# Back to Reality for Imitation Learning

Edward Johns
The Robot Learning Lab at Imperial College London

###### Abstract

Imitation learning, and robot learning in general, emerged due to
breakthroughs in machine learning, rather than breakthroughs in robotics. As
such, evaluation metrics for robot learning are deeply rooted in those for
machine learning, and focus primarily on data efficiency. We believe that a
better metric for real-world robot learning is time efficiency, which better
models the true cost to humans. This is a call to arms to the robot learning
community to develop our own evaluation metrics, tailored towards the long-
term goals of real-world robotics.

> Keywords: Imitation learning, Reinforcement learning, Evaluation, Benchmarks

##  1 Introduction

#### How do we define the term robot learning?

Let us consider two options. Firstly, the study of machine learning methods
for applications to robotics. And secondly, the study of methods to enable
robots to learn. The choice between these two can be distilled down to whether
robot learning is considered fundamentally to be a field within machine
learning, or a field within robotics, respectively. But whilst this
distinction may appear to be arbitrary, the implication is in fact profound,
for the following reason. It determines the evaluation metrics used by the
community to judge the performance of new methods, and therefore, it
determines the long-term impact that the field is shaping itself around. In
this paper, we argue that metrics which guide development of a technology
should be driven by the target application of that technology, irrespective of
the methods used by that technology. And therefore, we argue that robot
learning metrics should be driven by real-world robotics applications, rather
than the underlying machine learning algorithms. However, this is not what we
observe in the robot learning community today: evaluation is dominated by
traditional machine learning metrics. In this paper, we expose some problems
with this, which deserve foresight, attention, and debate.

#### Robot learning emerged from machine learning, not robotics.

Whilst the fields of both machine learning and robotics have existed for
decades, the field of robot learning is a more recent development, such as
with the arrival of our first CoRL in 2017. But its growth in popularity can
largely be attributed to breakthroughs in machine learning, and in particular,
breakthroughs in deep learning and reinforcement learning, rather than
breakthroughs in classical robotics. And during this emergence, the robot
learning community has brought with it evaluation metrics that were designed
by the machine learning community, for machine learning problems. For example,
a typical reinforcement learning paper published at a machine learning
conference today, will present graphs of Success Rate vs Number of Environment
Interactions. And we see these same graphs dominating papers published at
CoRL. Now, for the machine learning community, this metric is sensible: it
encourages data efficiency, which is arguably the most fundamental aim in
machine learning research. But is data efficiency the most fundamental aim in
robotics research?

##  2 Back to Reality

#### The curse of simulators.

To answer the above question, we now examine the differences between the
simulation benchmarks typically evaluated on in machine learning papers, and
the real-world environments of robotics applications. Our first observation is
that, whilst simulation benchmarks often assume prior existence of a task
definition, such as a reward function, in reality each new task must be
manually defined by the human teacher. And if we are aiming for robots to
learn from everyday humans in everyday environments, rather than from
engineers coding hand-crafted reward functions, then this should be an
intuitive and natural process, such as a human demonstration of the task. As
such, when we move away from simulators and into the real world, a large
number of robot learning problems can be considered as a form of imitation
learning. This is particularly the case for robot manipulation, where the huge
diversity of tasks makes it impractical to manually define each one in code.
Our second observation is that, whilst simulators facilitate rapid prototyping
and large-scale quantitative benchmarking, this comes at the cost of making
two problematic assumptions: (1) environment resetting comes for free, and (2)
low-dimensional ground-truth states (e.g. object poses) are readily available.
In real-world robotics, however, (1) automatic environment resetting requires
significant human supervision or construction of task-specific apparatus, and
(2) obtaining low-dimensional states requires additional development of a
state estimator for each new task.

#### From data efficiency to time efficiency.

The above observations lead to our main argument: data efficiency should not
be the most fundamental aim in robot learning research, and instead, we should
be optimising our methods for time efficiency. We define time efficiency as
how quickly a robot can learn a new task in a period of time, regardless of
the amount of data collected during that time, and we argue that time is a
much better approximation of the true cost to human teachers. Time has
historically been itself approximated by the amount of data required, but as
we have observed, this approximation is limited. If a method requires regular
environment resetting, then the additional time to reset the environment
should be a penalty, when that method is evaluated relative to other methods
that require less environment resetting. Similarly, if a method requires
object poses, then the additional time taken to train that pose estimator
should be accounted for. But this is not to say that data efficiency should
not be encouraged; data efficiency is always beneficial as long as it does not
come at the cost of time efficiency. Instead, we argue for a more
comprehensive metric than only data efficiency: the best imitation learning
method is one which optimally trades off all aspects that contribute to time
efficiency, and data efficiency is just one of those aspects. We believe that
this new metric will encourage researchers to develop methods which are more
useful to humans in practice, when moving away from the simulators and the
lab, and back into the real world.

#### Two flavours of time efficiency.

Imitation learning involves humans teaching robots. This takes time, but it
also takes effort, and an ideal imitation learning method would be one which
minimises both time and effort. However, these factors are not independent.
For example, a reinforcement learning method seeded with a larger number of
demonstrations may learn more quickly, but at the cost of more human effort.
As such, we now propose two flavours of the time efficiency metric.

Clock time efficiency. We define clock time efficiency as the speed at which a
robot learns a new task with respect to the amount of wall clock time. This
could be the metric of choice in a factory setting, for example, where an
engineer would like a robot to learn a new task as quickly as possible to then
begin a manufacturing job, even if the engineer needs to supervise the
learning throughout.

Human time efficiency. We define human time efficiency as the speed at which a
robot learns a new task with respect to the amount of time a human spends
supervising the learning. This could be the metric of choice in a domestic
setting, for example, where a consumer would like the robot to learn a new
task as autonomously as possible and with minimal human effort, even if the
robot takes a long time to learn.

Figure 1: A comparison of two hypothetical methods using data efficiency
(left) and time efficiency (right) metrics.

#### A choice of two futures.

Figure 1 left shows how robot learning methods are typically evaluated today.
According to the data efficiency metric, the red method is superior. But
hypothetically, let us now consider that the blue method requires no
environment resetting and learns end-to-end from images, whilst the red method
requires regular environment resetting and ground-truth object poses. Figure 1
right then shows how these two methods would be evaluated with a time
efficiency metric. Here, whilst the data efficiency of the blue method is
proportional to its time efficiency, the red method is significantly less time
efficient than it is data efficient. Today, we are being encouraged to develop
the red method. Through this paper, we aim to motivate the robot learning
community towards a future where we are instead developing the blue method.

##  3 Meta-analysis of CoRL 2020

#### Methodology.

Having established our criteria for good imitation learning metrics, we can
now take a look at evaluation methods used by the robot learning community
today. To do this, we performed a meta-analysis of imitation learning papers
published at CoRL 2020. To select these papers, we first searched for all
papers which contained either “imitation” or “demonstration” in the paper’s
abstract. From these, we inspected each paper and retained those which
actually involved providing a demonstration to learn a new task, which
resulted in 19 papers. We then analysed each paper to determine whether it
evaluated success rate as a function of either clock time or human time. We
also recorded whether a paper assumed ground-truth object poses, and whether
it required environment resetting. In some cases, the experiments themselves
did not require environment resetting but only because the tasks were simple
target reaching tasks, and in these cases we made a judgement as to whether
environment resetting would be required with more typical, everyday tasks. For
each paper, we also recorded whether any real-world experiments were done.

#### Results.

Table 1 shows the results of this meta-analysis. Papers are grouped into 8
groups, where all the papers in one group contained the same yes/no answers
across all 5 columns. The first observation we make is that not a single paper
evaluated the clock time efficiency or human time efficiency, with most
methods instead evaluating only data efficiency, and thus ignoring other real-
world costs associated with the method. The second observation we make is that
10 out of the 19 papers assumed access to ground-truth object poses, and 13
papers would require regular (and most likely manual) environment resetting in
real-world applications. Neither of these properties make for practical, real-
world imitation learning, and the utility of these methods is thus
artificially inflated due to simulation benchmarking. A final observation we
make is that 8 out of the 19 papers only included simulation experiments, an
unfortunate trend at all CoRL conferences which results in authors focussing
mainly on machine learning metrics, rather than those more appropriate for
real-world robotics. And often, those papers which did include real-world
experiments, required ground-truth object poses [1] or object-specific
keypoint detectors [2], but timing information was not provided for how to
train the relevant predictors for each new task.

# Papers | Clock Time Efficiency? | Human Time Efficiency? | Assumes Object Poses? | Environment Resetting? | Real-World?
---|---|---|---|---|---
4 [3, 4, 5, 6] | No | No | Yes | Yes | No
3 [7, 8, 1] | No | No | Yes | Yes | Yes
3 [9, 10, 11] | No | No | No | Yes | No
3 [12, 2, 13] | No | No | No | Yes | Yes
3 [14, 15, 16] | No | No | Yes | No | Yes
1 [17] | No | No | No | No | Yes
1 [18] | No | No | No | No | No
1 [19] | No | No | No | No | Yes
Table 1: Meta-analysis of 19 imitation learning papers at CoRL 2020.

##  4 Moving Forwards

#### Simulators are still important.

We have established our argument for the use of time efficiency instead of
data efficiency as an evaluation metric. But in practice, both clock time
efficiency and human time efficiency are difficult to objectively evaluate
with real-world experiments. For example, if the authors themselves are the
robot teachers, then there are likely to be implicit or explicit biases in the
time taken, depending on whether the authors’ own method, or a baseline, are
being evaluated. As such, we still recommend the use of simulation benchmarks
for large-scale objective evaluation, but with some significant changes that
will mitigate the aforementioned problems with these benchmarks. Specifically,
these changes should model any aspects of task learning which require time to
perform. For example, each time an environment would be manually reset, a
fixed amount of time should be added to both the method’s clock time and it’s
human time counters. As another example, each task can optionally be provided
with ground-truth object poses, but a fixed amount of time must then be added
to the method’s clock time counter, to model the user or robot collecting data
for an object pose estimator. Given benchmarks such as these, we believe that
the community will begin to develop methods which more optimally trade off all
the different aspects of real-world robot learning.

