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This is an exercise from Matsumura 2.4

$A$ is a non-trivial ring. If $A^n\cong A^m$, then $n=m$; prove this by reducing to the case of a field. (Note that there are counter-examples to this for non-commutative rings.)

What is the counter example and reason that I do not expect rank being well defined for the noncommutative case? Does it have anything to do with left ideal and right ideal might not be the same? So one may have one rank through left maximal ideal and another rank through the right maximal ideal?

user26857
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user45765
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  • A theorem by A. L. S. Corner states that, for every $k>0$, there exists a ring $R$ such that $R^m\cong R^n$ (as left $R$-modules) if and only if $m\equiv n\pmod{k}$. The case $k=1$ is not excluded (and it's the easiest to prove). – egreg May 20 '17 at 14:33

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The property that $A^m\cong A^n$ implies $m=n$ is known as invariant basis number (IBN). For a non-trivial commutative ring you can factor out by a maximal ideal. There are non-commutative rings without IBN. The classic example is the ring of linear maps from an infinite dimensional vector space to itself. In this case $A^2\cong A$.

Angina Seng
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