Semisimple Artinian rings as finite products

Theorem (Artin–Wedderburn, product form).
Let $R$ be a (unital) ring that is both semisimple and Artinian . Then there exist division rings $D_1,\dots,D_t$ and positive integers $n_1,\dots,n_t$ such that

Each factor $M_{n_i}(D_i)$ is a simple Artinian ring, hence (up to isomorphism) a matrix ring over a division ring .

Conversely, any finite product $\prod_i M_{n_i}(D_i)$ is semisimple Artinian.

Commutative specialization.
If $R$ is commutative and semisimple Artinian, then each $D_i$ must be a field and each $n_i=1$, so

Related knowls.

• Artin–Wedderburn theorem
• Simple Artinian rings are matrix rings
• Matrix ring

Examples

1. A single simple factor.
   For a field $k$, the ring $R=M_n(k)$ is semisimple Artinian, and the decomposition is just

   $$M_n(k)\cong M_n(k) \quad (t=1,\; D_1=k,\; n_1=n).$$

2. A product of two simple Artinian rings.
   $R=M_2(\mathbb{R})\times M_3(\mathbb{C})$ is semisimple Artinian with two factors.

3. A commutative example via Chinese remainder.
   If $\mathrm{char}(k)\neq 2$, then

   $$k[x]/(x^2-1)\;\cong\; k[x]/(x-1)\times k[x]/(x+1)\;\cong\; k\times k$$

   by the Chinese remainder theorem . This is semisimple Artinian (a product of fields).