Hyperelliptic $\mathcal {S}_7$-curves of prime conductor

Abstract

An abelian threefold $A_{/{\mathbb {Q}}}$ of prime conductor N is favorable if its 2-division field F is an ${\mathcal {S}}_7$-extension over ${\mathbb {Q}}$ with ramification index 7 over ${\mathbb {Q}}_2$. Let A be favorable and let B be a semistable abelian variety of conductor $N^d$ with B[2] filtered by d copies of A[2]. We obtain a class field theoretic criterion on F to guarantee that B is isogenous to $A^d$ and a fortiori, A is unique up to isogeny.

Appendix A. Conductors for Artin–Schreier-like extensions

We use the conventions of Serre [17, IV] for conductors. Let $G = {\text {Gal}}(L/K)$ be the Galois group of a finite extension of local fields of residue characteristic p. Write $G_j$ for the j-th ramification subgroup of G and $c_{L/K} = \max \{j \, \vert \, G_j \ne \{1\} \}$. The conductor exponent of L/K is given by

$$\begin{aligned} \begin{aligned} {\mathfrak {f}}(L/K) = 1 + \frac{1}{\vert {G_0} \vert } \left( \vert {G_1} \vert + \dots + \vert {G_{c_{L/K}}} \vert \right) . \end{aligned} \end{aligned}$$

If H is a subgroup of G, then $H_j = H \cap G_j$. In particular, if $G_1 \subseteq H$ and F is the fixed field of H, then $H_j = G_j$ for $j \ge 1$ and $c_{L/F} = c_{L/K}$, so

$$\begin{aligned} {\mathfrak {f}}(L/K) = 1 +\frac{1}{[{G_0}\!:\!{H_0}]} \left( {\mathfrak {f}}(L/F)-1 \right) . \end{aligned}$$

(A.1)

Assume from now on that K is an unramified extension of ${\mathbb {Q}}_p$ with ring of integers ${\mathcal {O}}_K$ and fixed algebraic closure $\overline{K}$. Finding the field of points of a group scheme ${\mathcal {V}}$ over ${\mathcal {O}}_K$ from the linear algebra of its associated Honda system often leads to rather complicated congruences. In §4.2, such congruences were simplified by a suitable change of variables, to arrive at a characteristic 0 version of Artin-Schreier theory. A key property is that distinct solutions to our Artin-Schreier-like equations differ by units, although the solutions themselves are not integral.

Let $q = p^n$ with $n \ge 1$ and let F be a finite extension of $K(\varvec{\mu }_{q-1})$ in $\overline{K}$ with ring of integers ${\mathcal {O}}_F$, maximal ideal $m_F$ and valuation $v_F$, normalized so that $v_F(p) = e_F$ is the ramification index of $F/{\mathbb {Q}}_p$.

Proposition A.2

Let $f(x) = x^q-x+C$, where $C = u w^{-p^m}$ with:

Let L be the splitting field of f. If ${\text {ord}}_p(C) > q/(1-q)$, then ${\text {Gal}}(L/F) \simeq {\mathbb {F}}_q$ is an elementary abelian p-group and L/F is totally ramified, with conductor exponent ${\mathfrak {f}}(L/F) = v_F(w)+1$.

Proof

For $\alpha$ and $\beta$ in L and an ideal ${\mathfrak {a}}$ of ${\mathcal {O}}_L$, write $\alpha =\beta + \mathrm{O}({\mathfrak {a}})$ if $\alpha -\beta$ is in ${\mathfrak {a}}$. To estimate interior terms of a binomial expansion, recall that if $1 \le j \le q-1$, then ${\text {ord}}_p \left( {\begin{array}{c}q\\ j\end{array}}\right) = n - {\text {ord}}_p{j} \ge 1$.

Fix a root $\theta$ of f in L, let $g(x) = f(x+\theta )$ and $M = F(\theta )$. By assumption, ${\text {ord}}_p(C) < 0$, so ${\text {ord}}_p(\theta ) = \frac{1}{q} {\text {ord}}_p(C) > \frac{1}{1-q}$. For $j = 1, \dots , q-1$, the coefficient of $x^j$ in $(x+\theta )^q$ is in the maximal ideal ${\mathfrak {m}}_M$. Indeed,

$$\begin{aligned} \textstyle {{\text {ord}}_p\left( \left( {\begin{array}{c}q\\ j\end{array}}\right) \theta ^j\right) \ge 1 + j {\text {ord}}_p(\theta ) \ge 1 + (q-1){\text {ord}}_p(\theta ) > 0}. \end{aligned}$$

Thus, $g(x) = x^q - x + h(x)$, where h(x) is in ${\mathfrak {m}}_M[x]$ and $h(0) = 0$.

By Hensel’s Lemma, every element of $\varvec{\mu }_{q-1}$ can be refined to a root of g in M, so $L = M$. Furthermore, for $\sigma$ in ${\text {Gal}}(L/F)$, the root $\sigma (\theta )$ of f has the form $\sigma (\theta ) = \theta + \zeta _\sigma + \mathrm{O}({\mathfrak {m}}_L)$, with $\zeta _1 = 0$ if $\sigma = 1$ is the identity and $\zeta _\sigma$ in $\varvec{\mu }_{q-1}$ otherwise. Pass to the residue field, to obtain an injective homomorphism ${\text {Gal}}(L/F) \rightarrow {\mathbb {F}}_q$ given by $\sigma \mapsto \zeta _\sigma \pmod {{\mathfrak {m}}_L}.$ Hence ${\text {Gal}}(L/F)$ is an elementary abelian p-group.

Since u is a unit in the unramified ring ${\mathcal {O}}_K$ with perfect residue field, there is a unit $u_1$ in ${\mathcal {O}}_K$ satisfying $u_1^{p^m} \equiv u \pmod {p}$. Set $t = \theta ^{p^{n-m}} + u_1 w^{-1}$. We claim that $t^{p^m} = \epsilon \theta$ for some unit $\epsilon$ in ${\mathcal {O}}_L$, thanks to the following estimates. Note first that

$$\begin{aligned} {\text {ord}}_p\left( \theta ^{p^{n-m}}\right) = p^{-m} {\text {ord}}_p(C)= -{\text {ord}}_p(w). \end{aligned}$$

If $m >0$ and $1 \le j \le p^m-1$, we have

$$\begin{aligned} \textstyle {{\text {ord}}_p\left( \left( {\begin{array}{c}p^m\\ j \,\end{array}}\right) (\theta ^{p^{n-m}})^j w^{-(p^m-j)}\right) \ge 1- p^m {\text {ord}}_p(w) = 1 + {\text {ord}}_p(C)} > {\text {ord}}_p(\theta ), \end{aligned}$$

where the last inequality follows from our lower bound on ${\text {ord}}_p(C)$. Hence

$$\begin{aligned} t^{p^m}= & {} \theta ^{q} + (u_1 w^{-1})^{p^m} + \lambda \theta = \theta - C + (u_1 w^{-1})^{p^m} + \lambda \theta \\= & {} \theta + \frac{u_1^{p^m} - u}{w^{p^m}} + \lambda \theta \end{aligned}$$

for some $\lambda \in {\mathfrak {m}}_L$. In addition,

$$\begin{aligned} \textstyle {{\text {ord}}_p\left( (u_1^{p^m} - u)w^{-p^m}\right) \ge {\text {ord}}_p\left( p w^{-p^m} \right) = 1 + {\text {ord}}_p(C) > {\text {ord}}_p(\theta )}, \end{aligned}$$

so $t^{p^m} = \epsilon \theta$ for some unit $\epsilon$ in L, as claimed. If $m = 0$ take $u_1 = u$, so $t = \theta$. Thus,

$$\begin{aligned} {\text {ord}}_p(t) = p^{-m} {\text {ord}}_p(\theta ) = - q^{-1} {\text {ord}}_p(w) \, \text { for all } m. \end{aligned}$$

Since $p \not \mid {\text {ord}}_p(w)$, the ramification index of L/F is a multiple of q. But $[{L}\!:\!{F}] \le q$, so L/F is totally ramified of degree q and $L = F(t)$. Similar estimates of interior terms give

$$\begin{aligned} \sigma (\theta ^{p^{n-m}}) = (\theta + \zeta _\sigma + \mathrm{O}({\mathfrak {m}}_L))^{p^{n-m}} = \theta ^{p^{n-m}} + \zeta _\sigma ^{p^{n-m}} + \mathrm{O}({\mathfrak {m}}_L). \end{aligned}$$

Then $\sigma (t) - t = \zeta _\sigma ^{p^{n-m}} + \mathrm{O}({\mathfrak {m}}_L)$, since $u_1w^{-1}$ is in F. If $e_L$ is the ramification index of $L/{\mathbb {Q}}_p$ and $v_L$ is the valuation on L satisfying $v_L(p) = e_L$, we find that

$$\begin{aligned} v_L(t) = e_L {\text {ord}}_p(t) = q e_F {\text {ord}}_p(t) = -e_F {\text {ord}}_p(w) = -v_F(w) \end{aligned}$$

is prime to p. Hence ${\mathfrak {f}}= v_F(w)+1$ by [5, Prop. C.5]. $\square$

Remark A.3

If $f(x) = x^q -x+C$, with C in ${\mathcal {O}}_F$, then Artin-Schreier theory over $k_F$ implies that the splitting field L of f(x) is unramified over F, of degree dividing p. Indeed, as in the proof above, the map ${\text {Gal}}(L/F) \rightarrow {\mathbb {F}}_q$ is injective, so ${\text {Gal}}(L/F)$ is an elementary p-group and it has order at most p, since unramified extensions are cyclic. In particular, if ${\text {ord}}_p(C) > 0$, then the elements of $\varvec{\mu }_{q-1}$ lead to roots of f by Hensel’s Lemma, so $L = F$.