The Stacks project

Proof. As the question is purely topological we may replace $X$ and $Y$ by their reductions. In particular this implies that $Y$ is integral, see Properties, Lemma 28.3.4. Let $X_\eta = X_{1, \eta } \cup \ldots \cup X_{n, \eta }$ be the decomposition of $X_\eta $ into connected components. Let $X_ i \subset X$ be the reduced closed subscheme whose generic fibre is $X_{i, \eta }$. Note that $Z_{i, j} = X_ i \cap X_ j$ is a closed subset of $X$ whose generic fibre $Z_{i, j, \eta }$ is empty. Hence after shrinking $Y$ we may assume that $Z_{i, j} = \emptyset $, see Lemma 37.24.1. After shrinking $Y$ some more we may assume that $X_ y = \bigcup X_{i, y}$ for $y \in Y$, see Lemma 37.24.5. Moreover, after shrinking $Y$ we may assume that each $X_ i \to Y$ is flat and of finite presentation, see Morphisms, Proposition 29.27.1. The morphisms $X_ i \to Y$ are open, see Morphisms, Lemma 29.25.10. Thus there exists an open neighbourhood $V$ of $\eta $ which is contained in $f(X_ i)$ for each $i$. For each $y \in V$ the schemes $X_{i, y}$ are nonempty closed subsets of $X_ y$, we have $X_ y = \bigcup X_{i, y}$ and the intersections $Z_{i, j, y} = X_{i, y} \cap X_{j, y}$ are empty! Clearly this implies that $X_ y$ has at least $n$ connected components. $\square$

Proof. This comes down to the statement that for $y' \in Y'$ with image $y \in Y$ the fibre $X'_{y'} = X_ y \times _ y y'$ is geometrically connected over $\kappa (y')$ if and only if $X_ y$ is geometrically connected over $\kappa (y)$. This follows from Varieties, Lemma 33.7.3. $\square$

Proof. Suppose that $y' \in Y'$ has image $y \in Y$. Suppose $K \supset \kappa (y)$ and $K' \supset \kappa (y')$ are separably closed extensions. Then we may choose a commutative diagram

of fields. The result follows as the morphisms of schemes

induce bijections between connected components, see Varieties, Lemma 33.7.6. $\square$

Proof. Choose a diagram

as in Lemma 37.24.8. Note that the generic fibre of $f'$ is geometrically connected (for example by Lemma 37.28.3). Suppose that the lemma holds for the morphism $f'$. This means that there exists a nonempty open $W \subset Y'$ such that every fibre of $X' \to Y'$ over $W$ is geometrically connected. Then, as $g$ is an open morphism by Morphisms, Lemma 29.36.13 all the fibres of $f$ at points of the nonempty open $V = g(W)$ are geometrically connected, see Lemma 37.28.3. In this way we see that we may assume that the irreducible components of the generic fibre $X_\eta $ are geometrically irreducible.

Let $Y'$ be the reduction of $Y$, and set $X' = Y' \times _ Y X$. Then it suffices to prove the lemma for the morphism $X' \to Y'$ (for example by Lemma 37.28.3 once again). Since the generic fibre of $X' \to Y'$ is the same as the generic fibre of $X \to Y$ we see that we may assume that $Y$ is irreducible and reduced (i.e., integral, see Properties, Lemma 28.3.4) and that the irreducible components of the generic fibre $X_\eta $ are geometrically irreducible.

At this point suppose that $X_\eta = X_{1, \eta } \bigcup \ldots \bigcup X_{n, \eta }$ is the decomposition of the generic fibre into (geometrically) irreducible components. Let $X_ i$ be the closure of $X_{i, \eta }$ in $X$. After shrinking $Y$ we may assume that $X = \bigcup X_ i$, see Lemma 37.24.5. Let $Z_{i, j} = X_ i \cap X_ j$. Let

where $(i, j) \in I$ if $Z_{i, j, \eta } = \emptyset $ and $(i, j) \in J$ if $Z_{i, j, \eta } \not= \emptyset $. After shrinking $Y$ we may assume that $Z_{i, j} = \emptyset $ for all $(i, j) \in I$, see Lemma 37.24.1. After shrinking $Y$ we obtain that $X_{i, y}$ is geometrically irreducible for each $i$ and all $y \in Y$, see Lemma 37.27.5. After shrinking $Y$ some more we achieve the situation where each $Z_{i, j} \to Y$ is flat and of finite presentation for all $(i, j) \in J$, see Morphisms, Proposition 29.27.1. This means that $f(Z_{i, j}) \subset Y$ is open, see Morphisms, Lemma 29.25.10. We claim that

