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Need to prove that simple objects in Quasi-coherent sheaves are isomorphic to structure sheaves of a closed point.

"Simple object" means that there is no non-trivial subobject.

I encountered this problem in Vakil's FOAG(April 1, 2023), 6.5.L. I have proved that if the simple sheaf is a simple module on every affine open set, we have the result. The approach is to notice that, under this assumption, the simple qcoh sheaf is supported at a closed point of the scheme X on every affine open set, then we can construct a skyscraper subsheaf.

But I don't know how to continue.

My attempt to prove the "fact" (the simple qcoh sheaf is a simple module on every affine open set) is as the following.

Suppose we have M on Spec A, M is an A-module, and since M is not simple, we have a submodule of M isomorphic to A/p, where p is a maximal ideal of A, but not necessarily a closed point.

I'm thinking of constructing a skyscraper at p with A/p. But since the closure of p is not necessarily inside Spec A, I can't do the zero extension and get an injection from the skyscraper into the quasi-coherent sheaf.

PS: the scheme X may not have any closed point. I think maybe there is a more subtle way of proving using the sheaf morphisms which I have not learned systematically.

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Aug 11, 2023 at 17:12
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    $\begingroup$ No idea why this question is closed. This question is fine for a starter user. $\endgroup$ Aug 12, 2023 at 0:32

1 Answer 1

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As you observe, it suffice to classify simple objects in $\mathrm{QCoh}(\mathrm{Spec} \ A)\simeq A-\mathrm{mod}$. Let $M\ne0$ be a simple $A$-module. Then for any $x\ne0\in M$, there is a sub-module $0\ne Ax\subset M$, so by simplicity, $M=Ax\simeq A/I$, where $I=\ker(x)=\{a\in A:ax=0\}$.

Now, $I$ must be maximal, since otherwise there is an ideal $A\supsetneq J\supsetneq I$, and $J/I\subsetneq A/I$ is a non-trivial sub-module. But $A/I$ with $I$ maximal is exactly the structure sheaf of the closed point $I\in \mathrm{Spec} \ A$.


For an arbitrary scheme $X$, let $\mathcal F\ne0\in\mathrm{QCoh}(X)$ be a simple object, so for some $x\in X$, $\mathcal F|_{\{x\}}\ne0$. Then for any closed sub-scheme $i_x\colon \{x\}\hookrightarrow X$, there is a surjection $\mathcal F\to i_{x*}\mathcal F|_{\{x\}}$, which must be an isomorphism. Then the same argument as in the affine case applies.

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  • $\begingroup$ Thank you! But I don't know how to apply it to general Qcoh(X), and that's the place I got stuck in. Could you please interpret it with more detail? $\endgroup$
    – Acoustica
    Aug 15, 2023 at 4:05
  • $\begingroup$ So this is the more advanced restriction map in the latter part of the book. I see, thank you so much! $\endgroup$
    – Acoustica
    Aug 15, 2023 at 16:54
  • $\begingroup$ Hello Kenta, for the single point {x}, how do you give it a scheme structure? Note that in the definition of a skyscraper sheaf, one can only give the one-point space a ringed topological space structure (x,O_{X,x}), but this is not a scheme. $\endgroup$
    – BenjaminY
    Oct 20, 2023 at 5:55
  • $\begingroup$ I'm giving {x} the structure of the spectrum of k(x), the residue field. $\endgroup$
    – Kenta S
    Oct 20, 2023 at 11:58
  • $\begingroup$ Thanks, you are absolutely right! But I still have several concerns. F|x is an O_{X,x}-module, not necessarily a k(x)-module; next, for x\to X to be a closed subscheme, x must be a closed point. But there are schemes without any closed point (in fact this phenomenon was my primary concern). $\endgroup$
    – BenjaminY
    Oct 28, 2023 at 4:09

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