# A positive proportion of quartic fields are not monogenic yet have no local
obstruction to being so

Levent Alpöge  ,  Manjul Bhargava  and  Ari Shnidman

###### Abstract.

We show that a positive proportion of quartic fields are not monogenic,
despite having no local obstruction to being monogenic. Our proof builds on
the corresponding result for cubic fields that we obtained in a previous work.
Along the way, we also prove that a positive proportion of quartic rings of
integers do not arise as the invariant order of an integral binary quartic
form despite having no local obstruction.

##  1\. Introduction

In a previous paper [1], we proved that a positive proportion of cubic fields
are not monogenic despite having no local obstruction to being monogenic. The
purpose of this paper is to prove the analogous result for quartic fields.

######  Theorem 1.

Let $r=0,\,2,$ or $4$. When isomorphism classes of quartic fields with $r$
real embeddings are ordered by absolute discriminant, a positive proportion
are not monogenic and yet have no local obstruction to being monogenic.

In other words, we prove that a positive proportion of quartic fields, up to
isomorphism, are non-monogenic for truly global reasons. Moreover, we exhibit
such positive proportions of non-monogenic quartic fields having each of the
three possible real signatures.

To state precisely what it means for a number field of degree $n$ to have “no
local obstruction to being monogenic”, recall the index form
$f_{K}:{\rm\mathcal{O}}_{K}/{\rm\mathbb{Z}}\to\wedge^{n}{\rm\mathcal{O}}_{K}$
of $K$ defined by $\alpha\mapsto
1\wedge\alpha\wedge\alpha^{2}\wedge\cdots\wedge\alpha^{n-1}$, which may be
viewed as a homogeneous form of degree $n\choose 2$ in $n-1$ variables. A
number field $K$ is monogenic if and only if $f_{K}$ represents $\pm 1$ over
${\rm\mathbb{Z}}$. We say that $K$ has no local obstruction to being monogenic
if $f_{K}$ represents $1$ over ${\rm\mathbb{Z}}_{p}$ for all primes $p$, or
represents $-1$ over ${\rm\mathbb{Z}}_{p}$ for all primes $p$. We note that
the index form of a number field $K$ always represents $+1$ and $-1$ over
${\rm\mathbb{R}}$ (see Lemma 13), and hence it is not necessary to consider
obstructions to monogenicity over ${\rm\mathbb{R}}$.

Our proof of Theorem 1 actually proves something stronger. Recall that a
homogeneous polynomial $F\in{\rm\mathbb{Z}}[x,y]$ of degree $n$ determines a
based ring $R_{F}$ of rank $n$ — called the invariant order of $F$ — that
satisfies ${\rm Disc}(R_{F})={\rm Disc}(F)$. If
$F(x,y)=\sum_{i=0}^{n}a_{i}x^{n-i}y^{i}$ is irreducible of degree $n$ and
$\theta$ is a root of $F(x,1)$, then a ${\rm\mathbb{Z}}$-basis for $R_{F}$ is
$\langle
1,a_{0}\theta,a_{0}\theta^{2}+a_{1}\theta,\ldots,\sum_{i=0}^{n-2}a_{i}\theta^{n-1-i}\rangle$;
see [9, 12]. In coordinate-free terms, $R_{F}$ is the ring of global functions
on the ${\rm\mathbb{Z}}$-scheme $\mathrm{Proj}\,{\rm\mathbb{Z}}[x,y]/(F)$; see
[14].

We say that a ring $R$ is binary if $R$ is isomorphic to the invariant order
of some homogeneous form $F(x,y)\in{\rm\mathbb{Z}}[x,y]$. A number field $K$
is binary if ${\rm\mathcal{O}}_{K}$ is. If $K$ is monogenic, i.e.,
${\rm\mathcal{O}}_{K}\simeq{\rm\mathbb{Z}}[x]/(f(x))$ for a monic integer
polynomial $f$, then $K$ is also binary, as it is isomorphic to the invariant
order of the homogenization $F(x,y)$ of $f$. However, there are many binary
number fields that are not monogenic. In fact, every ring of rank $n\leq 3$ is
binary, despite the fact that many cubic fields are not monogenic [1]. On the
other hand, for $n\geq 4$, it is natural to conjecture that, when ordered by
discriminant, $100\%$ of number fields of degree $n$ are not binary, though it
was not previously known that even a positive proportion of degree $n$ number
fields are not binary.

Our second main theorem states that a positive proportion of quartic fields
are not binary, despite having no local obstruction to being binary; thus they
are not binary for truly global reasons. In fact, we show that this holds even
if we restrict to quartic fields having any fixed real signature:

######  Theorem 2.

Let $r=0,\,2,$ or $4$. When isomorphism classes of quartic fields with $r$
real embeddings are ordered by absolute discriminant, a positive proportion
are not binary and yet have no local obstruction to being binary.

Just as for monogenicity, the condition that a rank $n$ ring $R$ has no local
obstruction to being binary is stronger than simply demanding that
$R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}$ is binary over
${\rm\mathbb{Z}}_{p}$ for all $p$. To state the condition, we use the notion
of an orientation of a ring $R$ of rank $n$ over a principal ideal domain $S$:
an orientation of $R$ is a generator $\omega$ of the $S$-module $\wedge^{n}R$.
An oriented ring of rank $n$ over $S$ is then a pair $(R,\omega)$, where $R$
is a ring of rank $n$ over $S$ and $\omega$ is an orientation of $R$. If
$F(x,y)=\sum_{i=0}^{n}a_{i}x^{n-i}y^{i}\in S[x,y]$ is a binary form, then the
ring $R_{F}$ naturally carries the orientation $\omega_{F}=1\wedge
a_{0}\theta\wedge(a_{0}\theta^{2}+a_{1}\theta)\wedge\cdots\wedge(a_{0}\theta^{n-1}+\cdots+a_{n-2}\theta)$.
An oriented ring $(R,\omega)$ of rank $n$ over $S$ is binary if there is an
isomorphism $R\simeq R_{F}$ such that $\omega\mapsto\omega_{F}$ under this
isomorphism.

If $(R,\omega)$ is an oriented ring of rank $n$, then $(R,\omega)$ has no
local obstruction to being binary if the oriented ring
$(R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p},\omega\otimes 1)$ of rank $n$
over ${\rm\mathbb{Z}}_{p}$ is binary for all primes $p$.111We view
$\omega\otimes 1\in(\wedge^{n}R)\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}$ as
an element of $\wedge^{n}(R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p})$ via
the canonical isomorphism. If $R$ is an (unoriented) ring of rank $n$, we say
that $R$ has no local obstruction to being binary if there exists an
orientation $\omega$ such that $(R,\omega)$ has no local obstruction to being
binary. Since, for a ring of rank $n$ over ${\rm\mathbb{Z}}$, there are
exactly two possible choices of orientation, this definition naturally extends
the earlier-stated notion for a ring of rank $n$ to have no local obstruction
to being monogenic.

### Geometric formulation

Theorems 1 and 2 can be phrased as failures of local-to-global principles for
certain embedding problems over ${\rm Spec}\,{\rm\mathbb{Z}}$. Indeed, a
number field $K$ is monogenic if and only if there is a closed immersion ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{A}}^{1}_{\rm\mathbb{Z}}={\rm
Spec}\,{\rm\mathbb{Z}}[x]$, as giving such a closed immersion is equivalent to
giving a surjection ${\rm\mathbb{Z}}[x]\to{\rm\mathcal{O}}_{K}$. Similarly, a
number field $K$ is binary if and only if there is a closed immersion ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{P}}^{1}_{\rm\mathbb{Z}}=\mathrm{Proj}\,{\rm\mathbb{Z}}[x,y]$
(see [14, Prop. 2.9]). If
$\omega\in\wedge^{n}_{\rm\mathbb{Z}}{\rm\mathcal{O}}_{K}$, then a closed
immersion ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{A}}^{1}_{\rm\mathbb{Z}}$
is $\omega$-oriented if $\omega=\bigwedge_{i=0}^{n-1}x^{i}$. Similarly, ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{P}}^{1}_{\rm\mathbb{Z}}$
is $\omega$-oriented if $\omega=a_{0}^{n-1}\bigwedge_{i=0}^{n-1}(x/y)^{i}$.

