# Artinian rings have finite length

In a recent question of mine here I asked whether it is true or not that Artinian (commutative) rings have finite length. I came up with a proof, and I want to know if it is valid.

So, I want to prove that Artinian rings have finite length (considered as a module over itself), which will follow directly from the following proposition.

$\textbf{Proposition}:$ If $R$ is an Artinian ring and $M$ a finite $R$-module, then $M$ has finite length.

$\textbf{Proof}:$ Every Artinian ring is also Noetherian and since $M$ is a finite $R$-module, there exists a chain of submodules $$0=M_0 \subset M_1 \subset \cdots \subset M_n=M$$ such that $M_i/M_{i-1} \cong A/P_i$, where $P_i \in \text{Spec}(A)$ for $i=1,...,n$. Also, $M_i/M_{i-1}$ is a simple module. To see this, suppose that $0 \not= N \subset M_i/M_{i-1}$ is a proper submodule; then $A/P_i$ contains a nonzero proper ideal. However, the ideals of $A/P_i$ are in one-to-one correspondence with ideals of $A$ containing $P_i$, and since $A$ is Artinian every prime ideal is also maximal, so this implies the existence of an ideal lying strictly between $P_i$ and $A$, contradicting the maximality of $P_i$.

Hence, we have found a Jordan-Hölder series for $M$, and since any two such series have the same length, we conclude that $M$ must have finite length. Thus the main result follows: every Artinian ring has finite length when considered as a module over itself.

Is this proof OK? I am a little unsure, I would like comments.

Your proof is missing the main point: when you say that there is a chain $$0=M_0 \subset M_1 \subset \cdots \subset M_n=M$$ with $M_i/M_{i-1}\cong A/P_i$ you have already shown that $M$ has finite length. Indeed, it is easy to show that if $X$ is a submodule of $Y$ and both $X$ and $Y/X$ have finite length, then also $X$ has finite length and go on by induction on $n$.

The point is to show that such a chain exists in the first place.

Five easy to prove facts, that make for a very easy proof that doesn't use maximal or prime ideals.

• Every module which is both artinian and noetherian has finite length.

• Any quotient of an artinian module is artinian

• Any quotient of a noetherian module is noetherian

• Any finite direct sum of artinian modules is artinian

• Any finite direct sum of noetherian modules is noetherian

A finitely generated module over an artinian (hence noetherian) ring $A$ is a quotient of $A^n$, for some integer $n$. Hence it is both artinian and noetherian.

Note that the commutativity of the ring is irrelevant: if a ring $R$ (with $1$) is right artinian (that is artinian as a right module over itself), then it is right noetherian (Hopkins-Levitzki theorem), so every finitely generated right $R$-module has finite length.

• Your answer is also missing a key point: the prime ideals $P_i$ are actually maximal. Jul 16 '15 at 18:04
• @user26857 That's irrelevant: if a ring $A$ is artinian, then also $A/I$ is artinian, for every ideal $I$. And the object of my comments after analyzing your proof is that it's not necessary to go with prime/maximal ideals. Jul 16 '15 at 18:05
• Then maybe you assume that artinian rings have finite length. (My proof???) Jul 16 '15 at 18:07
• @user26857 It's surely true, because artinian rings (with $1$) are noetherian. Jul 16 '15 at 18:07
• Now I'm completely lost in your arguments. Jul 16 '15 at 18:07

Let me explain what's going on here.

The OP shows us an alternative proof of the fact that finitely generated modules over artinian rings have finite length. The proof is based on a well known result about finitely generated modules over noetherian rings, that is, Theorem 6.4 from Matsumura's CRT which says the following:

If $A$ is a Noetherian ring and $M$ a finite $A$-module, then there exists a chain of submodules $$0=M_0 \subset M_1 \subset \cdots \subset M_n=M\qquad(*)$$ such that $M_i/M_{i-1} \cong A/P_i$, where $P_i \in \text{Spec}(A)$ for $i=1,...,n$.

Since $P_i$ are prime ideals in an Artinian ring they are maximal, so $A/P_i$ are fields. Then the chain $(*)$ is a composition series for $M$, and we are done.

To answer his question: the proof is a little longer that necessary, but it's OK.

• This is the kind of answer I was looking for, thanks for the comment above aswell. I will award you the bounty as soon as possible.
– user117449
Jul 16 '15 at 20:41
• Rather than “longer than necessary”, I'd say it's using a sledgehammer. Jul 16 '15 at 20:47