_nondegenerate_ if $R$ is nondegenerate. Let $\mathcal{F}^{x,y,xy}$ be the set
of isomorphism classes of nondegenerate such triples $(R,x,y)$.

We will embed $\mathcal{F}^{x,y,xy}$ in the space of pairs of integral ternary
quadratic forms, and use this embedding to define a height function on
$\mathcal{F}^{x,y,xy}$. In particular, consider the set of pairs of ternary
quadratic forms of the form888The $\ast$ symbol means any real number is
permitted to be in this entry, and $-$ means that this entry is determined by
symmetry.:

(34) $(A,B)=\begin{bmatrix}1&0&\ast\\\ -&0&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&0&\ast\\\ -&1&\ast\\\
-&-&\ast\end{bmatrix}.$

Similarly, now consider quadruples $(R,x,y,\varepsilon)$ where $x,y\in R$,
$\varepsilon=\pm 1$, and $1,x,y,\varepsilon xy$ is a basis of $R$. Again, we
say such a quadruple is _nondegenerate_ if $R$ is nondegenerate. We say two
such quadruples $(R,x,y,\varepsilon)$ and
$(R^{\prime},x^{\prime},y^{\prime},\varepsilon^{\prime})$ are isomorphic if
$\varepsilon=\varepsilon^{\prime}$ and there exists a ring isomorphism
$\alpha\colon R\rightarrow R^{\prime}$ such that $\alpha(x)=x^{\prime}$ and
$\alpha(y)=y^{\prime}$.

######  Theorem 5.8.

There is a canonical discriminant-preserving bijection between:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of}\\\
\text{nondegenerate quadruples $(R,x,y,\varepsilon)$}\\\ \text{where $x,y\in
R$ are such that }\\\ \text{$R=\mathbb{Z}\langle 1,x,y,xy\rangle$ and}\\\
\text{$\varphi(x+y)=\varphi(x)+\varphi(y)$}\text{ and $\epsilon=\pm
1$}\end{array}\right\\}\longleftrightarrow\left\\{\begin{array}[]{c}\text{nondegenerate
pairs of }\\\ \text{integral ternary quadratic forms}\\\ \text{of the form
\eqref{matrixform1xyxyresolventintro}}\end{array}\right\\}$

For a triple $T\in\mathcal{F}^{x,y,xy}$, we attach a pair of nondegenerate
ternary quadratic forms via the map above (setting $\varepsilon=1$); define
the _height_ to be

$H(T)\coloneqq\max\\{\operatorname{\lvert}a_{13}\operatorname{\rvert}^{2},\operatorname{\lvert}a_{23}\operatorname{\rvert}^{2},\operatorname{\lvert}b_{13}\operatorname{\rvert}^{2},\operatorname{\lvert}b_{23}\operatorname{\rvert}^{2},\operatorname{\lvert}a_{33}\operatorname{\rvert},\operatorname{\lvert}b_{33}\operatorname{\rvert}\\},$

where here $a_{ij},b_{ij}$ are the entries of $(A,B)$; see (36) for more
details999Another natural height function would be the
$H((R,x,y))\coloneqq\max\\{\operatorname{\lvert}x\operatorname{\rvert},\operatorname{\lvert}y\operatorname{\rvert}\\}$.
The author expects that with respect to this height function, the family
$\mathcal{F}^{x,y,xy}$ still corresponds to $(1/8,1/8,1/4,1/4,1/4)$..
Moreover, observe that the theorem above canonically attaches a cubic
resolvent ring to a triple $(R,x,y)\in\mathcal{F}^{x,y,x^{2}}$ via Bhargava’s
parametrization of quartic rings. Refer to the forward direction of the map
from Theorem 5.8 as $\psi$; via Bhargava’s parametrization of quartic rings
[BhargavaHCL3], it can be interpreted as map from such triples to pairs of
based quartic rings equipped with a based cubic resolvent ring.

######  Theorem 5.9.

The pair $(\mathcal{F}^{x,y,xy},H,\psi)$ corresponds to
$(1/8,1/8,1/4,1/4,1/4)$.

We now discuss the vertex $(1/10,1/5,1/5,1/5,3/10)$ of
$\operatorname{Poly}_{4}(S_{4})$. The coordinates of this vertex suggest
considering the family of quartic rings having a basis of the form
$1,x,y,x^{2}$. We consider a slight variant; consider triples $(R,x,y)$ where
$R$ is a quartic ring, $x\in R$, $y\in R/\mathbb{Z}$, and $x,y,x^{2}$ forms a
basis of $R/\mathbb{Z}$. We say two such triples $(R,x,y)$ and
$(R^{\prime},x^{\prime},y^{\prime})$ are _isomorphic_ if there exists a ring
isomorphism $\alpha\colon R\rightarrow R^{\prime}$ with the property that
$\alpha(x)=x^{\prime}$ and $\alpha(y)\equiv\alpha(y^{\prime})\ (\mathrm{mod}\
\mathbb{Z})$. We say such a triple is _nondegenerate_ if $R$ is nondegenerate.
Let $\mathcal{F}^{x,y,x^{2}}$ be the set of isomorphism classes of
nondegenerate such triples $(R,x,y)$. We employ the same strategy as
previously.

Consider the set of pairs of ternary quadratic forms of the form:

(35) $(A,B)=\begin{bmatrix}1&0&\ast\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&1/2&0\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix}.$

We consider quintuples $(R,x,y,\varepsilon,\nu)$ where $x\in R$, $y\in
R/\mathbb{Z}$, $\varepsilon=\pm 1$, $\nu\in\\{0,1\\}$, and $x,y,x^{2}$ is a
basis of $R/\mathbb{Z}$. We say such a quintuple is _nondegenerate_ if $R$ is
nondegenerate. We say two such quintuples $(R,x,y,\varepsilon,\nu)$ and
$(R^{\prime},x^{\prime},y^{\prime},\varepsilon^{\prime},\nu^{\prime})$ are
isomorphic if $\varepsilon=\varepsilon^{\prime}$, $\nu=\nu^{\prime}$, and
there exists a ring isomorphism $\alpha\colon R\rightarrow R^{\prime}$ such
that $\alpha(x)=x^{\prime}$ and $\alpha(y)=y^{\prime}\ (\mathrm{mod}\
\mathbb{Z})$.

######  Theorem 5.10.

There is a canonical discriminant-preserving bijection between:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of}\\\
\text{nondegenerate quintuples $(R,x,y,\epsilon,\nu)$}\\\ \text{where $x\in R$
and $y\in R/\mathbb{Z}$ are such that }\\\ \text{$x,y,x^{2}$ is a basis of
$R/\mathbb{Z}$}\text{ and $\varepsilon=\pm 1$ and
$\nu\in\\{0,1\\}$}\end{array}\right\\}\longleftrightarrow\left\\{\begin{array}[]{c}\text{nondegenerate
pairs of }\\\ \text{integral ternary quadratic forms}\\\ \text{of the form
\eqref{matrixform1xyx2intro}}\end{array}\right\\}$

For a triple $T\in\mathcal{F}^{x,y,x^{2}}$ we attach a nondegenerate pair of
integral ternary quadratic forms via the map above (setting $\varepsilon=1$
and $\nu=0$); define the _height_ to be

$H(T)\coloneqq\max\\{\operatorname{\lvert}a_{13}\operatorname{\rvert}^{2},\operatorname{\lvert}a_{22}\operatorname{\rvert},\operatorname{\lvert}a_{23}\operatorname{\rvert},\operatorname{\lvert}a_{33}\operatorname{\rvert},\operatorname{\lvert}b_{22}\operatorname{\rvert}^{2},\operatorname{\lvert}b_{23}\operatorname{\rvert}^{2},\operatorname{\lvert}b_{2}\operatorname{\rvert}^{2}\\}.$

As in the previous case, the theorem above canonically attaches a cubic
resolvent ring to a triple $(R,x,y)$ in $\mathcal{F}^{x,y,x^{2}}$; we will
prove that this is the _unique_ cubic resolvent ring of $R$. Refer to the
forward direction of map from the theorem above as $\psi$.

######  Theorem 5.11.

The pair $(\mathcal{F}^{x,y,x^{2}},H,\psi)$ corresponds to
$(1/10,1/5,1/5,1/5,3/10)$.

###  5.2. The parametrization of quartic rings

Our primary tool will be Bhargava’s parametrization of quartic rings, which we
now briefly summarize. Let
$V_{\mathbb{R}}=\mathbb{R}^{2}\otimes\operatorname{Sym}^{2}\mathbb{R}^{3}$. We
write an element of $(A,B)\in V_{\mathbb{R}}$ as a pair of $3\times 3$ real
symmetric matrices as follows:

(36) $2(A,B)=\Bigg{(}\begin{bmatrix}2a_{11}&a_{12}&a_{13}\\\
a_{12}&2a_{22}&a_{23}\\\
a_{13}&a_{23}&2a_{33}\end{bmatrix},\begin{bmatrix}2b_{11}&b_{12}&b_{13}\\\
b_{12}&2b_{22}&b_{23}\\\ b_{13}&b_{23}&2b_{33}\end{bmatrix}\Bigg{)}.$

An element $(A,B)$ is said to be _integral_ if $a_{ij},b_{ij}\in\mathbb{Z}$,
and the lattice of integral elements is denoted $V_{\mathbb{Z}}$. The group
$G(\mathbb{R})=\operatorname{GL}_{2}(\mathbb{R})\times\operatorname{GL}_{3}(\mathbb{R})$
acts on $V_{\mathbb{R}}$ in the following way; an element $(g_{2},g_{3})\in
G(\mathbb{R})$ sends $(A,B)$ to

$(rg_{3}Ag_{3}^{t}+sg_{3}Bg_{3}^{t},tg_{3}Ag_{3}^{t}+ug_{3}Bg_{3}^{t})$

where we write

$g_{2}=\begin{bmatrix}r&s\\\ t&u\end{bmatrix}.$

The subgroup $G(\mathbb{Z})$ acts on $V_{\mathbb{R}}$ and preserves the
lattice $V_{\mathbb{Z}}$. The action of $G_{\mathbb{Z}}$ on $V_{\mathbb{R}}$
has a unique polynomial invariant given by the formula
$\operatorname{Disc}(4\operatorname{Det}(Bx-Ay))$. This fundamental invariant
is known as the _discriminant_ and is denoted $\operatorname{Disc}(A,B)$. The
orbits of $G_{\mathbb{Z}}$ acting on $V_{\mathbb{Z}}$ parametrize quartic
rings equipped with a cubic resolvent ring; see [BhargavaHCL3] for more
details.

######  Theorem 5.12 (Bhargava, [BhargavaHCL3]).

There is a canonical, discriminant-preserving bijection between
$V_{\mathbb{Z}}$ and isomorphism classes of pairs $(R,C)$ where $R$ is a based
quartic ring and $C$ is a based cubic resolvent ring of $R$. Under the action
of $G_{\mathbb{Z}}$, this bijection descends to a bijection between the set of
$G_{\mathbb{Z}}$ orbits of $V_{\mathbb{Z}}$ and the set of isomorphism classes
of pairs $(R,C)$ where $R$ is a quartic ring and $C$ is a cubic resolvent ring
of $R$.

We now describe a few important details of this bijection. A cubic resolvent
ring $C$ of a quartic ring $R$ comes equipped with a quadratic map
$\varphi\colon R/\mathbb{Z}\rightarrow C/\mathbb{Z}$ called a _cubic resolvent
map_. Choosing bases for $R/\mathbb{Z}$ and $C/\mathbb{Z}$ gives rise to an
element of $V_{\mathbb{Z}}$. Conversely, for an element $(A,B)\in
V_{\mathbb{Z}}$, Bhargava constructed certain polynomials in the
$a_{ij},b_{ij}$ that form the structure coefficients for a basis of $R$.
Moreover, the binary cubic form $4\operatorname{Det}(Bx-Ay)$ via (8) gives
structure coefficients for a basis of $C$.

###  5.3. Reduction theory of quartic rings

We now fix some notation we use throughout this section. Unless otherwise
specified, $(A,B)$ will denote a nondegenerate element of $V_{\mathbb{Z}}$ and
$(R,C)$ will denote the corresponding quartic ring and cubic resolvent ring
respectively. Let
$\Delta=\operatorname{\lvert}\operatorname{Disc}(A,B)\operatorname{\rvert}$.
Analogously to the cubic case, in this section we develop the relationship
between:

  1. (1)

pairs $(R,C)$ equipped with “small” bases;

  2. (2)

and pairs $(A,B)$ with “small” coefficients.

Our proofs are almost identical to those in the cubic case.

######  Lemma 5.13.

Let $\\{1,v_{1},v_{2},v_{3}\\}$ and $\\{1,w_{1},w_{2}\\}$ be Minkowski bases
of any nondegenerate quartic ring $R$ and any cubic resolvent ring $C$ of $R$.
Let $(A,B)$ be the corresponding matrices given by the cubic resolvent map.
Then
$\operatorname{\lvert}a_{ij}\operatorname{\rvert}\ll\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}/\operatorname{\lvert}w_{1}\operatorname{\rvert}$
and
$\operatorname{\lvert}b_{ij}\operatorname{\rvert}\ll\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}/\operatorname{\lvert}w_{2}\operatorname{\rvert}$.

###### Proof.

Let $\sigma_{1},\dots,\sigma_{4}$ be the $4$ nonzero homomorphisms of $R$ into
the complex numbers. The cubic resolvent map $\varphi\colon
R/\mathbb{Z}\otimes R/\mathbb{Z}\rightarrow\frac{1}{2}C/\frac{1}{2}\mathbb{Z}$
is given by

$\varphi(x\otimes
y)\rightarrow\frac{1}{2}(\sigma_{1}(x)\sigma_{2}(y)+\sigma_{2}(x)\sigma_{1}(y)+\sigma_{3}(x)\sigma_{4}(y)+\sigma_{3}(y)\sigma_{4}(x))$

For an element $z\in C\otimes\mathbb{Q}$, let $\langle w_{i}\rangle z$ denote
the coefficient of $z$ with respect to the basis $\\{1,w_{1},w_{2}\\}$.
Because $\\{1,v_{1},v_{2},v_{3}\\}$ and $\\{1,w_{1},w_{2}\\}$ are Minkowski
bases,

$\displaystyle\operatorname{\lvert}a_{ij}\operatorname{\rvert}$
$\displaystyle\asymp\langle w_{1}\rangle(\varphi(v_{i}\otimes v_{j}))$
$\displaystyle\ll\frac{\operatorname{\lvert}\varphi(v_{i}\otimes
v_{j})\operatorname{\rvert}}{\operatorname{\lvert}w_{1}\operatorname{\rvert}}$
$\displaystyle\ll\frac{\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}}{\operatorname{\lvert}w_{1}\operatorname{\rvert}}.$

An analogous argument shows that
$\operatorname{\lvert}b_{ij}\operatorname{\rvert}\ll\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}/\operatorname{\lvert}w_{2}\operatorname{\rvert}$.
∎

For any $p=(p_{1},p_{2},p_{3},q_{1},q_{2})\in\mathbb{R}^{5}$ and any positive
real number $X$, define $B(p,X)$ to be the set of elements $(A,B)\in
V_{\mathbb{R}}$ such that $a_{ij}\leq X^{p_{i}+p_{j}-q_{1}}$ and $b_{ij}\leq
X^{p_{i}+p_{j}-q_{2}}$. Let
$p=(p_{1},p_{2},p_{3},q_{1},q_{2})\in\mathbb{R}^{5}$ denote a point. Moreover
$\epsilon$ will denote a positive real number, $M,D$ will denote real numbers
$\geq 1$, and $X$ will denote a real number $>1$. Recall that we say the pair
$(R,C)$ is $(\epsilon,X)$-close to $p$ if

$\displaystyle\max_{i=1,2,3}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(R)-p_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}$
$\displaystyle\max_{i=1,2}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(C)-q_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}.$

We say $((A,B),p)$ are $(\epsilon,X)$-close if the corresponding rings $(R,C)$
and $p$ are $(\epsilon,X)$-close. We say $(A,B)$ _comes from Minkowski bases_
if there exist Minkowski bases of the corresponding rings $(R,C)$ giving rise
to $(A,B)$.

######  Lemma 5.14.

There exists $M$ such that for every pair $(A,B)$, point $p$, and positive
real number $\epsilon$, the following statement holds: if $\Delta>1$ and
$((A,B),p)$ are $(\epsilon,\Delta)$-close and $(A,B)$ comes from Minkowski
bases, then $(A,B)\in Me^{3\epsilon}B(p,\Delta)$.

###### Proof.

Let $\\{1,v_{1},v_{2},v_{3}\\}$ and $\\{1,w_{1},w_{2}\\}$ be the Minkowski
bases of $(R,C)$ giving rise to $(A,B)$. We have

$\displaystyle\operatorname{\lvert}a_{i,j}\operatorname{\rvert}$
$\displaystyle\ll\frac{\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}}{\operatorname{\lvert}w_{1}\operatorname{\rvert}}$
by Lemma 5.13
$\displaystyle\asymp\frac{\lambda_{i}(R)\lambda_{j}(R)}{\lambda_{1}(C)}$
$\displaystyle\leq e^{3\epsilon}\Delta^{p_{i}+p_{j}-q_{1}}$
$\displaystyle\text{because $((A,B),p)$ is $(\epsilon,\Delta)$-close}.$

An analogous argument shows that
$\operatorname{\lvert}b_{i,j}\operatorname{\rvert}\ll
e^{3\epsilon}\Delta^{p_{i}+p_{j}-q_{2}}$. ∎

######  Lemma 5.15.

There exists $D$ such that for all $p$ satisfying (31), all $(A,B)$, and all
$M,X$ the following statement holds: if $(A,B)\in MB(p,X)$ then

$\\#\\{(A^{\prime},B^{\prime})\in DMB(p,X)\cap
V_{\mathbb{Z}}\mid\;(R,C)\simeq(R^{\prime},C^{\prime})\\}\geq\exp[X,(p_{3}-p_{2})+(p_{3}-p_{1})+(p_{2}-p_{1})+(q_{2}-q_{1})]\\}$

where here $(R^{\prime},C^{\prime})$ is the quartic ring and cubic resolvent
ring arising from the pair $(A^{\prime},B^{\prime})$.

###### Proof.

Recall that $V_{\mathbb{Z}}$ admits an action of
$G_{\mathbb{Z}}=\operatorname{GL}_{2}(\mathbb{Z})\times\operatorname{GL}_{3}(\mathbb{Z})$.
Act on $(A,B)$ by elements of $G_{\mathbb{Z}}$ of the form

$\Bigg{(}\begin{bmatrix}1&m\\\ 0&1\\\ \end{bmatrix},\begin{bmatrix}1&0&0\\\
n_{12}&1&0\\\ n_{13}&n_{23}&1\\\ \end{bmatrix}\Bigg{)}$

where $\operatorname{\lvert}n_{ij}\operatorname{\rvert}\leq\lceil
X^{p_{j}-p_{i}}\rceil$ and
$\operatorname{\lvert}m\operatorname{\rvert}\leq\lceil X^{q_{2}-q_{1}}\rceil$.
∎

For a pair $(A,B)$, let $\\{1,v_{1},v_{2},v_{3}\\}$ and $\\{1,w_{1},w_{2}\\}$
denote the explicit bases of $R$ and $C$ obtained from the structure
coefficients.

######  Lemma 5.16.

For every $p$ satisfying (31), every $(A,B)$, and every $M,X$ the following
statement holds: if $(A,B)\in MB(p,X)$, then
$\operatorname{\lvert}v_{i}\operatorname{\rvert}\ll_{M}X^{p_{i}}$ for
$i=1,2,3$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{M}X^{q_{i}}$ for $i=1,2$.

###### Proof.

The binary cubic form corresponding to the basis $\\{1,w_{1},w_{2}\\}$ is
given by $4\operatorname{Det}(Bx-Ay)$. Lemma 4.9 implies that
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{M}X^{q_{i}}$ for $i=1,2$.
The formulae in [BhargavaHCL3] of the explicit structure coefficients for the
basis $\\{1=v_{0},v_{1},v_{2},v_{3}\\}$ imply that if $(A,B)\in MB(p,X)$ then
the $v_{j}$-coefficient of $v_{i}^{2}$ is $\ll_{M}X^{2p_{i}-p_{j}}$. Let
$p_{0}=0$. Thus,

(37)
$\operatorname{\lvert}v_{i}\operatorname{\rvert}^{2}\ll_{M}\max_{j=0,1,2,3}\\{X^{2p_{i}-p_{j}}\operatorname{\lvert}v_{j}\operatorname{\rvert}\\}.$

Applying Lemma 11.2 completes the proof. ∎

######  Lemma 5.17.

For every $p$ satisfying (31) and every $(A,B)$, $X$, and $M$, the following
statement holds: if

$(A,B)\in MB(p,X)$

and $X/2\leq\Delta\leq X$, then there are
$\ll_{M}\exp[X,(p_{3}-p_{2})+(p_{3}-p_{1})+(p_{2}-p_{1})+(q_{2}-q_{1})]$
distinct elements $(A^{\prime},B^{\prime})\in MB(p,X)\cap V_{\mathbb{Z}}$ such
that $(R^{\prime},C^{\prime})\simeq(R,C)$.

###### Proof.

Let $\\{v_{0}=1,v_{1},v_{2},v_{3}\\}$ and $\\{w_{0}=1,w_{1},w_{2}\\}$ be the
explicit bases of $R$ and $C$ obtained from $(A,B)$. By Lemma 5.16, we have
$\operatorname{\lvert}v_{i}\operatorname{\rvert}\ll_{M}X^{p_{i}}$ for
$i=1,2,3$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{M}X^{q_{i}}$ for $i=1,2$.
Because

$X^{1/2}\asymp\Delta^{1/2}\ll\operatorname{\lvert}v_{1}\operatorname{\rvert}\operatorname{\lvert}v_{2}\operatorname{\rvert}\operatorname{\lvert}v_{3}\operatorname{\rvert}\ll_{M}X^{p_{1}+p_{2}+p_{3}}=X^{1/2}$

and

$X^{1/2}\asymp\Delta^{1/2}\ll\operatorname{\lvert}w_{1}\operatorname{\rvert}\operatorname{\lvert}w_{2}\operatorname{\rvert}\ll_{M}X^{q_{1}+q_{2}}=X^{1/2},$

we have $\operatorname{\lvert}v_{i}\operatorname{\rvert}\asymp_{M}X^{p_{i}}$
for $i=1,2,3$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\asymp_{M}X^{q_{i}}$ for
$i=1,2$.

Suppose $(A^{\prime},B^{\prime})\in MB(p,X)\cap V_{\mathbb{Z}}$ is such that
$(R^{\prime},C^{\prime})\simeq(R,C)$. Let
$\\{1,v^{\prime}_{1},v^{\prime}_{2},v^{\prime}_{3}\\}$ and
$\\{1,w^{\prime}_{1},w^{\prime}_{2}\\}$ denote the bases of $R^{\prime}$ and
$C^{\prime}$ given by $(A^{\prime},B^{\prime})$. By Lemma 5.16, we have
$\operatorname{\lvert}v^{\prime}_{i}\operatorname{\rvert}\ll_{M}X^{p_{i}}$ for
$i=1,2,3$ and
$\operatorname{\lvert}w^{\prime}_{i}\operatorname{\rvert}\ll_{M}X^{q_{i}}$ for
$i=1,2$. Moreover, write

$\displaystyle v^{\prime}_{i}$
$\displaystyle=c_{i0}+\sum_{j=0}^{3}c_{ij}v_{j}$ $\displaystyle
w^{\prime}_{i}$ $\displaystyle=d_{i0}+\sum_{j=0}^{2}d_{ij}w_{j}.$

The pair $(A^{\prime},B^{\prime})$ is determined by the choices of $c_{ij}$
for $1\leq i,j\leq 3$ and $d_{ij}$ for $1\leq i,j\leq 2$. We have:

$\displaystyle
X^{p_{i}}\asymp_{M}\operatorname{\lvert}v^{\prime}_{i}\operatorname{\rvert}$
$\displaystyle\asymp_{M}\max_{0\leq
j<4}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}\\}\asymp_{M}\max_{0\leq
j<n}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}X^{p_{j}}\\}$
$\displaystyle
X^{q_{i}}\asymp_{M}\operatorname{\lvert}w^{\prime}_{i}\operatorname{\rvert}$
$\displaystyle\asymp_{M}\max_{0\leq
j<3}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}\operatorname{\lvert}w_{j}\operatorname{\rvert}\\}\asymp_{M}\max_{0\leq
j<n}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}X^{q_{j}}\\}$

where $p_{0}=q_{0}=0$. If $j\leq i$, the number of choices for $c_{ij}$ and
$d_{ij}$ is $\ll_{M}1$. If $j>i$, the number of choices for $c_{ij}$ is
$\ll_{M}X^{p_{j}-p_{i}}$ and the number of choices for $d_{ij}$ is
$\ll_{M}X^{q_{j}-q_{i}}$. Hence, there are
$\ll_{M}\exp[X,(p_{3}-p_{2})+(p_{3}-p_{1})+(p_{2}-p_{1})+(q_{2}-q_{1})]$
distinct integral $(A^{\prime},B^{\prime})\in MB(p,X)$ such that
$(R^{\prime},C^{\prime})\simeq(R,C)$. ∎

######  Lemma 5.18.

For all $p$ satisfying (31) and all $(A,B)$, $M$, and $X$ the following
statement holds: if $(A,B)\in MB(p,X)$ and $\frac{X}{2}\leq\Delta$ then

$\lambda_{i}(R)\asymp_{M}\Delta^{p_{i}}.$

for $i=1,2,3$ and

$\lambda_{i}(C)\asymp_{M}\Delta^{q_{i}}.$

for $i=1,2$.

###### Proof.

Lemma 4.11 immediately implies that $\lambda_{i}(C)\asymp_{M}\Delta^{q_{i}}$
for $i=1,2$. We have

$\displaystyle\lambda_{1}(R)\lambda_{2}(R)\lambda_{3}(R)$
$\displaystyle\ll\operatorname{\lvert}v_{1}\operatorname{\rvert}\operatorname{\lvert}v_{2}\operatorname{\rvert}\operatorname{\lvert}v_{3}\operatorname{\rvert}$
by definition of successive minima
$\displaystyle\ll_{M}X^{p_{1}}X^{p_{2}}X^{p_{3}}$ by Lemma 5.16
$\displaystyle=X^{1/2}$ because $p_{1}+p_{2}+p_{3}=1/2$
$\displaystyle\asymp_{M}\Delta^{1/2}$ because $\Delta\geq X/2$ and $(A,B)\in
MB(p,X)$ $\displaystyle\asymp\lambda_{1}(R)\lambda_{2}(R)\lambda_{3}(R).$ by
Minkowski’s second theorem

Thus $\lambda_{i}(R)\asymp_{M}\Delta^{p_{i}}$ for $i=1,2,3$. ∎

###  5.4. Computing $\operatorname{Poly}_{4}(G)$

In this section, we will compute $\operatorname{Poly}_{4}(G)$ and
$\operatorname{Poly}_{4}$. Let
$p=(p_{1},p_{2},p_{3},q_{1},q_{2})\in\mathbb{R}^{5}$ be any point in
$\mathbb{R}^{5}$. Let $K$ denote a quartic étale algebra over $\mathbb{Q}$ and
$L$ denote the cubic resolvent algebra of $K$. Let $\varphi\colon
K/\mathbb{Q}\otimes K/\mathbb{Q}\rightarrow L/\mathbb{Q}$ denote the symmetric
bilinear form given by the cubic resolvent map. A _flag
$(\mathcal{F},\mathcal{G})$ of $(K,L)$_ is a pair of sequences
$\mathcal{F}=(F_{0},F_{1},F_{2},F_{3})$ and $\mathcal{G}=(G_{0},G_{1},G_{2})$
of $\mathbb{Q}$-vector spaces such that

$\mathbb{Q}=F_{0}\subset F_{1}\subset F_{2}\subset F_{3}=K$
$\mathbb{Q}=G_{0}\subset G_{1}\subset G_{2}=L$

and $\dim_{\mathbb{Q}}F_{i}=i+1$ and $\dim_{\mathbb{Q}}G_{i}=i+1$. For a
positive integer $n$, let $[n]$ refer to the set $\\{0,\dots,n-1\\}$.
Associate to a flag $(\mathcal{F},\mathcal{G})$ the function

$\displaystyle T_{\mathcal{F},\mathcal{G}}\colon[4]\times[4]$
$\displaystyle\longrightarrow[3]$ $\displaystyle(i\;\;,\;\;j)\;$
$\displaystyle\longmapsto\min\\{k\in[3]\mid\varphi\langle F_{i}\otimes
F_{j}\rangle\subseteq G_{k}\\}.$

######  Definition 5.19.

Let $P_{\mathcal{F},\mathcal{G}}$ be the set of
$p=(p_{1},p_{2},p_{3},q_{1},q_{2})\in\mathbb{R}^{5}$ satisfying (31) and the
following condition: for all $1\leq i,j\leq 3$, if
$T_{\mathcal{F},\mathcal{G}}(i,j)>0$ then

$q_{T_{\mathcal{F},\mathcal{G}}(i,j)}\leq p_{i}+p_{j}.$

The proof of the following theorem is almost exactly analogous to that of [me,
Theorem 1.1.2]; the reader is encouraged to refer to the proof of [me, Theorem
1.1.2] for more details. We will need the following definition.

######  Theorem 5.20.

We have
$\operatorname{Poly}_{4}(G)=\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$
where $(\mathcal{F},\mathcal{G})$ ranges across flags of pairs $(K,L)$ with
the property that $\operatorname{Gal}(K)\simeq G$ as a permutation group.

###### Proof.

We will first show that $\operatorname{Poly}_{4}(G)$ is contained in
$\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$. By
Minkowski’s second theorem and the definition of successive minima,
$\operatorname{Poly}_{4}(G)$ satisfies the linear inequalities given in (31).
Let $R$ be any nondegenerate quartic $G$-ring and let $C$ be a cubic resolvent
ring of $R$. Let $K=R\otimes\mathbb{Q}$ and let $L=C\otimes\mathbb{Q}$.
Picking Minkowski bases $\\{v_{0}=1,v_{1},v_{2},v_{3}\\}$ and
$\\{w_{0}=1,w_{1},w_{2}\\}$ of $R$ and $C$ respectively gives rise to a flag
$(\mathcal{F},\mathcal{G})$ of $(K,L)$ given by
$F_{i}\coloneqq\mathbb{Q}\langle v_{0},\dots,v_{i}\rangle$ and
$G_{i}\coloneqq\mathbb{Q}\langle w_{0},\dots,w_{i}\rangle$. For every $1\leq
i,j\leq 3$ such that $T_{\mathcal{F},\mathcal{G}}(i,j)>0$, we have

$\varphi\langle\mathbb{Z}\langle
v_{0},\dots,v_{i}\rangle\otimes\mathbb{Z}\langle
v_{0},\dots,v_{j}\rangle\rangle\subseteq\mathbb{Z}\langle
w_{0},\dots,w_{T_{\mathcal{F},\mathcal{G}}(i,j)}\rangle$

and

$\varphi\langle\mathbb{Z}\langle
v_{0},\dots,v_{i}\rangle\otimes\mathbb{Z}\langle
v_{0},\dots,v_{j}\rangle\rangle\not\subseteq\mathbb{Z}\langle
w_{0},\dots,w_{T_{\mathcal{F},\mathcal{G}}(i,j)-1}\rangle.$

Therefore, there exists $0\leq i^{\prime}\leq i$ and $0\leq j^{\prime}\leq j$
such that $\langle
w_{T_{\mathcal{F},\mathcal{G}}(i,j)}\rangle\varphi(v_{i^{\prime}}\otimes
v_{j^{\prime}})\neq 0$. So,

$\operatorname{\lvert}w_{T_{\mathcal{F},\mathcal{G}}(i,j)}\operatorname{\rvert}\ll\operatorname{\lvert}\varphi(v_{i^{\prime}}\otimes
v_{j^{\prime}})\operatorname{\rvert}\ll\operatorname{\lvert}v_{i^{\prime}}\operatorname{\rvert}\operatorname{\lvert}v_{j^{\prime}}\operatorname{\rvert}\ll\operatorname{\lvert}v_{i}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}.$

This implies that $\operatorname{Poly}_{4}(G)$ is contained in the set
$\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$ where
$(\mathcal{F},\mathcal{G})$ ranges across flags of $(K,L)$ with the property
that $\operatorname{Gal}(K)\simeq G$.

Now let $(\mathcal{F},\mathcal{G})$ be a flag of a pair $(K,L)$ with the
property that that $\operatorname{Gal}(K)\simeq G$. Choose bases
$\\{v_{0}=1,v_{1},v_{2},v_{3}\\}$ and $\\{w_{0}=1,w_{1},w_{2}\\}$ of $K$ and
$L$ giving rise to the flag $(\mathcal{F},\mathcal{G})$. Writing the symmetric
bilinear map $\varphi$ with respect to these bases gives a pair $(A,B)\in
V_{\mathbb{R}}$, which we write as

$2(A,B)=\Bigg{(}\begin{bmatrix}2a_{11}&a_{12}&a_{13}\\\
a_{12}&2a_{22}&a_{23}\\\
a_{13}&a_{23}&2a_{33}\end{bmatrix},\begin{bmatrix}2b_{11}&b_{12}&b_{13}\\\
b_{12}&2b_{22}&b_{23}\\\ b_{13}&b_{23}&2b_{33}\end{bmatrix}\Bigg{)}.$

For a positive integer $M$, define the pair of matrices $(A_{M},B_{M})$ to be

$2A_{M}=\Bigg{(}\begin{bmatrix}2a_{11}M^{p_{1}+p_{1}-q_{1}}&a_{12}M^{p_{1}+p_{2}-q_{1}}&a_{13}M^{p_{1}+p_{3}-q_{1}}\\\
a_{12}M^{p_{1}+p_{2}-q_{1}}&2a_{22}M^{p_{2}+p_{2}-q_{1}}&a_{23}M^{p_{2}+p_{2}-q_{1}}\\\
a_{13}M^{p_{1}+p_{3}-q_{1}}&a_{23}M^{p_{2}+p_{3}-q_{1}}&2a_{33}M^{p_{3}+p_{3}-q_{1}}\end{bmatrix}$
$2B_{M}=\begin{bmatrix}2b_{11}M^{p_{1}+p_{1}-q_{2}}&b_{12}M^{p_{1}+p_{2}-q_{2}}&b_{13}M^{p_{1}+p_{3}-q_{2}}\\\
b_{12}M^{p_{1}+p_{2}-q_{2}}&2b_{22}M^{p_{2}+p_{2}-q_{2}}&b_{23}M^{p_{2}+p_{3}-q_{2}}\\\
b_{13}M^{p_{1}+p_{3}-q_{2}}&b_{23}M^{p_{2}+p_{3}-q_{2}}&2b_{33}M^{p_{3}+p_{3}-q_{2}}\end{bmatrix}\Bigg{)}.$

For a rational point $p$ in the relative interior of
$P_{\mathcal{F},\mathcal{G}}$, one may check that there exists an infinite set
$\mathcal{M}\subseteq\mathbb{Z}_{>0}$ such that for all $M\in\mathcal{M}$, the
pair $(A_{M},B_{M})$ is contained in $V_{\mathbb{Z}}$; the proof is exactly
identical to that of LABEL:constructM. Each $M\in\mathcal{M}$ gives rise to a
distinct quartic ring and cubic resolvent ring $(R_{M},C_{M})$ with absolute
discriminant $\Delta_{M}$; for each $M\in\mathcal{M}$, write the bases of
$R_{M}$ and $C_{M}$ obtained from $(A_{M},B_{M})$ by
$\\{1,v_{1,M},\dots,v_{3,M}\\}$ and $\\{1,w_{1,M},w_{2,M}\\}$ respectively.
Then for $M,N\in\mathcal{M}$, we have:

$\displaystyle N\Delta_{M}$ $\displaystyle=M\Delta_{N}$ $\displaystyle
N^{p_{i}}v_{i,M}$ $\displaystyle=M^{p_{i}}v_{i,N}$ $\displaystyle
N^{q_{i}}w_{i,M}$ $\displaystyle=M^{q_{i}}w_{i,N}$

Therefore,

$\displaystyle\lim_{M\in\mathcal{M}}\Big{(}\log_{\Delta_{M}}\lambda_{1}(R_{M}),\dots,\log_{\Delta_{M}}\lambda_{3}(R_{M})\Big{)}$
$\displaystyle=(p_{1},p_{2},p_{3})$
$\displaystyle\lim_{M\in\mathcal{M}}\Big{(}\log_{\Delta_{M}}\lambda_{1}(C_{M}),\log_{\Delta_{M}}\lambda_{2}(C_{M})\Big{)}$
$\displaystyle=(q_{1},q_{2})$

Therefore, $p\in\operatorname{Poly}_{4}(G)$. Because
$\operatorname{Poly}_{4}(G)$ is defined to be a set of limit points, it must
be closed; therefore $P_{\mathcal{F},\mathcal{G}}$ is contained in
$\operatorname{Poly}_{4}(G)$, which completes the proof of the theorem. ∎

###### Proof of Theorem 5.1.

Theorem 5.20 expresses $\operatorname{Poly}_{4}(G)$ as a finite union of
polytopes. A computation shows that these polytopes all have dimension $3$. ∎

###### Proof of Theorem 5.4.

Let $p\in\operatorname{Poly}_{4}$ be any point. Suppose that
$q_{2}>p_{1}+p_{3}$. By Theorem 5.20,
$\operatorname{Poly}_{4}(G)=\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$
where $(\mathcal{F},\mathcal{G})$ ranges across flags of pairs $(K,L)$ where
$K$ is a quartic étale algebra. Choose such a flag $(\mathcal{F},\mathcal{G})$
such that $P_{\mathcal{F},\mathcal{G}}$ contains $p$. Choose bases of $K$ and
$L$ giving rise to that flag; in these bases, the cubic resolvent map is a
pair $(A,B)\in V_{\mathbb{Q}}$ of the form

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&0\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}.$

If $q_{2}>p_{2}+p_{3}$ then $b_{22}=b_{33}=0$. If $q_{1}>2p_{1}$ then
$a_{11}=0$. In what ensues, we will show that neither of these events occur.

Because $\operatorname{Disc}(A,B)\neq 0$, the binary cubic form
$\operatorname{Det}(Bx+Ay)$ does not have a double root, and hence
$\operatorname{Rank}(B)>1$. Therefore we cannot have both $b_{22}=0$ and
$b_{23}=0$. Now let $Q_{A}$ and $Q_{B}$ be the conics in $\mathbb{P}^{2}$
given by $A$ and $B$ respectively. Because $\operatorname{Disc}(A,B)\neq 0$,
the intersection $Q_{A}\cap Q_{B}$ is $4$ geometric points in $\mathbb{P}^{2}$
with the property that no $3$ are collinear. If $a_{11}=0$ then $[1\colon
0\colon 0]\in Q_{A}\cap Q_{B}$; however because $\operatorname{Rank}(B)=2$,
the conic $Q_{B}$ is a union of two lines intersecting at the point $[1\colon
0\colon 0]$, which forces three of the $4$ points in $Q_{A}\cap Q_{B}$ to be
collinear, contradiction. Hence, $a_{11}\neq 0$.

A computation shows that there exists $(A,B)\in V_{\mathbb{Q}}$ of the form

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&0\\\ -&0&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

generating a nondegenerate quartic ring. The pair $(A,B)$ gives rise to a flag
$(\mathcal{F},\mathcal{G})$ via the explicit structure coefficients. It is
easy to see that $P_{\mathcal{F},\mathcal{G}}$ is precisely the first polytope
in the theorem statement. Therefore, the first polytope specified in the
theorem statement is contained in $\operatorname{Poly}_{4}$ and the subset of
$\operatorname{Poly}_{4}$ with $q_{2}>p_{1}+p_{3}$ is contained in the first
polytope specified in the theorem statement.

Now suppose that $p\in\operatorname{Poly}_{4}(S_{4})$ is any point with
$q_{2}\leq p_{1}+p_{3}$ and $q_{1}>2p_{1}$. Let $(A,B)\in V_{\mathbb{Q}}$ be
the set of matrices arising from an appropriate flag generating a
nondegenerate quartic étale algebra; then $a_{11}=b_{11}=0$. If $q_{2}>2p_{2}$
then further $b_{12}=b_{22}=0$. So, the quadric $Q_{B}$ contains the line
$z=0$, where $x,y,z$ are the coordinates of $\mathbb{P}^{2}$; thus we may
suppose two distinct points on the line $z=0$ are contained in the
intersection $Q_{A}\cap Q_{B}$; hence $a_{12}\neq 0$ so $q_{1}\leq
p_{1}+p_{2}$. Therefore if $q_{2}\leq p_{1}+p_{3}$ and $q_{1}>2p_{1}$ then
$q_{1}\leq p_{1}+p_{2}$ or $q_{2}\leq 2p_{2}$. This shows that the part of
$\operatorname{Poly}_{4}(S_{4})$ not lying in the first polytope stated in the
theorem statement is contained in the union of the second polytope and third
polytope stated in the theorem statement. Similarly, a computation shows that
there exists $(A,B)\in V_{\mathbb{Q}}$ of the form

$(A,B)=\Bigg{(}\begin{bmatrix}0&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&*\\\ -&0&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

and

$(A,B)=\Bigg{(}\begin{bmatrix}0&0&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&*\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

generating a nondegenerate quartic ring, so Theorem 5.20 implies that the
second and third polytopes in the theorem statement are contained in
$\operatorname{Poly}_{4}$. ∎

###### Proof of Theorem 5.3.

As in the proof of Theorem 5.4, to show that $\operatorname{Poly}_{4}(D_{4})$
is contained in the given region, it suffices to show that for any element
$(A,B)\in V_{\mathbb{Q}}$ giving rise to a quartic étale algebra $K$, we have:

  1. (1)

if $a_{11}=b_{11}=0$, then $K$ is reducible;

  2. (2)

if $b_{11}=b_{12}=b_{22}=0$, then $K$ is reducible.

Suppose $K$ is a quartic étale algebra given by a pair $(A,B)\in
V_{\mathbb{Q}}$, and let $Q_{A}$ and $Q_{B}$ be the quadratic forms in
$\mathbb{P}^{2}_{\mathbb{Q}}$. Then

$K\simeq\Gamma(V(Q_{A},Q_{B})).$

If $a_{11}=b_{11}=0$ then $[1\colon 0\colon 0]$ is a common intersection of
the conics defined by $A$ and $B$, which implies that $K$ is reducible.
Similarly, if $b_{11}=b_{12}=b_{22}=0$ then the conic by $B$ in
$\mathbb{P}^{2}$ is of the form $axz+bxz+cz^{2}=z(ax+by+cz)$, where here
$x,y,z$ are the coordinates of $\mathbb{P}^{2}$; this implies that $K$ is
reducible. Therefore, the set $\operatorname{Poly}_{4}(D_{4})$ is contained in
the region specified in the theorem statement.

