Introduction
The zeros of classical orthogonal polynomials are among the most familiar algebraic numbers in study. Orthogonality places them on a real interval and makes them simple. However, their arithmetic is much less known.
An important class of orthogonal polynomials are Jacobi polynomials. More precisely, we use the normalization
P n ( α , β ) ( x ) = 2 − n ∑ j = 0 n ( n + α n − j ) ( n + β j ) ( x − 1 ) j ( x + 1 ) n − j , n ≥ 0 , (1) P_n^{(\alpha,\beta)}(x)=2^{-n}\sum_{j=0}^{n}\binom{n+\alpha}{n-j}\binom{n+\beta}{j}(x-1)^j(x+1)^{n-j},\qquad n\geq0, \tag*{(1)} P n ( α , β ) ( x ) = 2 − n j = 0 ∑ n ( n − j n + α ) ( j n + β ) ( x − 1 ) j ( x + 1 ) n − j , n ≥ 0 , (1) where ( z j ) = z ( z − 1 ) ⋯ ( z − j + 1 ) / j ! \binom{z}{j}=z(z-1)\cdots(z-j+1)/j! ( j z ) = z ( z − 1 ) ⋯ ( z − j + 1 ) / j ! ( z 0 ) = 1 \binom{z}{0}=1 ( 0 z ) = 1 α , β > − 1 \alpha,\beta>-1 α , β > − 1 n n n P n ( α , β ) ( 1 ) = ( n + α n ) P_n^{(\alpha,\beta)}(1)=\binom{n+\alpha}{n} P n ( α , β ) ( 1 ) = ( n n + α ) ( 1 − x ) α ( 1 + x ) β (1-x)^\alpha(1+x)^\beta ( 1 − x ) α ( 1 + x ) β ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) P n = P n ( 0 , 0 ) P_n=P_n^{(0,0)} P n = P n ( 0 , 0 ) P n ( 1 ) = 1 P_n(1)=1 P n ( 1 ) = 1 [26 Chapter IV] and [22 Chapter 18] .
For a nonconstant separable polynomial f ∈ K [ x ] f\in K[x] f ∈ K [ x ] Gal ( f / K ) \operatorname{Gal}(f/K) Gal ( f / K )
The irreducibility problem of Legendre polynomials goes back to Stieltjes in 19th century. In his letter to Hermite of 2 October 1890, he asked whether P 2 m P_{2m} P 2 m P 2 m + 1 / x P_{2m+1}/x P 2 m + 1 / x Q \mathbb{Q} Q [25 Letter 275] ; see also [7 §1] . The division by x x x [24 ] . For Legendre polynomials, Wahab obtained substantial infinite families of irreducible examples by Newton polygons [27 ] . Related questions for Bessel polynomials were settled through the work of Filaseta and Filaseta–Trifonov [13 , 14 ] .
Deng has recently proved Stieltjes’ conjecture in every degree [10 Theorem 1.1] . His argument compares the real size of a factor resultant with the size permitted by its p p p
Galois groups
The first family in our consideration comes from modular forms. Kaneko and Zagier constructed polynomials over Q \mathbb{Q} Q j j j [18 ] . After an affine change of variable and normalization, their Jacobi factors give the four families
P n ( λ / 3 , μ / 2 ) , λ , μ ∈ { − 1 , 1 } P_n^{(\lambda/3,\mu/2)}, \qquad\lambda,\mu\in\{-1,1\} P n ( λ /3 , μ /2 ) , λ , μ ∈ { − 1 , 1 } [8 Eq. (3)] . Thus their rational factorization describes the algebraic relations among the characteristic-zero lifts of these modular zeros.
Theorem 1.1 (The Mahlburg–Ono conjecture). For every n ≥ 1 n \ge1 n ≥ 1 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } P n ( λ / 3 , μ / 2 ) P_n^{(\lambda/3,\mu/2)} P n ( λ /3 , μ /2 ) Q \mathbb{Q} Q
Gal ( P n ( λ / 3 , μ / 2 ) / Q ) = S n . \operatorname{Gal}(P_n^{(\lambda/3,\mu/2)}/\mathbb{Q}) = S_n. Gal ( P n ( λ /3 , μ /2 ) / Q ) = S n . Mahlburg and Ono conjectured both assertions in 2004 and proved irreducibility in infinitely many degrees [19 p. 40 and Theorem 1.1] . Cullinan and Hajir proved further infinite families and established the implication from irreducibility to S n S_n S n ℓ ≤ 10 9 \ell\le10^9 ℓ ≤ 1 0 9 [8 Theorems 1.8–1.9] . Cullinan and Gajek-Leonard studied the associated Newton polygons [6 ] . Theorem 1.1 treats all four families in every positive degree. In § 6.4 we will give the precise normalization and explain the relation with modular differential equations. Appendix A gives stronger local irreducibility statements in prime-power degrees and in degrees 5 ⋅ 3 a 5 \cdot3^a 5 ⋅ 3 a
For Legendre polynomials, the symmetry of the zeros determines the largest possible group. Write
L 2 m + δ ( x ) = P 2 m + δ ( x ) / x δ , m ≥ 1 , δ ∈ { 0 , 1 } . L_{2m+\delta}(x) = P_{2m+\delta}(x)/x^\delta, \qquad m \ge1,\quad\delta\in\{0,1\}. L 2 m + δ ( x ) = P 2 m + δ ( x ) / x δ , m ≥ 1 , δ ∈ { 0 , 1 } . The 2 m 2m 2 m
W m = C 2 ≀ S m = F 2 m ⋊ S m . W_m = C_2 \wr S_m = \mathbb{F}_2^m \rtimes S_m. W m = C 2 ≀ S m = F 2 m ⋊ S m . Here S m S_m S m F 2 m \mathbb{F}_2^m F 2 m
Theorem 1.2 (The Legendre Galois groups). For every m ≥ 1 m \ge1 m ≥ 1 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 }
Gal ( P 2 m + δ ( x ) / x δ / Q ) = W m . \operatorname{Gal}(P_{2m+\delta}(x)/x^\delta/\mathbb{Q}) = W_m. Gal ( P 2 m + δ ( x ) / x δ / Q ) = W m . This proves both parts of [7 Conjecture 1.5] : the induced group on the pairs is S m S_m S m F 2 m \mathbb{F}_2^m F 2 m 10 9 10^9 1 0 9 [7 Theorem 1.9(a)] . Our argument removes that bound and determines the kernel in every degree.
Integral parameters
Now we consider parameters ( α , β ) (\alpha,\beta) ( α , β )
Theorem 1.3. Fix distinct nonnegative integers a , b a,b a , b N ( a , b ) N(a,b) N ( a , b ) n ≥ N ( a , b ) n \ge N(a,b) n ≥ N ( a , b ) P n ( a , b ) P_n^{(a,b)} P n ( a , b ) Q \mathbb{Q} Q S n S_n S n
Moreover, for every fixed number field E E E P n ( a , b ) P_n^{(a,b)} P n ( a , b ) E E E Q \mathbb{Q} Q n n n a , b , E a,b,E a , b , E E E E S n S_n S n
The restriction to sufficiently large degree is necessary, for example, take P 2 ( 4 , 1 ) ( x ) = 3 x ( 3 x + 2 ) P_2^{(4,1)}(x) = 3x(3x+2) P 2 ( 4 , 1 ) ( x ) = 3 x ( 3 x + 2 ) [9 ] have a different variation: the degree is fixed and a rational parameter varies.
Equal parameters connect the problem to derivatives. Define the Gegenbauer polynomials by
( 1 − 2 x t + t 2 ) − λ = ∑ j ≥ 0 C j λ ( x ) t j . (1-2xt+t^2)^{-\lambda} = \sum_{j \ge0} C_j^\lambda(x)t^j. ( 1 − 2 x t + t 2 ) − λ = j ≥ 0 ∑ C j λ ( x ) t j . For an integer r ≥ 0 r \ge0 r ≥ 0 C n r + 1 / 2 C_n^{r+1/2} C n r + 1/2 P n ( r , r ) P_n^{(r,r)} P n ( r , r ) [22 (18.7.1)] , and
d r d x r P n + r ( x ) = ( 2 r − 1 ) ! ! C n r + 1 / 2 ( x ) , ( 2 r − 1 ) ! ! = ∏ j = 1 r ( 2 j − 1 ) . (2) \frac{d^r}{dx^r}P_{n+r}(x)=(2r-1)!!\,C_n^{r+1/2}(x),\qquad(2r-1)!!=\prod_{j=1}^{r}(2j-1). \tag*{(2)} d x r d r P n + r ( x ) = ( 2 r − 1 )!! C n r + 1/2 ( x ) , ( 2 r − 1 )!! = j = 1 ∏ r ( 2 j − 1 ) . (2) The empty product is 1. A single superscript P ℓ ( r ) P_\ell^{(r)} P ℓ ( r )
F m , r , δ ( x ) = C 2 m + δ r + 1 / 2 ( x ) x δ , m ≥ 1 , δ ∈ { 0 , 1 } . (3) F_{m,r,\delta}(x)=\frac{C_{2m+\delta}^{r+1/2}(x)}{x^\delta},\qquad m\ge1,\quad\delta\in\{0,1\}. \tag*{(3)} F m , r , δ ( x ) = x δ C 2 m + δ r + 1/2 ( x ) , m ≥ 1 , δ ∈ { 0 , 1 } . (3) This polynomial has degree 2 m 2m 2 m x x x F m , r , δ F_{m,r,\delta} F m , r , δ
Theorem 1.4. For integers r ≥ 0 r \ge0 r ≥ 0 m ≥ 1 m \ge1 m ≥ 1 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } F m , r , δ F_{m,r,\delta} F m , r , δ Q \mathbb{Q} Q
m ≥ max { 100 r , 1000 } . m\ge\max\{100r,1000\}. m ≥ max { 100 r , 1000 } . There is also an effective absolute integer M 0 M_0 M 0 m ≥ max { 8 r , M 0 } m\ge\max\{8r,M_0\} m ≥ max { 8 r , M 0 }
The conclusion is uniform in the parameter: the derivative order may grow with m m m 0 ≤ r ≤ m / 8 0 \le r \le m/8 0 ≤ r ≤ m /8 m m m r = 0 r=0 r = 0 r = 1 r=1 r = 1 m m m
Theorem 1.5. For every m ≥ 1 m \ge1 m ≥ 1
P 2 m ( x ) , P 2 m + 1 ( x ) / x , P 2 m + 1 ′ ( x ) , P 2 m + 2 ′ ( x ) / x P_{2m}(x),\qquad P_{2m+1}(x)/x,\qquad P'_{2m+1}(x),\qquad P'_{2m+2}(x)/x P 2 m ( x ) , P 2 m + 1 ( x ) / x , P 2 m + 1 ′ ( x ) , P 2 m + 2 ′ ( x ) / x are irreducible over Q \mathbb{Q} Q
The first two assertions are Deng’s theorem [10 ] . The last two say that the nonzero critical points of each Legendre polynomial P ℓ P_\ell P ℓ ℓ ≥ 3 \ell\ge3 ℓ ≥ 3 Q \mathbb{Q} Q { P ℓ , P ℓ ′ : ℓ ≥ 1 } \{P_\ell,P'_\ell:\ell\ge1\} { P ℓ , P ℓ ′ : ℓ ≥ 1 } Corollary 6.2 ). This extends Deng’s common-zero result for the Legendre polynomials themselves [11 ] .
Rational parameters
We consider parameters ( α , β ) (\alpha,\beta) ( α , β )
Theorem 1.6. Fix α = a / u \alpha=a/u α = a / u β = b / v \beta=b/v β = b / v u , v > 0 u,v>0 u , v > 0 D D D α + β \alpha+\beta α + β
D is even , D ≥ max { u , v } , α , β ∉ { − 1 , − 2 , … } . D\ \text{is even},\qquad D\ge\max\{u,v\},\qquad\alpha,\beta\notin\{-1,-2,\ldots\}. D is even , D ≥ max { u , v } , α , β ∈ / { − 1 , − 2 , … } . For all sufficiently large n n n f n = P n ( α , β ) f_n=P_n^{(\alpha,\beta)} f n = P n ( α , β ) h n h_n h n M n > n / 2 M_n>n/2 M n > n /2
Gal ( h n / Q ) = S M n , n − M n = O α , β , A ( n ( log n ) A ) for every fixed A > 0. \operatorname{Gal}(h_n/\mathbb{Q})=S_{M_n},\qquad n-M_n=O_{\alpha,\beta,A}\left(\frac{n}{(\log n)^A}\right)\quad\text{for every fixed }A>0. Gal ( h n / Q ) = S M n , n − M n = O α , β , A ( ( log n ) A n ) for every fixed A > 0. Write f n = κ n h n g n f_n=\kappa_nh_ng_n f n = κ n h n g n g n g_n g n κ n ∈ Q × \kappa_n\in\mathbb{Q}^{\times} κ n ∈ Q × h n h_n h n g n g_n g n Q \mathbb{Q} Q
Gal ( f n / Q ) = S M n × Gal ( g n / Q ) . (4) \operatorname{Gal}(f_n/\mathbb{Q})=S_{M_n}\times\operatorname{Gal}(g_n/\mathbb{Q}). \tag*{(4)} Gal ( f n / Q ) = S M n × Gal ( g n / Q ) . (4) For every fixed c ∈ Q × c \in\mathbb{Q}^{\times} c ∈ Q × n n n h n ( 1 + c x 2 ) h_n(1+cx^2) h n ( 1 + c x 2 ) W M n W M_n W M n g n ( 1 + c x 2 ) g_n(1+cx^2) g n ( 1 + c x 2 )
Gal ( f n ( 1 + c x 2 ) / Q ) = W M n × Gal ( g n ( 1 + c x 2 ) / Q ) . (5) \operatorname{Gal}(f_n(1+cx^2)/\mathbb{Q}) = W M_n \times\operatorname{Gal}(g_n(1+cx^2)/\mathbb{Q}). \tag*{(5)} Gal ( f n ( 1 + c x 2 ) / Q ) = W M n × Gal ( g n ( 1 + c x 2 ) / Q ) . (5) The splitting fields of h n h_n h n h n ( 1 + c x 2 ) h_n(1+cx^2) h n ( 1 + c x 2 ) n n n
The threshold may depend on the fixed parameters, on c c c A A A Q \mathbb{Q} Q Q \mathbb{Q} Q
No orthogonality hypothesis is needed here. Some restriction on the parameters is necessary: the specialization α = β = − 1 / 2 \alpha=\beta=-1/2 α = β = − 1/2 [22 (18.7.3)] , whose zeros lie in cyclotomic fields and whose Galois groups are abelian. The small complementary factor g n g_n g n
The integral irreducibility theorem and the rational-parameter Galois theorem meet through a classical quadratic transformation. For δ = 0 , 1 \delta=0,1 δ = 0 , 1 F m , r , δ ( x ) F_{m,r,\delta}(x) F m , r , δ ( x )
P m ( δ − 1 / 2 , r ) ( 1 − 2 x 2 ) . P_m^{(\delta-1/2,r)}(1-2x^2). P m ( δ − 1/2 , r ) ( 1 − 2 x 2 ) . The parameters δ − 1 / 2 \delta-1/2 δ − 1/2 r r r D = 2 D=2 D = 2 F m , r , δ ( x ) = Q m , r , δ ( x 2 ) F_{m,r,\delta}(x)=Q_{m,r,\delta}(x^2) F m , r , δ ( x ) = Q m , r , δ ( x 2 ) Q m , r , δ ∈ Q [ t ] Q_{m,r,\delta}\in\mathbb{Q}[t] Q m , r , δ ∈ Q [ t ] m m m
Corollary 1.7. Fix an integer r ≥ 0 r\geq0 r ≥ 0 E E E m m m δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } F m , r , δ F_{m,r,\delta} F m , r , δ Q m , r , δ Q_{m,r,\delta} Q m , r , δ E E E
Gal ( F m , r , δ / E ) = W m , Gal ( Q m , r , δ / E ) = S m . \operatorname{Gal}(F_{m,r,\delta}/E)=W_m,\qquad\operatorname{Gal}(Q_{m,r,\delta}/E)=S_m. Gal ( F m , r , δ / E ) = W m , Gal ( Q m , r , δ / E ) = S m . Their splitting fields over Q \mathbb{Q} Q E E E Q \mathbb{Q} Q
The two groups describe the two natural sets of roots: S m S_m S m W m W_m W m r r r E E E r = 0 r=0 r = 0 E = Q E=\mathbb{Q} E = Q Theorem 1.2 gives every degree. For r = 1 r=1 r = 1
The resultant method
The idea is to estimate the resultant in two ways: the real geometry of the roots gives a lower bound, while their local arithmetic gives an upper bound. § 2 develops the common Jacobi form of this argument. It extends Deng’s comparison of real and p p p [10 §1.2] .
Fix rational parameters α , β > − 1 \alpha,\beta>-1 α , β > − 1 y n ( t ) = P n ( α , β ) ( 1 − 2 t ) y_n(t)=P_n^{(\alpha,\beta)}(1-2t) y n ( t ) = P n ( α , β ) ( 1 − 2 t ) Q \mathbb{Q} Q ( 0 , 1 ) (0,1) ( 0 , 1 ) H = κ y n H=\kappa y_n H = κ y n
H = A B , A , B ∈ Z [ t ] primitive , 1 ≤ d = deg A ≤ e = deg B . H=AB,\qquad A,B\in\mathbb{Z}[t]\text{ primitive},\qquad1\leq d=\deg A\leq e=\deg B. H = A B , A , B ∈ Z [ t ] primitive , 1 ≤ d = deg A ≤ e = deg B . The roots are simple by orthogonality [22 §18.2(vi)] , so the resultant R = Res ( A , B ) R=\operatorname{Res}(A,B) R = Res ( A , B )
log ∣ R ∣ = ∑ p v p ( R ) log p , (6) \log|R|=\sum_p v_p(R)\log p, \tag*{(6)} log ∣ R ∣ = p ∑ v p ( R ) log p , (6) where v p ( p ) = 1 v_p(p)=1 v p ( p ) = 1 d d d
The common input is an auxiliary polynomial constructed from orthogonality. Let I I I t α ( 1 − t ) β d t t^{\alpha}(1-t)^{\beta}\,\mathrm{d}t t α ( 1 − t ) β d t ( 0 , 1 ) (0,1) ( 0 , 1 ) I ( 1 ) = 1 I(1)=1 I ( 1 ) = 1
μ j = I ( t j ) = ( α + 1 ) j ( α + β + 2 ) j ∈ Q . \mu_j=I(t^j)=\frac{(\alpha+1)_j}{(\alpha+\beta+2)_j}\in\mathbb{Q}. μ j = I ( t j ) = ( α + β + 2 ) j ( α + 1 ) j ∈ Q . where ( u ) 0 = 1 (u)_0=1 ( u ) 0 = 1 ( u ) j = u ( u + 1 ) ⋯ ( u + j − 1 ) (u)_j=u(u+1)\cdots(u+j-1) ( u ) j = u ( u + 1 ) ⋯ ( u + j − 1 ) j ≥ 1 j\geq1 j ≥ 1 (9) . Define
Q H ( t ) = I s ( H ( t ) − H ( s ) t − s ) , σ ( t ) = t ( 1 − t ) . Q_H(t)=I_s\left(\frac{H(t)-H(s)}{t-s}\right),\qquad\sigma(t)=t(1-t). Q H ( t ) = I s ( t − s H ( t ) − H ( s ) ) , σ ( t ) = t ( 1 − t ) . The divided difference is a polynomial, so Q H ∈ Q [ t ] Q_H\in\mathbb{Q}[t] Q H ∈ Q [ t ] z z z H H H
σ ( z ) H ′ ( z ) Q H ( z ) = C H , C H = κ 2 C n ∈ Q > 0 . \sigma(z)H'(z)Q_H(z)=C_H,\qquad C_H=\kappa^2 C_n\in\mathbb{Q}_{>0}. σ ( z ) H ′ ( z ) Q H ( z ) = C H , C H = κ 2 C n ∈ Q > 0 . Here C n C_n C n (10) . Because the moments and the constant are rational, this identity can be used both over R \mathbb{R} R Q p \mathbb{Q}_p Q p
At the real place, the Gaussian quadrature weight ω z \omega_z ω z z z z Q H ( z ) = H ′ ( z ) ω z Q_H(z)=H'(z)\omega_z Q H ( z ) = H ′ ( z ) ω z [22 §3.5(v), equations (3.5.18) and (3.5.20.1)] . Since 0 < z ( 1 − z ) ≤ 1 / 4 0<z(1-z)\leq1/4 0 < z ( 1 − z ) ≤ 1/4 ∣ H ′ ( z ) ∣ ≥ 2 C H |H'(z)|\geq2\sqrt{C_H} ∣ H ′ ( z ) ∣ ≥ 2 C H
To pass from derivatives to the resultant, write a = lc A a=\operatorname{lc} A a = lc A ξ 1 , … , ξ d \xi_1,\ldots,\xi_d ξ 1 , … , ξ d A A A H = A B H=AB H = A B
∣ R ∣ = ∣ a ∣ e − d ∏ 1 ≤ i < j ≤ d ∣ ξ i − ξ j ∣ 2 ∏ i = 1 d ∣ H ′ ( ξ i ) ∣ . (7) |R|=\frac{|a|^{e-d}}{\prod_{1\leq i<j\leq d}|\xi_i-\xi_j|^2}\prod_{i=1}^{d}|H'(\xi_i)|. \tag*{(7)} ∣ R ∣ = ∏ 1 ≤ i < j ≤ d ∣ ξ i − ξ j ∣ 2 ∣ a ∣ e − d i = 1 ∏ d ∣ H ′ ( ξ i ) ∣. (7) See also [10 Lemma 5.3] . The denominator is the square of a Vandermonde determinant. The Chebyshev–Hadamard estimate of [10 Lemma 5.5] , recalled in Lemma 2.5, bounds it from above. Together with the derivative estimate and the information supplied by the endpoint values of A A A
At a finite prime, the same root identity controls how closely the roots can lie. The moment denominators bound the coefficients of Q H Q_H Q H v p ( H ′ ( z ) ) v_p(H'(z)) v p ( H ′ ( z )) z z z
v p ( R ) ≤ d E p , E p = v p ( C H ) − min 0 ≤ j ≤ 2 n − 1 v p ( μ j ) . v_p(R)\leq d\,\mathcal{E}_p,\qquad\mathcal{E}_p=v_p(C_H)-\min_{0\leq j\leq2n-1}v_p(\mu_j). v p ( R ) ≤ d E p , E p = v p ( C H ) − 0 ≤ j ≤ 2 n − 1 min v p ( μ j ) . The exact expression of v p ( R ) v_p(R) v p ( R ) (14) . This formulation also accounts for primes dividing the leading coefficient, at which some roots may have negative valuation.
A further step restricts the primes that can contribute. Reduction of the differential equation excludes multiple reduced roots away from certain distinguished points. Bounds for the endpoint, central, and leading coefficients then prevent the two factors from meeting at the remaining points of the projective line. The resulting exclusions are proved in (29) , (39) , and Lemma 4.4. They make the local sum small enough for the comparison to succeed in the ranges stated in our theorems.
The estimates needed to complete the argument depend on the parameters. For fixed unequal nonnegative integers a , b a,b a , b P 2 m + δ ( a , b ) ( 0 ) P_{2m+\delta}^{(a,b)}(0) P 2 m + δ ( a , b ) ( 0 ) δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } m m m d d d O a , b ( log n ) O_{a,b}(\log n) O a , b ( log n ) (31) . For equal parameters, we remove the zero forced by odd parity, as in (3) . The local bound then contains a factorial ratio whose valuations are retained with their signs; see (36) . Preserving this cancellation gives the sharper uniform estimates when the parameter grows proportionally to the degree (Proposition 3.6).
For the Kaneko–Zagier families, passing to the primitive polynomial is a substantial part of the argument. After denominators are cleared, the coefficients have a common divisor whose logarithm grows linearly with the degree; see (49) and Proposition 4.5. We estimate this content together with the moment denominators, counting divisibility at each prime power so as to preserve the cancellation between them (Lemma 4.2). At 2 and 3, Taylor recurrences from the differential equation control the separation of roots, while digit sums associated with the Newton polygons control the accumulated error (Proposition 4.6). Combining these estimates with the real bound gives the positive gap in (70) , uniformly in the degree of a possible factor. The explicit cutoff and the remaining finite checks are given in Propositions B.1 and B.2, completing the argument in every degree.
Arithmetic applications
To determine the Galois groups, we combine irreducibility with the ramification detected by Newton polygons. At suitable primes, the reciprocal polynomial has an Eisenstein factor, while squarefree reduction shows that the remaining roots lie in an unramified extension. In the tame cases, inertia therefore contains a single cycle fixing every other root (Lemma 5.1 and Proposition 5.2). On the transitive root set under consideration, a cycle moving more than half the roots forces every possible block size to divide its length (Lemma 5.3). The formula for the cycle length would then confine all the selected primes to one residue class for a larger modulus. Explicit estimates for primes in arithmetic progressions [3 Corollary 1.7] and the Brun–Titchmarsh inequality [21 Theorem 2, Eq. (1.10)] exclude such a block system; see Proposition 5.7 and its extension in § 5.5 . The action is therefore primitive. Since the constructed cycles leave at least three roots fixed, Jones’s theorem [17 Corollary 1.3] , recalled in Theorem 5.4, gives the alternating group. A cycle of even length then gives the symmetric group.
When the roots occur in opposite pairs, we first determine the permutation group on the pairs. A suitable power of an inertia element interchanges the two roots in an odd, nonempty proper set of pairs. Its conjugates generate all independent sign changes, giving the full wreath product (Lemma 5.5). The unequal integral case uses a different argument: wild inertia supplies a prime-length cycle, to which Jordan’s theorem applies [17 Theorem 1.1] , and an odd valuation of the discriminant supplies the odd permutation needed to obtain the symmetric group (Lemma 5.16).
The same ramification also controls intersections of splitting fields. Let L / Q L/\mathbb{Q} L / Q E ~ \widetilde{E} E E E E L ∩ E ~ L \cap\widetilde{E} L ∩ E Q \mathbb{Q} Q E ~ \widetilde{E} E Gal ( L / Q ) \operatorname{Gal}(L/\mathbb{Q}) Gal ( L / Q ) L ∩ E ~ = Q L \cap\widetilde{E} = \mathbb{Q} L ∩ E = Q § 5.7 . Since the selected primes tend to infinity with the degree, they eventually avoid the finite set of primes ramified in E ~ \widetilde{E} E
The modular origin of the Kaneko–Zagier families gives one reason to study these Galois groups. These Jacobi polynomials occur as polynomial factors of modular forms. When the corresponding modular index is prime, the affine normalization described in § 6.4 , together with the prescribed factors at j = 0 j=0 j = 0 j = 1728 j=1728 j = 1728 j j j [18 Theorem 1] . Theorem 1.1 determines the arithmetic of the characteristic-zero Jacobi factors in every positive degree.
A second application comes from spherical harmonics. The dependence on the polar angle of a mode of degree ℓ \ell ℓ r r r 0 ≤ r ≤ ℓ 0 \le r \le\ell 0 ≤ r ≤ ℓ
( sin θ ) r P ℓ ( r ) ( cos θ ) , (\sin\theta)^r P_{\ell}^{(r)}(\cos\theta), ( sin θ ) r P ℓ ( r ) ( cos θ ) , where P ℓ ( r ) P_{\ell}^{(r)} P ℓ ( r ) r r r [22 Eqs. 14.7.8 and 14.30.1] . The circular harmonic cone x 2 + y 2 − 2 z 2 = 0 x^2+y^2-2z^2=0 x 2 + y 2 − 2 z 2 = 0 cos θ = ± 1 / 3 \cos\theta= \pm1/\sqrt{3} cos θ = ± 1/ 3 3 x 2 − 1 3x^2-1 3 x 2 − 1 P ℓ ( r ) P_{\ell}^{(r)} P ℓ ( r ) Proposition 6.3 retains the arithmetic of this prescribed factor in the resultant comparison. It yields an effective absolute threshold beyond which, uniformly for 0 ≤ r ≤ ℓ 0 \le r \le\ell 0 ≤ r ≤ ℓ
P ℓ ( r ) ( 1 / 3 ) = 0 ⟹ 23 r > 4 ( ℓ − 1 ) P_{\ell}^{(r)}(1/\sqrt{3})=0 \quad\Longrightarrow\quad23r>4(\ell-1) P ℓ ( r ) ( 1/ 3 ) = 0 ⟹ 23 r > 4 ( ℓ − 1 ) (Corollary 6.5 ). This gives a simultaneous restriction on degree and order in Agranovsky’s question concerning homogeneous harmonic polynomials divisible by this cone [1 §6, Question 4] . It complements the theorem of Mangoubi and Weller Weiser [20 Theorem 2] : the order must be even, and each fixed order admits at most one degree. The new estimate restricts the possible modes without giving a complete classification.
The third application concerns hyperelliptic Jacobians. The roots of a separable polynomial f f f y 2 = f ( x ) y^2=f(x) y 2 = f ( x ) J f J_f J f f ∈ Q [ x ] f \in\mathbb{Q}[x] f ∈ Q [ x ] d ≥ 9 d \ge9 d ≥ 9 Gal ( f / Q ) = S d \operatorname{Gal}(f/\mathbb{Q})=S_d Gal ( f / Q ) = S d [28 Theorem 2.4] gives
End Q ( J f ) = Z . \operatorname{End}_{\mathbb{Q}}(J_f)=\mathbb{Z}. End Q ( J f ) = Z . His results also establish the Hodge and Tate conjectures on every power of J f J_f J f J f J_f J f [28 Theorems 8.1 and 9.1, and Remark 9.2] . Corollary 6.6 records these consequences for our families. In particular, Theorem 1.1 makes them available for the Jacobians associated with all four Kaneko–Zagier families in every degree at least 9.
Organization
Section 2 develops the common resultant comparison. Section 3 treats integral parameters and derivatives, and Section 4 treats Kaneko–Zagier irreducibility. Section 5 proves the Galois and splitting-field results. Section 6 gives the applications.
The appendices contain local irreducibility theorems, the explicit cutoff, and the finite calculations needed for the all-degree statements. The finite Galois inputs are the published conditional results in [7 , 8 ] , used with their stated degree and modular-index ranges.
Disclosure of AI assistance
This work was developed by the author with assistance from ChatGPT (OpenAI). The author developed the overall approach and methods, building on the prior work cited in the manuscript. During the research and preparation of the manuscript, ChatGPT was used to discuss and revise mathematical arguments, improve the exposition, and search for relevant literature. It also assisted with all computations presented in the appendices.
A common resultant comparison
We now establish the estimates used in the irreducibility proofs. The aim is to bound a factor resultant from below at the real place and from above through its prime divisors. Both estimates must be proportional to the smaller factor degree, so that the comparison excludes every proper factor at once.
The common input is a Jacobi identity obtained from orthogonality. We first use it to bound derivatives at the real roots and distances between roots over a p p p
For a prime p p p v p ( p ) = 1 v_p(p)=1 v p ( p ) = 1 Q ‾ p \overline{\mathbb{Q}}_p Q p v p ( 0 ) = + ∞ v_p(0)=+\infty v p ( 0 ) = + ∞ Z p [ t ] \mathbb{Z}_p[t] Z p [ t ] ( z ) j = z ( z + 1 ) ⋯ ( z + j − 1 ) (z)_j=z(z+1)\cdots(z+j-1) ( z ) j = z ( z + 1 ) ⋯ ( z + j − 1 ) ( z ) 0 = 1 (z)_0=1 ( z ) 0 = 1
The Jacobi identity
Fix real α , β > − 1 \alpha,\beta>-1 α , β > − 1 c = α + β c=\alpha+\beta c = α + β y n ( t ) = P n ( α , β ) ( 1 − 2 t ) y_n(t)=P_n^{(\alpha,\beta)}(1-2t) y n ( t ) = P n ( α , β ) ( 1 − 2 t ) n ≥ 1 n\ge1 n ≥ 1
y n ( t ) = ∑ j = 0 n ( − 1 ) j ( n + α n − j ) ( n + c + j j ) t j (8) y_n(t)=\sum_{j=0}^{n}(-1)^j\binom{n+\alpha}{n-j}\binom{n+c+j}{j}t^j \tag*{(8)} y n ( t ) = j = 0 ∑ n ( − 1 ) j ( n − j n + α ) ( j n + c + j ) t j (8) Let I I I
d ω ( t ) = t α ( 1 − t ) β B ( α + 1 , β + 1 ) d t , 0 < t < 1. d\omega(t)=\frac{t^\alpha(1-t)^\beta}{B(\alpha+1,\beta+1)}\,dt,\qquad0<t<1. d ω ( t ) = B ( α + 1 , β + 1 ) t α ( 1 − t ) β d t , 0 < t < 1. Here B ( s , t ) B(s,t) B ( s , t )
I ( t j ) = μ j = ( α + 1 ) j ( c + 2 ) j . (9) I(t^j)=\mu_j=\frac{(\alpha+1)_j}{(c+2)_j}. \tag*{(9)} I ( t j ) = μ j = ( c + 2 ) j ( α + 1 ) j . (9) The roots of y n y_n y n ( 0 , 1 ) (0,1) ( 0 , 1 ) I ( y n q ) = 0 I(y_nq)=0 I ( y n q ) = 0 deg q < n \deg q<n deg q < n [22 §§18.2–18.5] . Define
Q n ( t ) = I s ( y n ( t ) − y n ( s ) t − s ) , σ ( t ) = t ( 1 − t ) , ρ ( t ) = α − c t . Q_n(t)=I_s\left(\frac{y_n(t)-y_n(s)}{t-s}\right),\qquad\sigma(t)=t(1-t),\qquad\rho(t)=\alpha-ct. Q n ( t ) = I s ( t − s y n ( t ) − y n ( s ) ) , σ ( t ) = t ( 1 − t ) , ρ ( t ) = α − c t . The difference quotient is a polynomial of degree at most n − 1 n-1 n − 1 y n y_n y n I I I
Lemma 2.1. One has the polynomial identity
( c + 1 ) y n 2 + σ ( y n ′ Q n − y n Q n ′ ) + ρ y n Q n = C n , C n = ( α + 1 ) n ( β + 1 ) n n ! ( c + 2 ) n − 1 > 0. (10) (c+1)y_n^2+\sigma(y_n'Q_n-y_nQ_n')+\rho y_nQ_n=C_n,\qquad C_n=\frac{(\alpha+1)_n(\beta+1)_n}{n!(c+2)_{n-1}}>0. \tag*{(10)} ( c + 1 ) y n 2 + σ ( y n ′ Q n − y n Q n ′ ) + ρ y n Q n = C n , C n = n ! ( c + 2 ) n − 1 ( α + 1 ) n ( β + 1 ) n > 0. (10) At every root t i t_i t i y n y_n y n
∣ y n ′ ( t i ) ∣ ≥ 2 C n , C n ⟶ Γ ( c + 2 ) Γ ( α + 1 ) Γ ( β + 1 ) > 0. (11) \lvert y_n'(t_i)\rvert\ge2\sqrt{C_n},\qquad C_n\longrightarrow\frac{\Gamma(c+2)}{\Gamma(\alpha+1)\Gamma(\beta+1)}>0. \tag*{(11)} ∣ y n ′ ( t i )∣ ≥ 2 C n , C n ⟶ Γ ( α + 1 ) Γ ( β + 1 ) Γ ( c + 2 ) > 0. (11) Proof. Write τ = σ ′ + ρ = α + 1 − ( c + 2 ) t \tau=\sigma'+\rho=\alpha+1-(c+2)t τ = σ ′ + ρ = α + 1 − ( c + 2 ) t Λ = n ( n + c + 1 ) \Lambda=n(n+c+1) Λ = n ( n + c + 1 )
σ y n ′ ′ + τ y n ′ + Λ y n = 0 (12) \sigma y_n''+\tau y_n'+\Lambda y_n=0 \tag*{(12)} σ y n ′′ + τ y n ′ + Λ y n = 0 (12) Writing d ω = w ( t ) d t d\omega=w(t)\,dt d ω = w ( t ) d t ( σ w ) ′ = τ w (\sigma w)'=\tau w ( σ w ) ′ = τ w t > 1 t>1 t > 1
m ( t ) = I s ( ( t − s ) − 1 ) , q ( t ) = I s ( y n ( s ) / ( t − s ) ) . m(t)=I_s((t-s)^{-1}),\qquad q(t)=I_s(y_n(s)/(t-s)). m ( t ) = I s (( t − s ) − 1 ) , q ( t ) = I s ( y n ( s ) / ( t − s )) . Integration by parts gives
σ m ′ − ρ m = c + 1 , σ q ′ ′ + ( σ ′ − ρ ) q ′ + ( Λ + c ) q = 0. \sigma m'-\rho m=c+1,\qquad\sigma q''+(\sigma'-\rho)q'+(\Lambda+c)q=0. σ m ′ − ρ m = c + 1 , σ q ′′ + ( σ ′ − ρ ) q ′ + ( Λ + c ) q = 0. The endpoint terms vanish because α , β > − 1 \alpha,\beta>-1 α , β > − 1 σ w \sigma w σ w Q n = y n m − q Q_n=y_nm-q Q n = y n m − q
σ Q n ′ ′ + ( σ ′ − ρ ) Q n ′ + ( Λ + c ) Q n = 2 ( c + 1 ) y n ′ . \sigma Q_n''+(\sigma'-\rho)Q_n'+(\Lambda+c)Q_n=2(c+1)y_n'. σ Q n ′′ + ( σ ′ − ρ ) Q n ′ + ( Λ + c ) Q n = 2 ( c + 1 ) y n ′ . Differentiating the left side of (10) , and using these two differential equations, shows that it is constant. Orthogonality gives
q ( t ) = I ( y n 2 ) lc ( y n ) t − n − 1 + O ( t − n − 2 ) . q(t)=\frac{I(y_n^2)}{\operatorname{lc}(y_n)}t^{-n-1}+O(t^{-n-2}). q ( t ) = lc ( y n ) I ( y n 2 ) t − n − 1 + O ( t − n − 2 ) . Consequently the constant is ( 2 n + c + 1 ) I ( y n 2 ) (2n+c+1)I(y_n^2) ( 2 n + c + 1 ) I ( y n 2 ) C n C_n C n t > 1 t>1 t > 1
For a root t i t_i t i
ℓ i ( t ) = y n ( t ) ( t − t i ) y n ′ ( t ) . \ell_i(t)=\frac{y_n(t)}{(t-t_i)y_n'(t)}. ℓ i ( t ) = ( t − t i ) y n ′ ( t ) y n ( t ) . Since ℓ i 2 − ℓ i \ell_i^2-\ell_i ℓ i 2 − ℓ i y n y_n y n n − 2 n-2 n − 2
ω i : = I ( ℓ i ) = I ( ℓ i 2 ) > 0. \omega_i:=I(\ell_i)=I(\ell_i^2)>0. ω i := I ( ℓ i ) = I ( ℓ i 2 ) > 0. Also ∑ i ℓ i = 1 \sum_i\ell_i=1 ∑ i ℓ i = 1 ω i ≤ 1 \omega_i\le1 ω i ≤ 1 Q n Q_n Q n Q n ( t i ) = y n ′ ( t i ) ω i Q_n(t_i)=y_n'(t_i)\omega_i Q n ( t i ) = y n ′ ( t i ) ω i
C n = t i ( 1 − t i ) y n ′ ( t i ) 2 ω i ≤ 1 4 y n ′ ( t i ) 2 . C_n=t_i(1-t_i)y_n'(t_i)^2\omega_i\le\frac{1}{4}y_n'(t_i)^2. C n = t i ( 1 − t i ) y n ′ ( t i ) 2 ω i ≤ 4 1 y n ′ ( t i ) 2 . Finally, the stated limit follows by expressing the rising factorials in C n C_n C n
The numbers ω i \omega_i ω i deg q ≤ 2 n − 1 \deg q\le2n-1 deg q ≤ 2 n − 1 y n y_n y n q = y n s + r q=y_ns+r q = y n s + r deg s , deg r < n \deg s,\deg r<n deg s , deg r < n
I ( q ) = I ( r ) = ∑ i = 1 n ω i r ( t i ) = ∑ i = 1 n ω i q ( t i ) . I(q)=I(r)=\sum_{i=1}^{n}\omega_i r(t_i)=\sum_{i=1}^{n}\omega_i q(t_i). I ( q ) = I ( r ) = i = 1 ∑ n ω i r ( t i ) = i = 1 ∑ n ω i q ( t i ) . Thus a familiar formula for numerical integration supplies the derivative estimate needed for the arithmetic argument; see also [22 §3.5(v)] .
