ≡ X n − 1 ( X − n ) ( mod s + 1 ) A = b A=b A = b c = n = 2 b c=n=2b c = n = 2 b s + 1 ∣ b n s+1\mid b^n s + 1 ∣ b n log s < n log b < 2 n log n \log s<n\log b<2n\log n log s < n log b < 2 n log n
Laurent interpolation and cyclic intermediate fields
The imaginary quadratic case requires integral polynomials whose negative odd Laurent coefficients vanish and whose first positive odd coefficient is prescribed. We first prove the required surjectivity over the ground field; we then bound denominators and coefficient sizes. The next theorem gives a general criterion for the first step. At a fiber with a common ramification index q q q q q q
Let k k k C C C g g g T : C → P k 1 T:C\to\mathbb{P}^1_k T : C → P k 1 n = q b n=qb n = q b q , b ≥ 2 q,b\ge2 q , b ≥ 2
( T ) ∞ = q E , E = Q 1 + ⋯ + Q b , (T)_\infty=qE,\qquad E=Q_1+\cdots+Q_b, ( T ) ∞ = q E , E = Q 1 + ⋯ + Q b , with distinct points Q i Q_i Q i v i v_i v i T = v i − q T=v_i^{-q} T = v i − q K = k ( C ) K=k(C) K = k ( C ) F = k ( T ) F=k(T) F = k ( T )
L ( D E ) = { h ∈ K : ( h ) + D E ≥ 0 } ∪ { 0 } . L(DE)=\{h\in K:(h)+DE\ge0\}\cup\{0\}. L ( D E ) = { h ∈ K : ( h ) + D E ≥ 0 } ∪ { 0 } . For a finite branch value t t t e P e_P e P P P P t t t m ≥ 1 m\ge1 m ≥ 1
d m = ∑ t finite branch value max P ∣ t ⌊ m ( e P − 1 ) e P ⌋ . (96) d_m=\sum_{t\ \text{finite branch value}}\max_{P\mid t}\left\lfloor\frac{m(e_P-1)}{e_P}\right\rfloor. \tag*{(96)} d m = t finite branch value ∑ P ∣ t max ⌊ e P m ( e P − 1 ) ⌋ . (96) Theorem 4.21. Fix 1 ≤ a < q 1\le a<q 1 ≤ a < q e = q / gcd ( a , q ) e=q/\gcd(a,q) e = q / g cd( a , q ) D ≥ 0 D\ge0 D ≥ 0 q q q D b > 2 g − 2 Db>2g-2 D b > 2 g − 2
⌊ D + q + m q ⌋ > d m ( 1 ≤ m ≤ n − 1 , e ∤ m ) . (97) \left\lfloor\frac{D+q+m}{q}\right\rfloor>d_m\qquad(1\le m\le n-1,\ e\nmid m). \tag*{(97)} ⌊ q D + q + m ⌋ > d m ( 1 ≤ m ≤ n − 1 , e ∤ m ) . (97) If the coefficient map
J D , a : L ( D E ) ⟶ k b ( D / q + 1 ) , \mathcal{J}_{D,a}:L(DE)\longrightarrow k^{b(D/q+1)}, J D , a : L ( D E ) ⟶ k b ( D / q + 1 ) , h ⟼ ( [ v i − D + a + q j ] h ) 1 ≤ i ≤ b 0 ≤ j ≤ D / q (98) h\longmapsto\left([v_i^{-D+a+qj}]h\right)_{\substack{1\le i\le b\\0\le j\le D/q}} \tag*{(98)} h ⟼ ( [ v i − D + a + q j ] h ) 1 ≤ i ≤ b 0 ≤ j ≤ D / q (98) is not surjective, then there are intermediate fields
F ⊆ M ⊆ L ⊆ K with L / M cyclic of degree e . F\subseteq M\subseteq L\subseteq K\quad\text{with }L/M\text{ cyclic of degree }e. F ⊆ M ⊆ L ⊆ K with L / M cyclic of degree e . In particular, the map is surjective if the monodromy action of T T T
If B B B D = q B ( n − 2 ) D=qB(n-2) D = q B ( n − 2 ) 1 ≤ a < q 1\le a<q 1 ≤ a < q
Proof. We first describe the obstruction to prescribing Laurent coefficients. Let V V V v i − D , … , v i q − 1 v_i^{-D},\ldots,v_i^{q-1} v i − D , … , v i q − 1 1 ≤ i ≤ b 1\le i\le b 1 ≤ i ≤ b J : L ( D E ) → V \mathcal{J}:L(DE)\to V J : L ( D E ) → V L ( − q E ) = 0 L(-qE)=0 L ( − q E ) = 0 D b > 2 g − 2 Db>2g-2 D b > 2 g − 2
dim coker J = b ( D + q ) − ( D b + 1 − g ) = g + q b − 1. \dim\operatorname{coker}\mathcal{J}=b(D+q)-(Db+1-g)=g+qb-1. dim coker J = b ( D + q ) − ( D b + 1 − g ) = g + q b − 1. The same theorem gives dim H 0 ( C , Ω C ( q E ) ) = g + q b − 1 \dim H^0(C,\Omega_C(qE))=g+qb-1 dim H 0 ( C , Ω C ( q E )) = g + q b − 1 V V V
( h i ) i ⟼ ∑ i = 1 b Res Q i ( h i ω ) . (h_i)_i\longmapsto\sum_{i=1}^{b}\operatorname{Res}_{Q_i}(h_i\omega). ( h i ) i ⟼ i = 1 ∑ b Res Q i ( h i ω ) . This functional vanishes on global functions by the residue theorem. It determines ω \omega ω V V V ω \omega ω − q -q − q D − 1 D-1 D − 1 Q i Q_i Q i ω ∈ H 0 ( C , Ω C ( − D E ) ) = 0 \omega\in H^0(C,\Omega_C(-DE))=0 ω ∈ H 0 ( C , Ω C ( − D E )) = 0 H 0 ( C , Ω C ( q E ) ) H^0(C,\Omega_C(qE)) H 0 ( C , Ω C ( q E )) coker J \operatorname{coker}\mathcal{J} coker J
Suppose now that J D , a J_{D,a} J D , a V V V ω ∈ H 0 ( C , Ω C ( q E ) ) \omega\in H^0(C,\Omega_C(qE)) ω ∈ H 0 ( C , Ω C ( q E )) f = ω / d T ∈ K f=\omega/dT\in K f = ω / d T ∈ K Q i Q_i Q i d T = − q v i − q − 1 d v i dT=-qv_i^{-q-1}dv_i d T = − q v i − q − 1 d v i f = O ( v i ) f=O(v_i) f = O ( v i ) ω \omega ω D + q D+q D + q f f f − a -a − a q q q q q q ζ \zeta ζ
f ( ζ v i ) = ζ − a f ( v i ) + O ( v i D + q + 1 ) . (99) f(\zeta v_i)=\zeta^{-a}f(v_i)+O(v_i^{D+q+1}). \tag*{(99)} f ( ζ v i ) = ζ − a f ( v i ) + O ( v i D + q + 1 ) . (99) This is a relation between the local branches over T T T v i ↦ ζ v i v_i\mapsto\zeta v_i v i ↦ ζ v i C C C
Let 1 ≤ m ≤ n − 1 1\le m\le n-1 1 ≤ m ≤ n − 1 e ∤ m e\nmid m e ∤ m f = O ( v i ) f=O(v_i) f = O ( v i ) (99) to the m m m q q q
∑ j = 0 q − 1 f ( ζ j v i ) m = O ( v i D + q + m ) . \sum_{j=0}^{q-1}f(\zeta^jv_i)^m=O(v_i^{D+q+m}). j = 0 ∑ q − 1 f ( ζ j v i ) m = O ( v i D + q + m ) . The sum is a series in v i q = T − 1 v_i^q=T^{-1} v i q = T − 1 Tr K / F ( f m ) \operatorname{Tr}_{K/F}(f^m) Tr K / F ( f m ) ⌈ ( D + q + m ) / q ⌉ \lceil(D+q+m)/q\rceil ⌈( D + q + m ) / q ⌉
At a finite point P P P e P e_P e P ω \omega ω d T dT d T e P − 1 e_P-1 e P − 1 T − T ( P ) = u e P T-T(P)=u^{e_P} T − T ( P ) = u e P u u u e P e_P e P T ( P ) T(P) T ( P ) ⌊ m ( e P − 1 ) / e P ⌋ \lfloor m(e_P-1)/e_P\rfloor ⌊ m ( e P − 1 ) / e P ⌋ Tr K / F ( f m ) \operatorname{Tr}_{K/F}(f^m) Tr K / F ( f m ) d m d_m d m
Tr K / F ( f m ) = 0 ( 1 ≤ m ≤ n − 1 , e ∤ m ) . \operatorname{Tr}_{K/F}(f^m)=0\qquad(1\le m\le n-1,\ e\nmid m). Tr K / F ( f m ) = 0 ( 1 ≤ m ≤ n − 1 , e ∤ m ) . Since e ∣ n e\mid n e ∣ n
Norm K / F ( W − f ) ∈ F [ W e ] . \operatorname{Norm}_{K/F}(W-f)\in F[W^e]. Norm K / F ( W − f ) ∈ F [ W e ] . Let μ f \mu_f μ f f f f F F F μ f \mu_f μ f e e e f ≠ 0 f\ne0 f = 0 e e e μ f ∈ F [ W e ] \mu_f\in F[W^e] μ f ∈ F [ W e ]
Put L = F ( f ) L=F(f) L = F ( f ) M = F ( f e ) M=F(f^e) M = F ( f e ) e e e ξ \xi ξ f ↦ ξ f f\mapsto\xi f f ↦ ξ f F F F L L L M M M e e e [ L : M ] ≤ e [L:M]\le e [ L : M ] ≤ e f f f W e − f e W^e-f^e W e − f e M M M L / M L/M L / M e e e
If the monodromy is primitive, K / F K/F K / F Tr K / F ( f ) = 0 \operatorname{Tr}_{K/F}(f)=0 Tr K / F ( f ) = 0 f ∈ F f\in F f ∈ F L = K L=K L = K 2 ≤ e ≤ q < n 2\le e\le q<n 2 ≤ e ≤ q < n F ⊊ M ⊊ K F\subsetneq M\subsetneq K F ⊊ M ⊊ K
For the final assertion, note that d m ≤ B ( m − 1 ) ≤ B ( n − 2 ) d_m \le B(m-1) \le B(n-2) d m ≤ B ( m − 1 ) ≤ B ( n − 2 ) D = q B ( n − 2 ) D=qB(n-2) D = q B ( n − 2 ) (97) strictly larger than B ( n − 2 ) B(n-2) B ( n − 2 ) B ( n − 1 ) B(n-1) B ( n − 1 ) n − b n-b n − b
2 g − 2 ≤ B ( n − 1 ) − n − b < n B ( n − 2 ) = D b , 2g-2 \le B(n-1)-n-b < nB(n-2)=Db, 2 g − 2 ≤ B ( n − 1 ) − n − b < n B ( n − 2 ) = D b , as required. □
Failure does occur for covers with a cyclic intermediate extension. On C = P z 1 C=\mathbb{P}^{1}_{z} C = P z 1 U = z + z − 1 U=z+z^{-1} U = z + z − 1 T = U q T=U^{q} T = U q n = 2 q n=2q n = 2 q E = ( 0 ) + ( ∞ ) E=(0)+(\infty) E = ( 0 ) + ( ∞ ) v = U − 1 v=U^{-1} v = U − 1 E E E 1 ≤ a < q 1\le a<q 1 ≤ a < q
ω = d ( U a ) = − a v − a − 1 d v ∈ H 0 ( C , Ω C ( q E ) ) \omega=d(U^{a})=-av^{-a-1}dv\in H^{0}(C,\Omega_{C}(qE)) ω = d ( U a ) = − a v − a − 1 d v ∈ H 0 ( C , Ω C ( q E )) and the residue theorem give
[ v 0 a ] h + [ v ∞ a ] h = 0 ( h ∈ L ( D E ) ) . [v_{0}^{a}]h+[v_{\infty}^{a}]h=0 \qquad(h\in L(DE)). [ v 0 a ] h + [ v ∞ a ] h = 0 ( h ∈ L ( D E )) . Thus J D , a \mathcal{J}_{D,a} J D , a D D D q q q g 0 = gcd ( a , q ) g_{0}=\gcd(a,q) g 0 = g cd( a , q ) k ( U g 0 ) / k ( U q ) k(U^{g_{0}})/k(U^{q}) k ( U g 0 ) / k ( U q ) q / g 0 q/g_{0} q / g 0
For q = 2 q=2 q = 2 a = 1 a=1 a = 1 − D + 1 , − D + 3 , … , − 1 , 1 -D+1,-D+3,\ldots,-1,1 − D + 1 , − D + 3 , … , − 1 , 1 (97) becomes
D b > 2 g − 2 , ( D + m + 3 ) / 2 > d m ( m = 1 , 3 , … , n − 1 ) . (100) Db>2g-2,\qquad(D+m+3)/2>d_m\qquad(m=1,3,\ldots,n-1). \tag*{(100)} D b > 2 g − 2 , ( D + m + 3 ) /2 > d m ( m = 1 , 3 , … , n − 1 ) . (100) In this case the function f f f
f ( v i ) = O ( v i ) , f ( v i ) + f ( − v i ) = O ( v i D + 4 ) . (101) f(v_i)=O(v_i),\qquad f(v_i)+f(-v_i)=O(v_i^{D+4}). \tag*{(101)} f ( v i ) = O ( v i ) , f ( v i ) + f ( − v i ) = O ( v i D + 4 ) . (101) We use this case below. The theorem asserts a necessary field obstruction to failure of interpolation; the presence of a cyclic intermediate extension alone does not imply failure.
Separating branches at infinity
Write n = 2 b ≥ 4 n=2b\ge4 n = 2 b ≥ 4 C n ( Z , S ) = C n , S ( Z ) = H n , S ( Z − S ) C_n(Z,S)=C_{n,S}(Z)=H_{n,S}(Z-S) C n ( Z , S ) = C n , S ( Z ) = H n , S ( Z − S ) h n h_n h n (23) . At infinity set S = v − 2 S=v^{-2} S = v − 2 Z ( v , ζ ) = ζ / v + O ( 1 ) Z(v,\zeta)=\zeta/v+O(1) Z ( v , ζ ) = ζ / v + O ( 1 ) ζ \zeta ζ h n h_n h n Z ( − v , ζ ) = Z ( v , − ζ ) Z(-v,\zeta)=Z(v,-\zeta) Z ( − v , ζ ) = Z ( v , − ζ ) A − ⊂ Q [ T ] / ( h n ) \mathcal{A}^{-}\subset\mathbb{Q}[T]/(h_n) A − ⊂ Q [ T ] / ( h n ) b b b
We seek polynomials whose values on opposite branches have small differences at infinity. The following theorem supplies enough such polynomials to distinguish every selection of one branch from each pair. Its proof uses traces of odd powers and the generic symmetric group.
Theorem 4.22. Let D = 2 n 2 D=2n^{2} D = 2 n 2 P ( Z , S ) P(Z,S) P ( Z , S ) Z Z Z n n n D D D wt ( Z ) = 1 \operatorname{wt}(Z)=1 wt ( Z ) = 1 wt ( S ) = 2 \operatorname{wt}(S)=2 wt ( S ) = 2
v − D + 1 , v − D + 3 , … , v − 1 , v 1 v^{-D+1},v^{-D+3},\ldots,v^{-1},v^{1} v − D + 1 , v − D + 3 , … , v − 1 , v 1 in P ( Z ( v , ζ ) , v − 2 ) P(Z(v,\zeta),v^{-2}) P ( Z ( v , ζ ) , v − 2 ) ( A − ) D / 2 + 1 (\mathcal{A}^{-})^{D/2+1} ( A − ) D /2 + 1
In particular there exist b b b P 0 , … , P b − 1 P_{0},\ldots,P_{b-1} P 0 , … , P b − 1 v 1 v^{1} v 1 ζ , ζ 3 , … , ζ 2 b − 1 \zeta,\zeta^{3},\ldots,\zeta^{2b-1} ζ , ζ 3 , … , ζ 2 b − 1 Z Z Z
Proof. Let C C C C n ( Z , S ) = 0 C_n(Z,S)=0 C n ( Z , S ) = 0 E E E b = n / 2 b=n/2 b = n /2 ( S ) ∞ = 2 E (S)_{\infty}=2E ( S ) ∞ = 2 E S = 0 , − 1 , … , − ( n − 2 ) S=0,-1,\ldots,-(n-2) S = 0 , − 1 , … , − ( n − 2 ) X = Z − S = 0 X=Z-S=0 X = Z − S = 0 ∂ H n , S ( 0 ) / ∂ S = ( ( S ) n ) ′ ≠ 0 \partial H_{n,S}(0)/\partial S=((S)_n)'\ne0 ∂ H n , S ( 0 ) / ∂ S = (( S ) n ) ′ = 0 S n S_n S n S = 0 S=0 S = 0 ( n − 1 ) (n-1) ( n − 1 ) S = − 1 S=-1 S = − 1 S = − ( n − 2 ) S=-(n-2) S = − ( n − 2 ) S n S_n S n N = 2 , … , n N = 2,\ldots,n N = 2 , … , n g ( C ) = ( b − 1 ) 2 g(C) = (b-1)^2 g ( C ) = ( b − 1 ) 2 Theorem 4.21 is
d m = ∑ N = 2 n ⌊ m ( N − 1 ) N ⌋ ≤ ( n − 1 ) ( m − 1 ) . d_m = \sum_{N=2}^{n} \left\lfloor\frac{m(N-1)}{N} \right\rfloor\leq(n-1)(m-1). d m = N = 2 ∑ n ⌊ N m ( N − 1 ) ⌋ ≤ ( n − 1 ) ( m − 1 ) . The choice D = 2 n 2 D = 2n^2 D = 2 n 2 Theorem 4.21 with q = 2 q = 2 q = 2 a = 1 a = 1 a = 1 L ( D E ) L(DE) L ( D E )
The affine curve is smooth, so every function regular there belongs to its coordinate ring. Because C n C_n C n Z Z Z Z Z Z n n n L ( D E ) L(DE) L ( D E ) D D D Z Z Z S S S w > D w > D w > D p w ( T ) p_w(T) p w ( T ) Z = T / v Z = T/v Z = T / v S = v − 2 S = v^{-2} S = v − 2 deg p w < n \deg p_w < n deg p w < n p w p_w p w w w w w w w E E E p w p_w p w n n n h n h_n h n n n n C \mathbb{C} C Q [ T ] / ( h n ) \mathbb{Q}[T]/(h_n) Q [ T ] / ( h n ) C \mathbb{C} C Q \mathbb{Q} Q v v v P j P_j P j Z Z Z
Bounding the imaginary quadratic obstruction
We need quantitative bounds for this construction. Bounds for its denominators and coefficients gives a uniform upper bound for a parameter admitting an imaginary quadratic block field.
