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Suppose $R$ is a reduced noetherian ring, and $L$ is a finitely generated flat $R-$module such that $L_p\simeq R_p$ for every prime. Then how to prove $L$ is an invertible sheaf of $\operatorname{Spec} R$?(means we need to extend the isomorphism of stalks to a neighbourhood)

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    $\begingroup$ Here is how you could try to tackle this: Take generators $g_i$ for $L$, and look at their images in $R_p$. Then find an open affine neighbourhood $D(f)$ of $p$, where those images are defined, which gives a map $R_f^n \to R_f$. By localizing further, make sure this factors over $L_f \to R_f$, and gives an isomorphism. $\endgroup$ Sep 14 at 13:13
  • $\begingroup$ Check out 13.7.F & 13.7.K in Vakil: For nice enough settings, a module which is free at a point is free at a neighbourhood around that point (13.7.F) - and a module which has constant rank is locally free (13.7.K). $\endgroup$
    – Qi Zhu
    Sep 14 at 13:57
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There is always an evaluation homomorphism $$ev_L:L\otimes L^{*}\to R, \;\;\; (a,b)\mapsto b(a).$$ This evaluation is natural in the sense that if $f:M\to N$ is a module isomorphism, then there is a module isomorphism $$ev(f):M\otimes M^*\to N\otimes N^*,$$ which forms a commutative triangle with both evaluations (i.e. $ev_M\circ ev(f)=ev_N)$. We also note that the evaluation map $R_p\otimes R^*_p\to R_p$ is a isomorphism.

In particular, we see that after localization at $p$, the maps $(L\otimes L^*)_p=L_p\otimes L^*_p\to R_p$ is the composition of isomporphisms $L_p\otimes L^*_p\to R_p\otimes R_p^*\to R_p$ and so in particualr $ev_L$ is an isomorphism after any localization. As any homorphism that is an isomorphism after any localization is an isomorphism, we see that $ev_L$ is an isomorphism, so that $L$ is invertable.

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  • $\begingroup$ In the last step, how can we know $L$ is invertible? The definition I know is: $L$ is locally free of rank 1. But how to prove that(using $ev_L$ is an isomorphism)? $\endgroup$
    – Richard
    Sep 15 at 1:53
  • $\begingroup$ @Richard There are two standard definitions of invertable sheaves, that are equivalent. For a proof of this you can look at this post mathoverflow.net/questions/33489/… $\endgroup$
    – pax
    Sep 15 at 2:01

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