Bourbaki

Proposition 5

Tag 00I3, permanent. Cite it as [Bourbaki, Tag 00I3].

Algebra, VIII, p. 281, § 15. Brauer Groups

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

(i) The extension L is a splitting field for A.

(ii) There exists a central simple K-algebra of finite degree similar to A containing a maximal commutative subalgebra isomorphic to L.

Let us prove that (ii) implies (i); it suffices to consider the case when L is a maximal commutative subalgebra of A. Let ψ:A⊗KAo→\psi : A\otimes_KA^o\rightarrow EndK(A)_K(A) be the canonical isomorphism that sends a⊗a′a\otimes a' to the K-linear mapping x↦axa′x\mapsto axa' (VIII, p. 252, Theorem 1). We view A as a right vector space over L; then ψ\psi sends A⊗KLA\otimes_KL into the subspace EndL(A)_L(A) of EndK(A)_K(A) and induces, by restriction, an injective homomorphism of L-algebras ψ′:A⊗KL→\psi ': A\otimes_KL\rightarrow EndL(A)_L(A). Set n=[L:K]n= [L : K]. By VIII, p. 262, Proposition 3, we have [A:L]=n[A : L] =n, and therefore [A⊗KL:L]=[A:K]=n2[A\otimes_KL : L] = [A : K] =n^2 and [EndL(A):L]=n2_L(A) : L] =n^2. Consequently, ψ′\psi ' is an isomorphism, which proves (i).

The converse follows from Lemma 3 below.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.