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:
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 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
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 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 $(f')^{-1}(U') = U$). $\square$
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