# DENSITY OF $p$-ADIC POLYNOMIALS GENERATING EXTENSIONS WITH FIXED SPLITTING
TYPE

John Yin1 1Department of Mathematics at University of Wisconsin, Madison
<EMAIL_ADDRESS>
###### Abstract.

We prove that the density of polynomials $P(x)=\sum_{i=0}^{n}a_{n}x^{n}$ over
a local field $K$ generating an étale extension with specified splitting type
is a rational function in terms of the size of the residue field of $K$ in the
case where the splitting type is tame. Moreover, we give a computable
recursive formula for these densities and compute the asymptotics of this
density as the size of the residue field tends to infinity.

##  1\. Introduction

In [1], the authors conjectured that the probability that a random $p$-adic
polynomial $P(x)=c_{n}x^{n}+c_{n-1}x^{n-1}\dots+c_{0}$ with coefficients in
${\mathbb{Z}}_{p}$ generates an étale extension such that $p$ splits in a
specified way (e.g. ${\mathbb{Q}}_{p}[x]/f$ is a totally ramified extension,
unramified extension, a product of two extensions with specified residue
degrees) is a rational function $\rho$ of $p$, and moreover satisfies the
functional equation $\rho(p)=\rho(1/p)$. We will prove the rationality of
$\rho$ in the tamely ramified case. As we shall see, our methods are more
general and we will prove rationality of the density of polynomials satisfying
many other conditions. Moreover, we will give a recursive method to calculate
these densities.

We parametrize polynomials $P(x)$ of degree at most $n$ as above by
$\mathbb{Z}_{p}^{n+1}$ with the Haar measure normalized so that the total
measure of ${\mathbb{Z}}_{p}^{n+1}$ is $1$. Now given an étale extension $L$,
it is isomorphic to a product of field extensions $L_{i}$, so that
$L=\prod_{i=1}^{m}L_{i}$. We define the splitting type of $L$ to be the
multiset of pairs

$\sigma=\\{(e(L_{1}),f(L_{1})),(e(L_{2}),f(L_{2})),\dots,(e(L_{n}),f(L_{n}))\\}.$

This is written as $(f(L_{1})^{e(L_{1})},\dots,f(L_{m})^{e(L_{m})})$ in [1],
and we will follow their notation. Fixing such a splitting type $\sigma$, let
$U_{\sigma}(p)\subset\mathbb{Z}_{p}^{n+1}$ be the $p$-adic open subset of
polynomials with splitting type $\sigma$ and
$\rho(n,\sigma;p)=\mu_{\mathrm{Haar}}(U_{\sigma}(p))$. As a few sample
examples, [1] computes

(1.1) $\displaystyle\rho(2,(1^{1}1^{1});p)$ $\displaystyle=\frac{1}{2}$
$\displaystyle\rho(2,(2^{1});p)$
$\displaystyle=\frac{p^{2}-p+1}{2(p^{2}+p+1)}$ (1.2)
$\displaystyle\rho(3,(1^{1}1^{1}1^{1});p)$
$\displaystyle=\frac{p^{4}+2p^{2}+1}{6(p^{4}+p^{3}+p^{2}+p+1)}$
$\displaystyle\rho(3,(1^{1}2^{1});p)$
$\displaystyle=\frac{p^{4}+1}{2(p^{4}+p^{3}+p^{2}+p+1)}.$

Similarly, let $\alpha(n,\sigma;p)$ and $\beta(n,\sigma;p)$ be defined to be
the measure of $\\{P\in U_{\sigma(p)}:P\text{ is monic}\\}$ and $\\{P\in
U_{\sigma(p)}:P\text{ is monic and }P\equiv x^{n}\mod p\\}$, respectively.
Based on numerical evidence and some of their results, [1, Conjecture 1.2]
states the following conjecture.

######  Conjecture 1.1.

The densities $\rho(n,\sigma;p)$, $\alpha(n,\sigma;p)$, and
$\beta(n,\sigma;p)$ are rational functions in $p$ and satisfy

$\rho(n,\sigma;p^{-1})=\rho(n,\sigma;p).$
$\alpha(n,\sigma;p^{-1})=\beta(n,\sigma;p).$

In this paper, we will prove the rationality part of this conjecture for tame
extensions (3.1).

###  1.1. Overview of proof

One main ingredient is 3.4, proven in [3], which makes use of the $p$-adic
Kac-Rice formula to compute the average number of roots of a random polynomial
in a fixed open subset of ${\mathcal{O}}_{L}$ for a given étale extension $L$.
To use this formula, it turns out one needs to calculate the density of
elements $\alpha\in{\mathcal{O}}_{L}$ whose minimal polynomial
$P_{\alpha/{\mathbb{Q}}_{p}}$ over ${\mathbb{Q}}_{p}$ has a specified
valuation of polynomial discriminant $\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}}$.
That is, we need to calculate the density
$\mu_{L}(\\{\alpha\in{\mathcal{O}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$
for every $c$, where $\mu_{L}$ is the normalized Haar measure on
${\mathcal{O}}_{L}$. Calculating this density and summing over the étale
extensions $L$ is the main technical part of this paper, which we will explain
below. With this we obtain the average number of roots of a random polynomial
in extensions with prescribed splitting behavior. Finally, a couple of
transformations are done in Sections 3 and 6 to get from this to the density
of random polynomials generating an extension with prescribed splitting
behavior, which concludes the proof.

As an example of the process involved in the calculation of
$\mu_{L}(\\{\alpha\in{\mathcal{O}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$,
we fix a square root $p^{1/2}$ of $p$ and take $L$ to be
${\mathbb{Q}}_{p}[\zeta_{p^{3}-1},p^{1/2}]$, which is an extension of
${\mathbb{Q}}_{p}$ of ramification index $2$ and inertia degree $3$. Also let
${\mathfrak{m}}_{L}$ be the maximal ideal in ${\mathcal{O}}_{L}$. Elements of
$\alpha\in{\mathcal{O}}_{L}$ have a Teichmüller expansion of the form
$\alpha=\sum a_{i}p^{i/2}$, where $a_{i}$ is either a power of a
$(p^{3}-1)$-th root of unity or $0$. The $6$ Galois conjugates of $\alpha$ are
of the form $\sum a_{i}^{p^{j}}(-1)^{ib}p^{i/2}$ where $j\in\\{0,1,2\\}$ and
$b\in\\{0,1\\}$. With this expression of the Galois conjugates, one can
recursively calculate the valuation of the discriminant of the minimal
polynomial by using the definition
$\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}}=\prod_{\alpha^{\prime}\neq\alpha^{\prime\prime}\text{
Galois conjugates of }\alpha}(\alpha^{\prime}-\alpha^{\prime\prime})$. Here we
use the convention that in the case our polynomial is linear, the product is
empty and our discriminant is $1$. This recursion is somewhat combinatorially
complicated, but we will work out a few steps in the recursion for this $L$ to
give the reader an idea of the process.

The context of the following paragraphs is that we wish to calculate the
$p$-adic valuation of the product of pairwise differences between Galois
conjugates of $\alpha$ by analyzing its Teichmüller expansion
$\alpha=\sum_{i=0}^{\infty}a_{i}p^{i/2}$. We start out with considering the
lowest degree coefficient $a_{0}$. There are two cases. If
$a_{0}\in{\mathbb{Q}}_{p}$, which happens for exactly $p$ choices of $a_{0}$,
then the coefficient of $p^{0/2}$ in the Teichmüller expansion of the Galois
conjugates of $\alpha$ must all be $a_{0}$, and hence for these elements,
$v_{p}(P_{\alpha/{\mathbb{Q}}_{p}})=v_{p}(P_{\alpha-a_{0}})$, where $\alpha-
a_{0}\in{\mathfrak{m}}_{L}$. So, we just found that

$\mu_{L}(\\{\alpha\in{\mathcal{O}}_{L}:a_{0}\in{\mathbb{Q}}_{p},v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})=p\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\}).$

We now do the same in the other case. If $a_{0}\notin{\mathbb{Q}}_{p}$, which
happens for $p^{3}-p$ choices of $a_{0}$, then the coefficients of $p^{0/2}$
in the Teichmüller expansion of Galois conjugates of $\alpha$ are precisely
the Galois conjugates of $a_{0}$, of which there will be $3$. In fact, it
turns out that the $6$ Galois conjugates are neatly partitioned into $3$ sets
of $2$ Galois conjugates, with each set containing the three Galois conjugates
with a fixed $p^{0/2}$ coefficient. Now,
$v_{p}(a_{0}^{p^{j^{\prime}}}-a_{0}^{p^{j}})=0$ for distinct
$j,j^{\prime}\in\\{0,1,2\\}$ since $\alpha$ is a $p^{3}-1$-th root of unity.
In other words, for Galois conjugates $\alpha^{\prime},\alpha^{\prime\prime}$
in different partitions (i.e. with different $p^{0/2}$ coefficients), we have
$v_{p}(\alpha^{\prime}-\alpha^{\prime\prime})=0$. So, in the context of
finding the valuation of the product of pairwise differences of Galois
conjugates of $\alpha$, all that’s left is to calculate the valuation of the
product of $\alpha^{\prime}-\alpha^{\prime\prime}$, as
$\alpha^{\prime},\alpha^{\prime\prime}$ vary over pairs of Galois conjugates
of $\alpha$ with the same $p^{0/2}$-th coefficient. There are $3$ such pairs,
one coming from each element of the partition above. Now, note that the Galois
group acts transitively on the partitions themselves. In other words, the
valuation of the difference between each pair are the same. So, it suffices to
study one set in the partition, say the set $\\{\alpha,\alpha^{\prime}\\}$
whose coefficient of $p^{0/2}$ is $a_{0}$. Now we note that
$\alpha,\alpha^{\prime}$ are precisely the Galois conjugates of $\alpha$ over
${\mathbb{Q}}_{p}[a_{0}]$. Moreover, $\alpha-\alpha^{\prime}=(\alpha-
a_{0})-(\alpha^{\prime}-a_{0})$, and $\alpha-a_{0}$ lies in
${\mathfrak{m}}_{L}$. Thus, also we obtain

$\mu_{L}(\\{\alpha\in{\mathcal{O}}_{L}:a_{0}\notin{\mathbb{Q}}_{p},v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})=(p^{3}-p)\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}[a_{0}]}})=c\\}).$

The above two paragraphs conclude the first step of the recursion. The
combination of the two cases above reduced the calculation of
$\mu_{L}(\\{\alpha\in{\mathcal{O}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$
to the calculation of multiple densities of the form
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:v_{p}(\Delta_{P_{\alpha/K}})=c^{\prime}\\})$,
for some extension $K$ of ${\mathbb{Q}}_{p}$ and integer $c^{\prime}\leq c$.
As we shall see, this pattern continues. Although in the above cases,
$c^{\prime}$ happened to equal $c$, we do one more step in the recursion to
see that $c^{\prime}$ can be less than $c$. We continue from the first case,
where we ended up wanting to compute
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$.
Elements in $\alpha\in{\mathfrak{m}}_{L}$ have Teichmüller expansion of the
form $\sum_{i\geq 1}a_{i}p^{i/2}$. We again focus on the lowest degree term
$a_{1}p^{1/2}$. Now there are three cases. The first case is if $a_{1}=0$,
then we reduce to the calculation of
$\\{\alpha\in{\mathfrak{m}}_{L}^{2}:a_{1}=0,v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\}$.
The second case is if $a_{1}\in{\mathbb{Q}}_{p}\setminus 0$, which occurs for
a proportion of $p-1$ of all elements in ${\mathfrak{m}}_{L}$, then the Galois
conjugates are separated into $2$ partitions, each with $3$ conjugates with a
common leading term in the Teichmüller expansion $\sum_{i\geq 1}a_{i}p^{i/2}$
is $a_{1}p^{1/2}$ or $-a_{1}p^{1/2}$. Now for conjugates
$\alpha^{\prime},\alpha^{\prime\prime}$ belonging to two different partitions,
$v_{p}(\alpha^{\prime}-\alpha^{\prime\prime})=v_{p}(2a_{1}p^{1/2})=1/2$,
assuming $p\neq 2$. This is one example of where the tameness assumption comes
into play. There are $2\cdot 3\cdot 3=18$ such ordered pairs, giving a
contribution of $18/2$ to the $p$-adic valuation of the discriminant. Just as
we did in the previous paragraph, we can now focus on product of pairwise
differences of Galois conjugates $\alpha,\alpha^{\prime}$ in one partition,
say the one with leading term $a_{1}p^{1/2}$. Moreover, for Galois conjugates
in this partition, $\alpha-\alpha^{\prime}=(\alpha-
a_{1}p^{1/2})-(\alpha^{\prime}-a_{1}p^{1/2})$, and $\alpha-a_{1}p^{1/2}$ lies
in ${\mathfrak{m}}_{L}^{2}$. Now there are two partitions, with both of the
product of the pairwise differences of elements in each partition giving the
same contribution to the $p$-adic valuation of the discriminant. Thus, we have
that

$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:a_{1}\in{\mathbb{Q}}_{p}\setminus\\{0\\},v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})=(p-1)\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}[a_{1}p^{1/2}]}})=(c-9)/2\\}).$

Finally, we will not work out in full detail the third case, but one can show
with basic theory of local fields that if $a_{1}\notin{\mathbb{Q}}_{p}$, then
$a_{1}p^{1/2}$ will have $6$ conjugates, and in particular generate $L$
itself. So, the pairwise difference of Galois conjugates of $\alpha$ would all
have $p$-adic valuation of $1/2$, and we would actually obtain

$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:a_{1}\notin{\mathbb{Q}}_{p},v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})=(p^{3}-p)\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/L}})=(c-15)/6\\}).$

This concludes another step of the recursion. Note that this point, we have
reached a base case. Here we use the convention that the discriminant of a
linear polynomial is $1$, and so
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/L}})=(c-15)/6\\})$
is $p^{-6}$ if $(c-15)/6=0$, and $0$ otherwise.

More generally, given an integer $i$, a local field $K$, and an integer $c$,
and $L$ an extension of $K$, our recursion reduces the calculation of
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/K}})=c\\})$
to
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{i+1}:v_{p}(\Delta_{P_{\alpha/F}})=c^{\prime}\\})$
for some extension $F/K$ contained in $L$ and $c^{\prime}$ an integer at most
$c$. Now consider the poset

${\mathcal{P}}=\\{(i,c,F):i\in{\mathbb{Z}}_{\geq 0},c\in{\mathbb{Z}},K\subset
F\subset L\\}$

with ordering given by $(i^{\prime},c^{\prime},F^{\prime})\leq(i,c,F)$ if
$i^{\prime}\geq i$, $c^{\prime}\leq c$, and $F^{\prime}\supset F$. Associating
the set $\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/F}})=c\\}$
to each element of $(i,c,F)$ of ${\mathcal{P}}$, we see that the recursive
step outlined above is simply the process of moving down on this partial
order. Now, the only base cases we know is that we can find the measure of
$\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/F}})=c\\}$ when
$F=L$, or when $c<0$ since nothing in ${\mathcal{O}}_{L}$ can have
discriminant with negative valuation. Let $BC=\\{(i,c,F):F=L\text{ or
}c<0\\}\subset{\mathcal{P}}$ represent these base cases. Now, if
${\mathcal{P}}$ were bounded from below by $BC$, then this recursive step
would be enough for us to calculate all
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/K}})=c\\})$
from our base cases, but it’s not. Indeed, as we saw from the above paragraph,
in the first case where $a_{1}=0$, we merely reduced calculating
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$
to calculating
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$.
Actually, our recursive step will just keep pushing this along, and move from
$(i,c,{\mathbb{Q}}_{p})$ to $(i+1,c,{\mathbb{Q}}_{p})$ to
$(i+2,c,{\mathbb{Q}}_{p})$, and so on. So, we need a further input to close
off the recursion. The trick here is simply to notice that

(1.3)
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})=p^{-6}\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{2}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c-30\\})$

where we have used the fact that dividing by $p$ changes the $\mu_{L}$ measure
of a set by a factor of $p^{[L:{\mathbb{Q}}_{p}]}$, and also

$\prod_{\alpha^{\prime},\alpha^{\prime\prime}\text{ conjugates of
}\alpha}(\alpha^{\prime}/p-\alpha^{\prime\prime}/p)=p^{-30}\prod_{\alpha^{\prime},\alpha^{\prime\prime}\text{
conjugates of }\alpha}(\alpha^{\prime}-\alpha^{\prime\prime}).$

In terms of ${\mathcal{P}}$, we are essentially establishing an equivalence
relation $(i+e(L/F),c,F)\sim(i,c-\deg(L/F)(\deg(L/F)-1)i/e,{\mathbb{Q}}_{p})$.
Now the set ${\mathcal{P}}/\sim$ carries the induced partial order, which is
bounded below by base cases $BC/\sim$.

In summary, calculating
$\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/K}})=c\\}$ is done
with the following four steps.

  1. (1)

Take a general $\alpha\in{\mathfrak{m}}_{L}^{i}$ and consider the lowest term
$a_{i}p^{i/e(L)}$ in the Teichmüller expansion.

  2. (2)

For each choice of $a_{i}$ in the Teichmüller expansion, we get a partition of
the Galois conjugates of $\alpha$, with each set in the partition containing
Galois conjugates with the same leading term.

  3. (3)

Using the partition, reduce the calculation of terms like
$\mu(\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/K}})=c\\})$ to
the calculation of terms like
$\mu(\\{\alpha\in{\mathfrak{m}}_{L}^{i+1}:v_{p}(\Delta_{P_{\alpha/F}})=c^{\prime}\\})$

  4. (4)

When $i$ gets large use the trick similar to Equation 1.3 to reduce the
calculation
$\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/K}})=c\\}$ to
$\\{\alpha\in{\mathfrak{m}}_{L}^{i-e}:v_{p}(\Delta_{P_{\alpha/K}})=c^{\prime}\\}$
where $c^{\prime}<c$ when $i$ gets large.

  5. (5)

Repeat the above steps until we reach a base case.

The main idea and novelty of this paper is succinctly summarized above and
there are little additional conceptual difficulties. Much of the paper is
dedicated to the many technical and combinatorial details in these
calculations, including the following. Most of the notational baggage comes
from these technicalities.

  * •

We need to enumerate all extensions of a given splitting type. Fortunately,
this is not terrible for tame extensions, but it still takes some bookkeeping.

  * •

There are extensions like
$L={\mathbb{Q}}_{p}[\zeta_{p^{3}-1},(\zeta_{p^{3}-1}p)^{1/4}]$. So,
Teichmüller expansions of elements in ${\mathcal{O}}_{L}$ are of the form
$\sum a_{i}(\zeta_{p^{3}-1}p)^{1/4})^{i}\in{\mathcal{O}}_{L}$. Now although we
can still express the Galois conjugates of $\sum
a_{i}(\zeta_{p^{3}-1}p)^{1/4})^{i}\in{\mathcal{O}}_{L}$, and find its Galois
conjugates fairly easily, there are combinatorics involved in determining the
partitions of Galois conjugates arising from a given $a_{i}$, and determining
the number of $a_{i}$ giving a specified partition.

  * •

$L$ can be a product of field extensions rather than a single field extension.

  * •

We need to sum over all extensions $L$ with a given splitting type. We only
considered one $L$ above.

We remark that without summing over all extensions $L$ with a given splitting
type, we get only get “quasi-rationality” for $\rho$, where quasi-rationality
means that as a function of $p$, it is rational in arithmetic progressions
with a fixed modulus. Although we do not calculate it out in this paper,
following through with the same methods outlined above, we would see that the
proportion of polynomials in ${\mathbb{Z}}_{p}[x]$ of degree $9$ generating
${\mathbb{Q}}_{p}[\zeta_{p^{3}-1},p^{1/3}]$ is a quasi-rational function in
terms of $p$, so that its restriction to $p\equiv 0\mod 3$ is a rational
function, and its restriction to $p\equiv 1,2\mod 3$ is another rational
function.

###  1.2. Previous work

#### Densities of random polynomial with fixed splitting data

The study of statistics of roots of random polynomials dates back to the work
of Bloch and Pölya who proved asymptotic bounds on the number of expected
roots of polynomials with coefficients in $\\{-1,0,1\\}$. Then, Kac gives an
exact formula for the expected number of roots of polynomials with Gaussian
distributed coefficients. For a survey and related results, there is [4] and
[10].

The questions surrounding the expected number of roots of random polynomials
over $p$-adic fields were studied in [7],[2], and [9]. Following, there is the
paper [1] (the conjecture in which is precisely the one we attack in this
paper), which shows that the density of $p$-adic polynomials with fixed number
of roots is a rational function in $p$, and satisfies the functional equation
sending $p$ to $p^{-1}$. There is also the aforementioned paper [3], from
which we obtain a crucial formula for finding the average number of roots of a
random polynomial in ${\mathcal{O}}_{L}$ for a fixed extension $L$.

#### Carousels

Figure 1. This is a visualization of the main idea behind Proposition 4.2. It
is a calculation of the discriminant of the minimal polynomial over
${\mathbb{Q}}_{p}$ of
$(\alpha_{1},\alpha_{1}^{\prime})=(\zeta_{p^{3}-1}+p^{1/2},p^{1/2})$. Here
$\alpha_{2}=\zeta_{p^{3}-1}-p^{1/2}$,
$\alpha_{3}=\zeta_{p^{3}-1}^{p}+p^{1/2}$,
$\alpha_{4}=\zeta_{p^{3}-1}^{p}-p^{1/2}$,
$\alpha_{5}=\zeta_{p^{3}-1}^{p^{2}}+p^{1/2}$,
$\alpha_{6}=\zeta_{p^{3}-1}^{p^{2}}-p^{1/2}$, $\alpha_{2}^{\prime}=-p^{1/2}$.
The discriminant of $P_{(\alpha_{1},\alpha_{1}^{\prime})/{\mathbb{Q}}_{p}}$ is
the product of the pairwise differences of $\alpha_{i}$ and
$\alpha_{i}^{\prime}$. However, we can see clearly from the first terms of the
Teichmüller expansion of $\alpha_{i}$ and $\alpha_{i}^{\prime}$ the valuation
of some pairwise differences. This is visualized in the diagram in that for
any two elements not in the same inner circle, such as $\alpha_{1}$ and
$\alpha_{1}^{\prime}$, we know that $v_{p}(\alpha_{1}-\alpha_{1}^{\prime})=0$.
Thus, the calculation of the valuation of the discriminant of
$P_{(\alpha_{1},\alpha_{1}^{\prime})/{\mathbb{Q}}_{p}}$ is reduced to the
calculation of valuation of the pairwise products of elements in the inner
circles which is precisely the valuation of
$P_{\alpha_{1}/{\mathbb{Q}}_{p}[\zeta_{p^{3}-1}]}$ and
$P_{\alpha_{2}/{\mathbb{Q}}_{p}[0]}$.

In this paper, we make heavy use of the behavior of sets of Galois conjugates
under the Galois group. In particular, we consider the distances between
Galois conjugates. One may picture this as a carousel, as in 1. In the figure,
one sees that the conjugates $\alpha_{1}$ and $\alpha_{2}$ are closer to each
other than $\alpha_{1}$ and any other conjugate. As a result, there is a
subgroup of the Galois group stabilizing the set
$\\{\alpha_{1},\alpha_{2}\\}$, so pictorially, $\alpha_{1},\alpha_{2}$ are in
a carousel within a bigger carousel.

Other papers have made use of such structures. For example, in [5] and [6],
one sees that much information about the hyperelliptic curve $y^{2}=f(x)$ can
be obtained from the carousel of roots of $f(x)$. Note that in the above
paper, carousels are referred to as “cluster pictures.”

##  2\. Notation/Convention

We adopt a similar notation as in [3]. We fix a local field $K$ and an
algebraic closure $\bar{K}$. We equip $\bar{K}$ with the $p$-adic norm
normalized by $||p||=p^{-[K:{\mathbb{Q}}_{p}]}$. Let ${\mathcal{O}}_{K}$ be
the discrete valuation ring of $K$. Let ${\mathfrak{m}}_{K}$ be the maximal
ideal of ${\mathcal{O}}_{K}$. Let $\lambda_{K}$ be the associated Haar measure
on $K$. Let $e_{\operatorname{base}}$ be the ramification index of $K$, and
$f_{\operatorname{base}}$ be the inertia degree of $K$. Similarly, for an
étale extension $L=\prod_{i}L_{i}$, where $L_{i}$ are field extensions of $K$,
we denote ${\mathcal{O}}_{L}$ to be $\prod_{i}{\mathcal{O}}_{L_{i}}$ and
${\mathfrak{m}}_{L}$ to be $\prod_{i}{\mathfrak{m}}_{L_{i}}$.

Let $\Omega_{d}$ be the space of polynomials of degree at most $d$ with
coefficients in ${\mathcal{O}}_{K}$, which is set-theoretically identified
with ${\mathcal{O}}_{K}^{d+1}$. With this identification, we have a
probability measure $\mu_{r}$ on $\Omega_{r}$ associated to
$\lambda_{K}^{\otimes d+1}$ on ${\mathcal{O}}_{K}^{d+1}$. When we refer to a
random degree $d$ polynomial, we mean it in this sense. Also, for an étale
extension $E/K$, and a polynomial $f\in K[x]$, we call the finite set
$\operatorname{Hom}(K[x]/f,E)$ the roots of $f$ in $E$. Furthermore, if
$\varphi\in\operatorname{Hom}(K[x]/f,E)$ is surjective, we say $\varphi$ is
“new” in $E$. Note that this definition makes sense even for $f$ such that
$\deg(f)>\dim_{K}E$, but we will only concern ourselves with the case where
$\deg(f)=\dim_{K}E$.

Fix a positive integer $d$, and integers $e_{1},\dots,e_{m}$ and
$f_{1},\dots,f_{m}$ such that $d=\sum_{i}e_{i}f_{i}$. Let $\sigma$ be the
multiset of pairs $\\{(e_{1},f_{1}),\dots,(e_{m},f_{m})\\}$, which we will
write as $(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$. We call such symbols splitting
types. Let $L$ be an étale extension of $K$. We say $\pi_{K}$ has splitting
type $\sigma$ in $L$ if $\pi_{K}$ factors as $\prod_{i}\pi_{i}^{e_{i}}$ in
${\mathcal{O}}_{L}$, where $\pi_{i}$ are prime ideals of ${\mathcal{O}}_{L}$
having residue degree $f_{i}$. For any field $L$, we define $e(L)$ to be the
ramification index of $L$ over ${\mathbb{Q}}_{p}$, where $p$ is the
characteristic of the residue field of $L$, and $f(L)$ to be the inertia
degree of $L$. Similarly, we define $e(L/K)=e(L)/e(K)$ for any extension $L/K$
to be the relative ramification index of $L$ over $K$, and similarly for
$f(L/K)$. Finally, let $[m]$ denote the set $\\{1,\dots,m\\}$.

Let $q$ be the size of the residue field of $K$. Fix a field extension $L/K$
of inertia degree $f$. Let $\pi_{L}$ be a uniformizer of $L$. Then, we may
write elements of $L$ uniquely as $\sum_{i=-\infty}^{\infty}a_{i}\pi_{L}^{i}$,
where $a_{i}\in L$ is either a $q^{f}-1$-th root of unity or $0$, and there is
some cutoff $r\in{\mathbb{Z}}$ so that $a_{i}=0$ for $i<r$. This
representation of an element in $L$ is called a Teichmüller expansion.

##  3\. Main Results

For a splitting type $\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$, let
$\rho(\sigma,K)$ denote the density of polynomials $P\in{\mathcal{O}}_{K}[x]$
of degree $d=\sum_{i=1}^{m}e_{i}f_{i}$ such that $P$ generates an étale
extension $L$ over $K$ and $\pi_{K}$ has splitting type $\sigma$ in $L$. Let
$\alpha(\sigma,K)$ (resp. $\beta(\sigma,K)$) be the density of such
polynomials above with the added condition that $P$ is monic (resp. $P$ is
monic and $P(x)\mod\pi_{K}=x^{n}$. Our main result is the following.

######  Theorem 3.1.

Fix a splitting type $\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$.

  1. (1)

If $K,K^{\prime}$ are two local fields with residue characteristic $p$ such
that $\gcd(p,e_{i})=1$ for all $i\in[m]$, such that $e(K)=e(K^{\prime})$ and
$f(K)=f(K^{\prime})$, then $\rho(\sigma,K)=\rho(\sigma,K^{\prime})$. In
particular, we may replace the notation $\rho(\sigma,K)$ with
$\rho(\sigma,e,f;p)$ when $\gcd(p,e_{i})=1$ for all $i\in[m]$. Similarly,
$\alpha(\sigma,K)$ and $\beta(\sigma,K)$ may be replaced with
$\alpha(\sigma,e,f;p)$ and $\beta(\sigma,e,f;p)$.

  2. (2)

Fix $e$ and $f$ as well. Then, for primes $p$ such that $\gcd(p,e_{i})=1$ for
all $i\in[m]$, $\rho(\sigma,e,f;p)$, $\alpha(\sigma,e,f;p)$, and
$\beta(\sigma,e,f;p)$ are rational functions of $q=p^{f}$.

We note that it is conjectured in [1, Conjecture 1.2] that part 2 of this
theorem also holds for $p$ not satisfying the gcd condition, but we will not
prove that in this paper.

We also have an asymptotic for the large $q$ limit of $\rho$.

######  Theorem 3.2.

Let $q=p^{f}$. We have

$\rho(\sigma,e,f;p)\sim\frac{1}{\operatorname{perm}(\sigma)\prod_{i=1}^{m}f_{i}q^{\sum_{i=1}^{m}(e_{i}-1)f_{i}}}$

as $q\to\infty$.

More precisely, if one takes $p$ to infinity, $f$ to infinity, or both,

$\lim_{p^{f}\to\infty}\rho(\sigma,e,f;p)\operatorname{perm}(\sigma)\prod_{i=1}^{m}f_{i}q^{\sum_{i=1}^{m}(e_{i}-1)f_{i}}$

approaches $1$. Indeed, this agrees with the calculations 1.1 done in [1].

