Galois groups of Mori trinomials and hyperelliptic curves with big monodromy

Abstract

We compute the Galois groups for a certain class of polynomials over the the field of rational numbers that was introduced by Shigefumi Mori and study the monodromy of corresponding hyperelliptic jacobians.

1 Mori polynomials, their reductions and Galois groups

We write ${\mathbb Z},{\mathbb Q}$ and ${\mathbb C}$ for the ring of integers, the field of rational numbers and the field of complex numbers respectively. If $a$ and $b$ are nonzero integers then we write $(a,b)$ for its (positive) greatest common divisor. If $\ell$ is a prime then ${\mathbb F}_{\ell },{\mathbb Z}_{\ell }$ and ${\mathbb Q}_{\ell }$ stand for the prime finite field of characteristic $\ell$, the ring of ${\ell }$-adic integers and the field of ${\ell }$-adic numbers respectively.

We consider the subring ${\mathbb Z}[1/2] \subset {\mathbb Q}$ generated by $1/2$ over ${\mathbb Z}$. We have

$$\begin{aligned} {\mathbb Z}\subset {\mathbb Z}\biggl [\frac{1}{2}\biggr ] \subset {\mathbb Q}. \end{aligned}$$

If $\ell$ is an odd prime then the principal ideal $\ell {\mathbb Z}[1/2]$ is maximal in ${\mathbb Z}[1/2]$ and

$$\begin{aligned} {\mathbb Z}\biggl [\frac{1}{2}\biggr ]\bigg /\ell {\mathbb Z}\biggl [\frac{1}{2}\biggr ] ={\mathbb Z}/\ell {\mathbb Z}={\mathbb F}_{\ell }. \end{aligned}$$

If $K$ is a field then we write $\overline{K}$ for its algebraic closure and denote by $\mathrm {Gal}(K)$ its absolute Galois group $\mathrm{Aut}(\overline{K}/K)$. If $u(x)\in K[x]$ is a degree $n$ polynomial with coefficients in $K$ and without multiple roots then we write ${\mathfrak R}_u\subset \overline{K}$ for the $n$-element set of its roots, $K({\mathfrak R}_u)$ the splitting field of $u(x)$ and $\mathrm {Gal}(u/K)=\mathrm {Gal}(K({\mathfrak R}_u)/K)$ the Galois group of $u(x)$ viewed as a certain subgroup of the group $\mathrm {Perm}({\mathfrak R}_u) \cong {\mathbf S}_n$ of permutations of ${\mathfrak R}_u$. As usual, we write $\mathbf {A}_n$ for the alternating group, which is the only index $2$ subgroup in the full symmetric group ${\mathbf S}_n$.

1.1 Discriminants and alternating groups We write ${\Delta }(u)$ for the discriminant of $u$. We have

$$\begin{aligned} 0 \ne {\Delta }(u) \in K, \qquad \sqrt{{\Delta }(u)} \in K({\mathfrak R}_u). \end{aligned}$$

It is well known that

$$\begin{aligned} \mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )=\mathrm {Gal}(K({\mathfrak R}_u)/K)\cap \mathbf {A}_n \subset \mathbf {A}_n\subset {\mathbf S}_n=\mathrm {Perm}({\mathfrak R}_u). \end{aligned}$$

In particular, the permutation (sub)group $\mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )$ does not contain transpositions; ${\Delta }(u)$ is a square in $K$ if and only if $\mathrm {Gal}(u/K)$ lies in the alternating (sub)group $\mathbf {A}_n\subset {\mathbf S}_n$. On the other hand, if $\mathrm {Gal}(u/K)={\mathbf S}_n$ then $\mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )= \mathbf {A}_n$.

If $n$ is odd and $\mathrm {char}(K)\ne 2$ then we write $C_u$ for the genus $(n-1)/2$ hyperelliptic curve

$$\begin{aligned} C_u:y^2=u(x) \end{aligned}$$

and $J(C_u)$ for its jacobian, which is an $(n-1)/2$-dimensional abelian variety over $K$. We write $\mathrm{End}(J(C_u))$ for the ring of all $\overline{K}$-endomorphisms of $J(C_u)$ and $\mathrm {End}_K(J(C_u))$ for the (sub)ring of all its $K$-endomorphisms. We have

$$\begin{aligned} {\mathbb Z}\subset \mathrm {End}_K(J(C_u))\subset \mathrm{End}(J(C_u)). \end{aligned}$$

About 40 years ago Shigefumi Mori [8], Proposition 3, p. 107] observed that if $n=2g+1$ is odd and $\mathrm {Gal}(f/K)$ is a doubly transitive permutation group then $\mathrm {End}_K(J(C_u))={\mathbb Z}$. He constructed [8], Theorem 1, p. 105] explicit examples (in all dimensions $g$) of polynomials (actually, trinomials) $f(x)$ over ${\mathbb Q}$ such that $\mathrm {Gal}(f/{\mathbb Q})$ is doubly transitive and $\mathrm{End}(J(C_f))={\mathbb Z}$.

On the other hand, about 15 years ago the following assertion was proven by the author [17].

Theorem 1.2

Suppose that $\mathrm {char}(K)=0$ and $\mathrm {Gal}(u/K)={\mathbf S}_n$. Then $\mathrm{End}(J(C_u))={\mathbb Z}$.

The aim of this note is to prove that in Mori’s examples $\mathrm {Gal}(f/{\mathbb Q})={\mathbf S}_{2g+1}$. This gives another proof of the theorem of Mori [8], Theorem 1, p. 105]. Actually, we extend the class of Mori trinomials with $\mathrm{End}(J(C_f))={\mathbb Z}$, by dropping one of the congruence conditions imposed by Mori on the coefficients of $f(x)$. We also prove that the images of $\mathrm {Gal}({\mathbb Q})$ in the automorphism groups of Tate modules of $J(C_f)$ are almost as large as possible.

1.3 Mori trinomials  Throughout this paper, $g,p,b,c$ are integers that enjoy the following properties [8]:

• (i) The number $g$ is a positive integer and $p$ is an odd prime. In addition, there is a positive integer N such that $(p-1)^N/2^N$ is divisible by g. This means that every prime divisor of $g$ is also a divisor of $(p-1)/2$. This implies that $(p,g)=(p,2g)=1$. It follows that if $g$ is even then $p$ is congruent to $1$ modulo $4$.

• (ii) The residue $b~\mathrm{mod }~p$ is a primitive root of ${\mathbb F}_{p}={\mathbb Z}/p{\mathbb Z}$; in particular, $(b,p)=1$.

• (iii) The integer c is odd and $(b,c)=(b,2g{+}1)=(c,g)=1$. This implies that $(c,2g)=1$.

Mori [8] introduced and studied the monic degree $2g+1$ polynomial

$$\begin{aligned} f(x)=f_{g,p,b,c}(x)=x^{2g+1}-bx-\frac{pc}{4}\in {\mathbb Z}\biggl [\frac{1}{2}\biggr ][x]\subset {\mathbb Q}[x], \end{aligned}$$

which we call a Mori trinomial. He proved the following results [8], pp. 106–107].

Theorem 1.4

(Theorem of Mori) Let $f(x)=f_{g,p,b,c}(x)$ be a Mori trinomial. Then:

• (i) The polynomial $f(x)$ is irreducible over ${\mathbb Q}_2$ and therefore over ${\mathbb Q}$.

• (ii) The polynomial $f(x)~\mathrm{mod }~p \in {\mathbb F}_{p}[x]$ is a product $x (x^{2g}-b)$ of a linear factor $x$ and an irreducible $($over ${\mathbb F}_{p})$ degree $2g$ polynomial $x^{2g}-b$.

• (iii) Let $\mathrm {Gal}(f)$ be the Galois group of $f(x)$ over ${\mathbb Q}$ considered canonically as a $($transitive$)$ subgroup of the full symmetric group ${\mathbf S}_{2g+1}$. Then $\mathrm {Gal}(f)$ is a doubly transitive permutation group. More precisely, the transitive $\mathrm {Gal}(f)$ contains a permutation $\sigma$ that is a cycle of length $2g$.

• (iv) For each odd prime $\ell$ every root of the polynomial $f(x)~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ is either simple or double.

• (v) Let us consider the genus $g$ hyperelliptic curve

  $$\begin{aligned} C_f:y^2=f(x) \end{aligned}$$

  and its jacobian $J(C_f)$, which is a $g$-dimensional abelian variety over ${\mathbb Q}$. Assume additionally that $c$ is congruent to $-p$ modulo $4$. Then $C_f$ is a stable curve over ${\mathbb Z}$ and $J(C_f)$ has everywhere semistable reduction over ${\mathbb Z}$. In addition, $\mathrm{End}(J(C_f))={\mathbb Z}$.

Remark 1.5

• (I) The $2$-adic Newton polygon of Mori trinomial $f(x)$ consists of one segment that connects $(0,-2)$ and $(2g{+}1,0)$, which are its only integer points. Now the irreducibility of $f(x)$ follows from Eisenstein–Dumas Criterion [9], Corollary 3.6, p. 316], [4], p. 502]. It also follows that the field extension ${\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}$ is ramified at $2$.

• (II) If $g=1$ then $2g+1=3$ and the only doubly transitive subgroup of ${\mathbf S}_{3}$ is ${\mathbf S}_{3}$ itself. Concerning the double transitivity of the Galois group of trinomials of arbitrary degree, see [2], Theorem 4.2, p. 9 and Note 2, p. 10].

