Can a ring isomorphism change the structure of a module?

Question

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.

Answer 1

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 $M\ncong M'$.

Answer 2

I got a bit carried away writing answer to a similar question here, but the example of the group algebra of a finite abelian group is, I think, a useful one:

If $G$ a finite abelian group, then $\mathbb C[G]$, the complex group algebra of $G$, is a commutative semisimple $\mathbb C$-algebra, hence $\mathbb C[G] \cong \mathbb C^{|G|}$ as a $\mathbb C$-algebra, and the corresponding homomorphisms $\varphi_i\colon \mathbb C[G]\to \mathbb C$, for $1\leq i \leq |G|$, give the simple modules, which restricted to $G\subseteq \mathbb C[G]$ give homomorphisms $\phi_i\colon G \to \mathbb G_m$.

Multiplication gives $\hat{G} = \{\phi_i: 1\leq i \leq |G|\}$ a group structure, which is (non-canonically) isomorphic to $G$ -- any finite abelian group is isomorphic to a product of cyclic groups, and hence it suffices to verify the cyclic case, where if $G = \mathbb Z/n\mathbb Z$ then $\hat{G}= \{\phi_r \colon \mathbb Z/n\mathbb Z \to \mathbb C^\times\mid 0\leq r \leq n-1\}$, where $\phi_r(k) = \exp(2 \pi i rk/n)$, and $\phi_r = \phi_1^r$.

Now if $\phi\colon G\to G$ is an automorphism of $G$, then the induced action of $\phi^*$ on $\hat{G}$ is also a group homomorphism, and $\phi \mapsto \phi^*$ is isomorphism $\mathrm{Aut}(G)^{\mathrm{op}}\to \mathrm{Aut}(\hat{G})$. It follows that there are $\hat{G}^{\phi^*}$ simple modules which are isomorphic to their twists by $\phi^*$.

More concretely, if $G=\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z = \mathbb F_2^2$, then $\mathrm{Aut}(G) = \mathrm{GL}_2(\mathbb F_2) \cong S_3$. Then if, for example, $\phi = \left(\begin{array}{cc}0 & 1\\ 1 & 1\end{array}\right)$, it cyclically permutes the $3$ non-trivial elements of $G$ and thus the induced map on $\hat{G}$ correspondingly permutes the 3 non-trivial irreducible representations.