Projective root distances
A primitive polynomial over Z p \mathbb{Z}_p Z p p p p δ p ( z , w ) \delta_p(z,w) δ p ( z , w ) v p ( z − w ) v_p(z-w) v p ( z − w ) p p p
Lemma 2.2 (Projective moment bound). Let p p p H ∈ Z p [ t ] H\in\mathbb{Z}_p[t] H ∈ Z p [ t ] n ≥ 1 n\ge1 n ≥ 1 I I I Q p \mathbb{Q}_p Q p 2 n − 1 2n-1 2 n − 1 μ j = I ( t j ) \mu_j=I(t^j) μ j = I ( t j )
I ( H q ) = 0 ( deg q < n ) . I(Hq)=0\qquad(\deg q<n). I ( H q ) = 0 ( deg q < n ) . Put
Q H ( t ) = I s ( H ( t ) − H ( s ) t − s ) , σ ( t ) = t ( 1 − t ) . Q_H(t)=I_s\left(\frac{H(t)-H(s)}{t-s}\right),\qquad\sigma(t)=t(1-t). Q H ( t ) = I s ( t − s H ( t ) − H ( s ) ) , σ ( t ) = t ( 1 − t ) . Suppose that some C ∈ Q p × C\in\mathbb{Q}_p^\times C ∈ Q p ×
σ ( z ) H ′ ( z ) Q H ( z ) = C at every root z of H . \sigma(z)H'(z)Q_H(z)=C\qquad\text{at every root }z\text{ of }H. σ ( z ) H ′ ( z ) Q H ( z ) = C at every root z of H . Write
η p = min 0 ≤ j ≤ 2 n − 1 v p ( μ j ) , E p = v p ( C ) − η p . \eta_p=\min_{0\le j\le2n-1}v_p(\mu_j),\qquad\mathcal{E}_p=v_p(C)-\eta_p. η p = 0 ≤ j ≤ 2 n − 1 min v p ( μ j ) , E p = v p ( C ) − η p . For distinct roots put
δ p ( z , w ) = v p ( z − w ) − min { v p ( z ) , 0 } − min { v p ( w ) , 0 } . \delta_p(z,w)=v_p(z-w)-\min\{v_p(z),0\}-\min\{v_p(w),0\}. δ p ( z , w ) = v p ( z − w ) − min { v p ( z ) , 0 } − min { v p ( w ) , 0 } . Every root satisfies the projective row bound
∑ H ( w ) = 0 , w ≠ z δ p ( z , w ) ≤ E p . (13) \sum_{\substack{H(w)=0,\,w\ne z}}\delta_p(z,w)\le\mathcal{E}_p. \tag*{(13)} H ( w ) = 0 , w = z ∑ δ p ( z , w ) ≤ E p . (13) Consequently, for every primitive factorization H = A B H=AB H = A B Z p [ t ] \mathbb{Z}_p[t] Z p [ t ] k = deg A k=\deg A k = deg A R = Res ( A , B ) R=\operatorname{Res}(A,B) R = Res ( A , B )
v p ( R ) ≤ k E p . v_p(R)\le k\mathcal{E}_p. v p ( R ) ≤ k E p . Proof. The root identity implies that the roots are simple and avoid 0 , 1 0,1 0 , 1 δ p ( z , w ) \delta_p(z,w) δ p ( z , w ) F ∈ Z p [ t ] F \in\mathbb{Z}_p[t] F ∈ Z p [ t ]
v p ( lc F ) = − ∑ F ( w ) = 0 min { v p ( w ) , 0 } . v_p(\operatorname{lc} F)=-\sum_{F(w)=0}\min\{v_p(w),0\}. v p ( lc F ) = − F ( w ) = 0 ∑ min { v p ( w ) , 0 } . Thus the full projective row at an integral root z z z H H H v p ( H ′ ( z ) ) v_p(H'(z)) v p ( H ′ ( z )) z z z δ p \delta_p δ p H ^ ( X ) = X n H ( 1 / X ) \widehat{H}(X)=X^nH(1/X) H ( X ) = X n H ( 1/ X ) v p ( H ^ ′ ( z − 1 ) ) v_p(\widehat{H}'(z^{-1})) v p ( H ′ ( z − 1 ))
v p ( R ) = ∑ A ( z ) = 0 ∑ B ( w ) = 0 δ p ( z , w ) . (14) v_p(R)=\sum_{A(z)=0}\sum_{B(w)=0}\delta_p(z,w). \tag*{(14)} v p ( R ) = A ( z ) = 0 ∑ B ( w ) = 0 ∑ δ p ( z , w ) . (14) The coefficients of Q H Q_H Q H Z p \mathbb{Z}_p Z p v p ( Q H ( z ) ) ≥ η p v_p(Q_H(z))\geq\eta_p v p ( Q H ( z )) ≥ η p v p ( σ ( z ) ) ≥ 0 v_p(\sigma(z))\geq0 v p ( σ ( z )) ≥ 0 v p ( H ′ ( z ) ) ≤ E p v_p(H'(z))\leq\mathcal{E}_p v p ( H ′ ( z )) ≤ E p b = z − 1 b=z^{-1} b = z − 1 r = v p ( b ) > 0 r=v_p(b)>0 r = v p ( b ) > 0 I I I b b b
q ( t ) = H ( t ) 1 − b t q(t)=\frac{H(t)}{1-bt} q ( t ) = 1 − b t H ( t ) has integral coefficients in that field: if H j H_j H j q j q_j q j H H H q q q q j = H j + b q j − 1 q_j=H_j+bq_{j-1} q j = H j + b q j − 1 0 ≤ j < n 0\leq j<n 0 ≤ j < n q − 1 = 0 q_{-1}=0 q − 1 = 0 0 ≤ j < n 0\leq j<n 0 ≤ j < n I ( t j q ) = b I ( t j + 1 q ) I(t^jq)=bI(t^{j+1}q) I ( t j q ) = b I ( t j + 1 q )
I ( q ) = b n I ( t n q ) , Q ^ ( b ) = − b n I ( q ) = − b 2 n I ( t n q ) , Q ^ ( X ) = X n − 1 Q H ( 1 / X ) . I(q)=b^nI(t^nq),\qquad\widehat{Q}(b)=-b^nI(q)=-b^{2n}I(t^nq),\qquad\widehat{Q}(X)=X^{n-1}Q_H(1/X). I ( q ) = b n I ( t n q ) , Q ( b ) = − b n I ( q ) = − b 2 n I ( t n q ) , Q ( X ) = X n − 1 Q H ( 1/ X ) . Since deg ( t n q ) = 2 n − 1 \deg(t^nq)=2n-1 deg ( t n q ) = 2 n − 1 v p ( Q ^ ( b ) ) ≥ 2 n r + η p v_p(\widehat{Q}(b))\geq2nr+\eta_p v p ( Q ( b )) ≥ 2 n r + η p
( 1 − b ) H ^ ′ ( b ) Q ^ ( b ) = C b 2 n − 1 , (1-b)\widehat{H}'(b)\widehat{Q}(b)=Cb^{2n-1}, ( 1 − b ) H ′ ( b ) Q ( b ) = C b 2 n − 1 , so v p ( H ^ ′ ( b ) ) ≤ E p − r v_p(\widehat{H}'(b))\leq\mathcal{E}_p-r v p ( H ′ ( b )) ≤ E p − r k k k A A A (14) by k E p k\mathcal{E}_p k E p
The reflection t = − X t=-X t = − X σ ( X ) = X ( X + 1 ) \sigma(X)=X(X+1) σ ( X ) = X ( X + 1 )
Integral weights
For integral parameters, we move the endpoints to − 1 , 0 -1,0 − 1 , 0 0 0 0
Fix nonnegative integers a , b a,b a , b n ≥ 1 n\geq1 n ≥ 1
B = ( n + a a ) , S = 2 n + a + b + 1 , T = ∏ j = 1 a + b ( n + j ) , B=\binom{n+a}{a},\qquad S=2n+a+b+1,\qquad T=\prod_{j=1}^{a+b}(n+j), B = ( a n + a ) , S = 2 n + a + b + 1 , T = j = 1 ∏ a + b ( n + j ) , and write
J ( X ) = P n ( a , b ) ( 2 X + 1 ) = ∑ j = 0 n ( n + a n − j ) ( n + a + b + j j ) X j , Q = J / B . J(X)=P_n^{(a,b)}(2X+1)=\sum_{j=0}^{n}\binom{n+a}{n-j}\binom{n+a+b+j}{j}X^j,\qquad Q=J/B. J ( X ) = P n ( a , b ) ( 2 X + 1 ) = j = 0 ∑ n ( n − j n + a ) ( j n + a + b + j ) X j , Q = J / B . Thus J ∈ Z [ X ] J\in\mathbb{Z}[X] J ∈ Z [ X ] Q ( 0 ) = 1 Q(0)=1 Q ( 0 ) = 1 p 0 ( X ) = X ( X + 1 ) p_0(X)=X(X+1) p 0 ( X ) = X ( X + 1 )
I ( g ) = ∫ − 1 0 g ( t ) ( − t ) a ( t + 1 ) b d t , V ( X ) = I ( Q ( X ) − Q ( t ) X − t ) . I(g)=\int_{-1}^{0}g(t)(-t)^a(t+1)^b\,dt,\qquad V(X)=I\left(\frac{Q(X)-Q(t)}{X-t}\right). I ( g ) = ∫ − 1 0 g ( t ) ( − t ) a ( t + 1 ) b d t , V ( X ) = I ( X − t Q ( X ) − Q ( t ) ) . The quotient inside the integral is a polynomial. The moments of I I I I I I I I I I ( t j Q ) = 0 I(t^jQ)=0 I ( t j Q ) = 0 0 ≤ j < n 0 \le j < n 0 ≤ j < n
v p ( I ( t j ) ) ≥ − ⌊ log p ( j + a + b + 1 ) ⌋ ( j ≥ 0 ) . (15) v_p(I(t^j)) \ge-\left\lfloor\log_p(j+a+b+1)\right\rfloor\qquad(j \ge0). \tag*{(15)} v p ( I ( t j )) ≥ − ⌊ log p ( j + a + b + 1 ) ⌋ ( j ≥ 0 ) . (15) Indeed, the integral is an integral linear combination of reciprocals of integers at most j + a + b + 1 j+a+b+1 j + a + b + 1
Lemma 2.3. At every root α \alpha α Q Q Q
p 0 ( α ) Q ′ ( α ) V ( α ) = C n , C n = − ( a ! ) 2 n ! ( n + b ) ! ( n + a ) ! ( n + a + b ) ! . (16) p_0(\alpha)Q'(\alpha)V(\alpha)=C_n,\qquad C_n=-\frac{(a!)^2n!(n+b)!}{(n+a)!(n+a+b)!}. \tag*{(16)} p 0 ( α ) Q ′ ( α ) V ( α ) = C n , C n = − ( n + a )! ( n + a + b )! ( a ! ) 2 n ! ( n + b )! . (16) Proof. Apply Lemma 2.1 with parameters a , b a,b a , b t = − X t=-X t = − X B ( a + 1 , b + 1 ) B(a+1,b+1) B ( a + 1 , b + 1 ) Q ( X ) = y n ( − X ) / ( n + a a ) Q(X)=y_n(-X)/\binom{n+a}{a} Q ( X ) = y n ( − X ) / ( a n + a )
− B ( a + 1 , b + 1 ) ( a + 1 ) n ( b + 1 ) n ( n + a a ) 2 n ! ( a + b + 2 ) n − 1 = − ( a ! ) 2 n ! ( n + b ) ! ( n + a ) ! ( n + a + b ) ! . -\frac{B(a+1,b+1)(a+1)_n(b+1)_n}{\binom{n+a}{a}^2\,n!(a+b+2)_{n-1}}=-\frac{(a!)^2n!(n+b)!}{(n+a)!(n+a+b)!}. − ( a n + a ) 2 n ! ( a + b + 2 ) n − 1 B ( a + 1 , b + 1 ) ( a + 1 ) n ( b + 1 ) n = − ( n + a )! ( n + a + b )! ( a ! ) 2 n ! ( n + b )! . Lemma 2.4. Put
d p = − min j v p ( [ X j ] Q ) , L p = ⌊ log p S ⌋ . d_p=-\min_j v_p([X^j]Q),\qquad L_p=\left\lfloor\log_p S\right\rfloor. d p = − j min v p ([ X j ] Q ) , L p = ⌊ log p S ⌋ . For every root α \alpha α Q Q Q
∑ Q ( γ ) = 0 γ ≠ α δ p ( α , γ ) ≤ L p + 2 d p + v p ( C n ) . \sum_{\substack{Q(\gamma)=0\\\gamma\ne\alpha}}\delta_p(\alpha,\gamma)\le L_p+2d_p+v_p(C_n). Q ( γ ) = 0 γ = α ∑ δ p ( α , γ ) ≤ L p + 2 d p + v p ( C n ) . In particular, this sum is at most
D p = L p + v p ( T ) . (17) D_p=L_p+v_p(T). \tag*{(17)} D p = L p + v p ( T ) . (17) For a = b = r a=b=r a = b = r
D p = L p + v p ( ( n + r ) ! 2 n ! ( n + 2 r ) ! ) . (18) D_p=L_p+v_p\left(\frac{(n+r)!^2}{n!(n+2r)!}\right). \tag*{(18)} D p = L p + v p ( n ! ( n + 2 r )! ( n + r ) ! 2 ) . (18) Each bound applies at p = 2 p=2 p = 2
Proof. Since Q ( 0 ) = 1 Q(0)=1 Q ( 0 ) = 1 B Q ∈ Z [ X ] BQ\in\mathbb{Z}[X] B Q ∈ Z [ X ] 0 ≤ d p ≤ v p ( B ) 0\le d_p\le v_p(B) 0 ≤ d p ≤ v p ( B ) u = p d p Q u=p^{d_p}Q u = p d p Q p 2 d p C n p^{2d_p}C_n p 2 d p C n (15) , through degree 2 n − 1 2n-1 2 n − 1 − L p -L_p − L p U a = ∏ j = 1 a ( n + j ) U_a=\prod_{j=1}^{a}(n+j) U a = ∏ j = 1 a ( n + j ) V a = ∏ j = 1 a ( n + b + j ) V_a=\prod_{j=1}^{a}(n+b+j) V a = ∏ j = 1 a ( n + b + j )
2 d p + v p ( C n ) ≤ 2 v p ( B ) + v p ( C n ) = v p ( U a ) − v p ( V a ) ≤ v p ( T ) . 2d_p+v_p(C_n)\le2v_p(B)+v_p(C_n)=v_p(U_a)-v_p(V_a)\le v_p(T). 2 d p + v p ( C n ) ≤ 2 v p ( B ) + v p ( C n ) = v p ( U a ) − v p ( V a ) ≤ v p ( T ) . When a = b = r a=b=r a = b = r ( n + r ) ! 2 / ( n ! ( n + 2 r ) ! ) (n+r)!^2/(n!(n+2r)!) ( n + r ) ! 2 / ( n ! ( n + 2 r )!)
The real estimate and the prime 2
We use the Chebyshev–Hadamard estimate appearing in [10 Lemma 5.5] .
Lemma 2.5. For t 1 , … , t k ∈ [ 0 , 1 ] t_1,\ldots,t_k\in[0,1] t 1 , … , t k ∈ [ 0 , 1 ]
∏ i < j ∣ t i − t j ∣ 2 ≤ k k 2 − 2 ( k − 1 ) 2 . \prod_{i<j}|t_i-t_j|^2\le k^k2^{-2(k-1)^2}. i < j ∏ ∣ t i − t j ∣ 2 ≤ k k 2 − 2 ( k − 1 ) 2 . Proof. Let T j ( cos θ ) = cos ( j θ ) T_j(\cos\theta)=\cos(j\theta) T j ( cos θ ) = cos ( j θ ) j ≥ 1 j\ge1 j ≥ 1 q j ( t ) = 2 1 − 2 j T j ( 2 t − 1 ) q_j(t)=2^{1-2j}T_j(2t-1) q j ( t ) = 2 1 − 2 j T j ( 2 t − 1 ) 2 1 − 2 j 2^{1-2j} 2 1 − 2 j [ 0 , 1 ] [0,1] [ 0 , 1 ] q 0 = 1 q_0=1 q 0 = 1 ( q j − 1 ( t i ) ) i , j = 1 k (q_{j-1}(t_i))_{i,j=1}^k ( q j − 1 ( t i ) ) i , j = 1 k
k k / 2 ∏ j = 1 k − 1 2 1 − 2 j = k k / 2 2 − ( k − 1 ) 2 . k^{k/2}\prod_{j=1}^{k-1}2^{1-2j}=k^{k/2}2^{-(k-1)^2}. k k /2 j = 1 ∏ k − 1 2 1 − 2 j = k k /2 2 − ( k − 1 ) 2 . Squaring proves the assertion. □
We combine the derivative bound with this Vandermonde estimate. The values of an integral factor at the two endpoints also bound its leading coefficient from below. The next lemma records both contributions and separates the prime 2. We keep track of the powers of 2 introduced by x = 2 X + 1 x=2X+1 x = 2 X + 1
Lemma 2.6. Let h ∈ Z [ 1 / 2 ] [ x ] h\in\mathbb{Z}[1/2][x] h ∈ Z [ 1/2 ] [ x ] N N N ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) b h = h ( 1 ) b_h=h(1) b h = h ( 1 ) ∣ h ′ ( α ) ∣ > γ > 0 |h'(\alpha)|>\gamma>0 ∣ h ′ ( α ) ∣ > γ > 0 h ( 2 X + 1 ) h(2X+1) h ( 2 X + 1 ) D 2 D_2 D 2 κ > 0 \kappa>0 κ > 0 H = κ h H=\kappa h H = κh
H = F G , 1 ≤ d = deg F ≤ e = deg G , d + e = N , H=FG,\qquad1\le d=\deg F\le e=\deg G,\qquad d+e=N, H = F G , 1 ≤ d = deg F ≤ e = deg G , d + e = N , with F , G ∈ Z [ x ] F,G\in\mathbb{Z}[x] F , G ∈ Z [ x ]
R = Res ( F , G ) , R = ∣ R ∣ 2 v 2 ( R ) , u = lc F , c F = min j v 2 ( [ X j ] F ( 2 X + 1 ) ) , R=\operatorname{Res}(F,G),\qquad\mathcal{R}=\frac{|R|}{2^{v_2(R)}},\qquad u=\operatorname{lc} F,\qquad c_F=\min_j v_2([X^j]F(2X+1)), R = Res ( F , G ) , R = 2 v 2 ( R ) ∣ R ∣ , u = lc F , c F = j min v 2 ([ X j ] F ( 2 X + 1 )) , and V F = ∏ i < j ( α i − α j ) 2 V_F=\prod_{i<j}(\alpha_i-\alpha_j)^2 V F = ∏ i < j ( α i − α j ) 2 F F F u > 2 c F u>2^{c_F} u > 2 c F
log R d > e log 2 − log V F d + log γ b h − D 2 log 2 + e − d d log u 2 c F . (19) \frac{\log\mathcal{R}}{d}>e\log2-\frac{\log V_F}{d}+\log\frac{\gamma}{b_h}-D_2\log2+\frac{e-d}{d}\log\frac{u}{2^{c_F}}. \tag*{(19)} d log R > e log 2 − d log V F + log b h γ − D 2 log 2 + d e − d log 2 c F u . (19) In particular,
log R d > ( N − 3 ) log 2 − log d + log γ b h − D 2 log 2. (20) \frac{\log\mathcal{R}}{d}>(N-3)\log2-\log d+\log\frac{\gamma}{b_h}-D_2\log2. \tag*{(20)} d log R > ( N − 3 ) log 2 − log d + log b h γ − D 2 log 2. (20) Proof. The derivative and resultant formulas give
∣ R ∣ = u e − d V F ∏ F ( α ) = 0 ∣ H ′ ( α ) ∣ > u e − d V F ( κ γ ) d . |R|=\frac{u^{e-d}}{V_F}\prod_{F(\alpha)=0}|H'(\alpha)|>\frac{u^{e-d}}{V_F}(\kappa\gamma)^d. ∣ R ∣ = V F u e − d F ( α ) = 0 ∏ ∣ H ′ ( α ) ∣ > V F u e − d ( κγ ) d . Define c G c_G c G c F c_F c F
v 2 ( R ) ≤ e c F + d c G − d e + d D 2 . v_2(R)\le ec_F+dc_G-de+dD_2. v 2 ( R ) ≤ e c F + d c G − d e + d D 2 . Both F ( 1 ) F(1) F ( 1 ) F ( − 1 ) F(-1) F ( − 1 ) 2 c F 2^{c_F} 2 c F
u 2 = ∣ F ( 1 ) F ( − 1 ) ∣ ∏ i ( 1 − α i 2 ) > 2 2 c F . u^2=\frac{|F(1)F(-1)|}{\prod_i(1-\alpha_i^2)}>2^{2c_F}. u 2 = ∏ i ( 1 − α i 2 ) ∣ F ( 1 ) F ( − 1 ) ∣ > 2 2 c F . After subtracting the binary estimate, the first two displays yield the right side of (19) , with − log b h -\log b_h − log b h log κ − ( c F + c G ) log 2 \log\kappa-(c_F+c_G)\log2 log κ − ( c F + c G ) log 2 κ = 2 k / c \kappa=2^k/c κ = 2 k / c c c c h ∈ Z [ 1 / 2 ] [ x ] h\in\mathbb{Z}[1/2][x] h ∈ Z [ 1/2 ] [ x ] H H H c c c b h b_h b h t 2 t_2 t 2 h ( 2 X + 1 ) h(2X+1) h ( 2 X + 1 ) c F + c G = k + t 2 c_F+c_G=k+t_2 c F + c G = k + t 2 t 2 ≤ v 2 ( b h ) t_2\le v_2(b_h) t 2 ≤ v 2 ( b h )
log κ − ( c F + c G ) log 2 = − log c − t 2 log 2 ≥ − log b h . \log\kappa-(c_F+c_G)\log2=-\log c-t_2\log2\ge-\log b_h. log κ − ( c F + c G ) log 2 = − log c − t 2 log 2 ≥ − log b h . This proves (19) .
Finally, rescaling Lemma 2.5 from [ 0 , 1 ] [0,1] [ 0 , 1 ] [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
V F ≤ d d 2 − ( d − 1 ) ( d − 2 ) , − log V F d ≥ ( d − 3 ) log 2 − log d . V_F \le d^{d 2^{-(d-1)(d-2)}}, \qquad-\frac{\log V_F}{d} \ge(d-3)\log2-\log d. V F ≤ d d 2 − ( d − 1 ) ( d − 2 ) , − d log V F ≥ ( d − 3 ) log 2 − log d . The same inequality holds for d = 1 d=1 d = 1 (19) proves (20) . □
A bound for the prime powers
The local moment bound contains ⌊ log p S ⌋ \lfloor\log_p S\rfloor ⌊ log p S ⌋ B ≥ 1 B\ge1 B ≥ 1 S ≥ 1 S\ge1 S ≥ 1
Ψ B ( S ) = ∑ p ≤ B ⌊ log p S ⌋ log p . \Psi_B(S)=\sum_{p\le B}\lfloor\log_p S\rfloor\log p. Ψ B ( S ) = p ≤ B ∑ ⌊ log p S ⌋ log p . Lemma 2.7. If B ≥ 9 B\ge9 B ≥ 9 S ≤ 5 B S\le5B S ≤ 5 B
Ψ B ( S ) ≤ log lcm ( 1 , … , B ) + log lcm ( 1 , … , ⌊ S ⌋ ) < ( B + S ) log 3. (21) \Psi_B(S)\le\log\operatorname{lcm}(1,\ldots,B)+\log\operatorname{lcm}(1,\ldots,\lfloor\sqrt{S}\rfloor)<(B+\sqrt{S})\log3. \tag*{(21)} Ψ B ( S ) ≤ log lcm ( 1 , … , B ) + log lcm ( 1 , … , ⌊ S ⌋) < ( B + S ) log 3. (21) Proof. For p ≤ B p\le B p ≤ B l = ⌊ log p B ⌋ l=\lfloor\log_p B\rfloor l = ⌊ log p B ⌋ k = ⌊ log p S ⌋ k=\lfloor\log_p S\rfloor k = ⌊ log p S ⌋ B ≥ 9 B\ge9 B ≥ 9 p l ≥ 5 p^l\ge5 p l ≥ 5 p = 2 , 3 p=2,3 p = 2 , 3
S ≤ 5 B < 5 p l + 1 ≤ p 2 l + 1 , S\le5B<5p^{l+1}\le p^{2l+1}, S ≤ 5 B < 5 p l + 1 ≤ p 2 l + 1 , so k ≤ 2 l k\le2l k ≤ 2 l k ≤ l + ⌊ k / 2 ⌋ = l + ⌊ log p S ⌋ k\le l+\lfloor k/2\rfloor=l+\lfloor\log_p\sqrt{S}\rfloor k ≤ l + ⌊ k /2 ⌋ = l + ⌊ log p S ⌋ lcm ( 1 , … , t ) < 3 t \operatorname{lcm}(1,\ldots,t)<3^t lcm ( 1 , … , t ) < 3 t t ≥ 1 t\ge1 t ≥ 1 [16 ] . □
Integral parameters and derivatives
The local bounds of Section 2 involve primes up to 2 n + O ( 1 ) 2n+O(1) 2 n + O ( 1 ) m = ⌊ n / 2 ⌋ m=\lfloor n/2\rfloor m = ⌊ n /2 ⌋
For unequal integral parameters, the exceptional primes above m + O a , b ( 1 ) m+O_{a,b}(1) m + O a , b ( 1 ) O a , b ( log n ) O_{a,b}(\log n) O a , b ( log n ) m m m Theorems 1.3 and 1.4 , respectively.
Unequal parameters
By reflection, P n ( b , a ) ( − x ) = ( − 1 ) n P n ( a , b ) ( x ) P_n^{(b,a)}(-x)=(-1)^nP_n^{(a,b)}(x) P n ( b , a ) ( − x ) = ( − 1 ) n P n ( a , b ) ( x ) a > b ≥ 0 a>b\ge0 a > b ≥ 0 q = a − b q=a-b q = a − b s = a + b s=a+b s = a + b n = 2 m + δ n=2m+\delta n = 2 m + δ δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } b m = 4 − m ( 2 m m ) b_m=4^{-m}\binom{2m}{m} b m = 4 − m ( m 2 m ) ( − 1 ) m b m (-1)^m b_m ( − 1 ) m b m A δ ( m ) / D δ ( m ) A_\delta(m)/D_\delta(m) A δ ( m ) / D δ ( m ) a , b a,b a , b O a , b ( log n ) O_{a,b}(\log n) O a , b ( log n )
H δ = ⌊ q − δ 2 ⌋ , K δ = b + δ + H δ , (22) H_\delta=\left\lfloor\frac{q-\delta}{2}\right\rfloor,\qquad K_\delta=b+\delta+H_\delta, \tag*{(22)} H δ = ⌊ 2 q − δ ⌋ , K δ = b + δ + H δ , (22) D δ ( z ) = 2 δ ∏ j = 1 K δ ( z + j ) , (23) D_\delta(z)=2^\delta\prod_{j=1}^{K_\delta}(z+j), \tag*{(23)} D δ ( z ) = 2 δ j = 1 ∏ K δ ( z + j ) , (23) A δ ( z ) = ( ∏ j = 1 b + δ ( 2 z + j ) ) ∑ h = 0 H δ ( − 1 ) h ( q δ + 2 h ) z − h − ∏ j = b + δ + h + 1 K δ ( z + j ) . (24) A_\delta(z)=\left(\prod_{j=1}^{b+\delta}(2z+j)\right)\sum_{h=0}^{H_\delta}(-1)^h\binom{q}{\delta+2h}z^{-h}-\prod_{j=b+\delta+h+1}^{K_\delta}(z+j). \tag*{(24)} A δ ( z ) = ( j = 1 ∏ b + δ ( 2 z + j ) ) h = 0 ∑ H δ ( − 1 ) h ( δ + 2 h q ) z − h − j = b + δ + h + 1 ∏ K δ ( z + j ) . (24) where z h = z ( z − 1 ) ⋯ ( z − h + 1 ) z^h = z(z-1)\cdots(z-h+1) z h = z ( z − 1 ) ⋯ ( z − h + 1 ) z 0 = 1 z^0=1 z 0 = 1 A δ , D δ A_\delta,D_\delta A δ , D δ K δ K_\delta K δ A δ ( 0 ) > 0 A_\delta(0)>0 A δ ( 0 ) > 0 § 2 that
B = ( n + a a ) , S = 2 n + s + 1 , T = ∏ j = 1 s ( n + j ) . B=\binom{n+a}{a},\qquad S=2n+s+1,\qquad T=\prod_{j=1}^{s}(n+j). B = ( a n + a ) , S = 2 n + s + 1 , T = j = 1 ∏ s ( n + j ) . Put
M = m + s + 2 , c a , b = ( a + b + 1 ) a + b + 1 ( 2 a + 1 ) a + 1 / 2 ( 2 b + 1 ) b + 1 / 2 . (25) M=m+s+2,\qquad c_{a,b}=\frac{(a+b+1)^{a+b+1}}{(2a+1)^{a+1/2}(2b+1)^{b+1/2}}. \tag*{(25)} M = m + s + 2 , c a , b = ( 2 a + 1 ) a + 1/2 ( 2 b + 1 ) b + 1/2 ( a + b + 1 ) a + b + 1 . (25) Theorem 3.1. Suppose m ≥ 2 s + 10 m\geq2s+10 m ≥ 2 s + 10 A δ ( m ) ≠ 0 A_\delta(m)\ne0 A δ ( m ) = 0
n log 2 > ( M + S ) log 3 + 3 log T + log ∣ A δ ( m ) ∣ n\log2>(M+\sqrt{S})\log3+3\log T+\log\lvert A_\delta(m)\rvert n log 2 > ( M + S ) log 3 + 3 log T + log ∣ A δ ( m )∣ + log ( 8 ( m + 1 ) S B D δ ( m ) c a , b m ) , (26) +\log\left(\frac{8(m+1)SBD_\delta(m)}{c_{a,b}\sqrt{m}}\right), \tag*{(26)} + log ( c a , b m 8 ( m + 1 ) S B D δ ( m ) ) , (26) then P n ( a , b ) P_n^{(a,b)} P n ( a , b ) Q \mathbb{Q} Q
For fixed a , b a,b a , b ( n / 2 ) log 3 + O a , b ( n ) (n/2)\log3+O_{a,b}(\sqrt{n}) ( n /2 ) log 3 + O a , b ( n ) n log 2 n\log2 n log 2 log 2 − 1 2 log 3 \log2-\frac{1}{2}\log3 log 2 − 2 1 log 3
Central values and reduction
We first prove the central-value formula. Its exact numerator is needed to determine the exceptional primes; a lower bound for its absolute value will also give a lower bound for derivatives at roots.
Lemma 3.2. For m ≥ 0 m\geq0 m ≥ 0
P 2 m + δ ( a , b ) ( 0 ) = ( − 1 ) m b m A δ ( m ) D δ ( m ) . (27) P_{2m+\delta}^{(a,b)}(0)=(-1)^m b_m\frac{A_\delta(m)}{D_\delta(m)}. \tag*{(27)} P 2 m + δ ( a , b ) ( 0 ) = ( − 1 ) m b m D δ ( m ) A δ ( m ) . (27) If m ≥ 1 m\geq1 m ≥ 1 A δ ( m ) ≠ 0 A_\delta(m)\ne0 A δ ( m ) = 0
∣ P 2 m + δ ( a , b ) ( 0 ) ∣ ≥ 1 2 m D δ ( m ) . \left|P_{2m+\delta}^{(a,b)}(0)\right|\geq\frac{1}{2\sqrt{m}D_\delta(m)}. P 2 m + δ ( a , b ) ( 0 ) ≥ 2 m D δ ( m ) 1 . In particular, each parity has only finitely many zero central values.
Proof. Rodrigues’ formula [26 (4.3.1)] gives
P n ( a , b ) ( 0 ) = ( − 1 ) n 2 − n [ t n ] ( 1 − t ) n + a ( 1 + t ) n + b . P_n^{(a,b)}(0)=(-1)^n2^{-n}[t^n](1-t)^{n+a}(1+t)^{n+b}. P n ( a , b ) ( 0 ) = ( − 1 ) n 2 − n [ t n ] ( 1 − t ) n + a ( 1 + t ) n + b . Expanding ( 1 − t ) q ( 1 − t 2 ) n + b (1-t)^q(1-t^2)^{n+b} ( 1 − t ) q ( 1 − t 2 ) n + b
P n ( a , b ) ( 0 ) = ( − 1 ) m 2 − 2 m − δ ∑ h = 0 H δ ( − 1 ) h ( q δ + 2 h ) ( 2 m + δ + b m − h ) . P_n^{(a,b)}(0)=(-1)^m2^{-2m-\delta}\sum_{h=0}^{H_\delta}(-1)^h\binom{q}{\delta+2h}\binom{2m+\delta+b}{m-h}. P n ( a , b ) ( 0 ) = ( − 1 ) m 2 − 2 m − δ h = 0 ∑ H δ ( − 1 ) h ( δ + 2 h q ) ( m − h 2 m + δ + b ) . The identity
( 2 m + δ + b m − h ) ( 2 m m ) = ∏ j = 1 b + δ ( 2 m + j ) m h ∏ j = 1 b + δ + h ( m + j ) \frac{\binom{2m+\delta+b}{m-h}}{\binom{2m}{m}}=
\frac{\prod_{j=1}^{b+\delta}(2m+j)m^h}{\prod_{j=1}^{b+\delta+h}(m+j)} ( m 2 m ) ( m − h 2 m + δ + b ) = ∏ j = 1 b + δ + h ( m + j ) ∏ j = 1 b + δ ( 2 m + j ) m h proves (27) . It also holds for h > m h>m h > m m = 0 m=0 m = 0 h > 0 h>0 h > 0 A δ A_\delta A δ ∣ A δ ( m ) ∣ ≥ 1 \lvert A_\delta(m)\rvert\geq1 ∣ A δ ( m )∣ ≥ 1 b m ≥ 1 / ( 2 m ) b_m\geq1/(2\sqrt{m}) b m ≥ 1/ ( 2 m ) m ≥ 1 m\geq1 m ≥ 1 b 1 = 1 / 2 b_1=1/2 b 1 = 1/2 b m + 1 / b m = ( 2 m + 1 ) / ( 2 m + 2 ) b_{m+1}/b_m=(2m+1)/(2m+2) b m + 1 / b m = ( 2 m + 1 ) / ( 2 m + 2 )
:::
For example, when ( a , b ) = ( 1 , 0 ) (a,b)=(1,0) ( a , b ) = ( 1 , 0 )
A 0 ( z ) = D 0 ( z ) = 1 , A 1 ( z ) = 2 z + 1 , D 1 ( z ) = 2 ( z + 1 ) . A_0(z)=D_0(z)=1,\qquad A_1(z)=2z+1,\qquad D_1(z)=2(z+1). A 0 ( z ) = D 0 ( z ) = 1 , A 1 ( z ) = 2 z + 1 , D 1 ( z ) = 2 ( z + 1 ) . Thus the central value never vanishes in either parity in this case.
The next lemma controls roots whose reduction is nonzero. For an odd prime p p p p p p F p \mathbb{F}_p F p
Lemma 3.3. Suppose m ≥ 2 s + 10 m\ge2s+10 m ≥ 2 s + 10 p > M p>M p > M p ∤ T p\nmid T p ∤ T n = ϵ p + j n=\epsilon p+j n = ϵ p + j ϵ ∈ { 0 , 1 } \epsilon\in\{0,1\} ϵ ∈ { 0 , 1 } 0 ≤ j < p 0\le j<p 0 ≤ j < p j + s < p j+s<p j + s < p
P n ( a , b ) ‾ = x ϵ p P j ( a , b ) ‾ . \overline{P_n^{(a,b)}}=x^{\epsilon p}\overline{P_j^{(a,b)}}. P n ( a , b ) = x ϵ p P j ( a , b ) . Every nonzero root of this reduction is simple, and the values at 1 1 1 − 1 -1 − 1
Proof. We have n + s < 2 p n+s<2p n + s < 2 p j + s ≥ p j+s\ge p j + s ≥ p p p p n + 1 n+1 n + 1 n + s n+s n + s p ∤ T p\nmid T p ∤ T ϵ = 0 \epsilon=0 ϵ = 0 ϵ = 1 \epsilon=1 ϵ = 1 (1) leaves only the ranges 0 ≤ k ≤ j 0\le k\le j 0 ≤ k ≤ j p ≤ k ≤ p + j p\le k\le p+j p ≤ k ≤ p + j k ≥ p − a k\ge p-a k ≥ p − a k ≤ j + b k\le j+b k ≤ j + b j + s < p j+s<p j + s < p
2 − p { ( x + 1 ) p + ( x − 1 ) p } P j ( a , b ) ‾ = x p P j ( a , b ) ‾ . 2^{-p}\{(x+1)^p+(x-1)^p\}\overline{P_j^{(a,b)}}=x^p\overline{P_j^{(a,b)}}. 2 − p {( x + 1 ) p + ( x − 1 ) p } P j ( a , b ) = x p P j ( a , b ) . The endpoint values of P j ( a , b ) P_j^{(a,b)} P j ( a , b ) ( j + a a ) \binom{j+a}{a} ( a j + a ) ( − 1 ) j ( j + b b ) (-1)^j\binom{j+b}{b} ( − 1 ) j ( b j + b ) p p p
( 1 − x 2 ) y ′ ′ + { b − a − ( a + b + 2 ) x } y ′ + j ( j + s + 1 ) y = 0. (28) (1-x^2)y''+\{b-a-(a+b+2)x\}y'+j(j+s+1)y=0. \tag*{(28)} ( 1 − x 2 ) y ′′ + { b − a − ( a + b + 2 ) x } y ′ + j ( j + s + 1 ) y = 0. (28) A root away from the endpoints with multiplicity t ≥ 2 t\ge2 t ≥ 2 t ( t − 1 ) t(t-1) t ( t − 1 ) t < p t<p t < p x p x^p x p
Comparison and effectivity.
Proof of Theorem 3.1. Put f = P n ( a , b ) f=P_n^{(a,b)} f = P n ( a , b ) H = κ f = F G H=\kappa f=FG H = κ f = F G 1 ≤ d = deg F ≤ e = deg G 1\le d=\deg F\le e=\deg G 1 ≤ d = deg F ≤ e = deg G R = Res ( F , G ) R=\operatorname{Res}(F,G) R = Res ( F , G ) R = ∣ R ∣ / 2 v 2 ( R ) \mathcal{R}=|R|/2^{v_2(R)} R = ∣ R ∣/ 2 v 2 ( R )
We first estimate the odd part from above. For odd p p p x = 2 X + 1 x=2X+1 x = 2 X + 1 Lemma 2.4 gives v p ( R ) ≤ d ( L p + v p ( T ) ) v_p(R)\le d(L_p+v_p(T)) v p ( R ) ≤ d ( L p + v p ( T ))
p > M , p ∤ T A δ ( m ) ⟹ v p ( R ) = 0. (29) p>M,\quad p\nmid TA_\delta(m)\quad\Longrightarrow\quad v_p(R)=0. \tag*{(29)} p > M , p ∤ T A δ ( m ) ⟹ v p ( R ) = 0. (29) To see this, f ( 1 ) = B f(1)=B f ( 1 ) = B f f f p p p κ \kappa κ Lemma 3.3 makes every unit root simple modulo p p p p 2 > 2 n + s p^2>2n+s p 2 > 2 n + s 2 − n ( 2 n + s n ) 2^{-n}\binom{2n+s}{n} 2 − n ( n 2 n + s ) f ( 0 ) f(0) f ( 0 ) p > m + K δ p>m+K_\delta p > m + K δ p ∤ A δ ( m ) p\nmid A_\delta(m) p ∤ A δ ( m )
For a polynomial U U U
v p ( U ( 0 ) ) = ∑ α max { v p ( α ) , 0 } , v p ( lc U ) = − ∑ α min { v p ( α ) , 0 } . (30) v_p(U(0))=\sum_\alpha\max\{v_p(\alpha),0\},\qquad
v_p(\operatorname{lc}U)=-\sum_\alpha\min\{v_p(\alpha),0\}. \tag*{(30)} v p ( U ( 0 )) = α ∑ max { v p ( α ) , 0 } , v p ( lc U ) = − α ∑ min { v p ( α ) , 0 } . (30) For each primitive integral factor, both sums are nonnegative integers. A total of at most 1 therefore cannot be positive in both F F F G G G f f f (29) .
