[Math] Structure of finitely generated modules over local Artinian rings

Question

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"?

Answer 1

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.