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If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a nontrivial commutative ring, does it follow that $m \geq n$?

This is obviously true for vector spaces over a field, but how would one show this over just a commutative ring?

-----Edit

Is there any way to use the following?

If $\varphi : M \to M'$ is an epimorphism of left $S$-modules and $N$ is any right $S$-module then $id_N \otimes \varphi $ is an epimorphism.

darij grinberg
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user6560
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    @Chandru1: what's wrong with the question? It seems perfectly clear to me. (In fact it is a standard question: one of the exercises in Atiyah-Macdonald.) Hint: tensor with $R/\mathfrak{m}$, where $\mathfrak{m}$ is a maximal ideal of $R$. – Pete L. Clark Feb 03 '11 at 02:57
  • @Pete L.Clark: I can't understand as to what he wants to ask. I think that the title should be a part of the question. Since you know the subject you could understand it, since i am not much well versed it becomes hard for me to understand as to what he is asking. –  Feb 03 '11 at 03:01
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    @Chandru1: The question is: If $\varphi: R^{m} \to R^{n}$ is a epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? As Pete pointed out, this follows from the first sentence in the question by tensoring with $R / \mathfrak{m}$. – t.b. Feb 03 '11 at 03:05
  • It's worth noting that the result is false for noncommutative rings; in fact, for every positive integers $m,n$, there exist a noncommutative ring $R$ such that $R^s \cong R^t$ if and only if $s\geq m$ and $s\equiv t \pmod{n}$. – Arturo Magidin Feb 03 '11 at 03:27
  • Thank you. Sorry for the shorthand post. – user6560 Feb 03 '11 at 03:46
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    Google "invariant basis number" (IBN) for more than you wanted to know. – Bill Dubuque Feb 03 '11 at 03:50
  • Will this work? Since R has an IBN, choose a maximally linearly independent set in R^n, {N_i}. Then each of these is the image of some {M_i} under phi. Hence any linear combination r1N1+...+rkNk where k<=n. is the sum r1f(M1)+...+rkf(Mk) = f(r1M1+...+rkMk). The set {M_i} must also be linearly independent and must be at most a maximally linearly independent set, so it is the subset of some basis. – user6560 Feb 03 '11 at 03:58
  • @Chandru: I am not leaving an answer because it is a standard exercise, so I think it is more appropriate to leave a hint. The analogous statement for monomorphisms is much trickier -- someone asked me about this exercise a few years ago and I wasn't able to solve it -- but I found a nice treatment in one of Lam's books and posted it on several websites, most recently on MO. – Pete L. Clark Feb 03 '11 at 04:02
  • I understand, I am not asking to be given anything. I appreciate the hint. Tensors are still something I'm trying to understand yet so I tried to avoid them. I'll have to think about this more. – user6560 Feb 03 '11 at 04:05
  • @Pete: Ok, Pete –  Feb 03 '11 at 04:06
  • @Pete: That's a nice a simple proof. I was thinking about using determinants to show that a left inverse of a square matrix is also a right inverse, which implies the result. The matrix method does also work for monomorphisms though. – George Lowther Feb 03 '11 at 20:08
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    Commenting so that this will be visible at the top: This question is Problem 1.4 on the homework for Math 620, at the University of Buffalo. As user "Student" points out, every question user6560 has asked is a homework question from that course. http://www.math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf – David E Speyer Feb 13 '11 at 19:28

2 Answers2

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As mentioned in the comments to the question, it’s the first part of Exercise 2.11 in Atiyah-MacDonald, and I refer to the comments for an answer.

The second part of Exercise 2.11 (which is perhaps more interesting) has been the subject of this MO question.

I especially like Balazs Strenner’s answer.

