if}\;i>j\mbox{ and }j\leq k,\\\ \ell_{i,n+2-i}&\;{\rm
otherwise}.\end{array}\right.$

We define the auxiliary set ${\mathcal{Z}}^{*}$ by

${\mathcal{Z}}^{*}\,=\,\\{a_{ij}\mid i\leq j,\,i+j\leq
n\\}\cup\\{\ell_{ij}\mid i\leq j,\,i+j\leq
n+1\\}\,=\,{\mathcal{Z}}_{1}\backslash\\{\ell_{ij}\mid i>j,\,i+j\leq n+1\\}.$

Then, when restricted to ${\mathcal{Z}}^{*}\subset{\mathcal{Z}}_{1}$, the
functions $\pi_{k}$ are equal for every $k$.

###### Lemma 6.12

For any $k\in\\{0,\ldots,g\\}$, we have

$\Big{(}\prod_{\alpha\in{\mathcal{Z}}^{*}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Big{)}\,\delta(s)=1.$
(47)

Proof: We directly compute

$\begin{array}[]{rcl}\displaystyle\prod_{\alpha\in{\mathcal{Z}}^{*}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}&=&\displaystyle\Bigl{(}\prod_{\begin{subarray}{c}i\leq
j\\\
i+j<n+1\end{subarray}}\frac{w(a_{n+1-j,j})}{w(a_{ij})}\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}i\leq
j\\\
i+j<n+2\end{subarray}}\frac{w(\ell_{i,n+2-i})}{w(\ell_{ij})}\Bigr{)}\\\\[7.22743pt]
&=&\displaystyle\Bigl{(}\prod_{\begin{subarray}{c}i\leq j\\\
i+j<n+1\end{subarray}}\frac{t_{i}}{t_{n+1-j}}\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}i\leq
j\\\ i+j<n+2\end{subarray}}\frac{t_{j}}{t_{n+2-i}}\Bigr{)}\\\\[7.22743pt]
&=&\displaystyle\Bigl{(}\prod_{\begin{subarray}{c}i<r\\\ i+r\leq
n+1\end{subarray}}\frac{t_{i}}{t_{r}}\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}j<r\\\
j+r\geq n+2\end{subarray}}\frac{t_{j}}{t_{r}}\Bigr{)},\end{array}$

which is $\delta(s)^{-1}$. $\Box$

Fix a saturated set ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ such that
$a_{11}\in{\mathcal{Z}}$, $a_{g+1,g+1}\notin{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Let $k\in\\{1,\ldots,g\\}$ be the
largest integer such that $a_{kk}\in{\mathcal{Z}}$. Then we have the following
results.

###### Lemma 6.13

Let ${\mathcal{Z}}$ and $k$ be as above. Then for every
$\alpha\in{\mathcal{Z}}$, we have $\pi_{k}(\alpha)\notin{\mathcal{Z}}$. In
particular, for any $s\in T_{\mathcal{Z}}(L,Y)$, we have
$Yw(\pi_{k}(\alpha))\gg 1.$

Proof: Since $a_{n+1-j,j}\not\in{\mathcal{Z}}_{1}$ for any $j$ and
${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$, we have
$\pi_{k}(a_{ij})\not\in{\mathcal{Z}}$ for any $a_{ij}\in{\mathcal{Z}}$.
Moreover, since $a_{jj}\in{\mathcal{Z}}$ for every $j\leq k$, it follows from
Lemma 6.5 that $\ell_{n+1-j,j}\not\in{\mathcal{Z}}$. Furthermore,
$\ell_{i,n+2-i}\not\in{\mathcal{Z}}_{1}$. Hence
$\pi_{k}(\ell_{ij})\not\in{\mathcal{Z}}$ for any $\ell_{ij}\in{\mathcal{Z}}$.
$\Box$

###### Lemma 6.14

Let ${\mathcal{Z}}$ and $k$ be as above. Then, uniformly for $s\in
T_{\mathcal{Z}}(L,Y)$, we have

$\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Big{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\ i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Big{)}\,\delta(s)\ll
1.$ (48)

Proof: Since we have

$\frac{w(a_{n+1-j,j})}{w(a_{ij})}=\frac{t_{i}}{t_{n+1-j}},\quad\frac{w(\ell_{i,n+2-i})}{w(\ell_{ij})}=\frac{t_{j}}{t_{n+2-i}},\quad\frac{w(\ell_{n+1-j,j})}{w(\ell_{ij})}=\frac{L_{n+1-j}}{L_{i}},$

it follows that $w(\pi_{k}(\alpha))/w(\alpha)\gg 1$ for every $k$,
$\alpha\in{\mathcal{Z}}_{1}$, and $s\in T_{\mathcal{Z}}(L,Y)$. Thus, by adding
elements in ${\mathcal{Z}}_{1}$ to ${\mathcal{Z}}$, if necessary, we can
assume that ${\mathcal{Z}}$ is equal to

$\begin{array}[]{rcl}{\mathcal{Z}}_{0}&=&\bigl{\\{}a_{ij}:i\leq j,\,i\leq
k,\,i+j\leq n\bigr{\\}}\cup\bigl{\\{}\ell_{ij}:i>j>k,\,i+j\leq
n+1\bigr{\\}}\\\\[3.61371pt] &&\cup\,\bigl{\\{}\ell_{ij}:i>j,\,j\leq
k,\,i+j\leq n+1\bigr{\\}}\cup\bigl{\\{}\ell_{ij}:i\leq j,\,i+j\leq
n+1\bigr{\\}}.\end{array}$

Denote the four sets on the right hand side of the above equation as $S_{1}$,
$S_{2}$, $S_{3}$, and $S_{4}$, respectively. For an element $\ell_{ij}\in
S_{2}$, we have

$\frac{w(\pi_{k}(\ell_{ij}))}{w(\ell_{ij})}=\frac{w(\ell_{i,n+2-i})}{w(\ell_{ij})}=\frac{t_{j}}{t_{n+2-i}}=\frac{w(\pi_{k}(a_{j,i-1}))}{w(a_{j,i-1})}.$

Therefore,

$\begin{array}[]{rcl}\displaystyle\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}_{0}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Big{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Big{)}\,\delta(s)&=&\displaystyle\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}^{*}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Bigl{(}\prod_{\alpha\in
S_{3}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Big{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Big{)}\,\delta(s)\\\\[1.8063pt]
&=&\displaystyle\Bigl{(}\prod_{\begin{subarray}{c}i>j,\,j\leq k\\\ i+j\leq
n+1\end{subarray}}\frac{L_{n+1-j}}{L_{i}}\Bigr{)}\Big{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\ i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Big{)}\\\\[18.06749pt]
&=&\displaystyle\prod_{\begin{subarray}{c}1\leq i<j\leq n\\\ i+j>n+1\\\
j<n+1-k\end{subarray}}\frac{L_{i}}{L_{j}}\\\\[18.06749pt] &\leq&1,\end{array}$

where the second equality follows from Lemma 6.12, and the last inequality
follows because the $L_{i}$’s are nondecreasing. $\Box$

Proposition 6.3 and Lemmas 6.13 and 6.14 thus yield the bound

$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)\ll X^{n+1}\int_{1\ll
s_{1},\ldots,s_{n}\ll X^{\Theta}}d^{\times}s\ll_{\epsilon}X^{n+1+\epsilon}.$

We now work towards obtaining a power saving.

####  6.3.2 Strategy towards a power saving

In light of Proposition 6.3, it is enough to have a bound of the form

$Y^{-\\#{\mathcal{Z}}}w({\mathcal{Z}})^{-1}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\ i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)\ll
X^{-\delta}$ (49)

for some $\delta>0$, for all $s\in T_{\mathcal{Z}}(L,Y).$ By modifying
$\pi_{k}$ on a certain subset of ${\mathcal{Z}}$, we are able to obtain (49)
except for some $s\in T_{\mathcal{Z}}(L,Y)$ satisfying some special
conditions. We then consider the contribution from these special $s$ using a
different count for $\\#\bigl{(}(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}\bigr{)}$.

More precisely, let ${\mathcal{K}}_{1}:=\\{a_{1j}:1\leq j\leq g+2\\}$. Then
${\mathcal{K}}_{1}$ consists exactly of those $\alpha\in{\mathcal{K}}$ such
that the exponent of every $s_{i}$ is negative in $w(\alpha)$. As such, one
expects that the hardest case is when ${\mathcal{Z}}={\mathcal{K}}_{1}$. We
show first in Lemma 6.15 how to reduce to considering only
${\mathcal{Z}}\cap{\mathcal{K}}_{1}$.

###### Lemma 6.15

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. For any
${\mathcal{Z}}^{\prime}\subset{\mathcal{K}}_{1}$ and any $s\in T^{\prime}$, we
write

$I({\mathcal{Z}}^{\prime},s)=Y^{-\\#{\mathcal{Z}}^{\prime}}w({\mathcal{Z}}^{\prime})^{-1}\prod_{i=1}^{g}s_{i}^{-(n+1)(g+2)}\prod_{i=g+1}^{n-1}s_{i}^{-(n+1)(n-i)}.$

Then for any $s\in T_{\mathcal{Z}}(L,Y)$, we have

$Y^{-\\#{\mathcal{Z}}}w({\mathcal{Z}})^{-1}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\ i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)\ll
I({\mathcal{Z}}\cap{\mathcal{K}}_{1},s).$

We then prove in Lemma 6.16 the following bound for
$I({\mathcal{Z}}\cap{\mathcal{K}}_{1})$ when
${\mathcal{Z}}\cap{\mathcal{K}}_{1}$ is a proper subset of
${\mathcal{K}}_{1}$, which gives a bound of the form (49) when $s_{n}\ll
Y^{1/2-\delta}$.

###### Lemma 6.16

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{Z}}\cap{\mathcal{K}}_{1}\neq{\mathcal{K}}_{1}$. For any $s\in
T_{\mathcal{Z}}(L,Y)$, if $I({\mathcal{Z}}\cap{\mathcal{K}}_{1},s)\gg
Y^{-2\delta}$, then $s_{n}\gg Y^{1/2-\delta}.$

In the case where $s_{n}\gg Y^{1/2-\delta}$, the Haar measure turns out to be
very small so we may simply ignore the singularity condition of $B$ and prove
the following bound.

###### Lemma 6.17

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{Z}}\cap{\mathcal{K}}_{1}\neq{\mathcal{K}}_{1}$. Then for any $s\in
T_{\mathcal{Z}}(L,Y)$ with $s_{n}\gg Y^{1/2-\delta}$,

$\\#\bigl{(}(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap
W({\mathbb{Z}})\bigr{)}\,\delta(s)\ll
Y^{(\frac{n}{2}+2)(n+1)+n(2n^{2}+9n+9)\delta}.$

Therefore, by taking $\delta=(n-2)/(4n^{2}+14n+4)$, we obtain the following
result from Proposition 6.3 and Lemmas 6.15, 6.16, and 6.17:

###### Proposition 6.18

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{Z}}\cap{\mathcal{K}}_{1}\neq{\mathcal{K}}_{1}$. Then

$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)\ll
X^{n+1-\frac{n-2}{2(n+1)(n^{2}+7n+7)}}.$

We next handle the case ${\mathcal{K}}_{1}\subset{\mathcal{Z}}$. We give
necessary conditions in Lemma 6.19 on $s$ so that a bound of the form (49)
does not hold.

###### Lemma 6.19

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{K}}_{1}\subset{\mathcal{Z}}$. For any $s\in T_{\mathcal{Z}}(L,Y)$,
if $I({\mathcal{K}}_{1},s)\gg X^{-\delta}$, then

$\begin{array}[]{rcccl}\displaystyle
Y^{-\delta}&\ll&\displaystyle\frac{s_{i}}{s_{n-i}}&\ll&Y^{\delta},\quad\mbox{for
}\quad i=1,\ldots,g-1\\\\[3.61371pt] \displaystyle
Y^{1/2-(g/2)\delta}{\mathcal{R}}^{-1}&\ll&s_{g}&\ll&Y^{1/2+3g\delta}{\mathcal{R}}^{-1}\\\\[5.42003pt]
\displaystyle 1&\ll&s_{g+1}&\ll&Y^{\delta}\\\\[3.61371pt] \displaystyle
Y^{1/2-\delta}{\mathcal{R}}^{-1}&\ll&s_{g+2}&\ll&Y^{1/2+g\delta}{\mathcal{R}}^{-1}\end{array}$
(50)

where

${\mathcal{R}}=\prod_{i=g+3}^{n}s_{i}\ll Y^{1/2+3g\delta}.$

Note that the coefficients of $\delta$ in the exponents in the above bounds
are not optimal and are simply chosen to make the formula look nice. The
optimal coefficients can be obtained from the proof.

When $s$ satisfies (50), we give further conditions in Lemma 6.20 on $s$ so
that simply using the Haar measure and ignoring the singularity condition by
counting all symmetric matrices is not enough for a power saving.

###### Lemma 6.20

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{K}}_{1}\subset{\mathcal{Z}}$. For any $s\in T_{\mathcal{Z}}(L,Y)$,
if

$I({\mathcal{K}}_{1},s)\gg
X^{-\delta},\qquad\mbox{and}\qquad\\#\bigl{(}(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap W({\mathbb{Z}})\bigr{)}\,\delta(s)\gg X^{n+1-\delta},$

then

$s_{i}\ll Y^{258g^{3}\delta}\qquad\mbox{ for }\quad i=g+3,\ldots,n.$ (51)

To obtain a further saving, we need to use the $|q|$-invariant!

###### Lemma 6.21

Suppose $M>X^{\eta}$ where $\eta>0$ is some fixed constant. Let
${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$ , $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{K}}_{1}\subset{\mathcal{Z}}$. Then for
$\delta<\min(\eta,1)/(1355g^{6})$ and any $s\in T_{\mathcal{Z}}(L,Y)$ such
that (50) and (51) hold, we have

$\\#\bigl{(}(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}\bigr{)}\,\delta(s)\ll
X^{n+1+514g^{3}\delta-1/2}.$

Therefore, by taking $\delta=64\min(\eta,1)/(1355n^{6})$, we obtain the
following result from Proposition 6.3 and Lemmas 6.15, 6.19, 6.20, and 6.21.

