1 Introduction

Let A be an abelian variety over a field k of characteristic \(\ge 0\). The Mordell–Weil theorem asserts that if k is furthermore a number field, then the set of k-rational points A(k) on A is a finitely generated abelian group. The rank of this abelian group is referred to as the rank of the abelian variety A with notation \(\mathrm{rank}(A(k))\). One interesting problem in arithmetic geometry is to find abelian varieties with arbitrarily large ranks.

The present paper is a generalization of the work of Salami in [5] to arbitrary Galois coverings. More precisely, let \(n\in {\mathbb {N}}\) be a natural number such that the characteristic of the field k in the previous paragraph does not divide n and k contains a primitive n-th root of unity \(\xi \). Suppose there is an embedding \(G\hookrightarrow {{\,\mathrm{Aut}\,}}(A)\) with G a finite group with \(|G|=n\). Further, let X, Y be smooth projective varieties over k with function fields F, K respectively. We first define the notion of Prym variety associated to a G-Galois covering \(f:X\rightarrow Y\). We will show that this is a natural generalization of the notion of Prym variety in the case of coverings of curves. The group G is well-known to be also the Galois group of the Galois field extension F/K. By the results of [3, 4], the twist of A by the extension F/K is equivalent to a twist by the 1-cocyle \(a=(a_g)\in Z^1(G,{{\,\mathrm{Aut}\,}}(A))\) given by \(a_g=g\), where g is viewed both as a group element and as an automorphism of A corresponding to \(g\in G\) (in other words, we identify g with its image \(g\in G\hookrightarrow {{\,\mathrm{Aut}\,}}(A)\)). We prove a theorem which gives an isomorphism for \(A_a(K)\) in terms of the Prym variety introduced in Section 2, see Theorem 2.4. Then we consider the n-times product \(\prod _i f\) of a G-cover \(f:X\rightarrow Y\) with itself. We prove a result which describes the Prym variety of the self product of G-covers of varieties in terms of the Prym variety of the cover. We apply this to the case of an abelian cover \(C\rightarrow {\mathbb {P}}^1\). More precisely, we prove that \(\mathrm{rank}({{\,\mathrm{Jac}\,}}(C)_a(K))\) can be made arbitrarily large by taking n large enough.

2 The Prym variety of a Galois covering

Let V be a smooth projective variety over a field k. The Albanese variety of V is by definition the initial object for the morphisms from V to abelian varieties. In particular, if V is moreover defined over a field of characteristic zero (the classical case), it is the abelian variety

$$\begin{aligned} {{\,\mathrm{Alb}\,}}(V)=H^0(V,\Omega ^1_V)^*/H_1(V,{\mathbb {Z}}). \end{aligned}$$
(2.0.1)

Now let X, Y be smooth projective varieties with \(f:X\rightarrow Y\) a Galois covering with the Galois group G. Throughout this paper, we consider the situation where there is a finite group G with \(|G|=n\) that has a G-action on X such that \(Y:=X/G\) and \(f:X\rightarrow Y\) is the quotient map. The universal property of the Albanese variety implies that there is an induced action of G on \({{\,\mathrm{Alb}\,}}(X)\). We denote by \({({{\,\mathrm{Alb}\,}}(X)^G)}^0\) the largest abelian subvariety of \({{\,\mathrm{Alb}\,}}(X)\) fixed (pointwise) under this action. Equivalently, \({({{\,\mathrm{Alb}\,}}(X)^G)}^0\) is the connected component containing the identity of the subvariety \({{\,\mathrm{Alb}\,}}(X)^G\) of fixed points of \({{\,\mathrm{Alb}\,}}(X)\) under the action of G. In characteristic zero, the induced action of G on \({{\,\mathrm{Alb}\,}}(X)\) can be described through its action on \(H^0(X,\Omega ^1_X)\). With respect to this action, we write

$$\begin{aligned} H^0(X,\Omega ^1_X)^+= & {} H^0(X,\Omega ^1_X)^G(\cong H^0(Y,\Omega ^1_Y)) \text { and }\nonumber \\ H^0(X,\Omega ^1_X)^-= & {} H^0(X,\Omega ^1_X)/H^0(X,\Omega ^1_X)^+= \bigoplus \limits _{\chi \in {{\,\mathrm{Irr}\,}}(G)\setminus \{1\}}H^0({\widetilde{C}}, \omega _{{\widetilde{C}}})^{\chi }.\nonumber \\ \end{aligned}$$
(2.0.2)

Similarly, one defines \(H_1(X,{\mathbb {Z}})^-\). Here \({{\,\mathrm{Irr}\,}}(G)\) is the set of irreducible characters of G.

