# Picard group of product of spaces

Suppose $X,Y$ are varieties over an algebraically closed field $k$. Can we compute $\operatorname{Pic}(X \times_k Y)$ in terms of $\operatorname{Pic}(X),\operatorname{Pic}(Y)$? It seems that $\operatorname{Pic}(X \times_k Y) \cong \operatorname{Pic}(X) \times \operatorname{Pic}(Y)$ is not quite right, but I cannot figure out a counterexample. (I thought one might construct UFDs $A,B$, but their tensor product is not UFD).

• See this for example mathoverflow.net/questions/81611/… – Ehsan M. Kermani Sep 14 '13 at 3:44
• It may be a good idea to recall <b>everytime</b> what do you mean by an "algebraic variety". Taken in the broad sense, there is an easy counterexample: let $Y$ be two points and take $X$ with $\mathrm{Pic}(X)\ne\{1\}$. Then the LHS is $\mathrm{Pic}(X)\times \mathrm{Pic}(X)$, which is different from the RHS $\mathrm{Pic}(X)$. But even if you suppose $X, Y$ connected, there are still lot of counterexamples. Essentially there is almost never equality. – Cantlog Sep 20 '13 at 21:25

In some cases it is true:

If $X$ is a projective variety over an algebraically closed field $k$ such that $H^1(X,\mathcal{O}_X)=0$, and $T$ is a connected scheme of finite type over $k$, then $\mathrm{Pic}(X \times T) \cong \mathrm{Pic}(X) \times \mathrm{Pic}(T)$. This is exercise III.12.6. in Hartshorne's book.

Since $\mathrm{Pic}(\mathbb{A}^1)=0$, a special case of the question is the "homotopy invariance" $\mathrm{Pic}(X \times \mathbb{A}^1) \cong \mathbb{Pic}(X)$. This holds when $X$ is normal, but not in general (SE/432217).

• I wonder how user18119 there found the counterexample... – Cantlog Sep 14 '13 at 10:36
• I regret that user18119 = Qing Liu has gone away ... To your question: It's the simplest local non-normal ring. – Martin Brandenburg Sep 14 '13 at 10:43
• Since $H^1(X,\mathcal{O}_X)$ is the tangent space of $Pic(X)$ at the origin, the condition $H^1(X,\mathcal{O}_X)=0$ says that $X$ has discrete i.e. very small Picard group. That condition is equivalent, if $X$ is smooth over $\mathbb C$, to the first Betti number of $X$ being zero. And that last condition is close to $X$ being simply connected and thus having only trivial topological line bundles. I find these links with topology intriguing and fascinating. – Georges Elencwajg Sep 14 '13 at 21:09

Ischebeck has proved that given an algebraically closed field $k$ and two normal integral algebraic $k$-schemes $X,Y$ there is an exact sequence of groups $$0\to Pic(X)\times Pic (Y)\to Pic(X\times Y)\to Pic (k(X)\otimes _k k(Y))$$ Note that neither variety is supposed complete, nor affine, nor...
This is quite interesting because although other users have shown that the Picard group of the product of two varieties may be bigger than the product of that of the factors, Ischebeck gives a bound for the discrepancy.
In particular if one of the varieties, say $Y$, is rational, then the ring $k(X)\otimes _k k(Y)$ is a fraction ring $S^{-1}A$ of the polynomial ring $A=k(X)[T_1,...,T_n]$ over the field $k(X)$ and so is a UFD and thus has zero Picard group: $$Y\operatorname {rational}\implies Pic(X\times Y) =Pic(X)\times Pic (Y)$$
This is a vast generalization of $Pic( X \times\mathbb A^1) =Pic(X)$.

• Very nice! for a projective rational Y, does H1 of O_Y vanish ? – Cantlog Sep 14 '13 at 21:22
• Dear @Cantlog, yes, for a projective smooth variety $Y$, we have $H^1(Y,\mathcal O_Y)=0$. More generally, $H^1(Y,\mathcal O_Y)$ is a birational invariant of smooth projective varieties. – Georges Elencwajg Sep 14 '13 at 22:10
• @GeorgesElencwajg Thank you for your crystal clear explanation. I feel bad that I cannot accept more than one answer:( – Li Yutong Sep 15 '13 at 3:00
• Dear Li, don't feel bad: that the explanation was of some use to you is the only thing that matters. – Georges Elencwajg Sep 15 '13 at 7:21
• In my first commment above read "..,yes, for a rational projective smooth variety $Y$, we have $H^1(Y,\mathcal O_Y=0)$" The adjective "rational" was regrettably missing . – Georges Elencwajg Mar 27 '14 at 21:41

If $X$ and $Y$ are two curves, then $\mbox{Pic}(X\times Y)\simeq\mbox{Pic}(X)\times\mbox{Pic}(Y)\times\mbox{Hom}(J_X,J_Y)$, where $J_X$ and $J_Y$ denote the jacobian varieties of $X$ and $Y$, respectively. In particular, for example, if $X$ and $Y$ are two isogenous elliptic curves, then $\mbox{Hom}(J_X,J_Y)\neq0$.

• Dear @Robert, where does that interesting isomorphism on the first line come from? – Georges Elencwajg Sep 14 '13 at 21:12
• Dear @Georges, if $D$ is a prime divisor on $X\times Y$, then we can restrict the projections of $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$ to $D$ to get projections $\pi_{1,D}$ and $\pi_{2,D}$ and obtain a homomorphism $\phi_D:\mbox{Pic}(X)\to\mbox{Pic}(Y)$ by $L\mapsto (\pi_{2,D})_*(\pi_{1,D})^*L$. Restricting this to $\mbox{Pic^0(X)}$ gives us a morphism between the jacobians of $X$ and $Y$. Extending linearly to all divisors, we get a map $\mbox{Pic}(X\times Y)\to\mbox{Hom}(J_X,J_Y)$. One can show that this map is surjective, and the kernel consists of divisors coming from fibers. – rfauffar Sep 14 '13 at 22:44
• This kernel is then isomorphic to $\mbox{Pic}(X)\times\mbox{Pic(Y)}$, and the map actually splits, making this group a factor of $\mbox{Pic}(X\times Y)$. – rfauffar Sep 14 '13 at 22:46
• Thank you for this explanation, @Robert:+1. (Don't you mean $Hom$ in place of $End$ at the end of your answer?) – Georges Elencwajg Sep 14 '13 at 22:59
• Yes!! Thanks, I edited it. – rfauffar Sep 15 '13 at 4:57

This is to elaborate on Elencwajg's nice answer. Ischebeck's sequence holds under the following general hypotheses: $k$ is any field, $X$ and $Y$ are non-empty, normal, locally of finite type and geometrically integral $k$-schemes. The proof of exactness at the middle of the sequence is Ischebeck's, while exactness at the first nontrivial term follows from a Galois descent argument of Colliot-Thelene and Sansuc's which appears in their 1977 paper on $R$-equivalence on tori (see proof of Lemma 11 on pp.188-189). By the way, the preliminary results proved by Ischebeck's in order to derive his theorem are actually rediscoveries of (particular cases of) results of Raynaud which appear in the Errata list for EGA IV_4 (Err 53). When one of the factors, say $Y$, is rational over the separable closure of $k$ and has a $k$-rational point, Sansuc proved the formula $Pic\, X\oplus Pic\, Y\simeq Pic (X\times_{k}Y)$ in his 1981 Crelle paper using a method that requires smoothness and differs completely from Ischebeck's (note that, by Ischebeck and Raynaud, normality suffices). Let me note that neither Sansuc nor Colliot-Thelene-Sansuc (in the 1977 paper alluded to above) cite Ischebeck's paper. Sansuc's statement can be generalized by replacing the smoothness hypothesis by normality and the existence of a $k$-rational point by the condition that the separable indices of $X$ and $Y$ are coprime. The reliance on the existence of a $k$-rational point on $Y$ can in fact be removed without much effort directly from Sansuc's statement by using a standard restriction-corestriction argument, as pointed out to me by Mathieu Florence. Beyond the base field case, one might further ask for a relative version of the question posed by Li Yutong, i.e., let $f\colon X\to S$ be a morphism of schemes. Then what are the kernel and cokernel of the canonical homomorphism $Pic\, X\oplus Pic\, Y\to Pic (X\times_{S}Y)$? In this rarefied setting, assuming $S$ to be normal and integral (with function field $K$) seems reasonable (because of well-known counterexamples to many sorts of questions over non-normal bases). Under suitable conditions, there should exist an exact sequence of the sort $$Pic\,S\to Pic\, X\oplus Pic\, Y\to Pic (X\times_{S}Y)\to Pic\,(R(X_{K})\otimes_{K}R(Y_{K})),$$ where $R(X_{K}), R(Y_{K})$ are the rings of rational functions of the generic fibers. So, if one lets $S$ shrink to $Spec\, K$, then Ischebeck's sequence is recovered (by the way, if the reader suspects that the localization sequence for the Picard group is at play above, then he/she is right indeed...)

• Thank you for this extremely interesting comment, dear Cristian. I hope you will become a frequent contributor here: we need experts like you on this site! – Georges Elencwajg Feb 28 '16 at 8:42