###### Proposition 6.22

Suppose $M>X^{\eta}$ where $\eta>0$ is some fixed constant. Let
${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be saturated with
$a_{11}\in{\mathcal{Z}}$, $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$. Suppose
${\mathcal{K}}_{1}\subset{\mathcal{Z}}$. Then

$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)\ll
X^{n+1-64\min(\eta,1)/(1355n^{6})}.$

Theorem 6.11 then follows immediately from (38), Proposition 6.18, Proposition
6.22 and summing over the $O(1)$ different possible ${\mathcal{Z}}$’s and the
$O(Y^{\epsilon})$ different possible $L$’s.

####  6.3.3 Proofs of Lemmas 6.15, 6.16, 6.17, 6.19, 6.20 and 6.21.

We fix a saturated ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ with
$a_{11}\in{\mathcal{Z}}$, $a_{g+1,g+1}\not\in{\mathcal{Z}}$ and
$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$.

Proof of Lemma 6.15: Recall that ${\mathcal{K}}_{1}:=\\{a_{1j}:1\leq j\leq
g+2\\}$. Let $k\in\\{1,\ldots,g\\}$ be the largest integer such that
$a_{kk}\notin{\mathcal{Z}}$. Then, applying Lemma 6.14 to the saturated set
${\mathcal{Z}}\cup{\mathcal{K}}_{1}$, we have

$\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}\backslash{\mathcal{K}}_{1}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)\ll\prod_{\alpha\in{\mathcal{K}}_{1}}\frac{w(\alpha)}{w(\pi_{k}(\alpha))}=\prod_{i=1}^{g}s_{i}^{-(n+1)(g+2)}\prod_{i=g+1}^{n-1}s_{i}^{-(n+1)(n-i)}.$

Hence, by Lemma 6.13, we obtain for any $s\in T_{\mathcal{Z}}(L,Y)$,

$\begin{array}[]{rcl}\displaystyle
Y^{-\\#{\mathcal{Z}}}w({\mathcal{Z}})^{-1}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)&\ll&\displaystyle
Y^{-\\#{\mathcal{Z}}}w({\mathcal{Z}})^{-1}\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}\backslash{\mathcal{K}}_{1}}Yw(\pi_{k}(\alpha))\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)\\\\[14.45377pt]
&\ll&\displaystyle
Y^{-\\#({\mathcal{Z}}\cap{\mathcal{K}}_{1})}w({\mathcal{Z}}\cap{\mathcal{K}}_{1})^{-1}\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}\backslash{\mathcal{K}}_{1}}\frac{w(\pi_{k}(\alpha))}{w(\alpha)}\Bigr{)}\Bigl{(}\prod_{\begin{subarray}{c}1\leq
i<j\leq n\\\
i+j>n+1\end{subarray}}\frac{L_{i}}{L_{j}}\Bigr{)}\,\delta(s)\\\\[7.22743pt]
&=&\displaystyle
Y^{-\\#({\mathcal{Z}}\cap{\mathcal{K}}_{1})}w({\mathcal{Z}}\cap{\mathcal{K}}_{1})^{-1}\prod_{i=1}^{g}s_{i}^{-(n+1)(g+2)}\prod_{i=g+1}^{n-1}s_{i}^{-(n+1)(n-i)}\\\\[14.45377pt]
&=&\displaystyle I({\mathcal{Z}}\cap{\mathcal{K}}_{1},s),\end{array}$

as desired. $\Box$

Note that a direct computation yields

$I({\mathcal{K}}_{1},s)=\displaystyle
Y^{-(g+2)}\prod_{j=1}^{g+2}t_{n+1-j}t_{j}=Y^{-(g+2)}\frac{t_{g+1}t_{g+2}}{t_{n+1}}.$
(52)

Proof of Lemma 6.16: Since ${\mathcal{Z}}$ is saturated and
${\mathcal{Z}}\cap{\mathcal{K}}_{1}\neq{\mathcal{K}}_{1}$, we have
${\mathcal{Z}}\cap{\mathcal{K}}_{1}=\\{a_{11},\ldots,a_{1j}\\}$ for some
$j=1,\ldots,g+1$. Since $a_{g+1,g+1}$ and $a_{1,g+2}$ do not belong to
${\mathcal{Z}}$, we have for $s\in T_{\mathcal{Z}}(L,Y)$:

$\begin{array}[]{rcl}I(\\{a_{11},\ldots,a_{1,g+1}\\},s)&=&YI({\mathcal{K}}_{1},s)w(a_{1,g+2})\\\\[3.61371pt]
&\ll&Y^{g+1}I({\mathcal{K}}_{1},s)w(a_{g+1,g+1})w(a_{1,g+2})^{g}\\\\[1.8063pt]
&\ll&\displaystyle
Y^{-1}\frac{1}{t_{1}^{g}t_{g+1}t_{g+2}^{g-1}t_{n+1}}\\\\[10.84006pt]
&\ll&\displaystyle Y^{-1}s_{n}^{2},\end{array}$

since the powers of the $s_{i}$’s in the third line are negative for $i<n$.

Similarly, for any $j=1,\ldots,g$, we compute

$I(\\{a_{11},\ldots,a_{1j}\\},s)(Yw(a_{1,j+1}))^{j-1}\ll Y^{-1}s_{n}^{2},$

as desired. $\Box$

Proof of Lemma 6.17: Suppose now $s_{n}\gg Y^{1/2-\delta}$. First note, that
the inequality

$1\ll Y^{g+1}w(a_{1,n})w(a_{2,n-1})\cdots
w(a_{g+1,g+2})=Y^{g+1}\prod_{i=1}^{n}s_{i}^{-i}$ (53)

implies that we have

$\prod_{j=1}^{n-1}s_{j}^{j}\ll Y^{g+1}s_{n}^{-n}\ll Y^{n\delta}.$ (54)

Since each $s_{i}\gg 1$, we also have $s_{n}\ll Y^{1/2}$ by (54). Hence

$t_{1}^{-1}\ll t_{2}^{-1}\ll\ldots\ll
t_{n}^{-1}=s_{n}^{-1}\prod_{j=1}^{n-1}s_{j}^{j}\ll Y^{-1/2+(n+1)\delta};\quad
t_{n+1}^{-1}=s_{n}^{n}\prod_{j=1}^{n-1}s_{j}^{j}\ll Y^{n/2+n\delta}.$

Thus

$Yw(a_{ij})=Yw(b_{ij})=\frac{Y}{t_{i}t_{j}}\,\ll\,\begin{cases}Y^{(2n+2)\delta}&\mbox{
if }i\leq j\leq n,\\\ Y^{(n+1)/2+(2n+1)\delta}&\mbox{ if }i\leq n,\,j=n+1,\\\
Y^{n+1+2n\delta}&\mbox{ if }i=j=n+1.\end{cases}$

Multiplying these weights together and applying Proposition 4.4 gives the
estimate

$\\#\bigl{(}s(Y{\mathcal{D}})\times s(Y{\mathcal{D}})\cap
W_{n+1}({\mathbb{Z}})\bigr{)}\ll Y^{(n+2)(n+1)+2n(n+2)^{2}\delta}.$ (55)

Meanwhile, in this region where $s_{n}\geq X^{1/2-\delta}$, the quantity
$\delta(s)$ satisfies

$\delta(s)=\prod_{k=1}^{n}s_{k}^{-(n+1)k(n+1-k)}\ll s_{n}^{-n(n+1)}\ll
Y^{-n(n+1)/2+n(n+1)\delta}.$ (56)

Multiplying the bounds in (55) and (56) together yields

$\\#\bigl{(}(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap
W_{n+1}({\mathbb{Z}})\bigr{)}\,\delta(s)\ll
Y^{(\frac{n}{2}+2)(n+1)+n(2n^{2}+9n+9)\delta},$

as desired. $\Box$

Proof of Lemma 6.19: Suppose now
${\mathcal{K}}_{1}\subset{\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ and
$I({\mathcal{K}}_{1},s)\gg X^{-\delta}$ for some $s\in T_{\mathcal{Z}}(L,Y)$.
We prove first that for any $i=1,\ldots,g-1$, we have

$Y^{-\delta}\ll\frac{s_{i}}{s_{n-i}}\ll Y^{\delta}.$ (57)

Indeed, since $a_{j,n+1-j}\notin{\mathcal{Z}}$ for all $j$, we have from (52)
that, for any $k=1,\ldots,g$,

$\begin{array}[]{rcl}I({\mathcal{K}}_{1},s)&\ll&I({\mathcal{K}}_{1},s)Y^{g+2}w(a_{k,n+1-k})^{g}w(a_{g+1,g+2})^{2}\\\\[7.22743pt]
&\ll&\displaystyle\frac{t_{g+1}t_{g+2}}{t_{n+1}}\frac{1}{t_{j}^{g}t_{g+1}^{2}t_{g+2}^{2}t_{n-k+1}^{g}}\\\\[12.64746pt]
&=&\displaystyle\frac{t_{1}\cdots t_{g}}{t_{k}^{g}}\frac{t_{n}\cdots
t_{g+3}}{t_{n-k+1}^{g}}\\\\[12.64746pt]
&=&\displaystyle\frac{t_{1}}{t_{g}}\cdots\frac{t_{g-1}}{t_{g}}\left(\frac{t_{g}}{t_{k}}\right)^{g}\frac{t_{g+4}}{t_{g+3}}\cdots\frac{t_{n}}{t_{g+3}}\left(\frac{t_{g+3}}{t_{n-k+1}}\right)^{g}.\end{array}$

Hence

$\begin{array}[]{rcl}I({\mathcal{K}}_{1},s)&\ll&\left(s_{1}s_{2}^{2}\cdots
s_{g-1}^{g-1}(s_{k}s_{k+1}\cdots s_{g-1})^{-g}s_{n-1}^{-1}s_{n-2}^{-2}\cdots
s_{g+3}^{-(g-1)}(s_{g+3}s_{g+4}\cdots
s_{n-k})^{g})\right)^{n+1}\\\\[10.84006pt]
&=&\displaystyle\left(\prod_{i=1}^{g-1}\left(\frac{s_{i}}{s_{n-i}}\right)^{i(n+1)}\right)\left(\prod_{i=k}^{g-1}\left(\frac{s_{i}}{s_{n-i}}\right)^{-g(n+1)}\right).\end{array}$

Denote the product of the two factors in the final line by $J_{k}$. Then

$\prod_{k=1}^{g}J_{k}=1\qquad\mbox{and}\qquad\frac{J_{i+1}}{J_{i}}=\left(\frac{s_{i}}{s_{n-i}}\right)^{g(n+1)}\,\mbox{
for }i=1,\ldots,g-1.$

Since, by assumption, $I({\mathcal{K}}_{1},s)\gg X^{-\delta}$, we have
$J_{k}\gg Y^{-(n+1)\delta}$ for every $k=1,\ldots,g$. Therefore, for every
$i=1,\ldots,g-1$, we have

$\begin{array}[]{rcccccl}\displaystyle\frac{s_{i}}{s_{n-i}}&=&\displaystyle\left(\frac{J_{i+1}}{J_{i}}\right)^{\frac{1}{g(n+1)}}&=&\left(J_{1}\cdots
J_{i-1}\cdot J_{i+1}^{2}\cdot J_{i+2}\cdots
J_{g}\right)^{\frac{1}{g(n+1)}}&\gg&Y^{-\delta};\\\\[7.22743pt]
\displaystyle\frac{s_{i}}{s_{n-i}}&=&\displaystyle\left(\frac{J_{i}}{J_{i+1}}\right)^{-\frac{1}{g(n+1)}}&=&\left(J_{1}\cdots
J_{i-1}\cdot J_{i}^{2}\cdot J_{i+2}\cdots
J_{g}\right)^{-\frac{1}{g(n+1)}}&\ll&Y^{\delta}.\end{array}$

The claimed bound (57) follows.

By (53) and (57), we have

$s_{g}^{g}s_{g+1}^{g+1}s_{g+2}^{g+2}\ll
Y^{g+1}\prod_{i=1}^{g-1}\left(\frac{s_{i}}{s_{n-i}}\right)^{-i}\prod_{i=g+3}^{n}s_{i}^{-n}\ll
Y^{g+1+\frac{g(g-1)}{2}\delta}\cdot{\mathcal{R}}^{-n}$ (58)

where

${\mathcal{R}}=\prod_{i=g+3}^{n}s_{i}.$

We next prove the desired lower bounds:

$s_{g}\gg Y^{1/2-(g/2)\delta}{\mathcal{R}}^{-1};\quad s_{g+1}\gg 1;\quad
s_{g+2}\gg Y^{1/2-\delta}{\mathcal{R}}^{-1}.$ (59)

The bound $s_{g+1}\gg 1$ follows from the definition of $T^{\prime}$. For the
bounds on $s_{g}$ and $s_{g+2}$, we use the assumption that
$a_{g+1,g+1}\notin{\mathcal{Z}}$ and the computation of
$I({\mathcal{K}}_{1},s)$ in (52) to obtain

$\begin{array}[]{rcl}I({\mathcal{K}}_{1},s)&\ll&I({\mathcal{K}}_{1},s)Y^{1/2}w(a_{g+1,g+1})^{1/2}\\\\[3.61371pt]
&=&\displaystyle Y^{-(n+1)/2}\frac{t_{g+2}}{t_{n+1}}\\\\[10.84006pt]
&=&\displaystyle Y^{-(n+1)/2}s_{g+2}^{n+1}\,{\mathcal{R}}^{n+1},\end{array}$

which along with $I({\mathcal{K}}_{1},s)\gg Y^{-(n+1)\delta}$ implies the
desired lower bounds on $s_{g+2}$; and

$\begin{array}[]{rcl}I({\mathcal{K}}_{1},s)&\ll&\displaystyle
I({\mathcal{K}}_{1},s)Y^{2}w(a_{g+1,g+1})^{2}\\\\[1.8063pt] &=&\displaystyle
Y^{-g}\frac{t_{g+2}}{t_{g+1}^{3}t_{n+1}}\\\\[7.22743pt] &=&\displaystyle
Y^{-g}\Big{(}\prod_{i=1}^{g}s_{i}^{3i}\Big{)}\,s_{g+1}^{-3(g+2)}\Big{(}\prod_{i=g+2}^{n}s_{i}^{-2n-2+3i}\Big{)}\\\\[3.61371pt]
&\ll&\displaystyle
Y^{-g+\frac{3g(g-1)}{2}\delta}\Big{(}\prod_{i=1}^{g-1}s_{n-i}^{3i}\Big{)}s_{g}^{3g}s_{g+1}^{-3(g+2)}s_{g+2}^{-g}\Big{(}\prod_{i=g+3}^{n-1}s_{i}^{-2n-2+3i}\Big{)}s_{n}^{2g}\\\\[7.22743pt]
&\ll&\displaystyle
Y^{-g+\frac{3g(g-1)}{2}\delta}s_{g}^{3g}s_{g+1}^{-3(g+2)}s_{g+2}^{-g}\,{\mathcal{R}}^{2g}\\\\[7.22743pt]
&\ll&\displaystyle
Y^{-\frac{3g}{2}+\frac{g(3g-1)}{2}\delta}s_{g}^{3g}\,{\mathcal{R}}^{3g},\end{array}$

implying the desired lower bound on $s_{g}$, where in the last inequality we
used the already-established lower bounds on $s_{g+1}$ and $s_{g+2}$.

