Lemma 10.139.1. Given ring maps $A \to B \to C$ with $B \to C$ smooth, then the sequence
\[ 0 \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]
of Lemma 10.131.7 is exact.
10.139 Smoothness and differentials
Some results on differentials and smooth ring maps.
Proof. This follows from the more general Lemma 10.138.9 because a smooth ring map is formally smooth, see Proposition 10.138.13. But it also follows directly from Lemma 10.134.4 since $H_1(L_{C/B}) = 0$ is part of the definition of smoothness of $B \to C$. $\square$
Lemma 10.139.2. Let $A \to B \to C$ be ring maps with $A \to C$ smooth and $B \to C$ surjective with kernel $J \subset B$. Then the exact sequence of Lemma 10.131.9 is split exact.
\[ 0 \to J/J^2 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to 0 \]
Proof. This follows from the more general Lemma 10.138.10 because a smooth ring map is formally smooth, see Proposition 10.138.13. $\square$
Lemma 10.139.3. Let $A \to B \to C$ be ring maps. Assume $A \to C$ is surjective (so also $B \to C$ is) and $A \to B$ smooth. Denote $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(B \to C)$. Then the sequence of Lemma 10.134.7 is exact.
\[ 0 \to I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0 \]
Proof. This follows from the more general Lemma 10.138.11 because a smooth ring map is formally smooth, see Proposition 10.138.13. $\square$
Lemma 10.139.4. Let $\varphi : R \to S$ be a smooth ring map. Let $\sigma : S \to R$ be a left inverse to $\varphi $. Set $I = \mathop{\mathrm{Ker}}(\sigma )$. Then
$I/I^2$ is a finite locally free $R$-module, and
if $I/I^2$ is free, then $S^\wedge \cong R[[t_1, \ldots , t_ d]]$ as $R$-algebras, where $S^\wedge $ is the $I$-adic completion of $S$.
Proof. By Lemma 10.131.10 applied to $R \to S \to R$ we see that $I/I^2 = \Omega _{S/R} \otimes _{S, \sigma } R$. Since by definition of a smooth morphism the module $\Omega _{S/R}$ is finite locally free over $S$ we deduce that (1) holds. If $I/I^2$ is free, then choose $f_1, \ldots , f_ d \in I$ whose images in $I/I^2$ form an $R$-basis. Consider the $R$-algebra map defined by
\[ \Psi : R[[x_1, \ldots , x_ d]] \longrightarrow S^\wedge , \quad x_ i \longmapsto f_ i. \]
Denote $P = R[[x_1, \ldots , x_ d]]$ and $J = (x_1, \ldots , x_ d) \subset P$. We write $\Psi _ n : P/J^ n \to S/I^ n$ for the induced map of quotient rings. Note that $S/I^2 = \varphi (R) \oplus I/I^2$. Thus $\Psi _2$ is an isomorphism. Denote $\sigma _2 : S/I^2 \to P/J^2$ the inverse of $\Psi _2$. We will prove by induction on $n$ that for all $n > 2$ there exists an inverse $\sigma _ n : S/I^ n \to P/J^ n$ of $\Psi _ n$. Namely, as $S$ is formally smooth over $R$ (by Proposition 10.138.13) we see that in the solid diagram
\[ \xymatrix{ S \ar@{..>}[r] \ar[rd]_{\sigma _{n - 1}} & P/J^ n \ar[d] \\ & P/J^{n - 1} } \]
of $R$-algebras we can fill in the dotted arrow by some $R$-algebra map $\tau : S \to P/J^ n$ making the diagram commute. This induces an $R$-algebra map $\overline{\tau } : S/I^ n \to P/J^ n$ which is equal to $\sigma _{n - 1}$ modulo $J^{n - 1}$. By construction the map $\Psi _ n$ is surjective and now $\overline{\tau } \circ \Psi _ n$ is an $R$-algebra endomorphism of $P/J^ n$ which maps $x_ i$ to $x_ i + \delta _{i, n}$ with $\delta _{i, n} \in J^{n - 1}/J^ n$. It follows that $\Psi _ n$ is an isomorphism and hence it has an inverse $\sigma _ n$. This proves the lemma. $\square$
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