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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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    @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 $R$.2011-02-03
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    @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.2011-02-03
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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 $R / \mathfrak{m}$.2011-02-03
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    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 $s\equiv t \pmod{n}$.2011-02-03
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    Thank you. Sorry for the shorthand post.2011-02-03
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    Google **"invariant basis number" (IBN)** for more than you wanted to know.2011-02-03
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    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.2011-02-03
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    @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.2011-02-03
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    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.2011-02-03
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    @Pete: Ok, Pete2011-02-03
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    @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.2011-02-03
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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.pdf2011-02-13

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