Can infinitely many primes lie over a prime? 
Let $R \subset S$ be an integral extension of domains and $\mathfrak p \subset R$ a prime ideal. Can it be the case that there are infinitely many distinct primes ${\cal P} \subset S$ such that ${\cal P} \cap R=\mathfrak p$? 

Certainly this is impossible if $S$ is a Dedekind domain, because the primes lying over $\mathfrak p$ are the primes of $S$ occurring in the factorization of $\mathfrak p$ over $S$.  I don't have much of an intuition for integral ring extensions that aren't number fields, so past this I'm not particularly sure. 
 A: I think the answer is yes. 
Consider $\mathbb{Z} \subset S$, where $S$ is the ring of all the algebraic integers in $\mathbb{C}$. This is by definition integral, and for a fixed prime $p\in\mathbb Z$, one can find a sequence of Galois extensions $L_i/\mathbb{Q}$ such that $g_i(p) \to \infty$, where $g_i(p)$ is the number of primes of $L_i$ lying above $p$. (Using cyclotomic extensions suffice.) As each of these $g_i(p)$ primes can be further lifted up to distinct primes in $S$, this shows that there are infinitely many primes in $S$ lying above $p$.
Why cyclotomic extensions have arbitrarily large number of primes lying above a fixed $p$?
Consider $L = \mathbb{Q}(\zeta_n)$. Take $n$ that is not divisible by $p$, then $e(p) = 1$ in this extension. $f(p)$ is the order of Frobenius, which in this case is the smallest positive integer $f$ such that $n \mid p^f - 1$, and $g(p) = \phi(n)/f$. So everything would work if $n$ is large and $f$ is small. So take any $f$, and let $n = p^f - 1$, then $\frac{\phi(p^f-1)}{f} \to \infty$ as $f \to \infty$.