• (III) The additional congruence condition in Theorem 1.4 (v) guarantees that $C_f$ has stable (even good) reduction at $2$ [8], p. 106]. Mori’s proof of the last assertion of Theorem 1.4 (v) is based on results of [12] and the equality $\mathrm {End}_{{\mathbb Q}}(J(C_f))={\mathbb Z}$; the latter follows from the double transitivity of Galois groups of Mori trinomials.

Remark 1.6

Since a cycle of even length $2g$ is an odd permutation, it follows from Theorem 1.4 (iii) that $\mathrm {Gal}(f)$ is not contained in $\mathbf {A}_{2g+1}$. In other words, ${\Delta }(f)$ is not a square in ${\mathbb Q}$.

Our first main result is the following statement.

Theorem 1.7

Let $f(x)=f_{g,p,b,c}(x)$ be a Mori trinomial.

• (i) If $\ell$ is an odd prime then the polynomial $f(x)~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ has, at most, one double root and this root $($if exists$)$ lies in ${\mathbb F}_{\ell }$.

• (ii) There exists an odd prime $\ell \ne p$ such that $f(x)~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ has a double root $\overline{\alpha }\in {\mathbb F}_{\ell }$. All other roots of $f(x)~\mathrm{mod }~\ell$ $($in an algebraic closure of ${\mathbb F}_{\ell })$ are simple.

• (iii) The Galois group $\mathrm {Gal}(f)$ of $f(x)$ over ${\mathbb Q}$ coincides with the full symmetric group ${\mathbf S}_{2g+1}$. The Galois $($sub$)$group $\mathrm {Gal}\bigl ({\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )\bigr )$ coincides with the alternating group $\mathbf {A}_{2g+1}$.

• (iii’) The Galois extension ${\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )$ is ramified at all prime divisors of $2$. It is unramified at all prime divisors of every odd prime $\ell$.

• (iv) Suppose that $g>1$. Then $\mathrm{End}(J(C_f))={\mathbb Z}$.

Remark 1.8

Theorem 1.7 (iv) was proven by Mori under an additional assumption that $c$ is congruent to $-p$ modulo $4$, see Theorem 1.4 (v) above.

Remark 1.9

Thanks to Theorem 1.2, Theorem 1.7 (iv) follows readily from Theorem 1.7 (iii).

Remark 1.10

Let $g>1$ and suppose we know that $\mathrm {Gal}(f)$ contains a transposition. Now the double transitivity implies that $\mathrm {Gal}(f)$ coincides with ${\mathbf S}_{2g+1}$, see [15], Lemma 4.4.3, p. 40].

Let $K$ be a field of characteristic zero and $u(x) \in K[x]$ be a degree $2g+1$ polynomial without multiple roots. Then the jacobian $J(C_u)$ is a $g$-dimensional abelian variety over $K$. For every prime $\ell$ let $T_{\ell }(J(C_u))$ be the $\ell$-adic Tate module of $J(C_u)$, which is a free ${\mathbb Z}_{\ell }$-module of rank $2g$ provided with the canonical continuous action

$$\begin{aligned} \rho _{\ell ,u}:\mathrm {Gal}(K) \rightarrow \mathrm {Aut}_{{\mathbb Z}_{\ell }}\bigl (T_{\ell }(J(C_u))\bigr ) \end{aligned}$$

of $\mathrm {Gal}(K)$ [10, 14, 20]. There is a Riemann form

$$\begin{aligned} e_{\ell }:T_{\ell }(J(C_u)){\times } T_{\ell }(J(C_u)) \rightarrow {\mathbb Z}_{\ell } \end{aligned}$$

that corresponds to the canonical principal polarization on $J(C_u)$ ([10], Section 20], [21], Section 1]) and is a nondegenerate (even perfect) alternating ${\mathbb Z}_{\ell }$-bilinear form that satisfies

$$\begin{aligned} e_{\ell }(\sigma (x),\sigma (y))= \chi _{\ell }(\sigma )e_{\ell }(\sigma (x),\sigma (y)). \end{aligned}$$

This implies that the image

$$\begin{aligned} \rho _{\ell ,u}(\mathrm {Gal}(K))\subset \mathrm {Aut}_{{\mathbb Z}_{\ell }}\bigl (T_{\ell }(J(C_u))\bigr ) \end{aligned}$$

lies in the (sub)group

$$\begin{aligned} \mathrm {Gp}\bigl (T_{\ell }(J(C_u)),e_{\ell }\bigr ) \subset \mathrm {Aut}_{{\mathbb Z}_{\ell }}\bigl (T_{\ell }(J(C_u))\bigr ) \end{aligned}$$

of symplectic similitudes of $e_{\ell }$ [18, 19, 21].

Using results of Chris Hall [5] and the author [21], we deduce from Theorem 1.7 the following statement. (Compare it with [18], Theorem 2.5] and [19], Theorem 8.3].)

Theorem 1.11

Let $K={\mathbb Q}$ and $f(x)=f_{g,p,b,c}(x)\in {\mathbb Q}[x]$ be a Mori trinomial. Suppose that $g>1$. Then:

• (i) For all primes $\ell$ the image $\rho _{\ell ,f}(\mathrm {Gal}({\mathbb Q}))$ is an open subgroup of finite index in $\mathrm {Gp}\bigl (T_{\ell }(J(C_f)),e_{\ell }\bigr )$.

• (ii) Let $L$ be a number field and $\mathrm {Gal}(L)$ be its absolute Galois group, which we view as an open subgroup of finite index in $\mathrm {Gal}({\mathbb Q})$. Then for all but finitely many primes $\ell$ the image $\rho _{\ell ,f}(\mathrm {Gal}(L))$ coincides with $\mathrm {Gp}\bigl (T_{\ell }(J(C_f)),e_{\ell }\bigr )$.

The paper is organized as follows. In Sect. 2 we deduce Theorem 1.11 from Theorem 1.7. In Sect. 3 we discuss a certain class of trinomials that is related to Mori polynomials. Section 4 deals with discriminants of Mori polynomials. We prove Theorem 1.7 in Sect. 5.

2 Monodromy of hyperelliptic jacobians

Proof

of Theorem 1.11 (modulo Theorem 1.7) By Theorem 1.7 (iii), $\mathrm {Gal}(f/{\mathbb Q})$ coincides with the full symmetric group ${\mathbf S}_{2g+1}$. By Theorem 1.7 (iv), $\mathrm{End}(J(C_f))={\mathbb Z}$. It follows from Theorem 1.7 (i) that there is an odd prime $\ell$ such that $J(C_f)$ has at $\ell$ a semistable reduction with toric dimension $1$ [5]. Now the assertion (i) follows from [21], Theorem 4.3]. The assertion (ii) follows from [5], Theorem 1].$\square$

3 Reduction of certain trinomials

In order to prove Theorem 1.7 (i), we will use the following elementary statement that was inspired by [15], Remark 2, p. 42] and [8], p. 106].

Lemma 3.1

(key lemma) Let

$$\begin{aligned} u(x)=u_{n,B,C}(x)=x^n+Bx+C \in {\mathbb Z}[x] \end{aligned}$$

be a monic polynomial of degree $n>1$ such that $B\ne 0$ and $C\ne 0$.

• (I) If $u(x)$ has a multiple root then $n$ divides $B$ and $n-1$ divides $C$.

• (II) Let $\ell$ be a prime that enjoys the following properties:

  • (i) $(B,C)$ is not divisible by $\ell$,

  • (ii) $(n,B)$ is not divisible by $\ell$,

  • (iii) $(n-1,C)$ is not divisible by $\ell$.

  Suppose that $u(x)$ has no multiple roots. Let us consider the polynomial

  $$\begin{aligned} \overline{u}(x)=u(x) ~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]. \end{aligned}$$

  Then:

  • (a) $\overline{u}(x)$ has, at most, one multiple root in an algebraic closure of ${\mathbb F}_{\ell }$.

  • (b) If such a multiple root say, $\gamma$, does exist, then $\ell$ does not divide $n(n{-}1)BC$ and $\gamma$ is a double root of $\overline{u}(x)$. In addition, $\gamma$ is a nonzero element of ${\mathbb F}_{\ell }$.

  • (c) If such a multiple root does exist then either the field extension ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}$ is unramified at $\ell$ or a corresponding inertia subgroup at $\ell$ in

    $$\begin{aligned} \mathrm {Gal}({\mathbb Q}({\mathfrak R}_u)/{\mathbb Q})=\mathrm {Gal}(u/{\mathbb Q}) \subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

    is generated by a transposition. In both cases the Galois extension ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at all prime divisors of $\ell$.

Remark 3.2

The discriminant $\mathrm {Discr}(n,B,C)={\Delta }(u_{n,B,C})$ of $u_{n,B,C}(x)$ is given by the formula [3], Example 834]

$$\begin{aligned} \mathrm {Discr}(n,B,C)=(-1)^{n(n-1)/2} n^n C^{n-1}\!+ (-1)^{(n-1)(n-2)/2} (n-1)^{n-1} B^n\!. \end{aligned}$$

Remark 3.3

In the notation of Lemma 3.1, assume that $\overline{u}(x)$ has no multiple roots, i.e., ${\Delta }(u)$ is not divisible by $\ell$. Then obviously ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}$ is unramified at $\ell$. This implies that ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at all prime divisors of $\ell$.

Proof

of Lemma 3.1 (I)   Since $u(x)$ has a multiple root, its discriminant

$$\begin{aligned} {\Delta }(u)=(-1)^{n(n-1)/2} n^n C^{n-1}\!+ (-1)^{(n-1)(n-2)/2} (n-1)^{n-1} B^n=0. \end{aligned}$$

This implies that

$$\begin{aligned} n^n C^{n-1}= \pm (n-1)^{n-1} B^n\!. \end{aligned}$$

Since $n$ and $n-1$ are relatively prime, $n^n\,{\mid }\, B^n$ and $(n-1)^{n-1} \,{\mid }\, C^{n-1}$. This implies that $n\,{\mid }\, B$ and $(n-1)\,{\mid }\, C$.

