Right Artinian ring

A ring is right Artinian if its right ideals satisfy the descending chain condition, equivalently its right regular module is an Artinian module. Left and right chain conditions should be distinguished for a general ring.

Let $K/k$ be an infinite-dimensional field extension. For the triangular ring in the display, the right regular module has length three: $e_{1}T$ has its off-diagonal simple submodule and simple quotient, and $e_{2}T$ is simple. Thus $T$ is right Artinian. Its off-diagonal ideal is, on the left, the infinite-dimensional $k$-vector space $K$. Every $k$-subspace determines a left ideal, so an ascending sequence of finite-dimensional subspaces violates left Noetherianity. For example take $K=k(z)$ and the spans of $1,z,\dots ,z^m$.

The finite intersections of maximal right ideals of a right Artinian ring have a minimal member by the descending chain condition. Intersecting it with any other maximal right ideal changes nothing, so it equals the Jacobson radical. If that radical is zero, the diagonal map from the regular module into a finite direct sum of its simple quotients is injective. A submodule of a finite direct sum of simple modules is semisimple and has finite length. This proves semisimplicity of the regular module without assuming Noetherianity first.

A right Artinian ring is a right Noetherian ring. Its right regular module therefore has finite composition length. Its Jacobson radical is a nilpotent ideal, and its quotient by that Jacobson radical is a semisimple ring. In particular every finitely generated right module over such a ring has finite composition length.

In a right Artinian ring, the powers of its Jacobson radical stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne 0$, choose a minimal right ideal $K$ with $KI\ne 0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne 0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The unit criterion for the Jacobson radical makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of Nakayama lemma to an arbitrary Artinian module.