3.1. Proof of Propositions 2·1 and 2·2

By the Criterion of Néron-Ogg-Shafarevich (see [Reference Serre and Tate14, Section 2, Theorem 2(ii)]), if $\ell \neq p$ the Galois representation $\rho_\ell$ on $T_\ell \mathrm{Jac}(X)$ , restricted to inertia, has finite image, independent of $\ell$ . Moreover by Corollary 3 in the same paper, this image is isomorphic to $\mathrm{Gal}(K^{nr}(\mathrm{Jac}(X)[m])/K^{nr})$ for any $m \geq 3$ coprime to p, and $K^{nr}(\mathrm{Jac}(X)[m])$ is the minimal extension of $K^{nr}$ over which $\mathrm{Jac}(X)$ acquires good reduction. We can fix $m=4$ ; then by [Reference Yelton16, Theorem 1·1], we have that

$${K^{nr}}({\rm{Jac}}(X)[4]) = {K^{nr}}\left( {{{\left\{ {\sqrt {{\alpha _i} - {\alpha _j}} } \right\}}_{i,j \in \{ 1, \ldots ,p\} }}} \right).$$

We denote by $C_p \rtimes C_{2(p-1)}$ the semidirect product of $C_p$ and $C_{2(p-1)}$ which has the following presentation:

$$\left\langle {\sigma ,\tau |{\sigma ^p} = {\tau ^{2(p - 1)}} = 1,\tau \sigma {\tau ^{ - 1}} = {\sigma ^b}} \right\rangle ,$$

for some b coprime to p. The exact value of b will not be relevant for the rest of the paper. This group has a degree 2 quotient isomorphic to the Frobenius group $C_p \rtimes C_{p-1}$ , i.e. the group of affine transformations on the finite field with p elements, given by the non-abelian semidirect product $C_p \rtimes \mathrm{Aut}(C_p)$ . The extra $C_2$ contained in $C_p \rtimes C_{2(p-1)}$ is generated by $\tau^{p-1}$ . We will recall and use this notation in Section 3·2.

Lemma 3·2. With notations above, we have isomorphisms

$${\rho _\ell }({I_K}) \cong {\rm{Gal}}\left( {{K^{nr}}\left( {{{\left\{ {\sqrt {{\alpha _i} - {\alpha _j}} } \right\}}_{i,j}}} \right)/{K^{nr}}} \right) \cong {I_G}$$

and this group is isomorphic to a subgroup of $C_p \rtimes C_{2(p-1)}$ .

Proof. Recall that $I_G$ is the inertia subgroup of the Galois group of $K(\{\sqrt{\alpha_i-\alpha_j}\}_{i,j})/K$ , so we have that $K(\{\sqrt{\alpha_i-\alpha_j}\}_{i,j})K^{nr}/K^{nr}$ is also Galois with Galois group isomophic to $I_G$ .

Let $L=K^{nr}(\{\sqrt{\alpha_i-\alpha_j}\}_{i,j})$ and $L' = K^{nr}(\mathrm{Jac}(X)[2])$ , and consider the tower of field extensions $L/L'/K^{nr}$ . Notice that L ^′ is the splitting field of f over $K^{nr}$ , by [Reference Cornelissen3, Lemma 2·1].

For all i, $\alpha_i \in L'$ , therefore L is obtained from L ^′ by adjoining square roots of some elements of L ^′. So the Galois group of $L/L'$ is a direct product of some copies of $C_2$ . However, it is a totally ramified extension since $L' \supseteq K^{nr}$ , and it is tame since p is odd, therefore it must be cyclic. So $L/L'$ can only be trivial or quadratic.

Now let us consider $L'/K^{nr}$ . Since f is irreducible over $K^{nr}$ , we have that $\mathrm{Gal}(L'/K^{nr})$ has a cyclic subgroup of order p. Therefore $\mathrm{Gal}(L'/K^{nr})$ injects into $S_p$ , the group of permutations on p elements, and since p divides $|S_p|$ exactly once, necessarily the p-Sylow subgroup of $\mathrm{Gal}(L'/K^{nr})$ is isomorphic to $C_p$ . So, the wild inertia subgroup of $\mathrm{Gal}(L'/K^{nr})$ is isomorphic to $C_p$ , and the quotient by $C_p$ is the Galois group of the maximal tamely ramified subextension of $L'/K^{nr}$ , so it is cyclic. Now the image of it in $S_p$ is contained in the normaliser of $C_p$ , that is equal to $C_p \rtimes C_{p-1}$ . Therefore $\mathrm{Gal}(L'/K^{nr})$ injects into $C_p \rtimes C_{p-1}$ .

