Proof. If $L/F$ is an $S_3$ cubic extension with $\mathrm {Disc}(L/F)=n$ , then its associated quadratic resolvent $K/F$ has $ {\mathrm {Disc}}(K/F) = m$ and $m| {\mathrm {Disc}}(L/F)$ . By class field theory, quadratic extensions of F are in bijection to surjective homomorphisms from the idèle class group $C_F = \mathbb {A}^{\times }_F/F^{\times }$ to $C_2$ , where $\mathbb {A}^{\times }_F$ is the idèle group. We have the following exact sequence of $C_F$ ,

$$ \begin{align*}1\rightarrow O_F^{\times} \rightarrow \prod_v O_{v}^{\times}\rightarrow C_F \rightarrow {\mathrm{Cl}}_F\rightarrow 1,\end{align*} $$

where the product is over places v of F and $\mathcal {O}_v^\times $ is, at finite places, the group of elements of the completion $F_v$ with valuation $0$ , and, at infinite places, the group of nonzero elements of the completion $F_v$ . Since $ {\mathrm {Hom}}(\cdot , A)$ is left exact, we have

$$ \begin{align*}1\rightarrow {\mathrm{Hom}}({\mathrm{Cl}}_F, C_2 ) \rightarrow {\mathrm{Hom}}(C_F, C_2) \rightarrow {\mathrm{Hom}}( \prod_v O_{v}^{\times}, C_2).\end{align*} $$

Here $| {\mathrm {Hom}}( {\mathrm {Cl}}_F, C_2)| = h_2(F)$ . The global Artin map of class field theory maps $O_v^\times \subset C_F$ to the inertia group of $ {\mathrm {Gal}}(F^{ab}/F)$ , and the relative discriminant ideal of an extension only depends on the image of the inertia groups in the Galois group (e.g., since the discriminant is given by the Artin conductor of the permutation representation). Thus given a $\rho \in {\mathrm {Hom}}(C_F, C_2)$ , its image in $ {\mathrm {Hom}}( \prod _v O_{v}^{\times }, C_2)$ determines the relative discriminant ideal of the quadratic field corresponding to $\rho $ . Therefore for each fixed m, the number of possible $L/F$ with $ {\mathrm {Disc}}(L/F)=m$ is bounded by $h_2(F)$ . For each fixed n, there are at most $O_{[F:{\mathbb {Q}}],\epsilon }(n^{\epsilon })$ possible m, therefore there are at most $O_{[F:\mathbb {Q}],\epsilon }(n^\epsilon h_2(F))$ quadratic extensions $K/F$ with $\mathrm {Disc}(L/F)|n$ .

Similarly, for each fixed quadratic extension $K/F$ , any associated $S_3$ cubic extension $L/F$ has Galois closure $\tilde {L}/F$ such that $\tilde {L}/K$ has $ {\mathrm {Gal}}(\tilde {L}/K) = C_3$ . Thus, we can use class field theory again to determine that, for each fixed $K/F$ , the number of associated $S_3$ cubic extensions with $ {\mathrm {Disc}}(L/F)|n$ is

$$ \begin{align*}\ll_{[F:\mathbb{Q}],\epsilon} n^{\epsilon} h_3(K/F) \ll_{[F:\mathbb{Q}],\epsilon} n^{\epsilon}\mathrm{Disc}(K/F)^{1/2+\epsilon}\mathrm{Disc}(F)^{1/2+\epsilon} \ll_{[F:\mathbb{Q}],\epsilon} n^{1/2+\epsilon} \mathrm{Disc}(F)^{1/2+\epsilon},\end{align*} $$

by the trivial bound on $h_3(K/F)$ in Lemma 2.4. Altogether, the total number of $S_3$ -extensions is now $O_{[F:\mathbb {Q]},\epsilon }(n^{1/2+\epsilon } h_2(F) \mathrm {Disc}(F)^{1/2+\epsilon })$ , as claimed. As above, we can bound the number of $C_3$ -extensions $K/F$ by $O_{[F:\mathbb {Q}],\epsilon }(n^\epsilon h_3(F))$ and the lemma follows.

Proof. We provide references for these claims without regard to necessarily providing the original reference. The first claim follows from [Reference WrightWri85, Theorem 4.1]. The second through fourth primarily follow from [Reference Datskovsky and WrightDW86, Theorem 6.2], with the claim that $\hat \xi _{F,\alpha }(s) < \xi _{F,\alpha }(s)$ for real $s>1$ following from the absolute convergence of $\xi _{F,\alpha }(s)$ in this region. The fifth follows from a generalization of [Reference ShintaniShi72, Theorem 2.1.iii] as is explained on [Reference WrightWri82, pg. 96].

Proof. From the formulas given, it follows that each $c_{\alpha \beta }(s)$ has a zero of order $[F:\mathbb {Q}]$ at $s=1$ . The claim then follows from the functional equation for $\xi _{F,\alpha }(s)$ .

Proof. By the triangle inequality, it suffices to prove the lemma when $t=0$ . Assume $\sigma>3/2$ is given. With the notation of Lemma 3.9, for any cubic étale algebra A, we have $\zeta _A(\sigma ) \leq \zeta _F(\sigma )^3$ whenever $\sigma> 1$ . It follows that

(3.2) $$ \begin{align} \begin{aligned} \xi_{F,\alpha}(\sigma) & \leq \sum_{[L:F]\leq 3} \frac{1}{\mathrm{Disc}(L/F)^\sigma} \cdot \zeta_F(4\sigma) \zeta_F(6\sigma-1) \frac{\zeta_A(2\sigma)}{\zeta_A(4\sigma)} \\ & \leq \zeta_F(2\sigma)^3\zeta_F(4\sigma)\zeta_F(6\sigma-1) \sum_{[L:F]\leq 3} \frac{1}{\mathrm{Disc}(L/F)^\sigma}, \end{aligned} \end{align} $$

where the sum runs over isomorphism classes of field extensions $L/F$ of degree at most $3$ . The second inequality comes from $\zeta _A(4\sigma )\ge 1$ , valid for $\sigma>1$ . The factors of $\zeta _F(\sigma )$ may be bounded by suitable powers of the Riemann zeta function $\zeta (\sigma )$ , depending only on the degree of $F/\mathbb {Q}$ . So it suffices to understand the Dirichlet series of field discriminants. By Lemma 3.5, we then find that for any $\epsilon < \sigma - 3/2$ ,

$$\begin{align*}\sum_{L/F} \frac{1}{\mathrm{Disc}(L/F)^\sigma} \ll_{[F:\mathbb{Q}],\epsilon} h_2(F)\mathrm{Disc}(F)^{1/2+\epsilon} \sum_{n}\frac{n^{1/2+\epsilon}}{n^\sigma} \ll_{[F:{\mathbb{Q}}], \epsilon} \mathrm{Disc}(F)^{1/2+\epsilon}h_2(F) \zeta(\sigma-1/2-\epsilon), \end{align*}$$

yielding the lemma.