#### Visualising time efficiency.

In Figure 2, we present various options for visualising these metrics, on
hypothetical data. Figure 2(a) compares three different methods by plotting
success rates as a function of human time. Whilst Method D has the highest
human time efficiency, it is not necessarily the most data efficient.
Therefore, it would not necessarily be considered a desirable method if
traditional machine learning metrics were used. Figure 2(b) then presents a
more complete evaluation of one individual method. Here, each line represents
a particular success rate and trades off the amount of human time required,
and the overall clock time required. This is a useful visualisation for
choosing an operating point during deployment, where a user may have a
particular preference for minimising overall training time, compared to
minimising overall human effort or expertise. Figure 2(c) then shows the
trade-off between pre-training time and fine-tuning time across three
different methods. For example, meta-learning methods typically require a
large amount of pre-training, whereas behavioural cloning methods typically
learn from scratch. Here, each line represents one success rate (e.g. the
amount of time needed for 90 % success rate). This graph provides an intuitive
way to compare methods across entirely different families of imitation
learning approaches, which is typically not possible when considering only one
of these dimensions. Here, Method I outperforms both Methods G and H, and we
can see that Method H is faster than Method G at learning new tasks when there
is an abundance of pre-training time, whereas Method G is better when learning
from scratch.

(a)

(b)

(c)

Figure 2: Examples of how evaluations could be visualised using our proposed
clock time efficiency and human time efficiency metrics.

##  5 Conclusions

In this paper, we have argued that data efficiency is not the optimal
evaluation metric for robot learning, when the long-term goal is real-world
robotics. Instead, two alternative aspects should be considered: the overall
time taken for learning a new task, and the amount of time spent by the human
supervising the learning. Methods which are the most data efficient may not
necessarily be the most time efficient, and may sacrifice consideration of
real-world practicalities in an attempt to beat the state-of-the-art on
simulation benchmarks. Our intention through this paper, is to stimulate the
community to develop benchmarks which more accurately reflect the costs to
humans in real-world imitation learning. Re-thinking evaluation with these
real-world costs, outside of purely simulated environments, has inspired our
own recent work on Coarse-to-Fine Imitation Learning: from just a single human
demonstration, with just a single environment reset, and without any prior
object knowledge [20, 21].

### Acknowledgments

This work was supported by the Royal Academy of Engineering under the Research
Fellowship scheme.

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# Ring class fields and a result of Hasse

Ron Evans  Department of Mathematics, UCSD
La Jolla, CA 92093-0112<EMAIL_ADDRESS>https://mathweb.ucsd.edu/ revans ,
Mark Van Veen  2138 Edinburg Avenue
Cardiff by the Sea, CA 92007<EMAIL_ADDRESS>and  Franz Lemmermeyer
Mörikeweg 1
73489 Jagstzell, Germany<EMAIL_ADDRESS>https://www.mathi.uni-
heidelberg.de/ flemmermeyer

(Date: March 2024)

###### Abstract.

For squarefree $d>1$, let $M$ denote the ring class field for the order
$\mathbb{Z}[\sqrt{-3d}]$ in $F=\mathbb{Q}(\sqrt{-3d})$. Hasse proved that $3$
divides the class number of $F$ if and only if there exists a cubic extension
$E$ of $\mathbb{Q}$ such that $E$ and $F$ have the same discriminant. Define
the real cube roots $v=(a+b\sqrt{d})^{1/3}$ and
$v^{\prime}=(a-b\sqrt{d})^{1/3}$, where $a+b\sqrt{d}$ is the fundamental unit
in $\mathbb{Q}(\sqrt{d})$. We prove that $E$ can be taken as
$\mathbb{Q}(v+v^{\prime})$ if and only if $v\in M$. As byproducts of the
proof, we give explicit congruences for $a$ and $b$ which hold if and only if
$v\in M$, and we also show that the norm of the relative discriminant of
$F(v)/F$ lies in $\\{1,3^{6}\\}$ or $\\{3^{8},3^{18}\\}$ according as $v\in M$
or $v\notin M$. Some of the results above are extended for subsets of
$\mathbb{Q}(\sqrt{d})$ properly containing the fundamental units
$a+b\sqrt{d}$.

###### Key words and phrases:

ring class fields, fundamental units, class number, relative discriminants,
Artin symbol, cubic residuacity

###### 1991 Mathematics Subject Classification:

11R11, 11R27, 11R29, 11R37

##  1\. Introduction

Write $a+b\sqrt{d}$ for the fundamental unit in $\mathbb{Q}(d)$, where $d>1$
is squarefree. Define the real cube roots $v=(a+b\sqrt{d})^{1/3}$ and
$v^{\prime}=(a-b\sqrt{d})^{1/3}$. Note that $vv^{\prime}=\pm 1$. Write
$t=v+v^{\prime}$ and $F=\mathbb{Q}(\sqrt{-3d})$. Let $M$ denote the ring class
field for the order $\mathbb{Z}[\sqrt{-3d}]$ in $F$. For number field
extensions $K/k$, write $D(K/k)$ for the relative discriminant and $D(K)$ for
the discriminant.

In [5], Hasse proved that $3$ divides the class number of $F$ if and only if
there exists a cubic extension $E$ of $\mathbb{Q}$ such that $D(E)=D(F)$. When
$v\in M$, we prove in Theorems 1.1 and 1.2 below that $E$ can be taken to be
$\mathbb{Q}(t)$. Theorems 1.1 and 1.2 address the cases $3\nmid d$ and $3\mid
d$, respectively. They are proved in Sections 2 and 3.

######  Theorem 1.1.

Suppose that $3\nmid d$ and $v\in M$. Then the fields $E=\mathbb{Q}(t)$ and
$F$ have the same discriminant.

######  Theorem 1.2.

Suppose that $3\mid d$ and $v\in M$. Then the fields $E=\mathbb{Q}(t)$ and $F$
have the same discriminant.

Remarks (2A) and (3B) show that $v\in M$ implies that $3$ divides the class
number of $F$. However, the converse is false; the smallest counterexample
when $3\nmid d$ is $d=142$, while the smallest counterexample when $3\mid d$
is $d=786$.

When $v\notin M$, then in contrast with the theorems above, the discriminant
of $\mathbb{Q}(t)$ does not equal $D(F)$. In fact, when $v\notin M$, Theorems
1.3 and 1.4 show that $D(\mathbb{Q}(t))$ equals $9D(F)$ or $81D(F)$ according
as $3\nmid d$ or $3\mid d$. The proofs are given in Sections 4 and 5.

######  Theorem 1.3.

Suppose that $3\nmid d$ and $v\notin M$. Then the field $E=\mathbb{Q}(t)$ has
discriminant $9D(F)$.

######  Theorem 1.4.

Suppose that $3\mid d$ and $v\notin M$. Then the field $E=\mathbb{Q}(t)$ has
discriminant $81D(F)$.

The proofs will use the following notation: $w=(-1+\sqrt{-3})/2$,
$C=\mathbb{Q}(\sqrt{d})$, $B=\mathbb{Q}(\sqrt{-3},\sqrt{d})=F(\sqrt{d})=F(w)$,
$k=\mathbb{Q}(t)$, $K=\mathbb{Q}(v)=k(\sqrt{d})$, and
$L=F(v)=\mathbb{Q}(w,v)=K(w)$. The Galois extension $L/F$ is cyclic of degree
6. As a byproduct of the proofs, we have the following evaluation of $D(L/F)$;
for a generalization, see Conjecture 7.7.

######  Theorem 1.5.

When $v\in M$, the norm of the relative discriminant $D(L/F)$ equals $1$ or
$3^{6}$ according as $3\nmid d$ or $3\mid d$; and when $v\notin M$, the norm
of $D(L/F)$ equals $3^{8}$ or $3^{18}$ according as $3\nmid d$ or $3\mid d$.

###### Proof.

See Remarks (2A), (3D), (4F), and (5G), respectively. ∎

The inclusion $v\in M$ is connected to cubic residuacity of $u\pmod{p}$, where
$p$ is any prime of the form $p=x^{2}+3dy^{2}$. This is shown in Theorem 1.6
below.

When $d=2$, we have $\left(\displaystyle{\frac{2}{p}}\right)=1$. When $d$ is
odd, one of $d,p$ is $1\pmod{4}$, so that
$\left(\displaystyle{\frac{d}{p}}\right)=\left(\displaystyle{\frac{p}{d}}\right)=1$.
Thus $d$ is a square $\pmod{p}$, so that $u=a+b\sqrt{d}$ can be viewed as a
rational integer $u_{p}\pmod{p}$.

######  Theorem 1.6.

$v\in M$ if and only if $u_{p}$ is a cubic residue $\pmod{p}$.