Putting all this together, if f is irreducible over $K^{nr}$ then $\mathrm{Gal}(L/K^{nr})$ has at most order $2 \cdot p(p-1)$ , with wild inertia subgroup of order p and a cyclic quotient of order at most $2(p-1)$ , corresponding to the maximal tame subextension. Since $L'/K^{nr}$ is an intermediate subextension with Galois group isomorphic to a subgroup of the Frobenius group $C_p \rtimes C_{p-1}$ , this concludes the proof.

This lemma shows that, for a curve defined as in $({*})$ acquiring good reduction over a wildly ramified extension, the size of the image of inertia under $\rho_\ell$ is at most $2p(p-1)$ . In fact equality can be achieved, and in this case we say that the curve has maximal inertia image. We now complete the Proof of Proposition 2·1.

Proof of Proposition 2·1. In Lemma 3·2, we proved that $|\rho_\ell(I_K)|$ divides $2p(p-1)$ , and that (with the same notation used in the proof) $[L\,:\,L'] \leq 2$ and $[L'\,:\,K^{nr}] \leq p(p-1)$ . Clearly, the second inequality is an equality precisely when the Galois group of the splitting field of f over $K^{nr}$ is $C_p \rtimes C_{p-1}$ . Moreover, we have:

$$L = L'\left( {\sqrt {{\alpha _2} - {\alpha _1}} } \right) = {K^{nr}}\left( {{\alpha _1}, \ldots ,{\alpha _p},\sqrt {{\alpha _2} - {\alpha _1}} } \right),$$

in fact at most one of the elements $\sqrt{\alpha_i-\alpha_j}$ is sufficient to generate L over L ^′ and by Remark 3·1 any of these elements works as they all have the same valuation. More precisely, the extension $L/L'$ is quadratic if and only if $\alpha_2-\alpha_1$ is not a square in L ^′, or equivalently it has odd valuation. We denote by $v_L$ , $v_{L'}$ the normalised valuations on L and L ^′ respectively. Then $v_L(\sqrt{\alpha_2-\alpha_1})=[L\,:\,L']v_{L'}(\alpha_2-\alpha_1)/2$ and since by definition $\Delta = \prod_{i > j} (\alpha_i - \alpha_j)^2$ , we have:

$${v_L}(\Delta ) = \left( \matrix{ p \cr 2 \cr} \right)2{v_L}\left( {{\alpha _2} - {\alpha _1}} \right) = \left( \matrix{ p \cr 2 \cr} \right)4{v_L}\left( {\sqrt {{\alpha _2} - {\alpha _1}} } \right) = 2p(p - 1){v_L}\left( {\sqrt {{\alpha _2} - {\alpha _1}} } \right);$$

on the other hand since the valuations on $K^{nr}$ and K agree on the elements of K we have $v_L(\Delta)=[L\,:\,K^{nr}]v_K(\Delta)$ .

Suppose that $|\rho_{\ell}(I_K)|=2p(p-1)$ , so $[L\,:\,K^{nr}]=2p(p-1)$ . In particular, $[L\,:\,L']=2$ and by the observation above this means $v_{L'}(\alpha_2-\alpha_1)$ is odd. Then simplifying from the equalities above we obtain that $v_K(\Delta)=v_L(\sqrt{\alpha_2-\alpha_1}) = v_{L'}(\alpha_2-\alpha_1)$ is odd and, as we already noted, $\mathrm{Gal}(L'/K^{nr}) \cong C_p \rtimes C_{p-1}$ . Conversely suppose that $\mathrm{Gal}(L'/K^{nr}) \cong C_p \rtimes C_{p-1}$ and that $v_K(\Delta)$ is odd. Then comparing the two expressions for $v_L(\Delta)$ and using that $[L'\,:\,K^{nr}]=p(p-1)$ we obtain

$$\left[ {L{\mkern 1mu} :{\mkern 1mu} L'} \right]{v_K}(\Delta ) = 2 \cdot {1 \over 2}\left[ {L{\mkern 1mu} :{\mkern 1mu} L'} \right]{v_{L'}}\left( {{\alpha _2} - {\alpha _1}} \right),$$

so $v_{L'}(\alpha_2-\alpha_1)=v_K(\Delta)$ is odd, which implies $[L\,:\,L']=2$ and therefore $[L\,:\,K^{nr}]$ $=2p(p-1)$ .

