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
0 \to J/J^2 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to 0
of Lemma 10.131.9 is split exact.
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
0 \to I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0
of Lemma 10.134.7 is exact.
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.slogan 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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