Let \(g\in G\) be an element of the group G. We will denote also by g the automorphism of \({{\,\mathrm{Alb}\,}}(X)\) induced by g. Using this notation and following that of [8, Prop. 3.1] for the case of curves, we write \({{\,\mathrm{Nm}\,}}G:{{\,\mathrm{Alb}\,}}(X)\rightarrow {{\,\mathrm{Alb}\,}}(X)\) for the norm endomorphism of \({{\,\mathrm{Alb}\,}}(X)\) given by \({{\,\mathrm{Nm}\,}}G:=\sum _{g\in G}g\).

Now we are ready to define the notion of Prym variety for the covering f.

Definition 2.1

Let \(f:X\rightarrow Y\) be a G-Galois covering of smooth projective varieties over a field k. The Prym variety P(X/Y) associated with the covering f is defined to be

$$\begin{aligned} P(X/Y):=\frac{{{\,\mathrm{Alb}\,}}(X)}{{({{\,\mathrm{Alb}\,}}(X)^G)}^0}. \end{aligned}$$
(2.1.1)

For the classical case of Prym varieties of double coverings of curves, we refer to [1]. The following result gives an equivalent description of the Prym variety and shows furthermore that it coincides with the Prym variety of covers of curves (see [1, 6,7,8]) up to isogeny.

Proposition 2.2

With notation and assumptions of Definition 2.1, it holds that \(P(X/Y)=\frac{{{\,\mathrm{Alb}\,}}(X)}{{{\,\mathrm{Im}\,}}{{\,\mathrm{Nm}\,}}G}\). Furthermore it is isogeneous to the abelian variety \({(\ker {{\,\mathrm{Nm}\,}}G)}^0\). In particular, if the varieties are defined over a field of characteristic zero, then P(X/Y) is isogeneous to the abelian variety \({{\,\mathrm{Alb}\,}}(X)^-={(H^0(X,\Omega ^1_X)^-)}^*/H_1(X,{\mathbb {Z}})^-\).

Proof

The induced linear action on the tangent space of \({{\,\mathrm{Alb}\,}}(X)\) at the origin gives that \(\dim {{\,\mathrm{Alb}\,}}(X)=\dim {(\ker {{\,\mathrm{Nm}\,}}G)}^0+\dim {({{\,\mathrm{Alb}\,}}(X)^G)}^0\). Let \(Q\in {({{\,\mathrm{Alb}\,}}(X)^G)}^0\cap {(\ker {{\,\mathrm{Nm}\,}}G)}^0\). It follows that \(nQ=0\) in \({{\,\mathrm{Alb}\,}}(X)\), therefore \({({{\,\mathrm{Alb}\,}}(X)^G)}^0\cap {(\ker {{\,\mathrm{Nm}\,}}G)}^0\subseteq {{\,\mathrm{Alb}\,}}(X)[n]\). In particular, this intersection is finite so that \({{\,\mathrm{Alb}\,}}(X)\) is isogenous to the product \({({{\,\mathrm{Alb}\,}}(X)^G)}^0\times {(\ker {{\,\mathrm{Nm}\,}}G)}^0\). Since \((\sum _{g\in G}g)\cdot h=\sum _{g\in G}(g\cdot h)=\sum _{g\in G}g\) for every \(h\in G\), it follows that \({{\,\mathrm{Im}\,}}{{\,\mathrm{Nm}\,}}G\subseteq {({{\,\mathrm{Alb}\,}}(X)^G)}^0\). On the other hand, the above isogeny decomposition for \({{\,\mathrm{Alb}\,}}(X)\) implies that \(\dim {{\,\mathrm{Im}\,}}{{\,\mathrm{Nm}\,}}G=\dim {{\,\mathrm{Alb}\,}}(X)-\dim {(\ker {{\,\mathrm{Nm}\,}}G)}^0=\dim {({{\,\mathrm{Alb}\,}}(X)^G)}^0\). Hence \({{\,\mathrm{Im}\,}}{{\,\mathrm{Nm}\,}}G={({{\,\mathrm{Alb}\,}}(X)^G)}^0\). This completes the proof of the claims. \(\square \)

