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I want to prove (or to find a reference to) the following statement:

Statement:

Let $Z$ be an irreducible locally closed set (Zariski topology) of $\mathbb C^n$ and $\pi$ be a projection on the first $l$ coordinates (the values of $l$ is not important). Assume that there exists $d$ such that $\dim\pi^{-1}(\pi(z))\geq d$ for all $z\in Z$. Then $$ \dim \pi(Z)\leq \dim Z - d.\ \ \ \ \ \ (1) $$

I found several versions of the fiber dimension theorem, but they cannot be applied directly to my case since $Z$ is always assumed to be a variety.

For example, Theorem 3.7 on p.78 in Perrin D., Algebraic Geometry.. an introduction, Springer 2008 says (I omit statement 1))

Theorem 3.7

Let $\phi:\ X\rightarrow Y$ be a dominant morphism of irreducible algebraic varieties. Then there exists a non-empty open set $U\subset Y$ such that

a) $U\subset \phi(X)$

b) $\forall y\in U$, $\dim \phi^{-1}(y)=\dim X-\dim Y$.

Question: Is the following derivation of the statement above from Theorem 3.7 correct?

Since $Z$ is a locally closed set, there exist polynomials $p_1,\dots,p_k$ and $q_1,\dots,q_m$ in $n$ variables such that $$ Z=\{z\in\mathbb C^n:\ p_1(z)=\dots=p_k(z)=0\ \text{ and }\ q_1(z)\ne 0\ \text{ or }\ q_2(z)\ne 0\ \text{ or } \ \dots q_m(z)\ne 0\}. $$ Since $Z$ is irreducible, $m=1$, that is $$ Z=\{(z_1,\dots,z_n):\ p_1(z_1,\dots,z_n)=\dots=p_k(z_1,\dots,z_n)=0\ \text{ and }\ q(z_1,\dots,z_n)\ne 0\}. $$ We extend $Z$ to a Zariski closed subset $\widetilde{Z}\subset \mathbb C^{n+1}$: \begin{equation*} \begin{split} \widetilde{Z}=\{(z_1,\dots,z_n, z_{n+1}):\ &p_1(z_1,\dots,z_n)=\dots=p_k(z_1,\dots,z_n)=0\ \text{ and }\\ &z_{n+1}q(z_1,\dots,z_n)=1\}. \end{split} \end{equation*} It is clear $Z=\pi_{1,\dots,n}\widetilde{Z}$ (projection onto the first $n$ coordinates). On the other hand, $\widetilde{Z}$ is the image of $Z$ under the mapping $(z_1,\dots,z_n)\in Z\mapsto (z_1,\dots,z_n,\frac{1}{q(z_1,\dots,z_n)})\in\widehat{Z}$. Hence, $\dim Z=\dim\widetilde{Z}$. Thus, we can apply Theorem 3.7.

guest_09072015
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1 Answers1

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I recommend the following. Assume without loss of generality that $Z$ is irreducible, otherwise we can do this for a component of maximal dimension. There is an open subset $U\subseteq \mathbb C^n$ such that $Z$ is closed in $U$, by definition. There is some polynomial $f$ such that $D(f)=\{ x\in\mathbb C^n \mid f(x)\ne0 \}\subseteq U$ and $Z':=D(f)\cap Z\ne\emptyset$. Indeed, we choose $f\in I(\mathbb C^n\setminus U)\setminus I(Z)$ and this is possible because $Z\subseteq U$.

Then $Z'$ is a closed subset of $D(f)$ and an open subset of $Z$, in particular $Z'$ is dense in $Z$ and $\dim(Z')=\dim(Z)$ as well as $\dim(\pi(Z'))=\dim(\pi(Z))$, since $$\overline{\pi(Z)}=\overline{\pi(\overline{Z'})}=\overline{\pi(Z')}.$$ Since $D(f)$ is an affine variety though, the Theorem applies and you are done.

Jesko Hüttenhain
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  • Why $D(f)$ is an affine variety? $D(f)$ is an open set, isn't it? – guest_09072015 Jul 12 '15 at 19:53
  • It is open in $\mathbb C^n$, but it can be given the structure of an affine variety. Inside $\mathbb C^{n}\times\mathbb C$, it correponds to the points $(z,t)$ subject to the polynomial condition $f(z)\cdot t = 1$. – Jesko Hüttenhain Jul 12 '15 at 20:35
  • Thanks, I think I did something similar in my new solution (I have replaced the old solution by a new one). Is it true, that Theorem 3.7is valid if $X$ is an open subset (instead of irreducible algebraic variety)? – guest_09072015 Jul 12 '15 at 20:46
  • @guest_09072015: Yes, this is true, but simply because a nonempty, open subset of an irreducible algebraic variety is again an irreducible algebraic variety. It might not be affine, but it is a variety. – Jesko Hüttenhain Jul 13 '15 at 06:04
  • Thank you for your explanation. – guest_09072015 Jul 13 '15 at 16:13