We begin with a denominator estimate for the root expansions. The Hermite differential equation makes the estimate independent of a discriminant denominator.
Lemma 4.23. Let F n F_n F n h n h_n h n A = Z [ T ] / ( h n ) A = \mathbb{Z}[T]/(h_n) A = Z [ T ] / ( h n )
T ( v ) = T + ∑ k ≥ 1 a k v k ∈ ( A ⊗ Q ) [ [ v ] ] , F n ( T ( v ) , v ) = 0 , T(v) = T + \sum_{k \geq1} a_k v^k \in(A \otimes\mathbb{Q})[[v]], \qquad F_n(T(v),v) = 0, T ( v ) = T + k ≥ 1 ∑ a k v k ∈ ( A ⊗ Q ) [[ v ]] , F n ( T ( v ) , v ) = 0 , satisfies ( 3 k ) ! a k ∈ A (3k)!a_k \in A ( 3 k )! a k ∈ A k ≥ 1 k \geq1 k ≥ 1 ( 3 k ) ! [ v k ] T ( v ) j ∈ A (3k)![v^k]T(v)^j \in A ( 3 k )! [ v k ] T ( v ) j ∈ A j ≥ 0 j \geq0 j ≥ 0
Proof. All quotients in this proof are first interpreted in Q [ T ] / ( h n ) \mathbb{Q}[T]/(h_n) Q [ T ] / ( h n ) h n ′ h_n' h n ′ h n r / h n ′ h_n^r/h_n' h n r / h n ′ A r A_r A r
A 0 = 0 , A 1 = 1 , A r + 2 = − T A r + 1 + ( n − r ) A r , A_0 = 0, \qquad A_1 = 1, \qquad A_{r+2} = -T A_{r+1} + (n-r)A_r, A 0 = 0 , A 1 = 1 , A r + 2 = − T A r + 1 + ( n − r ) A r , using the Hermite differential equation. Thus these quotients do not introduce the denominators of a general inverse of h n ′ h_n' h n ′ F n ( T , v ) = ∑ l ≥ 0 F l ( T ) v l F_n(T,v) = \sum_{l \geq0} F_l(T)v^l F n ( T , v ) = ∑ l ≥ 0 F l ( T ) v l
F l ( j ) ( T ) = ∑ r 3 , r 4 , … ≥ 0 ∑ m ≥ 3 ( m − 2 ) r m = l h n ( M + j ) ( T ) ∏ m ≥ 3 m r m r m ! , M = ∑ m ≥ 3 m r m ≤ 3 l . F_l^{(j)}(T) =
\sum_{\substack{r_3,r_4,\ldots\geq0\\ \sum_{m\geq3}(m-2)r_m=l}}
\frac{h_n^{(M+j)}(T)}{\prod_{m\geq3}m^{r_m}r_m!},
\qquad
M = \sum_{m\geq3} mr_m \leq3l. F l ( j ) ( T ) = r 3 , r 4 , … ≥ 0 ∑ m ≥ 3 ( m − 2 ) r m = l ∑ ∏ m ≥ 3 m r m r m ! h n ( M + j ) ( T ) , M = m ≥ 3 ∑ m r m ≤ 3 l . The denominator in each summand divides M ! M! M ! ( 3 l ) ! (3l)! ( 3 l )!
( 3 l ) ! F l ( j ) / h n ′ ∈ A . (3l)!F_l^{(j)}/h_n' \in A. ( 3 l )! F l ( j ) / h n ′ ∈ A . This includes l = 0 l = 0 l = 0 k k k A ( v ) = ∑ i ≥ 1 a i v i A(v) = \sum_{i\geq1} a_i v^i A ( v ) = ∑ i ≥ 1 a i v i F n ( T + A ( v ) , v ) = 0 F_n(T + A(v),v)=0 F n ( T + A ( v ) , v ) = 0 Q [ T ] / ( h n ) [ [ v ] ] \mathbb{Q}[T]/(h_n)[[v]] Q [ T ] / ( h n ) [[ v ]] h n ′ h_n' h n ′ a k a_k a k a k a_k a k v k v^k v k
F l ( j ) ( T ) h n ′ ( T ) ∏ i ≥ 1 a i r i r i ! , j = ∑ i r i , l + ∑ i i r i = k , \frac{F_l^{(j)}(T)}{h_n'(T)}\prod_{i\geq1}\frac{a_i^{r_i}}{r_i!},\qquad j=\sum_i r_i,\qquad l+\sum_i ir_i=k, h n ′ ( T ) F l ( j ) ( T ) i ≥ 1 ∏ r i ! a i r i , j = i ∑ r i , l + i ∑ i r i = k , and uses only coefficients already covered by the induction. Its denominator divides
( 3 l ) ! ∏ i ≥ 1 ( ( 3 i ) ! ) r i ∏ i ≥ 1 r i ! . (3l)!\prod_{i\geq1}((3i)!)^{r_i}\prod_{i\geq1}r_i!. ( 3 l )! i ≥ 1 ∏ (( 3 i )! ) r i i ≥ 1 ∏ r i ! . This integer divides ( 3 k ) ! (3k)! ( 3 k )! 3 k 3k 3 k 3 l 3l 3 l r i r_i r i 3 i 3i 3 i i i i ( 3 k ) ! a k ∈ A (3k)!a_k\in\mathcal{A} ( 3 k )! a k ∈ A ( T + A ( v ) ) j (T+A(v))^j ( T + A ( v ) ) j ( 3 k ) ! [ v k ] T ( v ) j ∈ A (3k)![v^k]T(v)^j\in\mathcal{A} ( 3 k )! [ v k ] T ( v ) j ∈ A j ≥ 0 j\geq0 j ≥ 0
The following analytic estimates give a common disk for all root branches.
Lemma 4.24. Let n ≥ 4 n\geq4 n ≥ 4 F ( t , v ) = F n ( t , v ) F(t,v)=F_n(t,v) F ( t , v ) = F n ( t , v ) h = h n h=h_n h = h n ζ i \zeta_i ζ i h h h ∣ ζ i ∣ ≤ 2 n |\zeta_i|\leq2\sqrt{n} ∣ ζ i ∣ ≤ 2 n 1 / ( 6 n ) 1/(6\sqrt{n}) 1/ ( 6 n ) ∣ v ∣ ≤ r = ( 1000 n 2 ) − 1 |v|\leq r=(1000n^2)^{-1} ∣ v ∣ ≤ r = ( 1000 n 2 ) − 1 t i ( v ) t_i(v) t i ( v ) F F F ∣ t i ( v ) − ζ i ∣ ≤ 2 n ∣ v ∣ |t_i(v)-\zeta_i|\leq2n|v| ∣ t i ( v ) − ζ i ∣ ≤ 2 n ∣ v ∣ ∣ t i ( v ) ∣ < 3 n |t_i(v)|<3\sqrt{n} ∣ t i ( v ) ∣ < 3 n
Proof. The roots ζ 1 , … , ζ n \zeta_1,\ldots,\zeta_n ζ 1 , … , ζ n h h h ∣ ζ i ∣ ≤ 2 n |\zeta_i|\leq2\sqrt{n} ∣ ζ i ∣ ≤ 2 n j \sqrt{j} j − j -\sqrt{j} − j 1 ≤ j < n 1\leq j<n 1 ≤ j < n h h h i i i 2 n 2\sqrt{n} 2 n
The equation h ′ ′ + t h ′ − n h = 0 h''+th'-nh=0 h ′′ + t h ′ − nh = 0
∣ ζ i − ζ j ∣ > 1 6 n ( i ≠ j ) . (102) |\zeta_i-\zeta_j|>\frac{1}{6\sqrt{n}}\qquad(i\ne j). \tag*{(102)} ∣ ζ i − ζ j ∣ > 6 n 1 ( i = j ) . (102) Indeed, at a root put A k = h ( k ) ( ζ ) / h ′ ( ζ ) A_k=h^{(k)}(\zeta)/h'(\zeta) A k = h ( k ) ( ζ ) / h ′ ( ζ ) A k + 2 = − ζ A k + 1 + ( n − k ) A k A_{k+2}=-\zeta A_{k+1}+(n-k)A_k A k + 2 = − ζ A k + 1 + ( n − k ) A k A 0 = 0 A_0=0 A 0 = 0 A 1 = 1 A_1=1 A 1 = 1 ∣ A k ∣ ≤ ( 3 n ) k − 1 |A_k|\leq(3\sqrt{n})^{k-1} ∣ A k ∣ ≤ ( 3 n ) k − 1 0 < ∣ z ∣ ≤ 1 / ( 6 n ) 0<|z|\leq1/(6\sqrt{n}) 0 < ∣ z ∣ ≤ 1/ ( 6 n )
∣ h ( ζ + z ) h ′ ( ζ ) z − 1 ∣ ≤ ∑ k = 2 n 2 1 − k k ! < 2 ( e 1 / 2 − 3 / 2 ) < 1 , \left|\frac{h(\zeta+z)}{h'(\zeta)z}-1\right|\leq\sum_{k=2}^{n}\frac{2^{1-k}}{k!}<2(e^{1/2}-3/2)<1, h ′ ( ζ ) z h ( ζ + z ) − 1 ≤ k = 2 ∑ n k ! 2 1 − k < 2 ( e 1/2 − 3/2 ) < 1 , so the Taylor quotient cannot vanish.
Set r = ( 1000 n 2 ) − 1 r=(1000n^2)^{-1} r = ( 1000 n 2 ) − 1 F ( t , v ) F(t,v) F ( t , v ) v ( 2 j − n − 1 ) v(2j-n-1) v ( 2 j − n − 1 )
j 1 + ( n − j ) v 2 , − j 1 + ( n − j ) v 2 . \sqrt{j}\sqrt{1+(n-j)v^2},\qquad-\sqrt{j}\sqrt{1+(n-j)v^2}. j 1 + ( n − j ) v 2 , − j 1 + ( n − j ) v 2 . Here the square roots are the branches equal to 1 1 1 v = 0 v=0 v = 0 ∣ v ∣ ≤ r |v|\leq r ∣ v ∣ ≤ r v = 0 v=0 v = 0 2 n ∣ v ∣ 2n|v| 2 n ∣ v ∣ 1 / ( 18 n ) 1/(18\sqrt{n}) 1/ ( 18 n ) ζ i \zeta_i ζ i t i ( v ) t_i(v) t i ( v ) ∣ v ∣ ≤ r |v|\leq r ∣ v ∣ ≤ r
t i ( 0 ) = ζ i , ∣ t i ( v ) ∣ < 3 n . t_i(0)=\zeta_i,\qquad|t_i(v)|<3\sqrt{n}. t i ( 0 ) = ζ i , ∣ t i ( v ) ∣ < 3 n . The assertions remain valid on an open neighborhood of this closed disk, since all the preceding norm inequalities are strict.
The bound ∣ t i ( v ) − ζ i ∣ ≤ 2 n ∣ v ∣ |t_i(v)-\zeta_i|\leq2n|v| ∣ t i ( v ) − ζ i ∣ ≤ 2 n ∣ v ∣ v = 0 v=0 v = 0
Eliminating the odd monomials. The large jet system has a triangular part which can be removed before taking determinants. This reduces the number of rows from order n 3 n^3 n 3 n 2 n^2 n 2
Put n = 2 b n=2b n = 2 b D = 2 m ≥ n D=2m\ge n D = 2 m ≥ n u = v 2 u=v^2 u = v 2 A = Q [ T ] / ( h n ) \mathcal{A}=\mathbb{Q}[T]/(h_n) A = Q [ T ] / ( h n ) Q \mathbb{Q} Q T ( v ) : A → A [ [ v ] ] \mathcal{T}(v):\mathcal{A}\to\mathcal{A}[[v]] T ( v ) : A → A [[ v ]]
T ( v ) ( T j ) = Y ( v ) j ( m o d h n ) , 0 ≤ j < n , \mathcal{T}(v)(T^j)=Y(v)^j\pmod{h_n},\qquad0\le j<n, T ( v ) ( T j ) = Y ( v ) j ( mod h n ) , 0 ≤ j < n , where Y ( v ) = v Z ( v , T ) Y(v)=vZ(v,T) Y ( v ) = v Z ( v , T ) A = A + ⊕ A − \mathcal{A}=\mathcal{A}^{+}\oplus\mathcal{A}^{-} A = A + ⊕ A − 1 , T 2 , … , T 2 b − 2 1,T^2,\ldots,T^{2b-2} 1 , T 2 , … , T 2 b − 2 T , T 3 , … , T 2 b − 1 T,T^3,\ldots,T^{2b-1} T , T 3 , … , T 2 b − 1
T − − ( v ) = O ( u ) , T − + ( v ) = v B ( u ) , O ( 0 ) = I . \mathcal{T}_{--}(v)=\mathcal{O}(u),\qquad\mathcal{T}_{-+}(v)=v\mathcal{B}(u),\qquad\mathcal{O}(0)=I. T −− ( v ) = O ( u ) , T −+ ( v ) = v B ( u ) , O ( 0 ) = I . Define
R ( u ) = u O ( u ) − 1 B ( u ) = v O ( v 2 ) − 1 T − + ( v ) . (103) R(u)=u\mathcal{O}(u)^{-1}\mathcal{B}(u)=v\mathcal{O}(v^2)^{-1}\mathcal{T}_{-+}(v). \tag*{(103)} R ( u ) = u O ( u ) − 1 B ( u ) = v O ( v 2 ) − 1 T −+ ( v ) . (103) Its column corresponding to the constant polynomial is zero.
Lemma 4.25. Consider a polynomial P P P deg Z P < n \deg_Z P<n deg Z P < n 2 m 2m 2 m Z Z Z
A i ( u ) = p 2 i ( u − 1 ) u − i ( 1 ≤ i < b ) , A 0 = 0 , A_i(u)=p_{2i}(u^{-1})u^{-i}\qquad(1\le i<b),\qquad A_0=0, A i ( u ) = p 2 i ( u − 1 ) u − i ( 1 ≤ i < b ) , A 0 = 0 , and
B i i n ( u ) = p 2 i + 1 ( u − 1 ) u − i ( 0 ≤ i < b ) . B_i^{\mathrm{in}}(u)=p_{2i+1}(u^{-1})u^{-i}\qquad(0\le i<b). B i in ( u ) = p 2 i + 1 ( u − 1 ) u − i ( 0 ≤ i < b ) . Thus A i A_i A i u − m , … , u − i u^{-m},\ldots,u^{-i} u − m , … , u − i B i i n B_i^{\mathrm{in}} B i in u − ( m − 1 ) , … , u − i u^{-(m-1)},\ldots,u^{-i} u − ( m − 1 ) , … , u − i y ∈ Q b y\in\mathbb{Q}^b y ∈ Q b B i n B^{\mathrm{in}} B in P ( Z ( v , T ) , v − 2 ) P(Z(v,T),v^{-2}) P ( Z ( v , T ) , v − 2 ) v 1 v^1 v 1 y y y
[ u k ] ( R ( u ) A ( u ) ) i = 0 ( 1 − i ≤ k ≤ 0 ) , [ u 1 ] ( R ( u ) A ( u ) ) i = y i , 0 ≤ i < b . (104) [u^k](R(u)A(u))_i=0\quad(1-i\le k\le0),\qquad[u^1](R(u)A(u))_i=y_i,\quad0\le i<b. \tag*{(104)} [ u k ] ( R ( u ) A ( u ) ) i = 0 ( 1 − i ≤ k ≤ 0 ) , [ u 1 ] ( R ( u ) A ( u ) ) i = y i , 0 ≤ i < b . (104) When it exists, the odd part is unique and is obtained by taking the negative of the terms of ( R A ) i (RA)_i ( R A ) i − ( m − 1 ) , … , − i -(m-1),\ldots,-i − ( m − 1 ) , … , − i (104) has exactly b ( b + 1 ) / 2 b(b+1)/2 b ( b + 1 ) /2 D = 2 n 2 D=2n^2 D = 2 n 2
Proof. The odd part, in both the Laurent parameter and the Hermite variable, of the evaluated polynomial is
v ( B ( u ) A ( u ) + u − 1 O ( u ) B i n ( u ) ) = v − 1 O ( u ) ( B i n ( u ) + R ( u ) A ( u ) ) . v\bigl(\mathcal{B}(u)A(u)+u^{-1}\mathcal{O}(u)B^{\mathrm{in}}(u)\bigr)=v^{-1}\mathcal{O}(u)\bigl(B^{\mathrm{in}}(u)+R(u)A(u)\bigr). v ( B ( u ) A ( u ) + u − 1 O ( u ) B in ( u ) ) = v − 1 O ( u ) ( B in ( u ) + R ( u ) A ( u ) ) . It equals v y + O ( v 3 ) vy+O(v^3) v y + O ( v 3 )
B i n ( u ) + R ( u ) A ( u ) = u y + O ( u 2 ) , B^{\mathrm{in}}(u)+R(u)A(u)=uy+O(u^2), B in ( u ) + R ( u ) A ( u ) = u y + O ( u 2 ) , since O ( 0 ) = I \mathcal{O}(0)=I O ( 0 ) = I R R R u u u R A RA R A u − ( m − 1 ) u^{-(m-1)} u − ( m − 1 ) B i i n B^{\mathrm{in}}_i B i in u − i u^{-i} u − i (104) . There are i + 1 i+1 i + 1 i i i b ( b + 1 ) / 2 b(b+1)/2 b ( b + 1 ) /2
For D = 2 n 2 D=2n^2 D = 2 n 2 Theorem 4.22 is surjective. Multiplication by O ( u ) − 1 \mathcal{O}(u)^{-1} O ( u ) − 1 (104) , which is therefore surjective. □
The inverse matrix in (103) has a uniform analytic disk. Working in a Hermite quadrature norm avoids a loss caused by the condition number of a monomial Vandermonde matrix.
Lemma 4.26. Put r = ( 1000 n 2 ) − 1 r=(1000n^2)^{-1} r = ( 1000 n 2 ) − 1 R ( u ) R(u) R ( u ) ∣ u ∣ ≤ r 2 |u|\le r^2 ∣ u ∣ ≤ r 2
∥ O ( v 2 ) − 1 ∥ ≤ 2 ( ∣ v ∣ ≤ r ) . \lVert\mathcal{O}(v^2)^{-1}\rVert\le2\qquad(|v|\le r). ∥ O ( v 2 ) − 1 ∥ ≤ 2 ( ∣ v ∣ ≤ r ) . In the monomial bases, its entries also satisfy
∣ R i j ( u ) ∣ ≤ exp ( 6 n log n ) ( ∣ u ∣ ≤ r 2 ) . |R_{ij}(u)| \le\exp(6n\log n) \qquad(|u| \le r^2). ∣ R ij ( u ) ∣ ≤ exp ( 6 n log n ) ( ∣ u ∣ ≤ r 2 ) . Moreover every coefficient of u k u^k u k R ( u ) R(u) R ( u ) ( 6 k − 3 ) ! (6k-3)! ( 6 k − 3 )! k ≥ 1 k \ge1 k ≥ 1
Proof. Write the Hermite roots as ζ i = i η i \zeta_i=i\eta_i ζ i = i η i η i ∈ R \eta_i\in\mathbb{R} η i ∈ R n n n
∥ q ∥ 2 = 1 2 π ∫ R ∣ q ( i x ) ∣ 2 e − x 2 / 2 d x . \lVert q\rVert^2=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}|q(ix)|^2e^{-x^2/2}\,dx. ∥ q ∥ 2 = 2 π 1 ∫ R ∣ q ( i x ) ∣ 2 e − x 2 /2 d x . Gaussian quadrature at the n n n q ( ζ i ) q(\zeta_i) q ( ζ i )
The polynomials h k ( T ) / k ! h_k(T)/\sqrt{k!} h k ( T ) / k ! n n n n \sqrt{n} n k k k n k / 2 n^{k/2} n k /2
Lemma 4.24 gives root branches t i ( v ) t_i(v) t i ( v ) ∣ v ∣ ≤ r |v|\le r ∣ v ∣ ≤ r
δ i ( v ) = t i ( v ) − ζ i , max i ∣ δ i ( v ) ∣ ≤ 2 n ∣ v ∣ . \delta_i(v)=t_i(v)-\zeta_i,\qquad\max_i|\delta_i(v)|\le2n|v|. δ i ( v ) = t i ( v ) − ζ i , i max ∣ δ i ( v ) ∣ ≤ 2 n ∣ v ∣. Let D \mathcal{D} D
T ( v ) = I + ∑ k = 1 n − 1 diag ( δ i ( v ) k ) D k k ! . \mathcal{T}(v)=I+\sum_{k=1}^{n-1}\frac{\operatorname{diag}(\delta_i(v)^k)\mathcal{D}^k}{k!}. T ( v ) = I + k = 1 ∑ n − 1 k ! diag ( δ i ( v ) k ) D k . It follows that
∥ T ( v ) − I ∥ ≤ exp ( 2 n 3 / 2 ∣ v ∣ ) − 1 < 1 2 . \lVert\mathcal{T}(v)-I\rVert\le\exp(2n^{3/2}|v|)-1<\frac{1}{2}. ∥ T ( v ) − I ∥ ≤ exp ( 2 n 3/2 ∣ v ∣ ) − 1 < 2 1 . The projection onto odd polynomials is orthogonal because the quadrature weights are symmetric. Hence the same bound holds for O ( v 2 ) − I \mathcal{O}(v^2)-I O ( v 2 ) − I (103) proves the stated holomorphy of R R R
∥ R ( u ) ∥ ≤ 6 n 3 / 2 ∣ u ∣ ( ∣ u ∣ ≤ r 2 ) . \lVert R(u)\rVert\le6n^{3/2}|u|\qquad(|u|\le r^2). ∥ R ( u )∥ ≤ 6 n 3/2 ∣ u ∣ ( ∣ u ∣ ≤ r 2 ) . To pass to monomial coordinates, each input monomial has norm at most ( 2 n ) n / 2 (2n)^{n/2} ( 2 n ) n /2 h k / k ! h_k/\sqrt{k!} h k / k ! n ! \sqrt{n!} n ! n n ! n\sqrt{n!} n n ! exp ( 6 n log n ) \exp(6n\log n) exp ( 6 n log n ) n ≥ 4 n\ge4 n ≥ 4
By Lemma 4.23, every coefficient of v k v^k v k T ( v ) \mathcal{T}(v) T ( v ) ( 3 k ) ! (3k)! ( 3 k )! v = w 3 v=w^3 v = w 3 U ( w ) = ∑ j ≥ 0 U j w j / j ! U(w)=\sum_{j\ge0}U_jw^j/j! U ( w ) = ∑ j ≥ 0 U j w j / j ! U 0 = I U_0=I U 0 = I
V 0 = I , V j = − ∑ i = 1 j ( j i ) U i V j − i . V_0=I,\qquad V_j=-\sum_{i=1}^{j}\binom{j}{i}U_iV_{j-i}. V 0 = I , V j = − i = 1 ∑ j ( i j ) U i V j − i . They are therefore integral. Products have the same property by the binomial convolution formula. Applying these observations to the odd diagonal block gives the asserted integrality for its inverse and for its product with T − + \mathcal{T}_{-+} T −+ u k = v 2 k u^k=v^{2k} u k = v 2 k v 2 k − 1 v^{2k-1} v 2 k − 1 O ( v 2 ) − 1 T − + ( v ) \mathcal{O}(v^2)^{-1}\mathcal{T}_{-+}(v) O ( v 2 ) − 1 T −+ ( v ) ( 6 k − 3 ) ! (6k-3)! ( 6 k − 3 )!
The parameter bound. We construct integral polynomials with small differences on paired branches and apply them to the two blocks.
Proposition 4.27. Let n = 2 b ≥ 4 n=2b\ge4 n = 2 b ≥ 4 s ∈ Z > 0 s\in\mathbb{Z}_{>0} s ∈ Z > 0 H n , s H_{n,s} H n , s b b b
log s < 1000 n 4 log n . \log s<1000n^4\log n. log s < 1000 n 4 log n . Proof. Write C ( Z , S ) = C n , S ( Z ) C(Z,S)=C_{n,S}(Z) C ( Z , S ) = C n , S ( Z ) D = 2 n 2 D=2n^2 D = 2 n 2 F = F n F=F_n F = F n h = h n h=h_n h = h n (24) . The generating function
∑ m ≥ 0 C m ( Z , S ) w m m ! = exp ( Z w + S ∑ j ≥ 2 w j j ) \sum_{m\ge0} C_m(Z,S)\frac{w^m}{m!}=\exp\left(Zw+S\sum_{j\ge2}\frac{w^j}{j}\right) m ≥ 0 ∑ C m ( Z , S ) m ! w m = exp ( Z w + S j ≥ 2 ∑ j w j ) shows that C C C n n n 1 , 2 1,2 1 , 2 C ( 1 , 1 ) = n ! C(1,1)=n! C ( 1 , 1 ) = n ! F ∈ Z [ t , v ] F\in\mathbb{Z}[t,v] F ∈ Z [ t , v ]
We use the branches and radius r r r Lemma 4.24 .
We now apply the triangular elimination in Lemma 4.25 , with m = n 2 m=n^2 m = n 2 D = 2 m D=2m D = 2 m u = v 2 u=v^2 u = v 2 T ( v ) \mathcal{T}(v) T ( v ) t j ↦ t ( v ) j t^j\mapsto t(v)^j t j ↦ t ( v ) j Q [ t ] / ( h ) \mathbb{Q}[t]/(h) Q [ t ] / ( h )
O ( u ) = T − − ( v ) , R ( u ) = v O ( u ) − 1 T − + ( v ) . \mathcal{O}(u)=\mathcal{T}_{--}(v),\qquad\mathcal{R}(u)=v\mathcal{O}(u)^{-1}\mathcal{T}_{-+}(v). O ( u ) = T −− ( v ) , R ( u ) = v O ( u ) − 1 T −+ ( v ) . The latter is a power series in u u u Lemma 4.26 shows that its entries, in monomial coordinates, satisfy
∣ R i j ( u ) ∣ ≤ exp ( 6 n log n ) ( ∣ u ∣ = r 2 ) . \lvert R_{ij}(u)\rvert\le\exp(6n\log n)\qquad(\lvert u\rvert=r^2). ∣ R ij ( u )∣ ≤ exp ( 6 n log n ) (∣ u ∣ = r 2 ) . Consequently, all coefficients through degree m + 1 m+1 m + 1
exp ( 20 n 2 log n ) . (105) \exp(20n^2\log n). \tag*{(105)} exp ( 20 n 2 log n ) . (105) Indeed log r − 1 ≤ 7 log n \log r^{-1}\le7\log n log r − 1 ≤ 7 log n 6 n log n + 14 ( n 2 + 1 ) log n ≤ 20 n 2 log n 6n\log n+14(n^2+1)\log n\le20n^2\log n 6 n log n + 14 ( n 2 + 1 ) log n ≤ 20 n 2 log n Lemma 4.26 gives the denominator bound ( 6 k − 3 ) ! (6k-3)! ( 6 k − 3 )! u k u^k u k R \mathcal{R} R
q = ( 6 m + 3 ) ! q=(6m+3)! q = ( 6 m + 3 )! clears every coefficient used below and satisfies
log q ≤ 36 n 2 log n . (106) \log q\le36n^2\log n. \tag*{(106)} log q ≤ 36 n 2 log n . (106) The reduced map in Lemma 4.25 has
R = b ( b + 1 ) 2 ≤ n 2 R=\frac{b(b+1)}{2}\le n^2 R = 2 b ( b + 1 ) ≤ n 2 rows. It sends the coefficients of the even monomials Z 2 i S k Z^{2i}S^k Z 2 i S k D D D [ u l ] ( R E ) i [u^l](\mathcal{R}E)_i [ u l ] ( R E ) i 1 − i ≤ l ≤ 1 1-i\le l\le1 1 − i ≤ l ≤ 1 E i ( u ) = u − i p 2 i ( u − 1 ) E_i(u)=u^{-i}p_{2i}(u^{-1}) E i ( u ) = u − i p 2 i ( u − 1 ) E 0 = 0 E_0=0 E 0 = 0 A A A Theorem 4.22 and the triangular elimination. Let J J J (105) and (106) give
max ∣ ( q J ) i j ∣ ≤ exp ( 60 n 2 log n ) . (107) \max\lvert(qJ)_{ij}\rvert\le\exp(60n^2\log n). \tag*{(107)} max ∣( q J ) ij ∣ ≤ exp ( 60 n 2 log n ) . (107) Choose R R R q J qJ q J B B B 1 ≤ k ≤ b 1\le k\le b 1 ≤ k ≤ b r k r_k r k u 1 u^1 u 1 k k k E ( k ) E^{(k)} E ( k )
q adj ( B ) e r k q\operatorname{adj}(B)e_{r_k} q adj ( B ) e r k on these selected columns and zero on all other columns. Its entries are integers divisible by q q q J J J
J E ( k ) = q − 1 B ( q adj ( B ) e r k ) = det ( B ) e r k . JE^{(k)}=q^{-1}B\bigl(q\operatorname{adj}(B)e_{r_k}\bigr)=\det(B)e_{r_k}. J E ( k ) = q − 1 B ( q adj ( B ) e r k ) = det ( B ) e r k . Thus every residual coefficient vanishes except the prescribed one, whose value is the same nonzero integer c = det ( B ) c=\det(B) c = det ( B ) k k k (107) bound the even coefficients by exp ( 70 n 4 log n ) \exp(70n^{4}\log n) exp ( 70 n 4 log n )
The odd coefficients are obtained by canceling the allowed Laurent polynomial part of R E \mathcal{R}E R E Lemma 4.25 . This step introduces no denominators: each even coefficient is divisible by q q q q q q R \mathcal{R} R m + 1 m+1 m + 1 q = ( 6 m + 3 ) ! q=(6m+3)! q = ( 6 m + 3 )! u − j u^{-j} u − j i i i Z 2 i + 1 S j − i Z^{2i+1}S^{j-i} Z 2 i + 1 S j − i i ≤ j ≤ m − 1 i\leq j\leq m-1 i ≤ j ≤ m − 1
Each reconstructed coefficient is a sum of at most 2 n 3 2n^{3} 2 n 3 (105) . We have therefore constructed integer polynomials P 1 , … , P b P_1,\ldots,P_b P 1 , … , P b Z Z Z n n n D D D
[ v 1 ] P k ( t ( v ) / v , v − 2 ) = c t 2 k − 1 , c = det B ∈ Z ∖ { 0 } . (108) [v^1]P_k(t(v)/v,v^{-2})=ct^{2k-1},\qquad c=\det B\in\mathbb{Z}\setminus\{0\}. \tag*{(108)} [ v 1 ] P k ( t ( v ) / v , v − 2 ) = c t 2 k − 1 , c = det B ∈ Z ∖ { 0 } . (108) Their coefficients satisfy
H : = max k , j , l ∣ [ Z j S l ] P k ∣ ≤ exp ( 80 n 4 log n ) . (109) H:=\max_{k,j,l}\left|[Z^jS^l]P_k\right|\leq\exp(80n^{4}\log n). \tag*{(109)} H := k , j , l max [ Z j S l ] P k ≤ exp ( 80 n 4 log n ) . (109) Let ρ = r / 2 \rho=r/2 ρ = r /2
M = 2 n 3 H ( 3 n ) n ρ − D . M=2n^{3}H(3\sqrt{n})^{n}\rho^{-D}. M = 2 n 3 H ( 3 n ) n ρ − D . Then
M ≤ exp ( 90 n 4 log n ) . (110) M\leq\exp(90n^{4}\log n). \tag*{(110)} M ≤ exp ( 90 n 4 log n ) . (110) For a branch with initial root ζ \zeta ζ
G k , ζ ( v ) = 1 2 ( P k ( t ζ ( v ) / v , v − 2 ) − P k ( t ζ ( − v ) / ( − v ) , v − 2 ) ) . G_{k,\zeta}(v)=\frac{1}{2}\left(P_k(t_{\zeta}(v)/v,v^{-2})-P_k(t_{\zeta}(-v)/(-v),v^{-2})\right). G k , ζ ( v ) = 2 1 ( P k ( t ζ ( v ) / v , v − 2 ) − P k ( t ζ ( − v ) / ( − v ) , v − 2 ) ) . The parity of the equation shows that this is the odd Laurent part. Its negative powers have been canceled, so it is holomorphic at 0 0 0 ∣ v ∣ = ρ |v|=\rho ∣ v ∣ = ρ M M M (108) therefore give, for 0 < v ≤ ρ / 2 0<v\leq\rho/2 0 < v ≤ ρ /2
∣ G k , ζ ( v ) ∣ ≤ 2 M v / ρ , (111) |G_{k,\zeta}(v)|\leq2Mv/\rho, \tag*{(111)} ∣ G k , ζ ( v ) ∣ ≤ 2 M v / ρ , (111) ∣ G k , ζ ( v ) − c ζ 2 k − 1 v ∣ ≤ 2 M v 3 / ρ 3 . (112) |G_{k,\zeta}(v)-c\zeta^{2k-1}v|\leq2Mv^{3}/\rho^{3}. \tag*{(112)} ∣ G k , ζ ( v ) − c ζ 2 k − 1 v ∣ ≤ 2 M v 3 / ρ 3 . (112) Suppose now that the two-block field is imaginary quadratic, and assume, for a contradiction, that T = log s ≥ 1000 n 4 log n T=\log s\geq1000n^{4}\log n T = log s ≥ 1000 n 4 log n v = s − 1 / 2 < ρ / 2 v=s^{-1/2}<\rho/2 v = s − 1/2 < ρ /2 i λ 1 , … , i λ b i\lambda_1,\ldots,i\lambda_b i λ 1 , … , i λ b 0 < λ 1 < ⋯ < λ b 0<\lambda_1<\cdots<\lambda_b 0 < λ 1 < ⋯ < λ b i λ j i\lambda_j i λ j − i λ j -i\lambda_j − i λ j ϵ j ∈ { 1 , − 1 } \epsilon_j\in\{1,-1\} ϵ j ∈ { 1 , − 1 } P k P_k P k
Δ k = 2 ∑ j = 1 b ϵ j G k , i λ j ( v ) . \Delta_k=2\sum_{j=1}^{b}\epsilon_jG_{k,i\lambda_j}(v). Δ k = 2 j = 1 ∑ b ϵ j G k , i λ j ( v ) . The sum of P k P_k P k P k P_k P k Δ k = − Δ k \Delta_k=-\Delta_k Δ k = − Δ k Δ k ≠ 0 \Delta_k\ne0 Δ k = 0
N K / Q ( Δ k ) = − Δ k 2 = ∣ Δ k ∣ 2 ∈ Z > 0 , N_{K/\mathbb{Q}}(\Delta_k)=-\Delta_k^{2}=|\Delta_k|^{2}\in\mathbb{Z}_{>0}, N K / Q ( Δ k ) = − Δ k 2 = ∣ Δ k ∣ 2 ∈ Z > 0 , and therefore ∣ Δ k ∣ ≥ 1 |\Delta_k|\geq1 ∣ Δ k ∣ ≥ 1
Equations (110) and (111) imply ∣ Δ k ∣ ≤ 4 b M v / ρ < 1 |\Delta_k| \leq4bMv/\rho< 1 ∣ Δ k ∣ ≤ 4 b M v / ρ < 1 Δ k = 0 \Delta_k = 0 Δ k = 0 k k k (112) and ∣ c ∣ ≥ 1 |c| \geq1 ∣ c ∣ ≥ 1
∣ ∑ j = 1 b ϵ j ( i λ j ) 2 k − 1 ∣ ≤ 2 b M v 2 / ρ 3 ( 1 ≤ k ≤ b ) . (113) \left|\sum_{j=1}^{b}\epsilon_j(i\lambda_j)^{2k-1}\right| \leq2bMv^2/\rho^3 \qquad(1 \leq k \leq b). \tag*{(113)} j = 1 ∑ b ϵ j ( i λ j ) 2 k − 1 ≤ 2 b M v 2 / ρ 3 ( 1 ≤ k ≤ b ) . (113) These inequalities contradict an elementary Vandermonde bound. Indeed, (102) implies
λ j > 1 12 n , ∣ λ i 2 − λ j 2 ∣ > 1 36 n ( i ≠ j ) , λ j 2 ≤ 4 n . \lambda_j > \frac{1}{12\sqrt{n}}, \qquad|\lambda_i^2-\lambda_j^2| > \frac{1}{36n} \quad(i\ne j), \qquad\lambda_j^2 \leq4n. λ j > 12 n 1 , ∣ λ i 2 − λ j 2 ∣ > 36 n 1 ( i = j ) , λ j 2 ≤ 4 n . For the matrix V k j = λ j 2 k − 1 V_{kj}=\lambda_j^{2k-1} V k j = λ j 2 k − 1 λ j 2 \lambda_j^2 λ j 2
∥ V − 1 ∥ ∞ ≤ 12 b n ( 180 n 2 ) b − 1 . \|V^{-1}\|_\infty\leq12b\sqrt{n}(180n^2)^{b-1}. ∥ V − 1 ∥ ∞ ≤ 12 b n ( 180 n 2 ) b − 1 . Since ∥ ϵ ∥ ∞ = 1 \|\epsilon\|_\infty=1 ∥ ϵ ∥ ∞ = 1 (113) has absolute value at least
η = 1 12 b n ( 180 n 2 ) b − 1 . \eta= \frac{1}{12b\sqrt{n}(180n^2)^{b-1}}. η = 12 b n ( 180 n 2 ) b − 1 1 . But log η − 1 ≤ 5 n log n \log\eta^{-1} \leq5n\log n log η − 1 ≤ 5 n log n (110) , ρ − 1 = 2000 n 2 \rho^{-1}=2000n^2 ρ − 1 = 2000 n 2 T T T 2 b M v 2 / ρ 3 < η 2bMv^2/\rho^3 < \eta 2 b M v 2 / ρ 3 < η
Completion of the uniform Galois theorem
The preceding bounds reduce a possible failure of alternating containment to two blocks with n < log s < n 5 \sqrt{n}<\log s<n^5 n < log s < n 5
Theorem 4.28. There is an effectively computable absolute integer N G a l N_{\mathrm{Gal}} N Gal n ≥ N G a l n\geq N_{\mathrm{Gal}} n ≥ N Gal s s s H n , s H_{n,s} H n , s A n ≤ G n , s A_n\leq G_{n,s} A n ≤ G n , s
Proof. Choose an effective absolute threshold for the results used below, including Corollary 4.5 with κ = 6 \kappa=6 κ = 6 D = 5 D=5 D = 5
max { C 1 ( log n ) 9 , ( log n ) 6 } < n , 2 n log n < 1000 n 4 log n < n 5 , n 101 / 200 log n − log 2 > n , \max\{C_1(\log n)^9,(\log n)^6\}<\sqrt{n}, \qquad2n\log n<1000n^4\log n<n^5, \qquad n^{101/200}\log n-\log2>\sqrt{n}, max { C 1 ( log n ) 9 , ( log n ) 6 } < n , 2 n log n < 1000 n 4 log n < n 5 , n 101/200 log n − log 2 > n , where C 1 C_1 C 1 Proposition 4.16 .
Suppose H n , s H_{n,s} H n , s G n , s G_{n,s} G n , s A n A_n A n Proposition 4.8 gives s > n 3 / ( 10 log n ) > n s>n^3/(10\log n)>n s > n 3 / ( 10 log n ) > n Theorem 4.14 shows that the root action is imprimitive. Choose a partition into a a a b b b a b = n ab=n ab = n a , b ≥ 2 a,b\geq2 a , b ≥ 2
log ( s + n ) ≥ n 101 / 200 log n , \log(s+n)\geq n^{101/200}\log n, log ( s + n ) ≥ n 101/200 log n , so L = log s > n L=\log s>\sqrt{n} L = log s > n Lemma 4.15 excludes b = 2 b=2 b = 2 a ≥ 3 a\geq3 a ≥ 3 Proposition 4.16 gives L ≤ C 1 ( log n ) 9 < n L\leq C_1(\log n)^9<\sqrt{n} L ≤ C 1 ( log n ) 9 < n a = 2 a=2 a = 2
The corresponding block field is quadratic. If it is real, Lemma 4.20 gives L < 2 n log n L<2n\log n L < 2 n log n Proposition 4.27 gives L < 1000 n 4 log n L<1000n^4\log n L < 1000 n 4 log n
n 1 / 2 < L < n 5 , n^{1/2}<L<n^5, n 1/2 < L < n 5 , contradicting Corollary 4.5. Every threshold used here is effective and independent of s s s
Proof of Theorem 1.1. Take N = max { N 0 , N G a l } N=\max\{N_0,N_{\mathrm{Gal}}\} N = max { N 0 , N Gal } N 0 N_0 N 0 Proposition 3.21 and N G a l N_{\mathrm{Gal}} N Gal Theorem 4.28 . For n ≥ N n\geq N n ≥ N s s s s s s
An explicit description of the symmetric specializations
The square-discriminant classification of Banerjee–Filaseta–Finch–Leidy [2, Theorem 2.1 turns uniform alternating containment into a short description of every Galois group in sufficiently large degree. The classification of alternating Laguerre specializations was further developed by Banerjee–Bera [1 ] . The square-discriminant families used below are those of [2 ] ; the uniform theorem determines the symmetric group throughout their complement in the region α = − n − s \alpha=-n-s α = − n − s s > 0 s>0 s > 0
For k ≥ 0 k\ge0 k ≥ 0
a k = ( 2 + 3 ) k + ( 2 − 3 ) k 2 , b k = ( 4 + 15 ) k + ( 4 − 15 ) k 2 , c k = ( 1 + 2 ) k + ( 1 − 2 ) k 2 . a_k=\frac{(2+\sqrt{3})^k+(2-\sqrt{3})^k}{2},\qquad b_k=\frac{(4+\sqrt{15})^k+(4-\sqrt{15})^k}{2},\qquad c_k=\frac{(1+\sqrt{2})^k+(1-\sqrt{2})^k}{2}. a k = 2 ( 2 + 3 ) k + ( 2 − 3 ) k , b k = 2 ( 4 + 15 ) k + ( 4 − 15 ) k , c k = 2 ( 1 + 2 ) k + ( 1 − 2 ) k . Corollary 4.29. There is an effectively computable absolute integer N 1 N_1 N 1 n ≥ N 1 n\ge N_1 n ≥ N 1 s s s G n , s = A n G_{n,s}=A_n G n , s = A n G n , s = S n G_{n,s}=S_n G n , s = S n
s ∉ { 1 , 2 , 3 , 4 , 5 , 6 , n − 4 , n , n + 2 , n + 4 } ⟹ G n , s = S n . s\notin\{1,2,3,4,5,6,n-4,n,n+2,n+4\}\quad\Longrightarrow\quad G_{n,s}=S_n. s ∈ / { 1 , 2 , 3 , 4 , 5 , 6 , n − 4 , n , n + 2 , n + 4 } ⟹ G n , s = S n . Table 1. The negative-parameter part of the square-discriminant classification in [2 Section 2, items (v)–(vii) and (xx)–(xxx)] , after imposing α = − n − s \alpha=-n-s α = − n − s n ≥ N 1 n\ge N_1 n ≥ N 1
:::
Proof. The cited classification gives thirty explicit families and an effectively computable finite exceptional set A 0 A_0 A 0 N 1 N_1 N 1 A 0 A_0 A 0 α = − n − s < − n \alpha=-n-s<-n α = − n − s < − n [2 Section 2] . Substitution gives the ten rows of the table. The discriminant is nonzero for s > 0 s>0 s > 0 A n ⊆ G n , s A_n\subseteq G_{n,s} A n ⊆ G n , s A n A_n A n S n S_n S n
Effective exceptional specializations
Runge’s theorem gives effective specialization results when an intermediate cover has several closed points above infinity. Their number is the number of orbits of the arithmetic decomposition group on the corresponding coset space. We use this observation to turn the cycle type of geometric inertia into an effective criterion.
Proposition 5.1. Let K / Q ( T ) K/\mathbb{Q}(T) K / Q ( T ) G G G D ∞ ≤ G D_\infty\leq G D ∞ ≤ G T = ∞ T=\infty T = ∞ S f S_f S f H \mathcal{H} H G G G
# ( D ∞ \ G / H ) > ∣ S f ∣ + 1 ( H ∈ H ) . \#(D_\infty\backslash G/H)>|S_f|+1\qquad(H\in\mathcal{H}). # ( D ∞ \ G / H ) > ∣ S f ∣ + 1 ( H ∈ H ) . Then the parameters t ∈ Z [ S f − 1 ] t \in\mathbb{Z}[S_f^{-1}] t ∈ Z [ S f − 1 ] H \mathcal{H} H
A sufficient condition is [ G : H ] > ∣ D ∞ ∣ ( ∣ S f ∣ + 1 ) [G:H] > |D_\infty|(|S_f|+1) [ G : H ] > ∣ D ∞ ∣ ( ∣ S f ∣ + 1 ) H ∈ H H \in\mathcal{H} H ∈ H S f = ∅ S_f=\varnothing S f = ∅ H ∈ H H \in\mathcal{H} H ∈ H [ G : H ] [G:H] [ G : H ] ∣ D ∞ ∣ |D_\infty| ∣ D ∞ ∣
Proof. Let X H X_H X H K H K^H K H D ∞ \ G / H D_\infty\backslash G/H D ∞ \ G / H T T T ∣ S f ∣ + 1 |S_f|+1 ∣ S f ∣ + 1 Q \mathbb{Q} Q [34 Theorem 1.2] , with S = S f ∪ { ∞ } S=S_f\cup\{\infty\} S = S f ∪ { ∞ }
{ P ∈ X H ( Q ) : T ( P ) ∈ Z [ S f − 1 ] } . \{P \in X_H(\mathbb{Q}) : T(P) \in\mathbb{Z}[S_f^{-1}]\}. { P ∈ X H ( Q ) : T ( P ) ∈ Z [ S f − 1 ]} . At an unramified rational parameter, the specialization group is contained in a conjugate of H H H G / H G/H G / H X H X_H X H H \mathcal{H} H
Every D ∞ D_\infty D ∞ ∣ D ∞ ∣ |D_\infty| ∣ D ∞ ∣ S f = ∅ S_f=\varnothing S f = ∅ [ G : H ] ∣ ∣ D ∞ ∣ [G:H]\mid|D_\infty| [ G : H ] ∣ ∣ D ∞ ∣
Lemma 5.2. Let σ ∈ S n \sigma\in S_n σ ∈ S n a d a_d a d d d d a 1 a_1 a 1 m = ord ( σ ) m=\operatorname{ord}(\sigma) m = ord ( σ )
∣ N S n ( ⟨ σ ⟩ ) ∣ = φ ( m ) ∏ d ≥ 1 d a d a d ! . |N_{S_n}(\langle\sigma\rangle)|=\varphi(m)\prod_{d\geq1}d^{a_d}a_d!. ∣ N S n (⟨ σ ⟩) ∣ = φ ( m ) d ≥ 1 ∏ d a d a d ! . If p p p n / 2 < p < n n/2<p<n n /2 < p < n
p ∤ ∣ N S n ( ⟨ σ ⟩ ) ∣ ⟺ a p = 0 and a 1 < p . p\nmid|N_{S_n}(\langle\sigma\rangle)| \Longleftrightarrow a_p=0\ \text{and}\ a_1<p. p ∤ ∣ N S n (⟨ σ ⟩) ∣ ⟺ a p = 0 and a 1 < p . Proof. The centralizer has order ∏ d d a d a d ! \prod_d d^{a_d}a_d! ∏ d d a d a d ! σ ↦ σ u \sigma\mapsto\sigma^u σ ↦ σ u ⟨ σ ⟩ \langle\sigma\rangle ⟨ σ ⟩ d d d u u u d d d φ ( m ) \varphi(m) φ ( m )
We claim that p ∤ φ ( m ) p\nmid\varphi(m) p ∤ φ ( m ) p 2 ∣ m p^2\mid m p 2 ∣ m q ∣ m q\mid m q ∣ m p ∣ q − 1 p\mid q-1 p ∣ q − 1 p 2 > n p^2>n p 2 > n p p p q ≥ 2 p + 1 > n q\geq2p+1>n q ≥ 2 p + 1 > n m m m n n n p ∣ d p\mid d p ∣ d d ≤ n d\leq n d ≤ n d = p d=p d = p d ≥ 2 d\geq2 d ≥ 2 a d ≤ n / 2 < p a_d\leq n/2<p a d ≤ n /2 < p p p p a 1 ! a_1! a 1 !
Lemma 5.3. Let n / 2 < p ≤ n − 3 n/2<p\leq n-3 n /2 < p ≤ n − 3 H ≤ S n H\leq S_n H ≤ S n A n ≰ H A_n\nleq H A n ≰ H p ∤ ∣ H ∣ p\nmid|H| p ∤ ∣ H ∣ p ∣ [ S n : H ] p\mid[S_n:H] p ∣ [ S n : H ]
Proof. If p ∣ ∣ H ∣ p\mid|H| p ∣ ∣ H ∣ p p p 2 p > n 2p>n 2 p > n p p p a a a b b b a , b ≤ n / 2 < p a,b\leq n/2<p a , b ≤ n /2 < p p p p H H H [13 Theorem 3.3E] , with p ≤ n − 3 p\leq n-3 p ≤ n − 3 A n ≤ H A_n\leq H A n ≤ H
Theorem 5.4. Let F ( X , T ) ∈ Q ( T ) [ X ] F(X,T)\in\mathbb{Q}(T)[X] F ( X , T ) ∈ Q ( T ) [ X ] n ≥ 8 n\geq8 n ≥ 8 X X X S n S_n S n σ ∈ S n \sigma\in S_n σ ∈ S n
n / 2 < p ≤ n − 3 n/2<p\leq n-3 n /2 < p ≤ n − 3 such that σ \sigma σ p p p p p p t t t F ( X , t ) F(X,t) F ( X , t ) n n n Q \mathbb{Q} Q A n A_n A n
Proof. The arithmetic group is also S n S_n S n
D ∞ ≤ N S n ( ⟨ σ ⟩ ) . D_{\infty} \leq N_{S_n}(\langle\sigma\rangle). D ∞ ≤ N S n (⟨ σ ⟩) . Lemma 5.2 shows that p ∤ ∣ D ∞ ∣ p \nmid|D_{\infty}| p ∤ ∣ D ∞ ∣ p p p A n A_n A n □ \square □
Effective exceptional sets for the Laguerre family
Corollary 5.5. For every fixed n ≥ 5 n \geq5 n ≥ 5 s s s H n , s H_{n,s} H n , s A n ≰ G n , s A_n \not\leq G_{n,s} A n ≤ G n , s
Proof. The generic splitting field is a regular S n S_n S n Q ( s ) \mathbb{Q}(s) Q ( s ) s = 0 , − 1 , − ( n − 2 ) s = 0,-1,-(n-2) s = 0 , − 1 , − ( n − 2 ) n n n ( n − 1 ) (n-1) ( n − 1 ) [30 Lemma 15] . The first two give double transitivity, and the transposition then gives S n S_n S n 2 ⌊ n / 2 ⌋ 1 n m o d 2 2^{\lfloor n/2\rfloor}1^{n\bmod2} 2 ⌊ n /2 ⌋ 1 n mod 2 n ≥ 8 n \geq8 n ≥ 8 n / 2 < p ≤ n − 3 n/2 < p \leq n-3 n /2 < p ≤ n − 3 p p p p p p n ≥ 8 n \geq8 n ≥ 8 n ≥ 51 n \geq51 n ≥ 51 [39 ] , and for 8 ≤ n ≤ 50 8 \leq n \leq50 8 ≤ n ≤ 50 5 , 7 , 11 , 13 , 17 , 23 , 29 5,7,11,13,17,23,29 5 , 7 , 11 , 13 , 17 , 23 , 29
It remains to treat n = 5 , 6 , 7 n = 5,6,7 n = 5 , 6 , 7 σ \sigma σ C n = C S n ( σ ) C_n = C_{S_n}(\sigma) C n = C S n ( σ ) σ \sigma σ D ∞ ≤ C n D_{\infty} \leq C_n D ∞ ≤ C n
∣ C 5 ∣ = 8 , ∣ C 6 ∣ = ∣ C 7 ∣ = 48. |C_5| = 8,\qquad|C_6| = |C_7| = 48. ∣ C 5 ∣ = 8 , ∣ C 6 ∣ = ∣ C 7 ∣ = 48. The classification of transitive groups of these degrees [8 pp. 869–871] shows that a transitive subgroup omitting A 5 A_5 A 5 AGL 1 ( F 5 ) \operatorname{AGL}_1(\mathbb{F}_5) AGL 1 ( F 5 ) 6 6 6 S 5 S_5 S 5 A 7 A_7 A 7 AGL 1 ( F 7 ) \operatorname{AGL}_1(\mathbb{F}_7) AGL 1 ( F 7 ) GL 3 ( F 2 ) \operatorname{GL}_3(\mathbb{F}_2) GL 3 ( F 2 ) 120 120 120 30 30 30 S 7 S_7 S 7 ∣ C n ∣ |C_n| ∣ C n ∣ C n C_n C n
In degree six, an imprimitive subgroup is contained in a conjugate of S 3 ≀ S 2 S_3 \wr S_2 S 3 ≀ S 2 S 2 ≀ S 3 S_2 \wr S_3 S 2 ≀ S 3 10 10 10 15 15 15 48 48 48 A 6 A_6 A 6 M = PGL 2 ( F 5 ) M = \operatorname{PGL}_2(\mathbb{F}_5) M = PGL 2 ( F 5 ) P 1 ( F 5 ) \mathbb{P}^1(\mathbb{F}_5) P 1 ( F 5 ) z ↦ 2 / z z \mapsto2/z z ↦ 2/ z M M M 2 3 2^3 2 3 ( 0 ∞ ) ( 1 2 ) ( 3 4 ) (0\ \infty)(1\ 2)(3\ 4) ( 0 ∞ ) ( 1 2 ) ( 3 4 ) σ \sigma σ S 6 / M S_6/M S 6 / M S 6 S_6 S 6 S 6 / M S_6/M S 6 / M M M M S 6 S_6 S 6 C 6 C_6 C 6 C 6 C_6 C 6 σ \sigma σ C 6 C_6 C 6 S 6 / M S_6/M S 6 / M S 6 / H S_6/H S 6 / H H H H M M M
In all three degrees, D ∞ D_{\infty} D ∞ S n / H S_n/H S n / H H H H A n A_n A n □ \square □
The conclusion is effective in each fixed degree. The uniform Galois theorem supplies the separate bound on the degrees that can remain exceptional.
Corollary 5.6. The positive integral pairs ( n , s ) (n,s) ( n , s ) H n , s H_{n,s} H n , s A n ≰ G n , s A_n \not\leq G_{n,s} A n ≤ G n , s
Proof. There are no exceptions in degrees n ≤ 4 n \leq4 n ≤ 4 [29 Theorem 6.3(i)] . Theorem 1.1 bounds every remaining exceptional degree by an effective absolute constant. In each such degree, Proposition 2.4 bounds every reducible specialization by
s < ⌊ n 2 / 4 ⌋ n . s < \left\lfloor n^2/4 \right\rfloor^n. s < ⌊ n 2 /4 ⌋ n . Exact factorization determines those exceptions. For the irreducible specializations, Corollary 5.5 gives a finite effective list containing every exception to alternating containment. Exact Galois-group computation determines which entries are exceptions. □
This reduces Hajir’s two conjectures to an effective finite verification. It does not give a practical enumeration: the estimates obtained in the uniform proof are very large.