Remark 2.3

For curves, the Albanese variety coincides with the Jacobian variety. Hence if X and Y are curves, the Prym variety P(X/Y) coincides with the Prym variety for covers of curves, see [6, 7]. In this case, there are two fundamental homomorphisms: the norm homomorphism \({{\,\mathrm{Nm}\,}}_f:{{\,\mathrm{Jac}\,}}(X)\rightarrow {{\,\mathrm{Jac}\,}}(Y)\) and the pull-back homomorphism \(f^*:{{\,\mathrm{Jac}\,}}(Y)\rightarrow {{\,\mathrm{Jac}\,}}(X)\) and it holds that \({{\,\mathrm{Nm}\,}}G=f^*\circ {{\,\mathrm{Nm}\,}}_f\), so that \(P(X/Y)={(\ker {{\,\mathrm{Nm}\,}}_f)}^0={(\ker {{\,\mathrm{Nm}\,}}G)}^0\), see [8, Prop. 3.1].

We recall from [3, 4] the notion of twist of a smooth projective variety. Suppose \(k^{\prime }/k\) is a G-Galois extension of fields. Let \({{\mathcal {A}}}\) be a G-set and suppose E is another G-set which is also a left \({{\mathcal {A}}}\)-set. Let \(a=(a_g)\in Z^1(G,{{\,\mathrm{Aut}\,}}({{\mathcal {A}}}))\) be a 1-cocycle of \({{\mathcal {A}}}\). For any \(g\in G\) and \(x\in E\), we denote by \(^gx\) the left action of g on x. The G-set E with this action of G is denoted by \(E_a\) and is called the twist of E by a. If X is a projective smooth variety over k, we denote by \({{\,\mathrm{Aut}\,}}(X)\) the automorphism scheme of X and let \(a=(a_g)\in Z^1(G,{{\,\mathrm{Aut}\,}}(X))\). There exist a projective variety Y over k and a \(k^{\prime }\)-isomorphism \(f^{\prime }:X\otimes _k k^{\prime }\rightarrow Y\otimes _k k^{\prime }\) such that \(^g(f^{\prime })=f^{\prime }\circ a_g\) for every \(g\in G\). We denote the variety Y by \(X_a\) and call it the twist of X by a. Of course the two notions of twist introduced above are compatible in the sense that the map \(f^{\prime }:X(k^{\prime })\rightarrow Y(k^{\prime })\) induces an isomorphism between \(X(k^{\prime })_a\cong X_a(k^{\prime })(=Y(k^{\prime }))\).

In this paper, the above results will be applied to an abelian variety A and its twist \(A_a\) by the extension F/K or equivalently by the 1-cocycle \(a=(a_g)\in Z^1(G,{{\,\mathrm{Aut}\,}}(A))\) given by \(a_g=g\).

Theorem 2.4

Assume that there exists a k-rational point \(x_0\in X(k)\). Then there is an isomorphism of abelian groups

$$\begin{aligned} A_a(K)\cong \mathrm{Hom}_k(P(X/Y),A)\oplus A[n](k). \end{aligned}$$
(2.4.1)

If in addition P(X/Y) is k-isogenous to \(A^n\times B\), where \(n\in {\mathbb {N}}\) and B is an abelian variety over k that does not have any simple components k-isogenous to A, then \(\mathrm{rank}(A_a(K))=n\cdot \mathrm{rank}({{\,\mathrm{End}\,}}_k(A))\).