###### Proof.

Consider the principal prime ideal $\mathfrak{p}=(x+y\sqrt{-3d})$ in $F$ of
norm $p$. By Theorem 1.5, $\mathfrak{p}$ is unramified in $L$. Let $\sigma$
denote the Artin symbol
$\left(\displaystyle{\frac{L/F}{\mathfrak{p}}}\right)$, and let $\mathfrak{P}$
be a prime ideal in $L$ above $\mathfrak{p}$. Since $\sigma(v)\equiv
v^{p}\pmod{\mathfrak{P}}$, we see that $\sigma$ is trivial on $L$ iff and only
if $v^{p-1}\equiv 1\pmod{\mathfrak{P}}$. This last congruence is equivalent to
$u_{p}^{(p-1)/3}\equiv 1\pmod{p}$. Thus $u_{p}$ is a cubic residue $\pmod{p}$
if and only if $\sigma$ is trivial on $L$. It remains to show that $\sigma$ is
trivial on $L$ if and only if $L\subset M$. We know that $\sigma$ is trivial
on $L$ if and only if the primes $p$ split completely in $L$. By [3, Thm.
8.19], the primes $p$ split completely in $L$ if and only if $L\subset M$, so
the proof is complete. ∎

In Theorem 6.1, we give explicit criteria in terms of $a$ and $b$ for
$v=(a+b\sqrt{d})^{1/3}$ to lie in $M$, where $a+b\sqrt{d}$ is the fundamental
unit.

Theorem 6.2 relates the ring class fields $M$ and $M_{3}$ to ray class fields
of $F$, where $M_{3}$ denotes the ring class field of $F$ for the order
$\mathbb{Z}[\sqrt{-27d}]$.

In Section 7, we introduce a large class $S_{d}$ of integers $r+s\sqrt{d}$
with cubic norms, where $S_{d}$ properly contains the set of fundamental units
$a+b\sqrt{d}$. Under certain conditions, we extend Theorem 6.1 for elements in
$S_{d}$.

##  2\. Proof of Theorem 1.1

Assume throughout this section that $v\in M$. The proof of Theorem 1.1
utilizes the following five lemmas.

######  Lemma 2.1.

When $3\nmid d$, $L/F$ is unramified.

###### Proof.

The principal prime ideals in $F$ of norm $p=x^{2}+3dy^{2}$ are the ones in
$F$ that split completely in $M$. The same set of prime ideals split
completely in $F(w)$, as the corresponding Artin symbols for the extension
$F(w)/F$ act trivially on $w$, since $w^{p}=w$. Thus by [3, Thm. 8.19],
$L\subset M$.

We may assume that $d\equiv 1\pmod{8}$; otherwise, by [3, Thm. 7.24], $M$ is
the Hilbert class field of $F$ so that $L/F$ is unramified. Under this
assumption, the order $\mathbb{Z}[\sqrt{-3d}]$ has conductor $2$, so it
suffices to show that $2$ is unramified in $L$. Consider the tower
$\mathbb{Q}\subset C\subset K\subset L$. Clearly $C/\mathbb{Q}$ is unramified
at $2$. Also $K/C=\mathbb{Q}(v)/\mathbb{Q}(\sqrt{d})$ is unramified at $2$,
since the polynomial $x^{3}-v^{3}$ has discriminant $-27v^{6}$. Finally,
$L/K=K(w)/K$ is unramified at $2$, since the polynomial $x^{2}+x+1$ has
discriminant $-3$. ∎

######  Remark (2A).

Let $3\nmid d$ and $v\in M$. Since $L/F$ is a cyclic unramified extension of
degree $6$ when $3\nmid d$ by Lemma 2.1, the class number of $F$ is divisible
by $6$, and $D(L/F)$ has (absolute) norm $1$. This is in contrast with the
case $3\mid d$; see Remark (3D).

######  Lemma 2.2.

When $3\nmid d$, $D(L/K)$ has norm $9$.

###### Proof.

We need only examine the ramification at $3$, since the polynomial $x^{2}+x+1$
has discriminant $-3$. Note that $3$ ramifies in $F$ but there can be no
further ramification at $3$ in $L/F$ by Lemma 2.1. Thus in the factorization
of $(3)$ in $L$, every prime ideal occurs to the second power.

The minimal polynomial of $t=v+v^{\prime}$ is $x^{3}-3\epsilon x-2a$, where
$\epsilon$ is the norm of the fundamental unit $v^{3}=a+b\sqrt{d}$. This
polynomial has discriminant $-108db^{2}$, in which the exponent of the factor
$3$ is odd. By [7, Prop. 2.13], $3$ divides $D(k)$, so $3$ must ramify in $k$
and in $K$.

The factorization of $(3)$ in $C$ is either $\mathfrak{q}$ or
$\mathfrak{p}\mathfrak{p}^{\prime}$, where $\mathfrak{q}$ has norm $9$ and
$\mathfrak{p}$, $\mathfrak{p}^{\prime}$ each have norm $3$. Since $3$ ramifies
in $K$, we have in $K$ either the prime ideal factorization
$(\mathfrak{q})=\mathfrak{Q}_{1}^{2}\mathfrak{Q}$ where the prime factors have
norm $9$, or
$(\mathfrak{p}\mathfrak{p}^{\prime})=\mathfrak{P}\mathfrak{P}_{1}^{2}\mathfrak{P}^{\prime}\mathfrak{P}_{1}^{\prime
2}$ where the prime factors have norm $3$. In the first case, $\mathfrak{Q}$
is the factor that ramifies in $L$, while in the second case, $\mathfrak{P}$
and $\mathfrak{P}^{\prime}$ are the factors that ramify in $L$. Since the
ramification is tame, $D(L/K)$ equals $\mathfrak{Q}$ or
$\mathfrak{P}\mathfrak{P}^{\prime}$, and in either case $D(L/K)$ has norm 9. ∎

######  Lemma 2.3.

When $q\neq 3$ is a rational prime, $(q)\nmid D(K/k)$ in $k$.

###### Proof.

Suppose for the purpose of contradiction that $(q)$ divides $D(K/k)$. Since
$q$ divides the discriminant of the $k$-basis $\\{1,v\\}$ in $K$, $q$ must
divide

(2.1) $t^{2}-4=v^{2}+v^{\prime 2}-2.$

Replacing $v$ in (2.1) by its conjugate $wv$ in $L$, we see that in $L$, $q$
divides

(2.2) $w^{2}v^{2}+wv^{\prime 2}-2.$

Subtracting, we see that $q$ divides

(2.3) $(w^{2}-1)v^{2}+(w-1)v^{\prime 2}.$

Replacing $w$ by $w^{2}$ in (2.3), we see that $q$ divides

(2.4) $(w-1)v^{2}+(w^{2}-1)v^{\prime 2}.$

Multiplying (2.3) by $w+1$ and then subtracting from (2.4), we see that $q$
divides $3wv^{2}$, which is impossible, since $q\neq 3$. This contradiction
proves the lemma. ∎

######  Lemma 2.4.

Suppose that $3\nmid d$. The only rational primes that ramify in $L$ are the
ones that divide $D(F)$. If the rational prime $p$ divides $D(C)$, then each
prime ideal in the factorizations of $(p)$ in $K$ and $L$ occurs to the second
power.

###### Proof.

If the prime $p$ divides $D(F)$, then $p$ ramifies in $F$ with ramification
index $2$. Then since $L/F$ is unramified by Lemma 2.1, each prime ideal in
the factorization of $(p)$ in $L$ occurs to the second power. If $p$ does not
divide $D(F)$, then clearly $p$ does not ramify in $L$.

Next, suppose that $p$ divides $D(C)$ (so that $p\neq 3$). Then $p$ ramifies
in $C$ with ramification index $2$. Recall that $K/C$ is unramified at $p$,
since the polynomial $x^{3}-v^{3}$ has discriminant $-27v^{6}$. Thus each
prime ideal factor in the factorization of $(p)$ in $K$ occurs to the second
power. ∎

######  Lemma 2.5.

When $3\nmid d$, $D(K)=9D(C)^{3}$.

###### Proof.

By Lemma 2.1, $D(L)=D(F)^{6}$. Thus by Lemma 2.2,

(2.5) $D(K)^{2}=D(L)/9=D(F)^{6}/9.$

Since $D(K)$ is positive [7, Prop. 2.15],

(2.6) $D(K)=-D(F)^{3}/3=9D(C)^{3}.$

∎

We are now prepared for the proof of Theorem 1.1

###### Proof.

Consider the set $S$ of rational primes $p$ that divide the discriminant
$D(C)$. Note that $3\notin S$. By Lemma 2.4, each prime ideal in the
factorization of $(p)$ in $K$ (as well as in $L$) occurs to the second power.
Each prime ideal in the factorization of $(p)$ in $k$ must occur to either the
first or second power, and those occurring to the first power are exactly the
ones that ramify in $K$. Those that ramify tamely in $K$ are the only ones
that divide $D(K/k)$ to the first power [7, p. 260].

For $p\in S$, if every prime ideal factor of $(p)$ in $k$ were to occur to the
first power, then $(p)$ would divide $D(K/k)$, contradicting Lemma 2.3. Thus
$p$ ramifies in $k$, so there is a unique prime ideal $\mathfrak{p}$ in $k$
that divides $(p)$ to the first power. Note that $\mathfrak{p}$ has norm $p$.
If $p=2$, then $\mathfrak{p}^{e}$ exactly divides $D(K/k)$ for some $e\geq 2$
depending on $d$, while if $p>3$, then $\mathfrak{p}$ exactly divides
$D(K/k)$.

We proceed to say more about the value of $e$. To distinguish the prime ideal
$\mathfrak{p}$ in the case $p=2$, call it $\mathfrak{p}_{2}$. The discriminant
of the $k$-basis $\\{1,\sqrt{d}\\}$ for $K$ is $4d$, so that $D(K/k)$ divides
$4d$. First suppose that $d\equiv 3\pmod{4}$. Since $\mathfrak{p}_{2}$ divides
$(2)$ to the first power in $k$, $\mathfrak{p}_{2}$ divides $(4d)$ to the
second power. Thus $e\leq 2$ in this case, so that $e=2$. Next suppose that
$d\equiv 2\pmod{4}$. Then $8$ divides $4d$, so that $e\in\\{2,3\\}$ in this
case.

