Lemma 37.45.2. Let $h : Y \to S$ be a morphism of schemes. Let $s \in S$ be a point. Let $T \subset Y_ s$ be an open subscheme. Assume

$h$ is flat and of finite presentation,

all fibres of $h$ are geometrically reduced, and

$T$ is geometrically connected over $\kappa (s)$.

Then we can find an affine elementary étale neighbourhood $(S', s') \to (S, s)$ and a quasi-compact open $V \subset Y_{S'}$ such that

all fibres of $V \to S'$ are geometrically connected,

$V_{s'} = T \times _ s s'$.

**Proof.**
The problem is clearly local on $S$, hence we may replace $S$ by an affine open neighbourhood of $s$. The topology on $Y_ s$ is induced from the topology on $Y$, see Schemes, Lemma 26.18.5. Hence we can find a quasi-compact open $V \subset Y$ such that $V_ s = T$. The restriction of $h$ to $V$ is quasi-compact (as $S$ affine and $V$ quasi-compact), quasi-separated, locally of finite presentation, and flat hence flat of finite presentation. Thus after replacing $Y$ by $V$ we may assume, in addition to (1) and (2) that $Y_ s = T$ and $S$ affine.

Pick a closed point $y \in Y_ s$ such that $h$ is Cohen-Macaulay at $y$, see Lemma 37.21.7. By Lemma 37.22.4 there exists a diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & Y \ar[d] \\ & S } \]

such that $Z \to S$ is flat, locally of finite presentation, locally quasi-finite with $Z_ s = \{ y\} $. Apply Lemma 37.40.1 to find an elementary neighbourhood $(S', s') \to (S, s)$ and an open $Z' \subset Z_{S'} = S' \times _ S Z$ with $Z' \to S'$ finite with a unique point $z' \in Z'$ lying over $s$. Note that $Z' \to S'$ is also locally of finite presentation and flat (as an open of the base change of $Z \to S$), hence $Z' \to S'$ is finite locally free, see Morphisms, Lemma 29.48.2. Note that $Y_{S'} \to S'$ is flat and of finite presentation with geometrically reduced fibres as a base change of $h$. Also $Y_{s'} = Y_ s$ is geometrically connected. Apply Lemma 37.45.1 to $Z' \to Y_{S'}$ over $S'$ to get $V \subset Y_{S'}$ quasi-compact open satisfying (2) whose fibres over $S'$ are either empty or geometrically connected. As $V \to S'$ is open (Morphisms, Lemma 29.25.10), after replacing $S'$ by an affine open neighbourhood of $s'$ we may assume $V \to S'$ is surjective, whence (1) holds.
$\square$

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