The Stacks project

Proof. By Morphisms, Lemma 29.41.8 the morphism $f$ is quasi-compact. Hence the normalization $S'$ of $S$ in $X$ is defined (Morphisms, Definition 29.53.3) and we have the factorization $X \to S' \to S$. By Morphisms, Lemma 29.53.11 we have (2), (4), and (5). The morphism $f'$ is universally closed by Morphisms, Lemma 29.41.7. It is quasi-compact by Schemes, Lemma 26.21.14 and quasi-separated by Schemes, Lemma 26.21.13.

To show the remaining statements we may assume the base scheme $S$ is affine, say $S = \mathop{\mathrm{Spec}}(R)$. Then $S' = \mathop{\mathrm{Spec}}(A)$ with $A = \Gamma (X, \mathcal{O}_ X)$ an integral $R$-algebra. Thus it is clear that $f'_*\mathcal{O}_ X$ is $\mathcal{O}_{S'}$ (because $f'_*\mathcal{O}_ X$ is quasi-coherent, by Schemes, Lemma 26.24.1, 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 $V(I) \subset S' = \mathop{\mathrm{Spec}}(A)$. 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 ideal of $A$ and we conclude that $V(I) = S'$ as desired. $\square$

Proof. Recall that there are no specializations among the points of $\pi ^{-1}(\{ s\} )$, see Algebra, Lemma 10.36.20. As $f'$ is surjective, we find that $|X_ s| \to \pi ^{-1}(\{ s\} )$ is surjective. Observe that $X_ s$ is a quasi-separated scheme of finite type over a field (quasi-compactness was shown in the proof of the referenced lemma). Thus $X_ s$ is Noetherian (Morphisms, Lemma 29.15.6). A topological argument (omitted) now shows that $\pi ^{-1}(\{ s\} )$ is finite. For each $i$ we can pick a finite type point $x_ i \in X_ s$ mapping to $s_ i$ (Morphisms, Lemma 29.16.7). We conclude that $\kappa (s_ i)/\kappa (s)$ is finite: $x_ i$ can be represented by a morphism $\mathop{\mathrm{Spec}}(k_ i) \to X_ s$ of finite type (by our definition of finite type points) and hence $\mathop{\mathrm{Spec}}(k_ i) \to s = \mathop{\mathrm{Spec}}(\kappa (s))$ is of finite type (as a composition of finite type morphisms), hence $k_ i/\kappa (s)$ is finite (Morphisms, Lemma 29.16.1). $\square$

Proof. If $X_ s$ is geometrically connected, then any base change of it is connected. On the other hand, suppose that $X_ s$ is not geometrically connected. Then by Varieties, Lemma 33.7.11 we see that $X_ s \times _{\mathop{\mathrm{Spec}}(\kappa (s))} \mathop{\mathrm{Spec}}(k)$ is disconnected for some finite separable field extension $k/\kappa (s)$. By Lemma 37.35.2 there exists an affine étale neighbourhood $(U, u) \to (S, s)$ such that $\kappa (u)/\kappa (s)$ is identified with $k/\kappa (s)$. In this case $X_ u$ is disconnected. $\square$

Proof. Let $f = \pi \circ f'$ be the factorization of Lemma 37.53.1. Note that besides the conclusions of Lemma 37.53.1 we also have that $f'$ is separated (Schemes, Lemma 26.21.13) and finite type (Morphisms, Lemma 29.15.8). Hence $f'$ is proper. By Cohomology of Schemes, Proposition 30.19.1 we see that $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ S$-module. Hence we see that $\pi $ is finite, i.e., (2) holds.

This proves all but the most interesting assertion, namely that all the fibres of $f'$ are geometrically connected. It is clear from the discussion above that we may replace $S$ by $S'$, and we may therefore assume that $S$ is Noetherian, affine, $f : X \to S$ is proper, and $f_*\mathcal{O}_ X = \mathcal{O}_ S$. Let $s \in S$ be a point of $S$. We have to show that $X_ s$ is geometrically connected. By Lemma 37.53.3 we see that it suffices to show $X_ u$ is connected for every étale neighbourhood $(U, u) \to (S, s)$. We may assume $U$ is affine. Thus $U$ is Noetherian (Morphisms, Lemma 29.15.6), the base change $f_ U : X_ U \to U$ is proper (Morphisms, Lemma 29.41.5), and that also $(f_ U)_*\mathcal{O}_{X_ U} = \mathcal{O}_ U$ (Cohomology of Schemes, Lemma 30.5.2). Hence after replacing $(f : X \to S, s)$ by the base change $(f_ U : X_ U \to U, u)$ it suffices to prove that the fibre $X_ s$ is connected when $f_*\mathcal{O}_ X = \mathcal{O}_ S$. We can deduce this from Derived Categories of Schemes, Lemma 36.32.7 (by looking at idempotents in the structure sheaf of $X_ s$) but we will also give a direct argument below.

