The Stacks project

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. $\square$