# Two non-isomorphic $\mathbb Z[i]$-modules which are isomorphic as abelian groups.

I was asked the following question:

Give an example of two non-isomorphic $$\mathbb Z[i]$$-modules which are isomorphic as $$\mathbb Z$$-modules (abelian groups).

There are several similar questions on math.stackexchange, but I was not able to adapt answers for my problem.

I need a hint to start. Thanks!

• Try to show that $\mathbb Z[i]/\langle 2+i\rangle$ and $\mathbb Z[i]/\langle 2-i\rangle$ are not isomorphic $\mathbb Z[i]$-modules... though they are both cyclic groups of order $5$. (Hint: the prime $5\in\mathbb Z$ does not remain prime in the Euclidean ring $\mathbb Z[i]$, but factors as $5=(2+i)(2-i)$... Apr 5, 2020 at 23:53

So the question is actually: find an abelian group $$G$$ and two automorphisms $$\phi$$,$$\psi$$ with square $$-id$$ such that no automorphism of $$G$$ maps $$\phi$$ to $$\psi$$.
In other words, we want to find an abelian group $$G$$ and $$\phi,\psi \in Aut(G)$$ that are not in the same conjugacy class, with $$\phi^2=\psi^2=-1$$.
We are done as soon as $$Aut(G)$$ is abelian and we can choose $$\phi \neq \psi$$.
For instance, we can take $$\mathbb{Z}/5\mathbb{Z}$$ as its automorphism group is isomorphic to $$\mathbb{Z}/4\mathbb{Z}$$ so is abelian, and take $$\phi=2id$$, $$\psi=3id$$.
In terms of the original problem: define $$M_2$$ (resp. $$M_3$$) to be the $$\mathbb{Z}[i]$$ module $$\mathbb{F}_5$$ where $$i$$ acts by multiplication by $$2$$ (resp. $$3$$). So $$M_2$$ and $$M_3$$ are isomorphic over $$\mathbb{Z}$$, but they are not isomorphic over $$\mathbb{Z}[i]$$. Indeed, the ideal $$I$$ of elements acting as $$0$$ on $$M_k$$ is $$(5,i-k)$$, and $$(2-i)=(5,i-2) \neq (5,i-3)=(5,2+i)=(2+i)$$.