Can infinitely many primes lie over a prime?

Solution 1:

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 $f \to \infty$.