The Stacks project

Proof. By Morphisms of Spaces, Lemma 67.9.7 the morphism $f$ is quasi-compact. We just define $Y'$ as the normalization of $Y$ in $X$, so (5) and (2) hold automatically. By Morphisms of Spaces, Lemma 67.48.9 we see that (4) holds. The morphism $f'$ is universally closed by Morphisms of Spaces, Lemma 67.40.6. It is quasi-compact by Morphisms of Spaces, Lemma 67.8.9 and quasi-separated by Morphisms of Spaces, Lemma 67.4.10.

To show the remaining statements we may assume the base $Y$ is affine (as taking normalization commutes with étale localization). Say $Y = \mathop{\mathrm{Spec}}(R)$. Then $Y' = \mathop{\mathrm{Spec}}(A)$ with $A = \Gamma (X, \mathcal{O}_ X)$ an integral $R$-algebra. Thus it is clear that $f'_*\mathcal{O}_ X$ is $\mathcal{O}_{Y'}$ (because $f'_*\mathcal{O}_ X$ is quasi-coherent, by Morphisms of Spaces, Lemma 67.11.2, and hence equal to $\widetilde{A}$). This proves (3).

Let us show that $f'$ is surjective. As $f'$ is universally closed (see above) the image of $f'$ is a closed subset $V(I) \subset Y' = \mathop{\mathrm{Spec}}(A)$. Pick $h \in I$. Then $h|_ X = f^\sharp (h)$ is a global section of the structure sheaf of $X$ which vanishes at every point. As $X$ is quasi-compact this means that $h|_ X$ is a nilpotent section, i.e., $h^ n|X = 0$ for some $n > 0$. But $A = \Gamma (X, \mathcal{O}_ X)$, hence $h^ n = 0$. In other words $I$ is contained in the Jacobson radical of $A$ and we conclude that $V(I) = Y'$ as desired. $\square$

Proof. Recall that there are no specializations among the points of $\pi ^{-1}(\{ y\} )$, see Algebra, Lemma 10.36.20. As $f'$ is surjective, we find that $|X_ y| \to \pi ^{-1}(\{ y\} )$ is surjective. Observe that $X_ y$ is a quasi-separated algebraic space of finite type over a field (quasi-compactness was shown in the proof of the referenced lemma). Thus $|X_ y|$ is a Noetherian topological space (Morphisms of Spaces, Lemma 67.28.6). A topological argument (omitted) now shows that $\pi ^{-1}(\{ y\} )$ is finite. For each $i$ we can pick a finite type point $x_ i \in |X_ y|$ mapping to $y_ i$ (Morphisms of Spaces, Lemma 67.25.6). We conclude that $\kappa (y_ i)/\kappa (y)$ is finite: $x_ i$ can be represented by a morphism $\mathop{\mathrm{Spec}}(k_ i) \to X_ y$ of finite type (by our definition of finite type points) and hence $\mathop{\mathrm{Spec}}(k_ i) \to y = \mathop{\mathrm{Spec}}(\kappa (y))$ is of finite type (as a composition of finite type morphisms), hence $k_ i/\kappa (y)$ is finite (Morphisms, Lemma 29.16.1). $\square$

Proof. Since the category of étale neighbourhoods of $\overline{y}$ is cofiltered and contains a cofinal collection of schemes (Properties of Spaces, Lemma 66.19.3) we may replace $Y$ by one of these neighbourhoods and assume that $Y$ is a scheme. Let $y \in Y$ be the point corresponding to $\overline{y}$. Then $X_ y$ is geometrically connected over $\kappa (y)$ if and only if $X_{\overline{y}}$ is connected and if and only if $(X_ y)_{k'}$ is connected for every finite separable extension $k'$ of $\kappa (y)$. See Spaces over Fields, Section 72.12 and especially Lemma 72.12.8. By More on Morphisms, Lemma 37.35.2 there exists an affine étale neighbourhood $(V, v) \to (Y, y)$ such that $\kappa (s) \subset \kappa (u)$ is identified with $\kappa (s) \subset k'$ any given finite separable extension. The lemma follows. $\square$

Proof. Let $f = \pi \circ f'$ be the factorization of Lemma 76.36.1. Note that besides the conclusions of Lemma 76.36.1 we also have that $f'$ is separated (Morphisms of Spaces, Lemma 67.4.10) and finite type (Morphisms of Spaces, Lemma 67.23.6). Hence $f'$ is proper. By Cohomology of Spaces, Lemma 69.20.2 we see that $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ Y$-module. Hence we see that $\pi $ is finite, i.e., (2) holds.

