Bourbaki

Lemma 4

Tag 0D9I, permanent. Cite it as [Bourbaki, Tag 0D9I].

Commutative Algebra, chapter X, § 7. Algèbres lisses

Suppose that A is formally smooth over k for the J-adic topology. In order that A/JA/J be formally smooth over k, it is necessary and sufficient that the canonical mapping dˉ\bar{d} have an A-linear retraction.

It is already known that if A/JA/J is formally smooth over kk, the mapping dˉ\bar{d} admits an A-linear retraction (No. 2, Remark 1). Conversely, suppose that dˉ\bar{d} has an A-linear retraction. Let π:A/J2→A/J\pi : A/J^2 \to A/J be the canonical surjection; by Prop. 2 of No. 1, there exists a ring homomorphism h:A/J→A/J2h : A/J \to A/J^2 such that π∘h=IdA/J\pi \circ h = \mathrm{Id}_{A/J}. Let C be a kk-algebra, N an ideal of C which is square zero, and ρ:C→C/N\rho : C \to C/N the canonical surjection; endow C and C/N with the discrete topology. Let u:A/J→C/Nu : A/J \to C/N be a continuous homomorphism of kk-algebras. Since A is formally smooth over kk for the J-adic topology, there exists a homomorphism of kk-algebras v:A→Cv : A \to C making the diagram

A→vC↓↓ρA/J→uC/N \begin{array}{ccc} A & \xrightarrow{v} & C \\ \downarrow & & \downarrow \rho \\ A/J & \xrightarrow{u} & C/N \end{array}

commutative, where the vertical arrows represent the canonical surjections. One has v(J)⊂Nv(J) \subset \mathbf{N}, hence v(J2)⊂N2={0}v(J^2) \subset \mathbf{N}^2 = \{0\}, and vv defines, by passing to quotients, a homomorphism vˉ:A/J2→C\bar{v} : A/J^2 \to C which satisfies ρ∘vˉ=u∘π\rho \circ \bar{v} = u \circ \pi. Then vˉ∘h\bar{v} \circ h is a lifting of uu to CC.

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

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