We now launch into the preliminaries of the proof of these theorems. The goal
for the rest of this section to transform the problem of finding
$\rho(\sigma,K)$ to the problem of finding measures of sets of the form
$\mu_{L}(\\{\alpha\in{\mathfrak{m}}_{L}^{i}:v_{p}(\Delta_{P_{\alpha/{\mathbb{Q}}_{p}}})=c\\})$.
Let $L/K$ be an étale extension of degree $d$. For $U\subset L$ an open
subset, let $Z_{U,L}^{\operatorname{new}}:\Omega_{d}\to{\mathbb{Z}}$ be a
random variable sending $P$ to the number of new roots of $P$ in $U$. With
this notation, we have that

######  Lemma 3.3.

Fix a local field $K$ and a splitting type
$\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$. For $L=\prod_{i=1}^{m}L_{i}$,
and $I\subset[m]$ let $U_{L,I}=\prod_{i\in
I}{\mathfrak{m}}_{L_{i}}\times\prod_{i\in[m]\setminus
I}{\mathcal{O}}_{L_{i}}$. Then,

  1. (1)

$\rho(\sigma,K)=\sum_{\begin{subarray}{c}\textup{isom classes of }L,\\\
\pi_{K}\textup{ has splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]}{\operatorname{Aut}(L)}.$

  2. (2)

$\alpha(\sigma,K)=\sum_{\begin{subarray}{c}\textup{isom classes of }L,\\\
\pi_{K}\textup{ has splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{{\mathcal{O}}_{L},L}^{\operatorname{new}}]}{\operatorname{Aut}(L)}.$

  3. (3)

$\beta(\sigma,K)=\sum_{I\subsetneq[m]}(-1)^{|I|+1}\sum_{\begin{subarray}{c}\textup{isom
classes of }L,\\\ \pi_{K}\textup{ has splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{U_{L,I},L}^{\operatorname{new}}]}{\operatorname{Aut}(L)}.$

###### Proof.

We first partition degree $d$ polynomials $P$ by the isomorphism classes of
étale extensions that they generate. Thus

$\rho(\sigma,K)=\sum_{\begin{subarray}{c}\text{isom classes of }L,\\\
\pi_{K}\text{ has splitting type }\sigma\text{ in }L\end{subarray}}\rho(L,K),$

where $\rho(L,K)$ is the proportion of degree $d$ polynomials $P$ over $K$
such that $K[x]/P\cong L$. In other words, this is the proportion of
polynomials with a new root in $L$. Now, for any such $P$, it has
$\operatorname{Aut}(L)$ roots in $L$. Thus, $\operatorname{Aut}(L)\rho(L,K)$
is precisely the average number of new roots of a random degree $d$ polynomial
in $L$. Hence,

$\operatorname{Aut}(L)\rho(L,K)={\mathbb{E}}[Z_{L,L}^{\operatorname{new}}].$

The formula for $\alpha$ follow after noting that monic polynomials are
precisely the ones for which all roots lie in the ring of integers of the
associated étale extension. Now, monic polynomials $P(x)$ that is
$x^{n}\mod\pi_{K}$ are precisely the ones for which all roots lie in
${\mathcal{O}}_{L}$, with at least one root lies in ${\mathfrak{m}}_{L}$.
Thus, we obtain the formula for $\beta$ after applying inclusion exclusion. ∎

In the following theorem [3, Theorem 4.8], Caruso gives a formula for
${\mathbb{E}}[Z_{U,L}]$, for open subsets $U\subset{\mathcal{O}}_{L}$. Let
$w(d)=w_{f(K)}(d)=\frac{q^{d+1}-q^{d}}{q^{d+1}-1}$.

######  Theorem 3.4 (Theorem 4.5.5 [3]).

Let $K$ be a local field, and $L$ be an étale extension of degree $d$. Let
$U\subset{\mathcal{O}}_{L}$ be any open set. Then, the average number of new
roots of a random degree $d$ polynomial over $K$ in $L$ is

${\mathbb{E}}[Z_{U,L}^{\operatorname{new}}]=||D_{L}||w(d)\int_{\alpha\in
U}\frac{1}{[{\mathcal{O}}_{L}:{\mathcal{O}}_{K}[\alpha]]}d\lambda_{L}.$

Thus, we have turned a statement about polynomials over $K$ to a statement
about elements in étale extensions of $K$. We can apply the above formula
directly to calculate
${\mathbb{E}}[Z_{{\mathcal{O}}_{L},L}^{\operatorname{new}}]$ and
${\mathbb{E}}[Z_{U_{L},L}^{\operatorname{new}}]$ and thus obtain $\alpha$ and
$\beta$. However, since we require $U$ to be a subset of ${\mathcal{O}}_{L}$
in the above formula, we see that we cannot use this to calculate
${\mathbb{E}}[Z_{L,L}]$ and obtain $\rho$ directly. However, we will prove the
following in Section 6.

######  Theorem 3.5.

Let $K$ be a local field and let $q=p^{f(K)}$. Let $L$ be an étale extension
of degree $d$. So, $L=\prod_{i=1}^{m}L_{i}$, where $L_{i}$ are field
extensions of $K$. For any $A\subset[m]$, denote $L_{A}=\prod_{i\in A}L_{i}$
and $A^{c}=[m]\setminus A$. Then,

${\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]=\sum_{A\subset[m]}\frac{w([L:K])}{w([L_{A}:K])w([L_{A^{c}}:K])}{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}],$

where we make the convention that if $A$ is empty,
${\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]={\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A}},L_{A}}^{\operatorname{new}}]=1$
and $[L_{A}:K]=0$ (so that $w([L_{A}:K])=1$).

We are concerned mainly with tamely ramified extensions. Thus, with Theorem
3.5 and the following characterization of tamely ramified extensions, we can
transform our sum

$\sum_{\begin{subarray}{c}\text{isom classes of }L,\\\ \pi_{K}\text{ has
splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]}{\operatorname{Aut}(L)},$

which involves terms of the form ${\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]$,
to something that involves only terms of the form
${\mathbb{E}}[Z_{{\mathcal{O}}_{L},L}]$ and
${\mathbb{E}}[Z_{{\mathfrak{m}}_{L},L}]$, which we should be able to calculate
using Caruso’s formula (3.4). The theorem below is given in [8, Chapter 16].

######  Theorem 3.6.

Fix a uniformizer $\pi_{K}\in K$ and a positive integer $n$ coprime to $p$.
Then, there are $n$ tamely and totally ramified extensions of degree $n$ of
$K$ in $\bar{K}$. These extensions may be split into $\gcd(q-1,n)$ isomorphism
classes, each class isomorphic to $K[(\zeta_{q-1}^{j}\pi_{K})^{1/n}]$, for
some $0\leq j<\gcd(q-1,n)$.

Since every tamely ramified extension is an unramified extension followed by a
tamely and totally ramified extension, we see that every tamely ramified
extension $L$ with ramification index $e_{0}$ and inertia degree $f_{0}$ over
$K$ is of the form
$K[\zeta_{q^{f_{0}}-1},(\zeta_{q^{f_{0}}-1}^{j}\pi_{K})^{1/e_{0}}]$ for some
$0\leq j<\gcd(q^{f_{0})}-1,e_{0})$. We will now denote this field as $L_{j}$.
Thus, extensions $L/K$ with splitting type $\sigma$ are of the form
$\prod_{i}L_{j_{i}}$, where $0\leq j_{i}\leq\gcd(q^{f_{i}}-1,e_{i})$. This
also shows that
$|\operatorname{Aut}(L_{j_{i}}/K)|=f_{i}\gcd(q^{f_{i}}-1,e_{i})$.

Let

$\operatorname{perm}(\sigma)=\prod_{e,f}(\text{occurrences of }e^{f}\text{ in
}\sigma)!.$

Now we are ready to transform our formula in terms of ${\mathbb{E}}[Z_{L,L}]$,
to a formula in terms ${\mathbb{E}}[Z_{{\mathcal{O}}_{L},L}]$ and
${\mathbb{E}}[Z_{{\mathfrak{m}}_{L},L}]$. With Lemma 3.3, Theorem 3.5, and
Theorem 3.6, we have

######  Corollary 3.7.

Let $\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$ be a splitting type with
$p\nmid e_{i}$. Let $d=\sum e_{i}f_{i}$. Then,

$\rho(\sigma,K)=\frac{w(d)\sum_{A\subset[m]}N_{1,A}^{\prime}N_{2,A}^{\prime}}{\operatorname{perm}(\sigma)}$

where

$N_{1,A}^{\prime}=\frac{\sum_{\begin{subarray}{c}L_{A}=\prod_{i\in
A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]}{\prod_{i\in
A}f_{i}\gcd(q^{f_{i}}-1,e_{i})w([L_{A}:K])}$

and

$N_{2,A}^{\prime}=\frac{\sum_{\begin{subarray}{c}L_{A^{c}}=\prod_{i\notin
A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}]}{\prod_{i\notin
A}f_{i}\gcd(q^{f_{i}}-1,e_{i})w([L_{A^{c}}:K])}.$

###### Proof.

We first have by Lemma 3.3 that

$\rho(\sigma,K)=\sum_{\begin{subarray}{c}\text{isom classes of }L,\\\
\pi_{K}\text{ has splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]}{\operatorname{Aut}(L)}.$

Now as $j_{i}$ ranges in $[\gcd(q^{f_{i}}-1),e_{i})]$, $L=\prod_{i}L_{j_{i}}$
ranges among étale extensions of splitting type $\sigma$ over $K$. However,
this will hit the same extension
$\operatorname{perm}(\sigma)\left|\frac{\operatorname{Aut}(L)}{\prod_{i}\operatorname{Aut}(L_{j_{i}})}\right|^{-1}$
times. Thus,

$\displaystyle\sum_{\begin{subarray}{c}\text{isom classes of }L,\\\
\pi_{K}\text{ has splitting type }\sigma\text{ in
}L\end{subarray}}\frac{{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]}{\operatorname{Aut}(L)}$
$\displaystyle=\operatorname{perm}(\sigma)^{-1}\prod_{i}|\operatorname{Aut}(L_{j_{i}})|^{-1}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]$
$\displaystyle=\operatorname{perm}(\sigma)^{-1}\prod_{i}\frac{1}{f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]$

Now, by Theorem 3.5,

$\displaystyle\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]$
$\displaystyle=\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\sum_{A\subset[m]}\frac{w(d)}{w([L_{A}:K])w([L_{A^{c}}:K])}{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}]$
$\displaystyle=w(d)\sum_{A\subset[m]}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\frac{{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]}{w([L_{A}:K])}\frac{{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}]}{w([L_{A^{c}}:K])}$

where $L_{A}=\prod_{i\in A}L_{j_{i}}$ and $L_{A^{c}}=\prod_{i\in
A^{c}}L_{j_{i}}$. Thus, factoring the inner sum into a product, we get

$\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{L,L}^{\operatorname{new}}]=w(d)\sum_{A\subset[m]}\left(\sum_{\begin{subarray}{c}L_{A}=\prod_{i\in
A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\frac{{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]}{w([L_{A}:K])}\right)\left(\sum_{\begin{subarray}{c}L_{A^{c}}=\prod_{i\in
A^{c}}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\frac{{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}]}{w([L_{A^{c}}:K])}\right).$

∎

Although the formula in Corollary 3.7 seems complicated, one just needs to
note that since by definition $w(d)$ is a rational function of $q=p^{f(K)}$,
Theorem 3.1 for $\rho$ will be proven if we can show that

$\frac{1}{\prod_{i\in
A}f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L_{A}=\prod_{i\in
A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{{\mathcal{O}}_{L_{A}},L_{A}}^{\operatorname{new}}]$

and

$\frac{1}{\prod_{i\in
A^{c}}f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L_{A^{c}}=\prod_{i\in
A^{c}}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{{\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}}^{\operatorname{new}}]$

are rational functions in terms of $q$, and that these expressions depends on
$K$ through $e(K)$ and $f(K)$. Similarly, following the same process for
$\alpha$ and $\beta$, one simply needs to show that

$\prod_{i}\frac{1}{f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{{\mathcal{O}}_{L},L}^{\operatorname{new}}]$

and

$\prod_{i}\frac{1}{f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{U_{L,I},L}^{\operatorname{new}}]$

($U_{L,I}$ defined as in Lemma 3.3) are rational functions in terms of $q$,
and that these expressions depends on $K$ through $e(K)$ and $f(K)$. In fact,
we will show something more general, that this is true for

$\frac{1}{\prod_{i=1}^{m}f_{i}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}{\mathbb{E}}[Z_{\prod_{i=1}^{m}{\mathfrak{m}}_{L_{i}}^{b_{i}},\prod_{i}L_{j_{i}}}^{\operatorname{new}}]$

for any tamely ramified splitting type
$\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$ and nonnegative integers
$b_{1},\dots,b_{m}$.

Note that combining Theorem 3.4 with Corollary 3.7 gives us the formula

(3.1)
$\rho(\sigma,K)=\frac{w(d)\sum_{A\subset[m]}N_{1,A}N_{2,A}}{\operatorname{perm}(\sigma)q^{\sum_{i=1}^{m}(e_{i}-1)f_{i}}\prod_{i=1}^{m}f_{i}},$

where

$N_{1,A}=\frac{\sum_{\begin{subarray}{c}L_{A}=\prod_{i\in A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\int_{\alpha\in{\mathcal{O}}_{L_{A}}}\frac{1}{[{\mathcal{O}}_{L_{A}}:{\mathcal{O}}_{K}[\alpha]]}d\lambda_{L_{A}}}{\prod_{i\in
A}\gcd(q^{f_{i}}-1,e_{i})}$

and

$N_{2,A}=\frac{\sum_{\begin{subarray}{c}L_{A^{c}}=\prod_{i\notin
A}L_{j_{i}}\\\ j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\int_{\alpha\in
m_{L_{A^{c}}}}\frac{1}{[{\mathcal{O}}_{L_{A^{c}}}:{\mathcal{O}}_{K}[\alpha]]}d\lambda_{L_{A^{c}}}}{\prod_{i\notin
A}\gcd(q^{f_{i}}-1,e_{i})}.$

Finally, the integral in Caruso’s formula (3.4) involves terms of the form
$[{\mathcal{O}}_{L}:{\mathcal{O}}_{K}[\alpha]]$, which we will transform with
the following lemma.

######  Lemma 3.8.

Let $P$ be a monic polynomial in ${\mathcal{O}}_{K}[x]$, which generates an
étale extension $L=\prod_{i}L_{i}/K$. Let $\Delta_{P}$ be the polynomial
discriminant of $P$, and let $\Delta_{L}$ be the discriminant of the étale
extension $L$ (aka $N(\prod_{i}\pi_{L_{j_{i}}}^{e(L_{j_{i}})-1})$, where
$\pi_{L}\in{\mathcal{O}}_{L}$ is a uniformizer). Then,
$\pi_{K}^{2\log_{q}[{\mathcal{O}}_{L}:{\mathcal{O}}_{K}[x]/P]}\Delta_{L}=\Delta_{P}$.

Thus, combining Theorem 3.4 and Lemma 3.8, we can reduce our problem to
showing that

$\frac{q^{\frac{\sum(e_{i}-1)f_{i}}{2}}}{\prod_{i=1}^{m}\gcd(q^{f_{i}}-1,e_{i})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i}}-1,e_{i})]\end{subarray}}\int_{\alpha\in\prod_{i}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}}q^{-\frac{1}{2}v_{K}(\Delta_{P_{\alpha/K}})}d\lambda_{L}$

is a rational function in terms of $q$ and depends on $K$ only through $e(K)$
and $f(K)$. Sections 4 and 5 are dedicated to the calculation of this sum and
showing that it is rational function in terms of $q$.

##  4\. Generating Functions and Recurrence Relations

First, we will make a switch from relative notation to absolute notation. So,
let $\\{L_{i}\\}_{i=1}^{m}$ be field extensions of $K$. We define
$e_{i},f_{i}$ to be the absolute ramification index and inertia degree of
$L_{i}$. Now, for nonnegative integers $b_{i}$, we define

$a(\\{L_{i}\\}_{i=1}^{m},K,c,\\{b_{i}\\}_{i=1}^{m}):=\mu(\\{\alpha\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{i}}^{b_{i}}:v_{p}(\Delta_{P_{\alpha/K}})=c\\}).$

Also, fix two positive integers $e_{\operatorname{base}}$,
$f_{\operatorname{base}}$, representing the ramification index and inertia
degree of our base extension, which we will denote by $K$. Then, let
$\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$ be a tame splitting type, so that
$p\nmid e_{i}/e_{\operatorname{base}}$ for any $i$. Let
$e_{i,\operatorname{rel}}=e_{i}/e_{\operatorname{base}}$ and
$f_{i,\operatorname{rel}}=f_{i}/f_{\operatorname{base}}$. Then, we define

$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\}_{i=1}^{m},p)=\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]}a(\\{L_{j_{i}}\\},K,c,\\{b_{i}\\}_{i=1}^{m})$

where $K$ is any $p$-adic field with $e(K)=e_{\operatorname{base}}$ and
$f(K)=f_{\operatorname{base}}$. As the notation suggests, the value of this
expression will be shown to only depend on $K$ through $e(K)$ and $f(K)$. We
will often omit the dependence of $a$ on $p$.

######  Example 4.1.

As a base case, we can see that from the definition,

$a((e^{f}),e,f,c,\\{b\\})=a(\\{K\\},K,c,\\{b\\})=\begin{cases}p^{-bf}&c=0\\\
0&c\neq 0\end{cases},$

for any $j$ since the discriminants of all degree $1$ polynomials are all
$1=p^{0}$.

Our main proposition of this section, as well as the main technical part of
our paper is

######  Proposition 4.2.

Fix a tamely ramified splitting type
$\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$. Then, the definition of

$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$

doesn’t depend on the choice of $K$ (as long as $e(K)=e_{\operatorname{base}}$
and $f(K)=f_{\operatorname{base}}$). Moreover, we have the following
recurrence.

$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})=\sum_{\begin{subarray}{c}\textup{partitions
}\\{E_{l}\\}_{l=1}^{k}\text{ of }[m]\\\
\\{n_{O,l}\\}_{l=1}^{k}\in{\mathbb{Z}}_{\geq
1}^{k}\end{subarray}}P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$
$\sum_{\begin{subarray}{c}\\{c_{l}\\}_{l=1}^{k}\in{\mathbb{Z}}_{\geq 0}^{k}\\\
\sum
n_{O,l}c_{l}=r(c)\end{subarray}}\prod_{l}a(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}}).$

Where ${\mathbb{1}}_{I}(i)$ denotes the indicator function of $I$ and
$\sigma_{l}=(e_{i}^{f_{i}})_{i\in E_{l}}$.

The notation used in this proposition will be introduced shortly. The main
idea of this proposition is that we are expressing
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$ in
terms of

$a(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}}).$

After introducing the appropriate notations, it will be apparent that this
seemingly complicated term satisfies the four conditions:

  1. (1)

$\sigma_{l}\subset\sigma$,

  2. (2)

$h(\beta,n_{O,l})\geq 1$,

  3. (3)

$\frac{n_{O,l}}{h(\beta,n_{O,l})}\geq 1$,

  4. (4)

and $\min_{i}\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}}>\min_{i}\\{b_{i}\\}$.

By induction on degree, we have theoretically already calculated the term

$a(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}})$

whenever one of the first three conditions is satisfied strictly. Thus, this
proposition effectively reduces the calculation of
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$ to
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}+{\mathbb{1}}_{I}(i)\\})$.

Now we introduce the relevant terms in the proposition. Fix a tamely ramified
splitting type $(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$. Let
$d_{i}=e_{i,\operatorname{rel}}f_{i,\operatorname{rel}}$, and so
$d=\sum_{i}d_{i}$. Also, let
$\beta=\beta(\\{e_{i}\\},e_{\operatorname{base}},\\{b_{i}\\})=\min\frac{b_{i}}{e_{i,\operatorname{rel}}}$.
In other words, for any choices of $i\in[m]$ and
$j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]$, $\beta$ is the
smallest possible valuation of any element in
${\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}$. We will often omit the dependence of
$\beta$ on $\\{e_{i}\\},e_{\operatorname{base}},$ and $\\{b_{i}\\}$, when
these inputs are clear from context. Let
$I=I(\\{e_{i}\\},e_{\operatorname{base}},\\{b_{i}\\})=\\{i\in[m]:\frac{b_{i}}{e_{i,\operatorname{rel}}}=\beta$.
Let $\operatorname{denom}(\beta)$ denote the denominator of $\beta$ in its
lowest form. Note that if $\frac{a}{b}=\beta$, then
$\operatorname{denom}(\beta)=\frac{b}{\gcd(a,b)}$. Let $T_{f_{i},p}$ denote
the Teichmüller representatives (aka $p^{f_{i}}-1$ roots of unity and $0$).

We make a few more definitions which are more ad hoc, and so the reasons that
we have defined these terms as such may be clearer if one reads the proof of
Proposition 4.2, and refer to these definitions as needed. Fix a splitting
type $\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$, a partition
$\\{E_{l}\\}_{l=1}^{j}$ of $[m]$, an integer $n_{O,l}$ associated to each
element $E_{l}$ of the partition, and an integer $b_{i}$ for each $i\in[m]$.
Let

$r(c)=r(c,\\{E_{l}\\},\\{n_{O,l}\\})=c-\frac{\beta}{e_{\operatorname{base}}}(d(d-1)-\sum_{l}\sum_{i\in
E_{l}}d_{i}(\sum_{i\in E_{l}}(d_{i}/n_{O,l})-1)),$

and

$h(\beta,n_{O,l})=\begin{cases}\operatorname{denom}(\beta)&n_{O,l}\neq 1\\\
1&n_{O,l}=1\end{cases}.$

Now consider the set

$\prod_{i=1}^{m}[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}}]\times\prod_{i\in
I}T_{f_{i},p}.$

We consider three conditions on elements
$(\\{j_{i}\\}_{i=1}^{m},\\{a_{i,b_{i}}\\}_{i\in I})$ in this set:

  1. (1)

For all $i_{1},i_{2}\in I$, $i_{1},i_{2}$ belong to the same $E_{l}$ if and
only if $a_{i_{1},b_{i_{1}}}\pi_{L_{j_{i_{1}}}}^{b_{i_{1}}}$ and
$a_{i_{2},b_{i_{2}}}\pi_{L_{j_{i_{2}}}}^{b_{i_{2}}}$ are Galois conjugates.

  2. (2)

If $i\in I$, then $n_{O}(a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}})=n_{O,l}$.

  3. (3)

If $I\neq[m]$, then there is some $E_{l_{0}}$ containing all of $[m]\setminus
I$. Moreover, for any $i\in E_{l_{0}}\cap I$, $a_{i,b_{i}}=0$.

Let $P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ be
$\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}$ times
the number of elements in
$\prod_{i=1}^{m}[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}}]\times\prod_{i\in
I}T_{f_{i},p}$ satisfying these conditions. Note that if there is such an
$l_{0}$ as in the last condition, then unless $n_{O,l_{0}}=1$, the second and
third condition will contradict each other, and
$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ in this case would just be
$0$.

To give a bit more context to these notation, we will relate some of these
notations to the overview of proof given at the beginning.

Notation | Corresponding idea in overview
---|---
$\beta$ | lowest degree
$n_{O,l}$ | number of Galois conjugates of the lowest degree term
$\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in E_{l}}$ | going from $i$ to $i+1$
$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ |  number of choices of Teichmüller representatives $a_{i}$ giving a certain partition of Galois conjugates
$r(c)$ | $c^{\prime}$

###### Proof of Proposition 4.2.

First fix choices $\\{j_{i}\\}_{i\in[m]}$, and consider elements of
$L_{j_{i}}$ in power series expansion using Teichmüller representatives and
the uniformizer
$\pi_{L_{j_{i}}}=(\zeta_{p^{f_{ext}}-1}^{j_{i}}\pi_{K})^{1/e_{i,\operatorname{rel}}}$.
Then, the elements in ${\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}$ are of the form
$\alpha_{i}=\sum_{n=b_{i}}^{\infty}a_{i,n}\pi_{L_{j_{i}}}^{n}$. Our
proposition will be based on a rewriting of the following:

$\mu(\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:v_{p}(\Delta_{P_{(\alpha_{i})/K}})=c\\})$
$=\sum_{\begin{subarray}{c}i\in I\\\ a_{i,b_{i}}\in
T_{f_{i},p}\end{subarray}}\mu(\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:\text{
for all }i\in I,\alpha_{i}\in
a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}+\alpha_{i^{\prime}},\alpha_{i}^{\prime}\in{\mathfrak{m}}_{L_{j_{i}}}^{i+1},v_{p}(\Delta_{P_{(\alpha_{i})}})=c\\}.$

Now, let $(\alpha_{1},\dots,\alpha_{m})\in\prod_{i\in[m]}L_{j_{i}}$, then the
discriminant of its minimal polynomial $P_{(\alpha_{1},\dots,\alpha_{m})}$ is
equal to

$\prod_{(i,k)\neq(i^{\prime},k^{\prime}),k,k^{\prime}\in[d_{i}]}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}),$

where $\alpha_{i,k}$ are Galois conjugates of $\alpha_{i}$ over $K$. Our first
step is to calculate the $p$-adic valuation of minimal polynomials of elements
in

$\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:\text{
for all }i\in I,\alpha_{i}\in
a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}+\alpha_{i^{\prime}},\alpha_{i}^{\prime}\in{\mathfrak{m}}_{L_{j_{i}}}^{i+1}\\}$

in a clever way as in Figure 1. The next few paragraphs will explain in
rigorous detail the idea reflected in the figure.

Define

$C=C(\\{L_{i}\\},K,\\{b_{i}\\},\\{a_{i,b_{i}}\\}_{i\in I})$

to be the set of Galois conjugates of $a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}$, as
$i$ varies over $I$. If there is an $i$ so that
$b_{i}/e_{i,\operatorname{rel}}\neq\beta$, we also add $0$ to $C$. Define
$\bar{\gamma}$ to be the set of Galois conjugates of $\gamma$, and similarly
define $\bar{C}$ to be $C$ up to Galois conjugates. This way, for every $i,k$,
with $k\in[d_{i}]$,

$\alpha_{i,k}=\gamma_{i,k}+H.O.T.,$

for some $\gamma_{i,k}\in C$. Now we partition the set
$\\{(i,k):k\in[d_{i}]\\}$ with the equivalence relation
$(i,k)\sim(i^{\prime},k^{\prime})$ if
$\bar{\gamma}_{i,k}=\bar{\gamma}_{i^{\prime},k^{\prime}}$. Let $S_{\gamma}$
denote these partitions. Then,

(4.1)
$\prod_{(i,k)\neq(i^{\prime},k^{\prime}),k,k^{\prime}\in[d_{i}]}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})=\prod_{(i,k)\not\sim(i^{\prime},k^{\prime}),k,k^{\prime}\in[d_{i}]}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})\prod_{\gamma\in
C}\prod_{\begin{subarray}{c}(i,k),(i^{\prime},k^{\prime})\in S_{\gamma}\\\
(i,k)\neq(i^{\prime},k^{\prime})\end{subarray}}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}).$

In the case where $(i,k)\not\sim(i^{\prime},k^{\prime})$,
$\gamma_{i,k}\neq\gamma_{i^{\prime},k^{\prime}}$, and so if
$v_{K}(\gamma_{i,k}-\gamma_{i^{\prime},k})=\beta$, then since
$\alpha_{i,k}=\gamma_{i,k}+H.O.T.,$ we would also have that
$v_{K}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})=\beta$. Now, our
$\gamma$’s are either $0$ or of the form

$a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}=\zeta_{p^{f_{i}}-1}^{l_{i}}(\zeta_{p^{f_{i}}-1}^{j_{i}}\pi_{K})^{b_{i}/e_{i,\operatorname{rel}}}.$

Thus, $\gamma_{i,k}-\gamma_{i^{\prime},k^{\prime}}$ will be of the form

$\zeta_{e(p^{f}-1)}^{l}\pi_{K}^{\beta}-\zeta_{e^{\prime}(p^{f^{\prime}}-1)}^{l^{\prime}}\pi_{K}^{\beta}=\zeta_{e(p^{f}-1)}^{l}(1-\frac{\zeta_{e^{\prime}(p^{f^{\prime}}-1)}^{l^{\prime}}}{\zeta_{e(p^{f}-1)}^{l}})\pi_{K}^{\beta}$

and since $p\nmid e$ and $p\nmid e^{\prime}$, we see that the order of the
root of unity
$\frac{\zeta_{e^{\prime}(p^{f^{\prime}}-1)}^{l^{\prime}}}{\zeta_{e(p^{f}-1)}^{l}}$
is not divisible by $p$, and therefore

$v_{p}(\zeta_{e(p^{f}-1)}^{l}(1-\frac{\zeta_{e^{\prime}(p^{f^{\prime}}-1)}^{l^{\prime}}}{\zeta_{e(p^{f}-1)}^{l}})\pi_{K}^{\beta})=v_{K}(\pi_{K}^{\beta}),$

as desired. Now, the number of choices of pairs
$(i,k),(i^{\prime},k^{\prime})$ so that $(i,k)\not\sim(i^{\prime},k^{\prime})$
is equal to

$d(d-1)-\sum_{\gamma}|S_{\gamma}|(|S_{\gamma}|-1).$

Thus,

(4.2)
$v_{K}(\prod_{(i,k)\not\sim(i^{\prime},k^{\prime}),k,k^{\prime}\in[d_{i}]}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}))=\frac{1}{e_{\operatorname{base}}}(d(d-1)-\sum_{\gamma}|S_{\gamma}|(|S_{\gamma}|-1))\beta$

Now we compute the term

$\prod_{\gamma\in C}\prod_{\begin{subarray}{c}(i,k),(i^{\prime},k^{\prime})\in
S_{\gamma}\\\
(i,k)\neq(i^{\prime},k^{\prime})\end{subarray}}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}).$

We may express the inner product

$\prod_{\begin{subarray}{c}(i,k),(i^{\prime},k^{\prime})\in S_{\gamma}\\\
(i,k)\neq(i^{\prime},k^{\prime})\end{subarray}}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})$

as discriminant of some the minimal polynomial over $K[\gamma]$ of some
element in the étale extension of $K[\gamma]$. One can see that since
$S_{\gamma}$ consists of all pairs $(i,k)$ such that $\gamma_{i,k}=\gamma$,
this element may be constructed as the tuple consisting of $\alpha_{i,k_{i}}$,
as $i$ ranges between elements of $[m]$ such that $\gamma$ is a Galois
conjugate of $\gamma_{i}$, and $k_{i}$ is a choice of any Galois conjugate of
$\alpha_{i,k_{i}}$ so that $\gamma_{i,k_{i}}=\gamma$.

Now let $\bar{C}=C/\sim$, where $\gamma\sim\gamma^{\prime}$ if they are Galois
conjugates. Then, $\bar{C}$ gives a partition $\\{E_{\bar{\gamma}}\\}$ of
$[m]$ via the equivalence $i\sim i^{\prime}$ if
$\bar{\gamma}_{i}=\bar{\gamma}_{i^{\prime}}$. Thus, $E_{\bar{\gamma}}$ would
consist of all indices $i$ so that $\bar{\gamma}_{i}=\bar{\gamma}$. From now
on, we fix representatives $\gamma$ of $\bar{\gamma}$. With these
representatives, for any element $i\in E_{\bar{\gamma}}$, we may define
$\psi_{\gamma_{i}}\in Gal(K)$ be any element such that
$\psi_{\gamma_{i}}(\gamma_{i})=\gamma$. Now we can express what we shown in
the previous paragraph as the fact that

$\prod_{\begin{subarray}{c}(i,k),(i^{\prime},k^{\prime})\in S_{\gamma}\\\
(i,k)\neq(i^{\prime},k^{\prime})\end{subarray}}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})$

is equal to the discriminant of the minimal polynomial of
$(\psi_{\gamma_{i}}(\alpha_{i})-\gamma)_{i\in E_{\bar{\gamma}}}$ over
$K[\gamma]$. Thus, we have

$\displaystyle v_{p}(\prod_{\gamma\in
C}\prod_{\begin{subarray}{c}(i,k),(i^{\prime},k^{\prime})\in S_{\gamma}\\\
(i,k)\neq(i^{\prime},k^{\prime})\end{subarray}}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}))$
$\displaystyle=v_{p}(\prod_{\gamma\in
C}\Delta_{P_{(\psi_{\gamma_{i}}(\alpha_{i})-\gamma)}/K[\gamma]})$
$\displaystyle=v_{p}(\prod_{\bar{\gamma}\in\bar{C}}\prod_{\gamma\in\bar{\gamma}}\Delta_{P_{(\psi_{\gamma_{i}}(\alpha_{i})-\gamma)}/K[\gamma]})$
$\displaystyle=\sum_{\bar{\gamma}\in\bar{C}}\\#\bar{\gamma}v_{p}(\Delta_{P_{(\psi_{\gamma_{i}}(\alpha_{i})-\gamma)}/K[\gamma]}),$

where in RHS of the last equality, $\gamma$ is any chosen representative of
$\bar{\gamma}$.

