2 Conditions for $G_{2}(k)\simeq G$

At first we have to know when $G_{2}(k)\simeq Q_{2^{n}},D_{2^{n}}$ , or $\mathit{SD}_{2^{n}}$ . Moreover, the conditions should be able to be written by splitting of primes in a number field because we use the Chebotarev density theorem. The conditions are known by several authors and gathered (together with other groups) in the table in [Reference Benjamin, Lemmermeyer and Snyder1]. In this section, we outline the conditions and prove them partially.

The $2$ -rank of the narrow class group of a quadratic field is obtained by the genus theory:

The $4$ -rank can also be calculated by using the Rédei–Reichardt criterion [Reference Rédei and Reichardt14]:

Proposition 2. Let $F$ be a quadratic field and $l_{1},\ldots ,l_{t}$ be all the distinct prime divisors of $d_{F}$ . We define the Rédei matrix:

where $[l_{i},d_{F}/l_{j}]\in \mathbb{F}_{2}$ is defined as

Then the $4$ -rank $\dim _{\mathbb{F}_{2}}(\operatorname{Cl}_{2}^{+}(F)^{2}/\operatorname{Cl}_{2}^{+}(F)^{4})$ of $\operatorname{Cl}_{2}^{+}(F)$ equals to $t-1-\operatorname{rank}R_{F}$ .

From Propositions 1 and 2, one can show

The following conditions are shown by several authors, see [Reference Benjamin, Lemmermeyer and Snyder1]:

This proposition is shown by Kisilevsky [Reference Kisilevsky8] except for the conditions $2^{n}=4h_{2}(qr)$ and $2^{n}=4h_{2}(-pq)$ in (2b), (2c), and (2d). We give a proof of the conditions $2^{n}=4h_{2}(qr)$ and $2^{n}=4h_{2}(-pq)$ in (2b), (2c), and (2d) because no proof is given in [Reference Benjamin, Lemmermeyer and Snyder1].

Let $k=\mathbb{Q}(\sqrt{-pqr})$ where $p,q,r$ are distinct primes with $-p\equiv q\equiv r\equiv 1~(4)$ , and assume $h_{2}(k)=4$ . Then the Hilbert $2$ -class field $k^{1}$ is $\mathbb{Q}(\sqrt{-p},\sqrt{q},\sqrt{r})$ . Let $K$ be the maximal real subfield of $k^{1}$ , that is, $K=\mathbb{Q}(\sqrt{q},\sqrt{r})$ .

Lemma 1.

$$\begin{eqnarray}h_{2}(k^{1})={\textstyle \frac{1}{2}}h_{2}(K)h_{2}(-pq)h_{2}(-pr),\end{eqnarray}$$

where $h_{2}(m)=h_{2}(\mathbb{Q}(\sqrt{m}))$ for $m\in \mathbb{Z}$ .

Proof. From the relative class number formula [Reference Washington18, Chapter 4], one has

$$\begin{eqnarray}h(k^{1})/h(K)=wQ\mathop{\prod }_{\unicode[STIX]{x1D712}:\text{odd}}\biggl(-\frac{1}{2f_{\unicode[STIX]{x1D712}}}\mathop{\sum }_{a=1}^{f_{\unicode[STIX]{x1D712}}}\unicode[STIX]{x1D712}(a)a\biggr)\end{eqnarray}$$

where the product runs over the odd Dirichlet characters of $k^{1}$ and

1. (i) $h$ is the class number;

2. (ii) $w$ is the number of the roots of unity in $k^{1}$ ;

3. (iii) $Q=[U:\unicode[STIX]{x1D707}U_{K}]$ , where $U$ and $U_{K}$ are the unit groups of $k^{1}$ and $K$ , respectively, and $\unicode[STIX]{x1D707}$ is the group of the roots of unity in $k^{1}$ ;

4. (iv) $f_{\unicode[STIX]{x1D712}}$ is the conductor of $\unicode[STIX]{x1D712}$ .

Note that one can easily show

Similarly we have

for $m=-p,-pq,-pr,-pqr$ , where $h(m)=h(\mathbb{Q}(\sqrt{m})),\,w_{m}=w_{\mathbb{Q}(\sqrt{m})}$ , and $d_{m}=d_{\mathbb{Q}(\sqrt{m})}$ . Putting these formulas together we obtain

$$\begin{eqnarray}h(k^{1})/h(K)=wQ\frac{h(-p)h(-pq)h(-pr)h(k)}{w_{p}w_{pq}w_{pr}w_{k}}.\end{eqnarray}$$

Now $h(-p)$ is odd by Proposition 1 and $h_{2}(k)=4$ . Also we know that the $2$ -parts of $w,w_{p},w_{pq},w_{pr}$ , and $w_{k}$ are $2$ since $k^{1}/\mathbb{Q}$ is unramified at $2$ .

Hence it suffices to show $Q=1$ . By [Reference Washington18, Theorem 4.12], $Q=1$ if and only if $\unicode[STIX]{x1D700}/\overline{\unicode[STIX]{x1D700}}\in \unicode[STIX]{x1D707}^{2}$ for any $\unicode[STIX]{x1D700}\in U$ , where $\overline{\unicode[STIX]{x1D700}}$ means the conjugate of $\unicode[STIX]{x1D700}$ over $K$ . Suppose $\unicode[STIX]{x1D700}/\overline{\unicode[STIX]{x1D700}}\notin \unicode[STIX]{x1D707}^{2}$ for some $\unicode[STIX]{x1D700}\in U$ . It can be easily shown that $w=2$ or $6$ . Hence, $\unicode[STIX]{x1D700}/\overline{\unicode[STIX]{x1D700}}=-1$ or $(1\pm \sqrt{-3})/2$ . We let

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\left\{\begin{array}{@{}ll@{}}\displaystyle \unicode[STIX]{x1D700}\sqrt{-p}\quad & \displaystyle \text{if }\unicode[STIX]{x1D700}/\overline{\unicode[STIX]{x1D700}}=-1,\\ \displaystyle \unicode[STIX]{x1D700}\biggl(\frac{3\mp \sqrt{-3}}{2}\biggr)\quad & \displaystyle \text{if }\unicode[STIX]{x1D700}/\overline{\unicode[STIX]{x1D700}}=\frac{1\pm \sqrt{-3}}{2}\quad (p=3).\end{array}\right.\end{eqnarray}$$

We can check $\unicode[STIX]{x1D6FC}=\overline{\unicode[STIX]{x1D6FC}}$ so that $\unicode[STIX]{x1D6FC}\in K$ . Hence,

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}^{2}=N_{k^{1}/K}(\unicode[STIX]{x1D6FC})=pN_{k^{1}/K}(\unicode[STIX]{x1D700}).\end{eqnarray}$$

Since $N_{k^{1}/K}(\unicode[STIX]{x1D700})\in U_{K}$ , this implies that $p$ ramifies in $K/\mathbb{Q}$ , which is absurd.◻

Proof of Proposition 4.

