The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.135.10. Let $R \to S$ be a smooth ring map. There exists an open covering of $\mathop{\mathrm{Spec}}(S)$ by standard opens $D(g)$ such that each $S_ g$ is standard smooth over $R$. In particular $R \to S$ is syntomic.

Proof. Choose a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I = (f_1, \ldots , f_ m)$. For every subset $E \subset \{ 1, \ldots , m\} $ consider the open subset $U_ E$ where the classes $f_ e, e\in E$ freely generate the finite projective $S$-module $I/I^2$, see Lemma 10.78.3. We may cover $\mathop{\mathrm{Spec}}(S)$ by standard opens $D(g)$ each completely contained in one of the opens $U_ E$. For such a $g$ we look at the presentation

\[ \beta : R[x_1, \ldots , x_ n, x_{n + 1}] \longrightarrow S_ g \]

mapping $x_{n + 1}$ to $1/g$. Setting $J = \mathop{\mathrm{Ker}}(\beta )$ we use Lemma 10.132.12 to see that $J/J^2 \cong (I/I^2)_ g \oplus S_ g$ is free. We may and do replace $S$ by $S_ g$. Then using Lemma 10.134.6 we may assume we have a presentation $\alpha : R[x_1, \ldots , x_ n] \to S$ with kernel $I = (f_1, \ldots , f_ c)$ such that $I/I^2$ is free on the classes of $f_1, \ldots , f_ c$.

Using the presentation $\alpha $ obtained at the end of the previous paragraph, we more or less repeat this argument with the basis elements $\text{d}x_1, \ldots , \text{d}x_ n$ of $\Omega _{R[x_1, \ldots , x_ n]/R}$. Namely, for any subset $E \subset \{ 1, \ldots , n\} $ of cardinality $c$ we may consider the open subset $U_ E$ of $\mathop{\mathrm{Spec}}(S)$ where the differential of $\mathop{N\! L}\nolimits (\alpha )$ composed with the projection

\[ S^{\oplus c} \cong I/I^2 \longrightarrow \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S \longrightarrow \bigoplus \nolimits _{i \in E} S\text{d}x_ i \]

is an isomorphism. Again we may find a covering of $\mathop{\mathrm{Spec}}(S)$ by (finitely many) standard opens $D(g)$ such that each $D(g)$ is completely contained in one of the opens $U_ E$. By renumbering, we may assume $E = \{ 1, \ldots , c\} $. For a $g$ with $D(g) \subset U_ E$ we look at the presentation

\[ \beta : R[x_1, \ldots , x_ n, x_{n + 1}] \to S_ g \]

mapping $x_{n + 1}$ to $1/g$. Setting $J = \mathop{\mathrm{Ker}}(\beta )$ we conclude from Lemma 10.132.12 that $J = (f_1, \ldots , f_ c, fx_{n + 1} - 1)$ where $\alpha (f) = g$ and that the composition

\[ J/J^2 \longrightarrow \Omega _{R[x_1, \ldots , x_{n + 1}]/R} \otimes _{R[x_1, \ldots , x_{n + 1}]} S_ g \longrightarrow \bigoplus \nolimits _{i = 1}^ c S_ g\text{d}x_ i \oplus S_ g \text{d}x_{n + 1} \]

is an isomorphism. Reordering the coordinates as $x_1, \ldots , x_ c, x_{n + 1}, x_{c + 1}, \ldots , x_ n$ we conclude that $S_ g$ is standard smooth over $R$ as desired.

This finishes the proof as standard smooth algebras are syntomic (Lemmas 10.135.7 and 10.134.14) and being syntomic over $R$ is local on $S$ (Lemma 10.134.4). $\square$


Comments (4)

Comment #724 by Keenan Kidwell on

It seems like, at the end of the first paragraph, is implicitly being replaced by , because the result 07CF applies to , not (obviously) to itself. Maybe this should be made explicit?

Comment #2887 by Dario Weißmann on

I don't understand a part in the proof. Why do the cover ? I don't think this follows from Lemma 10.78.3. We only get that the are open. However we only need to be free on and this is exactly Lemma 10.77.2 (the finite locally free part).

The second time we (implicitly) apply 10.78.3 it works because stays injective upon tensoring with as is smooth. Maybe we could mention this?

Comment #2937 by on

These opens cover because to see this we can check over the residue fields of the primes and then we reduce to the statement: if generate a vector space , then a subset of is a basis for . OK?

Comment #2940 by Dario Weißmann on

Yes, of course. Thank you.


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