Bourbaki

Lemma 3

Tag 0D9H, permanent. Cite it as [Bourbaki, Tag 0D9H].

Commutative Algebra, chapter X, § 7. Algèbres lisses

Let K be an ideal of A containing J. Suppose that there exists an integer m such that J∩KmJ \cap K^m is contained in JK (this condition is satisfied if A is noetherian). If A/JA/J is formally smooth over k for the K/J-adic topology, the mapping dˉA/K:A/K⊗A/JJ/J2⟶A/K⊗AΩk(A)\bar{d}_{A/K} : A/K \otimes_{A/J} J/J^2 \longrightarrow A/K \otimes_A \Omega_k(A) has an A-linear retraction.

Let C denote the kk-algebra A/(JK+Km)A/(JK + K^m); the ideal N = (J+Km)/(JK+Km)(J + K^m)/(JK + K^m) of C is square zero and the quotient ring C/N identifies with A/(J+Km)A/(J + K^m). Endow C and C/N with the discrete topology, and A/J with the K/J-adic topology. The canonical homomorphism A/J→A/(J+Km)A/J \to A/(J + K^m) is continuous; it therefore has a lifting φ:A/J→A/(JK+Km)\varphi : A/J \to A/(JK + K^m).

The mapping a↦a1C−φ(a1A/J)a \mapsto a1_C - \varphi(a1_{A/J}) of A into N is then a kk-derivation (No. 1, Prop. 1), hence is written a↦u(da)a \mapsto u(da) with u∈HomA(Ωk(A),N)u \in \mathrm{Hom}_A(\Omega_k(A), N). But the assumption J∩Km⊂JKJ \cap K^m \subset JK implies J∩(JK+Km)=JKJ \cap (JK + K^m) = JK, so that the canonical mapping ψ:J/JK→N\psi : J/JK \to N is bijective; there therefore exists v∈HomA/K(A/K⊗AΩk(A),J/JK)v \in \mathrm{Hom}_{A/K}(A/K \otimes_A \Omega_k(A), J/JK) such that, for a in A, one has a1C=φ(a1A/J)+ψ(v(1⊗da))a1_C = \varphi(a1_{A/J}) + \psi(v(1 \otimes da)). Taking a in J, one sees that v(1⊗da)v(1 \otimes da) is equal to the class of a in J/JK. Since A/K⊗A/JJ/J2A/K \otimes_{A/J} J/J^2 identifies with J/JKJ/JK, vv is the required retraction.

The fact that the condition on K is satisfied when the algebra A is noetherian follows from III, § 3, No. 1, Cor. 2 of Prop. 1.

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

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