#  $S_{n}$-extensions with prescribed norms

Sebastian Monnet

###### Abstract.

Given a number field $k$, a finitely generated subgroup $\mathcal{A}\subseteq
k^{\times}$, and an integer $n\geq 3$, we study the distribution of
$S_{n}$-extensions of $k$ such that the elements of $\mathcal{A}$ are norms.
For $n\leq 5$, and conjecturally for $n\geq 6$, we show that the density of
such extensions is the product of so-called “local masses” at the places of
$k$. When $n$ is an odd prime, we give formulas for these local masses,
allowing us to express the aforementioned density as an explicit Euler
product. For $n=4$, we determine almost all of these masses exactly and give
an efficient algorithm for computing the rest, again yielding an explicit
Euler product.

###### Contents

  1. 1 Introduction
    1. 1.1 Qualitative results
    2. 1.2 Computing the masses $m_{\mathcal{A},\mathfrak{p}}$
    3. 1.3 Acknowledgements
  2. 2 Results for General $n$
    1. 2.1 Étale algebras, splitting symbols, and local conditions
    2. 2.2 Expressing our problem in terms of local conditions
  3. 3 Prime degree extensions
    1. 3.1 Finding $\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})$ when $p\neq\ell$
    2. 3.2 Finding $\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})$ when $p=\ell$
  4. 4 $S_{4}$-quartic extensions
    1. 4.1 $p$-adic fields for odd primes $p$
    2. 4.2 $2$-adic fields
    3. 4.3 Totally ramified $C_{4}$-extensions
  5. A Proof of Theorem 4.33
  6. B Explicit helper functions

##  1\. Introduction

Fix a number field $k$ and let $\mathcal{A}\subseteq k^{\times}$ be a finitely
generated subgroup. In [FLN22], Frei, Loughran, and Newton count abelian
extensions $K/k$ such that $\mathcal{A}\subseteq N_{K/k}K^{\times}$. The
natural follow-up is: what about non-abelian extensions? Counting non-abelian
number fields in general is the subject of Malle’s Conjecture ([Mal04,
Conjecture 1.1]), arguably the biggest open problem in arithmetic statistics,
and we do not address it here. Instead, we tackle one of the best-understood
classes of non-abelian number fields, $S_{n}$-$n$-ics.

An _$S_{n}$ -$n$-ic extension of $k$_ is a degree $n$ field extension $K/k$,
whose normal closure $\widetilde{K}$ has
$\operatorname{Gal}(\widetilde{K}/k)\cong S_{n}$. Given a real number $X$,
write $N_{k,n}(X)$ for the number of $S_{n}$-$n$-ics $K/k$ with
$N(\operatorname{disc}(K/k))\leq X$. The “Malle–Bhargava heuristics” give
conjectural asymptotics for $N_{k,n}(X)$, which have been proved for $n\leq 5$
by Bhargava and collaborators (see [Bha07] for the original heuristics and
[BSW15] for the proof when $n\leq 5$). The cases $n\geq 6$ are wide open and
very famous.

Again for a real number $X$, write $N_{k,n}(X;\mathcal{A})$ for the number of
$S_{n}$-$n$-ics $K/k$ such that $N(\operatorname{disc}(K/k))\leq X$ and
$\mathcal{A}\subseteq N_{K/k}K^{\times}$. We are primarily interested in the
leading coefficient,

$\lim_{X\to\infty}\frac{N_{k,n}(X;\mathcal{A})}{X},$

in the asymptotic expansion of $N_{k,n}(X;\mathcal{A})$, as well as the
proportion

$\lim_{X\to\infty}\frac{N_{k,n}(X;\mathcal{A})}{N_{k,n}(X)}$

of $S_{n}$-$n$-ics with $\mathcal{A}$ inside their norm groups.

Since $N_{k,n}(X)$ is only known for $n\leq 5$, barring major breakthroughs in
the field, we can only hope to solve our problem completely for $n=3,4,5$. It
is common for papers to focus only on one of these three cases, but we have
opted to address all three simultaneously. In fact, contingent on the
Malle–Bhargava heuristics, we extend our results to conjectures for $n\geq 6$.

###  1.1. Qualitative results

In Section 2, we explore our problem in as much generality as possible, namely
for all $n\geq 3$. As already discussed, our results are conjectural for
$n\geq 6$, since they assume the Malle–Bhargava heuristics. Specialising to
the known cases, we isolate the following headline result:

######  Theorem 1.1.

For $n=3,4,5$, we have

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

with equality if and only if $\mathcal{A}\subseteq k^{\times n}$.

######  Corollary 1.2.

Let $n\in\\{3,4,5\\}$. For every finitely generated subgroup
$\mathcal{A}\subseteq k^{\times}$, there are infinitely many $S_{n}$-$n$-ic
extensions $K/k$ with $\mathcal{A}\subseteq N_{K/k}K^{\times}$.

For arbitrary $n\geq 3$, the more general (conjectural) analogue of Theorem
1.1 is contained in Theorems 2.18 and 2.20.

Let us unpack the content of Theorem 1.1. It tells us that, for $n=3,4,5$, any
finitely generated subgroup $\mathcal{A}\subseteq k^{\times}$ is contained in
the norm group of a positive proportion of $S_{n}$-$n$-ic extensions.
Moreover, it tells us that $\mathcal{A}$ can only be contained in 100% of such
norm groups if it is contained in _every_ such norm group.

######  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. For general subgroups
$\mathcal{A}$, ours is the first proof we are aware of.

In order to prove Theorem 1.1, we expand the limit
$\lim_{X\to\infty}\frac{N_{n,k}(X;\mathcal{A})}{N_{n,k}(X)}$ as an Euler
product over all primes of the number field $k$. Throughout this paper, we use
the term _$p$ -adic field_ to mean a finite degree extension of
$\mathbb{Q}_{p}$, for some rational prime $p$. Given a $p$-adic field $F$, we
write $\mathbb{F}_{F}$ for its residue field. Let $F$ be a field that is
either $p$-adic or equal to $\mathbb{R}$ or $\mathbb{C}$. For such $F$, write
$\textnormal{\'{E}t}_{n/F}$ for the set of degree $n$ étale algebras over $F$.
For a subset $S\subseteq\textnormal{\'{E}t}_{n/F}$, the _mass_ $m(S)$ of $S$
is defined as follows:

  * •

If $F$ is $p$-adic then, writing $q=\\#\mathbb{F}_{F}$, we define

$m(S)=\frac{q-1}{q}\cdot\sum_{L\in
S}\frac{N(\operatorname{disc}(L/F))^{-1}}{\\#\operatorname{Aut}(L/F)}.$

  * •

If $F$ is $\mathbb{R}$ or $\mathbb{C}$, then we define

$m(S)=\sum_{L\in S}\frac{1}{\\#\operatorname{Aut}(L/F)}.$

Given a subgroup $\mathcal{A}\subseteq F^{\times}$, write
$\textnormal{\'{E}t}_{n/F}^{\mathcal{A}}$ for the set of
$L\in\textnormal{\'{E}t}_{n/F}$ with $\mathcal{A}\subseteq N_{L/F}L^{\times}$.
For each prime $\mathfrak{p}$ of $k$, we may view $\mathcal{A}$ as a subgroup
of $k_{\mathfrak{p}}^{\times}$, which allows us to write
$m_{\mathcal{A},\mathfrak{p}}$ for the mass

$m_{\mathcal{A},\mathfrak{p}}=m(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}}).$

######  Theorem 1.4.

For $n=3,4,5$, we have

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

and the same result holds for $n\geq 6$, subject to the Malle–Bhargava
heuristics.

###  1.2. Computing the masses $m_{\mathcal{A},\mathfrak{p}}$

Theorem 1.4 tells us that we can state the density
$\lim_{X\to\infty}\frac{N_{n,k}(X;\mathcal{A})}{N_{n,k}(X)}$ as an explicit
Euler product, if we can evaluate the masses $m_{\mathcal{A},\mathfrak{p}}$
for all primes $\mathfrak{p}$. As such, we dedicate much of this paper to a
careful study of these masses. The masses for infinite primes are the easiest
to state:

######  Theorem 1.5.

Let $\mathfrak{p}$ be an infinite prime of $k$ and let $f:k\to\mathbb{C}$ be
the corresponding embedding.

  * •

If $f$ is a real embedding, then

$m(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})=\begin{cases}\sum_{s=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{s!(n-2s)!2^{s}}\quad\text{if
$f(\alpha)>0$ for all $\alpha\in\mathcal{A}$},\\\
\sum_{s=0}^{\lceil\frac{n}{2}\rceil-1}\frac{1}{s!(n-2s)!2^{s}}\quad\text{otherwise}.\end{cases}$

  * •

If $f$ is a complex embedding, then

$m(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})=\frac{1}{n!}.$

Recall that we are most interested in the cases $n=3,4,5$, since these are the
$n$ for which the Malle–Bhargava heuristics have been proved. We do more than
just these cases, studying $m_{\mathcal{A},\mathfrak{p}}$ when $n=4$ and when
$n$ is an odd prime.

For $p$-adic fields $F$, the definition of mass has a normalising factor
$\frac{q-1}{q}$. To avoid this cumbersome notation, we define the _pre-mass_
of a set $S\subseteq\textnormal{\'{E}t}_{n/F}$ by

$\widetilde{m}(S)=\sum_{L\in
S}\frac{N(\operatorname{disc}(L/F))^{-1}}{\\#\operatorname{Aut}(L/F)}.$

Since it is trivial to compute the mass from the pre-mass, all results in the
current subsection will be about pre-mass.

Suppose that $F$ is $p$-adic. Let $L/F$ be an étale algebra, and write
$L=L_{1}\times\ldots\times L_{r}$, where each $L_{i}$ is a field extension of
$F$ with ramification index $e_{i}$ and inertia degree $f_{i}$. The _splitting
symbol_ $(L,F)$ of $L$ over $F$ is the symbol $(f_{1}^{e_{1}}\ldots
f_{r}^{e_{r}})$, which is defined up to permutation of indices (see Definition
2.3 and the subsequent discussion for a more rigorous definition). The
_degree_ of the splitting symbol $(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$ is
defined to be the quantity $\sum_{i}e_{i}f_{i}$. Given a degree $n$ splitting
symbol $\sigma$, write $\textnormal{\'{E}t}_{\sigma/F}^{\mathcal{A}}$ for the
set of $L\in\textnormal{\'{E}t}_{n/F}^{\mathcal{A}}$ with $(L,F)=\sigma$. When
$\sigma$ is a splitting symbol of the form $(f^{e})$, and for a finite group
$G$, write $\textnormal{\'{E}t}_{\sigma/F}^{G/F,\mathcal{A}}$ for the set of
field extensions $L\in\textnormal{\'{E}t}_{\sigma/F}^{\mathcal{A}}$ such that
the Galois closure $\widetilde{L}$ of $L$ over $F$ has
$\operatorname{Gal}(\widetilde{L}/F)\cong G$.

Our next result, Theorem 1.6, gives
$\widetilde{m}(\textnormal{\'{E}t}_{n/F}^{\mathcal{A}})$ for all $p$-adic
fields $F$, whenever $n$ is a rational prime. Consequently, for such $n$, it
allows us to write down all the masses $m_{\mathcal{A},\mathfrak{p}}$. In
particular, this covers the cases $n=3$ and $n=5$, two of the three we are
especially interested in. For a $p$-adic field $F$, a finitely generated
subgroup $\mathcal{A}\subseteq F^{\times}$, and a positive integer $n$, define

$\overline{\mathcal{A}}^{n}=\mathcal{A}F^{\times n}/F^{\times n}.$

For each nonnegative integer $t$, write

$\overline{\mathcal{A}}^{n}_{t}=\overline{\mathcal{A}}^{n}\cap\Big{(}U_{F}^{(t)}F^{\times
n}/F^{\times n}\Big{)},$

where $U_{F}^{(t)}$ is the $t^{\mathrm{th}}$ term of the unit filtration,
given by $U_{F}^{(t)}=1+\mathfrak{p}_{F}^{t}$.

######  Theorem 1.6.

Let $p$ be a rational prime, let $F$ be a $p$-adic field with residue field of
size $q$, and let $\ell$ be a rational prime. Let $\mathcal{A}\subseteq
F^{\times}$ be a finitely generated subgroup. Choose a minimal set of
generators $A_{0}\cup A_{1}$ for $\overline{\mathcal{A}}^{\ell}$ such that
$v_{F}(\alpha)=r$ for all $[\alpha]\in A_{r}$, for $r=0,1$.

  1. (1)

We have

$\widetilde{m}(\textnormal{\'{E}t}_{\ell/F}^{\mathcal{A}})=\widetilde{m}(\textnormal{\'{E}t}_{(\ell)/F}^{\mathcal{A}})+\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})+(1-\ell^{-1})+\sum_{d=1}^{\ell-2}\frac{\operatorname{Part}(d,\ell-d)}{q^{d}},$

where $\operatorname{Part}(d,m)$ is the partition function defined in Section
2.

  2. (2)

For all $p$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{(\ell)/F}^{\mathcal{A}})=\begin{cases}\frac{1}{\ell}\quad\text{if
$A_{1}=\varnothing$},\\\ 0\quad\text{otherwise}.\end{cases}$

  3. (3)

Suppose that $p\neq\ell$. Then we have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{\ell-1}}\quad\text{if
$q\not\equiv 1\pmod{\ell}$ or $\mathcal{A}\subseteq F^{\times\ell}$},\\\
\frac{1}{\ell q^{\ell-1}}\quad\text{else if
$A_{0}\subseteq\mathcal{O}_{F}^{\times\ell}$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times\ell}$ for all
$\alpha_{i},\alpha_{j}\in A_{1}$},\\\ 0\quad\text{otherwise}.\end{cases}$

  4. (4)

Suppose that $p=\ell$. We have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{\mathcal{A}})=\frac{1}{q^{p-1}}-\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})+\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}}).$

The two missing ingredients,
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})$ and
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})$, are as
follows. We have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})=\frac{q-1}{p-1}q^{-2}\Big{(}\mathbbm{1}_{e_{F}\geq
p-1}\cdot
A\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}+\mathbbm{1}_{(p-1)\nmid
e_{F}}\cdot
B\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}\Big{)}+\mathbbm{1}_{\mu_{p}\subseteq
F}\cdot q^{-(p-1)(e_{F}+1)},$

where $A$ and $B$ are the explicit functions defined in Appendix B, and

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})=\mathbbm{1}_{\mu_{p}\subseteq
F}\cdot\mathbbm{1}_{\mathcal{A}_{\frac{pe_{F}}{p-1}}=0}\cdot\frac{pq^{-(p-1)(e_{F}+1)}}{\\#\overline{\mathcal{A}}^{p}}+\frac{1}{(p-1)\\#\overline{\mathcal{A}}^{p}}\cdot
q^{-2}\cdot\sum_{c=1}^{\lceil\frac{pe_{F}}{p-1}\rceil}\frac{q\cdot\\#\overline{\mathcal{A}}^{p}_{c}-\\#\overline{\mathcal{A}}^{p}_{c-1}}{q^{(p-2)c+\lfloor\frac{c-2}{p}\rfloor}}.$

Since the sizes $\\#\overline{\mathcal{A}}^{p}_{c}$ can all be different, the
series for
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})$ does
not have a closed form. In the special case where $\mathcal{A}$ is generated
by one element, we can write down such a closed form, which is stated in
Theorem 3.33. Along the way, we find the exact number of extensions
$L\in\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}$ with each possible
discriminant, which we state in Theorem 3.23.

We now turn our attention to the final case of particular interest: $n=4$.
When $\mathfrak{p}$ is a prime of $k$ not lying over $2$, the masses
$\widetilde{m}(\textnormal{\'{E}t}_{4/k_{\mathfrak{p}}}^{\mathcal{A}})$ take a
relatively simple, explicit form, stated in Theorem 4.1.

So our problem is completely solved for $\mathfrak{p}\nmid 2$. Since there are
only finitely many primes $\mathfrak{p}\mid 2$, it suffices to have a
practicable method for computing
$\widetilde{m}(\textnormal{\'{E}t}_{4/k_{\mathfrak{p}}}^{\mathcal{A}})$ for
each such $\mathfrak{p}$. In principle, this can be done by brute force, since
there are finitely many quartic étale algebras of $k_{\mathfrak{p}}$, and we
can find all of them e.g. by adapting the methods of [PR01]. Unfortunately,
[Kra66, Theorem 2] tells us that $\\#\textnormal{\'{E}t}_{4/k_{\mathfrak{p}}}$
is on the order of $2^{3[k_{\mathfrak{p}}:\mathbb{Q}_{2}]}$, which becomes
very large very quickly, so the brute force approach is unfeasible in many
cases.

For $\mathfrak{p}\mid 2$, Theorem 4.1 tells us the value of
$\widetilde{m}(\textnormal{\'{E}t}_{\sigma/k_{\mathfrak{p}}}^{\mathcal{A}})$
for all $\sigma\not\in\\{(1^{2}1^{2}),(2^{2}),(1^{4})\\}$, so we only need to
address these three cases. For $\sigma=(1^{2}1^{2})$ and $\sigma=(2^{2})$, we
state the required masses in Theorems 4.7 and 4.8, respectively.

All that remains is to compute
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}}}^{\mathcal{A}})$
for primes $\mathfrak{p}\mid 2$. This final case is significantly harder,
because there is so much wild ramification at play. We now give a brief
overview of the results, which are stated in full technicality in Theorem 4.9.
We partition by Galois closure group $G$, considering the different
$\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}}}^{G/k_{\mathfrak{p}},\mathcal{A}}$
separately. For $G\in\\{A_{4},S_{4}\\}$, we can state the mass explicitly. For
other $G$, we will instead consider the quantities

$\\#\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}},m}^{G/k_{\mathfrak{p}},\mathcal{A}}=\\#\\{L\in\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}}}^{G/k_{\mathfrak{p}},\mathcal{A}}:v_{F}(d_{L/F})=m\\}$

for all $m$, where $d_{L/F}$ is the discriminant ideal of the extension $L/F$.
From these quantities, it is trivial to compute the mass by summing over $m$.
We give explicit expressions for
$\\#\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}},m}^{G/k_{\mathfrak{p}},\mathcal{A}}$
in the cases $G\in\\{D_{4},V_{4}\\}$. Finally, we give an explicit expression
for $\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}}$ whenever
$E\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}$ and $m_{2}$ is an integer.
This allows us to formulate an efficient algorithm for computing
$\\#\textnormal{\'{E}t}_{(1^{4})/k_{\mathfrak{p}},m}^{C_{4}/k_{\mathfrak{p}},\mathcal{A}}$
for each $m$.

Given a quadratic extension $E/F$ and $\alpha\in F^{\times}$, write
$t_{\mathrm{norm}}(F)$ for the time it takes to determine whether $\alpha\in
N_{E/F}E^{\times}$ and, if so, find an element $\widetilde{\alpha}\in
E^{\times}$ such that $N_{E/F}\widetilde{\alpha}\in\alpha F^{\times 2}$. This
can be done e.g. using Magma’s ‘NormEquation’ function [BCP97].

######  Theorem 1.7.

Let $F$ be a $2$-adic field with absolute ramification index $e_{F}$. Given
$E\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}$, there is an $O(e_{F}\cdot
t_{\mathrm{norm}}(F))$ algorithm for computing each quantity
$\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}},$ so we can
compute

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{C_{4}/F,\mathcal{A}})$

with time complexity $O(e_{F}\cdot t_{\mathrm{norm}}(F)\cdot
2^{[F:\mathbb{Q}_{2}]})$.

Since there are on the order of $2^{3[k_{\mathfrak{p}}:\mathbb{Q}_{2}]}$
quartic field extensions of $k_{\mathfrak{p}}$ (see [Kra66, Theorem 2]), a
brute force approach would have a time complexity of at least
$O(2^{3[k_{\mathfrak{p}}:\mathbb{Q}_{2}]})$, even if all quartic extensions
were already known. As such, the time complexity of our algorithm is a
significant improvement over the naïve baseline.

###  1.3. Acknowledgements

I would like to thank my supervisor, Rachel Newton, for suggesting the initial
research direction and for providing a great deal of support and encouragement
throughout. I am also grateful to Jiuya Wang, Ross Paterson, Ashvin
Swaminathan, and Brandon Alberts for helpful conversations.

This work was supported by the Engineering and Physical Sciences Research
Council [EP/S021590/1], via the EPSRC Centre for Doctoral Training in Geometry
and Number Theory (The London School of Geometry and Number Theory),
University College London.

##  2\. Results for General $n$

The main goal of this section is to prove Theorem 1.1, along with its
conjectural generalisations in Theorems 2.18 and 2.20.

###  2.1. Étale algebras, splitting symbols, and local conditions

For a local field $F$, write $\textnormal{\'{E}t}_{n/F}$ for the set of
isomorphism classes of degree $n$ étale algebras over $F$.

