Lemma 89.8.6. Let $R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda$. Then the induced morphism $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth if and only if $S$ is a power series ring over $R$.

Proof. Assume $S$ is a power series ring over $R$. Say $S = R[[x_1, \ldots , x_ n]]$. Smoothness of $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ means the following (see Remark 89.8.4): Given a surjective ring map $B \to A$ in $\mathcal{C}_\Lambda$, a ring map $R \to B$, a ring map $S \to A$ such that the solid diagram

$\xymatrix{ S \ar[r] \ar@{..>}[rd] & A \\ R \ar[u] \ar[r] & B \ar[u] }$

is commutative then a dotted arrow exists making the diagram commute. (Note the similarity with Algebra, Definition 10.138.1.) To construct the dotted arrow choose elements $b_ i \in B$ whose images in $A$ are equal to the images of $x_ i$ in $A$. Note that $b_ i \in \mathfrak m_ B$ as $x_ i$ maps to an element of $\mathfrak m_ A$. Hence there is a unique $R$-algebra map $R[[x_1, \ldots , x_ n]] \to B$ which maps $x_ i$ to $b_ i$ and which can serve as our dotted arrow.

Conversely, assume $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. Let $x_1, \ldots , x_ n \in S$ be elements whose images form a basis in the relative cotangent space $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ of $S$ over $R$. Set $T = R[[X_1, \ldots , X_ n]]$. Note that both

$S/(\mathfrak m_ R S + \mathfrak m_ S^2) \cong R/\mathfrak m_ R[x_1, \ldots , x_ n]/(x_ ix_ j)$

and

$T/(\mathfrak m_ R T + \mathfrak m_ T^2) \cong R/\mathfrak m_ R[X_1, \ldots , X_ n]/(X_ iX_ j).$

Let $S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the local $R$-algebra isomorphism given by mapping the class of $x_ i$ to the class of $X_ i$. Let $f_1 : S \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the composition $S \to S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$. The assumption that $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth means we can lift $f_1$ to a map $f_2 : S \to T/\mathfrak {m}_ T^2$, then to a map $f_3 : S \to T/\mathfrak {m}_ T^3$, and so on, for all $n \geq 1$. Thus we get an induced map $f : S \to T = \mathop{\mathrm{lim}}\nolimits T/\mathfrak m_ T^ n$ of local $R$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to \mathfrak m_ T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ of relative cotangent spaces. Hence $f$ is surjective by Lemma 89.4.2 (where we think of $f$ as a map in $\widehat{\mathcal{C}}_ R$). Choose preimages $y_ i \in S$ of $X_ i \in T$ under $f$. As $T$ is a power series ring over $R$ there exists a local $R$-algebra homomorphism $s : T \to S$ mapping $X_ i$ to $y_ i$. By construction $f \circ s = \text{id}$. Then $s$ is injective. But $s$ induces an isomorphism on relative cotangent spaces since $f$ does, so it is also surjective by Lemma 89.4.2 again. Hence $s$ and $f$ are isomorphisms. $\square$

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