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> 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 $\mathfrak{p} = 0$?
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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.



Is Krull dimension non-increasing along ring epimorphisms?
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 $g(k)=A$.