So far we have shown that

(2.7) $D(K/k)=\mathfrak{p}_{2}^{e}\prod_{3<p\in S}\mathfrak{p},$

where $\mathfrak{p}_{2}$ is to be interpreted as $1$ when $d\equiv 1\pmod{4}$.
(Note that no prime ideal above $3$ can occur in this product, since $3$ does
not divide $4d$.) Taking absolute norms on both sides of (2.7), we have

(2.8) $D(K)/D(k)^{2}=\begin{cases}d,\quad\quad&d\equiv 1\pmod{4}\\\
4d,\quad\quad&d\equiv 3\pmod{4}\\\ 2^{e-1}d,\quad\quad&d\equiv
2\pmod{4}.\end{cases}$

By Lemma 2.5,

(2.9) $D(K)=\begin{cases}9d^{3},\quad\quad&d\equiv 1\pmod{4}\\\ 9\cdot
2^{6}d^{3},\quad&d\equiv 2,3\pmod{4}.\end{cases}$

Thus

(2.10) $D(k)^{2}=\begin{cases}9d^{2},\quad\quad&d\equiv 1\pmod{4}\\\ 9\cdot
2^{4}d^{2},\quad\quad&d\equiv 3\pmod{4}\\\ 9\cdot
2^{7-e}d^{2},\quad\quad&d\equiv 2\pmod{4}.\end{cases}$

This shows that $e$ must be odd, so $e=3$. Finally, since $D(k)$ is negative
[7, Prop. 2.15], we obtain the desired result

(2.11) $D(k)=\begin{cases}-3d,\quad\quad&d\equiv 1\pmod{4}\\\
-12d,\quad&d\equiv 2,3\pmod{4}.\end{cases}$

∎

##  3\. Proof of Theorem 1.2

Assume throughout this section that $v\in M$. Write $d=3m$ so that
$F=\mathbb{Q}(\sqrt{-m})$. Note that the order
$\mathbb{Z}[\sqrt{-3d}]=\mathbb{Z}[3\sqrt{-m}]$ has conductor $3$ or $6$. The
proof of Theorem 1.2 utilizes the following two lemmas.

######  Lemma 3.1.

When $3\mid d$, the extension $F(t)/F$ is unramified.

###### Proof.

Let $J$ denote the ring class field for the order $\mathbb{Z}[\sqrt{-m}]$.
Note that this order has conductor $1$ or $2$. The formula for class numbers
of orders [3, Thm. 7.24] shows that the extension $M/J$ has degree $2$ or $4$.
Note that $t\in M$, since $v\in M$. We claim that moreover $t\in J$.

Since $t$ has degree $3$ over $\mathbb{Q}$ and $J(t)\subset M$, we have
$|J(t)/J|\leq 2$. Assume for the purpose of contradiction that equality holds.
The cubic minimal polynomial of $t$ over $\mathbb{Q}$ is divisible by the
quadratic minimal polynomial of $t$ over $J$. Therefore some conjugate of $t$,
and thus $t$ itself, lies in $J$. This contradiction proves the claim.

Consider the tower $F\subset F(w)\subset F(v)=L$. The extensions $F(w)/F$ and
$L/F(w)=\mathbb{Q}(w)(v)/\mathbb{Q}(w)(\sqrt{d})$ cannot ramify at any
rational prime other than $3$. Thus the same is true of the extension
$F(t)/F$. Since $F(t)\subset J$ and $3$ does not divide the conductor of
$\mathbb{Z}[\sqrt{-m}]$, the extension $F(t)/F$ must be unramified. ∎

######  Remark (3B).

By Lemma 3.1, $F(t)/F$ is a cyclic unramified cubic extension when $3\mid d$.
Thus $F(t)$ lies in the Hilbert class field of $F$, so that $3$ divides the
class number of $F$.

######  Remark (3C).

Consider the principal prime ideal $\mathfrak{p}=(x+y\sqrt{-m})$ in $F$ of
norm $p$. By Lemma 3.1, the Artin symbol for the extension $L/F$ fixes $t$, so
it maps $v$ to either $v$ or $v^{\prime}$. In the first case, $p\equiv
1\pmod{3}$ and $v^{p-1}\equiv 1\pmod{\mathfrak{P}}$, while in the second case,
$p\equiv-1\pmod{3}$ and $v^{p+1}\equiv 1\pmod{\mathfrak{P}}$, where
$\mathfrak{P}$ is a prime ideal of $L$ above $\mathfrak{p}$.

######  Remark (3D).

When $3\mid d$ and $v\in M$, $D(L/F)$ has norm $729$. To see this, first
observe that by Lemma 3.1, $F(t)/Q$ is unramified at $3$. Note that
$L=F(t)(\sqrt{-3})$. It follows that $D(L/F(t))=(3)$, since by [9, p. 685],
$D(L/F(t))$ is the product of the prime ideals in the factorization of $(3)$
in $F(t)$ which divide $(3)$ to the first power. Taking the norm, we obtain
$D(L)/D(F(t))^{2}=3^{6}$. By Lemma 3.1, $D(F(t))=D(F)^{3}$, so that
$D(L)/D(F)^{6}=3^{6}$. This proves that $D(L/F)$ has norm $729$.

######  Lemma 3.2.

Suppose that $3\mid d$. The only rational primes that ramify in $L$ are the
ones that divide $D(C)$. For a rational prime $p$ dividing $D(C)$, each prime
ideal in the factorizations of $(p)$ in $K$ and $L$ occurs to the second
power. Consequently the extension $K/C$ is unramified.

###### Proof.

If $p$ divides $D(C)$, then $p$ ramifies in $C$ with ramification index $2$.
First assume that $p\neq 3$. Then $L/C$ is unramified at $p$, since $K/C$ and
$L/K$ are unramified at $p$. Thus each prime ideal in the factorizations of
$(p)$ in $K$ and $L$ occurs to the second power. Next consider the case $p=3$.
We know that $3$ ramifies in $K$ and in $L$, because it ramifies in $C$. By
Lemma 3.1, $3$ does not ramify in $F(t)$. Since $F(t)$ is a subfield of $L$ of
index $2$, it follows that each prime ideal in the factorizations of $(3)$ in
$K$ and $L$ occurs to the second power. Thus when $p$ divides $D(C)$, there
can be no further ramification from $C$ to $L$, so that $L/C$ is unramified. ∎

We are now prepared for the proof of Theorem 1.2.

###### Proof.

Consider the set $S$ of rational primes $p$ that divide the discriminant
$D(C)$. Note that $3\in S$. By Lemma 3.2, each prime ideal in the
factorization of $(p)$ in $K$ occurs to the second power. Each prime ideal in
the factorization of $(p)$ in $k$ must occur to either the first or second
power, and those occurring to the first power are exactly the ones that ramify
in $K$.

Mimicking the proof of (2.7), we obtain

(3.1) $D(K/k)=3\mathfrak{p}_{2}^{e}\prod_{3<p\in S}\mathfrak{p}.$

The reason for the factor $3$ is as follows. The proof of Theorem 1.1 applies
here to show that all primes in $S$ other than $3$ ramify in $k$. However, $3$
does not ramify in $k$; this is due to Lemma 3.1, since $k\subset F(t)$.
Consequently each prime ideal in the factorization of $(3)$ in $k$ occurs to
the first power, so each ramifies tamely in $K$. This explains the factor $3$
in (3.1).

Taking absolute norms on both sides of (3.1), we have

(3.2) $D(K)/D(k)^{2}=\begin{cases}9d,\quad\quad&d\equiv 1\pmod{4}\\\ 4\cdot
9d,\quad\quad&d\equiv 3\pmod{4}\\\ 2^{e-1}9d,\quad\quad&d\equiv
2\pmod{4}.\end{cases}$

By Lemma 3.2, $D(K)=D(C)^{3}$ so that

(3.3) $D(K)=\begin{cases}d^{3},\quad\quad&d\equiv 1\pmod{4}\\\
2^{6}d^{3},\quad&d\equiv 2,3\pmod{4}.\end{cases}$

Thus

(3.4) $D(k)^{2}=\begin{cases}m^{2},\quad\quad&d\equiv 1\pmod{4}\\\
2^{4}m^{2},\quad\quad&d\equiv 3\pmod{4}\\\ 2^{7-e}m^{2},\quad\quad&d\equiv
2\pmod{4}.\end{cases}$

This shows that $e$ must be odd, so $e=3$. Finally, since $D(k)$ is negative,
we obtain the desired result

(3.5) $D(k)=\begin{cases}-m,\quad\quad&d\equiv 1\pmod{4}\\\ -4m,\quad&d\equiv
2,3\pmod{4}.\end{cases}$

∎

##  4\. Proof of Theorem 1.3

Assume throughout this section that $v\notin M$. The proof of Theorem 1.3
utilizes the following five lemmas.

######  Lemma 4.1.

Suppose that $3\nmid d$. If $D(C)$ is divisible by the prime $q$, then each
prime ideal in the factorizations of $(q)$ in $F(t)$, $K$, and $L$ occurs to
the second power. On the other hand, if $D(C)$ is not divisible by $q$ and
$q\neq 3$, then $q$ is unramified in $L$.

###### Proof.

First suppose that $q\nmid D(C)$ and $q\neq 3$. Then the extension
$C/\mathbb{Q}$ is unramified at $q$. The extension $K/C$ is also unramified at
$q$ since it can ramify only at $3$, due to the fact that $x^{3}-v^{3}$ has
discriminant $-27v^{6}$. Moreover, $L/K$ can ramify only at $3$, since
$L=K(w)$ and $x^{2}+x+1$ has discriminant $-3$. Thus $q$ is unramified in $L$.

