Simple Artinian Rings are Matrix Rings

Statement

A ring $R$ is simple Artinian if it is simple (no nontrivial two-sided ideals) and Artinian (descending chain condition on ideals).

Theorem (Wedderburn–Artin, simple case).
If $R$ is simple Artinian, then there exist an integer $n\ge 1$ and a division ring $D$ such that

the ring of $n\times n$ matrices over $D$ (see matrix ring ).

Moreover, $D$ can be taken to be (the opposite of) the endomorphism ring of a simple left $R$-module:

for a simple $R$-module $S$.

This is the “one simple component” case of the full Artin–Wedderburn theorem ; compare also semisimple Artinian = product of matrix rings .

Examples

1. Full matrix algebras over a field.
   For a field $k$, the ring $M_n(k)$ is simple Artinian.
   Here $D=k$, so the theorem recovers $R\cong M_n(k)$ itself.

2. Division rings.
   Any division ring $D$ is simple Artinian (it has no nontrivial ideals, and it is Artinian).
   This is the special case $n=1$: $D \cong M_1(D)$.

3. Endomorphism rings of finite-dimensional vector spaces.
   If $V$ is an $n$-dimensional vector space over a field $k$, then

   $$\mathrm{End}_k(V) \cong M_n(k).$$

   Since $M_n(k)$ is simple Artinian, so is $\mathrm{End}_k(V)$.

Remark (commutative case): If $R$ is commutative and simple Artinian, then the theorem forces $n=1$ and $D$ commutative, so $R$ is a field.