######  Definition 2.1.

Let $F$ be a $p$-adic field and let $S\subseteq\textnormal{\'{E}t}_{n/F}$. The
_pre-mass_ of $S$ is the quantity

$\widetilde{m}_{F}(S)=\sum_{L\in
S}\frac{\operatorname{Nm}(d_{L/F})^{-1}}{\\#\operatorname{Aut}(L/F)},$

and the _mass_ of $S$ is the quantity

$m_{F}(S)=\frac{q_{F}-1}{q_{F}}\cdot\widetilde{m}_{F}(S).$

######  Definition 2.2.

Let $F$ be either $\mathbb{R}$ or $\mathbb{C}$, and let
$S\subseteq\textnormal{\'{E}t}_{n/F}$. The _mass_ of $S$ is

$m_{F}(S)=\sum_{L\in S}\frac{1}{\\#\operatorname{Aut}(L/F)}.$

###### Proof of Theorem 1.5.

Suppose that $f$ is a real embedding. Then

$\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}=\\{\mathbb{R}^{r}\times\mathbb{C}^{s}:r+2s=n\text{
and $r,s\geq 0$}\\}.$

If $L=\mathbb{R}^{r}\times\mathbb{C}^{s}$, we have

$N_{L/k_{\mathfrak{p}}}L^{\times}=\begin{cases}\mathbb{R}^{\times}\quad\text{if
$r>0$},\\\ \mathbb{R}_{>0}\quad\text{if $r=0$},\end{cases}$

and

$\\#\operatorname{Aut}(L/k_{\mathfrak{p}})=2^{s}\cdot r!s!.$

The result for real $f$ follows. If $f$ is a complex embedding, then the only
element of $\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}$ is $\mathbb{C}^{n}$,
which has $n!$ automorphisms and norm group $\mathbb{C}^{\times}$, so we are
done. ∎

######  Definition 2.3.

Let $n$ be a positive integer. A _degree $n$ splitting symbol_ is a symbol
$(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$, where the $f_{i}$ and $e_{i}$ are
positive integers. We identify splitting symbols that are permutations of each
other, i.e. $(1^{2}2^{3})=(2^{3}1^{2})$. The superscripts are purely symbolic,
and do not represent exponentiation, so that for example $(1^{2})$ and
$(1^{3})$ are different splitting symbols. Finally, we suppress exponents with
value $1$, writing e.g. $(2)$ instead of $(2^{1})$. Write $\mathrm{Split}_{n}$
for the set of degree $n$ splitting symbols.

Let $F$ be a local field and let $L\in\textnormal{\'{E}t}_{n/F}$. Then we have

$L=L_{1}\times\ldots\times L_{r},$

for field extensions $L_{i}/F$ with inertia degree $f_{i}$ and ramification
index $e_{i}$. The _splitting symbol $(L,F)$ of $L$ over $F$_ is defined to be
the symbol

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

For a splitting symbol $\sigma\in\mathrm{Split}_{n}$, write
$\textnormal{\'{E}t}_{\sigma/F}$ for the set of
$L\in\textnormal{\'{E}t}_{n/F}$ with $(L,F)=\sigma$.

######  Definition 2.4.

We introduce our key terminology for local conditions.

  1. (1)

Let $k$ be a number field and let $\mathfrak{p}$ be a prime of $k$. A _degree
$n$ local condition at $\mathfrak{p}$_ is a set
$\Sigma_{\mathfrak{p}}\subseteq\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}$.

  2. (2)

A _degree $n$ collection of local conditions on $k$_ is a collection
$(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ of local conditions at each prime
$\mathfrak{p}$ of $k$.

  3. (3)

Let $\Sigma$ be a degree $n$ collection of local conditions on $k$. We call
$\Sigma$ _acceptable_ if, for all but finitely many $\mathfrak{p}$, the set
$\Sigma_{\mathfrak{p}}$ contains all étale algebras $L/k_{\mathfrak{p}}$ with
$v_{k_{\mathfrak{p}}}(d_{L/k_{\mathfrak{p}}})\leq 1$.

  4. (4)

Let $K/k$ be a finite field extension and let $\Sigma_{\mathfrak{p}}$ be a
local condition at the prime $\mathfrak{p}$ of $k$. The extension $K/k$
_satisfies_ the condition $\Sigma_{\mathfrak{p}}$ if the completion
$K_{\mathfrak{p}}:=K\otimes_{k}k_{\mathfrak{p}}$ is in
$\Sigma_{\mathfrak{p}}$.

  5. (5)

The extension $K/k$ _satisfies_ a collection
$\Sigma=(\Sigma_{\mathfrak{p}})_{\mathfrak{p}}$ of local conditions if it
satisfies $\Sigma_{\mathfrak{p}}$ for all $\mathfrak{p}$.

Write $\mathbf{Ext}_{k,n}$ for the set of isomorphism classes of degree $n$
extensions $K/k$ with Galois closure group $S_{n}$. For each real number $X$,
define

$\mathbf{Ext}_{k,n,\leq
X}=\\{K\in\mathbf{Ext}_{k,n}:\operatorname{Nm}(d_{K/k})\leq X\\}.$

For a system $\Sigma$ of local conditions, define

$N_{k,n}(X;\Sigma)=\\#\\{K\in\mathbf{Ext}_{k,n,\leq X}:\text{$K$ satisfies
$\Sigma$}\\}.$

######  Conjecture 2.5 (Malle–Bhargava heuristics).

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

$\lim_{X\to\infty}\frac{N_{k,n}(X;\Sigma)}{X}=\frac{1}{2}\operatorname{Res}_{s=1}\zeta_{k}(s)\prod_{\mathfrak{p}}m_{k_{\mathfrak{p}}}(\Sigma_{\mathfrak{p}}).$

######  Theorem 2.6.

Conjecture 2.5 is true for $n=2,3,4,5$.

###### Proof.

This is [BSW15, Theorem 2]. ∎

###  2.2. Expressing our problem in terms of local conditions

Let $F$ be a local field and let $\mathcal{A}\subseteq F^{\times}$ be a
finitely generated subgroup. Write $\textnormal{\'{E}t}_{n/F}^{\mathcal{A}}$
for the set of $L\in\textnormal{\'{E}t}_{n/F}$ with $\mathcal{A}\subseteq
N_{L/F}L^{\times}$. Define $\textnormal{\'{E}t}_{\sigma/F}^{\mathcal{A}}$
analogously.

Recall that in our original setting, $k$ is a number field and
$\mathcal{A}\subseteq k^{\times}$ is a finitely generated subgroup. For each
prime $\mathfrak{p}$ of $k$, we view $\mathcal{A}$ naturally as a subgroup of
$k_{\mathfrak{p}}^{\times}$, and therefore it makes sense to talk about the
set $\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}}$ of étale algebras
over $k_{\mathfrak{p}}$. In particular, we have a system of local conditions
$(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})_{\mathfrak{p}}$.

######  Theorem 2.7 (Hasse Norm Principle for $S_{n}$ $n$-ics).

Let $K/k$ be an $S_{n}$ $n$-ic extension. Then $\mathcal{A}\subseteq
N_{K/k}K^{\times}$ if and only if $K/k$ satisfies the system
$(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})_{\mathfrak{p}}$ of
local conditions.

###### Proof.

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

######  Lemma 2.8.

Let $E/F$ be a tamely ramified extension of $p$-adic fields, with ramification
index $e$ and inertia degree $f$. Then

$v_{F}(d_{E/F})=f(e-1).$

###### Proof.

This follows easily from [NS13, Theorem 2.6] and [NS13, Theorem 2.9]. ∎

######  Lemma 2.9.

If $n\geq 3$, then the system
$(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})_{\mathfrak{p}}$ of
local conditions is acceptable.

###### Proof.

Let $\mathfrak{p}$ be such that $\gcd(N(\mathfrak{p}),n)=1$ and $n\mid
v_{\mathfrak{p}}(\alpha)$ for all $\alpha\in\mathcal{A}$. Let
$L\in\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}$ such that
$v_{k_{\mathfrak{p}}}(d_{L/k_{\mathfrak{p}}})\leq 1$. Writing
$L=L_{1}\times\ldots\times L_{r}$ for field extensions
$L_{i}/k_{\mathfrak{p}}$, at most one of the extensions
$L_{i}/k_{\mathfrak{p}}$ is ramified, so either $L=L_{1}$ or
$L_{i}/k_{\mathfrak{p}}$ is unramified for some $i$. Suppose that $L=L_{1}$.
Let $(L,k_{\mathfrak{p}})=(f^{e})$. Since $N(\mathfrak{p})$ is coprime to $n$,
it follows that $L/k_{\mathfrak{p}}$ is tamely ramified, so Lemma 2.8 tells us
that $v_{K}(d_{L/k_{\mathfrak{p}}})=f(e-1)$, and therefore $f=1$ and $e\leq
2$, so $n\leq 2$, contradicting the assumption that $n\geq 3$. Therefore,
$L_{i}/k_{\mathfrak{p}}$ is unramified for some $i$, and therefore
$\mathcal{O}_{k_{\mathfrak{p}}}^{\times}\subseteq
N_{L/k_{\mathfrak{p}}}L^{\times}$, and it follows that
$L\in\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}}$. ∎

###### Proof of Theorem 1.4.

This is immediate from Conjecture 2.5, Theorems 2.6 and 2.7, and Lemma 2.9. ∎

Call a splitting symbol $(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$ _predictable_ if
the $e_{i}$ are mutually coprime. Write
$\textnormal{\'{E}t}_{n/F}^{\mathrm{pred}}$ for the set of
$L\in\textnormal{\'{E}t}_{n/F}$ such that $(L,F)$ is predictable. Call a
symbol _unpredictable_ if it is not predictable. Write
$\textnormal{\'{E}t}_{n/F}^{\mathrm{unpred}}$ for the set of
$L\in\textnormal{\'{E}t}_{n/F}$ such that $(L,F)$ is unpredictable. Moreover,
we call a symbol $(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$ _epimorphic_ if it is
predictable and the $f_{i}$ are mutually coprime. Write
$\textnormal{\'{E}t}_{n/F}^{\mathrm{epi}}$ for the set of
$L\in\textnormal{\'{E}t}_{n/F}$ with $(L,F)$ epimorphic. Write
$\mathrm{Split}_{n}^{\mathrm{pred}}$ and $\mathrm{Split}_{n}^{\mathrm{epi}}$
for the sets of predictable and epimorphic splitting symbols, respectively.

######  Lemma 2.10.

Let $F$ be a $p$-adic field and let $\sigma$ be a predictable splitting
symbol. Write $\sigma=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$, and let
$L\in\textnormal{\'{E}t}_{\sigma/F}$. Let $g=\gcd(f_{1},\ldots,f_{r})$. Then
we have

$N_{L/F}L^{\times}=\\{x\in F:g\mid v_{K}(x)\\}.$

In particular, if $L\in\textnormal{\'{E}t}_{n/F}^{\mathrm{epi}}$, then

$N_{L/F}L^{\times}=F^{\times}.$

###### Proof.

Write $L=L_{1}\times\ldots\times L_{r}$, where $L_{i}/F$ is a field extension
with ramification index $e_{i}$ and inertia degree $f_{i}$. For each $i$, let
$L_{i}^{\mathrm{ur}}$ be the maximal unramified subextension of $L_{i}/F$. By
considering the towers $L_{i}/L_{i}^{\mathrm{ur}}/F$, it is easy to see that
we have

$\mathcal{O}_{F}^{\times e_{i}}\subseteq N_{L_{i}/F}L_{i}^{\times}$

for each $i$. Moreover, for each $i$, there is some $x_{i}\in
N_{L_{i}/F}L_{i}^{\times}$ with $v_{F}(x_{i})=f_{i}$. Since the $e_{i}$ are
mutually coprime, there exist integers $r_{i}$ such that
$\sum_{i}r_{i}e_{i}=1$, so for each $\alpha\in\mathcal{O}_{F}^{\times}$, we
have

$\alpha=\prod_{i}(\alpha^{e_{i}})^{r_{i}}\in\prod_{i}N_{L_{i}/F}L_{i}^{\times}=N_{L/F}L^{\times}.$

Therefore, $\mathcal{O}_{F}^{\times}\subseteq N_{L/F}L^{\times}$. On the other
hand, there are integers $s_{i}$ such that $\sum_{i}s_{i}f_{i}=g$. Setting

$x=\prod_{i}x_{i}^{s_{i}}\in N_{L/F}L^{\times},$

we obtain $v_{K}(x)=g$, so

$\\{x\in F:g\mid v_{K}(x)\\}\subseteq N_{L/F}L^{\times}.$

Conversely, we have

$N_{L_{i}/F}L_{i}^{\times}\subseteq
N_{L_{i}^{\mathrm{ur}}/F}L_{i}^{\mathrm{ur},\times}=\\{x\in F:f_{i}\mid
v_{K}(x)\\}\subseteq\\{x\in F:g\mid v_{K}(x)\\}$

for each $i$, so we are done. ∎

######  Corollary 2.11.

Let $F$ be a $p$-adic field and let
$\mathcal{A}=\langle\alpha_{1},\ldots,\alpha_{d}\rangle$ be a finitely
generated subgroup. Suppose that $n\mid v_{F}(\alpha_{i})$ for each $i$. Then

$\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathrm{pred}}\subseteq\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}}.$

###### Proof.

This is immediate from Lemma 2.10. ∎

######  Lemma 2.12.

Let $n$ be any integer with $n\geq 3$. Let $F$ be a $p$-adic field with $p>n$,
and let
$L\in\textnormal{\'{E}t}_{n/F}\setminus\textnormal{\'{E}t}_{n/F}^{\mathrm{pred}}$.
Then $v_{F}(d_{L/F})\geq 2$.

###### Proof.

Let $(L,F)=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$. Since $p>n$, we have $p\nmid
e_{i}$ for each $i$, so by Lemma 2.8 we have

$v_{F}(d_{L/F})=\sum_{i}f_{i}(e_{i}-1).$

Since $(L,F)$ is not predictable, the $e_{i}$ are not mutually coprime, so
there is some $d\geq 2$ with $d\mid e_{i}$ for all $i$, so $e_{i}\geq 2$ for
each $i$. If any of the $e_{i}$ is at least $3$, then we are done. Suppose
that $e_{i}=2$ for every $i$. Then

$v_{F}(d_{L/F})=\sum_{i}f_{i}=\frac{n}{2}>1,$

so we are done. ∎

Let $d$ be a nonnegative integer and let $m$ be a positive integer. The
symmetric group $S_{m}$ acts by permutation of coordinates on the set
$\mathbb{Z}_{\geq 0}^{m}$ of nonnegative integers. We define a _partition of
$d$ into $m$ parts_ to be an equivalence class

$[(a_{i})]\in\mathbb{Z}_{\geq 0}^{m}/S_{m}$

such that $\sum_{i}a_{i}=d$. This is often referred to as a partition into “at
most” $m$ parts, but for our purposes it is more convenient to allow parts to
be zero. Write $\operatorname{Part}(d,m)$ for the number of partitions of $d$
into $m$ parts. Given a partition

$\pi=(a_{1},\ldots,a_{m})$

of $d$ into $m$ parts, write $\sum\pi$ for the sum $\sum_{i}a_{i}$.

Let $\sigma=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$ be a degree $n$ splitting
symbol with

$\sum_{i}f_{i}(e_{i}-1)=d.$

Then we obtain a partition

$\pi(\sigma):=(e_{1}-1,\ldots e_{1}-1,e_{2}-1,\ldots e_{2}-1,\ldots
e_{r}-1,\ldots,e_{r}-1),$

of $d$ into $n-d$ parts, where each term $e_{i}-1$ appears $f_{i}$ times. For
a splitting symbol $\sigma=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$, define

$d_{\sigma}:=\sum_{i}f_{i}(e_{i}-1)$

and

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

Note that $d_{\sigma}$ can be read off from $\pi(\sigma)$.

######  Lemma 2.13.

Let $F$ be a $p$-adic field for some rational prime $p$. Let $0\leq d\leq
n-1$, and let $\pi_{0}$ be a partition of $n$ into $n-d$ parts. We have

$\widetilde{m}(\\{L\in\textnormal{\'{E}t}_{n/F}:\pi(L,F)=\pi_{0}\\})=\frac{1}{q^{d}}.$

###### Proof.

In the case $F=\mathbb{Q}_{p}$, this is [Bha07, Proposition 2.2]. The appendix
of [Bha07] explains how to extend the proof to arbitrary $F$. ∎

######  Corollary 2.14.

Let $F$ be a $p$-adic field for some rational prime $p$, and let $0\leq d\leq
n-1$ be an integer. Then

$\widetilde{m}(\\{L\in\textnormal{\'{E}t}_{n/F}:d_{\sigma}=d\\})=\frac{\operatorname{Part}(d,n-d)}{q^{d}}.$

###### Proof.

This follows easily from Lemma 2.13. ∎

We also record the following lemma, which will be useful later:

######  Lemma 2.15.

Let $F$ be a $p$-adic field and let $\sigma\in\mathrm{Split}_{n}$. We have

$\widetilde{m}_{F}(\textnormal{\'{E}t}_{\sigma/F})=\frac{1}{d_{\sigma}}\cdot\frac{1}{\\#\operatorname{Aut}(\sigma)}.$

###### Proof.

This is [Bha07, Proposition 2.1]. ∎

######  Lemma 2.16.

For $n\geq 3$, we have

$\prod_{\mathfrak{p}}\frac{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})}{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})}>0.$

###### Proof.

Since

$\frac{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})}{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})}>0$

for each $\mathfrak{p}$, we may ignore any finite set of primes
$\mathfrak{p}$. Therefore, we may assume that $\mathfrak{p}$ is finite, we
have $v_{\mathfrak{p}}(\alpha_{i})=0$ for each $i$, and
$\operatorname{char}(\mathcal{O}_{k}/\mathfrak{p})>n$. Then all degree $n$
étale algebras $L/k_{\mathfrak{p}}$ are tamely ramified, so Lemma 2.8 tells us
that

$v_{k_{\mathfrak{p}}}(d_{L/k_{\mathfrak{p}}})=d_{(L,k_{\mathfrak{p}})},$

where $(L,k_{\mathfrak{p}})$ is the splitting symbol of $L/k_{\mathfrak{p}}$.
For $0\leq n\leq n-1$, let $a_{d}=\operatorname{Part}(d,n-d).$ Then Lemma 2.12
and Corollary 2.14 tell us that

$\widetilde{m}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathrm{pred}})\geq
a_{0}+a_{1}q^{-1}$

and

$\widetilde{m}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})=a_{0}+a_{1}q^{-1}+\ldots+a_{n-1}q^{-(n-1)}.$

It is easy to write down a positive real number $a$, independent of
$\mathfrak{p}$, such that

$1-aq^{-2}\leq\frac{\widetilde{m}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathrm{pred}})}{\widetilde{m}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})}\leq
1,$

and the result follows. ∎

######  Lemma 2.17.

Let $F$ be a $p$-adic field and let $n$ be any positive integer. Then

$\mathcal{O}_{F}^{\times}\cap\bigcap_{L\in\textnormal{\'{E}t}_{(1^{n})/F}}N_{L/F}L^{\times}=\mathcal{O}_{F}^{\times
n}.$

###### Proof.

By the structure theorem for finitely generated abelian groups, we may write

$\mathcal{O}_{F}^{\times}/\mathcal{O}_{F}^{\times
n}=\bigoplus_{i=1}^{r}(\mathbb{Z}/d_{i}\mathbb{Z})e_{i}$

for integers $d_{i}\geq 2$ with $d_{i}\mid n$. For each $i$, let

$A_{i}=\bigoplus_{j\neq i}(\mathbb{Z}/d_{j}\mathbb{Z})e_{j},$

so that $[\mathcal{O}_{F}^{\times}/\mathcal{O}_{F}^{\times
n}:A_{i}]=\frac{n}{d_{i}}$. Let $\pi_{F}$ be a uniformiser of $F$, and let
$E_{i}/F$ be the degree $d_{i}$ abelian extension with

$\Big{(}N_{E_{i}/F}E_{i}^{\times}\Big{)}/F^{\times n}=\langle
A_{i},\pi_{F}\rangle.$

Since $d_{i}\mid n$, there is a totally ramified extension $L_{i}/E_{i}$ of
degree $d_{i}$. We have

$\Big{(}N_{L_{i}/F}L_{i}^{\times}\Big{)}/F^{\times
n}\subseteq\Big{(}N_{E_{i}/F}E_{i}^{\times}\Big{)}/F^{\times n},$

so

$(N_{L_{i}/F}L_{i}^{\times}\cap\mathcal{O}_{F}^{\times})/\mathcal{O}_{F}^{\times
n}\subseteq A_{i}.$

The result follows since $\bigcap_{i}A_{i}=0$. ∎

######  Theorem 2.18.

Let $n\geq 3$. Assuming Conjecture 2.5 is true for $n$ (as is the case for
$n\leq 5$), we have

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

with equality if and only if

$\mathcal{A}\subseteq\ker\Big{(}k^{\times}\to\prod_{\mathfrak{p}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)}.$

###### Proof.

Theorem 1.4 and Lemma 2.16 tell us that

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

for all $\mathcal{A}$, so we need only determine when there is equality. If

$\mathcal{A}\subseteq\ker\Big{(}k^{\times}\to\prod_{\mathfrak{p}\in\Pi_{k}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)},$

then
$\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})=\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})$
for all $\mathfrak{p}$, so we have equality. Suppose conversely that we have
equality. Then

$\prod_{\mathfrak{p}}\frac{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})}{\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})}=1.$

Let $\mathfrak{p}$ be a finite prime. Then
$\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}})=\widetilde{m}_{\mathfrak{p}}(\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}})$,
and hence
$\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}^{\mathcal{A}}=\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}},$
so

$\mathcal{A}k_{\mathfrak{p}}^{\times
n}\subseteq\bigcap_{L\in\textnormal{\'{E}t}_{n/k_{\mathfrak{p}}}}N_{L/k_{\mathfrak{p}}}L^{\times}.$

Considering the unramified degree $n$ field extension $L/k_{\mathfrak{p}}$
tells us that $n\mid v_{\mathfrak{p}}(\alpha)$ for each
$\alpha\in\mathcal{A}$, and Lemma 2.17 tells us that
$(\mathcal{A}k_{\mathfrak{p}}^{\times
n})\cap\mathcal{O}_{k_{\mathfrak{p}}}^{\times}\subseteq\mathcal{O}_{k_{\mathfrak{p}}}^{\times
n}$. It follows that $\mathcal{A}\subseteq k_{\mathfrak{p}}^{\times n}.$ If
$\mathfrak{p}$ is an infinite prime, a similar argument shows that
$\mathcal{A}\subseteq k_{\mathfrak{p}}^{\times n}$. Therefore, we have shown
that

$\mathcal{A}\subseteq\ker\Big{(}k^{\times}\to\prod_{\mathfrak{p}\in\Pi_{k}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)},$

so we are done. ∎

######  Definition 2.19.

Given a number field $k$, we say that a positive integer $n$ is _power-
pathological in $k$_ if the following three statements are true:

  1. (1)

$n=2^{r}n^{\prime}$ for an odd integer $n^{\prime}$ and an integer $r\geq 3$.

