Can anyone point to me where I can find the proof that the Picard group of the product of two curves is isomorphic to the product of the Picard groups times the hom among the Jacobians? Does the result work on an arbitrary field? Thank you very much for your time and attention

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    $\begingroup$ Over $\mathbb{C}$, this is easy to see from Hodge theory and the Lefschetz $(1,1)$-theorem; one just computes with Kunneth, and uses that Hom of the Jacobians is the same as Hom of $H^1$ of the curves. The result is true over arbitrary fields, however, but requires more argument. I can write out more details if you'd like over arbitrary fields, though it's a bit exhausting. Have you come up with maps both ways yet? $\endgroup$ Commented Nov 18, 2014 at 2:22

1 Answer 1


It seems likely that you are assuming the curves are smooth and geometrically connected (e.g., you don't have in mind generalized Jacobians for singular curves), but you have omitted hypotheses on the ground field and have not indicated if you are assuming the existence of rational points.

Let $X$ and $Y$ be geometrically integral proper schemes over a field $k$. First assume there exist $x_0 \in X(k)$ and $y_0 \in Y(k)$, so the relative Picard functors ${\rm{Pic}}_{X/k}$ and ${\rm{Pic}}_{Y/k}$ respectively classify isomorphism classes of line bundles on $X$ and $Y$ trivialized along $x_0$ and $y_0$. Let $L$ be a line bundle on $X \times Y$, and define the line bundles $L_1 = (1_X \times y_0)^{\ast}(L)$ on $X$ and $L_2 = (x_0 \times 1_Y)^{\ast}(L)$ on $Y$, so the line bundle $$L' := L \otimes (p_X)^{\ast}(L_1)^{-1} \otimes (p_Y)^{\ast}(L_2)^{-1}$$ has trivial pullbacks along $x_0 \times Y$ and $X \times y_0$. Hence, $L'$ as line bundle on $X \times Y$ trivial along $x_0 \times Y$ is classified by a $k$-morphism $f_L:Y \rightarrow {\rm{Pic}}_{X/k}$ carrying $y_0$ to $0$. This map lands inside ${\rm{Pic}}^0_{X/k}$, and it is clear that $f_{L \otimes M} = f_L + f_M$ for $L$ and $M$ on $X \times Y$.

Assume now that $X$ and $Y$ are curves, so by the Albanese property of Jacobians we see that $f$ corresponds to a $k$-homomorphism $J_Y \rightarrow J_X$. This defines a homomorphism $${\rm{Pic}}(X \times Y) \rightarrow {\rm{Mor}}_k((Y,y_0), (J_X,0)) = {\rm{Hom}}_k(J_Y, J_X)$$ that admits a natural homomorphic section via the Picard functorialty of $J_X$ with respect to $(X, x_0)$ (essentially run parts of this construction in reverse).

The seesaw theorem implies that the kernel of this surjective map consists of exactly those $L$ for which $L'$ is trivial, which is to say it is the image of the visibly injective map $${\rm{Pic}}(X) \times {\rm{Pic}}(Y) \rightarrow {\rm{Pic}}(X \times Y).$$ Hence, we have built the desired isomorphism, but our construction rests on the choices of $x_0$ and $y_0$.

If these construction are carried out in a more careful scheme-theoretic manner over $k$-algebras uses the schematic form of the seesaw theorem, and keeps track of trivializations then this argument provides a left exact sequence of $k$-group schemes $$0 \rightarrow {\rm{Pic}}_{X/k} \times {\rm{Pic}}_{Y/k} \stackrel{j}{\rightarrow} {\rm{Pic}}_{(X \times Y)/k} \rightarrow \underline{\rm{Hom}}(J_Y,J_X)$$ where the final term of the \'etale Hom-scheme over $k$ and the induced map on geometric points is short exact. But the initial inclusion $j$ between smooth $k$-groups locally of finite type induces on Lie algebras the natural map ${\rm{H}}^1(X,O_X) \oplus {\rm{H}}^1(Y,O_Y) \rightarrow {\rm{H}}^1(X \times Y, O_{X \times Y})$ that is an isomorphism, so ${\rm{coker}}(j)$ is \'etale. Consideration of geometric points then shows that ${\rm{coker}}(j)$ maps isomorphically onto the Hom-scheme, so the left-exact sequence of $k$-groups is short exact.

What happens to the identification of the cokernel of $j$ with the Hom-scheme if we change $(x_0,y_0)$, using the identification of Jacobian as dual to Picard scheme (thereby suppressing its dependence on the base point, via Grothendieck's definition of the Picard functor through sheafification)? Since the entire construction can be carried out using sections after base change over any $k$-scheme, the dependence on $(x_0,y_0)$ is classified by a $k$-scheme morphism $$X \times Y \rightarrow \underline{\rm{Isom}}({\rm{coker}}(j), \underline{\rm{Hom}}(J_Y,J_X))$$ which must be constant since the target is etale.

Hence, the identification of the cokernel of $j$ with the Hom-scheme is independent of the choice of $(x_0,y_0)$, so in general without assuming such $k$-points to exist we can carry out the construction over a finite Galois extension of $k$ where rational points can be chosen and then use Galois descent to bring it down to $k$ (independently of all choices). This solves the original question in the form of a short exact sequence of smooth $k$-groups which is split-exact (hence also split on $k$-points) when $(x_0, y_0) \in (X \times Y)(k)$ is chosen. In the absence of such base points it isn't clear that one should really expect a description of ${\rm{Pic}}(X \times Y)$ in the form requested. (Presumably the case of curves of genus 0 or 1 without $k$-points may already exhibit some obstacles, but I am too lazy to sort that out at the moment.)

  • $\begingroup$ Thank you very much for taking the time to explain everything to me. You are correct, of course, about the hypothesis. I apologize for the sloppiness. In truth I was just looking for a reference... this is so much better. Thank you very much! $\endgroup$ Commented Nov 18, 2014 at 17:58
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    $\begingroup$ It seems that it might be a bit easier to keep track of things if one uses the Leray spectral sequence for the maps $X\times Y\to X\to *$. Then one gets an exact sequence $$0\to \text{Pic}(X)\to \text{Pic}(X\times Y)\to H^0(X, \text{Pic}(Y))\to \text{Br}(X)$$ and we may write $H^0(X, \text{Pic}(Y))$ as $\text{Hom}(\text{Pic}^1(X), \text{Pic}(Y))$. I haven't thought very hard about when the last map in this sequence is zero, though... $\endgroup$ Commented Nov 20, 2014 at 21:22

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