Proof

Note that the universal property of the Albanese variety together with the fact that a regular morphism of abelian varieties is obtained by a homomorphism followed by a translation (see, for example, [2, Prop. 1.2.1]) shows that

$$\begin{aligned} A(K)\cong \mathrm{Hom}_k({{\,\mathrm{Alb}\,}}(X),A)\oplus A(k). \end{aligned}$$
(2.4.2)

We assume that the Albanese map \(\iota _X:X\rightarrow {{\,\mathrm{Alb}\,}}(X)\) satisfies \(\iota _X(x_0)=0\) so that it is defined over k. Recall that [4, Prop. 1.1] shows that

$$\begin{aligned} A_a(K)\cong \{P\in A(F)\mid a_g\cdot ^g(P)=P \}. \end{aligned}$$

This implies in our particular case that for any \(g\in G\), viewed as remarked earlier also as an automorphism of \({{\,\mathrm{Alb}\,}}(X)\), we have that \((\alpha ,Q)\in A_a(K)\) if and only if \(g(\alpha \circ g,Q)=(\alpha ,Q)\) or equivalently \((\alpha \circ g,Q)=g^{-1}(\alpha ,Q)\). But this is the case if and only if \(\alpha \) annihilates \({{\,\mathrm{Im}\,}}{{\,\mathrm{Nm}\,}}G\) and \(Q\in A[n](k)\). Proposition 2.2 then shows that it must actually lie in P(X/Y), so we obtain the claimed isomorphism in (2.4.1).

If moreover P(X/Y) is isogenous to \(A^n\times B\) with n, A, and B as in the statement of the proposition, then

$$\begin{aligned} A_a(K)&\cong \mathrm{Hom}_k(P(X/Y),A)\oplus A[n](k)\\&\cong \mathrm{Hom}_k(A^n\times B,A)\oplus A[n](k)\\&\cong \mathrm{Hom}_k(A^n,A)\oplus \mathrm{Hom}_k(B,A)\oplus A[n](k)\\&\cong {{\,\mathrm{End}\,}}_k(A)^n\oplus \mathrm{Hom}_k(B,A)\oplus A[n](k). \end{aligned}$$

Which implies that as \({\mathbb {Z}}\)-modules, it holds that \(\mathrm{rank}(A_a(K))= n\cdot \mathrm{rank}({{\,\mathrm{End}\,}}_k(A))\). \(\square \)

Given a G-Galois covering \(f:X\rightarrow Y\), one can form the n-fold self product \(\prod _{i=1}^n f:\prod _{i=1}^n X\rightarrow \prod _{i=1}^n Y\) which is a \(\underbrace{G\times \cdots \times G}_{n-times}\)- Galois covering. Then the diagonal embedding \(G\hookrightarrow \prod _i G:=G\times \cdots \times G\) gives a subgroup of \(\prod _i G\) isomorphic to G. We denote this subgroup by \({\tilde{G}}\). This gives rise to an intermediate Galois covering \(f:\prod _{i=1}^n X\rightarrow (\prod _{i=1}^n X)/{\tilde{G}}\). Let us write \({\mathcal {X}}=\prod _{i=1}^n X\) and \({\mathcal {Y}}=(\prod _i X)/{\tilde{G}}\). We are interested in the Prym variety \(P({\mathcal {X}}/{\mathcal {Y}})\). In fact, we show

Proposition 2.5

With the above notation, there is an isogeny

$$\begin{aligned} P({\mathcal {X}}/{\mathcal {Y}})\sim _k \prod _i P(X_i/ Y_i). \end{aligned}$$
(2.5.1)

Proof

It suffices to only treat the case \(n=2\). The general case follows by an induction argument. So suppose \(n=2\) and denote the Galois group of the cover \({\mathcal {X}}/{\mathcal {Y}}\) by \({\tilde{G}}(\cong G)\). By Proposition 2.2, it suffices to show that there is a k-isogeny \({(\ker {{\,\mathrm{Nm}\,}}{\tilde{G}})}^0\sim _k{(\ker {{\,\mathrm{Nm}\,}}G)}^0\times {(\ker {{\,\mathrm{Nm}\,}}G)}^0\). In fact, we show that there is an isomorphism between these abelian varieties. Notice that there is an isomorphism

$$\begin{aligned} \beta :{{\,\mathrm{Alb}\,}}(X_1)\times {{\,\mathrm{Alb}\,}}(X_2)\xrightarrow {\sim }{{\,\mathrm{Alb}\,}}({\mathcal {X}}). \end{aligned}$$
(2.5.2)

