Theorem 2

Let k be a ring, A a k-algebra, and J an ideal of A such that the k-algebra A/J is formally smooth. Endow A with the J-adic topology. The following conditions are equivalent:

(i) the topological k-algebra A is formally smooth; (ii) the $A/J$-module $J/J^2$ is projective and the canonical homomorphism ($\S$ 5, n° 2)

is bijective; (iii) the $A/J$-module $J/J^2$ is projective and there exists an isomorphism of topological k-algebras of the separated completion of A onto the completed algebra of the graded algebra $S_{A/J}(J/J^2)$.

If A is noetherian, these conditions are also equivalent to:

(iv) the ideal J is completely secant.

Let us first observe that (iii) implies (i): indeed, under the hypotheses of (iii), the algebra $S_{A/J}(J/J^2)$, endowed with the topology associated with its graduation, is formally smooth over $A/J$ (No. 3, example 1), hence over k (No. 2, prop. 3, a)); assertion (i) then follows from prop. 3, c) of No. 2.

Let $\hat{A}$ denote the separated completion of the algebra A and $\hat{J}$ the separated completion of J. The canonical homomorphism $i : A \to \hat{A}$ induces an isomorphism $A/J \to \hat{A}/\hat{J}$ (III, § 2, No. 12, formula (21)). Let $\varphi : A/J \to \hat{A}$ be a lifting of this isomorphism (No. 4, Prop. 5). Let $\lambda : \hat{J} \to J/J^2$ denote the surjection deduced from the canonical isomorphism $J/J^2 \to \hat{J}/\hat{J}^2$ (III, § 2, No. 12, formula (21)). Let $a$ be an element of A, $\bar{a}$ its class in $A/J$, and $z$ an element of $\hat{J}$; one has $\varphi(\bar{a}) \equiv i(a) \pmod{\hat{J}}$, whence $\varphi(\bar{a})z \equiv i(a)z \pmod{\hat{J}^2}$ and $\lambda(\varphi(\bar{a})z) = \lambda(i(a)z) = \bar{a}\lambda(z)$. In other words, $\lambda$ is A/J-linear when $\hat{J}$ is endowed with the structure of an A/J-module deduced from $\varphi$.

Suppose that the homomorphism $\lambda$ admits an A/J-linear section $\sigma : J/J^2 \to \hat{J}$. Let S denote the graded k-algebra $S_{A/J}(J/J^2)$ and $\hat{S}$ its completion. Let

be the homomorphism of k-algebras such that $\theta(x) = \varphi(x)$ for $x$ in $S^0 = A/J$, and $\theta(x) = \sigma(x)$ for $x$ in $S^1 = J/J^2$. Since $\theta$ maps $S^1$ into $\hat{J}$, it maps $S^n$ into $\hat{J}^n$ and therefore extends to a continuous homomorphism $\hat{\theta} : \hat{S} \to \hat{A}$. The mapping $\mathrm{gr}_1(\theta) : J/J^2 \to \hat{J}/\hat{J}^2$ is the composite of $\sigma$ with the canonical surjection $\hat{J} \to \hat{J}/\hat{J}^2$; since $\sigma$ is a section of $\lambda$, $\mathrm{gr}_1(\theta)$ coincides with the canonical isomorphism of $J/J^2$ onto $\hat{J}/\hat{J}^2$. It follows that $\mathrm{gr}(\theta) : S \to \mathrm{gr}_{\hat{J}}(\hat{A})$ is the composite of the canonical surjection $\beta$ with the canonical isomorphism $\mathrm{gr}_J(A) \to \mathrm{gr}_{\hat{J}}(\hat{A})$ (III, § 2, No. 12, formula (22)).

Let us now prove the implication (ii) $\Rightarrow$ (iii). Under assumption (ii), the A/J-module $J/J^2$ is projective, and therefore $\lambda$ admits an A/J-linear section; the homomorphism $\hat{\theta} : \hat{S} \to \hat{A}$ associated with this section by the preceding construction induces by

Let us prove (i) ⇒ (ii). Suppose that the topological k-algebra A is formally smooth. Let us first prove that the A/J-module J/J^2 is projective. Let M be a Λ/J-module and $f : M \to J/J^2$ a surjective A/J-linear mapping; the question is to prove that f admits a Λ/J-linear section.

Let π denote the canonical surjection $A/J^2 \to A/J$. By Remark 1 of No. 2, there exists an isomorphism of k-algebras ψ : A/J ⊕ J/J^2 → A/J^2 such that π(ψ(y, z)) = y and ψ(0, z) = z for $y \in A/J$, $z \in J/J^2$. Consider the k-algebra (A/J) ⊕ M (No. 1, Example) and the mapping u : (A/J) ⊕ M → A/J^2 such that u(x, m) = ψ(x, f(m)). This is a surjective homomorphism of k-algebras, whose kernel is the submodule Ker f of M, and therefore is square zero. The canonical surjection ρ : A → A/J^2 is continuous; since the topological k-algebra A is formally smooth, there exists a homomorphism of k-algebras ŝ : A → (A/J)⊕M such that u◦ŝ = ρ. Since pr_1 = π◦ψ = π◦u, one has pr_1◦ŝ = π◦u◦ŝ = π◦ρ, so that pr_1◦ŝ is the canonical surjection of A onto A/J. Hence one has ŝ(J) ⊂ M and consequently ŝ(J^2) = 0, so that ŝ induces an A/J-linear mapping s : J/J^2 → M. One has u◦ŝ = ρ and pr_2◦ψ^{-1}◦u(y, m) = f(m) for $y \in A/J$ and $m \in M$. Let $x \in J$, and ū its class in J/J^2; one has f(s(ū)) = f(pr_2(ŝ(x))) = pr_2(ψ^{-1}(ū)) = ū. Thus s is a section of f.

It remains to prove that the homomorphism β is injective. Since the A/J-module J/J^2 is projective, λ admits an A/J-linear section; let θ : S → Ā denote the associated homomorphism. The homomorphism gr(θ) is identified with β. Let m be an integer; let Σ_m denote the graded k-algebra quotient of S by the ideal $\sum_{i>m} S^i$ and $\theta_m : \Sigma_m \to A/J^{m+1}$ the homomorphism deduced from θ. The composite of θ_m with the canonical surjection A/J^{m+1} → A/J is the canonical projection of Σ_m onto S^0 = A/J; hence the kernel of θ_m is a nilpotent two-sided ideal. By the example of No. 4, there exists a lifting ψ_m : A → Σ_m of the canonical surjection A → A/J^{m+1}. Since the composite of ψ_m with the canonical projection of Σ_m onto A/J is the canonical surjection, ψ_m(J) consists of elements of degree > 0. By passing to associated graded algebras, one deduces from ψ_m a graded k-linear mapping gr(ψ_m) : gr_J(A) → Σ_m such that gr_m(θ)◦gr_m(ψ_m) = Id_{J^m/J^{m+1}}. It follows that gr_m(θ), hence also β_m, is injective, which completes the proof of (ii).

Finally, when A is noetherian, conditions (ii) and (iv) are equivalent (§ 5, No. 2, Th. 1).