Applications and further questions
The preceding arguments have two kinds of applications. The finite-fiber and prime-valuation criteria apply to other polynomial families. The uniform Galois theorem also gives consequences for splitting fields, algebraic relations among roots, and products of hyperelliptic Jacobians.
An Appell family with two scales
6.1. An Appell family with two scales. We apply the finite-fiber criterion of § 2.3 to a convolution of two Laguerre families. This application uses the centered root strip, but does not require the uniform Galois theorem. For an integer c ≥ 2 c \ge2 c ≥ 2
P n , s ( c ) ( X ) = n ! [ t n ] e X t ( 1 − t ) − s ( 1 − c t ) − s , P n , s = P n , s ( 2 ) , κ c = 1 + c . (114) P_{n,s}^{(c)}(X) = n![t^n]e^{Xt}(1-t)^{-s}(1-ct)^{-s}, \qquad P_{n,s}=P_{n,s}^{(2)}, \qquad\kappa_c=1+c. \tag*{(114)} P n , s ( c ) ( X ) = n ! [ t n ] e X t ( 1 − t ) − s ( 1 − c t ) − s , P n , s = P n , s ( 2 ) , κ c = 1 + c . (114) For real s > 0 s>0 s > 0 E ( X + Y 1 + c Y 2 ) n \mathbb{E}(X+Y_1+cY_2)^n E ( X + Y 1 + c Y 2 ) n s s s X n − k X^{n-k} X n − k
∑ j = 0 k n ! ( n − k ) ! j ! ( k − j ) ! ( s ) j ( s ) k − j c k − j . \sum_{j=0}^{k}\frac{n!}{(n-k)!j!(k-j)!}(s)_j(s)_{k-j}c^{k-j}. j = 0 ∑ k ( n − k )! j ! ( k − j )! n ! ( s ) j ( s ) k − j c k − j . Thus P n , s ( c ) ∈ Z [ s , X ] P_{n,s}^{(c)}\in\mathbb{Z}[s,X] P n , s ( c ) ∈ Z [ s , X ] s s s k k k
Theorem 6.1. For every ε > 0 \varepsilon>0 ε > 0 N ( ε ) N(\varepsilon) N ( ε ) P n , s P_{n,s} P n , s Q \mathbb{Q} Q
n ≥ N ( ε ) , s ≥ 1 , log s ≥ n ε . n\ge N(\varepsilon), \qquad s\ge1, \qquad\log s\ge n^{\varepsilon}. n ≥ N ( ε ) , s ≥ 1 , log s ≥ n ε . The same argument permits rational parameters and a scale that grows with the degree.
Corollary 6.2. There is a finite effectively computable set E ⊂ Z ≥ 3 E\subset\mathbb{Z}_{\ge3} E ⊂ Z ≥ 3 0 < η < 1 / 3472 0<\eta<1/3472 0 < η < 1/3472 B ≥ 0 B\ge0 B ≥ 0 ε > 0 \varepsilon>0 ε > 0 N ( B , ε , η ) N(B,\varepsilon,\eta) N ( B , ε , η ) c c c P n , a / b ( c ) P_{n,a/b}^{(c)} P n , a / b ( c ) Q \mathbb{Q} Q
n ≥ N ( B , ε , η ) , 2 ≤ c ≤ n η , c ∈ Z ∖ E , n\ge N(B,\varepsilon,\eta), \qquad2\le c\le n^{\eta}, \qquad c\in\mathbb{Z}\setminus E, n ≥ N ( B , ε , η ) , 2 ≤ c ≤ n η , c ∈ Z ∖ E , a , b ∈ Z > 0 , gcd ( a , b ) = 1 , b ≤ n B , log a ≥ n ε . a,b\in\mathbb{Z}_{>0}, \qquad\gcd(a,b)=1, \qquad b\le n^B, \qquad\log a\ge n^\varepsilon. a , b ∈ Z > 0 , g cd( a , b ) = 1 , b ≤ n B , log a ≥ n ε . The set E E E 2 ∉ E 2\notin E 2 ∈ / E
Proof of the theorem and corollary. We check the two inputs to the finite-fiber criterion: a strip bound from polynomial convolution, and trace separation at four fibers. We then make both estimates uniform in c c c
Convolution and the root strip.
For monic polynomials p , q p,q p , q n n n q ( X ) = ∑ k = 0 n q k X n − k q(X)=\sum_{k=0}^{n}q_kX^{n-k} q ( X ) = ∑ k = 0 n q k X n − k
( p ⊞ n q ) ( X ) = ∑ k = 0 n ( n − k ) ! n ! q k p ( k ) ( X ) . (p\boxplus_n q)(X)=\sum_{k=0}^{n}\frac{(n-k)!}{n!}q_kp^{(k)}(X). ( p ⊞ n q ) ( X ) = k = 0 ∑ n n ! ( n − k )! q k p ( k ) ( X ) . This is Walsh’s additive convolution, also called symmetric additive finite free convolution; see [37 Section 1] . We need its preservation of half-planes, including for polynomials whose coefficients are not real.
Suppose the roots of p p p q q q
p ~ ( Z ) = i n p ( Z / i ) , q ~ ( Z ) = i n q ( Z / i ) \widetilde{p}(Z)=i^{n}p(Z/i),\qquad\widetilde{q}(Z)=i^{n}q(Z/i) p ( Z ) = i n p ( Z / i ) , q ( Z ) = i n q ( Z / i ) have their roots in the closed lower half-plane. Consider
T q ~ = ∑ k = 0 n ( n − k ) ! n ! q ~ k d k d X k . T_{\widetilde{q}}=\sum_{k=0}^{n}\frac{(n-k)!}{n!}\widetilde{q}_{k}\frac{d^{k}}{dX^{k}}. T q = k = 0 ∑ n n ! ( n − k )! q k d X k d k . Its algebraic symbol is T q ~ [ ( X + Y ) n ] = q ~ ( X + Y ) T_{\widetilde{q}}[(X+Y)^{n}]=\widetilde{q}(X+Y) T q [( X + Y ) n ] = q ( X + Y ) [5 Theorem 4, p. 471] , T q ~ p ~ T_{\widetilde{q}}\widetilde{p} T q p
( p ~ ⊞ n q ~ ) ( Z ) = i n ( p ⊞ n q ) ( Z / i ) , (\widetilde{p}\boxplus_n\widetilde{q})(Z)=i^{n}(p\boxplus_nq)(Z/i), ( p ⊞ n q ) ( Z ) = i n ( p ⊞ n q ) ( Z / i ) , p ⊞ n q p\boxplus_nq p ⊞ n q
For formal power series f , g f,g f , g
( n ! [ t n ] e X t f ( t ) ) ⊞ n ( n ! [ t n ] e X t g ( t ) ) = n ! [ t n ] e X t f ( t ) g ( t ) . \bigl(n![t^{n}]e^{Xt}f(t)\bigr)\boxplus_n\bigl(n![t^{n}]e^{Xt}g(t)\bigr)=n![t^{n}]e^{Xt}f(t)g(t). ( n ! [ t n ] e X t f ( t ) ) ⊞ n ( n ! [ t n ] e X t g ( t ) ) = n ! [ t n ] e X t f ( t ) g ( t ) . Consequently
P n , s ( c ) = H n , s ⊞ n ( c n H n , s ( X / c ) ) . P_{n,s}^{(c)}=H_{n,s}\boxplus_n\left(c^{n}H_{n,s}(X/c)\right). P n , s ( c ) = H n , s ⊞ n ( c n H n , s ( X / c ) ) . Proposition 2.2 and the half-plane bounds just proved show that, for every real s > 0 s>0 s > 0 α \alpha α
∣ Re ( α + κ c s ) ∣ ≤ κ c ( n − 1 ) . (115) \left|\operatorname{Re}(\alpha+\kappa_c s)\right|\leq\kappa_c(n-1). \tag*{(115)} ∣ Re ( α + κ c s ) ∣ ≤ κ c ( n − 1 ) . (115) Four fibers and the trace condition.
Introduce indeterminates N , C N,C N , C
f k ( N ) = N ( N − 1 ) ⋯ ( N − k + 1 ) , f 0 ( N ) = 1 , A C ( t ) = ( 1 − t ) ( 1 − C t ) . f_k(N)=N(N-1)\cdots(N-k+1),\qquad f_0(N)=1,\qquad A_C(t)=(1-t)(1-Ct). f k ( N ) = N ( N − 1 ) ⋯ ( N − k + 1 ) , f 0 ( N ) = 1 , A C ( t ) = ( 1 − t ) ( 1 − C t ) . For m = 1 , 2 , 3 m=1,2,3 m = 1 , 2 , 3
R m , C ( N , X ) = ∑ k = 0 2 m f k ( N ) [ t k ] A C ( t ) m X 2 m − k . R_{m,C}(N,X)=\sum_{k=0}^{2m}f_k(N)[t^k]A_C(t)^mX^{2m-k}. R m , C ( N , X ) = k = 0 ∑ 2 m f k ( N ) [ t k ] A C ( t ) m X 2 m − k . These belong to Z [ N , C , X ] \mathbb{Z}[N,C,X] Z [ N , C , X ] n ≥ 6 n\geq6 n ≥ 6
P n , 0 ( c ) = X n , P n , − m ( c ) = X n − 2 m R m , c ( n , X ) . P_{n,0}^{(c)}=X^n,\qquad P_{n,-m}^{(c)}=X^{n-2m}R_{m,c}(n,X). P n , 0 ( c ) = X n , P n , − m ( c ) = X n − 2 m R m , c ( n , X ) . Let U m , C ( N , Z ) U_{m,C}(N,Z) U m , C ( N , Z ) R m , C R_{m,C} R m , C U 0 , C ( N , Z ) = Z U_{0,C}(N,Z)=Z U 0 , C ( N , Z ) = Z
U m , C ∈ Z [ N , C , Z ] , deg Z U m , C = 2 2 m . U_{m,C}\in\mathbb{Z}[N,C,Z],\qquad\deg_ZU_{m,C}=2^{2m}. U m , C ∈ Z [ N , C , Z ] , deg Z U m , C = 2 2 m . The full subset has sum m ( 1 + C ) N m(1+C)N m ( 1 + C ) N m = 2 , 3 m=2,3 m = 2 , 3
V m , C ( N , Z ) = U m , C ( N , Z ) Z ( Z − m ( 1 + C ) N ) . V_{m,C}(N,Z)=\frac{U_{m,C}(N,Z)}{Z\left(Z-m(1+C)N\right)}. V m , C ( N , Z ) = Z ( Z − m ( 1 + C ) N ) U m , C ( N , Z ) . These are monic polynomials in Z [ N , C , Z ] \mathbb{Z}[N,C,Z] Z [ N , C , Z ] C = 2 C=2 C = 2 C C C R m , C R_{m,C} R m , C U m , C U_{m,C} U m , C V m , C V_{m,C} V m , C s s s d d d A = τ + κ c d s A=\tau+\kappa_cds A = τ + κ c d s
p ∣ s + m ⟹ p ∣ U m , c ( n , A + m κ c d ) , 0 ≤ m ≤ 3. (116) p\mid s+m\quad\Longrightarrow\quad p\mid U_{m,c}(n,A+m\kappa_cd),\qquad0\leq m\leq3. \tag*{(116)} p ∣ s + m ⟹ p ∣ U m , c ( n , A + m κ c d ) , 0 ≤ m ≤ 3. (116) The first residual polynomial is
R 1 , c ( n , X ) = X 2 − κ c n X + c n ( n − 1 ) . R_{1,c}(n,X)=X^2-\kappa_cnX+cn(n-1). R 1 , c ( n , X ) = X 2 − κ c n X + c n ( n − 1 ) . Its discriminant ( c − 1 ) 2 n 2 + 4 c n (c-1)^2n^2+4cn ( c − 1 ) 2 n 2 + 4 c n n ≥ 5 n\geq5 n ≥ 5 q 2 q^2 q 2
U = ( c − 1 ) 2 n + 2 c , V = ( c − 1 ) q U=(c-1)^2n+2c,\qquad V=(c-1)q U = ( c − 1 ) 2 n + 2 c , V = ( c − 1 ) q satisfy U > V ≥ 0 U > V \ge0 U > V ≥ 0 ( U − V ) ( U + V ) = 4 c 2 (U-V)(U+V)=4c^2 ( U − V ) ( U + V ) = 4 c 2 4 c 2 4c^2 4 c 2 4 c 2 + 1 4c^2+1 4 c 2 + 1 2 U ≤ 4 c 2 + 1 2U \le4c^2+1 2 U ≤ 4 c 2 + 1 U ≤ 2 c 2 U \le2c^2 U ≤ 2 c 2 n ≤ 2 c / ( c − 1 ) ≤ 4 n \le2c/(c-1) \le4 n ≤ 2 c / ( c − 1 ) ≤ 4 U 1 , c ( n , Z ) U_{1,c}(n,Z) U 1 , c ( n , Z ) 0 0 0 κ c n \kappa_c n κ c n
Thus fibers 0, 1 already give two nonzero evaluations unless the centered trace is 0 0 0 − κ c d -\kappa_c d − κ c d κ c ( n − d ) \kappa_c(n-d) κ c ( n − d ) d d d κ = 1 + C \kappa=1+C κ = 1 + C
D 0 ( N , C ) = Res D ( V 2 , C ( N , 2 κ D ) , V 3 , C ( N , 3 κ D ) ) , D_0(N,C)=\operatorname{Res}_D\bigl(V_{2,C}(N,2\kappa D),V_{3,C}(N,3\kappa D)\bigr), D 0 ( N , C ) = Res D ( V 2 , C ( N , 2 κ D ) , V 3 , C ( N , 3 κ D ) ) , D 1 ( N , C ) = Res D ( V 2 , C ( N , κ D ) , V 3 , C ( N , 2 κ D ) ) . D_1(N,C)=\operatorname{Res}_D\bigl(V_{2,C}(N,\kappa D),V_{3,C}(N,2\kappa D)\bigr). D 1 ( N , C ) = Res D ( V 2 , C ( N , κ D ) , V 3 , C ( N , 2 κ D ) ) . Both pairs have degrees 14, 62 in D D D κ \kappa κ N N N N N N C C C D i ( N ) = D i ( N , 2 ) D_i(N)=D_i(N,2) D i ( N ) = D i ( N , 2 )
D 0 ( 11 ) D 1 ( 11 ) ≠ 0. (117) D_0(11)D_1(11)\ne0. \tag*{(117)} D 0 ( 11 ) D 1 ( 11 ) = 0. (117) It therefore proves nonvanishing of the two resultants as polynomials, rather than merely a trace calculation in one degree.
Suppose n ≥ 6 n\ge6 n ≥ 6 c ≥ 2 c\ge2 c ≥ 2 D 0 ( n , c ) D 1 ( n , c ) ≠ 0 D_0(n,c)D_1(n,c)\ne0 D 0 ( n , c ) D 1 ( n , c ) = 0 z ∈ Q z\in\mathbb{Q} z ∈ Q 0 < d < n 0<d<n 0 < d < n
U m , c ( n , z + m κ c d ) , m = 0 , 1 , 2 , 3 , (118) U_{m,c}(n,z+m\kappa_c d),\qquad m=0,1,2,3, \tag*{(118)} U m , c ( n , z + m κ c d ) , m = 0 , 1 , 2 , 3 , (118) are nonzero. Outside the three exceptional traces above, this follows from fibers 0, 1. If z = 0 z=0 z = 0 0 < d < n 0<d<n 0 < d < n D 0 ( n , c ) ≠ 0 D_0(n,c)\ne0 D 0 ( n , c ) = 0 z = − κ c d z=-\kappa_c d z = − κ c d d = 0 d=0 d = 0 d = 2 n d=2n d = 2 n d = 3 n / 2 d=3n/2 d = 3 n /2 0 < d < n 0<d<n 0 < d < n D 1 ( n , c ) ≠ 0 D_1(n,c)\ne0 D 1 ( n , c ) = 0 z = κ c ( n − d ) z=\kappa_c(n-d) z = κ c ( n − d ) ( z , d ) (z,d) ( z , d ) ( − z , n − d ) (-z,n-d) ( − z , n − d )
U m , c ( n , m κ c n − Z ) = U m , c ( n , Z ) ( m ≥ 1 ) , U 0 , c ( n , − Z ) = − U 0 , c ( n , Z ) . U_{m,c}(n,m\kappa_c n-Z)=U_{m,c}(n,Z)\quad(m\ge1),\qquad U_{0,c}(n,-Z)=-U_{0,c}(n,Z). U m , c ( n , m κ c n − Z ) = U m , c ( n , Z ) ( m ≥ 1 ) , U 0 , c ( n , − Z ) = − U 0 , c ( n , Z ) . This replacement therefore preserves which evaluations vanish and reduces the last case to the preceding one.
