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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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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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    $\begingroup$ I don't understand. Do you mean that two finite dimensional real vector spaces of different dimensions are isomorphic as abelian groups? $\endgroup$ Commented Jul 23, 2017 at 3:24
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    $\begingroup$ Yes. That is precisely what I wrote. $\endgroup$ Commented Jul 23, 2017 at 5:00
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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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Here's another answer, motivated by a more recent question that asked for a more "concrete" example. Let $R$ be the polynomial ring $\mathbb Z[x]$. One $R$-module is just $R$ itself, and the other is the ideal $(2, x)$ in $R$. (This ideal consists of all integral polynomials with even constant coefficient.) These two $R$-modules are not isomorphic, as the first can be generated by a single element, namely 1, but the second cannot. However, an isomorphism from $R$ to $(2, x)$ as abelian groups is given by doubling the constant coefficient and leaving everything else the same.

One can construct similar examples using various other rings that have non-principal ideals, such as $k[x, y]$, $\mathbb Z[\sqrt{-5}]$, etc.

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  • $\begingroup$ This is a great example! Very simple and intuitive (at least to me). For the sake of generalizing it a little bit: You constructed a principal ring $R$ which is not PID, and then you found a non-principal ideal $I\subset R$ which has underlying additive group isomorphic to that of parent ring $R.$ Using this strategy -- for instance -- it should make sense that $(a,x)\subset\mathbb{Z}[x]$ would provide a sufficient counter-example to the original claim given any $a\in\mathbb{Z}^{>1}.$ Am I correct in saying this? $\endgroup$
    – JAG131
    Commented 14 hours ago

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