Next suppose that $q$ divides $D(C)$. Then $q$ ramifies in $C$ and in $F$, but
there can be no further ramification in the extension $L/C$, since $q\neq 3$.
Thus each prime ideal in the factorizations of $(q)$ in $F(t)$, $K$, and $L$
occurs to the second power. ∎

######  Lemma 4.2.

When $3\nmid d$, the extensions $L/F(t)$ and $B/F$ are unramified.

###### Proof.

Any prime ideal in $F(t)$ ramifying in $L=F(t,w)=F(t,\sqrt{d})$ would have to
divide both $F(t)$-basis discriminants $4d$ and $-3$, which is impossible. The
extension $B/F$ is unramified for the same reason. ∎

######  Lemma 4.3.

When $3\nmid d$, the extension $L/B$ ramifies only at $3$.

###### Proof.

In view of Lemma 4.1, we need only show that the extension $L/B$ is ramified.
Suppose for the purpose of contradiction that it isn’t. Then by Lemma 4.2, the
extension $L/F$ would be unramified. Consequently $L\subset M$, which
contradicts $v\notin M$. ∎

######  Lemma 4.4.

When $3\nmid d$, $D(L/B)$ has norm $3^{8}$ and $D(K/C)$ has norm $3^{6}$.

###### Proof.

We have the prime ideal factorization $(3)=(\sqrt{-3})^{2}$ in
$\mathbb{Q}(\sqrt{-3})$. In $B$, either $(\sqrt{-3})=\mathfrak{q}$ or
$(\sqrt{-3})=\mathfrak{p}\mathfrak{p}^{\prime}$, where $\mathfrak{q}$ has norm
$9$ and $\mathfrak{p},\mathfrak{p}^{\prime}$ have norm $3$. By Lemma 4.3,
these prime ideals ramify wildly in the cubic cyclic extension $L/B$, with
ramification index $3$. Thus for some integer $s\geq 3$,
$D(L/B)=\mathfrak{q}^{s}$ or $D(L/B)=(\mathfrak{p}\mathfrak{p}^{\prime})^{s}$.
Consequently $D(L/B)$ has norm $9^{s}$. Equivalently,

(4.1) $D(L)=9^{s}D(B)^{3}.$

We now know that the ramification index of $3$ in $L/\mathbb{Q}$ is divisible
by $3$. Therefore the ramification index in the cubic extension $K/C$ at $3$
must equal $3$. In $C$, either $(3)=\mathfrak{q}$ or
$(3)=\mathfrak{p}\mathfrak{p}^{\prime}$, where $\mathfrak{q}$ has norm $9$ and
$\mathfrak{p},\mathfrak{p}^{\prime}$ have norm $3$. Therefore we have wild
ramification in $K$ of the form $(3)=\mathfrak{Q}^{3}$ or
$(3)=(\mathfrak{P}\mathfrak{P}^{\prime})^{3}$, where $\mathfrak{Q}$ has norm
$9$ and $\mathfrak{P},\mathfrak{P}^{\prime}$ have norm $3$. It follows that
$D(K/C)$ equals $\mathfrak{q}^{r}$ or
$(\mathfrak{p}\mathfrak{p}^{\prime})^{r}$ for some integer $r\geq 3$. Thus
$D(K/C)$ has norm $9^{r}$ Moreover, since $D(K/C)$ divides $27v^{6}$, the norm
of $D(K/C)$ divides $27^{2}=9^{3}$.Thus $r=3$, so that $D(K/C)$ has the
desired norm $3^{6}$. Equivalently,

(4.2) $D(K)=3^{6}D(C)^{3}=-3^{3}D(F)^{3}.$

By Lemma 4.2, $D(B)=D(F)^{2}$. Thus, by (4.1),

(4.3) $D(L)=9^{s}D(F)^{6}.$

We next evaluate $D(L/K)$. Since $L=K(\sqrt{-3})$, $D(L/K)$ equals the product
of the prime ideals in the factorization of $(3)$ in $K$ that occur to odd
powers. Thus, in the notation above, $D(L/K)$ equals $\mathfrak{Q}$ or
$\mathfrak{P}\mathfrak{P}^{\prime}$. Consequently, $D(L/K)$ has norm $9$, so
that by (4.2),

(4.4) $D(L)=9D(K)^{2}=3^{8}D(F)^{6}.$

Comparing equations (4.3) and (4.4), we see that $s=4$. Therefore, by (4.1),
$D(L/B)$ has the desired norm $3^{8}$. ∎

######  Remark (4E).

Since each prime ideal factor of $D(L/B)$ occurs to the power $s=4$, the
corresponding higher ramification groups $G_{i}$ must be trivial for $i\geq
2$, in view of [7, p. 265]. This is because $s=4=(|G_{0}|-1)+(|G_{1}|-1)$,
where $G_{0}=G_{1}=\mathrm{Gal}(L/B)$.

######  Lemma 4.5.

When $3\nmid d$, $D(F(t))=3^{4}D(F)^{3}$.

###### Proof.

By (4.4), $D(L)=3^{8}D(F)^{6}$. By Lemma 4.2, $D(F(t))^{2}=D(L)$, so that
$D(F(t))=3^{4}D(F)^{3}$. ∎

######  Remark (4F).

When $3\nmid d$ and $v\notin M$, it follows from (4.4) that $D(L/F)$ has norm
$3^{8}$. This is in contrast with the case $3\mid d$; see Remark (5G).

We are now prepared for the proof of Theorem 1.3.

###### Proof.

Since $F(t)=k(\sqrt{-3d})$, the odd part of $D(F(t)/k)$ is the product of the
prime ideals of odd norm in the factorization of $(-3d)$ in $k$ which occur to
odd powers [9]. Let $p$ be any odd prime dividing $3d$. By [7, Prop. 2.13],
$p|D(k)$, since the minimal polynomial of $t$ over $\mathbb{Q}$ has
discriminant $-108db^{2}$, which $p$ divides to an odd power. Thus $p$
ramifies in $k$, so that exactly one prime ideal $\mathfrak{p}_{p}$ in the
factorization of $(p)$ in $k$ occurs to an odd power, where $\mathfrak{p}_{p}$
has norm $p$. Therefore the odd part of $D(F(t)/k)$ equals
$\prod\mathfrak{p}_{p}$, where $p$ runs through the odd prime factors of $3d$.

We must now consider the case where $2|D(C)$. By Lemma 4.1, each prime ideal
in the factorizations of $(2)$ in $F(t)$ and $K$ occurs to the second power.
Each prime ideal in the factorization of $(2)$ in $k$ must occur to the first
or second power, and those occurring to the first power are exactly the ones
that ramify in $F(t)$ and in $K$. If every prime ideal factor of $(2)$ in $k$
were to occur to the first power, then $(2)$ would divide $D(K/k)$,
contradicting Lemma 2.3. Thus exactly one prime ideal $\mathfrak{p}_{2}$ in
$k$ divides $(2)$ to the first power, and $\mathfrak{p}_{2}$ has norm $2$.
Then $\mathfrak{p}_{2}^{e}$ exactly divides $D(K/k)$ for some $e\geq 2$
depending on $d$.

The discriminant of the $k$-basis $\\{1,\sqrt{-3d}\\}$ for $F(t)$ is $-12d$,
so that $D(F(t)/k)$ divides $12d$. First suppose that $d\equiv 3\pmod{4}$.
Since $\mathfrak{p}_{2}$ divides $(2)$ to the first power in $k$,
$\mathfrak{p}_{2}$ divides $(12d)$ to the second power. Thus $e\leq 2$ in this
case, so that $e=2$. Next suppose that $d\equiv 2\pmod{4}$. Then $8$ divides
$12d$, so that $e\in\\{2,3\\}$ in this case.

So far we have shown that

(4.5) $D(F(t)/k)=\mathfrak{p}_{2}^{e}\prod_{2<p|3d}\mathfrak{p}_{p},$

where $\mathfrak{p}_{2}$ is to be interpreted as $1$ when $d\equiv 1\pmod{4}$.
Taking absolute norms on both sides of (4.5), we have

(4.6) $D(F(t))/D(k)^{2}=\begin{cases}-3d,\quad\quad&d\equiv 1\pmod{4}\\\
-12d,\quad\quad&d\equiv 3\pmod{4}\\\ -3d\cdot 2^{e-1},\quad\quad&d\equiv
2\pmod{4}.\end{cases}$

By (4.6) and Lemma 4.4,

(4.7) $D(k)^{2}=\begin{cases}3^{6}d^{2},\quad\quad&d\equiv 1\pmod{4}\\\
3^{6}\cdot 2^{4}d^{2},\quad\quad&d\equiv 3\pmod{4}\\\ 3^{6}\cdot
2^{7-e}d^{2},\quad\quad&d\equiv 2\pmod{4}.\end{cases}$

This shows that $e$ must be odd, so $e=3$. Finally, since $D(k)$ is negative
[7, Prop. 2.15], we obtain the desired result $D(k)=9D(F)$. ∎

##  5\. Proof of Theorem 1.4

Assume throughout this section that $v\notin M$ and $d=3m$. Then
$F=\mathbb{Q}(\sqrt{-m})$. Note that the order
$\mathbb{Z}[\sqrt{-3d}]=\mathbb{Z}[3\sqrt{-m}]$ has conductor $3$ or $6$. The
proof of Theorem 1.4 utilizes the following three lemmas.

