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This question is a follow-up to this question which I asked on MSE.

Let $f: X \rightarrow Y$ be a surjective morphism of schemes, and $\mathscr{F}$ a coherent sheaf on $Y$. Are there conditions we can put on $X$ and $Y$ that imply that the global pullback map $H^0(Y, \mathscr{F}) \rightarrow H^0(X, f^* \mathscr{F})$ is injective?

If $f$ is flat or $\mathscr{F}$ is locally free, this map is always injective. In addition, this answer to the linked question on MSE indicates a proof that, at least when the schemes are noetherian and we either assume that $Y$ is regular or that $X$ and $Y$ are both normal, the same is true when $f$ is any finite map (of course, by miracle flatness, this is only interesting when $X$ is not Cohen-Macaulay or $Y$ is not regular). In the comments and edits to this question, we found counterexamples when $X \rightarrow Y$ is the normalization of a curve and when $X$ is smooth but not connected. Both of these are, to some degree, related to the phenomenon of $f$ "breaking apart infinitesimal neighborhoods". (Both of the modules considered were infinitesimal thickenings of points).

Can this sort of behavior occur when $X, Y$ are connected smooth varieties (ideally over $\mathbb{C}$)? It seems to me like the natural place to find a counterexample is when $f$ is a birational morphism and $\mathscr{F}$ has support on the image of the exceptional locus, but I have not been able to construct one along these lines.

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  • $\begingroup$ What is a surjective morphism of schemes? $\endgroup$ – Qiaochu Yuan Nov 30 '16 at 6:48
  • $\begingroup$ Just a morphism which is surjective on the underlying topological spaces? E.g. the "faithful" part of "faithfully flat". (this is the Stacks project definition, which I would hope is also the one in EGA) I guess surjective on closed points is another reasonable criterion, but I'm only thinking about schemes that are of finite type over a field, so these should be equivalent (right?) $\endgroup$ – dorebell Nov 30 '16 at 6:54
  • $\begingroup$ If $f \colon X \to Y$ is birational with connected fibres, then $f_* \mathscr{O}_X = \mathscr{O}_Y$, hence by projection formula we have $$H^0(X, \, f^* \mathscr{F})=H^0(Y, \, f_*f^*\mathscr{F}) = H^0(Y, \, \mathscr{F} \otimes f_* \mathscr{O}_X) = H^0(Y, \, \mathscr{F} ).$$ On the other hand, Zariski's main theorem implies that if $Y$ is normal and $f \colon X \to Y$ is birational, then every fibre of $f$ is connected. So, if you want to look for a counterexample where $f$ is birational, you must take a non-normal $Y$. $\endgroup$ – Francesco Polizzi Nov 30 '16 at 7:08
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    $\begingroup$ Maybe this is a silly question, but I'm not seeing how you use the projection formula: don't you need $\mathscr{F}$ to be locally free? $\endgroup$ – dorebell Nov 30 '16 at 7:15
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    $\begingroup$ @dorebell It is true that if $X$ and $Y$ are finite type over a field $k$ then surjective on closed points implies surjective. Indeed, the image is constructible by Chevalley's theorem. So, it's complement is constructible. But, closed points are very dense, so if the complement was non-empty it would contain a closed point. $\endgroup$ – Alex Youcis Nov 30 '16 at 21:07
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Begin with $Y$ equal to the affine plane, $\text{Spec}\ R$, for $R=k[s,t]$. Let $\overline{f}:\overline{X}\to Y$ be the blowing up of $Y$ at the ideal $\mathfrak{m} = \langle s,t \rangle$. Define $I\subset \mathfrak{m}$ to be the ideal $\langle s,t^2 \rangle$. Define $\mathcal{G}$ to be $\widetilde{R/\mathfrak{m}}$, define $\mathcal{F}$ to be $\widetilde{R/I}$, and define $p:\mathcal{F}\to \mathcal{G}$ to be the natural surjection. The pullback $\overline{f}^*\mathcal{G}$ is the structure sheaf of the exceptional divisor $E$. The pullback $$\overline{f}^*p:\overline{f}^*\mathcal{F}\to \overline{f}^*\mathcal{G},$$ is an isomorphism except at a single point $q$ on $E$.

Define $X$ to be the open complement of $\{q\}$ in $\overline{X}$. Define $f:X\to Y$ to be the restriction of $\overline{f}$ to $X$. Since we remove a single closed point from the positive dimensional fiber $E$ of $\overline{f}$, the morphism $f$ is still surjective. On $X$, the natural homomorphism $$f^*p:f^*\mathcal{F}\to f^*\mathcal{G},$$ is an isomorphism. Therefore, the pullback map $H^0(Y,\mathcal{F}) \to H^0(X,f^*\mathcal{F})$ factors through the map $$H^0(p):H^0(Y,\mathcal{F})\to H^0(Y,\mathcal{G}).$$ This map has a one-dimensional kernel.

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