Proof. Let $\delta> 0$ be small. For $\sigma = 3/2+\delta $ , it follows from Lemma 3.10 that

$$\begin{align*}\xi_F(\sigma+it) \ll_{[F:\mathbb{Q}],\delta,\epsilon} {\mathrm{Disc}}(F)^{1/2+\epsilon} h_2(F). \end{align*}$$

For $s = \sigma + it$ with $\sigma = 3/2 + \delta $ , we apply the functional equation and Stirling’s formula to extract cancellation between the gamma factors and the $c_{\alpha \beta }(s)$ in order to find also

$$ \begin{align*} \xi_{F,\alpha}(1-s) &= \left(\frac{3^{(6s-2)}}{\pi^{4s}}\Gamma(s)^{2}\Gamma(s-1/6)\Gamma(s+1/6)\right)^{[F:\mathbb{Q}]} \mathrm{Disc}(F)^{4s-2} \sum_{\beta} c_{\alpha\beta}(s) \hat\xi_{F,\beta}(s) \\ &\ll_{[F:\mathbb{Q}],\delta,\epsilon} (1+|t|)^{(4+4\delta)[F:\mathbb{Q}]+\epsilon} \mathrm{Disc}(F)^{9/2+4\delta+\epsilon} h_2(F). \end{align*} $$

The general result follows upon using the Phragmen–Lindelöf principle [Reference Iwaniec and KowalskiIK04, Theorem 5.53] applied to the function $\frac {(s-1)(s-5/6)}{(s+1)^2} \xi _{F,\alpha }(s)$ and letting $\delta $ be sufficiently small. Specifically, we find for any $-1/2- \delta < \sigma < 3/2 + \delta $ that

$$\begin{align*}\xi_{F,\alpha}(\sigma+it) \ll_{[F:\mathbb{Q}],\delta,\epsilon} (1+|t|)^{2[F:\mathbb{Q}](\frac{3}{2}+\delta-\sigma)+\epsilon} {\mathrm{Disc}}(F)^{\frac{7}{2}+2\delta-2\sigma+\epsilon}h_2(F). \end{align*}$$

Upon choosing $\delta = \epsilon $ and then replacing $\epsilon $ in the equation above by $\epsilon /(1+2[F:\mathbb {Q}])$ , we obtain the required statement.

Proof of Proposition 3.6.

Let $\phi (x)$ be a positive smooth function satisfying $\phi (x)\geq 1$ whenever ${0\leq x \leq 1}$ and whose Mellin transform exhibits rapid decay away from the real axis; for example, we may take $\phi (x)=e^{1-x}$ . Letting R run over cubic rings over $\mathcal {O}_F$ , we have

$$\begin{align*}N_{\mathcal{O}_F,\alpha}(X) = \sum_{\substack{ 0<\mathrm{Disc}(R/\mathcal{O}_F) \leq X \\ \mathrm{sgn}(R) = \alpha}} 1 \ll \sum_{\substack{0<\mathrm{Disc}(R/\mathcal{O}_F) \\ \mathrm{sgn}(R) = \alpha}} \frac{\phi(\mathrm{Disc}(R/\mathcal{O}_F)/X)}{|\mathrm{Aut}_{\mathcal{O}_F}(R)|}. \end{align*}$$

This latter term may be evaluated via Perron’s formula and the full Shintani zeta function $\xi _{F,\alpha }(s)$ ,

$$\begin{align*}\sum_{\substack{0<\mathrm{Disc}(R/\mathcal{O}_F) \\ \mathrm{sgn}(R) = \alpha}} \frac{\phi(\mathrm{Disc}(R/\mathcal{O}_F)/X)}{|\mathrm{Aut}_{\mathcal{O}_F}(R)|} = \frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} \xi_{F,\alpha}(s) \Phi(s) X^s \,ds, \end{align*}$$

where $\Phi (s)$ is the Mellin transform of $\phi (x)$ . By Lemma 3.11, the exponential decay of $\Phi (s)$ off the real axis allows us to shift the contour to the line $\Re (s)=-1/2-\epsilon $ for any $\epsilon>0$ . Doing so, we must account for the poles of $\xi _{F,\alpha }(s)$ at $s=1$ and $s=5/6$ and of $\Phi (s)$ at $s=0$ . From Proposition 3.7, it follows that the contribution from the pole at $s=1$ subsumes that of the pole at $s=5/6$ . Moreover, Lemma 3.8 shows that in fact the integrand has no pole at $s=0$ owing to the zero of $\xi _{F,\alpha }(s)$ at $s=0$ . Altogether, the contribution from the poles is seen to be $O_{[F:\mathbb {Q}]}(X\cdot \mathrm {Res}_{s=1}\zeta _F(s))$ .

Finally, appealing to the functional equation and Lemma 3.10, we find

$$ \begin{align*} &\frac{1}{2\pi i} \int_{-1/2-\epsilon-i\infty}^{-1/2-\epsilon+i\infty} \xi_{F,\alpha}(s) \Phi(s) X^s \,ds \\& \quad = \frac{1}{2\pi i} \int_{3/2+\epsilon-i\infty}^{3/2+\epsilon+i\infty} \!\left(\frac{3^{(6s-2)}}{\pi^{4s}}\Gamma(s)^{2}\Gamma(s-1/6)\Gamma(s+1/6)\right)^{\![F:\mathbb{Q}]}\! \mathrm{Disc}(F)^{4s-2} \hat\xi(s) \Phi(1-s) X^{1-s} \,ds \\& \quad \ll_{[F:\mathbb{Q}],\epsilon} \mathrm{Disc}(F)^{9/2+\epsilon} h_2(F) X^{-1/2-\epsilon}, \end{align*} $$

where $\hat \xi (s) = \sum _{\beta }c_{\alpha \beta }(s)\hat \xi _{F,\beta }(s)$ .

We conclude that for any number field F and any $X \geq 1$ , we have

(3.3) $$ \begin{align} N_{\mathcal{O}_F,\alpha}(X) \ll_{[F:\mathbb{Q}],\epsilon} X \cdot \mathrm{Res}_{s=1} \zeta_F(s) + X^{-1/2-\epsilon} h_2(F) \mathrm{Disc}(F)^{9/2+\epsilon}. \end{align} $$

Using the well-known upper bound $\mathrm {Res}_{s=1} \zeta _F(s) \ll _{[F:\mathbb {Q}],\epsilon } \mathrm {Disc}(F)^\epsilon $ due to Landau (see also [Reference BrauerBra47, Lemma 3]), we conclude the proposition.

Proof. If $R \in \mathcal {R}(K)$ , then $\mathrm {Disc}(R/\mathcal {O}_F) = \mathrm {Disc}(K/F) [\mathcal {O}_A:R]^2$ . Thus, with the notation of Lemma 3.9, we have

$$\begin{align*}\sum_{R \in \mathcal{R}(K)} \frac{|\mathrm{Aut}_{F}(A)|^{-1}}{\mathrm{Disc}(R/\mathcal{O}_F)^s} = \mathrm{Disc}(K/F)^{-s} \cdot\left(\frac{1}{2}f_{F\times K}(s) + \sum_{\substack{[L:F] =3 \\ {\mathrm{disc}}(L/F) = {\mathrm{disc}}(K/F) \\ K \subseteq \widetilde{L}/F}} f_L(s) \right). \end{align*}$$

Note that the summation on the right is restricted to those cubic extensions $L/F$ with quadratic resolvent K and which satisfy $ {\mathrm {disc}}(L/F) = {\mathrm {disc}}(K/F)$ . For each such $L/F$ , by Lemma 3.9 we have

