Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V$ be an irreducible variety of dimension $k$ living in $\mathbb{F}^n$ for some algebraically closed field $\mathbb{F}$. Let $\pi: \mathbb{F}^n \to \mathbb{F}^m$ be projection onto the first $m$ coordinates. Suppose the dimension of the Zariski closure of $\pi(V)$ is equal to $j$. Is it true that for every $x \in \pi(V)$, the fiber $\pi^{-1}(x) \cap V$ has dimension $k-j$? If not, what is a counterexample?

share|cite|improve this question
This is not an appropriate question for MO. I'm getting more and more concerned with the number of such questions arising in here. – Anton Fonarev Apr 14 '12 at 23:51
No. You can use the (affine) blow up $V=\{xz=y\}$ projecting to the $xy$-plane. Compare the fibre over the origin with the other fibres. – Donu Arapura Apr 14 '12 at 23:53
Thanks! Even though the dimension of the fiber is not $k-j$ over all points, is there some sort of statement along the lines of saying that "most" of the fibers have dimension $k-j$? For example, in Donu's example, the dimension is large only for one point. – Alan Guo Apr 15 '12 at 2:01
Yes, there are such statements for flat maps, and a map between varieties is generically flat. This can be found in Hartshorne or Matsumura. (In the latter, see Section 15.) – Karl Schwede Apr 15 '12 at 3:33

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.