= Solution
A set $S$ of <prime numbers> has <Dirichlet density> $\delta$ if the following limit exists:
$$
\delta(S)=\lim_{s\downarrow1}\frac{\sum_{p\in S}p^{-s}}{\log(1/(s-1))}=\delta.
$$
Equivalently, the denominator can be $\sum_p p^{-s}$, since that sum is $\log(1/(s-1))+O(1)$. Adding or removing finitely many <prime numbers> leaves the <Dirichlet density> unchanged.
Put $\alpha=\sqrt[4]{2}>0$ and $L=\mathbb Q(\alpha,i)$. This is the <splitting field> of $X^4-2$, whose four <roots of a polynomial> are $\alpha,i\alpha,-\alpha,-i\alpha$. The polynomial is an <Eisenstein polynomial> at $2$, so $[\mathbb Q(\alpha):\mathbb Q]=4$. Since $\mathbb Q(\alpha)$ is contained in the <real numbers>, it does not contain $i$, giving $[L:\mathbb Q]=8$. Thus $L/\mathbb Q$ is a <Galois extension> of degree $8$. Its <Galois group> is the <dihedral group> of order $8$: the <field automorphisms> $r(\alpha)=i\alpha$, $r(i)=i$ and $s(\alpha)=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.
For an odd <prime number> $p$, the condition $p\equiv1\pmod4$ means that the <finite field> $\mathbb F_p$ contains all fourth <roots of unity>. Under this condition, $2$ is a <quartic residue> exactly when $X^4-2$ has a root in $\mathbb F_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over $\mathbb F_p$ gives both a fourth root of $2$ and a primitive fourth <root of unity>, so it also forces $p\equiv1\pmod4$.
The <polynomial discriminant> of $X^4-2$ is $-2^{11}$. Hence every odd <prime number> is unramified in $L$, and the root-splitting criterion is equivalent to its <Frobenius automorphism> acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the <Frobenius automorphism> being the identity, or to $p$ being a <completely split prime> of $L/\mathbb Q$. The <Chebotarev density theorem> says that the unramified <prime numbers> with <Frobenius conjugacy class> $C$ have <Dirichlet density> $|C|/|\operatorname{Gal}(L/\mathbb Q)|$. Here $C=\{1\}$, so
$$
\boxed{\delta(S)=\frac18.}
$$
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