Now we determine $n$ . We know that

$$\begin{eqnarray}2^{n}=\#G_{2}(k)=[k^{2}:k]=h_{2}(k)h_{2}(k^{1})=4h_{2}(k^{1}).\end{eqnarray}$$

From Lemma 2 we have $h_{2}(qr)=2h_{2}(K)$ . Hence from Lemma 1,

$$\begin{eqnarray}h_{2}(k^{1})={\textstyle \frac{1}{4}}h_{2}(qr)h_{2}(-pq)h_{2}(-pr).\end{eqnarray}$$

(2b), (2c)

Since $(p/q)=(p/r)=-1$ , we get $h_{2}(-pq)=h_{2}(-pr)=2$ from Proposition 1 (see Proposition 5 below). Hence $2^{n}=4h_{2}(k^{1})=4h_{2}(qr)$ .

(2d)

Similarly, we get $h_{2}(-pr)=2$ . Since $(q/r)=-1$ , we get $h_{2}^{+}(qr)=2$ from Proposition 1. Since $h_{2}(qr)=2h_{2}(K)$ is even, we have $h_{2}(qr)=2$ . Hence $2^{n}=4h_{2}(k^{1})=4h_{2}(-pq)$ .◻

Also the following conditions are known, see [Reference Yamamoto19]:

3 Densities for $Q_{8},Q_{16},D_{8}$ , and $\mathit{SD}_{16}$

We want to find the densities of quadratic fields which have maximal unramified $2$ -extensions with Galois groups isomorphic to $Q_{2^{n}},D_{2^{n}}$ , or $\mathit{SD}_{2^{n}}$ . In other words, we want to calculate the densities, $\unicode[STIX]{x1D6FF}(A(G)/B(V_{4}))$ for $G=Q_{2^{n}},D_{2^{n}},\mathit{SD}_{2^{n}}$ . For this purpose, we produce a multivariable version of the Chebotarev density theorem.

For an integer $n\geqslant 0$ , we let

$$\begin{eqnarray}\displaystyle P_{n} & = & \displaystyle \{\,p_{1}\cdots p_{n}\mid p_{1},\ldots ,p_{n}:\text{distinct primes}\,\}\!;\nonumber\\ \displaystyle P_{n}(x) & = & \displaystyle \{\,m\in P_{n}\mid m\leqslant x\,\}\!.\nonumber\end{eqnarray}$$

At first we calculate $\unicode[STIX]{x1D6FF}(A(G)/P_{3})$ . Note that in our definition of $\unicode[STIX]{x1D6FF}(A/B)$ , we do not assume $A\subseteq B$ . It is known [Reference Hardy and Wright7, Theorem 437] that for $n\geqslant 1$

(1) $$\begin{eqnarray}\#P_{n}(x)\sim \frac{x(\log \log x)^{n-1}}{(n-1)!\log x}.\end{eqnarray}$$

Suppose that $F_{a}$ and ${\mathcal{C}}_{a}$ are given for each square-free positive integer $a$ , where $F_{a}$ is a number field which is Galois over $\mathbb{Q}$ , and ${\mathcal{C}}_{a}$ is a conjugacy class in ${\mathcal{G}}_{a}=\operatorname{Gal}(F_{a}/\mathbb{Q})$ . For a prime number $p$ , we let $\unicode[STIX]{x1D711}_{a}(p)=1$ when $p$ does not ramify in $F_{a}$ and

and $\unicode[STIX]{x1D711}_{a}(p)=0$ otherwise. We denote for $n\geqslant 0$ ,

$$\begin{eqnarray}\displaystyle S_{n} & = & \displaystyle \left\{\hspace{2.22198pt}p_{1}\cdots p_{n}\in P_{n}\;|\;p_{1}<\cdots <p_{n},\mathop{\prod }_{k=1}^{n}\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})=1\hspace{2.22198pt}\right\}\!;\nonumber\\ \displaystyle S_{n}(x) & = & \displaystyle \{\,m\in S_{n}\mid m\leqslant x\,\}.\nonumber\end{eqnarray}$$

Note that the empty product means $1$ and $P_{0}=S_{0}=\{1\}$ .

Now we state our main theorem:

Theorem 2. Let $n$ be a positive integer. Assume that:

1. (i) the Generalized Riemann Hypothesis (GRH) holds for the Dedekind zeta function of $F_{a}$ for all $a\in S_{0}\cup \cdots \cup S_{n-1}$ ;

2. (ii) there are positive constants $c_{0}$ and $\unicode[STIX]{x1D700}$ depending only on $n$ , such that for $1\leqslant k\leqslant n-1$ and $a=p_{1}\cdots p_{k}\in S_{k}~(p_{1}<\cdots <p_{k})$ ,

   $$\begin{eqnarray}\log |d_{F_{a}}|\leqslant \frac{c_{0}\sqrt{p_{k}}}{(\log p_{k})^{1+\unicode[STIX]{x1D700}}}\qquad \text{and}\qquad [F_{a}:\mathbb{Q}]\leqslant \frac{c_{0}\sqrt{p_{k}}}{(\log p_{k})^{2+\unicode[STIX]{x1D700}}};\end{eqnarray}$$

3. (iii) for $1\leqslant k\leqslant n$ , there is a rational number $\unicode[STIX]{x1D6FF}_{k}$ such that for any $a\in S_{k-1}$ ,

   $$\begin{eqnarray}\frac{\#{\mathcal{C}}_{a}}{\#{\mathcal{G}}_{a}}=\unicode[STIX]{x1D6FF}_{k}.\end{eqnarray}$$

Then

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(S_{n}/P_{n})=\unicode[STIX]{x1D6FF}_{1}\cdots \unicode[STIX]{x1D6FF}_{n}.\end{eqnarray}$$

The proof of this theorem is given in the next section.

