Do localization and dual commute for locally free modules of rank $1$?

Question

Let $A$ be an integral domain and $M$, $N$ be finitely generated $A$-modules. I know from this topic that one cannot expect $\hom_A(M,N)_P \cong \hom_{A_P}(M_P,N_P)$ to be true in the general case (although I lack the background to fully grasp the given counterexample), but

what if I consider the localization of the dual module of $M$, i.e. $N = A$?

For my purposes, I may even assume that $M$ is locally free of rank $1$. If I didn't miscalculate, the natural map should be injective, but I fail to prove surjectivity. (A more abstract proof would be great, too!)

This MO question may be relevant. It appears that under your hypotheses, $M$ is f.g. and projective, and hence finitely presented. — Zhen Lin, Dec 09 '11 at 21:50

score 2 · Accepted Answer · answered Dec 09 '11 at 22:04

Yes, in the case you mention $Hom$ commutes with localization.

Reminder
For a locally free $A$-module $M$ of rank 1, the dual module $Hom_A(M,A)$ is the only module $L$ such that $M\otimes L=A$. [This is why such $M$ are also called invertible modules]

The rest is easy: localizing at $P$, you get $(M\otimes_A L)_P=M_P\otimes_{A_P} L_P=A_P$ and this proves, by the Reminder again, that $L_P=Hom_{A_P}(M_P,A_P)$.
This is what you wanted : $(Hom_A(M,A))_P=Hom_{A_P}(M_P,A_P)$

Do localization and dual commute for locally free modules of rank $1$?

1 Answers1