A computation shows that there exists $(A,B)\in V_{\mathbb{Q}}$ of the form

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&0\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

generating a $D_{4}$-quartic field, so Theorem 5.20 implies the region in the
theorem statement is contained in $\operatorname{Poly}_{4}(D_{4})$. ∎

###### Proof of Theorem 5.2.

Let $p\in\operatorname{Poly}_{4}(S_{4})$ be any point. The proof of Theorem
5.3 shows that $q_{1}\leq 2p_{1}$ and $q_{2}\leq 2p_{2}$. As in the proof of
Theorem 5.4, to show that $\operatorname{Poly}_{4}(S_{4})$ is contained in the
given region, it suffices to show that for any element $(A,B)\in
V_{\mathbb{Q}}$ giving rise to a quartic étale algebra $K$, we have that if
$b_{11}=b_{12}=b_{13}=0$ then $\operatorname{Gal}(K)\not\simeq S_{4}$. If
$b_{11}=b_{12}=b_{13}=0$ then $(1,0)$ is a root of the cubic resolvent
polynomial $\operatorname{Det}(By-Ax)$, which implies that the cubic resolvent
algebra of $K$ is reducible, which implies that
$\operatorname{Gal}(K)\not\simeq S_{4}$. Therefore, the set
$\operatorname{Poly}_{4}(S_{4})$ is contained in the region specified in the
theorem statement.

A computation shows that there exists $(A,B)\in V_{\mathbb{Q}}$ of the form

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&*\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

generating a $S_{4}$-quartic field, so Theorem 5.20 implies the region in the
theorem statement is contained in $\operatorname{Poly}_{4}(S_{4})$. ∎

###  5.5. Computing density functions of quartic rings

We now define some notation we use throughout this section to shorten our
formulae. For a point $p\in\mathbb{R}^{5}$, let

$f(p)\coloneqq 1+\sum_{1\leq i\leq j\leq 3,1\leq k\leq
2}\max\\{0,q_{k}-p_{i}-p_{j}\\}$

and

$g(p)\coloneqq(p_{3}-p_{2})+(p_{3}-p_{1})+(p_{2}-p_{1})+(q_{2}-q_{1}).$

As before, $\epsilon$ will denote a positive real number, $M,D$ will denote
real numbers $\geq 1$, $X$ will denote a real number $>1$, and $(A,B)$ will
denote a nondegenerate pair in $V_{\mathbb{Z}}$. As usual, $p$ will denote a
point in $\mathbb{R}^{5}$.

######  Lemma 5.21.

For any $M$, $X$, and $p$ satisfying (31), we have

$\exp[X,f(p)]\ll_{p}\\#\\{V_{\mathbb{Z}}\cap MB(p,X)\\}\ll
M^{12}\exp[X,f(p)].$

The first implicit constant is independent of $X$ and $M$ and the second
implicit constant is independent of $p$, $X$, and $M$.

###### Proof.

Let $S_{k}$ denote the set of pairs $(i,j)$ such that $q_{k}\geq p_{i}+p_{j}$
for $k=1,2$. Then for every $(i,j)\in S_{1}$ and every lattice point $(A,B)\in
MB(p,X)$, we have $\operatorname{\lvert}a_{ij}\operatorname{\rvert}\leq M$.
Similarly, for every $(i,j)\in S_{2}$ and every lattice point $(A,B)\in
MB(p,X)$, we have $\operatorname{\lvert}b_{ij}\operatorname{\rvert}\leq M$.
Let $\pi\colon V_{\mathbb{R}}\rightarrow\mathbb{R}^{12-\\#S_{1}-\\#S_{2}}$
denote the projection map away from $a_{ij}$ for every pair $(i,j)\in S_{1}$
and $b_{ij}$ for every pair $(i,j)\in S_{2}$. Then

$\displaystyle\\#\\{\pi(B(p,X))\cap\mathbb{Z}^{12-\\#S_{1}-\\#S_{2}}\\}$
$\displaystyle\leq\\#\\{MB(p,X)\cap V_{\mathbb{Z}}\\}$ $\displaystyle\leq
M^{\\#S_{1}+\\#S_{2}}\\#\\{\pi(MB(p,X))\cap\mathbb{Z}^{12-\\#S_{1}-\\#S_{2}}\\}.$

Davenport’s lemma implies that

$\displaystyle\\#\\{\pi(MB(p,X))\cap\mathbb{Z}^{12-\\#S_{1}-\\#S_{2}}\\}=\operatorname{Vol}\big{(}\pi(MB(p,X))\big{)}+O\Bigg{(}\frac{\operatorname{Vol}(\pi(MB(p,X)))}{Y}\Bigg{)}$

where

$Y=M\min_{(i,j,k)}\\{\exp[X,p_{i}+p_{j}-q_{k}]\;\mid\;(i,j)\notin S_{k}\\}$

and the implicit constant is independent of $p$, $M$, and $X$. To finish, a
computation shows that

$\operatorname{Vol}(\pi(MB(p,X)))=M^{12-\\#S_{1}-\\#S_{2}}\exp[X,f(p)].$

∎

######  Lemma 5.22.

We have

$\displaystyle\\#\mathcal{F}_{4}(p,\epsilon,X)\ll
e^{36\epsilon}\exp[X,f(p)-g(p)]$

where the implicit constant does not depend on $X$, $p$, or $\epsilon$.

###### Proof.

Let $S$ be the set of $(A,B)$ arising from Minkowski bases of pairs
$(R,C)\in\mathcal{F}_{4}(p,\epsilon,X)$. Lemma 5.14 implies that there exists
a constant $M$ such that $(A,B)\in Me^{3\epsilon}B(p,X)$ for all $(A,B)\in S$.
By Lemma 5.15, there exists $D\geq 1$, independent of $X$, $p$, $M$, and
$\epsilon$, such that

$\\#\\{(A^{\prime},B^{\prime})\in
DMe^{3\epsilon}B(p,X)\mid(R,C)\simeq(R^{\prime},C^{\prime})\\}\geq\exp[X,g(p)]$

for all elements of $S$. Therefore,

$\\#\mathcal{F}_{4}(p,\epsilon,X)\ll\frac{\\#\\{DMe^{3\epsilon}B(p,X)\cap
V_{\mathbb{Z}}\\}}{\exp[X,g(p)]}.$

Applying Lemma 5.21 to estimate the numerator completes the proof. ∎

######  Lemma 5.23.

There exists a constant $M$ such that for all $\epsilon>M$, we have:

  1. (1)

if $p\in\operatorname{Poly}_{4}$, then

$\displaystyle\\#\mathcal{F}_{4}(p,\epsilon,X)\gg_{p,\epsilon}\exp[X,f(p)-g(p)];$

  2. (2)

if $p\in\operatorname{Poly}_{4}(S_{4})$, then

$\displaystyle\\#\mathcal{F}(p,\epsilon,S_{4},X)\gg_{p,\epsilon}\exp[X,f(p)-g(p)];$

  3. (3)

if $p\in\operatorname{Poly}_{4}(D_{4})$, then

$\displaystyle\\#\mathcal{F}(p,\epsilon,D_{4},X)\gg_{p,\epsilon}\exp[X,1-g(p)+(q_{2}-2p_{1})+(q_{2}-p_{1}-p_{2})+(q_{2}-p_{1}-p_{3})].$

###### Proof.

We first prove $(1)$. Choose $p\in\operatorname{Poly}_{4}$. Lemma 5.18 shows
that there exists a constant $M$ such that for all $\epsilon>M$, if $(A,B)\in
B(p,X)\cap V_{\mathbb{Z}}$ and $X/2\leq\Delta$, then $(A,B)$ is
$(\epsilon,X)$-close to $p$. Let $S=\\{(A,B)\in B(p,X)\cap V_{\mathbb{Z}}\mid
X/2\leq\Delta\leq X\\}$. Therefore

$\displaystyle\\#\mathcal{F}_{4}(p,\epsilon,X)$
$\displaystyle\geq\frac{\\#S}{\max_{(A,B)\in
S}\\#\\{(A^{\prime},B^{\prime})\in
B(p,X)\mid(R^{\prime},C^{\prime})\simeq(R,C)\\}}$
$\displaystyle\asymp\frac{\\#\\{(A,B)\in B(p,X)\cap
V_{\mathbb{Z}}\\}}{\max_{(A,B)\in S}\\#\\{(A^{\prime},B^{\prime})\in
B(p,X)\mid(R^{\prime},C^{\prime})\simeq(R,C)\\}}$ $\displaystyle\text{by the
definition of $B(p,X)$}.$

Applying Lemma 5.21 to estimate the numerator and applying Lemma 5.17 to upper
bound the denominator completes the proof of $(1)$.

The argument above also proves $(2)$, with the additional input that there
exists $D\in\mathbb{R}_{\geq 1}$ such that a positive proportion, as
$X\rightarrow\infty$, of the lattice points in $DB(p,X)\cap V_{\mathbb{Z}}$
have Galois group $S_{4}$ and $\Delta\geq X/2$. In fact, we may replace the
$\Delta\geq X/2$ condition with $\Delta\geq X/M$ for arbitrarily large $M$.
Observe that

$\lim_{M\rightarrow\infty}\lim_{X\rightarrow\infty}\frac{\\#\\{(A,B)\in
DB(p,X)\cap V_{\mathbb{Z}}\;\mid\;\Delta>X/M\\}}{\\#(DB(p,X)\cap
V_{\mathbb{Z}})}=1.$

Thus it suffices to prove that that there exists $D\in\mathbb{R}_{\geq 1}$
such that a positive proportion of the lattice points in $DB(p,X)\cap
V_{\mathbb{Z}}$ have Galois group $S_{4}$ as $X\rightarrow\infty$. Consider
the subset of lattice points of $DB(p,X)$ such that $\Delta\equiv 5\
(\mathrm{mod}\ 25)$, the ring $R$ is irreducible modulo $7$, and the ring $C$
is irreducible modulo $11$; every element of this subset has Galois group
$S_{4}$. If this set is nonempty for large enough $X$, then it forms a
positive proportion of $DB(p,X)$. The nonemptiness follows from the fact that
there exists a pair of integral ternary quadratic forms

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&*\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

such that $\Delta=5\ (\mathrm{mod}\ 25)$, the ring $R$ is irreducible modulo
$7$, and the ring $C$ is irreducible modulo $11$. Choosing $D$ to be large
enough such that $(A,B)\in DB(p,X)$ completes the proof of this case.

Similarly, to prove $(3)$ we will show that there exists $D\in\mathbb{R}_{\geq
1}$ such that a positive proportion, as $X\rightarrow\infty$, of the lattice
points $(A,B)\in B(p,X)\cap V_{\mathbb{Z}}$ satisfying
$b_{11}=b_{12}=b_{13}=0$ have Galois group $D_{4}$. First, it is easy to see
that there exists a pair of integral ternary quadratic forms

$(A,B)=\Bigg{(}\begin{bmatrix}*&*&*\\\ -&*&*\\\
-&-&*\end{bmatrix},\begin{bmatrix}0&0&0\\\ -&*&*\\\
-&-&*\end{bmatrix}\Bigg{)}$

giving rise to a $D_{4}$-ring. Choose $D$ large enough so that $(A,B)\in
DB(p,X)$. Let $S_{k}=\\{(i,j)\colon q_{k}=p_{i}+p_{j}\\}$ for $k=1,2$. Let

$Z=\\{(A^{\prime},B^{\prime})\in V_{\mathbb{Z}}\mid\forall(i,j)\in
S_{1},\;\;a^{\prime}_{ij}=a_{ij}\text{ and }\forall(i,j)\in
S_{2},\;\;b^{\prime}_{ij}=b_{ij}\text{ and }b_{11}=b_{12}=b_{13}=0\\}.$

Observe that $Z\cap DB(p,X)$ contains a positive proportion of the lattice
points in $DB(p,X)$. Further, observe that a general pair of integral ternary
quadratic forms in $Z$ gives rise to a $D_{4}$-ring. Thus, by Hilbert’s
irreducibility theorem, the subset of $Z$ with Galois group $\neq D_{4}$ is a
thin set; an application of quantitative Hilbert irreducibility (specifically
Theorem 2.1 of [cohen]) shows that the density, as $X\rightarrow\infty$, of
lattice points in $Z$ with Galois group $\neq D_{4}$ is $0$. ∎

###### Proof of Theorem 1.5.

Clearly, the support of $d_{\epsilon,4}$ is contained in
$\operatorname{Poly}_{4}$. If $p\in\operatorname{Poly}_{4}$, then combining
Lemma 5.22 and Lemma 5.23 shows that there exists a constant $M$ such that for
all $\epsilon>M$, we have

$\displaystyle\\#\mathcal{F}_{4}(p,\epsilon,X)\asymp_{\epsilon,p}\exp[X,1-$
$\displaystyle(p_{3}-p_{2})-(p_{3}-p_{1})-(p_{2}-p_{1})-(q_{2}-q_{1})$
$\displaystyle+\sum_{i\leq j,k}\max\\{0,q_{k}-p_{i}-p_{j}\\};$

∎

###### Proof of Theorem 1.6.

The proof is analogous to that of Theorem 1.5. Combine Theorem 5.2, Lemma
5.22, and Lemma 5.23. ∎

###### Proof of Theorem 1.7.

Again, the proof is analogous to that of Theorem 1.5. Combine Theorem 5.3,
Lemma 5.22, and Lemma 5.23. ∎

###  5.6. The generic points of $D_{4}$ and $S_{4}$

In this section, we prove Theorem 5.5, first for $D_{4}$ and then for $S_{4}$.
To prove Theorem 5.5 for $D_{4}$, we will need the following lemma101010This
lemma can be strengthened significantly, but for our application this
statement suffices..

######  Lemma 5.24.

Choose $p\in\operatorname{Poly}_{4}(S_{4})$ such that $p_{i}+p_{j}-q_{k}>0$
for all $1\leq i,j\leq 3$ and $1\leq k\leq 2$. Then

$\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a $D_{4}$-ring, and $R$ is
$(\epsilon,X)$-close to
$p$}\\}\ll_{\epsilon,\gamma}\exp[X,1-\delta_{p}+\gamma]$

where $\delta_{p}=\frac{1}{2}\max_{1\leq i,j\leq 3,1\leq k\leq
2}\\{p_{i}+p_{j}-q_{i}\\}$. Moreover, the implicit constant is independent of
$p$.

###### Proof.

By Lemma 5.14, it suffices to show that for any $M\geq 1$, we have

$\\{(A,B)\in MB(p,X)\mid R\text{ is a $D_{4}$-ring and }\Delta\leq
X\\}\ll_{M,\gamma}X^{1-\delta_{p}+\gamma}.$

Observe that if $(A,B)$ gives rise to a $D_{4}$-ring, then the binary cubic
form $\operatorname{Det}(Ax+By)$ is reducible. We now use a version of
quantitative Hilbert irreducibility (specifically an easy generalization of
Theorem 2.1 of [cohen]) and Lemma 5.21 to conclude. ∎

###### Proof of Theorem 5.5 for $D_{4}$.

Let $p(D_{4})=(0,1/4,1/4,0,1/2)$. We prove (32) for $D_{4}$ by demonstrating
the following equivalent statement:

(38)
$\lim_{\epsilon\rightarrow\infty}\limsup_{X\rightarrow\infty}\frac{\\#\\{(R,C)\;\mid\;\Delta\leq
X\text{ and $R$ is a $D_{4}$-ring that is not $(\epsilon,X)$-close to
$p(D_{4})$}\\}}{\\#\\{(R,C)\;\mid\;\Delta\leq X\text{ and $R$ is a
$D_{4}$-ring}\\}}=0.$

It is easy to bound the denominator; when ordered by absolute discriminant,
$100\%$ of quadratic extensions of $\mathbb{Q}(i)$ are quartic
$D_{4}$-extensions, and there are $\asymp X$ quadratic extensions of
$\mathbb{Q}(i)$ of absolute discriminant $\leq X$. Hence

(39) $\\#\\{(R,C)\;\mid\;\Delta\leq X\text{ and $R$ is a $D_{4}$-ring}\\}\gg
X.$

We now focus our attention on bounding the numerator of (38). Let $\ell\leq k$
be positive integers whose values we fix their value later. Our proof strategy
is very similar to that of Corollary 4.5; we count the number of $D_{4}$-rings
of absolute discriminant $\leq X$ that are $(\epsilon/\ell,X)$-close to some
_other_ point $p^{\prime}$ on $\operatorname{Poly}_{4}(D_{4})$ with the
property that $p^{\prime}$ has distance at least $\epsilon/\log X$ from $p$.
We then sum over an appropriate set of points $p^{\prime}$; we evaluate this
sum by splitting it into a _main term_ and a _cusp_. We bound both the main
term and the cusp to conclude.

To this end, there exists a constant $M$ such that for all $\epsilon>M$, we
have:

$\displaystyle\\#\\{(R,C)\mid\Delta\leq X\text{, $R$ is a $D_{4}$-ring, and
$R$ is not $(\epsilon,X)$-close to $p(D_{4})$}\\}$
$\displaystyle\leq\operatorname{Main}+\operatorname{Cusp}$

where:

$\displaystyle\operatorname{Main}$
$\displaystyle\coloneqq\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a
$D_{4}$-ring, and $R$ is $(\epsilon/k,X)$-close to a point in}$
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\operatorname{Poly}_{4}(D_{4})\cap\\{p\in\mathbb{R}^{5}\mid\frac{\epsilon}{\log
X}\leq p_{1}\leq 1/100\\}$ $\displaystyle\operatorname{Cusp}$
$\displaystyle\coloneqq\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a
$D_{4}$-ring, and $R$ is $(\epsilon,X)$-close to a point in}$
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\operatorname{Poly}_{4}(D_{4})\cap\\{p\in\mathbb{R}^{5}\mid
p_{1}\geq 1/100\\}$

Therefore, to prove (38) it suffices to show that

(40)
$\lim_{\epsilon\rightarrow\infty}\limsup_{X\rightarrow\infty}\frac{\operatorname{Main}}{X}=\lim_{\epsilon\rightarrow\infty}\limsup_{X\rightarrow\infty}\frac{\operatorname{Cusp}}{X}=0.$

We begin by evaluating $\operatorname{Main}$. Choose values of $\ell$ and $k$
such that:

  1. (1)

$\ell>36$;

  2. (2)

and for any point $(p_{1},p_{2},p_{3},q_{1},q_{2})\in\operatorname{Poly}_{4}$
and any pair $(R,C)$, if

$\displaystyle\operatorname{\lvert}\log_{\Delta}\lambda_{1}(R)-p_{1}\operatorname{\rvert}$
$\displaystyle\leq\frac{\epsilon}{k\log\Delta}$
$\displaystyle\operatorname{\lvert}\log_{\Delta}\lambda_{2}(R)-p_{2}\operatorname{\rvert}$
$\displaystyle\leq\frac{\epsilon}{k\log\Delta}$
$\displaystyle\operatorname{\lvert}\log_{\Delta}\lambda_{1}(C)-q_{1}\operatorname{\rvert}$
$\displaystyle\leq\frac{\epsilon}{k\log\Delta},$

then $(R,C)$ is $(\epsilon/\ell,\Delta)$-close to
$(p_{1},p_{2},p_{3},q_{1},q_{2})$.

