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Let $M$ and $N$ be finitely generated modules over a local ring $A$ with residue field $k$ and $f:M\rightarrow N$ a A-homomorphism, such that the induced morphism

$M\otimes_{A}k \rightarrow N\otimes_{A}k$

is an isomorphism.

Can I conclude with Nakayama that already $f$ is an isomorphism?

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Take $M:=A$, $N:=k$. –  Pierre-Yves Gaillard Sep 7 '11 at 10:05

2 Answers 2

up vote 9 down vote accepted

This question is a little more subtle than it might look at first sight.

A) Consider the cokernel $Q$ of $f$ and the exact sequence $M\to N\to Q\to 0$.
It gives rise (by right-exactness of tensor product) to the exact sequence $\bar M \stackrel {\bar f}{\to} \bar N\to \bar Q\to 0$.
Since $\bar f:\bar M \to \bar N$ is an isomorphism by hypothesis , we have $\bar Q=0$, and since $Q$ is finitely generated as a quotient of $N$, we may apply Nakayama to deduce that $Q=0$ and so we are already half-way:
Yes, $f:M\to N$ is always surjective !

B) Analogously, we are led to consider the exact sequence $0\to K \to M\to N$ , with $K=Ker(f)$, and to tensor it with $ k \;$ so as to get the complex $0\to \bar K \to \bar M \stackrel {\bar f}{\to}\bar N$.
The first difficulty is that $K$ has no reason to be finitely generated. Never mind, assume $A$ noetherian and then $K$ will be finitely generated, right?
Right, but now comes the coup de grâce: although $\bar f:\bar M \to \bar N$ is an isomorphism, we cannot conclude that $\bar K$ is zero because the complex $0\to \bar K \to \bar M \to \bar N$ is not necessarily exact (since the tensor product is not left exact) and we cannot say that $\bar K$ is equal to $0=Ker(\bar f)$.
The analogy with A) has completely broken down!

And indeed, Pierre-Yves's counterexample shows that injectivity of $f$ cannot be deduced from the hypothesis that $\bar f$ is an isomorphism.
He considers the quotient map $A \to k$ , which is not injective if $A$ is not a field. However after tensorization with $k$, it becomes the isomorphism $ k \stackrel {\sim}{\to }k$.
We now understand why the conclusion is:
No, $f:M\to N$ is not always injective!

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Let $I$ be the augmentation ideal of $A$. Then $M\otimes _A k \cong M/MI$, i.e. a lot of information about $M$ is thrown away when you take this tensor product. You're asking "does $f$ inducing $M/MI\cong N/NI$ imply $f$ is an isomorphism $M \to N$? The answer here is definitely no (for general local rings - it's true for $A=k$ of course); Pierre-Yves gave a counterexample in the comments.

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