# A surjective homomorphism between finite free modules of the same rank

I know a proof of the following theorem using determinants. For some reason, I'd like to know a proof without using them.

Theorem Let $A$ be a commutative ring. Let $E$ and $F$ be finite free modules of the same rank over $A$. Let $f:E → F$ be a surjective $A$-homomorphism. Then $f$ is an isomorphism.

-
I think there is a similar result in Atiyah Macdonald chapter 2 no? The proof for it is using the Cayley Hamilton theorem, does that count as using determinants? –  user38268 Jun 3 '12 at 23:02
I cannot find the result in the book. Could you tel me the proposition No.? –  Makoto Kato Jun 3 '12 at 23:12
It's in the exercises exercise 11 of chapter 2. –  user38268 Jun 3 '12 at 23:13
Dear @Benjamin, Atiyah-Macdonald assume noetherianness. Actually you don't need noetherian, nor even freeness of the module. Only that the module be finitely generated: see here (where you can find a reference, Makoto) –  Georges Elencwajg Jun 3 '12 at 23:15
See M. Orzech, "Onto Endomorphisms are Isomorphisms", Amer. Math. Monthly 78 (1971), 357--362. To quote from the first section "We shall begin Section 3 by indicating several methods of approaching the proof of Vasconselos's theorem. Two of these methods, both known by Vasconselos, have in common the use of the theory of determinants over a commutative ring. We shall show that Theorem 1 can be proved without the use of determinants." –  KCd Jun 4 '12 at 1:06

You can show that every commutative ring is stably finite (see Lam's Lectures on Modules and Rings first 10 pages or so) which means that if $R^n\cong R^n\oplus N$, then $N=0$.
If you have a surjection $f:M\rightarrow M'$, then $M/\ker(f)\cong M'$, but $M'$ being projective implies that $0\rightarrow \ker(f)\rightarrow M\rightarrow M/\ker{f}\rightarrow 0$ splits, and so $M\cong \ker(f)\oplus M'$, whence $\ker{f}=\{0\}$.
By localizing at every maximal ideal of $A$, we can assume $A$ is a local ring. Let $I$ be the maximal ideal of $A$. Let $k = A/I$. Let $K = Ker(f)$. Since $0\rightarrow K\rightarrow E\rightarrow F\rightarrow 0$ splits, $0\rightarrow K\otimes k\rightarrow E\otimes k\rightarrow F\otimes k\rightarrow 0$ is exact. Hence $K\otimes k = 0$. By Nakayama's lemma, $K = 0$ as desired. –  Makoto Kato Jun 4 '12 at 19:15