EDIT. Here is Exercise 2.11 of Atiyah-MacDonald. Let $A$ be a nonzero commutative ring and $\phi:A^m\to A^n$ an $A$-linear map. Then

(a) $m\ge n$ if $\phi$ is surjective,

(b) $m\le n$ if $\phi$ is injective,

As pointed out in the comments to the question, there is an obvious proof of (a) [tensor with $A/\mathfrak m$, $\mathfrak m$ maximal]. My favorite proof of (b) is Strenner's one mentioned above. A natural question is: Can one use Strenner's argument to prove (a) and (b) at one go? I'll try to do that below.

Lemma. Let $B$ be a commutative ring, $A$ a subring, $b$ a nonzero element of $B$ which is integral over $A$ and which is not a zero divisor. Then there is a nonzero $a$ in $A$ and a monic $f$ in $A[X]$ such that $a=bf(b)$.

Proof. Let $g\in A[X]$ be a least degree monic polynomial annihilating $b$. Such exists because $b$ is integral over $A$. The constant term $a$ of $g$ is nonzero because $b$ is nonzero and not a zero divisor. QED

Assume (a) [resp. (b)] is false. Then there is an $n$ and a surjective [resp. injective] endomorphism $b$ of $A^n$ satisfying $b(e_n)=0$ [resp. $b(A^n)\subseteq A^{n-1}$], where $e_n$ is the last vector of the canonical basis and $A^{n-1}$ is the span of all the other vectors of this basis. Then $b$ is integral over $A$ by Cayley-Hamilton, and we get a contradiction by using the lemma (with $B:=A[b]$) and applying $a=bf(b)$ to $e_n$.

Community
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Pierre-Yves Gaillard
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  • @PierreYvesGaillard Is there a reason why $b(e_n)$ is in $A^{n-1}$ and not $A^n$? –  Apr 12 '12 at 13:55
  • Dear @BenjaminLim: We assume that (b) is false, which means (swapping $m$ and $n$) that there is an injective $\phi:A^n\to A^m$ with $n > m$. Putting $b:=i\circ\phi$, where $i$ is the natural injection $A^m\hookrightarrow A^n$, we get $be_j\in A^m\subset A^{n-1}$ for all $j$. – Pierre-Yves Gaillard Apr 12 '12 at 14:34
  • @RobertCardona - Which isomorphism are you referring to? – Pierre-Yves Gaillard Nov 06 '15 at 11:53
  • @RobertCardona - The assumption of Atiyah-MacDonald' exercise is that $A$ is a nonzero commutative ring and $\phi:A^m\to A^n$ is an $A$-linear map. (There is no $R$.) Sorry, I don't understand your answer. – Pierre-Yves Gaillard Nov 06 '15 at 15:42
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I am answering this question just because it took me a while to figure out a good proof, so it may help other people as well.

If $\varphi: R^m\rightarrow R^n $ is your epimorphism then you can construct the exact sequence $$\text{Ker}(\varphi) \xrightarrow{\ \ \iota \ \ }R^m \xrightarrow{\ \ \varphi \ \ }R^n\longrightarrow 0.\quad (*)$$

Let $\mathfrak{m}$ a maximal ideal of $R$. Therefore $R/\mathfrak{m}$ is a field and a $R$-module. Appling right-exactness to $(*)$ (Proposition 2.18, Introduction to Commutative Algebra - Michael Atiyah), we conclude that $$R/\mathfrak{m}\otimes_R\text{Ker}(\varphi) \xrightarrow{\ 1\otimes\iota \ }R/\mathfrak{m}\otimes_R R^m \xrightarrow{\ 1\otimes\varphi \ }R/\mathfrak{m}\otimes_R R^n\longrightarrow 0,$$ is an exact sequence.

Therefore $1\otimes\varphi$ is surjective. Since, $R/\mathfrak{m}\otimes_R R^m$ and $R/\mathfrak{m}\otimes_R R^n$ are vector spaces over $R/\mathfrak{m}$ and $1\otimes\varphi$ is a linear map between $R/\mathfrak{m}\otimes_R R^m$ and $R/\mathfrak{m}\otimes_R R^n$ over $R/\mathfrak{m}$, we have that $m\geq n.$

Matheus Manzatto
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