The terms v p ( T ) v_p(T) v p ( T ) log T \log T log T d d d p ≤ M p\le M p ≤ M Lemma 2.7 bounds the remaining contribution by ( M + S ) log 3 (M+\sqrt{S})\log3 ( M + S ) log 3 p > M p>M p > M p 2 > S p^2>S p 2 > S T A δ ( m ) TA_\delta(m) T A δ ( m )
log R d < ( M + S ) log 3 + 2 log T + log ∣ A δ ( m ) ∣ . (31) \frac{\log\mathcal{R}}{d}<(M+\sqrt{S})\log3+2\log T+\log|A_\delta(m)|. \tag*{(31)} d log R < ( M + S ) log 3 + 2 log T + log ∣ A δ ( m ) ∣. (31) We now need a lower bound for ∣ f ′ ( α ) ∣ |f'(\alpha)| ∣ f ′ ( α ) ∣ α \alpha α f f f λ = n ( n + s + 1 ) \lambda=n(n+s+1) λ = n ( n + s + 1 )
E ( x ) = ( 1 − x 2 ) f ′ ( x ) 2 + λ f ( x ) 2 . E(x)=(1-x^2)f'(x)^2+\lambda f(x)^2. E ( x ) = ( 1 − x 2 ) f ′ ( x ) 2 + λ f ( x ) 2 . At a root α \alpha α ( 1 − α 2 ) f ′ ( α ) 2 (1-\alpha^2)f'(\alpha)^2 ( 1 − α 2 ) f ′ ( α ) 2 E E E
E ′ ( x ) = 2 { q + ( s + 1 ) x } f ′ ( x ) 2 . E'(x)=2\{q+(s+1)x\}f'(x)^2. E ′ ( x ) = 2 { q + ( s + 1 ) x } f ′ ( x ) 2 . The minimum on [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] x 0 = − q / ( s + 1 ) x_0=-q/(s+1) x 0 = − q / ( s + 1 ) [ x 0 , 0 ] [x_0,0] [ x 0 , 0 ]
0 ≤ E ′ ( x ) E ( x ) ≤ 2 { q + ( s + 1 ) x } 1 − x 2 . 0\leq\frac{E'(x)}{E(x)}\leq\frac{2\{q+(s+1)x\}}{1-x^2}. 0 ≤ E ( x ) E ′ ( x ) ≤ 1 − x 2 2 { q + ( s + 1 ) x } . The integral of the last expression shows that E ( x 0 ) ≥ c a , b E ( 0 ) \sqrt{E(x_0)}\geq c_{a,b}\sqrt{E(0)} E ( x 0 ) ≥ c a , b E ( 0 ) (25) . Lemma 3.2 yields
E ( x 0 ) ≥ c a , b λ ∣ f ( 0 ) ∣ ≥ c a , b m D δ ( m ) = : γ . \sqrt{E(x_0)}\geq c_{a,b}\sqrt{\lambda}\,|f(0)|\geq\frac{c_{a,b}\sqrt{m}}{D_\delta(m)}=:\gamma. E ( x 0 ) ≥ c a , b λ ∣ f ( 0 ) ∣ ≥ D δ ( m ) c a , b m =: γ . At every root α \alpha α α ≠ 0 \alpha\ne0 α = 0 ∣ f ′ ( α ) ∣ > γ |f'(\alpha)|>\gamma ∣ f ′ ( α ) ∣ > γ b h = B b_h=B b h = B D 2 = D 2 D_2=\mathcal{D}_2 D 2 = D 2 d ≤ m + 1 d\leq m+1 d ≤ m + 1 D 2 log 2 ≤ log S + log T \mathcal{D}_2\log2\leq\log S+\log T D 2 log 2 ≤ log S + log T
log R d > n log 2 − log T − log ( 8 ( m + 1 ) S B D δ ( m ) c a , b m ) . (32) \frac{\log\mathcal{R}}{d}>n\log2-\log T-\log\left(\frac{8(m+1)SBD_\delta(m)}{c_{a,b}\sqrt{m}}\right). \tag*{(32)} d log R > n log 2 − log T − log ( c a , b m 8 ( m + 1 ) S B D δ ( m ) ) . (32) The two bounds contradict (26) . □
Irreducibility in Theorem 1.3 . Fix a > b a>b a > b A 0 ( m ) A_0(m) A 0 ( m ) A 1 ( m ) A_1(m) A 1 ( m )
∣ A δ ( m ) ∣ ≤ C ( m + 1 ) K δ , D δ ( m ) ≤ 2 ( m + K δ ) K δ , T ≤ ( n + s ) a , B ≤ ( n + a ) a . |A_\delta(m)|\leq C(m+1)^{K_\delta},\qquad D_\delta(m)\leq2(m+K_\delta)^{K_\delta},\qquad T\leq(n+s)^a,\qquad B\leq(n+a)^a. ∣ A δ ( m ) ∣ ≤ C ( m + 1 ) K δ , D δ ( m ) ≤ 2 ( m + K δ ) K δ , T ≤ ( n + s ) a , B ≤ ( n + a ) a . For sufficiently large m m m (26) is at least
m log ( 4 / 3 ) − C 1 m − C 2 log ( m + 1 ) − C 3 , m\log(4/3)-C_1\sqrt{m}-C_2\log(m+1)-C_3, m log ( 4/3 ) − C 1 m − C 2 log ( m + 1 ) − C 3 , with effective nonnegative constants depending on a , b a,b a , b m m m m m m b > a b>a b > a
Symmetric parameters
Equal parameters give an additional restriction on the reductions. At an odd prime, parity pairs every nonzero root with its distinct negative. A multiple root away from the central point and the endpoints must therefore occur in an opposite pair. The differential equation forces these multiplicities to be large, and the degree then excludes them at the primes under consideration. This is what removes the exceptional large primes present in the unequal-parameter argument.
Let r , m r,m r , m r ≥ 0 r\geq0 r ≥ 0 m ≥ 1 m\geq1 m ≥ 1 n = 2 m + δ n=2m+\delta n = 2 m + δ δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } (2) :
f ( x ) = P n + r ( r ) ( x ) = ( 2 r − 1 ) ! ! C n r + 1 / 2 ( x ) . f(x)=P_{n+r}^{(r)}(x)=(2r-1)!!\,C_n^{r+1/2}(x). f ( x ) = P n + r ( r ) ( x ) = ( 2 r − 1 )!! C n r + 1/2 ( x ) . Set
S = 2 n + 2 r + 1 , M = m + r , A = ( n + r ) ! 2 n ! ( n + 2 r ) ! , K = ( m + r r ) , B = ( n + r r ) , (33) S=2n+2r+1,\qquad M=m+r,\qquad A=\frac{(n+r)!^2}{n!(n+2r)!},\qquad K=\binom{m+r}{r},\qquad B=\binom{n+r}{r}, \tag*{(33)} S = 2 n + 2 r + 1 , M = m + r , A = n ! ( n + 2 r )! ( n + r ) ! 2 , K = ( r m + r ) , B = ( r n + r ) , (33) and put A ≤ m = ∏ p ≤ m p v p ( A ) A_{\leq m}=\prod_{p\leq m}p^{v_p(A)} A ≤ m = ∏ p ≤ m p v p ( A ) v p ( A ) v_p(A) v p ( A ) r r r A ≤ 1 A\leq1 A ≤ 1 A ≤ m A_{\le m} A ≤ m r = 1 r=1 r = 1 m = 4 m=4 m = 4 δ = 0 \delta=0 δ = 0 A = 9 / 10 A=9/10 A = 9/10 A ≤ m = 9 / 2 A_{\le m}=9/2 A ≤ m = 9/2
The exact criterion.
Theorem 3.4. Suppose m ≥ max { 2 r , 4 } m \ge\max\{2r,4\} m ≥ max { 2 r , 4 }
m log 4 > Ψ m ( S ) + log ( 8 m ) + log K + log A ≤ m , (34) m\log4 > \Psi_m(S) + \log(8\sqrt{m}) + \log K + \log A_{\le m}, \tag*{(34)} m log 4 > Ψ m ( S ) + log ( 8 m ) + log K + log A ≤ m , (34) then C 2 m + δ r + 1 / 2 ( x ) / x δ C_{2m+\delta}^{r+1/2}(x)/x^\delta C 2 m + δ r + 1/2 ( x ) / x δ Q \mathbb{Q} Q M ≥ 9 M \ge9 M ≥ 9
m log 4 > ( M + S ) log 3 + log ( 8 m ) + log K + 2 log B − log ( 2 r r ) . (35) m\log4 > (M+\sqrt{S})\log3 + \log(8\sqrt{m}) + \log K + 2\log B - \log\binom{2r}{r}. \tag*{(35)} m log 4 > ( M + S ) log 3 + log ( 8 m ) + log K + 2 log B − log ( r 2 r ) . (35) Proof. Apply Lemma 2.4 to Q ( X ) = B − 1 P n ( r , r ) ( 2 X + 1 ) Q(X)=B^{-1}P_n^{(r,r)}(2X+1) Q ( X ) = B − 1 P n ( r , r ) ( 2 X + 1 )
D p = L p + v p ( A ) , L p = ⌊ log p S ⌋ . (36) D_p=L_p+v_p(A), \qquad L_p=\lfloor\log_p S\rfloor. \tag*{(36)} D p = L p + v p ( A ) , L p = ⌊ log p S ⌋ . (36) This number is nonnegative, since it bounds a nonnegative row sum. Removing the zero forced by odd parity removes only nonnegative distances, so the same bound holds for h = f / x δ h=f/x^\delta h = f / x δ 2 X + 1 2X+1 2 X + 1
Differentiation of the Legendre coefficient formula gives
f ( x ) = 2 − n − r ∑ k = 0 m ( − 1 ) k ( 2 n + 2 r − 2 k ) ! k ! ( n + r − k ) ! ( n − 2 k ) ! x n − 2 k . (37) f(x)=2^{-n-r}\sum_{k=0}^{m}\frac{(-1)^k(2n+2r-2k)!}{k!(n+r-k)!(n-2k)!}x^{n-2k}. \tag*{(37)} f ( x ) = 2 − n − r k = 0 ∑ m k ! ( n + r − k )! ( n − 2 k )! ( − 1 ) k ( 2 n + 2 r − 2 k )! x n − 2 k . (37) Its nonzero central coefficient is
[ x δ ] f ( x ) = ( − 1 ) m 2 m m ! ( 2 m + 2 r + 2 δ − 1 ) ! ! . (38) [x^\delta]f(x)=\frac{(-1)^m}{2^m m!}(2m+2r+2\delta-1)!!. \tag*{(38)} [ x δ ] f ( x ) = 2 m m ! ( − 1 ) m ( 2 m + 2 r + 2 δ − 1 )!! . (38) The other roots are simple and lie in ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) κ > 0 \kappa>0 κ > 0 H = κ h H=\kappa h H = κh H = F G H=FG H = F G 1 ≤ d = deg F ≤ e = deg G 1\le d=\deg F\le e=\deg G 1 ≤ d = deg F ≤ e = deg G d + e = 2 m d+e=2m d + e = 2 m R = Res ( F , G ) R=\operatorname{Res}(F,G) R = Res ( F , G ) R = ∣ R ∣ / 2 v 2 ( R ) \mathcal{R}=|R|/2^{v_2(R)} R = ∣ R ∣/ 2 v 2 ( R )
We claim that
p > m ⟹ v p ( R ) = 0. (39) p>m \quad\Longrightarrow\quad v_p(R)=0. \tag*{(39)} p > m ⟹ v p ( R ) = 0. (39) Such a prime is odd, exceeds 2 r 2r 2 r p 2 > S p^2>S p 2 > S x δ H = κ f x^\delta H=\kappa f x δ H = κ f n n n n < 2 p n<2p n < 2 p 0 , 1 , − 1 0,1,-1 0 , 1 , − 1 p p p 2 p 2p 2 p
For the four remaining residue classes 0 , 1 , − 1 , ∞ 0,1,-1,\infty 0 , 1 , − 1 , ∞ (38) has valuation at most 1: its odd numerator factors are smaller than 3 p 3p 3 p p > m p>m p > m
lc ( f ) = ( 2 n + 2 r − 1 ) ! ! n ! . \operatorname{lc}(f)=\frac{(2n+2r-1)!!}{n!}. lc ( f ) = n ! ( 2 n + 2 r − 1 )!! . Its odd numerator factors are smaller than 5 p 5p 5 p p ≤ n p\le n p ≤ n p p p 3 p 3p 3 p n ! n! n ! p p p p > n p>n p > n 3 p 3p 3 p n ! n! n !
f ( 1 ) = ( n + 2 r ) ! 2 r r ! n ! = ( n + 2 r 2 r ) ( 2 r − 1 ) ! ! , ∣ f ( − 1 ) ∣ = f ( 1 ) . (40) f(1)=\frac{(n+2r)!}{2^r r!n!}=\binom{n+2r}{2r}(2r-1)!!,\qquad|f(-1)|=f(1). \tag*{(40)} f ( 1 ) = 2 r r ! n ! ( n + 2 r )! = ( 2 r n + 2 r ) ( 2 r − 1 )!! , ∣ f ( − 1 ) ∣ = f ( 1 ) . (40) The numerator product contains 2 r < p 2r<p 2 r < p p 2 p^2 p 2 (30) , and its translates at 1 and − 1 -1 − 1 F F F G G G (39) .
The row bound at the remaining odd primes gives
log R d ≤ Ψ m ( S ) + log A ≤ m − D 2 log 2. (41) \frac{\log R}{d} \le\Psi_m(S) + \log A_{\le m} - D_2 \log2. \tag*{(41)} d log R ≤ Ψ m ( S ) + log A ≤ m − D 2 log 2. (41) For the real estimate, the equation for f f f
( 1 − x 2 ) f ′ ′ − ( 2 r + 2 ) x f ′ + n ( n + 2 r + 1 ) f = 0. (1-x^2)f''-(2r+2)xf'+n(n+2r+1)f=0. ( 1 − x 2 ) f ′′ − ( 2 r + 2 ) x f ′ + n ( n + 2 r + 1 ) f = 0. With λ = n ( n + 2 r + 1 ) \lambda=n(n+2r+1) λ = n ( n + 2 r + 1 ) E = ( 1 − x 2 ) f ′ 2 + λ f 2 E=(1-x^2)f'^2+\lambda f^2 E = ( 1 − x 2 ) f ′2 + λ f 2 E ′ = 2 ( 2 r + 1 ) x f ′ 2 E'=2(2r+1)xf'^2 E ′ = 2 ( 2 r + 1 ) x f ′2 E ( x ) ≥ E ( 0 ) = γ 2 E(x)\ge E(0)=\gamma^2 E ( x ) ≥ E ( 0 ) = γ 2 γ 2 = λ f ( 0 ) 2 + f ′ ( 0 ) 2 > 0 \gamma^2=\lambda f(0)^2+f'(0)^2>0 γ 2 = λ f ( 0 ) 2 + f ′ ( 0 ) 2 > 0 α \alpha α ∣ h ′ ( α ) ∣ = ∣ f ′ ( α ) ∣ / ∣ α ∣ δ > γ |h'(\alpha)|=|f'(\alpha)|/|\alpha|^\delta>\gamma ∣ h ′ ( α ) ∣ = ∣ f ′ ( α ) ∣/∣ α ∣ δ > γ
γ f ( 1 ) = { 2 m ( 2 m + 2 r + 1 ) b m / K , δ = 0 , ( 2 m + 1 ) b m / K , δ = 1 , γ f ( 1 ) > m K . (42) \frac{\gamma}{f(1)}=
\begin{cases}
\sqrt{2m(2m+2r+1)}\,b_m/K, & \delta=0,\\
(2m+1)b_m/K, & \delta=1,
\end{cases}
\qquad
\frac{\gamma}{f(1)}>\frac{\sqrt{m}}{K}. \tag*{(42)} f ( 1 ) γ = { 2 m ( 2 m + 2 r + 1 ) b m / K , ( 2 m + 1 ) b m / K , δ = 0 , δ = 1 , f ( 1 ) γ > K m . (42) Lemma 2.6 , with N = 2 m N=2m N = 2 m d ≤ m d\le m d ≤ m
log R d > m log 4 − log ( 8 m ) − log K − D 2 log 2. (43) \frac{\log R}{d}>m\log4-\log(8\sqrt{m})-\log K-D_2\log2. \tag*{(43)} d log R > m log 4 − log ( 8 m ) − log K − D 2 log 2. (43) Comparing (41) and (43) proves (34) . In particular the binary contributions cancel exactly.
For the elementary criterion, retain the nonnegativity of D p D_p D p A A A ∏ j = 1 2 r ( n + j ) \prod_{j=1}^{2r}(n+j) ∏ j = 1 2 r ( n + j ) p > M p>M p > M 2 p > n + 2 r 2p>n+2r 2 p > n + 2 r ( n , n + 2 r ] (n,n+2r] ( n , n + 2 r ]
Ψ m ( S ) + log A ≤ m ≤ Ψ M ( S ) + ∑ n < p ≤ n + 2 r log p + log A . (44) \Psi_m(S)+\log A_{\le m}\le\Psi_M(S)+\sum_{n<p\le n+2r}\log p+\log A. \tag*{(44)} Ψ m ( S ) + log A ≤ m ≤ Ψ M ( S ) + n < p ≤ n + 2 r ∑ log p + log A . (44) The difference is a sum of nonnegative terms D p log p D_p\log p D p log p ( n , n + 2 r ] (n,n+2r] ( n , n + 2 r ] ( n + 2 r 2 r ) \binom{n+2r}{2r} ( 2 r n + 2 r )
log ( n + 2 r 2 r ) + log A = 2 log B − log ( 2 r r ) . \log\binom{n+2r}{2r}+\log A=2\log B-\log\binom{2r}{r}. log ( 2 r n + 2 r ) + log A = 2 log B − log ( r 2 r ) . Since S ≤ 5 M S\le5M S ≤ 5 M Lemma 2.7 proves (35) . □
Uniform linear ranges
The elementary criterion first gives an explicit range by bounding each additional term relative to m m m
The explicit assertion of Theorem 1.4 . Suppose m ≥ max { 100 r , 1000 } m\ge\max\{100r,1000\} m ≥ max { 100 r , 1000 } S ≤ 4.05 m S\le4.05m S ≤ 4.05 m ( u v ) ≤ ( e u / v ) v \binom{u}{v}\le(eu/v)^v ( v u ) ≤ ( e u / v ) v r r r
log K + 2 log B m ≤ 3 + log 101 + 2 log 201.1 100 < 0.1824. \frac{\log K+2\log B}{m}\le\frac{3+\log101+2\log201.1}{100}<0.1824. m log K + 2 log B ≤ 100 3 + log 101 + 2 log 201.1 < 0.1824. For r = 0 r=0 r = 0
( M − m ) log 3 m < 0.011 , S log 3 m < 1.1 ⋅ 2.1 31 < 0.0746 , log ( 8 m ) m < 0.006. \frac{(M-m)\log3}{m}<0.011,\qquad
\frac{\sqrt{S}\log3}{m}<\frac{1.1\cdot2.1}{31}<0.0746,\qquad
\frac{\log(8\sqrt{m})}{m}<0.006. m ( M − m ) log 3 < 0.011 , m S log 3 < 31 1.1 ⋅ 2.1 < 0.0746 , m log ( 8 m ) < 0.006. These bounds hold at m = 1000 m=1000 m = 1000 − log ( 2 r r ) -\log\binom{2r}{r} − log ( r 2 r ) (35) , the excess over m log 3 m\log3 m log 3 m m m
0.1824 + 0.011 + 0.0746 + 0.006 = 0.274 < 2 7 < log ( 4 / 3 ) . 0.1824+0.011+0.0746+0.006=0.274<\frac{2}{7}<\log(4/3). 0.1824 + 0.011 + 0.0746 + 0.006 = 0.274 < 7 2 < log ( 4/3 ) . The last inequality follows from the power series for 2 arctanh ( 1 / 7 ) 2\operatorname{arctanh}(1/7) 2 arctanh ( 1/7 )
When r r r m m m K K K A A A m m m r ≥ 1 r \ge1 r ≥ 1 t = m / r t=m/r t = m / r (34) , divided by m m m 1 + g ( t ) / t + o ( 1 ) 1+g(t)/t+o(1) 1 + g ( t ) / t + o ( 1 ) log 4 \log4 log 4 1 + g ( t ) / t < log 4 1+g(t)/t<\log4 1 + g ( t ) / t < log 4
g ( t ) = ( 4 t + 2 ) log ( t + 1 / 2 ) − 3 t log t − ( t + 1 ) log ( t + 1 ) . (45) g(t)=(4t+2)\log(t+1/2)-3t\log t-(t+1)\log(t+1). \tag*{(45)} g ( t ) = ( 4 t + 2 ) log ( t + 1/2 ) − 3 t log t − ( t + 1 ) log ( t + 1 ) . (45) Lemma 3.5. Uniformly for integers 1 ≤ r ≤ m / 2 1 \le r \le m/2 1 ≤ r ≤ m /2 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } m → ∞ m \to\infty m → ∞
Ψ m ( S ) + log K + log A ≤ m = m + r g ( m / r ) + o ( m ) . (46) \Psi_m(S)+\log K+\log A_{\le m}=m+r\,g(m/r)+o(m). \tag*{(46)} Ψ m ( S ) + log K + log A ≤ m = m + r g ( m / r ) + o ( m ) . (46) The error has an effective bound independent of r r r δ \delta δ g ( t ) / t g(t)/t g ( t ) / t t ≥ 2 t \ge2 t ≥ 2
Proof. The first prime powers contribute ϑ ( m ) = m + o ( m ) \vartheta(m)=m+o(m) ϑ ( m ) = m + o ( m ) O ( m log m ) = o ( m ) O(\sqrt{m}\log m)=o(m) O ( m log m ) = o ( m ) S ≤ 5 m + 3 S\le5m+3 S ≤ 5 m + 3 p > m p>m p > m p 2 > n + 2 r p^2>n+2r p 2 > n + 2 r m m m 3 p > n + 2 r 3p>n+2r 3 p > n + 2 r
A = ∏ j = 1 r ( n + j ) ∏ j = r + 1 2 r ( n + j ) A=\frac{\prod_{j=1}^{r}(n+j)}{\prod_{j=r+1}^{2r}(n+j)} A = ∏ j = r + 1 2 r ( n + j ) ∏ j = 1 r ( n + j ) are p p p 2 p 2p 2 p
log A − log A ≤ m = 2 ϑ ( n + r ) − ϑ ( n ) − ϑ ( n + 2 r ) + 2 ϑ ( ( n + r ) / 2 ) − ϑ ( n / 2 ) − ϑ ( ( n + 2 r ) / 2 ) . (47) \log A-\log A_{\le m}=2\vartheta(n+r)-\vartheta(n)-\vartheta(n+2r)+2\vartheta((n+r)/2)-\vartheta(n/2)-\vartheta((n+2r)/2). \tag*{(47)} log A − log A ≤ m = 2 ϑ ( n + r ) − ϑ ( n ) − ϑ ( n + 2 r ) + 2 ϑ (( n + r ) /2 ) − ϑ ( n /2 ) − ϑ (( n + 2 r ) /2 ) . (47) Every argument is between fixed positive multiples of m m m o ( m ) o(m) o ( m ) [12 Theorem 5.2] , makes this error effective.
The elementary factorial estimate log ( k ! ) = k log k − k + O ( log ( k + 1 ) ) \log(k!)=k\log k-k+O(\log(k+1)) log ( k !) = k log k − k + O ( log ( k + 1 )) log x \log x log x
log K + log A = r g ( m / r ) + O ( log m ) . \log K+\log A=r\,g(m/r)+O(\log m). log K + log A = r g ( m / r ) + O ( log m ) . Indeed, for t = m / r t=m/r t = m / r r r r
( t + 1 ) log ( t + 1 ) − t log t (t+1)\log(t+1)-t\log t ( t + 1 ) log ( t + 1 ) − t log t and
2 ( 2 t + 1 ) log ( 2 t + 1 ) − 2 t log ( 2 t ) − ( 2 t + 2 ) log ( 2 t + 2 ) . 2(2t+1)\log(2t+1)-2t\log(2t)-(2t+2)\log(2t+2). 2 ( 2 t + 1 ) log ( 2 t + 1 ) − 2 t log ( 2 t ) − ( 2 t + 2 ) log ( 2 t + 2 ) . Their sum is g ( t ) g(t) g ( t ) δ \delta δ O ( log m ) O(\log m) O ( log m ) (46) . Finally,
t g ′ ( t ) − g ( t ) = log t + 1 ( t + 1 / 2 ) 2 < 0 ( t ≥ 2 ) . tg'(t)-g(t)=\log\frac{t+1}{(t+1/2)^2}<0 \qquad(t\ge2). t g ′ ( t ) − g ( t ) = log ( t + 1/2 ) 2 t + 1 < 0 ( t ≥ 2 ) . Proposition 3.6. Let c ≥ 2 c\ge2 c ≥ 2
c ( log 4 − 1 ) > g ( c ) . (48) c(\log4-1)>g(c). \tag*{(48)} c ( log 4 − 1 ) > g ( c ) . (48) For all sufficiently large m m m 0 ≤ r ≤ m / c 0\le r\le m/c 0 ≤ r ≤ m / c C 2 m + δ r + 1 / 2 ( x ) / x δ C_{2m+\delta}^{r+1/2}(x)/x^\delta C 2 m + δ r + 1/2 ( x ) / x δ Q \mathbb{Q} Q c c c c = 8 c=8 c = 8
Proof. For r ≥ 1 r\ge1 r ≥ 1 t = m / r ≥ c t=m/r\ge c t = m / r ≥ c
Ψ m ( S ) + log K + log A ≤ m ≤ m ( 1 + g ( c ) c ) + o ( m ) . \Psi_m(S)+\log K+\log A_{\le m}\le m\left(1+\frac{g(c)}{c}\right)+o(m). Ψ m ( S ) + log K + log A ≤ m ≤ m ( 1 + c g ( c ) ) + o ( m ) . The strict margin in (48) absorbs this error and log ( 8 m ) \log(8\sqrt{m}) log ( 8 m ) Theorem 3.4 . For r = 0 r=0 r = 0 K = A = 1 K=A=1 K = A = 1 c c c
8 ( log 4 − 1 ) − g ( 8 ) > 0.0097. 8(\log4-1)-g(8)>0.0097. 8 ( log 4 − 1 ) − g ( 8 ) > 0.0097. An exact rational verification is recorded in Appendix C. This also proves the remaining assertion of Theorem 1.4 . □ \square □
Legendre polynomials and their first derivatives.
Proof of Theorem 1.5 . The assertions for P 2 m P_{2m} P 2 m P 2 m + 1 / x P_{2m+1}/x P 2 m + 1 / x [10 Theorem 1.1] . To see their relation to the present argument, specialize Theorem 3.4 to r = 0 r=0 r = 0 K = A = 1 K=A=1 K = A = 1 Lemma 2.7 gives the sufficient inequality
m log ( 4 / 3 ) > 4 m + 3 log 3 + log ( 8 m ) . m\log(4/3)>\sqrt{4m+3\log3+\log(8\sqrt{m})}. m log ( 4/3 ) > 4 m + 3 log 3 + log ( 8 m ) . This holds for m ≥ 100 m\ge100 m ≥ 100 m m m m = 100 m=100 m = 100 0.275 < 2 / 7 < log ( 4 / 3 ) 0.275<2/7<\log(4/3) 0.275 < 2/7 < log ( 4/3 )
For the first derivatives, take r = 1 r=1 r = 1 n = 2 m + δ n=2m+\delta n = 2 m + δ (33) become
K = m + 1 , A = n + 1 n + 2 , S = 2 n + 3. K=m+1,\qquad A=\frac{n+1}{n+2},\qquad S=2n+3. K = m + 1 , A = n + 2 n + 1 , S = 2 n + 3. Since A ≤ m ≤ n + 1 ≤ 2 m + 2 A\le m\le n+1\le2m+2 A ≤ m ≤ n + 1 ≤ 2 m + 2
m log ( 4 / 3 ) > 4 m + 5 log 3 + log ( 8 m ) + log ( 2 ( m + 1 ) 2 ) . m\log(4/3)>\sqrt{4m+5\log3+\log(8\sqrt{m})}+\log(2(m+1)^2). m log ( 4/3 ) > 4 m + 5 log 3 + log ( 8 m ) + log ( 2 ( m + 1 ) 2 ) . After division by m m m m = 200 m=200 m = 200 ( 29 ⋅ 1.1 + 5 + 12 ) / 200 = 0.2445 < 2 / 7 (29\cdot1.1+5+12)/200=0.2445<2/7 ( 29 ⋅ 1.1 + 5 + 12 ) /200 = 0.2445 < 2/7 m ≥ 200. m\ge200. m ≥ 200. Lemma C.2 treats 1 ≤ m ≤ 199 1\le m\le199 1 ≤ m ≤ 199 □ \square □
The Kaneko–Zagier families
We turn to irreducibility of the four Kaneko–Zagier families. The comparison of § 2 requires two further estimates here: one for the common divisor introduced by clearing denominators, and one for the clustering of roots at 2 2 2 3 3 3 § 8 makes the comparison explicit and completes the finite range.
Primitive normalization
In this section and Appendix § 8 , unless stated otherwise, n ≥ 1 n\ge1 n ≥ 1
α = λ / 3 , β = μ / 2 , λ , μ ∈ { − 1 , 1 } , c = α + β , y n ( t ) = P n ( α , β ) ( 1 − 2 t ) . \alpha=\lambda/3,\qquad\beta=\mu/2,\qquad\lambda,\mu\in\{-1,1\},\qquad c=\alpha+\beta,\qquad y_n(t)=P_n^{(\alpha,\beta)}(1-2t). α = λ /3 , β = μ /2 , λ , μ ∈ { − 1 , 1 } , c = α + β , y n ( t ) = P n ( α , β ) ( 1 − 2 t ) . All constants implicit in asymptotic notation are uniform in these four choices. The functional I I I μ j \mu_j μ j C n C_n C n (9) and (10) , evaluated at these parameters. Put
κ n = 2 n + v 2 ( n ! ) 3 n + v 3 ( n ! ) , g n = cont ( κ n y n ) , H n = κ n g n y n . (49) \kappa_n=2^{n+v_2(n!)}3^{n+v_3(n!)},\qquad g_n=\operatorname{cont}(\kappa_n y_n),\qquad H_n=\frac{\kappa_n}{g_n}y_n. \tag*{(49)} κ n = 2 n + v 2 ( n !) 3 n + v 3 ( n !) , g n = cont ( κ n y n ) , H n = g n κ n y n . (49) Here cont is the positive gcd of the integral coefficients.
The factor κ n \kappa_n κ n 2 2 2 3 3 3 g n g_n g n g n − 1 g_n^{-1} g n − 1 g n g_n g n
Lemma 4.1. The polynomial H n H_n H n ( g n , 6 ) = 1 (g_n,6)=1 ( g n , 6 ) = 1
log κ n = ( 2 log 2 + 3 2 log 3 ) n + O ( log n ) . \log\kappa_n=\left(2\log2+\frac{3}{2}\log3\right)n+O(\log n). log κ n = ( 2 log 2 + 2 3 log 3 ) n + O ( log n ) . Proof. For p ≥ 5 p\ge5 p ≥ 5 (8) are p p p − j − v 2 ( j ! ) -j-v_2(j!) − j − v 2 ( j !)
− n − v 3 ( j ! ) − v 3 ( ( n − j ) ! ) = − n − v 3 ( n ! ) + v 3 ( n j ) . -n-v_3(j!)-v_3((n-j)!)=-n-v_3(n!)+v_3\binom{n}{j}. − n − v 3 ( j !) − v 3 (( n − j )!) = − n − v 3 ( n !) + v 3 ( j n ) . The minimum at either prime is attained by the leading coefficient. This proves the integral and unit assertions. The last formula follows from v p ( n ! ) = ( n − s p ( n ) ) / ( p − 1 ) v_p(n!)=(n-s_p(n))/(p-1) v p ( n !) = ( n − s p ( n )) / ( p − 1 ) s p ( n ) s_p(n) s p ( n ) p p p n n n
The resultant away from 2 and 3
Suppose that
H n = A B , A , B ∈ Z [ t ] primitive , k = deg A ≤ e = deg B , H_n=AB,\qquad A,B\in\mathbb{Z}[t]\text{ primitive},\qquad k=\deg A\le e=\deg B, H n = A B , A , B ∈ Z [ t ] primitive , k = deg A ≤ e = deg B , and put
R = ∣ Res ( A , B ) ∣ > 0 , R ( 6 ) = R 2 v 2 ( R ) 3 v 3 ( R ) . (50) R=\lvert\operatorname{Res}(A,B)\rvert>0,\qquad R^{(6)}=\frac{R}{2^{v_2(R)}3^{v_3(R)}}. \tag*{(50)} R = ∣ Res ( A , B )∣ > 0 , R ( 6 ) = 2 v 2 ( R ) 3 v 3 ( R ) R . (50) The main estimates in this section are
log g n = ( c + o ( 1 ) ) n , log R ( 6 ) ≤ ( u + o ( 1 ) ) k n , (51) \log g_n=(c+o(1))n,\qquad\log R^{(6)}\le(u+o(1))kn, \tag*{(51)} log g n = ( c + o ( 1 )) n , log R ( 6 ) ≤ ( u + o ( 1 )) k n , (51) where
c = π 2 3 , u = 43 20 − π 3 . (52) c=\frac{\pi}{2\sqrt{3}},\qquad u=\frac{43}{20}-\frac{\pi}{\sqrt{3}}. \tag*{(52)} c = 2 3 π , u = 20 43 − 3 π . (52) A valuation calculation
The content and the moment denominators come from closely related products. Estimating them separately loses the cancellation needed in the resultant comparison. We instead count divisibility by each prime power q q q m q m_q m q e q e_q e q m q + e q ≤ 1 m_q+e_q\le1 m q + e q ≤ 1
Fix p ≥ 5 p\ge5 p ≥ 5 q = p h q=p^h q = p h a q , b q a_q,b_q a q , b q α , β \alpha,\beta α , β q q q
c q = ( a q + b q ) m o d q , d q = ⌊ a q + b q q ⌋ , s = n m o d q . c_q=(a_q+b_q)\bmod q,\qquad d_q=\left\lfloor\frac{a_q+b_q}{q}\right\rfloor,\qquad s=n\bmod q. c q = ( a q + b q ) mod q , d q = ⌊ q a q + b q ⌋ , s = n mod q . Thus 1 ≤ a q , b q ≤ q − 1 1\le a_q,b_q\le q-1 1 ≤ a q , b q ≤ q − 1 d q ∈ { 0 , 1 } d_q\in\{0,1\} d q ∈ { 0 , 1 }
E a = 1 s + a q ≥ q , E b = 1 s + b q ≥ q , E c = 1 s + c q ≥ q , E_a=\mathbf{1}_{s+a_q\ge q},\qquad E_b=\mathbf{1}_{s+b_q\ge q},\qquad E_c=\mathbf{1}_{s+c_q\ge q}, E a = 1 s + a q ≥ q , E b = 1 s + b q ≥ q , E c = 1 s + c q ≥ q , m q = 1 E a = E b = 1 , d q + E c = 1 , η q = 1 c q = q − 1 , b q ∗ = { 0 , c q = q − 1 , 1 − d q , c q ≠ q − 1 , m_q=\mathbf{1}_{E_a=E_b=1,\ d_q+E_c=1},\qquad\eta_q=\mathbf{1}_{c_q=q-1},\qquad
b_q^*=
\begin{cases}
0,&c_q=q-1,\\
1-d_q,&c_q\ne q-1,
\end{cases} m q = 1 E a = E b = 1 , d q + E c = 1 , η q = 1 c q = q − 1 , b q ∗ = { 0 , 1 − d q , c q = q − 1 , c q = q − 1 , and
e q = η q + E a + E b − E c + b q ∗ − 2 m q . (53) e_q=\eta_q+E_a+E_b-E_c+b_q^*-2m_q. \tag*{(53)} e q = η q + E a + E b − E c + b q ∗ − 2 m q . (53) The notation b q ∗ b_q^* b q ∗ b q b_q b q
Lemma 4.2. For every q = p h q=p^h q = p h p ≥ 5 p\ge5 p ≥ 5 e q ∈ { 0 , 1 } e_q\in\{0,1\} e q ∈ { 0 , 1 } m q + e q ≤ 1 m_q+e_q\le1 m q + e q ≤ 1 M = 12 n + 5 M=12n+5 M = 12 n + 5
E p ( n ) = ∑ h ≥ 1 p h ≤ M e p h , m p = v p ( g n ) , E_p(n)=\sum_{\substack{h\ge1\\p^h\le M}}e_{p^h},\qquad m_p=v_p(g_n), E p ( n ) = h ≥ 1 p h ≤ M ∑ e p h , m p = v p ( g n ) , one has
v p ( C n ) − 2 m p − min 0 ≤ j ≤ 2 n − 1 v p ( μ j ) ≤ E p ( n ) ≤ ⌊ log p M ⌋ . (54) v_p(C_n)-2m_p-\min_{0\le j\le2n-1}v_p(\mu_j)\le E_p(n)\le\lfloor\log_p M\rfloor. \tag*{(54)} v p ( C n ) − 2 m p − 0 ≤ j ≤ 2 n − 1 min v p ( μ j ) ≤ E p ( n ) ≤ ⌊ log p M ⌋ . (54) If p > M p>\sqrt{M} p > M m p = m p l e v m_p=m_p^{\mathrm{lev}} m p = m p lev m p l e v m_p^{\mathrm{lev}} m p lev m q m_q m q q = p q=p q = p
Proof. For a rational p p p z z z q = p h q=p^h q = p h v p ( j z ) v_p\binom{j}{z} v p ( z j )
1 j m o d q > z m o d q . \mathbf{1}_{j\bmod q>z\bmod q}. 1 j mod q > z mod q . This follows by counting multiples of q q q t j t^j t j (8) , with u = j m o d q u=j\bmod q u = j mod q
1 ( s − u m o d q ) > ( s + a q m o d q ) + 1 u + ( s + c q m o d q ) ≥ q . (55) \mathbf{1}_{(s-u\bmod q)>(s+ a_q\bmod q)}+\mathbf{1}_{u+(s+c_q\bmod q)\ge q}. \tag*{(55)} 1 ( s − u mod q ) > ( s + a q mod q ) + 1 u + ( s + c q mod q ) ≥ q . (55) Its minimum over residues u u u m q m_q m q E a = 0 E_a=0 E a = 0 u = 0 u=0 u = 0 E a = 1 E_a=1 E a = 1 [ q − a q , s ] [q-a_q,s] [ q − a q , s ] [ 0 , q − ( s + c q m o d q ) − 1 ] [0,q-(s+c_q\bmod q)-1] [ 0 , q − ( s + c q mod q ) − 1 ] E b = 1 E_b=1 E b = 1 d q + E c = 1 d_q+E_c=1 d q + E c = 1 0 ≤ j ≤ n 0\le j\le n 0 ≤ j ≤ n n < q n<q n < q [ q − a q , s ] [q-a_q,s] [ q − a q , s ] [ 0 , n ] [0,n] [ 0 , n ]
m p ≥ ∑ h ≥ 1 m p h . (56) m_p\ge\sum_{h\ge1}m_{p^h}. \tag*{(56)} m p ≥ h ≥ 1 ∑ m p h . (56) Counting the four products in C n C_n C n q q q η q + E a + E b − E c \eta_q+E_a+E_b-E_c η q + E a + E b − E c (9) , complete periods in the numerator and denominator cancel. If c q ≠ q − 1 c_q\ne q-1 c q = q − 1 q q q d q = 0 d_q=0 d q = 0 c q = q − 1 c_q=q-1 c q = q − 1 − b q ∗ -b_q^* − b q ∗
It remains to check (53) . If c q ≠ q − 1 c_q\ne q-1 c q = q − 1 d q = 0 d_q=0 d q = 0 E c ≥ max ( E a , E b ) E_c\ge\max(E_a,E_b) E c ≥ max ( E a , E b ) E a = E b = 1 E_a=E_b=1 E a = E b = 1 m q = 1 m_q=1 m q = 1 d q = 1 d_q=1 d q = 1 E c ≤ min ( E a , E b ) E_c\le\min(E_a,E_b) E c ≤ min ( E a , E b ) m q = 1 m_q=1 m q = 1 E a = E b = 1 E_a=E_b=1 E a = E b = 1 E c = 0 E_c=0 E c = 0 e q = 0 e_q=0 e q = 0 1 1 1 c q = q − 1 c_q=q-1 c q = q − 1 d q = 0 d_q=0 d q = 0 s = 0 s=0 s = 0 s ≥ 1 s\ge1 s ≥ 1 E a + E b − 21 E a = E b = 1 E_a+E_b-2\mathbf{1}_{E_a=E_b=1} E a + E b − 2 1 E a = E b = 1
All numerator factors relevant to the coefficients, C n C_n C n 2 n − 1 2n-1 2 n − 1 M M M (54) . For p > M p>\sqrt{M} p > M (55) is the actual content valuation. □
The moment bound alone allows primes as large as 12 n + 5 12n+5 12 n + 5 n ≥ 11 n\ge11 n ≥ 11 p > n p>n p > n p ≥ 5 p\ge5 p ≥ 5
Lemma 4.3. Let H ∈ Z p [ t ] H\in\mathbb{Z}_p[t] H ∈ Z p [ t ] n < p n<p n < p
t ( 1 − t ) H ′ ′ + τ ( t ) H ′ + Λ H = 0 , τ ∈ Z p [ t ] , Λ ∈ Z p . t(1-t)H''+\tau(t)H'+\Lambda H=0,\qquad\tau\in\mathbb{Z}_p[t],\qquad\Lambda\in\mathbb{Z}_p. t ( 1 − t ) H ′′ + τ ( t ) H ′ + Λ H = 0 , τ ∈ Z p [ t ] , Λ ∈ Z p . If
v p ( H ( 0 ) ) ≤ 1 , v p ( H ( 1 ) ) ≤ 1 , v p ( lc H ) ≤ 1 , v_p(H(0))\le1,\qquad v_p(H(1))\le1,\qquad v_p(\operatorname{lc} H)\le1, v p ( H ( 0 )) ≤ 1 , v p ( H ( 1 )) ≤ 1 , v p ( lc H ) ≤ 1 , then every primitive factorization H = A B H=AB H = A B Z p [ t ] \mathbb{Z}_p[t] Z p [ t ] v p ( Res ( A , B ) ) = 0 v_p(\operatorname{Res}(A,B))=0 v p ( Res ( A , B )) = 0
Proof. The nonzero reduction H ‾ \overline{H} H 0 , 1 0,1 0 , 1 ξ \xi ξ 2 ≤ d ≤ deg H ‾ < p 2\le d\le\deg\overline{H}<p 2 ≤ d ≤ deg H < p ξ ( 1 − ξ ) d ( d − 1 ) \xi(1-\xi)d(d-1) ξ ( 1 − ξ ) d ( d − 1 )
Homogenize A , B A,B A , B a = 0 a=0 a = 0 1 1 1 v p ( A ( a ) ) + v p ( B ( a ) ) ≥ 2 v_p(A(a))+v_p(B(a))\ge2 v p ( A ( a )) + v p ( B ( a )) ≥ 2 p p p
Lemma 4.4. For p ≥ 5 p\ge5 p ≥ 5
v p ( R ) ≤ k E p ( n ) . (57) v_p(R)\le kE_p(n). \tag*{(57)} v p ( R ) ≤ k E p ( n ) . (57) If n ≥ 11 n\ge11 n ≥ 11 p > n p>n p > n v p ( R ) = 0 v_p(R)=0 v p ( R ) = 0
Proof. Fix p ≥ 5 p \ge5 p ≥ 5 (10) allow Lemma 2.2 to be applied to H n H_n H n (9) and C = ( κ n / g n ) 2 C n C = (\kappa_n/g_n)^2 C_n C = ( κ n / g n ) 2 C n κ n \kappa_n κ n p p p Lemma 4.2 gives e p ≤ E p ( n ) e_p \le E_p(n) e p ≤ E p ( n ) (57) .
