# 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.

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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

## 1 Answer

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 $\ker{f}=\{0\}$.

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The Lam's book proves that every commutative ring is stably finite using determinants. Thanks, any way. – Makoto Kato Jun 4 '12 at 0:14
@MakotoKato Well rather than give up because the answer is not given to you on a platter, an idea might be to prove that without determinants... or if you are willing to prove for me that it is impossible not to use determinants, I would be interested in seeing that. – rschwieb Jun 4 '12 at 0:25
I've got an idea. I think we can assume that A is a local ring by localizing at every maximal ideal of the ring. – – Makoto Kato Jun 4 '12 at 1:14
For the record, the Lam's proof using determinants is exactly the same as I had in my mind when I asked the title question. – Makoto Kato Jun 4 '12 at 5:18
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