(II)   We have

$$\begin{aligned} \overline{u}(x)=x^n+\overline{B}x+\overline{C} \in {\mathbb F}_{\ell }[x] \end{aligned}$$

where

$$\begin{aligned} \overline{B}=B~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }, \qquad \overline{C}=C~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }. \end{aligned}$$

The condition (i) implies that either $\overline{B}\ne 0$ or $\overline{C} \ne 0$. The condition (ii) implies that if $\overline{B}= 0$ then $n \ne 0$ in ${\mathbb F}_{\ell }$. The condition (iii) implies that if $n-1=0$ in ${\mathbb F}_{\ell }$ then $\overline{C} \ne 0$ and $n \ne 0$ in ${\mathbb F}_{\ell }$. We have

$$\begin{aligned} {\Delta }(\overline{u})=(-1)^{n(n-1)/2} n^n \overline{C}^{\,n-1}\!+ (-1)^{(n-1)(n-2)/2} (n-1)^{n-1} \overline{B}^{\,n}=0 \end{aligned}$$

and therefore

$$\begin{aligned} n^n \overline{C}^{\,n-1}\!= \pm (n-1)^{n-1} \overline{B}^{\,n}\!. \end{aligned}$$

(1)

This implies that if $n-1=0$ in ${\mathbb F}_{\ell }$ then $\overline{C}=0$, which is not the case. This proves that $n-1\ne 0$ in ${\mathbb F}_{\ell }$. On the other hand, if $\overline{B}= 0$ then $\overline{C} \ne 0$ and $n \ne 0$ in ${\mathbb F}_{\ell }$. Then (1) implies that $\overline{C} = 0$ and we get a contradiction that proves that $\overline{B}\ne 0$. If $n=0$ in ${\mathbb F}_{\ell }$ then $n-1 \ne 0$ in ${\mathbb F}_{\ell }$ and (1) implies that $\overline{B}= 0$, which is not the case. The obtained contradiction proves that $n \ne 0$ in ${\mathbb F}_{\ell }$. If $\overline{C} = 0$ then (1) implies that $\overline{B}= 0$, which is not the case. This proves that $\ell$ does not divide $n(n-1)BC$.

The derivative of $\overline{u}(x)$ is $\overline{u}'(x)=n x^{n-1}+\overline{B}$. We have

$$\begin{aligned} x {\cdot } \overline{u}'(x)-n{\cdot } \overline{u}(x)=-(n-1)\overline{B}x-n\overline{C}. \end{aligned}$$

(2)

Suppose $\overline{u}(x)$ has a multiple root $\gamma$ in an algebraic closure of ${\mathbb F}_{\ell }$. Then

$$\begin{aligned} \overline{u}(\gamma )=0, \qquad \overline{u}'(\gamma )=0, \qquad n \cdot \gamma \cdot \overline{u}'(\gamma )-n{\cdot }\overline{u}(\gamma )=0. \end{aligned}$$

Using (2), we conclude that

$$\begin{aligned} 0=\gamma {\cdot }\overline{u}'(\gamma )-n{\cdot }\overline{u}(\gamma ) = -(n-1)\overline{B}\gamma -n\overline{C}, \qquad \gamma =-\frac{n\overline{C}}{(n-1)\overline{B}}\in {\mathbb F}_{\ell }. \end{aligned}$$

This implies that $\gamma \ne 0$.

Notice that the second derivative $\overline{u}''(x)=n(n-1)x^{n-2}$. This implies that

$$\begin{aligned} \overline{u}''(\gamma )=n(n-1)\gamma ^{n-2} \ne 0. \end{aligned}$$

It follows that $\gamma$ is a double root of $\overline{u}(x)$. This ends the proof of (a) and (b).

In order to prove (c), notice that there exists a monic degree $n-2$ polynomial $\overline{h}(x)\in {\mathbb F}_{\ell }[x]$ such that

$$\begin{aligned} \overline{u}(x)=(x-\gamma )^2 {\cdot } \overline{h}(x). \end{aligned}$$

Clearly, $\gamma$ is not a root of $\overline{h}(x)$ and therefore $\overline{h}(x)$ has no multiple roots and is relatively prime to $(x-\gamma )^2$.¹ By Hensel’s Lemma, there exist monic polynomials

$$\begin{aligned} h(x), v(x) \in {\mathbb Z}_{\ell }[x], \qquad \deg h=n-2,\quad \deg v=2 \end{aligned}$$

such that

$$\begin{aligned} u(x)=v(x) h(x) \end{aligned}$$

and

$$\begin{aligned} \overline{h}(x)=h(x)~\mathrm{mod }~\ell , \qquad (x-\gamma )^2=v(x) ~\mathrm{mod }~\ell . \end{aligned}$$

This implies that the splitting field ${\mathbb Q}_{\ell }({\mathfrak R}_h)$ of $h(x)$ (over ${\mathbb Q}_{\ell }$) is an unramified extension of ${\mathbb Q}_{\ell }$ while the splitting field ${\mathbb Q}_{\ell }({\mathfrak R}_u)$ of $u(x)$ (over ${\mathbb Q}_{\ell }$) is obtained from ${\mathbb Q}_{\ell }({\mathfrak R}_h)$ by adjoining to it two (distinct) roots say, $\alpha _1$ and $\alpha _2$ of quadratic $v(x)$. Clearly, ${\mathbb Q}_{\ell }({\mathfrak R}_u)$ either coincides with ${\mathbb Q}_{\ell }({\mathfrak R}_h)$ or with a certain quadratic extension of ${\mathbb Q}_{\ell }({\mathfrak R}_h)$, ramified or unramified. It follows that the inertia subgroup $I$ of

$$\begin{aligned} \mathrm {Gal}({\mathbb Q}_{\ell }({\mathfrak R}_u)/{\mathbb Q}_{\ell })\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

is either trivial or is generated by the transposition that permutes $\alpha _1$ and $\alpha _2$ (and leaves invariant every root of $h(x)$). In the former case ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}$ is unramified at $\ell$ while in the latter one an inertia subgroup in

$$\begin{aligned} \mathrm {Gal}({\mathbb Q}({\mathfrak R}_u)/{\mathbb Q})\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

that corresponds to $\ell$ is generated by a transposition. However, the permutation subgroup $\mathrm {Gal}\bigl ({\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )$ does not contain transpositions (see 1.1). This implies that ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at all prime divisors of $\ell$.$\square$

Example 3.4

Let us consider the polynomial

$$\begin{aligned} u(x)=u_{n,-1,-1}(x)=x^n-x-1\in {\mathbb Q}[x] \end{aligned}$$

over the field $K={\mathbb Q}$. Here $B=C=-1$ and the conditions of Lemma 3.1 hold for all primes $\ell$. It is known that $u(x)$ is irreducible [13], its Galois group over ${\mathbb Q}$ is ${\mathbf S}_n$ [11], Corollary 3, p. 233] and there exists a prime $\ell$ such that $u(x)~\mathrm{mod }~\ell$ acquires a multiple root [15], Remark 2, p. 42]. Clearly, the discriminant ${\Delta }(u)=\mathrm {Discr}(n,-1,-1)$ of $u(x)$ is an odd integer and therefore such an $\ell$ is odd. It follows from Lemma 3.1 that $u(x)~\mathrm{mod }~\ell$ has exactly one multiple root and its multiplicity is $2$.

Let $n=2g+1$ be an odd integer greater than or equal to $5$ and

$$\begin{aligned} u(x)=u_{2g+1,-1,-1}(x)=x^{2g+1}-x-1\in {\mathbb Q}[x]. \end{aligned}$$

Let us consider the $g$-dimensional jacobian $J(C_u)$ of the hyperelliptic curve $C_u:y^2=x^{2g+1}-x-1$. Since $\mathrm {Gal}(u/{\mathbb Q})={\mathbf S}_{2g+1}$, Theorem 1.2 tells us that $\mathrm{End}(J(C_u))={\mathbb Z}$. Now the same arguments as in Sect. 2 prove that

• (i) For all primes $\ell$ the image

  $$\begin{aligned} \rho _{\ell ,u}(\mathrm {Gal}({\mathbb Q}))\subset \mathrm {Gp}\bigl (T_{\ell }(J(C_u)),e_{\ell }\bigr ) \end{aligned}$$

  is an open subgroup of finite index in $\mathrm {Gp}\bigl (T_{\ell }(J(C_u)),e_{\ell }\bigr )$.

• (ii) Let $L$ be a number field and $\mathrm {Gal}(L)$ be its absolute Galois group, which we view as an open subgroup of finite index in $\mathrm {Gal}({\mathbb Q})$. Then for all but finitely many primes $\ell$ the image

  $$\begin{aligned} \rho _{\ell ,u}(\mathrm {Gal}(L))\subset \mathrm {Gp}\bigl (T_{\ell }(J(C_u)),e_{\ell }\bigr ) \end{aligned}$$

  coincides with $\mathrm {Gp}\bigl (T_{\ell }(J(C_u)),e_{\ell }\bigr )$.

Corollary 3.5

(Corollary to Lemma 3.1) Let

$$\begin{aligned} u(x)=u_{n,B,C}(x)=x^n+Bx+C \in {\mathbb Z}[x] \end{aligned}$$

be a monic polynomial of degree $n>1$ without multiple roots such that both $B$ and $C$ are nonzero integers that enjoy the following properties:

• $(B,C)$ is either $1$ or a power of $2$,

• $(n,B)$ is either $1$ or a power of $2$,

• $(n{-}1,C)$ is either $1$ or a power of $2$.

