Structure of finitely generated modules over local Artinian rings Let $(R,m)$ be an Artinian local ring with $m^2=0$. Let $M$ be a finitely generated $R$ module. Can we say anything about the structure of $M$? Perhaps to give a complete structure might be very difficult, but can we say anything "nice"?
 A: One case is really nice, and the other case is not so nice.
When $m$ is principal:
You can show that in fact $R$ has exactly three ideals $\{R,m,\{0\}\}$, and so it's an Artinian uniserial ring (a.k.a. an Artinian local, principal ideal ring or "special principal ring").  It's known that every module (f.g. or not) over an Artinian uniserial ring is a direct sum of cyclic submodules. That's pretty nice, IMHO.
When $m$ isn't principal:
At the very least you have that (for finitely generated $M$) $M$ has finite composition length (and is therefore Artinian and Noetherian,) and, that the socle $\operatorname{Soc}(M)$ is essential in $M$. Thus $M$ has to be a submodule $\operatorname{Soc}(M)\subseteq M\subseteq E(\operatorname{Soc}(M))$ where $E(-)$ denotes the injective hull.
What does the localness of $R$ buy us? Two things: one is that there is exactly one isotype of simple $R$ module, namely $R/m=S$. Secondly, since $R$ is Artinian, $R/m$ embeds in $R$ as an $R$ module, so it is actually a copy of an ideal of $R$. This means that $\operatorname{Soc}(M)\cong \oplus_{i=1}^nS$, and consequently $M\subseteq E(\operatorname{Soc}(M))\cong\oplus_{i=1}^nE(S)$. 
Since $E(S)$ is a indecomposable, we can say that $M$ is an essential submodule of a finite direct sum of copies of an indecomposible injective module. 
A: Let more generally $R$ be a Noetherian commutative ring with a maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=0$. Then every finitely generated $R$-module $M$ fits into a canonical exact sequence of finitely generated $R$-modules
$$0 \to \mathfrak{m} M \to M \to M/\mathfrak{m} M \to 0.$$
Since $\mathfrak{m} M$ and $M/\mathfrak{m} M$ are killed by $\mathfrak{m}$ (here we use $\mathfrak{m}^2=0$), these correspond to $R/\mathfrak{m}$-modules. Thus, there are unique natural numbers $r,s \in \mathbb{N}$ such that $\mathfrak{m} M \cong (R/\mathfrak{m})^s$ and $M/\mathfrak{m} M \cong (R/\mathfrak{m})^r$. Thus, $M$ is determined by these two natural numbers and the extension class
$$[M] \in \mathrm{Ext}^1_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s).$$
So we have to determine $\mathrm{Ext}^1_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s)$. In principle, because $\mathrm{Ext}$ is additive in both variables, we may restrict to $r=s=1$ here, but later I would like to describe the extensions explicitly for general $r,s$. And it's not harder at all to do the general case right away.
The long exact sequence associated to the short exact sequence $0 \to \mathfrak{m}^r \to R^r \to (R/\mathfrak{m})^r \to 0$ begins with:
$$0 \to \hom_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s) \to \hom_R(R^r,(R/\mathfrak{m})^s) \to \hom_R(\mathfrak{m}^r,(R/\mathfrak{m})^s) \to \mathrm{Ext}^1_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s) \to \mathrm{Ext}^1_R(R^r,(R/\mathfrak{m})^s)$$
Since $\mathrm{Ext}^1_R(R^r,(R/\mathfrak{m})^s)=0$ and $\hom_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s) \to \hom_R(R^r,(R/\mathfrak{m})^s)$ is an isomorphism, the sequence simplifies to
$$\mathrm{Ext}^1_R((R/\mathfrak{m})^r,(R/\mathfrak{m})^s) \cong \hom_R(\mathfrak{m}^r,(R/\mathfrak{m})^s).$$
Given a homomorphism $\delta : \mathfrak{m}^r \to (R/\mathfrak{m})^s$, the corresponding extension
$$0 \to (R/\mathfrak{m})^s \to E_\delta \to (R/\mathfrak{m})^r \to 0$$
is constructed as follows (see Weibel's Introduction to homological algebra, proof of Thm 3.4.3): We choose a pushout
$$\begin{array}{c} \mathfrak{m}^r & \xrightarrow{\subseteq} & R^r \\ \delta \downarrow ~~&& \downarrow \\ (R/\mathfrak{m})^s & \rightarrow & E_\delta. \end{array}$$
Explicitly, we have
$$E_\delta = (R^r \oplus (R/\mathfrak{m})^s) / \{(x,-\delta(x)) : x \in \mathfrak{m}^r\}.$$
The homomorphism $E_\delta \to (R/\mathfrak{m})^r$ is induced by $0 : (R/\mathfrak{m})^s \to (R/\mathfrak{m})^r$ and the projection $R^r \twoheadrightarrow (R/\mathfrak{m})^r$.
Therefore, every finitely generated $R$-module is isomorphic to $E_\delta$ for unique natural numbers $r,s$ and a unique homomorphism $\delta : \mathfrak{m}^r \to (R/\mathfrak{m})^s$. This finishes the classification.