(3.4) $$ \begin{align} f_L(s) =\zeta_F(4s) \zeta_F(6s-1) \frac{\zeta_L(2s)}{\zeta_L{(4s)}} = \zeta_F(2s) \zeta_F(6s-1)\frac{L(2s,\chi_{\widetilde{L}/K})}{L(4s,\chi_{\widetilde{L}/K})}, \end{align} $$

where $\widetilde {L}$ is the Galois closure of L over F, and $\chi _{\widetilde {L}/K}$ is one of the nontrivial cyclic cubic characters of the class group $\mathcal {C}= \mathrm {Cl}_K/\mathrm {Cl}_F$ that cuts out $\widetilde {L}/K$ . The second equality comes from the relation $\zeta _L(s)/\zeta _F(s)= L(s, \chi _{\widetilde {L}/K})$ . In fact, we have $L(s,\chi _{\widetilde {L}/K}) = L(s,\chi _{\widetilde {L}/K}^2) = L(s, \overline {\chi }_{\widetilde {L}/K})$ , since all are equal to the irreducible degree $2$ Artin L-function of $L/F$ . Thus, we obtain also

$$\begin{align*}f_L(s) = \zeta_F(2s) \zeta_F(6s-1)\frac{L(2s,\chi_{\widetilde{L}/K})}{L(4s,\chi_{\widetilde{L}/K}^2)}. \end{align*}$$

Meanwhile, the Dirichlet series for orders inside $K\times F$ is

(3.5) $$ \begin{align} f_{K\times F}(s) = \zeta_F(2s) \zeta_F(6s-1)\frac{\zeta_K(2s)}{\zeta_K(4s)}. \end{align} $$

Notice that $\zeta _K(s)$ is the L-function attached to the trivial character of $\mathcal {C}$ . Therefore,

$$ \begin{align*} \sum_{R \in \mathcal{R}(K)} \frac{|\mathrm{Aut}_{F}(A)|^{-1}}{\mathrm{Disc}(R/\mathcal{O}_F)^s} & = \frac{ \zeta_F(2s) \zeta_F(6s-1)}{2\mathrm{Disc}(K/F)^{s}} \sum_{\chi \in \widehat{\mathcal{C}}[3]} \frac{L(2s,\chi_{\widetilde{L}/K})}{L(4s,\chi_{\widetilde{L}/K}^2)} \\ &= \frac{ \zeta_F(2s) \zeta_F(6s-1)}{2\mathrm{Disc}(K/F)^{s}} \sum_{\mathfrak{a}\subseteq\mathcal{O}_K\,\text{sq. free}} \frac{1}{{\mathrm{Nm}}(\mathfrak{a})^{2s}} \sum_{\chi \in \widehat{\mathcal{C}}[3]}\chi(\mathfrak{a})\\ & =\frac{ h_3(K/F)\zeta_F(2s) \zeta_F(6s-1)}{2\mathrm{Disc}(K/F)^{s}} \sum_{\substack{\mathfrak{a}\subseteq \mathcal{O}_K\,\text{sq. free}\\ \mathfrak{a}\in 3{\mathrm{Cl}}_K + {\mathrm{Cl}}_F }} \frac{1}{{\mathrm{Nm}}(\mathfrak{a})^{2s}}, \end{align*} $$

by orthogonality of characters and the relation $|\widehat {\mathcal {C}}[3]| = h_3(K/F)$ .

Proof. Let $m \geq \frac {[F:\mathbb {Q}]}{2}+2$ be an integer. Then the number of ideals $\mathfrak {a} \subseteq \mathcal {O}_L$ with $ {\mathrm {Nm}}(\mathfrak {a}) \leq X$ satisfies

$$ \begin{align*} \#\{\mathfrak{a} \subseteq \mathcal{O}_L \! : \! {\mathrm{Nm}}(\mathfrak{a}) \leq X\} \geq \!\sum_{{\mathrm{Nm}}(\mathfrak{a}) \leq X} \!\left(1 - \frac{{\mathrm{Nm}}(\mathfrak{a})}{X}\right)^{\!\!m} \!\!= \frac{m!}{2\pi i} \int_{2-i\infty}^{2+i\infty} \zeta_F(s) X^s \frac{ds}{s(s+1)\dots(s+m)}. \end{align*} $$

The convexity bound for $\zeta _F(s)$ (see [Reference Iwaniec and KowalskiIK04, Equation (5.20)], for example) implies that when $\sigma :=\Re (s)$ lies in the critical strip $0 \leq \sigma \leq 1$ , we have

(3.6) $$ \begin{align} (s-1) \zeta_F(s) \ll_{[F:\mathbb{Q}],\epsilon} \mathrm{Disc}(F)^{(1-\sigma)/2+\epsilon} (1+|s|)^{\frac{[F:\mathbb{Q}](1-\sigma)}{2}+1+\epsilon}. \end{align} $$

Thus, the integral above converges absolutely when the line of integration is moved anywhere inside the critical strip. Taking it arbitrarily close to the line $\Re (s)=0$ , we conclude that

$$\begin{align*}\#\{\mathfrak{a} \subseteq \mathcal{O}_L \text{ ideal}: {\mathrm{Nm}}(\mathfrak{a}) \leq X\} \gg_{[F:\mathbb{Q}],\epsilon} X \cdot\mathrm{Res}_{s=1} \zeta_F(s) + O(X^\epsilon \mathrm{Disc}(F)^{\frac{1}{2}+\epsilon}). \end{align*}$$

The result follows.

Proof. Let Y be given by

(3.7) $$ \begin{align} Y := \max\left\{ { \mathrm{Disc}(F)^3 h_2(F)^{2/3}},\kappa_F^{-2} {X C^2\mathrm{Disc}(F)^{1+\epsilon}} \right\}, \end{align} $$

where C is the constant from Lemma 3.14. For each quadratic extension $K/F$ with discriminant at most X, we consider the contribution to $N_{\mathcal {O}_F,3}(Y)$ from the orders in the family $\mathcal {R}(K)$ from Lemma 3.13. For any $R \in \mathcal {R}(K)$ , any ring automorphism extends to an automorphism of the étale algebra $A = R \otimes _{\mathbb {Q}} F$ . Thus, $|\mathrm {Aut}_{\mathcal {O}_F} (R)| \leq |\mathrm {Aut}_F(A)|$ , and we obtain an inequality of generating functions,

$$\begin{align*}\sum_{R \in \mathcal{R}(K)} \frac{|\mathrm{Aut}_{\mathcal{O}_F} (R)|^{-1}}{\mathrm{Disc}(R/\mathcal{O}_F)^s} \geq \sum_{R \in \mathcal{R}(K)} \frac{|\mathrm{Aut}_{F}(A)|^{-1}}{\mathrm{Disc}(R/\mathcal{O}_F)^s}, \end{align*}$$

which is the Dirichlet series from Lemma 3.13. Appealing to Lemma 3.13, this series is in turn bounded below by

$$\begin{align*}\frac{h_3(K/F)}{2 \mathrm{Disc}(K/F)^{s}} \zeta_F(2s) \zeta_F(6s-1) \geq \frac{h_3(K/F)}{2 \mathrm{Disc}(K/F)^s} \zeta_F(2s), \end{align*}$$

since the summation over ideals $\mathfrak {a}$ has non-negative coefficients. It then follows from our assumption $Y\geq \kappa _F^{-2} {X C^2\mathrm {Disc}(F)^{1+\epsilon }} $ that we may appeal to Lemma 3.14 (with $X = Y^{1/2}/\mathrm {Disc}(K/F)^{1/2}$ ) to find that the contribution to $N_{\mathcal {O}_F,3}(Y)$ of orders arising in $\mathcal {R}(K)$ is

$$\begin{align*}\gg_{[F:\mathbb{Q}],\epsilon} \frac{h_3(K/F) Y^{1/2} \kappa_F }{\mathrm{Disc}(K/F)^{1/2}}. \end{align*}$$

Thus,

$$\begin{align*}N_{\mathcal{O}_F,3}(Y) \gg_{[F:\mathbb{Q}],\epsilon} \sum_{\substack{[K:F]=2 \\ X/2 \leq \mathrm{Disc}(K/F) \leq X}} \frac{h_3(K/F) Y^{1/2} \kappa_F }{\mathrm{Disc}(K/F)^{1/2}}, \end{align*}$$

and we conclude that

$$\begin{align*}\sum_{\substack{[K:F]=2 \\ X/2 \le \mathrm{Disc}(K/F) \leq X}} h_3(K/F) \ll_{[F:\mathbb{Q}],\epsilon} \frac{N_{\mathcal{O}_F,3}(Y) X^{1/2} }{Y^{1/2} \kappa_F}. \end{align*}$$

Finally, by Proposition 3.6, we have $N_{\mathcal {O}_F,3}(Y) \ll _{[F:\mathbb {Q}],\epsilon } \mathrm {Disc}(F)^{\epsilon /2} Y$ whenever $ Y\geq \mathrm {Disc}(F)^3 h_2(F)^{2/3},$ and the result follows using dyadic summation and the fact that $\kappa _F \ll _{[F:\mathbb {Q}],\epsilon } \mathrm {Disc}(F)^{\epsilon /2}$ .

Proof of Proposition 3.12.

Following the proof of Lemma $6.2$ in [Reference Datskovsky and WrightDW88], for a square-free ideal $\mathfrak {q}$ of F of norm $ {\mathrm {Nm}} \mathfrak {q} \leq X^{1/2}$ , the number of $S_3$ cubic fields $L/F$ with quadratic resolvent $K/F$ and $ {\mathrm {disc}}(L/F) = \mathfrak {q}^2 {\mathrm {disc}}(K/F)$ is bounded by $O(4^{\omega (\mathfrak {q})} h_3(K/F))$ , where $\omega (\mathfrak {q})$ denotes the number of prime ideals dividing $\mathfrak {q}$ . Thus, for fixed $\mathfrak {q}$ , the number of $S_3$ -extensions $L/F$ with $ {\mathrm {disc}}(L/F)\leq X$ whose discriminant is $\mathfrak {q}^2$ times that of their quadratic resolvent, may be bounded by a uniform constant times

$$\begin{align*}\sum_{\substack{[K:F] = 2 \\ {\mathrm{Disc}}(K/F) \leq \frac{X}{{\mathrm{Nm}}(\mathfrak{q})^2}}} \!\!\!\!\!4^{\omega(\mathfrak{q})} h_3(K/F) \ll_{[F:\mathbb{Q}],\epsilon} 4^{\omega(\mathfrak{q})} \mathrm{Disc}(F)^{1/2+\epsilon} \kappa_F^{-2} \left( \frac{\mathrm{Disc}(F) h_2(F)^{1/3}X^{1/2}}{{\mathrm{Nm}}(\mathfrak{q})} + \frac{X}{{\mathrm{Nm}}(\mathfrak{q})^{2}} \right), \end{align*}$$

the latter inequality provided by Proposition 3.15. Next we will sum these expressions over $\mathfrak {q}$ with norm at most $X^{1/2}$ in order to bound $N_F(S_3, X)$ . The Dirichlet series $\sum _{\mathfrak {q}} \frac {4^{\omega (q)}}{ {\mathrm {Nm}} (\mathfrak {q})^s}$ is bounded term by term by $\zeta (s)^{4[F:{\mathbb {Q}}]}$ . Thus it follows that

$$ \begin{align*} N_F(S_3, X) \ll_{[F:{\mathbb{Q}}],\epsilon} \mathrm{Disc}(F)^{3/2+\epsilon} \kappa_F^{-2} h_2(F)^{1/3} X^{1/2+\epsilon} + \mathrm{Disc}(F)^{1/2+\epsilon} \kappa_F^{-2} X. \end{align*} $$

If the second summand above is bigger than the first, then the proposition immediately follows. Otherwise, $X^{1/2-\epsilon }\leq \mathrm {Disc}(F) h_2(F)^{1/3}$ and the first summand is bigger. If also $\epsilon \leq 1/4$ , then $X\ll _{[F:{\mathbb {Q}}]} \mathrm {Disc}(F)^8$ and so the $X^\epsilon $ in the first term can be subsumed into the $\mathrm {Disc}(F)^\epsilon $ , and the proposition follows. The $\epsilon>1/4$ case of the proposition immediately follows from the $\epsilon \leq 1/4$ case.

Proof of Lemma 5.7.

Writing $D := {\mathrm {Disc}}(F_1)$ and $h:=h_2(F_1)$ , from Section 3 (and the translation from relative to absolute discriminant), we have

(5.5) $$ \begin{align} \sum_{\substack{F_2/ F_1\\ X/2\le {\mathrm{Disc}}(F_2)\le X}} h_3(F_2/F_1) \ll_{[k:{\mathbb{Q}}],\epsilon} h \cdot D^{\epsilon} \begin{cases} X^{3/2} D^{-5/2}, & \text{for all } D^2\leq X \leq D^3, \\ X^{1/2} D^{1/2} +X D^{-3/2}, & \text{for all } D^2\leq X, \\ X D^{-2}, & \text{if } X \geq D^5 h^{2/3}. \end{cases} \end{align} $$

Specifically, the first line follows from Proposition 3.4 and (3.8), the second from Proposition 3.15, and the third from the final case of Theorem 3.1.

We will use the first line of (5.5) for $ {\mathrm {Disc}}(F_1) \in (X^{1/3+\delta _0}, X^{1/2}]$ , the second line for $ {\mathrm {Disc}}(F_1) \in [Y_0, X^{1/3-\delta _0})$ , and the third line for $ {\mathrm {Disc}}(F_1) \in [Y, Y_0)$ , where we choose a value for $Y_0$ so that we can apply the third line to the range $[Y, Y_0)$ for every $F_1/k$ . In particular, if $C_{[K:{\mathbb {Q}}]}$ is such that we always have $h(K)^{2/3}\leq C_{[K:{\mathbb {Q}}]} {\mathrm {Disc}}(K)$ , we can choose $Y_0=C_{2[k:{\mathbb {Q}}]}^{-1/6}X^{1/6}$ . Applying (5.5) as described and applying Lemma 5.6 to bound h, we find

(5.6) $$ \begin{align} \begin{aligned} \sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{N}(X,Y)\\ X/2\le {\mathrm{Disc}}(F_2)\le X}} h_3(F_2/k) & \ll_{\epsilon',\epsilon, [k:{\mathbb{Q}}]} h_2(k)^{2} \cdot X^{3/2} \!\!\!\sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in (X^{1/3+\delta_0}, X^{1/2})}} \!\!\! h_3(F_1/k) {\mathrm{Disc}}(F_1)^{-5/2+\epsilon'}\\ & \quad + h_2(k)^{2} \cdot X^{1/2} \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in [Y_0,X^{1/3-\delta_0})}} h_3(F_1/k) {\mathrm{Disc}}(F_1)^{1/2+\epsilon'}\\ & \quad + h_2(k)^{2} \cdot X \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in [Y_0,X^{1/3-\delta_0})}} h_3(F_1/k) {\mathrm{Disc}}(F_1)^{-3/2+\epsilon'}\\ & \quad + h_2(k)^{2} \cdot X \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in [Y,Y_0)}} h_3(F_1/k) {\mathrm{Disc}}(F_1)^{-2+\epsilon} \\ & \ll_{\epsilon',\epsilon, [k:{\mathbb{Q}}]} {\mathrm{Disc}}(k)^{1/3+2\epsilon'} \cdot (X^{1-3\delta_0/2+\epsilon'} + X^{11/12+\epsilon'} + \frac{X}{Y^{1-\epsilon}}), \end{aligned} \end{align} $$

