Hopkins–Levitzki theorem

Remarks

Over any ring, condition 3 holds exactly when $M$ is both Noetherian and Artinian, so the point is that each chain condition alone gives the other. Choose $n$ with $J^n = 0$ and look at the filtration $M \supseteq MJ \supseteq MJ^2 \supseteq \cdots \supseteq MJ^n = 0$. Each factor $MJ^{i}/MJ^{i+1}$ is killed by $J$, so it is a module over the semisimple ring $R/J$, hence a semisimple module. A semisimple module satisfies one chain condition only when it is a finite direct sum of simple modules, and then it has finite length. So each factor has finite length, and so does $M$.

Main corollary: a right Artinian ring is semiprimary, because its Jacobson radical is nilpotent and $R/J$ is Artinian and semiprimitive, hence semisimple. So every right Artinian ring is right Noetherian, and the same holds on the left.

The hypothesis on $R$ cannot be dropped. The ring $\Z$ is not semiprimary, and the Prüfer group $\Z[1/p]/\Z$ is an Artinian $\Z$-module that is not Noetherian. Also $\Z$ itself is a Noetherian $\Z$-module that is not Artinian.

For commutative rings, the Akizuki–Hopkins–Levitzki theorem gives the sharper description of Artinian rings as the Noetherian rings in which every prime ideal is maximal.

Going deeper

For a student

The proof peels the module into finitely many layers using the powers of the radical. Each layer is a module over a semisimple ring, where both chain conditions just say "finitely many simple summands". This argument is the reason why right Artinian rings are right Noetherian, a fact used all through the structure theory of Artinian rings, together with the Artin–Wedderburn theorem. The converse fails: $\Z$ is Noetherian but not Artinian, so Artinian rings are a strictly smaller class.