######  Lemma 5.1.

Let $p$ be a prime dividing $D(F)$. Then each prime ideal in the
factorizations of $(p)$ in $F(t)$, $K$, and $L$ occurs to the second power.

###### Proof.

The prime $p$ ramifies in $C$ and in $F$ with ramification index $2$. Since
$L=K(w)$ and $x^{2}+x+1$ has discriminant $-3$ and $p\neq 3$, the extension
$L/K$ is unramified at $p$. Similarly, $K/C$ is unramified at $p$, since
$x^{3}-v^{3}$ has discriminant $-27v^{6}$. Thus each prime ideal in the
factorization of $(p)$ in $L$ occurs to the second power, and the result
follows. ∎

######  Lemma 5.2.

The extension $F(t)/F$ ramifies at $3$.

###### Proof.

Suppose for the purpose of contradiction that the lemma is false. Then by
Lemma 5.1, the cubic cyclic extension $F(t)/F$ is unramified. Consequently,
$F(t)$ is contained in the Hilbert class field of $F$. Thus $t$ is contained
in the ring class field $J$ for the order $\mathbb{Z}[\sqrt{-m}]$, so that the
Artin symbol in Remark (3C) fixes $t$. As noted in Remark (3C), the Artin map
fixes $v$ when $p=x^{2}+my^{2}\equiv 1\pmod{3}$, which occurs for example when
$3|y$. It follows that the primes $p=X^{2}+9mY^{2}$ split completely in $L$.
These are the primes that split completely in $M$. By [3, Thm. 8.19], we have
$L\subset M$, so that $v\in M$, as contradiction. ∎

######  Lemma 5.3.

The norm of $D(F(t)/F)$ is equal to $3^{8}$.

###### Proof.

In $F$, either $(3)=\mathfrak{q}$ or $(3)=\mathfrak{p}\mathfrak{p}^{\prime}$,
where $\mathfrak{q}$ has norm $9$ and $\mathfrak{p},\mathfrak{p}^{\prime}$
have norm $3$. By Lemma 5.2, these prime ideals ramify wildly in $F(t)$ with
ramification index $3$. Thus

(5.1) $D(F(t)/F)=\mathfrak{q}^{s}\ \mbox{or}\
D(F(t)/F)=(\mathfrak{p}\mathfrak{p}^{\prime})^{s}$

for some integer $s$ with $3\leq s\leq 5$ [7, pp. 260,262]. In either case,
$D(F(t)/F)$ has norm $9^{s}$. Since $F(t)/F$ is a cubic extension, $D(F(t)/F)$
equals the square of an ideal [7, p. 266]. Thus by (5.1), $s=4$, so
$D(F(t)/F)$ has norm $3^{8}$. ∎

We are now prepared for the proof of Theorem 1.4.

###### Proof.

Let $p$ be a prime dividing $D(F)$. By Lemma 5.1, each prime ideal in the
factorization of $(p)$ in $k$ must occur to the first or second power, and
those that occur to the first power are exactly the ones that ramify in $F(t)$
and in $K$. If every prime ideal factor of $(p)$ in $k$ were to occur to the
first power, then $(p)$ would divide $D(K/k)$, contradicting Lemma 2.3. Thus
$p$ ramifies in $k$, so that exactly one prime ideal $\mathfrak{p}_{p}$ in the
factorization of $(p)$ in $k$ occurs to an odd power, and $\mathfrak{p}_{p}$
has norm $p$. Since $F(t)=k(\sqrt{-m})$, the odd part of $D(F(t)/k)$ is the
product of the prime ideals of odd norm in the factorization of $(m)$ in $k$
which occur to odd powers [9]. Thus the odd part of $D(F(t)/k)$ equals
$\prod\mathfrak{p}_{p}$, where $p$ runs through the odd prime factors of $m$.

Consider now the case when $p=2$. Since $\mathfrak{p}_{2}$ ramifies wildly in
$F(t)$, $\mathfrak{p}_{2}^{e}$ exactly divides $D(F(t)/k)$ for some integer
$e\geq 2$. Arguing as in the paragraph above (2.7), we see that $e=2$ when
$d\equiv 3\pmod{4}$ and $e\in\\{2,3\\}$ when $d\equiv 2\pmod{4}$. Thus

(5.2) $D(F(t)/k)=\mathfrak{p}_{2}^{e}\prod_{2<p|m}\mathfrak{p}_{p},$

where $\mathfrak{p}_{2}$ is to be interpreted as $1$ when $d\equiv 1\pmod{4}$.

Taking absolute norms on both sides of (5.2), we have

(5.3) $D(F(t))/D(k)^{2}=\begin{cases}-m,\quad\quad&d\equiv 1\pmod{4}\\\
-4m,\quad\quad&d\equiv 3\pmod{4}\\\ -2^{e-1}m,\quad\quad&d\equiv
2\pmod{4}.\end{cases}$

By Lemma 5.3,

(5.4) $D(F(t))=3^{8}D(F)^{3}=\begin{cases}-m^{3}3^{8},\quad\quad&d\equiv
1\pmod{4}\\\ -m^{3}2^{6}3^{8},\quad&d\equiv 2,3\pmod{4}.\end{cases}$

By (5.3) and (5.4),

(5.5) $D(k)^{2}=\begin{cases}m^{2}3^{8},\quad\quad&d\equiv 1\pmod{4}\\\
2^{4}m^{2}3^{8},\quad\quad&d\equiv 3\pmod{4}\\\
2^{7-e}m^{2}3^{8},\quad\quad&d\equiv 2\pmod{4}.\end{cases}$

This shows that $e$ must be odd, so $e=3$. Finally, since $D(k)$ is negative
[7, Prop. 2.15], we obtain the desired result $D(k)=81D(F)$. ∎

######  Remark (5G).

When $3\mid d$ and $v\notin M$, $D(L/F)$ has norm $3^{18}$. To see this, note
that by the proof of Lemma 5.3, the prime ideal factorization of $(3)$ in
$F(t)$ is either $\mathfrak{Q}^{3}$ or
$(\mathfrak{P}\mathfrak{P}^{\prime})^{3}$, where $\mathfrak{Q}$ has norm $9$
and $\mathfrak{P},\mathfrak{P}^{\prime}$ have norm $3$. Since these prime
ideals occur to the odd power $3$, and since $L=F(t)(\sqrt{-3})$, it follows
from [9] that $D(L/F(t))$ equals $\mathfrak{Q}$ or
$\mathfrak{P}\mathfrak{P}^{\prime}$. In either case, $D(L/F(t))$ has norm $9$,
so that $D(L)=9D(F(t))^{2}$. By Lemma 5.3, $D(F(t))=3^{8}D(F)^{3}$. Combining
these last two equalities, we obtain the desired result $D(L)=3^{18}D(F)^{6}$.

##  6\. Ring class and ray class fields

Recall that $v$ is the cube root of the fundamental unit $a+b\sqrt{d}$. The
following theorem gives explicit criteria in terms of $a$ and $b$ for $v$ to
lie in the ring class field $M$.

######  Theorem 6.1.

We have $v\in M$ if and only if

(6.1) $a\equiv 0\pmod{9}\quad\mbox{or}\quad\begin{cases}a\equiv\pm
2\pmod{9}\quad&\ \mbox{when}\ \ a^{2}-db^{2}=-1\\\ a\equiv\pm
1\pmod{27}\quad&\ \mbox{when}\ \ a^{2}-db^{2}=+1.\end{cases}$

###### Proof.

By the proof of [6, Thm. 6], (6.1) holds if and only if $F(t)/F$ is
unramified. The result now follows from Lemmas 2.1, 3.1, 4.5, and 5.3. ∎

Let $M_{3}$ denote the ring class field of $F$ for the order
$\mathbb{Z}[\sqrt{-27d}]$. It follows from [3, Thm. 7.24] that $|M_{3}:M|=3$.
Consider the example $v=(5/2+\sqrt{21}/2)^{1/3}$. By Theorem 6.1, $v\notin M$,
since $a=5/2\equiv 16\not\equiv\pm 1\pmod{27}$. On the other hand, the $v$ in
this example does lie in the ring class field $M_{3}$. In fact, we conjecture
that $v\in M_{3}$ for every $v$. This needs proof only in the case $d\equiv
1\pmod{4}$, since it follows when $d\equiv 2,3\pmod{4}$ from general theorems
of Sun [8, Section 5]. For a generalization of this conjecture, see Conjecture
7.8.

Let $\mathcal{O}$ denote the order $\mathbb{Z}[\sqrt{-3d}]$ in $F$. Let
$F_{(N)}$ denote the ray class field of $F$ modulo $(N)$. Define the class
field $F_{(N),\mathcal{O}}$ as in [2, p. 853]. By [2, Thm. 1] with $n=3d$,

(6.2) $M=F_{(1),\mathcal{O}},\quad M_{3}=F_{(3),\mathcal{O}}.$

By [2, Thm. 1],

(6.3) $F_{(N)}\subset F_{(N),\mathcal{O}}\subset F_{(NT)},$

where $T$ denotes the conductor of the order $\mathcal{O}$. Theorem 6.2
relates the ring class fields $M$ and $M_{3}$ to ray class fields of $F$.

######  Remark (6H).

Analogous to (6.2), we have $M_{6}=F_{(6),\mathcal{O}}$, where $M_{6}$ is the
ring class field of $F$ for the order $\mathbb{Z}[\sqrt{-108d}]$.

