TITLE: Finite length modules over local rings
QUESTION [6 upvotes]: Let $A$ be a noetherian local ring and $M$ be an artinian and noetherian module over $A$.
Does one know a priori anything about the structure of $M$?
Furthermore: if one knows that the length of $M$ as $A$-module is $1$, i.e. $M$ is simple over $A$, can one conclude that $M$ is isomorphic to $A/\mathcal m$, where $\mathcal m$ is the maximal ideal of $A$?

REPLY [6 votes]: If the  module $M$ is artinian and noetherian, it has a composition series  $M=M_0\supsetneq ...\supsetneq M_n $ with $M_i/M_{i+1} $ a simple module , automatically isomorphic to $k=A/m$ as proved by Sebastian. 
This gives a structure theorem with the important caveat that these quotients do not determine the module $M$.   
For example, if your local ring is $A=\mathbb Z/p^3\; $ ($p\in \mathbb N$   prime) , the three $A$-modules $\mathbb Z/p^3,\; \mathbb Z/p^2\oplus   \mathbb Z/p,  \; \mathbb Z/p \oplus \mathbb Z/p\oplus \mathbb Z/p \;\;$  are non isomorphic but have the same associated graded module $k\oplus k \oplus k \quad (k=\mathbb Z/p)$