Suppose we're given a dominant morphism of irreducible algebraic varieties over an algebraically closed field $k$, $f:X \rightarrow Y$, then why do there exist subvarieties $Z$ of $X$ with $\dim Z=\dim Y$ and $f(Z)$ dense in $Y$? One idea I've heard is to resrict to the affine case, so $X=\text{Spec }A$ and $Y=\text{Spec }B$ and then consider $A\otimes \overline{K(B)}$, where $\overline{K(B)}$ is the algebraic closure of the function field of $B$. One can obtain sections of the inclusion of $\overline{K(B)}$ into $A\otimes \overline{K(B)}$ by considering $\overline{K(B)}$rational points in $X_{\overline{K(B)}}$ (this follows from the Nullstellensatz), which really lie in some finite extension of $L$ of $K(B)$. So we consider the integral closure $C$ of $B$ in $L$ and this gives us a finite morphism from $Y'=\text{Spec }C$ to $Y$, but I don't know how this eventually gives us an answer to the original question. If anyone knows how to continue this argument, or a nicer argument on how to get the result, please let me know.Thanks
2 Answers
More generally, let $f:X\to Y$ be a dominant morphism of finite type of schemes, with $Y$ irreducible. Let $y$ be the generic point of $Y$. Then the fiber $X_y$ is a nonempty scheme of finite type over $\kappa(y)$. Let $z\in X_y$ be a closed point. By Hilbert's Nullstellensatz, $\kappa(z)$ is a finite extension of $\kappa(y)$. Let $Z$ be the Zariski closure of $\{z\}$: this is an irreducible closed subset of $X$, which dominates $Y$ (the image contains the generic point) and the function field extension $\kappa(z)/\kappa(y)$ is finite, which implies $\dim Z=\dim Y$ if $Y$ is of finite type over a field.
This argument is of course close to the original idea. Sándor's proof is also OK but uses linear systems, hence works well only under quasiprojective assumptions (or after reducing to this case, e.g. by taking affine open sets).
Assume that $\dim X>\dim Y$, i.e., that the relative dimension of $f$ is positive. Let $H\subseteq X$ be a general member of a base point free linear system. Then $H$ does not contain any irreducible component of the general fiber of $f$ and hence the relative dimension of the restriction morphism $f_H:H\to Y$ is strictly less than the relative dimension of $X$. Since $\dim H=\dim X1$, this can only happen if the relative dimension of $f_H$ is exactly $1$ less than the relative dimension of $f$ and $f_H$ is dominant. Replace $X$ with $H$ and repeat as long as the initial assumption ($\dim X>\dim Y$) is satisfied. The process will stop when you reach an appropriate $Z$.

$\begingroup$ Re: Laurent's comment: I assumed, as stated in the question, that we have already restricted to the affine case. I probably should have said so. :) $\endgroup$ Jun 1, 2011 at 7:12