Now we shall calculate $\unicode[STIX]{x1D6FF}(A(G)/B(V_{4}))$ by using the theorem. Here we only calculate $\unicode[STIX]{x1D6FF}(A(D_{8})/B(V_{4}))$ . From Propositions 4 and 5, we know that $\mathbb{Q}(\sqrt{-m})$ has odd discriminant and $m\in A(D_{8})$ , if and only if, $m=pqr$ such that $p,q$ , and $r$ are distinct primes with $-p\equiv q\equiv r\equiv 1~(4)$ and

Let

$$\begin{eqnarray}\displaystyle A_{1} & = & \displaystyle \{\,pqr\in A(D_{8})\cap P_{3}\mid p<q<r,\,-p\equiv q\equiv r\equiv 1~(4)\,\}\!;\nonumber\\ \displaystyle A_{2} & = & \displaystyle \{\,pqr\in A(D_{8})\cap P_{3}\mid q<p<r,\,-p\equiv q\equiv r\equiv 1~(4)\,\}\!;\nonumber\\ \displaystyle A_{3} & = & \displaystyle \{\,pqr\in A(D_{8})\cap P_{3}\mid q<r<p,\,-p\equiv q\equiv r\equiv 1~(4)\,\}\!;\nonumber\\ \displaystyle A_{4} & = & \displaystyle \{\,m\in A(D_{8})\mid m\not \equiv -1~(4)\,\}.\nonumber\end{eqnarray}$$

Then clearly these are disjoint, and we know

$$\begin{eqnarray}A_{1}\cup A_{2}\cup A_{3}\cup A_{4}=A(D_{8})\end{eqnarray}$$

from Propositions 1 and 4. Note that if $m\in A_{4}$ then $\mathbb{Q}(\sqrt{-m})$ has even discriminant; therefore from Proposition 1, we can easily show that $m\in P_{2}$ or $m/2\in P_{2}$ . But we know $\unicode[STIX]{x1D6FF}(P_{2}/P_{3})=0$ from (1), which yields $\unicode[STIX]{x1D6FF}(A_{4}/P_{3})=0$ . Hence,

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(A(D_{8})/P_{3})=\unicode[STIX]{x1D6FF}(A_{1}/P_{3})+\unicode[STIX]{x1D6FF}(A_{2}/P_{3})+\unicode[STIX]{x1D6FF}(A_{3}/P_{3}).\end{eqnarray}$$

Now we see $\unicode[STIX]{x1D6FF}(A_{1}/P_{3})$ . We put:

1. (i) $F_{1}=\mathbb{Q}(\sqrt{-1}),{\mathcal{C}}_{1}=\{j\},\unicode[STIX]{x1D6FF}_{1}=\frac{1}{2}$ where  corresponds to $\mathbb{Q}$ ;

2. (ii) $F_{p}=\mathbb{Q}(\sqrt{-1},\sqrt{p}),{\mathcal{C}}_{p}=\{\unicode[STIX]{x1D70E}\},\unicode[STIX]{x1D6FF}_{2}=\frac{1}{4}$ for primes $p\equiv -1~(4)$ , where  corresponds to $\mathbb{Q}(\sqrt{-1})$ ;

3. (iii) $F_{pq}=\mathbb{Q}(\sqrt{-1},\sqrt{p},\sqrt{q},\sqrt{\unicode[STIX]{x1D700}_{q}}),{\mathcal{C}}_{pq}=\{\unicode[STIX]{x1D70F}\},\unicode[STIX]{x1D6FF}_{3}=\frac{1}{16}$ for primes $p<q,\,-p\equiv q\equiv 1~(4)$ , where $\unicode[STIX]{x1D700}_{q}$ is the fundamental unit of $\mathbb{Q}(\sqrt{q})$ , and  corresponds to $\mathbb{Q}(\sqrt{-1},\sqrt{q},\sqrt{p\unicode[STIX]{x1D700}_{q}})$ .

Assume that GRH holds for these fields. $F_{a}$ and ${\mathcal{C}}_{a}$ are unused for the other integers $a$ , so we may let them be anything.

Then we show $S_{3}=A_{1}$ . As is well-known that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{1}(p)=1 & \;\Longleftrightarrow \; & \displaystyle p\equiv 3~(4)\quad (\forall p:\text{prime});\nonumber\\ \displaystyle \unicode[STIX]{x1D711}_{p}(q)=1 & \;\Longleftrightarrow \; & \displaystyle q\equiv 1~(4)\text{ and }(p/q)=-1\quad (\forall p,q:\text{primes},p\equiv 3~(4)).\nonumber\end{eqnarray}$$

Now we consider $F_{pq}$ . Assume $p<q$ and $-p\equiv q\equiv 1~(4)$ . By Scholz’s reciprocity law [Reference Lemmermeyer12, Proposition 5.8], if $q\equiv r\equiv 1~(4)$ and $(q/r)=1$ ,

We know that

$$\begin{eqnarray}\displaystyle (p/r)=-1 & \;\Longleftrightarrow \; & \displaystyle r\text{ is inert in }\mathbb{Q}(\sqrt{p})/\mathbb{Q};\nonumber\\ \displaystyle (q/r)=+1 & \;\Longleftrightarrow \; & \displaystyle r\text{ splits in }\mathbb{Q}(\sqrt{q})/\mathbb{Q};\nonumber\\ \displaystyle (\unicode[STIX]{x1D700}_{q}/r)=-1 & \;\Longleftrightarrow \; & \displaystyle \text{the primes above }r\text{ are inert in }\mathbb{Q}(\sqrt{q},\sqrt{\unicode[STIX]{x1D700}_{q}})/\mathbb{Q}(\sqrt{q}).\nonumber\end{eqnarray}$$

$F_{pq}$ includes these fields. It is straightforward to show that $-(p/r)=(q/r)=-(\unicode[STIX]{x1D700}_{q}/r)=1$ if and only if the decomposition group of any prime ideal of $F_{pq}$ above $r$ is $\mathbb{Q}(\sqrt{-1},\sqrt{q},\sqrt{p\unicode[STIX]{x1D700}_{q}})$ . Hence, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{pq}(r)=1 & \;\Longleftrightarrow \; & \displaystyle r\equiv 1~(4)\text{ and }-(p/r)=(q/r)=-(q/r)_{4}(r/q)_{4}=1.\nonumber\\ \displaystyle & & \displaystyle (\forall p,q,r:\text{primes},p<q,-p\equiv q\equiv 1~(4))\nonumber\end{eqnarray}$$

Therefore, for any primes $p<q<r$ we get

$$\begin{eqnarray}\unicode[STIX]{x1D711}_{1}(p)\unicode[STIX]{x1D711}_{p}(q)\unicode[STIX]{x1D711}_{pq}(r)=1\;\Longleftrightarrow \;-p\equiv q\equiv r\equiv 1~(4)\text{ and }pqr\in A(D_{8}).\end{eqnarray}$$

This means $S_{3}=A_{1}$ by definition.

