Semisimple Artinian rings decompose as finite products

Let $R$ be a semisimple Artinian ring, not necessarily commutative. Then there exist simple Artinian rings $R_1,\dots,R_t$ such that

R \cong R_1 \times \cdots \times R_t .

Moreover, each $R_i$ is a matrix ring over a division ring: $R_i \cong M_{n_i}(D_i)$ . This refinement is exactly the content of the Artin–Wedderburn theorem, and the simple-factor description is packaged in simple Artinian matrix rings.

If $R$ is also commutative, then every simple Artinian factor must be a field (since a commutative division ring is a field, and commutativity forces $n_i=1$ ). Hence in the commutative case,

R \cong K_1 \times \cdots \times K_t

for fields $K_i$ .

Interpretation

The theorem expresses $R$ as a finite product of rings with no nontrivial two-sided ideals. In the commutative case, those simple factors are exactly fields.

Examples

Squarefree integers (commutative case). By the Chinese remainder theorem, $\mathbb Z/6\mathbb Z \cong \mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z$ . Both factors are fields, so $\mathbb Z/6\mathbb Z$ is a commutative semisimple Artinian ring.

Finite products of fields. For any field $k$ , the ring $k \times k \times k$ is semisimple Artinian (each factor is simple Artinian), and it is already presented in the product form of the theorem.

An Artinian ring that is not semisimple. The ring $\mathbb Z/4\mathbb Z$ is Artinian, but it is not semisimple: the class of $2$ is nonzero and nilpotent since $2^2=0$ mod $4$ . Equivalently, its Jacobson radical is nonzero, so it cannot be semisimple.