works, i.e., that $X_ y$ is geometrically connected for each $y \in V$. Namely, the fact that $X_\eta $ is connected implies that the equivalence relation generated by the pairs in $J$ has only one equivalence class. Now if $y \in V$ and $K \supset \kappa (y)$ is a separably closed extension, then the irreducible components of $(X_ y)_ K$ are the fibres $(X_{i, y})_ K$. Moreover, we see by construction and $y \in V$ that $(X_{i, y})_ K$ meets $(X_{j, y})_ K$ if and only if $(i, j) \in J$. Hence the remark on equivalence classes shows that $(X_ y)_ K$ is connected and we win. $\square$

Proof. Let $Z$ be the reduced induced scheme structure on $\overline{\{ y\} }$. Let $f_ Z : X_ Z \to Z$ be the base change of $f$. Clearly it suffices to prove the lemma for $f_ Z$ and the generic point of $Z$. Hence we may assume that $Y$ is an integral scheme, see Properties, Lemma 28.3.4. Our goal in this case is to produce a nonempty open $V \subset Y$ such that $n_{X/Y}|_ V$ is constant.

We apply Lemma 37.24.8 to $f : X \to Y$ and we get $g : Y' \to V \subset Y$. As $g : Y' \to V$ is surjective finite étale, in particular open (see Morphisms, Lemma 29.36.13), it suffices to prove that there exists an open $V' \subset Y'$ such that $n_{X'/Y'}|_{V'}$ is constant, see Lemma 37.27.3. Thus we see that we may assume that all irreducible components of the generic fibre $X_\eta $ are geometrically irreducible over $\kappa (\eta )$. By Varieties, Lemma 33.8.16 this implies that also the connected components of $X_\eta $ are geometrically connected.

At this point suppose that $X_\eta = X_{1, \eta } \bigcup \ldots \bigcup X_{n, \eta }$ is the decomposition of the generic fibre into (geometrically) connected components. In particular $n_{X/Y}(\eta ) = n$. Let $X_ i$ be the closure of $X_{i, \eta }$ in $X$. After shrinking $Y$ we may assume that $X = \bigcup X_ i$, see Lemma 37.24.5. After shrinking $Y$ some more we see that each fibre of $f$ has at least $n$ connected components, see Lemma 37.28.1. Hence $n_{X/Y}(y) \geq n$ for all $y \in Y$. After shrinking $Y$ some more we obtain that $X_{i, y}$ is geometrically connected for each $i$ and all $y \in Y$, see Lemma 37.28.4. Since $X_ y = \bigcup X_{i, y}$ this shows that $n_{X/Y}(y) \leq n$ and finishes the proof. $\square$

Proof. Fix $n$. Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E_ n \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation whose base change to $Y$ recovers $f$. By Lemma 37.28.3 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E_ n$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E_ n \cap Z$ either contains a nonempty open subset or is not dense in $Z$. Let $\xi \in Z$ be the generic point. Then Lemma 37.28.5 shows that $n_{X/Y}$ is constant in a neighbourhood of $\xi $ in $Z$. This clearly implies what we want. $\square$

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and let $\pi \in R$ be a uniformizer. To get a contradiction assume that $X_\eta $ is disconnected. This means there exists a nontrivial idempotent $e \in \Gamma (X_\eta , \mathcal{O}_{X_\eta })$. Let $U = \mathop{\mathrm{Spec}}(A)$ be any affine open in $X$. Note that $\pi $ is a nonzerodivisor on $A$ as $A$ is flat over $R$, see More on Algebra, Lemma 15.22.9 for example. Then $e|_{U_\eta }$ corresponds to an element $e \in A[1/\pi ]$. Let $z \in A$ be an element such that $e = z/\pi ^ n$ with $n \geq 0$ minimal. Note that $z^2 = \pi ^ nz$. This means that $z \bmod \pi A$ is nilpotent if $n > 0$. By assumption $A/\pi A$ is reduced, and hence minimality of $n$ implies $n = 0$. Thus we conclude that $e \in A$! In other words $e \in \Gamma (X, \mathcal{O}_ X)$. As $X$ is connected it follows that $e$ is a trivial idempotent which is a contradiction. $\square$