Proof that artinian ring implies noetherian ring I'm trying to understand the classic proof that an artinian ring is noetherian. I saw this post, but I'm having trouble with the last step. I'll copy it below.
Suppose $R$ is Artinian. Let $\mathscr{M}_1, \mathscr{M}_2, \cdots, \mathscr{M}_n$ be the maximal ideals in $R$. Since all prime ideals are maximal we have that $\bigcap_{i=1}^{n} \mathscr{M}_i$ is equal to the nilradical of $R$. Since in an Artinian ring, the nilradical is nilpotent, $\exists k > 0$ such that $(\bigcap_{i=1}^{n} \mathscr{M}_i)^k = 0$. Therefore, we have that
$$ (\mathscr{M}_1\mathscr{M}_2\cdots\mathscr{M}_n)^k = 0. $$
Consider the following descending chain of ideals:
$$ R \supseteq \mathscr{M}_1 \supseteq \mathscr{M}_1^2 \supseteq \cdots \supseteq  \mathscr{M}^k \supseteq  \mathscr{M}_1^k\mathscr{M}_2 \supseteq  \cdots \mathscr{M}_1^k\cdots \mathscr{M}_n^k  = 0.$$
Then we consider the quotients
$$ \mathscr{M}_1^k\mathscr{M}_2^k\cdots\mathscr{M}^{i}_{r} / \mathscr{M}_1^k\mathscr{M}_2^k\cdots\mathscr{M}^{i+1}_r$$
As $R/\mathscr{M_r}$-module (ie. as vector spaces). However, I'm having trouble to see why this implies that easy that $R$ is noetherian.
 A: Noting that each factor $m_1^k\cdots m_{r-1}^km_r^{i}/m_1^k\cdots m_{r-1}^km_r^{i+1}$ is an $R/m_r$ vector space, as $R$ is Artinian, we know that the above vector space is finite dimensional. Hence, the above vector space is also Noetherian $R/m_r$ module. This also shows that the above kind of each factor is Noetherian $R$ module.
Now, we use the lemma: if we have short exact sequence $0\to M'\to M\to M''\to 0$ of $R$ modules, then $M$ is Noetherian module iff $M',M''$ are Noetherian modules. A hint to prove this lemma is to consider pulling back (or pushing out) ascending chain of ideals through short exact sequence.
As we have this short exact sequence $0\to m_1^k\cdots m_n^k\to m_1^k\cdots m_{n-1}^km_n^{k-1}\to m_1^k\cdots m_{n-1}^k m_n^{k-1}/m_1^k\cdots m_n^k\to 0$ of $R$ modules, and we know $m_1^k\cdots m_n^k=0$, using the lemma yields that $m_1^k\cdots m_{n-1}^km_n^{k-1}$ is also Noetherian $R$ module. We can proceed in this manner and finally conclude that each factor $m_1^k\cdots m_r^km_{r+1}^i$ is Noetherian $R$ module. Then we have $0\to m_1\to R\to R/m_1\to 0$ with $m_1$ being Noetherian $R$ module. As $R/m_1$ is also Noetherian, the lemma tells $R$ is also Noetherian.
