Artinian reduced rings are semisimple

Remarks

The Jacobson radical $J$ of a left Artinian ring is nilpotent, so each element of $J$ is nilpotent. Since $R$ is reduced, $J = 0$. A left Artinian ring with zero Jacobson radical is semisimple, and the Artin–Wedderburn theorem gives $R \cong M_{n_1}(D_1) \times \cdots \times M_{n_k}(D_k)$. If some $n_i \ge 2$, the matrix unit $e_{12}$ in that factor is a nonzero element with square $0$, which is impossible; so every $n_i = 1$.

Example: $\Z/6\Z \cong \Z/2\Z \times \Z/3\Z$ is Artinian and reduced, and it is a product of two fields. The Artinian hypothesis is needed: $\Z$ is reduced but not semisimple, since the ideal $2\Z$ has no complement in $\Z$. Reducedness is needed: $\Z/4\Z$ is Artinian, but $2$ is a nonzero nilpotent element, and $\Z/4\Z$ is not semisimple.

Going deeper

For a student

The nilpotent elements are the only obstruction to semisimplicity for Artinian rings, because the radical of an Artinian ring is a nilpotent ideal. So for Artinian rings, "reduced" implies "semiprimitive", which implies "semisimple". In commutative algebra this is the statement that a reduced Artinian ring is a finite product of fields. It is used, for example, to describe reduced zero-dimensional Noetherian rings and finite étale algebras over a field.