  2. (2)

The extension $k(\mu_{2^{r}})/k$ is not cyclic.

  3. (3)

All primes $\mathfrak{p}$ of $k$ lying over $2$ decompose in
$k(\mu_{2^{r}})/k$.

######  Theorem 2.20 (Hasse principle for $n^{\mathrm{th}}$ powers).

Let $\Pi_{k}$ be the set of primes of $k$. We have

$\\#\ker\Big{(}k^{\times}/k^{\times
n}\to\prod_{\mathfrak{p}\in\Pi_{k}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)}=\begin{cases}1\quad\text{if $n$ is not power-pathological in
$k$},\\\ 2\quad\text{if $n$ is power-pathological in $k$}.\end{cases}$

###### Proof.

This is the special case $T=\Pi_{k}$ of [NSW00, Theorem 9.1.11]. ∎

######  Corollary 2.21.

If $n$ is not power-pathological in $k$, then

$\ker\Big{(}k^{\times}\to\prod_{\mathfrak{p}\in\Pi_{k}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)}=k^{\times n}.$

If $n$ is power-pathological in $k$, then

$\ker\Big{(}k^{\times}\to\prod_{\mathfrak{p}\in\Pi_{k}}k_{\mathfrak{p}}^{\times}/k_{\mathfrak{p}}^{\times
n}\Big{)}=k^{\times n}\cup uk^{\times n}$

for some $u\in k^{\times(n/2)}\setminus k^{\times n}$.

###### Proof of Theorem 1.1.

This is immediate from Theorems 2.18 and 2.20. ∎

######  Corollary 2.22.

If $n$ is not power-pathological in $k$, then $\mathcal{A}$ is in the norm
group of $100\%$ of $S_{n}$ $n$-ics if and only if $\mathcal{A}$ is in the
norm group of all $S_{n}$ $n$-ics.

##  3\. Prime degree extensions

The main purpose of this section is to prove Theorem 1.6. We start by fixing
some notation: Let $p$ and $\ell$ be rational primes; let $F$ be a $p$-adic
field with residue field of size $q$; let $\pi_{F}$ be a uniformiser of $F$;
let $\mathcal{A}\subseteq F^{\times}$ be a finitely generated subgroup.

######  Lemma 3.1.

We have

$\mathrm{Split}_{\ell}=\\{(1^{\ell}),(\ell)\\}\sqcup\mathrm{Split}_{\ell}^{\mathrm{epi}}.$

###### Proof.

Let $\sigma\in\mathrm{Split}_{\ell}\setminus\\{(1^{\ell}),(\ell)\\}$. Write
$\sigma=(f_{1}^{e_{1}}\ldots f_{r}^{e_{r}})$. Since $\ell$ is prime, we have
$r\geq 2$. Since $\ell=\sum_{i}f_{i}e_{i},$ the $e_{i}$ (respectively the
$f_{i}$) are mutually coprime, so
$\sigma\in\mathrm{Split}_{\ell}^{\mathrm{epi}}$. ∎

######  Corollary 3.2.

We have

$\widetilde{m}(\textnormal{\'{E}t}_{\ell/F}^{\mathcal{A}})=\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})+\widetilde{m}(\textnormal{\'{E}t}_{(\ell)/F}^{\mathcal{A}})+\widetilde{m}(\textnormal{\'{E}t}_{\ell/F}^{\mathrm{epi}}).$

###### Proof.

This follows easily from Lemmas 2.10 and 3.1. ∎

######  Lemma 3.3.

We have

$\widetilde{m}(\textnormal{\'{E}t}_{\ell/F}^{\mathrm{epi}})=(1-\frac{1}{\ell})+\sum_{d=1}^{\ell-2}\frac{\operatorname{Part}(d,\ell-d)}{q^{d}}.$

###### Proof.

Since $(1^{\ell})$ is the only degree $\ell$ splitting symbol $\sigma$ with
$d_{\sigma}=\ell-1$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{\ell/F}^{\mathrm{epi}})=\sum_{d=0}^{\ell-2}\widetilde{m}\big{(}\\{L\in\textnormal{\'{E}t}_{\ell/F}^{\mathrm{epi}}:d_{(L,F)}=d\\}\big{)}.$

Since $d_{(\ell)}=0$, it follows from Lemma 3.1 that

$\\{L\in\textnormal{\'{E}t}_{\ell/F}^{\mathrm{epi}}:d_{(L,F)}=d\\}=\\{L\in\textnormal{\'{E}t}_{\ell/F}:d_{(L,F)}=d\\}$

whenever $1\leq d\leq\ell-2$. The result follows from Corollary 2.14. ∎

###  3.1. Finding
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})$ when
$p\neq\ell$

######  Lemma 3.4.

Let $\zeta_{q-1}\in F$ be a primitive $(q-1)^{\mathrm{st}}$ root of unity, let
$e$ be a positive integer coprime to $q$, and define $g=\gcd(e,q-1)$. For
$0\leq j\leq g-1$, define

$L_{j}=F(\sqrt[e]{\zeta_{q-1}^{j}\pi_{F}}).$

Then we have

$\textnormal{\'{E}t}_{(1^{e})/F}=\\{L_{j}:j=0,1,\ldots,g-1\\}.$

###### Proof.

This is essentially [PR01, Theorem 7.2], where we replace the polynomial
$X^{e}+\zeta_{q-1}^{j}\pi_{F}$ with $X^{e}-\zeta_{q-1}^{j}\pi_{F}$. The
modifications to the proof are trivial. ∎

######  Lemma 3.5.

Suppose that $p\neq\ell$.

  1. (1)

If $\ell\mid q-1$, then Lemma 3.4 tells us that
$\textnormal{\'{E}t}_{(1^{\ell})/F}=\\{L_{0},\ldots,L_{\ell-1}\\}$, where
$L_{j}=F(\sqrt[\ell]{\zeta_{q-1}^{j}\pi_{F}})$. Then
$\operatorname{Aut}(L_{j}/F)\cong C_{\ell}$ for each $j$, and

$N_{L_{j}/F}L_{j}^{\times}=\\{u^{\ell}(\zeta_{q-1}^{j}\pi_{F})^{m}:u\in\mathcal{O}_{F}^{\times},k\in\mathbb{Z}\\}.$

  2. (2)

If $\ell\nmid q-1$, then Lemma 3.4 tells us that
$\textnormal{\'{E}t}_{(1^{\ell})/F}=\\{L\\}$, where
$L=F(\sqrt[\ell]{\pi_{F}})$. We have $\operatorname{Aut}(L/F)=1$ and
$N_{L/F}L^{\times}=F^{\times}$.

###### Proof.

We prove the statements one at a time.

  1. (1)

Suppose that $\ell\mid q-1$, so that $\zeta_{\ell}\in F^{\times}$. Since the
elements $\zeta_{\ell}^{j}\pi_{F}$ are all uniformisers of $F$, and $\pi_{F}$
is an arbitrary uniformiser, we only need to prove the result for
$L_{0}=F(\sqrt[\ell]{\pi_{F}})$. The element $\sqrt[\ell]{\pi_{F}}$ has
minimal polynomial $X^{\ell}-\pi_{F}$ over $F$, which splits in $L_{0}$, so
$L_{0}/F$ is Galois, hence cyclic. By class field theory, we have

$[F^{\times}:N_{L_{0}/F}L_{0}^{\times}]=\ell.$

Moreover, it is clear that

$\Big{\\{}u^{\ell}(\zeta_{q-1}^{j}\pi_{F})^{m}:u\in\mathcal{O}_{F}^{\times},m\in\mathbb{Z}\Big{\\}}\subseteq
N_{L_{j}/F}L_{j}^{\times}.$

By [Neu13, Proposition 3.7], we have

$[\mathcal{O}_{F}^{\times}:\mathcal{O}_{F}^{\times\ell}]=\\#\mu_{\ell}(F)=\ell.$

It follows that

$\big{[}F^{\times}:\big{\\{}u^{\ell}(\zeta_{q-1}^{j}\pi_{F})^{m}:u\in\mathcal{O}_{F}^{\times},m\in\mathbb{Z}\big{\\}}\big{]}=\ell,$

and the result follows.

  2. (2)

Since $\ell\nmid q-1$, we have $\zeta_{\ell}\not\in L$, so the polynomial
$X^{\ell}-\pi_{F}$ has only one root in $L$, and therefore $L/F$ is not
Galois, hence it has only one automorphism. The maximal abelian subextension
of $L/F$ is $F$, so it has trivial norm group.

∎

######  Lemma 3.6.

Suppose that $\ell\mid q-1$. Choose a minimal generating set $A_{0}\cup A_{1}$
for $\overline{\mathcal{A}}^{\ell}$ such that $v_{F}(\alpha)=r$ for all
$[\alpha]\in A_{r}$, for $r=0,1$. Then

$\\#\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}}=\begin{cases}\ell\quad\text{if
$\mathcal{A}\subseteq F^{\times\ell}$},\\\ 1\quad\text{else if
$A_{0}\subseteq\mathcal{O}_{F}^{\times\ell}$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times\ell}$ for all
$\alpha_{i},\alpha_{j}\in A_{1}$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

The classification of $\textnormal{\'{E}t}_{(1^{\ell})/F}$ follows from Lemma
3.4, and the case $\mathcal{A}\subseteq F^{\times\ell}$ is immediate. If
$\mathcal{A}\not\subseteq F^{\times\ell}$, then the result follows from Lemma
3.5. ∎

######  Corollary 3.7.

  1. (1)

Suppose that $\ell\nmid q-1$. Then

$\widetilde{m}_{F}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})=\frac{1}{q^{\ell-1}}.$

  2. (2)

Suppose that $\ell\mid q-1$. Then, in the language of Lemma 3.6, we have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{\ell-1}}\quad\text{if
$\mathcal{A}\subseteq F^{\times\ell}$}\\\ \frac{1}{\ell
q^{\ell-1}}\quad\text{else if $A_{0}\subseteq\mathcal{O}_{F}^{\times\ell}$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times\ell}$ for all
$\alpha_{i},\alpha_{j}\in A_{1}$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

This is immediate from Lemmas 3.5 and 3.6. ∎

###  3.2. Finding
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{\ell})/F}^{\mathcal{A}})$ when
$p=\ell$

######  Lemma 3.8.

We have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{\mathcal{A}})=\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F})-\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})+\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}}).$

###### Proof.

Let
$L\in\textnormal{\'{E}t}_{(1^{p})/F}\setminus\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}.$
Then the maximal abelian subextension of $L/F$ is $F$, so by class field
theory we have $N_{L/F}L^{\times}=F^{\times}$, and therefore

$\textnormal{\'{E}t}_{(1^{p})/F}^{\mathcal{A}}=\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}}\cup\textnormal{\'{E}t}_{(1^{p})/F}\setminus\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F},$

so the result follows. ∎

Let $L/F$ be an abelian extension of $p$-adic fields, and let
$G=\operatorname{Gal}(L/F)$. For a character $\chi:G\to\mathbb{C}^{\times}$,
the _Artin conductor_ of $\chi$ is defined to be

$\mathfrak{f}(\chi)=\sum_{i\geq 0}\frac{\lvert G_{i}\rvert}{\lvert
G_{0}\rvert}(\chi(1)-\chi(G_{i})),$

where $(G_{i})_{i}$ is the ramification filtration of $G$, defined by

$G_{i}=\Big{\\{}\varphi\in
G:v_{L}\Big{(}\frac{\varphi(\pi_{L})-\pi_{L}}{\pi_{L}}\Big{)}\geq i\Big{\\}},$

and $\chi(G_{i})$ is the average value of $\chi(g)$ as $g$ ranges over
$G_{i}$. Write $\operatorname{Irr}(G)$ for the set of irreducible characters
$\chi:G\to\mathbb{C}^{\times}$.

######  Theorem 3.9.

Let $L/F$ be an abelian extension of $p$-adic fields, and let
$G=\operatorname{Gal}(L/F)$. We have

  1. (1)

$\mathfrak{f}(L/F)=\max_{\chi\in\operatorname{Irr}(G)}\mathfrak{f}(\chi)$.

  2. (2)

$v_{F}(d_{L/F})=\sum_{\chi\in\operatorname{Irr}(G)}\mathfrak{f}(\chi)$.

###### Proof.

The first item is [AT68, Page 135, Corollary to Theorem 14]. The second is
[Keu23, Theorem 17.50]. ∎

######  Lemma 3.10.

Let $L/F$ be a $C_{p}$-extension of $p$-adic fields. Then

$\mathfrak{f}(L/F)=\frac{v_{F}(d_{L/F})}{p-1}.$

###### Proof.

Take $G=\operatorname{Gal}(L/F)=\langle g\rangle$, and write $c$ for the
unique integer with $G_{c-1}\neq G_{c}$. We have

$\operatorname{Irr}(G)=\\{\chi_{j}:j=0,1,2,\ldots,p-1\\},$

where $\chi_{j}(g)=\zeta_{p}^{j}$ for some fixed primitive $p^{\mathrm{th}}$
roots of unity $\zeta_{p}$. We have

$\mathfrak{f}(\chi_{j})=\begin{cases}0\quad\text{if $j=0$},\\\
c\quad\text{otherwise},\end{cases}$

so Theorem 3.9 tells us that $\mathfrak{f}(L/F)=c$ and
$v_{F}(d_{L/F})=(p-1)c$, and the result follows. ∎

######  Lemma 3.11.

There is a surjective $(p-1)$-to-$1$ map

$\operatorname{Epi}_{\mathbb{F}_{p}}\Big{(}F^{\times}/U_{F}^{(\lfloor\frac{m}{p-1}\rfloor)}F^{\times
p},\mathbb{F}_{p}\Big{)}\to\textnormal{\'{E}t}_{p/F,\leq m}^{C_{p}/F},$

where the $\mathbb{F}_{p}$-linear map
$\chi:F^{\times}/U_{F}^{(\lfloor\frac{m}{p-1}\rfloor)}F^{\times
p}\twoheadrightarrow\mathbb{F}_{p}$ maps to the unique abelian extension $L/F$
with

$N_{L/F}L^{\times}=\ker\Big{(}F^{\times}\to
F^{\times}/U_{F}^{(\lfloor\frac{m}{p-1}\rfloor)}F^{\times
p}\overset{\chi}{\to}\mathbb{F}_{p}\Big{)}.$

###### Proof.

This follows from Lemma 3.10 by basic class field theory. ∎

######  Corollary 3.12.

For every positive integer $m$, we have

$\\#\textnormal{\'{E}t}_{(p)/F,\leq
m}^{C_{p}/F}=\frac{1}{p-1}\Big{(}\\#\Big{(}F^{\times}/F^{\times
p}U_{F}^{(\lfloor\frac{m}{p-1}\rfloor)}\Big{)}-1\Big{)}.$

###### Proof.

This is immediate from Lemma 3.11. ∎

######  Lemma 3.13.

Let $u\in\mathcal{O}_{F}^{\times}$ and let $n$ be an integer with
$n>\frac{pe_{F}}{p-1}$. Then $U_{F}^{(n)}F^{\times p}=U_{F}^{(n+1)}F^{\times
p}$.

###### Proof.

Let $u\in U_{F}^{(n)}F^{\times p}$. Then $\frac{u}{a^{p}}\equiv
1\pmod{\mathfrak{p}_{F}^{n}}$ for some $a\in F^{\times}$, so there is some
$x\in\mathfrak{p}_{F}^{n}$ with

$\frac{u}{a^{p}}=1+x.$

Since $v_{F}(x)\geq n$ and $n>\frac{pe_{F}}{p-1}$, it is easy to see that

$\Big{(}1+\frac{x}{p}\Big{)}^{p}\equiv 1+x\pmod{\mathfrak{p}_{F}^{n+1}},$

and it follows that $u\in U_{F}^{(n+1)}F^{\times p}$, as required. ∎

######  Corollary 3.14.

We have $U_{F}^{(\lfloor\frac{pe_{F}}{p-1}\rfloor+1)}\subseteq F^{\times p}$.

###### Proof.

This follows from Lemma 3.13 by Hensel’s lemma. ∎

######  Lemma 3.15.

For $0\leq i\leq\lfloor\frac{pe_{F}}{p-1}\rfloor$, let
$W_{i}=U_{F}^{(i)}F^{\times p}/U_{F}^{(i+1)}F^{\times p}$. For each positive
integer $c$, we have an isomorphism

$F^{\times}/F^{\times p}U_{F}^{(c)}\cong\Big{(}F^{\times}/U_{F}^{(0)}F^{\times
p}\Big{)}\oplus\bigoplus_{i=0}^{\min\\{c-1,\lfloor\frac{pe_{F}}{p-1}\rfloor\\}}W_{i}.$

###### Proof.

Corollary 3.14 tells us that

$F^{\times}/F^{\times p}U_{F}^{(c)}=F^{\times}/F^{\times
p}U_{F}^{(\lfloor\frac{pe_{F}}{p-1}\rfloor+1)}$

whenever $c\geq\lfloor\frac{pe_{F}}{p-1}\rfloor+1$, so we only need to prove
the result for $c\leq\lfloor\frac{pe_{F}}{p-1}\rfloor+1$.

The left- and right-hand side have the same cardinality by definition of the
$W_{i}$. Moreover, both are $p$-torsion groups, hence $\mathbb{F}_{p}$-vector
spaces, which means they are isomorphic as groups. ∎

######  Lemma 3.16.

Let $0\leq i\leq\frac{pe_{F}}{p-1}$, and let $m\in U_{F}^{(i)}$.

  1. (1)

If $p\nmid i$, then $m\in U_{F}^{(i+1)}F^{\times p}$ if and only if $m\in
U_{F}^{(i+1)}$.

  2. (2)

If $p\mid i$, then $m\in U_{F}^{(i+1)}F^{\times p}$ if and only if there is
some $[x]\in\mathfrak{p}_{F}^{\frac{i}{p}}$ such that $(1+x)^{p}\equiv
m\pmod{\mathfrak{p}_{F}^{i+1}}$.

  3. (3)

If $p\mid i$ and $i<\frac{pe_{F}}{p-1}$, then we always have $m\in
U_{F}^{(i+1)}F^{\times p}$, and in particular

$m\equiv(1+\pi_{F}^{i/p}y)^{p}\pmod{\mathfrak{p}_{F}^{i+1}}$

where $[y]\in\mathcal{O}_{F}/\mathfrak{p}_{F}$ is the unique element with
$[y]^{p}=\Big{[}\frac{m-1}{\pi_{F}^{i}}\Big{]}.$

  4. (4)

If $i=\frac{pe_{F}}{p-1}$, then $m\in U_{F}^{(i+1)}F^{\times p}$ if and only
if
$\Big{[}\frac{m-1}{{\pi_{F}^{e_{F}/(p-1)}p}}\Big{]}\in\mathcal{O}_{F}/\mathfrak{p}_{F}$
is in the image of the map

$\mathcal{O}_{F}/\mathfrak{p}_{F}\to\mathcal{O}_{F}/\mathfrak{p}_{F},\quad
y\mapsto y+\frac{\pi_{F}^{e_{F}}}{p}y^{p}.$

###### Proof.

The case $i=0$ is trivial, so we assume that $i\geq 1$. For the first two
statements, the “if” directions are trivial, so we focus on the “only if”.

Suppose that $m\in U_{F}^{(i+1)}F^{\times p}\setminus U_{F}^{(i+1)}$, so that
$v_{F}(m-1)=i$. We will show that this implies $p\mid i$ and there is some
$x\in\mathfrak{p}_{F}^{\frac{i}{p}}$ with $(1+x)^{p}\equiv
m\pmod{\mathfrak{p}_{F}^{i+1}}$, proving the first two statements.

Since $m\in U_{F}^{(i+1)}F^{\times p}$, there is some $c\in F^{\times}$ such
that $\frac{m}{c^{p}}\equiv 1\pmod{\mathfrak{p}_{F}^{i+1}}$, so $v_{F}(c)=0$
and $m\equiv c^{p}\pmod{\mathfrak{p}_{F}^{i+1}}$. Write $c=1+x$ for
$x\in\mathcal{O}_{F}$, so that

$m-1\equiv\sum_{j=1}^{p-1}\binom{p}{j}x^{j}+x^{p}\pmod{\mathfrak{p}_{F}^{i+1}}.$

Since $v_{F}(m-1)>0$, we have $v_{F}(x)>0$. We have

$v_{F}\Big{(}\sum_{j=1}^{p-1}\binom{p}{j}x^{j}\Big{)}=e_{F}+v_{F}(x)$

and

$v_{F}(x^{p})=pv_{F}(x).$

Suppose for contradiction that $v_{F}(x)>\frac{e_{F}}{p-1}$. Then
$v_{F}(x)+e_{F}<pv_{F}(x)$, so

$i=v_{F}(m-1)=v_{F}(x)+e_{F}>\frac{pe_{F}}{p-1},$

which is impossible since by assumption $i\leq\frac{pe_{F}}{p-1}$.

Therefore, $v_{F}(x)\leq\frac{e_{F}}{p-1}$. We will consider the cases
$v_{F}(x)=\frac{e_{F}}{p-1}$ and $v_{F}(x)<\frac{e_{F}}{p-1}$ separately.
Suppose first that $v_{F}(x)=\frac{e_{F}}{p-1}$. Then
$pv_{F}(x)=e_{F}+v_{F}(x)=\frac{pe_{F}}{p-1}$, so

$i=v_{F}(m-1)\geq\frac{pe_{F}}{p-1},$

and therefore $i=\frac{pe_{F}}{p-1}$ and $v_{F}(x)\geq\frac{i}{p}$, as
required.