Suppose that the discriminant $D=\mathrm {Discr}(n,B,C)= 2^{2M}{\cdot } D_0$ where $M$ is a nonnegative integer and $D_0$ is an integer such that $D_0\equiv 1 ~\mathrm{mod }~4$. Assume also that $D$ is not a square. Then:

• (a) The quadratic extension ${\mathbb Q}\bigl (\sqrt{D}\bigr )/{\mathbb Q}$ is unramified at $2$. For all odd primes $\ell$ the Galois extension ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{D}\bigr )$ is unramified at every prime divisor of $\ell$.

• (b) There exists an odd prime $\ell$ that enjoys the following properties:

  • (i) $\ell$ divides $D_0$ and $u(x)~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ has exactly one multiple root and its multiplicity is $2$. In addition, this root lies in ${\mathbb F}_{\ell }$.

  • (ii) The field extension ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}$ is ramified at $\ell$ and the Galois group

    $$\begin{aligned} \mathrm {Gal}({\mathbb Q}({\mathfrak R}_u)/{\mathbb Q})=\mathrm {Gal}(u/{\mathbb Q})\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

    contains a transposition. In particular, if $\mathrm {Gal}(u/{\mathbb Q})$ is doubly transitive then

    $$\begin{aligned} \mathrm {Gal}(u/{\mathbb Q})=\mathrm {Perm}({\mathfrak R}_f)\cong {\mathbf S}_n \end{aligned}$$

    and

    $$\begin{aligned} \mathrm {Gal}\bigl ({\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}\bigl (\sqrt{D} \bigr )\bigr )=\mathbf {A}_n. \end{aligned}$$

Proof

Clearly, $D_0$ is not a square and

$$\begin{aligned} {\mathbb Q}\bigl (\sqrt{D}\bigr ) ={\mathbb Q}\bigl (\sqrt{D_0}\bigr ) \end{aligned}$$

is a quadratic field. Since $D_0$ is congruent to $1$ modulo $4$, the quadratic extension ${\mathbb Q}\bigl (\sqrt{D_0}\bigr )/{\mathbb Q}$ is unramified at $2$, which proves the first assertion of (a). The conditions of Lemma 3.1 (II) hold for all odd primes $\ell$. Now the second assertion of (a) follows from Remark 3.3 and Lemma 3.1 (II)(c).

Let us start to prove (b). There are nonzero integers $S$ and $S_0$ such that $D_0=S^2 S_0$ and $S_0$ is square-free. Clearly, both $S$ and $S_0$ are odd. Since

$$\begin{aligned} D=2^{2M}{\cdot } D_0=2^{2M}{\cdot } S^2 S_0=\bigl (2^M S\bigr )^2 S_0 \end{aligned}$$

is not a square, $S_0 \ne 1$. Since $S$ is odd, $S^2\equiv 1 ~\mathrm{mod }~4$. Since $D_0\equiv 1 ~\mathrm{mod }~4$, we obtain that $S_0\equiv 1 ~\mathrm{mod }~4$. It follows that $S_0 \ne -1$. We already know that $S_0 \ne 1$. This implies that there is a prime $\ell$ that divides $S_0$. Since $S_0$ is odd and square-free, $\ell$ is also odd and $\ell ^2$ does not divide $S_0$. Let $T$ be the nonnegative integer such that $\ell ^T{\mid \mid }\, S$. Then $\ell ^{2T+1}{\mid \mid }\, 2^{2M}S^2 S_0$, and therefore $\ell ^{2T+1}{\mid \mid }\, D$. This implies that the quadratic field extension ${\mathbb Q}\bigl (\sqrt{D}\bigr )/{\mathbb Q}$ is ramified at $\ell$. Since

$$\begin{aligned} {\mathbb Q}\subset {\mathbb Q}\bigl (\sqrt{D}\bigr )\subset {\mathbb Q}({\mathfrak R}_u), \end{aligned}$$

the field extension ${\mathbb Q}({\mathfrak R}_u)/{\mathbb Q}$ is also ramified at $\ell$. Since $\ell \,{\mid }\, D$, the polynomial $u(x) ~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ has a multiple root. Now the result follows from Lemma 3.1 combined with Remark 1.10. $\square$

4 Discriminants of Mori trinomials

Let

$$\begin{aligned} f(x)=f_{g,p,b,c}(x)=x^{2g+1}-bx-\frac{pc}{4} \end{aligned}$$

be a Mori trinomial. Following Mori [8], let us consider the polynomial

$$\begin{aligned} {\mathbf u}(x)=2^{2g+1} f\biggl (\frac{x}{2}\biggr )=x^{2g+1}-2^{2g}b x-2^{2g-1}pc=u_{n,B,C}(x) \in {\mathbb Z}[x]\subset {\mathbb Q}[x] \end{aligned}$$

with $n=2g+1$, $B=-2^{2g}b$, $C= -2^{2g-1}pc$.

Remark 4.1

• Clearly, $f(x)$ and ${\mathbf u}(x)$ have the same splitting field and Galois group. It is also clear that

  $$\begin{aligned} {\Delta }({\mathbf u})=2^{(2g+1)2g}{\cdot } {\Delta }(f)=\bigl [2^{(2g+1)g}\bigr ]^{2} {\cdot } {\Delta }(f). \end{aligned}$$

  In particular, ${\Delta }({\mathbf u})$ is not a square, thanks to Remark 1.6.

• By Theorem 1.4 (i, iii), the polynomial $f(x)$ is irreducible over ${\mathbb Q}$ and its Galois group is doubly transitive. This implies that ${\mathbf u}(x)$ is irreducible over ${\mathbb Q}$ and its Galois group over ${\mathbb Q}$ is also doubly transitive. $($See also Theorem 6.6 (i, ii) below.$)$

• For all $g$ the hyperelliptic curves $C_f$ and $C_{{\mathbf u}}$ are biregularly isomorphic over ${\mathbb Q}\bigl (\sqrt{2}\bigr )$. It follows that the jacobians $J(C_{{\mathbf u}})$ and $J(C_f)$ are also isomorphic over ${\mathbb Q}\bigl (\sqrt{2}\bigr )$. In particular, $\mathrm{End}(J(C_{{\mathbf u}}))=\mathrm{End}(J(C_f))$.

Clearly, the conditions of Lemma 3.1 hold for $u(x)={\mathbf u}(x)$ for all odd primes $\ell$. The discriminant ${\Delta }({\mathbf u})$ of ${\mathbf u}(x)$ coincides with

$$\begin{aligned} \mathrm {Discr}(n,B,C)=(-1)^{(2g+1)2g/2}&(2g{+}1)^{2g+1} \bigl [-2^{2g-1}pc\bigr ]^{2g}\\ {}&+ (-1)^{2g(2g-1)/2} (2g)^{2g} \bigl [-2^{2g}b\bigr ]^{2g+1}\!. \end{aligned}$$

It follows that

$$\begin{aligned} {\Delta }({\mathbf u})=(-1)^{g} 2^{2g(2g-1)}\bigl [(2g{+} 1)^{2g+1}(pc)^{2g}-2^{6g}g^{2g}b^{2g+1}\bigr ]. \end{aligned}$$

This implies that

$$\begin{aligned} {\Delta }({\mathbf u}) =2^{2[g(2g-1)]} D_0, \end{aligned}$$

(3)

where

$$\begin{aligned} D_0=(-1)^{g} \bigl \{(2g{+}1) \bigl [(2g{+}1)^{g} (pc)^{g}\bigr ]^{2}-2^{6g} g^{2g} b^{2g+1}\bigr \}. \end{aligned}$$

Clearly, $D_0$ is an odd integer that is not divisible by $p$. It is also clear that $D_0$ is congruent to $(-1)^{g}(2g{+}1)$ modulo $4$ (because every odd square is congruent to $1$ modulo $4$). This implies that

$$\begin{aligned} D_0 \equiv 1 ~\mathrm{mod }~4 \end{aligned}$$

(4)

for all $g$.

5 Proof of Theorem 1.7

Let us apply Lemma 3.1 (II) to

$$\begin{aligned} {\mathbf u}(x)=2^{2g+1} f\biggl (\frac{x}{2}\biggr )=x^{2g+1}-2^{2g}bx-2^{2g-1}pc. \end{aligned}$$

We obtain that for each odd prime $\ell$ the polynomial ${\mathbf u}(x)~\mathrm{mod }~\ell \in {\mathbb F}_{\ell }[x]$ has, at most, one multiple root; in addition, this root is double and lies in ${\mathbb F}_{\ell }$. Applying to ${\mathbf u}(x)$ Corollary 3.5 combined with formulas (3) and (4) of Sect. 4, we conclude that there exists an odd $\ell \ne p$ such that ${\mathbf u}(x)~\mathrm{mod }~\ell$ has exactly one multiple root; this root is double and lies in ${\mathbb F}_{\ell }$. In addition, $\mathrm {Gal}({\mathbf u}/{\mathbb Q})$ coincides with ${\mathbf S}_{2g+1}$, because it is doubly transitive. Now the assertions (i) and (ii) follow readily from the equality

$$\begin{aligned} f(x)~\mathrm{mod }~\ell =\frac{{\mathbf u}(2x)}{2^{2g+1}} ~\mathrm{mod }~\ell \end{aligned}$$

that holds for all odd primes $\ell$.

