The Stacks project

67.52 Zariski's Main Theorem (representable case)

This is the version you can prove using that normalization commutes with étale localization. Before we can prove more powerful versions (for non-representable morphisms) we need to develop more tools. See More on Morphisms of Spaces, Section 76.34.

Lemma 67.52.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is representable, of finite type, and separated. Let $Y'$ be the normalization of $Y$ in $X$. Picture:

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & Y' \ar[ld]^\nu \\ & Y & } \]

Then there exists an open subspace $U' \subset Y'$ such that

  1. $(f')^{-1}(U') \to U'$ is an isomorphism, and

  2. $(f')^{-1}(U') \subset X$ is the set of points at which $f$ is quasi-finite.

Proof. Let $W \to Y$ be a surjective étale morphism where $W$ is a scheme. Then $W \times _ Y X$ is a scheme as well. By Lemma 67.48.4 the algebraic space $W \times _ Y Y'$ is representable and is the normalization of the scheme $W$ in the scheme $W \times _ Y X$. Picture

\[ \xymatrix{ W \times _ Y X \ar[rd]_{(1, f)} \ar[rr]_{(1, f')} & & W \times _ Y Y' \ar[ld]^{(1, \nu )} \\ & W & } \]

By More on Morphisms, Lemma 37.43.1 the result of the lemma holds over $W$. Let $V' \subset W \times _ Y Y'$ be the open subscheme such that

  1. $(1, f')^{-1}(V') \to V'$ is an isomorphism, and

  2. $(1, f')^{-1}(V') \subset W \times _ Y X$ is the set of points at which $(1, f)$ is quasi-finite.

By Lemma 67.34.7 there is a maximal open set of points $U \subset X$ where $f$ is quasi-finite and $W \times _ Y U = (1, f')^{-1}(V')$. The morphism $f'|_ U : U \to Y'$ is an open immersion by Lemma 67.12.1 as its base change to $W$ is the isomorphism $(1, f')^{-1}(V') \to V'$ followed by the open immersion $V' \to W \times _ Y Y'$. Setting $U' = \mathop{\mathrm{Im}}(U \to Y')$ finishes the proof (omitted: the verification that $(f')^{-1}(U') = U$). $\square$

In the following lemma we can drop the assumption of being representable as we've shown that a locally quasi-finite separated morphism is representable.

Lemma 67.52.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-finite and separated. Let $Y'$ be the normalization of $Y$ in $X$. Picture:

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & Y' \ar[ld]^\nu \\ & Y & } \]

Then $f'$ is a quasi-compact open immersion and $\nu $ is integral. In particular $f$ is quasi-affine.

Proof. By Lemma 67.51.1 the morphism $f$ is representable. Hence we may apply Lemma 67.52.1. Thus 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 (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. $\square$


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