Now suppose n ≥ 11 n \ge11 n ≥ 11 p > n p > n p > n
y n ( 0 ) = ( α + 1 ) n n ! , y n ( 1 ) = ( − 1 ) n ( β + 1 ) n n ! , lc ( y n ) = ( − 1 ) n ( n + c + 1 ) n n ! y_n(0) = \frac{(\alpha+1)_n}{n!}, \qquad y_n(1) = (-1)^n\frac{(\beta+1)_n}{n!}, \qquad\operatorname{lc}(y_n) = (-1)^n\frac{(n+c+1)_n}{n!} y n ( 0 ) = n ! ( α + 1 ) n , y n ( 1 ) = ( − 1 ) n n ! ( β + 1 ) n , lc ( y n ) = ( − 1 ) n n ! ( n + c + 1 ) n has valuation at most one: n ! n! n ! n < p n < p n < p p p p 12 n + 5 < ( n + 1 ) 2 ≤ p 2 12n+5 < (n+1)^2 \le p^2 12 n + 5 < ( n + 1 ) 2 ≤ p 2 κ n \kappa_n κ n H n H_n H n (12) has p p p Lemma 4.3 gives v p ( R ) = 0 v_p(R) = 0 v p ( R ) = 0
The two asymptotic constants
For primes exceeding 12 n + 5 \sqrt{12n+5} 12 n + 5 n / p n/p n / p p p p n n n
Proposition 4.5. The estimates (51) hold uniformly in the four families and in the rational factorization.
Proof. For p > 12 n + 5 p > \sqrt{12n+5} p > 12 n + 5 p ≡ r ( m o d 6 ) p \equiv r \pmod{6} p ≡ r ( mod 6 ) r = 2 λ + 3 μ r = 2\lambda+ 3\mu r = 2 λ + 3 μ a p / p = 1 / 3 + O ( 1 / p ) a_p/p = 1/3 + O(1/p) a p / p = 1/3 + O ( 1/ p ) a p / p = 2 / 3 + O ( 1 / p ) a_p/p = 2/3 + O(1/p) a p / p = 2/3 + O ( 1/ p ) b p / p = 1 / 2 + O ( 1 / p ) b_p/p = 1/2 + O(1/p) b p / p = 1/2 + O ( 1/ p ) Lemma 4.2 gives the following intervals, up to endpoint changes of size O ( 1 / p ) O(1/p) O ( 1/ p )
Table 1.
Small primes and higher prime powers contribute O ( n log n ) O(\sqrt{n}\log n) O ( n log n ) v p ( g n ) ≤ v p ( lc ( y n ) ) ≤ ⌊ log p ( 12 n + 5 ) ⌋ v_p(g_n) \le v_p(\operatorname{lc}(y_n)) \le\lfloor\log_p(12n+5)\rfloor v p ( g n ) ≤ v p ( lc ( y n )) ≤ ⌊ log p ( 12 n + 5 )⌋
Apply the prime number theorem in the two classes modulo 6 to the intervals in the table; the fixed-modulus consequence of [3 Corollary 1.7] suffices. For each fixed bound on ℓ = ⌊ n / p ⌋ \ell= \lfloor n/p\rfloor ℓ = ⌊ n / p ⌋ O ( n / ℓ ) O(n/\ell) O ( n / ℓ ) o ( n ) o(n) o ( n ) p ≤ 12 n + 5 p \le\sqrt{12n+5} p ≤ 12 n + 5 O ( n ) O(\sqrt{n}) O ( n ) O ( 1 ) O(1) O ( 1 )
Lemma 4.4 permits restriction to p ≤ n p \le n p ≤ n
1 n ∑ 5 ≤ p ≤ n E p ( n ) log p ⟶ 1 2 ∑ ℓ ≥ 1 ( 1 ℓ − 1 ℓ + 1 / 6 + 1 ℓ + 1 / 3 − 1 ℓ + 2 / 3 + 1 ℓ + 5 / 6 − 1 ℓ + 1 ) = 43 20 − π 3 . \frac{1}{n}\sum_{5 \le p \le n} E_p(n)\log p \longrightarrow\frac{1}{2}\sum_{\ell\ge1}\left(\frac{1}{\ell}-\frac{1}{\ell+1/6}+\frac{1}{\ell+1/3}-\frac{1}{\ell+2/3}+\frac{1}{\ell+5/6}-\frac{1}{\ell+1}\right)=\frac{43}{20}-\frac{\pi}{\sqrt{3}}. n 1 5 ≤ p ≤ n ∑ E p ( n ) log p ⟶ 2 1 ℓ ≥ 1 ∑ ( ℓ 1 − ℓ + 1/6 1 + ℓ + 1/3 1 − ℓ + 2/3 1 + ℓ + 5/6 1 − ℓ + 1 1 ) = 20 43 − 3 π . For the content all primes are included, so the corresponding sum starts at zero:
log g n n ⟶ 1 2 ∑ ℓ ≥ 0 ( 1 ℓ + 2 / 3 − 1 ℓ + 1 + 1 ℓ + 1 / 2 − 1 ℓ + 5 / 6 ) = π 2 3 . \frac{\log g_n}{n} \longrightarrow\frac{1}{2}\sum_{\ell\ge0}\left(\frac{1}{\ell+2/3}-\frac{1}{\ell+1}+\frac{1}{\ell+1/2}-\frac{1}{\ell+5/6}\right)=\frac{\pi}{2\sqrt{3}}. n log g n ⟶ 2 1 ℓ ≥ 0 ∑ ( ℓ + 2/3 1 − ℓ + 1 1 + ℓ + 1/2 1 − ℓ + 5/6 1 ) = 2 3 π . These evaluations follow from the partial-fraction expansion of π cot ( π z ) \pi\cot(\pi z) π cot ( π z ) Γ \Gamma Γ [22 Eqs. 5.5.4 and 5.5.8] . □
The primes 2 and 3
The monic endpoint expansions place every root in the ball v 2 ( t ) ≥ 1 v_2(t) \ge1 v 2 ( t ) ≥ 1 v 3 ( 1 − t ) ≥ 1 v_3(1-t) \ge1 v 3 ( 1 − t ) ≥ 1
To sum the resulting errors, we use the Newton polygon. For a polynomial f ( X ) = ∑ j c j X j f(X) = \sum_j c_jX^j f ( X ) = ∑ j c j X j Q p \mathbb{Q}_p Q p ( j , v p ( c j ) ) (j,v_p(c_j)) ( j , v p ( c j )) − r -r − r d d d d d d Q p \mathbb{Q}_p Q p r r r
For the factorization H n = A B H_n = AB H n = A B
D 2 , A = v 2 ( A ( 0 ) ) , D 2 , B = v 2 ( B ( 0 ) ) , D 3 , A = v 3 ( A ( 1 ) ) , D 3 , B = v 3 ( B ( 1 ) ) . D_{2,A}=v_2(A(0)),\qquad D_{2,B}=v_2(B(0)),\qquad D_{3,A}=v_3(A(1)),\qquad D_{3,B}=v_3(B(1)). D 2 , A = v 2 ( A ( 0 )) , D 2 , B = v 2 ( B ( 0 )) , D 3 , A = v 3 ( A ( 1 )) , D 3 , B = v 3 ( B ( 1 )) . The leading coefficients of both factors are units at 2 and 3.
Proposition 4.6. Let s p ( m ) s_p(m) s p ( m ) p p p m m m δ = ( μ + 1 ) / 2 \delta=(\mu+1)/2 δ = ( μ + 1 ) /2
v 2 ( R ) ≤ min ( e D 2 , A , k D 2 , B ) + k e 2 + k 2 s 2 ( n ) , v_2(R) \le\min(eD_{2,A},kD_{2,B})+\frac{ke}{2}+\frac{k}{2}s_2(n), v 2 ( R ) ≤ min ( e D 2 , A , k D 2 , B ) + 2 k e + 2 k s 2 ( n ) , v 3 ( R ) ≤ min ( e D 3 , A , k D 3 , B ) + k e 3 + k 6 s 3 ( 2 n + δ ) . v_3(R) \le\min(eD_{3,A},kD_{3,B})+\frac{ke}{3}+\frac{k}{6}s_3(2n+\delta). v 3 ( R ) ≤ min ( e D 3 , A , k D 3 , B ) + 3 k e + 6 k s 3 ( 2 n + δ ) . These bounds hold for every factorization; no restriction on k k k O ( k log n ) O(k\log n) O ( k log n )
Proof. The proof has two parts. First, the Taylor coefficients at a root give a uniform upper bound for the extra closeness of another root with the same endpoint valuation. The error in that bound measures how far the root valuation lies below 2 at the prime 2, or below 3 / 2 3/2 3/2 O ( log n ) O(\log n) O ( log n )
At p = 2 p=2 p = 2 z = t z=t z = t b = α b=\alpha b = α p = 3 p=3 p = 3 z = 1 − t z=1-t z = 1 − t b = β b=\beta b = β P ( z ) P(z) P ( z )
σ P ′ ′ + τ P ′ + Λ P = 0 , σ = z ( 1 − z ) , τ = b + 1 − ( c + 2 ) z , Λ = n ( n + c + 1 ) . \sigma P''+\tau P'+\Lambda P=0,\qquad\sigma=z(1-z),\qquad\tau=b+1-(c+2)z,\qquad\Lambda=n(n+c+1). σ P ′′ + τ P ′ + Λ P = 0 , σ = z ( 1 − z ) , τ = b + 1 − ( c + 2 ) z , Λ = n ( n + c + 1 ) . Every root z z z v p ( z ) ≥ 1 v_p(z)\ge1 v p ( z ) ≥ 1
n ! P n ( a , b ) ( 2 X − 1 ) ( n + a + b + 1 ) n = ∑ j = 0 n ( − 1 ) n − j ( n j ) ( b + j + 1 ) n − j ( n + a + b + j + 1 ) n − j X j . (58) \frac{n!P_n^{(a,b)}(2X-1)}{(n+a+b+1)_n}
=\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}\frac{(b+j+1)_{n-j}}{(n+a+b+j+1)_{n-j}}X^j. \tag*{(58)} ( n + a + b + 1 ) n n ! P n ( a , b ) ( 2 X − 1 ) = j = 0 ∑ n ( − 1 ) n − j ( j n ) ( n + a + b + j + 1 ) n − j ( b + j + 1 ) n − j X j . (58) Here b ∈ Z p b\in\mathbb{Z}_p b ∈ Z p − 1 -1 − 1 n − j n-j n − j
Fix a root z z z r = v p ( z ) ≥ 1 r=v_p(z)\ge1 r = v p ( z ) ≥ 1
h = min ( 1 , r − 1 ) , u j = z j − 1 P ( j ) ( z ) P ′ ( z ) . h=\min(1,r-1),\qquad u_j=z^{j-1}\frac{P^{(j)}(z)}{P'(z)}. h = min ( 1 , r − 1 ) , u j = z j − 1 P ′ ( z ) P ( j ) ( z ) . Thus u 0 = 0 u_0=0 u 0 = 0 u 1 = 1 u_1=1 u 1 = 1 m m m
u m + 2 = − a m u m + 1 − b m u m , u_{m+2}=-a_mu_{m+1}-b_mu_m, u m + 2 = − a m u m + 1 − b m u m , a m = b + 1 + m − ( c + 2 + 2 m ) z 1 − z , b m = z ( n − m ) ( n + m + c + 1 ) 1 − z . (59) a_m=\frac{b+1+m-(c+2+2m)z}{1-z},\qquad b_m=\frac{z(n-m)(n+m+c+1)}{1-z}. \tag*{(59)} a m = 1 − z b + 1 + m − ( c + 2 + 2 m ) z , b m = 1 − z z ( n − m ) ( n + m + c + 1 ) . (59) Since v p ( c + 2 + 2 m ) = v p ( n + m + c + 1 ) = − 1 v_p(c+2+2m)=v_p(n+m+c+1)=-1 v p ( c + 2 + 2 m ) = v p ( n + m + c + 1 ) = − 1 v p ( b m ) ≥ r − 1 ≥ h v_p(b_m)\ge r-1\ge h v p ( b m ) ≥ r − 1 ≥ h v p ( a m ) ≥ 0 v_p(a_m)\ge0 v p ( a m ) ≥ 0 α + 1 \alpha+1 α + 1 Z 2 \mathbb{Z}_2 Z 2 v 2 ( a m ) ≥ h v_2(a_m)\ge h v 2 ( a m ) ≥ h m m m β + 1 \beta+1 β + 1 δ = 0 \delta=0 δ = 0 δ = 1 \delta= 1 δ = 1 v 3 ( a m ) ≥ h v_3(a_m) \ge h v 3 ( a m ) ≥ h m ≡ 1 − δ ( m o d 3 ) m \equiv1-\delta\pmod{3} m ≡ 1 − δ ( mod 3 ) (59) therefore gives
v 2 ( u j ) ≥ ⌊ j 2 ⌋ h , v 3 ( u j ) ≥ ⌊ j + δ 3 ⌋ h . (60) v_2(u_j) \ge\left\lfloor\frac{j}{2} \right\rfloor h,\qquad v_3(u_j) \ge\left\lfloor\frac{j+\delta}{3} \right\rfloor h. \tag*{(60)} v 2 ( u j ) ≥ ⌊ 2 j ⌋ h , v 3 ( u j ) ≥ ⌊ 3 j + δ ⌋ h . (60) For example, in the 3 3 3 h h h h h h
For p = 2 p=2 p = 2 3 3 3 j ≥ 2 j\ge2 j ≥ 2 0 ≤ h ≤ 1 0\le h\le1 0 ≤ h ≤ 1
v p ( j ! ) − ⌊ j / p ⌋ h j − 1 ≤ 1 p − 1 − h p . (61) \frac{v_p(j!)-\lfloor j/p\rfloor h}{j-1}\le\frac{1}{p-1}-\frac{h}{p}. \tag*{(61)} j − 1 v p ( j !) − ⌊ j / p ⌋ h ≤ p − 1 1 − p h . (61) To verify it, write j = p q + s j=pq+s j = pq + s 0 ≤ s < p 0\le s<p 0 ≤ s < p
1 − s p ( j ) p − 1 + h ( s − 1 ) p ≤ 0. \frac{1-s_p(j)}{p-1}+\frac{h(s-1)}{p}\le0. p − 1 1 − s p ( j ) + p h ( s − 1 ) ≤ 0. For s = 0 , 1 s=0,1 s = 0 , 1 s ≥ 2 s\ge2 s ≥ 2 s p ( j ) ≥ s s_p(j)\ge s s p ( j ) ≥ s h ≤ 1 h\le1 h ≤ 1 w ≠ z w\ne z w = z
P ( w ) ( w − z ) P ′ ( z ) = 1 + ∑ j = 2 n u j j ! ( w − z z ) j − 1 \frac{P(w)}{(w-z)P'(z)}=1+\sum_{j=2}^{n}\frac{u_j}{j!}\left(\frac{w-z}{z}\right)^{j-1} ( w − z ) P ′ ( z ) P ( w ) = 1 + j = 2 ∑ n j ! u j ( z w − z ) j − 1 cannot vanish when v p ( ( w − z ) / z ) > 1 / ( p − 1 ) − h / p v_p((w-z)/z)>1/(p-1)-h/p v p (( w − z ) / z ) > 1/ ( p − 1 ) − h / p ρ p = p / ( p − 1 ) \rho_p=p/(p-1) ρ p = p / ( p − 1 )
v p ( w − z ) − min ( v p ( w ) , v p ( z ) ) ≤ 1 p + ( ρ p − v p ( z ) ) + p if v p ( w ) = v p ( z ) . (62) v_p(w-z)-\min(v_p(w),v_p(z))\le\frac{1}{p}+\frac{(\rho_p-v_p(z))_+}{p}\qquad\text{if }v_p(w)=v_p(z). \tag*{(62)} v p ( w − z ) − min ( v p ( w ) , v p ( z )) ≤ p 1 + p ( ρ p − v p ( z ) ) + if v p ( w ) = v p ( z ) . (62) The excess is zero when the root valuations differ. Here ( x ) + = max ( x , 0 ) (x)_+=\max(x,0) ( x ) + = max ( x , 0 )
1 p − 1 − min ( 1 , r − 1 ) p ≤ 1 p + ( ρ p − r ) + p . \frac{1}{p-1}-\frac{\min(1,r-1)}{p}\le\frac{1}{p}+\frac{(\rho_p-r)_+}{p}. p − 1 1 − p min ( 1 , r − 1 ) ≤ p 1 + p ( ρ p − r ) + . It remains to control the total endpoint deficits. Let c j c_j c j y n ( t ) y_n(t) y n ( t ) 2 2 2
v 2 ( c j ) = 2 ( n − j ) + Ψ ( j ) , Ψ ( j ) = s 2 ( j ) − s 2 ( n ) + v 2 ( n + α n − j ) . v_2(c_j)=2(n-j)+\Psi(j),\qquad\Psi(j)=s_2(j)-s_2(n)+v_2\binom{n+\alpha}{n-j}. v 2 ( c j ) = 2 ( n − j ) + Ψ ( j ) , Ψ ( j ) = s 2 ( j ) − s 2 ( n ) + v 2 ( n − j n + α ) . The generalized binomial coefficient is 2 2 2 Ψ ( j ) ≥ − s 2 ( n ) \Psi(j)\ge-s_2(n) Ψ ( j ) ≥ − s 2 ( n ) Ψ ( n ) = 0 \Psi(n)=0 Ψ ( n ) = 0 W W W ( j , Ψ ( j ) ) (j,\Psi(j)) ( j , Ψ ( j ))
∑ i ( 2 − v 2 ( t i ) ) + = ∫ 0 n ( W ′ ( x ) ) + d x = W ( n ) − min W ≤ s 2 ( n ) . \sum_i(2-v_2(t_i))_+=\int_0^n(W'(x))_+\,dx=W(n)-\min W\le s_2(n). i ∑ ( 2 − v 2 ( t i ) ) + = ∫ 0 n ( W ′ ( x ) ) + d x = W ( n ) − min W ≤ s 2 ( n ) . The middle equality follows from convexity: a convex function first decreases and then increases.
At 3 3 3 z = 1 − t z=1-t z = 1 − t
3 2 ( n − j ) + 1 2 Φ ( j ) , Φ ( j ) = s 3 ( 2 j + δ ) + s 3 ( n − j ) − s 3 ( 2 n + δ ) . \frac{3}{2}(n-j)+\frac{1}{2}\Phi(j),\qquad\Phi(j)=s_3(2j+\delta)+s_3(n-j)-s_3(2n+\delta). 2 3 ( n − j ) + 2 1 Φ ( j ) , Φ ( j ) = s 3 ( 2 j + δ ) + s 3 ( n − j ) − s 3 ( 2 n + δ ) . Here Φ ( j ) ≥ − s 3 ( 2 n + δ ) \Phi(j)\ge-s_3(2n+\delta) Φ ( j ) ≥ − s 3 ( 2 n + δ ) Φ ( n ) = 0 \Phi(n)=0 Φ ( n ) = 0
∑ i ( 3 2 − v 3 ( 1 − t i ) ) + ≤ 1 2 s 3 ( 2 n + δ ) . \sum_i\left(\frac{3}{2}-v_3(1-t_i)\right)_+\le\frac{1}{2}s_3(2n+\delta). i ∑ ( 2 3 − v 3 ( 1 − t i ) ) + ≤ 2 1 s 3 ( 2 n + δ ) . The leading coefficients of A , B A,B A , B 2 , 3 2,3 2 , 3 e D p , A eD_{p,A} e D p , A k D p , B kD_{p,B} k D p , B (62) , using the root of B B B k e / p ke/p k e / p k / p k/p k / p B B B H n H_n H n
The archimedean comparison
The real roots lie in ( 0 , 1 ) (0,1) ( 0 , 1 ) 0 0 0 1 1 1 2 2 2 3 3 3
Write γ n = κ n / g n \gamma_n=\kappa_n/g_n γ n = κ n / g n
D 2 , n = n + ∑ j = 1 n v 2 ( 3 j + λ ) , T 2 , n = D 2 , n log 2 / n . (63) D_{2,n}=n+\sum_{j=1}^{n}v_2(3j+\lambda), \qquad T_{2,n}=D_{2,n}\log2/n. \tag*{(63)} D 2 , n = n + j = 1 ∑ n v 2 ( 3 j + λ ) , T 2 , n = D 2 , n log 2/ n . (63) D 3 , n = n + ∑ j = 1 n v 3 ( 2 j + μ ) , T 3 , n = D 3 , n log 3 / n . (64) D_{3,n}=n+\sum_{j=1}^{n}v_3(2j+\mu), \qquad T_{3,n}=D_{3,n}\log3/n. \tag*{(64)} D 3 , n = n + j = 1 ∑ n v 3 ( 2 j + μ ) , T 3 , n = D 3 , n log 3/ n . (64) These are the endpoint valuations of H n H_n H n
D 2 , n = 2 n + O ( log n ) , D 3 , n = 3 2 n + O ( log n ) . (65) D_{2,n}=2n+O(\log n), \qquad D_{3,n}=\frac{3}{2}n+O(\log n). \tag*{(65)} D 2 , n = 2 n + O ( log n ) , D 3 , n = 2 3 n + O ( log n ) . (65) Define
L n = log ( e T 2 , n + e T 3 , n ) , E = 1 2 log 2 + 1 3 log 3 , L_n=\log(e^{T_{2,n}}+e^{T_{3,n}}), \qquad E=\frac{1}{2}\log2+\frac{1}{3}\log3, L n = log ( e T 2 , n + e T 3 , n ) , E = 2 1 log 2 + 3 1 log 3 , J n = 1 2 s 2 ( n ) log 2 + 1 3 s 3 ( 2 n + δ ) log 3 , δ = ( μ + 1 ) / 2. (66) J_n=\frac{1}{2}s_2(n)\log2+\frac{1}{3}s_3(2n+\delta)\log3, \qquad\delta=(\mu+1)/2. \tag*{(66)} J n = 2 1 s 2 ( n ) log 2 + 3 1 s 3 ( 2 n + δ ) log 3 , δ = ( μ + 1 ) /2. (66) Proposition 4.7. For every factorization H n = A B H_n=AB H n = A B 1 ≤ k = deg A ≤ n / 2 1\leq k=\deg A\leq n/2 1 ≤ k = deg A ≤ n /2
log R ( 6 ) k ≥ log ( 2 γ n C n ) − J n + ( n − 2 k ) L n \frac{\log R^{(6)}}{k}\geq\log(2\gamma_n\sqrt{C_n})-J_n+(n-2k)L_n k log R ( 6 ) ≥ log ( 2 γ n C n ) − J n + ( n − 2 k ) L n − ( n − k ) ( T 2 , n + T 3 , n + E ) + 2 ( k − 1 ) 2 k log 2 − log k . (67) -(n-k)(T_{2,n}+T_{3,n}+E)+\frac{2(k-1)^2}{k}\log2-\log k. \tag*{(67)} − ( n − k ) ( T 2 , n + T 3 , n + E ) + k 2 ( k − 1 ) 2 log 2 − log k . (67) Consequently, uniformly in all such factorizations,
log R ( 6 ) ≥ ( l − o ( 1 ) ) k n , l = log ( 4 + 3 3 ) − 1 2 log 2 − 1 3 log 3 − π 2 3 . (68) \log R^{(6)}\geq(\mathfrak{l}-o(1))kn,\qquad\mathfrak{l}=\log(4+3\sqrt{3})-\frac{1}{2}\log2-\frac{1}{3}\log3-\frac{\pi}{2\sqrt{3}}. \tag*{(68)} log R ( 6 ) ≥ ( l − o ( 1 )) k n , l = log ( 4 + 3 3 ) − 2 1 log 2 − 3 1 log 3 − 2 3 π . (68) Proof. Let a = ∣ lc ( A ) ∣ a=|\operatorname{lc}(A)| a = ∣ lc ( A ) ∣ e = n − k e=n-k e = n − k t 1 , … , t k t_1,\ldots,t_k t 1 , … , t k A A A H n = A B H_n=AB H n = A B
R = a e − k ∏ i = 1 k ∣ H n ′ ( t i ) ∣ ∏ i < j ∣ t i − t j ∣ 2 . R=a^{e-k}\frac{\prod_{i=1}^{k}|H_n'(t_i)|}{\prod_{i<j}|t_i-t_j|^2}. R = a e − k ∏ i < j ∣ t i − t j ∣ 2 ∏ i = 1 k ∣ H n ′ ( t i ) ∣ . Lemmas 2.1 and 2.5 therefore give
R ≥ a n − 2 k ( 2 γ n C n ) k k − k 2 ( k − 1 ) 2 . R\geq a^{n-2k}(2\gamma_n\sqrt{C_n})^k k^{-k}2^{(k-1)^2}. R ≥ a n − 2 k ( 2 γ n C n ) k k − k 2 ( k − 1 ) 2 . By Hölder’s inequality,
( ∏ i t i ) 1 / k + ( ∏ i ( 1 − t i ) ) 1 / k ≤ 1. \left(\prod_i t_i\right)^{1/k}+\left(\prod_i(1-t_i)\right)^{1/k}\leq1. ( i ∏ t i ) 1/ k + ( i ∏ ( 1 − t i ) ) 1/ k ≤ 1. Hence, with U = D 2 , A log 2 / k U=D_{2,A}\log2/k U = D 2 , A log 2/ k V = D 3 , A log 3 / k V=D_{3,A}\log3/k V = D 3 , A log 3/ k
a 1 / k ≥ ∣ A ( 0 ) ∣ 1 / k + ∣ A ( 1 ) ∣ 1 / k ≥ e U + e V . (69) a^{1/k}\geq|A(0)|^{1/k}+|A(1)|^{1/k}\geq e^U+e^V. \tag*{(69)} a 1/ k ≥ ∣ A ( 0 ) ∣ 1/ k + ∣ A ( 1 ) ∣ 1/ k ≥ e U + e V . (69) Put x = k / n x=k/n x = k / n D p , n = D p , A + D p , B D_{p,n}=D_{p,A}+D_{p,B} D p , n = D p , A + D p , B
min ( e D 2 , A , k D 2 , B ) log 2 k n = min ( U , T 2 , n ) − x U , \frac{\min(e^{D_{2,A}},kD_{2,B})\log2}{kn}=\min(U,T_{2,n})-xU, k n min ( e D 2 , A , k D 2 , B ) log 2 = min ( U , T 2 , n ) − xU , and the analogous identity holds at 3. This exact expression allows either factor to contain an arbitrary share of the endpoint valuations. For 0 ≤ x ≤ 1 / 2 0 \le x \le1/2 0 ≤ x ≤ 1/2
( 1 − 2 x ) log ( e U + e V ) + x ( U + V ) − min ( U , T 2 ) − min ( V , T 3 ) (1-2x)\log(e^U+e^V)+x(U+V)-\min(U,T_2)-\min(V,T_3) ( 1 − 2 x ) log ( e U + e V ) + x ( U + V ) − min ( U , T 2 ) − min ( V , T 3 ) has its minimum at ( U , V ) = ( T 2 , T 3 ) (U,V)=(T_2,T_3) ( U , V ) = ( T 2 , T 3 ) V V V T 2 T_2 T 2
( 1 − 2 x ) e U e U + e V + x − 1 ≤ 0 , ( 1 − 2 x ) e U e U + e V + x ≥ 0. (1-2x)\frac{e^U}{e^U+e^V}+x-1\le0,\qquad(1-2x)\frac{e^U}{e^U+e^V}+x\ge0. ( 1 − 2 x ) e U + e V e U + x − 1 ≤ 0 , ( 1 − 2 x ) e U + e V e U + x ≥ 0. The same argument applies to V V V Proposition 4.6 and applying this minimum proves (67) .
To obtain the uniform asymptotic form, divide by n n n
2 ( k − 1 ) 2 k n log 2 = 2 x log 2 − 4 log 2 n + 2 log 2 k n . \frac{2(k-1)^2}{kn}\log2=2x\log2-\frac{4\log2}{n}+\frac{2\log2}{kn}. k n 2 ( k − 1 ) 2 log 2 = 2 x log 2 − n 4 log 2 + k n 2 log 2 . Since J n = O ( log n ) J_n=O(\log n) J n = O ( log n ) log k ≤ log n \log k\le\log n log k ≤ log n
log R k n ( 6 ) k n ≥ log γ n n + L n − T 2 , n − T 3 , n − E + x ( T 2 , n + T 3 , n + E − 2 L n + 2 log 2 ) − o ( 1 ) . \frac{\log R_{kn}^{(6)}}{kn}\ge\frac{\log\gamma_n}{n}+L_n-T_{2,n}-T_{3,n}-E+x(T_{2,n}+T_{3,n}+E-2L_n+2\log2)-o(1). k n log R k n ( 6 ) ≥ n log γ n + L n − T 2 , n − T 3 , n − E + x ( T 2 , n + T 3 , n + E − 2 L n + 2 log 2 ) − o ( 1 ) . with an error independent of k k k x x x L n ≤ max ( T 2 , n , T 3 , n ) + log 2 L_n\le\max(T_{2,n},T_{3,n})+\log2 L n ≤ max ( T 2 , n , T 3 , n ) + log 2 E − ∣ T 2 , n − T 3 , n ∣ E-\lvert T_{2,n}-T_{3,n}\rvert E − ∣ T 2 , n − T 3 , n ∣
5 2 log 2 − 7 6 log 3 = 1 6 log 2 15 3 7 > 0. \frac{5}{2}\log2-\frac{7}{6}\log3=\frac{1}{6}\log\frac{2^{15}}{3^7}>0. 2 5 log 2 − 6 7 log 3 = 6 1 log 3 7 2 15 > 0. We may discard this term. Now use Lemma 4.1 , Proposition 4.5, (11) , and (65) . The remaining expression tends to ℓ \ell ℓ (68) . □
The two uniform bounds already imply eventual irreducibility, since
ℓ − u = log ( 4 + 3 3 ) − 1 2 log 2 − 1 3 log 3 − 43 20 + π 2 3 > 1 4 . (70) \ell-u=\log(4+3\sqrt{3})-\frac{1}{2}\log2-\frac{1}{3}\log3-\frac{43}{20}+\frac{\pi}{2\sqrt{3}}>\frac{1}{4}. \tag*{(70)} ℓ − u = log ( 4 + 3 3 ) − 2 1 log 2 − 3 1 log 3 − 20 43 + 2 3 π > 4 1 . (70) For reference, ℓ = 0.5991078130 … \ell=0.5991078130\ldots ℓ = 0.5991078130 … u = 0.3362006358 … u=0.3362006358\ldots u = 0.3362006358 … 0.2629071772 … 0.2629071772\ldots 0.2629071772 …
1732 1000 < 3 < 1733 1000 , π > 157 50 , log 3 > 549 500 , log 2 < 7 10 . \frac{1732}{1000}<\sqrt{3}<\frac{1733}{1000},\qquad\pi>\frac{157}{50},\qquad\log3>\frac{549}{500},\qquad\log2<\frac{7}{10}. 1000 1732 < 3 < 1000 1733 , π > 50 157 , log 3 > 500 549 , log 2 < 10 7 . Since 4 + 3 3 > 9 ⋅ 2299 / 2250 4+3\sqrt{3}>9\cdot2299/2250 4 + 3 3 > 9 ⋅ 2299/2250 log ( 1 + t ) > t / ( 1 + t ) \log(1+t)>t/(1+t) log ( 1 + t ) > t / ( 1 + t ) t > 0 t>0 t > 0
ℓ − u > 5 ⋅ 549 3 ⋅ 500 − 7 20 − 43 20 + 49 2299 + 1570 1733 > 1 4 . \ell-u>\frac{5\cdot549}{3\cdot500}-\frac{7}{20}-\frac{43}{20}+\frac{49}{2299}+\frac{1570}{1733}>\frac{1}{4}. ℓ − u > 3 ⋅ 500 5 ⋅ 549 − 20 7 − 20 43 + 2299 49 + 1733 1570 > 4 1 . The logarithmic bounds follow, for example, from the positive-term series used in Appendix § 8 .§ 2 . Thus a proper factorization is impossible once n n n
An exact finite criterion
The resultant estimates give a finite criterion that does not require a factorization of the Jacobi polynomial. Use the quantities in (63) –(66) and the residue bounds of Lemma 4.2 , and put
S n = ∑ p p r i m e 5 ≤ p ≤ n E p ( n ) log p . S_n=\sum_{\substack{p\ \mathrm{prime}\\5\le p\le n}}E_p(n)\log p. S n = p prime 5 ≤ p ≤ n ∑ E p ( n ) log p . e n d s m i d = 1 {ends_mid=1} e n d s m i d = 1 For z ≥ 1 z \ge1 z ≥ 1
H n ( z ) = log ( 2 γ n C n ) − J n + ( n − 2 z ) L n − ( n − z ) ( T 2 , n + T 3 , n + E ) + 2 ( z − 1 ) 2 z log 2 − log z . (71) \mathcal{H}_n(z) = \log(2\gamma_n\sqrt{C_n}) - J_n + (n-2z)L_n - (n-z)(T_{2,n}+T_{3,n}+E) + \frac{2(z-1)^2}{z}\log2 - \log z. \tag*{(71)} H n ( z ) = log ( 2 γ n C n ) − J n + ( n − 2 z ) L n − ( n − z ) ( T 2 , n + T 3 , n + E ) + z 2 ( z − 1 ) 2 log 2 − log z . (71) Proposition 4.8. Let n ≥ 11 n \ge11 n ≥ 11
min 1 ≤ k ≤ ⌊ n / 2 ⌋ H n ( k ) > S n , \min_{1 \le k \le\lfloor n/2 \rfloor} \mathcal{H}_n(k) > S_n, 1 ≤ k ≤ ⌊ n /2 ⌋ min H n ( k ) > S n , then y n y_n y n Q \mathbb{Q} Q n / 2 n/2 n /2
Proof. A nontrivial factorization would give S n ≥ log R ( 6 ) / k ≥ H n ( k ) S_n \ge\log R^{(6)}/k \ge\mathcal{H}_n(k) S n ≥ log R ( 6 ) / k ≥ H n ( k ) (67) , a contradiction. □ \square □
The minimization is elementary. Set
a n = T 2 , n + T 3 , n + E − 2 L n + 2 log 2. a_n = T_{2,n} + T_{3,n} + E - 2L_n + 2\log2. a n = T 2 , n + T 3 , n + E − 2 L n + 2 log 2. The part of H n ( z ) \mathcal{H}_n(z) H n ( z ) z z z a n z + 2 log 2 / z − log z a_nz + 2\log2/z - \log z a n z + 2 log 2/ z − log z z > 0 z > 0 z > 0 a n ≤ 0 a_n \le0 a n ≤ 0 a n > 0 a_n > 0 a n > 0
q n = 1 + 1 + 8 a n log 2 2 a n . q_n = \frac{1+\sqrt{1+8a_n\log2}}{2a_n}. q n = 2 a n 1 + 1 + 8 a n log 2 . Thus on an interval of integer degrees it suffices to check the integers adjacent to the clipped value of q n q_n q n
The criterion depends only on endpoint valuations, content, and finite sums over primes. Appendix § 8 .§ 2 uses a slightly weaker form in which the possible factor degree has been eliminated and every comparison reduces to an inequality between linear forms in logarithms.
Completion in every degree
Completion in every degree. Appendix § 8 proves irreducibility for n ≥ 250 000 n \ge250\,000 n ≥ 250 000 1 ≤ n ≤ 8332 1 \le n \le8332 1 ≤ n ≤ 8332 [8 Remark 4.6] ; the conversion from their index to the degree is given in Appendix § 8 .§ 2 . For 8333 ≤ n ≤ 250 000 8333 \le n \le250\,000 8333 ≤ n ≤ 250 000
Irreducibility in Theorem 1.1 . Proposition B.1 treats n ≥ 250 000 n \ge250\,000 n ≥ 250 000 1 ≤ n ≤ 250 000 1 \le n \le250\,000 1 ≤ n ≤ 250 000 Theorem 5.12 . □ \square □
Galois groups and splitting fields
Irreducibility gives transitivity on the roots. The remaining task is to find enough permutations in the Galois group to determine it. We obtain these permutations from ramification at suitable primes. The first argument is common to the Legendre and Kaneko–Zagier families and extends to fixed rational parameters. For unequal integral parameters, a different local factorization gives a prime-length cycle. We also keep track of which roots each inertia element fixes. This support information will prove the splitting-field disjointness statements.
