Inverse problem of the field extension $K/\Bbb{Q}$ of given number of prime above $p$

Question

Let $g$ be an arbitrary fixed positive integer.

I want to find an example of $K/\Bbb{Q}$ : field extension and prime number $p$ such that the number of primes of ring of integers of $K$ above $p$ is exactly $g$.

How can I find such a $K$ and $p$ ?

Answer 1

It is a theorem that for every number field, infinitely many primes split completely in it. So let $K$ be an arbitrary number field with degree $g$ over $\mathbf Q$, e.g., $\mathbf Q(\sqrt[g]{2})$ and let $p$ be a prime that splits completely in it (such $p$ exist). Then there are $g$ primes in $\mathcal O_K$ lying over $p$.

Here is another approach that realizes $K$ inside some cyclotomic field. When $m \geq 1$ and $p \equiv 1 \bmod m$, $x^m - 1 \bmod p$ splits completely since $(x^m - 1) \mid (x^{p-1}-1)$ in $\mathbf Z[x]$. Thus the $m$th cyclotomic polynomial splits completely mod $p$, so $p$ splits completely in $\mathbf Q(\zeta_m)$. That implies $p$ splits completely in every subfield of $\mathbf Q(\zeta_m)$. We can find such $p$ because there are infiniely many primes $p \equiv 1 \bmod m$.

Now for your choice of $g$, let $m \geq 1$ satisfy $g \mid \varphi(m)$, e.g., $m = 2^g - 1$ since $2 \bmod m$ has order $g$. Then every prime $p \equiv 1 \bmod m$ splits completely in each subfield of $\mathbf Q(\zeta_m)$. Let's show there is a subfield of $\mathbf Q(\zeta_m)$ with degree $g$ over $\mathbf Q$. The Galois group of $\mathbf Q(\zeta_m)/\mathbf Q$ is $(\mathbf Z/m\mathbf Z)^\times$, which is abelian with order $\varphi(m)$. Abelian groups satisfy a converse to Lagrange's theorem: they have a subgroup with order equal to any divisor of the order of the group, and thus also with index equal to any divisor of the order of the group (see here). So $(\mathbf Z/m\mathbf Z)^\times$ has a subgroup $H$ with index $g$, and by Galois theory its fixed field inside $\mathbf Q(\zeta_m)$ has degree $g$ over $\mathbf Q$. Call that fixed field $K$. Then $[K:\mathbf Q] = g$ and every prime $p \equiv 1 \bmod m$ splits completely in $K$ and thus has $g$ primes over it in $K$.