Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 :-)

share|cite|improve this answer

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 $a\in k^\times.$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.