# A homomorphism between finite free modules over a local ring

The following lemma is stated in a book without a proof. How can this be proved?

Lemma Let $A$ be a local ring. Let $k$ be the residue field of $A$. Let $E$ and $F$ be finite free modules over $A$. Let $f:E → F$ be a $A$-homomorphism. Let $\bar{f}$$:$ $E\otimes k$ $→$ $F\otimes k$ be the homomorphism induced by $f$. Suppose $\bar{f}$ is an isomorphism. Then $f$ is also an isomorphism.

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## 1 Answer

Because $E$ and $F$ are finite free module, and $E/\mathfrak{m}E\cong F/\mathfrak{m}F$, where $\mathfrak{m}$ is the maximal ideal of $A$. So $E$ and $F$ have the same rank, say $n$. And by Nakayama theorem $E\to F$ is surjective. Hence $A^n\cong E\to F\cong A^n$ is surjective.

You need to show a surjective morphism between $A^n$ and $A^n$ is indeed bijective.

One easy way to see this is using linear algebra. Say $\phi$ is the surjective from $A^n\to A^n$, then we can find $\psi: A^n\to A^n$ such that $\phi\psi=1$, by theory of adjoint matrix or something like that (that is if a matrix has right inverse it itself is inverse), so $\psi \phi=1$.

Another way to see this is use Nakayama theorem again, by Nakayama's lemma, we can in fact show that a surjective endormorphism of an finite module $M$ is indeed bijective! (cf, Matsumura, commutative ring, ch 1,theorem ?).

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Thanks! Please let me wait for a few days before I accept your answer. – Makoto Kato May 20 '12 at 8:31