Theorem 6

Let K be a commutative field, and let A be a central simple A-algebra of finite degree over K. Let L be a separable extension of K of finite degree that is a splitting field for the algebra A. Then [L : K][A] is zero in Br(K).

There exists an extension M of L that is a Galois extension of K of finite degree (V, §10, No. 1, p. 57, Proposition 2). The class [A] of A in the Brauer group of K belongs to the subgroup Br(M$/K$). Let G be the Galois group of M over K, and let $\alpha$ be the image of [A] in $H^2(G,M^*)$ (VIII, p. 321, Corollary). Let H be the Galois group of M over L. Then H is a subgroup of index [L : K] in G (V, §10, No. 7, p. 68, Corollary 5). Since Res$^G_H(\alpha ) = \Phi_{M/L}(A_{(L)})$ (Proposition 14), we have Res$^G_H(\alpha ) = 0$. By Proposition 8 of VIII, p. 303, it follows that $[L : K]\alpha = 0$, and consequently [L : K][A] = 0.

Let K be a commutative field, and let A be a central simple algebra of finite degree over K. Then A is isomorphic to an algebra of the form $\mathbf{M}_n$(D), where D is a field with center K, and [A] = [D] in Br(K). The reduced degree of D depends only on A. We call this reduced degree the index of A. The index of A divides the reduced degree of A. The exponent of A is the order of the class of A in the Brauer group of K.