Now it is clear from the definition that $|S_{\gamma}|=\sum_{i\in
E_{\bar{\gamma}}}d_{i}/\\#\bar{\gamma}$. Thus,

$\frac{\beta}{e_{\operatorname{base}}}(d(d-1)-\sum_{\bar{\gamma}}|S_{\gamma}|(|S_{\gamma}|-1))=\frac{\beta}{e_{\operatorname{base}}}(d(d-1)-\sum_{\bar{\gamma}\in\bar{C}}\sum_{i\in
E_{\bar{\gamma}}}d_{i}(\sum_{i\in
E_{\bar{\gamma}}}(d_{i}/\\#\bar{\gamma})-1)).$

Thus, combining Equation 4.2 and the above paragraph, we get that

(4.3)
$v_{p}(\Delta_{P_{(\alpha_{i})/K}})=\frac{\beta}{e_{\operatorname{base}}}(d(d-1)-\sum_{\bar{\gamma}\in\bar{C}}\sum_{i\in
E_{\bar{\gamma}}}d_{i}(\sum_{i\in
E_{\bar{\gamma}}}(d_{i}/\\#\bar{\gamma})-1))+\sum_{\bar{\gamma}\in\bar{C}}\\#\bar{\gamma}v_{p}(\Delta_{P_{(\psi_{\gamma_{i}}(\alpha_{i})-\gamma)}/K[\gamma]}).$

This is the extent of the idea reflected in Figure 1, and this concludes our
first step. In summary, what we have done in Equation 4.3 is express the
discriminant of the minimal polynomial of an element $(\alpha_{i})$ in terms
of

  1. (1)

$e_{\operatorname{base}}$, $d_{i}$, $d$

  2. (2)

$\bar{C}(\\{L_{i}\\},K,\\{b_{i}\\},\\{a_{i,b_{i}}\\})$, and
$E_{\bar{\gamma}}$, $\\#\bar{\gamma}$, $\psi_{\gamma_{i}}$,

where the first item consist of constants that we defined prior to the proof,
and the second item consist of objects depending only on the choices of
$a_{i,b_{i}}$ and $L_{i}$.

As we noted at the beginning of this proof, we will rewrite the formula

$a(\\{L_{j_{i}}\\},K,c,\\{b_{i}\\})=\mu(\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:v_{p}(\Delta_{P_{(\alpha_{i})/K}})=c\\})$
$=\sum_{\begin{subarray}{c}i\in I\\\ a_{i,b_{i}}\in
T_{f_{i},p}\end{subarray}}\mu(\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:\text{
for all }i\in I,\alpha_{i}\in
a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}+\alpha_{i^{\prime}},\alpha_{i}^{\prime}\in{\mathfrak{m}}_{L_{j_{i}}}^{i+1},v_{p}(\Delta_{P_{(\alpha_{i})}})=c\\},$

and with Equation 4.3, we will now do so. So, we will continue to fix a choice
of $\\{j_{i}\\}_{i\in[m]}$. Now fix a choice of $(a_{i,b_{i}})_{i\in I}$.
Then, we consider the associated term

$\mu(\\{\\{\alpha_{i}\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}:\text{
for all }i\in I,\alpha_{i}\in
a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}+\alpha_{i^{\prime}},\alpha_{i}^{\prime}\in{\mathfrak{m}}_{L_{j_{i}}}^{i+1},v_{p}(\Delta_{P_{(\alpha_{i})}})=c\\}.$

We apply $(\psi_{\gamma_{i}})_{i=1}^{m}$ and get

$\mu(\\{\\{\psi_{\gamma_{i}}(\alpha_{i})\\}_{i\in[m]}\in\prod_{i\in[m]}{\mathfrak{m}}_{\psi_{\gamma_{i}}(L_{j_{i}})}^{b_{i}}:\text{
for all }i\in
I,\psi_{\gamma_{i}}(\alpha_{i})=\psi_{\gamma_{i}}(a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}})+\alpha_{i}^{\prime},\alpha_{i}^{\prime}\in{\mathfrak{m}}_{\psi_{\gamma_{i}}(L_{j_{i}})}^{i+1},v_{p}(\Delta_{P_{(\alpha_{i})}})=c\\}.$

Using Equation 4.3, we obtain that the above is

$\mu(\\{\\{\alpha_{i}^{\prime}\\}_{i\in[m]}\in\prod_{i\in
I}{\mathfrak{m}}_{\psi_{\gamma_{i}}(L_{j_{i}})}^{b_{i}+1}\prod_{i\notin
I}{\mathfrak{m}}_{\psi_{\gamma_{i}}(L_{j_{i}})}^{b_{i}}:\sum_{\bar{\gamma}\in\bar{C}}\\#\bar{\gamma}v_{p}(\Delta_{P_{\alpha_{i}^{\prime}}/K[\gamma]})=r(c)\\}$
$=\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}\mu(\\{\\{\alpha_{i}^{\prime}\\}_{i\in[m]}\in\prod_{i\in
E_{\bar{\gamma}}}{\mathfrak{m}}_{\psi_{\gamma_{i}}(L_{j_{i}})}^{b_{i}+{\mathbb{1}}_{I}(i)}:v_{p}(\Delta_{P_{\alpha_{i}^{\prime}}/K[\gamma]})=c_{\bar{\gamma}}\\}$
$=\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

Thus, we have

$a(\\{L_{j_{i}}\\},K,c,\\{b_{i}\\})=\sum_{\begin{subarray}{c}i\in I\\\
a_{i,b_{i}}\in
T_{f_{i},p}\end{subarray}}\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq
0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

With this, we may now calculate
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\}_{i=1}^{m})$.
We have

$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\}_{i=1}^{m})=\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]}a(\\{L_{j_{i}}\\},K,c,\\{b_{i}\\}_{i=1}^{m})$
$=\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]}\sum_{\begin{subarray}{c}i\in
I\\\ a_{i,b_{i}}\in T_{f_{i},p}\end{subarray}}$ (4.4)
$\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

At this point, we have something that somewhat resembles the proposition we
are proving. One annoying term in our sum is the
$\psi_{\gamma_{i}}(L_{j_{i}})$, but now we will fix this. It will turn out to
“equidistribute” nicely over extensions of $K[\gamma]$, as $j_{i}$ and
$a_{i,b_{i}}$ varies.

We now rearrange the sum. First fix any $\bar{C}_{0}$ and partition
$\\{E_{\bar{\gamma}}^{0}\\}_{\bar{\gamma}\in\bar{C}_{0}}$ that may show up in
the sum. Then, consider the set of pairs

$B=\\{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in
I}):\bar{C}_{0}=\bar{C}(\\{L_{i}\\},K,\\{b_{i}\\},\\{a_{i,b_{i}}\\}_{i\in
I}),E_{\bar{\gamma}}^{0}=E_{\bar{\gamma}}(\\{L_{i}\\},K,\\{b_{i}\\},\\{a_{i,b_{i}}\\}_{i\in
I})\\}.$

The reason to consider such sets $B$ is that for any
$(\\{j_{i}\\},\\{a_{i,b_{i}}\\})$ in this set, Equation 4.3 is expressed in
the same exact way. In other words,
$r(c)=r(c,\\{E_{\bar{\gamma}}^{0}\\}_{\bar{\gamma}\in\bar{C}_{0}},\\{\\#\bar{\gamma}\\}_{\bar{\gamma}\in\bar{C}_{0}})$
is constant on this set. Thus, we may swap the two outer sums as follows

$\sum_{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in I})\in
B}\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}})$
$=\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\sum_{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in
I})\in
B}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}),$

and note that the sum in Equation 4.4 is simply the sum of the above over all
choices of $B$. Let

$B_{\\{j_{i}^{0}\\}_{i\in[m]}}=\\{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in
I}):j_{i}=j_{i}^{0}\\}.$

Now we state a lemma that we will prove later.

######  Lemma 4.3.

There’s a bijection between $B_{\\{j_{i}^{0}\\}_{i\in[m]}}$ and
$B_{\\{j_{i}^{1}\\}_{i\in[m]}}$, for any choice of $j_{i}^{0}$ and $j_{i}^{1}$
such that for all $\bar{\gamma}\in\bar{C}_{0}$ and $i\in
E_{\bar{\gamma}}^{0}$, $L_{j_{i}}$ contains some element of $\bar{\gamma}$.

This lemma is the “equidistribution” mentioned earlier. A simple way to
express this lemma is the fact that as we vary
$\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in E_{\gamma}}$ up to isomorphism over
$(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in I})\in B$, any collections of
isomorphism classes of extensions $\\{L_{i}\\}_{i\in E_{\gamma}}$ of
$K[\gamma]$, such that $f(L_{i})=f_{i}$, $e(L_{i})=e_{i}$ shows up the same
number of times. In terms of our proof, we have

$\sum_{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in I})\in
B}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}})$
$=|B|\prod_{\bar{\gamma}\in\bar{C}}\sum_{j_{i}\in[\gcd(p^{f_{i}}-1,e(K[\gamma]))],i\in
E_{\bar{\gamma}}}a(\\{L_{j_{i}}\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

Now let $\sigma$ be partitioned by $\sigma_{\bar{\gamma}}$, where
$\sigma_{\bar{\gamma}}=(e_{i}^{f_{i}})_{i\in E_{\bar{\gamma}}}$. Now by the
induction hypothesis, in particular the dependence on choice of $K$ only via
$e(K)$ and $f(K)$, the very inner sum is just

$a(\sigma_{\bar{\gamma}},e(K[\gamma]),f(K[\gamma]),c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

If $\gamma\neq 0$, then we get

$e(K[\gamma])=e(K[\zeta\pi_{K}^{\beta}])=e(K)\operatorname{denom}(\beta)$

and

$f(K[\gamma])=f(K)n_{O}(\gamma)/\operatorname{denom}(\beta)$

where $\zeta$ is some root of unity with order coprime to $p$. Thus, what we
have obtained is that

$\sum_{(\\{j_{i}\\}_{i\in[m]},\\{a_{i,b_{i}}\\}_{i\in I})\in
B}\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}})$
$=|B|\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\sigma_{\bar{\gamma}},e(K)h(\beta,n_{O}(\gamma)),f(K)\frac{n_{O}(\gamma)}{h(\beta,n_{O}(\gamma))},c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}),$

where we recall
$r(c)=r(c,\\{E_{\bar{\gamma}}^{0}\\}_{\bar{\gamma}\in\bar{C}_{0}},\\{\\#\bar{\gamma}\\}_{\bar{\gamma}\in\bar{C}_{0}})$

Finally, we complete the rearrangement of the sum

$\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]}\sum_{\begin{subarray}{c}i\in
I\\\ a_{i,b_{i}}\in T_{f_{i},p}\end{subarray}}$
$\sum_{\begin{subarray}{c}c_{\bar{\gamma}}\in{\mathbb{Z}}_{\geq 0}\\\
\sum_{\bar{\gamma}\in\bar{C}}c_{\bar{\gamma}}\\#\bar{\gamma}=r(c)\end{subarray}}\prod_{\bar{\gamma}\in\bar{C}}a(\\{\psi_{\gamma_{i}}(L_{j_{i}})\\}_{i\in
E_{\bar{\gamma}}},K[\gamma],c_{\bar{\gamma}},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{\bar{\gamma}}}).$

In the above paragraph, we calculated a subsum after fixing $\bar{C}_{0}$ and
$\\{E^{0}_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}_{0}}$ and so we can sum
over all possible choices of $\bar{C}_{0}$ and
$\\{E^{0}_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}_{0}}$. However, we will
instead sum over something more coarse. Given a choice of $\bar{C}_{0}$ and
$\\{E^{0}_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}_{0}}$, we have in
particular a partition
$\\{E^{0}_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}_{0}}$ of $[m]$, along with
for each partition $E^{0}_{\bar{\gamma}}$, a number $\\#\bar{\gamma}$. This is
how we will rearrange our sum. Thus, we obtain

$\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{\begin{subarray}{c}\text{partitions
}\\{E_{l}\\}_{l=1}^{k}\text{ of }[m]\\\
\\{n_{O,l}\\}_{l=1}^{k}\in{\mathbb{Z}}_{\geq
1}^{k}\end{subarray}}\\#\\{(\\{j_{i}\\},\\{a_{i,b_{i}}\\}_{i\in
I}):\\{E_{l}\\}_{l}=\\{E_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}},\\{n_{O,l}\\}_{l}=\\{\\#\bar{\gamma}\\}_{\bar{\gamma}\in\bar{C}}\\}$
$\sum_{\begin{subarray}{c}\\{c_{l}\\}_{l=1}^{k}\in{\mathbb{Z}}_{\geq 0}^{k}\\\
\sum_{l}n_{O,l}c_{l}=r(c)\end{subarray}}\prod_{l}a(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}}).$

We are done after noting that the definition of
$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ is equivalent to

$\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\\#\\{(\\{j_{i}\\},\\{a_{i,b_{i}}\\}_{i\in
I}):\\{E_{l}\\}_{l}=\\{E_{\bar{\gamma}}\\}_{\bar{\gamma}\in\bar{C}},\\{n_{O,l}\\}_{l}=\\{\\#\bar{\gamma}\\}_{\bar{\gamma}}\in\bar{C}\\}.$

∎

###### proof of Lemma 4.3.

Without loss of generality, it suffices to show this when
$j_{i}^{0}=j_{i}^{1}$ except in the first coordinate, where $j_{0}^{0}\neq
j_{0}^{1}$. Now it’s clear that if $0\notin I$, we already have this
bijection, via just replacing $j_{0}^{0}$ by $j_{0}^{1}$ for each
$(\\{j_{i}\\},\\{a_{i,b_{i}}\\}_{i\in I})\in B_{\\{j_{i}^{0}\\}_{i=1}^{m}}$.
So, assume that $0\in I$. Now by assumption, $L_{j_{0}^{0}}$ and
$L_{j_{0}^{1}}$ both contain some conjugate of $\gamma\in C$. This conjugate
is of the form $a_{0,b_{0}}\pi_{L_{j_{0}^{0}}}^{b_{0}}$. Now, since the
uniformizer of $L_{j_{0}^{0}}$ is of the form

$\pi_{L_{j_{0}^{0}}}=(\zeta_{p^{f_{0}}-1}^{j_{0}^{0}}\pi_{K})^{1/e_{i,\operatorname{rel}}}=\zeta_{e_{i,\operatorname{rel}}(p^{f_{0}}-1)}^{j_{0}^{0}}\pi_{K}^{1/e_{i,\operatorname{rel}}},$

we have that

$\pi_{L_{j_{0}^{0}+\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})}n}}^{b_{0}}=\zeta_{p^{f_{0}}-1}^{s}\pi_{L_{j_{0}^{0}}}^{b_{0}}$

for some $s$ depending on $n$. It is also easy to see that
$j_{0}^{0}+\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})}n$
are the only such choices. In other words, the choices of $j_{0}$ so that
$L_{j_{0}}$ contains some conjugate $\gamma$ are precisely fields in the
isomorphism classes of
$L_{j_{0}^{0}+\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})}n}$,
which are of the form
$L_{j_{0}^{0}+\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})}n_{1}+(p^{f_{0}}-1)n_{2}}$.
Thus, we see that
$\gcd(\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})},p^{f_{0}}-1)|j_{0}^{1}-j_{0}^{0}$.