The isomorphism \(\beta \) is given as follows: Let \(j_i:X_i\rightarrow {\mathcal {X}}\), for \(i=1,2\), be the natural inclusions. Then \(\beta =\tilde{j_1}+\tilde{j_2}\), where \(\tilde{j_i}\) denotes the induced homomorphism \({{\,\mathrm{Alb}\,}}(X_i)\rightarrow {{\,\mathrm{Alb}\,}}({\mathcal {X}})\). This isomorphism is compatible with the action of \({\tilde{G}}\), namely, there is the following commutative diagram. From this, one deduces the isomorphism \(\ker {{\,\mathrm{Nm}\,}}{\tilde{G}}\xrightarrow {\sim }\ker {{\,\mathrm{Nm}\,}}G\times \ker {{\,\mathrm{Nm}\,}}G\) which implies the desired isomorphism. \(\square \)

Consider an abelian cover \(C\rightarrow {\mathbb {P}}^1\) with Galois group G. Consider the product \({\mathcal {C}}_n=\prod _{i=1}^n C\), i.e., the product of n copies of the same abelian cover in the above and let \({\tilde{G}}\) be the image of G under the diagonal embedding \(G\hookrightarrow \prod _{i=1}^n G\) as above. Set \({\mathcal {D}}_n={\mathcal {C}}_n/{\tilde{G}}\). By Proposition 2.5, we have that

$$\begin{aligned} P({\mathcal {C}}_n/{\mathcal {D}}_n)=\prod _i P(C/{\mathbb {P}}^1). \end{aligned}$$
(2.5.3)

By Remark 2.3, \(P(C/{\mathbb {P}}^1)={(\ker {{\,\mathrm{Nm}\,}}_f)}^0\). However, as \({{\,\mathrm{Jac}\,}}({\mathbb {P}}^1)=0\), it follows that \(P(C/{\mathbb {P}}^1)={{\,\mathrm{Jac}\,}}(C)\). Now Proposition 2.5 gives that

$$\begin{aligned} P({\mathcal {C}}_n/{\mathcal {D}}_n)=\prod _i P(C/{\mathbb {P}}^1)=({{\,\mathrm{Jac}\,}}(C))^n. \end{aligned}$$
(2.5.4)

Note that the function field K(C) of C is generated over the function field \(K({\mathbb {P}}^1)=K(z)\) of \({\mathbb {P}}^1\) by taking roots of (transcendental) elements of K(C), i.e., it is of the form \(K(z)(x_1^{1/m},\dots , x_r^{1/m})\). Then the function field \({\mathcal {L}}_n\) of \({\mathcal {C}}_n\) is \(K(z)(x_{i1}^{1/m},\dots , x_{ir}^{1/m}), i=1,\dots , n\). Let \(K=k({\mathcal {D}}_n)\) be the function field of \({\mathcal {D}}_n\).

We define the 1-cocycle \(Z^1(G,{{\,\mathrm{Aut}\,}}(C))\) by \(a_g=g\). Let \({{\,\mathrm{Jac}\,}}(C)_a\) be the twist corresponding to this 1-cocycle. By applying (2.4.1) and (2.5.4), it follows that

$$\begin{aligned} {{\,\mathrm{Jac}\,}}(C)_a(K)&\cong \mathrm{Hom}_k(P({\mathcal {C}}_n/{\mathcal {D}}_n),{{\,\mathrm{Jac}\,}}(C))\oplus {{\,\mathrm{Jac}\,}}(C)[n](k)\\&\cong \mathrm{Hom}_k(({{\,\mathrm{Jac}\,}}(C))^n,{{\,\mathrm{Jac}\,}}(C))\oplus {{\,\mathrm{Jac}\,}}(C)[n](k)\\&\cong {{\,\mathrm{End}\,}}_k({{\,\mathrm{Jac}\,}}(C))^n\oplus {{\,\mathrm{Jac}\,}}(C)[n](k). \end{aligned}$$

So that \(\mathrm{rank}({{\,\mathrm{Jac}\,}}(C)_a(K))= n\cdot \mathrm{rank}({{\,\mathrm{End}\,}}_k({{\,\mathrm{Jac}\,}}(C)))\). Let us see an example.