In this language, the Levi–Delone–Faddeev correspondence implies that every
cubic field $K$ admits exactly one closed immersion ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{P}}^{1}_{\rm\mathbb{Z}}$,
up to automorphisms of ${\rm\mathbb{P}}^{1}_{\rm\mathbb{Z}}$. However, [1,
Theorem 1] states that for a positive proportion of cubic fields $K$, there
does not exist an embedding of ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{A}}^{1}_{\rm\mathbb{Z}}$
even though there is no local obstruction to such an embedding. Similarly,
Theorems 1 and 2 of the current paper state that, for a positive proportion of
quartic fields $K$, there does not exist a closed immersion ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{A}}^{1}_{\rm\mathbb{Z}}$
(resp. ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{P}}^{1}_{\rm\mathbb{Z}}$),
despite there not being any local obstructions to the existence of such
embeddings. By having no local obstruction, we mean that, for some orientation
$\omega\in\wedge^{n}_{\rm\mathbb{Z}}{\rm\mathcal{O}}_{K}$, there are
$\omega$-oriented closed immersions ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{A}}^{1}_{{\rm\mathbb{Z}}_{p}}$
(resp. ${\rm
Spec}\,{\rm\mathcal{O}}_{K}\hookrightarrow{\rm\mathbb{P}}^{1}_{{\rm\mathbb{Z}}_{p}}$)
for all $p$.

### Methods

While Theorems 1 and 2 are logically independent of each other, we prove them
simultaneously by exhibiting a positive-density family of quartic fields that
have no local obstruction to being monogenic (hence also have no local
obstruction to being binary) and yet are not binary (hence are also not
monogenic). Our method relies on connecting the monogenicity of quartic rings
to that of their cubic resolvent rings. Indeed, if a quartic ring is
monogenic, then so is its cubic resolvent ring. More generally, if a quartic
ring is binary, then its cubic resolvent ring is monogenic [15].

It is thus natural to try and prove Theorems 1 and 2 by considering the family
$\\{K\\}$ of non-monogenic cubic fields from [1], and for each such $K$ to
construct a quartic field $L$ such that the cubic resolvent ring of
${\rm\mathcal{O}}_{L}$ is ${\rm\mathcal{O}}_{K}$. As such an $L$ would not be
binary and have ${\rm Disc}(L)={\rm Disc}(K)$, it would follow that a positive
proportion of quartic fields are not binary (and hence not monogenic) when
ordered by discriminant. There is even a natural candidate for $L$, namely the
quartic field $L_{u}$ (well defined up to conjugation) contained in the Galois
closure of $K(\sqrt{u})$, where $u\in{\rm\mathcal{O}}_{K}^{\times}$ is any
choice of non-square unit.

There are a few obstacles in carrying out this plan. First, one cannot
guarantee that ${\rm Disc}(L_{u})={\rm Disc}(K)$ or, equivalently, that the
cubic resolvent ring of ${\rm\mathcal{O}}_{L_{u}}$ is the full ring of
integers ${\rm\mathcal{O}}_{K}$. Indeed, given a cubic field $K$, there might
not be any quartic field $L$ with cubic resolvent field $K$ such that ${\rm
Disc}(L)={\rm Disc}(K)$. However, we have that ${\rm Disc}(L_{u})/{\rm
Disc}(K)$ is uniformly bounded; we thus refine our proof in [1] to prove that
a positive proportion of the (possibly non-maximal) cubic resolvent rings of
${\rm\mathcal{O}}_{L_{u}}$ are non-monogenic and yet have no local obstruction
to being monogenic.

A more serious obstacle that arises is that we must prove that a positive
proportion of the quartic fields $L_{u}$ themselves have no local obstruction
to being monogenic. Indeed, it turns out that if $Q$ is a quartic ring with
cubic resolvent ring $R$, then $Q$ can have a local obstruction to being
monogenic even if $R$ does not. Thus, we must demonstrate the solubility of a
positive proportion of the index form equations $f_{L_{u}}(x,y,z)=\pm 1$ over
${\rm\mathbb{Z}}_{p}$, where $f_{L_{u}}$ is the ternary sextic index form
corresponding to the quartic field $L_{u}$.

We carry this out in two steps. We first observe that if the cubic resolvent
of $Q$ is monogenic over ${\rm\mathbb{Z}}_{p}$, then ${\rm
Spec}\,Q\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}\simeq\mathrm{Proj}({\rm\mathbb{Z}}_{p}[x,y]/F(x,y))$
for some binary quartic form $F$ over ${\rm\mathbb{Z}}_{p}$. We then use a
recent result of Fess (Theorem 8) to relate the solubility of the index form
equation to the solubility of the more manageable equation $z^{2}=F(x,y)$.
Imposing appropriate local conditions within our construction then guarantees
that a positive proportion of the quartic fields $L_{u}$ have no local
obstruction to being monogenic.

After performing this local analysis, a suitable adaptation of the arguments
in [1] then shows that a positive proportion of quartic fields are not binary
despite having no local obstruction to being monogenic. Moreover, we exhibit a
positive proportion of such quartic fields with positive (resp. negative)
discriminant; in particular, the $r=2$ cases of Theorems 1 and 2 follow, since
a quartic field has negative discriminant if and only if it has exactly two
real embeddings.

Using this approach to produce a positive proportion of quartic fields having
each of the two remaining real signatures (cases $r=0$ and $r=4$ of Theorems 1
and 2) turns out to be surprisingly subtle, as we must produce positive
proportions of non-monogenic orders in cubic fields $K$ having totally
positive non-square units $u$ as well as units $u$ of mixed signature. On the
one hand, we can guarantee that roughly $50\%$ of the cubic fields in the
families constructed in [1] are not monogenic. On the other hand, results of
Bhargava-Varma [7] give lower bounds on the proportion of cubic fields in such
families having a unit $u$ of prescribed signature. For units of mixed
signature, which give rise to quartic fields of signature $(0,2)$, the lower
bound of $75\%$ in [7] is strong enough for us to conclude the existence of a
positive proportion of mixed-signature quartic fields that are not monogenic.
However, for totally positive units, which give rise to totally real quartic
fields, the lower bound of $50\%$ contained therein is not quite strong enough
to guarantee overlap with our subset of non-monogenic cubic orders. To
overcome this slight miss, we strengthen the method of [1] just enough to
force a bit of overlap between these two sets of density roughly $1/2$. Along
the way we appeal to, and prove a slight generalization of, results of
Klagsbrun [11] on the average size of the $3$-torsion subgroups of $S$-class
groups of quadratic fields (for a fixed set $S$ of primes); the generalization
allows the imposition of local conditions at finitely many primes not in $S$.

### Contents

This paper is organized as follows. In §2 and §3, we establish some facts
about counting cubic and quartic fields, and we discuss some of the key
properties of their index forms, including Fess’s Theorem. In §4, we construct
a positive-proportion family of quartic fields with no local obstruction to
being monogenic. In §5, we show that when we fix the sign of the discriminant,
many of these quartic fields have non-monogenic cubic resolvent rings, thereby
proving Theorems 1 and 2 in the case $r=2$. In §6, we exhibit a positive
proportion of quartic fields (with no local obstructions to being monogenic)
which are not binary and have signatures $(4,0)$ and $(0,2)$, respectively,
thereby completing the proof of Theorems 1 and 2 for all real signatures.
Finally, in §7, we establish a mild generalization of Klagsbrun’s results on
the average size of 3-torsion in $S$-class groups that allows for specified
local conditions at a finite set of primes not in $S$, which is utilized as an
ingredient in §6.

### Acknowledgments

The authors thank Dan Fess and Arul Shankar for many helpful conversations.
The first author was supported by NSF grant DMS-2002109. The second author was
supported by a Simons Investigator Grant and NSF grant DMS-1001828. The third
author was supported by the Israel Science Foundation (grant No. 2301/20).

