Existence of a prime number which splits in $\mathbb{Q}(\sqrt{m})$ and remains inert in $\mathbb{Q}(2^\frac{1}{3})$

Question

I was reading a paper on the simultaneous divisibility of class numbers of quadratic fields where I encountered this problem: For any non-zero, square-free integer $m$, there exist infinitely many primes $p$ in $\mathbb{Z}$ such that $p$ splits completely in $\mathbb{Q}(\sqrt{m})$ and stays inert in $\mathbb{Q}(2^\frac{1}{3})$.

At first I was trying to invoke the Chebotarev Density theorem into the picture but the later extension is not Galois. Could somebody tell me the density of such set of primes explicitly and how to compute it?

Answer 1

You could apply Chebotarev to $\Bbb Q(\sqrt{m},\sqrt{-3},2^{1/3})$ which is Galois with Galois group $S_2\times S_3$ of order $12$. I reckon the Frobeniuses of the $p$ you seek are of the form $(e,\sigma)$ where $e$ is the identity of $S_2$ and $\sigma$ is a $3$-cycle. There are two of these, so the Dirichlet density of your $p$ is $1/6$.

Answer 2

My thoughts :

Let $K= \mathbb{Q}(\sqrt{m},2^{1/3},\zeta_3)$,$[K:\mathbb{Q}] = 12$ the Galois closure of the two fields.

Find $\sigma_1 \in Gal(K/\mathbb{Q})$ such that $\sigma|_{\mathbb{Q}(2^{1/3},\zeta_3)}$ has order $3$ and $\sigma_1|_{\mathbb{Q}(\sqrt{m})}$ has order $1$. Also let $\rho \in Gal(K/\mathbb{Q})$ be the complex conjugaison and $\sigma_2 = \rho \sigma_1 \rho^{-1}$

For a prime number $p$, if the Frobenius of the two primes $\mathfrak{p}_1,\mathfrak{p}_2 = \rho(\mathfrak{p}_1) \in \mathcal{O}_K$ above $p$ are $\sigma_1,\sigma_2$,

Then the Frobenius of $\mathcal{P}_i=\mathfrak{p}_i \cap \mathcal{O}_{\mathbb{Q}(2^{1/3},\zeta_3)}$ is $\sigma_i|_{\mathbb{Q}(2^{1/3},\zeta_3)}$, of order $3$ so $N(\mathcal{P}_i) = p^3$ which guarantees $N(\mathcal{P}_i \cap \mathcal{O}_{\mathbb{Q}(2^{1/3})}) = p^3$ ie. $p$ is inert in $\mathcal{O}_{\mathbb{Q}(2^{1/3})}$.

Similarly the Frobenius of $\mathfrak{p}_i \cap \mathcal{O}_{\mathbb{Q}(\sqrt{m})}$ is $\sigma_i|_{\mathbb{Q}(\sqrt{m})}$, of order $1$ so $p$ splits completely in $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$.

Finally Chebotarev's density theorem guarantees the density of such prime numbers $p$ is $2/12$.