We will now explain the final inequality in more detail. From each of the four sums we can factor out $ {\mathrm {Disc}}(k)^{-1+\epsilon '}h_2(k)^{8/3}\ll _{[k:{\mathbb {Q}}],\epsilon '} {\mathrm {Disc}}(k)^{1/3+2\epsilon '}$ , by the trivial bound on $h_2(k)$ . Then the remaining factors are each of the form

(5.7) $$ \begin{align} X^{\gamma} \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in (A,B)}} \frac{h_3(F_1/k){\mathrm{Disc}}(k)^{1-\epsilon'}}{h_2(k)^{2/3}} {\mathrm{Disc}}(F_1)^{\beta} \end{align} $$

for some $\gamma , \beta , A, B$ . Corollary 3.2, translating to absolute discriminant, gives, for $B>1$ ,

(5.8) $$ \begin{align} \sum_{\substack{F_1/k\\ {\mathrm{Disc}}(F) <B}} \frac{h_3(F_1/k){\mathrm{Disc}}(k)^{1-\epsilon'}}{h_2(k)^{2/3}} \ll_{[k:{\mathbb{Q}}], \epsilon'} B. \end{align} $$

In the second sum, since $\beta>0$ , we can bound $ {\mathrm {Disc}}(F_1)^{1/2+\epsilon '}<X^{(1/3-\delta _0)(1/2+\epsilon ')}$ , and use the bound from (5.8) to find that the remaining factor is $\ll _{\epsilon ', [k:{\mathbb {Q}}]}X^{1-3\delta _0/2+\epsilon '}$ . In the first, third, and fourth sums, where $\beta <-1$ , we use partial summation and apply (5.8), to find that (5.7) is $ \ll _{\epsilon ', [k:{\mathbb {Q}}]} X^\gamma A^{\beta +1}. $ In particular the remaining factor in the first sum is $ \ll _{\epsilon ', [k:{\mathbb {Q}}]} X^{1-3\delta _0/2+\epsilon '}, $ the remaining factor in the third sum is $ \ll _{\epsilon ', [k:{\mathbb {Q}}]} XY_0^{-1/2+\epsilon '} \ll _{\epsilon ',\epsilon , [k:{\mathbb {Q}}]}X ^{11/12+\epsilon '}, $ and the remaining factor in the fourth sum is $ \ll _{\epsilon ',\epsilon , [k:{\mathbb {Q}}]} XY^{-1+\epsilon }. $

We now would like to choose $\delta ,\alpha $ and $\epsilon '$ such that the final bound in (5.6) is smaller than the bound we are trying to prove, that is, we want

$$ \begin{align*}{\mathrm{Disc}}(k)^{1/3+2\epsilon'} \cdot (X^{1-3\delta_0/2+\epsilon'} + X^{11/12+\epsilon'} + \frac{X}{Y^{1-\epsilon}}) \leq {\mathrm{Disc}}(k)^{\alpha}\Big (\frac{X}{Y^{1-\epsilon}} + X^{1-\delta} \Big). \end{align*} $$

We can choose, for example, $\alpha =2/5$ and $\delta =\min \{ \delta _0,1/15\}$ . Above if we take $\epsilon '=\min \{\delta _0/2,1/60\}$ , then the inequality above and the lemma follows.

Proof. By class field theory, the number $N_k(C_2, X)$ is bounded by the product of $h_2(k)$ and the number of integral ideals with bounded norm. The latter can be counted by integrating $\zeta _k(s)$ . If $a_n$ denotes the number of integral ideals of norm n, then by Perron’s formula we have

(5.10) $$ \begin{align} \sum_{n \leq X} a_n \leq \sum_{n=1}^\infty a_n e^{1-\frac{n}{X}} = \frac{e}{2\pi i} \int_{1+\epsilon-i\infty}^{1+\epsilon+i\infty} \zeta_k(s)\cdot \Gamma(s)\cdot X^s\,ds. \end{align} $$

Shifting the contour integral to $\Re (s)=1-\epsilon $ and using the convexity bound (3.6) on $\zeta _k(s)$ , we get the upper bound

(5.11) $$ \begin{align} \sum_{n \leq X} a_n = {\mathrm{Res}}(\zeta_k(s) \Gamma(s) X^s)_{s = 1} + O_{[k:\mathbb{Q}],\epsilon}({\mathrm{Disc}}(k)^{\epsilon/2} X^{1-\epsilon}) = O_{[k:\mathbb{Q}],\epsilon}({\mathrm{Disc}}(k)^{\epsilon} X), \end{align} $$

where the upper bound on the residue of the Dedekind zeta function comes from Landau.

Proof. We first give the idea of the proof in broad terms to motivate the notation that is to follow. Exploiting the factorization $h_3(F_2/k) = h_3(F_2/F_1) h_3(F_1/k)$ , Corollary 3.2 provides strong control on the average of $h_3(F_1/k)$ , while Lemma 2.1 provides a nontrivial bound on $h_3(F_2/F_1)$ if there are “enough” small primes that split in $F_2/F_1$ . Lemma 4.4 ensures there are enough small primes in k. Applying Lemma 4.3 twice, we see first that “most” extensions $F_1/k$ have roughly the same number of small primes as k, and then second that in “most” extensions $F_2/F_1$ for such $F_1$ , many of those small primes are split. This lets us apply Lemma 2.1, and yields the lemma.

To make this precise, and in preparation for applying Lemma 4.3, we let $\delta _1=1/20$ and $\epsilon _1=1/30$ . We set $c = \max \{c_{[k:{\mathbb {Q}}]}, c_{2[k:{\mathbb {Q}}]},1\}$ to be at least as large as the two constants given in Theorem 4.2 and $\sigma _1 = 1 - \epsilon _1/4c$ . Throughout the proof, we will make assumptions on X, including that X is at least any absolute constant necessary to apply the results we use, and then we will handle the remaining values of X at the end of the proof. Applying Lemma 4.3 the first time, we obtain an exceptional set $\mathcal {E}_0 =\mathcal {E}(k, X^{1/3+\delta _0}, \epsilon _1)$ . Then for each $F_1/k\notin \mathcal {E}_0$ , we apply Lemma 4.3 a second time to obtain an exceptional set $\mathcal {E}(F_1) = \mathcal {E}(F_1, X/ {\mathrm {Disc}}(F_1)^2, \epsilon _1)$ . By Lemma 4.3, both exceptional sets contain few elements:

(5.13) $$ \begin{align} |\mathcal{E}_0|\ll_{[k:{\mathbb{Q}}]} {\mathrm{Disc}}(k)^{\epsilon_1} X^{\epsilon_1} \quad \text{and} \quad |\mathcal{E}(F_1)|\ll_{[k:{\mathbb{Q}}]} {\mathrm{Disc}}(k)^{\epsilon_1} X^{\epsilon_1}. \end{align} $$