Set
D ( N , C ) = D 0 ( N , C ) D 1 ( N , C ) , E ( C ) = D ( 11 , C ) , E = { c ∈ Z ≥ 2 : E ( c ) = 0 } . \mathcal{D}(N,C)=D_0(N,C)D_1(N,C),\qquad E(C)=\mathcal{D}(11,C),\qquad\mathcal{E}=\{c\in\mathbb{Z}_{\ge2}:E(c)=0\}. D ( N , C ) = D 0 ( N , C ) D 1 ( N , C ) , E ( C ) = D ( 11 , C ) , E = { c ∈ Z ≥ 2 : E ( c ) = 0 } . By (6.4), E ( 2 ) ≠ 0 E(2)\ne0 E ( 2 ) = 0 E \mathcal{E} E E ⊂ Z ≥ 3 \mathcal{E}\subset\mathbb{Z}_{\ge3} E ⊂ Z ≥ 3
Put K = max { 1 , deg C D } K=\max\{1,\deg_C\mathcal{D}\} K = max { 1 , deg C D } K ≤ 3472 K\le3472 K ≤ 3472 d m = 2 2 m − 2 d_m=2^{2m}-2 d m = 2 2 m − 2 m = 2 , 3 m=2,3 m = 2 , 3 X 2 m − k X^{2m-k} X 2 m − k R m , C R_{m,C} R m , C C C C k k k Z d m − j Z^{d_m-j} Z d m − j V m , C V_{m,C} V m , C j j j k k k k k k C C C j j j Z = ℓ ( 1 + C ) D Z=\ell(1+C)D Z = ℓ ( 1 + C ) D D D D C C C d m d_m d m
deg C D i ≤ 62 ⋅ 14 + 14 ⋅ 62 = 1736 ( i = 0 , 1 ) , \deg_C D_i\le62\cdot14+14\cdot62=1736\qquad(i=0,1), deg C D i ≤ 62 ⋅ 14 + 14 ⋅ 62 = 1736 ( i = 0 , 1 ) , and hence K ≤ 3472 K\le3472 K ≤ 3472
There is an effective absolute C 0 ≥ 1 C_0\ge1 C 0 ≥ 1 D ( N , c ) \mathcal{D}(N,c) D ( N , c ) N N N C 0 ( 1 + c ) K C_0(1+c)^K C 0 ( 1 + c ) K c ∉ E c\notin\mathcal{E} c ∈ / E c c c 1 + C 0 ( 1 + c ) K 1+C_0(1+c)^K 1 + C 0 ( 1 + c ) K c ≤ n η c\le n^\eta c ≤ n η 0 < η < 1 / 3472 0<\eta<1/3472 0 < η < 1/3472
1 + C 0 ( 1 + c ) K ≤ 1 + 2 K C 0 n K η < n 1+C_0(1+c)^K\le1+2^K C_0n^{K\eta}<n 1 + C 0 ( 1 + c ) K ≤ 1 + 2 K C 0 n K η < n for all sufficiently large n n n η \eta η D 0 ( n , c ) D 1 ( n , c ) ≠ 0 D_0(n,c)D_1(n,c) \ne0 D 0 ( n , c ) D 1 ( n , c ) = 0
To finish, apply Corollary 2.14. For each sufficiently large n n n c = c ( n ) c=c(n) c = c ( n ) P n ( T , X ) = P n , T ( c ( n ) ) ( X ) P_n(T,X)=P_{n,T}^{(c(n))}(X) P n ( T , X ) = P n , T ( c ( n )) ( X ) c n ( T ) = − κ c ( n ) T c_n(T)=-\kappa_{c(n)}T c n ( T ) = − κ c ( n ) T (115) gives the required real-parameter bound, since κ c ( n − 1 ) ≤ 2 n 1 + η \kappa_c(n-1)\le2n^{1+\eta} κ c ( n − 1 ) ≤ 2 n 1 + η 6 6 6 n , c n,c n , c ∣ c n ( − m ) ∣ \lvert c_n(-m)\rvert ∣ c n ( − m )∣ n n n (118) supplies the trace condition. All constants are independent of the choice of c ( n ) c(n) c ( n ) N ( B , ε , η ) N(B,\varepsilon,\eta) N ( B , ε , η ) c = 2 c=2 c = 2 b = 1 b=1 b = 1 □ \square □
Further Appell families. Let F ( t ) ∈ Z [ t ] F(t)\in\mathbb{Z}[t] F ( t ) ∈ Z [ t ] F ( 0 ) = 1 F(0)=1 F ( 0 ) = 1 F ( t ) = ∏ j = 1 r ( 1 − c j t ) F(t)=\prod_{j=1}^r(1-c_jt) F ( t ) = ∏ j = 1 r ( 1 − c j t ) c j > 0 c_j>0 c j > 0 μ = ∑ j c j = − F ′ ( 0 ) ∈ Z > 0 \mu=\sum_j c_j=-F'(0)\in\mathbb{Z}_{>0} μ = ∑ j c j = − F ′ ( 0 ) ∈ Z > 0 Q n , s ( X ) = n ! [ t n ] e X t F ( t ) − s Q_{n,s}(X)=n![t^n]e^{Xt}F(t)^{-s} Q n , s ( X ) = n ! [ t n ] e X t F ( t ) − s
∣ Re ( α + μ s ) ∣ ≤ μ ( n − 1 ) \lvert\operatorname{Re}(\alpha+\mu s)\rvert\le\mu(n-1) ∣ Re ( α + μ s )∣ ≤ μ ( n − 1 ) for every root and every real s > 0 s>0 s > 0 F ( t ) − s = ∑ j ≥ 0 ( s ) j ( 1 − F ( t ) ) j / j ! F(t)^{-s}=\sum_{j\ge0}(s)_j(1-F(t))^j/j! F ( t ) − s = ∑ j ≥ 0 ( s ) j ( 1 − F ( t ) ) j / j ! Q n , s ∈ Z [ s , X ] Q_{n,s}\in\mathbb{Z}[s,X] Q n , s ∈ Z [ s , X ] X n − k X^{n-k} X n − k s s s k k k j ≤ k ≤ n j\le k\le n j ≤ k ≤ n n ! / ( ( n − k ) ! j ! ) n!/((n-k)!j!) n ! / (( n − k )! j !) s = − m s=-m s = − m r m rm r m n ≥ r m n\ge rm n ≥ r m n n n F F F
Generic irreducibility does not settle this verification. Indeed, Eisenstein at s = 0 s=0 s = 0 Q n , s ( X ) Q_{n,s}(X) Q n , s ( X ) s s s [ s ] Q n , s ( 0 ) = ( n − 1 ) ! ∑ j > 0 c j n > 0 [s]Q_{n,s}(0)=(n-1)!\sum_{j>0}c_j^n>0 [ s ] Q n , s ( 0 ) = ( n − 1 )! ∑ j > 0 c j n > 0
A deterministic application to trimmed exponentials
For positive integers n , u n,u n , u
E n , u ( X ) = ( u + n ) ! ∑ j = 0 n X j ( u + j ) ! = ∑ j = 0 n ( ∏ r = j + 1 n ( u + r ) ) X j . (119) E_{n,u}(X)=(u+n)!\sum_{j=0}^{n}\frac{X^j}{(u+j)!}=\sum_{j=0}^{n}\left(\prod_{r=j+1}^{n}(u+r)\right)X^j. \tag*{(119)} E n , u ( X ) = ( u + n )! j = 0 ∑ n ( u + j )! X j = j = 0 ∑ n ( r = j + 1 ∏ n ( u + r ) ) X j . (119) Bary-Soroker and Ben-Porath [3 Conjecture 1] conjectured that every such trimmed exponential is irreducible with Galois group S n S_n S n u > 1 u>1 u > 1 n ≥ 2 n\ge2 n ≥ 2 [3 Theorem 1] that, if k max ( x ) = x o ( 1 ) k_{\max}(x)=x^{o(1)} k m a x ( x ) = x o ( 1 ) m m m { 1 , … , x } \{1,\ldots,x\} { 1 , … , x } m > k max ( x ) m>k_{\max}(x) m > k m a x ( x ) E n , m − n E_{n,m-n} E n , m − n 8 < n < k max ( x ) 8<n<k_{\max}(x) 8 < n < k m a x ( x ) S n S_n S n S n S_n S n S n S_n S n
Local packets and the discriminant
Lemma 6.3. Let 1 ≤ N ≤ n 1\le N\le n 1 ≤ N ≤ n p > n p>n p > n e = v p ( u + N ) > 0 e=v_p(u+N)>0 e = v p ( u + N ) > 0 E n , u E_{n,u} E n , u ( 0 , e ) (0,e) ( 0 , e ) ( N , 0 ) (N,0) ( N , 0 )
v p ( Disc E n , u ) = e ( N − 1 ) . (120) v_p(\operatorname{Disc} E_{n,u})=e(N-1). \tag*{(120)} v p ( Disc E n , u ) = e ( N − 1 ) . (120) Its local inertia group is cyclic, and a generator has cycle type ( N / g ) g 1 n − N (N/g)^{g1^{n-N}} ( N / g ) g 1 n − N g = gcd ( N , e ) g=\gcd(N,e) g = g cd( N , e )
Proof. Only u + N u+N u + N u + 1 , … , u + n u+1,\ldots,u+n u + 1 , … , u + n p p p (119) therefore gives
v p ( [ X j ] E n , u ) = { e , 0 ≤ j < N , 0 , N ≤ j ≤ n . v_p([X^j]E_{n,u}) =
\begin{cases}
e, & 0 \le j < N,\\
0, & N \le j \le n.
\end{cases} v p ([ X j ] E n , u ) = { e , 0 , 0 ≤ j < N , N ≤ j ≤ n . This proves the Newton-polygon assertion. Put m = n − N m=n-N m = n − N e m ( X ) = ∑ j = 0 m X j / j ! e_m(X)=\sum_{j=0}^{m} X^j/j! e m ( X ) = ∑ j = 0 m X j / j ! u ≡ − N ( m o d p ) u\equiv-N\pmod p u ≡ − N ( mod p )
E ‾ n , u ( X ) = X N R m ( X ) , R m ( X ) = m ! e m ( X ) ∈ F p [ X ] . \overline{E}_{n,u}(X)=X^N R_m(X), \qquad R_m(X)=m!e_m(X)\in\mathbb{F}_p[X]. E n , u ( X ) = X N R m ( X ) , R m ( X ) = m ! e m ( X ) ∈ F p [ X ] . Here R m ( 0 ) = m ! ≠ 0 R_m(0)=m!\ne0 R m ( 0 ) = m ! = 0 R m − R m ′ = X m R_m-R'_m=X^m R m − R m ′ = X m R m R_m R m m = 0 m=0 m = 0 R m = 1 R_m=1 R m = 1 A ∈ Z p [ X ] A\in\mathbb{Z}_p[X] A ∈ Z p [ X ] N N N X N X^N X N A A A p e p^e p e e e e e / N e/N e / N 2 ( N 2 ) e / N = e ( N − 1 ) 2^{\binom{N}{2}e/N}=e(N-1) 2 ( 2 N ) e / N = e ( N − 1 ) (120) . □
Proposition 6.4. For each fixed η > 0 \eta>0 η > 0 n 0 ( η ) n_0(\eta) n 0 ( η ) E n , u E_{n,u} E n , u
n ≥ n 0 ( η ) , u ≥ ( gcd ( u , 2 ) − 1 + η ) n . n\ge n_0(\eta),\qquad u\ge\bigl(\gcd(u,2)-1+\eta\bigr)n. n ≥ n 0 ( η ) , u ≥ ( g cd( u , 2 ) − 1 + η ) n . In particular, u ≥ ( 1 + η ) n u\ge(1+\eta)n u ≥ ( 1 + η ) n u u u u ≥ η n u\ge\eta n u ≥ η n n n n u ≥ n 21 / 10 u\ge n^{21/10} u ≥ n 21/10 ⌊ n / 2 ⌋ / 102 \lfloor n/2\rfloor/102 ⌊ n /2 ⌋ /102 p > 3 ⌊ n / 2 ⌋ p>3\lfloor n/2\rfloor p > 3 ⌊ n /2 ⌋ Disc E n , u \operatorname{Disc}E_{n,u} Disc E n , u
Proof. Put k = ⌊ n / 2 ⌋ k=\lfloor n/2\rfloor k = ⌊ n /2 ⌋ h = gcd ( u , 2 ) h=\gcd(u,2) h = g cd( u , 2 ) A = ( u + 2 ) / h A=(u+2)/h A = ( u + 2 ) / h d = 2 / h d=2/h d = 2/ h d ∈ { 1 , 2 } d\in\{1,2\} d ∈ { 1 , 2 } gcd ( A , d ) = 1 \gcd(A,d)=1 g cd( A , d ) = 1
∏ j = 0 k − 1 ( A + j d ) = h − k ∏ j = 1 k ( u + 2 j ) . (121) \prod_{j=0}^{k-1}(A+jd)=h^{-k}\prod_{j=1}^{k}(u+2j). \tag*{(121)} j = 0 ∏ k − 1 ( A + j d ) = h − k j = 1 ∏ k ( u + 2 j ) . (121) If a prime p > n p>n p > n u + N u+N u + N N N N e = v p ( u + N ) e=v_p(u+N) e = v p ( u + N ) v p ( Disc E n , u ) = e ( N − 1 ) v_p(\operatorname{Disc}E_{n,u})=e(N-1) v p ( Disc E n , u ) = e ( N − 1 )
For the first assertion, set θ = min ( η , 1 ) / 4 \theta=\min(\eta,1)/4 θ = min ( η , 1 ) /4 C = 2 + θ C=2+\theta C = 2 + θ [23 Theorem 1.1] imply that, for fixed d d d k k k A A A
gcd ( A , d ) = 1 , A ≥ ( C − d + θ d ) k , \gcd(A,d)=1,\qquad A\ge(C-d+\theta d)k, g cd( A , d ) = 1 , A ≥ ( C − d + θ d ) k , the product in (121) has a prime p > C k p>Ck p > C k b y 2 by^2 b y 2 b b b C k Ck C k C ≥ d C\ge d C ≥ d 0 < θ < 1 0<\theta<1 0 < θ < 1 h = 2 h=2 h = 2
A ≥ ( 1 + η ) k ≥ ( 1 + 2 θ ) k = ( C − d + θ d ) k . A\ge(1+\eta)k\ge(1+2\theta)k=(C-d+\theta d)k. A ≥ ( 1 + η ) k ≥ ( 1 + 2 θ ) k = ( C − d + θ d ) k . If h = 1 h=1 h = 1 A ≥ 2 η k ≥ 3 θ k = ( C − d + θ d ) k A\ge2\eta k\ge3\theta k=(C-d+\theta d)k A ≥ 2 η k ≥ 3 θ k = ( C − d + θ d ) k k k k d = 1 d=1 d = 1 d = 2 d=2 d = 2 k > 1 / θ k>1/\theta k > 1/ θ p > C k > n p>Ck>n p > C k > n
For the quantitative assertion, u ≥ n 21 / 10 u\ge n^{21/10} u ≥ n 21/10 A ≥ k 21 / 10 A\ge k^{21/10} A ≥ k 21/10 C = 3 C=3 C = 3 r = 2 r=2 r = 2 ϵ = 1 / 10 \epsilon=1/10 ϵ = 1/10 δ = 1 / 102 \delta=1/102 δ = 1/102 k / 102 k/102 k /102 p > 3 k p>3k p > 3 k (121) . For n ≥ 5 n\ge5 n ≥ 5 n n n
Symmetric groups in a parameter window.
Theorem 6.5. Fix D > 0 D > 0 D > 0 κ > 5 \kappa> 5 κ > 5 N = N ( D , κ ) N = N(D,\kappa) N = N ( D , κ )
n ≥ N , u ≥ 1 , ( log n ) κ ≤ log u ≤ n D , n \ge N,\qquad u \ge1,\qquad(\log n)^\kappa\le\log u \le n^D, n ≥ N , u ≥ 1 , ( log n ) κ ≤ log u ≤ n D , irreducibility of E n , u E_{n,u} E n , u Q \mathbb{Q} Q
Gal ( E n , u / Q ) = S n . \operatorname{Gal}(E_{n,u}/\mathbb{Q}) = S_n. Gal ( E n , u / Q ) = S n . In particular, for fixed 0 < ε ≤ D 0 < \varepsilon\le D 0 < ε ≤ D n ε ≤ log u ≤ n D n^\varepsilon\le\log u \le n^D n ε ≤ log u ≤ n D n n n ε , D \varepsilon,D ε , D
Proof. For each prime n / 2 < P ≤ n − 3 n/2 < P \le n - 3 n /2 < P ≤ n − 3 u + P u + P u + P 2 n / ( 5 log n ) 2n/(5\log n) 2 n / ( 5 log n ) n n n p > n p > n p > n u + P u + P u + P s = u s = u s = u C = 1 C = 1 C = 1 c = 2 / 5 c = 2/5 c = 2/5 j P = P j_P = P j P = P E n , u E_{n,u} E n , u ⌊ C / c ⌋ + 3 = 5 \lfloor C/c\rfloor+ 3 = 5 ⌊ C / c ⌋ + 3 = 5 n n n log u ≥ ( log n ) κ \log u \ge(\log n)^\kappa log u ≥ ( log n ) κ u ≥ 2 n u \ge2n u ≥ 2 n η = 1 \eta= 1 η = 1 A n A_n A n S n S_n S n κ = 6 \kappa= 6 κ = 6 ( log n ) 6 ≤ n ε (\log n)^6 \le n^\varepsilon ( log n ) 6 ≤ n ε n n n
Corollary 6.6. Let n , u n,u n , u q 1 , … , q t > n q_1,\ldots,q_t > n q 1 , … , q t > n t ≥ 1 t \ge1 t ≥ 1
e i = v q i ( u + n ) > 0 , gcd ( n , e 1 , … , e t ) = 1. e_i = v_{q_i}(u+n) > 0,\qquad\gcd(n,e_1,\ldots,e_t) = 1. e i = v q i ( u + n ) > 0 , g cd( n , e 1 , … , e t ) = 1. Then E n , u E_{n,u} E n , u Q \mathbb{Q} Q n , u n,u n , u S n S_n S n q > n q > n q > n u + n u+n u + n gcd ( n , v q ( u + n ) ) = 1 \gcd(n,v_q(u+n)) = 1 g cd( n , v q ( u + n )) = 1 u + n u+n u + n
Proof. Every nonleading coefficient in (119) has q i q_i q i e i e_i e i 1 1 1 ( 0 , e i ) (0,e_i) ( 0 , e i ) ( n , 0 ) (n,0) ( n , 0 ) e i / n e_i/n e i / n d d d q i q_i q i d e i / n de_i/n d e i / n n / gcd ( n , e i ) n/\gcd(n,e_i) n / g cd( n , e i ) d d d i i i
lcm 1 ≤ i ≤ t n gcd ( n , e i ) = n gcd ( n , e 1 , … , e t ) = n . \operatorname{lcm}_{1\le i\le t}\frac{n}{\gcd(n,e_i)}
=\frac{n}{\gcd(n,e_1,\ldots,e_t)}=n. lcm 1 ≤ i ≤ t g cd( n , e i ) n = g cd( n , e 1 , … , e t ) n = n . There is therefore no proper rational factor. This argument works for all positive n , u n,u n , u S n S_n S n u + n u+n u + n
For example, if 6 ∣ n 6 \mid n 6 ∣ n u + n = q 1 2 q 2 3 u+n = q_1^2q_2^3 u + n = q 1 2 q 2 3 q 1 , q 2 > n q_1,q_2 > n q 1 , q 2 > n n n n S n S_n S n
Theorem 6.5 is uniform in every parameter in its window, conditional on irreducibility. Corollary 6.6 removes that condition on an explicit arithmetic subfamily. These statements do not settle the full conjecture of [3 ] : irreducibility outside the endpoint condition remains unproved here. The local Newton-polygon step is classical. Bary-Soroker and Ben-Porath already use an endpoint valuation-one prime for irreducibility and a full cycle, and an even-length cycle to exclude alternating groups [3 Propositions 5.1 and 5.2] . Theorem 4.3 supplies the prime valuations needed for alternating containment throughout the stated parameter window. The linear discriminant range uses [23 Theorem 1.1] , while Theorem 4.6 gives a positive proportion of odd discriminant valuations in the larger-parameter range. All these conclusions are deterministic.
