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I saw a statement in a paper like what follows:

Let $X=\text{Spec} A$ be an affine scheme, and let $\mathscr{F}$ be a sheaf of $\mathscr{O}_X$-modules on $X$. For each geometric point $x$ of $X$ we set $\mathscr{F}(x):=\mathscr{F}\otimes_A k(x)$.

My question is: what does $\mathscr{F}\otimes_A k(x)$ mean?

Thank you for your help!

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    $\begingroup$ This $\mathscr{F}(x)$ doesn't seem to be the stalk of $\mathscr{F}$ at the geometric/étale point $x \in X$. It looks to me that $\mathscr{F} \otimes_A k(x)$ here either means the tensor product over $A$ of the stalk at $x$ of $\mathscr{F}$ with the residue field $k(x)$ at $x$, or it is an abuse of notations of the tensor product of $\mathcal{O}_X$-modules of $\mathscr{F}$ with the skyscraper sheaf at $x$ with value $k(x)$. I honestly can not tell for sure without more context, but at least now I'm leaning more towards the first possibility. $\endgroup$
    – Dat Minh Ha
    Commented Jun 26, 2022 at 16:48
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    $\begingroup$ If $\mathcal F$ is a quasicoherent sheaf of $\mathcal O_X$-modules, we can abuse notation to view $\mathcal F$ as an $A$-module because of the equivalence of categories, and then $\mathcal F \otimes_A k(x)$ makes sense as a tensor product of $A$-modules over the ring $A$. But for a general sheaf, the definition one wants to use is $\mathcal F_x \otimes_{A_x} k(x)$, the tensor product of the Zariski stalk of $\mathcal F$ at $x$ over the Zariski local ring $A_x$. $\endgroup$
    – Will Sawin
    Commented Jun 26, 2022 at 18:24
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    $\begingroup$ Thank both you for the reply. I think it's more likely that Will's answer is true because $\mathscr{F}$ is locally free in the context. $\endgroup$
    – AAAS
    Commented Jun 26, 2022 at 20:10

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