# Hartshorne Chapter II exercise 5.7 on Invertible sheaves

I'm working on part c) which is to prove that for a Noetherian scheme $X$, a coherent sheaf $\mathscr{F}$ is invertible (locally free of rank 1) iff there exists a coherent sheaf $\mathscr{G}$ such that $\mathscr{F}\otimes_{O_X}\mathscr{G}=O_X$. I proved that if $\mathscr{F}$ is a coherent invertible sheaf then the natural evaluation map $\mathscr{F}\otimes_{O_X}\mathscr{F}^*\to O_X$ is an isomorphism (here $\mathscr{F}^*$ is the dual of $\mathscr{F}$).

For the converse I thought I'd try to use part b) which says that $\mathscr{F}$ is locally free iff it's stalks are all free. By taking an affine open set I can reduce this to showing that if $M,N$ are finitely generated modules over a noetherian local ring $R$ then $M\otimes_R N=R$ implies $M,N=R$. I'm not sure if this is the approach I should be taking or not, but I'm not sure how to proceed from here.

• That's a good approach. You might want to think about what happens when you tensor with the residue field. What does that imply about the number of generators you need? Oct 10 '14 at 18:35
• @zcn I would accept your comment as an answer if you want. Otherwise I'll just post it myself as an answer and accept.
– Seth
Oct 13 '14 at 21:04
• @Seth: I apologize for the delay - I have posted an answer now (and removed my earlier comment)
– zcn
Oct 20 '14 at 21:43

Proposition: Let $R$ be a local ring, and $M, N$ $R$-modules with $M \otimes_R N \cong R$. Then $M, N \cong R$.
Proof: By the answers to this question, $M, N$ are finitely generated and projective. Since $R$ is local, $M, N$ are in fact free, say $M \cong R^m, N \cong R^n$. Then $R \cong M \otimes N \cong R^m \otimes R^n \cong R^{mn}$, so $mn = 1 \implies m = n = 1$, i.e $M \cong N \cong R$ (since commutative rings have IBN).