Theorem 29.56.1 (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:
\xymatrix{ Y \ar[rd]_ f \ar[rr]_{f'} & & X' \ar[ld]^\nu \\ & X & }
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
Comments (0)
There are also: