The Stacks project

37.43 Application to the structure of finite type morphisms

The result in this section can be found in [GruRay]. Loosely stated it says that a finite type morphism is étale locally on the source and target the composition of a finite morphism by a smooth morphism with geometrically connected fibres of relative dimension equal to the fibre dimension of the original morphism.

Lemma 37.43.1. 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

\[ \xymatrix{ X \ar[dd] & X' \ar[l]^ g \ar[d]^\pi & x \ar@{|->}[dd] & x' \ar@{|->}[l] \ar@{|->}[d] \\ & Y \ar[d]^ h & & y \ar@{|->}[d] \\ S \ar@{=}[r] & S & s & s \ar@{=}[l] } \]

and a point $x' \in X'$ with $g(x') = x$ such that with $y = \pi (x')$ we have

  1. $h : Y \to S$ is smooth of relative dimension $n$,

  2. $g : (X', x') \to (X, x)$ is an elementary étale neighbourhood,

  3. $\pi $ is finite, and $\pi ^{-1}(\{ y\} ) = \{ x'\} $, and

  4. $\kappa (y)$ is a purely transcendental extension of $\kappa (s)$.

Moreover, if $f$ is locally of finite presentation then $\pi $ is of finite presentation.

Proof. The problem is local on $X$ and $S$, hence we may assume that $X$ and $S$ are affine. By Algebra, Lemma 10.125.3 after replacing $X$ by a standard open neighbourhood of $x$ in $X$ we may assume there is a factorization

\[ \xymatrix{ X \ar[r]^\pi & \mathbf{A}^ n_ S \ar[r] & S } \]

such that $\pi $ is quasi-finite and such that $\kappa (\pi (x))$ is purely transcendental over $\kappa (s)$. By Lemma 37.37.1 there exists an elementary étale neighbourhood

\[ (Y, y) \to (\mathbf{A}^ n_ S, \pi (x)) \]

and an open $X' \subset X \times _{\mathbf{A}^ n_ S} Y$ which contains a unique point $x'$ lying over $y$ such that $X' \to Y$ is finite. This proves (1) – (4) hold. For the final assertion, use Morphisms, Lemma 29.21.11. $\square$

slogan

Lemma 37.43.2. 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

  1. $h : Y' \to S'$ is smooth of relative dimension $n$,

  2. all fibres of $Y' \to S'$ are geometrically integral,

  3. $g : (X', x') \to (X, x)$ is an elementary étale neighbourhood,

  4. $\pi $ is finite, and $\pi ^{-1}(\{ y'\} ) = \{ x'\} $,

  5. $\kappa (y')$ is a purely transcendental extension of $\kappa (s')$, and

  6. $e : (S', s') \to (S, s)$ 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 37.43.1 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 37.18.3. At this point we have verified all assumption of Lemma 37.42.4 hold for the morphism $h : Y \to S$ and open $T \subset Y_ s$. As a result of applying Lemma 37.42.4 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. $\square$

Lemma 37.43.3. Assumption and notation as in Lemma 37.43.2. In addition to properties (1) – (6) we may also arrange it so that

  1. $S'$, $Y'$, $X'$ are affine.

Proof. Note that if $Y'$ is affine, then $X'$ is affine as $\pi $ is finite. Choose an affine open neighbourhood $U' \subset S'$ of $s'$. Choose an affine open neighbourhood $V' \subset h^{-1}(U')$ of $y'$. Let $W' = h(V')$. This is an open neighbourhood of $s'$ in $S'$, see Morphisms, Lemma 29.34.10, contained in $U'$. Choose an affine open neighbourhood $U'' \subset W'$ of $s'$. Then $h^{-1}(U'') \cap V'$ is affine because it is equal to $U'' \times _{U'} V'$. By construction $h^{-1}(U'') \cap V' \to U''$ is a surjective smooth morphism whose fibres are (nonempty) open subschemes of geometrically integral fibres of $Y' \to S'$, and hence geometrically integral. Thus we may replace $S'$ by $U''$ and $Y'$ by $h^{-1}(U'') \cap V'$. $\square$

The significance of the property $\pi ^{-1}(\{ y'\} ) = \{ x'\} $ is partially explained by the following lemma.

Lemma 37.43.4. Let $\pi : X \to Y$ be a finite morphism. Let $x \in X$ with $y = \pi (x)$ such that $\pi ^{-1}(\{ y\} ) = \{ x\} $. Then

  1. For every neighbourhood $U \subset X$ of $x$ in $X$, there exists a neighbourhood $V \subset Y$ of $y$ such that $\pi ^{-1}(V) \subset U$.

  2. The ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is finite.

  3. If $\pi $ is of finite presentation, then $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is of finite presentation.

  4. For any quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $\mathcal{F}_ x = \pi _*\mathcal{F}_ y$ as $\mathcal{O}_{Y, y}$-modules.

Proof. The first assertion is purely topological; use that $\pi $ is a continuous and closed map such that $\pi ^{-1}(\{ y\} ) = \{ x\} $. To prove the second and third parts we may assume $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $A \to B$ is a finite ring map and $y$ corresponds to a prime $\mathfrak p$ of $A$ such that there exists a unique prime $\mathfrak q$ of $B$ lying over $\mathfrak p$. Then $B_{\mathfrak q} = B_{\mathfrak p}$, see Algebra, Lemma 10.41.11. In other words, the map $A_{\mathfrak p} \to B_{\mathfrak q}$ is equal to the map $A_{\mathfrak p} \to B_{\mathfrak p}$ you get from localizing $A \to B$ at $\mathfrak p$. Thus (2) and (3) follow from simple properties of localization (some details omitted). For the final statement, suppose that $\mathcal{F} = \widetilde M$ for some $B$-module $M$. Then $\mathcal{F} = M_{\mathfrak q}$ and $\pi _*\mathcal{F}_ y = M_{\mathfrak p}$. By the above these localizations agree. Alternatively you can use part (1) and the definition of stalks to see that $\mathcal{F}_ x = \pi _*\mathcal{F}_ y$ directly. $\square$


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