Lemma 30.5.9. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $s \in S$ be a point which is not in the image of $f$. Then $s$ is not weakly associated to $f_*\mathcal{F}$.

Proof. The question is local so we may assume $X = \mathop{\mathrm{Spec}}(A)$. By Schemes, Lemma 25.24.1 the sheaf $f_*\mathcal{F}$ is quasi-coherent, say corresponding to the $A$-module $M$. Say $s$ corresponds to $\mathfrak p \subset A$. As $s$ is not in the image of $f$ we see that $X = \bigcup _{a \in \mathfrak p} f^{-1}D(a)$ is an open covering. Since $X$ is quasi-compact we can find $a_1, \ldots , a_ n \in \mathfrak p$ such that $X = f^{-1}D(a_1) \cup \ldots \cup f^{-1}D(a_ n)$. It follows that

$M \to M_{a_1} \oplus \ldots \oplus M_{a_ r}$

is injective. Hence for any nonzero element $m$ of the stalk $M_\mathfrak p$ there exists an $i$ such that $a_ i^ n m$ is nonzero for all $n \geq 0$. Thus $\mathfrak pA_\mathfrak p$ is not weakly associated to $M_\mathfrak p$. $\square$

Comment #4177 by Zhiyu Zhang on

"The question is local so we may assume $X=Spec(A)$", maybe you mean $S=Spec A$ ? Moreover, I think in order to have $X = \bigcup _{a \in \mathfrak p} f^{-1}D(a)$, we may need $s$ be a closed point, hence it's better to base chagne to $Spec O_{S,s}$ at first (as the question is local).

Comment #4178 by on

Good catch and thanks for explaining the fix! I will update the text later. Thanks again.

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