1
$\begingroup$

$\textbf{Question:}$ 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

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Batominovski Aug 22 '15 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.