#  $S_{4}$-quartics with Prescribed Norms

Sebastian Monnet

###### Abstract.

Given a number field $k$ and a finitely generated subgroup
$\mathcal{A}\subseteq k^{*}$, we study the distribution of $S_{4}$-quartic
extensions of $k$ such that the elements of $\mathcal{A}$ are norms. We show
that the density of such extensions is the product of so-called “local masses”
at every place of $k$. We give these local masses explicitly in almost all
cases and give an algorithm for computing the remaining cases.

##  1\. Introduction

Fix a number field $k$. An _$S_{4}$ -quartic extension of $k$_ is a quartic
field extension $K/k$, such that the normal closure $\widetilde{K}$ of $K$
over $k$ has $\operatorname{Gal}(\widetilde{K}/k)\cong S_{4}$. For a real
number $X>0$, write $N(X)$ for the number of isomorphism classes of
$S_{4}$-quartic extensions $K/k$ such that the norm of the relative
discriminant of $K$ over $k$ is at most $X$. Given a finitely generated
subgroup $\mathcal{A}$ of the unit group $k^{*}$, write $N(X;\mathcal{A})$ for
the number of such isomorphism classes with $\mathcal{A}\subseteq
N_{K/k}(K^{*})$.

######  Theorem 1.1.

For every finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, we have

$0<\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X)}\leq 1,$

with equality if and only if $\mathcal{A}\subseteq k^{*4}$.

######  Corollary 1.2.

For every finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, there are
infinitely many $S_{4}$-quartic extensions $K$ of $k$ with
$\mathcal{A}\subseteq N_{K/k}(K^{*})$.

Let us unpack the content of Theorem 1.1. It tells us that any finitely
generated subgroup $\mathcal{A}\subseteq k^{*}$ is contained in the norm group
of a positive proportion of $S_{4}$-quartic extensions. Moreover, it tells us
that $\mathcal{A}$ can only be contained in 100% of such norm groups if
$\mathcal{A}\subseteq k^{*4}$, in which case $\mathcal{A}\subseteq
N_{K/k}(K^{*})$ for every $S_{4}$-quartic extension $K$ of $k$. In other
words, if $\mathcal{A}\subseteq N_{K/k}(K^{*})$ for almost all $K$, then
$\mathcal{A}\subseteq N_{K/k}(K^{*})$ for all $K$.

######  Remark 1.3.

In the special case where $\mathcal{A}$ is generated by one element, Corollary
1.2 can be obtained by classical methods, using Hilbert’s irreducibility
theorem. See [FLN22, Example 1.13] for details.

In [FLN22], Frei, Loughran, and Newton considered the same question for
abelian extensions with a given Galois group. They used class field theory to
solve the problem. In our setting, the extensions are nonabelian, so we are
unable to use class field theory in the same way. Our approach is to use a
local-global principle called the Hasse norm principle to reduce the global
problem to a collection of local conditions. We then apply a result of
Bhargava, Shankar, and Wang to count the $S_{4}$-quartics satisfying this
collection of local conditions. Bhargava, Shankar, and Wang’s result is also
valid for $S_{3}$-cubics and $S_{5}$-quintics, and we expect that our methods
will generalise to such extensions.

To prove Theorem 1.1, we first obtain the following explicit formula for the
density of number fields in question.

######  Theorem 1.4.

Let $\mathcal{A}\subseteq k^{*}$ be a finitely generated subgroup. There exist
positive rational numbers $m_{\mathcal{A},\mathfrak{p}}$, indexed by the
primes $\mathfrak{p}$ of $k$, such that each $m_{\mathcal{A},\mathfrak{p}}$ is
defined explicitly in terms of $\mathcal{A}$ and $\mathfrak{p}$, and

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}=\frac{1}{2}\operatornamewithlimits{Res}_{s=1}\big{(}\zeta_{k}(s)\big{)}\prod_{\mathfrak{p}}m_{\mathcal{A},\mathfrak{p}}.$

For an element $\alpha\in k^{*}$, write $N(X;\alpha)$ to mean
$N(X;\langle\alpha\rangle)$, where $\langle\alpha\rangle$ is the subgroup of
$k^{*}$ generated by $\alpha$. Equivalently, $N(X;\alpha)$ is the number of
$S_{4}$-quartic extensions $K/k$ with appropriately bounded discriminant such
that $\alpha\in N_{K/k}(K^{*})$. Similarly, write $m_{\alpha,\mathfrak{p}}$
for the number $m_{\langle\alpha\rangle,\mathfrak{p}}$ from Theorem 1.1.

######  Theorem 1.5.

For $\alpha\in k^{*}$ and finite primes $\mathfrak{p}$ of $k$ with odd norm,
the values of $m_{\alpha,\mathfrak{p}}$ are given by Tables 1 and 2. If
$\mathfrak{p}$ is an infinite prime and $f:k\to\mathbb{C}$ is an embedding
corresponding to $\mathfrak{p}$, then

$m_{\alpha,\mathfrak{p}}=\begin{cases}\frac{5}{12}\quad\text{if $f$ is real
and $f(\alpha)>0$},\\\ \frac{7}{24}\quad\text{if $f$ is real and
$f(\alpha)<0$},\\\ \frac{1}{24}\quad\text{if $f$ is complex.}\end{cases}$

In the special case $k=\mathbb{Q}$ and $\mathfrak{p}=(2)$, the values of
$m_{\alpha,\mathfrak{p}}$ are given by Table 3.

We illustrate how to use our results, and in particular how to read Tables 1,
2, and 3, with an example.

######  Example 1.6.

We will give an exact expression for the proportion of $S_{4}$-quartic number
fields $K/\mathbb{Q}$ such that $2\in\mathbb{N}_{K/\mathbb{Q}}(K^{*})$. Since
$2>0$, we have

$m_{2,\infty}=m_{1,\infty}=\frac{5}{12}.$

In Table 3, we have $r=u=1$, so

$m_{2,2}=\frac{6523}{8192}.$

For an odd prime $p$, we know that $2$ is a quadratic residue if and only if
$p\equiv\pm 1\pmod{8}$. Therefore, if $p\equiv\pm 3\pmod{8}$, we have

$m_{2,p}=\frac{(4p^{2}+4p+5)(p-1)}{4p^{3}}.$

If $p\equiv 7\pmod{8}$, then Table 2 tells us that

$m_{2,p}=\frac{(p^{3}+p^{2}+2p+1)(p-1)}{p^{4}}.$

For $p\equiv 1\pmod{8}$, the situation is more complicated, and we have

$m_{2,p}=\begin{cases}\frac{(p^{3}+p^{2}+2p+1)(p-1)}{p^{4}}\quad\text{if $2$
is a quartic residue modulo $p$},\\\
\frac{(p^{2}+p+2)(p-1)}{p^{3}}\quad\text{otherwise}.\end{cases}$

We have

$m_{1,p}=\frac{(p^{3}+p^{2}+2p+1)(p-1)}{p^{4}},$

for all finite primes (including $2$), and therefore

$\frac{m_{2,p}}{m_{1,p}}=\begin{cases}1\quad\text{if $p=\infty$},\\\
\frac{6523}{8704}\quad\text{if $p=2$},\\\
\frac{4p^{3}+4p^{2}+5p}{4(p^{3}+p^{2}+2p+1)}\quad\text{if $p\equiv\pm
3\pmod{8}$},\\\ 1\quad\text{if $p\equiv 7\pmod{8}$},\\\ 1\quad\text{if
$p\equiv 1\pmod{8}$ and $2$ is a quartic residue modulo $p$},\\\
\frac{p^{3}+p^{2}+2p}{p^{3}+p^{2}+2p+1}\quad\text{if $p\equiv 1\pmod{8}$ and
$2$ is a quartic nonresidue modulo $p$}\end{cases}$

Finally then, the proportion of $S_{4}$-quartic extensions $K/\mathbb{Q}$ with
$2\in N_{K/\mathbb{Q}}(K^{*})$ is given by the Euler product

$\frac{1}{2}\cdot\frac{6523}{8704}\cdot\prod_{p\equiv\pm
3\pmod{8}}\Big{(}1-\frac{3p+4}{4p^{3}+4p^{2}+8p+4}\Big{)}\cdot\prod_{\begin{subarray}{c}p\equiv
1\pmod{8}\\\ \text{$2$ is a quartic nonresidue modulo
$p$}\end{subarray}}\Big{(}1-\frac{1}{p^{3}+p^{2}+2p+1}\Big{)},$

which is approximately equal to $33.06\%$.

###  1.1. Structure of the paper

In Section 2, we explain our approach to the problem. Our first step is to
introduce the _Hasse Norm Principle_ in Section 2.1, which lets us consider
the situation locally at each prime of $k$. Subsequently, in Section 2.2, we
state Bhargava, Shankar, and Wang’s counting result, which requires our local
conditions to be “acceptable”. In Section 2.3, we demonstrate that our
conditions are indeed acceptable, allowing us to prove Theorem 1.4.

The goal of Section 3 is to prove Theorem 1.5. That is, we compute the masses
$m_{\alpha,\mathfrak{p}}$ explicitly for $\alpha\in k^{*}$ and primes
$\mathfrak{p}$ that are either infinite or have odd norm, as well as in the
special case $k=\mathbb{Q}$ and $\mathfrak{p}=(2)$. In Section 3.1, we
introduce some notation and outline the strategy of Sections 3.2–3.4. Then, in
Section 3.2, we compute all possible norm groups of several classes of quartic
étale algebras over $k_{\mathfrak{p}}$, where $\mathfrak{p}$ is a finite prime
with odd norm. We apply these norm group computations in Section 3.3 to
compute all possible masses $m_{\alpha,\mathfrak{p}}$, for finite primes
$\mathfrak{p}$ with odd norm, as well as for infinite primes. In Section 3.4,
we present our MAGMA ([BCP97]) algorithm to compute masses in general and
obtain the values in Table 3. We then assemble our results to prove Theorem
1.5.

