# Finite length modules over local rings

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 $A$?

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Dear Cyril: +1! Nice question! If $A$ is any ring, and if $M$ is an $A$-module which is artinian and noetherian, then $M$ has finite length. In fact, any modular lattice which is artinian and noetherian has finite length. –  Pierre-Yves Gaillard Jan 3 '12 at 14:14
Given $A$ as above, an $A$-module $M$ is Artian and Noetherian iff it is finitely generated and killed by some power of the maximal ideal of $A$. –  user18119 Jan 3 '12 at 16:01

Your question about simple modules over $A$ can be answered affirmatively. One can see this as follows: take any non-zero $m \in M$ and consider the map of $A$-modules $f:A \to M$ given by $a \mapsto a \cdot m$. This map is non-zero as $1 \cdot m = m \neq 0$ and therefore it must be surjective as $M$ is simple. Therefore we have $M \cong A/\text{ker}(f)$. Next, note that $A/\text{ker}(f)$ can only be simple if $\text{ker}(f) = \mathfrak{m}$: indeed, if $\text{ker}(f) \subsetneq \mathfrak{m}$, then $\mathfrak{m}/\text{ker}(f)$ is a proper submodule of $A/\text{ker}(f)$. We conclude that $M \cong A/\mathfrak{m}$.

For your general question, I don't know if one can say anything more than that $M$ has finite length.

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Dear Sebastian: +1! Nice answer. –  Pierre-Yves Gaillard Jan 3 '12 at 13:50

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 $k\oplus k \oplus k \quad (k=\mathbb Z/p)$

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Dear Georges: +1 for an answer of the usual quality! –  Pierre-Yves Gaillard Jan 3 '12 at 13:56
Dear @Pierre-Yves: thank you, but I'm sure you could have given the same answer, or probably a better one! –  Georges Elencwajg Jan 3 '12 at 13:59