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)

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.