The Stacks project

Proof. As $f$ is smooth at $x$ we may assume, after replacing $X$ by an open neighbourhood of $x$ that $f$ is smooth. In particular we see that $f$ is flat and locally of finite presentation. By Lemma 37.23.1 we already know there exists an open neighbourhood $U \subset X$ of $x$ such that $h$ comes from $h \in \Gamma (U, \mathcal{O}_ U)$ and such that $D = V(h)$ is an effective Cartier divisor in $U$ with $x \in D$ and $D \to S$ flat and of finite presentation. By Morphisms, Lemma 29.32.15 we have a short exact sequence

where $i : D \to U$ is the closed immersion and $\mathcal{C}_{D/U}$ is the conormal sheaf of $D$ in $U$. As $D$ is an effective Cartier divisor cut out by $h \in \Gamma (U, \mathcal{O}_ U)$ we see that $\mathcal{C}_{D/U} = h \cdot \mathcal{O}_ S$. Since $U \to S$ is smooth the sheaf $\Omega _{U/S}$ is finite locally free, hence its pullback $i^*\Omega _{U/S}$ is finite locally free also. The first arrow of the sequence maps the free generator $h$ to the section $\text{d}h|_ D$ of $i^*\Omega _{U/S}$ which has nonzero value in the fibre $\Omega _{U/S, x} \otimes \kappa (x)$ by assumption. By right exactness of $\otimes \kappa (x)$ we conclude that

By Morphisms, Lemma 29.34.14 we see that $\Omega _{U/S, x} \otimes \kappa (x)$ can be generated by at most $\dim _ x(U_ s)$ elements. By the displayed formula we see that $\Omega _{D/S, x} \otimes \kappa (x)$ can be generated by at most $\dim _ x(U_ s) - 1$ elements. Note that $\dim _ x(D_ s) = \dim _ x(U_ s) - 1$ for example because $\dim (\mathcal{O}_{D_ s, x}) = \dim (\mathcal{O}_{U_ s, x}) - 1$ by Algebra, Lemma 10.60.13 (also $D_ s \subset U_ s$ is effective Cartier, see Divisors, Lemma 31.18.1) and then using Morphisms, Lemma 29.28.1. Thus we conclude that $\Omega _{D/S, x} \otimes \kappa (x)$ can be generated by at most $\dim _ x(D_ s)$ elements and we conclude that $D \to S$ is smooth at $x$ by Morphisms, Lemma 29.34.14 again. After shrinking $U$ we get that $D \to S$ is smooth and we win. $\square$

Proof. Write $k = \kappa (s)$ and $R = \mathcal{O}_{X_ s, x}$. Denote $\mathfrak m$ the maximal ideal of $R$ and $\kappa = R/\mathfrak m$ so that $\kappa = \kappa (x)$. As formation of modules of differentials commutes with localization (see Algebra, Lemma 10.131.8) we have $\Omega _{X_ s/s, x} = \Omega _{R/k}$. By Algebra, Lemma 10.131.9 there is an exact sequence

Hence if (2) holds, there exists an element $\overline{h} \in \mathfrak m$ such that $\text{d}\overline{h}$ is nonzero. Choose a lift $h \in \mathcal{O}_{X, x}$ of $\overline{h}$ and apply Lemma 37.38.1. $\square$

Proof. Write $k = \kappa (s)$ and $R = \mathcal{O}_{X_ s, x}$. Denote $\mathfrak m$ the maximal ideal of $R$ and $\kappa = R/\mathfrak m$ so that $\kappa = \kappa (x)$. As formation of modules of differentials commutes with localization (see Algebra, Lemma 10.131.8) we have $\Omega _{X_ s/s, x} = \Omega _{R/k}$. By assumption (2) and Algebra, Lemma 10.140.4 the map

is injective. Assumption (3) implies that $\mathfrak m/\mathfrak m^2 \not= 0$. Thus there exists an element $\overline{h} \in \mathfrak m$ such that $\text{d}\overline{h}$ is nonzero. Choose a lift $h \in \mathcal{O}_{X, x}$ of $\overline{h}$ and apply Lemma 37.38.1. $\square$

Proof. We may and do replace $S$ by an affine open neighbourhood of $s$. We may and do replace $X$ by an affine open neighbourhood of $x$ such that $X \to S$ is smooth. We will prove the lemma for smooth morphisms of affines by induction on $d = \dim _ x(X_ s)$.

