Jordan-Hölder factors of a finite length module

Question

Suppose one has a local ring $(A,\mathfrak{m})$ and a finite length $A$-module $M$ with $\operatorname{supp}(M) = \{\mathfrak{m}\}$. Does $M$ have a composition series consisting only of $A/\mathfrak{m}$'s?

Answer 1

Yes, this is true:

a) Since $M$ has finite length, it has by definition a composition series.
This means that there exists a non-refinable filtration $M_0=M \supsetneq M_1 \supsetneq...\supsetneq M_n=0$ , in other words a filtration where each $M_i/M_{i+1}$ is a simple $A$-module.

b) Now, a simple $A$-module $S$ is isomorphic to $A/{\mathfrak m}$ with ${\mathfrak m}\subset A$ maximal.
Indeed choose a non-zero element $s\in S$ (the zero module is not regarded as simple!) and consider the morphism $f:A\to S: a\mapsto as$. Its image is non-zero [because $f(1)=s\neq0$] hence it is $S$ since, by simplicity, $S$ is the only non-zero submodule of $S$ . So $f$ is surjective and $S\simeq A/Ker(f)$. But $A/Ker(f)$ can only be simple if $Ker(f)$ is a maximal ideal, so that $Ker(f)$ is a maximal ideal $\mathfrak m$ and, as announced, $S\simeq A/\mathfrak m$

c) Since your ring is local, $\mathfrak m$ is its only maximal ideal, and your question has an affirmative answer: $M$ has a composition series with quotients isomorphic to $A/\mathfrak m$. As Pierre-Yves very judiciously remarks, you don't need the hypothesis that the support of $M$ is $\lbrace\mathfrak m \rbrace$.

Acknowledgment
My former proof was much more complicated and needed $A$ noetherian. Any merit that the above proof might have should go to QiL and Pierre-Yves Gaillard (see their comments below).