Theorem 29.55.1 (Algebraic version of Zariski's Main Theorem). Let $f : Y \to X$ be an affine morphism of schemes. Assume $f$ is of finite type. Let $X'$ be the normalization of $X$ in $Y$. Picture:
Then there exists an open subscheme $U' \subset X'$ such that
$(f')^{-1}(U') \to U'$ is an isomorphism, and
$(f')^{-1}(U') \subset Y$ is the set of points at which $f$ is quasi-finite.
Proof. There is an immediate reduction to the case where $X$ and hence $Y$ are affine. Say $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $X' = \mathop{\mathrm{Spec}}(A')$, where $A'$ is the integral closure of $R$ in $A$, see Definitions 29.53.2 and 29.53.3. By Algebra, Theorem 10.123.12 for every $y \in Y$ at which $f$ is quasi-finite, there exists an open $U'_ y \subset X'$ such that $(f')^{-1}(U'_ y) \to U'_ y$ is an isomorphism. Set $U' = \bigcup U'_ y$ where $y \in Y$ ranges over all points where $f$ is quasi-finite. It remains to show that $f$ is quasi-finite at all points of $(f')^{-1}(U')$. If $y \in (f')^{-1}(U')$ with image $x \in X$, then we see that $Y_ x \to X'_ x$ is an isomorphism in a neighbourhood of $y$. Hence there is no point of $Y_ x$ which specializes to $y$, since this is true for $f'(y)$ in $X'_ x$, see Lemma 29.44.8. By Lemma 29.20.6 part (3) this implies $f$ is quasi-finite at $y$. $\square$
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