Library / Algebra / Chapter III / § 10
Proposition 4
Tag 066L, permanent. Cite it as [Bourbaki, Tag 066L].
Algebra, chapter III, § 10. Derivations
Let d : A → E d : A \to E d : A → E be an ε \varepsilon ε -derivation of degree δ \delta δ . Suppose that A has a unit element 1 and let x x x be a homogeneous element of A with an inverse x − 1 x^{-1} x − 1 in A. Then
d ( x − 1 ) = − ε δ , deg ( x ) x − 1 ( ( d x ) x − 1 ) = − ε δ , deg ( x ) ( x − 1 ( d x ) ) x − 1 .
d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}((dx)x^{-1}) = -\varepsilon_{\delta, \deg(x)} (x^{-1}(dx))x^{-1}.
d ( x − 1 ) = − ε δ , d e g ( x ) x − 1 (( d x ) x − 1 ) = − ε δ , d e g ( x ) ( x − 1 ( d x )) x − 1 .
We have d ( x x − 1 ) = d ( 1 ) = 0 d(xx^{-1}) = d(1) = 0 d ( x x − 1 ) = d ( 1 ) = 0 (Proposition 3), whence
( d x ) x − 1 + ε δ , deg ( x ) x ( d ( x − 1 ) ) = 0
(dx)x^{-1} + \varepsilon_{\delta, \deg(x)} x(d(x^{-1})) = 0
( d x ) x − 1 + ε δ , d e g ( x ) x ( d ( x − 1 )) = 0 which proves the first formula of (19). On the other hand,
x − 1 x^{-1} x − 1 is homogeneous of degree
− deg ( x ) -\deg(x) − deg ( x ) and
ε δ , deg ( x ) = ε δ , − deg ( x ) \varepsilon_{\delta, \deg(x)} = \varepsilon_{\delta, -\deg(x)} ε δ , d e g ( x ) = ε δ , − d e g ( x ) by formulae (1) of no. 1; writing
d ( x − 1 x ) = 0 d(x^{-1}x) = 0 d ( 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.
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Cites
Transcribed from the printed edition, 1998, Springer. Not translated.