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I'm looking for an example of of two isomorphic abelian groups, which are not isomorphic $R$ modules for some ring $R$.

I suppose we can just the same abelian group $M$ twice, and use a different operation $R\times M \rightarrow M$ so the $R$ modules aren't isomorphic. I can't think of such a group $M$ and ring $R$ to make this possible, though. Any ideas? Thanks.

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2 Answers

up vote 2 down vote accepted

All finite dimensional real vector spaces are isomorphic as abelian groups. You can even add the countably infinite dimensional ones.

Later. $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, so a direct sum of finitely many copies (or countable many) of $\mathbb R$ is isomorphic to $\mathbb R$ as a $\mathbb Q$-vector space, and therefore as an abelian group. This depends on vector spaces having bases (so on the axiom of choice, more or less) and on the fact that if $A$ is an infinite set then there is a bijection between $A\times\{0,1\}$ and $A$, which probably also depends on having choice at hand. Since we do have choice, there is no problem :-)

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The dual numbers over a field come to mind.

Let $k[\epsilon]=k[t]/(t^2)$ be considered as a $k[t]$-module. As a $k$-module (i.e. vector space), $k[\epsilon]\cong k^2.$ This implies that $k[\epsilon]\cong k^2$ as abelian groups. However, we can give $k^2$ the trivial $k[t]$-module structure whereby $t\cdot(a,b)=(0,0)$ for every $a,b\in k.$ This structure is different from the previous, since in $k[\epsilon]$ we have $t\cdot (a,0)=(0,a\epsilon)\neq (0,0)$ for $a\in k^\times.$

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