Theorem 7

Let A be a central simple K-algebra of finite degree and L a subalgebra of A. The following properties are equivalent:

(i) The algebra L is a maximal étale subalgebra of A.

(ii) There exist an extension $K'$ of K, an integer $n\geqslant 1$, and an isomorphism $\theta$ from $A_{(K')}$ to $\mathbf{M}_n(K')$ that sends $L_{(K')}$ to the set of diagonal matrices.

(iii) There exist $K',n$, and $\theta$ as in (ii), where the extension $K'$ is, moreover, assumed Galois and of finite degree.

It is clear that (iii) implies (ii).

If property (ii) holds, then $L_{(K')}$ is a maximal commutative subalgebra of $A_{(K')}$ (Proposition 6), and it is diagonalizable. The K-algebra L is then étale (V, §6, No. 3, p. 28, Definition 1), and it is a maximal commutative subalgebra of A (VIII, p. 261, Lemma 3, b)). We have proved that (ii) implies (i).

Suppose that property (i) holds. Since L is étale over K, by Proposition 7, there exists a Galois extension $K_1$ of K of finite degree such that the $K_1$-algebra $L_{(K_1)}$ is diagonalizable. The algebra A is central simple; by (VIII, p. 252, Theorem 1), there exist a Galois extension $K_2$, a vector space V of finite dimension $n$ over $K_2$, and an isomorphism $\theta$ from $A_{(K_2)}$ to End$_{K_2}(V)$. By Proposition 1 of V, p. 55, we may assume $K_1= K_2$. By Proposition 4, b) of VIII, p. 264 and Lemma 3, b) of VIII, p. 261$, L_{(K')}$ is a maximal commutative subalgebra of $A_{(K')}$, so $\theta (L_{(K')})$ is a maximal commutative subalgebra of End$_{K'}(V')$. Let us apply Proposition 6 to the diagonalizable algebra $\theta (L_{(K')}):$ there exists a basis $(e_1, . . . , e_n)$ of $V'$ over $K'$ such that $\theta (L_{(K')})$ consists of the endomorphisms of $V'$ with diagonal matrix with respect to this basis. So (i) implies (iii).