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Let $R$ be a commutative ring. We know that if $R^n\simeq R^m$ as $R$-modules for some positive integers $n,m$ then $n=m$. But is it still true when they are isomorphic as rings? Thanks!

Ergin Süer
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No. Just take a ring $A$ and $R=\prod_{i\in I}A$, where $I$ is a countable set of index.

Joe
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  • Do you want me to show that $R^n\simeq R^m$? There is nothing to prove... in this case $R^n$ is exactly equal to $R^m$. – Joe May 01 '15 at 21:00
  • Suppose $I=\Bbb N$, then $R^2=\prod_{i\in\Bbb N}A\times\prod_{i\in\Bbb N}A=\prod_{i\in\Bbb N}A=R$ because of some basic fact of how the cardinality of countable works – Joe May 01 '15 at 21:05
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    Two countable sets are in bijection, but not necessarily equal. – user26857 May 01 '15 at 21:06
  • @user26857: They are not equal but the sets are used as indexing sets. So it seems to me that the products of A indexed by these two sets will be equal. – Manos May 01 '15 at 21:07
  • Given an element in $\prod_{i\in\Bbb N}A\times\prod_{i\in\Bbb N}A$, you can find this very element in $\prod_{i\in\Bbb N}A$ and viceversa. So they are more than in bij, they are equal. What's wrong? – Joe May 01 '15 at 21:09
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    @Manos Let's start from scratch: if $I\sim J$, then $A^I=A^J$, where $A^I$ is the set of functions $f:I\to A$. How so? – user26857 May 01 '15 at 21:15
  • @Manos My point is that we have to define a ring isomorphism between $R$ and $R\times R$, instead of decreeing they are equal. – user26857 May 01 '15 at 21:17
  • @user26857: I see your point now. – Manos May 01 '15 at 21:20