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. $\square$

Comment #1163 by Hu Fei on

In the statement of Lemma 31.2.2, ''Let $\Spec(A) = U \subset X$ 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 $U$ is an open subset of a scheme $X$, then we endow it with the scheme structure inherited from $X$. Hence when we say "Let $U \subset X$ be an affine open", then we really mean "Let $U \subset X$ be an open such that with the scheme structure inherited from $X$ it is an affine scheme"...

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