The desired lower bounds for $s_{g}$, $s_{g+1}$, $s_{g+2}$ then follow by
combining the upper bound on $s_{g}^{g}s_{g+1}^{g+1}s_{g+2}^{g+2}$ in (58) and
the individual lower bounds on $s_{g}$, $s_{g+1}$, $s_{g+2}$ in (59). The
desired upper bound on ${\mathcal{R}}$ follows by comparing the upper bound on
$s_{g}$ and the trivial lower bound $s_{g}\gg 1$. $\Box$

Proof of Lemma 6.20: Suppose ${\mathcal{K}}_{1}\subset{\mathcal{Z}}$ and $s\in
T_{\mathcal{Z}}(L,Y)$ satisfies (50). Then

$t_{j}^{-1}\ll\begin{cases}\displaystyle
Y^{-(g+2)+5g^{2}\delta}\prod_{i=n-j+1}^{n}s_{i}^{n+1}&\mbox{ for
}j=1,\ldots,g,\\\ Y^{-1/2+20g^{2}\delta}&\mbox{ for }j=g+1,g+2,\\\
\displaystyle Y^{g+1+23g^{2}\delta}\prod_{i=j}^{n}s_{i}^{-(n+1)}&\mbox{ for
}j=g+3,\ldots,n+1,\end{cases}$ (60)

where the upper bound on ${\mathcal{R}}$ also gives $t_{j}^{-1}\ll
Y^{-1/2+20g^{2}\delta}$ for $j=1,\ldots,g$. For $i,j\leq g+2$, we have
$Yt_{i}^{-1}t_{j}^{-1}\ll Y^{40g^{2}\delta}$. For $j\geq g+3$ and $i\leq
n-j+1$, we have $Yt_{i}^{-1}t_{j}^{-1}\ll Y^{28g^{2}\delta}$. Using (60) for
the rest of the coordinates gives

$\\#\bigl{(}(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap
W({\mathbb{Z}})\bigr{)}\ll
Y^{(n+1)(g+1)(g+4)+972g^{4}\delta}\left(\prod_{i=1}^{g}s_{g+2+i}^{-2i(i+3)(n+1)}\right).$

The Haar measure satisfies the following bound:

$\delta(s)=\prod_{k=1}^{n}s_{k}^{-(n+1)k(n+1-k)}\ll
Y^{-(g^{2}+3g+1)(n+1)+55g^{4}\delta}\left(\prod_{i=1}^{g}s_{g+2+i}^{2i(i+2)(n+1)}\right).$

Hence

$\\#\bigl{(}(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap
W({\mathbb{Z}})\bigr{)}\,\delta(s)\ll
Y^{(n+1)^{2}+1027g^{4}\delta}\left(\prod_{i=1}^{g}s_{g+2+i}^{-2i(n+1)}\right).$
(61)

Suppose now $\\#\bigl{(}(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap
W({\mathbb{Z}})\bigr{)}\,\delta(s)\gg X^{n+1-\delta}$. Then, for any
$i=g+3,\ldots,n$,

$s_{i}\ll Y^{(1027g^{4}+2g+3)\delta/(2(i-g-2)(2g+3))}\ll
Y^{(1032g^{4}/(4g))\delta}=Y^{258g^{3}\delta},$

as desired. $\Box$

Proof of Lemma 6.21: Suppose $M>X^{\eta}$ where $\eta>0$ is some fixed
constant. Suppose $\delta<{\rm max}(\eta,1)/1355g^{6}$. Suppose
${\mathcal{K}}_{1}\subset{\mathcal{Z}}$ and $s\in T_{\mathcal{Z}}(L,Y)$
satisfies (50) and (51). We now impose the conditions $\det(B)=0$ and
$|q|(A,B)>M$ for any $(A,B)\in{{\mathcal{L}}(M)}$ to obtain a further saving
for $\\#\bigl{(}(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}\bigr{)}\,\delta(s)$.

The bound (51) on $s_{g+3},\ldots,s_{n}$ gives ${\mathcal{R}}\ll
Y^{258g^{4}\delta}.$ Hence

$t_{j}^{-1}\ll\begin{cases}\displaystyle Y^{-(g+2)+1295g^{5}\delta},&\mbox{
for }j=1,\ldots,g,\\\ Y^{-1/2+20g^{2}\delta},&\mbox{ for }j=g+1,g+2,\\\
\displaystyle Y^{g+1+23g^{2}\delta},&\mbox{ for }j=g+3,\ldots,n+1,\end{cases}$
(62)

thus improving (60). In this case,

$Yt_{g}^{-1}t_{g+2}^{-1}\ll Y^{-(n+1)/2+1315g^{5}\delta},\qquad
Yt_{g}^{-1}t_{n+1}^{-1}\ll Y^{1318g^{5}\delta},\vspace{-.05in}$
$Yt_{g+2}^{-1}t_{n+1}^{-1}\ll Y^{(n+1)/2+43g^{2}\delta},\qquad
Yt_{n+1}^{-2}\ll Y^{n+1+46g^{2}\delta}.$

Since $\delta<1/(1315g^{4})$, we may assume that every
$(A,B)\in(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap W({\mathbb{Z}})$
satisfies:

  * (a)

The top left $g\times(g+2)$-blocks of $A$ and $B$ are $0$.

  * (b)

The entries of the top right $g\times(g+1)$ blocks of $A$ and $B$ are
$O(Y^{1315g^{5}\delta})$.

  * (c)

The entries $a_{g+1,g+1}$, $a_{g+1,g+2}$, $a_{g+2,g+2}$, $b_{g+1,g+1}$,
$b_{g+1,g+2}$, and $b_{g+2,g+2}$ are $O(Y^{40g^{2}\delta})$.

  * (d)

The entries $a_{g+1,j}$, $a_{g+2,j}$, $b_{g+1,j}$ and $b_{g+2,j}$ are
$O(Y^{(n+1)/2+43g^{2}\delta})$ for $g+3\leq j\leq n+1$.

  * (e)

The entries $a_{ij}$ and $b_{ij}$ are $O(Y^{n+1+46g^{2}\delta})$ for $g+3\leq
i,j\leq n+1$.

Suppose now that $(A,B)$ is an element of $(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}$. Then $f_{A,B}=xg(x,y)$, where
$g(x,1)$ is a degree $n$ polynomial with Galois group $S_{n}$.

###### Lemma 6.23

Let $(A,B)$ be as above. If $b_{g+1,g+1}=b_{g+1,g+2}=b_{g+2,g+2}=0$, then

$|q|(A,B)\ll X^{1355g^{6}\delta}.$

Proof: Let $v=x_{0}e_{g+1}+y_{0}e_{g+2}$ where $x_{0},y_{0}\in K$ are not both
$0$, where $K$ is some quadratic extension of ${\mathbb{Q}}$, and where
$(x_{0},y_{0})$ is a solution to

$a_{g+1,g+1}x^{2}+a_{g+1,g+2}xy+a_{g+2,g+2}y^{2}=0.$ (63)

We claim first that $v$ can be chosen in ${\mathbb{Q}}^{n+1}$. Indeed,
$\text{Span}_{K}\\{e_{1},\ldots,e_{g},v\\}$ is a $(g+1)$-dimensional subspace
isotropic with respect to $A$ and $B$. Since $(A,B)$ is distinguished over
${\mathbb{Q}}$, the set of $(g+1)$-dimensional common isotropic subspaces
defined over any number field $L$ is in bijection with $J[2](L)$, where $J$ is
the Jacobian of the hyperelliptic curve $y^{2}=xg(x,1)$ (which has a rational
Weierstrass point at infinity), and $J[2](L)$ is in bijection with the
factorizations of $xg(x,1)$ over $L$. Since $g(x,1)$ has Galois group $S_{n}$,
it does not admit any factorization over any quadratic extension of
${\mathbb{Q}}$. Therefore, $J[2](K)=J[2]({\mathbb{Q}})$ with the only
nontrivial element corresponding to the factorization $x\cdot g(x,1)$. In
other words, $\text{Span}_{K}\\{e_{1},\ldots,e_{g},v\\}$ is defined over
${\mathbb{Q}}$ and we may choose $v$ to be in ${\mathbb{Q}}^{n+1}$. In
particular, the discriminant
$a_{g+1,g+2}^{2}-4a_{g+1,g+1}a_{g+2,g+2}\in{\mathbb{Z}}$ is a square.

If $a_{g+1,g+1}\neq 0$, let

$x_{1}=-a_{g+1,g+2}+\sqrt{a_{g+1,g+2}^{2}-4a_{g+1,g+1}a_{g+2,g+2}},\qquad
y_{1}=2a_{g+1,g+1}.$

If $a_{g+1,g+1}=0$, let $x_{1}=1,\,y_{1}=0$. Then $x_{1},y_{1}$ are integers
$\ll Y^{40g^{2}\delta}$, not both zero, and are solutions to (63). Let
$x_{0}=x_{1}/\gcd(x_{1},y_{1})$ and $y_{0}=y_{1}/\gcd(x_{1},y_{1})$. There
then exist integers $x_{2},y_{2}\ll Y^{40g^{2}\delta}$ such that

$\\{e_{1},\ldots,e_{g},x_{0}e_{g+1}+y_{0}e_{g+2},x_{2}e_{g+1}+y_{2}e_{g+2},e_{g+3},\ldots,e_{n}\\}$

forms an integral basis for ${\mathbb{Z}}^{n+1}$ such that the first $g+1$
vectors generate a primitive lattice isotropic with respect to $A$ and $B$,
and the first $g+2$ vectors generate a primitive lattice isotropic with
respect to $B$. That is, we compute the $|q|$-invariant of $(A,B)$ using this
basis. When so expressed, the top right $(g+1)\times(g+2)$ blocks of the Gram
matrices of $A$ and $B$ have the form

$A^{\rm top}=\begin{pmatrix}0&\flat&\cdots&\flat\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&\flat&\cdots&\flat\\\
\flat&*&\cdots&*\end{pmatrix},\qquad B^{\rm
top}=\begin{pmatrix}0&\flat&\cdots&\flat\\\ \vdots&\vdots&\ddots&\vdots\\\
0&\flat&\cdots&\flat\\\ 0&*&\cdots&*\end{pmatrix},$

where entries labeled “$0$” are $0$, entries labeled “$\flat$” are
$O(Y^{1355g^{5}\delta})$, and entries labeled “$*$” are
$O(Y^{(n+1)/2+83g^{2}\delta})$. Let $M_{1}$ denote the $(g+2)\times(g+2)$
matrix whose $i$th row consists of the coefficients of $\det(A_{i}x-B_{i}y)$,
where $A_{i}$ and $B_{i}$ are the $(g+1)\times(g+1)$ matrices formed by
removing the $i$-th columns from $A^{\rm top}$ and $B^{\rm top}$,
respectively. Then $M_{1}$ is of the form

$M_{1}=\begin{pmatrix}*&\cdots&*&*\\\ \sharp&\cdots&\sharp&0\\\
\vdots&\ddots&\vdots&\vdots\\\ \sharp&\cdots&\sharp&0\end{pmatrix},$

where entries labeled “$0$” are $0$, entries labeled “$\sharp$” are
$O(Y^{2710g^{6}\delta})$, and entries labeled “$*$” are
$O(Y^{(n+1)/2+1438g^{6}\delta})$, where the top right coefficient $m^{\prime}$
of $M_{1}$ is the determinant of the top right $(g+1)\times(g+1)$ block
$B^{\prime}$ of $B^{\rm top}$, up to sign. Thus

$|q|(A,B)=\frac{|Q|(A,B)}{|\det(B^{\prime})|}=\frac{|\det(M_{1})|}{|m^{\prime}|}=|\det(M_{1}^{\prime})|,$

where $M_{1}^{\prime}$ is the bottom left $(g+1)\times(g+1)$ block of $M_{1}$.
Since the coefficients of $M_{1}^{\prime}$ are $\ll Y^{2710g^{6}\delta}$, it
follows that $|q|(A,B)\ll X^{2710g^{6}(g+1)\delta/(n+1)}\ll
X^{1355g^{6}\delta}$. $\Box$

We now return to the proof of Lemma 6.21. For any
$(A,B)\in(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}$,
since $|q|(A,B)>M>X^{\eta}$, we may assume that $b_{g+1,g+1}$, $b_{g+1,g+2}$,
and $b_{g+2,g+2}$ are not all $0$ since $\delta<\eta/(1355g^{6})$.

