Lemma 37.37.6. Let $f : X \to S$ be a smooth morphism of schemes. Let $s \in S$ be a point in the image of $f$. Then there exists an étale neighbourhood $(S', s') \to (S, s)$ and a $S$-morphism $S' \to X$.

First proof of Lemma 37.37.6. 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 37.37.5 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 37.37.6. Pick a point $x \in X$ with $f(x) = s$. Choose a diagram

$\xymatrix{ X \ar[d] & U \ar[l] \ar[d] \ar[r]_-\pi & \mathbf{A}^ d_ V \ar[ld] \\ S & V \ar[l] }$

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

$R' = R[x_1, \ldots , x_ d]/(f_1, \ldots , f_ d)$

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 10.136.11 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 10.136.14. 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

$S'' = U \times _{\pi , \mathbf{A}^ d_ V, j} S'.$

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. $\square$

Comment #3822 by Johannes Anschuetz on

The "Y" in the diagram seems to mean "S".

Comment #4927 by robot0079 on

Here is a quick proof.

We can assume $X/S$ is surjective.

When S is a strict henselian local ring, this result follows from the fact that $X(S)\to X(k)$ is surjective, where k is residue field of S. We can prove this by factoring it into composition of etale and relative affine space morphism.

Now in general case. From above we see that $X(O_{S, \tilde{s}})$ is non empty, we deduce our result by limit preserving property of morphism of locally of finite presentation.

Comment #5195 by on

@#4927. Yes, this is an outline of a proof.

Comment #5970 by Dario Weißmann on

typo: W should be defined as $\pi(U)\cap ...$ instead of $\pi(V)\cap...$

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).