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Let us work in the "nice" situation where $X,Y,Z$ are smooth complex algebraic varieties, not necessarily compact. Assume that the fiber product $W:= X \times_Z Y$ is also smooth. What can we say about the Picard group of $W$?

More precisely, assume that $Pic(X)=Pic(Z)=0$.

1) Can we deduce that $Pic(W)=Pic(Y)$?

2) If 1) is not true in general, can we draw the conclusion if $Z$ is a point and thus $W=X \times Y$?

3) If 1) is not true in general, can we draw the conclusion if $Y \to Z$ is a finite étale cover?

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    $\begingroup$ I believe this is true if $Y$ is complete and $Z$ is a point, by the so-called Seesaw theorem (Chapter 2, Corollary 6 in Mumford's Abelian varieties). I think the argument should go through for more general $Z$, however. @Martin: There actually is a ``Theorem of the Square"; it's just a bit later in the same chapter of Mumford, though it's unrelated to the situation at hand. $\endgroup$
    – Daniel Litt
    Commented Nov 22, 2011 at 15:25
  • $\begingroup$ (That last was in reference to a comment of Martin Brandenburg's, which is now deleted, apparently.) $\endgroup$
    – Daniel Litt
    Commented Nov 22, 2011 at 16:48
  • $\begingroup$ Unfortunately this is not always true, even when $Y$ is complete and $Z$ is a point. In fact it can even fail when $Z = \textrm{Spec}(k)$ for an algebraically closed field $k$, $X=Y$, and $Y$ is smooth and projective over $Z$. $\endgroup$
    – Chuck Hague
    Commented Nov 22, 2011 at 16:51
  • $\begingroup$ (This is exercise IV.4.10 in Hartshorne). $\endgroup$
    – Chuck Hague
    Commented Nov 22, 2011 at 16:54
  • $\begingroup$ See also mathoverflow.net/questions/57830/… $\endgroup$
    – David E Speyer
    Commented Nov 22, 2011 at 19:35

2 Answers 2

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It can happen that $Pic(X) = Pic(Y) = Pic(Z) = 0$ but $Pic(W) \neq 0$!

For example, let $f: \mathcal{E} \to Z$ be a non-isotrivial family of elliptic curves, where $Z$ is a smooth rational curve. Then $Pic(\mathcal{E})$ is finitely generated, so by removing a finite number of (images of) sections of $f$ we obtain a surface $X \to Z$ such that $Pic(X) = 0$. Letting $Y = X$, I claim that $Pic(W) \neq 0$:

To see this, let $\Delta$ be the diagonal inside $W = X \times_Z X$ so $L :=\mathcal{O}(\Delta) \in Pic(W)$. The restriction of $L$ to any fibre of the map $W \to Z$ is nonzero since the cohomology class of the diagonal in $E \times E$, where $E$ is any elliptic curve, remains non-zero when restricted to $E' \times E'$ where $E' \subset E$ is any non-empty Zariski open subset. In particular, $L \neq 0$, so $Pic(W) \neq 0$.

3) is also false.

For example, let $Y \to Z$ be a finite etale cover such that $Z$ is a smooth rational curve and the genus of $\bar{Y}$, the smooth compactification of $Y$, is at least $2$. Let $X$ also be a rational curve and take any morphism $X \to Z$ of degree $> 1$. Then $W$ is a smooth curve with a map to $Y$ of degree $> 1$. Since $g(\bar{Y}) > 1$, at least one component of $\bar{W}$ has genus $> g(\bar{Y})$ or it has more than one component, so the cokernel of the induced map $Pic(\bar{Y}) \to Pic(\bar{W})$ is uncountable. Since the kernel of the map $Pic(\bar{W}) \to Pic(W)$ is finitely generated, it follows that the the map $Pic(Y) \to Pic(W)$ cannot be surjective.

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  • $\begingroup$ Thank you very much, your examples are both very nice and simple. The first example makes me believe that the question, as I posed it, is really wrong. This because $H^2(X \times Y)$ contains $H^1(X) \otimes H^1(Y)$, and the latters could make algebraic classes on $X \times Y$. Does this make sense? $\endgroup$
    – Calc
    Commented Nov 22, 2011 at 19:36
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    $\begingroup$ Exactly: for proper smooth connected $X$ and $Y$ over an algebraically closed field $Z$, $\mathop{\rm Pic}(X\times_Z Y)=\mathop{\rm Pic}(X)\oplus \mathop{\rm Pic}(Y)\oplus\mathop{\rm Hom}(A_X,P_Y)$, where $A_X$ is the Albanese variety of $X$ and $P_Y$ is the Picard variety of $Y$ (both being Abelian varieties of dimensions $h^1(X,\mathcal O_X)$ and $h^1(Y,\mathcal O_Y)$ respectively) $\endgroup$
    – ACL
    Commented Nov 22, 2011 at 23:03
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A weakening of 2) is an exercise (III.12.6) in Hartshorne: Let $X$ be an integral projective scheme over an algebraically closed field $k$ and assume that $H^1(X, \mathcal O_X) = 0$. Let $T$ be a connected scheme of finite type over $k$. Then $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T) $ under the obvious morphism. However, it is not true in general that $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T) $ for two arbitrary $k$-schemes $X$ and $T$, cf exercise IV.4.10 in Hartshorne.

In general, you might be interested in exercises III.12.4 and III.12.5 of Hartshorne as well; they give more results on Picard groups in this context.

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  • $\begingroup$ I was aware of IV 4.10, but I was still hoping to see an example where $Pic(X)=0$. I was not aware of III.12.4 and 12.5, thanks a lot!! $\endgroup$
    – Calc
    Commented Nov 22, 2011 at 19:39

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