Splitting fields and Jacobians
The uniform Galois theorem has consequences for relations among selected roots and for products of hyperelliptic Jacobians. The general criteria are due to Bary-Soroker–Kadets, Zarhin, and Lombardo; here they apply with arbitrary positive integral parameters. Increase N N N N ≥ 5 N \ge5 N ≥ 5 J n , s J_{n,s} J n , s Y 2 = H n , s ( X ) Y^2 = H_{n,s}(X) Y 2 = H n , s ( X )
Splitting fields of distinct degrees
Proposition 6.7. Let f 1 , … , f r ∈ Q [ X ] f_1,\ldots,f_r \in\mathbb{Q}[X] f 1 , … , f r ∈ Q [ X ] n i ≥ 5 n_i \ge5 n i ≥ 5 G i = A n i G_i = A_{n_i} G i = A n i S n i S_{n_i} S n i L i L_i L i δ i ∈ Q ∗ / Q ∗ 2 \delta_i \in\mathbb{Q}^{*}/\mathbb{Q}^{*2} δ i ∈ Q ∗ / Q ∗ 2 R = { ( e i ) ∈ F 2 r : ∏ i δ i e i = 1 } R = \{(e_i) \in\mathbb{F}_2^r : \prod_i \delta_i^{e_i} = 1\} R = {( e i ) ∈ F 2 r : ∏ i δ i e i = 1 }
Gal ( L 1 ⋯ L r / Q ) = { ( g i ) ∈ ∏ i G i : ∏ i sgn ( g i ) e i = 1 for every ( e i ) ∈ R } . \operatorname{Gal}(L_1\cdots L_r/\mathbb{Q}) =
\left\{(g_i) \in\prod_i G_i : \prod_i \operatorname{sgn}(g_i)^{e_i} = 1 \text{ for every } (e_i) \in R\right\}. Gal ( L 1 ⋯ L r / Q ) = { ( g i ) ∈ i ∏ G i : i ∏ sgn ( g i ) e i = 1 for every ( e i ) ∈ R } . If the δ i \delta_i δ i a a a [ L 1 ⋯ L r : Q ] = 2 a ∏ i ( n i ! / 2 ) [L_1\cdots L_r:\mathbb{Q}] = 2^a \prod_i(n_i!/2) [ L 1 ⋯ L r : Q ] = 2 a ∏ i ( n i ! /2 )
Proof. The compositum group G G G G i G_i G i G ′ G' G ′ A n i A_{n_i} A n i G ′ = ∏ i A n i G' = \prod_i A_{n_i} G ′ = ∏ i A n i Q ( δ 1 , … , δ r ) \mathbb{Q}(\sqrt{\delta_1},\ldots,\sqrt{\delta_r}) Q ( δ 1 , … , δ r ) R R R
Thus, for distinct n i ≥ N n_i \ge N n i ≥ N s i ≥ 1 s_i \ge1 s i ≥ 1 H n i , s i H_{n_i,s_i} H n i , s i
Algebraic relations among selected roots
The criterion of Bary-Soroker–Kadets [4 Definition 1.6 and Theorem 1.7] extends their application with fixed Laguerre parameters as follows.
Corollary 6.8. For any sequence of positive integers ( s n ) n ≥ 1 (s_n)_{n\ge1} ( s n ) n ≥ 1 S = { z ∈ Q ‾ : H n , s n ( z ) = 0 for some n ≥ 1 } S = \{z \in\overline{\mathbb{Q}} : H_{n,s_n}(z) = 0 \text{ for some } n \ge1\} S = { z ∈ Q : H n , s n ( z ) = 0 for some n ≥ 1 } m ≥ 1 m \ge1 m ≥ 1 B ⊂ S m B \subset S^m B ⊂ S m B B B A m \mathbb{A}^m A m X i = X j X_i = X_j X i = X j i ≠ j i \ne j i = j X i = c X_i = c X i = c c ∈ S c \in S c ∈ S S 2 S^2 S 2 S S S
Proof. For n ≥ N n \ge N n ≥ N n n n A n A_n A n S n S_n S n S S S
Products of Jacobians
Theorem 6.10. Let K K K f 1 , … , f r ∈ K [ X ] f_1,\ldots,f_r \in K[X] f 1 , … , f r ∈ K [ X ] n i ≥ 5 n_i \ge5 n i ≥ 5 Gal ( f i / K ) ⊇ A n i \operatorname{Gal}(f_i/K) \supseteq A_{n_i} Gal ( f i / K ) ⊇ A n i A i A_i A i y 2 = f i ( x ) y^2=f_i(x) y 2 = f i ( x ) B = ∏ i A i e i B = \prod_i A_i^{e_i} B = ∏ i A i e i e i ≥ 1 e_i \ge1 e i ≥ 1
End K ‾ ( A i ) = Z , Hom K ‾ ( A i , A j ) = 0 ( i ≠ j ) , End K ‾ ( B ) ≃ ∏ i M e i ( Z ) . \operatorname{End}_{\overline{K}}(A_i)=\mathbb{Z}, \qquad
\operatorname{Hom}_{\overline{K}}(A_i,A_j)=0\ (i\ne j), \qquad
\operatorname{End}_{\overline{K}}(B)\simeq\prod_i M_{e_i}(\mathbb{Z}). End K ( A i ) = Z , Hom K ( A i , A j ) = 0 ( i = j ) , End K ( B ) ≃ i ∏ M e i ( Z ) . Fix K ↪ C K \hookrightarrow\mathbb{C} K ↪ C ψ i \psi_i ψ i V i = H 1 ( A i ( C ) , Q ) V_i = H_1(A_i(\mathbb{C}), \mathbb{Q}) V i = H 1 ( A i ( C ) , Q )
Hg ( B ) = ∏ i Sp ( V i , ψ i ) , \operatorname{Hg}(B) = \prod_i \operatorname{Sp}(V_i,\psi_i), Hg ( B ) = i ∏ Sp ( V i , ψ i ) , MT ( B ) = G : = { ( u i ) ∈ ∏ i GSp ( V i , ψ i ) : ν ( u 1 ) = ⋯ = ν ( u r ) } , \operatorname{MT}(B) = \mathcal{G} := \left\{(u_i) \in\prod_i \operatorname{GSp}(V_i,\psi_i) : \nu(u_1) = \cdots= \nu(u_r)\right\}, MT ( B ) = G := { ( u i ) ∈ i ∏ GSp ( V i , ψ i ) : ν ( u 1 ) = ⋯ = ν ( u r ) } , where ν \nu ν ℓ \ell ℓ G Q ℓ \mathcal{G}_{\mathbb{Q}_{\ell}} G Q ℓ
Q ℓ Id ⊕ ⨁ i s p ( V ℓ ( A i ) , ψ i , ℓ ) , V ℓ ( A i ) = T ℓ ( A i ) ⊗ Q ℓ . \mathbb{Q}_{\ell}\operatorname{Id} \oplus\bigoplus_i \mathfrak{sp}(V_{\ell}(A_i),\psi_{i,\ell}), \qquad V_{\ell}(A_i) = T_{\ell}(A_i) \otimes\mathbb{Q}_{\ell}. Q ℓ Id ⊕ i ⨁ sp ( V ℓ ( A i ) , ψ i , ℓ ) , V ℓ ( A i ) = T ℓ ( A i ) ⊗ Q ℓ . The Mumford–Tate conjecture holds for B B B B B B B B B ℓ \ell ℓ K K K
Proof. The endomorphism and homomorphism assertions follow from [48 Theorem 1.1 and Corollary 1.4] , including adjacent degrees 2 g + 1 , 2 g + 2 2g+1, 2g+2 2 g + 1 , 2 g + 2 ℓ \ell ℓ [47 Theorem 2.6] . Lombardo’s product criterion [35 Theorem 4.1] now applies: the symplectic algebras are simple, their standard modules are irreducible, and every automorphism preserves the standard module. Indeed, a proper semisimple subdirect product would identify two simple factors, yielding an intertwiner of their Tate modules after a finite extension. Faltings’s homomorphism theorem [19 ] would contradict geometric Hom-vanishing. Thus the derived Lie algebra is the full direct sum. Polarizations force a common scalar line, and the cyclotomic character supplies that line.
The algebraic monodromy group lies in the connected group G Q ℓ \mathcal{G}_{\mathbb{Q}_{\ell}} G Q ℓ ℓ \ell ℓ MT ( B ) = G \operatorname{MT}(B) = \mathcal{G} MT ( B ) = G [47 proof of Theorem 9.1] . Repeated factors merely repeat the representations.
Finally, invariants of each symplectic factor are generated by pairwise contractions. In cohomology these are divisor classes, obtained from polarizations by pullback and addition between repeated copies; see [47 Theorems 8.1 and 9.1] . The same argument applies to every power. Finite extensions of K K K G = G m ⋅ ∏ i Sp ( V i ) \mathcal{G} = \mathbb{G}_{m} \cdot\prod_i \operatorname{Sp}(V_i) G = G m ⋅ ∏ i Sp ( V i )
Corollary 6.11. For pairwise distinct n i ≥ N n_i \ge N n i ≥ N s i , e i s_i,e_i s i , e i ∏ i J n i , s i e i \prod_i J_{n_i,s_i}^{e_i} ∏ i J n i , s i e i J 2 g + 1 , s J_{2g+1,s} J 2 g + 1 , s J 2 g + 2 , t J_{2g+2,t} J 2 g + 2 , t s , t ≥ 1 s,t \ge1 s , t ≥ 1 2 g + 1 ≥ N 2g+1 \ge N 2 g + 1 ≥ N
Proof. Apply Theorem 1.1. □
Further questions
Theorem 1.1 gives a uniform eventual form of Hajir’s conjectures. Corollary 5.6 reduces both all-degree conjectures to an effective finite verification. The remaining question is whether that finite exceptional set is empty. The current bounds do not make its enumeration practical. An all-degree proof needs estimates that remain useful when the degree is comparable to the constants in the local and logarithmic arguments. Equivalently, one may ask for irreducibility and alternating containment for every nonconstant numerator and denominator in the exponential Padé table.
The finite-fiber theorem suggests a broader irreducibility problem. For fixed positive integers c 1 , … , c r c_1,\ldots,c_r c 1 , … , c r
P n , s ( X ) = n ! [ z n ] e X z ∏ ν = 1 r ( 1 − c ν z ) − s . (122) P_{n,s}(X) = n![z^n]e^{Xz}\prod_{\nu=1}^{r}(1-c_\nu z)^{-s}. \tag*{(122)} P n , s ( X ) = n ! [ z n ] e X z ν = 1 ∏ r ( 1 − c ν z ) − s . (122) Does there exist an effective N ( c 1 , … , c r ) N(c_1,\ldots,c_r) N ( c 1 , … , c r ) P n , s P_{n,s} P n , s n ≥ N ( c 1 , … , c r ) n\ge N(c_1,\ldots,c_r) n ≥ N ( c 1 , … , c r ) s s s s = 0 s=0 s = 0
[ s ] P n , s ( 0 ) = ( n − 1 ) ! ∑ ν = 1 r c ν n ≠ 0. [s]P_{n,s}(0)=(n-1)!\sum_{\nu=1}^{r}c_\nu^n\ne0. [ s ] P n , s ( 0 ) = ( n − 1 )! ν = 1 ∑ r c ν n = 0. The specialization problem needs more information. For each fixed c ≥ 2 c\ge2 c ≥ 2 E \mathcal{E} E P n , s ( c ) P_{n,s}^{(c)} P n , s ( c ) log s ≥ n ε \log s\ge n^\varepsilon log s ≥ n ε ε > 0 \varepsilon>0 ε > 0 1 ≤ s ≤ exp ( n δ ) 1\le s\le\exp(n^\delta) 1 ≤ s ≤ exp ( n δ ) δ > 0 \delta>0 δ > 0 n − 2 j n-2j n − 2 j
Theorem 4.3 and its Newton-polygon consequence isolate a different question. For trimmed exponentials, the remaining irreducibility hypothesis is removed here by an endpoint valuation condition. A uniform irreducibility theorem without that condition would give the full symmetric group throughout the same logarithmic parameter range. The nonsquare-discriminant theorem already has no upper bound on u u u u ≥ ( gcd ( u , 2 ) − 1 + η ) n u\ge(\gcd(u,2)-1+\eta)n u ≥ ( g cd( u , 2 ) − 1 + η ) n [3 ] ; the current argument neither proves irreducibility for every endpoint nor covers every relation between u u u n n n ( log n ) κ (\log n)^\kappa ( log n ) κ κ > 5 \kappa>5 κ > 5 log n \log n log n
Finally, Theorem 4.21 is geometric, whereas the application to quadratic subfields also uses arithmetic bounds for the interpolants. It would be useful to obtain such bounds for other covers from explicit ramification and height data. The Hermite argument here uses integral derivative identities and Gaussian quadrature, so those bounds are not supplied by the geometric theorem alone. For higher cyclic degree, one must also control all relevant embeddings and the possible effect of units; a small value at one embedding no longer gives the imaginary-quadratic contradiction used in Section 4.4.4. A quantitative interpolation theorem with a second arithmetic application would substantially extend this part of the method.
