I'd like to find an example of a nonzero ring $R$ and positive integers $m,n$ with $m\neq n$ such that $R^m\cong R^n$ as $R$-modules.
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1Any conditions for the ring? Can it be the zero ring? – B.Swan Jul 15 '19 at 19:12
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@B.Swan nope, I reedit the question – yzll Jul 15 '19 at 19:20
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See example 1.4 here https://books.google.iq/books?id=6iLUBwAAQBAJ&lpg=PR1&dq=lectures%20on%20rings%20and%20modules&pg=PA4#v=onepage&q&f=false – Math137 Aug 02 '19 at 19:39
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For commutative rings with one this does not happen. For a non-commutative example take $R = \text{End}_K(K[X])$ for a field $K$. Then $R$ is a free $R$-module with basis $\{\text{id}_{K[X]}\}$. Now consider $$f_1(X^n) := \begin{cases} X^{n/2} &\mbox{if $n$ is even}, \\ 0 &\mbox{otherwise}\end{cases}$$ and $$f_2(X^n) := \begin{cases} 0 &\mbox{if $n$ is even}, \\ X^{(n-1)/2} &\mbox{otherwise}\end{cases}.$$ Then $\{f_1, f_2\}$ is also a basis of $R$, i.e. $R \simeq R^2$. Inductively we see $R \simeq R^n$.
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Why can't it happen in commutative rings with $1$? Let $V={f: \Bbb R \to \Bbb Q }$ with addition and multiplication defined pointwise. Isn't it the case that $V$ is a $\Bbb Q$-module and that $V \cong V^2$ as $\Bbb Q$-modules? – Robert Shore Jul 16 '19 at 22:42
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There is no integer $n$ with $V = \mathbb{Q}^n$. And the correct formulation would actually be: Over commutative rings with unity, the cardinality of a basis of a free module is well-defined. – Jul 17 '19 at 06:40