By Remark 4.1, $\mathrm {Gal}(f/{\mathbb Q})=\mathrm {Gal}({\mathbf u}/{\mathbb Q})$ and therefore also coincides with ${\mathbf S}_{2g+1}$, which implies (in light of 1.1) that $\mathrm {Gal}\bigl ({\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )\bigr )=\mathbf {A}_{2g+1}$. This proves (iii). Now Remark 1.9 implies that $\mathrm {End}(J(C_f))={\mathbb Z}$. This proves (iv). In order to prove (iii$'$), first notice that the Galois extension ${\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}$ is ramified at $2$, Remark 1.5 (I), while ${\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )={\mathbb Q}\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at $2$ over ${\mathbb Q}$ in light of formulas (3) and (4) in Sect. 4 (and Corollary 3.5 (a)). This implies that ${\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )$ is ramified at some prime divisor of $2$. Since all the field extensions involved are Galois, ${\mathbb Q}({\mathfrak R}_f)/{\mathbb Q}\bigl (\sqrt{{\Delta }(f)}\bigr )$ is actually ramified at all prime divisors of $2$. This proves the first assertion of (iii$'$). The second assertion of (iii$'$) follows from Corollary 3.5 (a). This proves (iii$'$).

6 Variants and complements

Throughout this section, $K$ is a number field. We write $\fancyscript{O}$ for the ring of integers in $K$. If ${\mathfrak b}$ is a maximal ideal in $\fancyscript{O}$ then we write $k({\mathfrak b})$ for the (finite) residue field $\fancyscript{O}/{\mathfrak b}$. As usual, we call $\mathrm {char}(k({\mathfrak b}))$ the residual characteristic of ${\mathfrak b}$. We write $K_{{\mathfrak b}}$ for the ${\mathfrak b}$-adic completion of $K$ and

$$\begin{aligned} \fancyscript{O}_{{\mathfrak b}}\subset K_{{\mathfrak b}} \end{aligned}$$

for the ring of ${\mathfrak b}$-adic integers in the field $K_{{\mathfrak b}}$. We consider the subring $\fancyscript{O}[1/2]\subset K$ generated by $1/2$ over $\fancyscript{O}$. We have

$$\begin{aligned} \fancyscript{O}\subset \fancyscript{O}\biggl [\frac{1}{2}\biggr ]\subset K. \end{aligned}$$

If ${\mathfrak b}\subset \fancyscript{O}$ is a maximal ideal in $\fancyscript{O}$ with odd residual characteristic then

$$\begin{aligned} \fancyscript{O}\subset \fancyscript{O}\biggl [\frac{1}{2}\biggr ]\subset \fancyscript{O}_{{\mathfrak b}}, \end{aligned}$$

the ideal ${\mathfrak b}\fancyscript{O}[1/2]$ is a maximal ideal in $\fancyscript{O}[1/2]$ and

$$\begin{aligned} k({\mathfrak b})=\fancyscript{O}/{\mathfrak b}=\fancyscript{O}\biggl [\frac{1}{2}\biggr ]\bigg /{\mathfrak b}\fancyscript{O}\biggl [\frac{1}{2}\biggr ]=\fancyscript{O}_{{\mathfrak b}}/{\mathfrak b}\fancyscript{O}_{{\mathfrak b}}. \end{aligned}$$

Lemma 3.1 (II) and its proof admit the following straightforward generalization.

Lemma 6.1

Let

$$\begin{aligned} u(x)=u_{n,B,C}(x)=x^n+Bx+C \in \fancyscript{O}[x] \end{aligned}$$

be a monic polynomial of degree $n>1$ such that $B\ne 0$ and $C\ne 0$. Let ${\mathfrak b}$ be a maximal ideal in $\fancyscript{O}$ that enjoys the following properties:

• (i) $B \fancyscript{O}+C\fancyscript{O}+ {\mathfrak b}=\fancyscript{O}$,

• (ii) $n\fancyscript{O}+B\fancyscript{O}+ {\mathfrak b}=\fancyscript{O}$,

• (iii) $(n{-}1)\fancyscript{O}+C\fancyscript{O}+ {\mathfrak b}=\fancyscript{O}$.

Suppose that $u(x)$ has no multiple roots. Let us consider the polynomial

$$\begin{aligned} \overline{u}(x)=u(x)~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b})[x]. \end{aligned}$$

Then:

• (a) $\overline{u}(x)$ has, at most, one multiple root in an algebraic closure of $k({\mathfrak b})$.

• (b) If such a multiple root say, $\gamma$, does exist, then $n(n-1)BC \not \in {\mathfrak b}$ and $\gamma$ is a double root of $\overline{u}(x)$. In addition, $\gamma$ is a nonzero element of $k({\mathfrak b})$.

• (c) If such a multiple root does exist then either the field extension $K({\mathfrak R}_u)/K$ is unramified at ${\mathfrak b}$ or a corresponding inertia subgroup at ${\mathfrak b}$ in

  $$\begin{aligned} \mathrm {Gal}(K({\mathfrak R}_u)/K)=\mathrm {Gal}(u/\mathrm {K}) \subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

  is generated by a transposition. In both cases the Galois extension $K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at all prime divisors of ${\mathfrak b}$.

Remark 6.2

In the notation of Lemma 6.1, suppose that $\overline{u}(x)$ has no multiple roots, i.e., ${\Delta }(u) \not \in {\mathfrak b}$. Then clearly the Galois extension $K({\mathfrak R}_u)/K$ is unramified at ${\mathfrak b}$.

Proof

We have

$$\begin{aligned} \overline{u}(x)=x^n+\overline{B}x+\overline{C} \in k({\mathfrak b})[x], \end{aligned}$$

where

$$\begin{aligned} \overline{B}=B~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b}), \qquad \overline{C}=C~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b}). \end{aligned}$$

The condition (i) implies that either $\overline{B}\ne 0$ or $\overline{C} \ne 0$. The condition (ii) implies that if $\overline{B}= 0$ then $n \ne 0$ in $k({\mathfrak b})$. It follows that if $\overline{B}= 0$ then $n\overline{C} \ne 0$.

The condition (iii) implies that if $n-1=0$ in $k({\mathfrak b})$ then $\overline{C} \ne 0$ (and, of course, $n \ne 0$ in $k({\mathfrak b})$). On the other hand, if $\overline{C}=0$ then $n-1 \ne 0$ in $k({\mathfrak b})$.

Suppose $\overline{u}(x)$ has a multiple root $\gamma$ in an algebraic closure of $k({\mathfrak b})$. Then as in the proof of Lemma 3.1 (II),

$$\begin{aligned} {\Delta }(\overline{u})=(-1)^{n(n-1)/2} n^n \overline{C}^{\,n-1}\!+ (-1)^{(n-1)(n-2)/2} (n-1)^{n-1} \overline{B}^{\,n}=0. \end{aligned}$$

This implies that

$$\begin{aligned} n^n \overline{C}^{\,n-1}\!= \pm (n-1)^{n-1} \overline{B}^{\,n}\!. \end{aligned}$$

(5)

This implies that if $n-1=0$ in $k({\mathfrak b})$ then $\overline{C}=0$, which is not the case. This proves that $n-1\ne 0$ in $k({\mathfrak b})$. On the other hand, if $\overline{B}= 0$ then $\overline{C} \ne 0$ and $n \ne 0$ in $k({\mathfrak b})$. Then (5) implies that $\overline{C} = 0$ and we get a contradiction that proves that $\overline{B}\ne 0$. If $n=0$ in $k({\mathfrak b})$ then $n-1 \ne 0$ in $k({\mathfrak b})$ and (5) implies that $\overline{B}= 0$, which is not the case. The obtained contradiction proves that $n \ne 0$ in $k({\mathfrak b})$. If $\overline{C} = 0$ then (5) implies that $\overline{B}= 0$, which is not the case. This proves that the maximal ideal ${\mathfrak b}$ does not contain $n(n-1)BC$.

On the other hand, we have as in the proof of Lemma 3.1 (II) that

$$\begin{aligned} x {\cdot } \overline{u}'(x)-n{\cdot }\overline{u}(x)=- (n-1)\overline{B}x-n\overline{C} \end{aligned}$$

and therefore $-(n-1)\overline{B}\gamma -n\overline{C}=0$. It follows that

$$\begin{aligned} \gamma =-\frac{n\overline{C}}{(n-1)\overline{B}} \end{aligned}$$

is a nonzero element of $k({\mathfrak b})$. The second derivative $\overline{u}''(x)=n(n-1)x^{n-2}$ and

$$\begin{aligned} \overline{u}''(\gamma )=n(n-1)\gamma ^{n-2} \ne 0. \end{aligned}$$

It follows that $\gamma$ is a double root of $\overline{u}(x)$. This proves (a) and (b).