We now consider the consequences of Lemma 4.3 for fields outside the exceptional sets. For $F_1/k \notin \mathcal {E}_0$ with $ {\mathrm {Disc}}(F_1) \in \mathfrak {C}(X)$ and $x =(X/(2 {\mathrm {Disc}}(F_1)^2))^{\delta _1} \leq X^{1/3+\delta _0}$ , by Lemma 4.3 we have, for X sufficiently large,

(5.14) $$ \begin{align} \pi_{F_1}(x/2) \ge \pi_k(x/2;F_1, e) \geq \frac{1}{8}\pi_k(x/4)-C_{[k:{\mathbb{Q}}],\epsilon_1}(x/2)^{\sigma_1} \log^2( X^{1/3+\delta_0}{\mathrm{Disc}}(k)). \end{align} $$

We next assume that $X\ge 2^{\frac {1}{1/3-2/100}} 4^{\frac {1}{(1/3-2/100)\delta _1}} {\mathrm {Disc}}(k)^A$ where $A =\frac { \max \{ \beta , 8c\gamma /\epsilon _1 \} }{(1/3-2/100) \delta _1} = A([k:{\mathbb {Q}}])$ , and $\gamma $ and $\beta $ are absolute constants for which Lemma 4.4 holds. This assumption ensures that $x/4 \geq {\mathrm {Disc}}(k)^\beta $ and that $ x^{\frac {1-\sigma _1}{2}}\geq {\mathrm {Disc}}(k)^\gamma $ . Let also $D_0$ be an absolute constant such that Lemma 4.4 holds for those $\gamma $ and $\beta $ . If $ {\mathrm {Disc}}(k) \geq D_0$ , we use Lemma 4.4, and for $X\ge {\mathrm {Disc}}(k)^A$ and X sufficiently large given $[k:{\mathbb {Q}}]$ , we obtain

(5.15) $$ \begin{align} \frac{1}{8}\pi_k(x/4)-C_{[k:{\mathbb{Q}}],\epsilon_1}(x/2)^{\sigma_1} \log^2( X^{1/3+\delta_0}{\mathrm{Disc}}(k)) \gg_{[k:{\mathbb{Q}}]} \frac{x}{{\mathrm{Disc}}(k)^{\gamma} \log x}. \end{align} $$

Here we used $x/4 \geq {\mathrm {Disc}}(k)^\beta $ to apply Lemma 4.4, and $ x^{\frac {1-\sigma _1}{2}}\geq {\mathrm {Disc}}(k)^\gamma $ as a somewhat generous assumption to ensure that the lower bound from Lemma 4.4 is sufficiently large to subtract off the term above and still, up to constants, obtain the same lower bound. We require that X is large enough so that (5.15) holds also for the finitely many fields k with $ {\mathrm {Disc}}(k) \leq D_0$ . Under these assumptions, we conclude for $F_1 \notin \mathcal {E}_0$ that

(5.16) $$ \begin{align} \pi_{F_1}(x/2) \gg_{[k:{\mathbb{Q}}]} \frac{x}{{\mathrm{Disc}}(k)^{\gamma} \log x}. \end{align} $$

Next, for $F_2/F_1\notin \mathcal {E}(F_1)$ , by Lemma 4.3 and (5.16), and the assumptions above on X, we have

(5.17) $$ \begin{align} \pi_{F_1}(x; F_2, e) \gg_{[k:{\mathbb{Q}}]} \frac{x}{ {\mathrm{Disc}}(k)^{\gamma} \log x}. \end{align} $$

We now consider the contribution to the summation from the various kinds of exceptional and nonexceptional fields. First, we consider the summation $F_2/F_1/k$ where $F_1/k\in \mathcal {E}_0$ . Using the trivial bound on the class group and the sparsity of exceptional fields along with Lemmas 5.8 and 5.6, we find

(5.18) $$ \begin{align} \begin{aligned} \sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)\\ X/2\le {\mathrm{Disc}}(F_2)\le X, F_1/k\in \mathcal{E}_0}} h_3(F_2/k) & \ll_{[k:{\mathbb{Q}}],\epsilon} X^{1/2+\epsilon} \sum_{\substack{F_1/k \in \mathcal{E}_0 \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)}} \sum_{\substack{F_2/ F_1 \\ X/2\le {\mathrm{Disc}}(F_2)\le X}} 1 \\ & \ll_{[k:{\mathbb{Q}}],\epsilon} h_2(k)^2 {\mathrm{Disc}}(k)^{\epsilon_1} \cdot X^{5/6+2\delta_0+\epsilon_1+\epsilon} \\ & \ll_{[k:{\mathbb{Q}}],\epsilon} {\mathrm{Disc}}(k)^{1+\epsilon_1+\epsilon} \cdot X^{5/6+2\delta_0+\epsilon_1+\epsilon}. \end{aligned} \end{align} $$

Second, we consider the contribution from those $F_2/F_1/k$ where $F_1 \notin \mathcal {E}_0$ but $F_2/F_1\in \mathcal {E}(F_1)$ . Using the trivial bound on relative class group $h_3(F_2/F_1)$ and an average bound on $h_3(F_1/k)$ , we find

(5.19) $$ \begin{align} \begin{aligned} \sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)\\ X/2\le {\mathrm{Disc}}(F_2)\le X\\ F_2/F_1\in \mathcal{E}(F_1)}} \!\!\!\! h_3(F_2/k) & \ll_{[k:{\mathbb{Q}}],\epsilon} \!\!\!\!\!\!\!\! \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)}} \!\!\!\! h_3(F_1/k) \!\!\!\! \sum_{\substack{F_2/ F_1 \\ X/2 \leq {\mathrm{Disc}}(F_2) \leq X \\ F_2/F_1\in \mathcal{E}(F_1)}} \!\!\!\!\!\! {\mathrm{Disc}}(F_2/F_1)^{1/2+\epsilon} {\mathrm{Disc}}(F_1)^{1/2+\epsilon} \\ & \ll_{[k:{\mathbb{Q}}],\epsilon} {\mathrm{Disc}}(k)^{\epsilon_1} X^{1/2+\epsilon_1+\epsilon} \sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)}} h_3(F_1/k) {\mathrm{Disc}}(F_1)^{-1/2+\epsilon}\\ & \ll_{[k:{\mathbb{Q}}],\epsilon} {\mathrm{Disc}}(k)^{4/3+\epsilon_1+\epsilon} \cdot X^{2/3+\delta_0/2+\epsilon_1+\epsilon}. \\ \end{aligned} \end{align} $$

Precisely, the first inequality follows from Lemma 2.4. The second inequality follows from the estimate $|\mathcal {E}(F_1)|\ll _{[k:{\mathbb {Q}}], \epsilon _1} {\mathrm {Disc}}(k)^{\epsilon _1} X^{\epsilon _1}$ from (5.13). The third inequality follows from Corollary 3.2 and partial summation.