Suppose instead that $v_{F}(x)<\frac{e_{F}}{p-1}$. Then
$pv_{F}(x)<v_{F}(x)+e_{F}$, so

$i=v_{F}(m-1)=pv_{F}(x),$

and therefore $p\mid i$ and $v_{F}(x)\geq\frac{i}{p}$. Thus we have proved
Statements $(1)$ and $(2)$.

Suppose that $p\mid i$ and $i<\frac{pe_{F}}{p-1}$. Let $y\in\mathcal{O}_{F}$
be such that $y^{p}\equiv\frac{m-1}{\pi_{F}^{i}}\pmod{\mathfrak{p}_{F}}$, and
set $x=\pi_{F}^{\frac{i}{p}}y$. Then we have

$(1+x)^{p}\equiv 1+x^{p}\equiv m\pmod{\mathfrak{p}_{F}^{i+1}},$

so Statement $(3)$ follows from Statement $(2)$.

Suppose that $i=\frac{pe_{F}}{p-1}$. Statement $(2)$ tells us that $m\in
U_{F}^{(i+1)}F^{\times p}$ if and only if there is some
$x\in\mathfrak{p}_{F}^{\frac{e_{F}}{p-1}}$ with

$m-1\equiv px+x^{p}\pmod{\mathfrak{p}_{F}^{\frac{pe_{F}}{p-1}+1}},$

and Statement $(4)$ follows easily. ∎

######  Lemma 3.17.

Suppose that $(p-1)\mid e_{F}$. Define $\varphi$ by

$\varphi:\mathcal{O}_{F}/\mathfrak{p}_{F}\to\mathcal{O}_{F}/\mathfrak{p}_{F},\quad
y\mapsto y+\frac{\pi_{F}^{e_{F}}}{p}y^{p}.$

We have

$\\#\operatorname{im}\varphi=\begin{cases}q/p\quad\text{if $\mu_{p}\subseteq
F$},\\\ q\quad\text{if $\mu_{p}\not\subseteq F$}.\end{cases}$

###### Proof.

The map is $\mathbb{F}_{p}$-linear, and its kernel consists of the roots of
the polynomial

$X\Big{(}X^{p-1}+\frac{p}{\pi_{F}^{e_{F}}}\Big{)}\in\mathbb{F}_{K}[X].$

Since $\mathbb{F}_{K}$ contains all $(p-1)^{\mathrm{st}}$ roots of unity, the
polynomial has either $1$ or $p$ roots. By Hensel’s lemma, a root of
$X^{p-1}+\frac{p}{\pi_{F}^{e_{F}}}$ in $\mathbb{F}_{K}$ lifts to a root in
$F$, which exists if and only if $-p\in F^{\times(p-1)}$. The result [Was97,
Lemma 14.6] states that
$\mathbb{Q}_{p}(\sqrt[p-1]{-p})=\mathbb{Q}_{p}(\zeta_{p})$, and the result
follows. ∎

We note the following algorithm as an immediate consequence of Lemmas 3.16 and
3.17.

######  Algorithm 3.18.

Input: $\alpha\in\mathcal{O}_{F}^{\times}$.

Output: Returns a pair $(i,\lambda)$. If $\alpha\in F^{\times p}$, then
$i=\infty$ and $\lambda\in\mathcal{O}_{F}^{\times}$ is such that
$\alpha\equiv\lambda^{p}\pmod{\mathfrak{p}_{F}^{\lfloor\frac{pe_{F}}{p-1}\rfloor+1}}$.
Otherwise, $i$ is the largest integer with $\alpha\in U_{F}^{(i)}F^{\times
p}$, and $\lambda\in\mathcal{O}_{F}^{\times}$ is an element with
$\alpha\equiv\lambda^{p}\pmod{\mathfrak{p}_{F}^{i}}$.

Algorithm:

  1. (1)

Set $m_{0}=\alpha$ and $\lambda_{0}=1$.

  2. (2)

For $0\leq i\leq\frac{pe_{F}}{p-1}$, do the following:

If $p\nmid i$, then:

     * •

If $m_{i}\equiv 1\pmod{\mathfrak{p}_{F}^{i+1}}$, then set $m_{i+1}=m_{i}$ and
$\lambda_{i+1}=\lambda_{i}$.

     * •

Otherwise, return $(i,\lambda_{i})$ and break the for loop.

If $p\mid i$ and $i<\frac{pe_{F}}{p-1}$, then:

     * •

Let $[y]\in\mathcal{O}_{F}/\mathfrak{p}_{F}$ be the unique element with
$[y]^{p}=\Big{[}\frac{m_{i}-1}{\pi_{F}^{i/p}}\Big{]}$, and set
$\lambda_{i+1}=\lambda_{i}(1+\pi_{F}^{i/p}y)$ and
$m_{i+1}=\frac{m_{i}}{(1+\pi_{F}^{i/p}y)^{p}}$.

If $i=\frac{pe_{F}}{p-1}$, then:

     * •

If
$\Big{[}\frac{m_{i}-1}{{\pi_{F}^{e_{F}/(p-1)}p}}\Big{]}\in\mathcal{O}_{F}/\mathfrak{p}_{F}$
is in the image of the map

$\mathcal{O}_{F}/\mathfrak{p}_{F}\to\mathcal{O}_{F}/\mathfrak{p}_{F},\quad
y\mapsto y+\frac{\pi_{F}^{e_{F}}}{p}y^{p},$

then take $y$ in the preimage and set
$\lambda_{i+1}=\lambda_{i}(1+\pi_{F}^{i/p}y)$ and
$m_{i+1}=\frac{m_{i}}{(1+\pi_{F}^{i/p})^{p}}$.

By Lemma 3.16, we see inductively that $m_{i}\equiv
1\pmod{\mathfrak{p}_{F}^{i}}$ and $\alpha=\lambda_{i}^{p}m_{i}$ for all $i$
that it is defined.

  3. (3)

If the for loop from the previous step finishes, then return
$(\infty,\frac{\lambda_{0}}{\lambda_{\frac{pe_{F}}{p-1}+1}})$.

######  Lemma 3.19.

Suppose that we have already computed a section of the map

$\mathcal{O}_{F}/\mathfrak{p}_{F}\to\mathcal{O}_{F}/\mathfrak{p}_{F},\quad
y\mapsto y+\frac{\pi_{F}^{e_{F}}}{p}y^{p}.$

Assume that arithmetic operations can be performed in $F$ with time complexity
$t(F)$, where $t(F)$ is some function of $F$. Then the time complexity of
Algorithm 3.18 is $O(e_{F}t(F))$.

###### Proof.

Steps $(1)$ and $(3)$ take $O(1)$ time. There are $\frac{pe_{F}}{p-1}+1$ steps
in the for loop, and each one just involves arithmetic in $F$, so Step $(2)$
has time complexity $O(e_{F}t(F))$. ∎

######  Corollary 3.20.

Let $i$ be an integer with $0\leq i\leq\frac{pe_{F}}{p-1}$. We have

$W_{i}\cong\begin{cases}\mathbb{F}_{F}\quad\text{if $i<\frac{pe_{F}}{p-1}$ and
$p\nmid i$},\\\ \mathbb{F}_{p}\quad\text{if $i=\frac{pe_{F}}{p-1}$ and
$\mu_{p}\subseteq F$},\\\ 1\quad\text{otherwise}.\end{cases}$

###### Proof.

Suppose first that $p\nmid i$. Then Lemma 3.16 tells us that the natural exact
sequence

$1\to U_{F}^{(i+1)}\to U_{F}^{(i)}\to W_{i}\to 1,$

is exact, so $W_{i}\cong\mathbb{F}_{F}$. If $p\mid i$ and
$i<\frac{pe_{F}}{p-1}$, then Lemma 3.16 tells us that $W_{i}=0$. Finally,
suppose that $i=\frac{pe_{F}}{p-1}$. Write $\varphi$ for the map

$\varphi:\mathcal{O}_{F}/\mathfrak{p}_{F}\to\mathcal{O}_{F}/\mathfrak{p}_{F},\quad
y\mapsto y+\frac{\pi_{F}^{e_{F}}}{p}y^{p}.$

By Lemma 3.16, we have an exact sequence

$1\to\operatorname{im}\varphi\to U_{F}^{(i)}/U_{F}^{(i+1)}\to W_{i}\to 1,$

where the map $\operatorname{im}\varphi\to U_{F}^{(i)}/U_{F}^{(i+1)}$ is given
by $x\mapsto 1+\pi_{F}^{\frac{e_{F}}{p-1}}px$. The result then follows by
Lemma 3.17. ∎

######  Corollary 3.21.

Let $t$ be a nonnegative integer. If $0\leq
t\leq\lceil\frac{pe_{F}}{p-1}\rceil$. We have

$\\#\Big{(}F^{\times}/U_{F}^{(t)}F^{\times
p}\Big{)}=pq^{t-1-\lfloor\frac{t-1}{p}\rfloor}.$

If $t>\lceil\frac{pe_{F}}{p-1}\rceil$, then

$\\#\Big{(}F^{\times}/U_{F}^{(t)}F^{\times
p}\Big{)}=\begin{cases}p^{2}q^{e_{F}}\quad\text{if $\mu_{p}\subseteq F$}\\\
pq^{e_{F}}\quad\text{otherwise}.\end{cases}$

###### Proof.

This is immediate from Lemma 3.15 and Corollary 3.20. ∎

######  Corollary 3.22.

Let $L\in\textnormal{\'{E}t}_{p/F}^{C_{p}/F}$. Then $v_{F}(d_{L/F})=(p-1)c$
for an integer $c$ with and $0\leq c\leq\lfloor\frac{pe_{F}}{p-1}\rfloor+1$.

For $1\leq c\leq\lceil\frac{pe_{F}}{p-1}\rceil$, we have

$\\#\textnormal{\'{E}t}_{p/F,\leq(p-1)c}^{C_{p}/F}=\frac{1}{p-1}\Big{(}pq^{c-1-\lfloor\frac{c-1}{p}\rfloor}-1\Big{)}.$

If $(p-1)\mid e_{F}$, we have

$\\#\textnormal{\'{E}t}_{p/F,\leq
pe_{F}+p-1}^{C_{p}/F}=\begin{cases}\frac{1}{p-1}\Big{(}p^{2}q^{e_{F}}-1\Big{)}\quad\text{if
$\mu_{p}\subseteq F$},\\\
\frac{1}{p-1}\Big{(}pq^{e_{F}}-1\Big{)}\quad\text{otherwise}.\end{cases}$

###### Proof.

This is immediate from Corollaries 3.12 and 3.21. ∎

######  Theorem 3.23.

If $\textnormal{\'{E}t}_{(1^{p})/F,m}^{C_{p}/F}$ is nonempty, then $m=(p-1)c$
for an integer $c$ with $1\leq c\leq\frac{pe_{F}}{p-1}+1$, and

$\\#\textnormal{\'{E}t}_{(1^{p})/F,m}^{C_{p}/F}=\begin{cases}\frac{p(q-1)}{p-1}\cdot
q^{c-2-\lfloor\frac{c-2}{p}\rfloor}\quad\text{if $c\not\equiv 1\pmod{p}$},\\\
pq^{e_{F}}\quad\text{if $c=\frac{pe_{F}}{p-1}+1$ and $\mu_{p}\subseteq F$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

This is immediate from Corollary 3.22. ∎

######  Lemma 3.24.

Let $t$ be a positive integer. Then we have

$\sum_{\begin{subarray}{c}1\leq c\leq t\\\ c\not\equiv
1\pmod{p}\end{subarray}}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}=\mathbbm{1}_{t\geq
p}\cdot A(t)+\mathbbm{1}_{t\not\equiv 0,1\pmod{p}}\cdot B(t),$

where $A(t)$ and $B(t)$ are as in Appendix B.

###### Proof.

Writing $a=\lfloor\frac{c-2}{p}\rfloor$, see that

$\sum_{\begin{subarray}{c}1\leq c\leq t\\\ c\not\equiv
1\pmod{p}\end{subarray}}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}$

is the sum of the following two quantities:

  1. (1)

$\mathbbm{1}_{t\geq
p}\sum_{a=0}^{\lfloor\frac{t}{p}\rfloor-1}\sum_{c=ap+2}^{(a+1)p}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}.$

  2. (2)

$\sum_{c=\lfloor\frac{t}{p}\rfloor
p+2}^{t}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}.$

If $p\geq 3$, these simplify respectively to:

  1. (1)

$q^{-p(p-2)}\cdot\frac{q^{(p-1)(p-2)}-1}{q^{p-2}-1}\cdot\frac{q^{-(p-1)^{2}(\lfloor\frac{t}{p}\rfloor)}-1}{q^{-(p-1)^{2}}-1}.$

  2. (2)

We have

$\mathbbm{1}_{t\geq\lfloor\frac{t}{p}\rfloor p+2}\cdot
q^{-(\lfloor\frac{t}{p}\rfloor)}\cdot\frac{q^{-(p-2)(t+1)}-q^{-(p-2)((\lfloor\frac{t}{p}\rfloor)p+2)}}{q^{-(p-2)}-1}.$

The case $p=2$ is similar. ∎

######  Corollary 3.25.

Recall the explicit rational functions $A(t)$ and $B(t)$ from Appendix B. We
have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})=\frac{q-1}{p-1}q^{-2}\Big{(}\mathbbm{1}_{e_{F}\geq
p-1}\cdot
A\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}+\mathbbm{1}_{(p-1)\nmid
e_{F}}\cdot
B\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}\Big{)}+\mathbbm{1}_{\mu_{p}\subseteq
F}\cdot q^{-(p-1)(e_{F}+1)}.$

###### Proof.

The mass $\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F})$ is the sum
of the following two quantities:

  1. (1)

$\frac{q-1}{p-1}\cdot q^{-2}\cdot\sum_{\begin{subarray}{c}1\leq
c<\frac{pe_{F}}{p-1}+1\\\ c\not\equiv
1\pmod{p}\end{subarray}}q^{-\big{(}(p-2)c+\lfloor\frac{c-2}{p}\rfloor\big{)}}.$

  2. (2)

$\mathbbm{1}_{\mu_{p}\subseteq F}\cdot q^{-(p-1)(e_{F}+1)}.$

For $c\in\mathbb{Z}$, we have $c<\frac{pe_{F}}{p-1}+1$ if and only if
$c\leq\lceil\frac{pe_{F}}{p-1}\rceil$. Setting
$t=\lceil\frac{pe_{F}}{p-1}\rceil$, Lemma 3.24 tells us that

$\sum_{\begin{subarray}{c}1\leq c<\frac{pe_{F}}{p-1}+1\\\ c\not\equiv
1\pmod{p}\end{subarray}}q^{-\big{(}(p-2)c+\lfloor\frac{c-2}{p}\rfloor\big{)}}=\mathbbm{1}_{\lceil\frac{pe_{F}}{p-1}\rceil\geq
p}\cdot
A\Big{(}\lceil\frac{pe_{F}}{p-1}\rceil\Big{)}+\mathbbm{1}_{\lceil\frac{pe_{F}}{p-1}\rceil\not\equiv
0,1\pmod{p}}\cdot B\Big{(}\lceil\frac{pe_{F}}{p-1}\rceil\Big{)}.$

It is easy to see that $\lceil\frac{pe_{F}}{p-1}\rceil\geq p$ if and only if
$e_{F}\geq p-1$ and $\lceil\frac{pe_{F}}{p-1}\rceil\not\equiv 0,1\pmod{p}$ if
and only if $(p-1)\nmid e_{F}$. ∎

Recall that we write

$\overline{\mathcal{A}}^{n}=\mathcal{A}F^{\times n}/F^{\times n}$

and

$\overline{\mathcal{A}}^{n}_{t}=\overline{\mathcal{A}}^{n}\cap\Big{(}U_{F}^{(t)}F^{\times
n}/F^{\times n}\Big{)}$

for each nonnegative integer $t$.

######  Lemma 3.26.

Let $c$ be an integer with $0\leq c\leq\frac{pe_{F}}{p-1}+1$. We have

$\\#\textnormal{\'{E}t}_{p/F,\leq(p-1)c}^{C_{p}/F,\mathcal{A}}=\frac{1}{p-1}\cdot\Big{(}\frac{\\#\overline{\mathcal{A}}^{p}_{c}}{\\#\overline{\mathcal{A}}^{p}}\Big{(}1+(p-1)\\#\textnormal{\'{E}t}_{p/F,\leq(p-1)c}^{C_{p}/F}\Big{)}-1\Big{)}.$

###### Proof.

By Lemma 3.11, we need to count $\mathbb{F}_{p}$-linear transformations

$\chi:F^{\times}/F^{\times p}\to\mathbb{F}_{p}$

such that

$\chi\big{(}\overline{\mathcal{A}}^{p}\big{)}=\chi\big{(}U_{F}^{(c)}F^{\times
p}/F^{\times p}\big{)}=0.$

In other words, we need to compute the size of the annihilator

$\Big{(}\overline{\mathcal{A}}^{p}+\Big{(}U_{F}^{(c)}F^{\times p}/F^{\times
p}\Big{)}\Big{)}^{\bot},$

and it is easy to see that

$\dim_{\mathbb{F}_{p}}\Big{(}\big{(}\overline{\mathcal{A}}^{p}+(U_{F}^{(c)}F^{\times
p}/F^{\times
p})\big{)}^{\bot}\Big{)}=\dim_{\mathbb{F}_{p}}\Big{(}F^{\times}/U_{F}^{(c)}F^{\times
p}\Big{)}-\dim_{\mathbb{F}_{p}}(\overline{\mathcal{A}}^{p}/\overline{\mathcal{A}}^{p}_{c}),$

so

$\\#\Big{(}\overline{\mathcal{A}}^{p}+(U_{F}^{(c)}F^{\times p}/F^{\times
p})\Big{)}^{\bot}=\frac{\\#\overline{\mathcal{A}}^{p}_{c}}{\\#\overline{\mathcal{A}}^{p}}\cdot\\#\Big{(}F^{\times}/U_{F}^{(c)}F^{\times
p}\Big{)}.$

The result then follows by Lemma 3.11. ∎

######  Corollary 3.27.

For $1\leq c\leq\frac{pe_{F}}{p-1}+1$, we have

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\mathcal{A}}=\begin{cases}\frac{p}{(p-1)\\#\overline{\mathcal{A}}^{p}}q^{c-2-\lfloor\frac{c-2}{p}\rfloor}\Big{(}q\cdot\\#\overline{\mathcal{A}}^{p}_{c}-\\#\overline{\mathcal{A}}^{p}_{c-1}\Big{)}\quad\text{if
$c\not\equiv 1\pmod{p}$},\\\
\frac{pq^{e_{F}}}{\\#\overline{\mathcal{A}}^{p}}\quad\text{if
$c=\frac{pe_{F}}{p-1}+1$ and $\mu_{p}\subseteq F$ and
$\overline{\mathcal{A}}^{p}_{\frac{pe_{F}}{p-1}}=0$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Let $1\leq c\leq\frac{pe_{F}}{p-1}+1$. If $c<\frac{pe_{F}}{p-1}+1$ and
$c\equiv 1\pmod{p}$ , then
$\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F}=\varnothing$ by Theorem 3.23, so

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\mathcal{A}}=0.$

Lemma 3.26 tells us that, for $1\leq c\leq\frac{pe_{F}}{p-1}+1$, we have

(1)
$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\mathcal{A}}=\frac{\\#\overline{\mathcal{A}}^{p}_{c}\\#\textnormal{\'{E}t}_{p/F,\leq(p-1)c}^{C_{p}/F}-\\#\overline{\mathcal{A}}^{p}_{c-1}\\#\textnormal{\'{E}t}_{p/F,\leq(p-1)(c-1)}^{C_{p}/F}}{\\#\overline{\mathcal{A}}^{p}}-\frac{\\#\overline{\mathcal{A}}^{p}_{c-1}-\\#\overline{\mathcal{A}}^{p}_{c}}{(p-1)\cdot\\#\overline{\mathcal{A}}^{p}}$

Suppose that $1\leq c\leq\frac{pe_{F}}{p-1}+1$ and $c\not\equiv 1\pmod{p}$.
Substituting Corollary 3.22 into Equation 1 gives

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\mathcal{A}}=\frac{p}{(p-1)\\#\overline{\mathcal{A}}^{p}}\cdot
q^{c-2-\lfloor\frac{c-2}{p}\rfloor}\Big{(}q\cdot\\#\overline{\mathcal{A}}^{p}_{c}-\\#\overline{\mathcal{A}}^{p}_{c-1}\Big{)}.$

Finally, consider $c=\frac{pe_{F}}{p-1}+1$. If $\mu_{p}\not\subseteq F$, then
Theorem 3.23 tells us that
$\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F}=\varnothing$, so

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\mathcal{A}}=0.$

Suppose that $\mu_{p}\subseteq F$. Substituting Corollary 3.22 into Equation 1
gives

$\\#\textnormal{\'{E}t}_{p/F,pe_{F}+p-1}^{C_{p}/F,\mathcal{A}}=\frac{pq^{e_{F}}}{(p-1)\\#\overline{\mathcal{A}}^{p}}\Big{(}\\#\overline{\mathcal{A}}^{p}_{c}\cdot
p-\\#\overline{\mathcal{A}}^{p}_{c-1}\Big{)},$

and the result follows from the fact that
$\\#\overline{\mathcal{A}}^{p}_{\frac{pe_{F}}{p-1}+1}=1$ and
$\\#\overline{\mathcal{A}}^{p}_{(\frac{pe_{F}}{p-1})}\subseteq
U_{F}^{\frac{pe_{F}}{p-1}}F^{\times p}/F^{\times p}\cong C_{p}$, where the
final isomorphism comes from Corollary 3.20. ∎

######  Corollary 3.28.

The mass
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})$ is the
sum of the following two quantities:

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})=\mathbbm{1}_{\mu_{p}\subseteq
F}\cdot\mathbbm{1}_{\overline{\mathcal{A}}^{p}_{\frac{pe_{F}}{p-1}}=0}\cdot\frac{pq^{-(p-1)(e_{F}+1)}}{\\#\overline{\mathcal{A}}^{p}}+\frac{1}{(p-1)\\#\overline{\mathcal{A}}^{p}}q^{-2}\sum_{c=1}^{\lceil\frac{pe_{F}}{p-1}\rceil}\frac{q\\#\overline{\mathcal{A}}^{p}_{c}-\\#\overline{\mathcal{A}}^{p}_{c-1}}{q^{(p-2)c+\lfloor\frac{c-2}{p}\rfloor}}.$

###### Proof of Theorem 1.6.

We prove the statements one by one.

  1. (1)

This is immediate from Lemmas 3.1 and 3.3.

  2. (2)

Let $L/F$ be the unique unramified extension of degree $\ell$. Clearly
$\\#\operatorname{Aut}(L/F)=\ell$ and $v_{F}(d_{L/F})=0$, and we have
$\mathcal{A}\subseteq N_{L/F}L^{\times}$ if and only if $A_{1}=\varnothing$.

  3. (3)

This is essentially Corollary 3.7.

  4. (4)

For all
$L\in\textnormal{\'{E}t}_{(1^{p})/F}\setminus\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}$,
we have $N_{L/F}L^{\times}=F^{\times}$, so

$\textnormal{\'{E}t}_{(1^{p})/F}^{\mathcal{A}}=\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}}\sqcup\Big{(}\textnormal{\'{E}t}_{(1^{p})/F}\setminus\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}\Big{)},$

and therefore

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{\mathcal{A}})=\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})+\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F})-\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}).$

Serre’s mass formula [Ser78, Theorem 2] tells us that tells us that

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F})=\frac{1}{q^{p-1}},$

and the first part of the statement follows. The rest of the theorem is given
by Corollaries 3.25 and 3.28.

∎

######  Definition 3.29.

Let $\alpha\in F^{\times}$. We define
$c_{\alpha}\in\mathbb{Z}\cup\\{\infty\\}$ as follows:

  * •

If $\alpha\in F^{\times p}$, then $c_{\alpha}=\infty$.

  * •

Otherwise, adopting the convention that $U_{F}^{(-1)}=F^{\times}$, we define
$c_{\alpha}$ to be the largest integer $c$ such that $\alpha\in
U_{F}^{(c)}F^{\times p}$.

######  Remark 3.30.

We can compute $c_{\alpha}$ with Algorithm 3.18. Lemma 3.19 tells us that this
computation has time complexity $O(e_{F}t(F))$, where $t(F)$ is the time taken
to perform elementary operations in $F$ (morally, we think of $O(t(F))$ as
being $O(1)$, but we choose to keep track of it since it depends on $F$).

######  Lemma 3.31.

Let $\alpha\in F^{\times}\setminus F^{\times p}$. Then
$c_{\alpha}\leq\lfloor\frac{pe_{F}}{p-1}\rfloor$. Moreover, if
$c_{\alpha}<\frac{pe_{F}}{p-1}$ then $p\nmid c_{\alpha}$, and if
$c_{\alpha}=\frac{pe_{F}}{p-1}$ then $\mu_{p}\subseteq F$.

###### Proof.

This follows from Corollary 3.20. ∎

######  Lemma 3.32.

Let $\alpha\in F^{\times}$ and let $c$ be an integer with $1\leq
c\leq\frac{pe_{F}}{p-1}+1$. Then we have

  1. (1)

If $c\leq c_{\alpha}$ and $c\leq\frac{pe_{F}}{p-1}+1$, then we have

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\alpha}=\begin{cases}\frac{p(q-1)}{p-1}q^{c-2-\lfloor\frac{c-2}{p}\rfloor}\quad\text{if
$c\not\equiv 1\pmod{p}$},\\\ pq^{e_{F}}\quad\text{if $c=\frac{pe_{F}}{p-1}+1$
and $\mu_{p}\subseteq F$},\\\ 0\quad\text{if $c\equiv 1\pmod{p}$}.\end{cases}$

  2. (2)

If $1\leq c_{\alpha}<\frac{pe_{F}}{p-1}$, then

$\\#\textnormal{\'{E}t}_{p/F,(p-1)(c_{\alpha}+1)}^{C_{p}/F,\alpha}=\frac{q-p}{p-1}q^{c_{\alpha}-1-\lfloor\frac{c_{\alpha}-1}{p}\rfloor}.$

  3. (3)

If $c_{\alpha}=\frac{pe_{F}}{p-1}$, then

$\\#\textnormal{\'{E}t}_{p/F,(p-1)(c_{\alpha}+1)}^{C_{p}/F,\alpha}=0.$

  4. (4)

If $c_{\alpha}+2\leq c\leq\frac{pe_{F}}{p-1}+1$, then

$\\#\textnormal{\'{E}t}_{p/F,(p-1)c}^{C_{p}/F,\alpha}=\begin{cases}\frac{(q-1)}{p-1}q^{c-2-\lfloor\frac{c-2}{p}\rfloor}\quad\text{if
$c\not\equiv 1\pmod{p}$},\\\ q^{e_{F}}\quad\text{if $c=\frac{pe_{F}}{p-1}+1$
and $\mu_{p}\subseteq F$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

This follows easily from Corollary 3.27. ∎

######  Theorem 3.33.

Consider the special case where $\mathcal{A}=\langle\alpha\rangle$ is
generated by one element $\alpha\in F^{\times}\setminus F^{\times p}$. If
$p\nmid v_{F}(\alpha)$, then

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})=\frac{1}{p}\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F}).$

If $p\mid v_{F}(\alpha)$, then
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\mathcal{A}})$ is the
sum of the following seven quantities:

  1. (1)

$\mathbbm{1}_{c_{\alpha}\geq p}\cdot\frac{q-1}{p-1}q^{-2}A(c_{\alpha}).$

  2. (2)

$\mathbbm{1}_{c_{\alpha}\not\equiv
0,1\pmod{p}}\cdot\frac{q-1}{p-1}q^{-2}B(c_{\alpha}).$

  3. (3)

$\mathbbm{1}_{c_{\alpha}<\frac{pe_{F}}{p-1}}\cdot\frac{q-p}{p-1}\cdot\frac{1}{p}\cdot
q^{-2-(p-2)(c_{\alpha}+1)-\lfloor\frac{c_{\alpha}-1}{p}\rfloor}.$

  4. (4)

$\mathbbm{1}_{c_{\alpha}<\frac{pe_{F}}{p-1}}\cdot\frac{q-1}{pq^{2}(p-1)}\cdot\Big{(}\mathbbm{1}_{e_{F}\geq
p-1}\cdot
A\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}-\mathbbm{1}_{c_{\alpha}\geq
p-2}\cdot A(c_{\alpha}+1)\Big{)}.$

  5. (5)

$\mathbbm{1}_{\lceil\frac{pe_{F}}{p-1}\rceil\not\equiv
0,1\pmod{p}}\cdot\frac{q-1}{pq^{2}}\cdot
B\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}.$

  6. (6)

$-\mathbbm{1}_{c_{\alpha}\not\equiv-1,0\pmod{p}}\cdot\frac{q-1}{pq^{2}}\cdot
B(c_{\alpha}+1).$

  7. (7)

$\mathbbm{1}_{\mu_{p}\subseteq F}\cdot\frac{1}{p}\cdot q^{-(p-1)e_{F}-p+1}.$

###### Proof.

Suppose first that $c_{\alpha}<\frac{pe_{F}}{p-1}$. Then the pre-mass is the
sum of the following four quantities:

  1. (1)

$\frac{q-1}{p-1}\cdot q^{-2}\sum_{\begin{subarray}{c}1\leq c\leq c_{\alpha}\\\
c\not\equiv 1\pmod{p}\end{subarray}}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}.$

  2. (2)

$\frac{q-p}{p-1}\cdot\frac{1}{p}\cdot
q^{-2-(p-2)(c_{\alpha}+1)-\lfloor\frac{c_{\alpha}-1}{p}\rfloor}.$

  3. (3)

$\frac{q-1}{p-1}\cdot\frac{1}{p}\cdot
q^{-2}\sum_{\begin{subarray}{c}c_{\alpha}+2\leq c<\frac{pe_{F}}{p-1}+1\\\
c\not\equiv 1\pmod{p}\end{subarray}}q^{-(p-2)c-\lfloor\frac{c-2}{p}\rfloor}.$

  4. (4)

$\mathbbm{1}_{\mu_{p}\subseteq F}\cdot\frac{1}{p}\cdot q^{-(p-1)e_{F}-p+1}.$

The first three rearrange to

  1. (1)

$\frac{q-1}{p-1}\cdot q^{-2}\cdot(\mathbbm{1}_{c_{\alpha}\geq p}\cdot
A(c_{\alpha})+\mathbbm{1}_{c_{\alpha}\not\equiv 0,1\pmod{p}}B(c_{\alpha})).$

  2. (2)

$\frac{q-p}{p-1}\cdot\frac{1}{p}\cdot
q^{-2-(p-2)(c_{\alpha}+1)-\lfloor\frac{c_{\alpha}-1}{p}\rfloor}.$

  3. (3)

$\displaystyle\frac{q-1}{p-1}\cdot\frac{1}{p}\cdot
q^{-2}\cdot\Big{(}\mathbbm{1}_{e_{F}\geq p-1}\cdot
A\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}$
$\displaystyle-\mathbbm{1}_{c_{\alpha}\geq p-1}A(c_{\alpha}+1)\Big{)}$
$\displaystyle+\mathbbm{1}_{\lceil\frac{pe_{F}}{p-1}\rceil\not\equiv
0,1\pmod{p}}B\Big{(}\Big{\lceil}\frac{pe_{F}}{p-1}\Big{\rceil}\Big{)}-\mathbbm{1}_{c_{\alpha}\not\equiv-1,0\pmod{p}}B(c_{\alpha}+1).$

The argument for $c_{\alpha}=\frac{pe_{F}}{p-1}$ is similar. ∎

######  Corollary 3.34.

Let $\alpha$ be as in Lemma 3.33. If $c_{\alpha}=\frac{pe_{F}}{p-1}$, then the
formula simplifies to

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{p})/F}^{C_{p}/F,\alpha})=\frac{q-1}{p-1}q^{-2}A\Big{(}\frac{pe_{F}}{p-1}\Big{)}+\frac{1}{p}q^{-(p-1)e_{F}-p+1}.$

##  4\. $S_{4}$-quartic extensions

Throughout Section 4, let $F$ be a $2$-adic field and let
$\mathcal{A}\subseteq F^{\times}$ be a finitely generated subgroup.

###  4.1. $p$-adic fields for odd primes $p$

The goal of this subsection is to prove the following theorem:

######  Theorem 4.1.

For all $p$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{4/F}^{\mathcal{A}})=\frac{5q^{2}+8q+8}{8q^{2}}+\sum_{\sigma\in\\{(4),(22),(1^{2}1^{2}),(2^{2}),(1^{4})\\}}\widetilde{m}(\textnormal{\'{E}t}_{\sigma/F}^{\mathcal{A}}).$

  1. (1)

For all $p$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{(4)/F}^{\mathcal{A}})=\begin{cases}\frac{1}{4}\quad\text{if
$4\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise},\end{cases}$

and

$\widetilde{m}(\textnormal{\'{E}t}_{(22)/F}^{\mathcal{A}})=\begin{cases}\frac{1}{8}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

  2. (2)

Suppose that $p\neq 2$. Let $\mathcal{A}F^{\times 2}/F^{\times 2}$ have a
minimal generating set $A_{0}\cup A_{1}$, where $v_{F}(\alpha)=r$ for all
$[\alpha]\in A_{r}$, for $r=0,1$. Then we have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 2}$},\\\ \frac{3}{8q^{2}}\quad\text{else if
$A_{0}\subseteq F^{\times 2}$ and $\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times
2}$ for all $\alpha_{i},\alpha_{j}\in A_{1}$},\\\
\frac{1}{4q^{2}}\quad\text{otherwise}.\end{cases}$

  3. (3)

Suppose that $p\neq 2$. Let $\mathcal{A}F^{\times 4}/F^{\times 4}$ have a
minimal generating set $A_{0}\cup A_{1}\cup A_{2}$, where $v_{F}(\alpha)=r$
for all $[\alpha]\in A_{r}$, for $r=0,1,2$. Then we have

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$A_{0}\subseteq F^{\times 2}$ and $A_{1}=A_{2}=0$},\\\
\frac{1}{4q^{2}}\quad\text{if $A_{0}\subseteq F^{\times 2}$ and
$A_{1}=\varnothing$ and $\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 2}$ for
all $\alpha_{i},\alpha_{j}\in A_{2}$},\\\ 0\quad\text{otherwise}.\end{cases}$

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

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 4}$},\\\ \frac{1}{2q^{3}}\quad\text{if
$A_{0}\subseteq F^{\times 4}$ and $A_{1}=\varnothing$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 4}$ whenever
$\alpha_{i},\alpha_{j}\in A_{2}$,}\\\ \frac{1}{4q^{3}}\quad\text{if
$A_{0}\subseteq F^{\times 4}$ and $A_{1}\neq\varnothing$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 4}$ whenever
$\alpha_{i},\alpha_{j}\in A_{1}$ or $\alpha_{i},\alpha_{j}\in A_{2}$},\\\
0\quad\text{otherwise}.\end{cases}$

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

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 2}$},\\\ \frac{1}{2q^{3}}\quad\text{if
$A_{0}\cup A_{2}\subseteq F^{\times 2}$ and $\frac{\alpha_{i}}{\alpha_{j}}\in
F^{\times 2}$ whenever $\alpha_{i},\alpha_{j}\in A_{1}$},\\\
0\quad\text{otherwise}.\end{cases}$

######  Lemma 4.2.

We have

$\mathrm{Split}_{4}^{\mathrm{epi}}=\\{(31),(21^{2}),(211),(1^{2}11),(1^{3}1),(1111)\\}$

and

$\mathrm{Split}_{4}=\\{(4),(22),(1^{2}1^{2}),(2^{2}),(1^{4})\\}\cup\mathrm{Split}_{4}^{\mathrm{epi}}.$

###### Proof.

The eleven possible splitting symbols are listed on [Bha04, Page 1353], and it
is clear that the epimorphic ones are as stated. ∎

######  Lemma 4.3.

For all $p$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{4/F}^{\mathrm{epi}})=\frac{5q^{2}+8q+8}{8q^{2}}.$

###### Proof.

This follows from Lemmas 2.15 and 4.2. ∎

######  Lemma 4.4.

For all $p$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{(4)/F}^{\mathcal{A}})=\begin{cases}\frac{1}{4}\quad\text{if
$4\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise},\end{cases}$

and

$\widetilde{m}(\textnormal{\'{E}t}_{(22)/F}^{\mathcal{A}})=\begin{cases}\frac{1}{8}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

This follows from basic class field theory. ∎

######  Lemma 4.5.

Let $\mathcal{A}F^{\times 2}/F^{\times 2}$ a minimal generating set $A_{0}\cup
A_{1}$, where $v_{F}(\alpha)=r$ for all $[\alpha]\in A_{r}$, for $r=0,1$. Then
we have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 2}$},\\\ \frac{3}{8q^{2}}\quad\text{else if
$A_{0}\subseteq\mathcal{O}_{F}^{\times 2}$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 2}$ for all
$\alpha_{i},\alpha_{j}\in A_{1}$},\\\
\frac{1}{4q^{2}}\quad\text{otherwise}.\end{cases}$

###### Proof.

By Lemma 3.4, we have

$\textnormal{\'{E}t}_{(1^{2}1^{2})/F}=\\{L_{0}\times L_{0},L_{0}\times
L_{1},L_{1}\times L_{1}\\},$

where

$L_{j}=F(\sqrt{\zeta_{q-1}^{j}\pi_{F}}),\quad j=0,1.$

Lemma 2.8 tells us that $v_{F}(d_{L_{j}/F})=1$ for each $j$. It follows that,
for $i,j\in\\{0,1\\}$, we have

$\widetilde{m}(\\{L_{i}\times
L_{j}\\})=\begin{cases}\frac{1}{8q^{2}}\quad\text{if $i=j$},\\\
\frac{1}{4q^{2}}\quad\text{if $i\neq j$}.\end{cases}$

The result then follows from the fact that

$N_{L_{j}/F}L_{j}^{\times}=\langle\mathcal{O}_{F}^{\times
2},-\zeta_{q-1}^{j}\pi_{F}\rangle.$

∎

######  Lemma 4.6.

Let $\mathcal{A}F^{\times 4}/F^{\times 4}$ have generators a minimal
generating set $A_{0}\cup A_{1}\cup A_{2}$, where $v_{F}(\alpha)=r$ for all
$[\alpha]\in A_{r}$, for $r=0,1,2$. Then we have

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{2q^{2}}\quad\text{if
$A_{0}\subseteq\mathcal{O}_{F}^{\times 2}$ and $A_{1}=A_{2}=\varnothing$},\\\
\frac{1}{4q^{2}}\quad\text{else if $A_{0}\subseteq\mathcal{O}_{F}^{\times 2}$
and $A_{1}=\varnothing$ and $\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 2}$
for all $\alpha_{i},\alpha_{j}\in A_{2}$},\\\
0\quad\text{otherwise}.\end{cases}$

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

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 4}$},\\\ \frac{1}{2q^{3}}\quad\text{else if
$A_{0}\subseteq\mathcal{O}_{F}^{\times 4}$ and $A_{1}=\varnothing$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 4}$ whenever
$\alpha_{i},\alpha_{j}\in A_{2}$},\\\ \frac{1}{4q^{3}}\quad\text{else if
$A_{0}\subseteq\mathcal{O}_{F}^{\times 4}$ and
$\frac{\alpha_{i}}{\alpha_{j}}\in F^{\times 4}$ whenever
$\alpha_{i},\alpha_{j}\in A_{1}$ or $\alpha_{i},\alpha_{j}\in A_{2}$},\\\
0\quad\text{otherwise}.\end{cases}$

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

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{\mathcal{A}})=\begin{cases}\frac{1}{q^{3}}\quad\text{if
$\mathcal{A}\subseteq F^{\times 2}$},\\\ \frac{1}{2q^{3}}\quad\text{else if
$A_{0}\cup A_{2}\subseteq F^{\times 2}$ and $\frac{\alpha_{i}}{\alpha_{j}}\in
F^{\times 2}$ whenever $\alpha_{i},\alpha_{j}\in A_{1}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Let $E/F$ be the quadratic unramified extension. Lemma 3.4 tells us that

$\textnormal{\'{E}t}_{(2^{2})/F}=\\{L_{0},L_{1}\\},$

where

$L_{j}=E\Big{(}\sqrt{\zeta_{q^{2}-1}^{j}\pi_{F}}\Big{)},$

for each $j$. On the other hand, we have two index $4$ subgroups

$\langle\mathcal{O}_{F}^{\times
2},\pi_{F}^{2}\rangle,\quad\langle\mathcal{O}_{F}^{\times
2},\zeta_{q-1}\pi_{F}^{2}\rangle$

of $F^{\times}$. Clearly the quartic abelian extensions corresponding to these
subgroups have splitting symbol $(2^{2})$, so they must be equal to $L_{0}$
and $L_{1}$ in some order, which implies that $L_{0}$ and $L_{1}$ are abelian
extensions of $F$. Lemma 2.8 tells us that $v_{F}(d_{L_{j}/F})=2$ for each
$j$, so $\widetilde{m}(\\{L_{j}\\})=\frac{1}{4q^{2}}$, and the formula for
$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{\mathcal{A}})$ follows.