Inertia cycles and permutation groups
We call a permutation a single d d d d d d d = 1 d = 1 d = 1 v p v_p v p
A local criterion
Under the hypotheses of the next lemma, the Newton polygon has a final edge of slope 1 / d 1/d 1/ d d d d − 1 / d -1/d − 1/ d
In a finite Galois extension of Q p \mathbb{Q}_p Q p d < p d < p d < p
Lemma 5.1. Let f ( X ) = ∑ j = 0 N a j X j ∈ Z p [ X ] f(X) = \sum_{j=0}^{N} a_jX^j \in\mathbb{Z}_p[X] f ( X ) = ∑ j = 0 N a j X j ∈ Z p [ X ] N N N a 0 a_0 a 0 v p ( a N ) = 1 v_p(a_N) = 1 v p ( a N ) = 1 f ‾ \overline{f} f N − d N-d N − d 1 ≤ d < p 1 \le d < p 1 ≤ d < p f f f Q p \mathbb{Q}_p Q p d d d N − d N-d N − d Q p \mathbb{Q}_p Q p
If p p p c ∈ Z p × c \in\mathbb{Z}_p^\times c ∈ Z p × 2 d < p 2d < p 2 d < p f ( c X 2 ) f(cX^2) f ( c X 2 ) 2 d 2d 2 d d d d d d d d d d
Proof. The normalized reciprocal
R ( Y ) = Y N f ( 1 / Y ) a 0 R(Y) = \frac{Y^N f(1/Y)}{a_0} R ( Y ) = a 0 Y N f ( 1/ Y ) is monic and integral. Its reduction is Y d B ‾ ( Y ) Y^d\overline{B}(Y) Y d B ( Y ) B ‾ ( 0 ) ≠ 0 \overline{B}(0) \ne0 B ( 0 ) = 0 B ‾ \overline{B} B R = A B R = AB R = A B A A A B B B deg A = d \deg A = d deg A = d A ≡ Y d ( m o d p ) A \equiv Y^d \pmod p A ≡ Y d ( mod p ) B ( 0 ) B(0) B ( 0 ) v p ( R ( 0 ) ) = 1 v_p(R(0)) = 1 v p ( R ( 0 )) = 1 A A A B B B
The polynomial A A A d d d p p p B B B S d S_d S d d < p d < p d < p d d d B = 1 B = 1 B = 1
The reduction of f ( c X 2 ) f(cX^2) f ( c X 2 ) f ‾ \overline{f} f p ≠ 2 p \ne2 p = 2 2 d 2d 2 d 2 d 2d 2 d τ \tau τ τ \tau τ τ d \tau^d τ d □ \square □
Jacobi coefficients and reduction
The normalization (1) satisfies
P n ( α , β ) ( 2 X + 1 ) = ∑ j = 0 n ( n + α n − j ) ( n + α + β + j j ) X j , (72) P_n^{(\alpha,\beta)}(2X+1)=\sum_{j=0}^{n}\binom{n+\alpha}{n-j}\binom{n+\alpha+\beta+j}{j}X^j, \tag*{(72)} P n ( α , β ) ( 2 X + 1 ) = j = 0 ∑ n ( n − j n + α ) ( j n + α + β + j ) X j , (72) P n ( α , β ) ( 1 ) = ( n + α n ) , P n ( α , β ) ( − 1 ) = ( − 1 ) n ( n + β n ) , (73) P_n^{(\alpha,\beta)}(1)=\binom{n+\alpha}{n}, \qquad P_n^{(\alpha,\beta)}(-1)=(-1)^n\binom{n+\beta}{n}, \tag*{(73)} P n ( α , β ) ( 1 ) = ( n n + α ) , P n ( α , β ) ( − 1 ) = ( − 1 ) n ( n n + β ) , (73) and its differential equation is
( 1 − X 2 ) y ′ ′ + [ β − α − ( α + β + 2 ) X ] y ′ + n ( n + α + β + 1 ) y = 0. (74) (1-X^2)y''+[\beta-\alpha-(\alpha+\beta+2)X]y'+n(n+\alpha+\beta+1)y=0. \tag*{(74)} ( 1 − X 2 ) y ′′ + [ β − α − ( α + β + 2 ) X ] y ′ + n ( n + α + β + 1 ) y = 0. (74) These identities follow from the binomial expansion and hold as polynomial identities in the parameters; see also [22 ] , Eqs. 18.5.7–18.5.8 and Table 18.8.1.
Write α = a / u \alpha=a/u α = a / u β = b / v \beta=b/v β = b / v a , b ∈ Z a,b \in\mathbb{Z} a , b ∈ Z u , v u,v u , v D D D
r = D ( α + β ) ∈ Z . (75) r = D(\alpha+\beta) \in\mathbb{Z}. \tag*{(75)} r = D ( α + β ) ∈ Z . (75) Equation (72) gives the integral normalization
K n α , β ( X ) : = u n n ! P n ( α , β ) ( 1 + 2 D u X ) = ∑ j = 0 n ( n j ) ∏ k = j + 1 n ( u k + a ) ∏ k = 1 j ( D n + r + D k ) X j . (76) \begin{aligned}
K_n^{\alpha,\beta}(X) := u^n n!P_n^{(\alpha,\beta)}\left(1+\frac{2D}{u}X\right) \\
&= \sum_{j=0}^{n}\binom{n}{j}\prod_{k=j+1}^{n}(uk+a)\prod_{k=1}^{j}(Dn+r+Dk)X^j. \tag*{(76)}
\end{aligned} K n α , β ( X ) := u n n ! P n ( α , β ) ( 1 + u 2 D X ) = j = 0 ∑ n ( j n ) k = j + 1 ∏ n ( u k + a ) k = 1 ∏ j ( D n + r + D k ) X j . (76) This normalization separates the factors involving α \alpha α α + β \alpha+\beta α + β p p p p p p
Proposition 5.2. Let p p p
p > n , p ∤ u v D , 0 < ∣ u k + a ∣ < p , 0 < ∣ v k + b ∣ < p ( 1 ≤ k ≤ n ) . (77) p>n,\qquad p\nmid uvD,\qquad0<|uk+a|<p,\qquad0<|vk+b|<p\quad(1\le k\le n). \tag*{(77)} p > n , p ∤ uv D , 0 < ∣ u k + a ∣ < p , 0 < ∣ v k + b ∣ < p ( 1 ≤ k ≤ n ) . (77) 0 < D ( n + 1 ) + r , 2 D n + r < 2 p , p ≡ r ( m o d D ) . (78) 0<D(n+1)+r,\qquad2Dn+r<2p,\qquad p\equiv r\pmod D. \tag*{(78)} 0 < D ( n + 1 ) + r , 2 D n + r < 2 p , p ≡ r ( mod D ) . (78) Suppose
1 ≤ k 0 : = p − r D − n ≤ n , d : = n − k 0 + 1. 1\le k_0:=\frac{p-r}{D}-n\le n,\qquad d:=n-k_0+1. 1 ≤ k 0 := D p − r − n ≤ n , d := n − k 0 + 1. Then an inertia subgroup at p p p Gal ( P n ( α , β ) / Q ) \operatorname{Gal}(P_n^{(\alpha,\beta)}/\mathbb{Q}) Gal ( P n ( α , β ) / Q ) d d d c ∈ Q × c\in\mathbb{Q}^{\times} c ∈ Q × v p ( c ) = 0 v_p(c)=0 v p ( c ) = 0 2 d < p 2d<p 2 d < p p p p Gal ( P n ( α , β ) ( 1 + c X 2 ) / Q ) \operatorname{Gal}(P_n^{(\alpha,\beta)}(1+cX^2)/\mathbb{Q}) Gal ( P n ( α , β ) ( 1 + c X 2 ) / Q ) 2 d 2d 2 d Lemma 5.1 .
Proof. In (76) , the binomial coefficients and the first product are p p p ( 0 , 2 p ) (0,2p) ( 0 , 2 p ) p p p k 0 k_0 k 0 p p p j < k 0 j<k_0 j < k 0 j ≥ k 0 j\ge k_0 j ≥ k 0 k 0 − 1 = n − d k_0-1=n-d k 0 − 1 = n − d
It remains to prove that this reduction is squarefree. The Jacobi polynomial and (74) are integral at p p p (73) are units by (77) . Its reduction has degree at most n < p n<p n < p z ≠ ± 1 z\ne\pm1 z = ± 1 h ≥ 2 h\ge2 h ≥ 2 y ˉ = ( X − z ) h g ( X ) \bar y=(X-z)^h g(X) y ˉ = ( X − z ) h g ( X ) g ( z ) ≠ 0 g(z)\ne0 g ( z ) = 0 2 ≤ h < p 2\le h<p 2 ≤ h < p ( X − z ) h − 2 (X-z)^{h-2} ( X − z ) h − 2
( 1 − z 2 ) h ( h − 1 ) g ( z ) ≠ 0 , (1-z^2)h(h-1)g(z)\ne0, ( 1 − z 2 ) h ( h − 1 ) g ( z ) = 0 , a contradiction. Thus the reduction is squarefree. The scalar and affine transformation in (76) are units at p p p K n α , β K_n^{\alpha,\beta} K n α , β
Lemma 5.1 now gives a single d d d
P n ( α , β ) ( 1 + c X 2 ) = ( u n n ! ) − 1 K n α , β ( u c 2 D X 2 ) P_n^{(\alpha,\beta)}(1+cX^2)=(u^n n!)^{-1}K_n^{\alpha,\beta}\left(\frac{uc}{2D}X^2\right) P n ( α , β ) ( 1 + c X 2 ) = ( u n n ! ) − 1 K n α , β ( 2 D u c X 2 ) and apply the second part of that lemma. □
Blocks and opposite pairs
A block system for a transitive group is a partition into equally sized sets permuted by the group. It is nontrivial when the block size is strictly between 1 1 1
A cycle moving more than half the roots places a simple restriction on a nontrivial block system: every block meeting the support of the cycle must lie entirely in that support. The block size must therefore divide the cycle length.
Lemma 5.3. Let a transitive group of degree N N N b b b d d d d > N / 2 d>N/2 d > N /2 b ∣ d b\mid d b ∣ d
Proof. We have b ≤ N / 2 < d b \le N/2 < d b ≤ N /2 < d d > b d > b d > b b ∣ d b \mid d b ∣ d
Once primitivity is known, a single cycle with at least three fixed points forces the alternating group. We use Jones’s theorem in this form; the cycle length need not be prime.
Theorem 5.4 (Jones [17 Corollary 1.3] ). A primitive permutation group of degree N N N A N A_N A N
This form of Jones’s theorem uses the classification of finite simple groups. The prime-length case used later is Jordan’s classical theorem.
For roots occurring in opposite pairs, the largest possible group has two parts: permutations of the pairs and interchanges within each pair. Even after the group on pairs is known to be S N S_N S N 2 d 2d 2 d d < N d < N d < N d d d
Lemma 5.5. Let Γ ≤ F 2 N ⋊ S N \Gamma\le\mathbb{F}_2^N \rtimes S_N Γ ≤ F 2 N ⋊ S N S N S_N S N Γ \Gamma Γ 2 d 2d 2 d d d d d d d 1 ≤ d < N 1 \le d < N 1 ≤ d < N Γ = F 2 N ⋊ S N \Gamma= \mathbb{F}_2^N \rtimes S_N Γ = F 2 N ⋊ S N
Proof. Let V V V d d d V V V d d d v v v S N S_N S N Γ \Gamma Γ V V V i i i j j j v + ( i j ) v = e i + e j v + (ij)v = e_i + e_j v + ( ij ) v = e i + e j V V V S N S_N S N e k + e l e_k + e_l e k + e l k ≠ l k \ne l k = l v v v F 2 N \mathbb{F}_2^N F 2 N V = F 2 N V = \mathbb{F}_2^N V = F 2 N
We will also use these local permutations to control common subfields. Restriction to a Galois subfield has a normal kernel. The next lemma shows that certain supported cycles cannot lie in a proper normal subgroup of the groups under consideration. Thus local elements that fix one field can force the entire group to fix its intersection with another field.
Lemma 5.6. Let N ≥ 3 N \ge3 N ≥ 3
A normal subgroup of S N S_N S N d d d 2 ≤ d < N 2 \le d < N 2 ≤ d < N S N S_N S N
A normal subgroup of C 2 ≀ S N C_2 \wr S_N C 2 ≀ S N 2 d 2d 2 d 1 ≤ d < N 1 \le d < N 1 ≤ d < N C 2 ≀ S N C_2 \wr S_N C 2 ≀ S N
Proof. For the first assertion, let H ◃ S N H \triangleleft S_N H ◃ S N σ \sigma σ i i i σ \sigma σ j j j
σ ( i j ) σ − 1 ( i j ) = ( σ ( i ) j ) ( i j ) ∈ H . \sigma(ij)\sigma^{-1}(ij) = (\sigma(i)j)(ij) \in H. σ ( ij ) σ − 1 ( ij ) = ( σ ( i ) j ) ( ij ) ∈ H . This is a 3-cycle. Its conjugates generate A N A_N A N σ \sigma σ H = S N H = S_N H = S N
For the second assertion, the image of the normal subgroup in S N S_N S N S N S_N S N d d d
An explicit prime-counting criterion
The local cycle lengths convert a possible block size b b b p p p D b Db D b
Proposition 5.7. Let D ∈ { 2 , 6 } D \in\{2,6\} D ∈ { 2 , 6 } r ∈ Z r \in\mathbb{Z} r ∈ Z ∣ r ∣ < D |r|<D ∣ r ∣ < D ( r , D ) = 1 (r,D)=1 ( r , D ) = 1 n ≥ 10 6 n \ge10^6 n ≥ 1 0 6
P D , r ( n ) = { p prime : D n + 10 D < p < 3 D n / 2 , p ≡ r ( m o d D ) } . \mathcal{P}_{D,r}(n)=\{p\ \text{prime}: Dn+10D<p<3Dn/2,\ p\equiv r\pmod{D}\}. P D , r ( n ) = { p prime : D n + 10 D < p < 3 D n /2 , p ≡ r ( mod D )} . Suppose that G ≤ S n G\le S_n G ≤ S n d p d_p d p p ∈ P D , r ( n ) p\in\mathcal{P}_{D,r}(n) p ∈ P D , r ( n )
d p = 2 n + 1 − p − r D . d_p=2n+1-\frac{p-r}{D}. d p = 2 n + 1 − D p − r . Then G = S n G=S_n G = S n
Proof. Write L = log n L=\log n L = log n c D = D / ( 2 φ ( D ) ) c_D=D/(2\varphi(D)) c D = D / ( 2 φ ( D )) φ \varphi φ c 2 = 1 c_2=1 c 2 = 1 c 6 = 3 / 2 c_6=3/2 c 6 = 3/2 π ( x ; q , a ) \pi(x;q,a) π ( x ; q , a ) x x x a a a q q q Li ( x ) = ∫ 2 x d t log t \operatorname{Li}(x)=\int_2^x \frac{\mathrm{d}t}{\log t} Li ( x ) = ∫ 2 x l o g t d t
∣ π ( x ; q , a ) − Li ( x ) φ ( q ) ∣ < 0.027 x log 2 x ( x ≥ 10 6 , 1 ≤ q ≤ 10 5 , ( a , q ) = 1 ) , (79) \left|\pi(x;q,a)-\frac{\operatorname{Li}(x)}{\varphi(q)}\right|<0.027\frac{x}{\log^2 x}\qquad(x\ge10^6,\ 1\le q\le10^5,\ (a,q)=1), \tag*{(79)} π ( x ; q , a ) − φ ( q ) Li ( x ) < 0.027 log 2 x x ( x ≥ 1 0 6 , 1 ≤ q ≤ 1 0 5 , ( a , q ) = 1 ) , (79) from [3 Corollary 1.7] . Only moduli at most 18 occur in applications of (79) below.
The interval ( D n , 3 D n / 2 ) (Dn,3Dn/2) ( D n , 3 D n /2 ) D n / 2 Dn/2 D n /2 10 D 10D 10 D
T : = ∣ P D , r ( n ) ∣ > D n 2 φ ( D ) ( L + log ( 3 D / 2 ) ) − 0.0675 D n L 2 − 10 D ≥ c D n L ( L L + log 9 − 0.27 L − 40 L n ) > 0.84 c D n L . (80) \begin{aligned}
T:=|\mathcal{P}_{D,r}(n)|>\frac{Dn}{2\varphi(D)(L+\log(3D/2))}-\frac{0.0675Dn}{L^2}-10D \\
&\ge c_D\frac{n}{L}\left(\frac{L}{L+\log9}-\frac{0.27}{L}-\frac{40L}{n}\right)>0.84c_D\frac{n}{L}. \tag*{(80)}
\end{aligned} T := ∣ P D , r ( n ) ∣ > 2 φ ( D ) ( L + log ( 3 D /2 )) D n − L 2 0.0675 D n − 10 D ≥ c D L n ( L + log 9 L − L 0.27 − n 40 L ) > 0.84 c D L n . (80) The expression in parentheses increases with n n n n = 10 6 n=10^6 n = 1 0 6 69 / 5 < log ( 10 6 ) < 691 / 50 69/5<\log(10^6)<691/50 69/5 < log ( 1 0 6 ) < 691/50 log 9 < 11 / 5 \log9<11/5 log 9 < 11/5
69 80 − 9 460 − 691 1250000 > 21 25 . \frac{69}{80}-\frac{9}{460}-\frac{691}{1250000}>\frac{21}{25}. 80 69 − 460 9 − 1250000 691 > 25 21 . For b = 2 , 3 b=2,3 b = 2 , 3 a a a D b Db D b
# { p : D n + 10 D < p < 3 D n / 2 , p ≡ a ( m o d D b ) } > n 2 ( L + log 9 ) − 0.405 n L 2 − 60 > 0.4 n L . (81) \#\{p:Dn+10D<p<3Dn/2,\ p\equiv a\pmod{Db}\}
>\frac{n}{2(L+\log9)}-\frac{0.405n}{L^2}-60>0.4\frac{n}{L}. \tag*{(81)} # { p : D n + 10 D < p < 3 D n /2 , p ≡ a ( mod D b )} > 2 ( L + log 9 ) n − L 2 0.405 n − 60 > 0.4 L n . (81) Indeed, D / ( 2 φ ( D b ) ) ≥ 1 / 2 D/(2\varphi(Db))\ge1/2 D / ( 2 φ ( D b )) ≥ 1/2 D ≤ 6 D\le6 D ≤ 6 L / n L/n L / n n n n 10 6 10^6 1 0 6
69 160 − 27 920 − 2073 2500000 > 2 5 . \frac{69}{160}-\frac{27}{920}-\frac{2073}{2500000}>\frac{2}{5}. 160 69 − 920 27 − 2500000 2073 > 5 2 . In particular, both reduced classes r r r r + D r+D r + D 2 D 2D 2 D
Suppose that G G G b b b 2 ≤ b ≤ n / 2 2\le b\le n/2 2 ≤ b ≤ n /2 b ∣ n b\mid n b ∣ n ∣ r ∣ < D |r|<D ∣ r ∣ < D
n / 2 < d p ≤ n − 9. n/2<d_p\le n-9. n /2 < d p ≤ n − 9. Lemma 5.3 gives b ∣ d p b\mid d_p b ∣ d p
p = D ( 2 n + 1 − d p ) + r ≡ D + r ( m o d D b ) . (82) p=D(2n+1-d_p)+r\equiv D+r\pmod{Db}. \tag*{(82)} p = D ( 2 n + 1 − d p ) + r ≡ D + r ( mod D b ) . (82) This class is reduced: otherwise every prime in it would divide D b ≤ D n / 2 Db\le Dn/2 D b ≤ D n /2 p > D n p>Dn p > D n
If b = 2 b=2 b = 2 3 3 3 D D D D b Db D b r r r φ ( D b ) / φ ( D ) \varphi(Db)/\varphi(D) φ ( D b ) / φ ( D ) 2 2 2 3 3 3 (82) ; the positive lower bound (81) is a contradiction.
If 4 ≤ b ≤ n 1 / 4 4\le b\le n^{1/4} 4 ≤ b ≤ n 1/4
φ ( D b ) ≥ 4 φ ( D ) . \varphi(Db)\ge4\varphi(D). φ ( D b ) ≥ 4 φ ( D ) . ::: For D = 2 D=2 D = 2 φ ( 2 b ) ≥ 4 \varphi(2b)\ge4 φ ( 2 b ) ≥ 4 D = 6 D=6 D = 6 φ ( 6 b ) / 2 ≥ φ ( 2 b ) \varphi(6b)/2\ge\varphi(2b) φ ( 6 b ) /2 ≥ φ ( 2 b )
π ( x + y ; q , a ) − π ( x ; q , a ) < 2 y φ ( q ) log ( y / q ) ( y > q , ( a , q ) = 1 ) ; (83) \pi(x+y;q,a)-\pi(x;q,a)<\frac{2y}{\varphi(q)\log(y/q)}\qquad(y>q,\ (a,q)=1); \tag*{(83)} π ( x + y ; q , a ) − π ( x ; q , a ) < φ ( q ) log ( y / q ) 2 y ( y > q , ( a , q ) = 1 ) ; (83) see [21 Theorem 2, Eq. (1.10)] . With y = D n / 2 y=Dn/2 y = D n /2 q = D b q=Db q = D b
T < D n φ ( D b ) log ( n / ( 2 b ) ) ≤ D n 4 φ ( D ) ( 3 L / 4 − log 2 ) = c D n L L 2 ( 3 L / 4 − log 2 ) < 0.75 c D n L , \begin{aligned}
T<\frac{Dn}{\varphi(Db)\log(n/(2b))} \\
&\le\frac{Dn}{4\varphi(D)(3L/4-\log2)}=c_D\frac{n}{L}\frac{L}{2(3L/4-\log2)}<0.75c_D\frac{n}{L},
\end{aligned} T < φ ( D b ) log ( n / ( 2 b )) D n ≤ 4 φ ( D ) ( 3 L /4 − log 2 ) D n = c D L n 2 ( 3 L /4 − log 2 ) L < 0.75 c D L n , contradicting (80) . The last normalized factor decreases with n n n 10 6 10^6 1 0 6 L > 69 / 5 L>69/5 L > 69/5 log 2 < 7 / 10 \log2<7/10 log 2 < 7/10 138 / 193 < 3 / 4 138/193<3/4 138/193 < 3/4
Finally, if b > n 1 / 4 b>n^{1/4} b > n 1/4 (82) gives
T ≤ n 2 b + 1 < 1 2 n 3 / 4 + 1 < 0.25 n L ≤ 0.25 c D n L , T\le\frac{n}{2b}+1<\frac{1}{2}n^{3/4}+1<0.25\frac{n}{L}\le0.25c_D\frac{n}{L}, T ≤ 2 b n + 1 < 2 1 n 3/4 + 1 < 0.25 L n ≤ 0.25 c D L n , again a contradiction. Here L / ( 2 n 1 / 4 ) + L / n L/(2n^{1/4})+L/n L / ( 2 n 1/4 ) + L / n n ≥ 10 6 n\ge10^6 n ≥ 1 0 6 n 1 / 4 > 158 / 5 n^{1/4}>158/5 n 1/4 > 158/5 L < 691 / 50 L<691/50 L < 691/50
691 3160 + 691 50000000 < 1 4 . \frac{691}{3160}+\frac{691}{50000000}<\frac{1}{4}. 3160 691 + 50000000 691 < 4 1 . Thus G G G
Every cycle under consideration is nontrivial and fixes at least nine points. Theorem 5.4 gives A n ≤ G A_n\le G A n ≤ G r + D r+D r + D 2 D 2D 2 D (81) . Then ( p − r ) / D (p-r)/D ( p − r ) / D d p d_p d p G = S n G=S_n G = S n
Remark 5.8. The strict comparisons needed in this proof follow from elementary monotonicity and the bounds
13.8 < log ( 10 6 ) < 13.82 , log 9 < 2.2 , log 2 < 0.7 , ( 10 6 ) 1 / 4 > 31.6. 13.8<\log(10^6)<13.82,\qquad\log9<2.2,\qquad\log2<0.7,\qquad(10^6)^{1/4}>31.6. 13.8 < log ( 1 0 6 ) < 13.82 , log 9 < 2.2 , log 2 < 0.7 , ( 1 0 6 ) 1/4 > 31.6. Thus the numerical threshold requires no additional finite computation.
The Legendre Galois group
We first replace each opposite pair of nonzero Legendre roots by its common square. The resulting degree-n n n T = − X 2 T=-X^2 T = − X 2
For ϵ = 2 δ − 1 \epsilon=2\delta-1 ϵ = 2 δ − 1
J n ϵ ( T ) = ∑ j = 0 n ( n j ) ∏ k = 1 n ( 2 j + 2 k + ϵ ) T j . (84) J_n^\epsilon(T)=\sum_{j=0}^{n}\binom{n}{j}\prod_{k=1}^{n}(2j+2k+\epsilon)T^j. \tag*{(84)} J n ϵ ( T ) = j = 0 ∑ n ( j n ) k = 1 ∏ n ( 2 j + 2 k + ϵ ) T j . (84) Coefficient comparison gives
J n ϵ ( − X 2 ) = ( − 2 ) n n ! L 2 n + δ ( X ) . (85) J_n^\epsilon(-X^2)=(-2)^n n!L_{2n+\delta}(X). \tag*{(85)} J n ϵ ( − X 2 ) = ( − 2 ) n n ! L 2 n + δ ( X ) . (85) Theorem 5.9. Let n ≥ 1 n\ge1 n ≥ 1 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } ϵ = 2 δ − 1 \epsilon=2\delta-1 ϵ = 2 δ − 1 J n ϵ J_n^\epsilon J n ϵ Q \mathbb{Q} Q
Gal ( J n ϵ / Q ) = S n , Gal ( L 2 n + δ / Q ) = W n . \operatorname{Gal}(J_n^\epsilon/\mathbb{Q})=S_n,\qquad\operatorname{Gal}(L_{2n+\delta}/\mathbb{Q})=W_n. Gal ( J n ϵ / Q ) = S n , Gal ( L 2 n + δ / Q ) = W n . In particular, J n ϵ J_n^\epsilon J n ϵ L 2 n + δ L_{2n+\delta} L 2 n + δ Q \mathbb{Q} Q
These are the two conditional Galois assertions in [7 Conjecture 1.5] . The irreducibility equivalence was proved in [7 Lemma 1.3] ; the proof below also establishes it directly.
The integral polynomial in (84) is the specialization of (76) with D = 2 D=2 D = 2
J n ϵ ( T ) = 2 n n ! P n ( ϵ / 2 , 0 ) ( 2 T + 1 ) . (86) J_n^\epsilon(T)=2^n n!P_n^{(\epsilon/2,0)}(2T+1). \tag*{(86)} J n ϵ ( T ) = 2 n n ! P n ( ϵ /2 , 0 ) ( 2 T + 1 ) . (86) Indeed, the two products in (76) concatenate to the product in (84) . Together with the classical quadratic transformation, this proves (85) ; see [7 Lemma 1.2] .
Lemma 5.10. Let N = 2 n + δ N=2n+\delta N = 2 n + δ ϵ = 2 δ − 1 \epsilon=2\delta-1 ϵ = 2 δ − 1 p p p N < p ≤ 4 n + ϵ N<p\le4n+\epsilon N < p ≤ 4 n + ϵ
d = 2 n + 1 − p − ϵ 2 . d=2n+1-\frac{p-\epsilon}{2}. d = 2 n + 1 − 2 p − ϵ . Then 1 ≤ d ≤ n 1\le d\le n 1 ≤ d ≤ n L N L_N L N 2 d 2d 2 d d d d d d d d d d
Proof. Apply Proposition 5.2 with ( a , u , b , v ) = ( ϵ , 2 , 0 , 1 ) (a,u,b,v)=(\epsilon,2,0,1) ( a , u , b , v ) = ( ϵ , 2 , 0 , 1 ) D = 2 D=2 D = 2 2 k + ϵ 2k+\epsilon 2 k + ϵ k k k p p p 2 n + ϵ + 2 k 2n+\epsilon+2k 2 n + ϵ + 2 k 4 n + ϵ < 2 p 4n+\epsilon<2p 4 n + ϵ < 2 p k 0 = ( p − ϵ ) / 2 − n k_0=(p-\epsilon)/2-n k 0 = ( p − ϵ ) /2 − n [ 1 , n ] [1,n] [ 1 , n ]
2 d = 4 n + 2 + ϵ − p < p . 2d=4n+2+\epsilon-p<p. 2 d = 4 n + 2 + ϵ − p < p . For δ = 0 \delta=0 δ = 0 p ≥ 2 n + 1 p\ge2n+1 p ≥ 2 n + 1 δ = 1 \delta=1 δ = 1 p ≥ 2 n + 3 p\ge2n+3 p ≥ 2 n + 3 c = − 2 c=-2 c = − 2 (86) , and then (85) . □
Lemma 5.11. For every n ≥ 4 n\ge4 n ≥ 4 ϵ ∈ { − 1 , 1 } \epsilon\in\{-1,1\} ϵ ∈ { − 1 , 1 } d d d 1 ≤ d < n 1\le d<n 1 ≤ d < n
Proof. Cullinan and Hajir [7 Lemma 2.2] prove that every real x ≥ 9 x\ge9 x ≥ 9 4 4 4 [ x , 2 x − 5 ] [x,2x-5] [ x , 2 x − 5 ] x = 2 n + 3 + ϵ x=2n+3+\epsilon x = 2 n + 3 + ϵ p ≡ ϵ ( m o d 4 ) p\equiv\epsilon\pmod{4} p ≡ ϵ ( mod 4 ) p p p p > x p>x p > x ϵ = − 1 \epsilon=-1 ϵ = − 1 4 n − 1 4n-1 4 n − 1 ϵ = 1 \epsilon=1 ϵ = 1 4 n + 3 4n+3 4 n + 3 p ≤ 4 n + 1 p\le4n+1 p ≤ 4 n + 1
2 n + 2 + ϵ < p ≤ 4 n + ϵ . 2n+2+\epsilon<p\le4n+\epsilon. 2 n + 2 + ϵ < p ≤ 4 n + ϵ . These inequalities give 1 ≤ d < n 1\le d<n 1 ≤ d < n ( p − ϵ ) / 2 (p-\epsilon)/2 ( p − ϵ ) /2 d d d
Proof of Theorem 5.9. Assume that J n ϵ J_n^\epsilon J n ϵ n ≥ 10 6 n\ge10^6 n ≥ 1 0 6 P 2 , ϵ ( n ) \mathcal{P}_{2,\epsilon}(n) P 2 , ϵ ( n ) Gal ( J n ϵ / Q ) = S n \operatorname{Gal}(J_n^\epsilon/\mathbb{Q})=S_n Gal ( J n ϵ / Q ) = S n n < 10 6 n<10^6 n < 1 0 6 [7 Theorem 1.9(a)] , which covers every n ≤ 10 9 n\le10^9 n ≤ 1 0 9 S n S_n S n
The nonzero Legendre roots are distinct and occur in n n n Γ \Gamma Γ F 2 n ⋊ S n \mathbb{F}_2^n\rtimes S_n F 2 n ⋊ S n Gal ( J n ϵ / Q ) = S n \operatorname{Gal}(J_n^\epsilon/\mathbb{Q})=S_n Gal ( J n ϵ / Q ) = S n n ≥ 4 n\ge4 n ≥ 4 Γ = F 2 n ⋊ S n \Gamma=\mathbb{F}_2^n\rtimes S_n Γ = F 2 n ⋊ S n
For completeness, the three small degrees can be treated directly. When n = 1 n=1 n = 1 3 X 2 − 1 3X^2-1 3 X 2 − 1 5 X 2 − 3 5X^2-3 5 X 2 − 3 C 2 C_2 C 2 ( n , ϵ ) = ( 2 , − 1 ) , ( 3 , − 1 ) , ( 3 , 1 ) (n,\epsilon)=(2,-1),(3,-1),(3,1) ( n , ϵ ) = ( 2 , − 1 ) , ( 3 , − 1 ) , ( 3 , 1 ) p = 7 , 11 , 13 p=7,11,13 p = 7 , 11 , 13 d = 1 d=1 d = 1 ( n , ϵ ) = ( 2 , 1 ) (n,\epsilon)=(2,1) ( n , ϵ ) = ( 2 , 1 ) 63 X 4 − 70 X 2 + 15 63X^4-70X^2+15 63 X 4 − 70 X 2 + 15
y ± = 35 ± 2 70 63 ∈ K = Q ( 70 ) , y + y − = 5 21 . y_{\pm}=\frac{35\pm2\sqrt{70}}{63}\in K=\mathbb{Q}(\sqrt{70}),\qquad y_+y_-=\frac{5}{21}. y ± = 63 35 ± 2 70 ∈ K = Q ( 70 ) , y + y − = 21 5 . Neither is a square in K K K 5 / 21 5/21 5/21 K K K q q q Q ( 70 ) \mathbb{Q}(\sqrt{70}) Q ( 70 ) q q q q / 70 q/70 q /70 5 / 21 5/21 5/21 1 / 294 1/294 1/294 4 4 4 K K K 8 8 8 Q \mathbb{Q} Q C 2 ≀ S 2 C_2\wr S_2 C 2 ≀ S 2
Finally, the irreducibility equivalence follows directly. A factorization of J n ϵ J_n^\epsilon J n ϵ J n ϵ ( − X 2 ) J_n^\epsilon(-X^2) J n ϵ ( − X 2 ) L 2 n + δ L_{2n+\delta} L 2 n + δ J n ϵ J_n^\epsilon J n ϵ 2 n 2n 2 n L 2 n + δ L_{2n+\delta} L 2 n + δ
Combining Theorem 5.9 with the irreducibility in Theorem 1.5 proves Theorem 1.2.
The Kaneko–Zagier Galois group
Theorem 5.12. For every n ≥ 1 n \ge1 n ≥ 1 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } P n ( λ / 3 , μ / 2 ) P_n^{(\lambda/3,\mu/2)} P n ( λ /3 , μ /2 ) Q \mathbb{Q} Q
Gal ( P n ( λ / 3 , μ / 2 ) / Q ) = S n . \operatorname{Gal}(P_n^{(\lambda/3,\mu/2)}/\mathbb{Q}) = S_n. Gal ( P n ( λ /3 , μ /2 ) / Q ) = S n . This extends the conditional finite verification of [8 Theorem 1.8] to every degree.
For λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } r = 2 λ + 3 μ r=2\lambda+3\mu r = 2 λ + 3 μ K n λ / 3 , μ / 2 K_n^{\lambda/3,\mu/2} K n λ /3 , μ /2 K n λ , μ K_n^{\lambda,\mu} K n λ , μ (76) , with D = 6 D=6 D = 6
K n λ , μ ( X ) = ∑ j = 0 n ( n j ) ∏ k = j + 1 n ( 3 k + λ ) ∏ k = 0 j ( 6 n + r + 6 k ) X j = 3 n n ! P n ( λ / 3 , μ / 2 ) ( 4 X + 1 ) . (87) K_n^{\lambda,\mu}(X)=\sum_{j=0}^{n}\binom{n}{j}\prod_{k=j+1}^{n}(3k+\lambda)\prod_{k=0}^{j}(6n+r+6k)X^j
=3^n n!P_n^{(\lambda/3,\mu/2)}(4X+1). \tag*{(87)} K n λ , μ ( X ) = j = 0 ∑ n ( j n ) k = j + 1 ∏ n ( 3 k + λ ) k = 0 ∏ j ( 6 n + r + 6 k ) X j = 3 n n ! P n ( λ /3 , μ /2 ) ( 4 X + 1 ) . (87) To compare with the published finite result, put ℓ = 12 n + 6 + r \ell=12n+6+r ℓ = 12 n + 6 + r F ~ ℓ − 1 \widetilde{F}_{\ell-1} F ℓ − 1 (106) in § 6.4 . Its normalization gives
F ~ ℓ − 1 ( Y ) = 576 n n ! K n λ , μ ( − Y / 3456 ) . (88) \widetilde{F}_{\ell-1}(Y)=\frac{576^n}{n!}K_n^{\lambda,\mu}(-Y/3456). \tag*{(88)} F ℓ − 1 ( Y ) = n ! 57 6 n K n λ , μ ( − Y /3456 ) . (88) These polynomials therefore have the same irreducibility behavior and isomorphic splitting fields over Q \mathbb{Q} Q
Proof of Theorem 5.12. Suppose P n ( λ / 3 , μ / 2 ) P_n^{(\lambda/3,\mu/2)} P n ( λ /3 , μ /2 ) n ≥ 10 6 n \ge10^6 n ≥ 1 0 6 p ∈ P 6 , r ( n ) p \in\mathcal{P}_{6,r}(n) p ∈ P 6 , r ( n )
6 n + 60 < p < 9 n , p ≡ r ( m o d 6 ) . 6n+60<p<9n,\qquad p\equiv r\pmod{6}. 6 n + 60 < p < 9 n , p ≡ r ( mod 6 ) . The small factors 3 k + λ 3k+\lambda 3 k + λ 2 k + μ 2k+\mu 2 k + μ p p p 12 n + r < 2 p 12n+r<2p 12 n + r < 2 p
1 ≤ p − r 6 − n ≤ n . 1\le\frac{p-r}{6}-n\le n. 1 ≤ 6 p − r − n ≤ n . Thus Proposition 5.2 supplies a single cycle of length
d p = 2 n + 1 − p − r 6 . d_p=2n+1-\frac{p-r}{6}. d p = 2 n + 1 − 6 p − r . Irreducibility makes the group transitive, and Proposition 5.7 proves that it is S n S_n S n
For 1 ≤ n < 10 6 1\le n<10^6 1 ≤ n < 1 0 6 (105) satisfies
ℓ ≤ 12 ( 999999 ) + 11 = 11 999 999 < 10 9 . \ell\le12(999999)+11=11\,999\,999<10^9. ℓ ≤ 12 ( 999999 ) + 11 = 11 999 999 < 1 0 9 . Cullinan and Hajir [8 Theorem 1.8] prove that, for every ℓ ≤ 10 9 \ell\le10^9 ℓ ≤ 1 0 9 6 6 6 F ~ ℓ − 1 \widetilde{F}_{\ell-1} F ℓ − 1 S n S_n S n (106) completes the proof in all degrees. □
Together with the irreducibility proved in § 4 , this completes Theorem 1.1 .
Fixed rational parameters
We now allow fixed rational parameters and do not assume irreducibility. Every cycle of length greater than n / 2 n/2 n /2 D n Dn D n n n n Theorem 1.6 . The remaining primes establish primitivity and the symmetric group on that factor. The prime-number-theorem estimates concern fixed moduli; the uniform Brun–Titchmarsh bound still controls possible growing block sizes.
The large factor and its splitting field.