Finally, Section 4 is dedicated to proving Theorem 1.1. Along the way, we
obtain an explicit finite bound on $\lim_{X\to\infty}\frac{N(X;\alpha)}{X}$
for $\alpha\in k^{*}$, in terms of $\alpha$ and $k$. This bound implies that
$\lim_{X\to\infty}\frac{N(X,\mathcal{A})}{X}$ is finite for any finitely
generated subgroup $\mathcal{A}\subseteq k^{*}$, which allows us to obtain an
explicit expression for the proportion
$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X)}$. By considering this explicit
expression, we prove Theorem 1.1.

###  1.2. Acknowledgements

I would like to thank my supervisor, Rachel Newton, for suggesting the project
and for her unwavering support throughout. Jiuya Wang’s suggestion to look at
[BSW15] was vital to the eventual approach, so I am very grateful to her as
well. Thanks also to Ross Patterson for suggesting I use MAGMA to verify my
results; my code did actually uncover a mistake, so the check was very useful.
Ashvin Swaminathan was very generous in explaining some details of Bhargava’s
parametrisation of quartic rings, for which I am also grateful. Of course,
none of this work would have been possible without my generous funding from
the ESPRC, University College London and the Heilbronn Institute for
Mathematical Research.

###  1.3. Notation

We fix the following notation:

  * •

$k$ is a number field,

  * •

$\mathfrak{p}$ is a prime of $k$,

  * •

$\mathbb{F}_{\mathfrak{p}}$ is the residue field
$\mathcal{O}_{K}/\mathfrak{p}$, in the case where $\mathfrak{p}$ is finite,

  * •

$N(\mathfrak{p})$ and $q$ both denote the cardinality of
$\mathbb{F}_{\mathfrak{p}}$, in the case where $\mathfrak{p}$ is finite,

  * •

$K$ is a finite-degree extension of $k$,

  * •

$\alpha$ is an element of $k^{*}$,

  * •

$\mathcal{A}$ is a finitely generated subgroup of $K^{*}$,

  * •

$\zeta_{m}$ is a primitive $m^{\mathrm{th}}$ root of unity, where $m$ is a
positive integer.

###  1.4. Detailed version of the paper

This version of the paper has been optimised for concision. In case the reader
is interested in more detailed proofs and worked examples, there is a much
longer draft available on the author’s website (www.smonnet.com).

##  2\. Reducing to Local Conditions

In this section, we break our global problem into infinitely many local
problems. In Section 2.1, we recall the Hasse Norm Principle, which says that
$\alpha$ is a norm globally if and only if it is a norm everywhere locally. In
Section 2.2, we recall a result of Bhargava, Shankar, and Wang, which allows
us to count number fields satisfying a given collection of local conditions.
In a probabilistic sense, this counting result essentially says that the local
conditions at each prime behave like independent random events. The result
also tells us how to compute the “probabilities” of these independent events.
There is one technical hitch, which is that this result only applies to so-
called _acceptable_ collections of local conditions. Therefore, we spend
Section 2.3 proving that our local conditions are indeed acceptable, so that
we can apply Bhargava, Shankar, and Wang’s result.

###  2.1. The Hasse Norm Principle

As usual, let $K/k$ be an extension of number fields and let $\mathfrak{p}$ be
a prime of $k$. Write $K_{\mathfrak{p}}$ for the étale
$k_{\mathfrak{p}}$-algebra

$K_{\mathfrak{p}}=K\otimes_{k}k_{\mathfrak{p}},$

which we call the _completion of $K$ at $\mathfrak{p}$_. In this paper, we
will refer to the following result as “the Hasse Norm Principle”. This
terminology is slightly nonstandard; in general, the Hasse Norm Principle is a
local-global principle that may or may not hold in a particular extension.
Theorem 2.1 says that this principle holds for all $S_{4}$-quartic extensions.

######  Theorem 2.1 (Hasse Norm Principle).

Let $K/k$ be an degree $n$ extension whose normal closure has Galois group
$S_{n}$, and let $\alpha\in k^{*}$. The following are equivalent:

  1. (1)

$\alpha\in N_{K_{\mathfrak{p}}/k_{\mathfrak{p}}}(K_{\mathfrak{p}}^{*})$ for
each prime $\mathfrak{p}$ of $k$,

  2. (2)

$\alpha\in N_{K/k}(K^{*})$.

###### Proof.

This is [Vos88, Corollary to Theorem 4]. ∎

######  Corollary 2.2.

Let $K/k$ be a degree $n$ extension whose normal closure has Galois group
$S_{n}$, and let $\mathcal{A}\subseteq k^{*}$ be a finitely generated
subgroup. The following are equivalent:

  1. (1)

$\mathcal{A}\subseteq
N_{K_{\mathfrak{p}}/k_{\mathfrak{p}}}(K_{\mathfrak{p}}^{*})$ for each prime
$\mathfrak{p}$ of $k$,

  2. (2)

$\mathcal{A}\subseteq N_{K/k}(K^{*})$.

Let $\mathfrak{p}$ be a prime of $k$. Define
$\Sigma_{\mathcal{A},\mathfrak{p}}$ to be the set of isomorphism classes of
quartic étale algebras $L/k_{\mathfrak{p}}$ with $\mathcal{A}\subseteq
N_{L/k_{\mathfrak{p}}}(L^{*})$. Then Corollary 2.2 tells us that an
$S_{4}$-quartic extension $K/k$ has $\mathcal{A}\subseteq N_{K/k}(K^{*})$ if
and only if $K_{\mathfrak{p}}\in\Sigma_{\mathcal{A},\mathfrak{p}}$ for all
$\mathfrak{p}$. Given $\alpha\in k^{*}$, write $\Sigma_{\alpha,\mathfrak{p}}$
for $\Sigma_{\langle\alpha\rangle,\mathfrak{p}}$, where $\langle\alpha\rangle$
is the subgroup of $k^{*}$ generated by $\alpha$.

###  2.2. Counting number fields using local conditions

The Hasse Norm Principle tells us that the number fields $K$ we are looking
for are precisely those satisfying a certain collection of local conditions,
in the sense of the following definition.

######  Definition 2.3.

Let $n$ be a positive integer. A _degree $n$ collection of local conditions on
$k$_ is a collection $(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$, where
$\mathfrak{p}$ ranges over primes of $k$, and each $\Sigma_{\mathfrak{p}}$ is
a set of isomorphism classes of étale $k_{\mathfrak{p}}$-algebras of degree
$n$.

######  Definition 2.4.

An extension $K/k$ _satisfies_ a collection of local conditions
$(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ if
$K_{\mathfrak{p}}\in\Sigma_{\mathfrak{p}}$ for all $\mathfrak{p}$.

######  Remark 2.5.

We will refer extensively to the discriminant
$\operatorname{Disc}(L/k_{\mathfrak{p}})$, where $L$ is an étale
$k_{\mathfrak{p}}$-algebra. This discriminant has the well-known property that

$\operatorname{Disc}\big{(}(L_{1}\times\ldots\times
L_{r})/k_{\mathfrak{p}}\big{)}=\prod_{i=1}^{r}\operatorname{Disc}(L_{i}/k_{\mathfrak{p}}),$

for étale $k_{\mathfrak{p}}$-algebras $L_{i}$.

######  Definition 2.6.

A _$p$ -adic field_ is a finite degree extension of the $p$-adic numbers
$\mathbb{Q}_{p}$.

######  Definition 2.7.

Let $F$ be a $p$-adic field with maximal ideal $P$ and residue field of size
$q$. Let $L$ be an étale algebra over $F$. Write _$\operatorname{Disc}(L/F)$_
for the relative discriminant of $L$ over $F$, and define
_$\operatorname{Disc}_{P}(L/F)$_ by

$\operatorname{Disc}_{P}(L/F)=q^{v_{P}(\operatorname{Disc}(L/F))}.$

######  Definition 2.8.

Following [BSW15], a degree $n$ collection of local conditions
$(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ is said to be _acceptable_ if, for
all but finitely many $\mathfrak{p}$, the set $\Sigma_{\mathfrak{p}}$ contains
every degree $n$ étale algebra $L/k_{\mathfrak{p}}$ with
$v_{\mathfrak{p}}(\operatorname{Disc}(L/k_{\mathfrak{p}}))<2$.

######  Definition 2.9.

Let $\Sigma=(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ be a degree $4$ collection
of local conditions on $k$. For a positive real number $X$, write
$N(X;\Sigma)$ for the number of $S_{4}$-quartic extensions of $k$ satisfying
$\Sigma$ such that the norm of the relative discriminant of $K/k$ is at most
$X$.

######  Definition 2.10.

Let $\mathfrak{p}$ be a prime of $k$ and let $\Sigma_{\mathfrak{p}}$ be a set
of isomorphism classes of étale $k_{\mathfrak{p}}$-algebras. Define the _mass
of $\Sigma_{\mathfrak{p}}$_ to be

$m_{\mathfrak{p}}(\Sigma_{\mathfrak{p}})=\begin{dcases}\frac{N(\mathfrak{p})-1}{N(\mathfrak{p})}\sum_{L\in\Sigma_{\mathfrak{p}}}\frac{1}{\operatorname{Disc}_{\mathfrak{p}}(L/k_{\mathfrak{p}})}\cdot\frac{1}{\\#\operatorname{Aut}(L/k_{\mathfrak{p}})}\quad\text{if
$\mathfrak{p}$ is finite},\\\
\sum_{L\in\Sigma_{\mathfrak{p}}}\frac{1}{\\#\operatorname{Aut}(L/k_{\mathfrak{p}})}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{
if $\mathfrak{p}$ is infinite}.\end{dcases}$

######  Theorem 2.11.

Let $\Sigma=(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ be an acceptable
collection of local conditions. Then

$\lim_{X\to\infty}\frac{N(X;\Sigma)}{X}=\frac{1}{2}\operatornamewithlimits{Res}_{s=1}\big{(}\zeta_{k}(s)\big{)}\prod_{\mathfrak{p}}m_{\mathfrak{p}}(\Sigma_{\mathfrak{p}}).$

###### Proof.

This is [BSW15, Theorem 2], in the case where $n=4$ and $F$ is a number field.
∎

###  2.3. Showing that our conditions are acceptable

In order to apply Theorem 2.11, we need to show that the collection
$(\Sigma_{\mathcal{A},\mathfrak{p}})_{\mathfrak{p}}$ of local conditions is
acceptable. We start by defining the “splitting symbol” of a prime in an
extension, which will allow us to consider extensions $K/k$ separately
depending on how the prime $\mathfrak{p}$ splits in $K$.

######  Definition 2.12.

Let $K/k$ be an extension of number fields and let $\mathfrak{p}$ be a prime
of $k$. Suppose that

$\mathfrak{p}\mathcal{O}_{K}=P_{1}^{e_{1}}\ldots P_{r}^{e_{r}},$

for distinct prime ideals $P_{i}$ of $\mathcal{O}_{K}$. For each $i$, let
$f_{i}$ be the inertia degree of $P_{i}$ over $\mathfrak{p}$. Then the
_splitting symbol of $\mathfrak{p}$ in $K$_ is

$(K,\mathfrak{p})=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}}).$

If $e_{i}=1$ for some $i$, we suppress the notation and write $f_{i}$ instead
of $f_{i}^{1}$. Note that this symbol is only well-defined up to permutation
of the indices, so we identify the symbols $(f_{1}^{e_{1}}\ldots
f_{r}^{e_{r}})$ and $(f_{\sigma(1)}^{e_{\sigma(1)}}\ldots
f_{\sigma(r)}^{e_{\sigma(r)}})$ for each permutation $\sigma$ of the set
$\\{1,2,\ldots,r\\}$. For example, we say that $(1^{2}2)=(21^{2})$.

We are interested in the completion $K_{\mathfrak{p}}$ of $K$ at
$\mathfrak{p}$, which is an étale $k_{\mathfrak{p}}$-algebra. There is a
natural notion of splitting symbol for étale $k_{\mathfrak{p}}$-algebras. We
first define this splitting symbol.

######  Definition 2.13.

Let $F$ be a $p$-adic field with maximal ideal $P$, and let $L/F$ be an étale
algebra. Then $L\cong L_{1}\times\ldots\times L_{r}$ for finite field
extensions $L_{i}/F$. Let $e_{i}$ and $f_{i}$ be respectively the ramification
index and inertia degree of $L_{i}$ over $F$. Then the _splitting symbol of
$L$_ is

$(L,P)=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}}).$

Again, we identify splitting symbols related by permutation, in the sense of
Definition 2.12.

It is easy to see that our two definitions of splitting symbol agree, in the
sense that

$(K_{\mathfrak{p}},\mathfrak{p})=(K,\mathfrak{p}),$

whenever $\mathfrak{p}$ is a finite prime of $k$. The following result is
standard algebraic number theory.

######  Lemma 2.14.

Let $E/F$ be an extension of $p$-adic fields, and let $P$ be the maximal ideal
of $\mathcal{O}_{F}$. Let $p$ be the residue characteristic of $F$. Write $e$
and $f$ for the ramification index and inertia degree, respectively, of $E/F$.
If $p\nmid e$, then

$v_{P}(\operatorname{Disc}(E/F))=f(e-1).$

######  Definition 2.15.

Let $\sigma$ be a splitting symbol. We say that $\sigma$ is _overramified_ if
it is one of $(1^{2}1^{2}),(2^{2})$, and $(1^{4})$.

######  Lemma 2.16.

Let $F$ be a $p$-adic field with maximal ideal $P$, such that the residue
characteristic is not $2$. Let $L/F$ be a quartic étale algebra. If
$v_{P}(\operatorname{Disc}(L/F))<2$, then the splitting symbol $(L,P)$ is not
overramified.

###### Proof.

It suffices to show that overramified extensions have
$v_{P}(\operatorname{Disc}(L/F))\geq 2$. This follows from Lemma 2.14 and
Remark 2.5, since the residue characteristic is odd, hence does not divide any
of the ramification indices in overramified symbols. ∎

We will use the following well-known fact.

######  Lemma 2.17.

Let $E/F$ be an unramified extension of $p$-adic fields. Then

$\mathcal{O}_{F}^{*}\subseteq N_{E/F}(E^{*}).$

###### Proof.

This is precisely the corollary on Page 50 of [Lan94]. ∎

######  Lemma 2.18.

Let $L/k_{\mathfrak{p}}$ be a quartic étale algebra such that

$(L,\mathfrak{p})\in\\{(1111),(112),(13),(1^{2}11),(1^{2}2),(1^{3}1)\\}.$

We have

$N_{L/k_{\mathfrak{p}}}(L^{*})=k_{\mathfrak{p}}^{*}.$

###### Proof.

Write $L=L_{1}\times\ldots\times L_{r}$ as usual. If $\sigma\neq(1^{2}2)$,
then one of the $L_{i}$ is equal to $k_{\mathfrak{p}}$, so
$N_{L/k_{\mathfrak{p}}}(L^{*})=k_{\mathfrak{p}}^{*}$. Suppose that
$\sigma=(1^{2}2)$. Then we may assume that $L_{1}/k_{\mathfrak{p}}$ is
quadratic and unramified, and $L_{2}/k_{\mathfrak{p}}$ is quadratic and
totally ramified. Since $L_{1}/k_{\mathfrak{p}}$ is unramified, Lemma 2.17
tells us that the group $N_{L_{1}/k_{\mathfrak{p}}}(L_{1}^{*})$ contains all
units of $\mathcal{O}_{k_{\mathfrak{p}}}$. Since $L_{2}/k_{\mathfrak{p}}$ is
totally ramified, $N_{L_{2}/k_{\mathfrak{p}}}(L_{2}^{*})$ contains an element
of valuation $1$. The result follows. ∎

######  Definition 2.19.

In light of Lemma 2.18, write _$\mathrm{triv}$_ for the set

$\mathrm{triv}=\\{(1111),(112),(13),(1^{2}11),(1^{3}1),(1^{2}2)\\}.$

We call the elements of this set the _trivial_ splitting symbols.

######  Remark 2.20.

We have chosen to call these symbols trivial because the problem we want to
solve, namely determining whether $\mathcal{A}\subseteq
N_{L/k_{\mathfrak{p}}}(L^{*})$, is trivial whenever
$(L,\mathfrak{p})\in\mathrm{triv}$, since the answer is always yes.

######  Lemma 2.21.

Let $\alpha\in k^{*}$, and let $\mathfrak{p}$ be a finite prime of $k$ such
that $N(\mathfrak{p})$ is odd and $4\mid v_{\mathfrak{p}}(\alpha)$. Then
$\Sigma_{\alpha,\mathfrak{p}}$ contains every quartic étale algebra
$L/k_{\mathfrak{p}}$ with
$v_{\mathfrak{p}}(\operatorname{Disc}(L/k_{\mathfrak{p}}))<2$.

###### Proof.

Let $L/k_{\mathfrak{p}}$ be a quartic étale algebra with
$v_{\mathfrak{p}}(\operatorname{Disc}(L/k_{\mathfrak{p}}))<2$. By Lemma 2.16,
the symbol $(L,\mathfrak{p})$ is not overramified, which means that it is
either trivial or one of $(22),(4)$. Suppose that $(L,\mathfrak{p})$ is
trivial. Then Lemma 2.18 tells us that that $\alpha\in
N_{L/k_{\mathfrak{p}}}(L^{*})$. Suppose instead that $(L,\mathfrak{p})$ is one
of $(22)$ and $(4)$. Then $N_{L/k_{\mathfrak{p}}}(L^{*})$ contains all units
by Lemma 2.17. Since $4\mid v_{\mathfrak{p}}(\alpha)$, it follows that
$\alpha\in N_{L/k_{\mathfrak{p}}}(L^{*})$. In either case, we have shown that
$\alpha\in N_{L/k_{\mathfrak{p}}}(L^{*})$, so indeed
$L\in\Sigma_{\alpha,\mathfrak{p}}$. ∎

The next result is immediate from the definitions of
$\Sigma_{\mathcal{A},\mathfrak{p}}$ and $\Sigma_{\alpha,\mathfrak{p}}$.

######  Lemma 2.22.

Let $\mathcal{A}\subseteq k^{*}$ be a finitely generated subgroup, and let
$\alpha_{1},\ldots,\alpha_{n}$ be a finite set of generators for
$\mathcal{A}$. Then

$\Sigma_{\mathcal{A},\mathfrak{p}}=\bigcap_{i=1}^{n}\Sigma_{\alpha_{i},\mathfrak{p}}.$

######  Lemma 2.23.

For each finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, the
collection of local conditions
$(\Sigma_{\mathcal{A},\mathfrak{p}})_{\mathfrak{p}}$ is acceptable.

###### Proof.

Take a finite generating set $\alpha_{1},\ldots,\alpha_{n}$ for $\mathcal{A}$,
and let $S$ be the set of primes $\mathfrak{p}$ of $k$ such that at least one
of the following holds:

  1. (1)

$N(\mathfrak{p})$ is even,

  2. (2)

$\mathfrak{p}$ is infinite,

  3. (3)

$4\nmid v_{\mathfrak{p}}(\alpha_{i})$ for some $i$.

By Lemma 2.21, for every prime $\mathfrak{p}$ not in $S$, each set
$\Sigma_{\alpha_{i},\mathfrak{p}}$ contains every quartic étale algebra
$L/k_{\mathfrak{p}}$ with
$v_{\mathfrak{p}}(\operatorname{Disc}(L/k_{\mathfrak{p}}))<2$. By Lemma 2.22,
it follows that, for every prime $\mathfrak{p}$ not in $S$, the set
$\Sigma_{\mathcal{A},\mathfrak{p}}$ contains every quartic étale algebra
$L/k_{\mathfrak{p}}$ with
$v_{\mathfrak{p}}(\operatorname{Disc}(L/k_{\mathfrak{p}}))<2$.

Since $S$ is finite, this means that the collection
$(\Sigma_{\mathcal{A},\mathfrak{p}})_{\mathfrak{p}}$ of local conditions is
acceptable. ∎

Write _$m_{\mathcal{A},\mathfrak{p}}$_ for the mass
$m_{\mathfrak{p}}(\Sigma_{\mathcal{A},\mathfrak{p}})$.

###### Proof of Theorem 1.4.

The equality is immediate from Theorem 2.11 and Lemma 2.23. The numbers
$m_{\mathcal{A},\mathfrak{p}}$ are positive since
$\Sigma_{\mathcal{A},\mathfrak{p}}$ contains every extension with trivial
splitting symbol, and they are rational by definition of masses in Definition
2.10. ∎

##  3\. Computation of Masses

In this section, given an $\alpha\in k^{*}$, we give explicit values for the
masses $m_{\alpha,\mathfrak{p}}$, whenever $N(\mathfrak{p})$ is odd. We also
present an algorithm for computing these masses in general, which we use to
calculate every possible $m_{\alpha,\mathfrak{p}}$ in the case $k=\mathbb{Q}$
and $\mathfrak{p}=(2)$. In Section 3.1, we define some notation and outline
our strategy for computing the masses $m_{\alpha,\mathfrak{p}}$. In Section
3.2, we compute explicitly the norm groups of all possible quartic étale
algebras over $k_{\mathfrak{p}}$, for $\mathfrak{p}$ not lying over $2$. In
Section 3.3, we compute all possible sets $\Sigma_{\alpha,\mathfrak{p}}$,
again for $\mathfrak{p}$ not lying over $2$, as well as their masses
$m_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$. Finally, in Section 3.4, we
describe our algorithm for computing
$m_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$ when $\mathfrak{p}$ lies over
$2$. We conclude the section by assembling the pieces to prove Theorem 1.5.

###  3.1. Strategy

We start by introducing some auxiliary notation.

######  Definition 3.1.

Let $\mathfrak{p}$ be a finite prime of $k$, and let $\Sigma_{\mathfrak{p}}$
be a set of quartic étale $k_{\mathfrak{p}}$-algebras. Define the _pre-mass of
$\Sigma_{\mathfrak{p}}$_, denoted
_$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\mathfrak{p}})$_ , by

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\mathfrak{p}})=\sum_{L\in\Sigma_{\mathfrak{p}}}\frac{1}{\operatorname{Disc}_{\mathfrak{p}}(L/k_{\mathfrak{p}})}\cdot\frac{1}{\\#\operatorname{Aut}(L/k_{\mathfrak{p}})}.$

The additional notation of pre-mass is perhaps a little cumbersome, but it is
a natural quantity to consider and makes the results of the current section
easier to state. Of course, the point is that

$m_{\mathfrak{p}}(\Sigma_{\mathfrak{p}})=\frac{N(\mathfrak{p})-1}{N(\mathfrak{p})}\cdot\widetilde{m}_{\mathfrak{p}}(\Sigma_{\mathfrak{p}})$

for $\mathfrak{p}$ and $\Sigma_{\mathfrak{p}}$ as in Definition 3.1. As such,
for finite primes $\mathfrak{p}$, we will compute
$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$, and this
information will immediately allow us to compute the values of
$m_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$.

######  Definition 3.2.

Recall that, for $\alpha\in k^{*}$, we defined $\Sigma_{\alpha,\mathfrak{p}}$
to be the set of isomorphism classes of quartic étale
$k_{\mathfrak{p}}$-algebras $L$ with $\alpha\in
N_{L/k_{\mathfrak{p}}}(L^{*})$. For a possible splitting symbol $\sigma$,
write _$\Sigma_{\alpha,\mathfrak{p}}^{\sigma}$_ for the set

$\Sigma_{\alpha,\mathfrak{p}}^{\sigma}=\\{L\in\Sigma_{\alpha,\mathfrak{p}}:(L,\mathfrak{p})=\sigma\\}.$

Since the sets $\Sigma_{\alpha,\mathfrak{p}}^{\sigma}$ partition
$\Sigma_{\alpha,\mathfrak{p}}$ we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})=\sum_{\sigma}\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{\sigma}).$

Therefore, it suffices to compute the values of
$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{\sigma})$ for each
$\sigma$. This is more tractable, since there are not many quartic étale
algebras with a given splitting symbol, so we can consider each one in turn.
In the current section, for each $\mathfrak{p}$ with $N(\mathfrak{p})$ odd, we
explicitly compute the sets $\Sigma_{\alpha,\mathfrak{p}}^{\sigma}$ along with
the norm group of each étale algebra in
$\Sigma_{\alpha,\mathfrak{p}}^{\sigma}$.

###  3.2. Norm group computations

Recall that we fixed a number field $k$ and a prime ideal $\mathfrak{p}$ of
$k$, and that we write $N(\mathfrak{p})$ or $q$ for the size of the residue
field $\mathcal{O}_{k}/\mathfrak{p}$.

For each $\sigma$, we compute the norm group of each étale
$k_{\mathfrak{p}}$-algebra with splitting symbol $\sigma$. Throughout this
subsection, $L$ will denote a quartic étale $k_{\mathfrak{p}}$-algebra, and
$\pi$ will be a uniformiser of $k_{\mathfrak{p}}$. The following lemma is
well-known class field theory.

######  Lemma 3.3.

Let $E/F$ be an extension of $p$-adic fields, and write $E^{\mathrm{ab}}$ for
the largest abelian extension of $F$ contained in $E$. Then

$[F^{*}:N_{E/F}(E^{*})]=[E^{\mathrm{ab}}:F].$

######  Lemma 3.4.

Let $F$ be a $p$-adic field with residue field of order $q$ and uniformiser
$\pi$, and let $e$ be a positive integer coprime111So that our extensions are
tamely ramified. to $q$. Let $g=\gcd(e,q-1)$. There are $g$ isomorphism
classes of totally ramified degree $e$ extensions of $F$, and they are given
by

$\frac{F[X]}{(X^{e}-\zeta_{q-1}^{j}\pi)},\quad j=0,1,\ldots,g-1,$

where $\zeta_{q-1}$ is a primitive $(q-1)^{\mathrm{st}}$ root of unity in $F$.

###### Proof.

This is almost the same as [PR01, Theorem 7.2]. The only difference is that we
replace the polynomial $X^{e}+\zeta_{q-1}^{j}\pi$ with
$X^{e}-\zeta_{q-1}^{j}\pi$. This modification does not affect the proof. ∎

######  Definition 3.5.

Retaining the notation of Lemma 3.4, we will write
_$F\Big{(}\sqrt[e]{\zeta_{q-1}^{j}\pi}\Big{)}$_ for the extension

$\frac{F[X]}{(X^{e}-\zeta_{q-1}^{j}\pi)},$

where formally

$\sqrt[e]{\zeta_{q-1}^{j}\pi}=X+(X^{e}-\zeta_{q-1}^{j}\pi).$

######  Lemma 3.6.

If $(L,\mathfrak{p})=(22)$, then

$N_{L/k_{\mathfrak{p}}}(L^{*})=\\{v\pi^{2m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},\quad
m\in\mathbb{Z}\\}.$

###### Proof.

We have $L=L_{1}\times L_{1}$, where $L_{1}$ is the unique unramified
quadratic extension of $k_{\mathfrak{p}}$. By Lemma 2.17,
$N_{L_{1}/k_{\mathfrak{p}}}(L_{1}^{*})$ contains
$\mathcal{O}_{k_{\mathfrak{p}}}^{*}$. We have
$N_{L_{1}/k_{\mathfrak{p}}}(\pi)=\pi^{2}$, so

$N_{L/k_{\mathfrak{p}}}(L^{*})=N_{L_{1}/k_{\mathfrak{p}}}(L_{1}^{*})\supseteq\\{v\pi^{2m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},\quad
m\in\mathbb{Z}\\},$

and we have equality by Lemma 3.3. ∎

######  Lemma 3.7.

If $(L,\mathfrak{p})=(4)$, then

$N_{L/k_{\mathfrak{p}}}(L^{*})=\\{v\pi^{4m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},\quad
m\in\mathbb{Z}\\}.$

###### Proof.

The proof is essentially the same as that of Lemma 3.6. ∎

######  Lemma 3.8.

If $N(\mathfrak{p})$ is odd and $(L,\mathfrak{p})=(1^{2}1^{2})$, then the
possibilities for $L$, and its norm group, are as follows.

$L$ | $N_{L/k_{\mathfrak{p}}}(L^{*})$
---|---
$k_{\mathfrak{p}}(\sqrt{\pi})\times k_{\mathfrak{p}}(\sqrt{\pi})$ | $\\{v^{2}(-\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$
$k_{\mathfrak{p}}(\sqrt{\pi})\times k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})$ | $k_{\mathfrak{p}}^{*}$
$k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})\times k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})$ | $\\{v^{2}(-\zeta_{q-1}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$

###### Proof.

Since $N(\mathfrak{p})$ is odd, Lemma 3.4 tells us that there are exactly two
totally ramified quadratic extensions of $k_{\mathfrak{p}}$, namely
$k_{\mathfrak{p}}(\sqrt{\pi})$ and $k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})$.
Therefore, the possibilities for $L$ are indeed those listed in the table. It
is easy to see that, for $j=0,1$, we have

$N_{k_{\mathfrak{p}}\big{(}\sqrt{\zeta_{q-1}^{j}\pi}\big{)}/k_{\mathfrak{p}}}\Big{(}k_{\mathfrak{p}}\Big{(}\sqrt{\zeta_{q-1}^{j}\pi}\Big{)}^{*}\Big{)}\supseteq\\{v^{2}(-\zeta_{q-1}^{j}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},\quad
m\in\mathbb{Z}\\},$

and Lemma 3.3 tells us that we have equality in both cases. The result then
follows from the fact that

$N_{L/k_{\mathfrak{p}}}(L^{*})=N_{L_{1}/k_{\mathfrak{p}}}(L_{1}^{*})\cdot
N_{L_{2}/k_{\mathfrak{p}}}(L_{2}^{*}),$

where $L_{1}$ and $L_{2}$ are the quadratic extensions of $k_{\mathfrak{p}}$
with $L=L_{1}\times L_{2}$. ∎

######  Lemma 3.9.

If $N(\mathfrak{p})$ is odd and $(L,\mathfrak{p})=(2^{2})$, then the
possibilities for $L$, and the corresponding groups
$N_{L/k_{\mathfrak{p}}}(L^{*})$, are given by the following table:

$L$ | $N_{L/k_{\mathfrak{p}}}(L^{*})$
---|---
$k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}},\sqrt{\pi})$ | $\\{v^{2}\pi^{2m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$
$k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}},\sqrt{\zeta_{q^{2}-1}\pi})$ | $\\{v^{2}\zeta_{q-1}^{m}\pi^{2m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$

###### Proof.

By applying Lemma 3.4 to the maximal unramified subextension

$L^{ur}=k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}})$

of $L$, we see that $L$ is one of

$L_{j}=L^{ur}\Big{(}\sqrt{\zeta_{q^{2}-1}^{j}\pi}\Big{)},\quad j=0,1.$

Every element of $\mathcal{O}_{k_{\mathfrak{p}}}^{*}$ is square in $L^{ur}$,
and it follows that $\mathcal{O}_{k_{\mathfrak{p}}}^{*2}\subseteq
N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})$ for each $j$. We have
$\pi^{2}\zeta_{q-1}^{j}\in N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})$ for $j=0,1$,
which means that

$\\{v^{2}\zeta_{q-1}^{jm}\pi^{2m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},\quad
m\in\mathbb{Z}\\}\subseteq N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*}),\quad j=0,1.$

Finally, Lemma 3.3 tells us that these inclusions are in fact equalities. ∎

######  Lemma 3.10.

Let $(L,\mathfrak{p})=(1^{4})$ and $q\equiv 1\pmod{4}$. Up to isomorphism,
there are four distinct possibilities for $L$, given by
$L_{j}=k_{\mathfrak{p}}\Big{(}\sqrt[4]{\zeta_{q-1}^{j}\pi}\Big{)}$ for
$j\in\\{0,1,2,3\\}$, and

$N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})=\\{v^{4}(-\zeta_{q-1}^{j}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}.$

###### Proof.

The classification of extensions comes from Lemma 3.4. Clearly
$N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})$ contains all fourth powers in
$k_{\mathfrak{p}}$, and we have
$N_{L_{j}/k_{\mathfrak{p}}}\Big{(}\sqrt[4]{\zeta_{q-1}^{j}\pi}\Big{)}=-\zeta_{q-1}^{j}\pi$,
so

$\\{v^{4}(-\zeta_{q-1}^{j}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}\subseteq
N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*}),$

and we get equality from Lemma 3.3. ∎

######  Lemma 3.11.

Let $(L,\mathfrak{p})=(1^{4})$ and $q\equiv 3\pmod{4}$. Up to isomorphism,
there are two distinct possibilities for $L$, given by
$L_{j}=k_{\mathfrak{p}}\Big{(}\sqrt[4]{\zeta_{q-1}^{j}\pi}\Big{)}$ for
$j\in\\{0,1\\}$, and

$\displaystyle N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})$
$\displaystyle=\\{v^{2}(-\zeta_{q-1}^{j}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$
$\displaystyle=\\{v^{4}(-\zeta_{q-1}^{j}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$

###### Proof.

The proof is essentially the same as that of Lemma 3.10, except that the
subgroup has index $2$, because the extensions are nonabelian. Since $q\equiv
3\pmod{4}$, we have
$\mathcal{O}_{k_{\mathfrak{p}}}^{*2}=\mathcal{O}_{k_{\mathfrak{p}}}^{*4}$. ∎

###  3.3. Mass computations

######  Definition 3.12.

Let $\sigma=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$ be the splitting symbol of an
étale algebra $L$ over a $p$-adic field $F$ with residue field of size $q$.
Define

$\operatorname{Disc}_{\mathfrak{p}}(\sigma)=q^{\sum f_{i}(e_{i}-1)}$

and

$\\#\operatorname{Aut}(\sigma)=\Big{(}\prod_{i}f_{i}\Big{)}\cdot\\#\\{\varphi\in
S_{r}:f_{\varphi(i)}^{e_{\varphi(i)}}=f_{i}^{e_{i}}\text{ for all $i$}\\}.$

Note that $\Sigma_{1,\mathfrak{p}}^{\sigma}$ is just the set of all étale
algebras with splitting symbol $\sigma$.

######  Lemma 3.13.

Let $\sigma$ be a splitting symbol of degree $n$, and let $\mathfrak{p}$ be
any finite prime of $k$. Then

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{1,\mathfrak{p}}^{\sigma})=\frac{1}{\operatorname{Disc}_{\mathfrak{p}}(\sigma)\\#\operatorname{Aut}(\sigma)}.$

###### Proof.

By definition of $\Sigma_{\alpha,\mathfrak{p}}^{\sigma}$ and
$\tilde{m}_{\mathfrak{p}}$, we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{1,\mathfrak{p}}^{\sigma})=\sum_{\begin{subarray}{c}[L:k_{\mathfrak{p}}]=n\text{
\'{e}tale}\\\
(L,\mathfrak{p})=\sigma\end{subarray}}\frac{1}{\operatorname{Disc}_{\mathfrak{p}}(L)}\cdot\frac{1}{\\#\operatorname{Aut}(L/k_{\mathfrak{p}})}.$

Therefore, the result is essentially [Bha07, Proposition 2.1], adapted to the
case of general $p$-adic fields. The appendix of [Bha07] explains how to make
this adaptation. ∎

######  Lemma 3.14.

Let $\mathfrak{p}$ be any finite prime of $k$. The pre-masses of the sets
$\Sigma_{1,\mathfrak{p}}^{\sigma}$ are given by the following table:

$\sigma$ | $\widetilde{m}_{\mathfrak{p}}(\Sigma_{1,\mathfrak{p}}^{\sigma})$
---|---
$(1111)$ | $\frac{1}{24}$
$(112)$ | $\frac{1}{4}$
$(13)$ | $\frac{1}{3}$
$(22)$ | $\frac{1}{8}$
$(4)$ | $\frac{1}{4}$
$(1^{2}11)$ | $\frac{1}{2\,q}$
$(1^{2}2)$ | $\frac{1}{2\,q}$
$(1^{2}1^{2})$ | $\frac{1}{2\,q^{2}}$
$(2^{2})$ | $\frac{1}{2\,q^{2}}$
$(1^{3}1)$ | $\frac{1}{q^{2}}$
$(1^{4})$ | $\frac{1}{q^{3}}$

###### Proof.

This follows immediately from Lemma 3.13. ∎

Write $\Sigma_{\mathfrak{p}}^{\mathrm{triv}}$ for the set of isomorphism
classes of quartic étale $k_{\mathfrak{p}}$-algebras with trivial splitting
symbols. It follows from Lemma 2.18 that
$\Sigma_{\mathfrak{p}}^{\mathrm{triv}}\subseteq\Sigma_{\mathcal{A},\mathfrak{p}}$
for all finitely generated subgroups $\mathcal{A}\subseteq k^{*}$.

######  Lemma 3.15.

For all finite primes $\mathfrak{p}$ of $k$, we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\mathfrak{p}}^{\mathrm{triv}})=\frac{5q^{2}+8q+8}{8q^{2}}.$

###### Proof.

This follows immediately from Lemma 3.14. ∎

We fix some notation for use in the following few lemmas. As always, let
$\alpha\in k^{*}$. Let $\mathfrak{p}$ be a finite prime of $k$, and let
$q=N(\mathfrak{p})$. Write $r=\operatorname{ord}_{\mathfrak{p}}(\alpha)$, and
fix an arbitrary element $\pi\in\mathfrak{p}\setminus\mathfrak{p}^{2}$. Then
$\alpha=u\pi^{r}$ for a unique $u\in k^{*}$, and moreover
$u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*}$. The notations $r,\pi,$ and $u$ will
be used without introduction for the rest of the current subsection.

######  Lemma 3.16.

For all finite primes $\mathfrak{p}$, we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(22)})=\begin{cases}\frac{1}{8}\quad\text{if
$2\mid r$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

Lemma 3.6 tells us that

$\Sigma_{\alpha,p}^{\sigma}=\begin{cases}\Sigma_{1,p}^{\sigma}\quad\text{if
$2\mid r$},\\\ \varnothing\quad\text{otherwise},\end{cases}$

so the result follows from Lemma 3.14. ∎

######  Lemma 3.17.

For all finite primes $\mathfrak{p}$, we have

$\tilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(4)})=\begin{cases}\frac{1}{4}\quad\text{if
$4\mid r$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

The proof is essentially the same as that of Lemma 3.16, using Lemmas 3.7 and
3.14. ∎

######  Lemma 3.18.

If $q$ is odd, then we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(1^{2}1^{2})})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$2\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\frac{1}{4q^{2}}\quad\text{if $2\mid r$ and
$u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\frac{3}{8q^{2}}\quad\text{if $2\nmid r$}.\end{cases}$

###### Proof.

Lemma 3.8 tells us that
$\Sigma_{1,\mathfrak{p}}^{(1^{2}1^{2})}=\\{L_{1},L_{2},L_{3}\\}$, where the
fields $L_{j}$ and their norm groups are given by the following table:

$j$ | $L_{j}$ | $N_{L_{j}/k_{\mathfrak{p}}}(L_{j}^{*})$
---|---|---
$1$ | $k_{\mathfrak{p}}(\sqrt{\pi})\times k_{\mathfrak{p}}(\sqrt{\pi})$ | $\\{v^{2}(-\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$
$2$ | $k_{\mathfrak{p}}(\sqrt{\pi})\times k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})$ | $k_{\mathfrak{p}}^{*}$
$3$ | $k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})\times k_{\mathfrak{p}}(\sqrt{\zeta_{q-1}\pi})$ | $\\{v^{2}(-\zeta_{q-1}\pi)^{m}:v\in\mathcal{O}_{k_{\mathfrak{p}}}^{*},m\in\mathbb{Z}\\}$

We have

$\\#\operatorname{Aut}(L_{j}/k_{\mathfrak{p}})=\begin{cases}8\quad\text{if
$j=1,3$},\\\ 4\quad\text{otherwise},\end{cases}$

and

$\operatorname{Disc}_{\mathfrak{p}}(L_{j})=q^{2}\quad\text{for all $j$},$

by Lemma 2.14. It follows that

$\widetilde{m}_{\mathfrak{p}}(\\{L_{j}\\})=\begin{cases}\frac{1}{8q^{2}}\quad\text{if
$j=1,3$},\\\ \frac{1}{4q^{2}}\quad\text{otherwise}.\end{cases}$

Some casework establishes that

$\Sigma_{\alpha,\mathfrak{p}}^{(1^{2}1^{2})}=\begin{cases}\\{L_{1},L_{2},L_{3}\\}\quad\text{if
$2\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\\{L_{2}\\}\quad\text{if $2\mid r$ and
$u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\ \\{L_{2},L_{j}\\}\text{ for
some $j\in\\{1,3\\}$}\quad\text{if $2\nmid r$}.\end{cases}$

The result then follows by adding together the relevant pre-masses
$\widetilde{m}_{\mathfrak{p}}(\\{L_{j}\\})$. ∎

The following three lemmas are proved in essentially the same way as Lemma
3.18, using Lemmas 3.9, 3.10, and 3.11 respectively for the norm groups.

######  Lemma 3.19.

If $q$ is odd, then we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(2^{2})})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$4\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\ 0\quad\text{if
$4\mid r$ and $u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\frac{1}{4q^{2}}\quad\text{if $r\equiv 2\pmod{4}$},\\\ 0\quad\text{if $2\nmid
r$}.\end{cases}$

######  Lemma 3.20.

If $q\equiv 1\pmod{4}$, then we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(1^{4})})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$4\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*4}$},\\\ 0\quad\text{if
$4\mid r$ and $u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*4}$},\\\
\frac{1}{2q^{3}}\quad\text{if $r\equiv 2\pmod{4}$ and
$u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\ 0\quad\text{if $r\equiv
2\pmod{4}$ and $u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\frac{1}{4q^{3}}\quad\text{if $2\nmid r$}.\end{cases}$

######  Lemma 3.21.

If $q\equiv 3\pmod{4}$, then we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{(1^{4})})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$2\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\ 0\quad\text{if
$2\mid r$ and $u\not\in\mathcal{O}_{k_{\mathfrak{p}}}^{*2}$},\\\
\frac{1}{2q^{3}}\quad\text{if $2\nmid r$}.\end{cases}$

We now have complete descriptions of the pre-masses
$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$, and hence the
masses $m_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$, for finite primes
$\mathfrak{p}$ not lying over $2$. For infinite primes, the masses are given
by the following two lemmas.

######  Lemma 3.22.

Let $f:k\to\mathbb{C}$ be a real embedding, and let $\mathfrak{p}_{f}$ be the
prime corresponding to $f$. Then

$m_{\mathfrak{p}_{f}}(\Sigma_{\alpha,\mathfrak{p}_{f}})=\begin{cases}\frac{5}{12}\quad\text{if
$f(\alpha)>0$},\\\ \frac{7}{24}\quad\text{if $f(\alpha)<0$}.\end{cases}$

###### Proof.

Since $f$ is a real embedding, we have $k_{\mathfrak{p}_{f}}\cong\mathbb{R}$.
The algebraic extensions of $\mathbb{R}$ are precisely $\mathbb{R}$ and
$\mathbb{C}$. The extension $\mathbb{C}/\mathbb{R}$ is totally ramified, so
the possible quartic splitting symbols are $(1111),(111^{2}),(1^{2}1^{2})$.
Each of these corresponds to a unique étale algebra, and the consequent norm
groups and masses are given by the following table.

$\sigma$ | $L$ | $N_{L/k_{\mathfrak{p}}}(L^{*})$ | $m_{\infty}(\\{L\\})$
---|---|---|---
$(1111)$ | $\mathbb{R}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ | $\mathbb{R}^{*}$ | $\frac{1}{24}$
$(111^{2})$ | $\mathbb{R}\times\mathbb{R}\times\mathbb{C}$ | $\mathbb{R}^{*}$ | $\frac{1}{4}$
$(1^{2}1^{2})$ | $\mathbb{C}\times\mathbb{C}$ | $\mathbb{R}_{>0}$ | $\frac{1}{8}$

The masses in the table are computed using Definition 2.10. ∎

######  Lemma 3.23.

Let $g:k\to\mathbb{C}$ be a complex embedding, and let $\mathfrak{p}_{g}$ be
the prime corresponding to $g$. Then

$m_{\mathfrak{p}_{g}}(\Sigma_{\alpha,\mathfrak{p}_{g}})=\frac{1}{24}.$

###### Proof.

Since $k_{\mathfrak{p}_{g}}\cong\mathbb{C}$ is algebraically closed, it has
only one quartic étale algebra, up to isomorphism, and that étale algebra is
isomorphic to $L=\mathbb{C}\times\mathbb{C}\times\mathbb{C}\times\mathbb{C}$.
We have

$\\#\operatorname{Aut}(L/\mathbb{C})=24,$

so

$m_{\mathfrak{p}_{g}}(\\{L\\})=\frac{1}{24}.$

We have $N_{L/\mathbb{C}}(L^{*})=\mathbb{C}^{*}$, so
$\Sigma_{\alpha,\mathfrak{p}_{g}}=L$ for all $\alpha$. ∎

######  Lemma 3.24.

Let $\alpha\in k^{*}$ and let $\mathfrak{p}$ be a finite prime of $k$ with
$N(\mathfrak{p})$ odd. The values of $m_{\alpha,\mathfrak{p}}$ are given by
Tables 1 and 2.

###### Proof.

Since

$m_{\alpha,\mathfrak{p}}=\sum_{\sigma}m_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}}^{\sigma}),$

the result follows easily from Lemmas 3.15-3.21. ∎

###  3.4. Primes lying over $2$

Let $\mathfrak{p}$ be a prime with even norm. The methods above do not apply,
because in the wildly ramified case there are many more extensions. However,
we have implemented an algorithm in MAGMA that computes
$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$ for any number
field $k$ and any finite prime $\mathfrak{p}$ of $k$. This has two
applications:

  1. (1)

The algorithm allows us to verify the results from Section 3.3 for specific
primes $\mathfrak{p}$ with odd norm. Simultaneously, agreement with our
results provides evidence that the algorithm works correctly.

  2. (2)

The algorithm allows us to compute explicit values for
$m_{\alpha,\mathfrak{p}}(\Sigma_{\alpha,\mathfrak{p}})$ for primes
$\mathfrak{p}$ lying over $2$. In particular, we used the algorithm to compute
the values in Table 3.

Our code is in the GitHub repository at https://github.com/Sebastian-
Monnet/S4-quartics-prescribed-norms. The most important function in our code
is
ComputePreMassOfNormSet. This takes as input a number alpha, a $p$-adic field
BaseField, and a list Symbols of splitting symbols. The function then
explicitly computes the set of quartic étale algebras of BaseField such that
alpha is a norm and returns the pre-mass of that set. In our notation, the
function computes

$\tilde{m}_{\mathfrak{p}}\Big{(}\bigcup_{\sigma\in
S}\Sigma_{\alpha,\mathfrak{p}}^{\sigma}\Big{)},$

given a set $S$ of splitting symbols.

The MAGMA file in the repository executes two functions: PrintPreMassesFor2()
and TestUpTo(20,3). The first of these functions prints the value of
$m_{2}(\Sigma_{\alpha,2})$ for $\alpha=2^{r}u$, as $r$ ranges over
$\\{0,1,2,3\\}$ and $u$ ranges over $\\{1,3,5,\ldots,15\\}$. We will see in
the proof of Lemma 3.26 that this is precisely the set of values we need to
compute in order to know $m_{\alpha,2}$ for all $\alpha\in\mathbb{Q}^{*}$. The
second function checks, for each prime $p\in\\{5,7,11,13,17,19\\}$, that the
values in Tables 1 and 2 are correct for all extensions
$k_{\mathfrak{p}}/\mathbb{Q}_{p}$ of degree at most $3$. The function returns
true if the results agree with those we have proved, and false otherwise.

######  Lemma 3.25.

Let $\mathfrak{p}$ lie over $2$ and let
$\alpha\in\mathcal{O}_{k_{\mathfrak{p}}}^{*}$. Suppose that $\alpha\equiv
1\pmod{8\pi}$, for a uniformiser $\pi$ of $k_{\mathfrak{p}}$. Then
$\alpha\in\mathcal{O}_{k_{\mathfrak{p}}}^{*4}$.

###### Proof.

Let $f(X)=(2X+1)^{4}-\alpha$. Then we have

$f(X)=16X^{4}+32X^{3}+24X^{2}+8X+1-\alpha,$

so $g(X):=\frac{1}{8}f(X)\in\mathcal{O}_{k_{\mathfrak{p}}}[X]$, and we have

$\lvert
g(0)\rvert_{\mathfrak{p}}=\Big{\lvert}\frac{1-\alpha}{8}\Big{\rvert}_{\mathfrak{p}}<1,$

whereas

$\lvert g^{\prime}(0)\rvert_{\mathfrak{p}}=\lvert 1\rvert_{\mathfrak{p}}=1.$

By Hensel’s Lemma, the polynomial $g(X)$ has a root in
$\mathcal{O}_{k_{\mathfrak{p}}}$, hence $f(X)$ does too. ∎

######  Lemma 3.26.

When $k=\mathbb{Q}$, the masses $m_{\alpha,2}$ are given by Table 3.

###### Proof.

Lemma 3.25 implies that a system of representatives for
$\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*4}$ is given by $2^{r}u$, as $r$ ranges
over $\\{0,1,2,3\\}$ and $u$ ranges over $\\{1,3,5,\ldots,15\\}$. Since
$\mathbb{Q}_{2}^{*4}\subseteq N_{L/\mathbb{Q}_{2}}(L^{*})$ for any quartic
étale algebra $L/\mathbb{Q}_{2}$, it suffices to compute

$m_{2}(\Sigma_{\alpha,2})$

for each such $\alpha=2^{r}u$. The corresponding pre-masses are outputted by
the code in our repository, and we obtain the masses by halving the pre-
masses. ∎

###### Proof of Theorem 1.5.

The result is immediate from Lemmas 3.24, 3.22, 3.23, and 3.26. ∎

##  4\. Proportion of Extensions with Prescribed Norms

The goal of this section is to prove Theorem 1.1. Recall that, given a finite
prime $\mathfrak{p}$ of $k$, we often write $q$ for the norm $N(\mathfrak{p})$
of $\mathfrak{p}$.

######  Lemma 4.1.

Let $\alpha\in k^{*}$, and let $\mathfrak{p}$ be a finite prime of $k$ such
that the following two conditions hold:

  1. (1)

$q$ is odd,

  2. (2)

$4\mid\operatorname{ord}_{\mathfrak{p}}(\alpha)$.

Then $m_{\alpha,\mathfrak{p}}<1+\frac{1}{q^{2}}$.

###### Proof.

This is immediate from Tables 1 and 2. ∎

Let $\alpha\in k^{*}$. In the following lemma, we write $S$ for the set of
primes $\mathfrak{p}$ of $k$ such that at least one of the following holds:

  1. (1)

$\mathfrak{p}$ is finite and $q$ is even,

  2. (2)

$\mathfrak{p}$ is finite and
$4\nmid\operatorname{ord}_{\mathfrak{p}}(\alpha)$,

  3. (3)

$\mathfrak{p}$ is infinite.

Then $S$ is a finite set, so we can explicitly compute the mass
$m_{\alpha,\mathfrak{p}}$ for every $\mathfrak{p}\in S$.

######  Lemma 4.2.

Let $\alpha\in k^{*}$, and let $S$ be the set of primes defined above. We have
an explicit finite bound

$\lim_{X\to\infty}\frac{N(X;\alpha)}{X}\leq\frac{1}{2}\operatornamewithlimits{Res}_{s=1}\zeta_{k}(s)\cdot\zeta_{k,S}(2)\cdot\prod_{\mathfrak{p}\in
S}m_{\alpha,\mathfrak{p}},$

where $\zeta_{k,S}(s)=\prod_{p\not\in S}(1-(N\mathfrak{p})^{-s})^{-1}$ is the
partial zeta function.

###### Proof.

It is easy to see that $S$ is a finite set, so we have

$\prod_{\mathfrak{p}\in S}m_{\alpha,\mathfrak{p}}<\infty.$

By Theorem 1.4, we have

$\lim_{X\to\infty}\frac{N(X;\alpha)}{X}=\frac{1}{2}\operatornamewithlimits{Res}_{s=1}\zeta_{k}(s)\prod_{\mathfrak{p}\in
S}m_{\alpha,\mathfrak{p}}\cdot\prod_{\mathfrak{p}\not\in
S}m_{\alpha,\mathfrak{p}},$

so it suffices to show that

$\prod_{\mathfrak{p}\not\in S}m_{\alpha,\mathfrak{p}}\leq\zeta_{k,S}(2).$

Note that $S$ is precisely the set of primes to which Lemma 4.1 does not
apply, so we have

$\displaystyle\prod_{\mathfrak{p}\not\in S}m_{\alpha,\mathfrak{p}}$
$\displaystyle\leq\prod_{\mathfrak{p}\not\in
S}\Big{(}1+\frac{1}{N(\mathfrak{p})^{2}}\Big{)}$
$\displaystyle\leq\prod_{\mathfrak{p}\not\in
S}\Big{(}1+\frac{1}{N(\mathfrak{p})^{2}}+\frac{1}{N(\mathfrak{p})^{4}}+\ldots\Big{)}$
$\displaystyle=\zeta_{k,S}(2).$

∎

######  Lemma 4.3.

Let $\mathcal{A}\subseteq k^{*}$ be a finitely generated subgroup and let
$\mathfrak{p}$ be any prime of $k$. Then $m_{\mathcal{A},\mathfrak{p}}>0$.

###### Proof.

If $\mathfrak{p}$ is archimedean, the result follows from Lemmas 3.22 and
3.23. If $\mathfrak{p}$ is finite, then

$m_{\mathfrak{p}}(\Sigma_{\mathcal{A},\mathfrak{p}})\geq
m_{\mathfrak{p}}(\Sigma_{\mathfrak{p}}^{\mathrm{triv}})=\frac{q-1}{q}\cdot\frac{5q^{2}+8q+8}{8q^{2}},$

by Lemma 3.15. ∎

######  Lemma 4.4.

For any finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, we have

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}>0.$

###### Proof.

Let $\alpha_{1},\ldots,\alpha_{n}$ be a finite generating set for
$\mathcal{A}$, so that
$\Sigma_{\mathcal{A},\mathfrak{p}}=\cap_{i}\Sigma_{\alpha_{i},\mathfrak{p}}$
by Lemma 2.22. Let $S$ be the set of primes of $k$ such that at least one of
the following conditions holds:

  1. (1)

$\mathfrak{p}$ is finite and $q$ is even,

  2. (2)

$\mathfrak{p}$ is finite and $q=3$,

  3. (3)

$4\nmid\operatorname{ord}_{\mathfrak{p}}(\alpha_{i})$ for some $i$,

  4. (4)

$\mathfrak{p}$ is infinite.