Now let $\gamma^{0}$ and $\gamma^{1}$ be the Galois conjugates belonging to
$L_{j_{0}^{0}}$ and $L_{j_{0}^{1}}$, respectively. We will construct a
bijection from $B_{\\{j_{i}^{0}\\}_{i\in[m]}}$ to
$B_{\\{j_{i}^{1}\\}_{i\in[m]}}$. This bijection will send the pair
$(\\{j_{i}^{0}\\},\\{a_{i,b_{i}}\\})$ to
$(\\{j_{i}^{1}\\},\\{h_{i}a_{i,b_{i}}\\})$, where $h_{i}=1$ except in the case
where $i=0$. Thus, we only need to define $h_{0}$. We will need to find a
$h_{0}$ so that $h_{0}a_{0,b_{0}}\pi_{L_{j_{0}^{1}}}^{b_{0}}$ is a Galois
conjugate of $a_{0,b_{0}}\pi_{L_{j_{0}^{0}}}^{b_{0}}$. Since the uniformizer
of $L_{j_{0}^{0}}$ is of the form

$\pi_{L_{j_{0}^{0}}}=(\zeta_{p^{f_{0}}-1}^{j_{0}^{0}}\pi_{K})^{1/e_{i,\operatorname{rel}}}=\zeta_{e_{i,\operatorname{rel}}(p^{f_{0}}-1)}^{j_{0}^{0}}\pi_{K}^{1/e_{i,\operatorname{rel}}},$

and similarly for $L_{j_{0}^{1}}$, we need to find $h_{0}\in T_{f_{0},p}$ so
that up to Galois conjugates,

$a_{0,b_{0}}\zeta_{e_{i,\operatorname{rel}}(p^{f_{0}}-1)}^{b_{0}j_{0}^{0}}\pi_{K}^{b_{0}/e_{i,\operatorname{rel}}}=h_{0}a_{0,b_{0}}\zeta_{e_{i,\operatorname{rel}}(p^{f_{0}}-1)}^{b_{0}j_{0}^{1}}\pi_{K}^{b_{0}/e_{i,\operatorname{rel}}}.$

Thus, it suffices to show that up to powers of
$\zeta_{e_{i,\operatorname{rel}}}^{b_{0}}$,
$\zeta_{e_{i,\operatorname{rel}}(p^{f_{0}}-1)}^{b_{0}j_{0}^{1}-b_{0}j_{0}^{0}}\in
T_{f_{0},p}$. In other words, we ask for the existence of $h_{0},h_{1}$ so
that

$b_{0}(j_{0}^{1}-j_{0}^{0})=b_{0}(p^{f_{0}}-1)h_{1}+e_{i,\operatorname{rel}}h_{0}\mod
e_{i,\operatorname{rel}}(p^{f_{0}}-1).$

Now since

$\gcd(\frac{e_{i,\operatorname{rel}}}{\gcd(b_{0},e_{i,\operatorname{rel}})},p^{f_{0}}-1)|j_{0}^{1}-j_{0}^{0},$

we see that $\gcd(e_{i,\operatorname{rel}},b_{0}(p^{f_{0}}-1))$ divides the
LHS. Since $e_{i,\operatorname{rel}}$ and $b_{0}(p^{f_{0}}-1)$ are the
coefficients of $h_{0},h_{1}$ on the RHS, we obtain the existence of such
$h_{0}$ and $h_{1}$.

∎

As mentioned after Proposition 4.2, this proposition essentially reduces the
calculation of
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$ to
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}+{\mathbb{1}}_{I}(i)\\})$.
However, repeatedly applying Proposition 4.2 will only keep increasing the
elements $\\{b_{i}\\}$. Thus, we prove the following lemma will allow us to
reduce the calculation of
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}+e_{i,\operatorname{rel}}\\})$
to $a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$,
which will be key in allowing us to close off the induction on degree, as we
will soon see.

######  Lemma 4.4.

We have the relation

$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}+e_{i,\operatorname{rel}}\\})=p^{-f_{\operatorname{base}}d}a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c-\frac{d(d-1)}{e_{\operatorname{base}}},\\{b_{i}\\}).$

###### Proof.

This comes down to the fact that in the definition of discriminant, we have

$\prod_{\alpha_{i}\neq\alpha_{j}}(\pi_{K}\alpha_{i}-\pi_{K}\alpha_{j})=\pi_{K}^{d(d-1)}\prod_{\alpha_{i}\neq\alpha_{j}}(\alpha_{i}-\alpha_{j}).$

The rest follows by using definition of
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\})$. ∎

###  4.1. Generating function

We can express the above recurrences nicely with a generating function. Let

$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\},p)(t)=\sum_{c=0}^{\infty}a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\},p)t^{c}.$

Similar to our notation for $a$, we will often omit the dependence of $G$ on
$p$.

As an analogous example to Example 4.1, we have

######  Example 4.5.

$G((e_{\operatorname{base}}^{f_{\operatorname{base}}}),e_{\operatorname{base}},f_{\operatorname{base}},\\{b\\})=p^{-bf}$.

This will be our base case, and we will inductively calculate
$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})(t)$
using Proposition 4.2 and Lemma 4.4.

To simplify our equations, define

$\operatorname{inc}^{1}(\\{b_{i}\\})=\operatorname{inc}(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})=\\{b_{i}+{\mathbb{1}}_{I}(i)\\},$

where we recall, $I$ consists of $i$ so that $b_{i}/e_{i,\operatorname{rel}}$
is minimum among such terms. Let
$\operatorname{inc}^{n}(\\{b_{i}\\})=\operatorname{inc}(\operatorname{inc}^{n-1}(\\{b_{i}\\}))$.
Define

$\displaystyle
H(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})$
$\displaystyle=\left(\sum_{\begin{subarray}{c}\text{partitions
}\\{E_{l}\\}_{l=1}^{k}\text{ of }[m]\\\
\\{n_{O,l}\\}_{l=1}^{k}\in{\mathbb{Z}}_{\geq 1}^{k}\\\
(\\{E_{l}\\},\\{n_{O,l}\\})\neq(\\{[m]\\},\\{1\\})\end{subarray}}P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)\right)$
$\displaystyle\times
t^{\frac{1}{e_{\operatorname{base}}}(d(d-1)-\sum_{l}\sum_{i\in
E_{l}}d_{i}(\sum_{i\in E_{l}}(d_{i}/n_{O,l})-1))\beta}$
$\displaystyle\times\prod_{l=1}^{k}G(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}})(t^{n_{O,l}})$

From now on, we will often abbreviate
$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})(t)$
with $G(\\{b_{i}\\})(t)$, and similarly for $H(\\{b_{i}\\})(t)$. With these
notations, we get the following as a reformulation of Proposition 4.2.

######  Corollary 4.6.

$G(\\{b_{i}\\})(t)=P(\sigma,\\{[m]\\},\\{b_{i}\\},\\{1\\},p)G(\operatorname{inc}(\\{b_{i}\\}))(t)+H(\\{b_{i}\\})(t)$

Also, as a reformulation of Lemma 4.4, we get that

######  Corollary 4.7.

$G(\\{b_{i}+e_{i,\operatorname{rel}}\\})(t)=p^{-f_{\operatorname{base}}d}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}G(\\{b_{i}\\})(t)$

Next, we will use these two reformulations to inductively calculate
$(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})$. Now
fix $\\{b_{i}\\}$ and let
$k_{lim}=\max_{i}\lceil\frac{b_{i}}{e_{i,\operatorname{rel}}}\rceil$. It’s not
hard to see that $\operatorname{inc}^{n}(\\{b_{i}\\})$ will eventually give us
the constant sequence $\\{k_{lim}e_{i,\operatorname{rel}}\\}$. Let
$n_{\\{b_{i}\\}}$ be the smallest $n$ so that
$\operatorname{inc}^{n}(\\{b_{i}\\})=\\{k_{lim}e_{i,\operatorname{rel}}\\}$.
Also, let $n_{rec}$ be the $n$ so that
$\operatorname{inc}^{n}(\\{0\\}_{i=1}^{m})=\\{e_{i,\operatorname{rel}}\\}$.
Then, we have the following formula.

######  Proposition 4.8.

$G(\\{b_{i}\\})(t)=\sum_{n=0}^{n_{\\{b_{i}\\}}-1}H(\operatorname{inc}^{n}(\\{b_{i}\\}))(t)+\left(p^{-f_{\operatorname{base}}d}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}\right)^{k_{lim}}G(\\{0\\}_{i=1}^{m})(t),$

where

$G(\\{0\\}_{i=1}^{m})(t)=\frac{1}{1-p^{f_{\operatorname{base}}(1-d)}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}}\left(H(\\{0\\}_{i=1}^{m})(t)+p^{f_{\operatorname{base}}}\sum_{n=1}^{n_{rec}-1}H(\operatorname{inc}^{n}(\\{0\\}_{i=1}^{m}))(t)\right).$

###### Proof.

First, we use Corollary 4.6 repeatedly and get that
$G(\\{b_{i}\\})=G(\\{k_{lim}e_{i,\operatorname{rel}}\\})+\sum_{n=0}^{n_{\\{b_{i}\\}}-1}H(\operatorname{inc}^{n}(\\{b_{i}\\}))(t)$,
since in these cases $P(\sigma,\\{[m]\\},\\{b_{i}\\},\\{1\\},p)=1$ by Theorem
7.3. Now by applying Corollary 4.7 $k_{lim}$ times to
$G(\\{k_{lim}e_{i,\operatorname{rel}}\\})$, we get that it suffices to show

$G(\\{0\\}_{i=1}^{m})(t)=\frac{1}{1-p^{f_{\operatorname{base}}(1-d)}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}}\left(H(\\{0\\}_{i=1}^{m})(t)+p^{f_{\operatorname{base}}}\sum_{n=1}^{n_{rec}-1}H(\operatorname{inc}^{n}(\\{0\\}_{i=1}^{m}))(t)\right).$

Indeed, applying Corollary 4.6 $n_{rec}$ number of times here, we get

$G(\\{0\\}_{i=1}^{m})(t)=H(\\{0\\}_{i=1}^{m})(t)+p^{f_{\operatorname{base}}}\sum_{n=1}^{n_{rec}-1}H(\operatorname{inc}^{n}(\\{0\\}_{i=1}^{m}))(t)+p^{f_{\operatorname{base}}}G(\\{e_{i,\operatorname{rel}}\\})(t),$

since
$P(\sigma,\\{[m]\\},\\{0\\}_{i=1}^{m},\\{1\\},p)=p^{f_{\operatorname{base}}}$.
Finally, since
$G(\\{e_{i,\operatorname{rel}}\\})(t)=p^{-f_{\operatorname{base}}d}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}G(\\{0\\}_{i=1}^{m})(t)$
by Corollary 4.7, we can solve for $G(\\{0\\}_{i=1}^{m})(t)$ and obtain

$G(\\{0\\}_{i=1}^{m})(t)=\frac{1}{1-p^{f_{\operatorname{base}}(1-d)}t^{\frac{d(d-1)}{e_{\operatorname{base}}}}}\left(H(\\{0\\}_{i=1}^{m})(t)+p^{f_{\operatorname{base}}}\sum_{n=1}^{n_{rec}-1}H(\operatorname{inc}^{n}(\\{0\\}_{i=1}^{m}))(t)\right).$

∎

##  5\. Proof of Main Result

For this section, we use the same notations as we did for the section 4. In
particular, we use absolute ramification indices and inertia degrees.

Let $q=p^{f_{\operatorname{base}}}$. Recall that we wish to show that given a
tamely ramified splitting type $\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$
over a base field with ramification index $e_{\operatorname{base}}$ and
inertia degree $f_{\operatorname{base}}$, that

$\frac{q^{\frac{\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}}{2}}}{\prod_{i=1}^{m}f_{i,\operatorname{rel}}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]-1\end{subarray}}\int_{\alpha\in\prod_{i}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}}q^{-\frac{1}{2}v_{K}(\Delta_{P_{\alpha/K}})}d\lambda_{L}$

is a rational function in terms of $q$ and depends on $K$ only through $e(K)$
and $f(K)$.

###### proof of Theorem 3.1.

Let $q=p^{f_{\operatorname{base}}}$. It’s clear from the definition that

$\frac{1}{\prod_{i=1}^{m}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}\sum_{\begin{subarray}{c}L=\prod_{i}L_{j_{i}}\\\
j_{i}\in[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})]\end{subarray}}\int_{\alpha\in\prod_{i}{\mathfrak{m}}_{L_{j_{i}}}^{b_{i}}}p^{-\frac{1}{2}v_{K}(\Delta_{P_{\alpha/K}})}d\lambda_{L}$

is precisely
$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})(q^{-e(K)/2})$.
Thus, part 1 of the theorem on the dependence on $K$ via $e(K)$ and $f(K)$ is
a corollary of Proposition 4.2, since $G$ is a generating function with
coefficients of the form
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\},p)$.

Now we start on part 2 of the theorem on the fact that it’s a rational
function in terms of $q$. We again have our base case (Example 4.5)
$G((e^{f}),e,f,b)=p^{-bf}=q^{-b}$, which is a rational function in terms of
$q$. Next, we use Proposition 4.8. Now, evaluating at
$t=q^{-e_{\operatorname{base}}/2}$, we obtain that

$q^{\frac{\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}}{2}}G(\\{b_{i}\\})(q^{-e_{\operatorname{base}}/2})$

is a rational function of $q$ as long as

$q^{\frac{\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}}{2}}H(\\{b_{i}\\})(q^{-e_{\operatorname{base}}}/2)$

are all rational functions. Unraveling the definition of $H(\\{b_{i}\\})(t)$,
we need to show that when evaluated $t=q^{-e_{\operatorname{base}}/2}$, the
product of the following three terms

  1. (1)

$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$

  2. (2)

$t^{\frac{1}{e_{\operatorname{base}}}(d(d-1)-\sum_{l}\sum_{i\in
E_{l}}d_{i}(\sum_{i\in E_{l}}(d_{i}/n_{O,l})-1))\beta}$

  3. (3)

$\prod_{l}G(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}})(t^{n_{O,l}})$

is a rational function in terms of $q^{1/2}$, and moreover the exponent of
this rational function has the same parity as
$\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}$.

For the first item, we have by Theorem 7.3 that it is a rational function in
terms of $q$. So, we need that the product of second and third item to be a
rational function in terms of $q^{1/2}$ after evaluating at
$t=q^{-e_{\operatorname{base}}/2}$, and the exponent of this rational function
has the same parity as
$\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}$. To do this, we
will evaluate at $t=q^{-e_{\operatorname{base}}}$, show that all powers of $q$
are integral, then track the parity of powers of $q$.

We consider now the second item. Evaluating at $q^{-e_{\operatorname{base}}}$
gives us an exponent of

$-(d(d-1)-\sum_{l}\sum_{i\in E_{l}}d_{i}(\sum_{i\in
E_{l}}(d_{i}/n_{O,l})-1))\beta).$

We rewrite this as

$-\sum_{l}\sum_{i\in E_{l}}d_{i}(d-\sum_{i\in E_{l}}(d_{i}/n_{O,l}))\beta$

and show each term is integral. Now $\operatorname{denom}(\beta)|n_{O,l}$ for
all $l$ except in the case where $\operatorname{denom}(\beta)\neq 1$ and
$n_{O,l}=1$. Since we must have $n_{O,l}|d_{i}$, to show integrality, we only
have to consider the case with $\operatorname{denom}(\beta)\neq 1$ and
$n_{O,l}=1$. In this case, $d-\sum_{i\in E_{l}}(d_{i}/n_{O,l}))$ is precisely
$\sum_{i\notin E_{l}}d_{i}$, and since $\operatorname{denom}(\beta)$ divides
every such $d_{i}$, we get that

$-\sum_{l}\sum_{i\in E_{l}}d_{i}(d-\sum_{i\in E_{l}}(d_{i}/n_{O,l}))\beta$

is indeed an integer.