Namely, we apply the theorem on formal functions, more precisely Cohomology of Schemes, Lemma 30.20.7. It tells us that

where $X_ n$ is the $n$th infinitesimal neighbourhood of $X_ s$. Since the underlying topological space of $X_ n$ is equal to that of $X_ s$ we see that if $X_ s = T_1 \amalg T_2$ is a disjoint union of nonempty open and closed subschemes, 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 $\mathcal{O}^\wedge _{S, s}$ is a local ring. Thus the assumption was wrong, i.e., $X_ s$ is connected, and we win. $\square$

Proof. We may apply Lemma 37.53.1 to get the morphism $f' : X \to S'$. Note that besides the conclusions of Lemma 37.53.1 we also have that $f'$ is separated (Schemes, Lemma 26.21.13) and finite type (Morphisms, Lemma 29.15.8). Hence $f'$ is proper. At this point we have proved all of the statements except for the statement that $f'$ has geometrically connected fibres.

We may assume that $S = \mathop{\mathrm{Spec}}(R)$ is affine. Set $R' = \Gamma (X, \mathcal{O}_ X)$. Then $S' = \mathop{\mathrm{Spec}}(R')$. Thus we may replace $S$ by $S'$ and assume that $S = \mathop{\mathrm{Spec}}(R)$ is affine $R = \Gamma (X, \mathcal{O}_ X)$. Next, let $s \in S$ be a point. Let $U \to S$ be an étale morphism of affine schemes and let $u \in U$ be a point mapping to $s$. Let $X_ U \to U$ be the base change of $X$. By Lemma 37.53.3 it suffices to show that the fibre of $X_ U \to U$ over $u$ is connected. By Cohomology of Schemes, Lemma 30.5.2 we see that $\Gamma (X_ U, \mathcal{O}_{X_ U}) = \Gamma (U, \mathcal{O}_ U)$. Hence we have to show: Given $S = \mathop{\mathrm{Spec}}(R)$ affine, $X \to S$ proper with $\Gamma (X, \mathcal{O}_ X) = R$ and $s \in S$ is a point, the fibre $X_ s$ is connected.

To do this it suffices to show that the only idempotents $e \in H^0(X_ s, \mathcal{O}_{X_ s})$ are $0$ and $1$ (we already know that $X_ s$ is nonempty by Lemma 37.53.1). By Derived Categories of Schemes, Lemma 36.32.7 after replacing $R$ by a principal localization we may assume $e$ is the image of an element of $R$. Since $R \to H^0(X_ s, \mathcal{O}_{X_ s})$ factors through $\kappa (s)$ we conclude. $\square$

Proof. Apply Theorem 37.53.5 to get a factorization $X \to S' \to S$. It is enough to show that $S' = S$. This will follow from Morphisms, Lemma 29.54.8. Namely, $S'$ is reduced because $X$ is reduced (Morphisms, Lemma 29.53.8). The morphism $S' \to S$ is integral by the theorem cited above. Every generic point of $S'$ lies over $\xi $ by Morphisms, Lemma 29.53.9 and assumption (5). On the other hand, since $S'$ is the relative spectrum of $f_*\mathcal{O}_ X$ we see that the scheme theoretic fibre $S'_\xi $ is the spectrum of $H^0(X_\xi , \mathcal{O})$ which is equal to $\kappa (\xi )$ by assumption. Hence $S'$ is an integral scheme with function field equal to the function field of $S$. This finishes the proof. $\square$

Proof. Let $s \in S$. Set $n = n_{X/S}(s)$. Note that $n < \infty $ as the geometric fibre of $X \to S$ at $s$ is a proper scheme over a field, hence Noetherian, hence has a finite number of connected components. We have to find an open neighbourhood $V$ of $s$ such that $n_{X/S}|_ V \geq n$. Let $X \to S' \to S$ be the Stein factorization as in Theorem 37.53.5. By Lemma 37.53.2 there are finitely many points $s'_1, \ldots , s'_ m \in S'$ lying over $s$ and the extensions $\kappa (s'_ i)/\kappa (s)$ are finite. Then Lemma 37.42.1 tells us that after replacing $S$ by an étale neighbourhood of $s$ we may assume $S' = V_1 \amalg \ldots \amalg V_ m$ as a scheme with $s'_ i \in V_ i$ and $\kappa (s'_ i)/\kappa (s)$ purely inseparable. Then the schemes $X_{s_ i'}$ are geometrically connected over $\kappa (s)$, hence $m = n$. The schemes $X_ i = (f')^{-1}(V_ i)$, $i = 1, \ldots , n$ are flat and of finite presentation over $S$. Hence the image of $X_ i \to S$ is open (Morphisms, Lemma 29.25.10). Thus in a neighbourhood of $s$ we see that $n_{X/S}$ is at least $n$. $\square$