##  2\. Preliminaries on cubic rings

We recall the families of discriminants that were constructed in [1]. Let
$p_{1},\ldots,p_{t}$ be primes satisfying $p_{i}\equiv 2\mkern
4.0mu({\operator@font mod}\mkern 6.0mu3)$, and let $n=3\prod_{i=1}^{t}p_{i}$.
Define

$\Sigma_{n}=\left\\{\scalebox{0.7}[1.0]{$-$}27dn^{2}\in{\rm\mathbb{Z}}\colon
d\mbox{ fundamental},\,\left(\frac{d}{7}\right)\neq 1,\mbox{ and
}\left(\frac{d}{p}\right)=1\,\mbox{ for all }\,p\mid n\right\\},$

and write $\Sigma_{n}^{\pm}$ for the subset of $D\in\Sigma_{n}$ such that $\pm
D>0$.

Then we find that there are many cubic fields per discriminant in
$\Sigma_{n}^{\pm}$, on average:

######  Proposition 3.

Let $\Sigma_{n}^{\pm}(X)=\Sigma_{n}^{\pm}\cap[-X,X]$, and let
$N(\Sigma_{n}^{\pm},X)$ be the number of cubic fields with discriminant in
$\Sigma_{n}^{\pm}(X)$. Then

$\displaystyle\lim_{X\to\infty}\dfrac{N(\Sigma_{n}^{\pm},X)}{\\#\Sigma_{n}^{\pm}(X)}$
$\displaystyle=(2\mp 1)2^{t},$

###### Proof.

This was stated without proof in [1], since the result was not used there. We
deduce the result from [6, Thm. 8]. To compute the local factors at $p\mid n$,
we note that for $p\neq 3$, there is a unique totally ramified cubic extension
of ${\rm\mathbb{Q}}_{p}$, and it is not Galois — and thus it has trivial
automorphism group — since $p\equiv 2\mkern 4.0mu({\operator@font mod}\mkern
6.0mu3)$. When $p=3$, we note that there are three cubic extensions of
${\rm\mathbb{Q}}_{3}$ of discriminant valuation 5, and they too have trivial
automorphism groups. Therefore, we have

$\displaystyle N(\Sigma_{n}^{+},X)$
$\displaystyle\sim\frac{X}{12}\left(1-3^{-1}\right)\left(3\cdot
3^{-5}\right)\left(27\cdot
7^{-2}\right)\prod_{p\,\nmid\,7n}\left(1-p^{-1}\right)\left(1+p^{-1}\right)\prod_{p\,\mid\,\frac{n}{3}}\left(1-p^{-1}\right)\left(p^{-2}\right),$
$\displaystyle N(\Sigma_{n}^{-},X)$
$\displaystyle\sim\frac{X}{4}\left(1-3^{-1}\right)\left(3\cdot
3^{-5}\right)\left(27\cdot
7^{-2}\right)\prod_{p\,\nmid\,7n}\left(1-p^{-1}\right)\left(1+p^{-1}\right)\prod_{p\,\mid\,\frac{n}{3}}\left(1-p^{-1}\right)\left(p^{-2}\right),$

while

$\\#\Sigma_{n}^{\pm}(X)\sim\frac{X}{2}\left(1-3^{-1}\right)\left(\textstyle\frac{1}{2}\cdot
3^{-5}\right)\left(27\cdot
7^{-2}\right)\prod_{p\,\nmid\,7n}\left(1-p^{-1}\right)\left(1+p^{-1}\right)\prod_{p\,\mid\,\frac{n}{3}}\left(1-p^{-1}\right)\left(\textstyle\frac{1}{2}\cdot
p^{-2}\right).$

the $\frac{1}{2}$’s in the final Euler factor are due to the Kronecker symbol
condition at each $p\mid n$. ∎

In fact, for many individual discriminants $D\in\Sigma_{n}$, we can prove that
this average gives a lower bound on the number of cubic fields of discriminant
$D$. Indeed, let $U_{n}^{+}$ (resp. $U_{n}^{-}$) be the subset of
$D\in\Sigma_{n}^{+}$ (resp. $\Sigma_{n}^{-}$) such that the $3$-torsion in the
class group of ${\rm\mathbb{Q}}(\sqrt{-3D})$ (resp.
${\rm\mathbb{Q}}(\sqrt{D})$) is trivial. The following is stated as [1, Prop.
20–21].

######  Proposition 4.

If $D\in U_{n}^{+}$ $($resp. $U_{n}^{-})$, then there are exactly $2^{t}$
$($resp. $3\cdot 2^{t})$ cubic fields of discriminant $D$.

We will also require a result of this type when $3$-part of the class group
has rank 1.

######  Proposition 5.

Let $D\in\Sigma_{n}^{+}$, let
$F={\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D})$, and factor
$(p_{i})={\rm\mathfrak{p}}_{i}\bar{\rm\mathfrak{p}}_{i}$ in
${\rm\mathcal{O}}_{F}$ for each $1\leq i\leq t$. If $\\#{\rm Cl}(F)[3]=3$ and
at least one $[{\rm\mathfrak{p}}_{i}]$ is not a cube in ${\rm Cl}(F)$, then
there are at least $3\cdot 2^{t-1}$ cubic fields $K$ having discriminant
either $D$ or $D/p_{i}^{2}$.

###### Proof.

Write $D=\scalebox{0.7}[1.0]{$-$}27dn^{2}$ and fix some $f\mid n$. By the
proof of [1, Prop. 20], the number of cubic fields of discriminant
$\scalebox{0.7}[1.0]{$-$}27df^{2}$ is half the number of pairs $(J,[I])$,
where $J\subseteq{\rm\mathcal{O}}_{F}$ is an ideal of norm $f$ and $[I]\in{\rm
Cl}({\rm\mathcal{O}}_{F})$ a cube root of $[J]$. Note that if $[J]$ is a cube
then, by hypothesis, it has exactly $3$ cube roots.

By assumption, there is an $i$ such that $[{\rm\mathfrak{p}}_{i}]$ spans the
one-dimensional ${\rm\mathbb{F}}_{3}$-vector space ${\rm Cl}(F)/3{\rm Cl}(F)$.
For each of the $2^{t}$ ideals $J_{1}$ of norm $n/p_{i}$, there is a unique
ideal $J_{2}$ of norm dividing $1$ or $p_{i}$ such that $[J_{1}]=[J_{2}]$
modulo cubes. The ideal $J:=\bar{J}_{1}J_{2}$ is therefore a cube in the class
group, and has norm either $n$ or $n/p_{i}$. Up to ${\rm
Gal}({\rm\mathcal{O}}_{F}/{\rm\mathbb{Z}})$-conjugacy, this gives
$\frac{1}{2}3\cdot 2^{t}=3\cdot 2^{t-1}$ pairs $(J,[I])$, and hence there are
at least that many cubic fields $K$ of discriminant $D$ or $D/p_{i}^{2}$. ∎

##  3\. Preliminaries on quartic rings and their index forms

To study local obstructions to monogenicity for quartic fields, we use a
parametrization of quartic rings and their cubic resolvent rings from [3] and
[13], which in particular will allow us to relate the index form of a quartic
ring to that of its cubic resolvent ring.

For a principal ideal domain $S$, let $V(S)$ denote the space of pairs $(A,B)$
of ternary quadratic forms with coefficients in $S$. A cubic (resp. quartic)
ring over $S$ is an $S$-algebra that is free of rank 3 (resp. 4) as an
$S$-module.

######  Theorem 6 ([3],[13]).

Let $S$ be a principal ideal domain. There is a bijection between pairs
$(A,B)\in V(S)$ and isomorphism classes of pairs
$((Q,\langle\alpha,\beta,\gamma\rangle),(C,\langle\omega,\theta\rangle))$,
where $Q$ is a quartic ring over $S$ with a chosen basis
$\langle\alpha,\beta,\gamma\rangle$ of $Q/S$ and $C$ is a cubic resolvent ring
of $Q$ with a chosen basis $\langle\omega,\theta\rangle$ of $C/S$.

A cubic resolvent ring $C$ of a quartic ring $Q$ over $S$ is a cubic ring over
$S$ that is equipped with an isomorphism $\xi:\wedge^{4}Q\to\wedge^{3}C$ and a
quadratic mapping $\phi:Q/S\to C/S$ satisfying $\xi(1\wedge x\wedge y\wedge
xy)=1\wedge\phi(x)\wedge\phi(y)$ for any $x,y\in Q/S$. These conditions also
imply that ${\rm Disc}(Q)={\rm Disc}(C)$, when computed with respect to bases
$\langle\alpha,\beta,\gamma\rangle$ and $\langle\omega,\theta\rangle$ of $Q$
and $C$ such that
$\xi(1\wedge\alpha\wedge\beta\wedge\gamma)=1\wedge\omega\wedge\theta$. By
construction, the bases in Theorem 6 are compatible in this way. See [3] and
[13] for details.