We now fix $b_{ij}$ for $1\leq i\leq g,$ $g+3\leq j\leq n+1$, and $i=g+1,g+2$,
$j=g+1,g+2$. We consider the number of pairs $(A,B)\in(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}$ with these prescribed coefficients
by viewing $\det(B)$ as a polynomial $F$ in $b_{ij}$ for $g+1\leq i\leq n+1$
and $g+3\leq j\leq n+1$. Note that all of these remaining coefficients have
range at least

$Y^{(n+1)/2+43g^{2}\delta}\prod_{i=g+3}^{n}s_{i}^{-(n+1)}.$

Hence, to complete the proof of Lemma 6.21, it remains to prove that $F$ is a
nonzero polynomial, for then we would have, using (61), that

$\displaystyle\\#\bigl{(}(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}\bigr{)}\,\delta(s)$
$\displaystyle\ll$ $\displaystyle
Y^{(n+1)^{2}+1027g^{4}\delta-(n+1)/2-43g^{2}\delta}\,\prod_{i=1}^{g}s_{g+2+i}^{-(2i-1)(n+1)}$
$\displaystyle\ll$ $\displaystyle X^{n+1+514g^{3}\delta-1/2}.$

We may assume that the top right $g\times(g+1)$ block of $B$ has full rank,
for otherwise the kernel of $B$ would be isotropic with respect to $A$ forcing
$\Delta(A,B)=0$ by Lemma 6.4. Hence we may also assume that the top right
$g\times(g+1)$ block of $B$ equals $(I_{g}\,\,0)$, where $I_{g}$ denotes the
$g\times g$ identity matrix. Then

$\det(B)=\det\begin{pmatrix}b_{g+1,g+1}&b_{g+1,g+2}&b_{g+1,n+1}\\\
b_{g+2,g+1}&b_{g+2,g+2}&b_{g+2,n+2}\\\
b_{n+1,g+1}&b_{n+1,g+2}&b_{n+1,n+1}\end{pmatrix}.$

Since

$\begin{pmatrix}b_{g+1,g+1}&b_{g+1,g+2}\\\
b_{g+2,g+1}&b_{g+2,g+2}\end{pmatrix}\neq 0,$

we see that $\det(B)$ is a nonzero polynomial in $b_{g+1,n+1}$, $b_{g+2,n+1}$,
and $b_{n+1,n+1}$. $\Box$

###  6.4 Bounding the number of distinguished elements in the deep cusp

In this subsection, we bound the number of elements with large $q$-invariant
that lie in the deep cusp.

###### Theorem 6.24

We have $\displaystyle{\mathcal{I}}_{X}^{\rm
dcusp}({{\mathcal{L}}(M)})=O\Bigl{(}\frac{X^{n+1+\frac{1}{2}\kappa}}{M}\log^{2n}X\Bigr{)}.$

Recall from (38) that

${\mathcal{I}}_{X}^{\rm
dcusp}({{\mathcal{L}}(M)})\ll\displaystyle\sum_{L}\sum_{{\mathcal{Z}}:a_{g+1,g+1}\in{\mathcal{Z}}}N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X);$

here the first sum is over $r$-tuples $L=(L_{1},\ldots,L_{r})$ with $L_{1}\leq
L_{2}\leq\cdots\leq L_{n}$ that partition the region
$\\{(\mu_{1},\ldots,\mu_{r})\in[Y^{-\Theta_{1}},Y^{\Theta_{2}}]^{r}:\mu_{1}\leq\ldots\leq\mu_{r}\\}$
into dyadic ranges, and the second sum is over saturated subsets
${\mathcal{Z}}$ of ${\mathcal{K}}\cup{\mathcal{M}}$, where

$N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)=\int_{T_{\mathcal{Z}}(L,Y)}\\#\bigl{\\{}(A,B)\in(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}:B\in{\mathcal{S}}(L,s)\bigr{\\}}\,\delta(s)d^{\times}s.$

The set ${\mathcal{S}}(L,s)$ is the union over $\Lambda\in\Sigma(L,s)$ of
$S(\Lambda)$, where $S(\Lambda)$ denotes the lattice of integral symmetric
matrices whose row space is contained in $\Lambda\otimes{\mathbb{R}}$, and
$\Sigma(L,s)$ denotes the set of primitive lattices
$\Lambda\in{\mathbb{Z}}^{n+1}$ of rank $n$ such that the successive minima
$\mu_{1},\ldots,\mu_{n}$ of $s^{-1}(\Lambda)$ satisfy
$L_{i}\leq\mu_{i}<2L_{i}$ for each $i\in\\{1,\ldots,n\\}$. Finally recall from
§6.1 and Proposition 5.4 that

$w(\ell_{g+1,g+1})^{2}=L_{g+1}^{2}t_{g+1}^{-2}\leq
c_{1}Yt_{g+1}^{-2}=c_{1}Yw(a_{g+1,g+1})<c_{1}c_{g+1,g+1}<c_{g+1}^{\prime\,\,2}$

for every $s\in T_{\mathcal{Z}}(Y,L)$. Hence we may assume that
$\ell_{g+1,g+1}\in{\mathcal{Z}}$.

The deep cusp contains $\asymp X^{n+1}$ elements, and we obtain a saving
because the elements we are counting have $q$-invariant greater than $M$. To
make use of this condition, we require an upper bound on the size of the
$|q|$-invariant of elements in $(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(1)}$. To accomplish this, we have the
following preliminary result.

###### Lemma 6.25

Let $(A,B)\in(Y{\mathcal{D}}\times Y{\mathcal{D}})\cap W_{0}({\mathbb{R}})$ be
such that $\Delta(A,B)>X^{2n-2-\kappa}$. Denote the top right
$(g+1)\times(g+2)$ block of $B$ by $B^{\rm top}$. Then

$\det(B^{\rm top}(B^{\rm top})^{t})\gg Y^{2(g+1)-(n+1)\kappa}.$

Proof: Let
$(A^{\prime},B^{\prime})=Y^{-1}(A,B)\in({\mathcal{D}}\times{\mathcal{D}})\cap
W_{0}({\mathbb{R}})$. Then it suffices to prove that

$\det(B^{\prime\,{\rm top}}(B^{\prime\,{\rm top}})^{t})\gg Y^{-(n+1)\kappa}.$

Since $|\Delta(A,B)|>X^{2n-2-\kappa}$, we have
$|\Delta(A^{\prime},B^{\prime})|>X^{-\kappa}$. By Proposition 3.6, there is a
polynomial $P\in{\mathbb{Z}}[W_{0}]$ such that

$\Delta(A^{\prime\prime},B^{\prime\prime})=P(A^{\prime\prime},B^{\prime\prime})\det(B^{\prime\prime\,{\rm
top}}(B^{\prime\prime\,{\rm top}})^{t})$

for any $(A^{\prime\prime},B^{\prime\prime})\in W_{0}({\mathbb{R}})$. Since
$(A^{\prime},B^{\prime})\in{\mathcal{D}}\times{\mathcal{D}}$, which is an
absolutely bounded region, we have $|P(A^{\prime},B^{\prime})|\ll 1$. Hence

$\det(B^{\prime\,{\rm top}}(B^{\prime\,{\rm
top}})^{t})=\frac{\Delta(A^{\prime},B^{\prime})}{P(A^{\prime},B^{\prime})}\gg
X^{-\kappa}$

as desired. $\Box$

Next, we have the following upper bound on the $|q|$-invariant.

###### Proposition 6.26

Let ${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ be a saturated set containing
$a_{g+1,g+1}$ and $\ell_{g+1,g+1}$. Let $L=(L_{1},\ldots,L_{n})$ be a sequence
of nondecreasing positive real numbers. Then for any $s\in
T_{\mathcal{Z}}(L,Y)$ and $(A,B)\in(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(1)}$, we have

$|q|(A,B)\ll Y^{(g+1)^{2}+\frac{n+1}{2}\kappa}\prod_{i=1}^{g+1}L_{i}.$ (64)

Proof: Suppose $(A,B)\in(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(1)}$. Since
$a_{g+1,g+1}\in{\mathcal{Z}}$, we have $(A,B)\in W_{0}({\mathbb{Z}})$. By
Lemma 6.4, $\ker(B)$ is $1$-dimensional and does not lie inside
$\text{Span}\\{e_{1},\ldots,e_{g+1}\\}$ as this $(g+1)$-plane is isotropic
with respect to $A$. Let
$w_{1}\in\text{Span}_{\mathbb{Z}}\\{e_{g+2},\ldots,e_{n+1}\\}$ be a primitive
vector so that $\\{e_{1},\ldots,e_{g+1},w_{1}\\}$ forms a basis for the
primitive lattice in
$\text{Span}_{\mathbb{R}}\\{e_{1},\ldots,e_{g+1}\\}+\ker(B).$ Complete $w_{1}$
to an integral basis $\\{w_{1},\ldots,w_{g+2}\\}$ for
$\text{Span}_{\mathbb{Z}}\\{e_{g+2},\ldots,e_{n+1}\\}$. We can now use the
integral basis

$\\{e_{1},\ldots,e_{g+1},w_{1},\ldots,w_{g+2}\\}$ (65)

of ${\mathbb{Z}}^{n+1}$ to compute the $|q|$-invariant of $(A,B)$, as the
first $g+1$ vectors generate a primitive lattice isotropic with respect to $A$
and $B$, and the first $g+2$ vectors generate a primitive lattice isotropic
with respect to $B$. Note also that with respect to the standard inner product
on ${\mathbb{R}}^{n+1}$, since
$w_{1}\in\text{Span}_{\mathbb{R}}\\{e_{1},\ldots,e_{g+1}\\}+\ker(B)$, we have

$w_{1}\perp\bigl{(}\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}\cap
C(B)\bigr{)}$ (66)

where $C(B)$ denotes the column space of $B$.

Let $A^{\prime}$ and $B^{\prime}$ be the Gram matrices of the quadratic forms
defined by $A$ and $B$ with respect to this new basis (65). Since the first
$g+1$ vectors of this basis are part of the standard basis, we see that
$(A,B)$ and $(A^{\prime},B^{\prime})$ are $G_{0}({\mathbb{Z}})$-equivalent,
where $G_{0}$ is defined in §3.1. Hence

$|Q|(A^{\prime},B^{\prime})=|Q|(A,B)\ll
Y^{(g+1)(g+2)}\prod_{k=1}^{g+1}t_{k}^{-1}.$

Let $B^{\prime\prime}$ denote the top right $(g+1)\times(g+1)$ block of
$B^{\prime}$. Then, by the definition of $q$, we have

$|q|(A,B)=|q|(A^{\prime},B^{\prime})=\frac{|Q|(A^{\prime},B^{\prime})}{|\det(B^{\prime\prime})|}\ll\frac{1}{|\det(B^{\prime\prime})|}Y^{(g+1)(g+2)}\prod_{k=1}^{g+1}t_{k}^{-1}.$
(67)

We now work towards proving a lower bound on $|\det(B^{\prime\prime})|$. Let
$p_{1}$ (resp., $p_{2}$) denote the projection of ${\mathbb{R}}^{n+1}$ onto
the first $g+1$ coefficients (resp., the last $g+2$ coefficients). Let
$B^{{\rm top}}$ (resp., $B^{\prime{\rm top}}$) denote the top right
$(g+1)\times(g+2)$ block of $B$ (resp., $B^{\prime}$). Then by (66), we have
$B^{\rm top}p_{2}(w_{1})=0$. Consider the following two $(g+2)\times(g+2)$
matrices in block form:

$B^{*}=\begin{pmatrix}B^{\rm top}\\\
p_{2}(w_{1})^{t}\end{pmatrix},\qquad\gamma=\begin{pmatrix}p_{2}(w_{1})&\cdots&p_{2}(w_{g+2})\end{pmatrix}.$

Then

$B^{*}\gamma=\begin{pmatrix}0&B^{\prime\prime}\\\ |w_{1}|^{2}&*\end{pmatrix}.$

Let $\Lambda_{2}$ denote the rank $g+1$ lattice in ${\mathbb{Z}}^{g+2}$
spanned by the rows of $B^{{\rm top}}$. Then
$|\det(B^{*})|=d(\Lambda_{2})|w_{1}|.$ Since
$\\{p_{2}(w_{1}),\ldots,p_{2}(w_{g+2})\\}$ is an integral basis for
${\mathbb{Z}}^{g+2}$, we have $\det\gamma=\pm 1$ and so

$|\det(B^{\prime\prime})|=\frac{|\det(B^{*})\det\gamma|}{|w_{1}|^{2}}=\frac{d(\Lambda_{2})|w_{1}|\cdot
1}{|w_{1}|^{2}}=\frac{d(\Lambda_{2})}{|w_{1}|}.$

We now use the fact that $B\in{\mathcal{S}}(L,s)$. This means that the row
span of $B$ lies in an $n$-dimensional primitive lattice
$\Lambda\subset{\mathbb{Z}}^{n+1}$ with basis of the form
$\\{s\ell_{1},\ldots,s\ell_{n}\\}$ where $L_{i}\leq|\ell_{i}|<2L_{i}$ and
$\\{\ell_{1},\ldots,\ell_{n}\\}$ are reduced. By assumption,
$\ell_{g+1,g+1}\in{\mathcal{Z}}$, and hence $\ell_{i,j}\in{\mathcal{Z}}$ for
all $i\leq g+1$ and $j\leq g+1$. Thus the first $g+1$ coefficients of
$s\ell_{1},\ldots,s\ell_{g+1}$ are all $0$, and
$\\{s\ell_{1},\ldots,s\ell_{g+1}\\}$ forms an integral basis of a primitive
lattice $\Lambda_{1}$ of rank $g+1$ in
$\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}$. By (66), $w_{1}$ is a
primitive vector in $\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}$
orthogonal to $\Lambda_{1}$. Hence

$|w_{1}|=d(\Lambda_{1}).$

By (66), we have $\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}\cap
C(B)\neq\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}$, and so

$\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}\cap
C(B)=\Lambda_{1}\otimes{\mathbb{R}}.$

In particular, since $\Lambda_{1}$ is primitive, the first $g+1$ columns of
$B$ belong to $\Lambda_{1}$. That is, there is a $(g+1)\times(g+1)$ matrix $C$
(with integer coefficients) such that

$B^{\rm top}=C\begin{pmatrix}p_{2}(s\ell_{1})^{t}\\\ \vdots\\\
p_{2}(s\ell_{g+1})\end{pmatrix}$

and so

$|\det(C)|=\frac{d(\Lambda_{2})}{d(p_{2}(\Lambda_{1}))}=\frac{d(\Lambda_{2})}{|w_{1}|}=|\det(B^{\prime\prime})|.$
(68)

To obtain a lower bound on $|\det(C)|$, we write

$s=\begin{pmatrix}A_{1}&0\\\ 0&A_{2}\end{pmatrix}\qquad\mbox{with}\qquad
A_{1}=\begin{pmatrix}t_{1}^{-1}&&\\\ &\ddots&\\\
&&t_{g+1}^{-1}\end{pmatrix},\quad A_{2}=\begin{pmatrix}t_{g+2}^{-1}&&\\\
&\ddots&\\\ &&t_{n+1}^{-1}\end{pmatrix}.$

Let $M^{\rm top}$ denote the $(g+1)\times(g+2)$ matrix with rows
$p_{2}(\ell_{1})^{t},\ldots,p_{2}(\ell_{g+1})^{t}$. Then

$CM^{\rm top}A_{2}=B^{{\rm top}}.$

Consider the pair $(A_{0},B_{0}):=s^{-1}(A,B)\in(Y{\mathcal{D}}\times
Y{\mathcal{D}})\cap W_{0,n+1}({\mathbb{R}})$ satisfying

