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 $kj$? If not, what is a counterexample?
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$\begingroup$ This is not an appropriate question for MO. I'm getting more and more concerned with the number of such questions arising in here. $\endgroup$– Anton FonarevApr 14 '12 at 23:51

5$\begingroup$ 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. $\endgroup$– Donu ArapuraApr 14 '12 at 23:53

1$\begingroup$ Thanks! Even though the dimension of the fiber is not $kj$ over all points, is there some sort of statement along the lines of saying that "most" of the fibers have dimension $kj$? For example, in Donu's example, the dimension is large only for one point. $\endgroup$– Alan GuoApr 15 '12 at 2:01

3$\begingroup$ 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.) $\endgroup$– Karl SchwedeApr 15 '12 at 3:33
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