For the third item, we have by induction hypothesis that evaluating

$G(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}})$

at $q^{e_{\operatorname{base}}n_{O,l}}$ gives us integer exponents in terms of
$q^{\frac{n_{O,l}}{h(\beta,n_{O,l})}}$.

Now we track the parity of the exponents. This is purely an exercise in
casework, unravelling definitions, and simplifying until we see that the
exponents have even parity. By inductive hypothesis, the parity of the
exponent of $q^{\frac{n_{O,l}}{h(\beta,n_{O,l})}}$ in

$G(\sigma_{l},e(K)h(\beta,n_{O,l}),f(K)\frac{n_{O,l}}{h(\beta,n_{O,l})},c_{l},\\{b_{i}+{\mathbb{1}}_{I}(i)\\}_{i\in
E_{l}})$

evaluated at $q^{e_{\operatorname{base}}n_{O,l}}$ is precisely the parity of
$\sum_{i\in
E_{l}}(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-1)\frac{f_{i,\operatorname{rel}}h(\beta,n_{O,l})}{n_{O,l}}$.
Thus, the parity of the exponent in terms of $q$ is the parity of

$\sum_{i\in
E_{l}}(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-1)f_{i,\operatorname{rel}}.$

Thus, we must show that

$\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}-\sum_{l}\sum_{i\in
E_{l}}d_{i}(d-\sum_{i\in E_{l}}(d_{i}/n_{O,l}))\beta+\sum_{l}\sum_{i\in
E_{l}}(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-1)f_{i,\operatorname{rel}}$

is even. We have a series of equalities modulo $2$.

$\sum(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}-\sum_{l}\sum_{i\in
E_{l}}d_{i}(d-\sum_{i\in E_{l}}(d_{i}/n_{O,l}))\beta+\sum_{l}\sum_{i\in
E_{l}}(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-1)f_{i,\operatorname{rel}}$
$=\sum_{l}\sum_{i\in E_{l}}d_{i}(d-\sum_{i\in
E_{l}}(d_{i}/n_{O,l}))\beta+\sum_{l}\sum_{i\in
E_{l}}(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-e_{i,\operatorname{rel}})f_{i,\operatorname{rel}}$
$=\sum_{l}\sum_{i\in
E_{l}}f_{i,\operatorname{rel}}(e_{i,\operatorname{rel}}(d-\sum_{i\in
E_{l}}(d_{i}/n_{O,l}))\beta+(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-e_{i,\operatorname{rel}})).$

First of all, modulo $2$, we may remove all $i$ such that
$f_{i,\operatorname{rel}}$ is even. So, we reduce to the case where
$f_{i,\operatorname{rel}}$ are all odd.

In the case where $f_{i,\operatorname{rel}}$ are all odd, we have that
$d_{i}\equiv e_{i,\operatorname{rel}}\mod 2$. So, we may rewrite our sum as

$\sum_{l}\sum_{i\in
E_{l}}e_{i,\operatorname{rel}}(\sum_{i}e_{i,\operatorname{rel}}-\sum_{i\in
E_{l}}(d_{i}/n_{O,l}))\beta+(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-e_{i,\operatorname{rel}})).$

Now for $n_{O,l}\neq 1$, it is actually the degree of a subextension
$K^{\prime}/K$ of some $L_{j_{i}}$, where $K^{\prime}$ is of the form
$K^{\prime}=K[a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}]$. With this notation,
$\operatorname{denom}(\beta)=e(K^{\prime}/K)$, and
$n_{O,l}=\operatorname{denom}(\beta)f(K^{\prime}/K)$. Now note that since
we’re in the case where $f_{i,\operatorname{rel}}$ are all odd, modulo $2$, we
can assume $n_{O,l}=\operatorname{denom}(\beta)$. Now suppose there are no $l$
such that $n_{O,l}=1$. Then, we may rewrite our sum again as

$\sum_{l}\sum_{i\in
E_{l}}(e_{i,\operatorname{rel}}(\sum_{i}e_{i,\operatorname{rel}}-\sum_{i\in
E_{l}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)})\beta+(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}})).$

Now let $\operatorname{num}(\beta)$ denote the numerator of $\beta$ in the
lowest form. Then, we get

$\operatorname{num}(\beta)\sum_{l}\sum_{i\in
E_{l}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}(\sum_{i}e_{i,\operatorname{rel}}-\sum_{i\in
E_{l}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)})+\sum_{i}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}}).$

Multiplying the first term out, we get

$\operatorname{num}(\beta)\left(\sum_{i}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}\sum_{i}e_{i,\operatorname{rel}}-\sum_{l}\left(\sum_{i\in
E_{l}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}\right)^{2}\right)+\sum_{i}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}}).$

But modulo $2$, $x^{2}=x$, and so our expression simplifies to

$\operatorname{num}(\beta)\left(\sum_{i}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}(\sum_{i}e_{i,\operatorname{rel}}-1)\right)+\sum_{i}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}}).$

Now we have two cases. If $\operatorname{denom}(\beta)$ were odd, then we
immediately see that both terms are $0\mod 2$. If
$\operatorname{denom}(\beta)$ were even, then $\operatorname{num}(\beta)$ must
be odd. In this case, our expression simplifies to

$\sum_{i}e_{i,\operatorname{rel}}(\sum_{i}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-1).$

But we know $\operatorname{denom}(\beta)|e_{i,\operatorname{rel}}$ for each
$i$, and so this term is even as well.

Now we are left with the case where there is some $l$ with $n_{O,l}=1$, say
$n_{O,1}$. All of the above reasoning still applies, and the term

$\sum_{i\in
E_{l}}e_{i,\operatorname{rel}}(\sum_{i}e_{i,\operatorname{rel}}-\sum_{i\in
E_{l}}(d_{i}/n_{O,l}))\beta+(\frac{e_{i,\operatorname{rel}}}{h(\beta,n_{O,l})}-e_{i,\operatorname{rel}})),$

where $l=1$ simplifies to

$\sum_{i\in E_{1}}e_{i,\operatorname{rel}}\sum_{i\notin
E_{1}}e_{i,\operatorname{rel}}\beta.$

Thus, we wish to show that

$\sum_{i\in E_{1}}e_{i,\operatorname{rel}}\sum_{i\notin
E_{1}}e_{i,\operatorname{rel}}\beta+\operatorname{num}(\beta)\left(\sum_{i\notin
E_{1}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}(\sum_{i}e_{i,\operatorname{rel}}-1)\right)+\sum_{i\notin
E_{1}}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}})$

is even. This expression is equivalent to the following

$\operatorname{num}(\beta)\left(\sum_{i\notin
E_{1}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}(\sum_{i}e_{i,\operatorname{rel}}-1+\sum_{i\in
E_{1}}e_{i,\operatorname{rel}})\right)+\sum_{i\notin
E_{1}}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}})$
$=\operatorname{num}(\beta)\left(\sum_{i\notin
E_{1}}\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}(\sum_{i\notin
E_{1}}e_{i,\operatorname{rel}}-1)\right)+\sum_{i\notin
E_{1}}(\frac{e_{i,\operatorname{rel}}}{\operatorname{denom}(\beta)}-e_{i,\operatorname{rel}}).$

As before, we separate into cases where $\operatorname{denom}(\beta)$ is odd
and $\operatorname{denom}(\beta)$ is even, and see that in both cases, this
expression is even. ∎

##  6\. Proof of Theorem 3.5

First, we setup and recall some notation. For $a$ a nonnegative integer, let
$R[x]_{a}^{1}$ denote the monic polynomials of degree $a$ over $R$. In the
case where $R={\mathcal{O}}_{K}$, we identify ${\mathcal{O}}_{K}[x]_{a}^{1}$
with ${\mathcal{O}}_{K}^{a}$ via recording the coefficients of every term
besides the $x^{a}$ term. Thus, we also get a measure on this space, which
we’ll call $\mu_{a}^{1}$ (recall that $\mu_{a}$ is the name of the measure on
$\Omega_{a}$). Let ${\mathcal{O}}_{K}[x]_{a}^{2}\subset\Omega_{a}$ denote the
polynomials $P$ of degree $a$ over ${\mathcal{O}}_{K}$ that reduce to
$\bar{P}=1$ in the residue field
$\kappa_{K}={\mathcal{O}}_{K}/{\mathfrak{m}}_{K}$. Let $q=p^{f}$, where $f$ is
the absolute inertia degree of $K$. As before, let
$w(d)=w_{f(K)}(d)=\frac{q^{d+1}-q^{d}}{q^{d+1}-1}$.

Let $L=\prod_{i=1}^{m}L_{i}$ be any étale extension of $K$. For $U\subset L$,
denote $\rho(U,K)$ to be the proportion of degree $[L:K]$ polynomials (in
$\Omega_{[L:K]}$ with a new root in $U$. We have by linearity of expectation
that

$\rho(L,K)=\sum_{A\subset[m]}\frac{1}{|\operatorname{Aut}(A)|}\rho(\prod_{i\in
A}{\mathcal{O}}_{L_{i}}\times\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)$

where $\operatorname{Aut}(A)$ is the number of $A^{\prime}$ so that
$L_{A}\cong L_{A^{\prime}}$. This is equal to
$\left|\frac{\operatorname{Aut}(L)}{\operatorname{Aut}(L_{A})\operatorname{Aut}(L_{A^{c}})}\right|$
(since $\operatorname{Aut}(L)$ acts transitively on the set of such
$A^{\prime}$, and the stabilizer of $A$ is precisely
$\operatorname{Aut}(L_{A})\times\operatorname{Aut}(L_{A^{c}})$). Now, by the
same argument in Lemma 3.3, we get that
$\rho(U,K)=\frac{{\mathbb{E}}[Z^{\operatorname{new}}_{U,L}]}{\operatorname{Aut}(L)}$.
Thus, it suffices to show that

$\rho(\prod_{i\in A}{\mathcal{O}}_{L_{i}}\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)=\frac{w([L:K])}{w([L_{A}:K])w([L_{A^{c}}:K])}\rho({\mathcal{O}}_{L_{A}},L_{A})\rho({\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}}).$

The central idea to this equivalence is given in [1, Lemma 2.9]. In the case
where $L$ is an honest field extension, the central idea manifests as the fact
that polynomials in $\Omega_{d}$ whose roots are in ${\mathfrak{m}}_{L}$ are
identified with the polynomials whose roots are in
$L\setminus{\mathcal{O}}_{L}$, via the reversal map $P\to x^{d}P(1/x)$, and so
since the reversal map in this case is just a permutation of the coordinates,
the measure of the two sets of polynomials should be the same.

######  Lemma 6.1 (Bhargava, Cremona, Fisher, Gajović).

Let $a\geq b\geq 0$. The multiplication map

$\psi:{\mathcal{O}}_{K}[x]_{b}^{1}\times{\mathcal{O}}_{K}[x]_{a-b}^{2}\to\\{P\in\Omega_{a}:\bar{P}\in\kappa_{K}[x]^{1}_{b}\\}$

is a measure preserving bijection.

Note that the authors proved this over the $p$-adics, but it extends to all
extensions as well. Applying this lemma, we get that

$\\{P_{1}\in{\mathcal{O}}_{K}[x]_{[L_{A}:K]}^{1}:P_{1}\text{ has a new root in
}{\mathcal{O}}_{L_{A}}\\}\times\\{P_{2}\in{\mathcal{O}}_{K}[x]_{[L_{A^{c}}:K]}^{2}:P_{2}\text{
has a new root in }\prod_{i\notin A}L_{i}\setminus{\mathcal{O}}_{L_{i}}\\},$

which is a subset of
${\mathcal{O}}_{K}[x]_{[L_{A}:K]}^{1}\times{\mathcal{O}}_{K}[x]_{[L_{A^{c}}:K]}^{2}$,
has the same measure as

$\\{P\in\Omega_{[L:K]}:\bar{P}\in\kappa_{K}[x]^{1}_{b},P\text{ has a new root
in }{\mathcal{O}}_{L_{A}}\times\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}}\\},$

which is a subset of
$\\{P\in\Omega_{[L:K]}:\bar{P}\in\kappa_{K}[x]^{1}_{[L_{A}:K]}\\}.$

Thus, it suffices for us to show the following

  1. (1)

$\rho({\mathcal{O}}_{L_{A}},K)=w([L_{A}:K])\mu_{[L_{A}:K]}^{1}(\\{P_{1}\in{\mathcal{O}}_{K}[x]_{[L_{A}:K]}^{1}:P_{1}\text{
has a new root in }{\mathcal{O}}_{L_{A}}\\}),$

  2. (2)

$\rho({\mathfrak{m}}_{L_{A^{c}}},L_{A^{c}})=qw([L_{A^{c}}:K]))\mu_{[L_{A^{c}}:K]}(\\{P_{2}\in{\mathcal{O}}_{K}[x]_{[L_{A^{c}}:K]}^{2}:P_{2}\text{
has a new root in }\prod_{i\notin A}L_{i}\setminus{\mathcal{O}}_{L_{i}}\\}),$

  3. (3)

and

$\rho(\prod_{i\in A}{\mathcal{O}}_{L_{i}}\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)$
$=qw([L:K])\mu_{[L:K]}(\\{P\in\Omega_{[L:K]}:\bar{P}\in\kappa_{K}[x]^{1}_{b},P\text{
has a new root in }{\mathcal{O}}_{L_{A}}\times\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}}\\}).$

We have a surjective map

$\varphi:\\{\sum_{i=0}^{[L_{A}:K]}a_{i}x^{i}\in\Omega_{[L_{A}:K]}:v_{p}(a_{[L_{A}:K]})\leq
v_{p}(a_{i}),i=0,\dots,[L_{A}:K]-1\\}\to{\mathcal{O}}_{K}[x]_{[L_{A}:K]}^{1}$

via dividing by the coefficient of $x^{[L_{A}:K]}$. One can check on basic
opens that the pushforward of $\mu_{[L_{A}:K]}$ via this map is precisely
$w([L_{A}:K])\mu_{[L_{A}:K]}^{1}$. Now,

$\varphi^{-1}(\\{P_{1}\in{\mathcal{O}}_{K}[x]_{[L_{A}:K]}^{1}:P_{1}\text{ has
a new root in }{\mathcal{O}}_{L_{A}}\\}),$

is precisely the set of which $\rho({\mathcal{O}}_{L_{A}},K)$ is the measure.
Hence, we have our first item.

The second and third item are similar, and so we will just prove the second
item. Let

$S_{\beta}=\\{P_{2}\in{\mathcal{O}}_{K}[x]_{[L_{A^{c}}:K]}:P_{2}\text{ has a
new root in }\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},\bar{P}_{2}=\beta\\},$

so that $S_{1}$ is our set on the RHS of item 2. Then, we see by multiplying
by units, that $\mu_{[L_{A}^{c}:K]}(S_{1})=\mu_{[L_{A}^{c}:K]}(S_{\beta})$ for
any $\beta\in\kappa_{K}\setminus\\{0\\}$. Now let

$S=\\{P_{2}\in{\mathcal{O}}_{K}[x]_{[L_{A^{c}}:K]}:P_{2}\text{ has a new root
in }\prod_{i\notin A}L_{i}\setminus{\mathcal{O}}_{L_{i}}\\},$

so that

$\rho(\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)=\mu_{[L_{A^{c}}:K]}(S).$

Then,
$S=\pi_{K}S\sqcup\bigsqcup_{\beta\in\kappa_{K}\setminus\\{0\\}}S_{\beta}$.
Hence,

$(1-q^{-([L_{A^{c}}:K]+1)})\rho(\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)=(q-1)\mu_{[L_{A^{c}}:K]}(S_{1}).$

Thus, we get

$\rho(\prod_{i\notin
A}L_{i}\setminus{\mathcal{O}}_{L_{i}},K)=qw([L_{A^{c}}:K])\mu_{[L_{A^{c}}:K]}(S_{1}),$

as desired.