The pair $(C,\langle\omega,\theta\rangle)$ is called a based cubic ring and
$(Q,\langle\alpha,\beta,\gamma\rangle)$ a based quartic ring. A complete
description of the construction of
$((Q,\langle\alpha,\beta,\gamma\rangle),(C,\langle\omega,\theta\rangle))$ from
$(A,B)$ can be found in [3, §3.2–3.3]. We denote the (based) quartic ring and
(based) cubic resolvent ring associated a pair $(A,B)\in V(S)$ by $Q(A,B)$ and
$R(A,B)$, respectively.

We view elements of $(A,B)\in V(S)$ as pairs of $3$-by-$3$ symmetric matrices;
we denote the associated quadratic forms by
$q_{A}(x_{1},x_{2},x_{3}):=\sum_{1\leq i<j\leq 3}a_{ij}x_{i}x_{j}$ and
$q_{B}(x_{1},x_{2},x_{3}):=\sum_{1\leq i<j\leq 3}b_{ij}x_{i}x_{j}$,
respectively, where $a_{ij},b_{ij}\in S$. The index forms of $Q(A,B)$ and
$R(A,B)$, in terms of $(A,B)$, are then given in the following proposition.

######  Proposition 7.

Let $(A,B)\in V(S)$. The index form $f_{R(A,B)}$ of $R(A,B)$ is given by

$f_{R(A,B)}(x,y)=4\cdot\det(Ax-By),$ (3.1)

and the index form $f_{Q(A,B)}$ of $Q(A,B)$ is given by

$f_{Q(A,B)}(x,y,z)=f_{R(A,B)}(q_{B}(x,y,z),\scalebox{0.7}[1.0]{$-$}q_{A}(x,y,z)).$
(3.2)

In particular, if a quartic ring over $S$ is monogenic then its cubic
resolvent ring is monogenic.

###### Proof.

See [3, Lemma $9$] and the surrounding discussion for the assertions on the
index forms. A monogenic quartic ring $Q$ over $S$ is primitive and thus, by
[3, Corollary $4$], it has a unique cubic resolvent ring $R$. Furthermore, if
$f_{Q}$ represents 1, then by (3.2), $f_{R}$ also represents 1! ∎

There is a natural (based) ring of rank $n$ associated to any binary $n$-ic
form over $S$ [9, 12, 14]. The quartic ring $Q_{F}$ over $S$ associated to a
binary quartic form $F(x,y)=ax^{4}+bx^{3}y+cx^{2}y^{2}+dxy^{3}+ey^{4}$ over
$S$ is given by $Q(A_{1},B_{F})$, where

$A_{1}=\left[\begin{array}[]{ccc}\phantom{0}&\phantom{0}&1/2\\\
\phantom{0}&-1&\phantom{0}\\\
1/2&\phantom{0}&\phantom{0}\end{array}\right],\qquad
B_{F}=\left[\begin{array}[]{ccc}a&b/2&0\\\ b/2&c&d/2\\\
0&d/2&e\end{array}\right];$

see Wood [15, 5].

######  Theorem 8 (Fess [10]).

Let $F$ be a binary quartic form over a principal ideal domain $S$, and
$Q_{F}$ the associated based quartic ring over $S$. Then the index form
$f_{Q_{F}}$ of $Q_{F}$ satisfies

$f_{Q_{F}}(x^{2},xy,y^{2})=F(x,y)^{3}.$ (3.3)

###### Proof.

We have $q_{A_{1}}(x^{2},xy,y^{2})=0$ and $q_{B_{F}}(x^{2},xy,y^{2})=F(x,y)$.
Thus, by (3.1) and (3.2), we see that
$f_{Q_{F}}(x^{2},xy,y^{2})=4\cdot\det(A_{1}\cdot F(x,y))=F(x,y)^{3}$. ∎

######  Corollary 9.

If $F$ represents a square in $S^{\times}$, then $f_{Q_{F}}(x,y,z)$ represents
$1$ over $S$.

###### Proof.

If $F(a,b)=c^{2}$ with $c\in S^{\times}$. Then $c^{-1}\cdot(a^{2},ab,b^{2})$
is a solution to $f_{Q_{F}}(x,y,z)=1$. ∎

######  Proposition 10.

Let $S={\rm\mathbb{Z}}$ or ${\rm\mathbb{Z}}_{p}$ for some $p$. Suppose
$((Q,\langle 1,\alpha,\beta,\gamma\rangle),(R,\langle
1,\omega,\theta\rangle))$ is a based quartic ring and a compatibly-based cubic
resolvent ring over $S$, such that the index form of $R$ with respect to this
basis represents $1$. Then there is a binary quartic form $F$ over $S$ and
compatible bases $\langle
1,\alpha^{\prime},\beta^{\prime},\gamma^{\prime}\rangle$ and $\langle
1,\omega^{\prime},\theta^{\prime}\rangle$ for $Q$ and $R$ such that

$1\wedge\alpha\wedge\beta\wedge\gamma=1\wedge\alpha^{\prime}\wedge\beta^{\prime}\wedge\gamma^{\prime},\quad
1\wedge\omega\wedge\theta=1\wedge\omega^{\prime}\wedge\theta^{\prime},$

and $((Q,\langle
1,\alpha^{\prime},\beta^{\prime},\gamma^{\prime}\rangle),(R,\langle
1,\omega^{\prime},\theta^{\prime}\rangle))$ corresponds to $(A_{1},B_{F})$ via
the correspondence of Theorem 6. In particular, the index forms $f_{Q}(x,y,z)$
and $f_{R}(x,y)$ associated to $(Q,\langle
1,\alpha^{\prime},\beta^{\prime},\gamma^{\prime}\rangle)$ and $(R,\langle
1,\omega^{\prime},\theta^{\prime}\rangle)$, respectively, satisfy

$f_{Q}(x,y,z)=f_{R}(q_{B_{F}}(x,y,z),-q_{A_{1}}(x,y,z)).$

###### Proof.

Let $\langle 1,\alpha,\beta,\gamma\rangle$ be an basis of $Q$ that is
compatible via $\xi$ with the basis $\langle 1,\theta,\theta^{2}\rangle$ of
$R$. Let $(A,B)\in V(S)$ be the corresponding pair under the bijection of
Theorem 6. Because $f_{R}(1,0)=1$, we see that $4\cdot\det(A)=1$. Over $S$,
there is a _unique_ ${\rm GL}_{3}(S)$-equivalence class of $A\in{\rm
Sym}^{2}(S)^{\ast}$ for which $4\cdot\det(A)=1$, namely the class of $A_{1}$.
Via a change of basis, we may assume $A=A_{1}$ and $b_{13}=0$. The desired
binary quartic form is then
$F(x,y):=b_{11}x^{4}+b_{12}x^{3}y+b_{22}x^{2}y^{2}+b_{23}xy^{3}+b_{33}y^{4}$,
with $b_{ij}$ the coefficients of $B$. ∎

######  Proposition 11.

Let $p$ be a prime. Let $F\in{\rm\mathbb{Z}}_{p}[x,y]$ be a binary quartic
form and $Q_{F}$ the corresponding quartic ring over ${\rm\mathbb{Z}}_{p}$.
Let $\varepsilon\in\\{\pm 1\\}$. If the curve $\varepsilon z^{2}=F(x,y)$ over
${\rm\mathbb{F}}_{p}$ has a smooth ${\rm\mathbb{F}}_{p}$-point with $z\neq 0$,
then the index form of $Q_{F}$ represents $\varepsilon$.

###### Proof.

Via Hensel lifting, we have $x,y\in{\rm\mathbb{Z}}_{p}$ and
$z\in{\rm\mathbb{Z}}_{p}^{\times}$ such that $\varepsilon z^{2}=F(x,y)$. Then,
by Theorem 8, we have
$I_{Q_{F}}(z^{-1}\cdot(x^{2},xy,y^{2}))=z^{-6}F(x,y)^{3}=\varepsilon^{3}=\varepsilon.$
∎

######  Lemma 12.

Let $F\in{\rm\mathbb{F}}_{p}[x,y]$ be a quartic form with splitting type
$(1^{3}1)$ or $(1^{2}2)$. Then the curve $z^{2}=F(x,y)$ has an
${\rm\mathbb{F}}_{p}$-point with $z\neq 0$. If $p>3$, then the same holds for
$F$ of type $(1^{2}11)$ as well.

###### Proof.

We first consider the case $(1^{3}1)$. More generally, if $L_{1}$ and $L_{2}$
are linearly independent linear forms and $n\geq 1$, then the equation
$1=L_{1}(v)^{n}L_{2}(v)$ has a solution $v\in{\rm\mathbb{F}}_{p}^{2}$. Indeed,
the map
$(L_{1}(v),L_{2}(v)):{\rm\mathbb{F}}_{p}^{2}\to{\rm\mathbb{F}}_{p}^{2}$ is
surjective by hypothesis, so $(1,1)$ is in its image.

If $F$ has type $(1^{2}11)$ or $(1^{2}2)$, then $F=L_{0}^{2}Q$ for some linear
form $L_{0}$ and quadratic form $Q$ with nonzero discriminant. The conic
$z^{2}=Q(x,y)$ has $p+1>2$ points over ${\rm\mathbb{F}}_{p}$. If $Q$ is
irreducible, then none of these points have $z=0$, and at most two of them
intersect the line $L_{0}=0$. Hence if $F$ has splitting type $(1^{2}2)$, then
$z^{2}=F(x,y)$ has an ${\rm\mathbb{F}}_{p}$-point with $z\neq 0$. If
$Q=L_{1}L_{2}$ is reducible, then the conic intersects $L_{0}L_{1}L_{2}=0$ in
at most 4 points. Thus, if $p>3$, then $p+1>4$ and there is at least one
solution to $z^{2}=F(x,y)$ with $z\neq 0$. ∎

Finally, we show that the index form of an order in a number field of any
degree always represents $1$ and $-1$ over ${\rm\mathbb{R}}$, and hence we
need not consider archimedean places when studying local obstructions to
monogenicity.

######  Lemma 13.

The index form $f_{\rm\mathcal{O}}$ of an order ${\rm\mathcal{O}}$ in a number
field $L\neq{\rm\mathbb{Q}}$ represents both $1$ and $-1$ over
${\rm\mathbb{R}}$.

###### Proof.

Let $n\geq 2$ be the degree of $L$, and let $\sigma_{1},\ldots,\sigma_{n}$
denote the distinct embeddings $L\to{\rm\mathbb{C}}$. Let
$1,\theta_{1},\ldots,\theta_{n-1}$ be a ${\rm\mathbb{Z}}$-basis for
${\rm\mathcal{O}}$. Then, over ${\rm\mathbb{C}}$, we have the factorization

$f_{\rm\mathcal{O}}(x_{1},\ldots,x_{n-1})=\pm\frac{1}{\sqrt{{\rm
Disc}({\rm\mathcal{O}})}}\prod_{1\leq i<j\leq
n}\left(\sum_{k=1}^{n-1}(\sigma_{i}(\theta_{k})-\sigma_{j}(\theta_{k}))x_{k}\right).$

It suffices to check that at least one of the linear factors of
$f_{\rm\mathcal{O}}$ above is in fact defined over ${\rm\mathbb{R}}$. Indeed,
if $L$ is totally real, then all factors are defined over ${\rm\mathbb{R}}$.
Otherwise, choose a factor corresponding to $(i,j)$ where $\sigma_{i}$ and
$\sigma_{j}$ are complex conjugate embeddings; then, upon removing a factor of
$\sqrt{-1}$, we see that this factor is defined over ${\rm\mathbb{R}}$. Since
$f_{{\rm\mathcal{O}}_{L}}$ has a real linear factor, it takes on all real
values, and in particular both $1$ and $-1$.∎

##  4\. Construction of quartic fields with no local obstruction to being
monogenic

We use the family of discriminants $\Sigma_{n}$ from Section 2 and assume that
$n$ is a multiple of $30=2\cdot 3\cdot 5$. We denote the family of cubic
fields having discriminants in $\Sigma_{n}$ by $F_{3}$.

For each $K\in F_{3}$, let $\mathcal{U}_{K}$ be a set of representatives for
the nonzero elements of the finite group
$\left({\rm\mathcal{O}}_{K}^{\times}/{\rm\mathcal{O}}_{K}^{\times
2}\right)_{{\rm Nm}=1}$ of norm-one square-classes in
${\rm\mathcal{O}}_{K}^{\times}$. This set has size 3 if ${\rm Disc}(K)>0$ and
size 1 if ${\rm Disc}(K)<0$. For each $u\in\mathcal{U}_{K}$, since $K$ is an
$S_{3}$-cubic field, the Galois closure of $K(\sqrt{u})$ is an $S_{4}$-field
[2, Lem. 4–5], and thus contains a quartic field $L_{u}$, unique up to
conjugacy. Consider the family $F_{4}=\\{L_{u}:K\in
F_{3},u\in\mathcal{U}_{K}\\}$ consisting of all of the quartic fields $L_{u}$
arising in this way from all $K\in F_{3}$.

######  Lemma 14.

Let $L\in F_{4}$, and let $K$ be its cubic resolvent field. Let $R$ be the
cubic resolvent ring of ${\rm\mathcal{O}}_{L}$. Then
$R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}\simeq{\rm\mathcal{O}}_{K}\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}$
for $p>2$. If $p=2$, then $R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{2}$ is
isomorphic to $R_{0}\simeq{\rm\mathbb{Z}}_{2}[\sqrt[3]{2}]$,
$R_{1}\simeq{\rm\mathbb{Z}}_{2}[\sqrt[3]{4}]$, or
$R_{3}\simeq{\rm\mathbb{Z}}_{2}[2(\sqrt[3]{2}+\sqrt[3]{4})]$, and hence the
index of $R$ in ${\rm\mathcal{O}}_{K}$ is equal to $2^{r}$ for $r=0$, $1$, or
$3$, respectively.

###### Proof.

Since the extension $K(\sqrt{u})/K$ is unramified away from $2$, the
discriminant ${\rm Disc}(L)$ agrees with ${\rm Disc}(K)$ up to powers of $2$.
By the definition of $\Sigma_{n}$, the cubic field $K$ is totally ramified at
$2$, and since the unique totally ramified cubic extension of
${\rm\mathbb{Q}}_{2}$ is $K_{2}:={\rm\mathbb{Q}}_{2}(\sqrt[3]{2})$, we have
$K\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{2}\simeq K_{2}$ and
${\rm\mathcal{O}}_{K_{2}}\simeq R_{0}$. The group of squareclasses of norm-one
units in $R_{0}^{\times}$ is of order 4. The trivial squareclass gives rise to
the quartic ${\rm\mathbb{Q}}_{2}$-algebra
$L_{0}=K_{2}\times{\rm\mathbb{Q}}_{2}$ having discriminant valuation $2$,
while the three nontrivial elements give rise to three distinct
$S_{4}$-quartic fields $L_{1}$, $L_{3}$, and $L_{3}^{\prime}$ over
${\rm\mathbb{Q}}_{2}$ with discriminant valuations $4$, $8$, and $8$,
respectively. A computation shows the cubic resolvent rings of $L_{0}$,
$L_{1}$, $L_{3}$, and $L_{3}^{\prime}$ to be $R_{0}$, $R_{1}$, $R_{3}$, and
$R_{3}$, respectively, which have the stated indices in $R_{0}$. ∎

######  Lemma 15.

Let $L\in F_{4}$, and let $R$ be the cubic resolvent ring of
${\rm\mathcal{O}}_{L}$. Then $R$ is has no local obstruction to being
monogenic.

