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Prove that if $M$ is an $R$-module of finite length, then $\operatorname{End}_R(M)$ is artinian.

We can derive that $M$ is both artinian and noetherian from that it has finite length, and its generating set is finite. But how do we get the properties of $\operatorname{End}_R(M)$, I have difficulty in understanding what is the inclusion relationship in $\operatorname{End}_R(M)$.

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You have got a good answer to your question. Is that hard to accept it? – user26857 Jan 12 '13 at 17:21
up vote 1 down vote accepted

Let $R$ be a ring, $M$ a f.g (finitely generated) $R$-module and $N$ an artinian $R$-module. Prove that $\operatorname{Hom}_R(M,N)$ is artinian.

Proof. Since $M$ is f.g., we have an exact sequence $R^n\to M\to0$. This implies the sequence $0\to \operatorname{Hom}(M,N)\to \operatorname{Hom}(R^n,N)=N^n$ is exact.

Now, $N$ is artinian $\Rightarrow$ $N^n$ is artinian. Since $\operatorname{Hom}_R(M,N)$ is a submodule of $N^n$, it must be artinian.

By this way, we also have $\operatorname{Hom}_R(M,N)$ has finite length in case $N$ has finite length.

We can also get this by an other way. Let $H=\operatorname{Hom}_R(M,N)$. We know that $\operatorname{Hom}_R(M,N)$ is artinian as $R$-module iff $\operatorname{Hom}_R(M,N)$ is artinian as $R/\operatorname{Ann}_R(H)$-module. Keep mind that $\operatorname{Ann}_R(M)\subset \operatorname{Ann}_R(H)$, we have an natural surjective $R/\operatorname{Ann}_R(M)\to R/\operatorname{Ann}_R(H)\to 0$. But since $M$ has finite length we have $R/\operatorname{Ann}_R(M)$ is artinian. The above exact sequence show that $R/\operatorname{Ann}_R(H)$ is artinian. Since $H$ is f.g. as $R/\operatorname{Ann}_R(H)$-module, it must be artinian.

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