Bourbaki

Theorem 1

Tag 00M1, permanent. Cite it as [Bourbaki, Tag 00M1].

Algebra, VIII, p. 367, § 19. Quaternion Algebras

Let D be an R-algebra of finite degree that is a field. Then D is isomorphic to R, C, or H.

Denote the center of D by Z, and let L be a maximal commutative subfield of D. We have [D:Z]=[L:Z]2[D : Z] = [L : Z]^2 by VIII, p. 265, Corollary 2; we also have [L:R]⩽2[L : R]\leqslant 2 because C is an algebraic closure of R. There are consequently three possible cases:

a) We have R = Z = L, so [D : Z] = 1 and D = R.

b) We have R≠ZR\not= Z and Z = L, so [D : Z] = 1 and D = L. In this case, D is isomorphic to C.

c) We have R = Z and [L : R] = 2, so [D : R] = 4. By Proposition 4 of VIII, p. 364, the R-algebra D is isomorphic to a quaternion algebra of type (α,0,γ)(\alpha ,0, \gamma ), where α\alpha and γ\gamma are nonzero elements of R. Let i∈Di\in D- Z be such that i2=αi^2=\alpha. We have α≠0\alpha \not= 0. If α>0\alpha >0, then there exists an a∈Ra\in R such that a2=αa^2=\alpha (VI, §2, No. 6, p. 26, Theorem 3); we then have (a−i)(a+i)=0(a-i)(a+i) = 0, which is absurd because D is a field. So we have α<0\alpha <0. The inequality γ<0\gamma <0 is shown analogously. There then exist elements aa and cc of R∗R^* such that α=−a2\alpha =-a^2 and γ=−c2\gamma =-c^2 (loc. cit.). The algebra D is therefore isomorphic to the quaternion algebra of type (−1,0,−1)(-1,0,-1) (VIII, p. 362, Remark 2), that is, to H.

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.