Proposition 7

Let $k_0$ be a ring, $k$ a $k_0$-algebra, $A$ a $k$-algebra, $m$ a maximal ideal of $A$. Suppose that $k$ and $A/m$ are formally smooth over $k_0$. In order that $A$ be formally smooth over $k$ for the $m$-adic topology, it is necessary and sufficient that the following two conditions be satisfied:

(i) the canonical homomorphism $S_{A/m}(m/m^2) \to \mathrm{gr}_m(A)$ is bijective; (ii) the $A/m$-linear mapping

deduced from the canonical mapping $k \to A$ is injective.

Let us denote by $d_k : k \to \Omega_{k_0}(k)$ and $d_A : A \to \Omega_{k_0}(A)$ the universal $k_0$-derivations.

Suppose first that $A$ is formally smooth over $k$ for the $m$-adic topology. Then $A$ is formally smooth over $k_0$ for the $m$-adic topology (No. 2, Prop. 3, a)), which is equivalent to (i) (No. 5, Theorem 2). Moreover, the $k_0$-derivation $\lambda \mapsto 1 \otimes d_k(\lambda)$ of $k$ into $A/m \otimes_k \Omega_{k_0}(k)$ can be extended to a $k_0$-derivation of $A$ into $A/m \otimes_k \Omega_{k_0}(k)$ (No. 2, Remark 2). There therefore exists an $A$-linear mapping $u : \Omega_{k_0}(A) \to A/m \otimes_k \Omega_{k_0}(k)$ such that $u(d_A(\lambda 1_A)) = 1 \otimes d_k(\lambda)$ for every $\lambda \in k$. The $A/m$-linear mapping $A/m \otimes_A \Omega_{k_0}(A) \longrightarrow A/m \otimes_k \Omega_{k_0}(k)$ deduced from $u$ is a retraction of $\omega$, which proves (ii).

Suppose conversely that conditions (i) and (ii) are satisfied. Then $A$ is formally smooth over $k_0$ for the $m$-adic topology (No. 5, Theorem 2) and the $A$-module $\Omega_{k_0}(A)$ is projective (No. 5, Corollary 3 of Theorem 2). Let us fix an integer $r \geqslant 0$ and consider the $A/m^r$-linear mapping

deduced from the canonical mapping $k \to A$. Let $(\lambda_i)_{i \in I}$ be a family of elements of $k$ such that the $d_k(\lambda_i)$ form a basis of the vector space $\Omega_{k_0}(k)$ over $k$; by (ii), the elements $1 \otimes d_A(\lambda_i 1_A)$ are linearly independent in $A/\mathfrak{m} \otimes_{A} \Omega_{k_0}(A)$. By II, § 3, No. 2, Cor. 1 and 2 of Prop. 5, the $1 \otimes d_A(\lambda_i 1_A)$ form a basis of a direct factor of the $A/\mathfrak{m}^r$-module $A/\mathfrak{m}^r \otimes_{A} \Omega_{k_0}(A)$. There therefore exists an $A/\mathfrak{m}^r$-linear mapping

such that $u_r(1 \otimes d_A(\lambda_i 1_A)) = 1 \otimes d_k(\lambda_i)$ for all $i$, hence $u_r \circ \omega_r = \mathrm{Id}$.

Let us now verify that $A$ is formally smooth over $k$ for the $\mathfrak{m}$-adic topology. Let $C$ be a $k$-algebra, $N$ a square-zero ideal of $C$, and $\pi : C \to C/N$ the canonical surjection; endow $C$ and $C/N$ with the discrete topology. Let $\varphi : A \to C/N$ be a continuous homomorphism of $k$-algebras. Since $A$ is formally smooth over $k_0$ for the $\mathfrak{m}$-adic topology, there exists a continuous homomorphism of $k_0$-algebras $\tilde{\varphi}_0 : A \to C$ such that $\pi \circ \tilde{\varphi}_0 = \varphi$. By Prop. 1 of No. 1, the homomorphisms of $k_0$-algebras $\tilde{\varphi} : A \to C$ such that $\pi \circ \tilde{\varphi} = \varphi$ are the mappings $x \mapsto v(d_A(x)) + \tilde{\varphi}_0(x)$, where $v$ runs through $\mathrm{Hom}_{A}(\Omega_{k_0}(A), N)$. It remains to choose $v$ in such a way that $\tilde{\varphi}$ is a homomorphism of $k$-algebras. The mapping $\lambda \mapsto \lambda 1_C - \tilde{\varphi}_0(\lambda 1_A)$ is a $k_0$-derivation of $k$ into $N$ (loc. cit.), and therefore can be written $h \circ d_k$ with $h \in \mathrm{Hom}_k(\Omega_{k_0}(k), N)$.

Choose an integer $r$ such that the kernel of $\varphi$ contains $\mathfrak{m}^r$. The $A$-module $N$ is annihilated by $\mathfrak{m}^r$, and it is enough to take for $v$ the composite of the sequence of homomorphisms

where $h'$ is deduced from $h$. In fact, for $\lambda \in k$ one has: