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 $J^ n$. 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)