Example 2.6

An abelian Galois cover \(C\rightarrow {\mathbb {P}}^{1}\) is determined by a collection of equations in the following way:

Consider an \(m\times s\) matrix \(A=(r_{ij})\) whose entries \(r_{ij}\) are in \({\mathbb {Z}}/N{\mathbb {Z}}\) for some \(N\ge 2\). Let \(\overline{k(z)}\) be the algebraic closure of k(z). For each \(i=1,...,m,\) choose a function \(w_{i}\in \overline{k(z)}\) with

$$\begin{aligned} w_{i}^{N}=h_i(z)=\prod _{l=1}^{s}(z-z_{l})^{{\widetilde{r}}_{li}}\text { for }i=1,\dots , m. \end{aligned}$$
(2.6.1)

Here \({\widetilde{r}}_{ij}\) is the lift of \(r_{ij}\) to \({\mathbb {Z}} \cap [0,N)\) and \(z_j\in k\) for \(j=1,2,\dots , s\). Notice that (2.6.1) gives in general only a singular affine curve and we take a smooth projective model associated to this affine curve. The Galois group \({\tilde{G}}\) of the covering is a subgroup of \(({{\mathbb {Z}}}/N{{\mathbb {Z}}})^m\). We will assume for simplicity that \({\tilde{G}}=({{\mathbb {Z}}}/N{{\mathbb {Z}}})^m\). This is the case if the rows of the matrix are linearly independent over \({{\mathbb {Z}}}/N{{\mathbb {Z}}}\), for example if N is a prime number. Now consider the n-fold product \({{\mathcal {C}}}_n\) as introduced above. Denoting the independent transcendental variables on the copy \(C_i\) of C by \(z_i\) and \(w_{ij}\) (\(j=1,\dots ,m\)), we see that the function field of \({{\mathcal {C}}}_n\) is \(L=k(z_1,\dots ,z_n,w_{ij})\) (\(i=1,\dots ,n, j=1,\dots ,m\)). The branch points of each copy \(C_i\) as in (2.6.1) will be denoted by \(z_{il}, l=1,\dots , s\). Then the function field \(K=k({{\mathcal {D}}}_n)\) of \({{\mathcal {D}}}_n\) is the set of \({\tilde{G}}\)-invariant elements of L under the action of \({\tilde{G}}\), i.e., \(K=k(z_1,\dots ,z_n,w_{ij})^{{\tilde{G}}}\). In this case, \(K=k(z_1,\dots ,z_n,w_{1j}^{N-1}w_{ij})\) (\(i\ne 1\)). Since \((w_{1j}^{N-1}w_{ij})^N=(\prod _{l=1}^{s}(z_1-z_{1l})^{{\widetilde{r}}_{lj}})^{N-1}\prod _{l=1}^{s}(z_i-z_{il})^{{\widetilde{r}}_{ij}}, j=1,\dots , m\), by setting \(W_{ij}=w_{1j}^{N-1}w_{ij}\), one obtains a new abelian cover with function field K such that the extension \(L=K(w_{1j}\mid j=1,\dots , m)\) is also abelian with Galois group \({\tilde{G}}\). So we can also write \(L=k(z_1)(u_1,\dots ,u_m)\) where \(u_i=\root N \of {h_i(z_1)}\text { for }i=1,\dots , m\) (\(h_i\) as in (2.6.1)). Suppose \(C_a\) is the twist of the abelian cover C by the extension \(L\mid K\). Arguing as in [4, Corollary 3.1], one sees that \(C_a\) can be defined by the equation \(\prod _{l=1}^{s}(z_1-z_{1l})^{{\widetilde{r}}_{ij}}W_i^N=\prod _{l=1}^{s}(z-z_{j})^{{\widetilde{r}}_{ij}}, i=1,\dots ,m\), or in the notation of (2.6.1), \(h_i(z_1)W_i^N=h_i(z)\). Let us show this in detail. Let D denote the curve over \(k(z_1)\) defined by the equation \(h_i(z_1)W_i^N=h_i(z), i=1,\dots ,m\), for each \(1\le i\le m\). Define \(f^{\prime }:C\otimes _k L\rightarrow D\otimes _K L\) by

