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

Since $\mathbb{C}[x]$ s noetherian we have that $\mathbb{C}[x]/(x^2)$ is too. And thus finitely generated $\mathbb{C}[\epsilon]$ modules are also finitely presented.

I'm not sure where to go from here. I imagine I need to use the fact that $\epsilon$ is not a unit.

Also, how would this problem be different if it were over $\mathbb{C}[i]$ instead. Obviously, here $i$ is a unit.

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Note that $\mathbb{C}[\epsilon]$ is also artinian. –  M Turgeon Apr 30 '12 at 0:52

2 Answers 2

up vote 6 down vote accepted

A module over the ring $R=\mathbb C[\epsilon]/(\epsilon^2)$ is the same thing as a pair $(M,f)$ with $M$ a complex vector space and $f:M\to M$ a linear map such that $f\circ f=0$. Indeed:

  • to each $R$-module $M$ we can assign the pair $(V,f)$ with $V=M$, the underlaying compex vector space of $M$, and $f:v\in V\mapsto \epsilon v\in V$,

  • and to each pair $(V,f)$ we can asign the $R$-module $M$ which coincides with $V$ as a complex vector space and where multiplication by $\epsilon$ is given by $\epsilon\cdot v=f(v)$ for all $v\in M$.

One can easily check that these two assignemnts are mutually inverse. To classify $R$-modules, then, it is enough to classify pairs $(V,f)$ as above and then translate the result using the second assignment.

The classification of finitely generated $R$-modules is therefore equivalent to the theory of Jordan canonical forms of nilpotent endomorphisms of vector spaces of nilpotentcy index $2$. The non-finitely generated case can in fact be dealt with in exactly the same way, and we this get the result rschwieb mentions that all modules are direct sum of f.g. ones.

The end result is as follows: suppose $M$ is an $R$-module; it is canonically a complex vector space, and we use this structure below. Pick a basis $B$ of the subspace $\epsilon M$, for each $b\in B$ let $b'\in M$ be an element such that $\epsilon b'=b$, let $B'=\{b':b\in B\}$ and let $C$ be a basis of subspace of $V$ complementary to the subspace generated by $B\cup B'$. Then you can check that $B\cup B'\cup C$ is a $\mathbb C$-basis of $M$, that the subspaces $F$ and $Z$ spanned by $B\cup B'$ and $C$, respectively, are an $R$-submodule, and that $M=F\oplus Z$ as $R$-modules.

Moreover, you can check that $Z$ is a direct sum of $|C|$ copies of the $1$-dimensional $R$-module $S=R/(\epsilon)$, and that $F$ is a direct sum of $|B|$ copies of the free $R$-module of rank one $R$.

Wrapping it up, we see that every $R$-module is a direct sum of some number of copies of the $1$-dimensional $R$-module $S$ and some number of copies of the free $R$-module $R$.

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Although very interesting, I'm not really familiar with this machinery. Could you explain more how this makes my problem easier/solves it? –  Steven-Owen Apr 30 '12 at 1:35
    
I've expanded on the reduction. On the other hand, if you are not familiar with the theory of Jordan canonical forms, I emphatically recommend that you do familiarise yourself with it! –  Mariano Suárez-Alvarez Apr 30 '12 at 1:40
    
Thank you so much for your help. Just to follow up and add some more concreteness to my understanding. $M$ is a complex vector space (whose dimension is not necessarily the same as the number of generators for the module $M$). If we hit $M$ with the endomorphism $f$, the image is the subspace $\epsilon M$. $B$ is then the basis for Im$(f)$. What is $B'$ be the basis for? It's not Ker$(f)$? I don't see why $C$ is not the null space. It seems like $B$ is the basis for everything that will died in one hit of $f$ and $B'$ is the basis for everything that will die in two hits. What else is there? –  Steven-Owen Apr 30 '12 at 2:07
    
I should specify, it seems as though $B$ is the basis for everything that will take exactly two hits of $f$ to die and $B'$ is a basis for everything that will take exactly one hit to die. Since Everything is either dead to begin with, or will die in at most two hits of $f$, it seems like $C$ is empty to me. –  Steven-Owen Apr 30 '12 at 2:25
    
Find an example of a module which is $1$-dimensional. What is $C$ then? –  Mariano Suárez-Alvarez Apr 30 '12 at 3:04

Looks like $R=\mathbb{C}[\epsilon]$ is an Artinian uniserial ring (or "special principal ideal ring") with ideal structure $R\supset (x)\supset (0)$.

It's known the finitely generated modules $^\dagger$ of these rings are direct sums of cyclic modules, and in this case there are only two possibilities that cyclic modules are isomorphic to: $R$ and $R/(x)$.

$^\dagger$ All modules actually, if I'm not mistaken.

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