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.
Best Answer
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.