We now define our appropriate set of points; they will form a grid whose
points are spaced $\frac{\epsilon}{k\log X}$ distance apart and lie in the
subset of $\operatorname{Poly}_{4}(D_{4})$ where $p_{1}\leq 1/100$. For any
integers $x$, $y$, and $z$, define the point

$p_{xyz}\coloneqq(p_{1}(x),p_{2}(x,y),p_{3}(x,y),q_{1}(x,y,z),q_{2}(x,y,z))$

via the following formulae:

$\displaystyle p_{1}(x)$ $\displaystyle\coloneqq\frac{\epsilon}{\log
X}+\frac{\epsilon x}{k\log X}$ $\displaystyle p_{2}(x,y)$
$\displaystyle\coloneqq\bigg{(}\frac{1}{4}-\frac{1}{2}p_{1}(x)\bigg{)}-\frac{\epsilon
y}{k\log X}$ $\displaystyle p_{3}(x,y)$ $\displaystyle\coloneqq
1/2-p_{1}(x)-p_{2}(x,y)$ $\displaystyle q_{1}(x,y,z)$ $\displaystyle\coloneqq
1/2-2p_{2}(x,y)+\frac{\epsilon z}{k\log X}$ $\displaystyle q_{2}(x,y,z)$
$\displaystyle\coloneqq 1/2-q_{1}(x,y,z)$

Let $M$ be the smallest positive integer such that

$p_{1}(M)\geq 1/100.$

For any nonnegative integer $x$, let $N_{x}$ be the smallest positive integer
such that

$p_{2}(x,N_{x})\leq\frac{1}{4}-p_{1}(x).$

For any nonnegative integers $x$ and $y$, let $L_{xy}$ be the smallest
positive integer such that

$q_{1}(x,y,L_{xy})\geq 2p_{1}(x).$

The values $M$, $N_{x}$, and $L_{xy}$ indicate where the grid meets the
boundaries of $\operatorname{Poly}_{4}(D_{4})$; see Theorem 5.3 for an
explicit description of the facets. Thus, we may write $\operatorname{Main}$
as the following triple sum111111This is essentially Riemann integration on
$\operatorname{Poly}_{4}(D_{4})$!:

(41)
$\operatorname{Main}\leq\sum_{x=0}^{M}\sum_{y=0}^{N_{x}}\sum_{z=0}^{L_{xy}}\\#\\{(R,C)\mid\Delta\leq
X\text{, $R$ is a $D_{4}$-ring, and $R$ is $(\epsilon/\ell,X)$-close to
$p_{xyz}$}\\}$

Without loss of generality, fix $\epsilon$ to be large enough so that
$d_{\epsilon/k,4}$ does not depend on $\epsilon$; for conciseness, let
$h=d_{\epsilon/k,4}$. We now bound the inner sum of (41); choose positive
integers $x$ and $y$ such that $0\leq x\leq M$ and $0\leq y\leq N_{x}$. Then:

$\displaystyle\sum_{z=0}^{L_{xy}}\\#\\{(R,C)\mid\Delta\leq X\text{, $R$ is a
$D_{4}$-ring, and $R$ is $(\epsilon/\ell,X)$-close to $p_{xyz}$}\\}$
$\displaystyle\ll\sum_{z=0}^{L_{xy}}e^{36\epsilon/\ell}\exp[X,h(p_{xyz})]$ by
Lemma 5.22
$\displaystyle=\sum_{z=0}^{L_{xy}}e^{36\epsilon/\ell}\exp[X,h(p_{xy0})]X^{-\epsilon
z/k\log X}$ by Theorem 1.5
$\displaystyle=e^{36\epsilon/\ell}\exp[X,h(p_{xy0})]\sum_{z=0}^{L_{xy}}e^{-\epsilon
z/k}$ $\displaystyle\leq
e^{36\epsilon/\ell}\exp[X,h(p_{xy0})]\frac{1}{1-e^{-\epsilon/k}}$

Moreover, the implicit constants are independent of $x$, $y$, and $\epsilon$.
We now bound the second sum, also using Lemma 5.22 and Theorem 1.5. Fix a
positive integer $0\leq x\leq M$; there exists a positive real number $s$ such
that:

$\displaystyle\sum_{y=0}^{N_{x}}\exp[X,h(p_{xy0})]$
$\displaystyle=\exp[X,h(p_{x00})]\sum_{y=0}^{N_{x}}X^{-\epsilon sy/k\log X}$
$\displaystyle\leq\exp[X,h(p_{x00})]\frac{1}{1-e^{-\epsilon s/k}}.$

We bound the outer sum in exactly the same way; again Lemma 5.22 implies that
there exists a positive real number $t$ such that:

$\displaystyle\sum_{x=0}^{M}\exp[X,h(p_{x00})]$
$\displaystyle=\exp[X,h(p_{000})]\sum_{z=1}^{M}X^{-\epsilon tx/k\log X}$
$\displaystyle\leq\exp[X,h(p_{000})]\frac{1}{1-e^{-\epsilon t/k}}.$

Thus,

$\displaystyle\frac{\operatorname{Main}}{X}$ $\displaystyle\ll
e^{36\epsilon/\ell}X^{-1}\exp[X,h(p_{000})]\frac{1}{1-e^{-\epsilon
t/k}}\frac{1}{1-e^{-\epsilon s/k}}\frac{1}{1-e^{\epsilon/k}}$
$\displaystyle=e^{36\epsilon/\ell}X^{-\epsilon/\log X}\frac{1}{1-e^{-\epsilon
t/k}}\frac{1}{1-e^{-\epsilon s/k}}\frac{1}{1-e^{-\epsilon/k}}$
$\displaystyle=e^{36\epsilon/\ell-\epsilon}\frac{1}{1-e^{-\epsilon
t/k}}\frac{1}{1-e^{-\epsilon s/k}}\frac{1}{1-e^{-\epsilon/k}}$

where the implicit constant is independent of $\epsilon$. Because $\ell>36$,
we have

$\lim_{\epsilon\rightarrow\infty}e^{36\epsilon/\ell-\epsilon}\frac{1}{1-e^{-\epsilon
t/k}}\frac{1}{1-e^{-\epsilon s/k}}\frac{1}{1-e^{\epsilon/k}}=0.$

So,

$\lim_{\epsilon\rightarrow\infty}\limsup_{X\rightarrow\infty}\frac{\operatorname{Main}}{X}=0.$

We now bound $\operatorname{Cusp}$. For $p\in\operatorname{Poly}_{4}$, Lemma
5.22 implies that:

(42) $\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a $D_{4}$-ring, and $R$ is
$(\epsilon,X)$-close to $p$}\\}\ll e^{36\epsilon}\exp[X,h(p)]$

where the implicit constant is independent of $\epsilon$ and $p$. Let $B$ be
an open ball of radius $1/100$ around $(1/6,1/6,1/6,1/4,1/4)$. For $p\in
B\cap\operatorname{Poly}_{4}(S_{4})$, Lemma 5.24 shows that

(43) $\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a $D_{4}$-ring, and $R$ is
$(\epsilon,X)$-close to
$p$}\\}\ll_{\epsilon,\gamma}\exp[X,1-\delta_{p}+\gamma]$

where $\delta_{p}=\frac{1}{2}\max_{1\leq i,j\leq 3,1\leq k\leq
2}\\{p_{i}+p_{j}-q_{i}\\}$ and the implicit constant is independent of $p$.
Combining (42) and (43), we conclude there exists a positive real number
$\eta<1$ such that for every $p\in\operatorname{Poly}_{4}(D_{4})$ with
$p_{1}\geq 1/100$, we have

$\\#\\{(R,C)\mid\;\;\Delta\leq X\text{, $R$ is a $D_{4}$-ring, and $R$ is
$(\epsilon,X)$-close to $p$}\\}\ll_{\epsilon}\exp[X,\eta],$

with an implicit constant independent of $p$. Summing over an appropriate grid
in $\operatorname{Poly}_{4}(D_{4})$, we see that:

$\operatorname{Cusp}\ll_{\epsilon}(\log X)^{3}\exp[X,\eta].$

Thus

$\lim_{\epsilon\rightarrow\infty}\lim_{X\rightarrow\infty}\frac{\operatorname{Cusp}}{X}=0.$

We now show (33), which translates to showing that

$\liminf_{X\rightarrow\infty}\frac{\\#\\{(R,C)\;\mid\;\Delta\leq X\text{ and
$(R,C)$ is a $D_{4}$-ring and is $(\epsilon,X)$-close to
$p(D_{4})$}\\}}{\\#\\{(R,C)\;\mid\;\Delta\leq X\text{ and $(R,C)$ is
$(\epsilon,X)$-close to $p(D_{4})$}\\}}>0.$

Combining Lemma 5.22 and Lemma 5.23 shows that the denominator is
$\asymp_{\epsilon}X$.

We now show that the numerator is $\gg X$. When ordered by absolute
discriminant, $100\%$ of quadratic extensions of $\mathbb{Q}(i)$ are quartic
$D_{4}$-extensions. Let $R$ be a maximal order in a quartic $D_{4}$-extension
containing $\mathbb{Q}(i)$. Then $i\in R$, so $\lambda_{1}(R)=1$; moreover,
let $v\in R\setminus\mathbb{Z}[i]$ be such that
$\operatorname{\lvert}v\operatorname{\rvert}=\lambda_{2}(R)$; then
$\operatorname{\lvert}iv\operatorname{\rvert}=\operatorname{\lvert}v\operatorname{\rvert}$,
so $\lambda_{3}(R)=\lambda_{2}(R)$. Moreover, $R$ has a unique cubic resolvent
ring $C$; let $\varphi\colon R/\mathbb{Z}\rightarrow C/\mathbb{Z}$ be the
cubic resolvent map. Then because $R$ is irreducible, $\varphi(i)$ must be a
nonzero element of $C/\mathbb{Z}$; moreover,
$\lambda_{1}(C)\leq\operatorname{\lvert}\varphi(i)\operatorname{\rvert}\ll 1$.
Therefore there exists a positive real number $M$ such that for all
$\epsilon>M$, every such ring $R$ is $(\epsilon,\Delta)$-close to $p(D_{4})$.
Observing that there are $\asymp X$ many quadratic extensions of
$\mathbb{Q}(i)$ of absolute discriminant $\leq X$ concludes the proof. ∎

To prove Theorem 5.5 when $G=S_{4}$, we appeal to upcoming work of H and
Vincent. Given a nondegenerate rank $n$ ring $R$, we consider the rank $(n-1)$
sublattice
$\mathcal{L}_{R}\coloneqq\\{x\in\mathbb{Z}+nR\;\mid\;\operatorname{Tr}(x)=0\\}$,
up to isometry. It is naturally seen as an element of the double coset space

$\mathcal{S}_{n-1}\coloneqq\operatorname{GL}_{n-1}(\mathbb{Z})\backslash\operatorname{GL}_{n-1}(\mathbb{R})/\operatorname{GO}_{n-1}(\mathbb{R}).$

The space $\mathcal{S}_{n-1}$ has a natural measure $\mu_{n-1}$, obtained from
the Haar measure on $\operatorname{GL}_{n-1}(\mathbb{R})$ and
$\operatorname{GO}_{n-1}(\mathbb{R})$. To a nondegenerate pair $(R,C)$ we
associate an element
$(\mathcal{L}_{R},\mathcal{L}_{C})\in\mathcal{S}_{3}\times\mathcal{S}_{2}$; we
call this the _shape_ of the pair $(R,C)$.

######  Theorem 5.25 (H, Vincent).

When ordered by absolute discriminant, the shapes of pairs $(R,C)$ where $R$
is an $S_{4}$-ring are equidistributed with respect to $\mu_{3}\times\mu_{2}$.

###### Proof of Theorem 5.5 when $G=S_{4}$.

Let $p(S_{4})=(1/6,1/6,1/6,1/4,1/4)$. We will first prove (32). For an integer
$n\geq 2$, an element $\mathcal{L}\in\mathcal{S}_{n}$, and an integer $0\leq
i<n$, define $\lambda_{i}(\mathcal{L})$ to be the $i$-th successive minimum of
a lattice representing $\mathcal{L}$ with determinant $1$. For a positive real
number $\epsilon$ and an integer $n\geq 2$, define

$\mathcal{S}_{n,\epsilon}\coloneqq\\{\mathcal{L}\in\mathcal{S}_{n}\mid\max_{1\leq
i<n}\\{\log\lambda_{i}(\mathcal{L})\\}\leq\epsilon\\}.$

Then

$\displaystyle\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{(R,C)\;\mid\;\Delta\leq
X\text{ and $R$ is a $S_{4}$-ring that is $(\epsilon,X)$-close to
$p(S_{4})$}\\}}{\\#\\{(R,C)\;\mid\;\Delta\leq X\text{ and $R$ is a
$S_{4}$-ring}\\}}$
$\displaystyle\geq\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{(R,C)\;\mid\Delta\leq
X\text{ and $R$ is a $S_{4}$-ring and
}(\mathcal{L}_{R},\mathcal{L}_{C})\in\mathcal{S}_{3,\epsilon}\times\mathcal{S}_{2,\epsilon}\\}}{\\#\\{(R,C)\;\mid\;\Delta\leq
X\text{ and $R$ is a $S_{4}$-ring}\\}}$
$\displaystyle=\lim_{\epsilon\rightarrow\infty}\frac{(\mu_{3}\times\mu_{2})(\mathcal{S}_{3,\epsilon}\times\mathcal{S}_{2,\epsilon})}{(\mu_{3}\times\mu_{2})(\mathcal{S}_{3}\times\mathcal{S}_{2})}$
$\displaystyle=1$

We prove (33) with the following observation: Lemma 5.22 and Lemma 5.23
combine to prove that there exists a positive real number $M$ such that for
all $\epsilon>M$, a positive proportion of rings in
$\mathcal{F}(p(S_{4}),\epsilon,X)$ as $X\rightarrow\infty$ are $S_{4}$-rings.
∎

###  5.7. Special families of quartic rings

###  5.8. Quartic rings with a basis of the form $1,x,y,xy$

The following index formula of Bhargava [BhargavaHCL3] is essential to our
study of quartic rings equipped with a basis of the form $1,x,y,xy$. For a
rank $n$ ring $R$ and $n$ elements $v_{1},\dots,v_{n}\in R$, define
$\operatorname{Ind}(v_{1},\dots,v_{n})$ to be the index of $\mathbb{Z}\langle
v_{1},\dots,v_{n}\rangle$ in $R$ if $v_{1},\dots,v_{n}$ are linearly
independent, and $0$ otherwise.

######  Proposition 5.26 (Bhargava [BhargavaHCL3], Lemma 9).

If $R$ is a quartic ring and $C$ is a cubic resolvent ring of $R$, then for
any $x,y\in R$,

$\operatorname{Ind}(1,x,y,xy)=\operatorname{Ind}(1,\varphi(x),\varphi(y)).$

The above proposition implies that any nondegenerate quartic ring with a basis
$1,x,y,xy$ has a _unique_ cubic resolvent ring, namely $\mathbb{Z}\langle
1,\varphi(x),\varphi(y)\rangle$. Suppose $R$ is a nondegenerate quartic ring
with a basis $1,x,y,xy$. Let $C$ be the unique cubic resolvent ring of $R$ and
equip $C/\mathbb{Z}$ with the basis $\varphi(x),\varphi(y)$. Choose a basis
$x,y,z$ for $R/\mathbb{Z}$. Then, the cubic resolvent map $\varphi\colon
R/\mathbb{Z}\rightarrow C/\mathbb{Z}$, expressed in these bases, has the form:

(44) $(A,B)=\begin{bmatrix}1&\ast&\ast\\\ -&0&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&\ast&\ast\\\ -&1&\ast\\\
-&-&\ast\end{bmatrix}$

Conversely, a nondegenerate element of $V(\mathbb{Z})$ of the form (44) gives
rise to the following data: a pair $(R,C)$, a basis $x,y,z$ of $R/\mathbb{Z}$,
and a basis $\varphi(x),\varphi(y)$ of $C/\mathbb{Z}$. Proposition 5.26
implies that

$\operatorname{Ind}(1,x,y,xy)=\operatorname{Ind}(1,\varphi(x),\varphi(y))=1,$

so $x,y,xy$ is also a basis of $R/\mathbb{Z}$. There are unique lifts $X,Y\in
R$ of $x,y\in R/\mathbb{Z}$ and a unique value $\varepsilon=\pm 1$ such that
$z\equiv\varepsilon XY\ (\mathrm{mod}\ \mathbb{Z})$. This discussion will
prove Theorem 5.27, which we state after making the following definitions.

Consider quadruples $(R,x,y,z)$ where $x,y,z\in R/\mathbb{Z}$ is a basis of
$R/\mathbb{Z}$ and $1,X,Y,XY$ is a basis of $R$ for any lift $X,Y\in R$ of
$x,y$. We say such a quadruple is _nondegenerate_ if $R$ is nondegenerate.
Moreover, we say two such quadruples $(R,x,y,z)$ and
$(R^{\prime},x^{\prime},y^{\prime},z^{\prime})$ are isomorphic if there exists
a ring isomorphism $\alpha\colon R\rightarrow R^{\prime}$ such that:

$\displaystyle\alpha(x)$ $\displaystyle\equiv x^{\prime}\ (\mathrm{mod}\
\mathbb{Z})$ $\displaystyle\alpha(y)$ $\displaystyle\equiv y^{\prime}\
(\mathrm{mod}\ \mathbb{Z})$ $\displaystyle\alpha(z)$ $\displaystyle\equiv
z^{\prime}\ (\mathrm{mod}\ \mathbb{Z})$

Similarly, now consider quadruples $(R,X,Y,\varepsilon)$ where $X,Y\in R$,
$\varepsilon=\pm 1$, and $1,X,Y,\varepsilon XY$ is a basis of $R$. Again, we
say such a quadruple is _nondegenerate_ if $R$ is nondegenerate. We say two
such quadruples $(R,X,Y,\varepsilon)$ and
$(R^{\prime},X^{\prime},Y^{\prime},\varepsilon^{\prime})$ are isomorphic if
$\varepsilon=\varepsilon^{\prime}$ and there exists a ring isomorphism
$\alpha\colon R\rightarrow R^{\prime}$ such that:

$\displaystyle\alpha(X)$ $\displaystyle=X^{\prime}$ $\displaystyle\alpha(Y)$
$\displaystyle=Y^{\prime}$

######  Theorem 5.27.

There is a canonical discriminant-preserving bijection between the following
three sets:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of nondegenerate
quadruples $(R,x,y,z)$ such that }\\\ \text{$x,y,z\in R/\mathbb{Z}$ and
$x,y,z$ is a basis of $R/\mathbb{Z}$ and $1,X,Y,XY$ }\\\ \text{is a basis of
$R$ for any lifts $X,Y\in R$ of $x,y\in R/\mathbb{Z}$}\end{array}\right\\}$
$\left\\{\begin{array}[]{c}\text{isomorphism classes of nondegenerate
quadruples $(R,X,Y,\varepsilon)$}\\\ \text{such that $X,Y\in R$,
$\varepsilon=\pm 1$, and $1,X,Y,\varepsilon XY$ is a basis of
$R$}\end{array}\right\\}$ $\left\\{\begin{array}[]{c}\text{nondegenerate pairs
of integral ternary }\\\ \text{quadratic forms of the form
\eqref{matrixform1xyxy}}\end{array}\right\\}$

Let $H$ be the subgroup of
$\operatorname{GL}_{2}(\mathbb{Z})\times\operatorname{GL}_{3}(\mathbb{Z})$
acting on pairs of matrices of the form (44); $H$ is generated by the
following matrices:

(45) $\begin{split}\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}&,\begin{bmatrix}\pm
1&0&0\\\ 0&\pm 1&0\\\ \ast&\ast&\pm 1\end{bmatrix}\\\ \begin{bmatrix}0&1\\\
1&0\end{bmatrix}&,\begin{bmatrix}0&1&0\\\ 1&0&0\\\
0&0&1\end{bmatrix}\end{split}$

Suppose $R$ is a nondegenerate quartic ring. Consider pairs $(x,y)\in R$ such
that $1,x,y,xy$ is a basis of $R$. Consider the equivalence relation on such
pairs generated by $(x,y)\sim(y,x)$ and $(x,y)\sim(\pm x+a,y)$ for any
$a\in\mathbb{Z}$. We define a _digenization of $R$_ to be such an equivalence
class $(x,y)$. A _digenized quartic ring_ is defined to be a nondegenerate
quartic ring equipped with a digenization121212The notion of digenized quartic
rings is a natural variant of the concept of a _monogenized_ quartic ring.. We
say two digenized quartic rings $(R,x,y)$ and
$(R^{\prime},x^{\prime},y^{\prime})$ are isomorphic if there exists a ring
isomorphism $\alpha\colon R\rightarrow R^{\prime}$ such that
$(\alpha(x),\alpha(y))\sim(x^{\prime},y^{\prime})$. Then we obtain:

######  Theorem 5.28.

There is a canonical discriminant-preserving bijection between:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of}\\\ \text{digenized
quartic rings}\\\
\end{array}\right\\}\longleftrightarrow\left\\{\begin{array}[]{c}\text{nondegenerate
orbit classes of}\\\ \text{pairs of integral ternary quadratic forms}\\\
\text{of the form \eqref{matrixform1xyxy}}\\\ \text{under the action of
$H$}\end{array}\right\\}$

A theorem of Bhargava [monogenizations] proves that a nondegenerate quartic
ring has at most $2760$ many monogenizations; it is natural to ask if the
number of digenizations of a quartic ring is uniformly bounded! However, it is
clear that a monogenic quartic ring has infinitely digenizations. It is
interesting to ask if there is a positive integer $N$ such that for every
nondegenerate quartic ring $R$:

  1. (1)

if $R$ has infinitely many digenizations, then it is monogenic;

  2. (2)

else, $R$ has at most $N$ digenizations.

We now consider matrices of the form (34), which we repeat here:

$(A,B)=\begin{bmatrix}1&0&\ast\\\ -&0&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&0&\ast\\\ -&1&\ast\\\
-&-&\ast\end{bmatrix}.$

We obtain the following theorem as a corollary of Theorem 5.27. See 5.8

Recall that $\mathcal{F}^{x,y,xy}$ is defined to be the set of isomorphism
classes of triples $(R,x,y)$ where $R$ is a nondegenerate quartic ring and
$x,y\in R$ are such that $1,x,y,xy$ is a basis of $R$ and
$\varphi(x+y)=\varphi(x)+\varphi(y)$. Let $W(\mathbb{Z})$ denote the set of
matrices of the form (34) and let $\psi$ be the map from triples
$(R,x,y)\in\mathcal{F}^{x,y,xy}$ to pairs of integral ternary quadratic forms
referred to in Theorem 5.8 given by setting $\varepsilon=1$. We now prove
Theorem 5.9. Let $p=(1/8,1/8,1/4,1/4,1/4)$.

