Bourbaki

Proposition 4

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

Algebra, chapter III, § 10. Derivations

Let d:A→Ed : A \to E be an ε\varepsilon-derivation of degree δ\delta. Suppose that A has a unit element 1 and let xx be a homogeneous element of A with an inverse x−1x^{-1} in A. Then

d(x−1)=−εδ,deg⁡(x)x−1((dx)x−1)=−εδ,deg⁡(x)(x−1(dx))x−1. d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}((dx)x^{-1}) = -\varepsilon_{\delta, \deg(x)} (x^{-1}(dx))x^{-1}.

We have d(xx−1)=d(1)=0d(xx^{-1}) = d(1) = 0 (Proposition 3), whence

(dx)x−1+εδ,deg⁡(x)x(d(x−1))=0 (dx)x^{-1} + \varepsilon_{\delta, \deg(x)} x(d(x^{-1})) = 0
which proves the first formula of (19). On the other hand, x−1x^{-1} is homogeneous of degree −deg⁡(x)-\deg(x) and εδ,deg⁡(x)=εδ,−deg⁡(x)\varepsilon_{\delta, \deg(x)} = \varepsilon_{\delta, -\deg(x)} by formulae (1) of no. 1; writing d(x−1x)=0d(x^{-1}x) = 0, the second formula of (19) is obtained similarly.

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

Cited by

Nothing.

Cites

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