The Stacks project

Proof. By Morphisms, Lemma 29.56.2 the subset $U \subset X$ of points where $f$ is quasi-finite is open. The lemma is equivalent to

1. $U' = f'(U) \subset S'$ is open,

2. $U = (f')^{-1}(U')$, and

3. $U \to U'$ is an isomorphism.

Let $x \in U$ be arbitrary. We claim there exists an open neighbourhood $f'(x) \in V \subset S'$ such that $(f')^{-1}V \to V$ is an isomorphism. We first prove the claim implies the lemma. Namely, then $(f')^{-1}V \cong V$ is both locally of finite type over $S$ (as an open subscheme of $X$) and for $v \in V$ the residue field extension $\kappa (v)/\kappa (\nu (v))$ is algebraic (as $V \subset S'$ and $S'$ is integral over $S$). Hence the fibres of $V \to S$ are discrete (Morphisms, Lemma 29.20.2) and $(f')^{-1}V \to S$ is locally quasi-finite (Morphisms, Lemma 29.20.8). This implies $(f')^{-1}V \subset U$ and $V \subset U'$. Since $x$ was arbitrary we see that (a), (b), and (c) are true.

Let $s = f(x)$. Let $(T, t) \to (S, s)$ be an elementary étale neighbourhood. Denote by a subscript ${}_ T$ the base change to $T$. Let $y = (x, t) \in X_ T$ be the unique point in the fibre $X_ t$ lying over $x$. Note that $U_ T \subset X_ T$ is the set of points where $f_ T$ is quasi-finite, see Morphisms, Lemma 29.20.13. Note that

is the normalization of $T$ in $X_ T$, see Lemma 37.19.2. Suppose that the claim holds for $y \in U_ T \subset X_ T \to S'_ T \to T$, i.e., suppose that we can find an open neighbourhood $f'_ T(y) \in V' \subset S'_ T$ such that $(f'_ T)^{-1}V' \to V'$ is an isomorphism. The morphism $S'_ T \to S'$ is étale hence the image $V \subset S'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_ T(y) \in V'$. Observe that

is a fibre square (as $S'_ T \times _{S'} X = X_ T$). Since the left vertical arrow is an isomorphism and $\{ V' \to V\} $ is a étale covering, we conclude that the right vertical arrow is an isomorphism by Descent, Lemma 35.23.17. In other words, the claim holds for $x \in U \subset X \to S' \to S$.

By the result of the previous paragraph we may replace $S$ by an elementary étale neighbourhood of $s = f(x)$ in order to prove the claim. Thus we may assume there is a decomposition

into open and closed subschemes where $V \to S$ is finite and $x \in V$, see Lemma 37.41.4. Since $X$ is a disjoint union of $V$ and $W$ over $S$ and since $V \to S$ is finite we see that the normalization of $S$ in $X$ is the morphism

where $W'$ is the normalization of $S$ in $W$, see Morphisms, Lemmas 29.53.10, 29.44.4, and 29.53.12. The claim follows and we win. $\square$

Proof. This follows from Lemma 37.43.1. Namely, by that lemma there exists an open subscheme $U' \subset S'$ such that $(f')^{-1}(U') = X$ and $X \to U'$ is an isomorphism. In other words, $f'$ is an open immersion. Note that $f'$ is quasi-compact as $f$ is quasi-compact and $\nu : S' \to S$ is separated (Schemes, Lemma 26.21.14). It follows that $f$ is quasi-affine by Morphisms, Lemma 29.13.3. $\square$

Proof. Let $X \to S' \to S$ be as in the conclusion of Lemma 37.43.2. By Properties, Lemma 28.22.13 we can write $\nu _*\mathcal{O}_{S'} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{A}_ i$ as a directed colimit of finite quasi-coherent $\mathcal{O}_ X$-algebras $\mathcal{A}_ i \subset \nu _*\mathcal{O}_{S'}$. Then $\pi _ i : T_ i = \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}_ i) \to S$ is a finite morphism for each $i$. Note that the transition morphisms $T_{i'} \to T_ i$ are affine and that $S' = \mathop{\mathrm{lim}}\nolimits T_ i$.

By Limits, Lemma 32.4.11 there exists an $i$ and a quasi-compact open $U_ i \subset T_ i$ whose inverse image in $S'$ equals $f'(X)$. For $i' \geq i$ let $U_{i'}$ be the inverse image of $U_ i$ in $T_{i'}$. Then $X \cong f'(X) = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i'}$, see Limits, Lemma 32.2.2. By Limits, Lemma 32.4.16 we see that $X \to U_{i'}$ is a closed immersion for some $i' \geq i$. (In fact $X \cong U_{i'}$ for sufficiently large $i'$ but we don't need this.) Hence $X \to T_{i'}$ is an immersion. By Morphisms, Lemma 29.3.2 we can factor this as $X \to T \to T_{i'}$ where the first arrow is an open immersion and the second a closed immersion. Thus we win. $\square$

Proof. By Limits, Lemma 32.9.8 we can write $T = \mathop{\mathrm{lim}}\nolimits T_ i$ where all $T_ i$ are finite and of finite presentation over $Y$ and the transition morphisms $T_{i'} \to T_ i$ are closed immersions. By Limits, Lemma 32.4.11 there exists an $i$ and an open subscheme $U_ i \subset T_ i$ whose inverse image in $T$ is $X$. By Limits, Lemma 32.4.16 we see that $X \cong U_ i$ for large enough $i$. Replacing $T$ by $T_ i$ finishes the proof. $\square$