This proves all but the most interesting assertion, namely that the geometric fibres of $f'$ are connected. It is clear from the discussion above that we may replace $Y$ by $Y'$. Then $Y$ is locally Noetherian, $f : X \to Y$ is proper, and $f_*\mathcal{O}_ X = \mathcal{O}_ Y$. Let $\overline{y}$ be a geometric point of $Y$. At this point we apply the theorem on formal functions, more precisely Cohomology of Spaces, Lemma 69.22.7. It tells us that

where $X_ n = \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}/\mathfrak m_{\overline{y}}^ n) \times _ Y X$. Note that $X_1 = X_{\overline{y}} \to X_ n$ is a (finite order) thickening and hence the underlying topological space of $X_ n$ is equal to that of $X_{\overline{y}}$. Thus, if $X_{\overline{y}} = T_1 \amalg T_2$ is a disjoint union of nonempty open and closed subspaces, then similarly $X_ n = T_{1, n} \amalg T_{2, n}$ for all $n$. And this in turn means $H^0(X_ n, \mathcal{O}_{X_ n})$ contains a nontrivial idempotent $e_{1, n}$, namely the function which is identically $1$ on $T_{1, n}$ and identically $0$ on $T_{2, n}$. It is clear that $e_{1, n + 1}$ restricts to $e_{1, n}$ on $X_ n$. Hence $e_1 = \mathop{\mathrm{lim}}\nolimits e_{1, n}$ is a nontrivial idempotent of the limit. This contradicts the fact that $\mathcal{O}^\wedge _{Y, \overline{y}}$ is a local ring. Thus the assumption was wrong, i.e., $X_{\overline{y}}$ is connected as desired. $\square$

Proof. We may apply Lemma 76.36.1 to get the morphism $f' : X \to Y'$. Note that besides the conclusions of Lemma 76.36.1 we also have that $f'$ is separated (Morphisms of Spaces, Lemma 67.4.10) and finite type (Morphisms of Spaces, Lemma 67.23.6). Hence $f'$ is proper. At this point we have proved all of the statements except for the statement that $f'$ has connected geometric fibres.

It is clear from the discussion that we may replace $Y$ by $Y'$. Then $f : X \to Y$ is proper and $f_*\mathcal{O}_ X = \mathcal{O}_ Y$. Note that these conditions are preserved under flat base change (Morphisms of Spaces, Lemma 67.40.3 and Cohomology of Spaces, Lemma 69.11.2). Let $\overline{y}$ be a geometric point of $Y$. By Lemma 76.36.3 and the remark just made we reduce to the case where $Y$ is a scheme, $y \in Y$ is a point, $f : X \to Y$ is a proper algebraic space over $Y$ with $f_*\mathcal{O}_ X = \mathcal{O}_ Y$, and we have to show the fibre $X_ y$ is connected. Replacing $Y$ by an affine neighbourhood of $y$ we may assume that $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Then $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ signifies that the ring map $R \to \Gamma (X, \mathcal{O}_ X)$ is bijective.

By Limits of Spaces, Lemma 70.12.2 we can write $(X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$ with $X_ i \to Y_ i$ proper and of finite presentation and $Y_ i$ Noetherian. For $i$ large enough $Y_ i$ is affine (Limits of Spaces, Lemma 70.5.10). Say $Y_ i = \mathop{\mathrm{Spec}}(R_ i)$. Let $R'_ i = \Gamma (X_ i, \mathcal{O}_{X_ i})$. Observe that we have ring maps $R_ i \to R_ i' \to R$. Namely, we have the first because $X_ i$ is an algebraic space over $R_ i$ and the second because we have $X \to X_ i$ and $R = \Gamma (X, \mathcal{O}_ X)$. Note that $R = \mathop{\mathrm{colim}}\nolimits R'_ i$ by Limits of Spaces, Lemma 70.5.6. Then

