# Can a ring isomorphism change the structure of a module?

Let $M$ be an $R$-module, where $R$ is a ring with unit. Given a ring automorphism $\phi: R \rightarrow R$, we can define a new $R$-module structure on $M$ by $r \cdot x = \phi(r) x$ for all $r \in R$, $x \in M$.

Can we find examples where this module induced by $\phi$ is not isomorphic to $M$? It seems that the two modules are at least in many ways similar, for example they have the same submodules.

I can see that they must be isomorphic if $M$ is free, since then any basis of $M$ is also a basis for the induced module. It is also clearly true for $\mathbb{Z}$-modules, since the only nontrivial ring automorphism of $\mathbb{Z}$ is $x \rightarrow -x$, and the inverse map is an automorphism of abelian groups.

• Note that if $\phi$ is an inner automorphism then you get an isomorphic module, but in general you don't. I can't think of an immediate example for rings but for groups (where modules are representations) an example is that $Spin(2n)$ has two non-isomorphic spin representations but are related by the outer automorphism of $Spin(2n)$. – Eric O. Korman Apr 10 '14 at 18:23
• Ok, I guess you can turn that into an example for a ring if you know about Clifford algebras: take the even Clifford algebra on $\mathbb R^{2n}$. Then the even and odd spinors are two non-isomorphic representations. – Eric O. Korman Apr 10 '14 at 18:26
• @EricO.Korman: Wouldn't an example for groups give an example for rings by using the group algebra? – spin Apr 10 '14 at 18:27
• Yes that should be true. Though the Clifford algebra is a simpler object then the group algebra of $SO(2n)$. – Eric O. Korman Apr 10 '14 at 18:32
• The similarity is that any ring automorphism of $R$ induces an automorphism of the category of $R$-modules, and automorphisms of categories preserve categorical properties (like the structure of the lattice of subobjects, etc.). – Qiaochu Yuan Apr 10 '14 at 19:16

Consider a field $F$ and the ring $R=F\times F$.
Let $M=\{0\}\times F$ with both the ordinary right $R$-module structure $(0,m)(r,s)=(0,ms)$, and let $M'$ be the same set with another $R$-module structure given by $(0,m)(r,s)=(0,mr)$.
This second structure is just given by the involution on $R$ given by $(r,s)\mapsto(s,r)$.
The annihilator in $R$ of $M$ is $F\times \{0\}$ while the annihilator in $R$ of $M'$ is $\{0\}\times F$. Since isomorphic modules share annihilators, this shows $M\ncong M'$.