Lemma 37.28.1. Let $f : X \to Y$ be a morphism of schemes. Assume $Y$ irreducible with generic point $\eta $ and $f$ of finite type. If $X_\eta $ has $n$ connected components, then there exists a nonempty open $V \subset Y$ such that for all $y \in V$ the fibre $X_ y$ has at least $n$ connected components.

## 37.28 Connected components of fibres

**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.
$\square$

Lemma 37.28.2. Let $f : X \to Y$ be a morphism of schemes. Let $g : Y' \to Y$ be any morphism, and denote $f' : X' \to Y'$ the base change of $f$. Then

**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.
$\square$

Lemma 37.28.3. Let $f : X \to Y$ be a morphism of schemes. Let

be the function which associates to $y \in Y$ the number of connected components of $(X_ y)_ K$ where $K$ is a separably closed extension of $\kappa (y)$. This is well defined and if $g : Y' \to Y$ is a morphism then

where $X' \to Y'$ is the base change of $f$.

**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. $\square$

Lemma 37.28.4. Let $f : X \to Y$ be a morphism of schemes. Assume

$Y$ is irreducible with generic point $\eta $,

$X_\eta $ is geometrically connected, and

$f$ is of finite type.

Then there exists a nonempty open subscheme $V \subset Y$ such that $X_ V \to V$ has geometrically connected fibres.

**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. $\square$

Lemma 37.28.5. Let $f : X \to Y$ be a morphism of schemes. Let $n_{X/Y}$ be the function on $Y$ counting the numbers of geometrically connected components of fibres of $f$ introduced in Lemma 37.28.3. Assume $f$ of finite type. Let $y \in Y$ be a point. Then there exists a nonempty open $V \subset \overline{\{ y\} }$ such that $n_{X/Y}|_ V$ is constant.

**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. $\square$

Lemma 37.28.6. Let $f : X \to Y$ be a morphism of schemes. Let $n_{X/Y}$ be the function on $Y$ counting the numbers of geometric connected components of fibres of $f$ introduced in Lemma 37.28.3. Assume $f$ of finite presentation. Then the level sets

of $n_{X/Y}$ are locally constructible in $Y$.

**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. $\square$

Lemma 37.28.7. Let $f : X \to S$ be a morphism of schemes. Assume that

$S$ is the spectrum of a discrete valuation ring,

$f$ is flat,

$X$ is connected,

the closed fibre $X_ s$ is reduced.

Then the generic fibre $X_\eta $ is connected.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #4059 by Name* on

Comment #4140 by Johan on