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Given two free modules $R^n$ and $R^m$ over a commutative ring $R$, it's true that $R^n \cong R^m$ if and only if $n=m$. However every proof I've seen of this fact assumes that we've first proven it when $R$ is a field. Although this is probably true (most of us see linear algebra before commutative algebra), this sort of proof is a bit unsatisfactory to me.

Is there a (nice) proof of $R^n \cong R^m$ if and only if $n=m$ that doesn't assume we know it's true when $R$ is a field?

$\bf{Edit}$ Though the linked question implies a proof that does not use the result for $R$ a field, such an implication doesn't mean that the questions are duplicates of one another.

Exit path
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  • This requires that $R$ is finitely generated, right? Like if I take $R$ to be the free commutative ring generated by a $a_0, a_1, \dots$, then $R \cong R^2$. I think. My algebra is very rusty :P – Alex Meiburg Aug 30 '16 at 23:13
  • @AlexMeiburg I'm pretty sure the ring would have to be non-commutative for something weird like that to happen, otherwise I'm ridiculously lost – Exit path Aug 30 '16 at 23:17
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    @AlexMeiburg $R\cong R^2$ as rings but not as $R$-modules. – stewbasic Aug 30 '16 at 23:33

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