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In the book FGA of AMS page 109 said, "the schematic suppor of F is... " (F is a sheaf) What that means? Thank you for help.

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  • $\begingroup$ That supp(F) is proper means that the topological space supp(F) together with an arbitrary subscheme structure is a proper scheme. $\endgroup$ Commented Aug 5, 2010 at 13:11

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Assume $\mathcal{F}$ is coherent on $X$. To complement the answers you have already got, there are two useful definitions of its schematic support: either use the annihilator ideal or the Fitting ideal. They have the same underlying reduced scheme, but in general different scheme structures. The annihilator is usually (always, I think) meant if nothing else is said.

The annihilator support $Z = V(\mathrm{Ann}(\mathcal{F}))$ can be viewed as the minimal closed subscheme $i\colon Z \subset X$ such that the natural map $\mathcal{F} \to i_*i^*\mathcal{F}$ is an isomorphism (more or less a tautology).

The Fitting ideal (locally the maximal minors of a free presentation) gives a subscheme that contains $V(\mathrm{Ann}(\mathcal{F}))$, but with a possibly thicker scheme structure. A feature is that it is compatible with pullback; the annihilator construction is not.

Example: a rank $r$ vector bundle $i_*E$ on a divisor $i\colon D\subset X$ has $D$ as annihilator support, and $rD$ as Fitting support.

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The scheme structure on the support of the coherent sheaf of $\mathcal{O}_X$-modules $\mathcal{F}$ is defined by the sheaf of ideals $Ann(\mathcal{F}) \subset \mathcal{O}_X$.

It is easy to see that the (topological space underlying the) locus defined by this sheaf of ideals is the same as the locus supporting $\mathcal{F}$. Indeed this is a local assertion, which we prove as follows.

Let $A$ be a ring, $M$ a finitely generated $A$-module, $P \subset A$ a prime. Then we must check that $M_P \neq 0$ if and only if $Ann(M) \subset P$.

Note that an element $m/1 \in M_P$ is $0$ if and only if there exists $a \notin P$ such that $a m = 0$, that is $Ann(m) \not \subset P$, and $M_P$ will be trivial if and only if $m/1 = 0$ for all $m$. Now use the fact that $$Ann(M) = \bigcap_{m \in M} Ann(m),$$ which can be written as a finite intersection, taking only the generators. Since $P$ is prime, it contains this intersection if and only if if contains at least one term $Ann(m)$, which means exactly that $M_P \neq 0$.

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    $\begingroup$ More generally, the closure of $supp(F)$ is $V(Ann(F))$ (also if $F$ is not coherent). $\endgroup$ Commented Aug 5, 2010 at 14:07
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A sheaf $\mathcal{F}$ doesn't just come with topological support on a scheme. If you look at the definition of support (Which should be available in any good book, none of which I have at hand at the moment) it should describe to you that the support is a subscheme of the ambient scheme, not just a subspace, and may really have non-variety structure on it.

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