######  Theorem 6.2.

If $3\nmid d$, then

(6.4) $\begin{cases}M_{3}=F_{(6)}\quad&\ \mbox{when}\ \ d\equiv 1\pmod{4}\\\
M_{3}=F_{(3)}\quad&\ \mbox{when}\ \ d\equiv 2,3\pmod{4}.\end{cases}$

If $3\mid d$, then

(6.5) $\begin{cases}M=F_{(6)}\quad&\ \mbox{when}\ \ d\equiv 1\pmod{4}\\\
M=F_{(3)}\quad&\ \mbox{when}\ \ d\equiv 2,3\pmod{4}.\end{cases}$

###### Proof.

Suppose first that $3\nmid d$ and $d\equiv 2,3\pmod{4}$, so that $T=1$. Then
by (6.2) and (6.3) with $N=3$, we have $M_{3}=F_{(3)}$. Next suppose that
$3\nmid d$ and $d\equiv 1\pmod{4}$, so that $T=2$. Then by (6.2) and (6.3)
with $N=3$, we have $M_{3}\subset F_{(6)}$. To show that $M_{3}=F_{(6)}$, it
suffices to show that

(6.6) $|M_{3}:F|=|F_{(6)}:F|.$

Using the formula $\phi((6))/2\ $ [1, p. 50] for the ray class number on the
right, and the formula [3, Thm. 7.24] for the ring class number on the left,
we see that both members of (6.6) are equal to

(6.7) $\begin{cases}3h_{F},\quad&\ \mbox{when}\ \ d\equiv 5\pmod{8}\\\
9h_{F},\quad&\ \mbox{when}\ \ d\equiv 1\pmod{8},\end{cases}$

where $h_{F}$ is the class number of $F$. This completes the proof of (6.4).

Now suppose that $d=3m$, so that $F=\mathbb{Q}(\sqrt{-m})$ and
$\mathcal{O}=\mathbb{Z}[3\sqrt{-m}]$. Let $g$ denote the ordered pair of
Kronecker symbols $g=((-m/3),(-m/2))$.

First assume that $d\equiv 1\pmod{4}$, so that $T=6$. Then by (6.2) and (6.3)
with $N=1$, we have $M\subset F_{(6)}$. To show that $M=F_{(6)}$, it suffices
to show that

(6.8) $|M:F|=|F_{(6)}:F|.$

Appealing again to [3] and [1], we see that both members of (6.8) are equal to

(6.9) $\begin{cases}4h_{F},\quad&\ \mbox{when}\ \ g=(-1,1)\\\ 12h_{F},\quad&\
\mbox{when}\ \ g=(-1,-1)\\\ 2h_{F},\quad&\ \mbox{when}\ \ g=(1,1)\\\
6h_{F},\quad&\ \mbox{when}\ \ g=(1,-1).\end{cases}$

Finally assume that $d\equiv 2,3\pmod{4}$, so that $T=3$. Then by (6.2) and
(6.3) with $N=1$, we have $M\subset F_{(3)}$. To show that $M=F_{(3)}$, it
suffices to show that

(6.10) $|M:F|=|F_{(3)}:F|.$

Again by [3] and [1], we see that both members of (6.10) are equal to

(6.11) $\begin{cases}2h_{F},\quad&\ \mbox{when}\ \ (-m/3)=1\\\ 4h_{F},\quad&\
\mbox{when}\ \ (-m/3)=-1.\end{cases}$

This completes the proof of (6.5). ∎

##  7\. Generalizations for integers $u\in\mathbb{Q}(d)$ with cubic norms

For each squarefree $d>1$, let $\mathcal{O}(d)$ denote the ring of integers in
$\mathbb{Q}(\sqrt{d})$, and write $f(d)$ for the fundamental unit in
$\mathcal{O}(d)$. Let $S_{d}$ denote the set of $u\in\mathcal{O}(d)$ for which
the norm of $u$ is a cube, $u$ is not the cube of an element in
$\mathcal{O}(d)$, and $u$ is not divisible in $\mathcal{O}(d)$ by the cube of
a rational prime. Denote the norm of $u$ by $N(u)=n^{3}$, and let $\nu$ denote
the real cube root $u^{1/3}$.

Let $S_{d}^{*}$ denote the subset of $u\in S_{d}$ such that $p|d$ for each
rational prime $p$ dividing $u$. For example, $S_{d}^{*}$ contains the
fundamental unit $f(d)$. More generally, whenever $\mu\in\mathcal{O}(d)$ is
not divisible by a rational prime, $S_{d}^{*}$ contains

(7.1) $\mu^{3}f(d)^{\pm 1}.$

Examples of elements in $S_{d}^{*}$ not of the form (7.1) are

(7.2) $17+2\sqrt{79},\ 13+\sqrt{142},\ 14+\sqrt{223},\ (11+\sqrt{229})/2,\
28+3\sqrt{235},\ 77+2\sqrt{254},$

whose norms are $-27,\ 27,-27,-27,-11^{3},17^{3}$, respectively.

Let $P_{d}$ denote the set of $\mu=a+b\sqrt{d}\in\mathcal{O}(d)$ with norm
$N(\mu)=n^{3}$ satisfying the following conditions:

$\displaystyle n\equiv 2,5,8\pmod{9}$ $\displaystyle\implies a\equiv 0,\pm
2,\pm 7,\pm 9,\pm 11\pmod{27};$ $\displaystyle n\equiv 1\pmod{9}$
$\displaystyle\implies a\equiv 0,\pm 1,\pm 9\pmod{27};$ $\displaystyle n\equiv
4\pmod{9}$ $\displaystyle\implies a\equiv 0,\pm 8,\pm 9\pmod{27};$
$\displaystyle n\equiv 7\pmod{9}$ $\displaystyle\implies a\equiv 0,\pm 9,\pm
10\pmod{27};$ $\displaystyle n\equiv 0\pmod{9}$ $\displaystyle\implies
a\equiv\pm 4,\pm 5,\pm 13\pmod{27};$ $\displaystyle n\equiv\pm 3\pmod{9}$
$\displaystyle\implies a\equiv 0,\pm 4,\pm 5,\pm 9,\pm 13\pmod{27}.$

When $u$ is a fundamental unit (so that $n=\pm 1$), it follows from Theorem
6.1 that $v\in M$ if and only if $u\in P_{d}$. Theorem 7.3 shows that this
equivalence holds for a more general set of $u$ under the condition that the
class number $h(d)$ is not a multiple of $3$. We conjecture that the condition
on the class number can be dropped. The proof of Theorem 7.3 depends on the
following two lemmas.

######  Lemma 7.1.

Let $u\in S_{d}$. Then the principal ideal $(u)$ factors in $\mathcal{O}(d)$
as

(7.3) $(u)=\mathfrak{A}^{3}q\mathfrak{Q},$

where $\mathfrak{A}$ and $\mathfrak{Q}$ are ideals, and $q$ is a squarefree
integer with $q=N(\mathfrak{Q})$, where each rational prime factor of $q$
splits in $\mathcal{O}(d)$.

###### Proof.

Since $n^{3}=N(u)=uu^{\prime}$, the norm of each prime ideal factor of $(u)$
is a prime dividing $n$. For such a prime $p$, we proceed to investigate the
“$p$-part” of the ideal factorization of $(u)$.

First suppose that $p$ does not split in $\mathcal{O}(d)$. Let $\mathfrak{p}$
be a prime ideal factor of $(p)$ in $\mathcal{O}(d)$. For some $e\geq 1$,
$\mathfrak{p}^{e}||(u)$. Since $\mathfrak{p}^{2e}||(uu^{\prime})=(n^{3})$, we
have $3|e$. If $p$ were inert, then $p^{3}|u$, contradicting the definition of
$S_{d}$. Thus $(p)=\mathfrak{p}^{2}$. Since $\mathfrak{p}^{e}||(u)$, we
conclude that the $p$-part of $(u)$ is a cube.

Next suppose that $p$ splits, and write
$(p)=\mathfrak{p}\mathfrak{p}^{\prime}$. Without loss of generality,
$\mathfrak{p}^{e}||(u)$ for some $e\geq 2$. If
$\mathfrak{p}^{\prime}\nmid(u)$, then $p^{e}||n^{3}$; consequently $3|e$ and
once again the $p$-part of $(u)$ is a cube. It remains to consider the case
where $\mathfrak{p}^{\prime}\mid(u)$. In this case the power of $p$ in the
prime factorization of $u$ is $p^{i}$, where $i\in\\{1,2\\}$, and we have
$\mathfrak{p}^{e-i}||(u/p^{i})$. By taking the norm,
$p^{e-i}||(n^{3}/p^{2i})$, so that $3|(e+i)$. Therefore
$\mathfrak{p}^{i+3k}||(u/p^{i})$ for some $k\geq 0$, so that the $p$-part of
$(u)$ is $p^{i}\mathfrak{p}^{i}$ times a cube. Taking the product of all the
$p$-parts, we obtain (7.3), where $\mathfrak{Q}$ is the product of the
$\mathfrak{p}^{i}$ in the $p$-parts and $q=N(\mathfrak{Q})$ is the product of
the $p^{i}$ in the $p$-parts. We can arrange the product so that each $i$ is
equal to $1$, as follows. If a $p$-part equals
$p^{2}\mathfrak{p}^{2}=\mathfrak{p}^{\prime 2}\mathfrak{p}^{4}$, then pull out
$\mathfrak{p}^{3}$ and let that cube be absorbed in $\mathfrak{A}^{3}$. What’s
left is $p\mathfrak{p}^{\prime}$. The resulting $q$ in (7.3) can thus be taken
to be squarefree. ∎

Let $R_{d}$ be the set of $\beta\in\mathcal{O}(d)$ having the form
$\beta=nr+ns\sqrt{d}$, where $r+s\sqrt{d}$ has norm $n$ (so that
$N(u)=n^{3}$). For example, $\beta=(31+155\sqrt{5})/2\in R_{5}$ with
$N(\beta)=-31^{3}$. An example of an element of $S_{d}$ that is not in $R_{d}$
is $u=1376+387\sqrt{79}$, for which $N(u)=-215^{3}$.