Finally, we estimate the discriminant of $F_{a}$ . Clearly $|d_{F_{1}}|=4$ . Now $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-p})$ are linearly disjoint over $\mathbb{Q}$ , and their discriminants are coprime. This yields [Reference Lang11, Chapter III, Proposition 17]

$$\begin{eqnarray}d_{F_{p}}=d_{\mathbb{Q}(\sqrt{-1})}^{2}d_{\mathbb{ Q}(\sqrt{-p})}^{2}=16p^{2}.\end{eqnarray}$$

For $F_{pq}$ , we first let $L=\mathbb{Q}(\sqrt{-1},\sqrt{-p},\sqrt{q})$ and we deduce as above

$$\begin{eqnarray}d_{L}=d_{\mathbb{Q}(\sqrt{-1})}^{4}d_{\mathbb{ Q}(\sqrt{-p})}^{4}d_{\mathbb{ Q}(\sqrt{q})}^{4}=2^{8}p^{4}q^{4}.\end{eqnarray}$$

Then we use the formula:

$$\begin{eqnarray}d_{F_{pq}}\mathbb{Z}=N_{L/\mathbb{Q}}(\mathfrak{d}_{F_{pq}/L})\cdot d_{L}^{[F_{pq}:L]}\mathbb{Z}\end{eqnarray}$$

where $\mathfrak{d}_{F_{pq}/L}$ is the relative discriminant. Also we use the fact that $\mathfrak{d}_{F_{pq}/L}$ divides the discriminant of the minimal polynomial of $\sqrt{\unicode[STIX]{x1D700}_{q}}$ over $L$ in ${\mathcal{O}}_{L}$ , the ring of integers in $L$ . Hence, we have

$$\begin{eqnarray}\displaystyle & \displaystyle \mathfrak{d}_{F_{pq}/L}\mid 4\unicode[STIX]{x1D700}_{q}{\mathcal{O}}_{L}=4{\mathcal{O}}_{L}; & \displaystyle \nonumber\\ \displaystyle & \displaystyle \therefore ~N_{L/\mathbb{Q}}(\mathfrak{d}_{F_{pq}/L})\mid 4^{8}\mathbb{Z}; & \displaystyle \nonumber\\ \displaystyle & \displaystyle \therefore ~d_{F_{pq}}\mid 4^{8}\cdot (2^{8}p^{4}q^{4})^{2}=2^{32}p^{8}q^{8}. & \displaystyle \nonumber\end{eqnarray}$$

In any cases, $F_{a}$ satisfies the discriminant condition of Theorem 2. For the facts about discriminants used above, see [Reference Fröhlich and Taylor3, Section III.2].

At this time we can apply Theorem 2, which yields

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(A_{1}/P_{3})=\unicode[STIX]{x1D6FF}(S_{3}/P_{3})=\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}\unicode[STIX]{x1D6FF}_{3}={\textstyle \frac{1}{128}}.\end{eqnarray}$$

Similarly as above, we can also deduce $\unicode[STIX]{x1D6FF}(A_{2}/P_{3})=1/128$ by using

$$\begin{eqnarray}F_{1}=\mathbb{Q}(\sqrt{-1}),\qquad F_{q}=\mathbb{Q}(\sqrt{-1},\sqrt{q}),\qquad F_{qp}=\mathbb{Q}(\sqrt{-1},\sqrt{p},\sqrt{q},\sqrt{\unicode[STIX]{x1D700}_{q}}),\end{eqnarray}$$

and $\unicode[STIX]{x1D6FF}(A_{3}/P_{3})=1/128$ by using

$$\begin{eqnarray}F_{1}=\mathbb{Q}(\sqrt{-1}),\qquad F_{q}=\mathbb{Q}(\sqrt{-1},\sqrt{q},\sqrt{\unicode[STIX]{x1D700}_{q}}),\qquad F_{qr}=\mathbb{Q}(\sqrt{-1},\sqrt{q},\sqrt{r}).\end{eqnarray}$$

Therefore, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(A(D_{8})/P_{3})={\textstyle \frac{3}{128}}.\end{eqnarray}$$

It is known in [Reference Gerth6] that $\unicode[STIX]{x1D6FF}(B(V_{4})/P_{3})=7/32$ ; hence, we conclude

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(A(D_{8})/B(V_{4}))={\textstyle \frac{3}{28}}.\end{eqnarray}$$

Also we can calculate densities for $Q_{8},Q_{16}$ , and $\mathit{SD}_{16}$ in the same way. One can easily find the appropriate $F_{a}$ and ${\mathcal{C}}_{a}$ by using the following fact: If $q,r$ are odd primes with $r\equiv 1~(4)$ and $(q/r)=1$ then

$$\begin{eqnarray}\displaystyle (q/r)_{4}=-1 & \;\Longleftrightarrow \; & \displaystyle \text{the primes above }q\text{ are inert in }\mathbb{Q}(\unicode[STIX]{x1D701}_{r})_{4}/\mathbb{Q}(\sqrt{r});\nonumber\\ \displaystyle (q/r)_{4}=-1 & \;\Longleftrightarrow \; & \displaystyle \text{the primes above }r\text{ are inert in }\mathbb{Q}(\sqrt[4]{q})/\mathbb{Q}(\sqrt{q}),\nonumber\end{eqnarray}$$

where $\mathbb{Q}(\unicode[STIX]{x1D701}_{r})_{4}$ is the quartic subfield of the $r$ th cyclotomic field.

Table 1. Densities.

Note that if the condition for $m\in A(G)$ can be written by using only quadratic residue symbols, we do not need GRH since we can use the same way as Gerth.

4 Proof of Theorem 2

The following argument is basically from Gerth [Reference Gerth5]. However, we cannot use the same way to estimate error terms because the quartic residue symbol $(p/\cdot )_{4}$ cannot be considered as Dirichlet characters. Hence, we shall use the effective Chebotarev density theorem.

In what follows, we use the following notations:

1. (i) $\sum _{a<p\leqslant b}$ means a single summation on $p$ only over primes.

2. (ii) For a multivariable function $f$ , $O(f)$ means a term whose absolute value is always at most $c|f|$ for some constant $c$ depending only on $n$ and $\unicode[STIX]{x1D700}$ .

We shall use the following formulas [Reference Koch.9, Section 27.1], [Reference Hardy and Wright7, Chapter XXII]:

(2) $$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{p\leqslant y}1=\frac{y}{\log y}+O\biggl(\frac{y}{(\log y)^{2}}\biggr); & \displaystyle\end{eqnarray}$$

(3) $$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{p\leqslant y}\frac{1}{p}=\log \log y+O(1); & \displaystyle\end{eqnarray}$$

(4) $$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{p\leqslant y}\frac{\log p}{p}=\log y+O(1). & \displaystyle\end{eqnarray}$$

We may assume $n\geqslant 2$ . It suffices to show

$$\begin{eqnarray}\#S_{n}(x)=\unicode[STIX]{x1D6FF}_{1}\cdots \unicode[STIX]{x1D6FF}_{n}\#P_{n}(x)+o(\#P_{n}(x)).\end{eqnarray}$$

As in [Reference Gerth5], we write

$$\begin{eqnarray}\displaystyle \#S_{n}(x) & = & \displaystyle \mathop{\sum }_{1<p_{1}\leqslant x^{1/n}}\unicode[STIX]{x1D711}_{1}(p_{1})\mathop{\sum }_{p_{1}<p_{2}\leqslant (x/p_{1})^{1/(n-1)}}\unicode[STIX]{x1D711}_{p_{1}}(p_{2})\cdots \nonumber\\ \displaystyle & & \displaystyle \mathop{\sum }_{p_{n-2}<p_{n-1}\leqslant (x/(p_{1}\cdots p_{n-2}))^{1/2}}\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-2}}(p_{n-1})\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{p_{n-1}<p_{n}\leqslant x/(p_{1}\cdots p_{n-1})}\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n}).\nonumber\end{eqnarray}$$

Note that if $p_{1},\ldots ,p_{k}$ are in each interval, one has

$$\begin{eqnarray}x^{1/n}\leqslant \biggl(\frac{x}{p_{1}\cdots p_{k}}\biggr)^{1/(n-k)}.\end{eqnarray}$$

In particular,

(5) $$\begin{eqnarray}\frac{1}{\log (x/p_{1}\cdots p_{k})}=O\biggl(\frac{1}{\log x}\biggr).\end{eqnarray}$$

An elementary calculation shows

(6) $$\begin{eqnarray}\displaystyle \#S_{n}(x) & = & \displaystyle \mathop{\sum }_{p_{1}}\unicode[STIX]{x1D711}_{1}(p_{1})\mathop{\sum }_{p_{2}}\unicode[STIX]{x1D711}_{p_{1}}(p_{2})\cdots \mathop{\sum }_{p_{n}}\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{p_{1}}\unicode[STIX]{x1D6FF}_{1}\mathop{\sum }_{p_{2}}\unicode[STIX]{x1D6FF}_{2}\cdots \mathop{\sum }_{p_{n}}\unicode[STIX]{x1D6FF}_{n}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{p_{1}}(\unicode[STIX]{x1D711}_{1}(p_{1})-\unicode[STIX]{x1D6FF}_{1})\mathop{\sum }_{p_{2}}\unicode[STIX]{x1D6FF}_{2}\cdots \mathop{\sum }_{p_{n}}\unicode[STIX]{x1D6FF}_{n}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{p_{1}}\unicode[STIX]{x1D711}_{1}(p_{1})\mathop{\sum }_{p_{2}}(\unicode[STIX]{x1D711}_{p_{1}}(p_{2})-\unicode[STIX]{x1D6FF}_{2})\cdots \mathop{\sum }_{p_{n}}\unicode[STIX]{x1D6FF}_{n}\nonumber\\ \displaystyle & & \displaystyle +\,\cdots \nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{p_{1}}\unicode[STIX]{x1D711}_{1}(p_{1})\mathop{\sum }_{p_{2}}\unicode[STIX]{x1D711}_{p_{1}}(p_{2})\cdots \mathop{\sum }_{p_{n}}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n})-\unicode[STIX]{x1D6FF}_{n}).\end{eqnarray}$$

The first term in the right-hand side of (6) equals to $\unicode[STIX]{x1D6FF}_{1}\cdots \unicode[STIX]{x1D6FF}_{n}\#P_{n}(x)$ . So we have to estimate the remaining terms.

We need the following estimation:

Lemma 3. Let $1\leqslant k\leqslant n$ and $a=p_{1}\cdots p_{k-1}\in P_{k-1}~(p_{1}<\cdots <p_{k-1})$ . Then for $y\geqslant p_{k-1}$ ,

$$\begin{eqnarray}\biggl|\mathop{\sum }_{p\leqslant y}(\unicode[STIX]{x1D711}_{a}(p)-\unicode[STIX]{x1D6FF}_{k})\biggr|=O\biggl(\frac{y}{(\log y)^{1+\unicode[STIX]{x1D700}}}\biggr).\end{eqnarray}$$

Proof. From the definition of $\unicode[STIX]{x1D711}_{a}$ , we know

Also we note that $\unicode[STIX]{x1D70B}(y)=\sum _{p\leqslant y}1$ is the usual prime-counting function and $\unicode[STIX]{x1D6FF}_{k}=\#{\mathcal{C}}_{a}/\#{\mathcal{G}}_{a}$ .

From the effective version of the Chebotarev density theorem [Reference Lagarias and Odlyzko10, Reference Serre15], we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70B}_{{\mathcal{C}}_{a}}(y,F_{a}/\mathbb{Q}) & = & \displaystyle \frac{\#{\mathcal{C}}_{a}}{\#{\mathcal{G}}_{a}}(\operatorname{Li}(y)+O(\sqrt{y}([F_{a}:\mathbb{Q}]\log y+\log |d_{F_{a}}|)));\nonumber\\ \displaystyle \unicode[STIX]{x1D70B}(y) & = & \displaystyle \operatorname{Li}(y)+O(\sqrt{y}\log y),\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\operatorname{Li}(y)=\int _{2}^{y}\frac{dt}{\log t}.\end{eqnarray}$$

Note that GRH for the Dedekind zeta function of $F_{a}$ implies the usual Riemann hypothesis. Our claim follows from these equations and our assumptions for $|d_{F_{a}}|$ and $[F_{a}:\mathbb{Q}]$ .◻

We consider the last term of (6). Since $x/p_{1}\cdots p_{n-1}\geqslant p_{n-1}$ , we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|\mathop{\sum }_{p_{1}}\unicode[STIX]{x1D711}_{1}(p_{1})\mathop{\sum }_{p_{2}}\unicode[STIX]{x1D711}_{p_{1}}(p_{2})\cdots \mathop{\sum }_{p_{n}}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n})-\unicode[STIX]{x1D6FF}_{n})\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,\mathop{\sum }_{p_{1}}\mathop{\sum }_{p_{2}}\cdots \mathop{\sum }_{p_{n-1}}\biggl|\mathop{\sum }_{1<p_{n}\leqslant x/(p_{1}\cdots p_{n-1})}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n})-\unicode[STIX]{x1D6FF}_{n})\nonumber\\ \displaystyle & & \displaystyle \qquad -\mathop{\sum }_{1<p_{n}\leqslant p_{n-1}}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{n-1}}(p_{n})-\unicode[STIX]{x1D6FF}_{n})\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{p_{1}}\mathop{\sum }_{p_{2}}\cdots \mathop{\sum }_{p_{n-1}}O\biggl(\frac{x}{p_{1}\cdots p_{n-1}(\log (x/(p_{1}\cdots p_{n-1})))^{1+\unicode[STIX]{x1D700}}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\frac{p_{n-1}}{(\log p_{n-1})^{1+\unicode[STIX]{x1D700}}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{x}{(\log x)^{1+\unicode[STIX]{x1D700}}}\biggr)\mathop{\sum }_{p_{1}\leqslant x}\frac{1}{p_{1}}\mathop{\sum }_{p_{2}\leqslant x}\frac{1}{p_{2}}\cdots \mathop{\sum }_{p_{n-1}\leqslant x}\frac{1}{p_{n-1}}\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{x(\log \log x)^{n-1}}{(\log x)^{1+\unicode[STIX]{x1D700}}}\biggr)=o(\#P_{n}(x))\nonumber\end{eqnarray}$$

from Lemmas 3 and (5).

