# Isomorphism classes of $\mathbb{Z}[i]$ modules.

$\textbf{Question:}$ How many isomorphism classes of $\mathbb{Z}[i]$-modules with exactly $5$ elements are there?

$\textbf{My Attempt:}$ Since $\mathbb{Z}[i]$ is a P.I.D and any module with $5$ elements is finitely generated we can use the structure theorem. In the case of a finite abelian group, the isomorphism classes determined by the prime factorization of the order and then listing all invariant factors.

In this case we have that $5 = (2-i)(2+i)$, but I don't know how to use the ideal $(2-i)$ and $(2+i)$ in the structure theorem...

Any help working this problem or showing an example of a similar problem is appreciate.d

As you suggested, both of those are primes/irreducible in the PID $R=\mathbb{Z}[i]$, so $M \cong R/(a_{1}+ib_{1}) \oplus R/(a_{2}+ib_{2}) \oplus R/(a_{3}+ib_{3}) \cdots$ where $a_{1}+ib_{1} \vert a_{2}+ib_{2} \vert a_{3}+ib_{3} \cdots$. Now, the order of such module, is $(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})\cdots$. Now, since it is equal to $5$, you can conclude that $a_{2}+ib_{2}$ and everything coming after are units, and so one must have $a_{1}^{2}+b_{1}^{2}=5$. This being said, so $(a_{1},b_{1})=(1,2)$ or $(-1,2)$ or $(1,-2)$ or $(-1,-2)$, or $(2,1)$, or $(2,-1)$, and so on. Notice that many solutions will get you the same ideal! I think you will only have two ideals, the ones you mentionned! And so there are only two types of $\mathbb{Z}[i]$ modules of order $5$, $\mathbb{Z}[i]/(2-i)$ and $\mathbb{Z}[i]/(2+i)$. Those are not isomorphic as $\mathbb{Z}[i]$-modules, as they are cyclic and have different annihilators!
• Your solution doesn't cover the trivial module $M=\mathbb{Z}/5\mathbb{Z}$ where $x\cdot v=0_M$ for all $x\in\mathbb{Z}[\text{i}]$ and $v\in M$. However, this is only one extra module that is forgotten. – Batominovski Aug 22 '15 at 8:52