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This must be very easy, but I'm somehow unable to see it:

Let $M$ be a module of finite length over a noetherian local ring $A$ with residue field $k$.

If $M$ is nonzero, then there exists a surjective $A-$module homomorphism

$M \rightarrow k$.

How do I get it?

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2 Answers 2

up vote 4 down vote accepted

Let $M$ be a nonzero module over a commutative ring $A$.

If $M$ contains a maximal proper submodule $N$, then $M/N$ is isomorphic to $A/\mathfrak m$, for some maximal ideal $\mathfrak m$ of $A$.

If $M$ is finitely generated, it contains a maximal proper submodule.

If $M$ has finite length, it is finitely generated.

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At one end of a composition series for $M$, there is a step $N \subset M$ where $M/N$ is simple. If $x$ is a non-zero element of $M/N$, then define a map $A \to M/N$ by $a \mapsto ax$. What does the kernel have to be? [I don't see any use for Noetherian here.]

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