Whilst doing exercise $11.4.B$ of Ravi Vakil's "Foundations of Algebraic Geometry", I got stuck with the following problem (although I think that many of the hypotheses are unnecessary and a more general statement can be proved by reducing to something like this case):
Let $X=\rm{Spec}(A)$ and $Y=\rm{Spec}(B)$ be irreducible affine $k-$varieties of dimension $m,n$ respectively and $\pi:X\rightarrow Y$ be a dominant morphism. Then for any $q \in Y$, any irreducible component of the fibre $\pi^{-1}(q)$ has dimension at least $m-n$.
I know that for irreducible varieties codimension is the difference of dimensions, and that for $p \in X$ with $\pi(p) = q$, $\rm{codim}_Xp\leq \rm{codim}_Yq + \rm{codim}_{\pi^{-1}(q)}p$. If we take $p$ corresponding to the generic point of an irreducible component of $\pi^{-1}(q)$ then, putting these together gives that $m-n \leq \rm{dim}\bar{\{p\}}-\rm{dim}\bar{\{q\}}$ where the closures are taken in $X$ and $Y$ respectively. I'm having trouble showing that the right hand side is a lower bound for the dimension of the closure of $p$ in $\pi^{-1}(q)$, which is the dimension of the irreducible component that we want a lower bound for.
Algebraically this amounts to proving: given a prime $\mathfrak{q}$ of $B$ and a prime $\mathfrak{p}$ of $A$ lying over $\mathfrak{q}$, there is a chain of primes going up from $\mathfrak{p}$ that all lie over $\mathfrak{q}$, of length at least $\rm{dim}(A/\mathfrak{p})-\rm{dim}(B/\mathfrak{q})$. This is easy to prove if $\mathfrak{q}$ is maximal (since then any maximal length chain of primes over $\mathfrak{p}$ will work), otherwise this difference just gives a lower bound for the number of times two consecutive primes in any maximal chain over $\mathfrak{p}$ pull back to the same prime in $B$.
Edit: A more general inequality than the one I gave in the second paragraph gives that for any $p$ mapping to $q$, $m-n \leq \rm{dim}\bar{p}-\rm{dim}\bar{q}+\rm{codim}_{\pi^{-1}(q)}p$. In particular, as suggested by Hoot in the comments, given a particular irreducible component of the fibre, one can take $p$ to correspond to a closed point lying in that component and no other, so that it's codimension is the dimension of that component. It then remains to show that $\rm{dim}\bar{\{p\}}-\rm{dim}\bar{\{q\}} \leq 0$.
As in the question, it suffices to prove that for any $q \in Y$ and $p\in X$ corresponding to an irreducible component of the fibre, that $\rm{dim}_X\bar{\{p\}}-\rm{dim}_Y\bar{\{q\}} = \rm{dim}_{\pi^{-1}(q)}\bar{\{p\}}$. This is in fact true for all $p$ in the fibre:
By construction $\pi^{-1}(q)$ is a finite type $\kappa_q$-scheme, and so we can compute the dimension of irreducible components by computing the transcendence degree of the residue field at the generic points over $\kappa_q$. But for any $p \in \pi^{-1}(q)$, we have that the map of local rings $\mathcal{O}_{X,p}\rightarrow \mathcal{O}_{\pi^{-1}(q),p}$ is surjective, and so the residue field of $p$ in $\pi^{-1}(q)$ is isomorphic to the residue field in $X$. Thus we have that $\rm{dim}_{\pi^{-1}(q)}\bar{{\{p\}}} = \rm{tr.deg}_{\kappa_q}(\kappa_p)=\rm{tr.deg}_{k}(\kappa_p)-\rm{tr.deg}_{k}(\kappa_q)=\rm{dim}_X\bar{\{p\}}-\rm{dim}_Y\bar{\{q\}}$ as desired.