To conclude this subsection, we prove Proposition 2·2.

Proof of Proposition 2·2. Since $p-1$ is even, it follows from assumption (ii) that $v_K(\Delta)$ is odd. So, in order to prove Proposition 2·2 it suffices to prove that, if f is irreducible over K and $(v_K(\Delta),p-1)=1$ then $\mathrm{Gal}(L'/K^{nr}) \cong C_p \rtimes C_{p-1}$ , where L ^′ is as in the proofs of Lemma 3·2 and Proposition 2·1.

Let us denote by M the splitting field of f over K. First of all, since f is irreducible over K, then $\mathrm{Gal}(M/K)$ contains a subgroup H of order p. We observe that H is a p-Sylow subgroup of $\mathrm{Gal}(M/K)$ , since this group has a natural injection into $S_p$ , whose p-Sylow subgroups have order p. Therefore p divides exactly $[M\,:\,K]$ . Let U be the maximal unramified extension of K contained in M. If $p \mid [M\,:\,U]$ then H is the first ramification group of $\mathrm{Gal}(M/K)$ (see [Reference Greenberg and Serre11, Chapter IV, Section 1]). Otherwise, let $\alpha_1$ be any root of f in M and consider $K'=K(\alpha_1)$ , which has (prime) degree p over K. Since $p \nmid [M\,:\,U]$ , we have $K \subseteq K' \subseteq U$ . Now, $U/K$ is cyclic, therefore every subextension of it is Galois. In particular, $K'/K$ is Galois, thus K ^′ coincides with the splitting field of f, hence $K'=U=M$ . In both cases, H is normal in $\mathrm{Gal}(M/K)$ .

Now consider the element $\sqrt[p-1]{\Delta}$ . Using the expression of $\Delta$ in terms of the roots $\alpha_1,\dots,\alpha_p$ of f and the fact that $\alpha_i-\alpha_j$ all have the same valuation, we can prove that $\sqrt[p-1]{\Delta} \in L'$ . More precisely, we have

$$\root {p - 1} \of \Delta = {({\alpha _2} - {\alpha _1})^p}\root {p - 1} \of u ,\quad {\rm{where }} u = \prod\limits_{i > j} {{{\left( {{{{\alpha _i} - {\alpha _j}} \over {{\alpha _2} - {\alpha _1}}}} \right)}^2}} ;$$

note that u is an element of L ^′ with valuation 0 and so its $(p-1)$ -th root gives an unramified hence trivial extension of L ^′. Since $(v_K(\Delta),p-1)=1$ , we also have that $[K^{nr}(\sqrt[p-1]{\Delta})\,:\,K^{nr}]=p-1$ . Therefore $p-1$ divides both $[L'\,:\,K^{nr}]$ and $[M\,:\,K]$ . Now the inertia subgroup of $\mathrm{Gal}(M/K)$ is isomorphic to $\mathrm{Gal}(L'/K^{nr})$ , and it is normal with cyclic quotient. Therefore it must contain the subgroup H of $\mathrm{Gal}(M/K)$ isomorphic to $C_p$ . This proves that p and $p-1$ divide $[L'\,:\,K^{nr}]$ and as in the proof of Lemma 3·2 we conclude that $\mathrm{Gal}(L'/K^{nr}) \cong C_p \rtimes C_{p-1}$ .

3.2. The irreducible representations of the group $C_p \rtimes C_{2(p-1)}$

We now want to describe the representation induced from $\rho_\ell$ on $I_G$ . In order to do it, we make a digression on the irreducible representations of the group $C_p \rtimes C_{2(p-1)}$ .

Let $\sigma,\tau \in C_p \rtimes C_{2(p-1)}$ be as in the presentation given in Section 3.1, and denote by $\nu$ the element $\tau^{p-1}$ , that generates the extra $C_2$ contained in $C_p \rtimes C_{2(p-1)}$ . Note that $\nu$ is the only element of the subgroup $C_{2(p-1)}$ of $C_p \rtimes C_{2(p-1)}$ (except the identity) that commutes with $\sigma$ . The group $C_p \rtimes C_{2(p-1)}$ is the semidirect product of two abelian groups, $A= C_p$ and $H=C_{2(p-1)}$ , so we are in the setting of [Reference Serre15, Section 8·2]. Consider a set of representatives for the orbits of H in the group of characters of A. This set consists of two elements only, namely the trivial representation $\textbf{1}$ and a non-trivial character $\eta$ . Let $H_{\textbf{1}}$ (resp. $H_\eta$ ) denote the subgroup of H consisting of the elements that stabilise $\textbf{1}$ (resp. $\eta$ ). Then $H_{\textbf{1}} = H$ and $H_\eta=\langle \nu \rangle \cong C_2$ . Now for any irreducible representation $\xi$ of $H_{\bullet}$ we obtain a representation of G given by $\mathrm{Ind}_{A H_{\bullet}}^{C_p\rtimes C_{2(p-1)}} \bullet \otimes \xi$ . By [Reference Serre15, Section 8·2, Proposition 25] the representations obtained in this way are exactly all the irreducible representations of $C_p \rtimes C_{2(p-1)}$ . This proves the following lemma.

Out of these two $(p-1)$ -dimensional irreducible representations, exactly one is faithful. Since $H_\eta \cong C_2$ , the representation $\xi$ needed for the construction described above is either the trivial representation of $H_\eta$ , or the representation sgn, defined by $\mathrm{sgn}(\nu)=-1$ . So we obtain the two representations $\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta$ and $\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta \otimes \mathrm{sgn}$ .

1. (i) The representation $\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta$ is not faithful. In fact,

   \begin{align*}\mathrm{tr} \left(\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta\right) (1)=\mathrm{tr} \left(\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta\right) (\nu)= p-1.\end{align*}

2. (ii) The representation $\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta \otimes \mathrm{sgn}$ is faithful. In fact we have, for $s \in C_p\rtimes C_{2(p-1)}$ and for $t_1,\dots,t_{p-1}$ a set of representatives for $C_p\rtimes C_{2(p-1)}/C_{2p}$ :

   $${\rm{tr}}\left( {{\rm{Ind}}_{{C_{2p}}}^{{C_p} \rtimes {C_{2(p - 1)}}}\eta \otimes {\rm{sgn}}} \right)(s) = \sum\limits_{i{\kern 1pt} :{\kern 1pt} {t_i}st_i^{ - 1} \in {C_{2p}}} \eta \left( {{t_i}st_i^{ - 1}} \right){\rm{sgn}}(s).$$
   For the terms occurring in this sum (which are at most $p-1$ ) we have that $\eta(t_i s t_i^{-1})$ is some root of unity, and it is 1 if and only if $s=1$ . So for $s \neq 1$ we have a sum of at most $p-1$ roots of unity, different from 1, and therefore $\mathrm{tr} (\mathrm{Ind}_{C_{2p}}^{C_p\rtimes C_{2(p-1)}} \eta \otimes \mathrm{sgn}) (s) \neq p-1$ , or equivalently the representation is faithful.

Now we can determine the representation $\rho_\ell \big|_{I_K}$ . Since it factors through $I_G$ , which is isomorphic to $C_p \rtimes C_{2(p-1)}$ , we identify it with a representation of the finite group $C_p\rtimes C_{2(p-1)}$ , and as such, it is faithful. Furthermore, the discussion above implies the following lemma.

Proof. Suppose that $\rho_{\ell} \big|_{I_K}$ is reducible. Then, since it has dimension equal to $2g=p-1$ , it is the sum of $p-1$ one-dimensional representations, but in this case the image would be abelian. However, this representation factors through $I_G$ and it is faithful as a $I_G$ -representation, so since $I_G$ is non-abelian we have a contradiction. Therefore $\rho_\ell \big|_{I_K}$ must be irreducible.

Note that as a consequence of this, the representation $\rho_\ell$ is also irreducible. This proves the following result.

Remark 3·6. By [Reference Dokchitser, Dokchitser, Maistret and Morgan7, Theorem 10·1], the representation given by the action of $I_K$ on the first étale cohomology group $H^1_{\acute{e}t}(X \times_K \overline K,\mathbb{Q}_\ell)$ is given by

$${\rho ^*} = \gamma \otimes ({{\mathbb{Q}}_\ell }[R] \ominus {\bf{1}}),$$

where $\gamma$ is a certain character of order $2(p-1)$ . This is immediate from op. cit. since the cluster picture of X only contains the cluster R described in Remark 3·1.

