Bourbaki

Remark

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Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

(1) Let pp be a prime number such that p⋅1=0p \cdot 1 = 0 in KK. The binomial formula and the congruences (pi)≡0(modp)\binom{p}{i} \equiv 0 \pmod{p} for 1≤i≤p−11 \leq i \leq p - 1 imply that

P(E) is stable under the mapping x↦xpx \mapsto x^p.

(2) By definition, the diagram

0⟶P(E)⟶E+→c+E+⊗E+ 0 \longrightarrow P(E) \longrightarrow E^+ \xrightarrow{c^+} E^+ \otimes E^+

is an exact sequence. If K′K' is a commutative ring and ρ:K→K′\rho : K \to K' a ring homomorphism, ρ∗(E)=E⊗KK′\rho^*(E) = E \otimes_K K' is a K′K'-bigebra and the inclusion P(E)→EP(E) \to E defines a homomorphism of Lie K′K'-algebras

α:P(E)⊗KK′→P(E⊗KK′). \alpha : P(E) \otimes_K K' \to P(E \otimes_K K').

If K′K' is flat over KK (Commutative Algebra, Chapter I, § 2, no. 3, Definition 2), it follows from loc. cit. that the diagram

0⟶P(E)⊗KK′⟶E+⊗KK′→c+⊗KIdK′(E+⊗KK′)⊗K′(E+⊗KK′) 0 \longrightarrow P(E) \otimes_K K' \longrightarrow E^+ \otimes_K K' \xrightarrow{c^+ \otimes_K \mathrm{Id}_{K'}} (E^+ \otimes_K K') \otimes_{K'} (E^+ \otimes_K K')

is an exact sequence, which implies that α\alpha is an isomorphism.

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

  • Commutative Algebra, Chapter I, § 2, no. 3, Definition 2 not yet on this site: AC I
Transcribed from the printed volume. Not translated.