$|\Delta(A_{0},B_{0})|=|\Delta(A,B)|>X^{2n-2-\kappa}$

since $(A,B)\in{{\mathcal{L}}(1)}$. The top right $(g+1)\times(g+2)$ block
$B_{0}^{\rm top}$ of $B_{0}$ satisfies

$A_{1}B_{0}^{\rm top}A_{2}=B^{\rm top},$

and so

$CM^{\rm top}=A_{1}B_{0}^{\rm top}.$

The rows of $M^{\rm top}$ form a reduced basis for a lattice
$\Lambda_{3}\subset{\mathbb{Z}}^{g+2}$ with
$L_{i}\leq|p_{2}(\ell_{i})|<2L_{i}$. Thus

$\det(B^{\prime\prime})^{2}=\det(C)^{2}=\frac{\det(A_{1}B_{0}^{\rm
top}(B_{0}^{\rm top})^{t}A_{1})}{\det(M^{\rm top}(M^{\rm
top})^{t})}\gg\frac{t_{1}^{-2}\cdots t_{g+1}^{-2}}{L_{1}^{2}\cdots
L_{g+1}^{2}}\det(B_{0}^{\rm top}(B_{0}^{\rm top})^{t}).$ (69)

By Equations (67) and (67),

$|q|(A,B)\ll Y^{(g+1)(g+2)}\frac{L_{1}\cdots L_{g}}{\sqrt{\det(B_{0}^{\rm
top}(B_{0}^{\rm top})^{t})}}.$

The result now follows from Lemma 6.25. $\Box$

Proof of Theorem 6.24: We write

${\mathcal{I}}_{X}^{\rm
dcusp}({{\mathcal{L}}(M)})\ll\sum_{L}\sum_{\begin{subarray}{c}{\mathcal{Z}}\\\
a_{g+1,g+1}\in{\mathcal{Z}}\\\
\ell_{g+1,g+1}\in{\mathcal{Z}}\end{subarray}}N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X),$
(70)

and obtain upper bounds on $N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)$ for each
${\mathcal{Z}}\subset{\mathcal{Z}}_{1}$ with
$a_{g+1,g+1},\ell_{g+1,g+1}\in{\mathcal{Z}}$. Fix such a set ${\mathcal{Z}}$
with $N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)>0$ and an element $s\in
T_{\mathcal{Z}}(L,Y)$. Then

$(s(Y{\mathcal{D}})\times
s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}\subset(s(Y{\mathcal{D}})\cap
S({\mathbb{Z}}))\times(s(Y{\mathcal{D}})\cap{\mathcal{S}}(L,s).$

We begin by bounding the number of elements in $\\#(s(Y{\mathcal{D}})\cap
S({\mathbb{Z}}))$. Let ${\mathcal{K}}_{\rm dist}:=\\{a_{ij}\mid 1\leq i\leq
j\leq g+1\\}$. By assumption, ${\mathcal{K}}_{\rm dist}$ is a subset of
${\mathcal{Z}}\cap{\mathcal{K}}$. Define
$\pi_{\mathcal{K}}:{\mathcal{Z}}_{1}\cap{\mathcal{K}}\rightarrow{\mathcal{K}}\backslash{\mathcal{Z}}_{1}$
by

$\pi_{\mathcal{K}}(a_{ij}):=a_{n+1-j,j}.$

This agrees with the $\pi_{k}$ as defined in §6.3.1 when restricted to
${\mathcal{K}}$. For any $\alpha\in{\mathcal{Z}}_{1}\cap{\mathcal{K}}$, we
have $Yw(\pi_{\mathcal{K}}(\alpha))\gg 1$ and $w(\pi_{\mathcal{K}}(\alpha))\gg
w(\alpha)$. For any
$a_{ij}\in({\mathcal{Z}}_{1}\cap{\mathcal{K}})\backslash{\mathcal{K}}_{\rm
dist}$, we have $i<g+1<j$. Thus

$\prod_{\alpha\in({\mathcal{Z}}\cap{\mathcal{K}})\backslash{\mathcal{K}}_{\rm
dist}}\frac{w(\pi_{\mathcal{K}}(\alpha))}{w(\alpha)}\ll\prod_{\alpha\in({\mathcal{Z}}_{1}\cap{\mathcal{K}})\backslash{\mathcal{K}}_{\rm
dist}}\frac{w(\pi_{\mathcal{K}}(\alpha))}{w(\alpha)}=\prod_{\begin{subarray}{c}1\leq
i<g+1<j\\\ i+j\leq n\end{subarray}}\frac{t_{i}}{t_{n+1-j}}=\prod_{1\leq i\leq
j\leq g+1}\frac{t_{i}}{t_{j}}.$

Therefore,

$\displaystyle\\#\bigl{(}s(Y{\mathcal{D}})\cap S({\mathbb{Z}})\bigr{)}$
$\displaystyle\ll$ $\displaystyle
Y^{(n+1)(n+2)/2-\\#({\mathcal{Z}}\cap{\mathcal{K}})}\prod_{\alpha\in{\mathcal{Z}}\cap{\mathcal{K}}}\frac{1}{w(\alpha)}$
(71) $\displaystyle\ll$ $\displaystyle
Y^{(n+1)(n+2)/2-\\#({\mathcal{Z}}\cap{\mathcal{K}})}\Big{(}\prod_{\alpha\in{\mathcal{K}}_{\rm
dist}}\frac{1}{w(\alpha)}\Big{)}\Big{(}\prod_{\alpha\in({\mathcal{Z}}\cap{\mathcal{K}})\backslash{\mathcal{K}}_{\rm
dist}}\frac{Yw(\pi_{q}(\alpha))}{w(\alpha)}\Big{)}$ $\displaystyle\ll$
$\displaystyle\frac{Y^{(n+1)(n+2)/2}}{Y^{\\#{\mathcal{K}}_{\rm
dist}}}\Big{(}\prod_{1\leq i\leq j\leq
g+1}t_{i}t_{j}\Big{)}\Big{(}\prod_{1\leq i\leq j\leq
g+1}\frac{t_{i}}{t_{j}}\Big{)}$ $\displaystyle=$
$\displaystyle\frac{Y^{(n+1)(n+2)/2}}{Y^{(g+1)(g+2)/2}}(t_{1}\cdots
t_{g+1})^{g+2}\Big{(}\prod_{1\leq i\leq j\leq g+1}\frac{t_{i}}{t_{j}}\Big{)}.$

We now obtain an upper bound on
$\\#(s(Y{\mathcal{D}})\cap{\mathcal{S}}(L,s))$. Recall that

$\\#\bigl{(}s(Y{\mathcal{D}})\cap{\mathcal{S}}(L,s)\bigr{)}=\sum_{\Lambda\in\Sigma(L,s)}\\#\big{(}Y{\mathcal{D}}\cap
s^{-1}S(\Lambda)\big{)}.$ (72)

Let $\Lambda\in\Sigma(L,s)$ be a lattice such that $s^{-1}(\Lambda)$ has
reduced basis $\\{\ell_{1},\ldots,\ell_{n}\\}$ with
$L_{i}\leq|\ell_{i}|<2L_{i}$ for each $i=1,\ldots,n$. Suppose there exists
$(A,B)\in(s(Y{\mathcal{D}})\times s(Y{\mathcal{D}}))\cap{{\mathcal{L}}(M)}$
with $B\in s(Y{\mathcal{D}})\cap S(\Lambda)$. By Proposition 6.26,

$M\ll Y^{(g+1)^{2}+\frac{n+1}{2}\kappa}\prod_{i=1}^{g+1}L_{i}.$ (73)

Recall also from the proof of Proposition 6.26 that

$\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}\cap
C(B)=\text{Span}_{\mathbb{R}}\\{s\ell_{1},\ldots,s\ell_{g+1}\\}.$

Hence

$\text{Span}_{\mathbb{R}}\\{e_{g+2},\ldots,e_{n+1}\\}\cap\text{Span}_{\mathbb{R}}\\{s\ell_{g+2},\ldots,s\ell_{n}\\}=\\{0\\}.$

It follows that the set $\\{p_{1}(s\ell_{g+2}),\ldots,p_{1}(s\ell_{n})\\}$,
and thus the set $\\{p_{1}(\ell_{g+2}),\ldots,p_{1}(\ell_{n})\\}$, are both
linearly independent. There then exist vectors
$v_{g+2},\ldots,v_{n}\in\text{Span}_{\mathbb{R}}\\{e_{1},\ldots,e_{g+1}\\}$
such that

$\begin{pmatrix}v_{g+2}&\cdots&v_{n}\end{pmatrix}^{t}\begin{pmatrix}\ell_{g+2}&\cdots&\ell_{n}\end{pmatrix}=I_{g+1}$

is the identity matrix. Let $B^{\prime}\in s(Y{\mathcal{D}})\cap S(\Lambda)$
be any element and write

$s^{-1}B^{\prime}=\sum_{1\leq i\leq j\leq n}\beta_{ij}\ell_{i}\ast\ell_{j},$

where $\ell_{i}\ast\ell_{j}$ is as defined in (21). Then for $g+2\leq i\leq
j\leq n$, since $v_{i},v_{j}\perp\ell_{1},\ldots,\ell_{g+2}$, we have

$v_{i}^{t}(s^{-1}B^{\prime})v_{j}=\begin{cases}2\beta_{ij}&\mbox{if }i\neq
j,\\\ \beta_{ii}&\mbox{if }i=j\end{cases}.$

Since the top left $(g+1)\times(g+1)$ block of $B^{\prime}\in
s(Y{\mathcal{D}})\cap S(\Lambda)$ is $0$, the same is true for
$s^{-1}B^{\prime}$. Hence $\beta_{ij}=0$ whenever $g+2\leq i\leq j\leq n$. In
other words,

$Y{\mathcal{D}}\cap
s^{-1}S(\Lambda)\subset\mbox{Span}_{\mathbb{Z}}\\{\ell_{i}\ast\ell_{j}\mid
1\leq i\leq j\leq n\mbox{ and }i\leq g+1\\}.$

By Proposition 5.2, we have

$\displaystyle\\#(Y{\mathcal{D}}\cap s^{-1}S(\Lambda))$ $\displaystyle\ll$
$\displaystyle\displaystyle\prod_{\begin{subarray}{c}1\leq i\leq j\leq n\\\
i\leq g+1\\\ L_{i}L_{j}\ll Y\end{subarray}}\frac{Y}{L_{i}L_{j}}$ (74)
$\displaystyle\ll$
$\displaystyle\displaystyle\Bigl{(}\prod_{\begin{subarray}{c}1\leq i\leq j\leq
n\\\ i\leq
g+1\end{subarray}}\frac{Y}{L_{i}L_{j}}\Bigr{)}\prod_{\begin{subarray}{c}1\leq
i\leq j\leq n\\\ i\leq g+1\\\ L_{i}L_{j}\gg
Y\end{subarray}}\Bigl{(}\frac{L_{i}L_{j}}{Y}\frac{Y}{L_{n+1-j}{L_{j}}}\Bigr{)}$
$\displaystyle\ll$ $\displaystyle\displaystyle\frac{Y^{n(n+1)/2}}{(L_{1}\cdots
L_{n})^{n+1}}\frac{(L_{g+2}\cdots
L_{n})^{g+2}}{Y^{(g+1)(g+2)/2}}\prod_{\begin{subarray}{c}1\leq i\leq j\leq
n\\\ i\leq g+1\end{subarray}}\frac{L_{i}}{L_{n+1-j}}$ $\displaystyle\ll$
$\displaystyle\frac{Y^{n(n+1)/2}}{(L_{1}\cdots
L_{n})^{n+1}}\frac{(L_{g+2}\cdots L_{n})^{g+2}}{Y^{(g+1)(g+2)/2}}\prod_{1\leq
i<j\leq g+1}\frac{L_{j}}{L_{i}},$

where the second bound follows since $L_{n+1-j}L_{j}\ll Y$ for all $j$ by
Proposition 5.4; and the last bound follows because the map from
$\\{(i,j):1\leq i\leq j\leq n\mbox{ and }i\leq g+1\\}$ to $\\{(k,\ell)\\}$
sending $(i,j)$ to $(n+1-j,i)$ is one-to-one with its image contained within
the set of pairs $(k,\ell)$ with $k<\ell\leq g+1$, and because the $L_{i}$’s
are nondecreasing.

To obtain a bound on the size of $\Sigma(L,s)$, we use (41):

$\\#\Sigma(L,s)\ll(L_{1}L_{2}\cdots L_{n})^{n+1}\Bigl{(}\prod_{1\leq i<j\leq
n}\frac{L_{i}}{L_{j}}\Bigr{)}\Bigl{(}\prod_{\alpha\in{\mathcal{Z}}\cap{\mathcal{M}}}\frac{1}{w(\alpha)}\Bigr{)}.$

Let ${\mathcal{M}}_{\rm dist}=\\{\ell_{i,j}\mid 1\leq i\leq g+1,\,1\leq j\leq
g+1\\}$. Recall that elements $\ell_{i,j}\in{\mathcal{Z}}_{1}$ satisfy
$i+j\leq n+1$. Hence for any
$\ell_{ij}\in({\mathcal{Z}}_{1}\cap{\mathcal{M}})\backslash{\mathcal{M}}_{\rm
dist}$, exactly one of $i$ and $j$ is $\leq g+1$. Define

$\begin{array}[]{rcl}\pi_{\mathcal{M}}&:&({\mathcal{Z}}_{1}\cap{\mathcal{M}})\backslash{\mathcal{M}}_{\rm
dist}\to{\mathcal{M}}\\\\[7.22743pt]
\pi_{\mathcal{M}}(\ell_{i,j})&=&\displaystyle\left\\{\begin{array}[]{rl}\ell_{i,n+2-i}&\;{\rm
if}\;i\leq g+1;\\\ \ell_{n+1-j,j}&\;{\rm if}\;j\leq
g+1.\end{array}\right.\end{array}$

We claim that the image of $\pi_{\mathcal{M}}$ is disjoint from
${\mathcal{Z}}$. Indeed, when $i\leq g+1$, we have
$\pi_{\mathcal{M}}(\ell_{i,j})\not\in{\mathcal{Z}}_{1}$, and when $j\leq g+1$,
we have $\pi_{\mathcal{M}}(\ell_{i,j})\not\in{\mathcal{Z}}$ by Lemma 6.5 and
the fact that $a_{g+1,g+1}\in{\mathcal{Z}}$. Thus
$w(\pi_{\mathcal{M}}(\alpha))\gg 1$ and $w(\pi_{\mathcal{M}}(\alpha))\gg
w(\alpha)$ for every
$\alpha\in({\mathcal{Z}}_{1}\cap{\mathcal{M}})\backslash{\mathcal{M}}_{\rm
dist}$. It follows that

$\begin{array}[]{rcl}\displaystyle\prod_{\alpha\in{\mathcal{Z}}_{\mathcal{M}}}\frac{1}{w(\alpha)}&\ll&\displaystyle\Bigl{(}\prod_{\ell\in{\mathcal{M}}_{\rm
dist}}\frac{1}{w(\ell)}\Bigr{)}\Bigl{(}\prod_{({\mathcal{Z}}_{1}\cap{\mathcal{M}})\backslash{\mathcal{M}}_{\rm
dist}}\frac{w(\pi(\ell))}{w(\ell)}\Bigr{)}\\\\[14.45377pt]
&\ll&\displaystyle\frac{(t_{1}\cdots t_{g+1})^{g+1}}{(L_{1}\cdots
L_{g+1})^{g+1}}\Bigl{(}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}\Bigr{)}\Bigl{(}\prod_{g+2\leq i<j\leq
n+1}\frac{L_{j}}{L_{i}}\Bigr{)},\end{array}$

so that

$\\#\Sigma(L,s)\ll(L_{1}\cdots L_{n})^{n+1}\frac{(t_{1}\cdots
t_{g+1})^{g+1}}{(L_{1}\cdots L_{g+1})^{g+1}}\Bigl{(}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}\Bigr{)}\Bigl{(}\prod_{i=1}^{g+1}\prod_{j=i}^{n}\frac{L_{i}}{L_{j}}\Bigr{)}.$
(75)

Combining (72), (74), and (75) and the identity

$\Bigl{(}\prod_{1\leq i<j\leq
g+1}\frac{L_{j}}{L_{i}}\Bigr{)}\Bigl{(}\prod_{i=1}^{g+1}\prod_{j=i}^{n}\frac{L_{i}}{L_{j}}\Bigr{)}=\frac{(L_{1}\cdots
L_{g+1})^{g+1}}{(L_{g+2}\cdots L_{n})^{g+1}}$

now yields

$\displaystyle\\#\bigl{(}s(Y{\mathcal{D}})\cup{\mathcal{S}}(L,s)\bigr{)}$
$\displaystyle\ll$ $\displaystyle\frac{Y^{n(n+1)/2}}{(L_{1}\cdots
L_{n})^{n+1}}\frac{(L_{g+2}\cdots
L_{n})^{g+2}}{Y^{(g+1)(g+2)/2}}\Bigl{(}\prod_{1\leq i<j\leq
g+1}\frac{L_{j}}{L_{i}}\Bigr{)}$ (76) $\displaystyle\cdot\,\,(L_{1}\cdots
L_{n})^{n+1}\frac{(t_{1}\cdots t_{g+1})^{g+1}}{(L_{1}\cdots
L_{g+1})^{g+1}}\Bigl{(}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}\Bigr{)}\Bigl{(}\prod_{i=1}^{g+1}\prod_{j=i}^{n}\frac{L_{i}}{L_{j}}\Bigr{)}$
$\displaystyle=$
$\displaystyle\frac{Y^{n(n+1)/2}}{Y^{(g+1)(g+2)/2}}(t_{1}\cdots
t_{g+1})^{g+1}(L_{g+2}\cdots L_{n})\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}$ $\displaystyle\ll$
$\displaystyle\frac{Y^{n(n+1)/2}}{Y^{(g+1)(g+2)/2}}(t_{1}\cdots
t_{g+1})^{g+1}(L_{g+2}\cdots L_{n})\Bigl{(}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}\Bigr{)}\Bigl{(}\prod_{j=g+2}^{n}\frac{Y}{L_{j}L_{n+1-j}}\Bigr{)}$
$\displaystyle=$
$\displaystyle\frac{Y^{n(n+1)/2+g+1}}{Y^{(g+1)(g+2)/2}}\frac{(t_{1}\cdots
t_{g+1})^{g+1}}{L_{1}\cdots L_{g+1}}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}},$

where the fourth line follows since $L_{n+1-j}L_{j}\ll Y$ for all $j$ by
Proposition 5.4. Finally, note that

$\begin{array}[]{rcl}\displaystyle(t_{1}\cdots t_{g+1})^{2g+3}\prod_{1\leq
i<j\leq g+1}\frac{t_{i}}{t_{j}}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}&=&\displaystyle\frac{(t_{1}\cdots
t_{g+1})^{g+2}}{(t_{g+2}\cdots t_{n+1})^{g+1}}\prod_{1\leq i<j\leq
g+1}\frac{t_{i}}{t_{j}}\prod_{g+2\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}\\\\[14.45377pt] &=&\displaystyle\prod_{1\leq i<j\leq
n+1}\frac{t_{i}}{t_{j}}=\delta(s)^{-1}.\end{array}$

Therefore, combining (71), (73) and (76) gives

$\begin{array}[]{rcl}N({{\mathcal{L}}(M)},L,{\mathcal{Z}},X)&\ll&\displaystyle\int_{T_{\mathcal{Z}}(L,Y)}\\#\bigl{(}s(Y{\mathcal{D}})\cap
S({\mathbb{Z}})\bigr{)}\cdot\\#\bigl{(}s(Y{\mathcal{D}})\cap{\mathcal{S}}(L,s)\bigr{)}\,\delta(s)d^{\times}s\\\\[14.45377pt]
&\ll&\displaystyle\frac{Y^{(n+1)^{2}-(g+1)^{2}}}{L_{1}\cdots
L_{g+1}}\int_{T_{\mathcal{Z}}(L,Y)}d^{\times}s\\\\[14.45377pt]
&\ll&\displaystyle\frac{X^{(n+1)+\frac{1}{2}\kappa}}{M}\log^{n}Y.\end{array}$

Theorem 6.24 now follows immediately from (70) by summing over the $O(1)$
different possible ${\mathcal{Z}}$’s and the $O(\log^{n}Y)$ different possible
$L$’s. $\Box$

###  6.5 Proof of the main uniformity estimates

Proof of Theorem 5: Case (a) of Theorem 5 follows from an application of the
quantitative version of the Ekedahl geometric sieve developed in [3, Theorem
3.3]. Case (b) follows from (13), (15), and Theorem 4.1.

We now use the results of this section to prove the most intricate case,
namely, Case (c). For $m\leq X^{1/2}$, we have the immediate bound

$\\#\\{f\in{\mathcal{W}}_{m}:H(f)<X\\}\ll X^{n+1}/m^{2}.$

Using this bound for $m\leq X^{1/2}$, we may assume that $M>X^{1/2}$. We first
note that from (33), (35), and Lemmas 6.1 and 6.2, we have

$\\#\bigcup_{\begin{subarray}{c}m>M\\\
m\;\mathrm{squarefree}\end{subarray}}\\{f\in{\mathcal{W}}_{m}^{\rm{(2)}}:H(f)<X\\}\ll{\mathcal{I}}_{X}({\mathcal{L}}(\sqrt{M}))+\frac{X^{n+1}}{\sqrt{M}}+X^{n}+X^{n+1-\frac{\kappa}{2n-2}}.$

Applying Theorems 6.6, 6.11, and 6.24, we obtain

$\begin{array}[]{rcl}\displaystyle{\mathcal{I}}_{X}({\mathcal{L}}(\sqrt{M}))&=&\displaystyle{\mathcal{I}}_{X}^{\rm
main}({\mathcal{L}}(\sqrt{M}))+{\mathcal{I}}_{X}^{\rm
scusp}({\mathcal{L}}(\sqrt{M}))+{\mathcal{I}}_{X}^{\rm
dcusp}({\mathcal{L}}(\sqrt{M}))\\\\[3.61371pt] &\ll&\displaystyle
X^{n+1-1/(10n)}+X^{n+1-1/(88n^{6})}+\frac{X^{n+1+\frac{1}{2}\kappa}}{\sqrt{M}}\log^{2n}X.\end{array}$

Setting $\kappa=(2n-2)/(88n^{6})$ yields the desired result. $\Box$.

Theorem 5 has the following immediate consequence. For a positive squarefree
integer $m$, let
${\mathcal{W}}_{m}:={\mathcal{W}}_{m}^{\rm{(1)}}\cup{\mathcal{W}}_{m}^{\rm{(2)}}$.

###### Corollary 6.27

For a positive integer $N\geq 3$, and positive real numbers $M$ and $X$, we
have

$\sum_{\begin{subarray}{c}m>M\\\ m{\rm
squarefree}\end{subarray}}\\#\\{f\in{\mathcal{W}}_{m}:H(f)<X\\}\ll_{\epsilon}\frac{X^{n+1+\xi_{n}+\epsilon}}{M^{\delta_{n}}}+X^{n+1-{\eta_{n}}+\epsilon},$

where $\delta_{n}=1/2,\,\xi_{n}=0,\,\eta_{n}=1/(5n)$ when $n$ is odd and
$\delta_{n}=1/3,\,\xi_{n}=1/(88n^{5}),\,\eta_{n}=1/(88n^{6})$ when $n$ is
even.

Proof: Suppose $f\in{\mathcal{W}}_{m}$ for some $m>M$. When $n$ is odd, this
implies that $f$ belongs to either ${\mathcal{W}}_{m_{1}}^{(1)}$ or
${\mathcal{W}}_{m_{1}}^{(2)}$ for some $m_{1}>\sqrt{M}$. When $n$ is even, $f$
belongs either to ${\mathcal{W}}_{m_{2}}^{(1)}$ for $m_{2}>M^{1/3}$ or to
${\mathcal{W}}_{m_{3}}^{(2)}$ for $m_{3}>M^{2/3}$. A direct application of
Theorem 5 now yields the result. $\Box$

##  7 Proofs of the main results

We begin by proving a more general form of Theorem 6. Let $N$ be a positive
squarefree integer, and for each $p\mid N$, let $\Sigma_{p}\subset
V_{n}({\mathbb{Z}}/p^{2}{\mathbb{Z}})$ be a nonempty subset. Denote the
collection $(\Sigma_{p})_{p\mid N}$ by $\Sigma$. Let $V_{n}(\Sigma)$ be the
set of all $f\in V_{n}({\mathbb{Z}})$ such that the reduction of $f$ modulo
$p^{2}$ lies in $\Sigma_{p}$ for all $p\mid N$. For $p\mid N$, let
$\alpha_{n}(\Sigma,p)$ (resp., $\beta_{n}(\Sigma,p)$) denote the density of
elements $f\in V_{n}({\mathbb{Z}})$ such that $p^{2}\nmid\Delta(f)$ (resp.,
$R_{f}$ is maximal at $p$) and such that the reduction of $f$ modulo $p^{2}$
lies in $\Sigma_{p}$. For $p\nmid N$, simply set
$\alpha_{n}(\Sigma,p)=\alpha_{n}(p)$ and $\beta_{n}(\Sigma,p)=\beta_{n}(p)$.
Finally, define

$\alpha_{n}(\Sigma)=\prod_{p}\alpha_{n}(\Sigma,p);\quad\quad\beta_{n}(\Sigma)=\prod_{p}\beta_{n}(\Sigma,p).$

We are now ready to carry out our sieve.

###### Theorem 7.1

We have,

$\begin{array}[]{ccl}\displaystyle\\#\\{f\in V_{n}(\Sigma):H(f)<X\mbox{ and
$\Delta(f)$
squarefree}\\}&\\!\\!=\\!\\!&\alpha_{n}(\Sigma)(2X)^{n+1}+O_{\epsilon}(\mathcal{E}(X,N,\epsilon)),\\\\[8.31143pt]
\displaystyle\\#\\{f\in V_{n}(\Sigma):H(f)<X\mbox{ and $R_{f}$
maximal}\\}&\\!\\!=\\!\\!&\beta_{n}(\Sigma)(2X)^{n+1}+O_{\epsilon}(\mathcal{E}(X,N,\epsilon))\end{array}$

where the error term is given by

$\mathcal{E}(X,N,\epsilon):=X^{n+1-\eta_{n}+\epsilon}+N^{2}X^{n+3(\eta_{n}+\xi_{n})+\epsilon}+N^{2n+2}X^{(6n+3)(\eta_{n}+\xi_{n})+\epsilon},$

where $\eta_{n}=1/(5n),\,\xi_{n}=0$ when $n$ is odd, and
$\eta_{n}=1/(88n^{6}),\,\xi_{n}=1/(88n^{5})$ when $n$ is even.

Proof: For any squarefree integer $m$ that is relatively prime to $N$, let
${\mathcal{W}}_{m}(\Sigma)$ denote the set of elements $f\in V({\mathbb{Z}})$
such that $m^{2}\mid\Delta(f)$, and such that the reduction of $f$ modulo
$p^{2}$ belongs to $\Sigma_{p}$ for every $p\mid N$. Note that
${\mathcal{W}}_{m}(\Sigma)$ is a union of

$\gamma(\Sigma,N,m):=N^{2n+2}m^{2n+2}\prod_{p\mid
m}\Bigl{(}\frac{\\#\Sigma_{p}}{p^{2n+2}}-\alpha_{n}(\Sigma,p)\Bigr{)}=O_{\epsilon}(N^{2n+2}m^{2n+\epsilon})$

translates of $m^{2}N^{2}V({\mathbb{Z}})$. By inclusion-exclusion and
Corollary 6.27, we have for any $M>0$,

$\displaystyle\\#\\{f\in V_{n}(\Sigma)\colon H(f)<X\mbox{ and $\Delta(f)$
squarefree}\\}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}(m,N)=1\\\ m\leq
M\end{subarray}}\mu(m)\\#\\{f\in{\mathcal{W}}_{m}(\Sigma)\colon
H(f)<X\\}+O_{\epsilon}\Bigl{(}\frac{X^{n+1+\xi_{n}+\epsilon}}{M^{\delta_{n}}}+X^{n+1-{\eta_{n}}+\epsilon}\Bigr{)}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}(m,N)=1\\\ m\leq
M\end{subarray}}\mu(m)\gamma(\Sigma,N,m)\Bigl{(}\frac{2X}{N^{2}m^{2}}+O(1)\Bigr{)}^{n+1}+O_{\epsilon}\Bigl{(}\frac{X^{n+1+\xi_{n}+\epsilon}}{M^{\delta_{n}}}+X^{n+1-{\eta_{n}}+\epsilon}\Bigr{)}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}(m,N)=1\\\ m\leq
M\end{subarray}}\Biggl{(}(2X)^{n+1}\mu(m)\prod_{p\mid
m}\Bigl{(}\frac{\\#\Sigma_{p}}{p^{2n+2}}-\alpha_{n}(\Sigma,p)\Bigr{)}+O\Bigl{(}N^{2}X^{n}m^{\epsilon}+N^{2n+2}m^{2n+\epsilon}\Bigr{)}\Biggr{)}$
$\displaystyle+\,O_{\epsilon}\Bigl{(}\frac{X^{n+1+\xi_{n}+\epsilon}}{M^{\delta_{n}}}+X^{n+1-{\eta_{n}}+\epsilon}\Bigr{)}$
$\displaystyle=$
$\displaystyle(2X)^{n+1}\alpha_{n}(\Sigma)+O\Bigl{(}\frac{X^{n+1}}{M^{1-\epsilon}}+N^{2}M^{1+\epsilon}X^{n}+N^{2n+2}M^{2n+1+\epsilon}+\frac{X^{n+1+\xi_{n}+\epsilon}}{M^{\delta_{n}}}+X^{n+1-{\eta_{n}}+\epsilon}\Bigr{)}.$

Recalling that $\delta_{n}=1/2$ or $1/3$, we may take
$M=X^{3\eta_{n}+3\xi_{n}}$ to obtain the first claim in Theorem 7.1. The
second claim follows identically. $\Box$

Taking $N=1$ in Theorem 7.1 yields Theorem 6. Theorems 1 and 2 are then
immediate consequences of Theorem 6.

Next, we prove lower bounds on the number of $S_{n}$-fields having bounded
discriminant. Let $f(x,y)=a_{0}x^{n}+a_{1}x^{n-1}y+\cdots+a_{n}y^{n}$ be a
real binary $n$-ic form with $a_{0}\neq 0$ and nonzero discriminant. Let
$\theta$ be the image of $x$ in ${\mathbb{R}}[x]/(f(x,1))$, and write $R_{f}$
for the lattice spanned by

$1,\quad\zeta_{1}=a_{0}\theta,\quad\zeta_{2}=a_{0}\theta^{2}+a_{1}\theta,\quad\ldots,\quad\zeta_{n-1}=a_{0}\theta^{n-1}+\cdots+a_{n-1}$

in ${\mathbb{R}}[x]/(f(x,1))$. Here we identify ${\mathbb{R}}[x]/(f(x,1))$
with ${\mathbb{R}}^{n}$ via its real and complex embeddings and by identifying
${\mathbb{C}}={\mathbb{R}}\oplus i{\mathbb{R}}$ with ${\mathbb{R}}^{2}$.

We say that $f(x,y)$ is Minkowski-reduced if the basis
$\\{1,\zeta_{1},\ldots,\zeta_{n-1}\\}$ of $R_{f}$ is Minkowski-reduced. We say
that $f(x,y)$, or its ${\rm SL}_{2}({\mathbb{Z}})$-orbit, is quasi-reduced if
there exists $\gamma\in{\rm SL}_{2}({\mathbb{Z}})$ such that $\gamma.f$ is
Minkowski-reduced. We add the prefix “strongly” if the relevant lattice has a
unique Minkowski-reduced basis. The relevance of being strongly quasi-reduced
is contained in the following lemma.

###### Lemma 7.2

Let $n\geq 3$ and let $f(x,y)$ and $f^{*}(x,y)$ be strongly quasi-reduced
integral binary $n$-ic forms. Suppose the corresponding rank-$n$ rings $R_{f}$
and $R_{f^{*}}$ are isomorphic. Then $f(x,y)$ and $f^{*}(x,y)$ are ${\rm
SL}_{2}({\mathbb{Z}})$-equivalent.

Proof: It suffices to assume $f(x,y)=a_{0}x^{n}+\cdots+a_{n}y^{n}$ and
$f^{*}(x,y)=a_{0}^{*}x^{n}+\cdots+a_{n}^{*}y^{n}$ are strongly Minkowski-
reduced with $R_{f}\simeq R_{f^{*}}$. We show $f(x,y)=f^{*}(x,y)$. Let
$\phi:R_{f}\rightarrow R_{f^{*}}$ be a ring isomorphism. By the uniqueness of
Minkowski-reduced bases, $\phi$ must map the basis elements
$1,\zeta_{1},\ldots,\zeta_{n-1}$ for $R_{f}$ to the corresponding basis
elements $1,\zeta^{*}_{1},\ldots,\zeta^{*}_{n-1}$ for $R_{f^{*}}$. Let
$\theta$ denote the image of $x$ in ${\mathbb{Q}}[x]/(f(x,1))$ and
$\theta^{*}$ the image of $x$ in ${\mathbb{Q}}[x]/(f^{*}(x,1))$. Then
$\phi(a_{0}\theta)=a_{0}^{*}\theta^{*}$ and

$a_{0}^{*}\theta^{*2}+a_{1}^{*}\theta^{*}=\phi(a_{0}\theta^{2}+a_{1}\theta)=(a_{0}^{*2}/a_{0})\theta^{*2}+(a_{1}a_{0}^{*}/a_{0})\theta^{*}.$

Since $\theta^{*}$ and $\theta^{*2}$ are linearly independent, we have
$a_{0}=a_{0}^{*}$, $a_{1}=a_{1}^{*}$, and $\phi(\theta)=\theta^{*}$, where we
extend $\phi$ naturally to
$R_{f}\otimes{\mathbb{Q}}={\mathbb{Q}}[x]/(f(x,1))$. Then since
$\phi(\zeta_{n-1})=\zeta^{*}_{n-1}$, we have $a_{i}=a_{i}^{*}$ for
$i=0,\ldots,n-2$. Finally $\phi(-a_{n-1}\theta-
a_{n})=\phi(\theta\zeta_{n-1})=\phi(\theta^{*}\zeta_{n-1}^{*})=-a_{n-1}^{*}\theta^{*}-a_{n}^{*}$.
Hence $a_{n-1}=a^{*}_{n-1}$ and $a_{n}=a^{*}_{n}$. $\Box$

Proof of Theorem 3: The condition of being strongly quasi-reduced is open in
$V_{n}({\mathbb{R}})$. Therefore, given a strongly quasi-reduced element $f\in
V_{n}({\mathbb{R}})$, there exists an open neighbourhood ${\mathcal{B}}$ of
$f$ in which every element is strongly quasi-reduced. Moreover, since the
action of ${\rm SL}_{2}({\mathbb{Z}})$ on $V_{n}({\mathbb{R}})$ is discrete,
we may ensure that no two elements of ${\mathcal{B}}$ are ${\rm
SL}_{2}({\mathbb{Z}})$-equivalent. We may further scale ${\mathcal{B}}$ in
order to assume that every element in ${\mathcal{B}}$ has discriminant bounded
by $1$.

Consider the set ${\mathcal{B}}_{X}:=X^{1/(2n-2)}\cdot{\mathcal{B}}$. No two
elements in it are ${\rm SL}_{2}({\mathbb{Z}})$-equivalent and every element
in it is strongly quasi-reduced. Therefore the rings corresponding to any two
elements in ${\mathcal{B}}_{X}$ are nonisomorphic. On the other hand, applying
Theorem 7.1, we see that $\gg X^{(n+1)/(2n-2)}$ integral elements in
${\mathcal{B}}_{X}$ have discriminant less than $X$ and correspond to maximal
orders in degree-$n$ number fields. Since these rings are pairwise
nonisomorphic, so are their fields of fractions. Hence we have constructed
$\gg X^{(n+1)/(2n-2)}$ nonisomorphic degree-$n$ number fields of absolute
discriminant less than $X$. Restricting to counting forms that have squarefree
discriminant yields $\gg X^{(n+1)/(2n-2)}$ nonisomorphic $S_{n}$-number
fields. $\Box$

We note that Theorem 7.1 also allows us to construct $\gg X^{(n+1)/(2n-2)}$
$S_{n}$-number fields satisfying any finite set of splitting conditions.

##  Appendix A Computations of the local densities
$\alpha_{n}(p),\beta_{n}(p)$

Let $n\geq 2$ be a fixed integer. For a prime $p$, let $\alpha_{n}(p)$ denote
the density of the set of binary $n$-ic forms having discriminant indivisible
by $p^{2}$, and let $\beta_{n}(p)$ denote the density of binary $n$-ic forms
whose associated rank-$n$ rings are maximal at $p$. In this section, we
compute $\alpha_{n}(p)$ and $\beta_{n}(p)$ for all integers $n\geq 2$ and all
primes $p$.

###### Proposition A.1

We have $\alpha_{2}(2)=1/2$ and $\alpha_{n}(2)=3/8$ for $n\geq 3$. For odd
primes $p$, we have

$\alpha_{n}(p)=\begin{cases}\displaystyle\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}1+\frac{1}{p}-\frac{1}{p^{3}}\Bigr{)}&\mbox{
if }n=2,\\\\[10.84006pt]
\displaystyle\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}\Bigl{(}1+\frac{1}{p}\Bigr{)}^{2}&\mbox{
if }n=3,\\\\[10.84006pt]
\displaystyle\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}\Bigl{(}1+\frac{2}{p}-\frac{2}{p^{4}}+\frac{1}{p^{5}}\Bigr{)}&\mbox{
if }n=4,\\\\[10.84006pt]
\displaystyle\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}\Bigl{(}1+\frac{1}{p}\Bigr{)}\Bigl{(}1+\frac{1}{p}-\frac{1}{p^{2}}\Bigr{)}&\mbox{
if }n\geq 5.\end{cases}$

Proof: For $j\geq 0$, $n\geq 1$, and $p$ prime, we let $\nu_{j}(n,p)$ denote
the density within monic degree-$n$ integer polynomials of the set of those
whose discriminants have $p$-adic valuation $j$. Then $\nu_{0}(n,p)$ and
$\nu_{1}(n,p)$ are computed in [1, Proposition 6.4 and Theorem 6.8]:

$\displaystyle\nu_{0}(n,p)$ $\displaystyle=$
$\displaystyle\begin{cases}1&\mbox{ if }n=1;\\\ 1-p^{-1}&\mbox{ if }n\geq
2.\end{cases}$ $\displaystyle\nu_{1}(n,p)$ $\displaystyle=$
$\displaystyle\begin{cases}0&\mbox{ if }p=2\mbox{ or }n=1;\\\
p^{-1}(1-p^{-1})&\mbox{ if }n=2,p\neq 2;\\\ p^{-1}(1-p^{-1})^{2}&\mbox{ if
}n=3,p\neq 2;\\\ (1-p^{-1})^{2}(1-(-p)^{-n})(1+p)^{-1}&\mbox{ if }n\geq
4,p\neq 2.\end{cases}$

To compute the densities $\alpha_{n}(p)$, we partition the set of integral
binary $n$-ic forms $f(x,y)=a_{0}x^{n}+a_{1}x^{n-1}y+\cdots+a_{n}y^{n}$ whose
discriminants are not divisible by $p^{2}$ into three subsets, and compute
each of their densities. For any binary form $f(x,y)$ in ${\mathbb{Z}}[x,y]$
or in $({\mathbb{Z}}/p^{2}{\mathbb{Z}})[x,y]$, we write $\bar{f}(x,y)$ for its
reduction modulo $p$.

Subset 1: The set of $f(x,y)$ with $p\nmid a_{0}$ and $p^{2}\nmid\Delta(f)$.
Here, for any fixed leading coefficient $a_{0}\not\equiv 0\pmod{p}$, the
density of $f(x,y)$ having discriminant indivisible by $p^{2}$ is simply given
by $\nu_{0}(n,p)+\nu_{1}(n,p)$. Therefore, the $p$-adic density of this subset
is equal to

$\Bigl{(}1-\frac{1}{p}\Bigr{)}\big{(}\nu_{0}(n,p)+\nu_{1}(n,p)\big{)}.$

Subset 2: The set of $f(x,y)$ with $p\mid a_{0}$, $p\nmid a_{1}$, and
$p^{2}\nmid\Delta(f)$. In this case, we begin by proving that the density of
elements $f$ with fixed $a_{0}$ and $a_{1}$ and with $p^{2}\nmid\Delta(f)$ is
the same as the density of binary $(n-1)$-ic forms $g$, with fixed leading
coefficient $a_{1}$ such that $p^{2}\nmid\Delta(g)$. Indeed, given any
$(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}$, we write

$\begin{array}[]{ccrcc}f_{a_{2},\ldots,a_{n}}(x,y)&=&a_{0}x^{n}+a_{1}x^{n-1}y+a_{2}x^{n-2}y^{2}+\cdots+a_{n}y^{n}&\;\,\in&({\mathbb{Z}}/p^{2}{\mathbb{Z}})[x,y],\\\\[1.8063pt]
g_{a_{2},\ldots,a_{n}}(x,y)&=&\,\\!a_{1}x^{n-1}\\!+\,a_{2}x^{n-2}y\;+\,\\!\cdots\,\\!+a_{n}y^{n-1}\\!\\!\\!\\!\\!\\!&\;\,\in&({\mathbb{Z}}/p^{2}{\mathbb{Z}})[x,y].\end{array}$

Define

$\displaystyle S_{f}^{(1)}$ $\displaystyle=$
$\displaystyle\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\colon
p^{2}\mbox{ strongly divides }\Delta(f_{a_{2},\ldots,a_{n}})\\},$
$\displaystyle S_{f}^{(2)}$ $\displaystyle=$
$\displaystyle\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\colon
p^{2}\mbox{ weakly divides }\Delta(f_{a_{2},\ldots,a_{n}})\\},$ $\displaystyle
S_{g}^{(1)}$ $\displaystyle=$
$\displaystyle\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\colon
p^{2}\mbox{ strongly divides }\Delta(g_{a_{2},\ldots,a_{n}})\\},$
$\displaystyle S_{g}^{(2)}$ $\displaystyle=$
$\displaystyle\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\colon
p^{2}\mbox{ weakly divides }\Delta(g_{a_{2},\ldots,a_{n}})\\}.$

