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In old exams, I've encountered the two following similar exercises:

(1) Classify, up to isomorphism, all the unitary $\mathbb{Z}[i]$-modules with 10 elements.

(2) Classify, up to isomorphism, all the unitary $\mathbb{Z}_2[x]$-modules with 4 elements.

My thoughts on (1):

I use the following fact: "An abelian group $V$ admits an unitary $\mathbb{Z}[i]$-module structure iff there exists a group homomorphism $\varphi:V\to V$ such that $\varphi\circ\varphi=-\mathrm{Id}$". First of all, as an abelian group of order 10 $V$ must be $\mathbb{Z}_{10}$. Then, I look at $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_{10},\mathbb{Z}_{10})\cong \mathbb{Z}_{10}$ and I found that $n^2\equiv-1\,(10)\Rightarrow n\equiv 3,7$ so there are two $\mathbb{Z}[i]$-modules with 10 elements.

PROBLEM: How to distinguish if that two structures are isomorphic or not? Anyway, I'm following the correct approach? I thought of using fundamental theorem of modules over PID but I stuck following that way.

My thoughts on (2):

I know that the unitary $\mathbb{Z}_2[x]$-modules correspond with the vector spaces over $\mathbb{Z}_2$ with a fixed linear transformation $T$. We have to look at two non-isomorphic cases:

$V=\mathbb{Z}_2\oplus\mathbb{Z}_2$, then the representation of $T$ must be a matrix $\begin{pmatrix} a&b\\ c&d \end{pmatrix}$ with $a,b,c,d\in\mathbb{Z}_2$ and we can classify the structure modules by looking at Rational Canonical Form: if $m_T$ denotes the minimal polynomial of $T$ then $\mathrm{deg}\, m_T=1,2$. If it's 1 then $T=\lambda\mathrm{Id}$, i.e. the null matrix or the identity matrix. If it's 2 then $T$ is similar to $\begin{pmatrix} 0& -a_0\\ 1& -a_1\end{pmatrix}$ then there are 3 structure modules in this case. Is it ok??

PROBLEM: If $V=\mathbb{Z}_4$ I don't know how to approach the classification in this case.

I apprecciate any hints/suggestions/comments, sorry for the long question. Thanks!

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    $\begingroup$ $\mathbb{Z}_4$ is not a vector space over $\mathbb{Z}_2$. $\endgroup$ – egreg Jan 25 at 21:07
  • $\begingroup$ I supposed it but I didn't know to prove it! $\endgroup$ – Ale Tolcachier Jan 25 at 21:09
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    $\begingroup$ If $V$ is a vector space over $\mathbb{Z}_2$, then $x+x=0$, for every $x\in V$. $\endgroup$ – egreg Jan 25 at 21:10