Now, the representation $\rho_\ell\big|_{I_K}$ is dual to $\rho^*$ , by [Reference Milne4, Theorem 15·1], but since by [Reference Serre and Tate14, Theorem 2 (ii)] its character has integer values, it is in fact isomorphic to it.

4.1. The action of Frobenius

Suppose first that $n=f_{K/\mathbb{Q}_p}$ is even. Then $\mathrm{Frob}_F$ is central in $\mathrm{Gal}(L/K^{nr})$ (recall $L=F^{nr}$ ) and, since $\rho_\ell$ is irreducible, by Schur’s Lemma $\rho_\ell(\mathrm{Frob}_F)$ is a scalar matrix. Let $\lambda \in \overline{\mathbb{Q}}_\ell$ be such that $\rho_\ell(\mathrm{Frob}_F)=\lambda \mathrm{id}$ . Then we know $\det(\rho_\ell(\mathrm{Frob}_F))= |k|^g=p^{ng}$ and so $\lambda^{2g}=p^{ng}$ . On the other hand since $\rho_\ell(\mathrm{Frob}_F)$ is a scalar matrix then its characteristic polynomial is precisely $(T-\lambda)^{2g}$ , and by [Reference Deligne5, Theorem 1·6] it has integral coefficients, so $\lambda \in \mathbb{Z}$ and in particular $\lambda \in \{\pm p^{n/2}\}$ . Finally, since $F/K$ is Galois, we have $F=K(\mathrm{Jac}(X)[4])$ , so $\rho_\ell(\mathrm{Frob}_F)$ acts trivially modulo 4 and $\lambda \equiv 1 \pmod 4$ . Hence

\begin{align*}\lambda = \left(\left(\dfrac{-1}{p}\right) p\right)^{f_{K/\mathbb{Q}_p}/2}. \end{align*}

Suppose now that $n=f_{K/\mathbb{Q}_p}$ is odd. In general, $\rho_\ell(\mathrm{Frob}_F)$ acts as the nth power of the linear operator obtained for $n=1$ , so we may first assume that $n=1$ . Then the square of $\mathrm{Frob}_F$ is central in $\mathrm{Gal}(L/K^{nr})$ , hence the minimal polynomial of $\rho_\ell(\mathrm{Frob}_F)$ is of the form $T^2-\mu$ . As a consequence of the Weil Conjectures (see again [Reference Deligne5, Theorem 1·6]) we also have that the trace of $\rho_\ell(\mathrm{Frob}_F)$ is given by $p+1-|\tilde{X}_F(\mathbb{F}_p)|$ where $\tilde{X}_F$ is the reduction modulo p of $X \times_K F$ . Since for each $x \in \mathbb{F}_p$ , $x^p-x=0$ , we have precisely p affine points on $\tilde{X}_F$ , so $\mathrm{tr}(\rho_\ell(\mathrm{Frob}_F))=0$ . Therefore the characteristic polynomial of $\rho_\ell(\mathrm{Frob}_F)$ has g roots equal to $\sqrt{\mu}$ and g roots equal to $-\sqrt{\mu}$ , hence it is $(T^2-\mu)^g$ , and again it has constant term equal to $p^g$ and integer coefficients, hence $\mu \in \{\pm p\}$ . As in the previous case, we have $\mu \equiv 1 \pmod 4$ and so $\mu = \Big(\frac{-1}{p}\Big)p$ . Putting all this together, the eigenvalues of $\rho_\ell(\mathrm{Frob}_F)$ for generic odd $f_{K/\mathbb{Q}_p}$ are

\begin{align*}\pm \sqrt{\left(\dfrac{-1}{p}\right)p}^{\,f_{K/\mathbb{Q}_p}}, \end{align*}

each occurring g times.

Now we can prove Theorem 2·3 in the case of even $f_{K/\mathbb{Q}_p}$ .

Proof of Theorem 2·3 for even inertia degree. Since $f_{K/\mathbb{Q}_p}$ is even, we know $F/K$ is Galois with Galois group isomorphic to its inertia subgroup. We furthermore have

\begin{align*}\mathrm{Gal}(L/K)=\mathrm{Gal}(F/K) \times \mathrm{Gal} (K^{nr}/K),\end{align*}

since $L=F^{nr}=F K^{nr}$ . If we define $\chi$ as in the statement of Theorem 2·3 we have that $\rho_\ell(\mathrm{Frob}_F) = \chi(\mathrm{Frob}_F) \mathrm{id} = \chi(\mathrm{Frob}_K) \mathrm{id}$ and therefore if we let $\psi = \rho_\ell \otimes \chi^{-1}$ , then $\psi$ factors through $\mathrm{Gal}(F/K)$ which is isomorphic to $I_G$ , and as a representation of this group it is irreducible, faithful and $(p-1)$ -dimensional. By Lemma 3·5, there exists a unique such representation.

5.1. The irreducible representations of the group G

A set of representatives for the orbits of H in the group of characters of A consists, as in Section 3, only of the two elements $\textbf{1}$ and $\eta$ , for $\eta$ any non-trivial character. It is easy to check that, with the same notation as in Section 3, $H_{\textbf{1}}=H$ and $H_{\eta} = \langle \phi, \nu\rangle \cong C_2^2$ . All the representations of $H_{\textbf{1}}$ give rise to a representation of G of the same dimension. Now $H_{\textbf{1}} \cong C_{2(p-1)} \rtimes C_2$ is itself a semidirect product of two abelian subgroups, so using [Reference Serre15, Proposition 25] we have that all its irreducible representation have dimension dividing the order of the second subgroup, that is either 2 or 1. However $\psi$ is irreducible of dimension $p-1$ (since $\rho_\ell$ is), so unless $p=3$ it cannot arise from such a representation. For $p=3$ we need a more direct approach, and this case is dealt with in [Reference Coppola1, Section 3], so we can assume $p \neq 3$ .

Now let us consider the representations arising from $H_\eta$ . Since this group is abelian, it only has 1-dimensional irreducible representations, namely those given by the following characters:

\begin{align*} \begin{array}{c|r@{\quad}r@{\quad}r@{\quad}r} \rm class& 1 & \nu & \phi & \nu \phi\cr \hline \xi_{1}&1&1&1&1\cr \xi_{2}&1&1&-1&-1\cr \xi_{3}&1&-1&1&-1\cr \xi_{4}&1&-1&-1&1\cr \end{array}\end{align*}

The irreducible representations arising from these four representations are

\begin{align*}\mathrm{Ind}_{C_p \times C_2^2}^G \xi_j \otimes \eta \end{align*}

for $j \in \{1,\dots,4\}$ (note that the subgroup of G isomorphic to $C_p \rtimes C_2^2$ is in fact a direct product). In particular these representations have dimension equal to $[G\,:\,C_p \times C_2^2]=p-1$ . Following the same proof as in Lemma 3·5 we have that only the representations arising from $\xi_3$ and $\xi_4$ are faithful, so $\psi$ is one of these two.

Let $\sigma,\tau$ be the generators of $I_G$ , as in the previous sections.

Proof. First of all, it is easy to check that $C_p \times C_2^2$ is a normal subgroup of G. Moreover a set of representatives for $G/(C_p \times C_2^2)$ is given by $\tau,\tau^2,\dots,\tau^{p-1}$ . We have

\begin{align*}\mathrm{tr} \left(\psi_j (\sigma \phi)\right)= \sum_{i=1}^{p-1} \left(\xi_{j+2} \otimes \eta\right) \left(\tau^i\sigma \phi\tau^{-i}\right)= \sum_{i=1}^{p-1} \xi_{j+2} \left(\tau^i \phi \tau^{-i}\right) \eta \left(\tau^{i} \sigma \tau^{-i}\right).\end{align*}

By the relation $\phi \tau \phi = \tau^p$ we deduce $\tau^2 \phi = \phi \tau^2$ ; so if i is even then $\xi_{j+2} (\tau^i \phi \tau^{-i}) = \xi_{j+2}(\phi)$ , and if i is odd then $\xi_{j+2} (\tau^i \phi \tau^{-i}) = \xi_{j+2}(\phi \nu) = - \xi_{j+2}(\phi)$ (recall that $\nu = \tau^{p-1}$ ). On the other hand, since $\tau \sigma \tau^{-1} = \sigma^b$ , then $\eta(\tau^i \sigma \tau^{-i})$ varies among all the powers of $\eta(\sigma)$ , which is a primitive pth root of unity; without loss of generality we can assume it is $e^{2 \pi i/p}$ , seen as a complex number. Note that

