Is Krull dimension non-increasing along ring epimorphisms? 
Let $f \colon R \to S$ be an epimorphism of commutative rings, where $R$ and $S$ are integral domains. Suppose that $\mathfrak{p} \subset S$ is a prime such that $f^{-1}(\mathfrak{p}) = 0$.  Does it follow that $\mathfrak{p} = 0$?

If answer is "yes", then it follows that for any epimorphism of commutative rings $R \to S$, strictly increasing chains of prime ideals in $S$ lift to strictly increasing chains in $R$; hence, Krull dimension is non-increasing along ring epimorphisms.
All of the above hold for quotients and localizations, which are the only examples of ring epimorphisms that come readily to my mind.
 A: Let's make several reductions.  First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $S$ at the zero ideal of $R$; it is easy to check that this is an epimorphism via the universal property of localization.
But as George Lowther points out, an epimorphism from a field to an integral domain must be surjective.  To see this, assume to the contrary that $K\to S'$ is not surjective; thus $K'=\operatorname{Frac}(S')$ is not equal to $K$.  But $S'\to K'$ is an epimorphism, so composing, we've reduced to the case of a map between two fields $f: K\to K'$.  But $K'$ admits many embeddings into (say) its algebraic closure which agree on $K$, contradicting that $f$ was an epimorphism.
EDIT:  As the commenters point out, the algebraic closure doesn't quite work in the case $K'/K$ is purely inseparable, but this case is not difficult; see George Lowther's answer for an easy general argument.
A: First, pick a maximal chain of prime ideals in $S$ and mod out by the minimal one.  Now $S$ is an integral domain of the same dimension.  Similarly, you might as well assume $f$ is injective, since that can only decrease the Krull dimension of $R$.  
So, now, we have a map, which must induce an isomorphism on fraction fields, and both algebras inject into their fraction fields.  Now, take an ideal $I\subset S$ such that $I\cap R=0$, and let $s\neq 0$ be an element of $I$.  Then $s=r'/r''$ for $r',r''\in R$.  Thus $sr''\in R\cap I$, and we have arrived at a contradiction.
A: Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram.
$$
\begin{array}{ccc}
R&\stackrel{f}{\rightarrow}&S\\\\
\downarrow\scriptstyle{}&&\downarrow\scriptstyle{}\\\\
k&\stackrel{g}{\rightarrow}&S_{\mathfrak{p}}
\end{array}
$$
However, $f$ and the localization $S\to S_{\mathfrak{p}}$ are epimorphisms, so $g$ is an epimorphism with domain a field. This means that it is surjective, so $S_{\mathfrak{p}}$ is a field, and $\mathfrak{p}=0$.
To see that an epimorphism $g\colon k\to A$ of commutative rings with domain a field $k$ is surjective, consider the morphisms $u,v\colon A\to A\otimes_k A$ given by $u(a)=a\otimes1$ and $v(a)=1\otimes a$. Then, $u\circ g=v\circ g$ and, from the definition of epimorphism, $u=v$, in which case $g(k)=A$.