Lemma 5.13. Fix an integer q ≥ 2 q \ge2 q ≥ 2 A > 0 A > 0 A > 0 a a a q q q x x x
( x , x + x ( log x ) A ) \left(x,x+\frac{x}{(\log x)^A}\right) ( x , x + ( log x ) A x ) contains a prime congruent to a a a q q q a a a
Proof. Choose a fixed power Q = q m ≥ 10 5 Q=q^m \ge10^5 Q = q m ≥ 1 0 5 a a a q q q Q Q Q [3 ] , with Z = A + 1 Z=A+1 Z = A + 1
π ( t ; Q , a ) = Li ( t ) φ ( Q ) + O q , A ( t ( log t ) A + 2 ) ( t → ∞ ) . \pi(t;Q,a)=\frac{\operatorname{Li}(t)}{\varphi(Q)}+O_{q,A}\left(\frac{t}{(\log t)^{A+2}}\right)\qquad(t\to\infty). π ( t ; Q , a ) = φ ( Q ) Li ( t ) + O q , A ( ( log t ) A + 2 t ) ( t → ∞ ) . Indeed, for each fixed Z Z Z [3 Eq. (6.23)] can be chosen sufficiently large; the resulting threshold is fixed and uniform in the reduced classes. Set H = x / ( log x ) A H=x/(\log x)^A H = x / ( log x ) A
π ( x + H ; Q , a ) − π ( x ; Q , a ) = ( 1 φ ( Q ) + o q , A ( 1 ) ) x ( log x ) A + 1 > 0 \pi(x+H;Q,a)-\pi(x;Q,a)=\left(\frac{1}{\varphi(Q)}+o_{q,A}(1)\right)\frac{x}{(\log x)^{A+1}}>0 π ( x + H ; Q , a ) − π ( x ; Q , a ) = ( φ ( Q ) 1 + o q , A ( 1 ) ) ( log x ) A + 1 x > 0 for all sufficiently large x x x a a a q q q
Proof of Theorem 1.6. Write α = a / u \alpha=a/u α = a / u β = b / v \beta=b/v β = b / v α + β = r / D \alpha+\beta=r/D α + β = r / D
( r , D ) = 1 , D ≥ max { u , v } , D ≥ 2 is even . (r,D)=1,\qquad D\ge\max\{u,v\},\qquad D\ge2\text{ is even}. ( r , D ) = 1 , D ≥ max { u , v } , D ≥ 2 is even . In particular, α + β = r / D \alpha+\beta=r/D α + β = r / D P n ( α , β ) P_n^{(\alpha,\beta)} P n ( α , β ) n n n (74) , over a field of characteristic zero, then also shows that all its roots are simple.
Choose a fixed constant
B > ∣ r ∣ + ∣ a ∣ + ∣ b ∣ + 4 D + 10 , B>|r|+|a|+|b|+4D+10, B > ∣ r ∣ + ∣ a ∣ + ∣ b ∣ + 4 D + 10 , and consider primes in
p ≡ r ( m o d D ) , D n + B < p < 3 D n / 2 − B . (89) p\equiv r\pmod D,\qquad Dn+B<p<3Dn/2-B. \tag*{(89)} p ≡ r ( mod D ) , D n + B < p < 3 D n /2 − B . (89) For all sufficiently large n n n D n + B Dn+B D n + B D n + r + D k Dn+r+Dk D n + r + D k ( 0 , 2 p ) (0,2p) ( 0 , 2 p ) k 0 = ( p − r ) / D − n k_0=(p-r)/D-n k 0 = ( p − r ) / D − n 1 ≤ k 0 ≤ n 1\le k_0\le n 1 ≤ k 0 ≤ n
d p = 2 n + 1 − p − r D , n / 2 < d p ≤ n − 4. (90) d_p=2n+1-\frac{p-r}{D},\qquad n/2<d_p\le n-4. \tag*{(90)} d p = 2 n + 1 − D p − r , n /2 < d p ≤ n − 4. (90) The large orbit. Every cycle in (90) has support larger than n / 2 n/2 n /2 O O O A > 0 A>0 A > 0 q = D q=D q = D x = D n + B x=Dn+B x = D n + B (89) with
p = D n + O α , β , A ( n ( log n ) A ) . p=Dn+O_{\alpha,\beta,A}\left(\frac{n}{(\log n)^A}\right). p = D n + O α , β , A ( ( log n ) A n ) . The associated cycle has length n − O α , β , A ( n / ( log n ) A ) n-O_{\alpha,\beta,A}(n/(\log n)^A) n − O α , β , A ( n / ( log n ) A )
0 ≤ n − M = O α , β , A ( n ( log n ) A ) , M : = ∣ O ∣ , 0\le n-M=O_{\alpha,\beta,A}\left(\frac{n}{(\log n)^A}\right),\qquad M:=|O|, 0 ≤ n − M = O α , β , A ( ( log n ) A n ) , M := ∣ O ∣ , for every fixed A > 0 A>0 A > 0 O O O A A A A A A h n h_n h n
Primitivity on the large orbit. Let G O G_O G O O O O s s s 2 ≤ s ≤ M / 2 2 \le s \le M/2 2 ≤ s ≤ M /2 d p > n / 2 ≥ M / 2 d_p > n/2 \ge M/2 d p > n /2 ≥ M /2 s ∣ d p s \mid d_p s ∣ d p (89) , and hence
p ≡ D ( 2 n + 1 ) + r ( m o d D s ) . (91) p \equiv D(2n+1)+r \pmod{Ds}. \tag*{(91)} p ≡ D ( 2 n + 1 ) + r ( mod D s ) . (91) Here s s s M M M n n n (91) . Let T n T_n T n (89) . The prime number theorem in arithmetic progressions, with fixed modulus D D D
T n ∼ D 2 φ ( D ) n log n . (92) T_n \sim\frac{D}{2\varphi(D)}\frac{n}{\log n}. \tag*{(92)} T n ∼ 2 φ ( D ) D log n n . (92) Choose a fixed integer s 0 s_0 s 0 φ ( D s ) ≥ 4 φ ( D ) \varphi(Ds) \ge4\varphi(D) φ ( D s ) ≥ 4 φ ( D ) s ≥ s 0 s \ge s_0 s ≥ s 0 φ ( q ) ≥ q / 2 \varphi(q) \ge\sqrt{q/2} φ ( q ) ≥ q /2 s 0 ≤ s ≤ n 1 / 4 s_0 \le s \le n^{1/4} s 0 ≤ s ≤ n 1/4 (83) applied to (91) gives
T n < D n φ ( D s ) log ( n / ( 2 s ) ) ≤ D n 4 φ ( D ) ( 3 4 log n − log 2 ) ∼ D 3 φ ( D ) n log n , T_n < \frac{Dn}{\varphi(Ds)\log(n/(2s))} \le\frac{Dn}{4\varphi(D)(\frac{3}{4}\log n-\log2)} \sim\frac{D}{3\varphi(D)}\frac{n}{\log n}, T n < φ ( D s ) log ( n / ( 2 s )) D n ≤ 4 φ ( D ) ( 4 3 log n − log 2 ) D n ∼ 3 φ ( D ) D log n n , contradicting (92) . If s > n 1 / 4 s > n^{1/4} s > n 1/4
T n ≤ n 2 s + 1 ≤ 1 2 n 3 / 4 + 1 = o ( n / log n ) , T_n \le\frac{n}{2s}+1 \le\frac{1}{2}n^{3/4}+1=o(n/\log n), T n ≤ 2 s n + 1 ≤ 2 1 n 3/4 + 1 = o ( n / log n ) , another contradiction. In both cases a nonreduced forced class is already impossible, since its primes would divide D s ≤ D n / 2 Ds \le Dn/2 D s ≤ D n /2
It remains to exclude the finitely many values 2 ≤ s < s 0 2 \le s < s_0 2 ≤ s < s 0 D s Ds D s D D D D D D φ ( D s ) / φ ( D ) ≥ 2 \varphi(Ds)/\varphi(D) \ge2 φ ( D s ) / φ ( D ) ≥ 2 D D D 2 2 2 2 2 2 D s Ds D s r r r n n n (91) , even though that forced class may vary with n n n G O G_O G O
The symmetric group and the remaining roots. Both classes r r r r + D r+D r + D 2 D 2D 2 D
7 D n / 5 < p < 3 D n / 2 − B . (93) 7Dn/5 < p < 3Dn/2-B. \tag*{(93)} 7 D n /5 < p < 3 D n /2 − B . (93) The prime number theorem in the fixed class modulo 2 D 2D 2 D n n n d p ≤ 3 n / 5 + O ( 1 ) d_p \le3n/5+O(1) d p ≤ 3 n /5 + O ( 1 ) M = n − o ( n ) M=n-o(n) M = n − o ( n ) O O O A M ≤ G O A_M \le G_O A M ≤ G O ( p − r ) / D (p-r)/D ( p − r ) / D d p d_p d p G O = S M = Gal ( h n / Q ) G_O=S_M=\operatorname{Gal}(h_n/\mathbb{Q}) G O = S M = Gal ( h n / Q )
Write f n = κ n h n g n f_n=\kappa_nh_ng_n f n = κ n h n g n Gal ( f n / Q ) \operatorname{Gal}(f_n/\mathbb{Q}) Gal ( f n / Q ) S M × Gal ( g n / Q ) S_M\times\operatorname{Gal}(g_n/\mathbb{Q}) S M × Gal ( g n / Q ) g n g_n g n S M S_M S M S M S_M S M O O O S M S_M S M (4) and the linear disjointness.
Quadratic substitution. Fix c ∈ Q × c\in\mathbb{Q}^{\times} c ∈ Q × n n n (89) satisfy v p ( u c / ( 2 D ) ) = 0 v_p(uc/(2D))=0 v p ( u c / ( 2 D )) = 0 2 d p ≤ 2 n < p 2d_p\le2n<p 2 d p ≤ 2 n < p 2 d p 2d_p 2 d p P n ( α , β ) ( 1 + c X 2 ) P_n^{(\alpha,\beta)}(1+cX^2) P n ( α , β ) ( 1 + c X 2 ) d p d_p d p O O O
The roots of h n ( 1 + c X 2 ) h_n(1+cX^2) h n ( 1 + c X 2 ) O O O h n ( 1 ) ≠ 0 h_n(1)\ne0 h n ( 1 ) = 0 S M S_M S M (93) and in the class r r r 2 D 2D 2 D d p d_p d p 1 < d p < M 1<d_p<M 1 < d p < M 2 d p 2d_p 2 d p W M W_M W M h n ( 1 + c X 2 ) h_n(1+cX^2) h n ( 1 + c X 2 )
Finally, the full splitting group after quadratic substitution embeds in
W M × Gal ( g n ( 1 + c X 2 ) / Q ) W_M\times\operatorname{Gal}(g_n(1+cX^2)/\mathbb{Q}) W M × Gal ( g n ( 1 + c X 2 ) / Q ) and projects onto each factor. As above, surjectivity makes the kernel on the second factor normal in W M W_M W M 2 d p 2d_p 2 d p g n g_n g n f n f_n f n M = n M=n M = n g n = 1 g_n=1 g n = 1 Corollary 5.14 below.
Fixed number fields and symmetric parameters
The selected ramified primes tend to infinity with the degree, whereas a fixed number field is ramified at only finitely many primes. To use this observation, pass to the Galois closure of the fixed field. At every sufficiently large selected prime, inertia fixes its intersection with the splitting field. This intersection is Galois over Q \mathbb{Q} Q Q \mathbb{Q} Q
Corollary 5.14. Under the hypotheses of Theorem 1.6 , fix a number field E E E c ∈ Q × c \in\mathbb{Q}^{\times} c ∈ Q × n n n h n h_n h n h n ( 1 + c X 2 ) h_n(1+cX^2) h n ( 1 + c X 2 ) E E E Q \mathbb{Q} Q E E E S M n S_{M_n} S M n W M n W_{M_n} W M n
Proof. Let E ′ E' E ′ E E E Q \mathbb{Q} Q n n n E ′ E' E ′ K n K_n K n h n h_n h n h n ( 1 + c X 2 ) h_n(1+cX^2) h n ( 1 + c X 2 ) F n = K n ∩ E ′ F_n=K_n\cap E' F n = K n ∩ E ′ K n K_n K n F n F_n F n Q \mathbb{Q} Q F n / Q F_n/\mathbb{Q} F n / Q p p p p p p Gal ( K n / Q ) \operatorname{Gal}(K_n/\mathbb{Q}) Gal ( K n / Q ) F n F_n F n Gal ( K n / F n ) \operatorname{Gal}(K_n/F_n) Gal ( K n / F n ) h n h_n h n h n ( 1 + c X 2 ) h_n(1+cX^2) h n ( 1 + c X 2 ) 2 d p 2d_p 2 d p 1 ≤ d p < M n 1\le d_p<M_n 1 ≤ d p < M n Gal ( K n / F n ) = Gal ( K n / Q ) \operatorname{Gal}(K_n/F_n)=\operatorname{Gal}(K_n/\mathbb{Q}) Gal ( K n / F n ) = Gal ( K n / Q ) F n = Q F_n=\mathbb{Q} F n = Q K n K_n K n E ′ E' E ′ Q \mathbb{Q} Q E E E E ′ E' E ′ E E E
The denominator hypotheses have separate roles in the proof. The inequality D ≥ max { u , v } D\ge\max\{u,v\} D ≥ max { u , v } D n D_n D n D D D D D D r r r r + D r+D r + D 2 D 2D 2 D
Corollary 5.15. Fix a ∈ Q a\in\mathbb{Q} a ∈ Q a > − 1 a>-1 a > − 1 v v v 4 4 4 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } Q m a , δ Q_m^{a,\delta} Q m a , δ
P 2 m + δ ( a , a ) ( X ) X δ = Q m a , δ ( X 2 ) . \frac{P_{2m+\delta}^{(a,a)}(X)}{X^\delta}=Q_m^{a,\delta}(X^2). X δ P 2 m + δ ( a , a ) ( X ) = Q m a , δ ( X 2 ) . For all sufficiently large m m m Q m a , δ Q_m^{a,\delta} Q m a , δ
Gal ( Q m a , δ / Q ) = S m , Gal ( P 2 m + δ ( a , a ) ( X ) / X δ / Q ) = C 2 ≀ S m . \operatorname{Gal}(Q_m^{a,\delta}/\mathbb{Q})=S_m,\qquad
\operatorname{Gal}\left(P_{2m+\delta}^{(a,a)}(X)/X^\delta/\mathbb{Q}\right)=C_2\wr S_m. Gal ( Q m a , δ / Q ) = S m , Gal ( P 2 m + δ ( a , a ) ( X ) / X δ / Q ) = C 2 ≀ S m . Proof. The quadratic transformations Eqs. 18.7.13–18.7.14, followed by Jacobi symmetry, identify P 2 m + δ ( a , a ) ( X ) / X δ P_{2m+\delta}^{(a,a)}(X)/X^\delta P 2 m + δ ( a , a ) ( X ) / X δ
P m ( δ − 1 / 2 , a ) ( 1 − 2 X 2 ) . P_m^{(\delta-1/2,a)}(1-2X^2). P m ( δ − 1/2 , a ) ( 1 − 2 X 2 ) . The reduced denominator of δ − 1 / 2 + a \delta-1/2+a δ − 1/2 + a 2 v 2v 2 v v v v v v v 4 ∣ v 4\mid v 4 ∣ v max { 2 , v } \max\{2,v\} max { 2 , v } Theorem 1.6 , with c = − 2 c=-2 c = − 2 T ↦ 1 − 2 T T\mapsto1-2T T ↦ 1 − 2 T
These are symmetric Jacobi families, related to Gegenbauer polynomials with parameter a + 1 / 2 a+1/2 a + 1/2 a = 0 a=0 a = 0 Theorem 5.9 supplies every degree. For nonnegative integral parameters, Theorem 1.4 supplies the required eventual irreducibility, as used below. Unconditionally, Q m a , δ Q_m^{a,\delta} Q m a , δ m / 2 m/2 m /2 m − O a , A ( m / ( log m ) A ) m - O_{a,A}(m/(\log m)^A) m − O a , A ( m / ( log m ) A ) A > 0 A > 0 A > 0 T = X 2 T = X^2 T = X 2
Unequal integral parameters
Return to the notation of § 3.1 , with a > b ≥ 0 a > b \ge0 a > b ≥ 0 s = a + b s = a + b s = a + b n = 2 m + δ n = 2m + \delta n = 2 m + δ p ∈ ( 3 n / 4 , 7 n / 8 ] p \in(3n/4, 7n/8] p ∈ ( 3 n /4 , 7 n /8 ] p p p x p x^p x p p p p p p p n < 2 p n < 2p n < 2 p p p p p p p 1 − δ 1 - \delta 1 − δ
C δ ( z ) = 2 K 1 − δ A 1 − δ ( z + δ − 1 2 ) , W δ ( z ) = A δ ( z ) C δ ( z ) . (94) C_\delta(z) = 2^{K_{1-\delta}} A_{1-\delta}\left(z + \delta- \frac{1}{2}\right), \qquad W_\delta(z) = A_\delta(z) C_\delta(z). \tag*{(94)} C δ ( z ) = 2 K 1 − δ A 1 − δ ( z + δ − 2 1 ) , W δ ( z ) = A δ ( z ) C δ ( z ) . (94) Since deg A ε ≤ K ε \deg A_\varepsilon\le K_\varepsilon deg A ε ≤ K ε K 0 + K 1 = s K_0 + K_1 = s K 0 + K 1 = s deg W δ ≤ s \deg W_\delta\le s deg W δ ≤ s
Lemma 5.16. Suppose m ≥ 2 s + 10 m \ge2s + 10 m ≥ 2 s + 10 P n ( a , b ) P_n^{(a,b)} P n ( a , b ) Q \mathbb{Q} Q p p p
3 n 4 < p ≤ 7 n 8 , p ∤ W δ ( m ) , (95) \frac{3n}{4} < p \le\frac{7n}{8}, \qquad p \nmid W_\delta(m), \tag*{(95)} 4 3 n < p ≤ 8 7 n , p ∤ W δ ( m ) , (95) then Gal ( P n ( a , b ) / Q ) = S n \operatorname{Gal}(P_n^{(a,b)}/\mathbb{Q}) = S_n Gal ( P n ( a , b ) / Q ) = S n
Proof. Put f = P n ( a , b ) f = P_n^{(a,b)} f = P n ( a , b ) j = n − p j = n - p j = n − p
p > M , p < n + s < 2 p < 2 n + s < 3 p , 2 n + s < p 2 . p > M, \qquad p < n + s < 2p < 2n + s < 3p, \qquad2n + s < p^2. p > M , p < n + s < 2 p < 2 n + s < 3 p , 2 n + s < p 2 . In particular, p p p p ∤ T p \nmid T p ∤ T 2 − n ( 2 n + s n ) 2^{-n}\binom{2n+s}{n} 2 − n ( n 2 n + s ) p p p 2 − 1 − 1 = 0 2 - 1 - 1 = 0 2 − 1 − 1 = 0 Lemma 3.3 gives
f ‾ = x p P n ( a , b ) ‾ j . (96) \overline{f} = x^p \overline{P_n^{(a,b)}}_j. \tag*{(96)} f = x p P n ( a , b ) j . (96) The residual factor has degree j < p j < p j < p
To check its value at zero, write j = 2 k + ( 1 − δ ) j = 2k + (1 - \delta) j = 2 k + ( 1 − δ )
k = m + δ − p + 1 2 ≡ m + δ − 1 2 ( m o d p ) . k = m + \delta- \frac{p+1}{2} \equiv m + \delta- \frac{1}{2} \pmod p. k = m + δ − 2 p + 1 ≡ m + δ − 2 1 ( mod p ) . Here 2 k < p 2k < p 2 k < p k + K 1 − δ < p k + K_{1-\delta} < p k + K 1 − δ < p P j ( a , b ) ( 0 ) P_j^{(a,b)}(0) P j ( a , b ) ( 0 ) p ∤ C δ ( m ) p \nmid C_\delta(m) p ∤ C δ ( m ) p ∤ A 1 − δ ( k ) p \nmid A_{1-\delta}(k) p ∤ A 1 − δ ( k ) P j ( a , b ) ( 0 ) P_j^{(a,b)}(0) P j ( a , b ) ( 0 ) m < p ≤ 2 m m < p \le2m m < p ≤ 2 m m + K δ < p m + K_\delta< p m + K δ < p p ∤ A δ ( m ) p \nmid A_\delta(m) p ∤ A δ ( m )
v p ( f ( 0 ) ) = v p ( ( 2 m m ) ) = 1. (97) v_p(f(0)) = v_p\left(\binom{2m}{m}\right) = 1. \tag*{(97)} v p ( f ( 0 )) = v p ( ( m 2 m ) ) = 1. (97) After dividing f f f P j ( a , b ) P_j^{(a,b)} P j ( a , b ) (96) is a product of coprime monic polynomials. Hensel’s lemma lifts them to monic polynomials of degrees p p p j j j j j j p p p x p x^p x p Q p \mathbb{Q}_p Q p p p p f f f p p p p p p p p p n < 2 p n < 2p n < 2 p p p p
Irreducibility gives transitivity. In a nontrivial system of blocks, both the block size and the number of blocks are at most n / 2 < p n/2 < p n /2 < p p p p p ≤ n − 3 p \le n - 3 p ≤ n − 3 A n A_n A n [17 Theorem 1.1] .
Finally, the Jacobi discriminant formula is
disc P n ( a , b ) = 2 − n ( n − 1 ) ∏ h = 1 n h h − 2 n + 2 ( h + a ) h − 1 ( h + b ) h − 1 ( n + h + s ) n − h ; (98) \operatorname{disc} P_n^{(a,b)} = 2^{-n(n-1)} \prod_{h=1}^{n} h^{h-2n+2}(h+a)^{h-1}(h+b)^{h-1}(n+h+s)^{n-h}; \tag*{(98)} disc P n ( a , b ) = 2 − n ( n − 1 ) h = 1 ∏ n h h − 2 n + 2 ( h + a ) h − 1 ( h + b ) h − 1 ( n + h + s ) n − h ; (98) see [26 Theorem 6.71] or [22 (18.16.19)] . The multiples of p p p h = p h=p h = p p − a p-a p − a p − b p-b p − b 2 p − n − s 2p-n-s 2 p − n − s [ 1 , n ] [1,n] [ 1 , n ] p 2 p^2 p 2
v p ( disc f ) = ( p − 2 n + 2 ) + ( p − a − 1 ) + ( p − b − 1 ) + ( 2 n + s − 2 p ) = p . v_p(\operatorname{disc} f) = (p - 2n + 2) + (p - a - 1) + (p - b - 1) + (2n + s - 2p) = p. v p ( disc f ) = ( p − 2 n + 2 ) + ( p − a − 1 ) + ( p − b − 1 ) + ( 2 n + s − 2 p ) = p . The discriminant is not a rational square, so the Galois group is S n S_n S n
The required prime can be found effectively because a fixed polynomial value cannot contain all primes in a proportional interval.
Proposition 5.17. Suppose the hypotheses of Theorem 3.1 hold. If also
n ≥ 5 000 000 , W δ ( m ) ≠ 0 , log ∣ W δ ( m ) ∣ < n / 10 , (99) n \ge5\,000\,000,\qquad W_\delta(m) \ne0,\qquad\log|W_\delta(m)| < n/10, \tag*{(99)} n ≥ 5 000 000 , W δ ( m ) = 0 , log ∣ W δ ( m ) ∣ < n /10 , (99) then Gal ( P n ( a , b ) / Q ) = S n \operatorname{Gal}(P_n^{(a,b)}/\mathbb{Q}) = S_n Gal ( P n ( a , b ) / Q ) = S n
Proof. Let ϑ ( x ) = ∑ p ≤ x log p \vartheta(x) = \sum_{p \le x} \log p ϑ ( x ) = ∑ p ≤ x log p [12 Theorem 5.2] gives
∣ ϑ ( x ) − x ∣ < 0.2 x ( log x ) 2 ( x > 3 594 641 ) . |\vartheta(x) - x| < \frac{0.2x}{(\log x)^2}\qquad(x > 3\,594\,641). ∣ ϑ ( x ) − x ∣ < ( log x ) 2 0.2 x ( x > 3 594 641 ) . In particular, 0.99 x < ϑ ( x ) < 1.01 x 0.99x < \vartheta(x) < 1.01x 0.99 x < ϑ ( x ) < 1.01 x
ϑ ( 7 n / 8 ) − ϑ ( 3 n / 4 ) > ( 0.99 7 8 − 1.01 3 4 ) n = 0.10875 n > n / 10. \vartheta(7n/8) - \vartheta(3n/4) > \left(0.99\frac{7}{8} - 1.01\frac{3}{4}\right)n = 0.10875n > n/10. ϑ ( 7 n /8 ) − ϑ ( 3 n /4 ) > ( 0.99 8 7 − 1.01 4 3 ) n = 0.10875 n > n /10. If every prime in this interval divided W δ ( m ) W_\delta(m) W δ ( m ) log ∣ W δ ( m ) ∣ \log|W_\delta(m)| log ∣ W δ ( m ) ∣ (95) therefore exists, and Lemma 5.16 applies. □
The rational Galois assertion of Theorem 1.3 . For each parity, a root bound excludes the zeros of the fixed polynomial W δ W_\delta W δ ∣ W δ ( m ) ∣ ≤ C ( m + 1 ) s |W_\delta(m)| \le C(m+1)^s ∣ W δ ( m ) ∣ ≤ C ( m + 1 ) s C C C (99) holds for all sufficiently large m m m Theorem 3.1 , and use reflection when b > a b>a b > a
The fixed-field assertion of Theorem 1.3 . Let E ′ E' E ′ E E E Q \mathbb{Q} Q Lemma 5.16 , for all sufficiently large n n n f = P n ( a , b ) f=P_n^{(a,b)} f = P n ( a , b ) S n S_n S n p p p
3 n 4 < p ≤ 7 n 8 , v p ( disc f ) = p . \frac{3n}{4} < p \le\frac{7n}{8},\qquad v_p(\operatorname{disc} f)=p. 4 3 n < p ≤ 8 7 n , v p ( disc f ) = p . Enlarge the effective threshold so that n ≥ 5 n \ge5 n ≥ 5 p p p E ′ E' E ′
Let K K K f f f L = K ∩ E ′ L=K\cap E' L = K ∩ E ′ L / Q L/\mathbb{Q} L / Q p p p A n A_n A n Q ( disc f ) \mathbb{Q}(\sqrt{\operatorname{disc} f}) Q ( disc f ) v p ( disc f ) v_p(\operatorname{disc} f) v p ( disc f ) p p p Gal ( K / Q ) = S n \operatorname{Gal}(K/\mathbb{Q})=S_n Gal ( K / Q ) = S n L L L Gal ( K / L ) \operatorname{Gal}(K/L) Gal ( K / L ) S n S_n S n A n A_n A n n ≥ 5 n\ge5 n ≥ 5 Gal ( K / L ) = S n \operatorname{Gal}(K/L)=S_n Gal ( K / L ) = S n L = Q L=\mathbb{Q} L = Q K K K E ′ E' E ′ K K K E E E E E E S n S_n S n
For effectivity, choose a computable R ≥ 1 R \ge1 R ≥ 1 E ′ E' E ′ 3 n / 4 > R 3n/4 > R 3 n /4 > R ∣ Disc ( E ) ∣ |\operatorname{Disc}(E)| ∣ Disc ( E ) ∣ E = Q E = \mathbb{Q} E = Q
Fixed derivatives
The quadratic transformation now brings the symmetric integral families within the denominator hypotheses of the rational-parameter theorem, and their established irreducibility makes its large factor the whole quotient polynomial.
Proof of Corollary 1.7. Put J m = P m ( δ − 1 / 2 , r ) J_m = P_m^{(\delta-1/2,r)} J m = P m ( δ − 1/2 , r ) [22 Eqs. 18.7.15–18.7.16] , followed by Jacobi reflection, give
F m , r , δ ( X ) = ( − 1 ) m 2 δ ( r + 1 / 2 ) m + δ ( δ + 1 / 2 ) m J m ( 1 − 2 X 2 ) . (100) F_{m,r,\delta}(X) = (-1)^m 2^\delta\frac{(r+1/2)_{m+\delta}}{(\delta+1/2)_m} J_m(1-2X^2). \tag*{(100)} F m , r , δ ( X ) = ( − 1 ) m 2 δ ( δ + 1/2 ) m ( r + 1/2 ) m + δ J m ( 1 − 2 X 2 ) . (100) The scalar is a nonzero rational number. By Theorem 1.4 , F m , r , δ F_{m,r,\delta} F m , r , δ Q \mathbb{Q} Q m m m J m J_m J m J m J_m J m J m ( 1 − 2 X 2 ) J_m(1-2X^2) J m ( 1 − 2 X 2 )
The parameters δ − 1 / 2 \delta- 1/2 δ − 1/2 r r r D = 2 D = 2 D = 2 Theorem 1.6 and Corollary 5.14 apply, with c = − 2 c = -2 c = − 2 J m J_m J m J m ( 1 − 2 X 2 ) J_m(1-2X^2) J m ( 1 − 2 X 2 ) E E E m m m E E E S m S_m S m C 2 ≀ S m C_2 \wr S_m C 2 ≀ S m (100) identifies the second polynomial with F m , r , δ F_{m,r,\delta} F m , r , δ Q m , r , δ Q_{m,r,\delta} Q m , r , δ J m ( 1 − 2 T ) J_m(1-2T) J m ( 1 − 2 T ) T ↦ 1 − 2 T T \mapsto1-2T T ↦ 1 − 2 T E E E δ = 0 , 1 \delta= 0,1 δ = 0 , 1
Finally, P 2 m + δ + r ( r ) = ( 2 r − 1 ) ! ! ( 2 m + δ r + 1 / 2 ) P_{2m+\delta+r}^{(r)} = (2r-1)!!\binom{2m+\delta}{r+1/2} P 2 m + δ + r ( r ) = ( 2 r − 1 )!! ( r + 1/2 2 m + δ ) ( − 1 ) ! ! = 1 (-1)!! = 1 ( − 1 )!! = 1 X δ X^\delta X δ F m , r , δ F_{m,r,\delta} F m , r , δ
Applications and further questions
The results have consequences for both individual algebraic zeros and geometric objects defined by the polynomials. Irreducibility controls common and prescribed zeros; the full symmetric group provides the additional information needed for the Jacobians below.
Integral parameters and common roots
Negative integral parameters contribute endpoint factors. Removing them extends Theorem 1.3 to all fixed integral parameters whose absolute values differ.
Corollary 6.1. Fix integers a , b a,b a , b ∣ a ∣ ≠ ∣ b ∣ |a| \ne|b| ∣ a ∣ = ∣ b ∣ a − = max { − a , 0 } a_- = \max\{-a,0\} a − = max { − a , 0 } b − = max { − b , 0 } b_- = \max\{-b,0\} b − = max { − b , 0 } n n n a , b a,b a , b
P n ( a , b ) ( x ) ( x − 1 ) a − ( x + 1 ) b − \frac{P_n^{(a,b)}(x)}{(x-1)^{a_-}(x+1)^{b_-}} ( x − 1 ) a − ( x + 1 ) b − P n ( a , b ) ( x ) is irreducible over Q \mathbb{Q} Q S n − a − − b − S_{n-a_--b_-} S n − a − − b −
Proof. For A , b ≥ 0 A,b \ge0 A , b ≥ 0 n ≥ A n \ge A n ≥ A
P n ( − A , b ) ( x ) = ( n − A ) ! ( n + b ) ! n ! ( n − A + b ) ! ( x − 1 2 ) A P n − A ( A , b ) ( x ) . P_n^{(-A,b)}(x) = \frac{(n-A)!(n+b)!}{n!(n-A+b)!}\left(\frac{x-1}{2}\right)^A P_{n-A}^{(A,b)}(x). P n ( − A , b ) ( x ) = n ! ( n − A + b )! ( n − A )! ( n + b )! ( 2 x − 1 ) A P n − A ( A , b ) ( x ) . Reflection treats a negative second parameter. For A , B ≥ 0 A,B \ge0 A , B ≥ 0 n ≥ A + B n \ge A+B n ≥ A + B
P n ( − A , − B ) ( x ) = ( x − 1 2 ) A ( x + 1 2 ) B P n − A − B ( A , B ) ( x ) . P_n^{(-A,-B)}(x) = \left(\frac{x-1}{2}\right)^A \left(\frac{x+1}{2}\right)^B P_{n-A-B}^{(A,B)}(x). P n ( − A , − B ) ( x ) = ( 2 x − 1 ) A ( 2 x + 1 ) B P n − A − B ( A , B ) ( x ) . The remaining factor is a nonzero rational multiple of the Jacobi polynomial with parameters ∣ a ∣ , ∣ b ∣ |a|, |b| ∣ a ∣ , ∣ b ∣ Theorem 1.3 applies. □
Irreducibility also controls how the root sets vary with the degree. If two rational polynomials share an algebraic root, its minimal polynomial divides both. Thus an irreducible member cannot meet the root set of a lower-degree member. For the Legendre polynomials and their derivatives, the only additional comparison is between the four reduced polynomials of the same degree.
Corollary 6.2. For fixed distinct nonnegative integers a , b a, b a , b P n ( a , b ) P_n^{(a,b)} P n ( a , b ) { P ℓ , P ℓ ′ : ℓ ≥ 1 } \{P_\ell, P_\ell' : \ell\ge1\} { P ℓ , P ℓ ′ : ℓ ≥ 1 }
Proof. In the asymmetric case, once the larger degree is at least N ( a , b ) N(a,b) N ( a , b ) Theorem 1.5 settles different positive reduced degrees. At reduced degree 2 m 2m 2 m
P 2 m , P 2 m + 1 / x , P 2 m + 1 ′ , P 2 m + 2 ′ / x . P_{2m},\quad P_{2m+1}/x,\quad P'_{2m+1},\quad P'_{2m+2}/x. P 2 m , P 2 m + 1 / x , P 2 m + 1 ′ , P 2 m + 2 ′ / x . Their absolute ratios of the coefficient of x 2 m − 2 x^{2m-2} x 2 m − 2
m ( 2 m − 1 ) 4 m + 1 < m ( 2 m − 1 ) 4 m − 1 < m ( 2 m + 1 ) 4 m + 3 < m ( 2 m + 1 ) 4 m + 1 . \frac{m(2m-1)}{4m+1}
<
\frac{m(2m-1)}{4m-1}
<
\frac{m(2m+1)}{4m+3}
<
\frac{m(2m+1)}{4m+1}. 4 m + 1 m ( 2 m − 1 ) < 4 m − 1 m ( 2 m − 1 ) < 4 m + 3 m ( 2 m + 1 ) < 4 m + 1 m ( 2 m + 1 ) . They are therefore pairwise nonproportional. Irreducibility again excludes a common root. Reduced constants have no roots. □
The same argument gives a degree bound for any prescribed algebraic number ξ \xi ξ
P n ( a , b ) ( ξ ) ≠ 0 for n > max { N ( a , b ) , [ Q ( ξ ) : Q ] } . P_n^{(a,b)}(\xi) \ne0 \quad\text{for } n > \max\{N(a,b), [\mathbb{Q}(\xi):\mathbb{Q}]\}. P n ( a , b ) ( ξ ) = 0 for n > max { N ( a , b ) , [ Q ( ξ ) : Q ]} . Excluding a prescribed factor
The proof of irreducibility must allow every possible factor. When a candidate factor F F F F F F
Proposition 6.3. Let F ∈ Z [ x ] F \in\mathbb{Z}[x] F ∈ Z [ x ] d ≥ 1 d \ge1 d ≥ 1 u u u ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) F ( 0 ) ≠ 0 F(0) \ne0 F ( 0 ) = 0
c F = min j v 2 ( [ X j ] F ( 2 X + 1 ) ) , β ( F ) = log 4 + 2 d log u 2 c F . c_F = \min_j v_2([X^j]F(2X+1)),\qquad\beta(F) = \log4 + \frac{2}{d}\log\frac{u}{2^{c_F}}. c F = j min v 2 ([ X j ] F ( 2 X + 1 )) , β ( F ) = log 4 + d 2 log 2 c F u . Let c ≥ 2 c \ge2 c ≥ 2
c ( β ( F ) − 1 ) > g ( c ) . (101) c(\beta(F)-1)>g(c). \tag*{(101)} c ( β ( F ) − 1 ) > g ( c ) . (101) For all sufficiently large m m m 0 ≤ r ≤ m / c 0 \le r \le m/c 0 ≤ r ≤ m / c δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 }
F ∤ C 2 m + δ r + 1 / 2 ( x ) / x δ . F \nmid C_{2m+\delta}^{r+1/2}(x)/x^\delta. F ∤ C 2 m + δ r + 1/2 ( x ) / x δ . For rational c c c F , c F,c F , c
Proof. Suppose F ∣ H F \mid H F ∣ H H H H Theorem 3.4 . For large m m m H = F G H=FG H = F G e = 2 m − d ≥ d e=2m-d\ge d e = 2 m − d ≥ d z = u / 2 c F > 1 z=u/2^{c_F}>1 z = u / 2 c F > 1
V F = ∏ i < j ( α i − α j ) 2 , C F = d log 2 + 2 log z + 1 d log V F . V_F=\prod_{i<j}(\alpha_i-\alpha_j)^2,\qquad C_F=d\log2+2\log z+\frac{1}{d}\log V_F. V F = i < j ∏ ( α i − α j ) 2 , C F = d log 2 + 2 log z + d 1 log V F . Use (19) and (42) , without the Chebyshev bound. They give
log R d > m β ( F ) + 1 2 log m − log K − C F − D 2 log 2. (102) \frac{\log R}{d}>m\beta(F)+\frac{1}{2}\log m-\log K-C_F-D_2\log2. \tag*{(102)} d log R > m β ( F ) + 2 1 log m − log K − C F − D 2 log 2. (102) Comparison with (41) forces
m β ( F ) < Ψ m ( S ) + log K + log A ≤ m + C F − 1 2 log m . (103) m\beta(F) < \Psi_m(S) + \log K + \log A_{\leq m} + C_F - \frac{1}{2}\log m. \tag*{(103)} m β ( F ) < Ψ m ( S ) + log K + log A ≤ m + C F − 2 1 log m . (103) For r ≥ 1 r \geq1 r ≥ 1 m ( 1 + g ( c ) / c ) + o F ( m ) m(1+g(c)/c)+o_F(m) m ( 1 + g ( c ) / c ) + o F ( m ) r = 0 r=0 r = 0 K = A = 1 K=A=1 K = A = 1
Render [23 Theorem 6] proved that P n ( r , r ) ( b / c ) ≠ 0 P_n^{(r,r)}(\sqrt{b/c}) \ne0 P n ( r , r ) ( b / c ) = 0 b , c b,c b , c b ∉ { 1 , 3 } b \notin\{1,3\} b ∈ / { 1 , 3 } n , r ≥ 0 n,r \geq0 n , r ≥ 0 ± 1 \pm1 ± 1 m ≥ 2 r m \geq2r m ≥ 2 r m m m u / 2 c F u/2^{c_F} u / 2 c F
Corollary 6.4. There is an effective absolute integer M 1 M_1 M 1 m ≥ M 1 m \geq M_1 m ≥ M 1 0 ≤ r ≤ m / 2 0 \leq r \leq m/2 0 ≤ r ≤ m /2 C 2 m r + 1 / 2 C_{2m}^{r+1/2} C 2 m r + 1/2 C 2 m + 1 r + 1 / 2 / x C_{2m+1}^{r+1/2}/x C 2 m + 1 r + 1/2 / x
Proof. A nonzero rational root in ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) F = u x + v F=ux+v F = ux + v 0 < ∣ v ∣ < u 0<|v|<u 0 < ∣ v ∣ < u gcd ( u , v ) = 1 \gcd(u,v)=1 g cd( u , v ) = 1 u , v u,v u , v c F = 1 c_F=1 c F = 1 u ≥ 3 u\geq3 u ≥ 3 c F = 0 c_F=0 c F = 0 u ≥ 2 u\geq2 u ≥ 2 z = u / 2 c F ≥ 3 / 2 z=u/2^{c_F}\geq3/2 z = u / 2 c F ≥ 3/2 d = 1 d=1 d = 1 V F = 1 V_F=1 V F = 1
m log 9 + 1 2 log m − log K − log ( 9 / 2 ) − D 2 log 2. m\log9+\frac{1}{2}\log m-\log K-\log(9/2)-D_2\log2. m log 9 + 2 1 log m − log K − log ( 9/2 ) − D 2 log 2. Indeed, its z z z ( 2 m − 2 ) log z (2m-2)\log z ( 2 m − 2 ) log z
2 ( log 9 − 1 ) − g ( 2 ) = log 3 72 2 16 5 10 − 2 > 0. 2(\log9-1)-g(2)=\log\frac{3^{72}2^{16}}{5^{10}}-2>0. 2 ( log 9 − 1 ) − g ( 2 ) = log 5 10 3 72 2 16 − 2 > 0. The last inequality follows from e < 3 e<3 e < 3 3 5 2 16 > 5 10 3^5 2^{16}>5^{10} 3 5 2 16 > 5 10
Vanishing modes on a circular cone
In spherical coordinates ( ρ , θ , φ ) (\rho,\theta,\varphi) ( ρ , θ , φ ) ℓ \ell ℓ 0 ≤ r ≤ ℓ 0\leq r\leq\ell 0 ≤ r ≤ ℓ
ρ ℓ e i r φ ( sin θ ) r P ℓ ( r ) ( cos θ ) ; \rho^\ell e^{ir\varphi}(\sin\theta)^r P_\ell^{(r)}(\cos\theta); ρ ℓ e i r φ ( sin θ ) r P ℓ ( r ) ( cos θ ) ; see [22 Eqs. 14.7.8 and 14.30.1, and §14.30(iv)] . For α = cos θ ∈ ( − 1 , 1 ) \alpha=\cos\theta\in(-1,1) α = cos θ ∈ ( − 1 , 1 ) P ℓ ( r ) ( α ) = 0 P_\ell^{(r)}(\alpha)=0 P ℓ ( r ) ( α ) = 0 x 2 + y 2 − 2 z 2 = 0 x^2+y^2-2z^2=0 x 2 + y 2 − 2 z 2 = 0 α = ± 1 / 3 \alpha=\pm1/\sqrt{3} α = ± 1/ 3 3 x 2 − 1 3x^2-1 3 x 2 − 1
Corollary 6.5. For every integer D ≥ 1 D\geq1 D ≥ 1 L D L_D L D α ∈ ( − 1 , 1 ) \alpha\in(-1,1) α ∈ ( − 1 , 1 ) D D D
ℓ ≥ L D , 0 ≤ r ≤ ℓ , P ℓ ( r ) ( α ) = 0 ⟹ 17 r > ℓ − 1. \ell\geq L_D,\quad0\leq r\leq\ell,\quad P_\ell^{(r)}(\alpha)=0
\quad\Longrightarrow\quad17r>\ell-1. ℓ ≥ L D , 0 ≤ r ≤ ℓ , P ℓ ( r ) ( α ) = 0 ⟹ 17 r > ℓ − 1. For α = 1 / 3 \alpha=1/\sqrt{3} α = 1/ 3
23 r > 4 ( ℓ − 1 ) (104) 23r>4(\ell-1) \tag*{(104)} 23 r > 4 ( ℓ − 1 ) (104) holds for all sufficiently large ℓ \ell ℓ α ∈ ( − 1 , 1 ) \alpha\in(-1,1) α ∈ ( − 1 , 1 ) 5 r > ℓ − 1 5r>\ell-1 5 r > ℓ − 1 ℓ \ell ℓ
Proof. Write ℓ − r = 2 m + δ \ell-r=2m+\delta ℓ − r = 2 m + δ 17 r ≤ ℓ − 1 17r\leq\ell-1 17 r ≤ ℓ − 1 m ≥ 8 r m\geq8r m ≥ 8 r m ≥ 8 ( ℓ − 1 ) / 17 m\geq8(\ell-1)/17 m ≥ 8 ( ℓ − 1 ) /17 L D L_D L D ℓ ≥ L D \ell\geq L_D ℓ ≥ L D m ≥ M 0 m\geq M_0 m ≥ M 0 2 m > D 2m>D 2 m > D P ℓ ( r ) x δ P_\ell^{(r)}x^\delta P ℓ ( r ) x δ 2 m 2m 2 m α \alpha α [ Q ( α ) : Q ] ≤ D [\mathbb{Q}(\alpha):\mathbb{Q}]\leq D [ Q ( α ) : Q ] ≤ D
For α = 1 / 3 \alpha= 1/\sqrt{3} α = 1/ 3 F = 3 x 2 − 1 F = 3x^2 - 1 F = 3 x 2 − 1 d = 2 d = 2 d = 2 u = 3 u = 3 u = 3 c F = 1 c_F = 1 c F = 1 F ( 2 X + 1 ) = 12 X 2 + 12 X + 2 F(2X + 1) = 12X^2 + 12X + 2 F ( 2 X + 1 ) = 12 X 2 + 12 X + 2 β ( F ) = log 6 \beta(F) = \log6 β ( F ) = log 6 c = 19 / 8 c = 19/8 c = 19/8
19 8 ( log 6 − 1 ) − g ( 19 / 8 ) > 0.0042 ; \frac{19}{8}(\log6 - 1) - g(19/8) > 0.0042; 8 19 ( log 6 − 1 ) − g ( 19/8 ) > 0.0042 ; an exact rational verification is given in Appendix C. If 23 r ≤ 4 ( ℓ − 1 ) 23r \le4(\ell- 1) 23 r ≤ 4 ( ℓ − 1 ) m ≥ 19 r / 8 m \ge19r/8 m ≥ 19 r /8 ℓ \ell ℓ 5 r ≤ ℓ − 1 5r \le\ell- 1 5 r ≤ ℓ − 1 m ≥ 2 r m \ge2r m ≥ 2 r Corollary 6.4 applies, uniformly in the rational root. □
For this cone, Mangoubi and Weller Weiser show that r r r r r r ℓ \ell ℓ [20 Theorem 2] . The bound (104) restricts the two variables simultaneously: in large degree, a vanishing mode must have order greater than 4 ( ℓ − 1 ) / 23 4(\ell- 1)/23 4 ( ℓ − 1 ) /23 [1 §6, Question 4] remains open. The distinct generic existence problem of Agranovsky–Krasnov [2 ] was solved by Greilhuber [15 ] .