The case $d = 0$. In this case we show that we may take $Z$ to be an open neighbourhood of $x$. Namely, if $d = 0$, then $X \to S$ is quasi-finite at $x$, see Morphisms, Lemma 29.29.5. Hence there exists an affine open neighbourhood $U \subset X$ such that $U \to S$ is quasi-finite, see Morphisms, Lemma 29.56.2. Thus after replacing $X$ by $U$ we see that $X$ is quasi-finite and smooth over $S$, hence smooth of relative dimension $0$ over $S$, hence étale over $S$. Moreover, the fibre $X_ s$ is a finite discrete set. Hence after replacing $X$ by a further affine open neighbourhood of $X$ we see that $f^{-1}(\{ s\} ) = \{ x\} $ (because the topology on $X_ s$ is induced from the topology on $X$, see Schemes, Lemma 26.18.5). This proves the lemma in this case.

Next, assume $d > 0$. Note that because $x$ is a closed point of its fibre the extension $\kappa (x)/\kappa (s)$ is finite (by the Hilbert Nullstellensatz, see Morphisms, Lemma 29.20.3). Thus we see $\Omega _{\kappa (x)/\kappa (s)} = 0$ as this holds for algebraic separable field extensions. Thus we may apply Lemma 37.38.2 to find a diagram

with $x \in D$. Note that $\dim _ x(D_ s) = \dim _ x(X_ s) - 1$ for example because $\dim (\mathcal{O}_{D_ s, x}) = \dim (\mathcal{O}_{X_ s, x}) - 1$ by Algebra, Lemma 10.60.13 (also $D_ s \subset X_ s$ is effective Cartier, see Divisors, Lemma 31.18.1) and then using Morphisms, Lemma 29.28.1. Thus the morphism $D \to S$ is smooth with $\dim _ x(D_ s) = \dim _ x(X_ s) - 1 = d - 1$. By induction hypothesis we can find an immersion $Z \to D$ as desired, which finishes the proof. $\square$

First proof of Lemma 37.38.6. By assumption $X_ s \not= \emptyset $. By Varieties, Lemma 33.25.6 there exists a closed point $x \in X_ s$ such that $\kappa (x)$ is a finite separable field extension of $\kappa (s)$. Hence by Lemma 37.38.5 there exists an immersion $Z \to X$ such that $Z \to S$ is étale and such that $x \in Z$. Take $(S' , s') = (Z, x)$. $\square$

Second proof of Lemma 37.38.6. Pick a point $x \in X$ with $f(x) = s$. Choose a diagram

with $\pi $ étale, $x \in U$ and $V = \mathop{\mathrm{Spec}}(R)$ affine, see Morphisms, Lemma 29.36.20. In particular $s \in V$. The morphism $\pi : U \to \mathbf{A}^ d_ V$ is open, see Morphisms, Lemma 29.36.13. Thus $W = \pi (U) \cap \mathbf{A}^ d_ s$ is a nonempty open subset of $\mathbf{A}^ d_ s$. Let $w \in W$ be a point with $\kappa (s) \subset \kappa (w)$ finite separable, see Varieties, Lemma 33.25.5. By Algebra, Lemma 10.114.1 there exist $d$ elements $\overline{f}_1, \ldots , \overline{f}_ d \in \kappa (s)[x_1, \ldots , x_ d]$ which generate the maximal ideal corresponding to $w$ in $\kappa (s)[x_1, \ldots , x_ d]$. After replacing $R$ by a principal localization we may assume there are $f_1, \ldots , f_ d \in R[x_1, \ldots , x_ d]$ which map to $\overline{f}_1, \ldots , \overline{f}_ d \in \kappa (s)[x_1, \ldots , x_ d]$. Consider the $R$-algebra