###### Proof of Theorem 5.9.

Let $T=(R,x,y)\in\mathcal{F}^{x,y,xy}$ be any such triple of absolute
discriminant $\Delta$ and let $C$ be the unique cubic resolvent ring. A
computation shows that $H(T)\leq X^{1/4}$ if and only if $\psi(T)\in B(p,X)$;
thus for any real numbers $M\geq 1$ and $X>1$, an easy generalization of Lemma
5.18 implies that if $\Delta\geq X/M$ and $H(T)^{4}\leq X$, then

$\displaystyle\lambda_{i}(R)$ $\displaystyle\asymp_{M}\Delta^{p_{i}}$
$\displaystyle\lambda_{i}(C)$ $\displaystyle\asymp_{M}\Delta^{q_{i}}.$

So, for each $M\geq 1$, there exists a positive real number $\epsilon_{M}$
such that if $\Delta\geq X/M$ and $H(T)^{4}\leq X$, we have that $\psi(T)$ is
$(\epsilon_{M},X)$-close to $p$. Moreover, we may suppose
$\lim_{M\rightarrow\infty}\epsilon_{M}=\infty$. We now prove (32). We have:

$\displaystyle\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq
X\text{ and $\psi(T)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq X\\}}$
$\displaystyle=\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq
X^{1/4}\text{ and $\psi(T)$ is $(\epsilon,X^{1/4})$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq X^{1/4}\\}}$
$\displaystyle=\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq
X\text{ and $\psi(T)$ is $(4\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq X\\}}$
$\displaystyle=\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq
X\text{ and $\psi(T)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq X\\}}$
$\displaystyle\geq\lim_{M\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq
X\text{ and $\Delta\geq
X/M$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq X\\}}.$

Davenport’s lemma gives:

$\displaystyle{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq X\\}}$
$\displaystyle=\frac{1}{2}\\#\\{(A,B)\in W(\mathbb{Z})\cap
B(p,X)\mid\operatorname{Disc}(A,B)\neq 0\\}$
$\displaystyle=\frac{X}{2}\operatorname{Vol}\\{(A,B)\in W(\mathbb{R})\cap
B(p,1)\\}+O(X^{7/8}).$

We now compute the numerator: Let

$g_{X}=\Bigg{(}\begin{bmatrix}X^{1/4}&0\\\
0&X^{1/4}\end{bmatrix},\begin{bmatrix}X^{-1/8}&0&0\\\ 0&X^{-1/8}&0\\\
0&0&X^{-1/4}\end{bmatrix}\Bigg{)}.$

The element $g_{X}$ maps $B(p,X)$ to $B(p,1)$. By Davenport’s lemma, we have

$\displaystyle\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)^{4}\leq X\text{ and
}\Delta\geq X/M\\}$ $\displaystyle=\frac{1}{2}\\#\\{(A,B)\in W(\mathbb{Z})\cap
B(p,X)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
X/M\\}$ $\displaystyle=\frac{1}{2}\\#\\{(A,B)\in W(\mathbb{Z})\cap
B(p,X)\;\mid\;\operatorname{\lvert}\operatorname{disc}(g(A,B))\operatorname{\rvert}\geq
1/M\\}$ $\displaystyle=\frac{1}{2}X\operatorname{Vol}\\{(A,B)\in
W(\mathbb{R})\cap
B(p,1)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
1/M\\}+O(X^{7/8}).$

Note that $W(\mathbb{R})$ lies on an affine linear subspace, so the “volume”
above is taken relatively. Now observe that

$\lim_{M\rightarrow\infty}\operatorname{Vol}\\{(A,B)\in W(\mathbb{R})\cap
B(p,1)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
1/M\\}=1$

so

$\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq
X\text{ and $\psi(T)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,xy}\;\mid\;H(T)\leq X\\}}=1.$

We obtain Condition (7) via the proof of Lemma 5.22, using the additional
observation that a positive proportion (depending on $\epsilon$) of lattice
points in $B(p,X)$ lie on the affine linear subspace defined by
$a_{11}=b_{22}=1$ and $a_{12}=b_{12}=a_{22}=b_{11}=0$. ∎

###  5.9. Quartic rings with a basis of the form $1,x,y,x^{2}$

As in the case of quartic rings with a basis of the form $1,x,y,xy$, our
proofs crucially rely on an _index formula_.

######  Proposition 5.29.

If $R$ is a quartic ring and $C$ is a cubic resolvent ring of $R$, then for
any $x,y\in R$,

$\operatorname{Ind}(1,x,y,x^{2})=\operatorname{Ind}(1,\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y)).$

###### Proof.

The proof proceeds exactly as that of Proposition 5.26. Namely, the assertion
of the lemma is equivalent to the following identity:

$\begin{array}[]{|c c c c|}1&1&1&1\\\
x&x^{\prime}&x^{\prime\prime}&x^{\prime\prime\prime}\\\
y&y^{\prime}&y^{\prime\prime}&y^{\prime\prime\prime}\\\
x^{2}&(x^{\prime})^{2}&(x^{\prime\prime})^{2}&(x^{\prime\prime\prime})^{2}\end{array}=\begin{array}[]{|c
c c|}1&1&1\\\
xx^{\prime}+x^{\prime\prime}x^{\prime\prime\prime}&xx^{\prime\prime}+x^{\prime}x^{\prime\prime\prime}&xx^{\prime\prime\prime}+x^{\prime}x^{\prime\prime}\\\
f&g&h\end{array}$

where

$\displaystyle f$
$\displaystyle=(x+y)(x^{\prime}+y^{\prime})+(x^{\prime\prime}+y^{\prime\prime})(x^{\prime\prime\prime}+y^{\prime\prime\prime})-(xx^{\prime}+x^{\prime\prime}x^{\prime\prime\prime})-(yy^{\prime}+y^{\prime\prime}y^{\prime\prime\prime})$
$\displaystyle g$
$\displaystyle=(x+y)(x^{\prime\prime}+y^{\prime\prime})+(x^{\prime}+y^{\prime})(x^{\prime\prime\prime}+y^{\prime\prime\prime})-(xx^{\prime\prime}+x^{\prime}x^{\prime\prime\prime})-(yy^{\prime\prime}+y^{\prime}y^{\prime\prime\prime})$
$\displaystyle h$
$\displaystyle=(x+y)(x^{\prime\prime\prime}+y^{\prime\prime\prime})+(x^{\prime}+y^{\prime})(x^{\prime\prime}+y^{\prime\prime})-(xx^{\prime\prime\prime}+x^{\prime}x^{\prime\prime})-(yy^{\prime\prime\prime}+y^{\prime}y^{\prime\prime}).$

The identity may be verified by direct calculation. ∎

From the above proposition, we see that any nondegenerate quartic ring with a
basis of the form $1,x,y,x^{2}$ has a _unique_ cubic resolvent ring, namely
the ring spanned by the vectors
$1,\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y)$. Let $R$ be a nondegenerate
quartic ring equipped with the basis $1,x,y,x^{2}$ and let $C$ be the unique
cubic resolvent ring; equip $C/\mathbb{Z}$ with the basis
$\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y)$. Choose a basis $x,y,z$ for
$R/\mathbb{Z}$. Then, the cubic resolvent map $\varphi\colon
R/\mathbb{Z}\rightarrow C/\mathbb{Z}$, expressed in these bases, has the form:

(46) $(A,B)=\begin{bmatrix}1&0&\ast\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&1/2&\ast\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix}.$

Conversely, a nondegenerate element of $V(\mathbb{Z})$ of the form (46) gives
rise to the following data: a pair $(R,C)$, a basis $x,y,z$ of $R/\mathbb{Z}$,
and a basis $\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y)$ of $C/\mathbb{Z}$.
Proposition 5.29 implies that

$\operatorname{Ind}(1,x,y,x^{2})=\operatorname{Ind}(1,\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y))=1$

so $1,X,Y,X^{2}$ is a basis of $R$ for any lifts $X,Y\in R$ of $x,y\in
R/\mathbb{Z}$. This discussion proves the following theorem.

######  Theorem 5.30.

There is a canonical discriminant-preserving bijection between:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of}\\\
\text{nondegenerate quadruples $(R,x,y,z)$}\\\ \text{where $x,y,z\in
R/\mathbb{Z}$ are such that }\\\ \text{$1,X,Y,X^{2}$ is a basis of $R$}\\\
\text{for any lifts $X,Y\in R$ of $x,y\in
R/\mathbb{Z}$}\end{array}\right\\}\longleftrightarrow\left\\{\begin{array}[]{c}\text{nondegenerate
pairs of}\\\ \text{integral ternary quadratic forms}\\\ \text{of the form
\eqref{matrixform1xyx2}}\end{array}\right\\}$

Let $H$ be the largest subgroup of
$\operatorname{GL}_{2}(\mathbb{Z})\times\operatorname{GL}_{3}(\mathbb{Z})$
acting on matrices of the form (46); $H$ is generated by of elements of the
form:

(47) $\begin{split}\begin{bmatrix}1&-2u\\\
0&1\end{bmatrix}&,\begin{bmatrix}1&0&0\\\ u&1&0\\\ \ast&\ast&1\end{bmatrix}\\\
\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}&,\begin{bmatrix}1&0&0\\\ 0&1&0\\\
0&0&\pm 1\end{bmatrix}\\\ \begin{bmatrix}1&0\\\
0&-1\end{bmatrix}&,\begin{bmatrix}-1&0&0\\\ 0&-1&0\\\
0&0&1\end{bmatrix}\end{split}$

Suppose $R$ is a nondegenerate quartic ring. Consider pairs $(x,y)\in R$ such
that $1,x,y,x^{2}$ is a basis of $R$. Consider the equivalence relation on all
pairs generated by $(x,y)\sim(\pm x+a,b+cx+\pm y)$ for integers
$a,b,c\in\mathbb{Z}$. Now consider triples $(R,x,y)$ where $x\in
R/\mathbb{Z}$, $y\in R/\mathbb{Z}\langle 1,x\rangle$ are such that
$R=\mathbb{Z}\langle 1,x,y,x^{2}\rangle$. We say the triple is _nondegenerate_
if $R$ is nondegenerate. We say two such triples $(R,x,y)$ and
$(R^{\prime},x^{\prime},y)$ are isomorphic if there exists a ring isomorphism
$\alpha\colon R\rightarrow R^{\prime}$ such that $\alpha(x)\equiv x^{\prime}\
(\mathrm{mod}\ \mathbb{Z})$ and $\alpha(y)\equiv y^{\prime}\ (\mathrm{mod}\
\mathbb{Z}\langle 1,x\rangle)$. We obtain:

######  Theorem 5.31.

There is a canonical discriminant preserving bijection between:

$\left\\{\begin{array}[]{c}\text{isomorphism classes of}\\\
\text{nondegenerate triples $(R,x,y)$}\\\ \text{where $x\in R/\mathbb{Z}$,
$y\in R/\mathbb{Z}\langle 1,x\rangle$ are such that }\\\
\text{$R=\mathbb{Z}\langle 1,X,Y,X^{2}\rangle$ for any lifts}\\\ \text{$X,Y\in
R$ of
$x,y$}\end{array}\right\\}\longleftrightarrow\left\\{\begin{array}[]{c}\text{nondegenerate
orbit classes of}\\\ \text{pairs of integral ternary quadratic forms}\\\
\text{of the form \eqref{matrixform1xyx2}}\\\ \text{under the action of
$H$}\end{array}\right\\}$

We now restrict our attention to matrices of the form (35), which we repeat
here:

$(A,B)=\begin{bmatrix}1&0&\ast\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix},\begin{bmatrix}0&1/2&0\\\ -&\ast&\ast\\\
-&-&\ast\end{bmatrix}.$

We consider quintuples $(R,x,y,\varepsilon,\nu)$ where $x\in R$, $y\in
R/\mathbb{Z}$, $\varepsilon=\pm 1$, $\nu\in\\{0,1\\}$, and $x,y,x^{2}$ is a
basis of $R/\mathbb{Z}$. We such a quintuple is _nondegenerate_ if $R$ is
nondegenerate. We say two such quintuples $(R,x,y,\varepsilon,\nu)$ and
$(R^{\prime},x^{\prime},y^{\prime},\varepsilon^{\prime},\nu^{\prime})$ are
isomorphic if $\varepsilon=\varepsilon^{\prime}$, $\nu=\nu^{\prime}$, and
there exists a ring isomorphism $\alpha\colon R\rightarrow R^{\prime}$ such
that $\alpha(x)=x^{\prime}$ and $\alpha(y)=y^{\prime}\ (\mathrm{mod}\
\mathbb{Z})$.

See 5.10

###### Proof.

Given such a nondegenerate pair of integral ternary quadratic forms, Theorem
5.30 gives a nondegenerate quartic ring with a basis $x,y,z$ of $R/\mathbb{Z}$
with the property that $1,X,Y,X^{2}$ is a basis of $R$ for any lifts $X,Y\in
R$ of $x,y\in R/\mathbb{Z}$. There is a unique lift $X\in R$ of $x$ and a
unique value $\varepsilon=\pm 1$ and $\nu\in\\{0,1\\}$ such that
$z\equiv\varepsilon X^{2}+\nu X\ (\mathrm{mod}\ \mathbb{Z}\langle
1,y\rangle)$. We take the quintuple $(R,X,y,\varepsilon,\nu)$. Conversely,
given such a quintuple $(R,x,y,\epsilon,\nu)$, there exists a unique integer
$a$ such that

$\varphi(x+(\varepsilon x^{2}+\nu x+ay))-\varphi(\varepsilon x^{2}+\nu
x+ay)\equiv 0\ (\mathrm{mod}\ \mathbb{Z}\langle 1,\varphi(x)\rangle).$

Equip $R/\mathbb{Z}$ with the basis $x,y,\varepsilon x^{2}+\nu x+ay$. Let $C$
be the unique cubic resolvent ring of $R$ and equip $C/\mathbb{Z}$ with the
basis $1,\varphi(x),\varphi(x+y)-\varphi(x)-\varphi(y)$; taking the cubic
resolvent map with respect to these bases gives an element of (35). A
computation shows that these two maps are inverse to each other. ∎

Recall that $\mathcal{F}^{x,y,x^{2}}$ is defined to be the set of isomorphism
classes of triples $(R,x,y)$ where $R$ is a nondegenerate quartic ring and
$x\in R$ and $y\in R/\mathbb{Z}$ is such that $x,y,x^{2}$ forms a basis of
$R/\mathbb{Z}$. Let $W(\mathbb{Z})$ denote the set of matrices of the form
(35) and $\psi$ be the map from triples $(R,x,y)\in\mathcal{F}^{x,y,x^{2}}$ to
pairs of integral ternary quadratic forms referred to in Theorem 5.10 given by
setting $\varepsilon=1$ and $\nu=0$. We now prove Theorem 5.11. Let
$p=(1/10,1/5,1/5,1/5,3/10)$.

###### Proof of Theorem 5.11.

The proof is almost identical to that of Theorem 5.9. Let
$T=(R,x,y)\in\mathcal{F}^{x,y,x^{2}}$ be any triple of absolute discriminant
$\Delta$ and let $C$ be the unique cubic resolvent ring. A computation shows
that $H(T)\leq X^{1/5}$ if and only if $\psi(T)\in B(p,X)$; thus for any
positive real numbers $M$ and $X$, Lemma 5.18 implies that if $\Delta\geq
X/M$, then

$\displaystyle\lambda_{i}(R)$ $\displaystyle\asymp_{M}\Delta^{p_{i}}$
$\displaystyle\lambda_{i}(C)$ $\displaystyle\asymp_{M}\Delta^{q_{i}}.$

So, for each $M$ there exists a constant $\epsilon_{M}$ such that if
$\Delta\geq X/M$ and $H(T)^{5}\leq X$, we have that $T$ is
$(\epsilon_{M},X)$-close to $p$. Moreover, we may suppose
$\lim_{M\rightarrow\infty}\epsilon_{M}=\infty$. We first prove (32).

$\displaystyle\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)\leq
X\text{ and $\psi(f)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)\leq X\\}}$
$\displaystyle\geq\lim_{M\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)^{5}\leq
X\text{ and $\Delta\geq
X/M$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)^{5}\leq X\\}}.$

Davenport’s lemma gives:

$\displaystyle{\\#\\{T\in\mathcal{F}^{x,y,s^{2}}\;\mid\;H(T)^{5}\leq X\\}}$
$\displaystyle\asymp\frac{1}{4}\\#\\{(A,B)\in W(\mathbb{Z})\cap
B(p,X)\mid\operatorname{Disc}(A,B)\neq 0\\}$
$\displaystyle=\frac{X}{4}\operatorname{Vol}\\{(A,B)\in W(\mathbb{R})\cap
B(p,1)\\}+O(X^{9/10}).$

We now compute the numerator. Let

$g_{X}=\Bigg{(}\begin{bmatrix}X^{1/5}&0\\\
0&X^{3/10}\end{bmatrix},\begin{bmatrix}X^{-1/10}&0&0\\\ 0&X^{-1/5}&0\\\
0&0&X^{-1/5}\end{bmatrix}\Bigg{)}.$

By Davenport’s lemma, we have

$\displaystyle\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)^{5}\leq X\text{
and }\Delta\geq X/M\\}$ $\displaystyle\asymp\frac{1}{4}\\#\\{(A,B)\in
W(\mathbb{Z})\cap
B(p,X)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
X/M\\}$ $\displaystyle=\frac{1}{4}\\#\\{(A,B)\in W(\mathbb{Z})\cap
B(p,X)\;\mid\;\operatorname{\lvert}\operatorname{disc}(g(A,B))\operatorname{\rvert}\geq
1/M\\}$ $\displaystyle=\frac{1}{4}X\operatorname{Vol}\\{(A,B)\in
W(\mathbb{R})\cap
B(p,1)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
1/M\\}+O(X^{9/10}).$

Now observe that

$\lim_{M\rightarrow\infty}\operatorname{Vol}\\{(A,B)\in W(\mathbb{R})\cap
B(p,1)\;\mid\;\operatorname{\lvert}\operatorname{disc}(A,B)\operatorname{\rvert}\geq
1/M\\}=1$

so

$\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)\leq
X\text{ and $\psi(f)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{T\in\mathcal{F}^{x,y,x^{2}}\;\mid\;H(T)\leq X\\}}=1$

We now prove Condition (33). Observe that a positive proportion of lattice
points in $B(p,X)$ satisfy the conditions $a_{11}=b_{12}=1$ and
$a_{12}=b_{11}=0$. Moreover, there exists $C$ such that for every lattice
point in $B(p,X)$ with $a_{11}=b_{12}=1$ and $a_{12}=b_{11}=0$, there exists a
$\operatorname{GL}_{2}(\mathbb{Z})$-translate with $a_{11}=b_{12}=1$ and
$a_{12}=b_{11}=b_{13}=0$ in $CB(p,X)$. Thus, a positive proportion of lattice
points in $CB(p,X)$ arise from elements of $\mathcal{F}^{1,x,y,x^{2}}$;
combining this observation with the proof of Lemma 5.22 shows that for
$\epsilon$ sufficiently large, we have:

$\\#(\mathcal{F}(p,\epsilon,X)\cap\psi(\mathcal{F}^{1,x,y,x^{2}}))\ll_{\epsilon}X^{4/5}$

Conversely, Lemma 5.23 shows that

$\\#\mathcal{F}(p,\epsilon,X)\gg X^{4/5}.$

∎

###  5.10. Binary quartic forms

Let
$f=f_{0}x^{4}+f_{1}x^{3}y+f_{2}x^{2}y^{2}+f_{3}xy^{3}+f_{4}y^{4}\in\operatorname{Sym}^{4}\mathbb{Z}^{2}$
be a binary quartic form. Wood defines a map
$\operatorname{Sym}^{4}\mathbb{Z}^{2}\rightarrow\mathbb{Z}^{2}\otimes\operatorname{Sym}^{2}\mathbb{Z}^{3}$
which canonically attaches to a binary quartic form a based quartic ring
equipped with a cubic resolvent ring ([woodquartic]). In this thesis we use a
slight variant of Wood’s map131313Our map and the map of Wood send a form $f$
to isomorphic quartic rings and cubic resolvent rings. The only difference is
a choice of basis of the quartic ring and cubic resolvent ring.. In
particular, we send a form $f$ to the following element of
$\mathbb{Z}^{2}\otimes\operatorname{Sym}^{2}\mathbb{Z}^{3}$:

$\begin{bmatrix}f_{0}&f_{1}/2&0\\\ -&f_{2}&f_{3}/2\\\
-&-&f_{4}\end{bmatrix},\begin{bmatrix}0&0&1/2\\\ -&1&0\\\ -&-&0\end{bmatrix}.$

Let $p=(1/6,1/6,1/6,1/6,1/3)$ and let $\psi$ denote the map from
$\operatorname{Sym}^{4}\mathbb{Z}^{2}\rightarrow\mathbb{Z}^{2}\otimes\operatorname{Sym}^{2}\mathbb{Z}^{3}$.
Observe that for any form $f\in\operatorname{Sym}^{4}\mathbb{Z}^{2}$, we have
$H_{c}(f)\leq X^{1/6}$ if and only if $\psi(f)\in B(p,X)$.

###### Proof of Theorem 5.7.

Again, the proof is almost identical to that of Theorem 5.9. Let $M$ be any
constant. Let $f\in\mathcal{F}_{4}^{\operatorname{bin}}$ and let $(R,C)$ be
the corresponding rings. By Lemma 5.18, if $\Delta\geq X/M$ and $\psi(f)\in
B(p,X)$, then

$\displaystyle\lambda_{i}(R)$ $\displaystyle\asymp_{M}\Delta^{p_{i}}$
$\displaystyle\lambda_{i}(C)$ $\displaystyle\asymp_{M}\Delta^{q_{i}}.$

To prove (32), it suffices to show that

$\lim_{M\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{f\in\mathcal{F}_{4}^{\operatorname{bin}}\;\mid\;H_{c}(f)\leq
X\text{ and }\Delta\geq
X^{6}/M\\}}{\\#\\{f\in\mathcal{F}_{4}^{\operatorname{bin}}\;\mid\;H_{c}(f)\leq
X\\}}=1.$

The denominator is clearly $X^{5}+o(X^{5})$. By Davenport’s lemma, we have:

$\displaystyle\\#\\{f\in\mathcal{F}_{4}^{\operatorname{bin}}\;\mid\;H_{c}(f)\leq
X\text{ and }\Delta>X^{6}/M\\}$
$\displaystyle=\\#\\{(f_{0},\dots,f_{4})\in\mathbb{Z}^{5}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}\\}\leq
X\text{ and
}\operatorname{\lvert}\operatorname{disc}(f_{0},\dots,f_{4})\operatorname{\rvert}>X^{6}/M\\}$
$\displaystyle=\operatorname{Vol}\\{(f_{0},\dots,f_{4})\in\mathbb{R}^{5}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}\\}\leq
X\text{ and
}\operatorname{\lvert}\operatorname{disc}(f_{0},\dots,f_{4})\operatorname{\rvert}>X^{6}/M\\}+O(X^{4})$
$\displaystyle=X^{5}\operatorname{Vol}\\{(f_{0},\dots,f_{5})\in\mathbb{R}^{5}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}\\}\leq
1\text{ and
}\operatorname{\lvert}\operatorname{disc}(f_{0},\dots,f_{4})\operatorname{\rvert}>1/M\\}+O(X^{4})$

We now prove (33). Observe that by Lemma 5.22, there exists a constant $M$
such that for all $\epsilon>M$, there are $\ll_{\epsilon}X^{5/6}$ isomorphism
classes of tuples $(R,C)$ such that $\Delta\leq X$ and $(R,C)$ is
$(\epsilon,X)$-close to $p$. Theorem 4.28 shows that there are at most
$2^{80}$ $\operatorname{GL}_{2}(\mathbb{Z})$-equivalence classes of binary
quartic forms giving rise to the same quartic ring. Therefore, it suffices to
show that there are $\gg X^{5/6}$
$\operatorname{GL}_{2}(\mathbb{Z})$-equivalence classes of binary quartics
such that $\Delta\leq X$ and the corresponding rings $(R,C)$ are
$(\epsilon,X)$-close to $p$; this is precisely the statement of Theorem 1.4
when $r=(1/24,1/24)$. ∎

###  5.11. Monic quartic forms

Let $p=(1/12,1/6,1/4,1/6,1/3)$. Observe that for any monic integral quartic
form $f$, we have $H_{r}(f)\leq X^{1/12}$ if and only if $H_{r}(f)\in B(p,X)$.
Let $\mathcal{F}_{4}^{\operatorname{mon}}$ denote the set of monic quartic
forms with integral coefficients and nonzero discriminant.

