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Given two smooth elliptic curves $C_1$ and $C_2$ over $\mathbb{C}$. Assume they are not isogenous. I'm interested in the structure of $Pic(A)$ and $Pic^{0}(A)$ for $A:=C_1 \times C_2$.

Reading Birkenhake/Lange - Complex Abelian Varieties, i think this has to do with correspondences of curves. Since an elliptic curve is its own Jacobian and the two curves are not isogenous, we have $Hom(C_1,C_2)=0$. So the space of correspondences $Corr(C_1,C_2)$ is trivial, i.e. every line bundle $L$ on $A$ is of the form $L=q^{\*}M\otimes p^{\*}N$, where q and p are the projections on the factors and $M$ and $N$ are line bundles on the factors. This implies $Pic(A)=Pic(C_1)\times Pic(C_2)$.

Does this impliy $Pic^{0}(A)=Pic^{0}(C_1)\times Pic^{0}(C_2)$? That is, is the Picard variety of $A$ isomorphic to $A$ in this case?

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    $\begingroup$ Yes, any product $A$ of elliptic cures is principally polarizable(e.g. by the product polarization), hence isomorphic to its Picard variety. $\endgroup$
    – Pete L. Clark
    Commented Oct 7, 2010 at 18:01

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Yes: in fact $Pic^0(C_1\times C_2)=Pic^0(C_1)\times Pic^0(C_2)$ for any pair of curves. The fact that $C_1$ and $C_2$ are not isogenous in your case only affects the Neron-Severi group $Pic/Pic^0$ of $C_1\times C_2$, exactly for the reasons you describe.

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  • $\begingroup$ Interesting, can you give an explanation or a reference why this is true for all curves? I see that we have $Pic^{0}(C_1)\times Pic^{0}(C_2) \subset Pic^{0}(C_1\times C_2)$, via the pullbacks coming from the projections. But i don't see the other inclusion. $\endgroup$
    – TonyS
    Commented Oct 7, 2010 at 17:33
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    $\begingroup$ It comes out of the theory of correspondences. See Weil's classic "Courbes algebriques et les varietes qui s'en deduisent", or p.153 of Zariski's book on algebraic surfaces (2nd ed.) $\endgroup$
    – Tony Scholl
    Commented Oct 7, 2010 at 18:38
  • $\begingroup$ Thank you. I will try to get a look in that book and that article. $\endgroup$
    – TonyS
    Commented Oct 7, 2010 at 20:24
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Am I missing something? Doesn't this hold for arbitrary smooth pairs of varieties using the Kunneth decomposition?

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  • $\begingroup$ Not for the Picard groups. See Ex. 4.10, chap. IV in Hartshorne. $\endgroup$
    – Damian Rössler
    Commented Nov 1, 2021 at 10:23

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