Next we see the middle terms in the right-hand side of (6). We first claim

(7) $$\begin{eqnarray}\mathop{\sum }_{p_{n-1}<p_{n}\leqslant x/(p_{1}\cdots p_{n-1})}1=\frac{x}{p_{1}\cdots p_{n-1}\log x}+O\biggl(\frac{x\log p_{n-1}}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr).\end{eqnarray}$$

In fact, from (2) we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{1<p_{n}\leqslant (x/(p_{1}\cdots p_{n-1}))}1=\frac{x}{p_{1}\cdots p_{n-1}\log (x/(p_{1}\cdots p_{n-1}))}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O\biggl(\frac{x}{p_{1}\cdots p_{n-1}(\log (x/(p_{1}\cdots p_{n-1})))^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{x}{p_{1}\cdots p_{n-1}\log x}+\frac{x\log (p_{1}\cdots p_{n-1})}{p_{1}\cdots p_{n-1}\log (x/(p_{1}\cdots p_{n-1}))\log x}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O\biggl(\frac{x}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{x}{p_{1}\cdots p_{n-1}\log x}+O\biggl(\frac{x\log p_{n-1}}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr),\nonumber\end{eqnarray}$$

and from

$$\begin{eqnarray}O\biggl(\frac{p_{n-1}}{(\log p_{n-1})^{2}}\biggr)=O\biggl(\frac{x/(p_{1}\cdots p_{n-1})}{(\log (x/(p_{1}\cdots p_{n-1})))^{2}}\biggr)=O\biggl(\frac{x}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr)\end{eqnarray}$$

we have

$$\begin{eqnarray}\mathop{\sum }_{1<p_{n}\leqslant p_{n-1}}1=O\biggl(\frac{p_{n-1}}{\log p_{n-1}}\biggr)=O\biggl(\frac{x\log p_{n-1}}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr).\end{eqnarray}$$

Hence, (7) follows. Then we can ignore the error term of (7) since

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p_{1}}\mathop{\sum }_{p_{2}}\cdots \mathop{\sum }_{p_{n-1}}O\biggl(\frac{x\log p_{n-1}}{p_{1}\cdots p_{n-1}(\log x)^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{x}{(\log x)^{2}}\biggr)\mathop{\sum }_{p_{1}\leqslant x}\frac{1}{p_{1}}\mathop{\sum }_{p_{2}\leqslant x}\frac{1}{p_{2}}\cdots \mathop{\sum }_{p_{n-1}\leqslant x}\frac{\log p_{n-1}}{p_{n-1}}\nonumber\\ \displaystyle & = & \displaystyle O\biggl(\frac{x(\log \log x)^{n-2}}{\log x}\biggr)=o(\#P_{n}(x))\nonumber\end{eqnarray}$$

from (3) and (4). Thus, we have to estimate

(8) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p_{1}}\unicode[STIX]{x1D711}_{1}(p_{1})\cdots \mathop{\sum }_{p_{k}}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})-\unicode[STIX]{x1D6FF}_{k})\mathop{\sum }_{p_{k+1}}\unicode[STIX]{x1D6FF}_{k+1}\cdots \nonumber\\ \displaystyle & & \displaystyle \quad \mathop{\sum }_{p_{n-1}}\unicode[STIX]{x1D6FF}_{n-1}\frac{\unicode[STIX]{x1D6FF}_{n}x}{p_{1}\cdots p_{n-1}\log x}\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{x}{\log x}\biggr)\mathop{\sum }_{p_{1}}\frac{1}{p_{1}}\cdots \mathop{\sum }_{p_{k_{1}}}\frac{1}{p_{k-1}}\nonumber\\ \displaystyle & & \displaystyle \quad \biggl|\mathop{\sum }_{p_{k}}\frac{\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})-\unicode[STIX]{x1D6FF}_{k}}{p_{k}}\mathop{\sum }_{p_{k+1}}\frac{1}{p_{k+1}}\cdots \mathop{\sum }_{p_{n-1}}\frac{1}{p_{n-1}}\biggr|\end{eqnarray}$$

for $1\leqslant k\leqslant n-1$ .We write

$$\begin{eqnarray}s_{k}=\mathop{\sum }_{p_{k}<p_{k+1}\leqslant (x/(p_{1}\cdots p_{k}))^{1/(n-k)}}\frac{1}{p_{k+1}}\cdots \mathop{\sum }_{p_{n-2}<p_{n-1}\leqslant (x/(p_{1}\cdots p_{n-2}))^{1/2}}\frac{1}{p_{n-1}}\end{eqnarray}$$

for $1\leqslant k\leqslant n-2$ and $s_{n-1}=1$ . Then we prove

(9) $$\begin{eqnarray}s_{k}=\frac{(\log \log x-\log \log p_{k})^{n-k-1}}{(n-k-1)!}+O((\log \log x)^{n-k-2})\end{eqnarray}$$

for $1\leqslant k\leqslant n-1$ . If $k=n-1$ it is obvious, and if $k=n-2$ it follows from (3) and (5). Suppose $2\leqslant k\leqslant n-2$ . Put

$$\begin{eqnarray}f_{1}(t)=\frac{(\log \log x-\log \log t)^{n-k-1}}{(n-k-1)!}.\end{eqnarray}$$

Suppose now $s_{k}=f_{1}(p_{k})+O((\log \log x)^{n-k-2})$ . Let

$$\begin{eqnarray}y=\biggl(\frac{x}{p_{1}\cdots p_{k-1}}\biggr)^{1/(n-k+1)}\end{eqnarray}$$

then from (5),

$$\begin{eqnarray}\log \log y=\log \log x+O(1),\qquad f_{1}(y)=O(1).\end{eqnarray}$$

Let $C_{1}(t)=\sum _{p\leqslant t}1/p$ . Then from (3)

$$\begin{eqnarray}C_{1}(t)=\log \log t+O(1).\end{eqnarray}$$

Next we quote Abel’s summation formula [Reference Hardy and Wright7, Theorem 421]:

