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Possible Duplicate:
Atiyah-MacDonald, exercise 2.11

The following question came up during tea today.

Let $R$ be a commutative ring with an identity and let $M \subset R^n$ be a submodule. Assume that $M \cong R^k$ for some $k$. Question : Must $k \leq n$?

If $R$ is a domain, then this is obvious. The obvious approach to proving the general result then is to mod out by the radical of $R$. If the resulting map $M / \text{rad}(R) M \rightarrow (R / \text{rad}(R))^n$ were injective, then we'd be done. However, I can't seem to prove this injectivity (I'm not even totally convinced that it's true).

Thank you for any help!

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    $\begingroup$ This is one of the early exercises in Atiyah-Macdonald. The answer is "yes" but the proof is tricky and there's not enough room in this comment box for it ;-) I think that if you google around for solutions to all the exercises in Atiyah-Macdonald then you will find a document that looks promising but which contains an incorrect proof. I remember when I did this question finding the notion of Euler characteristic very helpful, which I learnt from one of the later chapters of Matsumura! It has been suggested that A-M might have put the question in in error, not realising how tricky it was. $\endgroup$
    – Kevin Buzzard
    Commented Jul 7, 2010 at 6:29
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    $\begingroup$ Even though Robin has already answered the question I'm still going to point out that this is a duplicate of mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/… , something I discovered after Robin had posted his answer. I'm voting to close, in a nice way. $\endgroup$
    – Kevin Buzzard
    Commented Jul 7, 2010 at 6:50
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    $\begingroup$ We have to assume $R \neq 0$. ;-) $\endgroup$
    – Martin Brandenburg
    Commented Jul 7, 2010 at 8:12
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    $\begingroup$ If I understand correctly, this question will be deleted. In my opinion, this would be unfortunate because the answer to it are (I think) of at least as high quality as the answers to the "Atiyah-MacDonald, exercise 2.11" question (mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11). The best would be of course to append the answers to this question to the answers to the other question. But if this is too complicated, it would be better to reopen this question. $\endgroup$
    – Pierre-Yves Gaillard
    Commented Jul 8, 2010 at 18:00

3 Answers 3

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This reduces to the question: is there an $R$-module injection from $R^{n+1}$ to $R^n$. This is a matrix question: is there a nonzero nullvector for an $n$-by-$n+1$ matrix $M$.

Clearly $M$ has a nullvector formed by the $n$-by-$n$ minors, the trouble is that it could be zero. In that case we need to show that an $n$-by-$n$ matrix $N$ with zero determinant has a nonzero nullvector.

Let $r$ be the determinantal rank of $N$: the size of the largest nonzero subdeterminant of $N$. Then $r < n$. Let's assume the top left $r$ by $r$ submatrix of $N$ has nonzero determinant. Let $N'$ be the top left $r+1$-by-$r+1$ submatrix of $N$. Then the adjugate of $N'$ has a nonzero row. Fill this out to a row vector of length $n$ by adding zeros. Then this is a nullvector of $N$.

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    $\begingroup$ Robin, this question is a duplicate ;-) grin but I only just discovered the source. mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/… . Having discovered this I'm voting to close. While you were answering the question I was grepping the database dumps :-/ $\endgroup$
    – Kevin Buzzard
    Commented Jul 7, 2010 at 6:50
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    $\begingroup$ Blimey Kevin, you're a hard man :-) Anyway I think my proof is nicer than those in the cited thread. OK, my argument is more-or-less the same as Anton's there, except I don't worry about annihilators as they seem unnecessary here :-) $\endgroup$
    – Robin Chapman
    Commented Jul 7, 2010 at 8:34
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    $\begingroup$ I really like the second proof in Pete Clark's answer. I found it a few days ago by googling for the paper mentioned in mathoverflow.net/questions/30066/… (dealing with a related question); it's Cor 2.7.4 in Antoine Chambert-Loir, ALGÈBRE COMMUTATIVE, Cours à l’Université de Rennes 1 (2006–2007) $\endgroup$
    – Victor Protsak
    Commented Jul 7, 2010 at 22:18
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For a proof using multilinear algebra, see Corollary 5.11 at

http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf

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  • $\begingroup$ Beautiful. I liked Chapman's proof a lot and felt the urge to translate it into coordinate-free language. Now I don't have to. $\endgroup$
    – Tom Goodwillie
    Commented Jul 7, 2010 at 20:56
  • $\begingroup$ That is, I wanted to see an argument explicitly using exterior powers instead of determinants. $\endgroup$
    – Tom Goodwillie
    Commented Jul 8, 2010 at 0:25
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Here's a proof by Karl Dahlke:

Math Reference: A Free Submodule Embeds

The result can be generalized to infinite ranks.

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  • $\begingroup$ I didn't read the details of the proof carefully (so I might be misinterpreting things), but at least at the beginning he claims to only be considering domains. $\endgroup$
    – Andy Putman
    Commented Jul 7, 2010 at 14:18
  • $\begingroup$ Andy, the link I posted does not assume the underlying ring is a domain. $\endgroup$
    – KConrad
    Commented Jul 8, 2010 at 3:09
  • $\begingroup$ @KConrad : Yes, the proof at the link you posted is very nice! I was just pointing out that this answer does not appear to answer the question posed by the OP. $\endgroup$
    – Andy Putman
    Commented Jul 8, 2010 at 3:19
  • $\begingroup$ @Andy: Read carefully. Karl generalizes step by step. $\endgroup$
    – Martin Brandenburg
    Commented Jul 8, 2010 at 9:55

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