Bourbaki

Proposition 8

Tag 066Q, permanent. Cite it as [Bourbaki, Tag 066Q].

Algebra, chapter III, § 10. Derivations

Let AA be a graded K-algebra.

(i) For every ε\varepsilon-derivation d:A→Ad : A \to A and every homogeneous element aa of AA,

(24)(24)
[d,adε(a)]ε=adε(da). [d, \mathrm{ad}_\varepsilon(a)]_\varepsilon = \mathrm{ad}_\varepsilon(da).

(ii) If the algebra AA is associative, adε(a)\mathrm{ad}_\varepsilon(a) is an ε\varepsilon-derivation of AA of degree deg⁡(a)\deg(a).

(i) Suppose that dd is of degree δ\delta, let α=deg⁡a\alpha = \deg a and write f=[d,adε(a)]εf = [d, \mathrm{ad}_\varepsilon(a)]_\varepsilon. For every homogeneous element x∈Ax \in A of degree ξ\xi, 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. f(x) = d(ax - \varepsilon(\alpha, \xi) xa) - \varepsilon_{\delta, \alpha}(a(dx)) - \varepsilon_{\alpha, \delta+\xi}(dx)a \\ = (da)x + \varepsilon_{\delta, \alpha} a(dx) - \varepsilon_{\alpha, \xi}(dx)a - \varepsilon_{\delta+\alpha, \xi} x(da) \\ - \varepsilon_{\delta, \alpha} a(dx) + \varepsilon_{\alpha, \xi}(dx)a \\ = (da)x - \varepsilon_{\delta+\alpha, \xi} x(da) = [da, x]_\varepsilon.

(ii) For all xx homogeneous of degree ξ\xi and all yy homogeneous of degree η\eta in AA,

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) \mathrm{ad}_\varepsilon(a)(xy) = a(xy) - \varepsilon_{\alpha, \xi+\eta}(xy)a \\ = (ax - \varepsilon_{\alpha, \xi} xa)y + \varepsilon_{\alpha, \xi} x(ay - \varepsilon_{\alpha, \eta} ya) \\ = \mathrm{ad}_\varepsilon(a)(x).y + \varepsilon_{\alpha, \xi} x.\mathrm{ad}_\varepsilon(a)(y)
taking account of (1) and the associativity of AA.

When AA is associative, adε(a)\mathrm{ad}_\varepsilon(a) is called the inner ε\varepsilon-derivation of AA defined by aa.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

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Transcribed from the printed edition, 1998, Springer. Not translated.