Lemma 3.4 tells us that

$\textnormal{\'{E}t}_{(1^{4})/F}=\begin{cases}\\{L_{0},L_{1},L_{2},L_{3}\\}\quad\text{if
$q\equiv 1\pmod{4}$},\\\ \\{L_{0},L_{1}\\}\quad\text{if $q\equiv
3\pmod{4}$},\end{cases}$

where

$L_{j}=F\Big{(}\sqrt[4]{\zeta_{q-1}^{j}\pi_{F}}\Big{)}.$

Suppose first that $q\equiv 1\pmod{4}$. Then the minimal polynomial of
$\sqrt[4]{\pi_{F}}$ over $F$ splits in $L_{0}$, so $L_{0}/F$ is Galois. Since
$\pi_{F}$ is an arbitrary choice of uniformiser, it follows that $L_{j}/F$ is
Galois for each $j$, and therefore

$N_{L_{j}/F}L_{j}^{\times}=\langle\mathcal{O}_{F}^{\times
4},-\zeta_{q-1}^{j}\pi_{F}\rangle.$

Lemma 2.8 tells us that $\widetilde{m}(\\{L_{j}\\})=\frac{1}{4q^{3}}$ for each
$j$, and the result follows. Finally, suppose that $q\equiv 3\pmod{4}$. Then
$\mu_{4}\not\subseteq F$, so $\sqrt[4]{\pi_{F}}$ has only two conjugates in
$L_{0}$, and therefore $L_{0}$ is non-Galois with a maximal abelian
subextension $F(\sqrt{\pi_{F}})$, so

$N_{L_{0}/F}L_{0}^{\times}=\langle\mathcal{O}_{F}^{\times 2},-\pi_{F}\rangle,$

and similarly $L_{1}/F$ is non-Galois and

$N_{L_{1}/F}L_{1}^{\times}=\langle\mathcal{O}_{F}^{\times
2},-\zeta_{q-1}\pi_{F}\rangle.$

By Lemma 2.8, we have $v_{F}(d_{L_{j}/F})=3$ for each $j$, and the result
follows. ∎

###### Proof of Theorem 4.1.

This is immediate from Lemmas 4.2 to 4.6 inclusive. ∎

###  4.2. $2$-adic fields

Let $F$ be a $2$-adic field. By Theorem 4.1, in order to compute
$\widetilde{m}(\textnormal{\'{E}t}_{4/F}^{\mathcal{A}})$, all we need is to
compute $\widetilde{m}(\textnormal{\'{E}t}_{\sigma/F}^{\mathcal{A}})$ for each
$\sigma\in\\{(1^{2}1^{2}),(2^{2}),(1^{4})\\}$. We state the first two of these
quantities in the next two theorems:

######  Theorem 4.7.

The pre-mass
$\widetilde{m}(\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}})$ is equal
to the sum of the following quantities:

  1. (1)

$\frac{1}{8}\cdot\sum_{4\leq m\leq
4e_{F}+2}q^{-m}\\#\textnormal{\'{E}t}_{(1^{2})/F,\frac{m}{2}}^{\mathcal{A}}.$

  2. (2)

$\frac{1}{4}\cdot\sum_{4\leq m\leq 4e_{F}+2}q^{-m}N_{(1^{2}1^{2})}^{\neq}(m).$

###### Proof.

Postponed. ∎

######  Theorem 4.8.

If $2\nmid v_{F}(\alpha)$ for some $\alpha\in\mathcal{A}$, then
$\textnormal{\'{E}t}_{(2^{2})/F}^{\mathcal{A}}=\varnothing$. Assume that
$2\mid v_{F}(\alpha)$. Then the following four statements are true:

  1. (1)

For $G\in\\{S_{4},A_{4}\\}$, we have
$\textnormal{\'{E}t}_{(2^{2})/F}^{G/F}=\varnothing$, so

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{G/F,\mathcal{A}})=0.$

  2. (2)

We have

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F,\mathcal{A}})=\frac{1}{2}\cdot\Big{(}q^{-2}-q^{-2e_{F}-2}-\frac{1}{q^{2}+q+1}\Big{(}q^{-1}-q^{-3e_{F}-1}\Big{)}+q^{-3e_{F}-2}(q^{e_{F}}-1)\Big{)}.$

  3. (3)

We have

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{V_{4}/F,\mathcal{A}})=\frac{1}{4\\#\overline{\mathcal{A}}^{2}}\Big{(}\mathbbm{1}_{\overline{\mathcal{A}}^{2}_{2e_{F}}=0}\cdot
q^{-3e_{F}-2}+\sum_{c=1}^{e_{F}}q^{-3c-1}(q\cdot\\#\overline{\mathcal{A}}^{2}_{2c}-\\#\overline{\mathcal{A}}^{2}_{2c-1})\Big{)}.$

  4. (4)

For $\alpha\in F^{\times}$, write $N_{\alpha}$ for the subgroup
$\Big{(}N_{F(\sqrt{\alpha})/F}F(\sqrt{\alpha})^{\times}\Big{)}/F^{\times
2}\subseteq F^{\times}/F^{\times 2}$. Define

$N^{\mathcal{A}}=\Big{(}\bigcap_{\begin{subarray}{c}[\alpha]\in\overline{\mathcal{A}}^{4}\\\
v_{F}(\alpha)\equiv
0\pmod{4}\end{subarray}}N_{\alpha}\Big{)}\setminus\Big{(}\bigcup_{\begin{subarray}{c}[\alpha]\in\overline{\mathcal{A}}^{4}\\\
v_{F}(\alpha)\equiv 2\pmod{4}\end{subarray}}N_{\alpha}\Big{)}.$

For each integer $m$ with $0\leq m\leq 2e_{F}$, write

$N^{\mathcal{A}}_{c}=N^{\mathcal{A}}\cap\Big{(}U_{F}^{(2e_{F}-c)}F^{\times
2}/F^{\times 2}\Big{)}.$

Then

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{C_{4}/F,\mathcal{A}})=\begin{cases}\frac{1}{8}\Big{(}\sum_{c=1}^{e_{F}}q^{-4c}(\\#N_{2c}^{\mathcal{A}}-\\#N_{2c-1}^{\mathcal{A}})+q^{-4e_{F}-2}(\\#N^{\mathcal{A}}-N_{2e_{F}}^{\mathcal{A}})\Big{)}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Postponed. ∎

For $\sigma=(1^{4})$, the actual mass is too cumbersome to state in closed
form, so we instead study the quantities
$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{G/F,\mathcal{A}}$ for each $G$. For
$G\neq C_{4}$, we state these quantities explicitly. For $G=C_{4}$, we give a
formula for $\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}}$,
whenever $E$ is a totally ramified quadratic extension of $F$. Using this
formula, we give an algorithm for computing the quantities
$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{C_{4}/F,\mathcal{A}}$.

######  Theorem 4.9.

The following four statements are true:

  1. (1)

For $G\in\\{S_{4},A_{4}\\}$, we have

$\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{G/F,\mathcal{A}})=\widetilde{m}(\textnormal{\'{E}t}_{(1^{4})/F}^{G/F}),$

which is known from [Mon24, Corollaries 1.6 and 1.7].

  2. (2)

We have

$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}=\frac{1}{2}\sum_{0<m_{1}<m/2}\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}\cdot\Big{(}N^{C_{2}}(m_{1})-N^{C_{4}}(m_{1},m_{2})-N^{V_{4}}(m_{1},m_{2})\Big{)},$

which we can compute explicitly using Lemma 3.26.

  3. (3)

The quantity $\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}}$ is
the sum of the following two quantities:

    1. (a)

$\frac{1}{2}\cdot\sum_{\begin{subarray}{c}m_{1}<m_{2}\\\
m_{1}+2m_{2}=m\end{subarray}}(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}})(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}^{\mathcal{A}}).$

    2. (b)

$\mathbbm{1}_{3\mid m}\cdot\frac{1}{3(\\#\overline{\mathcal{A}}^{2})^{2}}\cdot
q^{\frac{m}{6}-4}(q\\#\overline{\mathcal{A}}^{2}_{m/3}-\\#\overline{\mathcal{A}}^{2}_{m/3-1})(q\\#\overline{\mathcal{A}}^{2}_{m/3}-2\\#\overline{\mathcal{A}}^{2}_{m/3-1}).$

This expression is explicit by Lemma 3.26.

  4. (4)

Let $E\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}$ and set
$m_{1}=v_{F}(d_{E/F})$. If $m_{1}\leq e_{F}$, then let $\omega\in E^{\times}$
be such that
$E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E,3m_{1}-2}^{C_{4}/F}$. If
$m_{1}>e_{F}$, then let $\omega\in E^{\times}$ be such that
$E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E}^{C_{4}/F}$. Define

$N_{\omega}=\Big{(}N_{E(\sqrt{\omega})/F}E(\sqrt{\omega})^{\times}\Big{)}/F^{\times
4}\subseteq F^{\times}/F^{\times 4}.$

For $\alpha\in F^{\times}$, write

$N_{\alpha}=\Big{(}N_{F(\sqrt{\alpha})/F}F(\sqrt{\alpha})^{\times}\Big{)}/F^{\times
2}\subseteq F^{\times}/F^{\times 2}.$

Define

$N_{\omega}^{\mathcal{A}}=\Big{(}\bigcap_{[\alpha]\in\overline{\mathcal{A}}^{4}\cap
N_{\omega}}N_{\alpha}\Big{)}\setminus\Big{(}\bigcup_{[\alpha]\in\overline{\mathcal{A}}^{4}\setminus
N_{\omega}}N_{\alpha}\Big{)}$

and

$N^{\mathcal{A}}_{\omega,c}=N_{\omega}^{\mathcal{A}}\cap\Big{(}U_{F}^{(2e_{F}-c)}F^{\times
2}/F^{\times 2}\Big{)},$

for each integer $c$ with $0\leq c\leq 2e_{F}$. If $m_{1}>e_{F}$, then

$\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}}=\begin{cases}\frac{1}{2}\cdot\\#N_{\omega}^{\mathcal{A}}\quad\text{if
$m_{2}=m_{1}+2e_{F}$},\\\ 0\quad\text{otherwise}.\end{cases}$

Suppose instead that $m_{1}\leq e_{F}$. For even integers $m_{2}$, define

$c(m_{2})=2\Big{\lfloor}\frac{m_{1}+2\lfloor\frac{m_{2}}{2}\rfloor}{4}\Big{\rfloor}.$

Then we have

$\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}}=\begin{cases}\frac{1}{2}\cdot\\#N_{\omega,c(m_{2})}^{\mathcal{A}}\quad\text{if
$m_{2}=3m_{1}-2$},\\\
\frac{1}{2}\cdot\Big{(}\\#N_{\omega,c(m_{2})}^{\mathcal{A}}-\\#N_{\omega,c(m_{2}-2)}^{\mathcal{A}}\Big{)}\quad\text{if
$3m_{1}\leq m_{2}\leq 4e_{F}-m_{1}+1$},\\\
\frac{1}{2}\cdot\Big{(}\\#N^{\mathcal{A}}_{\omega}-\\#N^{\mathcal{A}}_{\omega,c(m_{2}-2)}\Big{)}\quad\text{if
$m_{2}=4e_{F}-m_{1}+2$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

Postponed. ∎

######  Lemma 4.10.

We have

$\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}}=\\{L_{1}\times
L_{2}\in\textnormal{\'{E}t}_{(1^{2}1^{2})/F}:L_{1}\not\cong
L_{2}\\}\cup\\{L_{1}\times
L_{1}:L_{1}\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}\\}.$

###### Proof.

It is clear that the left-hand side is contained inside the right-hand side,
and also that

$\\{L_{1}\times
L_{1}:L_{1}\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}\\}\subseteq\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}}.$

Finally, for distinct elements
$L_{1},L_{2}\in\textnormal{\'{E}t}_{(1^{2})/F}$, the norm groups
$N_{L_{i}/F}L_{i}^{\times}$ are distinct index $2$ subgroups of $F^{\times}$,
and therefore $\operatorname{Nm}(L_{1}\times L_{2})=F^{\times}$, so
$L_{1}\times L_{2}\in\textnormal{\'{E}t}_{(1^{2}1^{2})/F}^{\mathcal{A}}$. ∎

######  Lemma 4.11.

Let $m$ be an integer. We have

$\\#\\{L_{1}\times
L_{2}\in\textnormal{\'{E}t}_{(1^{2}1^{2})/F,m}:L_{1}\not\cong
L_{2}\\}=N_{(1^{2}1^{2})}^{\neq}(m)$

###### Proof.

Theorem 3.23 tells us that, for all $m_{1}$, we have

$\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}=\begin{cases}2(q-1)q^{\frac{m_{1}}{2}-1}\quad\text{if
$m_{1}$ is even with $2\leq m_{1}\leq 2e_{F}$},\\\ 2q^{e_{F}}\quad\text{if
$m_{1}=2e_{F}+1$},\\\ 0\quad\text{otherwise}.\end{cases}$

We will use this fact without reference for the rest of this proof.

For any $m$, the number we are looking for is equal to

(2)
$\frac{1}{2}\cdot\Big{(}\sum_{\begin{subarray}{c}m_{1}+m_{2}=m\end{subarray}}\Big{(}\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}\cdot\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}\Big{)}-\\#\textnormal{\'{E}t}_{(1^{2})/F,m/2}\Big{)}.$

It is easy to see this is $0$ unless one of the following is true:

  * •

$2\leq m\leq 4e_{F}$ and $m$ is a multiple of $4$.

  * •

$2e_{F}+3\leq m\leq 4e_{F}+1$ and $m$ is odd.

  * •

$m=4e_{F}+2$.

The result follows by considering these cases separately. ∎

###### Proof of Theorem 4.7.

This is immediate from Lemmas 4.10 and 4.11. ∎

######  Lemma 4.12.

We have

$\\#\textnormal{\'{E}t}_{(2^{2})/F,m}^{V_{4}/F,\mathcal{A}}=\begin{cases}\frac{1}{2}\\#\textnormal{\'{E}t}_{(1^{2})/F,m/2}^{\mathcal{A}}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Write $E_{\mathrm{ur}}$ for the unramified quadratic extension of $F$. Let
$L\in\textnormal{\'{E}t}_{(2^{2})/F,m}^{V_{4}/F}$. Then there are exactly two
elements $E\in\textnormal{\'{E}t}_{(1^{2})/F,m/2}$ such that
$L=E_{\mathrm{ur}}E$. It follows that

$\operatorname{Nm}L=\operatorname{Nm}E_{ur}\cap\operatorname{Nm}E=\\{x\in\operatorname{Nm}E:2\mid
v_{F}(x)\\}.$

Therefore, if there is some $x\in\mathcal{A}$ with $2\nmid v_{F}(x)$, then
$\textnormal{\'{E}t}_{(2^{2})/F}^{V_{4}/F,\mathcal{A}}=\varnothing$. On the
other hand, if $2\mid v_{F}(x)$ for all $x\in\mathcal{A}$, then we have a
$2$-to-$1$ surjection

$\textnormal{\'{E}t}_{(1^{2})/F,m/2}^{\mathcal{A}}\to\textnormal{\'{E}t}_{(2^{2})/F,m}^{V_{4}/F,\mathcal{A}},\quad
E\mapsto E_{\mathrm{ur}}E,$

and the result follows. ∎

######  Lemma 4.13.

We have

$\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F,\mathcal{A}}=\begin{cases}\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
\varnothing\quad\text{otherwise}.\end{cases}$

###### Proof.

Let $E/F$ be the unramified quadratic extension. By class field theory, every
$L\in\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F}$ has
$N_{L/F}L^{\times}=N_{E/F}E^{\times}$. The result follows. ∎

######  Lemma 4.14.

Let $E/F$ be the unique unramified quadratic extension. Then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,m}^{D_{4}/F}=\frac{1}{2}\cdot\Big{(}\\#\textnormal{\'{E}t}_{(1^{2})/E,m/2}-\\#\textnormal{\'{E}t}_{(1^{2})/E,m/2}^{C_{4}/F}-\\#\textnormal{\'{E}t}_{(1^{2})/E,m/2}^{V_{4}/F}\Big{)}.$

###### Proof.

It is easy to see that there is a well-defined surjection

$\textnormal{\'{E}t}_{(1^{2})/E,\frac{m}{2}}\setminus\Big{(}\textnormal{\'{E}t}_{(1^{2})/E,\frac{m}{2}}^{C_{4}/F}\cup\textnormal{\'{E}t}_{(1^{2})/E,\frac{m}{2}}^{V_{4}/F}\Big{)}\to\textnormal{\'{E}t}_{(2^{2})/F,m}^{D_{4}/F}.$

Moreover, [Mon24, Lemma 5.1, Part (2)] tells us that this surjection is
$2$-to-$1$. ∎

######  Lemma 4.15.

Let $E\in\textnormal{\'{E}t}_{2/F}$ and let $m_{2}$ be an integer such that
$\textnormal{\'{E}t}_{2/E,\leq m_{2}}^{C_{4}/F}$ is nonempty. Let $\omega\in
E^{\times}$ be such that $E(\sqrt{\omega})\in\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F}$. If $m_{2}\leq 2e_{E}$, then we have a $2$-to-$1$ surjection

$\Big{(}U_{E}^{(2e_{E}-2\lfloor\frac{m_{2}}{2}\rfloor)}E^{\times 2}\cap
F^{\times}\Big{)}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F},\quad t\mapsto E(\sqrt{\omega t}).$

If $m_{2}>2e_{E}$, then we have a $2$-to-$1$ surjection

$F^{\times}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F},\quad t\mapsto E(\sqrt{\omega t}).$

###### Proof.

By [CDO05, Proposition 1.2], there is a $2$-to-$1$ surjection

$F^{\times}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E}^{C_{4}/F},\quad t\mapsto
E(\sqrt{t\omega}),$

and the result follows from [Mon24, Lemma 4.6]. ∎

######  Lemma 4.16.

Let $E/F$ be the unique unramified quadratic extension. For all $m$, we ave

$\\#\textnormal{\'{E}t}_{(1^{2})/E,m}^{C_{4}/F}=\frac{1}{2}\\#\textnormal{\'{E}t}_{(1^{2})/F,m}$

###### Proof.

If $\textnormal{\'{E}t}_{2/E,m}^{C_{4}/F}\neq\varnothing$, then either $m$ is
even with $0\leq m\leq 2e_{F}$, or $m=2e_{F}+1$. Suppose that $m$ is an even
integer with $0\leq m\leq 2e_{F}$. Let $\omega\in E^{\times}$ be such that
$E(\sqrt{\omega})/E$ is unramified, hence a $C_{4}$-extension of $F$. By111The
statement in [CDO05] is for number fields modulo prime ideals, but exactly the
same proof works for $p$-adic fields. [CDO05, Proposition 3.6], we have

$U_{E}^{(2e_{E}-m)}E^{\times 2}\cap F^{\times}=U_{F}^{(2e_{F}-m)}F^{\times
2},$

so Lemma 4.15 gives a $2$-to-$1$ surjection

$U_{F}^{(2e_{F}-m)}F^{\times 2}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E,\leq
m}^{C_{4}/F},\quad t\mapsto E(\sqrt{\omega t}).$

But [Mon24, Lemma 4.6] tells us that, for $u\in F^{\times}/F^{\times 2}$, we
have

$v_{F}(d_{F(\sqrt{u})/F})\leq m\iff u\in U_{F}^{(2e_{F}-m)}F^{\times 2},$

so

$\\#\textnormal{\'{E}t}_{2/F,\leq m}=\\#(U_{F}^{(2e_{F}-m)}F^{\times
2}/F^{\times 2})-1,$

so

(3) $\\#\textnormal{\'{E}t}_{2/E,\leq
m}^{C_{4}/F}=\frac{1}{2}\\#\textnormal{\'{E}t}_{2/F,\leq m}+\frac{1}{2}.$

The result for $2\leq m\leq 2e_{F}$ follows from Equation 3. Similarly, we
have a $2$-to-$1$ surjection

$\\{x\in F^{\times}/F^{\times
2}:v_{F}(x)=1\\}\to\textnormal{\'{E}t}_{(1^{2})/E,2e_{F}+1}^{C_{4}/F},\quad
x\mapsto E(\sqrt{\omega x}),$

so the result for $m=2e_{F}+1$ follows from the fact that
$\\#(F^{\times}/F^{\times 2})=4q^{e_{F}}$, by [Neu13, Proposition 3.7]. ∎

######  Lemma 4.17.

If $\textnormal{\'{E}t}_{(2^{2})/F,m}^{D_{4}/F,\mathcal{A}}$ is nonempty, then
either $m$ is a multiple of $4$ with $4\leq m\leq 4e_{F}$, or $m=4e_{F}+2$. If
$m$ is a multiple of $4$ with $4\leq m\leq 4e_{F}$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,m}^{D_{4}/F,\mathcal{A}}=\begin{cases}(q-1)\Big{(}(q+1)q^{\frac{m}{2}-2}-q^{\frac{m}{4}-1}\Big{)}\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

If $m=4e_{F}+2$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,m}^{D_{4}/F,\mathcal{A}}=\begin{cases}q^{e_{F}}(q^{e_{F}}-1)\quad\text{if
$2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Lemma 4.12 tells us $\\#\textnormal{\'{E}t}_{(2^{2})/E,m/2}$ by the tower law
for discriminant and [Mon24, Lemma 5.1, Part (2)]. The result then follows
easily from Theorem 3.23 and Lemmas 4.13, 4.14, and 4.16. ∎

######  Lemma 4.18.

If $2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$, then

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F,\mathcal{A}})=\frac{1}{2}\cdot\Big{(}q^{-2}-q^{-2e_{F}-2}-\frac{1}{q^{2}+q+1}\Big{(}q^{-1}-q^{-3e_{F}-1}\Big{)}+q^{-3e_{F}-2}(q^{e_{F}}-1)\Big{)}.$

###### Proof.

Assume that $2\mid v_{F}(\alpha)$ for all $\alpha\in\mathcal{A}$. Then

$\widetilde{m}(\textnormal{\'{E}t}_{(2^{2})/F}^{D_{4}/F,\mathcal{A}})=\frac{1}{2}\cdot\Big{(}(q^{2}-1)q^{-2}\sum_{k=1}^{e_{F}}q^{-2k}-(q-1)q^{-1}\sum_{k=1}^{e_{F}}q^{-3k}+q^{-3e_{F}-2}(q^{e_{F}}-1)\Big{)}.$

For $a>0$, we have

$\sum_{k=1}^{e_{F}}q^{-ak}=q^{-ae_{F}}\cdot\frac{q^{ae_{F}}-1}{q^{a}-1},$

and the result follows. ∎

######  Lemma 4.19.

Let $M/L/K$ be a tower of quadratic extensions of $p$-adic fields. Suppose
that $M/K$ is Galois and let $\beta\in L^{\times}$. Let $\alpha=N_{L/K}\beta$.
Then

$\alpha\in N_{M/K}M^{\times}\iff\beta\in N_{M/L}M^{\times}.$

###### Proof.

The $(\Leftarrow)$ direction is obvious. For $(\Rightarrow)$, suppose that
$\alpha\in N_{M/K}M^{\times}$. Then $\alpha=N_{L/K}\theta$ for some $\theta\in
N_{M/L}M^{\times}$. Since $N_{L/K}(\theta/\beta)=1$, Hilbert’s theorem 90
tells us that $\theta=\beta\frac{x}{\bar{x}}$ for some $x\in L^{\times}$.
Writing $M=L(\sqrt{d})$ for $d\in L^{\times}$, we have

$\displaystyle(\beta,d)_{L}$
$\displaystyle=(d,\theta)_{L}(d,x)_{L}(d,\bar{x})_{L}$
$\displaystyle=(d,x)_{L}(\bar{d},x)_{L}$ $\displaystyle=1,$

where the final equality comes from the fact that $M/K$ is Galois, so
$L(\sqrt{d})=L(\sqrt{\bar{d}})=M$. ∎

######  Lemma 4.20.

Let $E$ be the unramified quadratic extension of $F$, and let $[\omega]\in
E^{\times}/E^{\times 2}$ be the unique element such that $E(\sqrt{\omega})$ is
unramified and quadratic. Let $\alpha\in F^{\times}$ and $u\in
F^{\times}/F^{\times 2}$. The following three statements are true:

  1. (1)

If $v_{F}(\alpha)$ is odd, then
$\alpha\not\in\operatorname{Nm}(E(\sqrt{u\omega}))$.