In order to prove (c), notice that as in the proof of Lemma 3.1 (II)(c), there exists a monic degree $n-2$ polynomial $\overline{h}(x)\in k({\mathfrak b})[x]$ such that

$$\begin{aligned} \overline{u}(x)=(x-\gamma )^2 {\cdot } \overline{h}(x) \end{aligned}$$

and $\overline{h}(x)$ and $(x-\gamma )^2$ are relatively prime. By Hensel’s Lemma, there exist monic polynomials

$$\begin{aligned} h(x), v(x) \in \fancyscript{O}_{{\mathfrak b}}[x], \qquad \deg h=n-2,\quad \deg v=2 \end{aligned}$$

such that

$$\begin{aligned} u(x)=v(x) h(x) \end{aligned}$$

and

$$\begin{aligned} \overline{h}(x)=h(x)~\mathrm{mod }~{\mathfrak b}, \qquad (x-\gamma )^2=v(x) ~\mathrm{mod }~{\mathfrak b}. \end{aligned}$$

This implies that the splitting field $K_{{\mathfrak b}}({\mathfrak R}_h)$ of $h(x)$ (over $K_{{\mathfrak b}}$) is an unramified extension of $\mathrm {K}_{{\mathfrak b}}$ while the splitting field $\mathrm {K}_{{\mathfrak b}}({\mathfrak R}_u)$ of $u(x)$ (over $\mathrm {K}_{{\mathfrak b}}$) is obtained from $\mathrm {K}_{{\mathfrak b}}({\mathfrak R}_h)$ by adjoining to it two (distinct) roots say, $\alpha _1$ and $\alpha _2$ of quadratic $v(x)$. The field $K_{{\mathfrak b}}({\mathfrak R}_u)$ coincides either with $K_{{\mathfrak b}}({\mathfrak R}_h)$ or with a certain quadratic extension of $\mathrm {K}_{{\mathfrak b}}({\mathfrak R}_h)$, ramified or unramified. It follows that the inertia subgroup $I$ of

$$\begin{aligned} \mathrm {Gal}(K_{{\mathfrak b}}({\mathfrak R}_u)/K_{{\mathfrak b}})\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

is either trivial or is generated by the transposition that permutes $\alpha _1$ and $\alpha _2$ (and leaves invariant every root of $h(x)$). In the former case $K({\mathfrak R}_u)/K$ is unramified at ${\mathfrak b}$ while in the latter one an inertia subgroup in

$$\begin{aligned} \mathrm {Gal}(K({\mathfrak R}_u)/K)\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

that corresponds to ${\mathfrak b}$ is generated by a transposition. In both cases the Galois (sub)group $\mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )$ does not contain transpositions (see 1.1). This implies that $K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at all prime divisors of ${\mathfrak b}$. $\square$

Corollary 3.5 admits the following partial generalization.

Lemma 6.3

Let $K$ be a number field and $\fancyscript{O}$ be its ring of integers. Let

$$\begin{aligned} u(x)=u_{n,B,C}(x)=x^n+Bx+C \in \fancyscript{O}[x] \end{aligned}$$

be a monic polynomial without multiple roots of degree $n>1$ such that both $B$ and $C$ are not zeros. Suppose that there is a nonnegative integer $N$ such that

$$\begin{aligned} 2^N \fancyscript{O}\subset B \fancyscript{O}+C\fancyscript{O}, \qquad 2^N \fancyscript{O}\subset n\fancyscript{O}+B\fancyscript{O}, \qquad 2^N \fancyscript{O}\subset (n-1)\fancyscript{O}+C\fancyscript{O}. \end{aligned}$$

Suppose that there is a nonnegative integer $M$ such that the discriminant $D={\Delta }(u)=2^{2M}{\cdot } D_0$ with $D_0 \in \fancyscript{O}$. Assume also that $D, D_0$ and $K$ enjoy the following properties:

• (i) $D$ is not a square in $K$ and $D_0-1 \in 4\fancyscript{O}$.

• (ii) The class number of $K$ is odd $($e.g., $\fancyscript{O}$ is a principal ideal domain$)$.

• (iii) Either $K$ is totally imaginary, i.e., it does not admit an embedding into the field of real numbers or $K$ is totally real and $D_0$ is totally positive.

Then:

• (a) The quadratic extension $K\bigl (\sqrt{{\Delta }(u)}\bigr )/K$ is unramified at every prime divisor of $2$. The Galois extension $K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at every prime ideal ${\mathfrak b}$ of odd residual characteristic.

• (b) There exists a maximal ideal ${\mathfrak b}\subset \fancyscript{O}$ with residue field $k({\mathfrak b})$ of odd characteristic that enjoys the following properties:

  • $D_0 \in {\mathfrak b}$, the polynomial $u(x) {\mathfrak b}{~\mathrm{mod }~} \in k({\mathfrak b})[x]$ has exactly one multiple root and its multiplicity is $2$. In addition, this root lies in $k({\mathfrak b})$.

  • The field extension $K({\mathfrak R}_u)/K$ is ramified at ${\mathfrak b}$ and the Galois group

    $$\begin{aligned} \mathrm {Gal}(K({\mathfrak R}_u)/K)=\mathrm {Gal}(u/\mathrm {K})\subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

    contains a transposition. In particular, if $\mathrm {Gal}(u/\mathrm {K})$ is doubly transitive then

    $$\begin{aligned} \mathrm {Gal}(u/\mathrm {K})=\mathrm {Perm}({\mathfrak R}_f)\cong {\mathbf S}_n \end{aligned}$$

    and

    $$\begin{aligned} \mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl ( \sqrt{{\Delta }(u)}\bigr )\bigr )=\mathbf {A}_n. \end{aligned}$$

Proof

Let us prove (a). Clearly,

$$\begin{aligned} E=K\bigl (\sqrt{D_0}\bigr ) =K\bigl (\sqrt{D} \bigr )=K\bigl (\sqrt{{\Delta }(u)} \bigr )\subset K({\mathfrak R}_u) \end{aligned}$$

is a quadratic extension of $K$. Notice that $\theta =\bigl (1{+}\sqrt{D_0}\bigr )/2 \in E$ is a root of the quadratic equation

$$\begin{aligned} v_2(x)=x^2-x +\frac{1-D_0}{4} \in \fancyscript{O}[x] \end{aligned}$$

and therefore is an algebraic integer. In addition, $E=K(\theta )$.

If a maximal ideal ${\mathfrak b}_2$ in $\fancyscript{O}$ has residual characteristic $2$ then the quadratic polynomial

$$\begin{aligned} v_2(x)~\mathrm{mod }~{\mathfrak b}_2 =x^2-x +\biggl (\frac{1-D_0}{4}\biggr )~\mathrm{mod }~{\mathfrak b}_2 \in k({\mathfrak b}_2)[x] \end{aligned}$$

has no multiple roots, because its derivative is a nonzero constant $-1$. This implies that $E/K$ is unramified at all prime divisors of $2$. On the other hand, the conditions of Lemma 6.1 hold for all maximal ideals ${\mathfrak b}$ of $\fancyscript{O}$ with odd residual characteristic. Now Remark 6.2 and Lemma 6.1 (c) imply that the Galois extension $K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )$ is unramified at every ${\mathfrak b}$ of odd residual characteristic. This proves (a).

In order to prove (b), notice that the condition (iii) implies that either all archimedean places of both $E$ and $K$ are complex or all archimedean places of both $E$ and $K$ are real. This implies that $E/K$ is unramified at all infinite primes. Since the class number of $K$ is odd, the classical results about Hilbert class fields [6], Chapter 2, Section 1.2] imply that there is a maximal ideal ${\mathfrak b}\subset \fancyscript{O}$ such that $E/K=K\bigl (\sqrt{D}\bigr )/K$ is ramified at ${\mathfrak b}$. Since $E/K$ is unramified at all prime divisors of $2$, the residual characteristic of ${\mathfrak b}$ is odd, i.e., $2\not \in {\mathfrak b}$. This implies that

$$\begin{aligned} {\Delta }(u) =D \in {\mathfrak b}. \end{aligned}$$

Since $D=2^{2M}{\cdot } D_0$ and ${\mathfrak b}$ is a prime (actually, maximal) ideal in $\fancyscript{O}$, we have $D_0 \in {\mathfrak b}$. It also follows that

$$\begin{aligned} u(x) ~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b})[x] \end{aligned}$$

has a multiple root. Now we are in a position to apply Lemma 6.1. Since $K({\mathfrak R}_u)\supset E$, the field extension $K({\mathfrak R}_u)/K$ is ramified at ${\mathfrak b}$. Applying Lemma 6.1, we conclude that $u(x) ~\mathrm{mod }~{\mathfrak b}$ has exactly one multiple root, this root is double and lies in $k({\mathfrak b})$. In addition,

$$\begin{aligned} \mathrm {Gal}(K({\mathfrak R}_u)/K) \subset \mathrm {Perm}({\mathfrak R}_u) \end{aligned}$$

contains a transposition. This implies that if $\mathrm {Gal}(K({\mathfrak R}_u)/K)$ is doubly transitive then $\mathrm {Gal}(K({\mathfrak R}_u)/K)$ coincides with $\mathrm {Perm}({\mathfrak R}_u)\cong {\mathbf S}_n$. Of course, this implies that $\mathrm {Gal}\bigl (K({\mathfrak R}_u)/K\bigl (\sqrt{{\Delta }(u)}\bigr )\bigr )=\mathbf {A}_n$. $\square$

6.4 Generalized Mori quadruples   Let us consider a quadruple $(g, {\mathfrak p}, {\mathbf b},{\mathbf c})$ where $g$ is a positive integer, ${\mathfrak p}$ is a maximal ideal in $\fancyscript{O}$ while ${\mathbf b}$ and ${\mathbf c}$ are elements of $\fancyscript{O}$ that enjoy the following properties:

• The residue field $k({\mathfrak p})=\fancyscript{O}/{\mathfrak p}$ is a finite field of odd characteristic. If $q$ is the cardinality of $k({\mathfrak p})$ then every prime divisor of $g$ is also a divisor of $(q-1)/2$. In particular, if $g$ is even then $q-1$ is divisible by $4$.