Proof. By Lemma 37.53.7 the function $n_{X/S}$ is lower semincontinuous. For $s \in S$ consider the $\kappa (s)$-algebra

By Varieties, Lemma 33.9.3 and the fact that $X_ s$ is geometrically reduced $A$ is finite product of finite separable extensions of $\kappa (s)$. Hence $A \otimes _{\kappa (s)} \kappa (\overline{s})$ is a product of $\beta _0(s) = \dim _{\kappa (s)} H^0(E \otimes ^\mathbf {L} \kappa (s))$ copies of $\kappa (\overline{s})$. Thus $X_{\overline{s}}$ has $\beta _0(s) = \dim _{\kappa (s)} A$ connected components. In other words, we have $n_{X/S} = \beta _0$ as functions on $S$. Thus $n_{X/S}$ is upper semi-continuous by Derived Categories of Schemes, Lemma 36.32.1. This 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 Chow's lemma (in the form of Limits, Lemma 32.12.1) 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 $X'_0$. Suppose the lemma holds for $(X', Y')$. Let $W' \subset X'$ be the open and closed subscheme proper over $A$ such that $Y' = W'_0$. By Morphisms, Lemma 29.41.7 $W = \varphi (W') \subset X$ and $Q = \varphi (X' \setminus W') \subset X$ are closed subsets and by Morphisms, Lemma 29.41.9 $W$ is proper over $A$. The image of $W \cap Q$ in $\mathop{\mathrm{Spec}}(A)$ is closed. Since $(A, I)$ is henselian, if $W \cap Q$ is nonempty, then we find that $W \cap Q$ has a point lying over $\mathop{\mathrm{Spec}}(A/I)$. This is impossible as $W'_0 = Y' = \varphi ^{-1}(Y)$. We conclude that $W$ is an open and closed subscheme of $X$ proper over $A$ with $W_0 = Y$. Thus we reduce to the case described in the next paragraph.

Assume there exists an immersion $j : X \to \mathbf{P}^ n_ A$ over $A$. Let $\overline{X}$ be the scheme theoretic image of $j$. Since $j$ is a quasi-compact morphism (Schemes, Lemma 26.21.14) we see that $j : X \to \overline{X}$ is an open immersion (Morphisms, Lemma 29.7.7). Hence the base change $j_0 : X_0 \to \overline{X}_0$ is an open immersion as well. Thus $j_0(Y) \subset \overline{X}_0$ is open. It is also closed by Morphisms, Lemma 29.41.7. Suppose that the lemma holds for $(\overline{X}, j_0(Y))$. Let $\overline{W} \subset \overline{X}$ be the corresponding open and closed subscheme proper over $A$ such that $j_0(Y) = \overline{W}_0$. Then $T = \overline{W} \setminus j(X)$ is closed in $\overline{W}$, hence has closed image in $\mathop{\mathrm{Spec}}(A)$ by properness of $\overline{W}$ over $A$. Since $(A, I)$ is henselian, we find that if $T$ is nonempty, then there is a point of $T$ mapping into $\mathop{\mathrm{Spec}}(A/I)$. This is impossible because $j_0(Y) = \overline{W}_0$ is contained in $j(X)$. Hence $\overline{W}$ is contained in $j(X)$ and we can set $W \subset X$ equal to the unique open and closed subscheme mapping isomorphically to $\overline{W}$ via $j$. Thus we reduce to the case described in the next paragraph.

Assume $X \subset \mathbf{P}^ n_ A$ is a closed subscheme. Then $X \to \mathop{\mathrm{Spec}}(A)$ is a proper morphism. Let $Z = X_0 \setminus Y$. This is an open and closed subscheme of $X_0$ and $X_0 = Y \amalg Z$. Let $X \to X' \to \mathop{\mathrm{Spec}}(A)$ be the Stein factorization as in Theorem 37.53.5. Let $Y' \subset X'_0$ and $Z' \subset X'_0$ be the images of $Y$ and $Z$. Since the fibres of $X \to Z$ are geometrically connected, we see that $Y' \cap Z' = \emptyset $. Hence $X'_0 = Y' \amalg Z'$ as $X \to X'$ is surjective. Since $X' \to \mathop{\mathrm{Spec}}(A)$ is integral, we see that $X'$ is the spectrum of an $A$-algebra integral over $A$. Recall that open and closed subsets of spectra correspond $1$-to-$1$ with idempotents in the corresponding ring, see Algebra, Lemma 10.21.3. Hence by More on Algebra, Lemma 15.11.6 we see that we may write $X' = W' \amalg V'$ with $W'$ and $V'$ open and closed and with $Y' = W'_0$ and $Z' = V'_0$. Let $W$ be the inverse image in $X$ to finish the proof. $\square$