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

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.

Solution 1:

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 $R$.

Solution 2:

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.