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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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    Commenting so that this will be visible at the top: This question is P$r$oblem 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.pdf2011-02-13

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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 $e_n$.

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    @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.2015-11-06