###### Proof of Theorem 5.6.

Again, the proof is almost identical to that of Theorem 5.9. Let $M$ be any
constant. For any $f\in\mathcal{F}_{4}^{\operatorname{mon}}$ let $(R,C)$ be
the corresponding rings. By Lemma 5.18, if $\Delta\geq X/M$ and $\psi(f)\in
B(p,X)$, then

$\displaystyle\lambda_{i}(R)$ $\displaystyle\asymp_{M}\Delta^{p_{i}}$
$\displaystyle\lambda_{i}(C)$ $\displaystyle\asymp_{M}\Delta^{q_{i}}.$

To prove (32), it suffices to show that

$\lim_{M\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{f\in\mathcal{F}_{4}^{\operatorname{mon}}\;\mid\;H_{r}(f)\leq
X\text{ and }\Delta\geq
X^{12}/M\\}}{\\#\\{f\in\mathcal{F}_{4}^{\operatorname{mon}}\;\mid\;H_{r}(f)\leq
X\\}}=1.$

The denominator is clearly $X^{10}+o(X^{10})$. By Davenport’s lemma, we have:

$\displaystyle\\#\\{f\in\mathcal{F}_{4}^{\operatorname{mon}}\;\mid\;H_{r}(f)\leq
X\text{ and }\Delta>X^{12}/M\\}$
$\displaystyle=\\#\\{(f_{1},\dots,f_{4})\in\mathbb{Z}^{4}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}^{1/i}\\}\leq
X\text{ and
}\operatorname{\lvert}\operatorname{disc}(1,f_{1},\dots,f_{4})\operatorname{\rvert}>X^{12}/M\\}$
$\displaystyle=\operatorname{Vol}\\{(f_{1},\dots,f_{4})\in\mathbb{R}^{4}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}^{1/i}\\}\leq
X\text{ and
}\operatorname{\lvert}\operatorname{disc}(1,f_{1},\dots,f_{4})\operatorname{\rvert}>X^{12}/M\\}+O(X^{9})$
$\displaystyle=X^{10}\operatorname{Vol}\\{(f_{1},\dots,f_{4})\in\mathbb{R}^{4}\;\mid\;\max\\{\operatorname{\lvert}f_{i}\operatorname{\rvert}\\}\leq
1\text{ and
}\operatorname{\lvert}\operatorname{disc}(1,f_{1},\dots,f_{4})\operatorname{\rvert}>1/M\\}+O(X^{9})$

We now prove (33). Observe that by Lemma 5.22, there exists a constant $M$
such that for all $\epsilon>M$, there are $\ll_{\epsilon}X^{3/4}$ isomorphism
classes of tuples $(R,C)$ such that $\Delta\leq X$ and $(R,C)$ is
$(\epsilon,X)$-close to $p$. Theorem 4.28 shows that there are at most
$2^{80}$ $\operatorname{GL}_{2}(\mathbb{Z})$-equivalence classes of binary
quartic forms giving rise to the same quartic ring. Therefore, it suffices to
show that there are $\gg X^{3/4}$
$\operatorname{GL}_{2}(\mathbb{Z})$-equivalence classes of binary quartics
such that $\Delta\leq X$, the binary quartic represents $1$, and the
corresponding rings $(R,C)$ are $(\epsilon,X)$-close to $p$; this follows from
Theorem 1.4 when $r=(0,1/12)$. ∎

###  5.12. Pictures of $\operatorname{Poly}_{4}$,
$\operatorname{Poly}_{4}(S_{4})$, and $\operatorname{Poly}_{4}(D_{4})$

Figure 9. $\operatorname{Poly}_{4}$ projected onto $p_{1}$, $p_{2}$, and
$q_{1}$. The color indicates the value of the density function
$d_{\epsilon,4}$; blue is the highest value.

Figure 10. A rotated version of the previous image.

Figure 11. $\operatorname{Poly}_{4}(D_{4})$ projected onto $p_{1}$, $p_{2}$,
and $q_{1}$. The color indicates the value of the lower bound on the density
function $d_{\epsilon,D_{4}}$ given in Theorem 1.7; blue is the highest value.

Figure 12. A rotated version of the previous image.

Figure 13. $\operatorname{Poly}_{4}(S_{4})$ projected onto $p_{1}$, $p_{2}$,
and $q_{1}$. The color indicates the value of the density function
$d_{\epsilon,S_{4}}$; blue is the highest value.

Figure 14. A rotated version of the previous image.

Figure 15. $\operatorname{Poly}_{4}(S_{4})$ projected onto $p_{1}$, $p_{2}$,
and $q_{1}$. The colored regions are the regions on which $d_{\epsilon,S_{4}}$
is linear. The vertices are labelled by the family of rings “corresponding” to
them.

Figure 16. A rotated version of the previous image.

##  6\. Counting quintic rings by successive minima

We now investigate density functions of quintic rings. In this section, let
$R$ denote a nondegenerate quintic ring with absolute discriminant $\Delta$
and let $S$ denote a sextic resolvent ring of $R$.

###  6.1. Statement of results

####  6.1.1. Computing $\operatorname{Poly}_{5}(G)$

######  Theorem 6.1.

For every permutation group $G\subseteq S_{5}$, the set
$\operatorname{Poly}_{5}(G)$ is a finite union of polytopes of dimension $7$.

In fact, the facets of these polytopes can be defined quite explicitly; see
Theorem 8.2 for an explicit description of these polytopes. Theorem 6.1 is a
natural generalization of [me, Theorem 1.1.2] in the quintic case. We now
explicitly describe this finite union of polytopes when $G=S_{5}$. Let

$p=(p_{1},\dots,p_{4},q_{1},\dots,q_{5})\in\mathbb{R}^{9}$

refer to a point of $\mathbb{R}^{9}$. By the definition of successive minima
and Minkowski’s theorem, the set $\operatorname{Poly}_{5}(G)$ satisfies the
linear inequalities

(48) $0\leq p_{1}\leq\dots\leq p_{4}$, |  $1/2=\sum_{i}p_{i}$, |  $0\leq q_{1}\leq\dots\leq q_{5}$, | and $3/2=\sum_{i}q_{i}$.
---|---|---|---

######  Theorem 6.2.

The set $\operatorname{Poly}_{5}(S_{5})$ is the set of points
$p\in\mathbb{R}^{9}$ such that (48) and

$1/2+p_{1}\geq q_{4}+q_{1}$, |  $1/2+p_{1}\geq q_{3}+q_{2}$, |  $1/2+p_{2}\geq q_{1}+q_{5}$, |  $1/2+p_{2}\geq q_{2}+q_{4}$,
---|---|---|---
$1/2+p_{3}\geq q_{5}+q_{2}$, |  $1/2+p_{3}\geq q_{3}+q_{4}$, |  and $1/2+p_{4}\geq q_{5}+q_{3}$. |

Moreover, $\operatorname{Poly}_{5}(S_{5})=\cup_{G}\operatorname{Poly}_{5}(G)$
as $G$ ranges across transitive subgroups of $S_{5}$.

Thus, we see that $\operatorname{Poly}_{5}(S_{5})$ is in fact a single
polytope! We now make a slight detour and discuss successive minima of
resolvent rings. A sextic resolvent ring of a quintic $S_{5}$-ring is a sextic
$\psi(S_{5})$-ring, where $\psi$ is a transitive embedding $\psi\colon
S_{5}\xhookrightarrow{}S_{6}$; the embedding $\psi$ is unique up to outer
automorphism. It is a natural question to ask if the successive minima of
sextic resolvent rings satisfy the same constraints as those of sextic
$\psi(S_{5})$-rings. More precisely, is the set of limit points of the
multiset

$\\{(\log_{\operatorname{\lvert}\operatorname{Disc}(S)\operatorname{\rvert}}\lambda_{1}(S),\dots,\log_{\operatorname{\lvert}\operatorname{Disc}(S)\operatorname{\rvert}}\lambda_{5}(S))\\}\subseteq\mathbb{R}^{5}$

as $S$ ranges across sextic resolvent rings of quintic $S_{5}$-rings equal to
set of limit points of the multiset

$\\{(\log_{\operatorname{\lvert}\operatorname{Disc}(S)\operatorname{\rvert}}\lambda_{1}(S),\dots,\log_{\operatorname{\lvert}\operatorname{Disc}(S)\operatorname{\rvert}}\lambda_{5}(S))\\}\subseteq\mathbb{R}^{5}$

as $S$ ranges across sextic $\psi(S_{5})$-rings? Let $R_{1}$ denote the first
set of limit points and let $R_{2}$ denote the second.

######  Corollary 6.3.

$R_{1}$ is strictly contained in $R_{2}$.

###### Proof.

The first region $R_{1}$ is $1/3$ times the projection of
$\operatorname{Poly}_{5}(S_{5})$ onto the last $5$ coordinates of
$\mathbb{R}^{9}$. By LABEL:onlyifthmshorten, the second region $R_{2}$ is
contained in of the set of points $q=(q_{1},\dots,q_{5})\in\mathbb{R}^{5}$
such that:

  1. (1)

$1/2=q_{1}+\dots+q_{5}$;

  2. (2)

$0\leq q_{1}\leq\dots\leq q_{5}$;

  3. (3)

and $q_{i+j}\leq q_{i}+q_{j}$ for all $1\leq i\leq j<i+j<6$.

Notice that this region is precisely $\operatorname{Len}(6)$; by [me, Theorem
1.1.2], $\operatorname{Len}(6)$ is contained in $R_{2}$, as
$\operatorname{Len}(6)$ is the polytope of the flag generated by a monogenic
basis of number field. In particular, the second region contains the point
$\frac{1}{30}(1,2,3,4,5)$ but the first does not. ∎

####  6.1.2. Density functions of quintic rings

Let $p=(p_{1},\dots,p_{4},q_{1},\dots,q_{5})\in\mathbb{R}^{9}$ denote a point.
For any $\epsilon\in\mathbb{R}_{>0}$, $X\in\mathbb{R}_{>1}$, and permutation
group $G\subseteq S_{5}$, define $\mathcal{F}(p,\epsilon,G,X)$ to be the set
of isomorphism classes of pairs $(R,S)$ such that $R$ is a $G$-ring,
$\Delta\leq X$, and

(49)
$\displaystyle\max_{i=1,2,3,4}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(R)-p_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}$ (50)
$\displaystyle\max_{i=1,2,3,4,5}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(S)-q_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}.$

Define the density function
$d_{\epsilon,G}\colon\mathbb{R}^{9}\rightarrow\mathbb{R}\cup\\{\ast\\}$ which
assigns to $p\in\mathbb{R}^{9}$ the value

$\lim_{X\rightarrow\infty}\frac{\log(\\#\mathcal{F}(p,\epsilon,G,X)+1)}{\log
X}$

if it exists and else assigns $\ast$. Similarly, let
$\mathcal{F}(p,\epsilon,X)$ be the set of isomorphism classes of pairs $(R,S)$
such that $\Delta\leq X$, (49), and (50). Let $d_{\epsilon,5}$ denote the
corresponding density function. We now compute the total density function for
quintic rings and the $S_{5}$-density function. In this section, we prove
Theorem 1.8 and Theorem 1.9.

####  6.1.3. Special families of quintic rings

Again, we see that quintic rings with extremal successsive minima tend to also
be extremal in an _algebraic_ sense. In the quartic case, we considered
vertices of $\operatorname{Poly}_{4}(S_{4})$ and showed that many of these
vertices corresponded to interesting families of quintic rings. The polytope
$\operatorname{Poly}_{5}(S_{5})$ is significantly more complicated than
$\operatorname{Poly}_{4}(S_{4})$; the polytope
$\operatorname{Poly}_{5}(S_{5})$ has $18$ vertices, and the author believes
that many of them likely give rise to interesting and natural families of
quartic rings. In this section, we discuss the following $3$ of these
vertices:

$\displaystyle(1/8,1/8,1/8,1/8,3/10,3/10,3/10,3/10,3/10)$
$\displaystyle(1/8,1/8,1/8,1/8,1/4,1/4,1/4,3/8,3/8)$
$\displaystyle(1/20,2/20,3/20,4/20,4/20,5/20,6/20,7/20,8/20)$

We’ll show that the first “corresponds” to generic $S_{5}$-rings, the second
“corresponds” to binary quintic forms, and the third “corresponds” to
monogenic quintic rings. We now formalize our definition of “correspond”.

We say a pair $(R,S)$ is $(\epsilon,X)$-close to $p$ if

(51) $\displaystyle\max_{1\leq i\leq
4}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(R)-p_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}$ (52) $\displaystyle\max_{1\leq
i\leq
5}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(S)-q_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}.$

Let $\mathcal{F}$ be an infinite set equipped with a map
$H\colon\mathcal{F}\rightarrow\mathbb{R}_{>0}$ such that for all $X$, we have
$\\#\\{f\in\mathcal{F}\mid H(f)\leq X\\}<\infty$. We call $H$ the _height
function_ of $\mathcal{F}$. Further suppose that $\psi$ is a map from
$\mathcal{F}$ to isomorphism classes of nondegenerate pairs of quintic rings
equipped with a sextic resolvent. For a point $p\in\operatorname{Poly}_{5}$,
we say the triple _$(\mathcal{F},H,\psi)$ corresponds to $p$_ if

(53)
$\lim_{\epsilon\rightarrow\infty}\liminf_{X\rightarrow\infty}\frac{\\#\\{f\in\mathcal{F}\;\mid\;H(f)\leq
X\text{ and $\psi(f)$ is $(\epsilon,X)$-close to
$p$}\\}}{\\#\\{f\;\mid\;H(f)\leq X\\}}=1$

and there exists a constant $C$ such that for all $\epsilon>C$, we have

(54)
$\liminf_{X\rightarrow\infty}\frac{\\#\\{(R,S)\in\mathcal{F}(p,\epsilon,X)\;\mid\;(R,S)\in\psi(\mathcal{F})\\}}{\\#\mathcal{F}(p,\epsilon,X)}>0.$

When $H$ or $\psi$ are implicit, we simply write that $\mathcal{F}$
corresponds to $p$. The following is a corollary of upcoming work of H.

######  Corollary 6.4.

The set of isomorphism classes of nondegenerate $S_{5}$-rings, where $H$ is
the absolute discriminant, corresponds to

$(1/8,1/8,1/8,1/8,3/10,3/10,3/10,3/10,3/10).$

The root height of a monic degree $n$ polynomial
$f(x)=x^{n}+a_{1}x^{2}+\dots+a_{n}\in\mathbb{Z}[x]$ is defined to be
$H_{r}(f)\coloneqq\max_{i}\\{\operatorname{\lvert}a_{i}\operatorname{\rvert}^{1/i}\\}$.
Given a binary $n$-ic form $f(x,y)=a_{0}x^{n}+\dots+a_{n}y^{n}$, define the
_coefficient height of $f$_ to be
$H_{c}(f)\coloneqq\max_{i}\\{\operatorname{\lvert}a_{i}\operatorname{\rvert}\\}$.
Fess [fess] constructed a map from
$\operatorname{Sym}^{5}\mathbb{Z}^{2}\rightarrow\mathbb{Z}^{4}\otimes\wedge^{2}\mathbb{Z}^{5}$;
via Bhargava’s parametrization of quintic rings [BhargavaHCL4], this map
canonically associates to an integral binary quintic form the data of a
quintic ring and a sextic resolvent ring. Fess’ construction is compatible
with that of Birch and Merriman [birch]; both constructions give isomorphic
quintic rings from a binary quintic form. Let
$\mathcal{F}_{5}^{\operatorname{bin}}$ be the set of binary quintic forms with
integral coefficients and nonzero discriminant.

######  Theorem 6.5.

Then $(\mathcal{F}_{5}^{\operatorname{bin}},H_{c})$ corresponds to

$(1/8,1/8,1/8,1/8,1/4,1/4,1/4,3/8,3/8).$

As in the case of monogenic cubics, monogenic quartics, and quartic rings
coming from binary quartic forms (Corollary 4.4, Theorem 5.6, Theorem 5.7),
there is a good reason to expect that $\mathcal{F}_{5}^{\operatorname{bin}}$
(equipped with the coefficient height) corresponds to the point above; this
reason comes from Fess’ characterization of pairs $(R,S)$ arising from binary
quintic forms. For a rank $n$ ring $R$, let
$\widetilde{R}\coloneqq\operatorname{Hom}(R/\mathbb{Z},\mathbb{Z})$. Recall
that a pair $(R,S)$ comes equipped with a _sextic resolvent map_
$\varphi\colon\widetilde{S}\otimes\widetilde{S}\rightarrow\widetilde{R}$. For
an basis $\\{v_{0}=1,v_{1},\dots,v_{n-1}\\}$ of a nondegenerate rank $n$ ring,
let $\\{v^{*}_{0},v^{*}_{1},\dots,v^{*}_{n-1}\\}$ denote the dual basis with
respect to the trace pairing.

######  Theorem 6.6 (Fess [fess], Theorem 5.1.4).

Let $(R,S)$ be a nondegenerate pair. There exists
$f\in\operatorname{Sym}^{5}\mathbb{Z}^{2}$ giving rise to the pair $(R,S)$ if
and only if $S$ has a basis $\\{1,w_{1},\dots,w_{5}\\}$ such that

$(w_{1},w_{2},w_{3})\equiv
8\operatorname{Disc}(R)((w^{*}_{5})^{2},2w^{*}_{4}w^{*}_{5},(w^{*}_{4})^{2})\
(\mathrm{mod}\ \mathbb{Q})$

and $\varphi(w_{4}^{*},w_{5}^{*})=0$.

From this classification, one might naively guess the following: if a pair
$(R,S)$ arises from a nondegenerate binary quintic form and
$\\{1,w_{1},\dots,w_{5}\\}$ is a Minkowski basis of $S$, then it is likely
that

$\displaystyle\operatorname{\lvert}w_{1}\operatorname{\rvert}$
$\displaystyle\asymp\Delta\operatorname{\lvert}w_{5}\operatorname{\rvert}^{-2}$
$\displaystyle\operatorname{\lvert}w_{2}\operatorname{\rvert}$
$\displaystyle\asymp\Delta\operatorname{\lvert}w_{4}\operatorname{\rvert}^{-1}\operatorname{\lvert}w_{5}\operatorname{\rvert}^{-1}$
$\displaystyle\operatorname{\lvert}w_{3}\operatorname{\rvert}$
$\displaystyle\asymp\Delta\operatorname{\lvert}w_{4}\operatorname{\rvert}^{-2}.$

For lack of other information, one might also guess that
$\operatorname{\lvert}w_{4}\operatorname{\rvert}$ is of similar size to
$\operatorname{\lvert}w_{5}\operatorname{\rvert}$ and that the successive
minima of the quintic ring $R$ are roughly equal as well. Theorem 6.5
formalizes this heuristic. A similar heuristic informs the monogenic case. Let
$\mathcal{F}_{5}^{\operatorname{mon}}$ be the set of binary quintic forms with
integral coefficients and nonzero discriminant.

