Lemma 31.5.9. 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. Consider the base change $f' : X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ of $f$ by the morphism $g : \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \to S$ and denote $g' : X' \to X$ the other projection. Then

$(f_*\mathcal{F})_ s = (g^*f_*\mathcal{F})_ s = (f'_*(g')^*\mathcal{F})_ s$

The first equality because $g$ induces an isomorphism on local rings at $s$ and the second by flat base change (Cohomology of Schemes, Lemma 30.5.2). Of course $s \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is not in the image of $f'$. Thus we may assume $S$ is the spectrum of a local ring $(A, \mathfrak m)$ and $s$ corresponds to $\mathfrak m$. By Schemes, Lemma 26.24.1 the sheaf $f_*\mathcal{F}$ is quasi-coherent, say corresponding to the $A$-module $M$. As $s$ is not in the image of $f$ we see that $X = \bigcup _{a \in \mathfrak m} f^{-1}D(a)$ is an open covering. Since $X$ is quasi-compact we can find $a_1, \ldots , a_ n \in \mathfrak m$ 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 $\mathfrak m$ is not weakly associated to $M$. $\square$

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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