is commutative with $Y'_ i = \mathop{\mathrm{Spec}}(R'_ i)$. Let $y'_ i \in Y'_ i$ be the image of $y$. We have $X_ y = \mathop{\mathrm{lim}}\nolimits X_{i, y'_ i}$ because $X = \mathop{\mathrm{lim}}\nolimits X_ i$, $Y = \mathop{\mathrm{lim}}\nolimits Y'_ i$, and $\kappa (y) = \mathop{\mathrm{colim}}\nolimits \kappa (y'_ i)$. Now let $X_ y = U \amalg V$ with $U$ and $V$ open and closed. Then $U, V$ are the inverse images of opens $U_ i, V_ i$ in $X_{i, y'_ i}$ (Limits of Spaces, Lemma 70.5.7). By Theorem 76.36.4 the fibres of $X_ i \to Y'_ i$ are connected, hence either $U$ or $V$ is empty. This finishes the proof. $\square$

Proof. Apply Theorem 76.36.5 to get a factorization $X \to Y' \to Y$. It is enough to show that $Y' = Y$. It suffices to show that $Y' \times _ Y V \to V$ is an isomorphism, where $V \to Y$ is an étale morphism and $V$ an affine integral scheme, see Spaces over Fields, Lemma 72.4.5. The formation of $Y'$ commutes with étale base change, see Morphisms of Spaces, Lemma 67.48.4. The generic points of $X \times _ Y V$ lie over the generic points of $X$ (Decent Spaces, Lemma 68.20.1) hence map to the generic point of $V$ by assumption (5). Moreover, condition (6) is preserved under the base change by $V \to Y$, for example by flat base change (Cohomology of Spaces, Lemma 69.11.2). Thus it suffices to prove the lemma in case $Y$ is a normal integral affine scheme.

Assume $Y$ is a normal integral affine scheme. We will show $Y' \to Y$ is an isomorphism by an application of Morphisms, Lemma 29.54.8. Namely, $Y'$ is reduced because $X$ is reduced (Morphisms of Spaces, Lemma 67.48.6). The morphism $Y' \to Y$ is integral by the theorem cited above. Since $Y$ is decent and $X \to Y$ is separated, we see that $X$ is decent too; to see this use Decent Spaces, Lemmas 68.17.2 and 68.17.5. By assumption (5), Morphisms of Spaces, Lemma 67.48.7, and Decent Spaces, Lemma 68.20.1 we see that every generic point of an irreducible component of $|Y'|$ maps to $\xi $. On the other hand, since $Y'$ is the relative spectrum of $f_*\mathcal{O}_ X$ we see that the scheme theoretic fibre $Y'_\xi $ is the spectrum of $H^0(X_\xi , \mathcal{O})$ which is equal to $\kappa (\xi )$ by assumption. Hence $Y'$ is an integral scheme with function field equal to the function field of $Y$. This finishes the proof. $\square$

Proof. The question is étale local on $Y$, hence we may and do assume $Y$ is an affine scheme. Let $y \in Y$. Set $n = n_{X/S}(y)$. Note that $n < \infty $ as the geometric fibre of $X \to Y$ at $y$ is a proper algebraic space over a field, hence Noetherian, hence has a finite number of connected components. We have to find an open neighbourhood $V$ of $y$ such that $n_{X/S}|_ V \geq n$. Let $X \to Y' \to Y$ be the Stein factorization as in Theorem 76.36.5. By Lemma 76.36.2 there are finitely many points $y'_1, \ldots , y'_ m \in Y'$ lying over $y$ and the extensions $\kappa (y'_ i)/\kappa (y)$ are finite. More on Morphisms, Lemma 37.42.1 tells us that after replacing $Y$ by an étale neighbourhood of $y$ we may assume $Y' = V_1 \amalg \ldots \amalg V_ m$ as a scheme with $y'_ i \in V_ i$ and $\kappa (y'_ i)/\kappa (y)$ purely inseparable. Then the algebraic spaces $X_{y_ i'}$ are geometrically connected over $\kappa (y)$, hence $m = n$. The algebraic spaces $X_ i = (f')^{-1}(V_ i)$, $i = 1, \ldots , n$ are flat and of finite presentation over $Y$. Hence the image of $X_ i \to Y$ is open (Morphisms of Spaces, Lemma 67.30.6). Thus in a neighbourhood of $y$ we see that $n_{X/Y}$ is at least $n$. $\square$

Proof. By Lemma 76.36.7 the function $n_{X/Y}$ is lower semincontinuous. Thus it suffices to show it is upper semi-continuous. To do this we may work étale locally on $Y$, hence we may assume $Y$ is an affine scheme. For $y \in Y$ consider the $\kappa (y)$-algebra

By Spaces over Fields, Lemma 72.14.3 and the fact that $X_ y$ is geometrically reduced $A$ is finite product of finite separable extensions of $\kappa (y)$. Hence $A \otimes _{\kappa (y)} \kappa (\overline{y})$ is a product of $\beta _0(y) = \dim _{\kappa (y)} A$ copies of $\kappa (\overline{y})$. Thus $X_{\overline{y}}$ has $\beta _0(y)$ connected components. In other words, we have $n_{X/S} = \beta _0$ as functions on $Y$. Thus $n_{X/Y}$ is upper semi-continuous by Derived Categories of Spaces, Lemma 75.26.2. This finishes the proof. $\square$

Proof. Formation of the Stein factorization commutes with flat base change, see Lemma 76.36.1. Thus we may work étale locally on $Y$ and we may assume $Y$ is an affine scheme. Then $Y'$ is an affine scheme and $Y' \to Y$ is integral.

Let $y \in Y$. Set $n$ be the number of connected components of the geometric fibre $X_{\overline{y}}$. Note that $n < \infty $ as the geometric fibre of $X \to Y$ at $y$ is a proper algebraic space over a field, hence Noetherian, hence has a finite number of connected components. By Lemma 76.36.2 there are finitely many points $y'_1, \ldots , y'_ m \in Y'$ lying over $y$ and for each $i$ we can pick a finite type point $x_ i \in |X_ y|$ mapping to $y'_ i$ the extension $\kappa (y'_ i)/\kappa (y)$ is finite. Thus More on Morphisms, Lemma 37.42.1 tells us that after replacing $Y$ by an étale neighbourhood of $y$ we may assume $Y' = V_1 \amalg \ldots \amalg V_ m$ as a scheme with $y'_ i \in V_ i$ and $\kappa (y'_ i)/\kappa (y)$ purely inseparable. In this case the algebraic spaces $X_{y_ i'}$ are geometrically connected over $\kappa (y)$, hence $m = n$. The algebraic spaces $X_ i = (f')^{-1}(V_ i)$, $i = 1, \ldots , n$ are proper, flat, of finite presentation, with geometrically reduced fibres over $Y$. It suffices to prove the lemma for each of the morphisms $X_ i \to Y$. This reduces us to the case where $X_{\overline{y}}$ is connected.

Assume that $X_{\overline{y}}$ is connected. By Lemma 76.36.8 we see that $X \to Y$ has geometrically connected fibres in a neighbourhood of $y$. Thus we may assume the fibres of $X \to Y$ are geometrically connected. Then $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ by Derived Categories of Spaces, Lemma 75.26.8 which finishes the proof. $\square$

Proof. We will denote $T \mapsto T_0$ the base change by $\mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A)$. By a weak version of Chow's lemma (in the form of Cohomology of Spaces, Lemma 69.18.1) there exists a surjective proper morphism $\varphi : X' \to X$ such that $X'$ admits an immersion into $\mathbf{P}^ n_ A$. Set $Y' = \varphi ^{-1}(Y)$. This is an open and closed subscheme of $X'_0$. The lemma holds for $(X', Y')$ by More on Morphisms, Lemma 37.53.9. Let $W' \subset X'$ be the open and closed subscheme proper over $A$ such that $Y' = W'_0$. By Morphisms of Spaces, Lemma 67.40.6 $Q_1 = \varphi (|W'|) \subset |X|$ and $Q_2 = \varphi (|X' \setminus W'|) \subset |X|$ are closed subsets and by Morphisms of Spaces, Lemma 67.40.7 any closed subspace structure on $Q_1$ is proper over $A$. The image of $Q_1 \cap Q_2$ in $\mathop{\mathrm{Spec}}(A)$ is closed. Since $(A, I)$ is henselian, if $Q_1 \cap Q_2$ is nonempty, then we find that $Q_1 \cap Q_2$ has a point lying over $\mathop{\mathrm{Spec}}(A/I)$. This is impossible as $W'_0 = Y' = \varphi ^{-1}(Y)$. We conclude that $Q_1$ is open and closed in $|X|$. Let $W \subset X$ be the corresponding open and closed subspace. Then $W$ is proper over $A$ with $W_0 = Y$. $\square$