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For a commutative ring $A$, we call a finite family of localizations $A \to A_{S_i}$ (where $S_i$ are some subsets of $A$) a covering if the canonical morphism $A \to \prod A_{S_i}$ is an effective monomorphism. That is, this arrow is an equalizer to two arrows $\prod\limits_{i} A_{S_i} \rightrightarrows \prod\limits_{i, j} A_{S_i \cup S_j}$ (the usual condition from the definition of a sheaf).

Is it correct that the coverings are stable under pushouts? That is, is it true that for a homomorphism $f \colon A \to B$ the induced family of localizations $B \to B_{f(S_i)}$ will be a covering?

If so, where can I read the proof?

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    $\begingroup$ Some comments on terminology: by definition, a map $A \to B$ in a category $\mathscr C$ is an effective monomorphism (not epimorphism) if $A \to B \rightrightarrows B \amalg_A B$ is an equaliser diagram. The pushout in commutative rings is the tensor product, but it is not clear to me that the natural map $(\prod_i A_{S_i}) \otimes_A (\prod_j A_{S_j}) \to \prod_{i,j} A_{S_i S_j}$ is always injective (this fails for general modules). So I think the name 'effective monomorphism' is not entirely the right one here. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 9, 2023 at 15:15
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    $\begingroup$ Likewise, it is not clear that either variant coincides with the geometric condition that $\coprod_i \operatorname{Spec} A_{S_i} \to \operatorname{Spec} A$ is an effective epimorphism. But all three coincide when the index set is finite, and in this case the condition is stable under base change (for instance since they are then pro-étale covers, so in particular fpqc covers, hence universal effective epimorphisms). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 9, 2023 at 15:26
  • $\begingroup$ @R.vanDobbendeBruyn Of course, "effective monomorphism", thank you. I was just switching between algebra and geometry and accidentally made a typo. $\endgroup$
    – Arshak Aivazian
    Commented Dec 9, 2023 at 16:59
  • $\begingroup$ In my condition, the index set is finite. Thank you very much for your comments! $\endgroup$
    – Arshak Aivazian
    Commented Dec 9, 2023 at 17:01

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