The root-modulus representation
We prove Theorem 2.5 by continuing the roots through the upper half of the parameter plane. At each parameter 0 , − 1 , … , 1 − n 0,-1,\ldots,1-n 0 , − 1 , … , 1 − n
Lemma A.1. Let 1 − n < s < 0 1-n<s<0 1 − n < s < 0 H n , s H_{n,s} H n , s n + s − 1 n+s-1 n + s − 1
Proof. The matrix argument in Lemma 2.1 remains valid for the upper bound even when D 0 D_0 D 0 x x x z 2 = x z^2=x z 2 = x ℜ z ≠ 0 \Re z\ne0 ℜ z = 0 ∣ x ∣ = v ∗ D 0 v < n + s − 1 |x|=v^*D_0v<n+s-1 ∣ x ∣ = v ∗ D 0 v < n + s − 1
Put A = n + s − 1 ∈ ( 0 , n − 1 ) A=n+s-1\in(0,n-1) A = n + s − 1 ∈ ( 0 , n − 1 ) κ = n − A / 2 > A / 2 \kappa=n-A/2>A/2 κ = n − A /2 > A /2 ρ 1 < ρ 2 \rho_1 < \rho_2 ρ 1 < ρ 2 A A A u ˉ \bar{u} u ˉ
sin θ ∫ ρ 1 ρ 2 ( κ ρ − cos θ 2 ) ∣ u ∣ 2 d ρ = 0. \sin\theta\int_{\rho_1}^{\rho_2}\left(\frac{\kappa}{\rho}-\frac{\cos\theta}{2}\right)|u|^2\,d\rho=0. sin θ ∫ ρ 1 ρ 2 ( ρ κ − 2 cos θ ) ∣ u ∣ 2 d ρ = 0. The parenthesized function is strictly positive, since ρ < A < 2 κ \rho< A < 2\kappa ρ < A < 2 κ u u u − ∫ ∣ u ′ ∣ 2 + ∫ V ∣ u ∣ 2 = 0 -\int|u'|^2+\int V|u|^2=0 − ∫ ∣ u ′ ∣ 2 + ∫ V ∣ u ∣ 2 = 0 V < 0 V<0 V < 0
Proof of Theorem 2.5. Continuation and boundary labels. The roots are simple and nonzero when ℑ s > 0 \Im s>0 ℑ s > 0 n n n { 0 , − 1 , … , 1 − n } \{0,-1,\ldots,1-n\} { 0 , − 1 , … , 1 − n }
We describe their boundary labels explicitly. For 0 ≤ m ≤ n − 1 0\le m\le n-1 0 ≤ m ≤ n − 1 N = n − m N=n-m N = n − m
H n , − m ( X ) = X N J m , n ( X ) , J m , n ( 0 ) = ( − 1 ) m n ! N ! . (123) H_{n,-m}(X)=X^N J_{m,n}(X),\qquad J_{m,n}(0)=(-1)^m\frac{n!}{N!}. \tag*{(123)} H n , − m ( X ) = X N J m , n ( X ) , J m , n ( 0 ) = ( − 1 ) m N ! n ! . (123) The m m m J m , n J_{m,n} J m , n
J m , n ( X ) = ( − 1 ) m e X X − N d m d X m ( e − X X n ) . J_{m,n}(X)=(-1)^m e^X X^{-N}\frac{d^m}{dX^m}\left(e^{-X}X^n\right). J m , n ( X ) = ( − 1 ) m e X X − N d X m d m ( e − X X n ) . Integration by parts shows that this monic degree-m m m X N e − X X^N e^{-X} X N e − X ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) m m m J m , n J_{m,n} J m , n m m m m = 0 m=0 m = 0
Write s = − m + δ s=-m+\delta s = − m + δ N N N
X N = − c m δ ( 1 + o ( 1 ) ) , c m = m ! ( N − 1 ) ! N ! n ! > 0. (124) X^N=-c_m\delta(1+o(1)),\qquad c_m=\frac{m!(N-1)!N!}{n!}>0. \tag*{(124)} X N = − c m δ ( 1 + o ( 1 )) , c m = n ! m ! ( N − 1 )! N ! > 0. (124) To see this, balance the coefficient J m , n ( 0 ) J_{m,n}(0) J m , n ( 0 ) X N X^N X N H n , − m + δ ( 0 ) = ( − 1 ) m m ! ( N − 1 ) ! δ + O ( δ 2 ) H_{n,-m+\delta}(0)=(-1)^m m!(N-1)!\delta+O(\delta^2) H n , − m + δ ( 0 ) = ( − 1 ) m m ! ( N − 1 )! δ + O ( δ 2 ) ρ = δ 1 / N \rho=\delta^{1/N} ρ = δ 1/ N ρ − N H n , − m + ρ N ( ρ w ) \rho^{-N}H_{n,-m+\rho^N}(\rho w) ρ − N H n , − m + ρ N ( ρw ) ρ = 0 \rho=0 ρ = 0 J m , n ( 0 ) ( w N + c m ) J_{m,n}(0)(w^N+c_m) J m , n ( 0 ) ( w N + c m ) 0 ≤ arg δ ≤ π 0\le\arg\delta\le\pi 0 ≤ arg δ ≤ π δ 1 / N \delta^{1/N} δ 1/ N − m -m − m ( 2 j + 1 ) π / N (2j+1)\pi/N ( 2 j + 1 ) π / N 0 ≤ j < N 0\le j<N 0 ≤ j < N π / N \pi/N π / N 2 π 2\pi 2 π − m -m − m δ = − r N \delta=-r^N δ = − r N X = r w X=rw X = r w r > 0 r>0 r > 0 w w w
On ( − m , − m + 1 ) (-m,-m+1) ( − m , − m + 1 ) 1 ≤ m ≤ n − 1 1\le m\le n-1 1 ≤ m ≤ n − 1 ( s ) m (s)_m ( s ) m m m m m m m m m m m m m
Near s = 0 + s=0+ s = 0 + x 0 , … , x n − 1 x_0,\ldots,x_{n-1} x 0 , … , x n − 1 ( 0 , 2 π ) (0,2\pi) ( 0 , 2 π ) s > 0 s>0 s > 0 j j j n − 1 − j n-1-j n − 1 − j 2 π 2\pi 2 π
These observations give an induction across 0 , − 1 , … , 1 − n 0,-1,\ldots,1-n 0 , − 1 , … , 1 − n ( − m , − m + 1 ) (-m,-m+1) ( − m , − m + 1 ) N = n − m N=n-m N = n − m θ r = ℑ log x r \theta_r=\Im\log x_r θ r = ℑ log x r
0 < θ 0 < ⋯ < θ N − 1 < 2 π , θ r + θ N − 1 − r = 2 π ( 0 ≤ r < N ) , 0<\theta_0<\cdots<\theta_{N-1}<2\pi,\qquad\theta_r+\theta_{N-1-r}=2\pi\qquad(0\le r<N), 0 < θ 0 < ⋯ < θ N − 1 < 2 π , θ r + θ N − 1 − r = 2 π ( 0 ≤ r < N ) , θ N = ⋯ = θ n − 1 = 2 π . (125) \theta_N=\cdots=\theta_{n-1}=2\pi. \tag*{(125)} θ N = ⋯ = θ n − 1 = 2 π . (125) Indeed, the m m m 2 π 2\pi 2 π s = − m s=-m s = − m ( 2 r + 1 ) π / N (2r+1)\pi/N ( 2 r + 1 ) π / N 2 ( r + 1 ) π / N 2(r+1)\pi/N 2 ( r + 1 ) π / N N − 1 N-1 N − 1 s = 1 − n s=1-n s = 1 − n 2 π 2\pi 2 π s < 1 − n s<1-n s < 1 − n
Continuing the modulus. For an original conjugate pair j , k = n − 1 − j j,k=n-1-j j , k = n − 1 − j j < k j<k j < k
q ( s ) = exp ( log x j ( s ) + log x k ( s ) − 2 π i 2 ) . (126) q(s)=\exp\left(\frac{\log x_j(s)+\log x_k(s)-2\pi i}{2}\right). \tag*{(126)} q ( s ) = exp ( 2 log x j ( s ) + log x k ( s ) − 2 π i ) . (126) On s > 0 s>0 s > 0 ∣ x j ( s ) ∣ |x_j(s)| ∣ x j ( s ) ∣ x j , x k x_j,x_k x j , x k [ 2 π , 4 π ] [2\pi,4\pi] [ 2 π , 4 π ] N N N θ j + θ N − 1 − j = 2 π \theta_j+\theta_{N-1-j}=2\pi θ j + θ N − 1 − j = 2 π k = n − 1 − j ≥ N − 1 − j k=n-1-j\ge N-1-j k = n − 1 − j ≥ N − 1 − j θ k ≥ θ N − 1 − j \theta_k\ge\theta_{N-1-j} θ k ≥ θ N − 1 − j 2 π 2\pi 2 π 2 π 2\pi 2 π q q q [ 0 , π ] [0,\pi] [ 0 , π ]
For odd n n n j = ( n − 1 ) / 2 j=(n-1)/2 j = ( n − 1 ) /2 [ π , 2 π ) [\pi,2\pi) [ π , 2 π ) 2 π 2\pi 2 π q = − x j q=-x_j q = − x j q q q
The expansion at infinity. Put z = s − 1 / 2 z=s^{-1/2} z = s − 1/2 − π / 2 ≤ arg z ≤ 0 -\pi/2\le\arg z\le0 − π /2 ≤ arg z ≤ 0 s s s F n ( w , z ) F_n(w,z) F n ( w , z ) h n ( w ) h_n(w) h n ( w ) w j ( z ) w_j(z) w j ( z )
For positive z z z w w w F n ( − w , − z ) = ( − 1 ) n F n ( w , z ) F_n(-w,-z)=(-1)^nF_n(w,z) F n ( − w , − z ) = ( − 1 ) n F n ( w , z ) w k ( z ) = − w j ( − z ) w_k(z)=-w_j(-z) w k ( z ) = − w j ( − z ) x j ( z ) = z − 1 w j ( z ) − z − 2 x_j(z)=z^{-1}w_j(z)-z^{-2} x j ( z ) = z − 1 w j ( z ) − z − 2 x k ( z ) = x j ( − z ) x_k(z)=x_j(-z) x k ( z ) = x j ( − z ) z − 4 = s 2 z^{-4}=s^2 z − 4 = s 2 s s s s + θ + O ( s − 1 ) s+\theta+O(s^{-1}) s + θ + O ( s − 1 ) θ ∈ R \theta\in\mathbb{R} θ ∈ R w w w z z z q = − x j q=-x_j q = − x j
The germs and the already continued functions agree on the positive axis and extend across it; the identity theorem identifies them in the exterior upper half-plane. Convergence of the full germ expansions gives, uniformly in that closed half-plane as ∣ s ∣ → ∞ |s|\to\infty ∣ s ∣ → ∞
q ( s ) = s + θ + O ( s − 1 ) , θ ∈ R . (127) q(s)=s+\theta+O(s^{-1}),\qquad\theta\in\mathbb{R}. \tag*{(127)} q ( s ) = s + θ + O ( s − 1 ) , θ ∈ R . (127) Cauchy’s formula and the measure. The root functions have convergent expansions in ( s + m ) 1 / N (s+m)^{1/N} ( s + m ) 1/ N − m -m − m q q q q q q q ( s ) = O ( ∣ s ∣ 1 / n ) q(s) = O(|s|^{1/n}) q ( s ) = O ( ∣ s ∣ 1/ n )
The boundary values of q q q s > 0 s > 0 s > 0 s < 1 − n s < 1-n s < 1 − n C ∖ [ 1 − n , 0 ] \mathbb{C} \setminus[1-n,0] C ∖ [ 1 − n , 0 ] ψ ( t ) = ℑ q ( − t + i 0 ) \psi(t) = \Im q(-t+i0) ψ ( t ) = ℑ q ( − t + i 0 ) ψ ≥ 0 \psi\ge0 ψ ≥ 0 ψ \psi ψ ψ ( t ) = O ( t 1 / n ) \psi(t) = O(t^{1/n}) ψ ( t ) = O ( t 1/ n ) t ↓ 0 t \downarrow0 t ↓ 0
Apply Cauchy’s formula to q ( s ) − s − θ q(s)-s-\theta q ( s ) − s − θ 2 i ψ ( t ) 2i\psi(t) 2 i ψ ( t )
q ( s ) = s + θ − 1 π ∫ 0 n − 1 ψ ( t ) s + t d t ( s ∈ C ∖ [ 1 − n , 0 ] ) . (128) q(s) = s+\theta-\frac{1}{\pi}\int_{0}^{n-1}\frac{\psi(t)}{s+t}\,dt \qquad(s \in\mathbb{C}\setminus[1-n,0]). \tag*{(128)} q ( s ) = s + θ − π 1 ∫ 0 n − 1 s + t ψ ( t ) d t ( s ∈ C ∖ [ 1 − n , 0 ]) . (128) Because ψ ( t ) / t = O ( t 1 / n − 1 ) \psi(t)/t = O(t^{1/n-1}) ψ ( t ) / t = O ( t 1/ n − 1 ) s ↓ 0 s \downarrow0 s ↓ 0
θ = 1 π ∫ 0 n − 1 ψ ( t ) t d t . \theta= \frac{1}{\pi}\int_{0}^{n-1}\frac{\psi(t)}{t}\,dt. θ = π 1 ∫ 0 n − 1 t ψ ( t ) d t . Hence (22) holds with the finite nonnegative measure
d μ ( t ) = ψ ( t ) π t d t . (129) d\mu(t) = \frac{\psi(t)}{\pi t}\,dt. \tag*{(129)} d μ ( t ) = π t ψ ( t ) d t . (129) There are no endpoint atoms: the formula is obtained from the full contour, and every indentation integral tends to zero. If two finite measures on [ 0 , n − 1 ] [0,n-1] [ 0 , n − 1 ] s s s C [ 0 , n − 1 ] C[0,n-1] C [ 0 , n − 1 ] s / ( s + t ) s/(s+t) s / ( s + t ) s > 0 s>0 s > 0
Proposition A.2. Fix n ≥ 2 n \ge2 n ≥ 2 0 ≤ j ≤ ⌊ ( n − 1 ) / 2 ⌋ 0 \le j \le\lfloor(n-1)/2\rfloor 0 ≤ j ≤ ⌊( n − 1 ) /2 ⌋
α j ( s ) = ( ( n − 1 ) ! s ) 1 / n exp ( ( 2 j + 1 ) π i n ) + o ( s 1 / n ) ( s ↓ 0 ) , \alpha_j(s) = ((n-1)!s)^{1/n}\exp\left(\frac{(2j+1)\pi i}{n}\right)+o(s^{1/n}) \qquad(s \downarrow0), α j ( s ) = (( n − 1 )! s ) 1/ n exp ( n ( 2 j + 1 ) π i ) + o ( s 1/ n ) ( s ↓ 0 ) , and continue it along the positive real axis. Put q j ( s ) = ∣ α j ( s ) ∣ q_j(s) = |\alpha_j(s)| q j ( s ) = ∣ α j ( s ) ∣ L j = n − 1 − j L_j = n-1-j L j = n − 1 − j
q j ( s ) = s + ∫ s s + t d μ j ( t ) q_j(s) = s+\int\frac{s}{s+t}\,d\mu_j(t) q j ( s ) = s + ∫ s + t s d μ j ( t ) has support exactly [ 0 , L j ] [0,L_j] [ 0 , L j ] ( m , m + 1 ) (m,m+1) ( m , m + 1 ) 0 ≤ m < L j 0 \le m < L_j 0 ≤ m < L j t > L j t>L_j t > L j
M j = n − 1 3 + ξ j 2 6 , α j ( s ) = − s + i ξ j s + O ( 1 ) ( s ⟶ ∞ ) , M_j = \frac{n-1}{3}+\frac{\xi_j^2}{6}, \qquad\alpha_j(s)=-s+i\xi_j\sqrt{s}+O(1) \qquad(s\longrightarrow\infty), M j = 3 n − 1 + 6 ξ j 2 , α j ( s ) = − s + i ξ j s + O ( 1 ) ( s ⟶ ∞ ) , then
M j s s + L j < q j ( s ) − s < M j ( s > 0 ) . (130) M_j\frac{s}{s+L_j}<q_j(s)-s<M_j \qquad(s>0). \tag*{(130)} M j s + L j s < q j ( s ) − s < M j ( s > 0 ) . (130) For odd n n n j = ( n − 1 ) / 2 j=(n-1)/2 j = ( n − 1 ) /2 ξ j = 0 \xi_j=0 ξ j = 0 L j = ( n − 1 ) / 2 L_j=(n-1)/2 L j = ( n − 1 ) /2
Proof. On ( − m , − m + 1 ) (-m,-m+1) ( − m , − m + 1 ) 1 ≤ m ≤ n − 1 1 \le m \le n-1 1 ≤ m ≤ n − 1 N = n − m N=n-m N = n − m j < k = n − 1 − j j<k=n-1-j j < k = n − 1 − j j , k < N j,k<N j , k < N k > N − 1 − j k>N-1-j k > N − 1 − j m > 0 m>0 m > 0
2 π < θ j + θ k < 4 π . 2\pi<\theta_j+\theta_k<4\pi. 2 π < θ j + θ k < 4 π . If j < N ≤ k j<N\le k j < N ≤ k θ j + 2 π \theta_j+2\pi θ j + 2 π N ≤ j N\le j N ≤ j . T h e h o l o m o r p h i c m o d u l u s b r a n c h h a s c o n t i n u o u s a r g u m e n t . The holomorphic modulus branch has continuous argument . T h e h o l o m or p hi c m o d u l u s b r an c hha sco n t in u o u s a r g u m e n t m ≤ n − 1 − j m \le n - 1 - j m ≤ n − 1 − j m ≥ n − j m \ge n - j m ≥ n − j s = − t s = -t s = − t 0 < t < L j 0 < t < L_j 0 < t < L j t > L j t > L_j t > L j
If n n n j = ( n − 1 ) / 2 j = (n - 1)/2 j = ( n − 1 ) /2 q j = − α j q_j = -\alpha_j q j = − α j j < N j < N j < N j > ( N − 1 ) / 2 j > (N - 1)/2 j > ( N − 1 ) /2 π < θ j < 2 π \pi< \theta_j < 2\pi π < θ j < 2 π N ≤ j N \le j N ≤ j 2 π 2\pi 2 π t = j = L j t = j = L_j t = j = L j
In either case the representation obtained from Cauchy’s formula is
d μ j ( t ) = Im q j ( − t + i 0 ) π t d t . d\mu_j(t) = \frac{\operatorname{Im} q_j(-t + i0)}{\pi t}\,dt. d μ j ( t ) = π t Im q j ( − t + i 0 ) d t . The density is positive on every indicated open interval. It vanishes past L j L_j L j [ 0 , L j ] [0,L_j] [ 0 , L j ]
s s + L j ≤ s s + t ≤ 1 \frac{s}{s+L_j} \le\frac{s}{s+t} \le1 s + L j s ≤ s + t s ≤ 1 and using the already established mass μ j ( [ 0 , L j ] ) = M j \mu_j([0,L_j]) = M_j μ j ([ 0 , L j ]) = M j ( 0 , L j ) (0,L_j) ( 0 , L j ) □ \square □
Appendix B. An exact certificate for the Appell family
The two resultants D i ( N ) = D i ( N , 2 ) D_i(N) = D_i(N,2) D i ( N ) = D i ( N , 2 ) N N N D 0 ( 11 ) D 1 ( 11 ) ≠ 0 D_0(11)D_1(11) \ne0 D 0 ( 11 ) D 1 ( 11 ) = 0 F 479 \mathbb{F}_{479} F 479 D 0 , D 1 D_0,D_1 D 0 , D 1 N N N N = 11 N = 11 N = 11
R 2 ( 11 , X ) = X 4 − 66 X 3 + 1430 X 2 − 11880 X + 31680 , R_2(11,X) = X^4 - 66X^3 + 1430X^2 - 11880X + 31680, R 2 ( 11 , X ) = X 4 − 66 X 3 + 1430 X 2 − 11880 X + 31680 , R 3 ( 11 , X ) = X 6 − 99 X 5 + 3630 X 4 − 62370 X 3 + 522720 X 2 − 1995840 X + 2661120. R_3(11,X) = X^6 - 99X^5 + 3630X^4 - 62370X^3 + 522720X^2 - 1995840X + 2661120. R 3 ( 11 , X ) = X 6 − 99 X 5 + 3630 X 4 − 62370 X 3 + 522720 X 2 − 1995840 X + 2661120. The first polynomial has four distinct roots
{ 72 , 180 , 353 , 419 } ⊂ F 479 . \{72, 180, 353, 419\} \subset\mathbb{F}_{479}. { 72 , 180 , 353 , 419 } ⊂ F 479 . Its nonempty proper subsets have fourteen distinct sums, namely
S = { 12 , 54 , 72 , 120 , 126 , 180 , 192 , 252 , 293 , 353 , 365 , 419 , 425 , 473 } . S = \{12, 54, 72, 120, 126, 180, 192, 252, 293, 353, 365, 419, 425, 473\}. S = { 12 , 54 , 72 , 120 , 126 , 180 , 192 , 252 , 293 , 353 , 365 , 419 , 425 , 473 } . These are the roots of V 2 ( 11 , Z ) V_2(11,Z) V 2 ( 11 , Z ) D 0 ( 11 ) D_0(11) D 0 ( 11 ) D 1 ( 11 ) D_1(11) D 1 ( 11 ) z / 6 z/6 z /6 z / 3 z/3 z /3 z ∈ S z \in S z ∈ S V 3 V_3 V 3 V 3 ( 11 , ⋅ ) V_3(11,\mathord\cdot) V 3 ( 11 , ⋅ )
Every entry is nonzero. Each specialized pair is therefore relatively prime. The four leading coefficients are 6 14 6^{14} 6 14 9 62 9^{62} 9 62 3 14 3^{14} 3 14 6 62 6^{62} 6 62
The table can be reproduced using matrices of order at most 20. Let C C C
X 6 + 380 X 5 + 277 X 4 + 379 X 3 + 131 X 2 + 153 X + 275 over F 479 . X^6 + 380X^5 + 277X^4 + 379X^3 + 131X^2 + 153X + 275 \quad\text{over } \mathbb{F}_{479}. X 6 + 380 X 5 + 277 X 4 + 379 X 3 + 131 X 2 + 153 X + 275 over F 479 . For 1 ≤ k ≤ 5 1 \le k \le5 1 ≤ k ≤ 5 C [ k ] C^{[k]} C [ k ] ⋀ k F 479 6 \bigwedge^k \mathbb{F}_{479}^6 ⋀ k F 479 6
C [ k ] ( v 1 ∧ ⋯ ∧ v k ) = ∑ j = 1 k v 1 ∧ ⋯ ∧ C v j ∧ ⋯ ∧ v k . C^{[k]}(v_1 \wedge\cdots\wedge v_k) = \sum_{j=1}^{k} v_1 \wedge\cdots\wedge Cv_j \wedge\cdots\wedge v_k. C [ k ] ( v 1 ∧ ⋯ ∧ v k ) = j = 1 ∑ k v 1 ∧ ⋯ ∧ C v j ∧ ⋯ ∧ v k . The eigenvalues of this additive exterior-power operator are the k k k C C C
V 3 ( 11 , Z ) = ∏ k = 1 5 det ( Z I − C [ k ] ) . V_3(11,Z) = \prod_{k=1}^{5} \det(ZI-C^{[k]}). V 3 ( 11 , Z ) = k = 1 ∏ 5 det ( Z I − C [ k ] ) . Triangularization proves the identity without a separability assumption.
