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$R$ is Noetherian, $f:R^m \rightarrow R^n$ is surjective. Is $\ker(f)$ free?

lnth
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    To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles May 28 '13 at 23:38
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    $\ker(f)$ is locally free of rank $m-n$, but not free in general. Try to play around with $m=2,n=1$. – Martin Brandenburg May 29 '13 at 00:10

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Let $K = \operatorname{ker} f$. We have a short exact sequence

$0 \rightarrow K \rightarrow R^m \rightarrow R^n \rightarrow 0$.

Since $R^n$ is free, it is projective, so the sequence splits: $R^m \cong K \oplus R^n$. Thus $K$ is finitely generated projective, which over certain rings already implies it must be free: e.g. this is true if $R$ is a PID.

Going deeper, the condition $R^m \cong K \oplus R^n$ says precisely that $K$ is stably free. Such modules need not be free in general, but to give non-free examples is a subtle matter. (For the cognoscenti: stably free modules are precisely the modules which algebraic K-theory cannot distinguish from free modules.)

A discussion of stably free modules -- including some examples of non-free ones -- can be found in $\S$ 6.5 of these notes.

Community
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Pete L. Clark
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    Keith Conrad also gives an example of a stably free but not free module in this note: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/stablyfree.pdf – Qiaochu Yuan May 29 '13 at 03:19
  • @Qiaochu: In fact my notes explicitly reference this "blurb" of Keith's. It is certainly highly recommended. – Pete L. Clark May 29 '13 at 03:20
  • I am little confused, when $m=n$ then $K=0$ and this means, for any ring $R$, any surjective $R$-homomorphism $R^n\rightarrow R^n$ is an isomorphism? – lnth May 29 '13 at 06:22
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    In fact, every surjective endomorphism of a finitely generated module over a commutative(!) ring is an isomorphism. – Martin Brandenburg May 29 '13 at 10:37
  • @lnth: Yes, that's right: this is an exercise in Section 6.5.2 of the notes. The result Martin mentions is Theorem 3.43 in the notes. (The fact that Martin's name appears in the paragraph before the statement of this theorem suggests that we've had a similar conversation at some point in the past!) – Pete L. Clark May 30 '13 at 02:47