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In II.5.2.4 of Hartshorne (the example) Hartshorne remarks:

If $Y$ is a closed subscheme of a scheme $X$, then the sheaf $\mathcal{O}_X| _Y$ is not in general quasi-coherent on $Y$. In fact, it is not even a sheaf of $\mathcal{O}_Y$ modules in general.

I would like to see a concrete example of this. I know that such examples must exist, since we have the later proposition:

Let $X$ be an affine scheme, $O \to \mathcal{F'} \to \mathcal{F} \to \mathcal{F''} \to 0$ an exact sequence of sheaves of $\mathcal{O}_X$ modules, and $\mathcal{F'}$ is quasi-coherent. Then the exact sequence $$0 \to \Gamma(X, \mathcal{F'}) \to \Gamma(X, \mathcal{F}) \to \Gamma(X, \mathcal{F''}) \to 0$$ is exact.

I have examples where the contrapositive of this theorem holds (i.e. exactness fails to induce exactness on global sections) - in fact I think there are some exercises to previous sections that fashion us with such examples. However, I would like a concrete example illustrating the failure without the need of any additional tools. I suppose I could simply chase through a proof of the contrapositive with a particular example, but ideally I would an explicit calculation.

This isn't homework - it's going in my notes for an oral exam on algebraic geometry, and I like to have concrete examples I can use to calculate, since my instructor likes to see explicit examples (and I do too).

A. Thomas Yerger
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    Do you have a description of $O_X|_Y$ in simple cases, like when $X$ is the line and $Y$ the origin? – Mariano Suárez-Álvarez Feb 26 '18 at 01:13
  • For points, the inverse image functor just returns the stalk, so that's easy enough. But this functor is not so good for things more complicated than points :( Also I don't think these give counter examples, unless I screwed up my calculation. – A. Thomas Yerger Feb 26 '18 at 01:16
  • @MarianoSuárez-Álvarez It is possible I have hopelessly misunderstood your comment or the definitions of these notions, but I am convinced the situation with $X$ the line and $Y$ a point is not a counter-example. Was it meant to imply something else? – A. Thomas Yerger Feb 26 '18 at 07:50

1 Answers1

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Here is an example. Consider $X = \operatorname{Spec}(\mathbf{Z})$, and let $Y = \{(p)\} \cong \operatorname{Spec}(\mathbf{Z}/p\mathbf{Z})$ for a prime $p \in \mathbf{Z}$. Then, the restriction $\mathcal{O}_X\rvert_Y$ is the constant sheaf on $Y$ with value $\mathbf{Z}_{(p)}$, since restricting to a point is the same as taking the stalk at that point. This group $\mathbf{Z}_{(p)}$ cannot be given the structure of a $\mathbf{Z}/p\mathbf{Z} = \Gamma(Y,\mathcal{O}_Y)$-module, since adding any non-zero element of $\mathbf{Z}_{(p)}$ to itself $p$ times does not yield zero.

Remark. Let $i\colon Y \hookrightarrow X$ be a closed immersion of ringed spaces. One potentially confusing thing is that Hartshorne consistently denotes $\mathscr{F}\rvert_Y := i^{-1}\mathscr{F}$, while one often sees $\mathscr{F}\rvert_Y := i^*\mathscr{F}$ in the literature.

Takumi Murayama
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