Lemma 10.137.10 (00TA)—The Stacks project

Lemma 10.137.10. 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 10.79.4. 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 10.134.12 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 10.136.6 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 10.134.12 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 10.137.7 and 10.136.13) and being syntomic over $R$ is local on $S$ (Lemma 10.136.4). $\square$

Comments (4)

Comment #724 by Keenan Kidwell on June 21, 2014 at 21:02

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

Comment #2887 by Dario Weißmann on October 07, 2017 at 12:57

I don't understand a part in the proof. Why do the $U_E$ cover $\text{Spec}(S)$ ? I don't think this follows from Lemma 10.78.3. We only get that the $U_E$ are open. However we only need $I/I^2$ to be free on $D(g)$ 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 $I/I^2\to \bigoplus Sdx_i$ stays injective upon tensoring with $\kappa(\mathfrak{p})$ as $R\to S$ is smooth. Maybe we could mention this?

Comment #2937 by Johan on October 10, 2017 at 02:18

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 $v_1, \ldots, v_m$ generate a vector space $V$ , then a subset of $\{v_1, \ldots, v_m\}$ is a basis for $V$ . OK?

Comment #2940 by Dario Weißmann on October 10, 2017 at 10:23

Yes, of course. Thank you.

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