Proposition 6. Let $\{{c_{m}\}}_{m\in \mathbb{N}}$ be a sequence of numbers. Put

$$\begin{eqnarray}C(t)=\mathop{\sum }_{m\leqslant t}c_{m}.\end{eqnarray}$$

Let $f(t)$ be a function which is continuously differentiable for $t\geqslant 1$ . Then

$$\begin{eqnarray}\mathop{\sum }_{x<m\leqslant y}c_{m}f(m)=C(y)f(y)-C(x)f(x)-\int _{x}^{y}C(t)f^{\prime }(t)\,dt.\end{eqnarray}$$

Using this proposition we have

$$\begin{eqnarray}\displaystyle s_{k-1} & = & \displaystyle \mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}\frac{s_{k}}{p_{k}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}\frac{1}{p_{k}}f_{1}(p_{k})+O((\log \log x)^{n-k-2})\mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}\frac{1}{p_{k}}\nonumber\\ \displaystyle & = & \displaystyle C_{1}(y)f_{1}(y)-C_{1}(p_{k-1})f_{1}(p_{k-1})\nonumber\\ \displaystyle & & \displaystyle -\,\int _{p_{k-1}}^{y}C_{1}(t)f_{1}^{\prime }(t)\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle (\log \log x+O(1))O(1)-(\log \log p_{k-1}+O(1))f_{1}(p_{k-1})\nonumber\\ \displaystyle & & \displaystyle +\,\int _{p_{k-1}}^{y}(\log \log t+O(1))|f_{1}^{\prime }(t)|\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle -f_{1}(p_{k-1})\log \log p_{k-1}+(\log \log x+O(1))\int _{p_{k-1}}^{y}|f_{1}^{\prime }(t)|\,dt\nonumber\\ \displaystyle & & \displaystyle -\,\int _{p_{k-1}}^{y}(\log \log x-\log \log t)|f_{1}^{\prime }(t)|\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle -f_{1}(p_{k-1})\log \log p_{k-1}-(\log \log x+O(1))\int _{p_{k-1}}^{y}f_{1}^{\prime }(t)\,dt\nonumber\\ \displaystyle & & \displaystyle -\,\int _{p_{k-1}}^{y}\frac{(\log \log x-\log \log t)^{n-k-1}}{(n-k-2)!\,t\log t}\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle -f_{1}(p_{k-1})\log \log p_{k-1}-(\log \log x+O(1))[f_{1}(t)]_{p_{k-1}}^{y}\nonumber\\ \displaystyle & & \displaystyle -\,\biggl[-\frac{(\log \log x-\log \log t)^{n-k}}{(n-k-2)!(n-k)}\biggr]_{p_{k-1}}^{y}+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle -f_{1}(p_{k-1})\log \log p_{k-1}+f_{1}(p_{k-1})\log \log x\nonumber\\ \displaystyle & & \displaystyle -\,\frac{(\log \log x-\log \log p_{k-1})^{n-k}}{(n-k-2)!(n-k)}+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle \frac{(\log \log x-\log \log p_{k-1})^{n-k}}{(n-k-2)!}\biggl(\frac{1}{n-k-1}-\frac{1}{n-k}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & = & \displaystyle \frac{(\log \log x-\log \log p_{k-1})^{n-k}}{(n-k)!}+O((\log \log x)^{n-k-1}).\nonumber\end{eqnarray}$$

Hence we have proved (9) by induction.

Next we claim

(10) $$\begin{eqnarray}\mathop{\sum }_{\quad p_{k-1}<p_{k}\leqslant (x/(p_{1}\cdots p_{k-1}))^{1/(n-k+1)}}\frac{\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})-\unicode[STIX]{x1D6FF}_{k}}{p_{k}}s_{k}=O((\log \log x)^{n-k-1})\end{eqnarray}$$

for $1\leqslant k\leqslant n-1$ . In fact, let

$$\begin{eqnarray}\displaystyle y & = & \displaystyle \biggl(\frac{x}{p_{1}\cdots p_{k-1}}\biggr)^{1/(n-k+1)};\nonumber\\ \displaystyle f_{2}(t) & = & \displaystyle \frac{(\log \log x-\log \log t)^{n-k-1}}{(n-k-1)!\,t};\nonumber\\ \displaystyle C_{2}(t) & = & \displaystyle \mathop{\sum }_{p\leqslant t}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p)-\unicode[STIX]{x1D6FF}_{k}).\nonumber\end{eqnarray}$$

Then from (9) we have

$$\begin{eqnarray}\frac{s_{k}}{p_{k}}=f_{2}(p_{k})+O\biggl(\frac{(\log \log x)^{n-k-2}}{p_{k}}\biggr),\end{eqnarray}$$

and from Lemma 3,

$$\begin{eqnarray}C_{2}(t)=O\biggl(\frac{t}{(\log t)^{1+\unicode[STIX]{x1D700}}}\biggr)\end{eqnarray}$$

for $t\geqslant p_{k-1}$ . Similarly as before, by Proposition 6,

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}\frac{\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})-\unicode[STIX]{x1D6FF}_{k}}{p_{k}}s_{k}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}(\unicode[STIX]{x1D711}_{p_{1}\cdots p_{k-1}}(p_{k})-\unicode[STIX]{x1D6FF}_{k})f_{2}(p_{k})+O((\log \log x)^{n-k-2})\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{p_{k-1}<p_{k}\leqslant y}\frac{1}{p_{k}}=C_{2}(y)f_{2}(y)-C_{2}(p_{k-1})f_{2}(p_{k-1})\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\int _{p_{k-1}}^{y}C_{2}(t)f_{2}^{\prime }(t)\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{y}{(\log y)^{1+\unicode[STIX]{x1D700}}}\biggr)O\biggl(\frac{1}{y}\biggr)+O\biggl(\frac{p_{k-1}}{(\log p_{k-1})^{1+\unicode[STIX]{x1D700}}}\biggr)O\biggl(\frac{(\log \log x)^{n-k-1}}{p_{k-1}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{p_{k-1}}^{y}O\biggl(\frac{t}{(\log t)^{1+\unicode[STIX]{x1D700}}}\biggr)|f_{2}^{\prime }(t)|\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{p_{k-1}}^{y}O\biggl(\frac{t}{(\log t)^{1+\unicode[STIX]{x1D700}}}\biggr)O\biggl(\frac{(\log \log x)^{n-k-1}}{t^{2}}\biggr)\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =O((\log \log x)^{n-k-1})\int _{p_{k-1}}^{y}\frac{1}{t(\log t)^{1+\unicode[STIX]{x1D700}}}\,dt+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =O((\log \log x)^{n-k-1})\biggl[-\frac{1}{(\log t)^{\unicode[STIX]{x1D700}}}\biggr]_{p_{k-1}}^{y}+O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =O((\log \log x)^{n-k-1}).\nonumber\end{eqnarray}$$

Thus (10) follows.

