$\textbf{Question:}$ Show the existence of two $\mathbb{Z}[i]$ modules $M$, $N$ such that both $M$ and $N$ have $13$ elements bot $M \not \cong N$ as $\mathbb{Z}[i]$ modules.

$\textbf{My Attempt:}$ Since $13=(2+3i)(2-3i)$ or $13=(3+2i)(3-2i)$, $13$ is not prime in $\mathbb{Z}[i]$. Anyway, the factors are associates of each other so I think I can write, by the Structure Theorem, $$ \mathbb{Z}[i] / (13) \cong \mathbb{Z}[i] / (2 + 3i) + \mathbb{Z}[i] / (3+2i)$$

I know of the existence of the theorem that if $a,b$ are relatively prime then $\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z} / (a^2 + b^2)$. However, in the notes I am self-studying from this theorem is not given.

How can I finish my proof without this theorem?


What is the cardinality of $\mathbb{Z}[i]/(13)$? What does that tell you about the cardinality of $\mathbb{Z}[i]/(2+3i)$ and $\mathbb{Z}[i]/(3+2i)$?

  • $\begingroup$ I'm going to use student logic here that $ \mid \mathbb{Z}_{13}[i] \mid = 13^2$ since that gives me the result and is pretty easy now that you mention it. $\endgroup$ – Yuugi Jul 29 '15 at 0:50

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