and set $S' = \mathop{\mathrm{Spec}}(R')$. By construction we have a closed immersion $j : S' \to \mathbf{A}^ d_ V$ over $V$. By construction the fibre of $S' \to V$ over $s$ is a single point $s'$ whose residue field is finite separable over $\kappa (s)$. Let $\mathfrak q' \subset R'$ be the corresponding prime. By Algebra, Lemma 10.136.10 we see that $(R')_ g$ is a relative global complete intersection over $R$ for some $g \in R'$, $g \not\in \mathfrak q$. Thus $S' \to V$ is flat and of finite presentation in a neighbourhood of $s'$, see Algebra, Lemma 10.136.13. By construction the scheme theoretic fibre of $S' \to V$ over $s$ is $\mathop{\mathrm{Spec}}(\kappa (s'))$. Hence it follows from Morphisms, Lemma 29.36.15 that $S' \to S$ is étale at $s'$. Set

By construction there exists a point $s'' \in S''$ which maps to $s'$ via the projection $p : S'' \to S'$. Note that $p$ is étale as the base change of the étale morphism $\pi $, see Morphisms, Lemma 29.36.4. Choose a small affine neighbourhood $S''' \subset S''$ of $s''$ which maps into the nonempty open neighbourhood of $s' \in S'$ where the morphism $S' \to S$ is étale. Then the étale neighbourhood $(S''', s'') \to (S, s)$ is a solution to the problem posed by the lemma. $\square$

Proof. For every $s \in S$ there exists an $i \in I$ such that $s$ is in the image of $S_ i \to S$. By Lemma 37.38.6 we can find an étale morphism $g_ s : T_ s \to S$ such that $s \in g_ s(T_ s)$ and such that $g_ s$ factors through $S_ i \to S$. Hence $\{ T_ s \to S\} $ is an étale covering of $S$ that refines $\mathcal{U}$. $\square$

Proof. Let $s \in S$. By Varieties, Lemma 33.25.6 the set of closed points $x \in X_ s$ such that $\kappa (x)/\kappa (s)$ is separable is dense in $X_ s$. Thus it suffices to construct an étale morphism $U \to X$ with $x$ in the image such that $U \to S$ factors in the manner described in the lemma. To do this, choose an immersion $Z \to X$ passing through $x$ such that $Z \to S$ is étale (Lemma 37.38.5). After replacing $S$ by $Z$ and $X$ by $Z \times _ S X$ we see that we may assume $X \to S$ has a section $\sigma : S \to X$ with $\sigma (s) = x$. Then we can first replace $S$ by an affine open neighbourhood of $s$ and next replace $X$ by an affine open neighbourhood of $x$. Then finally, we consider the subset $X^0 \subset X$ of Section 37.29. By Lemmas 37.29.6 and 37.29.4 this is a retrocompact open subscheme containing $\sigma $ such that the fibres $X^0 \to S$ are geometrically connected. If $X^0$ is not affine, then we choose an affine open $U \subset X^0$ containing $x$. Since $X^0 \to S$ is smooth, the image of $U$ is open. Choose an affine open neighbourhood $V \subset S$ of $s$ contained in $\sigma ^{-1}(U)$ and in the image of $U \to S$. Finally, the reader sees that $U \cap f^{-1}(V) \to V$ has all the desired properties. For example $U \cap f^{-1}(V)$ is equal to $U \times _ S V$ is affine as a fibre product of affine schemes. Also, the geometric fibres of $U \cap f^{-1}(V) \to V$ are nonempty open subschemes of the irreducible fibres of $X^0 \to S$ and hence connected. Some details omitted. $\square$