Recall that $p^{2}$ strongly divides the discriminant of $f$ if and only if
$\bar{f}(x,y)$ has a factor of the form $h(x,y)^{3}$ for some linear form $h$
or a factor of the form $j(x,y)^{2}$ where $j$ is a binary form of degree at
least 2. Since $\overline{f_{a_{2},\ldots,a_{n}}}(x,y)\equiv
y\,\overline{g_{a_{2},\ldots,a_{n}}}(x,y)$, and since $y$ does not divide
$\overline{g_{a_{2},\ldots,a_{n}}}(x,y)$, we see that
$\overline{f_{a_{2},\ldots,a_{n}}}(x,y)$ admits such a factor if and only if
$\overline{g_{a_{2},\ldots,a_{n}}}(x,y)$ does. Hence
$S_{f}^{(1)}=S_{g}^{(1)}$. On the other hand, we have

$\displaystyle\\#S_{f}^{(2)}$ $\displaystyle=$
$\displaystyle\\#\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\backslash
S_{f}^{(1)}\colon\exists
r\in{\mathbb{Z}}/p{\mathbb{Z}},(x-r)^{2}\mid\overline{f_{a_{2},\ldots,a_{n}}}(x,1),p^{2}\mid
f_{a_{2},\ldots,a_{n}}(r,1)\\}$ $\displaystyle=$
$\displaystyle\textstyle\frac{1}{p}\\#\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\backslash
S_{f}^{(1)}\colon\exists
r\in{\mathbb{Z}}/p{\mathbb{Z}},(x-r)^{2}\mid\overline{f_{a_{2},\ldots,a_{n}}}(x,1),p\mid
f_{a_{2},\ldots,a_{n}}(r,1)\\}$ $\displaystyle=$
$\displaystyle\textstyle\frac{1}{p}\\#\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\backslash
S_{g}^{(1)}\colon\exists
r\in{\mathbb{Z}}/p{\mathbb{Z}},(x-r)^{2}\mid\overline{g_{a_{2},\ldots,a_{n}}}(x,1),p\mid
g_{a_{2},\ldots,a_{n}}(r,1)\\}$ $\displaystyle=$
$\displaystyle\\#\\{(a_{2},\ldots,a_{n})\in({\mathbb{Z}}/p^{2}{\mathbb{Z}})^{n-1}\backslash
S_{g}^{(1)}\colon\exists
r\in{\mathbb{Z}}/p{\mathbb{Z}},(x-r)^{2}\mid\overline{g_{a_{2},\ldots,a_{n}}}(x,1),p^{2}\mid
g_{a_{2},\ldots,a_{n}}(r,1)\\}$ $\displaystyle=$
$\displaystyle\\#S_{g}^{(2)}.$

The density $(\\#S_{g}^{(1)}+\\#S_{g}^{(2)})/p^{2(n-1)}$ is
$\nu_{0}(n-1,p)+\nu_{1}(n-1,p)$. Taking into account that $p\mid a_{0}$ and
$p\nmid a_{1}$, we see that the density of this second subset is

$\frac{1}{p}\Bigl{(}1-\frac{1}{p}\Bigr{)}\big{(}\nu_{0}(n-1,p)+\nu_{1}(n-1,p)\big{)}.$

Subset 3: The set of $f(x,y)$ with $p\mid a_{0}$, $p\mid a_{1}$, and
$p^{2}\nmid\Delta(f)$. Note that we already have $p\mid\Delta(f)$ in this
case. To ensure that $p^{2}\nmid\Delta(f)$, we must have $p>2$, $p^{2}\nmid
a_{0}$, and $p\nmid a_{2}$. Indeed, if $p=2$, then since $2\mid\Delta(f)$, we
have $4\mid\Delta(f)$; if $p^{2}\mid a_{0}$, then $p^{2}$ (weakly) divides
$\Delta(f)$; and if $p\mid a_{2}$, then $y^{3}\mid\bar{f}$ and so $p^{2}$
(strongly) divides $\Delta(f)$. As polynomials in $a_{0},\ldots,a_{n}$, we
have

$\Delta(a_{0}x^{n}+\cdots+a_{n})\equiv-4a_{0}a_{2}^{3}\Delta(a_{2}x^{n-2}+\cdots+a_{n})\pmod{a_{0}^{2},a_{0}a_{1},a_{1}^{2}}.$

Hence if $p>2$, $p^{2}\nmid a_{0}$, and $p\nmid a_{2}$, then
$p^{2}\nmid\Delta(f)$ if and only if
$p\nmid\Delta(a_{2}x^{n-2}+a_{3}x^{n-3}y+\cdots a_{n}y^{n-2})$. Hence the
density of this third subset is

$\frac{1}{p^{2}}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}\nu_{0}(n-2,p).\\\ $

Adding together these three densities yields the proposition. $\Box$

Next, we compute the value of $\beta_{n}(p)$ for integers $n\geq 2$ and primes
$p$.

###### Proposition A.2

We have

$\beta_{n}(p)=\begin{cases}\displaystyle\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}1+\frac{1}{p}-\frac{1}{p^{3}}\Bigr{)}&\mbox{
if }n=2;\\\\[9.03374pt]
\displaystyle\Bigl{(}1-\frac{1}{p^{2}}\Bigr{)}\Bigl{(}1-\frac{1}{p^{3}}\Bigr{)}&\mbox{
if }n\geq 3.\end{cases}$

Proof: The density of monic degree-$n$ integer polynomials that are maximal at
$p$ was computed in [1, Proposition 3.5] to be $1-p^{-2}$ for all $n\geq 2$
and all primes $p$.

We compute $\beta_{n}(p)$ by working over ${\mathbb{Z}}_{p}$. Fix a binary
$n$-ic form $f(x,y)\in V_{n}({\mathbb{Z}}_{p})$. Suppose $f(x,y)$ (mod $p$)
factors as $y^{k}g(x,y)$, where $g(x,y)$ is a binary $(n-k)$-ic form over
${\mathbb{F}}_{p}$ with nonzero $x^{n-k}$-term for some
$k\in\\{0,\ldots,n\\}$. Then, by Hensel’s lemma, $f(x,y)$ factors as
$h_{1}(x,y)h_{2}(x,y)$ where $h_{1}(x,y)\in{\mathbb{Z}}_{p}[x]$ is a binary
$k$-ic form such that $h_{1}(x,y)$ (mod $p$) is $y^{k}$ and
$h_{2}(x,y)\in{\mathbb{Z}}_{p}[x]$ is a binary $(n-k)$-ic such that
$h_{2}(x,y)$ (mod $p$) is $g(x,y)$. By scaling $h_{1}$ and $h_{2}$, we may
further assume that the leading coefficient of $h_{2}(x,y)$ is $1$.

Since $h_{1}(x,y)$ and $h_{2}(x,y)$ share no common factors (mod $p$), the
rank-$n$ ring over ${\mathbb{Z}}_{p}$ associated to $f(x,y)$ is isomorphic to
the product of the rings associated to $h_{1}(x,y)$ and $h_{2}(x,y)$. Since
$h_{1}(x,y)$ reduces to a unit times $y^{k}$ modulo $p$, the rank-$k$ ring
associated to $h_{1}(x,y)$ is always maximal when $k\leq 1$ and is maximal
when $k\geq 2$ if and only if $p^{2}$ does not divide the $x^{k}$-coefficient.
On the other hand, $h_{2}(x,y)$ is monic, and so the probability that it is
maximal is exactly $1-p^{-2}$ when $n-k\geq 2$, and $1$ when $n-k=1$. When
$k=n$, $f(x,y)$ is a multiple of $p$ and is automatically nonmaximal. Summing
over $k$, we have for $n\geq 3$,

$\displaystyle\beta_{n}(p)$ $\displaystyle=$
$\displaystyle\sum_{k=0}^{1}\frac{1}{p^{k}}\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}1-\frac{1}{p^{2}}\Bigr{)}+\sum_{k=2}^{n-2}\frac{1}{p^{k}}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}\Bigl{(}1-\frac{1}{p^{2}}\Bigr{)}+\sum_{k=n-1}^{n}\frac{1}{p^{k}}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}$
$\displaystyle=$
$\displaystyle\Bigl{(}1-\frac{1}{p^{2}}\Bigr{)}\Bigl{(}1-\frac{1}{p^{3}}\Bigr{)}.$

When $n=2$, we have

$\beta_{2}(p)=\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}1-\frac{1}{p^{2}}\Bigr{)}+\frac{1}{p}\Bigl{(}1-\frac{1}{p}\Bigr{)}+\frac{1}{p^{2}}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2}=\Bigl{(}1-\frac{1}{p}\Bigr{)}\Bigl{(}1+\frac{1}{p}-\frac{1}{p^{3}}\Bigr{)}.$

This concludes the proof of Proposition A.2. $\Box$

### Acknowledgments

The first-named author was supported by a Simons Investigator Grant and NSF
Grant DMS-1001828. The second-named author was supported by an NSERC Discovery
Grant and Sloan Research Fellowship. The third-named author was supported by
an NSERC Discovery Grant.

## References

  * [1] A. Ash, J. Brakenhoff, and T. Zarrabi, Equality of Polynomial and Field Discriminants, Experiment. Math. 16 (2007), 367–374.
  * [2] M. Bhargava, The density of discriminants of quintic rings and fields, Ann. of Math. $($2$)$ 172 (2010), no. 3, 1559–1591.
  * [3] M. Bhargava, The geometric sieve and the density of squarefree values of polynomial discriminants and other invariant polynomials, http://arxiv.org/abs/1402.0031v1.
  * [4] M. Bhargava, Most hyperelliptic curves over ${\mathbb{Q}}$ have no rational points, https://arxiv.org/abs/1308.0395v1.
  * [5] M. Bhargava, Galois groups of random integer polynomials and van der Waerden’s Conjecture, https://arxiv.org/abs/2111.06507v1.
  * [6] M. Bhargava, B. Gross. and X. Wang, A positive proportion of locally soluble hyperelliptic curves over ${\mathbb{Q}}$ have no point over any odd degree extension, J. Amer. Math. Soc. 30 (2017), 451–493.
  * [7] M. Bhargava, B. Gross. and X. Wang, Arithmetic invariant theory II: Pure inner forms and obstructions to the existence of orbits, Representations of reductive groups, Progr. Math., 312, Birkhäuser/Springer, Cham., 2015, 139–171.
  * [8] M. Bhargava and W. Ho, On average sizes of Selmer groups and ranks in families of elliptic curves having marked points, preprint.
  * [9] M. Bhargava and A. Shankar, Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves, Ann. of Math. $($2$)$, 181 (2015), no. 1, 191–242.
  * [10] M. Bhargava, A. Shankar and X. Wang, Squarefree values of polynomial discriminants I, Invent. Math. 228 (2022), no. 3, 1037–1073.
  * [11] B. J. Birch and J. R. Merriman, Finiteness theorems for binary forms with given discriminant, Proc. London Math. Soc. $($3$)$, 24, (1972), 385–394.
  * [12] H. Davenport, On a principle of Lipschitz, J. London Math. Soc., 26, (1951), 179–183.
  * [13] H. Davenport and H. Heilbronn, On the density of discriminants of cubic fields II, Proc. Roy. Soc. London Ser. A 322 (1971), no. 1551, 405–420.
  * [14] T. Ekedahl, An infinite version of the Chinese remainder theorem, Comment. Math. Univ. St. Paul. 40 (1991), 53–59.
  * [15] A. Eskin and Y. Katznelson, Singular symmetric matrices, Duke Math J.  79, No. 2 (1995), 515–547.
  * [16] I. M. Gelfond, M. M. Kapranov and A. V. Zelevinsky, Discriminants, Resultants and Multidimensional Determinants, Springer Science+Business Media, New York, 1994.
  * [17] W. Ho, A. Shankar and I. Varma, Odd degree number fields with odd class number, Duke Math. J., 167 (2018), no. 5, 995–1047.
  * [18] K. S. Kedlaya, A construction of polynomials with squarefree discriminants, Proc. Amer. Math. Soc. 140 (2012), 3025–3033.
  * [19] J. Nakagawa, Binary forms and orders of algebraic number fields, Invent. Math. 97 (1989), no. 2, 219–235.
  * [20] J. Nakagawa, Erratum: “Binary forms and orders of algebraic number fields” [Invent. Math. 97 (1989), no. 2, 219–235], Invent. Math. 105 (1991), no. 2, 443.
  * [21] J. Nakagawa, Binary forms and unramified $A_{n}$-extensions of quadratic fields, J. Reine Angew. Math. 406, (1990), 167–178.
  * [22] J. Nakagawa, Correction to the paper: “Binary forms and unramified $A_{n}$-extensions of quadratic fields” [J. Reine Angew. Math. 406 (1990), 167–178], J. Reine Angew. Math. 413, (1991), 220.
  * [23] B. Poonen, Bertini theorems over finite fields, Ann. of Math. (2) 160 (2004), no. 3, 1099–1127.
  * [24] G. C. Sanjaya and X. Wang. On the squarefree values of $a^{4}+b^{3}$. Math. Ann.  To appear.
  * [25] M. Sato and T. Kimura, A classification of irreducible prehomogeneous vector spaces and their relative invariants, Nagoya Math. J. 65 (1977), 1–155.
  * [26] W. Schmidt, Asymptotic formulae for point lattices of bounded discriminant and subspaces of bounded height, Duke Math J.  35 (1968), 327–339.
  * [27] A. N. Shankar, A. Shankar, and X. Wang, Large families of elliptic curves ordered by conductor, Compos. Math. 157 (2021), no. 7, 1538–1583.
  * [28] A. Shankar and J. Tsimerman, Counting $S_{5}$-fields with a power saving error term, Forum Math. Sigma 2 (2014), e13, 8 pp.
  * [29] K. Uchida, Unramified extensions of quadratic number fields, II, Tohoku Math. J. 22 (1970), 220–224.
  * [30] M. M. Wood, Rings and ideals parameterized by binary $n$-ic forms, J. Lond. Math. Soc. $($2$)$, 83, (2011), no. 1, 208–231.
  * [31] M. M. Wood, Parametrization of ideal classes in rings associated to binary forms, J. Reine Angew. Math. 689 (2014), 169–199.
  * [32] Y. Yamamoto, On unramified Galois extensions of quadratic number fields, Osaka J. Math. 7 (1970), 57–76.
  * [33] K. Yamamura, On unramified Galois extensions of real quadratic number fields, Osaka J. Math. 23 (1986), no. 2, 471–478.