# Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?

If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a 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?

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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.

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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 3 '11 at 2:57
@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 3 '11 at 3:05
@Pete: You could have put that as an answer. – anonymous Feb 3 '11 at 3:09
Google "invariant basis number" (IBN) for more than you wanted to know. – Bill Dubuque Feb 3 '11 at 3:50
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. math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf – David Speyer Feb 13 '11 at 19:28

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 $e_n$.

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@PierreYvesGaillard Is there a reason why $b(e_n)$ is in $A^{n-1}$ and not $A^n$? – user38268 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 6 '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 6 '15 at 15:42