# Subvarieties that dominate a dominant morphism

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

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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.
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 $\dim H=\dim X-1$, 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 $Z$.