A surjective homomorphism between finite free modules of the same rank

Question

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? — , Jun 03 '12 at 23:02
I cannot find the result in the book. Could you tel me the proposition No.? — Makoto Kato, Jun 03 '12 at 23:12
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 03 '12 at 23:15
@GeorgesElencwajg I believe in chapter 2 they haven't covered the noetherian condition yet no? — , Jun 03 '12 at 23:16
@George I know Vasconcelos's theorem and I think it's somewhat tricky. Since the title theorem assumes freeness of modules, I guess there is a simpler proof(actualy there is at least such one using determinants). — Makoto Kato, Jun 03 '12 at 23:24
Dear @Benjamin, well, no, they haven't covered the noetherian condition yet in Chapter 2. But on the other hand I have just checked that exercise 11 doesn't address Makoto's question ! — Georges Elencwajg, Jun 03 '12 at 23:24
Dear Makoto, if you know Vasconcelos's theorem, I have nothing to add: sorry. — Georges Elencwajg, Jun 03 '12 at 23:26
@Benjamin I think exercise 11 of chapter 2 is different from the title theorem, no? — Makoto Kato, Jun 03 '12 at 23:38
@George Not at all. Sorry I didn't mention the Vasconcelos's theorem in the first place. — Makoto Kato, Jun 03 '12 at 23:40
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 04 '12 at 01:06
@George Atiyah-Macdonald's Exercise 6.1 is good enough for me. Thanks. — Makoto Kato, Jun 04 '12 at 08:01
@KCd I like Orzech's proof(though there are a few typos). It is conceptual enough. Since the module is finitely generated, it can be reduced to a Noetherian module case which is Atiyah-Macdonald's Exercise 6.1. — Makoto Kato, Jun 14 '12 at 22:22

score 9 · Accepted Answer · answered Jun 03 '12 at 23:42

9

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

answered Jun 03 '12 at 23:42

rschwieb

153,510

The Lam's book proves that every commutative ring is stably finite using determinants. Thanks, any way. – Makoto Kato Jun 04 '12 at 00:14
@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 04 '12 at 00:25
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 04 '12 at 01:14
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 04 '12 at 05:18
9

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 04 '12 at 19:15
@MakotoKato For the non-psychic editors' benefit, it would be good to include what you had in mind when you wrote the question :) I'm glad you found an approach you like! – rschwieb Jun 04 '12 at 21:16

Mike Pierce · Answer 2 · 2018-09-01T03:38:09.417

This answer is not complete. See the comments below.

The modules $E$ and $F$ being free of finite rank $n$ over $A$ means that they each have a finite basis over $A$. Take $y \in F$, and since $f$ is surjective some $x \in E$ maps to $y$. Pick a basis for $\langle e_1, \dots, e_n \rangle$ of $E$ over $A$, so $x = a_1e_1 + \dotsb + a_ne_n$ for some $a_i \in A$. Then for our arbitrary element $y \in F$, $$ y = f(a_1e_1 + \dotsb + a_ne_n) = a_1f(e_1) + \dotsb + a_nf(e_n) \, $$ so $\langle f(e_1),\dotsc, f(e_n)\rangle$ generates $F$. Since $F$ has the same rank as $E$, these generators must form a basis (this needs to be proven. See darij grinberg's comment below). Since these generators form a basis $$ 0 = f(\alpha_1e_1 + \dotsb + \alpha_ne_n) = \alpha_1f(e_1) + \dotsb + \alpha_nf(e_n) $$ only when the $\alpha_i$ are all zero, so $f$ is injective and hence an isomorphism. ${_\square}$

I don't see why we need $A$ to be a commutative ring. Since we're specifying that $E$ and $F$ have the same rank I assume they have the invariant dimension property. Otherwise commutivity would imply this. Also we're only talking about a single map $f \colon E \to F$ and don't need to talk about the module structure on $\mathrm{Hom}_R(E,F)$, for which we need $R$ to be commutative.

Also I've seen it asked as an exercise, is this still true if we assume $f$ is injective instead of surjective? The answer is no, looking at the counterexample $\mathbf{Z} \to \mathbf{Z}$ where $1 \mapsto 2$, regarding $\mathbf{Z}$ as a rank $1$ free module over itself.

"Since F has the same rank as E, these generators must form a basis": not obvious, and only true for commutative rings. — darij grinberg, Aug 31 '18 at 19:06
@darijgrinberg I see that's not obvious now. But it is true though, right? The elements $f(e_i)$ actually are linearly independent? Do you know a way to proceed with this line of proof, or would you recommend just accepting the approach in rschwieb's answer as the most reasonable way? Also, I still don't see what noncommutativity breaks here. Do you know an illuminating counterexample to this where $R$ is noncommutative? Thank you. :) — Mike Pierce, Sep 01 '18 at 03:37
It is true, but it is equivalent to the original question. In particular, it does not generalise to noncommutative rings that don't have the invariant basis property. — darij grinberg, Sep 01 '18 at 09:44

A surjective homomorphism between finite free modules of the same rank

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