Theorem 1 (Wedderburn)

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

(i) The A-module $A_s$ is semisimple.

(ii) For every left ideal $\mathfrak{a}$ of A, there exists a left ideal $\mathfrak{b}$ of A such that $A_s$ is the direct sum of $\mathfrak{a}$ and $\mathfrak{b}$.

(iii) The ring A is left Artinian, and the $(A,A)$-bimodule $_sA_d$ is semisimple.

(iv) The ring A is isomorphic to the product of a finite family of simple rings.

(v) There exist an integer $s\geqslant$ 0, fields $D_1, . . . ,D_s$, and integers $r_1\geqslant 1, . . .,r_s\geqslant 1$ such that the ring A is isomorphic to the product of the matrix rings $\mathbf{M}_{r_i}(D_i)$.

(vi) The ring A is left Artinian, and there exists a faithful and semisimple left A-module.

The equivalence of (i) and (ii) follows from Corollary 2 of VIII, p. 56, and that of (iv) and (v) follows from Theorem 1 of VIII, p. 120.

The A-module $A_s$ is finitely generated. If it is semisimple, then it is left Artinian (VIII, p. 71, Proposition 10). Since the A-module $A_s$ is faithful, this proves that (i) implies (vi). Conversely, suppose that property (vi) holds; let M be a faithful semisimple A-module. There exists an integer $m\geqslant 1$ such that $A_s$ is isomorphic to a submodule of $M^m$ (VIII, p. 50, Proposition 5, a)); since M is semisimple, the same holds for $A_s$. We have therefore proved the equivalence of (i) and (vi).

Let us prove that (i) implies (iii). Suppose that the ring A has property (i); we have already observed that A is then left Artinian. Now, the endomorphisms of the left A-module $A_s$ are the right multiplications by the elements of A. That the $(A$, A)-bimodule $_sA_d$ is semisimple then follows from Proposition 6 of VIII, p. 86.

Let us show that (iii) implies (iv). Suppose that the $(A$, A)-bimodule $_sA_d$ is semisimple. It is finitely generated, so there exists a finite family $(\mathfrak{a}_i)_{i\in I}$ of simple $(A$, A)-sub-bimodules with direct sum $_sA_d$. In other words, the $\mathfrak{a}_i$ are nonzero two-sided ideals of A, the additive group of A is the direct sum of the $\mathfrak{a}_i$, and for every $i\in I$, every two-sided ideal of A contained in $\mathfrak{a}_i$ is equal to 0 or $\mathfrak{a}_i$. Set $\mathfrak{b}_i=\sum_{j\not=i}\mathfrak{a}_j$ for every $i\in I$; this is a two-sided ideal of A. The mapping $a\mapsto (a+\mathfrak{b}_i)_{i\in I}$ is an isomorphism from the ring A to the product of the rings $A/\mathfrak{b}_i$. The $(A$, A)-bimodules $\mathfrak{a}_i$ and $A/\mathfrak{b}_i$ are isomorphic, so every two-sided ideal of $A/\mathfrak{b}_i$ is equal to 0 or $A/\mathfrak{b}_i$. If the ring A is left Artinian, then the same holds for the rings $A/\mathfrak{b}_i$, which are therefore simple (VIII, p. 120, Definition 1).

Finally, let us prove that (iv) implies (i). Suppose that A is the product of a finite family $(A_i)_{i\in I}$ of simple rings. Denote by $\pi_i$ the projection with index $i$ from A to $A_i$ and by $M_i$ the A-module with underlying additive group $A_i$ and law of action $(a, x)\mapsto \pi_i(a)x$. Since the ring $A_i$ is simple, the $A_i$-module $(A_i)_s$ is semisimple, so the A-module $M_i$ is semisimple. Since the A-module $A_s$ is nothing but the product $\prod_{i\in I}M_i$, it is semisimple.