Therefore, (8) becomes

$$\begin{eqnarray}\displaystyle & & \displaystyle O\biggl(\frac{x}{\log x}\biggr)\mathop{\sum }_{p_{1}}\frac{1}{p_{1}}\cdots \mathop{\sum }_{p_{k-1}}\frac{1}{p_{k-1}}O((\log \log x)^{n-k-1})\nonumber\\ \displaystyle & & \displaystyle \quad =O\biggl(\frac{x(\log \log x)^{n-2}}{\log x}\biggr)=o(\#P_{n}(x)).\nonumber\end{eqnarray}$$

Hence we have completed the proof.

5 Densities for the remaining groups

We cannot get the densities for the remaining groups because we do not know any conditions for them in which Theorem 2 can be applied. We want to know about the conditions $h_{2}(-pq)=2^{n},h_{2}(qr)=2^{n}$ or $\mathbf{N}\unicode[STIX]{x1D700}_{qr}=1$ in detail.

Cohn and Lagarias [Reference Cohn and Lagarias2] considered $\operatorname{Cl}_{2}(\mathbb{Q}(\sqrt{dp}))$ as $p$ varies where $d\not \equiv 2~(4)$ is an integer. They conjectured that the splitting of primes in a number field determines the structure of $\operatorname{Cl}_{2}(\mathbb{Q}(\sqrt{dp}))$ , and they also considered numerically. Here we consider $h_{2}(dp)$ and $h_{2}^{+}(dp)$ where $d$ is an odd prime discriminant.

We suggest the following conjecture:

Note that $h_{2}(pq)=h_{2}^{+}(pq)$ if and only if $\mathbf{N}\unicode[STIX]{x1D700}_{pq}=-1$ .

Stevenhagen [Reference Stevenhagen16] showed that the narrow $8$ -ranks of quadratic fields are determined by number fields. Hence the cases $n\leqslant 2$ in both 1 and 2 of Conjecture 1 are known to be true. Unfortunately however, Milovic [Reference Milovic13] gives some evidence against Conjecture 1 for $n\geqslant 3$ .

We calculated the numbers of $pq<10^{7}$ ( $p,q:$ primes) with $-p\equiv q\equiv 1~(4),h_{2}(-pq)=2^{n}$ ; and also with $p\equiv q\equiv 1~(4)$ and $h_{2}(pq)h_{2}^{+}(pq)/2=2^{n}$ for each $n\in \mathbb{N}$ (using SageMath 6.10); see Table 2. Note that both $p$ and $q$ vary.

Table 2. Numbers of $pq<10^{7}$ with $h_{2}(-pq)=2^{n}$ and with $h_{2}(pq)h_{2}^{+}(pq)/2=2^{n}$ .

Table 3. Conjectured densities.

This conjecture yields the densities for the remaining groups:

Theorem 4. Assume that Conjecture 1 holds. In addition, we assume that each $F_{p}$ in Conjecture 1 satisfies

$$\begin{eqnarray}\log |d_{F_{p}}|=O\biggl(\frac{\sqrt{p}}{(\log p)^{1+\unicode[STIX]{x1D700}}}\biggr),\qquad [F_{p}:\mathbb{Q}]=O\biggl(\frac{\sqrt{p}}{(\log p)^{2+\unicode[STIX]{x1D700}}}\biggr).\end{eqnarray}$$

Moreover we assume GRH.

Then the results in Table 3 hold.

Proof. From our assumptions, we can apply Theorem 2. We omit the details.

Note that $\unicode[STIX]{x1D6FF}(\cdot /B(V_{4}))$ is not countably additive in general. In this case, however, we can calculate the densities of $A_{Q}=\bigcup _{n=3}^{\infty }A(Q_{2^{n}})$ and $A_{D}=\bigcup _{n=3}^{\infty }A(D_{2^{n}})$ . In fact, for $N,x\in \mathbb{N}$

$$\begin{eqnarray}\mathop{\sum }_{n=3}^{N}\#A(Q_{2^{n}})(x)=\#\mathop{\bigcup }_{n=3}^{N}A(Q_{2^{n}})(x)\leqslant \#A_{Q}(x)\end{eqnarray}$$

where $A(x)=\{\,m\in A\,|\,m\leqslant x\,\}$ for $A\subseteq \mathbb{N}$ . Hence,

$$\begin{eqnarray}\mathop{\sum }_{n=3}^{N}\unicode[STIX]{x1D6FF}(A(Q_{2^{n}})/B(V_{4}))=\mathop{\sum }_{n=3}^{N}\lim _{x\rightarrow \infty }\frac{\#A(Q_{2^{n}})(x)}{\#B(V_{4})(x)}\leqslant \liminf _{x\rightarrow \infty }\frac{\#A_{Q}(x)}{\#B(V_{4})(x)}\end{eqnarray}$$

for $N\in \mathbb{N}$ . Hence,

$$\begin{eqnarray}\frac{2}{7}=\mathop{\sum }_{n=3}^{\infty }\unicode[STIX]{x1D6FF}(A(Q_{2^{n}})/B(V_{4}))\leqslant \liminf _{x\rightarrow \infty }\frac{\#A_{Q}(x)}{\#B(V_{4})(x)}.\end{eqnarray}$$

Similarly,

$$\begin{eqnarray}\frac{1}{7}\leqslant \liminf _{x\rightarrow \infty }\frac{\#A_{D}(x)}{\#B(V_{4})(x)}.\end{eqnarray}$$

On the other hand, since

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(A_{Q}\cup A_{D}/B(V_{4}))=1-\unicode[STIX]{x1D6FF}(A(V_{4})/B(V_{4}))-\unicode[STIX]{x1D6FF}\biggl(\mathop{\bigcup }_{n\geqslant 4}A(\mathit{SD}_{2^{n}})/B(V_{4})\biggr)=\frac{3}{7},\end{eqnarray}$$

we have

$$\begin{eqnarray}\displaystyle \limsup _{x\rightarrow \infty }\frac{\#A_{Q}(x)}{\#B(V_{4})(x)} & = & \displaystyle \limsup _{x\rightarrow \infty }\biggl(\frac{\#(A_{Q}(x)\cup A_{D}(x))}{\#B(V_{4})(x)}-\frac{\#A_{D}(x)}{\#B(V_{4})(x)}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \lim _{x\rightarrow \infty }\frac{\#(A_{Q}(x)\cup A_{D}(x))}{\#B(V_{4})(x)}-\liminf _{x\rightarrow \infty }\frac{\#A_{D}(x)}{\#B(V_{4})(x)}\leqslant \frac{2}{7}.\nonumber\end{eqnarray}$$

Putting these together, we obtain that all these inequalities are equalities and $\liminf =\limsup$ . It is similar for $A_{D}$ .◻