# Classifying finitely generated modules over complex quotient polynomial ring

Please help demystify the classification of finite modules over PIDs for me. I want to decompose a finitely generated module over a PID into its elementary divisors or its invariant factors (see Dummit & Foote, Theorems 12.5 and 12.6 if you're not familiar).

Here's an example from Artin:

Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2 = 0$.

How does one start this process in the above simple case? I understand much of the proofs in the text, but can't seem to work out a simple example like this one.

I see the same textbook problem here: Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2=0$

But don't believe they've completed the step of decomposing into its elementary divisors or invariant factors. Can someone help with this particular step?

• $\mathbb{C}[x]/\langle x^2\rangle$ is not a PID. It's not even a domain. – lhf Dec 11 '17 at 14:08

A module over the ring $R=\mathbb{C}[x]/\langle x^2\rangle$ is a complex vector space $V$ with a linear operator $T$ such that $T^2=0$.

If $V$ is finitely generated over $R$, then $V$ is finite dimensional over $\mathbb{C}$. (Why?)

So, the question really is:

Classify finite dimensional complex vector spaces with a linear operator $T$ such that $T^2=0$.

This question is now in a familiar form for which you can use the classification of finitely generated modules over PIDs.

The invariant factor decomposition will be $T \mid T \cdots \mid T^2 \cdots \mid T^2$. The number of factors of each type will determine the module. This is of course the same as a decomposition of the matrix of $T$ into a block for $\ker T$ plus some $2 \times 2$ Jordan blocks.

• Hey there @lhf. Thanks so much for the start. Now, I'm wondering how to decompose into elementary divisors or invariant factors. How does one proceed? – anonymous Dec 11 '17 at 19:34
• @anonymous, it all depends on whether the minimal polynomial of $T$ is $X$ or $X^2$. – lhf Dec 11 '17 at 22:44
• Let's assume it is $X^2$ (or $X$ if you prefer). Now what? – anonymous Dec 12 '17 at 3:31