Bourbaki

Exercise 2, § 10

Algebra, III, 645

Let K be a commutative field of characteristic ≠2\neq 2, A and B two K-algebras and α\alpha a K-linear mapping of A into B; an α\alpha-derivation of A into B is a K-linear mapping dd of A into B satisfying the relation

d(xy)=(dx)α(y)+α(x)(dy)for x,y in A, d(xy) = (dx)\alpha(y) + \alpha(x)(dy) \quad \text{for } x, y \text{ in } A,

in other words, a derivation on the sense of Definition 1 with A=A′=A′′A = A' = A'', B=B′=B′′B = B' = B'', d=d′=d′′d = d' = d'', μ\mu multiplication in A, λ1\lambda_1 the mapping (z,x)↦zα(x)(z, x) \mapsto z\alpha(x) of B×AB \times A into B and λ2\lambda_2 the mapping (x,z)↦α(x)z(x, z) \mapsto \alpha(x)z of A×BA \times B into B.

Show that if B is a commutative field, an extension of K, and d≠0d \neq 0, there exists b≠0b \neq 0 in B such that

α(xy)=α(x)α(y)+b(dx)(dy)for x,y in A. \alpha(xy) = \alpha(x)\alpha(y) + b(dx)(dy) \quad \text{for } x, y \text{ in } A.

If b=c2b = c^2 with c∈B,c≠0c \in B, c \neq 0, there exist two homomorphisms f,gf, g of A into B such that α(x)=(f(x)+g(x))/2\alpha(x) = (f(x) + g(x))/2, dx=(f(x)−g(x))/2cdx = (f(x) - g(x))/2c. Otherwise, there exists a quadratic algebra (§2§ 2, no. 3) B' over B, a μ∈B′\mu \in B' such that μ2=c\mu^2 = c, and a homomorphism ff of A into B' such that α(x)=(f(x)+f(x)‾)/2\alpha(x) = (f(x) + \overline{f(x)})/2, dx=(f(x)−f(x)‾)/2dx = (f(x) - \overline{f(x)})/2.

Solution

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Transcribed from the printed volume. Not translated.