• The residue ${\mathbf b}~\mathrm{mod }~{\mathfrak p}$ is a primitive element of $k({\mathfrak p})$, i.e., it has multiplicative order $q-1$. In particular,

  $$\begin{aligned} {\mathbf b}\fancyscript{O}+{\mathfrak p}=\fancyscript{O}. \end{aligned}$$

  The conditions (i) and (ii) imply that for each prime divisor $d$ of $g$ the residue ${\mathbf b}~\mathrm{mod }~{\mathfrak p}$ is not a $d$th power in $k({\mathfrak p})$. Since $q-1$ is even, ${\mathbf b}~\mathrm{mod }~{\mathfrak p}$ is not a square in $k({\mathfrak p})$. So, if $d$ is a prime divisor of $2g$ then ${\mathbf b}~\mathrm{mod }~{\mathfrak p}$ is not a $d$th power in $k({\mathfrak p})$. If $2g$ is divisible by $4$ then $g$ is even and $q-1$ is divisible by $4$, i.e., $-1$ is a square in $k({\mathfrak p})$. It follows that $-4 {\mathbf b}~\mathrm{mod }~{\mathfrak p}$ is not a square in $k({\mathfrak p})$. Thanks to [7], Thereom 9.1, Chapter VI, Section 9], the last two assertions imply that the polynomial

  $$\begin{aligned} x^{2g}-{\mathbf b}~\mathrm{mod }~{\mathfrak p}\in k({\mathfrak p})[x] \end{aligned}$$

  is irreducible over $k({\mathfrak p})$. This implies that its Galois group over $($the finite field$)$ $k({\mathfrak p})$ is an order $2g$ cyclic group.

• ${\mathbf c}\in {\mathfrak p}$, ${\mathbf c}-1 \in 2 \fancyscript{O}$ and

  $$\begin{aligned} \fancyscript{O}= {\mathbf b}\fancyscript{O}+ {\mathbf c}\fancyscript{O}= {\mathbf b}\fancyscript{O}+(2g{+}1)\fancyscript{O}= 2g \fancyscript{O}+{\mathbf c}\fancyscript{O}. \end{aligned}$$

We call such a quadruple a generalized Mori quadruple (in $K$).

Example 6.5

Suppose that $K$ and $g$ are given. By Dirichlet’s Theorem about primes in arithmetic progressions, there is a prime $p$ that does not divide $2g+1$ and is congruent to $1$ modulo $2g$. (In fact, there are infinitely many such primes.) Clearly, $p$ is odd. Let us choose a maximal ideal ${\mathfrak p}$ of $\fancyscript{O}$ that contains $p$ and denote by $q$ the cardinality of the finite residue field $k({\mathfrak p})$. Then $\mathrm {char}(k({\mathfrak p}))=p$ and $q$ is a power of $p$. This implies that $q-1$ is divisible by $p-1$ and therefore is divisible by $2g$. Let us choose a generator $\widetilde{{\mathbf b}}\in k({\mathfrak p})$ of the multiplicative cyclic group $k({\mathfrak p})^{*}$. Let $r$ be a nonzero integer that is relatively prime to $2g+1$. (E.g., $r=\pm 1, \pm 2$.) Using Chinese Remainder Theorem, one may find ${\mathbf b}\in \fancyscript{O}$ such that

$$\begin{aligned} {\mathbf b}~\mathrm{mod }~{\mathfrak p}=\widetilde{{\mathbf b}}, \qquad {\mathbf b}-r \in (2g{+}1)\fancyscript{O}. \end{aligned}$$

(Clearly, ${\mathbf b}\not \in {\mathfrak p}$.) Now the same theorem allows us to find ${\mathbf c}\in {\mathfrak p}\subset \fancyscript{O}$ such that ${\mathbf c}-1 \in 2g{\mathbf b}\fancyscript{O}$. Then $(g, {\mathfrak p}, {\mathbf b},{\mathbf c})$ is a generalized Mori quadruple in $K$.

Let us consider the polynomials

$$\begin{aligned} F(x)=F_{g, {\mathfrak p}, {\mathbf b},{\mathbf c}}(x)=x^{2g+1}-{\mathbf b}x - \frac{{\mathbf c}}{4} \in \fancyscript{O}\biggl [\frac{1}{2}\biggr ][x]\subset K[x] \end{aligned}$$

and

$$\begin{aligned} U(x)=2^{2g+1} F\biggl (\frac{x}{2}\biggr )=x^{2g+1}-2^{2g}{\mathbf b}x-2^{2g-1}{\mathbf c}\in \fancyscript{O}[x]\subset K[x]. \end{aligned}$$

Theorem 6.6

Let $(g, {\mathfrak p}, {\mathbf b},{\mathbf c})$ be a generalized Mori quadruple in $K$. Assume also that there exists a maximal ideal ${\mathfrak b}_2\subset \fancyscript{O}$ of residual characteristic $2$ such that the ramification index $e({\mathfrak b}_2)$ of ${\mathfrak b}_2$ $($over $2)$ in $K/{\mathbb Q}$ is relatively prime to $2g+1$. Then:

• (i) The polynomial $F(x)=F_{g, {\mathfrak p}, {\mathbf b},{\mathbf c}}(x)\in K[x]$ is irreducible over $K_{{\mathfrak b}_2}$ and therefore over $K$. In addition, the Galois extension $K({\mathfrak R}_{F})/K$ is ramified at ${\mathfrak b}_2$.

• (ii) The transitive Galois group

  $$\begin{aligned} \mathrm {Gal}(F/\mathrm {K})=\mathrm {Gal}(K({\mathfrak R}_{F})/K)\subset \mathrm {Perm}({\mathfrak R}_{F})={\mathbf S}_{2g+1} \end{aligned}$$

  contains a cycle of length $2g$. In particular, $\mathrm {Gal}(F/\mathrm {K})$ is doubly transitive and is not contained in $\mathbf {A}_{2g+1}$, and ${\Delta }(F)$ is not a square in $K$.

• (iii) Assume that $K$ is a totally imaginary number field with odd class number. Then $\mathrm {Gal}(F/K)=\mathrm {Perm}({\mathfrak R}_{F})$. If, in addition, $g>1$ then $\mathrm{End}(J(C_F))={\mathbb Z}$.

• (iv) Assume that $K$ is a totally imaginary number field with odd class number and $g>1$. Then:

  • For all primes $\ell$ the image $\rho _{\ell ,F}(\mathrm {Gal}(K))$ is an open subgroup of finite index in $\mathrm {Gp}\bigl (T_{\ell }(J(C_F)),e_{\ell }\bigr )$.

  • Let $L$ be a number field that contains $K$ and $\mathrm {Gal}(L)$ be the absolute Galois group of $L$, which we view as an open subgroup of finite index in $\mathrm {Gal}(L)$. Then for all but finitely many primes $\ell$ the image $\rho _{\ell ,F}(\mathrm {Gal}(L))$ coincides with $\mathrm {Gp}\bigl (T_{\ell }(J(C_F)),e_{\ell }\bigr )$.

Remark 6.7

If $K$ is a quadratic field then for every maximal ideal ${\mathfrak b}_2 \subset \fancyscript{O}$ (with residual characteristic $2$) the ramification index $e({\mathfrak b}_2)$ of ${\mathfrak b}_2$ in $K/{\mathbb Q}$ is either $1$ or $2$: in both cases it is relatively prime to odd $2g+1$. This implies that if $K$ is an imaginary quadratic field with odd class number then all conclusions of Theorem 6.6 hold for every generalized Mori quadruple $(g, {\mathfrak p}, {\mathbf b},{\mathbf c})$. In particular, the Galois extension $K({\mathfrak R}_F)/K$ is ramified at every ${\mathfrak b}_2$.

One may find the list of imaginary quadratic fields with small, ${\le }23$, odd class number in [1], pp. 322–324]; see also [16], Table 4, p. 936].

Proof

of Theorem 6.6 The ${\mathfrak b}_2$-adic Newton polygon of $F(x)$ consists of one segment that connects the points $(0,-2e({\mathfrak b}_2))$ and $(2g{+}1,0)$, which are its only integer points, because $e({\mathfrak b}_2)$ and $2g+1$ are relatively prime and therefore $2e({\mathfrak b}_2)$ and $2g+1$ are relatively prime. Now the irreducibility of $F(x)$ over $K_{{\mathfrak b}_2}$ follows from Eisenstein–Dumas Criterion [9], Corollary 3.6, p. 316], [4], p. 502]. This proves (i). It also proves that the Galois extension $K({\mathfrak R}_F)/K$ is ramified at ${\mathfrak b}_2$.

In order to prove (ii), let us consider the reduction

$$\begin{aligned} \widetilde{F}(x)=F(x)~\mathrm{mod }~{\mathfrak p}\fancyscript{O}\biggl [\frac{1}{2}\biggr ]=x^{2g+1}-\widetilde{{\mathbf b}} x \in k({\mathfrak p})[x] \end{aligned}$$

where $\widetilde{{\mathbf b}}={\mathbf b}~\mathrm{mod }~{\mathfrak p}\in k({\mathfrak p})$. So,

$$\begin{aligned} \widetilde{F}(x)=x\bigl (x^{2g}-\widetilde{{\mathbf b}}\bigr ) \in k({\mathfrak p})[x]. \end{aligned}$$

We have already seen in 6.4 that $x^{2g}-\widetilde{{\mathbf b}}$ is irreducible over $k({\mathfrak p})$ and its Galois group is an order $2g$ cyclic group. We also know that $\widetilde{{\mathbf b}} \ne 0$ and therefore the polynomials $x$ and $x^{2g}-\widetilde{{\mathbf b}}$ are relatively prime. This implies that $K({\mathfrak R}_F)/K$ is unramified at ${\mathfrak p}$ and a corresponding Frobenius element in $\mathrm {Gal}(K({\mathfrak R}_F)/K)\subset \mathrm {Perm}({\mathfrak R}_F)$ is a cycle of length $2g$. This proves (ii). (Compare with arguments on [8], p. 107].)

The map $\alpha \mapsto 2\alpha$ is a $\mathrm {Gal}(K)$-equivariant bijection between the sets of roots ${\mathfrak R}_F$ and ${\mathfrak R}_U$, which induces a group isomorphism between permutation groups $\mathrm {Gal}({\mathfrak R}_F)\subset \mathrm {Perm}({\mathfrak R}_F)$ and $\mathrm {Gal}({\mathfrak R}_U)\subset \mathrm {Perm}({\mathfrak R}_U)$. In particular, the double transitivity of $\mathrm {Gal}({\mathfrak R}_F)$ implies the double transitivity of $\mathrm {Gal}({\mathfrak R}_U)$. On the other hand,

$$\begin{aligned} {\Delta }(U)=2^{(2g+1)2g}{\Delta }(F)=\bigl [2^{(2g+1)g}\bigr ]^{2}{\Delta }(F). \end{aligned}$$

This implies that ${\Delta }(U)$ is not a square in $K$ as well. The discriminant ${\Delta }(U)$ is given by the formula, Remark 3.2,

We have $D=2^{2M} D_0$, where $M=g(2g-1)$ is a positive integer and

$$\begin{aligned} D_0=(-1)^{g}\bigl [(2g+1)^{2g+1}{\mathbf c}^{2g}-2^{6g}g^{2g}{\mathfrak b}^{2g+1}\bigr ] \in \fancyscript{O}. \end{aligned}$$

Since ${\mathbf c}-1 \in 2\fancyscript{O}$, we have ${\mathbf c}^2-1 \in 4\fancyscript{O}$ and

$$\begin{aligned} D_0 \equiv (-1)^{g} (2g+1)^{2g+1} ~\mathrm{mod }~4\fancyscript{O}. \end{aligned}$$

Since $(2g{+}1)^{2g}=\bigl [(2g{+}1)^2\bigr ]^{g} \equiv 1 ~\mathrm{mod }~4$, we conclude $D_0 \equiv (-1)^{g} (2g{+}1) \mod 4 \fancyscript{O}$. This implies that

$$\begin{aligned} D_0-1 \in 4\fancyscript{O}. \end{aligned}$$

Applying Lemma 6.3 to

$$\begin{aligned} n=2g+1,\qquad B=-2^{2g}{\mathbf b},\qquad C= -2^{2g-1}{\mathbf c},\\ u(x)=U(x),\qquad M=g(2g{-}1),\qquad N=2g, \end{aligned}$$

we conclude that doubly transitive $\mathrm {Gal}(U/K)$ coincides with $\mathrm {Perm}({\mathfrak R}_U)$ and therefore $\mathrm {Gal}(F/K)$ coincides with $\mathrm {Perm}({\mathfrak R}_F)\cong {\mathbf S}_{2g+1}$. If $g>1$ then Theorem 1.2 tells us that $\mathrm{End}(J(C_F))={\mathbb Z}$. This proves (iii). We also obtain that there exists a maximal ideal ${\mathfrak b}\subset \fancyscript{O}$ with odd residual characteristic such that $U(x) ~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b})[x]$ has exactly one multiple root, this root is double and lies in $k({\mathfrak b})$. Since

$$\begin{aligned} F(x)=\frac{U(2x)}{2^{2g+1}}, \end{aligned}$$

we obtain that

$$\begin{aligned} F(x)~\mathrm{mod }~{\mathfrak b}\fancyscript{O}\biggl [\frac{1}{2}\biggr ] =\frac{U(2x)}{2^{2g+1}}~\mathrm{mod }~{\mathfrak b}\in k({\mathfrak b})[x]. \end{aligned}$$

This implies that the polynomial $F(x)~\mathrm{mod }~{\mathfrak b}\fancyscript{O}[1/2] \in k({\mathfrak b})[x]$ has exactly one multiple root, this root is double and lies in $k({\mathfrak b})$. The properties of $F(x) ~\mathrm{mod }~{\mathfrak b}\fancyscript{O}[1/2]$ imply that $J(C_F)$ has a semistable reduction at ${\mathfrak b}$ with toric dimension $1$. Now it follows from [21], Theorem 4.3] that for for all primes $\ell$ the image $\rho _{\ell ,F}(\mathrm {Gal}(K))$ is an open subgroup of finite index in $\mathrm {Gp}\bigl (T_{\ell }(J(C_F)),e_{\ell }\bigr )$. It follows from [5], Theorem 1] that if $L$ is a number field containing $K$ then for all but finitely many primes $\ell$ the image $\rho _{\ell ,F}(\mathrm {Gal}(L))$ coincides with $\mathrm {Gp}(T_{\ell }(J(C_F)),e_{\ell })$. This proves (iv).$\square$

Corollary 6.8

We keep the notation of Theorem 6.6. Let $K$ be an imaginary quadratic field with odd class number. Let $(g, {\mathfrak p}, {\mathbf b},{\mathbf c})$ be a generalized Mori quadruple in $K$ and $F(x)=F_{g, {\mathfrak p}, {\mathbf b},{\mathbf c}}(x) \in K[x]$. Then

$$\begin{aligned} \mathrm {Gal}\bigl (K({\mathfrak R}_F)/K\bigl ( \sqrt{{\Delta }(F)}\bigr )\bigr )=\mathbf {A}_{2g+1} \end{aligned}$$

and the Galois extension $K({\mathfrak R}_F)/K\bigl (\sqrt{{\Delta }(F)}\bigr )$ is unramified everywhere outside $2$ and ramified at all prime divisors of $2$.

Proof

As above, let us consider the polynomial

$$\begin{aligned} U(x)=2^{2g+1} F\biggl (\frac{x}{2}\biggr )=x^{2g+1}-2^{2g}{\mathbf b}x-2^{2g-1}{\mathbf c}\in \fancyscript{O}[x]\subset K[x]. \end{aligned}$$

We have $K({\mathfrak R}_F)=K({\mathfrak R}_U)$, $K\bigl (\sqrt{{\Delta }(F)}\bigr )=K\bigl (\sqrt{{\Delta }(U)}\bigr )$. Since

$$\begin{aligned} {\mathbf S}_{2g+1}=\mathrm {Perm}({\mathfrak R}_U)=\mathrm {Gal}(U/K) =\mathrm {Gal}(K({\mathfrak R}_U)/K), \end{aligned}$$

we have

$$\begin{aligned} \mathrm {Gal}\bigl (K({\mathfrak R}_U)/K\bigl ( \sqrt{{\Delta }(U)}\bigr )\bigr )=\mathbf {A}_{2g+1}. \end{aligned}$$

It follows from Remark 6.7 that the Galois extension $K({\mathfrak R}_U)/K$ is ramified at every prime divisor of $2$ (in $K$). On the other hand, Lemma 6.3 (a) (applied to $u(x)=U(x)$) tells us that the quadratic extension $K\bigl (\sqrt{{\Delta }(U)} \bigr )/K$ is unramified at every prime divisor of $2$ (in $K$). Since all the field extensions involved are Galois, $K({\mathfrak R}_U)/K\bigl ( \sqrt{{\Delta }(U)}\bigr )$ is ramified at every prime divisor of $2$ $\bigl ($in $K\bigl (\sqrt{{\Delta }(U)}\bigr )\bigr )$.

Since $K$ is purely imaginary, $K\bigl (\sqrt{{\Delta }(U)}\bigr )$ is also purely imaginary and therefore (its every field extension, including) $K({\mathfrak R}_U)$ is unramified at all infinite places $\bigl ($in $K\bigl (\sqrt{{\Delta }(U)}\bigr )\bigr )$.

Remark 6.2 and Lemma 6.3 (a) (applied to $u(x)=U(x)$) imply that the field extension $K({\mathfrak R}_U)/K\bigl (\sqrt{{\Delta }(U)}\bigr )$ is unramified at all maximal ideals ${\mathfrak b}$ in $\fancyscript{O}$ with odd residual characteristic.$\square$

7 Corrigendum to [20]

• Page 660, the 6th displayed formula: insert $\subset$ between $\mathrm {End}_{\mathrm {Gal}(K)}V_{\ell }(X)$ and $\mathrm {End}_{{\mathbb Q}_{\ell }}V_{\ell }(X)$.

• Page 662, Theorem 2.6, line 3: $r_1$ should be $r_2$.

• Page 664, Remark 2.16: The reference to [23, Theorem 1.5] should be replaced by [23, Theorem 1].

• Page 664, Theorem 2.20: The following additional condition on $\ell$ was inadvertently omitted:

  “ (iii) If C is the center of $\mathrm {End}(X)$ then $C/\ell C$ is the center of $\mathrm{End}(X) / \ell \mathrm{End}(X)$.”

  In addition, “be” on the last line should be “is”.

• Page 666, Theorem. 3.3, line 2: $\ell$ should be assumed to be in $P$, i.e. one should read “Then for all but finitely many $\ell \in P\dots$”. In addition, $X_n$ should be $X_{\ell }$ throughout lines 3–6.

• Page 668, Lemma 3.9, line 1: $\mathrm {Isog}_P$ should be $\mathrm {Is}_P$.

• Page 668, Theorem 3.10, line 1: replace $\mathrm {Isog}_P((X {\times } X^t)^8, K, 1)$ by $\mathrm {Is}_P((X {\times } X^t)^4, K, 1)$.

• Page 670, Section 5.1, the first displayed formula: $t$ should be $g$.

• Page 672, line 9: $X'_{\ell }$ should be $X_{\ell }$.

(The author is grateful to Kestutis Cesnavicius for sending this list of typos.)