Let A be a graded K-algebra.
(i) For every ε-derivation d:A→A and every homogeneous element a of A,
[d,adε(a)]ε=adε(da).
(ii) If the algebra A is associative, adε(a) is an ε-derivation of A of degree deg(a).
(i) Suppose that d is of degree δ, let α=dega and write f=[d,adε(a)]ε. For every homogeneous element x∈A of degree ξ, we have, by (1) (no. 1),
f(x)=d(ax−ε(α,ξ)xa)−εδ,α(a(dx))−εα,δ+ξ(dx)a=(da)x+εδ,αa(dx)−εα,ξ(dx)a−εδ+α,ξx(da)−εδ,αa(dx)+εα,ξ(dx)a=(da)x−εδ+α,ξx(da)=[da,x]ε.
(ii) For all x homogeneous of degree ξ and all y homogeneous of degree η in A,
adε(a)(xy)=a(xy)−εα,ξ+η(xy)a=(ax−εα,ξxa)y+εα,ξx(ay−εα,ηya)=adε(a)(x).y+εα,ξx.adε(a)(y) taking account of (1) and the associativity of
A.
When A is associative, adε(a) is called the inner ε-derivation of A defined by a.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.