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? – Zhen Lin 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).