We provide a computer-generated proof via Mathematica for the following lemma.

######  Lemma 7.2.

Let $\beta,\gamma\in R_{d}$, $\mu\in\mathcal{O}(d)$, with norms $N(\mu)$ and
$N(\gamma)$ both nonzero modulo $3$. Then

(7.4) $\beta\in P_{d}\Longleftrightarrow\beta\mu^{3}\in P_{d}.$

and

(7.5) $(\beta\in P_{d}\ \mbox{and}\ \gamma\in
P_{d})\Longrightarrow\beta\gamma\in P_{d}.$

###### Proof.

Write $\mu=x+y\sqrt{d}$ and $\beta=nr+ns\sqrt{d}$ with $n=r^{2}-ds^{2}$, so
that $N(\beta)=n^{3}$. Create in Mathematica the master set of $9,565,938$
quintuples $(r,s,x,y,d)$ modulo $27$ for which $x^{2}-dy^{2}$ is nonzero
modulo $3$. Compute the subset of the master set for which
$(nr+ns\sqrt{d})(x+y\sqrt{d})^{3}\in P_{d}$. This turns out to be exactly the
same subset for which $\beta=nr+ns\sqrt{d}\in P_{d}$, thereby proving (7.4).

Next, write $\gamma=mx+my\sqrt{d}$ with $m=x^{2}-dy^{2}$, so that
$N(\gamma)=m^{3}$. Using the same master set, compute the subset for which
$(nr+ns\sqrt{d})(mx+my\sqrt{d})\in P_{d}$. This turns out to contain the
subset for which both of these factors lie in $P_{d}$, thus proving (7.5). The
details of the Mathematica proof are given in [4]. ∎

For $u\in S_{d}$, recall that $\nu$ denotes the real cube root $u^{1/3}$, $n$
denotes the cube root of the norm of $u$, and $v$ denotes the real cube root
of the fundamental unit $f(d)$.

######  Theorem 7.3.

Let $u\in S_{d}^{*}$ have nonzero norm modulo $3$. Assume that $3\nmid h(d)$.
Then $\nu\in M$ if and only if $u\in P_{d}$.

###### Proof.

By definition of $S_{d}^{*}$, any rational prime dividing $u$ must ramify in
$\mathcal{O}(d)$. Thus, by Lemma 7.1, we have $(u)=\mathfrak{A}^{3}$ for some
ideal $\mathfrak{A}$ in $\mathcal{O}(d)$. Since $3\nmid h(d)$ and both of
$\mathfrak{A}^{3}$ , $\mathfrak{A}^{h(d)}$ are principal, it follows that
$\mathfrak{A}$ is principal. This shows that $(u)=(\mu^{3})$ for some
$\mu\in\mathcal{O}(d)$, so that $u$ has the form (7.1). Without loss of
generality, we will assume the plus sign in (7.1), and we write
$u=f(d)(x+y\sqrt{d})^{3}$, where $(x+y\sqrt{d})$ is an element of
$\mathcal{O}(d)$ whose norm is not divisible by $3$.

Clearly $\nu=v(x+y\sqrt{d})\in M$ if and only if $v\in M$. As noted above
Lemma 7.1, $v\in M$ if and only if $f(d)\in P_{d}$. Thus $\nu\in M$ if and
only if $f(d)\in P_{d}$. It remains to prove that

(7.6) $f(d)\in P_{d}\Longleftrightarrow f(d)(x+y\sqrt{d})^{3}\in P_{d}.$

This follows from the special case $\beta=f(d)$ of Lemma 7.2. ∎

Let $M_{e}$ denote the ring class field for the order
$\mathbb{Z}[e\sqrt{-3d}]$ in $F=\mathbb{Q}(\sqrt{-3d})$. In particular,
$M_{1}=M$. For $u\in\mathcal{O}(d)$, let $c=c(u)$ be the product of the
distinct rational primes which divide $u$ but not $d$. Note that for $u\in
S_{d}$, we have $c(u)=1$ if and only if $u\in S_{d}^{*}$.

Theorem 7.5 below extends Theorem 7.3. Just as for Theorem 7.3, we conjecture
that the condition on the class number can be dropped. The proof of Theorem
7.5 is conditional on the following conjecture.

######  Conjecture 7.4.

Let $\beta\in\mathcal{O}(d)$ have a cubic norm with $3\nmid N(\beta)$. Then
$\beta^{1/3}\in M_{c}$ if and only if $\beta\in P_{d}$, where $c=c(\beta)$. In
particular, this equivalence holds for every $\beta\in R_{d}$ with $3\nmid
N(\beta)$.

######  Remark (7I).

Let $\beta,\gamma\in R_{d}$ with $3\nmid N(\beta\gamma)$. If $\beta^{1/3}\in
M_{c(\beta)}$ and $\gamma^{1/3}\in M_{c(\gamma)}$, then by definition of
$M_{c(u)}$, we have $(\beta\gamma)^{1/3}\in M_{c(\beta\gamma)}$. This
demonstrates that (7.5) is consistent with Conjecture 7.4.

######  Theorem 7.5.

Let $u\in S_{d}$ have nonzero norm modulo $3$. Assume that Conjecture 7.4
holds and that $3\nmid h(d)$. Then $\nu\in M_{c}$ if and only if $u\in P_{d}$,
where $c=c(u)$.

###### Proof.

Let $T\in\\{h(d),2h(d)\\}$ be chosen such that $3\mid(T-1)$. By (7.3),

(7.7) $(u^{T})=(\mathfrak{A}^{T})^{3}q^{T}\mathfrak{Q}^{T}.$

Since $\mathfrak{Q}^{T}$ is a principal ideal of norm $q^{T}$, we have
$(q^{T}\mathfrak{Q}^{T})=(\gamma)$ with $\gamma\in R_{d}$. By (7.7), $q=c$,
where $c=c(u)=c(\gamma)$. Since $\mathfrak{A}^{T}$ is principal, we have
$u^{T}=\mu_{0}^{3}\beta$ for some $\mu_{0}\in\mathcal{O}(d)$ and $\beta\in
R_{d}$. Therefore $u^{3k}u=\mu_{0}^{3}\beta$ for some $k$. Multiplying by the
conjugate $u^{\prime 3k}$, we obtain

(7.8) $j^{3}u=\mu^{3}\beta$

for some $\mu\in\mathcal{O}(d)$, where $j=N(u)^{k}$ is not divisible by 3.
Consequently

(7.9) $j\nu=\mu\beta^{1/3}.$

Since $j,\mu\in M\subset M_{c}$,

(7.10) $\nu\in M_{c}\Longleftrightarrow\beta^{1/3}\in
M_{c}\Longleftrightarrow\beta\in P_{d},$

where the first equivalence follows from (7.9) and the second follows from
Conjecture 7.4. By Lemma 7.2 and (7.8),

(7.11) $\beta\in P_{d}\Longleftrightarrow\mu^{3}\beta\in
P_{d}\Longleftrightarrow j^{3}u\in P_{d}\Longleftrightarrow u\in P_{d}.$

The result now follows from (7.10) and (7.11). ∎

Some numerical examples supporting Theorem 7.5 with $3\mid h(d)$ are given in
the last section of [4].

######  Theorem 7.6.

Let $u\in\mathcal{O}(d)$ have a cubic norm and write $\nu=u^{1/3}$. Then
$\nu\in M_{c}$ if and only if $\nu\in M_{c^{\prime}}$, where
$c^{\prime}=c^{\prime}(u)$ denotes the odd part of $c=c(u)$.

###### Proof.

It suffices to prove that $\nu\in M_{c}$ implies $\nu\in M_{c^{\prime}}$ for
even $c$. Assume for the purpose of contradiction that $\nu\in M_{c}$ but
$\nu\notin M_{c^{\prime}}$. By [3, Thm. 7.24], $M_{c}$ has degree $2$ over
$M_{c^{\prime}}$. Thus the minimal polynomial of $\nu$ over $M_{c^{\prime}}$
is quadratic. Since this quadratic polynomial must divide the cubic polynomial
$x^{3}-u=x^{3}-\nu^{3}$ over $M_{c^{\prime}}$, it follows that this cubic
polynomial has a linear factor over $M_{c^{\prime}}$. Since the cube roots of
unity lie in $M$, we obtain the desired contradiction $\nu\in M_{c^{\prime}}$.
∎

We close this section with two conjectures. When $u$ is a fundamental unit,
Conjecture 7.7 reduces to Theorem 1.5, while Conjecture 7.8 reduces to the
conjecture mentioned below Theorem 6.1. Extensive numerical evidence for
Conjectures 7.4, 7.7, and 7.8 is given in [4].

######  Conjecture 7.7.

Let $u\in S_{d}$ and let $c^{\prime}=c^{\prime}(u)$ denote the odd part of
$c=c(u)$. When $\nu\in M_{c}$, the norm of $D(F(\nu)/F)c^{-4}$ equals $1$ or
$3^{6}$ according as $3\nmid d$ or $3\mid d$, and when $\nu\notin M_{c}$, the
norm of $D(F(\nu)/F)c^{-4}$ equals $3^{8}$ or $3^{18}$ according as $3\nmid d$
or $3\mid d$, except that when $d\equiv 3\pmod{4}$, each $c^{-4}$ is to be
replaced by $c^{\prime-4}$.

######  Conjecture 7.8.

For every $u\in S_{d}$, we have $\nu\in M_{3c}$, where $c=c(u)$.

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