Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume $\phi:X\rightarrow Y$ is a dominating regular map of affine varieties. Mumford claim that for all $y\in Y$, the dimension of components of $\phi^{-1}(y)$ is at least $r-s$. His proof used the following fact:

Choose the smallest $k$ such that there exist a regular map $$\psi:X\rightarrow \mathbb{C}^{k}$$ for which $\{x\}$ is a component of $\psi^{-1}(0)$. Let $U$ be a Zariski open set containing $x$ such that $\psi^{-1}(0)\cap U=\{x\}$. In this case, noticed $\psi$ is dominating. In fact, if $Z=\overline{\psi(X)}$ had dimension $l<k$, choose $l$ polynomial functions $h_{1},...h_{l}$ on $Z$ such that $(0)$ is the component of $Z\cap V(h_{1},...,h_{l})$ and replace $\psi$ by $H\circ \psi$. Here $$H:\mathbb{C}^{k}\rightarrow \mathbb{C}^{l}$$ being defined by $(h_{1},...,h_{l})$. Thus $\psi$ is dominating and if $Y_{1},...,Y_{k}$ are coordinates in $\mathbb{C}^{k}$, $\psi^{*}Y_{1},...\psi^{*}Y_{k}$ are independent transcendentals in $\mathbb{C}(X)$.

I cannot follow the underlined part. From Algebraic Geometry, I, page 45.

share|improve this question
What isbetween the two sentences you boldened is the proof of the fact that the map psi is dominant. Where exactly do you have a problem? –  Mariano Suárez-Alvarez Nov 17 '12 at 6:16
I do not see this as a proof. The $H\circ \psi$ may not be dominant. –  Bombyx mori Nov 17 '12 at 6:22
I see. You are implying because the choice of $h_{i}$, the map $H\circ \psi$'s image must have dimension at least $l$, which implies its closure is $\mathbb{C}^{l}$. –  Bombyx mori Nov 17 '12 at 6:29
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.