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It is possible to have a ring $A$, with identity, such that $A^m\cong A^n$ as left $A$-modules and with $m\neq n$?

I have that for commutative rings this is not possible, but I have no idea of the proof.

user26857
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zacarias
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    These rings for which this not possible are known as IBN rings (Invariant Basis Number rings) and include (subrings of) skewfields. See for instance the answer to this question. There an example of a ring that does not have this property. – Bernard Sep 27 '15 at 12:38

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Tensor each side with $A/m$, where $m \in Spec(A).$ By noting that $A/m \otimes A^n \simeq A^n/mA^n $ is a $n$-dimensional vector space over $A/m$, compare the dimensions on both sides.

Hmm.
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