Bourbaki

Exercise 2, § 19

Tag 00M6, permanent. Cite it as [Bourbaki, Tag 00M6].

Algebra, VIII, p. 369

Bourbaki marks this one as harder than the rest.

a) Let A be a simple ring of finite rank over its center K, and let ss be an involutive antiautomorphism of A. Let s′s' be an involutive antiautomorphism of A that coincides with ss on K. Prove that there exists an element aa of A such that we have s′(x)=as(x)a−1s'(x) =as(x)a^{-1} for every x∈Ax\in A and s(a)=as(a) =a or s(a)=−as(a) =-a (apply the Skolem–Noether theorem and Theorem 3 of V, §11, No. 6, p. 85). Also prove the converse.

b) Suppose that the restriction of ss to K is not the identity; let K0K_0 be the subfield of K consisting of the elements fixed by ss. Prove that the set A0A_0 (resp. A1A_1) of elements aa of A such that s(a)=as(a) =a (resp. s(a)=−as(a) =-a) is a K0K_0-linear subspace of A of dimension [A : K] and that A0A_0 is a K0K_0-structure on A (use V, §10, No. 4, p. 63, Proposition 7).

c) Suppose that the restriction of ss to K is the identity; set [A:K]=m2[A : K] =m^2. Prove that A0A_0 is a K0K_0-linear subspace of A of dimension m(m+1)/2m(m+ 1)/2 or m(m−1)/2m(m-1)/2 (reduce to the case when A is a matrix algebra over K and use a)).

Solution

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Cites

  • V, §10, No. 4, p. 63, Proposition 7 not in this corpus
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