# Describe all $R$-modules of low order (up to isomorphism) [duplicate]

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!

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