Picard groups of (fiber) products - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:27:17Z http://mathoverflow.net/feeds/question/81611 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81611/picard-groups-of-fiber-products Picard groups of (fiber) products Calc 2011-11-22T14:34:10Z 2011-11-22T16:38:08Z <p>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 $W$? </p> <p>More precisely, assume that $Pic(X)=Pic(Z)=0$.</p> <p>1) Can we deduce that $Pic(W)=Pic(Y)$?</p> <p>2) If 1) is not true in general, can we draw the conclusion if $Z$ is a point and thus $W=X \times Y$?</p> <p>3) If 1) is not true in general, can we draw the conclusion if $Y \to Z$ is a finite étale cover?</p> http://mathoverflow.net/questions/81611/picard-groups-of-fiber-products/81617#81617 Answer by ulrich for Picard groups of (fiber) products ulrich 2011-11-22T15:31:50Z 2011-11-22T16:35:16Z <p>It can happen that $Pic(X) = Pic(Y) = Pic(Z) = 0$ but $Pic(W) \neq 0$!</p> <p>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 $Pic(W) \neq 0$:</p> <p>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 $Pic(W) \neq 0$.</p> <p>3) is also false. </p> <p>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. </p> http://mathoverflow.net/questions/81611/picard-groups-of-fiber-products/81624#81624 Answer by Chuck Hague for Picard groups of (fiber) products Chuck Hague 2011-11-22T16:38:08Z 2011-11-22T16:38:08Z <p>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.</p> <p>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.</p>