For comparison, the associated Legendre equation gives the opposite constraint
r 2 ≤ 2 3 ℓ ( ℓ + 1 ) ( r ≥ 1 ) r^2 \le\frac{2}{3}\ell(\ell+ 1) \qquad(r \ge1) r 2 ≤ 3 2 ℓ ( ℓ + 1 ) ( r ≥ 1 ) when P ℓ ( r ) ( 1 / 3 ) = 0 P_{\ell}^{(r)}(1/\sqrt{3}) = 0 P ℓ ( r ) ( 1/ 3 ) = 0 y ( x ) = ( 1 − x 2 ) r / 2 P ℓ ( r ) ( x ) y(x) = (1 - x^2)^{r/2}P_{\ell}^{(r)}(x) y ( x ) = ( 1 − x 2 ) r /2 P ℓ ( r ) ( x ) [ 1 / 3 , 1 ] [1/\sqrt{3},1] [ 1/ 3 , 1 ]
ℓ ( ℓ + 1 ) ∫ 1 / 3 1 y 2 = ∫ 1 / 3 1 ( ( 1 − x 2 ) ( y ′ ) 2 + r 2 y 2 1 − x 2 ) > 3 r 2 2 ∫ 1 / 3 1 y 2 . \ell(\ell+ 1)\int_{1/\sqrt{3}}^1 y^2 = \int_{1/\sqrt{3}}^1 \left((1 - x^2)(y')^2 + \frac{r^2y^2}{1 - x^2}\right) > \frac{3r^2}{2}\int_{1/\sqrt{3}}^1 y^2. ℓ ( ℓ + 1 ) ∫ 1/ 3 1 y 2 = ∫ 1/ 3 1 ( ( 1 − x 2 ) ( y ′ ) 2 + 1 − x 2 r 2 y 2 ) > 2 3 r 2 ∫ 1/ 3 1 y 2 . Both endpoint terms vanish. Thus the necessary conditions still leave a nonempty linear range of derivative orders.
Modular interpretations
The fractional parameters have a modular origin. Kaneko and Zagier’s characteristic-zero polynomials occur in modular-form solutions. At prime indices, adjoining the prescribed factors at 0 and 1728 gives a polynomial whose reduction is the supersingular j j j [18 Theorem 1] . Theorem 1.1 determines the arithmetic of each positive-degree Jacobi factor: its roots form one Galois orbit over Q \mathbb{Q} Q
For n ≥ 0 n \ge0 n ≥ 0 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 }
r = 2 λ + 3 μ , e = 6 + r , ℓ = 12 n + e , (105) r = 2\lambda+ 3\mu,\qquad e = 6 + r,\qquad\ell= 12n + e, \tag*{(105)} r = 2 λ + 3 μ , e = 6 + r , ℓ = 12 n + e , (105) and define
F ~ ℓ − 1 ( Y ) = 1728 n P n ( λ / 3 , μ / 2 ) ( 1 − Y / 864 ) . (106) \widetilde{F}_{\ell-1}(Y) = 1728^n P_n^{(\lambda/3,\mu/2)}(1 - Y/864). \tag*{(106)} F ℓ − 1 ( Y ) = 172 8 n P n ( λ /3 , μ /2 ) ( 1 − Y /864 ) . (106) Here e ∈ { 1 , 5 , 7 , 11 } e \in\{1,5,7,11\} e ∈ { 1 , 5 , 7 , 11 }
A n , r = ( − 1 ) n ( 2 n + r / 6 n ) . (107) A_{n,r} = (-1)^n \binom{2n + r/6}{n}. \tag*{(107)} A n , r = ( − 1 ) n ( n 2 n + r /6 ) . (107) The affine substitution and nonzero rational scalar preserve irreducibility and splitting fields. Put a = ( λ + 1 ) / 2 a = (\lambda+ 1)/2 a = ( λ + 1 ) /2 b = ( μ + 1 ) / 2 b = (\mu+ 1)/2 b = ( μ + 1 ) /2 k = ℓ − 1 = 12 n + 4 a + 6 b k = \ell- 1 = 12n + 4a + 6b k = ℓ − 1 = 12 n + 4 a + 6 b
F k ( z ) = Δ ( z ) n E 4 ( z ) a E 6 ( z ) b F ~ k ( j ( z ) ) , F_k(z) = \Delta(z)^n E_4(z)^a E_6(z)^b \widetilde{F}_k(j(z)), F k ( z ) = Δ ( z ) n E 4 ( z ) a E 6 ( z ) b F k ( j ( z )) , where E 4 , E 6 E_4,E_6 E 4 , E 6 Δ \Delta Δ j = E 4 3 / Δ j = E_4^3/\Delta j = E 4 3 /Δ A n , r A_{n,r} A n , r ℓ ≥ 5 \ell\ge5 ℓ ≥ 5
Y a ( Y − 1728 ) b F ~ ℓ − 1 ( Y ) Y^a(Y - 1728)^b\widetilde{F}_{\ell-1}(Y) Y a ( Y − 1728 ) b F ℓ − 1 ( Y ) has ℓ \ell ℓ j j j F ℓ \mathbb{F}_{\ell} F ℓ [18 Theorem 1] . Indeed, 2 n + r / 6 = ℓ / 6 − 1 2n+r/6=\ell/6-1 2 n + r /6 = ℓ /6 − 1 A n , r ≡ 1 ( m o d ℓ ) A_{n,r}\equiv1\pmod{\ell} A n , r ≡ 1 ( mod ℓ ) [4 Theorem 4.2] .
The Legendre family has a related classical supersingular interpretation. For an odd prime p = 2 m + 1 p=2m+1 p = 2 m + 1
( 1 − z ) m P m ( 1 + z 1 − z ) = ∑ j = 0 m ( m j ) 2 z j (1-z)^mP_m\left(\frac{1+z}{1-z}\right)=\sum_{j=0}^{m}\binom{m}{j}^{2}z^j ( 1 − z ) m P m ( 1 − z 1 + z ) = j = 0 ∑ m ( j m ) 2 z j The reduction is a Hasse polynomial for y 2 = x ( x − 1 ) ( x − z ) y^2=x(x-1)(x-z) y 2 = x ( x − 1 ) ( x − z ) z ≠ 0 , 1 z\ne0,1 z = 0 , 1 [5 §1] .
Hyperelliptic Jacobians
For the curve y 2 = f ( x ) y^2=f(x) y 2 = f ( x ) f f f 2 2 2 Z \mathbb{Z} Z Sp \operatorname{Sp} Sp ℓ \ell ℓ Q ℓ Id ⊕ s p \mathbb{Q}_{\ell}\operatorname{Id}\oplus\mathfrak{sp} Q ℓ Id ⊕ sp [28 ] . They also determine the Hodge and Tate classes on all powers. The following corollary collects these consequences under one Galois-group hypothesis.
Corollary 6.6. Let f ∈ Q [ x ] f\in\mathbb{Q}[x] f ∈ Q [ x ] d ≥ 9 d\ge9 d ≥ 9 Gal ( f / Q ) = S d \operatorname{Gal}(f/\mathbb{Q})=S_d Gal ( f / Q ) = S d J f J_f J f y 2 = f ( x ) y^2=f(x) y 2 = f ( x )
dim J f = ⌊ d − 1 2 ⌋ , End Q ( J f ) = Z . \dim J_f=\left\lfloor\frac{d-1}{2}\right\rfloor,\qquad\operatorname{End}_{\mathbb{Q}}(J_f)=\mathbb{Z}. dim J f = ⌊ 2 d − 1 ⌋ , End Q ( J f ) = Z . For V = H 1 ( J f ( C ) , Q ) V=H_1(J_f(\mathbb{C}),\mathbb{Q}) V = H 1 ( J f ( C ) , Q ) ψ \psi ψ
Hdg ( J f ) = Sp ( V , ψ ) , MT ( J f ) = GSp ( V , ψ ) . \operatorname{Hdg}(J_f)=\operatorname{Sp}(V,\psi),\qquad\operatorname{MT}(J_f)=\operatorname{GSp}(V,\psi). Hdg ( J f ) = Sp ( V , ψ ) , MT ( J f ) = GSp ( V , ψ ) . For every number field K ⊂ Q ‾ K\subset\overline{\mathbb{Q}} K ⊂ Q ℓ \ell ℓ V ℓ = T ℓ ( J f ) ⊗ Z ℓ Q ℓ V_\ell=T_\ell(J_f)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell V ℓ = T ℓ ( J f ) ⊗ Z ℓ Q ℓ ψ ℓ \psi_\ell ψ ℓ Q ℓ ( 1 ) \mathbb{Q}_\ell(1) Q ℓ ( 1 ) Gal ( Q ‾ / K ) \operatorname{Gal}(\overline{\mathbb{Q}}/K) Gal ( Q / K ) V ℓ V_\ell V ℓ
Q ℓ Id ⊕ s p ( V ℓ , ψ ℓ ) . \mathbb{Q}_\ell\operatorname{Id}\oplus\mathfrak{sp}(V_\ell,\psi_\ell). Q ℓ Id ⊕ sp ( V ℓ , ψ ℓ ) . Every rational Hodge class on every power of J f J_f J f h ≥ 1 h\ge1 h ≥ 1 c ≥ 0 c\ge0 c ≥ 0
CH c ( ( J f h ) K ) ⊗ Q ℓ ⟶ H e ˊ t 2 c ( ( J f h ) Q ‾ , Q ℓ ( c ) ) Gal ( Q ‾ / K ) \operatorname{CH}^{c}((J_f^h)_K)\otimes\mathbb{Q}_\ell\longrightarrow H^{2c}_{\mathrm{\acute et}}((J_f^h)_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell(c))^{\operatorname{Gal}(\overline{\mathbb{Q}}/K)} CH c (( J f h ) K ) ⊗ Q ℓ ⟶ H e ˊ t 2 c (( J f h ) Q , Q ℓ ( c ) ) Gal ( Q / K ) is surjective. The Hodge, Tate and Mumford–Tate conjectures therefore hold for J f J_f J f
Proof. These are consequences of Zarhin’s theorems [28 ] : Theorem 2.4 gives the geometric endomorphism ring, Theorem 2.6 gives the scalar plus symplectic ℓ \ell ℓ
For the Kaneko–Zagier family, take f = P d ( ℓ / 3 , μ / 2 ) f=P_d^{(\ell/3,\mu/2)} f = P d ( ℓ /3 , μ /2 ) a , b a,b a , b f = P d ( a , b ) f=P_d^{(a,b)} f = P d ( a , b ) d ≥ max { 9 , N ( a , b ) } d\ge\max\{9,N(a,b)\} d ≥ max { 9 , N ( a , b )} h n h_n h n d = M n d=M_n d = M n
For the derivative families, fix r r r f = Q m , r , δ f=Q_{m,r,\delta} f = Q m , r , δ m ≥ 9 m\ge9 m ≥ 9 t = x 2 t=x^2 t = x 2
{ y 2 = F m , r , δ ( x ) } ⟶ { y 2 = Q m , r , δ ( t ) } , ( x , y ) ⟼ ( x 2 , y ) . \{y^2=F_{m,r,\delta}(x)\}\longrightarrow\{y^2=Q_{m,r,\delta}(t)\},\qquad(x,y)\longmapsto(x^2,y). { y 2 = F m , r , δ ( x )} ⟶ { y 2 = Q m , r , δ ( t )} , ( x , y ) ⟼ ( x 2 , y ) . This map extends to a degree-two morphism of the smooth projective models. It is the curve on the right whose defining polynomial has group S m S_m S m F m , r , δ F_{m,r,\delta} F m , r , δ W m W_m W m
Further questions
The fixed rational-parameter theorem leaves a precise irreducibility problem.
Question 6.7. Suppose α , β > − 1 \alpha,\beta>-1 α , β > − 1 Theorem 1.6 . Is P n ( α , β ) P_n^{(\alpha,\beta)} P n ( α , β ) Q \mathbb{Q} Q n n n
The theorem bounds the complementary degree by O α , β , A ( n / ( log n ) A ) O_{\alpha,\beta,A}(n/(\log n)^A) O α , β , A ( n / ( log n ) A ) A > 0 A>0 A > 0
For integral parameters, the exceptional degrees form an effective finite set. Classifying them calls for more information about the central numerators and small factors. In the symmetric case, all-degree irreducibility cannot hold for every fixed r r r
C 2 7 / 2 ( x ) = 7 2 ( 3 x − 1 ) ( 3 x + 1 ) . C_{2}^{7/2}(x)=\frac{7}{2}(3x-1)(3x+1). C 2 7/2 ( x ) = 2 7 ( 3 x − 1 ) ( 3 x + 1 ) . The cases r = 0 , 1 r=0,1 r = 0 , 1 r r r m ≥ 8 r m\geq8r m ≥ 8 r Corollary 1.7 settles eventual maximality for each fixed r r r r r r
The prescribed-factor estimate suggests a related question: how does the arithmetic of a minimal polynomial constrain the occurrence of its roots in varying Jacobi families? The cone x 2 + y 2 − 2 z 2 = 0 x^2+y^2-2z^2=0 x 2 + y 2 − 2 z 2 = 0 23 r > 4 ( ℓ − 1 ) 23r>4(\ell-1) 23 r > 4 ( ℓ − 1 )
Finally, the local results in Appendix § 7 come from a dilation of digit-sum Newton polygons. A general description of when multiplication of the degree by a denominator prime dilates the lower hull could yield further local irreducibility sequences, including cases where the global resultant comparison gives no contradiction.
Local irreducibility
Rational irreducibility does not in general imply irreducibility over a completion. The following result gives the latter conclusion in two infinite sequences of degrees.
Theorem A.1. For every integer a ≥ 0 a\geq0 a ≥ 0 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } P 3 a ( λ / 3 , μ / 2 ) P_{3^a}^{(\lambda/3,\mu/2)} P 3 a ( λ /3 , μ /2 ) Q 3 \mathbb{Q}_3 Q 3 P 5 ⋅ 3 a ( λ / 3 , μ / 2 ) P_{5\cdot3^a}^{(\lambda/3,\mu/2)} P 5 ⋅ 3 a ( λ /3 , μ /2 ) Q 3 \mathbb{Q}_3 Q 3 μ = − 1 \mu=-1 μ = − 1 a ≥ 2 a\geq2 a ≥ 2 μ = 1 \mu=1 μ = 1 1 , 1 , 3 1,1,3 1 , 1 , 3 a = 0 a=0 a = 0 2 , 13 2,13 2 , 13 a = 1 a=1 a = 1
The first assertion follows from a general prime-power criterion. The second follows from a dilation identity for the lower convex hull of a digit-sum function.
Irreducibility in prime-power degrees
A denominator in one parameter, together with an exact valuation in the other, forces a single Newton edge.
Theorem A.2. Let p p p a ≥ 1 a\geq1 a ≥ 1 n = p a n=p^a n = p a α , β ∈ Q \alpha,\beta\in\mathbb{Q} α , β ∈ Q
v p ( α ) = − e < 0 , v p ( β ) ≥ 0 , v p ( ( β + 1 ) n ) = v p ( n ! ) = n − 1 p − 1 . v_p(\alpha)=-e<0,\qquad v_p(\beta)\geq0,\qquad v_p((\beta+1)_n)=v_p(n!)=\frac{n-1}{p-1}. v p ( α ) = − e < 0 , v p ( β ) ≥ 0 , v p (( β + 1 ) n ) = v p ( n !) = p − 1 n − 1 . Then P n ( α , β ) P_n^{(\alpha,\beta)} P n ( α , β ) Q p \mathbb{Q}_p Q p
n ! P n ( α , β ) ( 2 X − 1 ) ( n + α + β + 1 ) n \frac{n!P_n^{(\alpha,\beta)}(2X-1)}{(n+\alpha+\beta+1)_n} ( n + α + β + 1 ) n n ! P n ( α , β ) ( 2 X − 1 ) has a single Newton edge joining
( 0 , e n + n − 1 p − 1 ) and ( n , 0 ) , \left(0,en+\frac{n-1}{p-1}\right)\quad\text{and}\quad(n,0), ( 0 , e n + p − 1 n − 1 ) and ( n , 0 ) , with every intermediate coefficient point strictly above the edge.
Proof. Every denominator factor in (58) has valuation − e -e − e h j = v p ( ( β + 1 ) j ) h_j=v_p((\beta+1)_j) h j = v p (( β + 1 ) j ) h 0 = 0 h_0=0 h 0 = 0 1 ≤ i ≤ a 1\leq i\leq a 1 ≤ i ≤ a n / p i n/p^i n / p i β + 1 , … , β + n \beta+1,\ldots,\beta+n β + 1 , … , β + n p i Z p p^i\mathbb{Z}_p p i Z p h n h_n h n p a + 1 Z p p^{a+1}\mathbb{Z}_p p a + 1 Z p 1 ≤ j < n 1\leq j<n 1 ≤ j < n t = v p ( j ) t=v_p(j) t = v p ( j )
h j ≤ ∑ i = 1 a ⌊ j p i ⌋ = v p ( j ! ) + a − t , v p ( ( n j ) ) = a − t . h_j\leq\sum_{i=1}^{a}\left\lfloor\frac{j}{p^i}\right\rfloor=v_p(j!)+a-t,\qquad v_p\left(\binom{n}{j}\right)=a-t. h j ≤ i = 1 ∑ a ⌊ p i j ⌋ = v p ( j !) + a − t , v p ( ( j n ) ) = a − t . If b j b_j b j X j X^j X j
v p ( b j ) = e ( n − j ) + v p ( ( n j ) ) + h n − h j ≥ e ( n − j ) + h n − v p ( j ! ) . v_p(b_j)=e(n-j)+v_p\left(\binom{n}{j}\right)+h_n-h_j\geq e(n-j)+h_n-v_p(j!). v p ( b j ) = e ( n − j ) + v p ( ( j n ) ) + h n − h j ≥ e ( n − j ) + h n − v p ( j !) . Writing H = e n + ( n − 1 ) / ( p − 1 ) H=en+(n-1)/(p-1) H = e n + ( n − 1 ) / ( p − 1 )
v p ( b j ) − H ( 1 − j n ) ≥ s p ( j ) − j / n p − 1 > 0. v_p(b_j)-H\left(1-\frac{j}{n}\right)\geq\frac{s_p(j)-j/n}{p-1}>0. v p ( b j ) − H ( 1 − n j ) ≥ p − 1 s p ( j ) − j / n > 0. The endpoint valuations are H H H H ≡ 1 ( m o d p ) H\equiv1\pmod p H ≡ 1 ( mod p ) ( H , n ) = 1 (H,n)=1 ( H , n ) = 1 H / n H/n H / n Q p \mathbb{Q}_p Q p d d d d H / n ∈ Z dH/n\in\mathbb{Z} d H / n ∈ Z n ∣ d n\mid d n ∣ d
Corollary A.3. Let p p p a ≥ 1 a\geq1 a ≥ 1 α ∈ Q \alpha\in\mathbb{Q} α ∈ Q v p ( α ) < 0 v_p(\alpha)<0 v p ( α ) < 0 P p a ( α , − 1 / 2 ) P_{p^a}^{(\alpha,-1/2)} P p a ( α , − 1/2 ) P p a ( α , 1 / 2 ) P_{p^a}^{(\alpha,1/2)} P p a ( α , 1/2 ) Q p \mathbb{Q}_p Q p a ≥ 1 a\geq1 a ≥ 1 P 3 a ( λ / 3 , μ / 2 ) P_{3^a}^{(\lambda/3,\mu/2)} P 3 a ( λ /3 , μ /2 ) Q 3 \mathbb{Q}_3 Q 3
Proof. For n = p a n=p^a n = p a p p p
( 1 / 2 ) n = ( 2 n ) ! 4 n n ! , ( 3 / 2 ) n = ( 2 n + 1 ) ! 4 n n ! , (1/2)_n=\frac{(2n)!}{4^n n!},\qquad(3/2)_n=\frac{(2n+1)!}{4^n n!}, ( 1/2 ) n = 4 n n ! ( 2 n )! , ( 3/2 ) n = 4 n n ! ( 2 n + 1 )! , and v p ( ( 2 n ) ! ) = v p ( ( 2 n + 1 ) ! ) = 2 ( n − 1 ) / ( p − 1 ) v_p((2n)!)=v_p((2n+1)!)=2(n-1)/(p-1) v p (( 2 n )!) = v p (( 2 n + 1 )!) = 2 ( n − 1 ) / ( p − 1 )
A dilation law and the degrees 5 times powers of 3
Fix n ≥ 1 n \ge1 n ≥ 1 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } v 3 ( 3 ) = 1 v_{3}(3)=1 v 3 ( 3 ) = 1 s 3 ( r ) s_{3}(r) s 3 ( r ) r r r
Legendre’s formula v 3 ( k ! ) = ( k − s 3 ( k ) ) / 2 v_{3}(k!)=(k-s_{3}(k))/2 v 3 ( k !) = ( k − s 3 ( k )) /2
For δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 }
Φ n , δ ( j ) = s 3 ( 2 j + δ ) + s 3 ( n − j ) − s 3 ( 2 n + δ ) ( 0 ≤ j ≤ n ) , \Phi_{n,\delta}(j)=s_{3}(2j+\delta)+s_{3}(n-j)-s_{3}(2n+\delta) \qquad(0\le j\le n), Φ n , δ ( j ) = s 3 ( 2 j + δ ) + s 3 ( n − j ) − s 3 ( 2 n + δ ) ( 0 ≤ j ≤ n ) , and let H n , δ H_{n,\delta} H n , δ [ 0 , n ] [0,n] [ 0 , n ]
T n λ , μ ( X ) = n ! ( n + λ / 3 + μ / 2 + 1 ) n P n ( λ / 3 , μ / 2 ) ( 2 X − 1 ) T_{n}^{\lambda,\mu}(X)=\frac{n!}{(n+\lambda/3+\mu/2+1)_{n}}P_{n}^{(\lambda/3,\mu/2)}(2X-1) T n λ , μ ( X ) = ( n + λ /3 + μ /2 + 1 ) n n ! P n ( λ /3 , μ /2 ) ( 2 X − 1 ) has coefficients
b j = ( − 1 ) n − j ( n j ) ∏ r = j + 1 n 3 ( 2 r + μ ) 6 ( n + r ) + 2 λ + 3 μ . b_j=(-1)^{n-j}\binom{n}{j}\prod_{r=j+1}^{n}\frac{3(2r+\mu)}{6(n+r)+2\lambda+3\mu}. b j = ( − 1 ) n − j ( j n ) r = j + 1 ∏ n 6 ( n + r ) + 2 λ + 3 μ 3 ( 2 r + μ ) . Since each denominator in this product is a 3-adic unit,
v 3 ( b j ) = n − j + v 3 ( ( 2 n + δ ) ! ) − v 3 ( ( 2 j + δ ) ! ) − v 3 ( ( n − j ) ! ) = 3 2 ( n − j ) + 1 2 Φ n , δ ( j ) , δ = μ + 1 2 . (108) v_{3}(b_j)=n-j+v_{3}((2n+\delta)!)-v_{3}((2j+\delta)!)-v_{3}((n-j)!)=\frac{3}{2}(n-j)+\frac{1}{2}\Phi_{n,\delta}(j),\qquad\delta=\frac{\mu+1}{2}. \tag*{(108)} v 3 ( b j ) = n − j + v 3 (( 2 n + δ )!) − v 3 (( 2 j + δ )!) − v 3 (( n − j )!) = 2 3 ( n − j ) + 2 1 Φ n , δ ( j ) , δ = 2 μ + 1 . (108) Thus its Newton polygon is obtained from H n , δ H_{n,\delta} H n , δ λ \lambda λ
Lemma A.5. For every integer m ≥ 1 m\ge1 m ≥ 1
H 3 m , 0 ( x ) = H m , 0 ( x / 3 ) ( 0 ≤ x ≤ 3 m ) , H 9 m , 1 ( x ) = H m , 0 ( x / 9 ) ( 0 ≤ x ≤ 9 m ) . H_{3m,0}(x)=H_{m,0}(x/3)\qquad(0\le x\le3m),\qquad H_{9m,1}(x)=H_{m,0}(x/9)\qquad(0\le x\le9m). H 3 m , 0 ( x ) = H m , 0 ( x /3 ) ( 0 ≤ x ≤ 3 m ) , H 9 m , 1 ( x ) = H m , 0 ( x /9 ) ( 0 ≤ x ≤ 9 m ) . All points whose abscissae are not divisible by 3, respectively by 9, lie strictly above the indicated hull.
Proof. First, all slopes of H m , 0 H_{m,0} H m , 0 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
Φ m , 0 ( j ) − Φ m , 0 ( 0 ) = s 3 ( 2 j ) + s 3 ( m − j ) − s 3 ( m ) ≥ − s 3 ( j ) ≥ − j . \Phi_{m,0}(j)-\Phi_{m,0}(0)=s_{3}(2j)+s_{3}(m-j)-s_{3}(m)\ge-s_{3}(j)\ge-j. Φ m , 0 ( j ) − Φ m , 0 ( 0 ) = s 3 ( 2 j ) + s 3 ( m − j ) − s 3 ( m ) ≥ − s 3 ( j ) ≥ − j . Writing t = m − j t=m-j t = m − j
Φ m , 0 ( j ) ≥ s 3 ( t ) − s 3 ( 2 t ) ≥ − s 3 ( t ) ≥ − t . \Phi_{m,0}(j)\ge s_{3}(t)-s_{3}(2t)\ge-s_{3}(t)\ge-t. Φ m , 0 ( j ) ≥ s 3 ( t ) − s 3 ( 2 t ) ≥ − s 3 ( t ) ≥ − t . These inequalities bound the first and last slopes of the convex hull, and hence every slope, as asserted.
For 0 ≤ k < m 0\le k<m 0 ≤ k < m u = v 3 ( m − k ) u=v_{3}(m-k) u = v 3 ( m − k ) w = v 3 ( k + 1 ) w=v_{3}(k+1) w = v 3 ( k + 1 ) s 3 ( r − 1 ) = s 3 ( r ) − 1 + 2 v 3 ( r ) s_{3}(r-1)=s_{3}(r)-1+2v_{3}(r) s 3 ( r − 1 ) = s 3 ( r ) − 1 + 2 v 3 ( r ) r ≥ 1 r\ge1 r ≥ 1
Φ 3 m , 0 ( 3 k ) = Φ m , 0 ( k ) , Φ 3 m , 0 ( 3 k + 1 ) = Φ m , 0 ( k ) + 3 + 2 u , Φ 3 m , 0 ( 3 k + 2 ) = Φ m , 0 ( k + 1 ) + 1 + 2 w . \begin{aligned}
\Phi_{3m,0}(3k)&=\Phi_{m,0}(k),\\
\Phi_{3m,0}(3k+1)&=\Phi_{m,0}(k)+3+2u,\\
\Phi_{3m,0}(3k+2)&=\Phi_{m,0}(k+1)+1+2w.
\end{aligned} Φ 3 m , 0 ( 3 k ) Φ 3 m , 0 ( 3 k + 1 ) Φ 3 m , 0 ( 3 k + 2 ) = Φ m , 0 ( k ) , = Φ m , 0 ( k ) + 3 + 2 u , = Φ m , 0 ( k + 1 ) + 1 + 2 w . The endpoint identity also holds at k = m k=m k = m H m , 0 ( x / 3 ) H_{m,0}(x/3) H m , 0 ( x /3 ) [ − 1 / 3 , 1 / 3 ] [-1/3,1/3] [ − 1/3 , 1/3 ]
For the second identity, the grid points satisfy Φ 9 m , 1 ( 9 k ) = Φ m , 0 ( k ) \Phi_{9m,1}(9k)=\Phi_{m,0}(k) Φ 9 m , 1 ( 9 k ) = Φ m , 0 ( k ) 0 ≤ k ≤ m 0\le k\le m 0 ≤ k ≤ m
r 1 2 3 Φ 9 m , 1 ( 9 k + r ) − Φ m , 0 ( k ) − 2 u 3 4 3 \begin{array}{c|ccc}
r&1&2&3\\
\Phi_{9m,1}(9k+r)-\Phi_{m,0}(k)-2u&3&4&3
\end{array} r Φ 9 m , 1 ( 9 k + r ) − Φ m , 0 ( k ) − 2 u 1 3 2 4 3 3 r 4 5 6 7 8 Φ 9 m , 1 ( 9 k + r ) − Φ m , 0 ( k + 1 ) − 2 w 1 2 1 2 3 \begin{array}{c|ccccc}
r&4&5&6&7&8\\
\Phi_{9m,1}(9k+r)-\Phi_{m,0}(k+1)-2w&1&2&1&2&3
\end{array} r Φ 9 m , 1 ( 9 k + r ) − Φ m , 0 ( k + 1 ) − 2 w 4 1 5 2 6 1 7 2 8 3 They follow by separating the last two base-3 digits and using the same identity for s 3 ( r − 1 ) s_{3}(r-1) s 3 ( r − 1 ) [ − 1 / 9 , 1 / 9 ] [-1/9,1/9] [ − 1/9 , 1/9 ] 1 ≤ r ≤ 3 1 \le r \le3 1 ≤ r ≤ 3 3 − r / 9 3-r/9 3 − r /9 4 ≤ r ≤ 8 4 \le r \le8 4 ≤ r ≤ 8 1 − ( 9 − r ) / 9 1-(9-r)/9 1 − ( 9 − r ) /9
Completion of the proof of Theorem A.1. Corollary A.3 treats the degrees 3 a 3^{a} 3 a n = 5 ⋅ 3 a n=5\cdot3^{a} n = 5 ⋅ 3 a Φ 5 , 0 ( j ) \Phi_{5,0}(j) Φ 5 , 0 ( j )
1 , 2 , 1 , 2 , 3 , 0. 1,\ 2,\ 1,\ 2,\ 3,\ 0. 1 , 2 , 1 , 2 , 3 , 0. Their lower hull is the segment from ( 0 , 1 ) (0,1) ( 0 , 1 ) ( 5 , 0 ) (5,0) ( 5 , 0 ) n = 5 ⋅ 3 a n=5\cdot3^{a} n = 5 ⋅ 3 a μ = − 1 \mu=-1 μ = − 1 μ = 1 \mu=1 μ = 1 a ≥ 2 a\ge2 a ≥ 2
( 0 , ( 3 n + 1 ) / 2 ) − ( n , 0 ) , (0,(3n+1)/2)-(n,0), ( 0 , ( 3 n + 1 ) /2 ) − ( n , 0 ) , with every intermediate coefficient point strictly above it. Since n n n
gcd ( n , 3 n + 1 2 ) = 1. \gcd\left(n,\frac{3n+1}{2}\right)=1. g cd( n , 2 3 n + 1 ) = 1. The constant term of a monic factor of degree d d d Q 3 \mathbb{Q}_{3} Q 3 d ( 3 n + 1 ) / ( 2 n ) d(3n+1)/(2n) d ( 3 n + 1 ) / ( 2 n ) n ∣ d n\mid d n ∣ d μ = 1 \mu=1 μ = 1 λ \lambda λ
Table 1 (PDF p. 49).
Each side has relatively prime height and horizontal length. The Newton-polygon factorization therefore has one irreducible factor of each listed degree. Both polynomials are reducible, which proves the converse and the asserted factor degrees. □
The explicit cutoff and finite verification
The positive gap in § 4.5 already gives eventual irreducibility. Here we keep enough of the error terms to obtain a numerical cutoff. The useful quantity is S n + log g n S_n+\log g_n S n + log g n n ≥ 250 000 n\ge250\,000 n ≥ 250 000
An explicit cutoff
B.1. An explicit cutoff. Let α = λ / 3 \alpha=\lambda/3 α = λ /3 β = μ / 2 \beta=\mu/2 β = μ /2 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } H n H_n H n g n g_n g n S n S_n S n L n L_n L n J n J_n J n T 2 = T 2 , n T_2=T_{2,n} T 2 = T 2 , n T 3 = T 3 , n T_3=T_{3,n} T 3 = T 3 , n
N 0 = 250 000 , r = 2 λ + 3 μ , δ = μ + 1 2 , E = 1 2 log 2 + 1 3 log 3 , L ∗ = log ( 4 + 3 3 ) . N_0=250\,000,\qquad r=2\lambda+3\mu,\qquad\delta=\frac{\mu+1}{2},\qquad E=\frac{1}{2}\log2+\frac{1}{3}\log3,\qquad L_*=\log(4+3\sqrt{3}). N 0 = 250 000 , r = 2 λ + 3 μ , δ = 2 μ + 1 , E = 2 1 log 2 + 3 1 log 3 , L ∗ = log ( 4 + 3 3 ) . Proposition B.1. For every n ≥ 250 000 n\ge250\,000 n ≥ 250 000 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } P n ( λ / 3 , μ / 2 ) P_n^{(\lambda/3,\mu/2)} P n ( λ /3 , μ /2 ) Q \mathbb{Q} Q
We prove the proposition in three steps. We use the notation
θ ( x ; q , a ) = ∑ p ≤ x prime p ≡ a ( m o d q ) log p , θ ( x ) = ∑ p ≤ x prime log p , ψ ( x ) = ∑ p h ≤ x log p . \theta(x;q,a)=\sum_{\substack{p\le x\ \text{prime}\\ p\equiv a\pmod q}}\log p,\qquad
\theta(x)=\sum_{p\le x\ \text{prime}}\log p,\qquad
\psi(x)=\sum_{p^h\le x}\log p. θ ( x ; q , a ) = p ≤ x prime p ≡ a ( mod q ) ∑ log p , θ ( x ) = p ≤ x prime ∑ log p , ψ ( x ) = p h ≤ x ∑ log p . The necessary archimedean inequality
For every proper factorization with smaller degree 1 ≤ k ≤ n / 2 1 \le k \le n/2 1 ≤ k ≤ n /2
S n + log g n ≥ n ( L ∗ − E ) − 6 log ( 12 n ) ( n ≥ N 0 ) . (109) S_n+\log g_n \ge n(L_* - E)-6\log(12n) \qquad(n \ge N_0). \tag*{(109)} S n + log g n ≥ n ( L ∗ − E ) − 6 log ( 12 n ) ( n ≥ N 0 ) . (109) Put
b n = v 2 ( n + α n ) . b_n=v_2\binom{n+\alpha}{n}. b n = v 2 ( n n + α ) . The level-by-level binomial valuation formula gives 0 ≤ b n ≤ log 2 ( 3 n + 1 ) 0 \le b_n \le\log_2(3n+1) 0 ≤ b n ≤ log 2 ( 3 n + 1 )
D 2 , n = 2 n − s 2 ( n ) + b n , D 3 , n = 3 n 2 + δ + s 3 ( n ) − s 3 ( 2 n + δ ) 2 . D_{2,n}=2n-s_2(n)+b_n,\qquad D_{3,n}=\frac{3n}{2}+\frac{\delta+s_3(n)-s_3(2n+\delta)}{2}. D 2 , n = 2 n − s 2 ( n ) + b n , D 3 , n = 2 3 n + 2 δ + s 3 ( n ) − s 3 ( 2 n + δ ) . Also
log κ n = ( 2 n − s 2 ( n ) ) log 2 + ( 3 n / 2 − s 3 ( n ) / 2 ) log 3. \log\kappa_n=(2n-s_2(n))\log2+(3n/2-s_3(n)/2)\log3. log κ n = ( 2 n − s 2 ( n )) log 2 + ( 3 n /2 − s 3 ( n ) /2 ) log 3. Using s p ( m ) ≤ ( p − 1 ) ( 1 + log p m ) s_p(m)\le(p-1)(1+\log_p m) s p ( m ) ≤ ( p − 1 ) ( 1 + log p m )
n L n ≥ n L ∗ − log ( 6 n + 3 ) nL_n\ge nL_*-\log(6n+3) n L n ≥ n L ∗ − log ( 6 n + 3 ) and
log κ n − n ( T 2 + T 3 ) ≥ − log ( 3 n + 1 ) − 2 log ( 3 n ) − 1 2 log 3. \log\kappa_n-n(T_2+T_3)\ge-\log(3n+1)-2\log(3n)-\frac{1}{2}\log3. log κ n − n ( T 2 + T 3 ) ≥ − log ( 3 n + 1 ) − 2 log ( 3 n ) − 2 1 log 3. Furthermore,
J n ≤ 1 2 log ( 2 n ) + 1 3 log ( 6 n + 3 ) . J_n\le\frac{1}{2}\log(2n)+\frac{1}{3}\log(6n+3). J n ≤ 2 1 log ( 2 n ) + 3 1 log ( 6 n + 3 ) . The Wronskian constant satisfies C n ≥ 1 / 3 C_n\ge1/3 C n ≥ 1/3 C 1 = ( α + 1 ) ( β + 1 ) ≥ 1 / 3 C_1=(\alpha+1)(\beta+1)\ge1/3 C 1 = ( α + 1 ) ( β + 1 ) ≥ 1/3
C m + 1 C m = 1 + α β ( m + 1 ) ( m + c + 1 ) , c = α + β . \frac{C_{m+1}}{C_m}=1+\frac{\alpha\beta}{(m+1)(m+c+1)},\qquad c=\alpha+\beta. C m C m + 1 = 1 + ( m + 1 ) ( m + c + 1 ) α β , c = α + β . If α β > 0 \alpha\beta>0 α β > 0 α β = − 1 / 6 \alpha\beta=-1/6 α β = − 1/6 C 1 ≥ 2 / 3 C_1\ge2/3 C 1 ≥ 2/3 m ≥ 1 m\ge1 m ≥ 1
C m + 1 C m ≥ 1 − 1 6 m ( m + 1 ) . \frac{C_{m+1}}{C_m}\ge1-\frac{1}{6m(m+1)}. C m C m + 1 ≥ 1 − 6 m ( m + 1 ) 1 . The product is at least 1 − ∑ m ≥ 1 1 / ( 6 m ( m + 1 ) ) = 5 / 6 1-\sum_{m\ge1}1/(6m(m+1))=5/6 1 − ∑ m ≥ 1 1/ ( 6 m ( m + 1 )) = 5/6 C n ≥ 5 / 9 C_n\ge5/9 C n ≥ 5/9 2 C n ≥ 1 2\sqrt{C_n}\ge1 2 C n ≥ 1
Let a n = T 2 + T 3 + E − 2 L n + 2 log 2 a_n=T_2+T_3+E-2L_n+2\log2 a n = T 2 + T 3 + E − 2 L n + 2 log 2 n ≥ N 0 n\ge N_0 n ≥ N 0
∣ T 2 − T 3 ∣ ≤ 3 2 log 3 − 2 log 2 + 2 log ( 9 n ) n < 27 100 + 30 250000 < 3 10 . |T_2-T_3|\le\frac{3}{2}\log3-2\log2+\frac{2\log(9n)}{n}<\frac{27}{100}+\frac{30}{250000}<\frac{3}{10}. ∣ T 2 − T 3 ∣ ≤ 2 3 log 3 − 2 log 2 + n 2 log ( 9 n ) < 100 27 + 250000 30 < 10 3 . Here we used log 3 < 1.1 \log3<1.1 log 3 < 1.1 log 2 > 0.69 \log2>0.69 log 2 > 0.69 log ( 9 N 0 ) < 15 \log(9N_0)<15 log ( 9 N 0 ) < 15 log ( 9 n ) / n \log(9n)/n log ( 9 n ) / n log cosh u ≤ u 2 / 2 \log\cosh u\le u^2/2 log cosh u ≤ u 2 /2
a n = E − 2 log cosh ( ( T 2 − T 3 ) / 2 ) > 1 2 . a_n=E-2\log\cosh((T_2-T_3)/2)>\frac{1}{2}. a n = E − 2 log cosh (( T 2 − T 3 ) /2 ) > 2 1 . Consequently a n k + 2 log 2 / k − log k > 0 a_nk+2\log2/k-\log k>0 a n k + 2 log 2/ k − log k > 0 k > 0 k>0 k > 0 k / 2 − log k ≥ 1 − log 2 > 0 k/2-\log k\ge1-\log2>0 k /2 − log k ≥ 1 − log 2 > 0
S n + log g n ≥ log κ n − J n + n ( L n − T 2 − T 3 − E ) − 4 log 2 ≥ n ( L ∗ − E ) − { 4 3 log ( 6 n + 3 ) + log ( 3 n + 1 ) + 2 log ( 3 n ) + 1 2 log ( 2 n ) + 1 2 log 3 + 4 log 2 } . \begin{aligned}
S_n+\log g_n\ge\log\kappa_n-J_n+n(L_n-T_2-T_3-E)-4\log2 \\
&\ge n(L_*-E)
-\left\{\frac{4}{3}\log(6n+3)+\log(3n+1)+2\log(3n)+\frac{1}{2}\log(2n)+\frac{1}{2}\log3+4\log2\right\}.
\end{aligned} S n + log g n ≥ log κ n − J n + n ( L n − T 2 − T 3 − E ) − 4 log 2 ≥ n ( L ∗ − E ) − { 3 4 log ( 6 n + 3 ) + log ( 3 n + 1 ) + 2 log ( 3 n ) + 2 1 log ( 2 n ) + 2 1 log 3 + 4 log 2 } . The expression in braces is at most 6 log ( 12 n ) 6\log(12n) 6 log ( 12 n ) n ≥ 2 n\ge2 n ≥ 2 n = 2 n=2 n = 2 7 / ( 6 n ) 7/(6n) 7/ ( 6 n )
Content and the good-prime sum
Put M = 12 n + 5 M=12n+5 M = 12 n + 5
H ( M ) = ∑ p p r i m e , h ≥ 2 p h ≤ M log p . H(M)=\sum_{\substack{p\ \mathrm{prime},\ h\ge2\\p^h\le M}}\log p. H ( M ) = p prime , h ≥ 2 p h ≤ M ∑ log p . For each prime p ≥ 5 p\ge5 p ≥ 5 m p ( 1 ) m_p^{(1)} m p ( 1 )
v p ( g n ) ≤ m p ( 1 ) + 2 # { h ≥ 2 : p h ≤ M } . v_p(g_n)\le m_p^{(1)}+2\#\{h\ge2:p^h\le M\}. v p ( g n ) ≤ m p ( 1 ) + 2# { h ≥ 2 : p h ≤ M } . Since each exponent e p h e_{p^h} e p h
S n + log g n ≤ T 0 ( n ) + 3 H ( M ) , T 0 ( n ) = ∑ p ≥ 5 ( m p ( 1 ) + 1 p ≤ n e p ) log p . (110) S_n+\log g_n\le T_0(n)+3H(M),\qquad T_0(n)=\sum_{p\ge5}\left(m_p^{(1)}+\mathbf{1}_{p\le n}e_p\right)\log p. \tag*{(110)} S n + log g n ≤ T 0 ( n ) + 3 H ( M ) , T 0 ( n ) = p ≥ 5 ∑ ( m p ( 1 ) + 1 p ≤ n e p ) log p . (110) The two indicators m p ( 1 ) , e p m_p^{(1)},e_p m p ( 1 ) , e p T 0 T_0 T 0
For p ≥ 13 p\ge13 p ≥ 13 p ≡ 3 + 2 λ ( m o d 6 ) p\equiv3+2\lambda\pmod6 p ≡ 3 + 2 λ ( mod 6 ) p ≡ 3 − 2 λ ( m o d 6 ) p\equiv3-2\lambda\pmod6 p ≡ 3 − 2 λ ( mod 6 ) p ≤ n p\le n p ≤ n ℓ = ⌊ n / p ⌋ ≥ 1 \ell=\lfloor n/p\rfloor\ge1 ℓ = ⌊ n / p ⌋ ≥ 1
first class ( 6 n + r 6 ℓ + 1 , n ℓ ] and ( n ℓ + 1 , 2 n + μ 2 ℓ + 1 ] , second class ( n ℓ + 1 , 3 n + λ 3 ℓ + 1 ] . \begin{aligned}
\text{first class}\quad&\left(\frac{6n+r}{6\ell+1},\frac{n}{\ell}\right]\quad\text{and}\quad\left(\frac{n}{\ell+1},\frac{2n+\mu}{2\ell+1}\right],\\
\text{second class}\quad&\left(\frac{n}{\ell+1},\frac{3n+\lambda}{3\ell+1}\right].
\end{aligned} first class second class ( 6 ℓ + 1 6 n + r , ℓ n ] and ( ℓ + 1 n , 2 ℓ + 1 2 n + μ ] , ( ℓ + 1 n , 3 ℓ + 1 3 n + λ ] . For p > n p>n p > n
( n , 3 n + λ 2 ] in the first class , ( 6 n + r 5 , 2 n + μ ] in the second class . \left(n,\frac{3n+\lambda}{2}\right]\quad\text{in the first class},\qquad\left(\frac{6n+r}{5},2n+\mu\right]\quad\text{in the second class}. ( n , 2 3 n + λ ] in the first class , ( 5 6 n + r , 2 n + μ ] in the second class . These formulas follow by substituting a p = ( p + λ ) / 3 a_p=(p+\lambda)/3 a p = ( p + λ ) /3 ( 2 p + λ ) / 3 (2p+\lambda)/3 ( 2 p + λ ) /3 b p = ( p + μ ) / 2 b_p=(p+\mu)/2 b p = ( p + μ ) /2 p ≤ n p\le n p ≤ n 1 + E b − E c 1+E_b-E_c 1 + E b − E c E a E_a E a 5 , 7 , 11 5,7,11 5 , 7 , 11 log 385 \log385 log 385
We retain the first six groups of intervals. They contain enough of the main term for the required contradiction, while all discarded primes are at most n / 7 n/7 n /7 θ ( n / 7 ) \theta(n/7) θ ( n /7 ) n n n
Q 6 = 13 20 + 1 2 ∑ ℓ = 1 6 { 1 ℓ − 6 6 ℓ + 1 + 2 2 ℓ + 1 − 1 ℓ + 1 + 3 3 ℓ + 1 − 1 ℓ + 1 } = 180556377763 157066509600 < 23 20 . \begin{aligned}
Q_6
&=\frac{13}{20}+\frac{1}{2}\sum_{\ell=1}^{6}\left\{\frac{1}{\ell}-\frac{6}{6\ell+1}+\frac{2}{2\ell+1}-\frac{1}{\ell+1}+\frac{3}{3\ell+1}-\frac{1}{\ell+1}\right\}\\
&=\frac{180556377763}{157066509600}<\frac{23}{20}.
\end{aligned} Q 6 = 20 13 + 2 1 ℓ = 1 ∑ 6 { ℓ 1 − 6 ℓ + 1 6 + 2 ℓ + 1 2 − ℓ + 1 1 + 3 ℓ + 1 3 − ℓ + 1 1 } = 157066509600 180556377763 < 20 23 . The constant shifts in the interval lengths contribute at most
5 4 + 5 6 ∑ ℓ = 1 6 1 ℓ < 5. \frac{5}{4}+\frac{5}{6}\sum_{\ell=1}^{6}\frac{1}{\ell}<5. 4 5 + 6 5 ℓ = 1 ∑ 6 ℓ 1 < 5. Together with the small primes, this is less than 12.
We use two published estimates of Bennett, Martin, O’Bryant, and Rechnitzer [3 Corollary 1.7 and Theorem 1.9] : for a ∈ { 1 , 5 } a\in\{1,5\} a ∈ { 1 , 5 }
∣ θ ( x ; 6 , a ) − x / 2 ∣ ≤ 2.072 x ( 1 ≤ x ≤ 4 ⋅ 10 13 ) , \left|\theta(x;6,a)-x/2\right|\le2.072\sqrt{x}\qquad(1\le x\le4\cdot10^{13}), ∣ θ ( x ; 6 , a ) − x /2 ∣ ≤ 2.072 x ( 1 ≤ x ≤ 4 ⋅ 1 0 13 ) , ∣ θ ( x ; 6 , a ) − x / 2 ∣ < 0.024 x / log x ( x ≥ 10 6 ) . \left|\theta(x;6,a)-x/2\right|<0.024x/\log x\qquad(x\ge10^6). ∣ θ ( x ; 6 , a ) − x /2 ∣ < 0.024 x / log x ( x ≥ 1 0 6 ) . The range 4 ⋅ 10 13 4 \cdot10^{13} 4 ⋅ 1 0 13 x 2 ( 6 ) = x 2 ( 3 ) x_2(6)=x_2(3) x 2 ( 6 ) = x 2 ( 3 ) N 0 ≤ n ≤ 10 13 N_0 \le n \le10^{13} N 0 ≤ n ≤ 1 0 13 2 n + 1 < 4 ⋅ 10 13 2n+1<4\cdot10^{13} 2 n + 1 < 4 ⋅ 1 0 13 1 ≤ ℓ ≤ 6 1\le\ell\le6 1 ≤ ℓ ≤ 6 ( n + 1 ) / ℓ (n+1)/\ell ( n + 1 ) / ℓ 2 n + 1 2n+1 2 n + 1
2.072 { 6 n + 1 ∑ ℓ = 1 6 ℓ − 1 / 2 + 4 2 n + 1 } < 61 n . 2.072\left\{6\sqrt{n+1}\sum_{\ell=1}^{6}\ell^{-1/2}+4\sqrt{2n+1}\right\}<61\sqrt{n}. 2.072 { 6 n + 1 ℓ = 1 ∑ 6 ℓ − 1/2 + 4 2 n + 1 } < 61 n . To obtain the last inequality, use ∑ ℓ = 1 6 ℓ − 1 / 2 ≤ 2 6 − 1 < 3.9 \sum_{\ell=1}^{6}\ell^{-1/2}\le2\sqrt{6}-1<3.9 ∑ ℓ = 1 6 ℓ − 1/2 ≤ 2 6 − 1 < 3.9 1 + 1 / N 0 < 1.000002 \sqrt{1+1/N_0}<1.000002 1 + 1/ N 0 < 1.000002 2 + 1 / N 0 < 1.415 \sqrt{2+1/N_0}<1.415 2 + 1/ N 0 < 1.415 n ≥ 10 13 n\ge10^{13} n ≥ 1 0 13 n / 7 n/7 n /7 ∑ ℓ = 1 6 ℓ − 1 = 49 / 20 \sum_{\ell=1}^{6}\ell^{-1}=49/20 ∑ ℓ = 1 6 ℓ − 1 = 49/20
6 ( n + 1 ) 49 20 + 4 ( 2 n + 1 ) < 23 n . 6(n+1)\frac{49}{20}+4(2n+1)<23n. 6 ( n + 1 ) 20 49 + 4 ( 2 n + 1 ) < 23 n . Using log ( 10 13 / 7 ) > 27 \log(10^{13}/7)>27 log ( 1 0 13 /7 ) > 27
0.024 23 n 27 < 21 1000 n . 0.024\frac{23n}{27}<\frac{21}{1000}n. 0.024 27 23 n < 1000 21 n . Hanson’s elementary theorem [16 Theorem 1] states lcm ( 1 , … , N ) < 3 N \operatorname{lcm}(1,\ldots,N)<3^N lcm ( 1 , … , N ) < 3 N θ ( x ) ≤ ψ ( x ) < x log 3 \theta(x)\le\psi(x)<x\log3 θ ( x ) ≤ ψ ( x ) < x log 3 x ≥ 1 x\ge1 x ≥ 1
T 0 ( n ) < ( 23 20 + log 3 7 ) n + 12 + { 61 n , N 0 ≤ n ≤ 10 13 , 21 1000 n , n > 10 13 . (111) T_0(n)<\left(\frac{23}{20}+\frac{\log3}{7}\right)n+12+
\begin{cases}
61\sqrt{n}, & N_0\le n\le10^{13},\\
\frac{21}{1000}n, & n>10^{13}.
\end{cases} \tag*{(111)} T 0 ( n ) < ( 20 23 + 7 log 3 ) n + 12 + { 61 n , 1000 21 n , N 0 ≤ n ≤ 1 0 13 , n > 1 0 13 . (111) Hanson’s estimate also gives
H ( M ) ≤ log 3 { M + ( log 2 M ) M 1 / 3 } < 11 n ( n ≥ N 0 ) . H(M)\le\log3\left\{\sqrt{M}+(\log_2 M)M^{1/3}\right\}<11\sqrt{n}\qquad(n\ge N_0). H ( M ) ≤ log 3 { M + ( log 2 M ) M 1/3 } < 11 n ( n ≥ N 0 ) . For an explicit verification, use M ≤ 12.1 n M\le12.1n M ≤ 12.1 n n − 1 / 6 log ( 12.1 n ) n^{-1/6}\log(12.1n) n − 1/6 log ( 12.1 n ) n ≥ N 0 n\ge N_0 n ≥ N 0
12.1 < 3.5 , 12.1 1 / 3 < 2.3 , N 0 − 1 / 6 < 0.127 , log ( 12.1 N 0 ) < 15 , log 2 > 0.69 , log 3 < 1.1 \sqrt{12.1}<3.5,\qquad12.1^{1/3}<2.3,\qquad N_0^{-1/6}<0.127,\qquad\log(12.1N_0)<15,\qquad\log2>0.69,\qquad\log3<1.1 12.1 < 3.5 , 12. 1 1/3 < 2.3 , N 0 − 1/6 < 0.127 , log ( 12.1 N 0 ) < 15 , log 2 > 0.69 , log 3 < 1.1 give
H ( M ) n < 1.1 { 3.5 + 2.3 ⋅ 0.127 ⋅ 15 / 0.69 } = 10.835 < 11. \frac{H(M)}{\sqrt{n}}<1.1\{3.5+2.3\cdot0.127\cdot15/0.69\}=10.835<11. n H ( M ) < 1.1 { 3.5 + 2.3 ⋅ 0.127 ⋅ 15/0.69 } = 10.835 < 11. Combining this with (B.2) and (B.3), and rounding the first square-root coefficient upward, gives
S n + log g n < ( 23 20 + log 3 7 ) n + 12 + { 95 n , N 0 ≤ n ≤ 10 13 , 21 1000 n + 33 n , n > 10 13 . (112) S_n+\log g_n<\left(\frac{23}{20}+\frac{\log3}{7}\right)n+12+
\begin{cases}
95\sqrt{n}, & N_0\le n\le10^{13},\\
\frac{21}{1000}n+33\sqrt{n}, & n>10^{13}.
\end{cases} \tag*{(112)} S n + log g n < ( 20 23 + 7 log 3 ) n + 12 + { 95 n , 1000 21 n + 33 n , N 0 ≤ n ≤ 1 0 13 , n > 1 0 13 . (112) The contradiction
We first record the lower bound L ∗ − E > 3 / 2 L_* - E>3/2 L ∗ − E > 3/2 3 > 1.73 \sqrt{3}>1.73 3 > 1.73 4 + 3 3 > 9 ⋅ 919 / 900 4+3\sqrt{3}>9\cdot919/900 4 + 3 3 > 9 ⋅ 919/900 log 3 > 1.098 \log3>1.098 log 3 > 1.098 log 2 < 0.7 \log2<0.7 log 2 < 0.7 log ( 1 + t ) > t / ( 1 + t ) \log(1+t)>t/(1+t) log ( 1 + t ) > t / ( 1 + t ) t > 0 t>0 t > 0
L ∗ − E > 5 3 log 3 − 1 2 log 2 + log 919 900 > 5 549 3 500 − 7 20 + 19 919 = 34478 22975 > 3 2 . L_*-E>\frac{5}{3}\log3-\frac{1}{2}\log2+\log\frac{919}{900}>\frac{5\,549}{3\,500}-\frac{7}{20}+\frac{19}{919}=\frac{34478}{22975}>\frac{3}{2}. L ∗ − E > 3 5 log 3 − 2 1 log 2 + log 900 919 > 3 500 5 549 − 20 7 + 919 19 = 22975 34478 > 2 3 . For N 0 ≤ n ≤ 10 13 N_0\le n\le10^{13} N 0 ≤ n ≤ 1 0 13 log ( 12 n ) / n \log(12n)/n log ( 12 n ) / n log ( 12 N 0 ) < 15 \log(12N_0)<15 log ( 12 N 0 ) < 15
23 20 + log 3 7 + 95 n + 12 + 6 log ( 12 n ) n < 23 20 + 11 70 + 19 100 + 102 250000 = 1310357 875000 < 3 2 . \frac{23}{20}+\frac{\log3}{7}+\frac{95}{\sqrt{n}}+\frac{12+6\log(12n)}{n}<\frac{23}{20}+\frac{11}{70}+\frac{19}{100}+\frac{102}{250000}=\frac{1310357}{875000}<\frac{3}{2}. 20 23 + 7 log 3 + n 95 + n 12 + 6 log ( 12 n ) < 20 23 + 70 11 + 100 19 + 250000 102 = 875000 1310357 < 2 3 . For n > 10 13 n>10^{13} n > 1 0 13
33 n + 12 + 6 log ( 12 n ) n < 33 10 6 + 210 10 13 < 1 1000 , \frac{33}{\sqrt{n}}+\frac{12+6\log(12n)}{n}<\frac{33}{10^6}+\frac{210}{10^{13}}<\frac{1}{1000}, n 33 + n 12 + 6 log ( 12 n ) < 1 0 6 33 + 1 0 13 210 < 1000 1 , where log ( 12 ⋅ 10 13 ) < 33 \log(12 \cdot10^{13}) < 33 log ( 12 ⋅ 1 0 13 ) < 33
23 20 + log 3 7 + 21 1000 + 33 n + 12 + 6 log ( 12 n ) n < 23 20 + 11 70 + 22 1000 < 4 3 < 3 2 . \frac{23}{20}+\frac{\log3}{7}+\frac{21}{1000}+\frac{33}{\sqrt{n}}+\frac{12+6\log(12n)}{n}<\frac{23}{20}+\frac{11}{70}+\frac{22}{1000}<\frac{4}{3}<\frac{3}{2}. 20 23 + 7 log 3 + 1000 21 + n 33 + n 12 + 6 log ( 12 n ) < 20 23 + 70 11 + 1000 22 < 3 4 < 2 3 . Both ranges contradict (B.1). This proves irreducibility for every n ≥ 250000 n \ge250000 n ≥ 250000
The finite range
B.2. The finite range. We use a published computation for the small degrees. Cullinan and Hajir [8 Remark 4.6] checked the irreducibility of K ℓ K_\ell K ℓ 13 ≤ ℓ ≤ 10 5 13 \le\ell\le10^5 13 ≤ ℓ ≤ 1 0 5
ℓ = 12 n + 6 + 2 λ + 3 μ , K ℓ ( X ) = 3 n n ! P n ( λ / 3 , μ / 2 ) ( 4 X + 1 ) . \ell=12n+6+2\lambda+3\mu,\qquad K_\ell(X)=3^n n!P_n^{(\lambda/3,\mu/2)}(4X+1). ℓ = 12 n + 6 + 2 λ + 3 μ , K ℓ ( X ) = 3 n n ! P n ( λ /3 , μ /2 ) ( 4 X + 1 ) . Consequently their result covers all four parameter choices for 1 ≤ n ≤ 8332 1 \le n \le8332 1 ≤ n ≤ 8332 12 ⋅ 8332 + 11 = 99995 12\cdot8332+11=99995 12 ⋅ 8332 + 11 = 99995 [8 Theorem 1.8] .
Proposition B.2. For every 1 ≤ n ≤ 250000 1 \le n \le250000 1 ≤ n ≤ 250000 λ , μ ∈ { − 1 , 1 } \lambda,\mu\in\{-1,1\} λ , μ ∈ { − 1 , 1 } P n ( λ / 3 , μ / 2 ) P_n^{(\lambda/3,\mu/2)} P n ( λ /3 , μ /2 ) Q \mathbb{Q} Q
Proof. It remains to treat 8333 ≤ n ≤ 250000 8333 \le n \le250000 8333 ≤ n ≤ 250000 T 0 ( n ) T_0(n) T 0 ( n )
Q ( n ) = ∑ p ≥ 5 prime , h ≥ 2 p h ≤ 12 n + 5 log p . Q(n)=\sum_{\substack{p\ge5\ \text{prime},\ h\ge2\\ p^h\le12n+5}}\log p. Q ( n ) = p ≥ 5 prime , h ≥ 2 p h ≤ 12 n + 5 ∑ log p . The same content and moment calculation gives
S n + log g n ≤ T 0 ( n ) + 3 Q ( n ) . (113) S_n+\log g_n\le T_0(n)+3Q(n). \tag*{(113)} S n + log g n ≤ T 0 ( n ) + 3 Q ( n ) . (113) Write T 2 = T 2 , n T_2=T_{2,n} T 2 = T 2 , n T 3 = T 3 , n T_3=T_{3,n} T 3 = T 3 , n L = L n L=L_n L = L n J = J n J=J_n J = J n C n ≥ 1 / 3 C_n\ge1/3 C n ≥ 1/3 1 ≤ k ≤ n / 2 1\le k\le n/2 1 ≤ k ≤ n /2
log R ( 6 ) k ≥ log κ n − log g n − J + n ( L − T 2 − T 3 − E ) − 4 log 2 \frac{\log R^{(6)}}{k}\ge\log\kappa_n-\log g_n-J+n(L-T_2-T_3-E)-4\log2 k log R ( 6 ) ≥ log κ n − log g n − J + n ( L − T 2 − T 3 − E ) − 4 log 2 + a k + 2 log 2 k − log k , a = T 2 + T 3 + E − 2 L + 2 log 2. {}+ak+\frac{2\log2}{k}-\log k,\qquad a=T_2+T_3+E-2L+2\log2. + ak + k 2 log 2 − log k , a = T 2 + T 3 + E − 2 L + 2 log 2. If ∣ T 2 − T 3 ∣ ≤ 9 / 10 \lvert T_2-T_3\rvert\le9/10 ∣ T 2 − T 3 ∣ ≤ 9/10
a = E − 2 log cosh ( ( T 2 − T 3 ) / 2 ) ≥ E − 81 400 > 1 2 . a=E-2\log\cosh((T_2-T_3)/2)\ge E-\frac{81}{400}>\frac{1}{2}. a = E − 2 log cosh (( T 2 − T 3 ) /2 ) ≥ E − 400 81 > 2 1 . Thus the last three terms in the second line have positive sum, because k / 2 − log k ≥ 1 − log 2 > 0 k/2-\log k\ge1-\log2>0 k /2 − log k ≥ 1 − log 2 > 0 7 / 16 7/16 7/16
L ≥ 7 16 T 2 + 9 16 T 3 + 4 log 2 − 7 16 log 7 − 9 8 log 3. L\ge\frac{7}{16}T_2+\frac{9}{16}T_3+4\log2-\frac{7}{16}\log7-\frac{9}{8}\log3. L ≥ 16 7 T 2 + 16 9 T 3 + 4 log 2 − 16 7 log 7 − 8 9 log 3. Define the integers
c 2 = 216 n + 48 v 2 ( n ! ) − 24 s 2 ( n ) − 27 D 2 , n − 192 , c_2=216n+48v_2(n!)-24s_2(n)-27D_{2,n}-192, c 2 = 216 n + 48 v 2 ( n !) − 24 s 2 ( n ) − 27 D 2 , n − 192 , c 3 = 48 v 3 ( n ! ) − 22 n − 21 D 3 , n − 8 s 3 ( 2 n + δ ) , c_3=48v_3(n!)-22n-21D_{3,n}-8s_3(2n+\delta), c 3 = 48 v 3 ( n !) − 22 n − 21 D 3 , n − 8 s 3 ( 2 n + δ ) , c 7 = − 21 n , δ = μ + 1 2 , c_7=-21n,\qquad\delta=\frac{\mu+1}{2}, c 7 = − 21 n , δ = 2 μ + 1 , and the real number
Δ n = c 2 log 2 + c 3 log 3 + c 7 log 7 48 − T 0 ( n ) − 3 Q ( n ) . \Delta_n=\frac{c_2\log2+c_3\log3+c_7\log7}{48}-T_0(n)-3Q(n). Δ n = 48 c 2 log 2 + c 3 log 3 + c 7 log 7 − T 0 ( n ) − 3 Q ( n ) . Combining the preceding bounds shows that the two conditions
∣ D 2 , n log 2 − D 3 , n log 3 ∣ ≤ 9 10 n , Δ n > 0 (114) \left|D_{2,n}\log2-D_{3,n}\log3\right|\leq\frac{9}{10}n,\qquad\Delta_n>0 \tag*{(114)} ∣ D 2 , n log 2 − D 3 , n log 3 ∣ ≤ 10 9 n , Δ n > 0 (114) exclude every proper factorization.
For each of the four parameter pairs and every integer 8333 ≤ n ≤ 250 000 8333\leq n\leq250\,000 8333 ≤ n ≤ 250 000
Δ n > n 1000 . \Delta_n>\frac{n}{1000}. Δ n > 1000 n . The calculation uses only the residue indicators of Lemma 4.2 , digit sums, and rational bounds for logarithms. For completeness, these bounds follow by truncating
log 1 + t 1 − t = 2 ∑ j = 0 r − 1 t 2 j + 1 2 j + 1 + E r ( t ) , 0 ≤ E r ( t ) ≤ 2 t 2 r + 1 ( 2 r + 1 ) ( 1 − t 2 ) ( 0 ≤ t ≤ 1 / 3 ) . \log\frac{1+t}{1-t}=2\sum_{j=0}^{r-1}\frac{t^{2j+1}}{2j+1}+\mathcal{E}_r(t),\qquad0\leq\mathcal{E}_r(t)\leq\frac{2t^{2r+1}}{(2r+1)(1-t^2)}\qquad(0\leq t\leq1/3). log 1 − t 1 + t = 2 j = 0 ∑ r − 1 2 j + 1 t 2 j + 1 + E r ( t ) , 0 ≤ E r ( t ) ≤ ( 2 r + 1 ) ( 1 − t 2 ) 2 t 2 r + 1 ( 0 ≤ t ≤ 1/3 ) . Rescaling an integer by a power of 2 2 2
Appendix C. The finite first-derivative calculation
We give the remaining finite verification for the first derivatives, and record elementary rational bounds for three logarithmic constants. The coefficient formulas and prime bounds below specify the calculation completely.
A factor-degree criterion
Lemma C.1. Let H ∈ Z [ x ] H\in\mathbb{Z}[x] H ∈ Z [ x ] D D D p p p D D D D p D_p D p Σ p \Sigma_p Σ p
{ 1 , … , ⌊ D / 2 ⌋ } ∩ ⋂ p ∈ P Σ p = ∅ \{1,\ldots,\lfloor D/2\rfloor\}\cap\bigcap_{p\in\mathcal{P}}\Sigma_p=\varnothing { 1 , … , ⌊ D /2 ⌋} ∩ p ∈ P ⋂ Σ p = ∅ for a finite set P \mathcal{P} P H H H Q \mathbb{Q} Q
Proof. By Gauss’s lemma, a nontrivial rational factorization gives an integral factor of degree 1 ≤ k ≤ ⌊ D / 2 ⌋ 1\leq k\leq\lfloor D/2\rfloor 1 ≤ k ≤ ⌊ D /2 ⌋ H H H k ∈ Σ p k\in\Sigma_p k ∈ Σ p p ∈ P p\in\mathcal{P} p ∈ P
The finite ranges
For n = 2 m + δ n=2m+\delta n = 2 m + δ δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } H n H_n H n
∑ k = 0 m ( − 1 ) k ( 2 n + 2 − 2 k ) ! k ! ( n + 1 − k ) ! ( n − 2 k ) ! x n − 2 k − δ . (115) \sum_{k=0}^{m}\frac{(-1)^k(2n+2-2k)!}{k!(n+1-k)!(n-2k)!}x^{n-2k-\delta}. \tag*{(115)} k = 0 ∑ m k ! ( n + 1 − k )! ( n − 2 k )! ( − 1 ) k ( 2 n + 2 − 2 k )! x n − 2 k − δ . (115) Thus H n H_n H n P n + 1 ′ ( x ) / x δ P'_{n+1}(x)/x^\delta P n + 1 ′ ( x ) / x δ 2 m 2m 2 m
U = ∏ p ≤ m p v p ( n + 1 ) , V = ∏ p ≤ m p v p ( n + 2 ) , Z = ∏ p ≤ m p ⌊ log p ( 2 n + 3 ) ⌋ . U=\prod_{p\leq m}p^{v_p(n+1)},\qquad V=\prod_{p\leq m}p^{v_p(n+2)},\qquad Z=\prod_{p\leq m}p^{\lfloor\log_p(2n+3)\rfloor}. U = p ≤ m ∏ p v p ( n + 1 ) , V = p ≤ m ∏ p v p ( n + 2 ) , Z = p ≤ m ∏ p ⌊ l o g p ( 2 n + 3 )⌋ . For m ≥ 4 m\geq4 m ≥ 4 Theorem 3.4 is exactly
16 m V 2 > 64 m ( m + 1 ) 2 Z 2 U 2 . (116) 16^mV^2>64m(m+1)^2Z^2U^2. \tag*{(116)} 1 6 m V 2 > 64 m ( m + 1 ) 2 Z 2 U 2 . (116) Indeed, K = m + 1 K=m+1 K = m + 1 A ≤ m = U / V A_{\leq m}=U/V A ≤ m = U / V
Lemma C.2. For 1 ≤ m ≤ 199 1\leq m\leq199 1 ≤ m ≤ 199 δ ∈ { 0 , 1 } \delta\in\{0,1\} δ ∈ { 0 , 1 } H 2 m + δ H_{2m+\delta} H 2 m + δ Q \mathbb{Q} Q
Proof. For each parity, the indices m < 4 m < 4 m < 4 4 ≤ m ≤ 199 4 \le m \le199 4 ≤ m ≤ 199
E 0 = { 1 , … , 25 } ∪ { 28 , 30 , 31 , 32 , 34 , 37 , 40 , 42 , 43 , 45 , 46 , 47 } , E_0 = \{1,\ldots,25\} \cup\{28,30,31,32,34,37,40,42,43,45,46,47\}, E 0 = { 1 , … , 25 } ∪ { 28 , 30 , 31 , 32 , 34 , 37 , 40 , 42 , 43 , 45 , 46 , 47 } , E 1 = { 1 , … , 26 } ∪ { 29 , 31 , 32 , 33 , 34 , 35 , 37 , 38 , 41 , 43 , 44 , 45 , 47 , 49 , 50 , 53 , 73 } . E_1 = \{1,\ldots,26\} \cup\{29,31,32,33,34,35,37,38,41,43,44,45,47,49,50,53,73\}. E 1 = { 1 , … , 26 } ∪ { 29 , 31 , 32 , 33 , 34 , 35 , 37 , 38 , 41 , 43 , 44 , 45 , 47 , 49 , 50 , 53 , 73 } . The other 318 pairs satisfy the strict integer inequality. For the 37 + 43 = 80 37+43=80 37 + 43 = 80 m ∈ E δ m\in E_\delta m ∈ E δ Lemma C.1 to the full-degree, squarefree reductions at primes p ≤ 317 p\le317 p ≤ 317
Here is a complete procedure for this finite calculation. The coefficients are obtained from (C.1) by dividing by their positive gcd. The integers U , V , Z U,V,Z U , V , Z H n H_n H n { 1 , … , m } \{1,\ldots,m\} { 1 , … , m } gcd ( H ‾ n , H ‾ n ′ ) ≠ 1 \gcd(\overline{H}_n,\overline{H}'_n)\ne1 g cd( H n , H n ′ ) = 1 D p D_p D p
Make f = H ‾ n / lc ( H ‾ n ) f=\overline{H}_n/\operatorname{lc}(\overline{H}_n) f = H n / lc ( H n ) R = f R=f R = f h = x h=x h = x j = 1 , 2 , … j=1,2,\ldots j = 1 , 2 , … 2 j ≤ deg R 2j\le\deg R 2 j ≤ deg R h h h h p m o d R h^p\bmod R h p mod R G = gcd ( R , h − x ) G=\gcd(R,h-x) G = g cd( R , h − x ) j j j deg ( G ) / j \deg(G)/j deg ( G ) / j R R R R / G R/G R / G h h h R R R R R R D p D_p D p x p j − x x^{p^j}-x x p j − x j j j R R R j j j deg R < 2 j \deg R<2j deg R < 2 j
To compute Σ p \Sigma_p Σ p { 0 } \{0\} { 0 } d d d D p D_p D p d d d Σ p \Sigma_p Σ p Lemma C.1 proves the remaining 80 cases and completes the proof. □
Rational logarithm bounds
For 0 < z < 1 0<z<1 0 < z < 1
T ( z ) = 2 ∑ j = 0 4 z 2 j + 1 2 j + 1 , U ∗ ( z ) = T ( z ) + 2 z 11 11 ( 1 − z 2 ) . T(z)=2\sum_{j=0}^{4}\frac{z^{2j+1}}{2j+1},\qquad U_*(z)=T(z)+\frac{2z^{11}}{11(1-z^2)}. T ( z ) = 2 j = 0 ∑ 4 2 j + 1 z 2 j + 1 , U ∗ ( z ) = T ( z ) + 11 ( 1 − z 2 ) 2 z 11 . The power series for the inverse hyperbolic tangent gives
T ( z ) < log 1 + z 1 − z < U ∗ ( z ) . T(z)<\log\frac{1+z}{1-z}<U_*(z). T ( z ) < log 1 − z 1 + z < U ∗ ( z ) . The two margins used in the paper have the exact expressions
8 ( log 4 − 1 ) − g ( 8 ) = 13 log 2 + 9 log ( 9 / 8 ) − 34 log ( 17 / 16 ) − 8 , 8(\log4-1)-g(8)=13\log2+9\log(9/8)-34\log(17/16)-8, 8 ( log 4 − 1 ) − g ( 8 ) = 13 log 2 + 9 log ( 9/8 ) − 34 log ( 17/16 ) − 8 , 8 [ 19 8 ( log 6 − 1 ) − g ( 19 / 8 ) ] = 3 log 2 + 8 log ( 3 / 2 ) + 57 log ( 19 / 16 ) + 92 log ( 24 / 23 ) − 19. 8\left[\frac{19}{8}(\log6-1)-g(19/8)\right]
=3\log2+8\log(3/2)+57\log(19/16)
+92\log(24/23)-19. 8 [ 8 19 ( log 6 − 1 ) − g ( 19/8 ) ] = 3 log 2 + 8 log ( 3/2 ) + 57 log ( 19/16 ) + 92 log ( 24/23 ) − 19. Substitution of the rational bounds gives
13 T ( 1 / 3 ) + 9 T ( 1 / 17 ) − 34 U ∗ ( 1 / 33 ) − 8 > 97 10000 13T(1/3)+9T(1/17)-34U_*(1/33)-8>\frac{97}{10000} 13 T ( 1/3 ) + 9 T ( 1/17 ) − 34 U ∗ ( 1/33 ) − 8 > 10000 97 and
3 T ( 1 / 3 ) + 8 T ( 1 / 5 ) + 57 T ( 3 / 35 ) + 92 T ( 1 / 47 ) − 19 > 42 1250 . 3T(1/3)+8T(1/5)+57T(3/35)+92T(1/47)-19>\frac{42}{1250}. 3 T ( 1/3 ) + 8 T ( 1/5 ) + 57 T ( 3/35 ) + 92 T ( 1/47 ) − 19 > 1250 42 . Both are direct comparisons of rational numbers; the second right side is 8 ⋅ 21 / 5000 8\cdot21/5000 8 ⋅ 21/5000
2 ( log 9 − 1 ) − g ( 2 ) = log ( 3 72 2 16 5 10 ) − 2 > 0 , 2(\log9-1)-g(2)=\log\left(\frac{3^{72}2^{16}}{5^{10}}\right)-2>0, 2 ( log 9 − 1 ) − g ( 2 ) = log ( 5 10 3 72 2 16 ) − 2 > 0 , because e < 3 e<3 e < 3 3 5 2 16 > 5 10 3^52^{16}>5^{10} 3 5 2 16 > 5 10
