The Stacks project

Proof. By Morphisms of Spaces, Lemma 67.34.7 there is an open subspace $U \subset X$ corresponding to the points of $|X|$ where $f$ is quasi-finite. We have to prove

1. the image of $|U| \to |Y'|$ is $|U'|$ for some open subspace $U'$ of $Y'$,

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

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

Since formation of $U$ commutes with arbitrary base change (Morphisms of Spaces, Lemma 67.34.7), since formation of the normalization $Y'$ commutes with smooth base change (Lemma 76.25.2), since étale morphisms are open, and since “being an isomorphism” is fpqc local on the base (Descent on Spaces, Lemma 74.11.15), it suffices to prove (a), (b), (c) étale locally on $Y$ (some details omitted). Thus we may assume $Y$ is an affine scheme. This implies that $Y'$ is an (affine) scheme as well.

Let $x \in |U|$. Claim: there exists an open neighbourhood $f'(x) \in V \subset Y'$ 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 a scheme (as an open of $Y'$), locally of finite type over $Y$ (as an open subspace of $X$), and for $v \in V$ the residue field extension $\kappa (v)/\kappa (\nu (v))$ is algebraic (as $V \subset Y'$ and $Y'$ is integral over $Y$). Hence the fibres of $V \to Y$ are discrete (Morphisms, Lemma 29.20.2) and $(f')^{-1}V \to Y$ 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 $y = f(x) \in |Y|$. Let $(T, t) \to (Y, y)$ be an étale morphism of pointed schemes. Denote by a subscript ${}_ T$ the base change to $T$. Let $z \in X_ T$ be a 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 of Spaces, Lemma 67.34.7. Note that

is the normalization of $T$ in $X_ T$, see Lemma 76.25.2. Suppose that the claim holds for $z \in U_ T \subset X_ T \to Y'_ T \to T$, i.e., suppose that we can find an open neighbourhood $f'_ T(z) \in V' \subset Y'_ T$ such that $(f'_ T)^{-1}V' \to V'$ is an isomorphism. The morphism $Y'_ T \to Y'$ is étale hence the image $V \subset Y'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_ T(z) \in V'$. Observe that

is a fibre square (as $Y'_ T \times _{Y'} X = X_ T$). 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 on Spaces, Lemma 74.11.15. In other words, the claim holds for $x \in U \subset X \to Y' \to Y$.

By the result of the previous paragraph to prove the claim for $x \in |U|$, we may replace $Y$ by an étale neighbourhood $T$ of $y = f(x)$ and $x$ by any point lying over $x$ in $T \times _ Y X$. Thus we may assume there is a decomposition

into open and closed subspaces where $V \to Y$ is finite and $x \in V$, see Lemma 76.33.1. Since $X$ is a disjoint union of $V$ and $W$ over $Y$ and since $V \to Y$ is finite we see that the normalization of $Y$ in $X$ is the morphism

where $W'$ is the normalization of $Y$ in $W$, see Morphisms of Spaces, Lemmas 67.48.8, 67.45.6, and 67.48.10. The claim follows and we win. $\square$

Proof. This follows from Lemma 76.34.1. Namely, by that lemma there exists an open subspace $U' \subset Y'$ 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 $\nu : Y' \to Y$ is separated (Morphisms of Spaces, Lemma 67.8.9). Hence for every affine scheme $Z$ and morphism $Z \to Y$ the fibre product $Z \times _ Y X$ is a quasi-compact open subscheme of the affine scheme $Z \times _ Y Y'$. Hence $f$ is quasi-affine by definition. $\square$

Proof. Let $X \to Y' \to Y$ be as in the conclusion of Lemma 76.34.2. By Limits of Spaces, Lemma 70.9.7 we can write $\nu _*\mathcal{O}_{Y'} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{A}_ i$ as a directed colimit of finite quasi-coherent $\mathcal{O}_ X$-algebras $\mathcal{A}_ i \subset \nu _*\mathcal{O}_{Y'}$. Then $\pi _ i : T_ i = \underline{\mathop{\mathrm{Spec}}}_ Y(\mathcal{A}_ i) \to Y$ is a finite morphism for each $i$. Note that the transition morphisms $T_{i'} \to T_ i$ are affine and that $Y' = \mathop{\mathrm{lim}}\nolimits T_ i$.

By Limits of Spaces, Lemma 70.5.7 there exists an $i$ and a quasi-compact open $U_ i \subset T_ i$ whose inverse image in $Y'$ 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 of Spaces, Lemma 70.4.1. By Limits of Spaces, Lemma 70.5.12 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 of Spaces, Lemma 67.12.6 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 of Spaces, Lemma 70.11.3 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 of Spaces, Lemma 70.5.7 there exists an $i$ and an open subscheme $U_ i \subset T_ i$ whose inverse image in $T$ is $X$. By Limits of Spaces, Lemma 70.5.12 we see that $X \cong U_ i$ for large enough $i$. Replacing $T$ by $T_ i$ finishes the proof. $\square$