Support of a quasi-coherent sheaf on an affine scheme

Question

Let $A$ be a commutative ring, $X = \text{Spec }A$, and let $M$ be a $A$-module. The $\mathcal{F} = \tilde M$ is a sheaf on $X$.

Exercise II.5.6 in Hartshorne's Algebraic Geometry states that if $A$ is noetherian and $M$ is finitely generated, then $\text{Supp } \mathcal{F} = V(\text{Ann} M)$.

I proved this, but it feels suspicious that I did not use that $A$ should be noetherian. This is my proof:

Let $m_1,\dotsc,m_n$ be generators of $M$. Then $\text{Ann }M = \bigcap _i\text{Ann } m_i$. Also the $m_i$ generate the stalks of $\mathcal{F}$, so $\mathcal{F}_x = 0$ if and only if $(m_i)_x = 0$ for all $i = 1,\dotsc, n$. Hence $\text{Supp }\mathcal{F} = \bigcup_i \text{Supp }m_i = \bigcup_i V(\text{Ann } m_i) = V(\bigcap_i \text{Ann }m_i) = V(\text{Ann }M).$

Did I make any mistake, or can the noetherian hypothesis be omitted here?

Answer 1

You are right, the Noetherian hypothesis is not needed here. Actually, the following is true

Let $(X, \mathcal{O}_{X})$ be a scheme and $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_{X}$-module of finite type (i.e. a module such that for any point $p \in X$ there exists an open neighbourhood $U$ of $p$ and a surjection $\mathcal{O}_U^n \rightarrow \mathcal{F} \vert_U$), then the underlying topological space of $V(\text{Ann} \mathcal{F})$ is $\text{Supp} \mathcal{F}$.

As the question is local, the proof is exactly the one you did (you can find it in Ulrich Görtz, Torsten Wedhorn, Algebraic geometry I, Proposition 7.35).

In my opinion, the reason why the exercise has been put in that way is that if the scheme is noetherian we have the equivalence

$$\textit{quasi-coherent + finite type} \iff \textit{coherent}$$

Answer 2

$\def\sI{\mathcal{I}} \def\sO{\mathcal{O}} \def\supp{\operatorname{supp}} \def\ann{\operatorname{Ann}} \def\sF{\mathcal{F}}$Here's a general proof.

Definition. Let $X$ be a ringed space. Let $\sI\subset\sO_X$ be an ideal sheaf. The vanishing locus of $\sI$ is $$ V(\mathcal{I})=\supp(\sO_X/\sI). $$

$V(\sI)$ is a closed subset [01AV].

The annihilator sheaf $\ann_{\sO_X}\sF$ is defined over an arbitrary ringed space $X$ and for any sheaf of $\sO_X$-modules $\sF$ [0H2G]. In the case $X$ is a scheme and $\sF$ is finite, it holds $(\ann_{\sO_X}\sF)|_U\cong\widetilde{\ann_AM}$, where $\operatorname{Spec}A=U\subset X$ is open affine and $M=\sF(U)$ [ref], so the result follows from:

Lemma. Let $X$ be a ringed space and let $\sF$ be an $\sO_X$-module. Then $$ \supp\sF\subset V(\ann\sF), $$ with equality if $\sF$ is finite.

Proof. Let $x\in X$. In general, we have an inclusion of ideals of $\sO_{X,x}$: $$ \tag{1}\label{cont} (\ann_{\sO_X}\sF)_x\subset\ann_{\sO_{X,x}}\sF_x. $$ Thus $\sF_x\neq 0$ implies $(\ann_{\sO_X}\sF)_x\neq\sO_{X,x}$, i.e., we have the desired containment. The converse follows from the fact that \eqref{cont} is an equality if $\sF$ is finite [ref] $\square$.

If $X=\operatorname{Spec} A$ is an affine scheme and $M$ is an $A$-module, by the containment $(2)$ here, it follows that $\supp\widetilde{M}\subset V(\ann_{\sO_X}\widetilde{M})\subset V(\widetilde{\ann_AM})$. Thus $$ \supp M\subset V(\ann_AM) $$ always holds.