# Just how many primes can lie over a prime?

Let $R \subset S$ be an integral extensions 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.

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– Martin Brandenburg Feb 6 at 18:48

I think the answer is yes. Consider $\mathbb{Z} \to S$, the ring of all the algebraic integers in $\mathbb{C}$. This is by definition integral, and for a fixed prime $p$, 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 $n | p^f - 1$, 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 $f \to \infty$.
 I was thinking about $\mathbb Z$ into the algebraic integers. Is there an easy reason, or do you have a reference for why Cyclotomic extensions work? So for instance we could replace the algebraic integers, by integral closure of $\mathbb Z$ in $\mathbb Q^{ab}$? (Also I was asking if it was possible for an infinite number of primes to lie over, so the answer is yes). – JSchlather Feb 6 at 18:35 @JacobSchlather Oh, sorry for mixing up the yes's and no's. I will add some details about cyclotomic extensions part. – Sanchez Feb 6 at 18:36 What do you mean by "noetherian over"? – Martin Brandenburg Feb 6 at 18:44 @MartinBrandenburg, $S$ is a noetherian $R$-module. – Sanchez Feb 6 at 18:47 @MartinBrandenburg, actually that was a clearly wrong remark, not sure what I was thinking. – Sanchez Feb 6 at 18:52