###### Proof.

By Lemma 14, $R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{2}$ is monogenic, and
hence the index form $f_{R}$ is a primitive binary cubic form. Since ${\rm
Disc}(f_{R})$ is divisible by $2$, [1, Lem. 9] implies that $f_{R}$ represents
both $+1$ and $\scalebox{0.7}[1.0]{$-$}1$ over ${\rm\mathbb{Z}}_{2}$. For
$p>2$, the ring $R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}$ is maximal, and
the proof is exactly as in [1, Lem. 18]. ∎

######  Proposition 16.

If $L\in F_{4}$, then $L$ has no local obstruction to being monogenic.

###### Proof.

Let $R$ be the cubic resolvent ring of ${\rm\mathcal{O}}_{L}$, and set
$K=R\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Q}}$.

First assume $p>2$. Since $R$ is locally unobstructed (Lemma 15), we have
${\rm\mathcal{O}}_{L}\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}\simeq Q_{F}$
for some binary quartic form $F\in{\rm\mathbb{Z}}_{p}[x,y]$, and
$f_{{\rm\mathcal{O}}_{L}}(x,y,z)=f_{Q_{F}}((x,y,z)\cdot g)$ for some $g\in{\rm
SL}_{3}({\rm\mathbb{Z}}_{p})$ by Proposition 10. We will show that both
$f_{Q_{F}}(x,y,z)=1$ and $f_{Q_{F}}(x,y,z)=-1$ have
${\rm\mathbb{Z}}_{p}$-solutions. By Proposition 11, it suffices to show that
each curve $C_{F,\varepsilon}:\varepsilon z^{2}=F(x,y)$ with
$\epsilon\in\\{\pm 1\\}$ has an ${\rm\mathbb{F}}_{p}$-point with $z\neq 0$.
Because
$Q_{F}\simeq{\rm\mathcal{O}}_{L}\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{p}$ is
maximal, the splitting type of $F$ modulo $p$ matches the splitting type of
$p$ in $L$, and we break into cases based on whether $p$ divides ${\rm
Disc}(K)=\scalebox{0.7}[1.0]{$-$}27dn^{2}$ or not.

If $p\nmid dn$, then $p\nmid{\rm Disc}(F)$ and, as long as $p>7$, the
Hasse–Weil bound implies that $C_{F,\varepsilon}$ has an
${\rm\mathbb{F}}_{p}$-point with $z\neq 0$. For $p\mid d$, the splitting type
of $p$ in $K$ is $(1^{2}1)$, and so in $L$ it is $(1^{2}2)$ or $(1^{2}11)$.
Thus $C_{F,\varepsilon}$ has an ${\rm\mathbb{F}}_{p}$-point with $z\neq 0$
thanks to Lemma 12; note that $p>3$ when $p\mid d$, since we have insisted
that $6\mid n$. If $p=7$ and $p\nmid d$, the splitting type of $p$ in $K$ is
$(12)$ because $\left(\frac{d}{7}\right)\neq 1$. Hence the splitting type of
$p$ in $L$ is either $(4)$ or $(112)$, and so $F$ has at most $2$ roots over
${\rm\mathbb{F}}_{p}$. Thus the Hasse–Weil bound implies that
$C_{F,\varepsilon}$ has an ${\rm\mathbb{F}}_{7}$-point with $z\neq 0$. For
$p\mid n$ (which includes $p=3$ and $p=5$, by assumption), the splitting type
of $p$ in $K$ is $(1^{3})$, so that the splitting type of $p$ in $L$ is
$(1^{3}1)$. Hence the splitting type of $F$ modulo $p$ is $(1^{3}1)$ and Lemma
12 implies $C_{F,\varepsilon}$ has an ${\rm\mathbb{F}}_{p}$-point with $z\neq
0$.

Next assume $p=2$. We prove that the index form of ${\rm\mathcal{O}}_{L}$
represents $+1$ or $\scalebox{0.7}[1.0]{$-$}1$ over ${\rm\mathbb{Z}}_{2}$.
Since $2$ is totally ramified in $K$, we have
$K\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{2}\simeq{\rm\mathbb{Q}}_{2}(\sqrt[3]{2})$.
Hence either
$L\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{2}\simeq{\rm\mathbb{Q}}_{2}(\sqrt[3]{2})\times{\rm\mathbb{Q}}_{2}$,
implying that the splitting type of $2$ in $L$ is $(1^{3}1)$, or else
$L\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{2}$ is a field in which $2$ splits
as $(1^{4})$.

In the first case, we conclude by the same argument that we used to deal with
the primes $p\mid n$ with $p>2$. In the second case,
$L\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{2}$ is an $S_{4}$-quartic field. As
mentioned earlier, there are three quartic $S_{4}$-extensions of
${\rm\mathbb{Q}}_{2}$. Explicitly, we have
${\rm\mathcal{O}}_{L}\otimes_{\rm\mathbb{Z}}{\rm\mathbb{Z}}_{2}\cong{\rm\mathbb{Z}}_{2}[x]/(g)$
for some $g\in\\{x^{4}+2x+2,x^{4}+4x+2,x^{4}+4x^{2}+4x+2\\}$. Therefore, there
exists $u\in{\rm\mathbb{Z}}_{2}^{\times}$ such that
$f_{{\rm\mathcal{O}}_{L}}(x,y,z)=u\cdot f_{Q_{g}}(x,y,z)$. Thus it suffices to
check that, for all three quartic forms $g$ and all units
$u\in{\rm\mathbb{Z}}_{2}^{\times}$, the equation $f_{Q_{g}}(x,y,z)=1$ or
$f_{Q_{g}}(x,y,z)=\scalebox{0.7}[1.0]{$-$}1$ has a solution. Since we may
scale $u$ by an element of
$({\rm\mathbb{Z}}_{2}^{\times})^{6}=({\rm\mathbb{Z}}_{2}^{\times})^{2}=1+8{\rm\mathbb{Z}}_{2}$,
we need only verify this statement modulo $8$, which is a quick check. ∎

##  5\. Proof of Theorems 1 and 2: the case $r=2$

We prove that a positive proportion of positive discriminant (resp. negative
discriminant) quartic fields are not even binary, despite having no local
obstruction to being monogenic. The result for negative discriminants
immediately implies the case $r\\!=\\!2$ of Theorems 1 and 2.

For each choice of sign $\pm$, write $F_{3}^{\pm}$ and $F_{4}^{\pm}$ for the
subsets of $F_{3}$ and $F_{4}$ having discriminants of sign $\pm$. Recall from
Section 2 the subset $U_{n}^{\pm}\subset\Sigma_{n}^{\pm}$. By Proposition 4,
if $D\in U_{n}^{\pm}$, then the number of $L\in F_{4}$ having cubic resolvent
field $K$ of discriminant $D$ is

$\bigl{[}\\#\bigl{(}{\rm\mathcal{O}}_{K}^{\times}/{\rm\mathcal{O}}_{K}^{\times
2}\bigr{)}_{{\rm Nm}=1}-1\bigr{]}\cdot(2\mp 1)2^{t}=3\cdot 2^{t}.$

For $D\in\Sigma_{n}$, let $B_{D}$ denote the number of binary fields $L\in
F_{4}$ having cubic resolvent field of discriminant $D$.

For each nonzero $D\in{\rm\mathbb{Z}}$, we define the elliptic curve
$E^{D}\colon y^{2}=4x^{3}+D$. If $R_{L}$ is the cubic resolvent ring of
${\rm\mathcal{O}}_{L}$, then $f_{R_{L}}$ gives rise to two nonzero elements
$\alpha_{L}\in{\rm Sel}_{3}(E^{{\rm Disc}(L)})$ by Lemma 15 and the
construction of [1, §3.2]. Moreover, $\alpha_{L}=\pm\alpha_{L^{\prime}}$ if
and only if $R_{L}\cong R_{L^{\prime}}$, as in the proof of [1, Lem. 15]. If
$L$ is binary, then $R_{L}$ is monogenic by Proposition 10, and hence
$\pm\alpha_{L}$ both lie in $E^{{\rm Disc}(L)}({\rm\mathbb{Q}})/3E^{{\rm
Disc}(L)}({\rm\mathbb{Q}})$. Thus, by Lemma 14,

$9+2B_{D}\leq
3(3^{\mathop{\mathrm{rk}}{E^{D}}}+3^{\mathop{\mathrm{rk}}{E^{4D}}}+3^{\mathop{\mathrm{rk}}{E^{64D}}}).$

