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A question about following statement from Martin Olsson's book on Stacks. In the proof of Proposition 13.2.9. (p 269) is claimed that certain sheaf $K$ on a nodal curve $C$ is invertible it suffice to check it etale locally.

I don't get this part.
So far I understand correctly the notion of invertible sheaves makes sense on any site: Say $S$ is a site, $\mathcal{O}$ be a sheaf of commutative rings on $S$. Just as in the case of a scheme with the Zariski topology, we can define an invertible sheaf on $S$ to be a sheaf (wrt $S$'s Grothendieck topology) of $\mathcal{O}$-modules $\mathcal{L}$ such that for any element $U \in S$ there exists a covering $\{U_i \to U\}$ such that the restriction of $\mathcal{L}$ to the localized site $C/U_i$ is isomorphic to the restriction of $\mathcal{C}$ (viewed as an $\mathcal{C}$-module).

Now if we have a sheaf $K$ on $C$ for which we can show that it is etale locally invertible, then it would be an invertible sheaf in sense about on etale site of $C$. But I assume that above Olsson wants $K$ to be actually Zariski invertible sheaf.

Therefore I not understand how invertibility of $K$ as Zariski sheaf can be checked etale locally as claimed. Is it something what work only for curves, as intuitively, it seems that Zariski invertibility appears to much stronger/restrictive then etale invertibility, isn't it?

If I misread, any idea what Olsson means there?

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You're not misreading anything. It's just something that doesn't follow directly from the definitions and has to be checked using an argument.

A sheaf is invertible if and only if the stalk everywhere is a free module of rank $1$ for the local ring. By taking stalks, this reduces to the claim that if $R \to S$ is an étale morphism of local rings, $M$ is an $R$-module, and $M \otimes_R S$ is free of rank $1$ then $M $ is free of rank $1$. This follows from Nakayama's lemma: For $\mathfrak m$ the maximal ideal of $R$, we have $ M/\mathfrak m M$ a vector space over $R/\mathfrak m R$ which must have rank $1$ since $$(M/\mathfrak m M) \otimes_{R/\mathfrak m R} (S/\mathfrak m S) = (M \otimes_R S) \otimes_S (S/\mathfrak m S)$$ is free of rank $1$ over $S/\mathfrak m S$. So we can choose a generator for this space and then lift it to an element of $M$, which must be a generator, giving a surjection $R \to M$. For $f$ in the kernel of this surjection, $M$ is $f$-torsion so $M\otimes_R S$ is $f$-torsion which means the image of $f$ is zero in $S$ which means $f=0$ since étale morphisms of local rings are injective.

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  • $\begingroup$ Following your argumentation it seems to be nothing étale , fppf, descent etc specific. More precisely, it seems that to check that sheaf $K$ is invertible on $C$ we just need to find a surjective map $f: X \to C$ with property that all corresp maps at stalks $O_{C, f(x)} \to O_{X,x}$ are (just set theoretically) injective and $f^*K$ is invertible on $X$, right? $\endgroup$ Commented Oct 28, 2024 at 20:46
  • $\begingroup$ @user267839 Yes, that sounds right. $\endgroup$ Commented Oct 28, 2024 at 20:47

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