Lemma 37.27.4. Let $f : X \to Y$ be a morphism of schemes. Assume

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

2. $X_\eta$ is geometrically connected, and

3. $f$ 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

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_ V \ar[r] \ar[d] & X \ar[d]^ f \\ Y' \ar[r]^ g & V \ar[r] & Y }$

as in Lemma 37.23.8. Note that the generic fibre of $f'$ is geometrically connected (for example by Lemma 37.27.3). 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 37.27.3. 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 37.27.3 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 37.23.5. Let $Z_{i, j} = X_ i \cap X_ j$. Let

$\{ 1, \ldots , n\} \times \{ 1, \ldots , n\} = I \amalg J$

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 37.23.1. After shrinking $Y$ we obtain that $X_{i, y}$ is geometrically irreducible for each $i$ and all $y \in Y$, see Lemma 37.26.5. 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

$V = \bigcap \nolimits _{(i, j) \in J} f(Z_{i, j})$

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. $\square$

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