Since $g(x):=x^{\log_{3}{2}}$ is a subadditive function, i.e., $g\\!\left(\sum
a_{i}\right)\leq\sum g(a_{i})$, we have

$(9+2B_{D})^{\log_{3}{2}}\leq
2(2^{\mathop{\mathrm{rk}}{E^{D}}}+2^{\mathop{\mathrm{rk}}{E^{4D}}}+2^{\mathop{\mathrm{rk}}{E^{64D}}}).$

Denote by $\mu_{n,\pm}$ the lower density of $U_{n}^{\pm}$ in
$\Sigma_{n}^{\pm}$, which is at least $1/2$ by [8]. We have

$\mu_{n,\pm}\cdot{\rm avg}_{D\in
U_{n}^{\pm}}(9+2B_{D})^{\log_{3}{2}}+(1-\mu_{n,\pm})\cdot
3\>\leq\>2(3+3+3)\>=\>18,$

by [1, Thm. 13]. Therefore,

${\rm avg}_{D\in U_{n}^{\pm}}(9+2B_{D})^{\log_{3}{2}}\leq
3+15\mu_{n,\pm}^{-1}\leq 33.$

Given the constraints (a) $B_{D}\in[0,3\cdot 2^{t}]$ for all $D\in
U_{n}^{\pm}$, and (b) ${\rm avg}_{D\in U_{n}^{\pm}}(9+2B_{D})^{\log_{3}2}\leq
33$, the maximum of ${\rm avg}_{D\in U_{n}^{\pm}}(9+2B_{D})$ is achieved, for
$t\geq 4$, when $B_{D}$ attains its maximum value $3\cdot 2^{t}$ for a density
of $24/((9+3\cdot 2^{t+1})^{\log_{3}2}-1)$ of $D\in U_{n}^{\pm}$, and
$B_{D}=0$ otherwise. It follows that a positive lower density of at least
$1-O(2^{-t\log_{3}2})$ quartic fields $L$ with cubic resolvent field of
discriminant $D\in U_{n}^{\pm}$ are non-monogenic.

Thus if $t\geq 5$, the number of $L\in F_{4}$ which are not binary and are of
absolute discriminant at most $X$ is $\gg X$. By Proposition 16, these quartic
fields are all locally unobstructed to being monogenic. By [4], they give a
positive proportion of all quartic fields when ordered by absolute
discriminant, completing the proof. $\;\Box$

##  6\. Proof of Theorems 1 and 2: the cases $r=0$ and $r=4$

We first treat the case $r=0$. We use a result of the second author and Varma
[7, Cor. 10], which states that at least $\frac{3}{4}$ of cubic fields in
$F^{+}_{3}$ have a unit of mixed signature. By Proposition 3, the proportion
of cubic fields in $F_{3}^{+}$ with discriminant in $U_{n}^{+}$ is
$\mu_{n}^{+}\geq 1/2$. Hence, at least $\frac{1}{2}$ of cubic fields with
${\rm Disc}(K)\in U_{n}^{+}$ have a unit of mixed signature. For such cubic
fields, two of the three corresponding quartic fields in $F_{4}^{+}$ are
totally complex. On the other hand, for $t\gg 1$, the proof of Theorem 1 shows
that a proportion of more than $1-O(2^{-t\log_{3}{2}})$ of quartic fields in
$F_{4}^{+}$ with cubic resolvent discriminant in $U_{n}^{+}$ are not binary.
It follows that at least
$\frac{2}{3}\cdot\left(1-\frac{1}{2}-O(2^{-t\log_{3}{2}})\right)=\frac{1}{3}-O(2^{-t\log_{3}{2}})$
of the quartic fields produced in the proof of Section 5 are totally complex,
proving Theorem 1 in the case $r=0$. (In fact, $t=7$ suffices to make the
latter density positive.)

Finally, we turn to the totally real case $r=4$. Let
$\widetilde{\Sigma}_{n}^{+}$ be the set of integers of the form $D$ or
$D/p_{i}^{2}$, for some $D\in\Sigma_{n}^{+}$ and some $1\leq i\leq t$. Let
$U_{n,1}^{+}$ be the subset of $D\in\Sigma_{n}^{+}$ such that $\\#{\rm
Cl}({\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D}))[3]=3$ and such that
not all of the $[{\rm\mathfrak{p}}_{i}]$ are cubes in ${\rm
Cl}({\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D}))$. Let $\mu_{n,1}^{+}$
be the lower relative density of $U_{n,1}^{+}$ in $\Sigma_{n}^{+}$. Write also
$U_{n,\leq 1}^{+}:=U_{n}^{+}\cup U_{n,1}^{+}$ and $\mu_{n,\leq
1}^{+}:=\mu_{n}^{+}+\mu_{n,1}^{+}$.

######  Lemma 17.

We have $\mu_{n,\leq 1}^{+}\geq\frac{7}{8}-\frac{1}{2}\cdot 3^{-t}$.

###### Proof.

Let $\widetilde{U}_{n,1}^{+}\supset U_{n,1}^{+}$ be the subset of
$D\in\Sigma_{n}^{+}$ such that $\\#{\rm
Cl}({\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D}))[3]=3$, and let
$\tilde{\mu}_{n,1}$ be the lower relative density of $\widetilde{U}_{n,1}^{+}$
in $\Sigma_{n}^{+}$. Then
$\mu_{n,1}^{+}\leq\tilde{\mu}_{n,1}^{+}\leq\mu_{n,1}^{+}+\frac{1}{2}\cdot
3^{-t}$ by Corollary 19 in §7. By [8, Cor. $4$], we have ${\rm
avg}_{D\in\Sigma_{n}^{+}}\\#{\rm
Cl}({\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D}))[3]=2$, and hence

$(\mu_{n}^{+}+\tilde{\mu}_{n,1}^{+})\cdot
1+(1-\mu_{n}^{+}-\tilde{\mu}_{n,1}^{+})\cdot 9\leq 2.$

We conclude that $\mu_{n}^{+}+\tilde{\mu}_{n,1}^{+}\geq\frac{7}{8}$, and
therefore $\mu_{n}^{+}+\mu_{n,1}^{+}\geq\frac{7}{8}-\frac{1}{2}\cdot 3^{-t}$.
∎

By Proposition 3, the total number of cubic fields with discriminant in
$\widetilde{\Sigma}_{n}^{+}$ and corresponding $D\in\Sigma_{n}^{+}$ such that
$D<X$, is

$\sim\left(2^{t}+t\cdot 2^{t-1}\right)X=2^{t-1}(t+2)X.$ (6.1)

On the other hand, the total number of cubic fields with discriminant in
$\widetilde{\Sigma}_{n}^{+}$, with corresponding $D\in U_{n,\leq 1}^{+}$ such
that $D<X$, is at least

$\displaystyle\sim\left(\mu_{n}^{+}(t\cdot 2^{t-1}+2^{t})+\mu_{n,1}^{+}(3\cdot
2^{t-1})\right)X$
$\displaystyle\geq\left(\mu_{n}^{+}(t-1)2^{t-1}+(\mu_{n}^{+}+\mu_{n,1}^{+})(3\cdot
2^{t-1})\right)X$
$\displaystyle\geq\left(\textstyle\frac{1}{2}(t-1)2^{t-1}+(\textstyle\frac{7}{8}-\textstyle\frac{1}{2}\cdot
3^{-t})(3\cdot 2^{t-1})\right)X$ (6.2)

by Propositions 4-5 and Lemma 17. The lower density of (6) in (6.1) is thus at
least

$\frac{\textstyle\frac{1}{2}(t-1)2^{t-1}+(\textstyle\frac{7}{8}-\textstyle\frac{1}{2}\cdot
3^{-t})(3\cdot 2^{t-1})}{2^{t-1}(t+2)}=\frac{1}{2}+\frac{9-4\cdot
3^{-t+1}}{8(t+2)}.$

as $X\to\infty$.

