Prym variety associated to an étale cover of degree 2 of an hyperelliptic curve. 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! 
 A: 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 $\mu$.
A: 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.
