A morphism of finite type is, in étale neighbourhoods, finite over a smooth morphism.
Lemma 37.47.2. Let $f : X \to S$ be a morphism. Let $x \in X$ and set $s = f(x)$. Assume that $f$ is locally of finite type and that $n = \dim _ x(X_ s)$. Then there exists a commutative diagram
\[ \xymatrix{ X \ar[dd] & X' \ar[l]^ g \ar[d]^\pi & x \ar@{|->}[dd] & x' \ar@{|->}[l] \ar@{|->}[d] \\ & Y' \ar[d]^ h & & y' \ar@{|->}[d] \\ S & S' \ar[l]_ e & s & s' \ar@{|->}[l] } \]
and a point $x' \in X'$ with $g(x') = x$ such that with $y' = \pi (x')$, $s' = h(y')$ we have
$h : Y' \to S'$ is smooth of relative dimension $n$,
all fibres of $Y' \to S'$ are geometrically integral,
$g : (X', x') \to (X, x)$ is an elementary étale neighbourhood,
$\pi $ is finite, and $\pi ^{-1}(\{ y'\} ) = \{ x'\} $,
$\kappa (y')$ is a purely transcendental extension of $\kappa (s')$, and
$e : (S', s') \to (S, s)$ is an elementary étale neighbourhood.
Moreover, if $f$ is locally of finite presentation, then $\pi $ is of finite presentation.
Proof.
The question is local on $S$, hence we may replace $S$ by an affine open neighbourhood of $s$. Next, we apply Lemma 37.47.1 to get a commutative diagram
\[ \xymatrix{ X \ar[dd] & X' \ar[l]^ g \ar[d]^\pi & x \ar@{|->}[dd] & x' \ar@{|->}[l] \ar@{|->}[d] \\ & Y \ar[d]^ h & & y \ar@{|->}[d] \\ S \ar@{=}[r] & S & s & s \ar@{=}[l] } \]
where $h$ is smooth of relative dimension $n$ and $\kappa (y)$ is a purely transcendental extension of $\kappa (s)$. Since the question is local on $X$ also, we may replace $Y$ by an affine neighbourhood of $y$ (and $X'$ by the inverse image of this under $\pi $). As $S$ is affine this guarantees that $Y \to S$ is quasi-compact, separated and smooth, in particular of finite presentation. Let $T$ be the connected component of $Y_ s$ containing $y$. As $Y_ s$ is Noetherian we see that $T$ is open. We also see that $T$ is geometrically connected over $\kappa (s)$ by Varieties, Lemma 33.7.14. Since $T$ is also smooth over $\kappa (s)$ it is geometrically normal, see Varieties, Lemma 33.25.4. We conclude that $T$ is geometrically irreducible over $\kappa (s)$ (as a connected Noetherian normal scheme is irreducible, see Properties, Lemma 28.7.6). Finally, note that the smooth morphism $h$ is normal by Lemma 37.20.3. At this point we have verified all assumption of Lemma 37.46.4 hold for the morphism $h : Y \to S$ and open $T \subset Y_ s$. As a result of applying Lemma 37.46.4 we obtain $e : S' \to S$, $s' \in S'$, $Y'$ as in the commutative diagram
\[ \xymatrix{ X \ar[dd] & X' \ar[l]^ g \ar[d]^\pi & X' \times _ Y Y' \ar[l] \ar[d] & x \ar@{|->}[dd] & x' \ar@{|->}[l] \ar@{|->}[d] & (x', s') \ar@{|->}[l] \ar@{|->}[d] \\ & Y \ar[d]^ h & Y' \ar[d] \ar[l] & & y \ar@{|->}[d] & (y, s') \ar@{|->}[l] \ar@{|->}[d] \\ S \ar@{=}[r] & S & S' \ar[l]_ e & s & s \ar@{=}[l] & s' \ar@{|->}[l] } \]
where $e : (S', s') \to (S, s)$ is an elementary étale neighbourhood, and where $Y' \subset Y_{S'}$ is an open neighbourhood all of whose fibres over $S'$ are geometrically irreducible, such that $Y'_{s'} = T$ via the identification $Y_ s = Y_{S', s'}$. Let $(y, s') \in Y'$ be the point corresponding to $y \in T$; this is also the unique point of $Y \times _ S S'$ lying over $y$ with residue field equal to $\kappa (y)$ which maps to $s'$ in $S'$. Similarly, let $(x', s') \in X' \times _ Y Y' \subset X' \times _ S S'$ be the unique point over $x'$ with residue field equal to $\kappa (x')$ lying over $s'$. Then the outer part of this diagram is a solution to the problem posed in the lemma. Some minor details omitted.
$\square$
Comments (1)
Comment #1127 by Simon Pepin Lehalleur on