## 66.52 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 75.34.

Lemma 66.52.1. 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

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

$(f')^{-1}(U') \subset X$ 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 66.48.4 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

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

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

By Lemma 66.34.7 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 66.12.1 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 66.52.2. 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 66.51.1 the morphism $f$ is representable. Hence we may apply Lemma 66.52.1. 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 66.8.9). 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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