We may now conclude as in Section 5. For $D\in\Sigma_{n}^{+}$, let $B_{D}$ be
the number of monogenic quartic fields with cubic resolvent field of
discriminant $D$ or $D/p_{i}^{2}$ for some $i$. Then

$9(t+1)+2B_{D}\leq
3\biggl{(}3^{\mathop{\mathrm{rk}}{E^{D}}}+3^{\mathop{\mathrm{rk}}{E^{4D}}}+3^{\mathop{\mathrm{rk}}{E^{64D}}}+\sum_{i=1}^{t}\Bigl{(}3^{\mathop{\mathrm{rk}}{E^{D/p_{i}^{2}}}}+3^{\mathop{\mathrm{rk}}{E^{4D/p_{i}^{2}}}}+3^{\mathop{\mathrm{rk}}{E^{64D/p_{i}^{2}}}}\Bigr{)}\biggr{)}$

and so

$(9(t+1)+2B_{D})^{\log_{3}2}\leq
2\biggl{(}2^{\mathop{\mathrm{rk}}{E^{D}}}+2^{\mathop{\mathrm{rk}}{E^{4D}}}+2^{\mathop{\mathrm{rk}}{E^{64D}}}+\sum_{i=1}^{t}\Bigl{(}2^{\mathop{\mathrm{rk}}{E^{D/p_{i}^{2}}}}+2^{\mathop{\mathrm{rk}}{E^{4D/p_{i}^{2}}}}+2^{\mathop{\mathrm{rk}}{E^{64D/p_{i}^{2}}}}\Bigr{)}\biggr{)}.$

By Theorem [1, Thm. 13], we have

$\mu_{n,\leq 1}^{+}\cdot{\rm avg}_{D\in U_{n,\leq
1}^{+}}(9(t+1)+2B_{D})^{\log_{3}{2}}+(1-\mu^{+}_{n,\leq
1})\cdot(3(t+1))\>\leq\>18(t+1).$

Hence, using Lemma 17, we have

${\rm avg}_{D\in U_{n,\leq
1}^{+}}(9(t+1)+2B_{D})^{\log_{3}{2}}\>\leq\>(t+1)(15\cdot(\mu_{n,\leq
1}^{+})^{-1}+3)\>\leq\>21(t+1).$

For $D\in U_{n,\leq 1}^{+}$, the maximum possible value of $B_{D}$ is $3\cdot
2^{t}+t\cdot 3\cdot 2^{t-1}=3\cdot 2^{t-1}(t+2)$. Since (a) $B_{D}\in[0,3\cdot
2^{t-1}(t+2)]$ for all $D\in U_{n,\leq 1}$, and (b) ${\rm avg}_{D\in U_{n,\leq
1}}(9(t+1)+2B_{D})^{\log_{3}2}\leq 21(t+1),$ the maximum of ${\rm avg}_{D\in
U^{+}_{n,\leq 1}}(9(t+1)+2B_{D})$ is achieved when $B_{D}$ takes its maximum
value $3\cdot 2^{t-1}(t+2)$ for a density of
$21(t+1)(3(t+2)2^{t-1})^{-\log_{3}2}$ of $D\in U_{n,\leq 1}$, and $B_{D}=0$
otherwise.

We conclude that the total number of binary quartic fields with cubic
resolvent field of discriminant $D$ or $D/p_{i}^{2}$ for $D\in U_{n,\leq
1}^{+}$ and $D<X$ is $O(t^{2-\log_{3}{2}}\cdot 2^{t\cdot(1-\log_{3}{2})}X)$.
Since, for each $D\in U_{n}^{+}\subset U_{n,\leq 1}^{+}$, there are at least
$2^{t-1}(t+2)$ quartic fields with cubic resolvent field of discriminant $D$
or $D/p_{i}^{2}$ by Proposition 4, it follows that, ordering by the
corresponding $D\in U_{n,\leq 1}^{+}$, the probability that such a quartic
field is binary is $O(t^{1-\log_{3}{2}}\cdot 2^{-t\log_{3}{2}})$.

By [7, Cor. 9], at least $\frac{1}{2}$ of all cubic fields with discriminant
in $\widetilde{\Sigma}_{n}^{+}$ have a totally positive non-square unit. Hence
a proportion of at least

$\frac{1}{2}+\frac{9-4\cdot 3^{-t+1}}{8(t+2)}-\frac{1}{2}\>=\>\frac{9-4\cdot
3^{-t+1}}{8(t+2)}$

of the cubic fields with discriminant in $U_{n,\leq 1}^{+}$ have a totally
positive non-square unit, and hence give rise to a totally real quartic field.
The proportion of these which are not binary is at least
$\frac{9}{8(t+3)}-O(t^{1-\log_{3}{2}}\cdot 2^{-t\log_{3}{2}})$, which is
positive for $t\gg 1$.

##  7\. An auxiliary result on the average number of $3$-torsion elements in
$S$-class groups

We state a slight generalization of a result of Klagsbrun [11] on the average
number of $3$-torsion elements in the $S$-class groups of quadratic fields,
that allows specified local conditions at a finite set of primes not in $S$,
and which was used in the proof of Theorem 1.

For each prime $p$, let $\Sigma_{p}$ be a set of isomorphism classes of étale
quadratic algebras over ${\rm\mathbb{Q}}_{p}$. The collection
$(\Sigma_{p})_{p}$ is acceptable if, for all but finitely many $p$, the set
$\Sigma_{p}$ contains all étale quadratic algebras over ${\rm\mathbb{Q}}_{p}$.
Let $\Sigma$ denote the set of quadratic fields $F$, up to isomorphism, such
that $F\otimes_{\rm\mathbb{Q}}{\rm\mathbb{Q}}_{p}\in\Sigma_{p}$ for each $p$.

######  Theorem 18.

Let $S$ be a finite set of primes, and let $\Sigma=(\Sigma_{p})_{p}$ be any
acceptable set of quadratic fields defined by local conditions such that, for
$p\in S$, the set
$\Sigma_{p}:=\\{{\rm\mathbb{Q}}_{p}\times{\rm\mathbb{Q}}_{p}\\}$ consists of
just the unique split étale quadratic extension of ${\rm\mathbb{Q}}_{p}$.
Then, when quadratic fields with discriminants in $\Sigma$ are ordered by
absolute discriminant:

  1. $(i)$

the average size of ${\rm Cl}(F)_{S}[3]$, as $F$ ranges over imaginary
quadratic fields in $\Sigma$, is equal to $1+3^{-|S|};$ and

  2. $(ii)$

the average size of ${\rm Cl}(F)_{S}[3]$, as $F$ ranges over real quadratic
fields in $\Sigma$, is equal to $1+3^{-|S|-1}$.

###### Proof.

The proof proceeds exactly as in [11], but using the proof method of [8] to
impose additional local conditions at primes not in $S$. All Euler factors at
primes $p\in S$ are identical to those in the proof of [11, Theorem 2], while
all Euler factors at primes $p\notin S$ also remain unchanged due to the
identical cancellation as explained in [8, Proof of Corollary 4, §5.4]. Thus
the averages in [11, Theorem 2] remain the same even upon the imposition of
local conditions at additional primes not in $S$. ∎

######  Corollary 19.

Write $F={\rm\mathbb{Q}}(\sqrt{\scalebox{0.7}[1.0]{$-$}3D})$. The proportion
of $D\in\Sigma_{n}^{\pm}$ such that for each $1\leq i\leq t$, the primes
$[{\rm\mathfrak{p}}_{i}]$ are cubes in ${\rm Cl}(F)$, and such that $\\#{\rm
Cl}(F)[3]=3$, is at most $\frac{1}{2}\cdot 3^{-t}$.

###### Proof.

Let $S:=\\{p_{1},\ldots,p_{t}\\}$. Since $|{\rm Cl}(F)_{S}[3]|=|{\rm
Cl}(F)[3]|=3$ and ${\rm avg}_{D\in\Sigma_{n}^{\pm}}|{\rm
Cl}({\rm\mathbb{Q}}(\sqrt{D})_{S}[3]|$ is at most $1+3^{-t}$ by Theorem 18, it
follows that

$\mathrm{Prob}_{D\in\Sigma_{n}^{\pm}}(\\#{\rm Cl}(F)_{S}[3]=\\#{\rm
Cl}(F)[3]=3)\>\leq\>{\rm avg}_{D\in\Sigma_{n}^{\pm}}\,\frac{|{\rm
Cl}(F)_{S}[3]|-1}{2}\>\leq\>\frac{1}{2}\cdot 3^{-t}.$

∎

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