\begin{align*}\left(\dfrac{b^i}{p}\right)=({-}1)^i = \dfrac{\xi_{j+2} \left(\tau^i \phi \tau^{-i}\right)}{\xi_{j+2}(\phi)},\end{align*}

therefore

\begin{align*}\mathrm{tr} \psi_j (\sigma \phi) = \sum_{i=1}^{p-1} ({-}1)^i \xi_{j+2}(\phi) \eta(\sigma)^{b^i} = \xi_{j+2} (\phi) \sum_{a=1}^{p-1} \left(\dfrac{a}{p}\right) \left(e^{2 \pi i/p}\right)^a = \xi_{j+2} (\phi) \sqrt{\left(\dfrac{-1}{p}\right) p},\end{align*}

where the last equality follows from the Gauss summation formula.

5.2. The proof of Theorem 2·3

Proof. Let $\beta(x,y) = (x',y')$ be the change of variables described in the proof of Lemma 4·1, let “red” be the reduction map: $X (\overline{K}) \rightarrow \tilde{X}(\overline{k})$ and “lift” be any section of it (which exists by smoothness of $\tilde{X}$ ). Then we can compute the action of a Galois automorphism $\gamma$ on the reduced curve $\tilde{X}(\overline{k})$ via the composition $\mathrm{red} \circ \beta \circ \gamma \circ \beta^{-1} \circ \mathrm{lift}$ . In particular we will do it for $\gamma = \sigma \mathrm{Frob}_K$ ; then we have that

\begin{align*}\mathrm{tr} (\rho_\ell (\sigma \mathrm{Frob}_K)) = |k| + 1 - A, \end{align*}

where A is the number of points on the reduced curve fixed by the map $\mathrm{red} \circ \beta \circ \sigma \mathrm{Frob}_K \circ \beta^{-1} \circ \mathrm{lift}$ constructed above (see [Reference Dokchitser, Dokchitser and Morgan8, Theorem 1·5 and Remark 1·7] and [Reference Dokchitser and Dokchitser9, Section 6·5]).

Let $(\tilde x, \tilde y ) \in \tilde X(\overline{k})$ , then since $\sigma$ is a wild inertia element we have

\begin{align*} & \left(\tilde{x},\tilde{y}\right) \xrightarrow{\mathrm{lift}} \left(x,y\right) \xrightarrow{\beta^{-1}} \left(x\!\left(\alpha_2-\alpha_1\right) +\alpha_1,y \left(\sqrt{\alpha_2-\alpha_1}\right)^p\right) \\[4pt] & \xrightarrow{\sigma \mathrm{Frob}_K} \left(\sigma\!\left(\mathrm{Frob}_K(x)\right) \sigma\!\left(\alpha_2-\alpha_1\right) + \alpha_2, \sigma\!\left(\mathrm{Frob}_K(y)\right)\left(\sigma\!\left(\sqrt{\alpha_2-\alpha_1}\right)\right)^p\right)\\[4pt] & \xrightarrow{\beta} \left(\dfrac{\sigma\!\left(\mathrm{Frob}_K(x)\right) \sigma\!\left(\alpha_2-\alpha_1\right) + \alpha_2-\alpha_1}{\alpha_2-\alpha_1}, \sigma(\mathrm{Frob}_K(y))\dfrac{\left(\sigma\!\left(\sqrt{\alpha_2-\alpha_1}\right)\right)^p}{\left(\sqrt{\alpha_2-\alpha_1}\right)^p}\right)\\[4pt] & =\left( \sigma\!\left(\mathrm{Frob}_K(x)\right) \dfrac{\sigma\!\left(\alpha_2-\alpha_1\right)}{\alpha_2-\alpha_1}+1, \sigma\!\left(\mathrm{Frob}_K(y)\right)\dfrac{\left(\sigma\!\left(\sqrt{\alpha_2-\alpha_1}\right)\right)^p}{\left(\sqrt{\alpha_2-\alpha_1}\right)^p}\right) \xrightarrow{\mathrm{red}} \left(\tilde{x}^{|k|}+1,\tilde{y}^{|k|}\right). \end{align*}

