3
$\begingroup$

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!

$\endgroup$
1
  • 3
    $\begingroup$ 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 $5=(2+i)(2-i)$... $\endgroup$ Apr 5, 2020 at 23:53

1 Answer 1

4
$\begingroup$

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 $(2-i)=(5,i-2) \neq (5,i-3)=(5,2+i)=(2+i)$.

$\endgroup$
2
  • $\begingroup$ What do you mean by "the question is actually..."? It sounds like you're implying that the poster asked the wrong question $\endgroup$ Apr 6, 2020 at 0:38
  • $\begingroup$ That might have been unclear — I am merely reformulating. $\endgroup$
    – Aphelli
    Apr 6, 2020 at 6:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .