5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ That supp(F) is proper means that the topological space supp(F) together with an arbitrary subscheme structure is a proper scheme. $\endgroup$ Aug 5, 2010 at 13:11

3 Answers 3

5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $M_P \neq 0$.

$\endgroup$
1
  • 2
    $\begingroup$ More generally, the closure of $supp(F)$ is $V(Ann(F))$ (also if $F$ is not coherent). $\endgroup$ Aug 5, 2010 at 14:07
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.