  2. (2)

If $v_{F}(\alpha)\equiv 2\pmod{4}$, then
$\alpha\in\operatorname{Nm}(E(\sqrt{u\omega}))$ if and only if
$u\not\in\operatorname{Nm}F(\sqrt{\alpha})$.

  3. (3)

If $v_{F}(\alpha)\equiv 0\pmod{4}$, then
$\alpha\in\operatorname{Nm}(E(\sqrt{u\omega}))$ if and only if
$u\in\operatorname{Nm}F(\sqrt{\alpha})$.

###### Proof.

Write $L=E(\sqrt{u\omega})$. If $v_{F}(\alpha)$ is odd, then clearly
$\alpha\not\in N_{L/F}L^{\times}$, so assume that $v_{F}(\alpha)$ is even.
Then $\alpha=N_{E/F}\beta$ for some $\beta\in E^{\times}$. By Lemma 4.19, we
have

$\displaystyle\alpha\in N_{L/F}L^{\times}$ $\displaystyle\iff\beta\in
N_{L/E}L^{\times}$ $\displaystyle\iff(\beta,u\omega)_{E}=1$
$\displaystyle\iff(\beta,u)_{E}=(\beta,\omega)_{E}$
$\displaystyle\iff(\alpha,u)_{F}=\begin{cases}1\quad\text{if
$v_{E}(\beta)\equiv 0\pmod{2}$},\\\ -1\quad\text{if $v_{E}(\beta)\equiv
1\pmod{2}$},\end{cases}$
$\displaystyle\iff(\alpha,u)_{F}=\begin{cases}1\quad\text{if
$v_{F}(\alpha)\equiv 0\pmod{4}$},\\\ -1\quad\text{if $v_{F}(\alpha)\equiv
2\pmod{4}$},\end{cases}$

where the final equivalence comes from the fact that
$v_{F}(\alpha)=2v_{E}(\beta)$. ∎

For $\alpha\in F^{\times}$, write $N_{\alpha}$ for the subgroup

$N_{\alpha}=\Big{(}N_{F(\sqrt{\alpha})/F}F(\sqrt{\alpha})^{\times}\Big{)}/F^{\times
2}.$

######  Lemma 4.21.

Let $m$ be a nonnegative integer. Let

$N^{\mathcal{A}}=\Big{(}\bigcap_{\begin{subarray}{c}[\alpha]\in\overline{\mathcal{A}}^{4}\\\
v_{F}(\alpha)\equiv
0\pmod{4}\end{subarray}}N_{\alpha}\Big{)}\setminus\Big{(}\bigcup_{\begin{subarray}{c}[\alpha]\in\overline{\mathcal{A}}^{4}\\\
v_{F}(\alpha)\equiv 2\pmod{4}\end{subarray}}N_{\alpha}\Big{)}\subseteq
F^{\times}/F^{\times 2},$

and define

$N^{\mathcal{A}}_{c}=N^{\mathcal{A}}\cap(U_{F}^{(2e_{F}-c)}F^{\times
2}/F^{\times 2}).$

If $m\leq 4e_{F}+1$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,\leq
m}^{C_{4}/F,\mathcal{A}}=\begin{cases}\frac{1}{2}\cdot\\#N^{\mathcal{A}}_{2\lfloor{\frac{m}{4}\rfloor}}-1\quad\text{if
$v_{F}(\alpha)\equiv 0\pmod{4}$ for all
$[\alpha]\in\overline{\mathcal{A}}^{4}$},\\\
\frac{1}{2}\cdot\\#N^{\mathcal{A}}_{2\lfloor{\frac{m}{4}\rfloor}}\quad\text{else
if $2\mid v_{F}(\alpha)$ for all $[\alpha]\in\overline{\mathcal{A}}^{4}$},\\\
0\quad\text{otherwise}.\end{cases}$

If $m\geq 4e_{F}+2$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,\leq
m}^{C_{4}/F,\mathcal{A}}=\begin{cases}\frac{1}{2}\cdot\\#N^{\mathcal{A}}-1\quad\text{if
$v_{F}(\alpha)\equiv 0\pmod{4}$ for all
$[\alpha]\in\overline{\mathcal{A}}^{4}$},\\\
\frac{1}{2}\cdot\\#N^{\mathcal{A}}\quad\text{else if $2\mid v_{F}(\alpha)$ for
all $[\alpha]\in\overline{\mathcal{A}}^{4}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

Clearly, if $2\nmid v_{F}(\alpha)$ for any $\alpha\in\mathcal{A}$, then
$\textnormal{\'{E}t}_{(2^{2})/F,\leq m}^{C_{4}/F,\mathcal{A}}=\varnothing$.
Therefore, we will assume that $2\mid v_{F}(\alpha)$ for all
$\alpha\in\mathcal{A}$. By [Mon24, Lemma 5.1], the natural map

$\textnormal{\'{E}t}_{(1^{2})/E,\leq\lfloor\frac{m}{2}\rfloor}^{C_{4}/F}\to\textnormal{\'{E}t}_{(2^{2})/F,\leq
m}^{C_{4}/F}$

is a bijection. Let $m_{2}=\lfloor\frac{m}{2}\rfloor$. Suppose that $m\leq
4e_{F}+1$. Then Lemmas 4.15 and 4.20, along with [CDO05, Proposition 3.6],
give us a $2$-to-$1$ surjection

$N^{\mathcal{A}}_{2\lfloor\frac{m_{2}}{2}\rfloor}\to\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F,\mathcal{A}},\quad t\mapsto E(\sqrt{\omega t}),$

where $E(\sqrt{\omega})/E$ is unramified and quadratic. The result follows,
since $E(\sqrt{\omega})\in\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F,\mathcal{A}}$ if and only if $v_{F}(\alpha)\equiv 0\pmod{4}$
for all $[\alpha]\in\overline{\mathcal{A}}^{4}$. The argument for $m\geq
4e_{F}+2$ is similar but easier, so we omit it. ∎

###### Proof of Theorem 4.8.

We address the statements one by one.

  1. (1)

An $S_{4}$\- or $A_{4}$-quartic extension has no intermediate fields, so it
cannot have splitting symbol $(2^{2})$, since that would imply an unramified
quadratic intermediate field.

  2. (2)

This is precisely Lemma 4.18.

  3. (3)

Let $E^{\mathrm{ur}}$ be the unramified quadratic extensions of $F$. For each
$m$, there is a $2$-to-$1$ surjection

$\textnormal{\'{E}t}_{(1^{2})/F,\frac{m}{2}}^{\mathcal{A}}\to\textnormal{\'{E}t}_{(2^{2})/F,m}^{V_{4}/F,\mathcal{A}},\quad
E\mapsto EE^{\mathrm{ur}},$

and the result follows from Corollary 3.27.

  4. (4)

This follows from Lemma 4.21.

∎

######  Lemma 4.22.

Let $\alpha\in F^{\times}\setminus F^{\times 2}$ and
$d_{\alpha}=v_{F}(d_{F(\sqrt{\alpha})/F})$. Let $c$ be a nonnegative integer.
We have

$\\#N_{c}^{\mathcal{A}}=\begin{cases}q^{\lfloor\frac{c+1}{2}\rfloor}\quad\text{if
$c>2e_{F}-d_{\alpha}$},\\\ 2q^{\lfloor\frac{c+1}{2}\rfloor}\quad\text{if
$c\leq 2e_{F}-d_{\alpha}$ and $v_{F}(\alpha)\equiv 0\pmod{4}$},\\\
0\quad\text{otherwise}.\end{cases}$

###### Proof.

By Lemma 3.10, we have $U_{F}^{(2e_{F}-c)}F^{\times 2}/F^{\times 2}\subseteq
N_{\alpha}$ if and only if $c\leq 2e_{F}-d_{\alpha}$. By elementary linear
algebra, it follows that

$\\#N_{c}^{\mathcal{A}}=\begin{cases}\frac{1}{2}\cdot\\#\Big{(}U_{F}^{(2e_{F}-c)}F^{\times
2}/F^{\times 2}\Big{)}\quad\text{if $c>2e_{F}-d_{\alpha}$},\\\
\\#\Big{(}U_{F}^{(2e_{F}-c)}F^{\times 2}/F^{\times 2}\Big{)}\quad\text{if
$c\leq 2e_{F}-d_{\alpha}$ and $v_{F}(\alpha)\equiv 0\pmod{4}$},\\\
0\quad\text{otherwise}.\end{cases}$

The result follows by Corollary 3.21. ∎

######  Corollary 4.23.

Let $\alpha\in F^{\times}\setminus F^{\times 2}$, and let $m$ be an integer.
If $\textnormal{\'{E}t}_{(2^{2})/F,m}^{C_{4}/F,\langle\alpha\rangle}$ is
nonempty, $v_{F}(\alpha)$ is even and either $m=4e_{F}+2$, or $4\leq m\leq
4e_{F}$ and $m$ is a multiple of $4$. If $v_{F}(\alpha)$ is even and $m$ is a
multiple of $4$ with $4\leq m\leq 4e_{F}$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,m}^{C_{4}/F,\langle\alpha\rangle}=\begin{cases}\frac{1}{2}q^{\frac{m}{4}-1}(q-1)\quad\text{if
$m>4e_{F}-2d_{\alpha}+4$},\\\ \frac{1}{2}q^{\frac{m}{4}-1}(q-2)\quad\text{if
$m=4e_{F}-2d_{\alpha}+4$},\\\ q^{\frac{m}{4}-1}(q-1)\quad\text{if
$m<4e_{F}-2d_{\alpha}+4$}.\end{cases}$

If $v_{F}(\alpha)$ is even and $m=4e_{F}+2$, then

$\\#\textnormal{\'{E}t}_{(2^{2})/F,4e_{F}+2}^{C_{4}/F,\langle\alpha\rangle}=\begin{cases}\frac{1}{2}q^{e_{F}}\quad\text{if
$d_{\alpha}>0$},\\\ q^{e_{F}}\quad\text{if $v_{F}(\alpha)\equiv 2\pmod{4}$ and
$d_{\alpha}=0$},\\\ 0\quad\text{otherwise.}\end{cases}$

###### Proof.

This follows from Lemmas 4.21 and 4.22. ∎

######  Lemma 4.24.

For $G\in\\{S_{4},A_{4}\\}$, we have

$\textnormal{\'{E}t}_{(1^{4})/F,m}^{G/F,\mathcal{A}}=\textnormal{\'{E}t}_{(1^{4})/F,m}^{G/F}.$

###### Proof.

Let $L\in\textnormal{\'{E}t}_{4/F}^{G/F}$, for $G\in\\{S_{4},A_{4}\\}$. Then
$L/F$ has no intermediate fields, so class field theory tells us that
$N_{L/F}L^{\times}=F^{\times}$. ∎

######  Lemma 4.25.

There is a natural $2$-to-$1$ surjection

$\bigsqcup_{\begin{subarray}{c}2m_{1}+m_{2}=m\\\
m_{1},m_{2}>0\end{subarray}}\bigsqcup_{E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}}\\{L\in\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}:L/F\text{
not Galois}\\}\to\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}.$

###### Proof.

An element of the left-hand side may be written as a pair $(E,L)$, where
$E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}$ and
$L\in\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}$. It is easy to see that there is a
well-defined map

$\Phi:\bigsqcup_{\begin{subarray}{c}2m_{1}+m_{2}=m\\\
m_{1},m_{2}>0\end{subarray}}\bigsqcup_{E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}}\\{L\in\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}:L/K\text{
not
Galois}\\}\to\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}},\quad(L,E)\mapsto
L.$

Let $L\in\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}$. Since $L/F$
has Galois closure group $D_{4}$, it has a unique quadratic intermediate field
$E$. It is easy to see that the pair $(E,L)$ is in the domain of $\Phi$, so
$\Phi$ is surjective. Let
$L\in\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}$. It is easy to
see that

$\Phi^{-1}(L)=\\{(E,E(\sqrt{\alpha})),(E,E(\sqrt{\bar{\alpha}}))\\},$

where $E/F$ is the unique quadratic subextension of $L/F$, and
$L=E(\sqrt{\alpha})$. Since $L/F$ is non-Galois, we have
$E(\sqrt{\alpha})\not\cong E(\sqrt{\bar{\alpha}})$ as extensions of $E$, and
therefore the preimage $\Phi^{-1}(L)$ has exactly two elements, so we are
done. ∎

######  Corollary 4.26.

We have

$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}=\frac{1}{2}\sum_{\begin{subarray}{c}2m_{1}+m_{2}=m\\\
m_{1},m_{2}>0\end{subarray}}\sum_{E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}}\Big{(}\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}-\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F}-\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{V_{4}/F}.\Big{)}$

######  Lemma 4.27.

Let $E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}$ for some integer $m_{1}$. Then
for each $m_{2}$ we have

$\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{V_{4}/F}=\begin{cases}2q^{\frac{m_{2}}{2}-1}(q-1)\quad\text{if
$m_{2}<m_{1}$ and $m_{2}$ is even},\\\ q^{\frac{m_{2}}{2}-1}(q-2)\quad\text{if
$m_{2}=m_{1}$ and $m_{1}$ is even},\\\
q^{\frac{m_{1}+m_{2}}{4}-1}(q-1)\quad\text{if $m_{2}>m_{1}$ and $m_{1}\equiv
m_{2}\pmod{4}$},\\\ 0\quad\text{otherwise}.\end{cases}$

###### Proof.

Lemma 3.10 tells us that for any quadratic extension $E^{\prime}/F$, the
corresponding character $\chi:F^{\times}/F^{\times 2}\to\mathbb{F}_{2}$ has
conductor $v_{F}(d_{E^{\prime}/F})$.

Suppose that $L/F$ is a $V_{4}$-extension with intermediate fields
$E_{1},E_{2},E_{3}$. Then Theorem 3.9 tells us that

$v_{F}(d_{L/F})=\sum_{i=1}^{3}v_{F}(d_{E_{i}/F}).$

Suppose that $m_{2}<m_{1}$. Then using the two facts above, it is easy to see
that there is a bijection

$\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}\to\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{V_{4}/F},\quad
E^{\prime}\mapsto EE^{\prime}.$

If $m_{2}>m_{1}$, then similarly we obtain a $2$-to-$1$ surjection

$\textnormal{\'{E}t}_{(1^{2})/F,\frac{m_{1}+m_{2}}{2}}\to\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{V_{4}/F},\quad
E\mapsto EE^{\prime}.$

Finally, suppose that $m_{1}=m_{2}$. Suppose that
$L\in\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{V_{4}/F}$ and $L=EE^{\prime}$ for
some quadratic extension $E^{\prime}/F$. Let
$\chi,\chi^{\prime}:F^{\times}/F^{\times 2}\to\mathbb{F}_{2}$ be the quadratic
characters associated to $E$ and $E^{\prime}$ respectively. Then we have

$m_{1}+\mathfrak{f}(\chi^{\prime})+\mathfrak{f}(\chi\chi^{\prime})=v_{F}(d_{L/F})=2m_{1}+m_{2}=3m_{1},$

so

$\mathfrak{f}(\chi^{\prime})+\mathfrak{f}(\chi\chi^{\prime})=2m_{1}.$

If $\mathfrak{f}(\chi^{\prime})\neq m_{1}$, then

$\mathfrak{f}(\chi\chi^{\prime})=\max\\{m_{1},\mathfrak{f}(\chi^{\prime})\\},$

so

$\mathfrak{f}(\chi^{\prime})+\max\\{m_{1},\mathfrak{f}(\chi^{\prime})\\}=2m_{1},$

which is impossible. It follows that $\mathfrak{f}(\chi^{\prime})=m_{1}$, so

$\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{1}}^{V_{4}/F}=\frac{1}{2}\cdot\\#\Big{\\{}\chi^{\prime}:F^{\times}/U_{F}^{(m_{1})}F^{\times
2}\to\mathbb{F}_{2}\text{ }\Big{|}\text{
}\chi^{\prime}|_{W_{m_{1}-1}}\not\in\\{0,\chi|_{W_{m_{1}-1}}\\}\Big{\\}}.$

If $m_{1}=2e_{F}+1$, then $W_{m_{1}-1}\cong C_{2}$, so
$\textnormal{\'{E}t}_{(1^{2})/E,m_{1}}^{V_{4}/F}=\varnothing$. If $m_{1}$ is
even with $2\leq m_{1}\leq 2e_{F}$, we have $\\#W_{m_{1}-1}=q$, so there are
$q-2$ possible restrictions $\chi^{\prime}|_{W_{m_{1}-1}}$. Each such
restriction lifts to

$\\#(F^{\times}/U_{F}^{(m_{1}-2)}F^{\times 2})=2q^{\frac{m_{1}}{2}-1}$

characters, so we have

$\\#\Big{\\{}\chi^{\prime}:F^{\times}/U_{F}^{(m_{1})}F^{\times
2}\to\mathbb{F}_{2}\text{ }\Big{|}\text{
}\chi^{\prime}|_{W_{m_{1}-1}}\not\in\\{0,\chi|_{W_{m_{1}-1}}\\}\Big{\\}}=2q^{\frac{m_{1}}{2}-1}(q-2).$

∎

######  Corollary 4.28.

We have

$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{D_{4}/F,\mathcal{A}}=\frac{1}{2}\sum_{0<m_{1}<m/2}\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}\cdot\Big{(}N^{C_{2}}(m_{1})-N^{C_{4}}(m_{1},m_{2})-N^{V_{4}}(m_{1},m_{2})\Big{)},$

which we can compute explicitly by Lemma 3.26.

###### Proof.

This is immediate from Corollary 4.26 and Lemma 4.27. ∎

######  Lemma 4.29.

The number $\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}}$ is the
sum of the following two quantities:

  1. (1)

$\frac{1}{2}\cdot\sum_{\begin{subarray}{c}m_{1}<m_{2}\\\
m_{1}+2m_{2}=m\end{subarray}}(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}})(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}^{\mathcal{A}}).$

  2. (2)

$\mathbbm{1}_{3\mid m}\cdot\frac{1}{3(\\#\overline{\mathcal{A}}^{2})^{2}}\cdot
q^{\frac{m}{6}-4}(q\\#\overline{\mathcal{A}}^{2}_{m/3}-\\#\overline{\mathcal{A}}^{2}_{m/3-1})(q\\#\overline{\mathcal{A}}^{2}_{m/3}-2\\#\overline{\mathcal{A}}^{2}_{m/3-1}).$

###### Proof.

Define the map

$\Phi:\bigsqcup_{\begin{subarray}{c}m_{1}<m_{2}\\\
m_{1}+2m_{2}=m\end{subarray}}\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}\times\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}^{\mathcal{A}}\to\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}},\quad(E_{1},E_{2})\mapsto
E_{1}E_{2}.$

It is easy to see that this map is well-defined and $2$-to-$1$, so

$\\#\operatorname{im}\Phi=\frac{1}{2}\cdot\sum_{\begin{subarray}{c}m_{1}<m_{2}\\\
m_{1}+2m_{2}=m\end{subarray}}(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}})(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}^{\mathcal{A}}).$

If $3\nmid m$, then $\Phi$ is surjective, so we are done. Suppose that $3\mid
m$. Let $S$ be the set of
$L\in\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}}$ such that every
intermediate quadratic field $E$ of $L$ has $v_{F}(d_{E/F})=m/3$. Then

$\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}}=S\sqcup\operatorname{im}\Phi,$

so

$\\#\textnormal{\'{E}t}_{(1^{4})/F,m}^{V_{4}/F,\mathcal{A}}=\\#S+\frac{1}{2}\cdot\sum_{\begin{subarray}{c}m_{1}<m_{2}\\\
m_{1}+2m_{2}=m\end{subarray}}(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}})(\\#\textnormal{\'{E}t}_{(1^{2})/F,m_{2}}^{\mathcal{A}}).$

Let $\Sigma$ be the set of pairs $(\chi_{1},\chi_{2})$, where:

  1. (1)

$\chi_{1}$ and $\chi_{2}$ are quadratic characters
$\chi_{i}:F^{\times}/U_{F}^{(m/3)}F^{\times 2}\to\mathbb{F}_{2}$.

  2. (2)

The restrictions of $\chi_{1}$ and $\chi_{2}$ to $U_{F}^{(m/3-1)}F^{\times
2}/U_{F}^{(m/3)}F^{\times 2}$ are nonzero and distinct.

  3. (3)

For $i=1,2$, we have

$\chi_{i}\Big{(}\mathcal{A}F^{\times 2}U_{F}^{(m/3)}/F^{\times
2}U_{F}^{(m/3)}\Big{)}=0.$

Then there is a natural $6$-to-$1$ surjection $\Sigma\to S,$ so

$\\#S=\frac{1}{6}\cdot\\#\Sigma.$

These conditions look daunting, but they actually constitute a simple linear
algebra problem. To emphasise this simplicity, we define:

  1. (1)

$V=F^{\times}/U_{F}^{(m/3)}F^{\times 2}.$

  2. (2)

$M=U_{F}^{(m/3-1)}F^{\times 2}/U_{F}^{(m/3)}F^{\times 2}.$

  3. (3)

$N=\mathcal{A}U_{F}^{(m/3)}F^{\times 2}/U_{F}^{(m/3)}F^{\times 2}.$

Then $V$ is an $\mathbb{F}_{2}$-vector space, $M$ and $N$ are subspaces of
$V$, and we are looking for pairs of linear transformations
$\chi_{1},\chi_{2}:V\to\mathbb{F}_{2}$ such that:

  1. (1)

$\chi_{i}(N)=0$ for $i=1,2$.

  2. (2)

The restrictions $\chi_{1}|_{M}$ and $\chi_{2}|_{M}$ are nonzero and distinct.

