Suppose $C$ is a hyperelliptic curve. Then the set of two-torsion points on the jacobians is generated by the set of difference of Weierstrass points.
Suppose $C'$ is another hyperelliptic curve. Is it correct that the set of two-torsion points on the fibre product $C\times _{{\mathbb P}^1} C'$ is generated by pullback of the two-torsion line bundles on $C$ and $C'$.
Using Riemann-Hurwitz formula, there seems to be some relation coming. Is it possible to obtain the relation. Thanks.
No. First of all, you must assume that the images in $\mathbb{P}^1$ of the Weierstrass points of $C$ and $C'$ are distinct, otherwise the fiber product is not smooth.
Assuming that, $C\times _{\mathbb{P}^1}C'$ has a fixed point free involution $\tau $, given by $\tau (x,y)=(\sigma x, \sigma ' y)$, where $\sigma $ and $\sigma '$ are the hyperelliptic involutions. The pull back of 2-torsion points in $JC$ and $JC'$ are invariant under $\tau $, hence the subgroup of $J(C\times _{\mathbb{P}^1}C')$ they generate is invariant under $\tau $, and therefore much smaller than the full 2-torsion subgroup. In fact it is the 2-torsion subgroup of the Prym variety associated to $(C\times _{\mathbb{P}^1}C',\tau )$; this situation is thoroughly analyzed in Mumford's Prym varieties I (in Contributions to analysis, pp. 325-350. Academic Press, New York, 1974).