Here we use the following facts: since $\sigma$ belongs to $I_K$ , for every algebraic integer $x \in F$ , x and $\sigma(x)$ reduce to the same element of $\overline{k}$ . If in addition $x \neq 0$ then, since $\sigma$ is wild, $\sigma(x)/x$ reduces to 1 (again by [Reference Coppola1, Lemma 3·3]). Finally, by definition $\mathrm{Frob}_K(x)$ reduces to $\tilde{x}^{|k|}$ . We deduce that A is equal to the number of solutions (including the point at infinity) of the following system of equations:

(5.1) $$\left\{ {\matrix{ {x = {x^{|k|}} + 1} \cr {y = {y^{|k|}}} \cr {{y^2} = {x^p} - x;} \cr } } \right.$$

As in Section 4.1 let $n = f_{K/\mathbb{Q}_p}$ , so $|k|=p^n$ . If $n=1$ , then this system has 0 affine solutions if $p \equiv 3 \pmod 4$ and 2p affine solutions if $p \equiv 1 \pmod 4$ , so $A=1$ or $2p+1$ respectively. Therefore

$${\rm{tr}}({\rho _\ell }(\sigma {\rm{Fro}}{{\rm{b}}_K})) = - \left( {{{ - 1} \over p}} \right)p,$$

and so $\mathrm{tr} (\psi(\sigma \phi))=\dfrac{\mathrm{tr} (\rho_\ell(\sigma \mathrm{Frob}_K))}{\chi(\mathrm{Frob}_K)}=\dfrac{-\left(\dfrac{-1}{p}\right)p}{\sqrt{\left(\dfrac{-1}{p}\right)p}} = - \sqrt{\left(\dfrac{-1}{p}\right) p}$ .

For general odd n, we obtain the same system of equations independently of the curve X we use, as long as it is defined over a p-adic field K with $f_{K/\mathbb{Q}_p}=n$ and it satisfies the conditions of Theorem 2·3. Let $X/\mathbb{Q}_p \,:\, y^2= x^p - p$ as in Remark 2·4, and let $X_K$ be the base change of X to the field K given by the unique unramified extension of $\mathbb{Q}_p$ of degree n. The polynomial $x^p-p$ is irreducible over K, as any root gives a ramified extension, and $v_{K}(\Delta)=v_{\mathbb{Q}_p}(\Delta)=2p-1$ is coprime to $p-1$ . Let $\rho^{\prime}_\ell$ be the $\ell$ -adic Galois representation attached to $\mathrm{Jac}(X)$ and let $\rho_\ell$ be the $\ell$ -adic Galois representation attached to $\mathrm{Jac}(X_K)$ ; then:

1. (i) $\rho^{\prime}_\ell$ and $\rho_\ell$ have the same restriction to the inertia subgroup;

2. (ii) $\rho_\ell(\mathrm{Frob}_K)$ acts as the nth power of $\rho^{\prime}_\ell (\mathrm{Frob}_{\mathbb{Q}_p})$ .

So $\rho_\ell(\sigma)\rho_\ell(\mathrm{Frob}_K) =\rho^{\prime}_\ell (\sigma)\rho^{\prime}_\ell (\mathrm{Frob}_{\mathbb{Q}_p})^n$ . Notice that, by Section 4.1, since $n-1$ is even, we have that $\rho^{\prime}_\ell (\mathrm{Frob}_{\mathbb{Q}_p})^{n-1}$ is the scalar matrix with eigenvalue $\left(\Big(\frac{-1}{p}\Big)p\right)^{(n-1)/2}$ . Therefore

\begin{align*}\mathrm{tr} (\rho_\ell (\sigma \mathrm{Frob}_K)) = \left(\left(\dfrac{-1}{p}\right)p\right)^{(n-1)/2} \mathrm{tr} (\rho^{\prime}_\ell (\sigma \mathrm{Frob}_{\mathbb{Q}_p})) = -\left(\left(\dfrac{-1}{p}\right)p\right)^{(n+1)/2}.\end{align*}

We conclude, since

\begin{align*}\mathrm{tr} (\psi(\sigma \phi))=\dfrac{\mathrm{tr} (\rho_\ell(\sigma \mathrm{Frob}_K))}{\chi(\mathrm{Frob}_K)}=\dfrac{-\left(\left(\dfrac{-1}{p}\right)p\right)^{(n+1)/2}}{\sqrt{\left(\dfrac{-1}{p}\right)p}^{\,n}} = - \sqrt{\left(\dfrac{-1}{p}\right) p}.\end{align*}