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I have trouble determining some Dirichlet densities which I will denote as $\delta$.

I want to compute the Dirichlet densities of the following set of prime numbers:

  • The set $P_1$ of prime numbers $p$ for which $2$ is a square modulo $p$;

  • The set $P_2$ of prime numbers $p$ for which $2$ is a cube modulo $p$;

  • The set $P_3$ of prime numbers $p$ for which $2$ is a fourth power modulo $p$.

The first one is easy to compute. Consider the number field $L = \mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ which is a Galois extension. By some basic algebraic number theory, $p$ splits completely in $L$ if and only if $2$ is a square modulo $p$ which is exactly the set $P_1$. Hence by an important theorem about Dirichlet densities and completely splitting prime ideals, $\delta(P_1) = \frac{1}{[L : \mathbb{Q}]} = \frac{1}{2}$.

I have trouble with computing the other two cases. For the third set, I know for sure that $\delta(P_3) \leq \delta(P_1)$ since $P_3 \subseteq P_1$. Can anyone please help me?

I'm only familiar with the notion of Dirichlet densities and the Frobenius Density Theorem for Abelian number fields. I'm not familiar with the Chebotarev Density Theorem.

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The simplest density theorem is actually Kronecker's (the Frobenius density theorem is usually derived from Kronecker's), according to which primes that split completely in a anormal extension of degree $n$ have density $\frac1n$. The primes splitting completely in ${\mathbb Q}(\sqrt{-3},\sqrt[3]{2})$ are those for which $p \equiv 1 \bmod 3$ and $x^3 \equiv 2 \bmod p$ is solvable. Since $2$ is a cube for every prime $p \equiv 2 \bmod 3$ the density of such primes is $\frac12 + \frac16 = \frac23$.

For the third case, consider ${\mathbb Q}(i,\sqrt[4]{2})$.

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