Theorem 2

Let K be a commutative field and A a K-algebra. The following properties are equivalent:

(i) The K-algebra A is absolutely semisimple.

(ii) The K-algebra $B = A\otimes_KA^o$ is semisimple.

(iii) The B-module A is projective.

(iv) There exists an element $e$ of the $(A,A)$-bimodule B satisfying $\varepsilon (e) = 1$ and $ae=ea$ for every $a\in A$.

Suppose that the algebra A is absolutely semisimple, hence semisimple. Then the algebra $A^o$ is semisimple (VIII, p. 137, Proposition 2), and it follows from Corollary 1 of VIII, p. 234 that $B = A\otimes_KA^o$ is a semisimple K-algebra. Therefore, (i) implies (ii).

Since every module over a semisimple ring is projective (VIII, p. 138, Proposition 4), (ii) implies (iii).

The equivalence of (iii) and (iv) follows from Proposition 5. To complete the proof, let us show that (iv) implies (i). Let $e=\sum^r_{i=1}a_i\otimes a'_i$ be an element of B satisfying the conditions in item (ii) of Proposition 5. Let L be an extension of the field K; we must prove that the ring $A_{(L)}$ is semisimple or, equivalently, that every $A_{(L)}$-module is semisimple (VIII, p. 138, Proposition 4). Let M be an $A_{(L)}$-module, and let N be a submodule of M; we view M as a left $(A$, L)-bimodule and N as a sub-bimodule (III, §4, No. 3, p. 466). Since L is a field, there exists an L-linear projector $u$ in M with image N. Since the homotheties $a_M$ associated with the elements $a$ of A are L-linear, there exists a unique group homomorphism from $A\otimes_KA^o$ to End$_L(M)$ that sends an element $a\otimes a'$ to the L-linear mapping $x\mapsto au(a'x)$. We denote the image of $e$ by this homomorphism by $v$; it follows from Remark 2 that $v$ is an $A_{(L)}$-linear projector with image N. The kernel of $v$ is an $A_{(L)}$-submodule of M, supplementary to N. By Corollary 2 of VIII, p. 56, the $A_{(L)}$-module M is semisimple.