Lemma 38.2.9. Let $S$ be a scheme and $s \in S$ a point. Denote $\mathcal{O}_{S, s}^ h$ (resp. $\mathcal{O}_{S, s}^{sh}$) the henselization (resp. strict henselization), see Algebra, Definition 10.155.3. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ S$-module. The following are equivalent
$s$ is a weakly associated point of $\mathcal{F}$,
$\mathfrak m_ s$ is a weakly associated prime of $\mathcal{F}_ s$,
$\mathfrak m_ s^ h$ is a weakly associated prime of $\mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S, s}^ h$, and
$\mathfrak m_ s^{sh}$ is a weakly associated prime of $\mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S, s}^{sh}$.
Proof.
The equivalence of (1) and (2) is the definition, see Divisors, Definition 31.5.1. The implications (2) $\Rightarrow $ (3) $\Rightarrow $ (4) follows from Divisors, Lemma 31.6.4 applied to the flat (More on Algebra, Lemma 15.45.1) morphisms
\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^{sh}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^ h) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \]
and the closed points. To prove (4) $\Rightarrow $ (2) we may replace $S$ by an affine neighbourhood. Suppose that $x \in \mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S, s}^{sh}$ is an element whose annihilator has radical equal to $\mathfrak m_ s^{sh}$. (See Algebra, Lemma 10.66.2.) Since $\mathcal{O}_{S, s}^{sh}$ is equal to the limit of $\mathcal{O}_{U, u}$ over étale neighbourhoods $f : (U, u) \to (S, s)$ by Algebra, Lemma 10.155.11 we may assume that $x$ is the image of some $x' \in \mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{U, u}$. The local ring map $\mathcal{O}_{U, u} \to \mathcal{O}_{S, s}^{sh}$ is faithfully flat (as it is the strict henselization), hence universally injective (Algebra, Lemma 10.82.11). It follows that the annihilator of $x'$ is the inverse image of the annihilator of $x$. Hence the radical of this annihilator is equal to $\mathfrak m_ u$. Thus $u$ is a weakly associated point of $f^*\mathcal{F}$. By Lemma 38.2.8 we see that $s$ is a weakly associated point of $\mathcal{F}$.
$\square$
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