$$\begin{aligned}&(z(z_1,u), w_1(z_1,u),\dots ,w_i(z_1,u),\dots ,w_m(z_1,u))\\&\quad \mapsto (z(z_1,u),w_1(z_1,u)/u_1,\dots ,w_i(z_1,u)/u_i,\dots ,w_m(z_1,u)/u_m), \end{aligned}$$

where we use \((z_1,u)\) as an abbreviation for the coordinate \((z_1,u_1,\dots ,u_m)\) in \(L=k(z_1)(u_1,\dots ,u_m)\) as mentioned above. Suppose \({\tilde{G}}=\langle g_1\rangle \times \cdots \times \langle g_m\rangle \). The action of \({\tilde{G}}\) on L is by letting \(g_i\) act as \(g_i\cdot (z_1,u)=(z_1,u_1,\dots , \xi u_i,\dots ,u_m)\), where \(\xi \) is an N-th primitive root of unity, whose existence was assumed from the beginning of this paper in Section 1. With the above notation, we have

$$\begin{aligned}&^{g_i}f^{\prime }(z(z_1,u), w_1(z_1,u),\dots ,w_m(z_1,u))\\&\quad = {}^{g_i}(f^{\prime }(^{{g_i}^{N-1}}(z(z_1,u), w_1(z_1,u),\dots ,w_m(z_1,u))))\\&\quad = {}^{g_i}(f^{\prime }(z(z_1,u_1,\dots , \xi ^{N-1} u_i,\dots ,u_m),\dots ,w_m(z_1,u_1,\dots , \xi ^{N-1} u_i,\dots ,u_m)))\\&\quad = {}^{g_i}(z(z_1,u_1,\dots , \xi ^{N-1} u_i,\dots ,u_m),\dots ,w_m(z_1,u_1,\dots , \xi ^{N-1} u_i,\dots ,u_m)/u_m)\\&\quad = (z(z_1,u), \xi w_1(z_1,u)/u_1,\dots ,\xi w_m(z_1,u)/u_m). \end{aligned}$$

Note that the first equality is due to the fact that \({g_i}^{N}=1\). On the other hand, we have

$$\begin{aligned}&f^{\prime }\circ a_{g_i}(z(z_1,u), w_1(z_1,u),\dots ,w_m(z_1,u))\\&\quad = f^{\prime }(z(z_1,u), \xi w_1(z_1,u),\dots ,\xi w_i(z_1,u),\dots ,\xi w_m(z_1,u))\\&\quad = (z(z_1,u), \xi w_1(z_1,u)/u_1,\dots ,\xi w_m(z_1,u)/u_m). \end{aligned}$$

Therefore we have the equality \(^{g_i}f^{\prime }=f^{\prime }\circ a_{g_i}\) for every \(i=1,\dots ,m\). Since the \(g_i\) are generators of G, it holds that \(^{g}f^{\prime }=f^{\prime }\circ a_{g}\) for every \(g\in G\) which implies the isomorphism \(D\cong C_a\) because of the uniqueness of the twist by [3, Prop. 2.6]. So \(C_a\) contains the rational points \(P_1=(z_1,1), P_i=(z_{i+1,j},W_{ij}/\prod _{l=1}^{s}(z_1-z_{1j})^{{\widetilde{r}}_{ij}})\) (\(1\le i\le n-1\)). By the above computations, we see that

$$\begin{aligned} {{\,\mathrm{Jac}\,}}(C)_a(K)\cong {{\,\mathrm{End}\,}}_k({{\,\mathrm{Jac}\,}}(C))^n\oplus {{\,\mathrm{Jac}\,}}(C)[n](k). \end{aligned}$$

Denote by \(Q_i\) the image of \(P_i\) via the Abel–Jacobi embedding \(C_a\hookrightarrow J(C_a)\). Then the \(Q_i\) are among the generators of \({{\,\mathrm{Jac}\,}}(C)_a(K)\).