It is easy to see that $S$ is a finite set. Moreover, if $\mathfrak{p}\not\in
S$, then

$\Sigma_{\mathcal{A},\mathfrak{p}}\supseteq\Sigma_{\mathfrak{p}}^{\mathrm{triv}}\cup\Sigma_{1,\mathfrak{p}}^{(22)}\cup\Sigma_{1,\mathfrak{p}}^{(4)},$

by Lemmas 2.18, 3.6, and 3.7. Lemmas 3.15, 3.16, and 3.17 tell us that, for
all finite $\mathfrak{p}$, we have

$\widetilde{m}_{\mathfrak{p}}(\Sigma_{\mathfrak{p}}^{\mathrm{triv}}\cup\Sigma_{1,\mathfrak{p}}^{(22)}\cup\Sigma_{1,\mathfrak{p}}^{(4)})=\frac{q^{2}+q+1}{q^{2}}.$

It follows that

$m_{\mathfrak{p}}(\Sigma_{\mathfrak{p}}^{\mathrm{triv}}\cup\Sigma_{1,\mathfrak{p}}^{(22)}\cup\Sigma_{1,\mathfrak{p}}^{(4)})=1-\frac{1}{q^{3}}.$

Taking the product over all finite primes $\mathfrak{p}$ of $k$, we have

$\prod_{\mathfrak{p}\nmid\infty}\Big{(}1-\frac{1}{N(\mathfrak{p})^{3}}\Big{)}=\zeta_{k}(3)^{-1},$

which is greater than $0$. Since $S$ is finite, it follows that

$\prod_{\mathfrak{p}\not\in
S}m_{\mathcal{A},\mathfrak{p}}\geq\prod_{\mathfrak{p}\not\in
S}\Big{(}1-\frac{1}{N(\mathfrak{p})^{3}}\Big{)}>0.$

Since $S$ is finite, Lemma 4.3 implies that

$\prod_{\mathfrak{p}\in S}m_{\mathcal{A},\mathfrak{p}}>0.$

The result now follows from Theorem 1.4. ∎

######  Lemma 4.5.

For any finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, we have

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}<\infty.$

###### Proof.

This follows immediately from Lemma 4.2, since $N(X;\mathcal{A})\leq
N(X;\alpha)$ for all $\alpha\in\mathcal{A}$. ∎

The following lemma is elementary.

######  Lemma 4.6 (Elementary analysis lemma).

Let $(a_{n})$ and $(b_{n})$ be sequences of real numbers, and suppose that
$A=\prod_{n=1}^{\infty}a_{n}$ and $B=\prod_{n=1}^{\infty}b_{n}$ are
conditionally convergent products. Then the product

$\prod_{n=1}^{\infty}\frac{a_{n}}{b_{n}}$

is conditionally convergent to $\frac{A}{B}$.

######  Lemma 4.7.

We have

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X)}=\prod_{\mathfrak{p}}\frac{m_{\mathcal{A},\mathfrak{p}}}{m_{1,\mathfrak{p}}}.$

###### Proof.

This follows immediately from Theorem 1.4 and Lemma 4.6, whose hypotheses are
satisfied by Lemmas 4.5 and 4.4. ∎

###### Proof of Theorem 1.1.

For each finitely generated subgroup $\mathcal{A}\subseteq k^{*}$, Lemmas 4.4
and 4.5 tell us that

$0<\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}<\infty.$

It follows by the algebra of limits that

$\Bigg{(}\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}\Bigg{)}/\Bigg{(}\lim_{X\to\infty}\frac{N(X;\\{1\\})}{X}\Bigg{)}=\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X;\\{1\\})}.$

Lemma 4.4 tells us that the left-hand side is a ratio of two positive numbers,
so we have

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X;\\{1\\})}>0.$

Clearly

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X;\\{1\\})}\leq 1,$

with equality if and only if

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{X}=\lim_{X\to\infty}\frac{N(X;\\{1\\})}{X}.$

By Lemma 4.7, this equality holds if and only if
$m_{\mathcal{A},\mathfrak{p}}=m_{1,\mathfrak{p}}$ for all $\mathfrak{p}$, and
this is the case if and only if
$\Sigma_{\mathcal{A},\mathfrak{p}}=\Sigma_{1,\mathfrak{p}}$ for all
$\mathfrak{p}$, hence if and only if $\mathcal{A}\subseteq
N_{L/k_{\mathfrak{p}}}(L^{*})$ for every $\mathfrak{p}$ and every quartic
étale algebra $L/k_{\mathfrak{p}}$.

Suppose that this is the case. Taking $L$ to be the unramified quartic
extension of $k_{\mathfrak{p}}$ shows that $4\mid v_{\mathfrak{p}}(\alpha)$
for all $\alpha\in\mathcal{A}$ and for each prime $\mathfrak{p}$ of $k$. Let
$T$ be the set of primes $\mathfrak{p}$ of $k$ with $N(\mathfrak{p})$ odd. Let
$\alpha\in\mathcal{A}$ and let $\mathfrak{p}\in T$. Write $\pi$ for a
uniformiser of $\mathcal{O}_{k_{\mathfrak{p}}}$, so that $\alpha=\pi^{r}u$,
where $4\mid r$ and $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*}$. Taking $L$ to be
a totally ramified quartic field extension of $k_{\mathfrak{p}}$, Lemmas 3.10
and 3.11 tell us that $u\in\mathcal{O}_{k_{\mathfrak{p}}}^{*4}$. Therefore,
$\alpha$ is in $k_{\mathfrak{p}}^{*4}$ for all $\mathfrak{p}\in T$, so [NSW00,
Theorem 9.1.11] tells us that $\alpha\in k^{*4}$. We have shown that

$\lim_{X\to\infty}\frac{N(X;\mathcal{A})}{N(X;\\{1\\})}=1\implies\mathcal{A}\subseteq
k^{*4}.$

The converse implication is trivial, so we are done. ∎

##  Appendix A Tables of Masses

To use Tables 1 and 2, let $r=\operatorname{ord}_{\mathfrak{p}}(\alpha)$ and
choose $\pi\in\mathfrak{p}\setminus\mathfrak{p}^{2}$. Then set
$u=\pi^{-r}\alpha$. Let

$\bar{r}=r\pmod{4},\quad\bar{u}=u\pmod{\mathfrak{p}}\in\mathbb{F}_{\mathfrak{p}}.$

Then $m_{\alpha,\mathfrak{p}}$ is the entry in the table corresponding to
$\bar{r}$ and the status of $\bar{u}$ as a fourth power or a square.

To use Table 3, write $\alpha=2^{r}u$ for an odd integer $u$. Let

$\bar{r}=r\pmod{4},\quad\bar{u}=u\pmod{16}.$

Then $m_{\alpha,2}$ is the corresponding entry of the table.

Table 1. $q\equiv 1\pmod{4}$ $\bar{u}$ $\bar{r}$ | 0 | 1 | 2 | 3
---|---|---|---|---
$\mathbb{F}_{\mathfrak{p}}^{*4}$ | $\frac{{\left(q^{3}+q^{2}+2\,q+1\right)}{\left(q-1\right)}}{q^{4}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$ | $\frac{{\left(q^{2}+q+2\right)}{\left(3\,q+1\right)}{\left(q-1\right)}}{4\,q^{4}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$
$\mathbb{F}_{\mathfrak{p}}^{*2}\setminus\mathbb{F}_{\mathfrak{p}}^{*4}$ | $\frac{{\left(q^{2}+q+2\right)}{\left(q-1\right)}}{q^{3}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$ | $\frac{{\left(q^{2}+q+2\right)}{\left(3\,q+1\right)}{\left(q-1\right)}}{4\,q^{4}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$
$\mathbb{F}_{\mathfrak{p}}^{*}\setminus\mathbb{F}_{\mathfrak{p}}^{*2}$ | $\frac{{\left(4\,q^{2}+4\,q+5\right)}{\left(q-1\right)}}{4\,q^{3}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$ | $\frac{{\left(3\,q^{2}+4\,q+6\right)}{\left(q-1\right)}}{4\,q^{3}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+2\right)}{\left(q-1\right)}}{8\,q^{4}}$
Table 2. $q\equiv 3\pmod{4}$ $\bar{u}$ $\bar{r}$ | 0 | 1 | 2 | 3
---|---|---|---|---
$\mathbb{F}_{\mathfrak{p}}^{*2}$ | $\frac{{\left(q^{3}+q^{2}+2\,q+1\right)}{\left(q-1\right)}}{q^{4}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+4\right)}{\left(q-1\right)}}{8\,q^{4}}$ | $\frac{{\left(3\,q^{3}+4\,q^{2}+7\,q+4\right)}{\left(q-1\right)}}{4\,q^{4}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+4\right)}{\left(q-1\right)}}{8\,q^{4}}$
$\mathbb{F}_{\mathfrak{p}}^{*}\setminus\mathbb{F}_{\mathfrak{p}}^{*2}$ | $\frac{{\left(4\,q^{2}+4\,q+5\right)}{\left(q-1\right)}}{4\,q^{3}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+4\right)}{\left(q-1\right)}}{8\,q^{4}}$ | $\frac{{\left(3\,q^{2}+4\,q+6\right)}{\left(q-1\right)}}{4\,q^{3}}$ | $\frac{{\left(5\,q^{3}+8\,q^{2}+11\,q+4\right)}{\left(q-1\right)}}{8\,q^{4}}$
Table 3. Table defining $m_{\alpha,2}$. $\bar{u}$ $\bar{r}$ | 0 | 1 | 2 | 3
---|---|---|---|---
1 | $17/16$ | $6523/8192$ | $3791/4096$ | $6523/8192$
3 | $65/64$ | $6523/8192$ | $115/128$ | $6523/8192$
5 | $535/512$ | $6523/8192$ | $469/512$ | $6523/8192$
7 | $4159/4096$ | $6523/8192$ | $3679/4096$ | $6523/8192$
9 | $2175/2048$ | $6523/8192$ | $3791/4096$ | $6523/8192$
11 | $65/64$ | $6523/8192$ | $115/128$ | $6523/8192$
13 | $535/512$ | $6523/8192$ | $469/512$ | $6523/8192$
15 | $4159/4096$ | $6523/8192$ | $3679/4096$ | $6523/8192$

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