In view of this question, I have an additional question.

The situation is as follows. Let $C$ be the hyperelliptic curve over $\mathbb{C}$, which is given on an affine by the equation $y^2 = x^5 +1 = \prod_{i=1}^{5} (x-\alpha_{i})$. Etale Galois covers of $C$ with group the cyclic group of order 2, are in bijection with the 2-torsion points of the Jacobian $J(C)$ of $C$, see Hartshorne exercise 2.7 of Chpt. 4. The difference between two Weierstrass-points gives rise to a 2-torsion point on the Jacobian, see also question. Indeed, for example the function $z^{2} - \frac{(x-\alpha_{1})}{(x-\alpha_{2})} \in K[z]$, where $K$ is the function field of $C$, defines a quadratic extension of the function field $K$, which gives an etale cover of degree 2 of $C$, which we will denote by $f: C' \rightarrow C$. This gives the so-called norm map on Jacobians $f_{*}: J(C') \rightarrow J(C)$. If I understand it correctly, the Prym variety $P$ with respect to $f$ then is defined to be the connected component of the identity of the kernel of $f_{*}$.

My question is: what is the Prym variety $P$ in this case? Due to the Riemann-Hurwitz formula, I believe that $P$ must be 1-dimensional. I am actually not even sure what exactly the map $f$ is. Thanks in advance!

  • $\begingroup$ I answered your question in a fairly abstract way; are you looking for something more explicit? $\endgroup$
    – rfauffar
    Mar 14, 2014 at 13:38
  • $\begingroup$ Thank you for your answer and for your reference Robert! It is always useful to have different realizations of one object. But indeed, I am actually looking for ways to obtain a more explicit description of the Prym variety in this case, i.e., which elliptic curve it is. Do you maybe have any ideas? $\endgroup$ Mar 14, 2014 at 13:50
  • $\begingroup$ Dear Adam, in this case I'm not quite sure how to find the equation of the elliptic curve explicitly. $\endgroup$
    – rfauffar
    Mar 15, 2014 at 15:29
  • $\begingroup$ Dear Robert, thank you for your reply! Do you maybe know what $C'$ is? Because then I can try to count the number of points on $C$ and $C'$ over finite fields, and determine the zeta functions of the curves. In this way I think I can determine what the zeta function of the Prym variety I'm looking for should be. $\endgroup$ Mar 16, 2014 at 13:29
  • $\begingroup$ Well, $C'$ is just defined by your extension field $K(z^2-\frac{(x-\alpha_1)}{x-\alpha_2})$. $\endgroup$
    – rfauffar
    Mar 17, 2014 at 0:22

2 Answers 2


I know this question is old (came across it on a google search for something else) but the answer you want is in the (mostly expository) paper: http://arxiv.org/abs/alg-geom/9206008

In short, a genus 2 curve has six branch points for the hyperelliptic map, and a choice of two of them (unordered) is equivalent to a choice of two Weierstrass points, which is the same as the data of a point of order two/double cover of the curve.

The elliptic curve that is the Prym of that double cover is the one that is branched over $\mathbb{P}^1$ at the remaining 4 branch points.

As for a concrete way to picture $f$, the point of order two on the Jacobian is a line bundle $\mu$ that squares to the trivial line bundle $\mathscr{O}_C$, the isomorphism $\mu^{\otimes 2}\to \mathscr{O}_C$ gives an $\mathscr{O}_C$-algebra structure to the sheaf $\mathscr{O}_C\oplus\mu$, and $C'$ in your notation is the relative Spec of this algebra, and the map $f$ is projection.

Another way to construct $C'$ is to look locally at solutions to $z^2=1$ inside the total space of the line bundle $\mu$.

  • 1
    $\begingroup$ Thank you for your answer! The article that you mention is very useful. $\endgroup$ Jan 26, 2015 at 19:15

If $f:C'\to C$ is a morphism of smooth projective curves, then you can define in general the Prym variety defined by $f$ to be the connected component of $\ker f_*$. Recall that on a polarized abelian variety $(A,\Theta)$, if $X\leq A$ is an abelian subvariety, then we can canonically associate to it its complementary abelian variety $Y$ that has the property that the addition map $X\times Y\to A$ is an isogeny. It turns out that the Prym variety associated to $f$ is precisely the complementary abelian sub variety of $f^*(J_C)$, where $J_C$ denotes the Jacobian of $C$.

In your case, by Riemann-Hurwitz you have that the genus of $C'$ is 3. Therefore, since $f^*J_C$ is 2-dimensional, we have that the complementary abelian subvariety, or in other words the Prym variety of the covering, has dimension 1.

Another way to see it in your case is that $C$ is the quotient of $C'$ by an involution $\sigma\in\mbox{Aut}(C')$. You can then see that $\sigma$ induces an automorphism of the polarized Jacobian $(J_{C'},\Theta_{C'})$ and your Prym variety is precisely the connected component of $\ker(\mbox{id}+\sigma)$.

If you're willing to read in Spanish, the notes http://rhidalgo.mat.utfsm.cl/files/varf4.pdf are very good and explain this sort of thing starting on page 26.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .