# Nakayama and locally free sheaves

Let $S$ be a noetherian scheme and $F,E$ be two coherent sheaves on $S$ with $E$ locally free.

Suppose we have a morphisme $f : F \to E$ such that $f_s : F \otimes \kappa(s) \to E \otimes \kappa(s)$ is an isomorphism for all $s \in S$ then is it true that $f$ is an isomorphism ?

Nakayama implies it is surjective but I don't see injectivity ?

• Note that the noetherian hypothesis is not necessary in your question nor in Moos's excellent answer. It may seem like nitpicking but I think it is healthy to remember that Nakayama has nothing to do with noetherianness, contrary to a popular misconception. My experience is that the less hypotheses the better in order to remember and understand a theorem, even if the hypotheses per se are practically always fulfilled – Georges Elencwajg Mar 30 '16 at 10:56
• Thanks ! In fact I only added Noetherian to be sure but it's true that I always get confused about the noetherian hypothesis with Nakayama :) – A.champolion Mar 30 '16 at 11:23
• One often uses Nakayama for some ideal of a ring, without any further knowledge about the ideal. Then it is pretty good to be in the noetherian case :) But yes, in general, we only need finite generation to apply Nakayama. – MooS Mar 30 '16 at 13:07

$(R,\mathfrak m)$ is a noetherian local ring and $F,E$ two finitely generated $R$-modules with $E$ free. The map $f: F \to E$ is an isomorphism after tensoring with $R/\mathfrak m$, hence it is surjective by Nakayama.
But now, note that $E$ is free, in particular projective, hence the map splits, say by $g: E \to F$. After tensoring with $R/\mathfrak m$, $g \otimes R/\mathfrak m$ is an isomorphism, since it is the splitting of the isomorphism $f \otimes R/\mathfrak m$.
Again, by Nakayama, we deduce that $g$ is surjective. $g$ was a priori injective, hence $g$ is an isomorphism, thus $f$ is an isomorphism, too.