The Stacks project

Proof. In this proof we will use that flat, finite presentation, finite locally free are properties that are preserved under base change and composition. We will also use that a finite locally free morphism is both open and closed. You can find these facts as Morphisms, Lemmas 29.25.8, 29.21.4, 29.48.4, 29.25.6, 29.21.3, 29.48.3, 29.25.10, and 29.44.11.

Note that $X_ Z \to Z$ is flat morphism of finite presentation which has a section $s$ coming from $\sigma $. Let $X_ Z^0$ denote the subset of $X_ Z$ defined in Situation 37.29.1. By Lemma 37.29.6 it is an open subset of $X_ Z$.

The pullback $X_{Z \times _ Y Z}$ of $X$ to $Z \times _ Y Z$ comes equipped with two sections $s_0, s_1$, namely the base changes of $s$ by $\text{pr}_0, \text{pr}_1 : Z \times _ Y Z \to Z$. The construction of Situation 37.29.1 gives two subsets $(X_{Z \times _ Y Z})_{s_0}^0$ and $(X_{Z \times _ Y Z})_{s_1}^0$. By Lemma 37.29.2 these are the inverse images of $X_ Z^0$ under the morphisms $1_ X \times \text{pr}_0, 1_ X \times \text{pr}_1 : X_{Z \times _ Y Z} \to X_ Z$. In particular these subsets are open.

Let $(Z \times _ Y Z)_ y = \{ z_1, \ldots , z_ n\} $. As $X_ y$ is geometrically connected, we see that the fibres of $(X_{Z \times _ Y Z})_{s_0}^0$ and $(X_{Z \times _ Y Z})_{s_1}^0$ over each $z_ i$ agree (being equal to the whole fibre). Another way to say this is that

Since the sets $(X_{Z \times _ Y Z})_{s_0}^0$ and $(X_{Z \times _ Y Z})_{s_1}^0$ are open in $X_{Z \times _ Y Z}$ there exists an open neighbourhood $W \subset Z \times _ Y Z$ of $(Z \times _ Y Z)_ y$ such that

Then it follows directly from the construction in Situation 37.29.1 that

where $p : X_{Z \times _ Y Z} \to Z \times _ W Z$ is the projection. Because $Z \times _ Y Z \to Y$ is finite locally free, hence open and closed, there exists an affine open neighbourhood $V \subset Y$ of $y$ such that $q^{-1}(V) \subset W$, where $q : Z \times _ Y Z \to Y$ is the structure morphism. To prove the lemma we may replace $Y$ by $V$. After we do this we see that $X_ Z^0 \subset Y_ Z$ is an open such that

This means that the image $X^0 \subset X$ of $X_ Z^0$ is an open such that $(X_ Z \to X)^{-1}(X^0) = X_ Z^0$, see Descent, Lemma 35.13.6. Finally, $X^0$ is quasi-compact because $X_ Z^0$ is quasi-compact by Lemma 37.29.4 (use that at this point $Y$ is affine, hence $X$ is quasi-compact and quasi-separated, hence locally constructible is the same as constructible and in particular quasi-compact; details omitted). In this way we see that $X^0$ has all the desired properties. $\square$

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 37.22.7. By Lemma 37.23.4 there exists a diagram

such that $Z \to S$ is flat, locally of finite presentation, locally quasi-finite with $Z_ s = \{ y\} $. Apply Lemma 37.41.1 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 37.46.1 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. $\square$

Proof. Pick a point $x \in X$ with image $s \in S$. We will produce a diagram

and points $s' \in S'$, $x' \in X'$, $y \in S' \times _ S X$ such that $x'$ maps to $x$, $(S', s') \to (S, s)$ is an étale neighbourhood, $(X', x') \to (S' \times _ S X, y)$ is an étale neighbourhood¹, and $X' \to S'$ has geometrically connected fibres. If we can do this for every $x \in X$, then the lemma follows (with members of the covering given by the collection of étale morphisms $X' \to X$ so produced). The first step is the replace $X$ and $S$ by affine open neighbourhoods of $x$ and $s$ which reduces us to the case that $X$ and $S$ are affine (and hence $f$ of finite presentation).

Choose a separable algebraic extension $\overline{k}$ of $\kappa (s)$. Denote $X_{\overline{k}}$ the base change of $X_ s$. Choose a point $\overline{x}$ in $X_{\overline{k}}$ mapping to $x \in X_ s$. Choose a connected quasi-compact open neighbourhood $\overline{V} \subset X_{\overline{k}}$ of $\overline{x}$. (This is possible because any scheme locally of finite type over a field is locally connected as a locally Noetherian topological space.) By Varieties, Lemma 33.7.9 we can find a finite separable extension $k'/\kappa (s)$ and a quasi-compact open $V' \subset X_{k'}$ whose base change is $\overline{V}$. In particular $V'$ is geometrically connected over $k'$, see Varieties, Lemma 33.7.7. By Lemma 37.35.2 we can find an étale neighbourhood $(S', s') \to (S, s)$ such that $\kappa (s')$ is isomorphic to $k'$ as an extension of $\kappa (s)$. Denote $x' \in (S' \times _ S X)_{s'} = X_{k'}$ the image of $\overline{x}$. Thus after replacing $(S, s)$ by $(S', s')$ and $(X, x)$ by $(S' \times _ S X, x')$ we reduce to the case handled in the next paragrah.

Assume there is a quasi-compact open $V \subset X_ s$ which contains $x$ and is geometrically irreducible. Then we can apply Lemma 37.46.2 to find an affine étale neighbourhood $(S', s') \to (S, s)$ and a quasi-compact open $X' \subset S' \times _ S X$ such that $X' \to S'$ has geometrically connected fibres and such that $X'$ contains a point mapping to $x$. This finishes the proof. $\square$

Proof. Apply Lemma 37.46.2 to 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 and $V_{s'} = T \times _ s s'$. As $V$ is an open of the base change of $h$ all fibres of $V \to S'$ are geometrically normal, see Lemma 37.20.2. In particular, they are geometrically reduced. To finish the proof we have to show they are geometrically irreducible. But, if $t \in S'$ then $V_ t$ is of finite type over $\kappa (t)$ and hence $V_ t \times _{\kappa (t)} \overline{\kappa (t)}$ is of finite type over $\overline{\kappa (t)}$ hence Noetherian. By choice of $S' \to S$ the scheme $V_ t \times _{\kappa (t)} \overline{\kappa (t)}$ is connected. Hence $V_ t \times _{\kappa (t)} \overline{\kappa (t)}$ is irreducible by Properties, Lemma 28.7.6 and we win. $\square$