These correspond bijectively to pairs $\chi_{1},\chi_{2}:V/N\to\mathbb{F}_{2}$
such that the restrictions $\chi_{1}|_{(M+N)/N}$ and $\chi_{2}|_{(M+N)/N}$ are
nonzero and distinct. There are thus

$\Big{(}\\#\Big{(}\frac{M+N}{N}\Big{)}-1\Big{)}\Big{(}\\#\Big{(}\frac{M+N}{N}\Big{)}-2\Big{)}$

possibilities for the pair $(\chi_{1}|_{(M+N)/N},\chi_{2}|_{(M+N)/N})$. Each
of these lifts to

$\Big{(}\\#\Big{(}\frac{V}{M+N}\Big{)}\Big{)}^{2}$

pairs $(\chi_{1},\chi_{2})$, so we have

$\\#\Sigma=\Big{(}\\#\Big{(}\frac{V}{M+N}\Big{)}\Big{)}^{2}\cdot\Big{(}\\#\Big{(}\frac{M+N}{N}\Big{)}-1\Big{)}\Big{(}\\#\Big{(}\frac{M+N}{N}\Big{)}-2\Big{)}.$

We evaluate the sizes of the relevant vector spaces.

  1. (1)

Corollary 3.21 tells us that

$\\#V=2q^{\frac{m}{6}}.$

  2. (2)

We have

$N=\frac{\mathcal{A}\cdot U_{F}^{(m/3)}F^{\times 2}}{U_{F}^{(m/3)}F^{\times
2}}\cong\frac{\overline{\mathcal{A}}^{2}}{\overline{\mathcal{A}}^{2}_{m/3}},$

so

$\\#N=\frac{\\#\overline{\mathcal{A}}^{2}}{\\#\overline{\mathcal{A}}^{2}_{m/3}}.$

  3. (3)

We have

$M+N=\frac{\mathcal{A}F^{\times 2}U_{F}^{(m/3-1)}}{F^{\times
2}U_{F}^{(m/3)}},$

so

$\\#(M+N)=\frac{\\#(\mathcal{A}F^{\times 2}U_{F}^{(m/3-1)})}{\\#(F^{\times
2}U_{F}^{(m/3-1)})}\cdot\\#W_{m/3-1}.$

We also have

$\frac{\mathcal{A}F^{\times 2}U_{F}^{(m/3-1)}}{F^{\times
2}U_{F}^{(m/3-1)}}\cong\frac{\overline{\mathcal{A}}^{2}}{\overline{\mathcal{A}}^{2}_{m/3-1}},$

so

$\\#(M+N)=q\cdot\frac{\\#\overline{\mathcal{A}}^{2}}{\\#\overline{\mathcal{A}}_{m/3-1}^{2}}.$

It follows that

$\\#\Sigma=\frac{4q^{\frac{m}{3}-2}}{(\\#\overline{\mathcal{A}}^{2})^{2}}\cdot(q\\#\overline{\mathcal{A}}^{2}_{m/3}-\\#\overline{\mathcal{A}}^{2}_{m/3-1})(q\\#\overline{\mathcal{A}}^{2}_{m/3}-2\\#\overline{\mathcal{A}}^{2}_{m/3-1}).$

∎

###  4.3. Totally ramified $C_{4}$-extensions

Let $\alpha\in F^{\times}$. Write $m_{\alpha}=v_{F}(d_{F(\sqrt{\alpha})/F})$
and

$N_{\alpha}=\Big{(}N_{F(\sqrt{\alpha})/F}F(\sqrt{\alpha})^{\times}\Big{)}/F^{\times
2}\subseteq F^{\times}/F^{\times 2}.$

For $\omega\in E^{\times}$, define

$N_{\omega}=\Big{(}N_{E(\sqrt{\omega})/F}E(\sqrt{\omega})^{\times}\Big{)}/F^{\times
4}\subseteq F^{\times}/F^{\times 4}.$

Define

$N_{\omega}^{\mathcal{A}}=\Big{(}\bigcap_{[\alpha]\in\overline{\mathcal{A}}^{4}\cap
N_{\omega}}N_{\alpha}\Big{)}\setminus\Big{(}\bigcup_{[\alpha]\in\overline{\mathcal{A}}^{4}\setminus
N_{\omega}}N_{\alpha}\Big{)}\subseteq F^{\times}/F^{\times 2}$

and

$N_{\omega,c}^{\mathcal{A}}=N_{\omega}^{\mathcal{A}}\cap\Big{(}U_{F}^{(2e_{F}-c)}F^{\times
2}/F^{\times 2}\Big{)}.$

######  Lemma 4.30.

Let $E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}$ for a positive integer
$m_{1}$. Let $t$ be an integer with $1\leq t\leq 2e_{F}-\frac{m_{1}}{2}$. We
have

$(E^{\times 2}U_{E}^{(2t)}\cap F^{\times})/F^{\times
2}=\begin{cases}F^{\times}/F^{\times 2}\quad\text{if
$t\leq\frac{m_{1}}{2}-1$},\\\
U_{F}^{\Big{(}2\Big{\lceil}\frac{t-\frac{m_{1}}{2}}{2}\Big{\rceil}+1\Big{)}}F^{\times
2}/F^{\times 2}\quad\text{if $t\geq\frac{m_{1}}{2}$}.\end{cases}$

###### Proof.

The result [CDO05, Proposition 3.11] essentially tells us that

$\begin{cases}F^{\times}/F^{\times 2}\quad\text{if
$t\leq\frac{m_{1}}{2}-1$},\\\
U_{F}^{\Big{(}2\Big{\lceil}\frac{t-\frac{m_{1}}{2}}{2}\Big{\rceil}\Big{)}}F^{\times
2}/F^{\times 2}\quad\text{if $t\geq\frac{m_{1}}{2}$}.\end{cases}$

The result then follows from Corollary 3.20. ∎

######  Lemma 4.31.

Let $E\in\textnormal{\'{E}t}_{(1^{2})/F}^{\mathcal{A}}$. If $-1\not\in
N_{E/F}E^{\times}$, then

$\textnormal{\'{E}t}_{(1^{2})/E}^{C_{4}/F}=\varnothing.$

For the rest of the lemma, assume that $-1\in N_{E/F}E^{\times}$. Let
$m_{1}=v_{F}(d_{E/F})$, and let $m_{2}$ be an integer. Let

$c(m_{2})=2\big{\lfloor}\frac{m_{1}+2\lfloor\frac{m_{2}}{2}\rfloor}{4}\big{\rfloor}.$

If $m_{1}\leq e_{F}$, then let $\omega\in E^{\times}$ such that
$E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E,3m_{1}-2}^{C_{4}/F}$. Then
we have

$\\#\textnormal{\'{E}t}_{(1^{2})/E,\leq
m_{2}}^{C_{4}/F,\mathcal{A}}=\begin{cases}0\quad\text{if $m_{2}<3m_{1}-2$},\\\
\frac{1}{2}\cdot\\#N^{\mathcal{A}}_{\omega,c(m_{2})}\quad\text{if
$3m_{1}-2\leq m_{2}\leq 4e_{F}-m_{1}+1$},\\\
\frac{1}{2}\cdot\\#N^{\mathcal{A}}_{\omega}\quad\text{if $m_{2}\geq
4e_{F}-m_{1}+2$}.\end{cases}$

If $m_{1}>e_{F}$, then let $\omega\in E^{\times}$ such that
$E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E}^{C_{4}/F}$. Then we have

$\textnormal{\'{E}t}_{(1^{2})/E,\leq
m_{2}}^{C_{4}/F,\mathcal{A}}=\begin{cases}0\quad\text{if
$m_{2}<m_{1}+2e_{F}$},\\\ \frac{1}{2}\\#N^{\mathcal{A}}_{\omega}\quad\text{if
$m_{2}\geq m_{1}+2e_{F}$}.\end{cases}$

###### Proof.

If $-1\not\in N_{E/F}E^{\times}$, then
$\textnormal{\'{E}t}_{(1^{2})/E}^{C_{4}/F}=\varnothing$ by [Mon24, Corollary
4.9]. Suppose that $-1\in N_{E/F}E^{\times}$. Then [Mon24, Corollary 4.9] and
[Mon24, Lemma 4.21] tell us that $\textnormal{\'{E}t}_{(1^{2})/E,\leq
m_{2}}^{C_{4}/F}$ is nonempty if and only if

$m_{2}\geq\begin{cases}3m_{1}-2\quad\text{if $m_{1}\leq e_{F}$,}\\\
2e_{F}+m_{1}\quad\text{if $m_{1}>e_{F}$}.\end{cases}$

Suppose that $m_{1}\leq e_{F}$ and $m_{2}\geq 3m_{1}-2$. If $m_{2}\leq
4e_{F}-m_{1}+1$, then Lemmas 4.15 and 4.30 give a $2$-to-$1$ surjection

$\Big{(}U_{F}^{\Big{(}2e_{F}-2\Big{\lfloor}\frac{m_{1}+2\lfloor\frac{m_{2}}{2}\rfloor}{4}\Big{\rfloor}\Big{)}}F^{\times
2}\Big{)}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E,\leq m_{2}}^{C_{4}/F},\quad
t\mapsto E(\sqrt{\omega t}).$

If $m_{2}\geq 4e_{F}-m_{1}+2$, then similarly we have a $2$-to-$1$ surjection

$F^{\times}/F^{\times 2}\to\textnormal{\'{E}t}_{2/E,\leq
m_{2}}^{C_{4}/F},\quad t\mapsto E(\sqrt{\omega t}).$

Let $\alpha\in\mathcal{A}$, and let $\widetilde{\alpha}\in E^{\times}$ be such
that $\alpha=N_{E/F}\widetilde{\alpha}$. By Lemma 4.19, for each $t\in
F^{\times}$, we have

$\displaystyle\alpha\in N_{E(\sqrt{\omega t})/E}E(\sqrt{\omega t})^{\times}$
$\displaystyle\iff(\widetilde{\alpha},\omega t)_{E}=1$
$\displaystyle\iff(\widetilde{\alpha},\omega)_{E}=(\widetilde{\alpha},t)_{E}$
$\displaystyle\iff(\alpha,t)_{F}=(\widetilde{\alpha},\omega)_{E}$
$\displaystyle\iff\begin{cases}t\in N_{\alpha}\quad\text{if $\alpha\in
N_{\omega}$},\\\ t\not\in N_{\alpha}\quad\text{if $\alpha\not\in
N_{\omega}$}.\end{cases}$

It follows that $\mathcal{A}\subseteq N_{E(\sqrt{\omega t})/E}E(\sqrt{\omega
t})^{\times}$ if and only if $t\in N^{\mathcal{A}}_{\omega}$, and the result
follows. The argument for $m_{1}>e_{F}$ is similar but easier, so we omit it.
∎

###### Proof of Theorem 4.9.

This follows easily from Lemma 4.24, Corollary 4.26, Lemma 4.29, and Lemma
4.31

∎

######  Algorithm 4.32.

Let $E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}$ for even $m_{1}\leq e_{F}$.

  1. (1)

Take $d\in F^{\times}$ such that $E=F(\sqrt{d})$ and $v_{F}(d)=m_{1}$.

  2. (2)

Take $a,b\in F^{\times}$ with $d=a^{2}+4b$, such that $v_{F}(a)=m_{1}/2$ and
$v_{F}(b)=1$.

  3. (3)

Define $\rho=\frac{a+\sqrt{d}}{2}$, which is a uniformiser of $E$.

  4. (4)

Choose $\omega\in E^{\times}$ such that $N_{E/F}\omega\in dF^{\times 2}$ and
$v_{E}(\omega)=0$.

  5. (5)

Take $\lambda\in F^{\times}$ such that

$N_{E/F}\omega\equiv\lambda^{2}\pmod{\mathfrak{p}_{F}^{2e_{F}+1-m_{1}}}.$

  6. (6)

Define

$\omega_{1}=\begin{cases}\omega\quad\text{if $v_{E}(\omega-\lambda)\geq
m_{1}$},\\\ \frac{\omega b}{\rho^{2}}\quad\text{if
$v_{E}(\omega-\lambda)=m_{1}-1$}.\end{cases}$

  7. (7)

Define

$\omega_{2}=\begin{cases}\omega_{1}\quad\text{if
$v_{E}(\omega_{1}-\bar{\omega}_{1})=2m_{1}$},\\\
\omega_{1}(1+\rho)^{2}\quad\text{if
$v_{E}(\omega_{1}-\bar{\omega}_{1})>2m_{1}$}\end{cases}$

and

$\lambda_{2}=\begin{cases}\lambda\quad\text{if
$v_{E}(\omega_{1}-\bar{\omega}_{1})=2m_{1}$},\\\ \lambda(1+a-b)\quad\text{if
$v_{E}(\omega_{1}-\bar{\omega}_{1})>2m_{1}$}.\end{cases}$

  8. (8)

Write $\omega_{2}=r_{2}+s_{2}\rho$ and define

$q=\frac{r_{2}-\lambda_{2}}{s_{2}},\quad n=\frac{q^{2}+b}{r_{2}}.$

  9. (9)

Output $\frac{\omega_{2}n}{(q+\rho)^{2}}.$

######  Theorem 4.33.

Let $E,m_{1}$ be as in Algorithm 4.32.

  1. (1)

All the steps of the algorithm are well-defined.

  2. (2)

Let $\omega$ be the output of the algorithm. Then
$E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E,3m_{1}-2}^{C_{4}/F}$.

###### Proof.

This is essentially [CDO05, Proposition 3.15]. We rewrite their proof using
our notation in Appendix A. ∎

######  Theorem 4.34.

The time complexity of Algorithm 4.32 is $O(e_{F}\cdot t_{\mathrm{norm}}(F))$.

###### Proof.

We compute the time complexity of each step of the algorithm, one by one.

  1. (1)

Clearly this has time complexity $O(t(F))$.

  2. (2)

We need to solve the congurence $\frac{X^{2}}{d}-1\equiv
0\pmod{\mathfrak{p}_{F}^{2e_{F}+1-m_{1}}}$. Using Algorithm 3.18, we can do
this with time complexity $O(e_{F}t(F))$.

  3. (3)

This is clearly $O(t(F))$.

  4. (4)

Using Hensel approximation, we may solve the congruence
$\frac{X^{2}+1}{d}-1\equiv 0\pmod{\mathfrak{p}_{F}^{2e_{F}+1}}$ with time
complexity $O(t_{\mathrm{norm}}(F))$. Let $x$ be a solution to this
congruence. By Corollary 3.14, it follows that $dF^{\times
2}=(x^{2}+1)F^{\times 2}$. But

$x^{2}+1=N_{F(\sqrt{d})/F}\Big{(}\frac{x^{2}+1}{x}+\frac{\sqrt{d}}{x}\Big{)},$

so we may set $\omega=\frac{x^{2}+1}{x}+\frac{\sqrt{d}}{x}$. Therefore this
step has time complexity $O((2e_{F}+1)t(F))$.

  5. (5)

Using Algorithm 3.18, we can find $\lambda$ with time complexity
$O(e_{F}t(F))$.

  6. (6)-(9)

The remaining steps are all just computations in $F$, so they have complexity
$O(t(F))$.

∎

######  Lemma 4.35.

Assume that we have already computed the norm groups
$N_{F(\sqrt{-1})/F}F(\sqrt{-1})^{\times}$ and
$N_{F(\sqrt{\mathcal{A}})/F}F(\sqrt{\mathcal{A}})^{\times}$. Let
$E\in\textnormal{\'{E}t}_{(1^{2})/F}$. We can compute
$\\#\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\mathcal{A}}$ with time
complexity $O(t_{\mathrm{norm}}(F))$.

###### Proof.

Using our pre-computed norm groups, we can check quickly whether $-1\in
N_{E/F}E^{\times}$ and $\mathcal{A}\subseteq N_{E/F}E^{\times}$. If $-1\not\in
N_{E/F}E^{\times}$, then $\textnormal{\'{E}t}_{2/E}^{C_{4}/F}$ is empty, so we
are done. Similarly, if $\mathcal{A}\not\subseteq N_{E/F}E^{\times}$, then
$\textnormal{\'{E}t}_{2/E}^{\mathcal{A}}$ is empty, so we are done.

Suppose that $-1\in N_{E/F}E^{\times}$ and $\mathcal{A}\subseteq
N_{E/F}E^{\times}$. Let $m_{1}=v_{F}(d_{E/F})$. If $m_{1}\leq e_{F}$, then
Theorems 4.33 and 4.34 tell us that we can compute $\omega\in E^{\times}$ such
that $E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E,3m_{1}-2}^{C_{4}/F}$
with time complexity $O(t_{\mathrm{norm}}(F))$. If $m_{1}>e_{F}$, then let
$d\in F^{\times}$ be such that $E=F(\sqrt{d})$, and, again with time
complexity $O(t_{\mathrm{norm}}(F))$, let $\omega\in E^{\times}$ be such that
$N_{E/F}\omega\in dF^{\times 2}$. In that case, [Mon24, Lemma 4.8] tells us
that $E(\sqrt{\omega})\in\textnormal{\'{E}t}_{(1^{2})/E}^{C_{4}/F}$. Using
standard algorithms for computing norm groups, it is routine to find the
$\mathbb{F}_{2}$-vector space $N_{\omega}^{\mathcal{A}}$, and the result
follows by Lemma 4.31. ∎

###### Proof of Theorem 1.7.

There is a natural bijection

$\bigsqcup_{2m_{1}+m_{2}=m}\bigsqcup_{E\in\textnormal{\'{E}t}_{(1^{2})/F,m_{1}}^{\mathcal{A}}}\textnormal{\'{E}t}_{(1^{2})/E,m_{2}}^{C_{4}/F,\langle\alpha\rangle}\longleftrightarrow\textnormal{\'{E}t}_{(1^{2})/F,m}^{C_{4}/F,\mathcal{A}}.$

Since $\\#\textnormal{\'{E}t}_{(1^{2})/F}=4q^{e_{F}}-2$, the result follows
from Lemma 4.35. ∎

##  Appendix A Proof of Theorem 4.33

Let $E,m_{1}$ be as in the setup of the algorithm. In this appendix, we will
step through the algorithm, showing that each stage is well-defined, and
eventually proving that the output has the desired property.

  1. (1)

Start by taking any $d$ with $E=F(\sqrt{d})$. Since $v_{K}(d_{E/F})\leq
e_{F}$, [Mon24, Lemma 4.6] tells us that $v_{F}(d)$ is even, which means we
can multiply by some even power of $\pi_{F}$ to get $v_{F}(d)=m_{1}$.

  2. (2)

[Mon24, Lemma 4.6] tells us that there is some $a\in F^{\times}$ such that
$\frac{d}{a^{2}}\equiv 1\pmod{\mathfrak{p}_{F}^{2e_{F}+1-m_{1}}}$, and
moreover that there is no such $a$ for any higher power of $\mathfrak{p}_{F}$.
This implies that

$v_{F}(\frac{d}{a^{2}}-1)=2e_{F}+1-m_{1},$

so

$v_{F}(d-a^{2})=2e_{F}+1.$

Setting $b=d-a^{2}$, we obtain $a,b$ as required.

  3. (3)

It is easy to see that $\rho^{2}-a\rho-b=0$, so the minimal polynomial of
$\rho$ over $F$ is Eisenstein, and therefore $\rho$ is a uniformiser.

  4. (4)

Since $-1\equiv 1\pmod{\mathfrak{p}_{F}^{e_{F}}}$, we have
$v_{F}(d_{F(\sqrt{-1})/F})\leq e_{F}$, so
$U_{F}^{(e_{F})}\subseteq\operatorname{Nm}F(\sqrt{-1})$ so

$\frac{d}{a^{2}}\in U_{F}^{(2e_{F}+1-m_{1})}\subseteq
U_{F}^{(e_{F})}\subseteq\operatorname{Nm}F(\sqrt{-1}),$

which implies that $(-1,d)_{F}=1$, and therefore [Mon24, Corollary 4.9] tells
us that we may choose $\omega\in E^{\times}$ with $N_{E/F}\omega\in dF^{\times
2}$. Moreover, we may ensure that $v_{E}(\omega)=0$, since $v_{F}(d)$ is even.

  5. (5)

We know that $N_{E/F}\omega=dx^{2}$ for some $x\in F^{\times}$ with
$v_{F}(x)=-m_{1}/2$. Setting $\lambda=ax$, it is easy to see that
$N_{E/F}\omega\equiv\lambda^{2}\pmod{\mathfrak{p}_{F}^{2e_{F}+1-m_{1}}}$.

  6. (6)

######  Lemma A.1.

For all $x\in\mathcal{O}_{E}$, we have
$v_{F}(\operatorname{Tr}_{E/F}x)\geq\frac{m_{1}}{2}$.

######  Lemma A.2.

We have

$v_{E}(\omega-\lambda)\geq m_{1}-1.$

###### Proof.

Let $\gamma=\omega-\lambda$. Define the sequence $(a_{n})_{n\geq 0}$ as
follows. Set $a_{0}=0$, and for each $n\geq 0$, define

$a_{n+1}=\min\\{\lfloor\frac{a_{n}}{2}\rfloor+\frac{m_{1}}{2},2e_{F}+1-m_{1}\\}.$

We claim that $v_{E}(\gamma)\geq a_{n}$ for all $n$. The base case $n=0$ is
clear. Suppose that $v_{E}(\gamma)\geq a_{n}$ for some $n$. Then
$\gamma/\pi_{F}^{\lfloor\frac{a_{n}}{2}\rfloor}\in\mathcal{O}_{E}$, so it
follows from Lemma A.1 that

$v_{F}(\operatorname{Tr}_{E/F}\gamma)\geq\frac{m_{1}}{2}+\lfloor\frac{a_{n}}{2}\rfloor.$

Since
$N_{E/F}\omega\equiv\lambda^{2}\pmod{\mathfrak{p}_{F}^{2e_{F}+1-m_{1}}}$, we
have