We now consider the contribution from those $F_2/F_1/k$ where neither $F_1/k \notin \mathcal {E}_0$ nor $F_2/F_1 \notin \mathcal {E}(F_1)$ . Using the relative Ellenberg–Venkatesh method developed in Section 2 and existence of small split primes for nonexceptional fields developed in Section 4 and made explicit in (5.17), we find,

(5.20) $$ \begin{align} \begin{aligned} &\sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)\\ X/2\le {\mathrm{Disc}}(F_2)\le X \\ F_1 \notin \mathcal{E}_0, F_2/F_1\notin \mathcal{E}(F_1)}} h_3(F_2/k) \\&\quad\quad \ll_{[k:{\mathbb{Q}}], \epsilon} {\mathrm{Disc}}(k)^{\gamma} \cdot \!\!\!\!\!\!\!\!\sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)}} \!\!\!\!\! h_3(F_1/k) \left(\sum_{\substack{F_2/ F_1 \\ F_2\notin \mathcal{E}(F_1)\\ X/2\le {\mathrm{Disc}}(F_2)\le X }} {\mathrm{Disc}}(F_2/F_1)^{1/2-\delta_1+\epsilon} {\mathrm{Disc}}(F_1)^{1/2+\epsilon} \!\!\right) \\&\quad\quad \ll_{[k:{\mathbb{Q}}], \epsilon} {\mathrm{Disc}}(k)^{\gamma} \cdot h_2(k)^2 \!\cdot\! \left(\sum_{\substack{F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X)}} h_3(F_1/k) \cdot (\frac{X}{{\mathrm{Disc}}(F_1)^2})^{3/2-\delta_1+\epsilon}\cdot {\mathrm{Disc}}(F_1)^{1/2+\epsilon}\!\right) \\ & \quad\quad \ll_{[k:{\mathbb{Q}}],\epsilon} {\mathrm{Disc}}(k)^{\gamma+7/3+\epsilon} X^{1 - \frac{\delta_1}{3}+\delta_0(\frac{3}{2}-2\delta_1)+\epsilon}.\\ \end{aligned} \end{align} $$

Precisely, the first inequality comes from Lemma 2.1 with $\theta =\delta _1=1/20$ and and (5.17) with ${x= {\mathrm {Disc}}(F_2/F_1)^{\delta _1}}$ . The second inequality follows from Lemma 5.8 and Lemma 5.6. We apply Corollary 3.2 and partial summation for the last inequality.

Finally, in the complementary case that X is not sufficiently large, we may assume that $X \ll _{[k:\mathbb {Q}]} {\mathrm {Disc}}(k)^A$ , so that by applying the trivial bound $h_3(F_2/k) \ll _{[k:\mathbb {Q}],\epsilon } X^{1/2+\epsilon }$ , we find

(5.21) $$ \begin{align} \begin{aligned} \sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X) \\ X/2\le {\mathrm{Disc}}(F_2)\le X}} h_3(F_2/k) & \ll_{[k:\mathbb{Q}],\epsilon} X^{1/2+\epsilon} \cdot \sum_{\substack{F_2/ F_1/k \\ {\mathrm{Disc}}(F_1) \in \mathfrak{C}(X) \\ X/2\le {\mathrm{Disc}}(F_2)\le X}} 1 \\ & \ll_{[k:{\mathbb{Q}}], \epsilon} X^{7/6+\delta_0 + \epsilon} h_2(k)^3 {\mathrm{Disc}}(k)^{-2+\epsilon} \\ & \ll_{[k:{\mathbb{Q}}],\epsilon} {\mathrm{Disc}}(k)^{(7/6+\delta_0+\epsilon)A-1/2+\epsilon}. \end{aligned} \end{align} $$

Here the second inequality follows from Lemma 5.8 twice, and the last inequality follows from the bound on X.

So to deduce the statement of the lemma, we note we have four upper bounds of the form $ {\mathrm {Disc}}(k)^{a_i} X^{b_i+\epsilon }$ . We can take, for example, $\epsilon =1/1000$ , and note that each of the $b_i$ is less than $998/1000$ . Then we can take $\delta =1/1000$ , and $\alpha $ can be taken to be $\max _i{a_i}$ , which depends on $[k:{\mathbb {Q}}]$ .

Proof of Theorem 5.1.

The statement for $m = 2$ follows from the combination of Lemmas 5.7 and 5.9. Now assume for some i that we can prove the lemma for every $m\le i$ . Our goal is to prove the lemma for $m = i+1$ .

Firstly, we treat the summation when $ {\mathrm {Disc}}(F_i) \in \mathfrak {N}(X,Y)=[Y, X^{1/3-\delta _0}) \cup (X^{1/3+\delta _0}, X^{1/2}]$ , the noncritical range. We use an argument similar to that in Lemma 5.7 where the noncritical range is treated for $m=2$ . For any $0< \delta _0 < 1/6$ and $Y \geq 1$ , using the results on counting cubic fields in Section 3, as collected in (5.5), and the induction hypothesis in the form of Corollary 5.5, we find

(5.22) $$ \begin{align} \begin{aligned} \sum_{\substack{F_{i+1}/F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_i) \in \mathfrak{N}(X,Y) \\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X}} h_3(F_{i+1}/k) &= \sum_{\substack{F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_i) \in \mathfrak{N}(X,Y)}} h_3(F_i/k) \sum_{\substack{F_{i+1}/F_i \\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X }} h_3(F_{i+1}/F_i) \\ &\ll_{[k:{\mathbb{Q}}],i,\epsilon',\epsilon} {\mathrm{Disc}}(k)^{\alpha_i+2^{i-1}+\epsilon} \Big(X^{1-\frac{3\delta_0}{2}+\epsilon'} + X^{11/12+\epsilon'}+\frac{X}{Y^{1-\epsilon}}\Big) .\\ \end{aligned} \end{align} $$

The inequality above follows as in (5.6), with $F_i$ in place of $F_1$ , using the cases of (5.5) in three ranges of $ {\mathrm {Disc}}(F_i)$ followed by partial summation with Corollary 5.5 (in place of Corollary 3.2), and the $h_2(k)^2$ factor from bounding $h_2(F_1)$ is replaced by $h_2(k)^{2^i}$ in the bound for $h_2(F_i)$ , using Lemma 5.6.

We now treat the contribution from towers with $ {\mathrm {Disc}}(F_i)\in \mathfrak {C}(X) = [X^{1/3-\delta _0},X^{1/3+\delta _0}]$ . We separate the discussion into two cases depending on the size of $ {\mathrm {Disc}}(F_{i-1})$ . Fix $1 \leq Y_2 \leq X^{\frac {1}{6}-\frac {\delta _0}{2}}$ . Then, we find

(5.23) $$ \begin{align} \begin{aligned} \sum_{\substack{F_{i+1}/F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\ge Y_2 \\ {\mathrm{Disc}}(F_i) \in \mathfrak{C}(X) \\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X}} \!\!\!\!h_3(F_{i+1}/k) & = \sum_{\substack{F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\ge Y_2 \\ {\mathrm{Disc}}(F_i) \in \mathfrak{C}(X)}} h_3(F_{i}/k) \sum_{\substack{F_{i+1}/F_i\\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X}} h_3(F_{i+1}/F_i)\\ & \ll_{[k:{\mathbb{Q}}],i, \epsilon} h_2(k)^{2^i}\!\!\!\!\!\! \sum_{\substack{F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\ge Y_2 \\ {\mathrm{Disc}}(F_i) \in \mathfrak{C}(X)}} \!\!\!\!\!\! h_3(F_{i}/k)\! \cdot\! {\mathrm{Disc}}(F_i)^{1/2+\epsilon}\! \cdot\! (\frac{X}{{\mathrm{Disc}}(F_{i})^2})^{3/2+\epsilon}\\ & \ll_{[k:{\mathbb{Q}}], i,\epsilon} h_2(k)^{2^i} {\mathrm{Disc}}(k)^{\alpha_i}\cdot X^{3/2+\epsilon} \!\cdot\! \Big( \frac{ X^{-\frac{1}{2}+\frac{3\delta_0}{2}} }{Y_2^{1-\epsilon}} + X^{-(\frac{1}{3}-\delta_0)(\frac{3}{2}+\delta_i)}\Big)\\ & \ll_{[k:{\mathbb{Q}}], i,\epsilon} {\mathrm{Disc}}(k)^{\alpha_i+2^{i-1}+\epsilon} \cdot \Big( \frac{ X^{1+{3\delta_0}/{2}+\epsilon} }{Y_2^{1-\epsilon}} + X^{ 1+3\delta_0/2-\delta_i/3+\delta_0\delta_i+\epsilon }\Big). \end{aligned} \end{align} $$

Precisely, the first inequality follows from (3.8), Proposition 3.4, and Lemma 5.6. The second inequality follows from partial summation, and the induction hypothesis with $m=i$ applied to $F_i/\cdots /F_1/k$ with $Y_2$ being the lower bound for $ {\mathrm {Disc}}(F_{i-1})$ . The third inequality follows from the trivial bound on $h_2(k)$ .

Finally we consider those towers with $ {\mathrm {Disc}}(F_{i-1}) \le Y_2$ . Let $\delta ^{\prime }_2$ and $\alpha ^{\prime }_2$ be the constants associated to $[F_{i-1}:{\mathbb {Q}}]$ and $m =2$ . We have

(5.24) $$ \begin{align} \begin{aligned} \sum_{\substack{F_{i+1}/F_i/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\le Y_2 \\ {\mathrm{Disc}}(F_i) \in \mathfrak{C}(X) \\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X}} \!\!\!\!h_3(F_{i+1}/k) &= \sum_{\substack{F_{i-1}/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\le Y_2}} h_3(F_{i-1}/k) \sum_{\substack{F_{i+1}/F_i/F_{i-1}\\ {\mathrm{Disc}}(F_i) \in \mathfrak{C}(X)\\ X/2\le {\mathrm{Disc}}(F_{i+1})\le X}} h_3(F_{i+1}/F_{i-1})\\ & \ll_{ [k:{\mathbb{Q}}],i, \epsilon} \!\!\!\!\sum_{\substack{F_{i-1}/\cdots/F_1/k \\ {\mathrm{Disc}}(F_{i-1})\le Y_2}} \!\!\!\!h_3(F_{i-1}/k) \Big( X^{2/3+\delta_0+\epsilon} + X^{1-\delta^{\prime}_2} \Big) \cdot {\mathrm{Disc}}(F_{i-1})^{\alpha^{\prime}_2} \\ & \ll_{[k:{\mathbb{Q}}],i} {\mathrm{Disc}}(k)^{\alpha_{i-1}} X^{\max\{{3/4+\delta_0, 1-\delta^{\prime}_2}\}} Y_2^{1+\alpha^{\prime}_2}.\\ \end{aligned} \end{align} $$

Here precisely, the first inequality follows from the induction hypothesis with $m=2$ and base field $k = F_{i-1}$ . The second inequality follows from the induction hypothesis with $m=i-1$ and taking $\epsilon = 1/12$ .

If we let $Y_2=X^b$ , then to obtain the desired bound, (5.23) requires us to choose $\delta _0$ small, given b and $\delta _i$ , and (5.24) requires us to take b small given $\delta _2'$ and $\alpha _2'$ (as long as we have taken $\delta _0$ absolutely sufficiently small). Since we can do this by choosing b and then $\delta _0$ , we will be able to obtain the theorem. In particular, making the convenient but nonoptimal choices $b=\min \{\delta ^{\prime }_2/(2(1+\alpha ^{\prime }_2)),1/(8(1+\alpha ^{\prime }_2))\}$ and $Y_2 = X^{b}$ and $\delta _0 = \min \{\delta _i/9, b/2,1/16\}$ , we conclude that the statement of the theorem holds for $m=i+1$ with $ \alpha _{i+1} = \max \{\alpha _i+2^{i}, \alpha _{i-1}\} \quad \text {and} \quad \delta _{i+1} = \min \{\frac {\delta ^{\prime }_2}{2}, \frac {b}{8},\frac {\delta _i}{36},\frac {1}{32}\}. $

Proof. We proceed by induction over m. For $m = 1$ , we know from class field theory (see the proof of Lemma 3.5) that the number of relative quadratic extensions $K/k$ with $ {\mathrm {Disc}}(K/k) = D$ is at most $O_{[k:{\mathbb {Q}}], \epsilon }(h_2(k)\cdot D^{\epsilon }) = O_{[k:{\mathbb {Q}}], \epsilon }( D^{\epsilon } {\mathrm {Disc}}(k)^{1/2+\epsilon })$ . Now assuming the statement is true for $m =i$ , we will prove that it also holds for $m = i+1$ . By Lemma 5.4, it suffices to count quadratic extensions of degree $2^i 2$ -extensions $F/k$ with $ {\mathrm {Disc}}(F/k)^2| D$ . The number of such F is bounded by $O_{[k:{\mathbb {Q}}], \epsilon , i}( D^{\epsilon } {\mathrm {Disc}}(k)^{\alpha _i})$ by the induction hypothesis. For each such F, the number of quadratic relative extensions $K/F$ with $ {\mathrm {Disc}}(K/F) = D/ {\mathrm {Disc}}(F/k)^2$ is bounded by $O_{[F:{\mathbb {Q}}], \epsilon }(h_2(F) \cdot D^{\epsilon })$ from class field theory. Applying Lemma 5.6 and summation over all divisors of D, we obtain the upper bound $O_{[k:{\mathbb {Q}}], i+1, \epsilon }( D^{\epsilon } {\mathrm {Disc}}(k)^{\alpha _{i+1}})$ , where $\alpha _{i+1}$ can be taken as $\alpha _{i}+ 2^{i-1}$ .

Proof. Notice that when G does not contain a transposition, the discriminant ideal $ {\mathrm {disc}}(K/k)$ must have exponent at least $2$ at every ramified prime, that is, it is powerful. (The exponent of a prime in $ {\mathrm {disc}}(K/k)$ can be computed from the exponent in the different of a prime above in the Galois closure, which is the Artin conductor of the permutation representation [Reference NeukirchNeu99, VII.11.8]. In particular, the exponent of a ramified prime in $ {\mathrm {disc}}(K/k)$ is at least $2^m$ minus the number of orbits of the inertia subgroup, and the number of orbits is $2^m - 1$ if and only if the inertia subgroup is cyclic and generated by a transposition.) Thus, its norm, $ {\mathrm {Disc}}(K/k)$ , must be a powerful integer, and there are $O(X^{1/2})$ powerful integers below X. The result now follows from Lemma 6.3.

Proof. A G-extension K can be constructed as a relative quadratic extension over a degree $2^{m-1} 2$ -extension F with $ {\mathrm {Gal}}(F/k) =: H\subset S_{2^{m-1}}$ by Lemma 5.4. We fix $\delta _1=1/24$ and $\epsilon _1=1/4$ . Let $\gamma $ , $\beta $ and $D_0$ be absolute constants allowable in Lemma 4.4. Let $Y = X^{\delta '}$ for some $\delta '$ that we will choose sufficiently small in terms of m and $[k:{\mathbb {Q}}]$ . Then by Theorem 5.1 and dyadic summation, taking $\delta '\leq \delta _m$ , we have