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As stated in the title, I am trying to prove that (I am assuming $R$ is a commutative ring with unity);

Let $M$ be an $R$-module. Then $M$ is artinian and noetherian if and only if $\ell \left( M\right)<\infty$.

Assume that $M$ has finite length $\ell \left( M\right) = n$. Take any (possibly infinite) chain of sub-modules of $M$, denote it by $\mathcal{N}$. Assume for contradiction that $\mathcal{N}$ has $m > n$ proper non-trivial elements in the chain, by Jordan-Holder theorem, this series of sub-modules can be refined to a composition series which has to be of length at least $m$. This is a contradiction because all composition series have the same length $\ell \left( M\right)$. $\mathcal{N}$ was an arbitrary chain of modules (in particular, increasing or decreasing chains), hence every chain of sub-modules stabilizes, which proves $M$ is artinian and noetherian.

I am stuck proving the other implication and would appreciate any help in doing so (I would also love to get improvement suggestions for my proof attempt :). Thank you!

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I will use the following definition:

A $R$-module $M$ is Noetherian iff every nonempty set of submodules contains a maximal element with respect to inclusion. (It is equivalent to saying it satisfies the ascending chain condition)

$(\Longrightarrow)$ Consider the set of all proper submodules of $M$. Since $M$ is Noetherian, there exists a maximal element, call it $M_1$, so $M/M_1$ is simple. Then, consider the set of all proper submodules of $M_1$, we can find a submodule $M_2$ such that $M_1/M_2$ is a simple module. Keep doing this and suppose we have found submodules $$M=M_0\supseteq M_1\supseteq M_2\supseteq \cdots \supseteq M_k$$ with each $M_i/M_{i+1}$ a simple module for $0\leq i\leq k-1$, then if $M_k\neq \{0\}$, there exists a submodule $M_{k+1}$ of $M_k$ such that $M_k/M_{k+1}$ is simple. Either $M$ has a composition series of finite length or we get a descending chain that never terminates. Since $M$ is Artinian, the latter case is excluded, i.e., $M$ has a composition series of finite length. $\square$

Your idea of the proof of $(\Longleftarrow)$ is fine. Actually, since $\ell(M)<\infty,$ we can conclude directly that $M$ is Artinian by definition.

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  • $\begingroup$ would you mind explaining why "we can find a submodule $M_{2}$ so $ M_{1}/M_{2}$ is simple"? I guess that to some extent you are using the claim that a sub-module of a noetherian and artinian module is such, right? $\endgroup$ Commented Nov 19 at 11:12
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    $\begingroup$ @Shavit Yes. For example, if $M$ is noetherian, then $M_1$ is also noetherian since for any ascending chain in $M_1$, it is also an ascending chain in $M$ and thus stable. $\endgroup$ Commented Nov 19 at 11:14

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