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I am reading Ravi Vakil's Foundations of Algebraic Geometry, in the paragraph before theorem 18.1.4, Vakil mentions this:

For any coherent sheaf $\mathscr{F}$ on $\mathbb{P}_A^n$, we can find a surjection $\mathscr{O}(m)^{\oplus j}\to\mathscr{F}$, which yields the exact sequence $$0\to \mathscr{G} \to \mathscr{O}(m)^{\oplus j}\to\mathscr{F} \to 0$$ for some coherent sheaf $\mathscr{G}$.

Here, $A$ is a ring, and $\mathscr{O}(m)$ is the degree $m$ invertible sheaf on $\mathbb{P}^n_A$

And in the proof of theorem 18.1.4(ii), which says that $H^i(X,\mathscr{F}(m))=0$ for a coherent sheaf $\mathscr{F}$on the projective $A$-scheme $X$ and for all $i>0$, Vakil has used this result.

And my question is, why should the sheaf $\mathscr{G}$ be coherent? From the coherence of $\mathscr{F}$, I can only see that $\mathscr{G}$ is finitely generated.


As Takumi Murayama mentioned, I need the coherence of $\mathscr{O}(m)$, which may not be true even $A$ is coherent itself. Actually I need this "result" to prove the following, which is the exercise 18.1.A in Vakil's note:

Assuming $A$ is a coherent ring, and $\mathscr{F}$ is a coherent sheaf on the projective $A$-scheme $\mathbb{P}^n_A$, show that $H^i(\mathbb{P}^n_A,\mathscr{F})$ is a coherent $A$-module.

My way to do this exercise: using the exact sequence of quasicoherent sheaf $$0\to \mathscr{G} \to \mathscr{O}(m)^{\oplus j}\to\mathscr{F} \to 0$$ we have an exact sequence of $A$-modules: $$H^n(\mathbb{P}^n_A,\mathscr{G})\to H^n(\mathbb{P}^n_A,\mathscr{O}(m)^{\oplus j}) \to H^n(\mathbb{P}^n_A,\mathscr{F})\to 0$$ Since $H^n(\mathbb{P}^n_A,\mathscr{O}(m)^{\oplus j})$ is finitely generated, so is $H^n(\mathbb{P}^n_A,\mathscr{F})$. And since $\mathscr{G}$ is finitely generated, we can used a similar argument to show that $H^n(\mathbb{P}^n_A,\mathscr{G})$ is finitely generated. Therefore $H^n(\mathbb{P}^n_A,\mathscr{F})$ is finitely presented, which is the same as coherent since $A$ is coherent over itself. Now, we just use the coherence of $\mathscr{F}$ to show coherence of $H^n(\mathbb{P}^n_A,\mathscr{F})$, if we know the coherence of $\mathscr{G}$, we can show the coherence of $H^n(\mathbb{P}^n_A,\mathscr{F})$. If we repeat the arguments, we can show that $H^{n-1}(\mathbb{P}^n_A,\mathscr{F})$ and $H^{n-1}(\mathbb{P}^n_A,\mathscr{G})$ are coherent, and similar the other cohomology groups.

In my proof, I assume that $\mathscr{G}$ is coherent, so is there any way to show the coherence of $H^i(\mathbb{P}^n_A,\mathscr{F})$ with the "result" I mentioned above?

Any help or hints are apprecaited, thank you so much.

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  • $\begingroup$ It's a general fact that the kernel of a morphism of coherent $\mathcal{O}_X$-modules is in fact coherent. In Vakil's notes, this is Exercise 13.8.G. $\endgroup$ – Takumi Murayama Jan 29 '17 at 5:32
  • $\begingroup$ To apply ex13.8.G, we need to assume $\mathscr{O}(m)$ is a coherent sheaf, right? $\endgroup$ – chan kifung Jan 29 '17 at 5:38
  • $\begingroup$ Did you end up solving the exercise? $\endgroup$ – Takumi Murayama Feb 1 '17 at 16:20
  • $\begingroup$ still can't solve the the problem without assuming$ /mathscr{G}$ is coherent $\endgroup$ – chan kifung Feb 1 '17 at 16:25
  • $\begingroup$ I came back to this today for whatever reason, and it looks like Exercise 18.1.A has not been in Vakil's notes for some time. He also said that he wanted to think more about this particular exercise here. $\endgroup$ – Takumi Murayama Nov 27 '17 at 19:27
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It's a general fact that the kernel of a morphism of coherent $\mathcal{O}_X$-modules is in fact coherent. In Vakil's notes, this is Exercise 13.8.G. Thus, it suffices to note that $\mathcal{O}_{\mathbf{P}^n_A}(m)$ is coherent for any $m$. In Theorem 18.1.4, you assume that $A$ is noetherian, so it suffices to note that $\mathcal{O}_{\mathbf{P}^n_A}(m)$ is locally free, hence locally isomorphic to the structure sheaf, which is coherent by noetherianity.

On the other hand, if $A$ is not noetherian, funny things can occur: even if $A$ is coherent, $A[x]$ may not be coherent [Soublin, Prop. 18]. In this case, $\mathcal{O}_{\mathbf{P}^1_A}(m)$ would not be coherent, so you are right to be worried in the non-noetherian setting!

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  • $\begingroup$ However, Vakil mentions that to prove part (ii) of the theorem, we do not need Noetherian hypothesis. Also, in the excercise 18.1 A after this theorem, Vakil asks us to prove part (i) assuming only that $A$ is a coherent over itself, I think I also need coherence of $\mathscr{G}$ to prove it. $\endgroup$ – chan kifung Jan 29 '17 at 6:33
  • $\begingroup$ I have added my way to do exercise 18.1A in this post, maybe there is some other way to avoid assuming $\mathscr{G}$ is coherent? $\endgroup$ – chan kifung Jan 29 '17 at 7:08

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