# Surjectivity implies injectivity of finitely generated modules, localization?

The following problem is canonical:

Suppose $A$ is a commutative unitary ring, and $M$ is a finitely generated module over $A$. If an endomorphism $f\colon M\to M$ is surjective, then it's also injective.

It appears in Eisenbud's Commutative Algebra, or another MSE post. There're two canonical ways to crack it. However, I'm considering another method, but get stuck at some point. I wonder whether my method could be completed. I want some help on this.

First, note that injectivity and surjectivity are local properties, we can assume that $(A,m)$ is a local ring and $k=A/m$ is the residue field of $m$. Now we tensor the following exact sequence $$0\to N=\ker f\to M\xrightarrow f M\to 0$$ and obtain a new exact sequence $$N\otimes_A k\to M\otimes_A k\xrightarrow{f\otimes1}M\otimes_A k\to 0$$ Since $M$ is finitely generated, $M\otimes_A k$ is finite dimensional over $k$, $f\otimes1$ is surjective therefore an isomorphism, thus the image of $N\otimes_A k$ in $M\otimes_A k$ is zero, therefore $N\subseteq mM$. However, I don't know how to proceed then. I want to take advantage of Nakayama's lemma.

Well, as user26857 said, there's no evidence that $N$ is immediately finitely-generated, so even if $N\otimes_A k=0$, $N$ needn't be $0$. I don't know whether $N\subseteq mM$ could be used to lift the inverse map of $f\otimes1$, etc. The next step is still unclear.

Note that if $f\otimes1$ is surjective, by Nakayama's lemma, $f$ is also surjective, so the reduction is somewhat safe. Note also that $M\otimes_A k$ is also fiber of the module $M$ over $m$, and the pattern is quite similar to inverse function theorem in smooth category, which lifts a linear isomorphism to a local diffeomorphism.

Any help? Thanks!

Remark: For completeness, here's the proof on Eisenbud's book:

We may regard $M$ as a module over $A[X]$, letting $X$ act by $X.m=f(m)$ for $m\in M$. If we set $I=(X)$, then since $f$ is surjective, $IM=M$ and by a version of Nakayama's lemma, there exists $g\in A[X]$ such that $(1-g(X)X)M=0$, therefore $1-g(f)f=0$, thus $g(f)$ is the inverse to $f$.

If somebody could point out a nice geometric viewpoint of the preceding proof, it could also be a good answer.

• @user26857 Maybe $N$ isn't essential. – Yai0Phah Jul 22 '14 at 14:25
• I doubt that the proof works. Base change to the residue field loses too much information. One has to use other arguments. At the moment I have 4 proofs for the statement. For example, the usual arguments let us reduce to the case that $A$ is of finite type and hence noetherian. But then $M$ is noetherian, hence Hopfian. – Martin Brandenburg Jul 22 '14 at 17:43
• @MartinBrandenburg I didn't understand your point, but by Nakayama, the surjectivity of $M\otimes_k k\xrightarrow{f\otimes1}M\otimes_k k$ implies the surjectivity of $M\xrightarrow fM$. See, for example, Atiyah & Macdonald, Ex 2.10. – Yai0Phah Jul 23 '14 at 10:12
• Of course I know that. But you want to show that $f$ is injective. And for this the residue field is not enough. – Martin Brandenburg Jul 23 '14 at 11:42
• @MartinBrandenburg Yeah, it might be frustrating. Incidentally, is there any geometric viewpoint of viewing $M$ as an $A[X]$ module by action $X.m=f(m)$? – Yai0Phah Jul 23 '14 at 11:57