Proposition 3

Let A be a commutative ring. The following properties are equivalent:

(i) The ring A is semisimple.

(ii) The ring A is Artinian and reduced (V, §6, No. 7, p. 34).

(iii) The ring A is isomorphic to the product of a finite family of commutative fields.

Commutative simple rings are commutative fields (VIII, p. 120, Remark 1). So (i) is equivalent to (iii).

It is clear that (iii) implies (ii). Conversely, suppose that the ring A is Artinian and reduced. The intersection of the set of prime ideals of A consists of the nilpotent elements of A (V, §15, No. 1, p. 118, Proposition 2), hence is reduced to 0 because A is reduced. By VIII, p. 2, there then exist distinct prime ideals $\mathfrak{p}_1, . . . ,\mathfrak{p}_r$ of A such that we have $\mathfrak{p}_1\cap \cdots \cap \mathfrak{p}_r= 0$. By the corollary of VIII, p. 8, each of the prime ideals $\mathfrak{p}_i$ of the Artinian ring A is maximal; we therefore have $\mathfrak{p}_i+\mathfrak{p}_j= A$ whenever $i$ and $j$ are distinct. By Proposition 9 of I, §8, No. 11, p. 110, the canonical homomorphism from A to the ring $\prod^r_{i=1}(A/\mathfrak{p}_i)$ is an isomorphism. The ring $A/\mathfrak{p}_i$ is a field for every $i$, and (ii) therefore implies (iii).

A commutative algebra of finite degree over a field is an Artinian commutative ring. Proposition 3 therefore generalizes Proposition 5 of V, §6, No. 7, p. 34.