# Finite maps to normal varieties have fibers with bounded number of points

Let $$f\colon X\rightarrow Y$$ be a dominant, finite, and proper map of normal varieties of degree $$d$$ over an algebraically closed field $$k$$. Let $$y\in Y$$ be any closed point.

Question. Is it true that $$\# f^{-1}(y)\le d$$? (counting points without multiplicity)

Normality of $$Y$$ is necessary as is shown by the resolution of the nodal cubic. Probably normality of $$X$$ is unnecessary. This is easy to show in the case of flat maps, but it seems strangely annoying to prove in the case of normal varieties. Ideally, it would be a straightforward application of Zariski's main theorem.

• Note that the result is also false if you count lengths; for example $k[x^2,xy,y^2] \subseteq k[x,y]$ has length $3$ at the origin. I feel that I've read about a positive answer somewhere for the cardinality question as stated, but I cannot reproduce it... Jul 17, 2020 at 1:02
• Have you consulted Shafarevich, Basic Algebraic Geometry 1, second edition, the section II.6.3 in chapter II on ramification, Thm. 3, page 143; or in the earlier one volume version, Grundlehren der Math . Wissenschaften, Band 213, Thm. 6, section II.5.3, page 116? He assumes only f finite and dominant, and Y normal. In Mumford's yellow book, p. 53, for the complex projective case, he assumes the map projective and quasi finite, and Y topologically unibranch. Jul 17, 2020 at 4:34
• Let $R$ be the complete local ring of $Y$ at $y$, and let $R \to S$ be the base change of $X \to Y$, so $S$ is the complete semilocal ring of $X$ at $f^{-1}(y)$. Then $R \to S$ is a finite morphism, so $S = \prod_{i=1}^r S_i$ is a product of finite local $R$-algebras with $r = \#f^{-1}(y)$. Each $S_i$ is flat over $X$, so contains a point over the generic point of $X$. As $Y$ is normal, $R$ is a domain whose unique generic point lives over that of $Y$. So each $S_i$ contains a point over the generic point of $R$, and thus $r \leq \mathrm{deg}(S \otimes K(R)/K(R)) = d$. Jul 17, 2020 at 13:39
• @Anonymous Hmmm... It seems that when you assert that each $S_i$ is flat over $X$ ($R$?) then you would know that $X\rightarrow Y$ is flat in a neighborhood of $y\in Y$. I am not interested in this case... Maybe I am missing something? Jul 17, 2020 at 15:22
• @roysmith Great! Whenever I look in Shafarevich I am always satisfied. If you post this as an answer I will accept it. Jul 17, 2020 at 15:23