######  Theorem 6.7.

Then $(\mathcal{F}_{5}^{\operatorname{mon}},H_{r})$ corresponds to

$(1/20,2/20,3/20,4/20,4/20,5/20,6/20,7/20,8/20).$

Observe that, as case of Theorem 6.5, the vertex

$(1/20,2/20,3/20,4/20,4/20,5/20,6/20,7/20,8/20)$

satisfies the conditions:

$\displaystyle q_{1}$ $\displaystyle=1-2q_{5}$ $\displaystyle q_{2}$
$\displaystyle=1-q_{4}-q_{5}$ $\displaystyle q_{3}$ $\displaystyle=1-2q_{4}$

##  7\. The parametrization of quintic rings

Our primary tool will be Bhargava’s parametrization of quintic rings; we now
briefly summarize this parametrization. Let
$V_{\mathbb{R}}=\mathbb{R}^{4}\otimes\wedge^{2}\mathbb{R}^{5}$. We write an
element $\mathcal{A}=(A_{1},A_{2},A_{3},A_{4})$ as a quadruple of $5\times 5$
real alternating matrices as follows:

$A_{k}=\begin{bmatrix}0&a^{k}_{12}&a^{k}_{13}&a^{k}_{14}&a^{k}_{15}\\\
-a^{k}_{12}&0&a^{k}_{23}&a^{k}_{24}&a^{k}_{25}\\\
-a^{k}_{13}&-&0&a^{k}_{34}&a^{k}_{35}\\\
-a^{k}_{14}&-a^{k}_{24}&-a^{k}_{34}&0&a^{k}_{45}\\\
-a^{k}_{15}&-a^{k}_{25}&-a^{k}_{35}&-a^{k}_{45}&0\end{bmatrix}.$

An element $\mathcal{A}$ is said to be _integral_ if
$a^{k}_{ij}\in\mathbb{Z}$, and the lattice of integral elements is denoted
$V_{\mathbb{Z}}$. The group
$\operatorname{GL}_{4}(\mathbb{R})\times\operatorname{GL}_{5}(\mathbb{R})$
acts on $V_{\mathbb{R}}$ in a natural way; an element $(g_{4},g_{5})$ sends
$\mathcal{A}$ to

$g_{4}\begin{bmatrix}g_{5}A_{1}g_{5}^{t}\\\ g_{5}A_{2}g_{5}^{t}\\\
g_{5}A_{3}g_{5}^{t}\\\ g_{5}A_{4}g_{5}^{t}\end{bmatrix}.$

The subgroup
$G(\mathbb{Z})=\operatorname{GL}_{4}(\mathbb{Z})\times\operatorname{GL}_{5}(\mathbb{Z})$
acts on $V_{\mathbb{R}}$ and preserves the lattice $V_{\mathbb{Z}}$. The
action of $G_{\mathbb{Z}}$ on $V_{\mathbb{Z}}$ has a unique polynomial
invariant, denoted $\operatorname{Disc}(\mathcal{A})$. The orbits of
$G_{\mathbb{Z}}$ acting on $V_{\mathbb{Z}}$ parametrize quintic rings equipped
with a sextic resolvent ring; see [BhargavaHCL4] for more details.

######  Theorem 7.1 (Bhargava, [BhargavaHCL4]).

There is a canonical, discriminant-preserving bijection between
$V_{\mathbb{Z}}$ and isomorphism classes of pairs $(R,S)$ where $R$ is a based
quintic ring and $S$ is a based sextic resolvent ring of $R$. Under the action
of $G_{\mathbb{Z}}$, this bijection descends to a bijection between the set of
$G_{\mathbb{Z}}$ orbits of $V_{\mathbb{Z}}$ and the set of isomorphism classes
of pairs $(R,S)$ where $R$ is a quintic ring and $S$ is a sextic resolvent
ring of $R$.

We now describe a few important details of this bijection. Let $\widetilde{R}$
and $\widetilde{S}$ denote $\operatorname{Hom}(R/\mathbb{Z},\mathbb{Z})$ and
$\operatorname{Hom}(S/\mathbb{Z},\mathbb{Z})$ respectively. A sextic resolvent
ring $S$ of a quintic ring comes equipped with an alternating map
$\widetilde{S}\otimes\widetilde{S}\rightarrow\widetilde{R}$. Choosing bases
for $R/\mathbb{Z}$ and $S/\mathbb{Z}$ gives rise to an element of
$V_{\mathbb{Z}}$. Conversely, given an element $\mathcal{A}\in
V_{\mathbb{Z}}$, we obtain bases for $R$ and $S$ via certain explicit
polynomials in the $a^{k}_{ij}$ that form the structure coefficients for these
bases.

###  7.1. Reduction theory of quintic rings

We now fix some notation we use throughout this section. Unless otherwise
specified, let $\mathcal{A}=(A_{1},A_{2},A_{3},A_{4})$ denote an element of
$V_{\mathbb{Z}}$ and let $(R,S)$ denote the corresponding quintic ring and
sextic resolvent ring. Let
$\Delta=\operatorname{\lvert}\operatorname{Disc}(\mathcal{A})\operatorname{\rvert}=\operatorname{\lvert}\operatorname{Disc}(R)\operatorname{\rvert}$.
For a basis $\\{v_{0}=1,v_{1},\dots,v_{n-1}\\}$ of a nondegenerate rank $n$
ring $R$, let $\\{v^{*}_{0},v^{*}_{1},\dots,v^{*}_{n-1}\\}$ denote the dual
basis with respect to the trace pairing and let
$R^{*}=\operatorname{Hom}(R,\mathbb{Z})$. Unless otherwise specified,
$\\{1=v_{0},\dots,v_{4}\\}$ and $\\{1=w_{0},\dots,w_{5}\\}$ will denote bases
of the quintic ring $R$ and the sextic resolvent ring $S$ respectively. Given
a pair $(R,S)$ and bases $\\{1=v_{0},\dots,v_{4}\\}$ and
$\\{1=w_{0},\dots,w_{5}\\}$ respectively, we obtain a tensor $\mathcal{A}$ by
writing the resolvent
$\varphi\colon\widetilde{S}\otimes\widetilde{S}\rightarrow\widetilde{R}$ map
in terms of the bases $\\{v^{*}_{4},\dots,v^{*}_{1}\\}$ and
$\\{w^{*}_{5},\dots,w^{*}_{1}\\}$ of $\widetilde{R}$ and $\widetilde{S}$
respectively141414This ordering is very important! We encourage the reader to
keep this ordering of the basis elements in mind throughout this section.. In
this section, we develop the relationship between:

  1. (1)

pairs $(R,S)$ equipped with “small” bases;

  2. (2)

and tensors $\mathcal{A}$ with “small” coefficients.

We will need to relate the lengths of $\\{v^{*}_{1},\dots,v^{*}_{4}\\}$ and
$\\{w^{*}_{1},\dots,w^{*}_{5}\\}$ to the lengths of
$\\{1,v_{1},\dots,v_{4}\\}$ and $\\{1,w_{1},\dots,w_{5}\\}$, when the latter
two bases are “short”; the content of the following lemma consists of the
relationship between these two lengths.

######  Lemma 7.2.

Let $R$ be any nondegenerate rank $n$ ring and let
$\\{1=v_{0},\dots,v_{n-1}\\}$ be a basis of $R$. Let $\theta$ be a positive
real number such that the angle between any $v_{i}$ and $v_{j}$ in
$\mathbb{R}^{n}$ is bounded below by $\theta$ in the Minkowski embedding of
$R$. For all $0\leq i\leq n-1$, we have

$\operatorname{\lvert}v_{i}\operatorname{\rvert}\asymp_{n,\theta}\lambda_{i}(R)\asymp_{n,\theta}\lambda_{n-1-i}(R^{*})^{-1}\asymp_{n,\theta}\operatorname{\lvert}v_{i}^{*}\operatorname{\rvert}^{-1}.$

###### Proof.

Let $r$ be the number of nonzero homomorphisms of $R$ into $\mathbb{R}$ and
let $s$ be the number of complex conjugate pairs of nonzero homomorphisms of
$R$ into $\mathbb{C}$ whose image does not lie in $\mathbb{R}$. The Minkowski
embedding of $R$ into $\mathbb{R}^{n}$ is given via the composition of
$R\rightarrow\mathbb{R}^{r}\oplus\mathbb{C}^{s}$ with the map

$\displaystyle\mathbb{R}^{r}\oplus\mathbb{C}^{s}$
$\displaystyle\rightarrow\mathbb{R}^{n}$
$\displaystyle(y_{1},\dots,y_{r},z_{1},\dots,z_{s})$
$\displaystyle\rightarrow(y_{1},\dots,y_{r},\operatorname{Re}(z_{1}),\operatorname{Im}(z_{1}),\dots,\operatorname{Re}(z_{s}),\operatorname{Im}(z_{s})).$

Because all norms on finite dimensional vector spaces are equivalent, it
suffices to suppose that all lengths are taken with respect to the Euclidean
norm on $\mathbb{R}^{n}$, i.e. let
$\operatorname{\lvert}\cdot\operatorname{\rvert}$ denote the Euclidean norm.
Note that $v_{0},\dots,v_{n-1}$ form a set of linearly independent vectors in
$\mathbb{R}^{n}$ via the Minkowski embedding; let
$\\{v_{0}^{\vee},\dots,v_{n-1}^{\vee}\\}$ denote the dual basis with respect
to the standard Euclidean norm. We will show that the length of $v_{i}^{*}$ is
comparable to the length of $v_{i}^{\vee}$. Afterwards, we compute the lengths
of $v_{i}^{\vee}$.

On $\mathbb{R}^{n}$, the trace form is given by the quadratic form

$q(x_{1},\dots,x_{n})=\sum_{1\leq i\leq r}x_{i}^{2}+\sum_{1\leq i\leq
s}(2x_{r+2i-1}^{2}-2x_{r+2i}^{2}).$

Let $\sigma\colon\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be the map sending

$(y_{1},\dots,y_{r},z_{1},\dots,z_{s})\rightarrow(y_{1},\dots,y_{r},\frac{1}{2}\overline{z_{1}},\dots,\frac{1}{2}\overline{z_{s}}).$

For $0\leq i\leq n-1$ write $v_{i}=(v_{1,i},\dots,v_{n,i})$ and
$v^{\vee}_{i}=(v^{\vee}_{1,i},\dots,v^{\vee}_{n,i})$.

$\displaystyle q(\sigma(v_{i}^{\vee}),v_{j})$ $\displaystyle=\sum_{1\leq k\leq
r}v^{\vee}_{k,i}v_{k,j}+\sum_{1\leq k\leq
s}(2\frac{1}{2}v^{\vee}_{r+2k-1,i}v_{r+2k-1,j}-2\frac{-1}{2}v^{\vee}_{r+2k,i}v_{r+2k,j})$
$\displaystyle=\sum_{1\leq k\leq r}v^{\vee}_{k,i}v_{k,j}+\sum_{1\leq k\leq
s}(v^{\vee}_{r+2k-1,i}v_{r+2k-1,j}+v^{\vee}_{r+2k,i}v_{r+2k,j})$
$\displaystyle=\delta_{ij}$

Thus, $\sigma(v_{i}^{\vee})=v_{i}^{*}$ for all $0\leq i\leq n-1$. Moreover,
the angle between $v_{i}^{\vee}$ and $v_{i}$ is $\leq c_{\theta}\pi/2$ where
$0\leq c_{\theta}<1$ and $c_{\theta}$ is dependent only on $\theta$. Hence

$\displaystyle 1$ $\displaystyle=v_{i}^{\vee}\cdot v_{i}$ because
$v_{i}^{\vee}\cdot v_{j}=\delta_{ij}$
$\displaystyle\asymp_{\theta}\operatorname{\lvert}v_{i}^{\vee}\operatorname{\rvert}\operatorname{\lvert}v_{i}\operatorname{\rvert}$
because of the angle bound $c_{\theta}$
$\displaystyle=\operatorname{\lvert}\sigma^{-1}(v_{i}^{*})\operatorname{\rvert}\operatorname{\lvert}v_{i}\operatorname{\rvert}$
because $\sigma(v_{i}^{\vee})=v_{i}^{*}$
$\displaystyle\asymp_{n}\operatorname{\lvert}v_{i}^{*}\operatorname{\rvert}\operatorname{\lvert}v_{i}\operatorname{\rvert}.$

Moreover,

$\operatorname{Covol}(R)^{-1}\asymp_{n,\theta}\prod_{i=0}^{n-1}\operatorname{\lvert}v_{i}\operatorname{\rvert}^{-1}\asymp_{n,\theta}\prod_{i=0}^{n-1}\operatorname{\lvert}v_{i}^{*}\operatorname{\rvert}\gg_{n}\prod_{i=0}^{n-1}\lambda_{i}(R^{*})\asymp_{n}\operatorname{Covol}(R^{*})\asymp_{n}\operatorname{Covol}(R)^{-1}.$

Hence,
$\operatorname{\lvert}v_{i}^{*}\operatorname{\rvert}\asymp_{n,\theta}\lambda_{n-1-i}(R^{\vee})$,
which completes the proof. ∎

We now show that if the sextic resolvent map is expressed with respect to
“short” bases of $R$ and $S$, then the coefficients of the map are “small”. To
do so, we will need the following explicit description of the sextic resolvent
map. Let $\sigma_{1},\dots,\sigma_{6}$ be the $6$ nonzero homomorphisms of $S$
into the complex numbers. The sextic resolvent map
$\varphi\colon\widetilde{S}\otimes\widetilde{S}\rightarrow\widetilde{R}$ is
given by the formula:

$\varphi(u\otimes
v)=\frac{\sqrt{\operatorname{Disc}(S)}}{48\operatorname{Disc}(R)}\;\;\begin{array}[]{|c
c c|}1&1&1\\\
\sigma_{1}(u)+\sigma_{2}(u)&\sigma_{3}(u)+\sigma_{6}(u)&\sigma_{4}(u)+\sigma_{5}(u)\\\
\sigma_{1}(v)+\sigma_{2}(v)&\sigma_{3}(v)+\sigma_{6}(v)&\sigma_{4}(v)+\sigma_{5}(v)\end{array}$

######  Lemma 7.3.

If $\\{1,v_{1},\dots,v_{4}\\}$ and $\\{1,w_{1},\dots,w_{5}\\}$ are Minkowski
bases of $R$ and $S$ respectively, then for any $1\leq i,j\leq 4$ and any
$1\leq k\leq 5$, we have:

$\operatorname{\lvert}a^{k}_{ij}\operatorname{\rvert}\ll\frac{\Delta^{1/2}\operatorname{\lvert}v_{5-k}\operatorname{\rvert}}{\operatorname{\lvert}w_{6-i}\operatorname{\rvert}\operatorname{\lvert}w_{6-j}\operatorname{\rvert}}$

###### Proof.

From the definition of the sextic resolvent map, we have that:

$\displaystyle\operatorname{\lvert}a^{k}_{ij}\operatorname{\rvert}$
$\displaystyle\ll\frac{\operatorname{\lvert}\varphi(w^{*}_{6-i}\otimes
w^{*}_{6-j})\operatorname{\rvert}}{\operatorname{\lvert}v^{*}_{5-k}\operatorname{\rvert}}$
$\displaystyle\ll\frac{\Delta^{1/2}\operatorname{\lvert}w^{*}_{6-i}\operatorname{\rvert}\operatorname{\lvert}w^{*}_{6-j}\operatorname{\rvert}}{\operatorname{\lvert}v^{*}_{5-k}\operatorname{\rvert}}$
by the explicit description of $\varphi$
$\displaystyle\ll\frac{\Delta^{1/2}\operatorname{\lvert}v_{5-k}\operatorname{\rvert}}{\operatorname{\lvert}w_{6-i}\operatorname{\rvert}\operatorname{\lvert}w_{6-j}\operatorname{\rvert}}$
by Lemma 7.2

∎

Let $p=(p_{1},\dots,p_{4},q_{1},\dots,q_{5})\in\mathbb{R}^{9}$ be any point.
Let $B(p,X)$ denote the set of elements $\mathcal{A}\in V_{\mathbb{R}}$ such
that

$\operatorname{\lvert}a^{k}_{ij}\operatorname{\rvert}\leq\exp[X,1/2+p_{5-k}-q_{6-i}-q_{6-j}].$

We now fix more notation we use throughout this section. Let $\epsilon$ denote
a positive real number, let $C,D$ denote real numbers $\geq 1$, and let $X$
denote a real number $>1$. Recall that we say the pair $(R,S)$ is
$(\epsilon,X)$-close to $p$ if:

$\displaystyle\max_{1\leq i\leq
4}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(R)-p_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}$ $\displaystyle\max_{1\leq i\leq
5}\\{\operatorname{\lvert}\log_{\Delta}\lambda_{i}(S)-q_{i}\operatorname{\rvert}\\}$
$\displaystyle\leq\frac{\epsilon}{\log X}$

For an integral element $\mathcal{A}$, we say $(\mathcal{A},p)$ are
$(\epsilon,X)$-close if the corresponding pair $(R,S)$ is $(\epsilon,X)$-close
to $p$. We say an integral element $\mathcal{A}$ _comes from Minkowski bases_
if there exist Minkowski bases of the corresponding rings giving rise to
$\mathcal{A}$.

######  Lemma 7.4.

There exists $C$ such that for all $\epsilon$, $\mathcal{A}$ and $p$ the
following statement holds: if $\Delta>1$ and $(\mathcal{A},p)$ are
$(\epsilon,\Delta)$-close and $\mathcal{A}$ comes from Minkowski bases, then
$\mathcal{A}\in Ce^{3\epsilon}B(p,\Delta)$.

###### Proof.

Let $\\{1,v_{1},\dots,v_{4}\\}$ and $\\{1,w_{1},\dots,w_{5}\\}$ be the
Minkowski bases of $(R,S)$ giving rise to $\mathcal{A}$. We have:

$\displaystyle\operatorname{\lvert}a^{k}_{ij}\operatorname{\rvert}$
$\displaystyle\ll\frac{\Delta^{1/2}\operatorname{\lvert}v_{5-k}\operatorname{\rvert}}{\operatorname{\lvert}w_{6-i}\operatorname{\rvert}\operatorname{\lvert}w_{6-j}\operatorname{\rvert}}$
by Lemma 7.3
$\displaystyle\asymp\frac{\Delta^{1/2}\lambda_{5-k}(R)}{\lambda_{6-i}(S)\lambda_{6-j}(S)}$
by Lemma 7.2 $\displaystyle\leq
e^{3\epsilon}\exp[X,1/2+p_{5-k}-q_{6-i}-q_{6-j}]$ because $(\mathcal{A},p)$ is
$(\epsilon,\Delta)$-close

∎

######  Lemma 7.5.

There exists $D$ such that for all $p$, $X$, $C$ and $\mathcal{A}$, the
following statement holds: if $p$ satisfies (48) and $\mathcal{A}\in CB(p,X)$
then

$\\#\\{\mathcal{A}^{\prime}\in DCB(p,X)\cap
V_{\mathbb{Z}}\mid\;(R,S)\simeq(R^{\prime},S^{\prime})\\}\geq\exp[X,\sum_{1\leq
i<j\leq 4}(p_{j}-p_{i})+\sum_{1\leq i<j\leq 5}(q_{j}-q_{i})]\\}.$

where here $(R^{\prime},S^{\prime})$ denotes the quintic ring and sextic
resolvent ring arising from tensor $\mathcal{A}^{\prime}$.

###### Proof.

Recall that $V_{\mathbb{Z}}$ admits an action of
$G_{\mathbb{Z}}=\operatorname{GL}_{4}(\mathbb{Z})\times\operatorname{GL}_{5}(\mathbb{Z})$.
Act by elements of $G_{\mathbb{Z}}$ of the form

$\Bigg{(}\begin{bmatrix}1&n_{12}&n_{13}&n_{14}\\\ 0&1&n_{23}&n_{24}\\\
0&0&1&n_{34}\\\ 0&0&0&1\\\ \end{bmatrix},\begin{bmatrix}1&0&0&0&0\\\
m_{12}&1&0&0&0\\\ m_{13}&m_{23}&1&0&0\\\ m_{14}&m_{24}&m_{34}&1&0\\\
m_{15}&m_{25}&m_{35}&m_{45}&1\\\ \end{bmatrix}\Bigg{)}$

where
$\operatorname{\lvert}n_{ij}\operatorname{\rvert}\leq\lceil\exp[X,p_{5-i}-p_{5-j}]\rceil$
and
$\operatorname{\lvert}m_{ij}\operatorname{\rvert}\leq\lceil\exp[X,q_{6-i}-q_{6-j}]\rceil$.
∎

Choose $\mathcal{A}\in V_{\mathbb{Z}}$ and let
$\\{v_{0}=1,v_{1},\dots,v_{4}\\}$ and $\\{w_{0}=1,w_{1},\dots,w_{5}\\}$ be the
explicit bases of $R$ and $S$ obtained from $\mathcal{A}$ via the explicit
multiplication coefficients in [BhargavaHCL4].

######  Lemma 7.6.

For all $\mathcal{A}$, $C$, $p$, and $X$, the following statement holds: if
$p$ satisfies (48) and $\mathcal{A}\in CB(p,X)$, then
$\operatorname{\lvert}v_{i}\operatorname{\rvert}\ll_{C}X^{p_{i}}$ for
$i=1,2,3,4$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{C}X^{q_{i}}$ for
$i=1,2,3,4,5$.

###### Proof.

Using the fact that $\mathcal{A}\in B(p,X)$, we may bound the size of the
multiplication coefficients given in [BhargavaHCL4]. We obtain:

$\operatorname{\lvert}v_{i}\operatorname{\rvert}^{2}\ll_{C}\max_{0\leq j\leq
4}\\{X^{2p_{i}-p_{j}}\operatorname{\lvert}v_{j}\operatorname{\rvert}\\}$
$\operatorname{\lvert}w_{i}\operatorname{\rvert}^{2}\ll_{C}\max_{0\leq j\leq
5}\\{X^{2q_{i}-q_{j}}\operatorname{\lvert}w_{j}\operatorname{\rvert}\\}$

Applying Lemma 11.2 completes the proof. ∎

######  Lemma 7.7.

For all $p$ satisfying (48), all $X,C$, and all $\mathcal{A}$, the following
statement holds: if $\mathcal{A}\in V_{\mathbb{Z}}\cap CB(p,X)$ and
$X/2\leq\Delta\leq X$, then there are

$\\#\\{\mathcal{A}^{\prime}\in CB(p,X)\cap
V_{\mathbb{Z}}\mid\;(R,S)\simeq(R^{\prime},S^{\prime})\\}\ll_{C}\exp[X,\sum_{1\leq
i<j\leq 4}(p_{j}-p_{i})+\sum_{1\leq i<j\leq 5}(q_{j}-q_{i})].$

###### Proof.

Let $\\{v_{0}=1,v_{1},v_{2},v_{3},v_{4}\\}$ and
$\\{w_{0}=1,w_{1},\dots,w_{5}\\}$ be the explicit bases of $R$ and $S$
obtained from $\mathcal{A}$. By Lemma 7.6, we have
$\operatorname{\lvert}v_{i}\operatorname{\rvert}\ll_{C}X^{p_{i}}$ for
$i=1,2,3,4$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{C}X^{q_{i}}$ for
$i=1,2,3,4,5$. Because

$X^{1/2}\asymp\Delta^{1/2}\ll\prod_{i=1}^{4}\operatorname{\lvert}v_{i}\operatorname{\rvert}\ll_{C}X^{\sum_{i=1}^{4}p_{i}}=X^{1/2}$

and

$X^{3/2}\asymp\Delta^{3/2}\ll\prod_{i=1}^{5}\operatorname{\lvert}w_{i}\operatorname{\rvert}\ll_{C}X^{\sum_{i=1}^{5}q_{i}}=X^{3/2}$

we have $\operatorname{\lvert}v_{i}\operatorname{\rvert}\asymp_{C}X^{p_{i}}$
for $i=1,2,3,4$ and
$\operatorname{\lvert}w_{i}\operatorname{\rvert}\asymp_{C}X^{q_{i}}$ for
$i=1,2,3,4,5$.

Suppose $\mathcal{A}^{\prime}\in CB(p,X)\cap V_{\mathbb{Z}}$ is such that
$(R^{\prime},S^{\prime})\simeq(R,S)$. Let
$\\{1,v^{\prime}_{1},\dots,v^{\prime}_{4}\\}$ and
$\\{1,w^{\prime}_{1},\dots,w^{\prime}_{5}\\}$ denote the bases of $R^{\prime}$
and $S^{\prime}$ given by $\mathcal{A}^{\prime}$. By Lemma 7.6, we have
$\operatorname{\lvert}v^{\prime}_{i}\operatorname{\rvert}\asymp_{C}X^{p_{i}}$
for $i=1,2,3,4$ and
$\operatorname{\lvert}w^{\prime}_{i}\operatorname{\rvert}\asymp_{C}X^{q_{i}}$
for $i=1,2,3,4,5$. Moreover, write

$\displaystyle v^{\prime}_{i}$
$\displaystyle=c_{i0}+\sum_{j=1}^{4}c_{ij}v_{j}$ $\displaystyle
w^{\prime}_{i}$ $\displaystyle=d_{i0}+\sum_{j=1}^{5}d_{ij}w_{j}.$

The tensor $\mathcal{A}^{\prime}$ is determined by the choices of $c_{ij}$ and
$d_{ij}$. Letting $p_{0}=q_{0}=0$, we have:

$\displaystyle X^{p_{i}}$
$\displaystyle\asymp_{C}\operatorname{\lvert}v^{\prime}_{i}\operatorname{\rvert}\asymp_{C}\max_{0\leq
j<5}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}\operatorname{\lvert}v_{j}\operatorname{\rvert}\\}\asymp_{C}\max\\{0\leq
j<5\\}\\{\operatorname{\lvert}c_{ij}\operatorname{\rvert}X^{p_{j}}\\}$
$\displaystyle X^{q_{i}}$
$\displaystyle\asymp_{C}\operatorname{\lvert}w^{\prime}_{i}\operatorname{\rvert}\asymp_{C}\max_{0\leq
j<6}\\{\operatorname{\lvert}d_{ij}\operatorname{\rvert}\operatorname{\lvert}w_{j}\operatorname{\rvert}\\}\asymp_{C}\max\\{0\leq
j<6\\}\\{\operatorname{\lvert}d_{ij}\operatorname{\rvert}X^{q_{j}}\\}$

Thus if $j\leq i$, the number of choices for $c_{ij}$ and $d_{ij}$ is
$\ll_{C}1$. If $j>i$, the number of choices for $c_{ij}$ is
$\ll_{C}X^{p_{j}-p_{i}}$ and the number of choices for $d_{ij}$ is
$\ll_{C}X^{q_{j}-q_{i}}$. Hence, there are $\ll_{C}\exp[X,\sum_{1\leq i<j\leq
4}(p_{j}-p_{i})+\sum_{1\leq i<j\leq 5}(q_{j}-q_{i})]$ distinct integral
$\mathcal{A}^{\prime}\in CB(p,X)$ such that
$(R^{\prime},S^{\prime})\simeq(R,S)$. ∎

######  Lemma 7.8.

For all $p$ satisfying (48), all $X,C$, and all $\mathcal{A}$, the following
statement holds: if $\mathcal{A}\in V_{\mathbb{Z}}\cap CB(p,X)$ and
$\frac{X}{2}\leq\Delta$ then

$\lambda_{i}(R)\asymp_{C}\Delta^{p_{i}}$

for $i=1,2,3,4$ and

$\lambda_{i}(C)\asymp_{C}\Delta^{q_{i}}$

for $i=1,2,3,4,5$.

###### Proof.

We have

$\displaystyle\prod_{i=1}^{4}\lambda_{i}(R)$
$\displaystyle\ll\prod_{i=1}^{4}\operatorname{\lvert}v_{i}\operatorname{\rvert}$
by definition of successive minima
$\displaystyle\ll_{C}\prod_{i=1}^{4}X^{p_{i}}$ by Lemma 7.6
$\displaystyle=X^{1/2}$ because $p_{1}+p_{2}+p_{3}+p_{4}=1/2$
$\displaystyle\asymp\Delta^{1/2}$ because $\Delta\geq X/2$
$\displaystyle\asymp\prod_{i=1}^{4}\lambda_{i}(R).$ by Minkowski’s second
theorem

Thus $\lambda_{i}(R)\asymp_{C}\Delta^{p_{i}}$ for $i=1,2,3,4$. The same
argument shows, for the sextic resolvent, that
$\lambda_{i}(S)\asymp_{C}\Delta^{q_{i}}$ for $i=1,2,3,4,5$. ∎

##  8\. Computing $\operatorname{Poly}_{5}(G)$

Let $K$ denote a quintic étale algebra over $\mathbb{Q}$ and $L$ denote the
sextic resolvent algebra of $K$. Let $K_{0}$ denote the subspace of $K$ with
trace $0$, and let $L_{0}$ denote the subspace of $L$ with trace $0$. Let
$\varphi\colon L_{0}\otimes L_{0}\rightarrow K_{0}$ denote the alternating
bilinear form given by the sextic resolvent map. A _flag
$(\mathcal{F},\mathcal{G})$ of $(K,L)$_ is a pair of sequences
$\mathcal{F}=(F_{0},\dots,F_{5})$ and $\mathcal{G}=(G_{0},\dots,G_{4})$ of
$\mathbb{Q}$-vector spaces such that

$\\{0\\}=F_{0}\subset F_{1}\subset F_{2}\subset\dots\subset F_{5}=L_{0}$
$\\{0\\}=G_{0}\subset G_{1}\subset G_{1}\subset\dots\subset G_{4}=K_{0}$

and $\dim_{\mathbb{Q}}F_{i}=i$ and $\dim_{\mathbb{Q}}G_{i}=i$. Recall that for
an integer $n>1$, the set $[n]=\\{0,\dots,n-1\\}$. Associate to the flag
$(\mathcal{F},\mathcal{G})$ the function

$\displaystyle T_{\mathcal{F},\mathcal{G}}\colon[6]\times[6]$
$\displaystyle\longrightarrow[5]$ $\displaystyle(i\;\;,\;\;j)\;$
$\displaystyle\longmapsto\min\\{k\in[5]\mid\varphi\langle F_{i}\otimes
F_{j}\rangle\subseteq G_{k}\\}.$

######  Definition 8.1.

The polytope $P_{\mathcal{F},\mathcal{G}}$ is the set of
$p=(p_{1},\dots,p_{4},q_{1},\dots,q_{5})\in\mathbb{R}^{9}$ satisfying (48) and
the following condition: for all $1\leq i\leq j\leq 5$, if
$T_{\mathcal{F},\mathcal{G}}(i,j)>0$ then

$p_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}+1/2\geq q_{6-i}+q_{6-j}.$

The proof of the following theorem is almost exactly analogous to that of
Theorem 5.20.

######  Theorem 8.2.

We have
$\operatorname{Poly}_{5}(G)=\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$
where $(\mathcal{F},\mathcal{G})$ ranges across flags of pairs $(K,L)$ with
the property that $\operatorname{Gal}(K)\simeq G$ as a permutation group.

###### Proof.

We first show that $\operatorname{Poly}_{5}(G)$ is contained in
$\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$. By
Minkowski’s second theorem and the definition of successive minima,
$\operatorname{Poly}_{5}(G)$ satisfies the linear inequalities given in (48).
Let $R$ be any nondegenerate quintic $G$-ring and let $S$ be a sextic
resolvent ring of $G$. Let $K=R\otimes\mathbb{Q}$ and let
$L=S\otimes\mathbb{Q}$. Picking Minkowski bases $\\{v_{0}=1,\dots,v_{4}\\}$
and $\\{w_{0}=1,\dots,w_{5}\\}$ of $R$ and $S$ gives rise to dual bases
$\\{v^{*}_{0},\dots,v^{*}_{4}\\}$ and $\\{w^{*}_{0},\dots,w^{*}_{5}\\}$ of $R$
and $S$ respectively. These give rise to a flag $(\mathcal{F},\mathcal{G})$ of
$(K,L)$ given by $F_{0}=G_{0}\coloneqq\\{0\\}$ and

$\displaystyle F_{i}$ $\displaystyle\coloneqq\mathbb{Q}\langle
w^{*}_{5},\dots,w^{*}_{6-i}\rangle$ $\displaystyle G_{i}$
$\displaystyle\coloneqq\mathbb{Q}\langle v^{*}_{4},\dots,v^{*}_{5-i}\rangle.$

Choose $1\leq i\leq j\leq 5$ and suppose $T_{\mathcal{F},\mathcal{G}}(i,j)>0$.
Then if $T_{\mathcal{F},\mathcal{G}}(i,j)=1$ then

$0\neq\varphi\langle\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-i}\rangle\otimes\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-j}\rangle\rangle\subseteq\mathbb{Z}\langle
v^{*}_{4},\dots,v^{*}_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}\rangle.$

If $T_{\mathcal{F},\mathcal{G}}(i,j)>1$ then

$\varphi\langle\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-i}\rangle\otimes\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-j}\rangle\rangle\subseteq\mathbb{Z}\langle
v^{*}_{4},\dots,v^{*}_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}\rangle$

and

$\varphi\langle\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-i}\rangle\otimes\mathbb{Z}\langle
w^{*}_{5},\dots,w^{*}_{6-j}\rangle\rangle\not\subseteq\mathbb{Z}\langle
v^{*}_{4},\dots,v^{*}_{5-T_{\mathcal{F},\mathcal{G}}(i,j)+1}\rangle$

In either case we obtain:

$\operatorname{\lvert}v^{*}_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}\operatorname{\rvert}\ll\operatorname{\lvert}w^{*}_{6-i}\operatorname{\rvert}\operatorname{\lvert}w^{*}_{6-j}\operatorname{\rvert}\Delta^{1/2}$

which implies by Lemma 7.2 that

$\operatorname{\lvert}v_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}\operatorname{\rvert}^{-1}\ll\operatorname{\lvert}w_{6-i}\operatorname{\rvert}^{-1}\operatorname{\lvert}w_{6-j}\operatorname{\rvert}^{-1}\Delta^{1/2}.$

Multiplying out the inverses, we obtain

$\operatorname{\lvert}w_{6-i}\operatorname{\rvert}\operatorname{\lvert}w_{6-j}\operatorname{\rvert}\ll\operatorname{\lvert}v_{5-T_{\mathcal{F},\mathcal{G}}(i,j)}\operatorname{\rvert}\Delta^{1/2}.$

So, $\operatorname{Poly}_{5}(G)$ is contained in the set
$\bigcup_{(\mathcal{F},\mathcal{G})}P_{\mathcal{F},\mathcal{G}}$ where
$(\mathcal{F},\mathcal{G})$ ranges across flags of $(K,L)$ with the property
that $\operatorname{Gal}(K)\simeq G$.

Now let $(\mathcal{F},\mathcal{G})$ be a flag of a pair $(K,L)$ with the
property that $\operatorname{Gal}(K)\simeq G$. Choose bases
$\\{v_{0}=1,v_{1},\dots,v_{4}\\}$ and $\\{w_{0}=1,w_{1},\dots,w_{5}\\}$ of $K$
and $L$ giving rise to bases $\\{v^{*}_{4},\dots,v^{*}_{1}\\}$ and
$\\{w^{*}_{5},\dots,w^{*}_{1}\\}$ of $K_{0}$ and $L_{0}$. These bases give
rise to a flag $(\mathcal{F},\mathcal{G})$ of $(K,L)$. Moreover, the dual
bases give rise to a tensor $\mathcal{A}\in V_{\mathbb{R}}$. For a positive
integer $M$, define
$g_{M}\in\operatorname{GL}_{4}(\mathbb{R})\times\operatorname{GL}_{5}(\mathbb{R})$
to be

$g_{M}=\Bigg{(}\begin{bmatrix}M^{p_{4}}&0&0&0\\\ 0&M^{p_{3}}&0&0\\\
0&0&M^{p_{2}}&0\\\ 0&0&0&M^{p_{1}}\\\
\end{bmatrix},\begin{bmatrix}M^{-q_{5}}&0&0&0&0\\\ 0&M^{-q_{4}}&0&0&0\\\
0&0&M^{-q_{3}}&0&0\\\ 0&0&0&M^{-q_{2}}&0\\\ 0&0&0&0&M^{-q_{1}}\\\
\end{bmatrix}\Bigg{)}$

Define $\mathcal{A}_{M}=\sqrt{M}g_{M}\mathcal{A}$. For a rational point $p$ in
the relative interior of $P_{\mathcal{F},\mathcal{G}}$, one may check that
there exists an infinite set $\mathcal{M}\subseteq\mathbb{Z}_{>0}$ such that
for all $M\in\mathcal{M}$ the tensor $\mathcal{A}_{M}\in V_{\mathbb{Z}}$. Each
$M\in\mathcal{M}$ gives rise to a distinct quintic ring and sextic resolvent
ring $(R_{M},S_{M})$ with absolute discriminant $\Delta_{M}$; for each
$M\in\mathcal{M}$, write the bases of $R_{M}$ and $S_{M}$ obtained from
$\mathcal{A}_{M}$ by $\\{1,v_{1,M},\dots,v_{4,M}\\}$ and
$\\{1,w_{1,M},\dots,w_{5,M}\\}$ respectively. Then for $M,N\in\mathcal{M}$, we
have:

$\displaystyle N\Delta_{M}$ $\displaystyle=M\Delta_{N}$ $\displaystyle
N^{p_{i}}v_{i,M}$ $\displaystyle=M^{p_{i}}v_{i,N}$ $\displaystyle
N^{q_{i}}w_{i,M}$ $\displaystyle=M^{q_{i}}w_{i,N}$

Therefore,

$\displaystyle\lim_{M\in\mathcal{M}}\Big{(}\log_{\Delta_{M}}\lambda_{1}(R_{M}),\dots,\log_{\Delta_{M}}\lambda_{4}(R_{M})\Big{)}$
$\displaystyle=(p_{1},p_{2},p_{3},p_{4})$
$\displaystyle\lim_{M\in\mathcal{M}}\Big{(}\log_{\Delta_{M}}\lambda_{1}(S_{M}),\dots,\log_{\Delta_{M}}\lambda_{5}(S_{M})\Big{)}$
$\displaystyle=(q_{1},q_{2},q_{3},q_{4},q_{5})$

Therefore, $p\in\operatorname{Poly}_{5}(G)$. Because
$\operatorname{Poly}_{5}(G)$ is defined to be a set of limit points, it must
be closed; therefore $P_{\mathcal{F},\mathcal{G}}$ is contained in
$\operatorname{Poly}_{5}(G)$, which completes the proof of the theorem. ∎

###### Proof of Theorem 6.1.

Theorem 8.2 expresses $\operatorname{Poly}_{5}(G)$ as a finite union of
polytopes. A computation shows that these polytopes have dimension $7$. ∎

To compute $\operatorname{Poly}_{5}(S_{5})$, we will need the following lemma.

######  Lemma 8.3 (Bhargava [Bhquintic], Lemma 10).

Let $\mathcal{A}\in V_{\mathbb{Q}}$ be an element such that all the variables
in one of the following sets vanish:

  1. (1)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{14},a^{4}_{15},a^{4}_{23},a^{4}_{24},a^{4}_{25}\\}$

  2. (2)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{14},a^{4}_{23},a^{4}_{24},a^{4}_{34}\\}$

  3. (3)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{14},a^{4}_{15}\\}\cup\\{a^{3}_{12},a^{3}_{13},a^{3}_{14},a^{3}_{15}\\}$

  4. (4)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{14},a^{4}_{23},a^{4}_{24}\\}\cup\\{a^{3}_{12},a^{3}_{13},a^{3}_{14},a^{3}_{23},a^{3}_{24}\\}$

  5. (5)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{14}\\}\cup\\{a^{3}_{12},a^{3}_{13},a^{3}_{14}\\}\cup\\{a^{2}_{12},a^{2}_{13},a^{2}_{14}\\}$

  6. (6)

$\\{a^{4}_{12},a^{4}_{13},a^{4}_{23}\\}\cup\\{a^{3}_{12},a^{3}_{13},a^{3}_{23}\\}\cup\\{a^{2}_{12},a^{2}_{13},a^{2}_{23}\\}$

  7. (7)

$\\{a^{4}_{12},a^{4}_{13}\\}\cup\\{a^{3}_{12},a^{3}_{13}\\}\cup\\{a^{2}_{12},a^{2}_{13}\\}\cup\\{a^{1}_{12},a^{1}_{13}\\}$

Then $\mathcal{A}$ is reducible.

###### Proof of Theorem 6.2.

Let $(\mathcal{F},\mathcal{G})$ be any flag of a pair $(K,L)$ such that
$\operatorname{Gal}(K)$ is a transitive subgroup of $S_{5}$. Then the seven
statements of Lemma 8.3 imply that:

  1. (1)

$T_{\mathcal{F},\mathcal{G}}(2,5)=4$

  2. (2)

$T_{\mathcal{F},\mathcal{G}}(3,4)=4$

  3. (3)

$T_{\mathcal{F},\mathcal{G}}(1,5)\geq 3$

  4. (4)

$T_{\mathcal{F},\mathcal{G}}(2,4)\geq 3$

  5. (5)

$T_{\mathcal{F},\mathcal{G}}(1,4)\geq 2$

  6. (6)

$T_{\mathcal{F},\mathcal{G}}(2,3)\geq 2$

  7. (7)

$T_{\mathcal{F},\mathcal{G}}(1,3)\geq 1$

The $7$ statements above imply that $P_{\mathcal{F},\mathcal{G}}$ satisfies
the linear inequalities:

  1. (1)

$p_{1}+1/2\geq q_{4}+q_{1}$

  2. (2)

$p_{1}+1/2\geq q_{3}+q_{2}$

  3. (3)

$p_{2}+1/2\geq q_{5}+q_{1}$

  4. (4)

$p_{2}+1/2\geq q_{4}+q_{2}$

  5. (5)

$p_{3}+1/2\geq q_{5}+q_{2}$

  6. (6)

$p_{3}+1/2\geq q_{4}+q_{3}$

  7. (7)

$p_{4}+1/2\geq q_{5}+q_{3}$

In total, this proves that $\operatorname{Poly}_{5}(\operatorname{Gal}(K))$ is
contained in the region stated in the theorem statement. The converse is shown
via computer calculation by producing an example of
$\mathcal{A}=(A_{1},A_{2},A_{3},A_{4})\in V_{\mathbb{Q}}$ of the form

$\Bigg{(}\begin{bmatrix}-&0&*&*&*\\\ -&-&*&*&*\\\ -&-&-&*&*\\\ -&-&-&-&*\\\
-&-&-&-&-\\\ \end{bmatrix},\begin{bmatrix}-&0&0&*&*\\\ -&-&*&*&*\\\
-&-&-&*&*\\\ -&-&-&-&*\\\ -&-&-&-&-\\\
\end{bmatrix},\begin{bmatrix}-&0&0&0&*\\\ -&-&0&*&*\\\ -&-&-&*&*\\\
-&-&-&-&*\\\ -&-&-&-&-\\\ \end{bmatrix},\begin{bmatrix}-&0&0&0&0\\\
-&-&0&0&*\\\ -&-&-&*&*\\\ -&-&-&-&*\\\ -&-&-&-&-\\\ \end{bmatrix}\Bigg{)}$

defining an $S_{5}$-quintic field. ∎

##  9\. Computing density functions of quintic rings

To make our formulas shorter, we use the following notation in this section.
Define:

$f(p)\coloneqq 1+\sum_{1\leq i<j\leq 5,1\leq k\leq
4}\max\\{0,-1/2+q_{i}+q_{j}-p_{k}\\}$ $g(p)\coloneqq\sum_{1\leq i<j\leq
4}(p_{j}-p_{i})+\sum_{1\leq i<j\leq 5}(q_{j}-q_{i})$

######  Lemma 9.1.

For any $C,X$ and $p\in\mathbb{R}^{5}$ satisfying (48), we have

$\\{V_{\mathbb{Z}}\cap CB(p,X)\\}\asymp_{C}\exp[X,f(p)].$

###### Proof.

The argument is identical to that of Lemma 5.21. ∎

######  Lemma 9.2.

For any $\epsilon$, $X$, and $p\in\operatorname{Poly}_{5}$, we have

$\\#\mathcal{F}_{5}(p,\epsilon,X)\ll_{\epsilon}\exp[X,f(p)-g(p)].$

###### Proof.

The proof is again completely identical to that of Lemma 5.22. ∎

######  Lemma 9.3.

There is a constant $C$ such that for all $\epsilon>C$ we have: