The Stacks project

Lemma 31.2.2. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open, and set $M = \Gamma (U, \mathcal{F})$. Let $x \in U$, and let $\mathfrak p \subset A$ be the corresponding prime.

  1. If $\mathfrak p$ is associated to $M$, then $x$ is associated to $\mathcal{F}$.

  2. If $\mathfrak p$ is finitely generated, then the converse holds as well.

In particular, if $X$ is locally Noetherian, then the equivalence

\[ \mathfrak p \in \text{Ass}(M) \Leftrightarrow x \in \text{Ass}(\mathcal{F}) \]

holds for all pairs $(\mathfrak p, x)$ as above.

Proof. This follows from Algebra, Lemma 10.63.15. But we can also argue directly as follows. Suppose $\mathfrak p$ is associated to $M$. Then there exists an $m \in M$ whose annihilator is $\mathfrak p$. Since localization is exact we see that $\mathfrak pA_{\mathfrak p}$ is the annihilator of $m/1 \in M_{\mathfrak p}$. Since $M_{\mathfrak p} = \mathcal{F}_ x$ (Schemes, Lemma 26.5.4) we conclude that $x$ is associated to $\mathcal{F}$.

Conversely, assume that $x$ is associated to $\mathcal{F}$, and $\mathfrak p$ is finitely generated. As $x$ is associated to $\mathcal{F}$ there exists an element $m' \in M_{\mathfrak p}$ whose annihilator is $\mathfrak pA_{\mathfrak p}$. Write $m' = m/f$ for some $f \in A$, $f \not\in \mathfrak p$. The annihilator $I$ of $m$ is an ideal of $A$ such that $IA_{\mathfrak p} = \mathfrak pA_{\mathfrak p}$. Hence $I \subset \mathfrak p$, and $(\mathfrak p/I)_{\mathfrak p} = 0$. Since $\mathfrak p$ is finitely generated, there exists a $g \in A$, $g \not\in \mathfrak p$ such that $g(\mathfrak p/I) = 0$. Hence the annihilator of $gm$ is $\mathfrak p$ and we win.

If $X$ is locally Noetherian, then $A$ is Noetherian (Properties, Lemma 28.5.2) and $\mathfrak p$ is always finitely generated. $\square$

Comments (2)

Comment #1163 by Hu Fei on

In the statement of Lemma 31.2.2, ''Let be an affine open'' should add ''subscheme''

Comment #1180 by on

Well, I am going to leave it as is, because our convention is that if is an open subset of a scheme , then we endow it with the scheme structure inherited from . Hence when we say "Let be an affine open", then we really mean "Let be an open such that with the scheme structure inherited from it is an affine scheme"...

There are also:

  • 2 comment(s) on Section 31.2: Associated points

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