##  7\. Computation of $P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$

In this section, our goal is to calculate
$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$, which was defined to be
$\frac{1}{\prod_{i\in[m]}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})}$ times
the number of elements

$(\\{j_{i}\\}_{i=1}^{m},\\{a_{i,b_{i}}\\}_{i\in
I})\in\prod_{i=1}^{m}[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}}]\times\prod_{i\in
I}T_{f_{i},p}$

satisfying the following three conditions:

  1. (1)

For all $i_{1},i_{2}\in I$, $i_{1},i_{2}$ belong to the same $E_{l}$ if and
only if $a_{i_{1},b_{i_{1}}}\pi_{L_{j_{i_{1}}}}^{b_{i_{1}}}$ and
$a_{i_{2},b_{i_{2}}}\pi_{L_{j_{i_{2}}}}^{b_{i_{2}}}$ are Galois conjugates.

  2. (2)

If $i\in I$, then $n_{O}(a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}})=n_{O,l}$.

  3. (3)

If $I\neq[m]$, then there is some $E_{l_{0}}$ containing all of $[m]\setminus
I$. Moreover, for any $i\in E_{l_{0}}\cap I$, $a_{i,b_{i}}=0$, and so in
particular, $n_{O,l_{0}}=1$.

Now suppose $P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ is nonzero.
Then, first of all, since for $i\in I$, we have that
$e(K[a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}]/K)$ is $\operatorname{denom}(\beta)$
unless $a_{i,b_{i}}=0$, we must have that unless $n_{O,l}=1$ and
$\operatorname{denom}(\beta)\neq 1$ (in which case
$a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}=0$), $\operatorname{denom}(\beta)|n_{O,l}$
for all $l$ and
$\frac{n_{O,l}}{\operatorname{denom}(\beta)}|f_{i,\operatorname{rel}}$ for all
$i\in E_{l}$. Second of all, we must have that for any given $k$, the number
of elements $E_{l}$ of the partition such that $n_{O,l}=k$ is at most the
number of possible Galois orbits of size $k$ among elements of the form
$a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}}$. We will shortly see that the number of
such Galois orbits is
$P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$, where

$P(f_{\operatorname{base}},k^{\prime},p)=\frac{1}{k^{\prime}}((p^{f_{\operatorname{base}}k^{\prime}}-1)-\sum_{\ell_{1}\text{
prime},\ell_{1}|k^{\prime}}(p^{f_{\operatorname{base}}k^{\prime}/\ell_{1}}-1)+\sum_{\ell_{1}\neq\ell_{2}\text{
prime},\ell_{i}|k^{\prime}}(p^{\frac{f_{\operatorname{base}}k^{\prime}}{\ell_{1}\ell_{2}}}-1)-\dots).$

Thus, $P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=0$ unless all of
following conditions are satisfied:

  1. (1)

For all $l$ such that $n_{O,l}\neq 1$, $\operatorname{denom}(\beta)|n_{O,l}$.

  2. (2)

For all $i,l$ such that $i\in E_{l}$ and $n_{O,l}\neq 1$,
$\frac{n_{O,l}}{\operatorname{denom}(\beta)}|f_{i,\operatorname{rel}}$.

  3. (3)

$\\#\\{l:n_{O,l}=k\\}\leq
P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$.

  4. (4)

If $\operatorname{denom}(\beta)\neq 1$, then there is at most one $l_{0}$ so
that $n_{O,l_{0}}=1$.

  5. (5)

If $I\neq[m]$, then there is some $l_{0}$ such that $E_{l_{0}}$ contains all
of $[m]\setminus I$.

For the rest of this section, assume all of these conditions hold. Let $l_{0}$
be defined as in the conditions above. Now let $F_{k}=\\{i\in[m]:\exists
l,n_{O,l}=k,i\in E_{l}\\}.$ Let
$P_{k}(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$ be the number of
elements

$(\\{j_{i}\\}_{i\in F_{k}},\\{a_{i,b_{i}}\\}_{i\in I\cap F_{k}})\in\prod_{i\in
F_{k}}[\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}}]\times\prod_{i\in F_{k}\cap
I}T_{f_{i},p}$

satisfying the following conditions:

  1. (1)

For all $i_{1},i_{2}\in F_{k}\cap I$, $i_{1},i_{2}$ belong to the same $E_{l}$
if and only if $a_{i_{1},b_{i_{1}}}\pi_{L_{j_{i_{1}}}}^{b_{i_{1}}}$ and
$a_{i_{2},b_{i_{2}}}\pi_{L_{j_{i_{2}}}}^{b_{i_{2}}}$ are Galois conjugates.

  2. (2)

If $i\in I$, then $n_{O}(a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}})=k$.

  3. (3)

If $k=1$ and $I\neq[m]$, then for any $i\in E_{l_{0}}\cap I$, $a_{i,b_{i}}=0$.

Since these conditions are independent as $k$ varies, we see that

$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=\prod_{k}P_{k}(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p),$

as $k$ varies over integers for which $F_{k}$ is nonempty. Thus, it suffices
to calculate $P_{k}$. Now

$P_{k}(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=P((e_{i}^{f_{i}})_{i\in
F_{k}},\\{E_{l}\\}_{l,n_{O,l}=k},\\{b_{i}\\}_{i\in
F_{k}},\\{k\\}_{l,n_{O,l}=k},p).$

Thus, it suffices for us to calculate
$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)$, in the case where all
$n_{O,l}$ are equal to $k$, which we will now proceed to do in the case where
$k\neq 1$. The case where $k=1$ is similar, but just requires more casework.

Now we start with the simplest case, where $\sigma$ is of the form
$(e_{1}^{f_{1}})$. Thus, $m=1$ and we have that the only partition of $[1]$
consist of one set $E_{1}=\\{1\\}$.

######  Lemma 7.1.

Suppose $\operatorname{denom}(\beta)|k$ (here
$\beta=\min_{i}\frac{b_{i}}{e_{i,\operatorname{rel}}}=\frac{b_{1}}{e_{1,\operatorname{rel}}}$,
and so
$\operatorname{denom}(\beta)=\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}$),
then

$\\#\\{(a_{1,b_{1}},j_{1})\in(T_{f_{1},p}\setminus\\{0\\})\times[\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})],n_{O}(a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}})=k\\}$
$=\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\frac{k}{\operatorname{denom}(\beta)}P(f_{\operatorname{base}},\frac{k}{\operatorname{denom}(\beta)},p).$

Hence, in this simple case, we have that

$P((e_{1}^{f_{1}}),\\{E_{1}\\},\\{b_{1}\\},\\{k\\},p)=\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\frac{k}{\operatorname{denom}(\beta)}P(f_{\operatorname{base}},\frac{k}{\operatorname{denom}(\beta)},p)$

if $k\neq 1$ (since $a_{1,b_{1}}$ cannot be $0$ if $n_{O,1}=k\neq 1$). Next,
we do the case where $\sigma=(e_{1}^{f_{1}},e_{2}^{f_{2}})$, and the general
case will follow by similar reasoning. This case comes down to the following.

######  Lemma 7.2.

Suppose $b_{1}/e_{1,\operatorname{rel}}=b_{2}/e_{2,\operatorname{rel}}$ (and
so
$\beta=\frac{b_{1}}{e_{1,\operatorname{rel}}}=\frac{b_{2}}{e_{2,\operatorname{rel}}}$
and
$\operatorname{denom}(\beta)=\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}=\frac{e_{2,\operatorname{rel}}}{\gcd(b_{2},e_{2,\operatorname{rel}})}$).
Given an element $a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}}$ with $k$ Galois
conjugates, the number of elements $a_{2,b_{2}}\pi_{L_{j_{2}}}^{b_{2}}$ such
that it is a Galois conjugate of $a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}}$ is
$\frac{k}{\operatorname{denom}(\beta)}\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})$.

First of all, note that combining Lemmas 7.1 and 7.2 shows that the number of
Galois orbits of size $k$ among elements of the form
$a_{1}\pi_{L_{1}}^{b_{1}}$ is precisely
$P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$ (here
$\beta=\frac{b_{1}}{e_{1,\operatorname{rel}}}$) by setting $b_{1}=b_{2}$ and
$e_{1,\operatorname{rel}}=e_{2,\operatorname{rel}}$ in Lemma 7.2. In addition,
Lemma 7.2 shows that after fixing a Galois orbit of size $k$ among elements of
the form $a_{1}\pi_{L_{1}}^{b_{1}}$, there are
$\frac{k}{\operatorname{denom}(\beta)}\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})$
choices of $a_{1},j_{1}$ such that $a_{1}\pi_{L_{1}}^{b_{1}}$ belongs in the
chosen Galois orbit.

Now, by Lemmas 7.1 and 7.2, if $k\neq 1$, we get a total of

$\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\left(\frac{k}{\operatorname{denom}(\beta)}\right)^{2}P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$

as our count for

$P((e_{1}^{f_{1}},e_{2}^{f_{2}}),\\{\\{1,2\\}\\},\\{b_{1},b_{2}\\},\\{k\\},p).$

Similarly,

$P((e_{1}^{f_{1}},e_{2}^{f_{2}}),\\{\\{1\\},\\{2\\}\\},\\{b_{1},b_{2}\\},\\{k,k\\},p)$

is

$\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})\left(\frac{k}{\operatorname{denom}(\beta)}\right)^{2}(P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)^{2}-P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)).$

One may interpret this as follows. For

$P((e_{1}^{f_{1}},e_{2}^{f_{2}}),\\{\\{1,2\\}\\},\\{b_{1},b_{2}\\},\\{k\\},p),$

we need to choose an orbit of size $k$ among elements of the form
$a_{i}\pi_{L_{j}}^{b_{i}}$, and there are
$P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$ many of such
orbits, as mentioned in the above paragraph. Now, having chosen an orbit, we
need to choose Galois conjugates $a_{1}\pi_{L_{j_{1}}}^{\beta}$ and
$a_{2}\pi_{L_{j_{2}}}^{\beta}$ in that orbit, where $(a_{1},j_{1})\in
T_{f_{1},p}\times[\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}}]$ and
$(a_{2},j_{2})\in
T_{f_{2},p}\times[\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}}]$. The number of
ways we can choose $(a_{i},j_{i})$ is
$\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})\frac{k}{\operatorname{denom}(\beta)}$.
Multiplying these together gives us the desired count. Similarly, for
$P((e_{1}^{f_{1}},e_{2}^{f_{2}}),\\{\\{1,2\\}\\},\\{b_{1},b_{2}\\},\\{k\\},p)$,
we need to do the exact same thing, except we need to choose two different
orbits of size $k$ (in order), and there are

$\frac{P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)!}{(P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)-2)!}=P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)^{2}-P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$

many choices.

We may apply the same reasoning to the general case of
$\sigma=(e_{1}^{f_{1}},\dots,e_{m}^{f_{m}})$. In this case, we obtain that if
$k\neq 1$ and we have $M$ elements in our partition $\\{E_{l}\\}_{l=1}^{M}$,

$P(\sigma,\\{E_{l}\\}_{l=1}^{M},\\{b_{i}\\}_{i=1}^{m},\\{k\\}_{l=1}^{M},p)=\left(\prod_{i=1}^{m}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})\right)\left(\frac{k}{\operatorname{denom}(\beta)}\right)^{m}\frac{P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)!}{(P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)-M)!}.$

The calculation of
$P(\sigma,\\{E_{l}\\}_{l=1}^{M},\\{b_{i}\\}_{i=1}^{m},\\{1\\}_{l=1}^{M},p)$ is
similar. In this case we also have to consider the cases of whether $[m]\neq
I$, and so there is some $i$ so that $\beta<b_{i}/e_{i}$. Let
$\\{E_{l}\\}_{l=1}^{k}$ be the partition we desire. Now if
$\operatorname{denom}(\beta)\neq 1$, then actually there can only be one
partition, as there’s only one element of the form
$a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}/e_{i,\operatorname{rel}}}$, where
$a_{i,b_{i}}\in T_{f_{i},p}$, that belongs to $K$, which is when
$a_{i,b_{i}}=0$. Thus, in this case we obtain
$\prod_{i=1}^{m}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})$ total choices,
coming from the choices of $j_{i}$. Next, if $\operatorname{denom}(\beta)=1$,
and so $\beta$ is an integer, then we get something more interesting. In this
case,
$\pi_{L_{j_{i}}}^{b_{i}/e_{i,\operatorname{rel}}}=\zeta_{p^{f_{i}}-1}^{j_{i}\beta}\pi_{K}^{\beta}$.
Since we’re considering elements of the form
$a_{i,b_{i}}\pi_{L_{j_{i}}}^{b_{i}/e_{i,\operatorname{rel}}}$, and
$\zeta_{p^{f_{i}}-1}^{j_{i}\beta}\in T_{f_{i},p}$, we can ignore this factor
by shifting each choice of $a_{i,b_{i}}$ by
$\zeta_{p^{f_{i}}-1}^{j_{i}\beta}$. Thus, the choices of $j_{i}$ are
independent from the choices of $a_{i,b_{i}}$. Hence, we just need to make
choices for $a_{i,b_{i}}$. First, since we expect $a_{i,b_{i}}\pi_{K}^{\beta}$
to have size $1$ Galois orbit over $K$, it is actually in $K$, and so we must
have that $a_{i,b_{i}}\in T_{f_{\operatorname{base}},p}$. Second, we need that
$i_{1}$ and $i_{2}$ belong to the same partition if and only if
$a_{i_{1},b_{i_{1}}}=a_{i_{2},b_{i_{2}}}$. Thus, if $I=[m]$, then we get a
total of

$\frac{p^{f_{\operatorname{base}}}!}{(p^{f_{\operatorname{base}}}-M)!}$

choices of $\\{a_{i,b_{i}}\\}_{i\in I}$. If $I\neq[m]$, we are forced to
choose $0$ to represent one of our Galois orbits, and so we get

$\frac{(p^{f_{\operatorname{base}}}-1)!}{(p^{f_{\operatorname{base}}}-1-(M-1))!}$

choices of $\\{a_{i,b_{i}}\\}_{i\in I}$. Thus,

$P(\sigma,\\{E_{l}\\}_{l=1}^{M},\\{b_{i}\\}_{i=1}^{m},\\{1\\}_{l=1}^{M},p)=\prod_{i=1}^{m}\gcd(p^{f_{i}}-1,e_{i,\operatorname{rel}})\begin{cases}1&\operatorname{denom}(\beta)\neq
1\\\
\frac{p^{f_{\operatorname{base}}}!}{(p^{f_{\operatorname{base}}}-M)!}&\operatorname{denom}(\beta)=1,I=[m]\\\
\frac{(p^{f_{\operatorname{base}}}-1)!}{((p^{f_{\operatorname{base}}}-1)-(M-1))!}&\operatorname{denom}(\beta)=1,I\neq[m]\end{cases}.$

Now with the definition $F_{k}=\\{i\in[m]:\exists l,n_{O,l}=k,i\in E_{l}\\}$
recall that we had

$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=\prod_{k}P_{k}(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p),$

as $k$ varies over integers for which $F_{k}$ is nonempty and

$P_{k}(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=P((e_{i}^{f_{i}})_{i\in
F_{k}},\\{E_{l}\\}_{l,n_{O,l}=k},\\{b_{i}\\}_{i\in
F_{k}},\\{k\\}_{l,n_{O,l}=k},p),$

the RHS of which we just calculated. Thus, we obtain the following theorem,
which is the main theorem of this section.

######  Theorem 7.3.

$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=0$ unless all of the
following are satisfied:

  1. (1)

For all $l$ such that $n_{O,l}\neq 1$, $\operatorname{denom}(\beta)|n_{O,l}$.

  2. (2)

For all $i,l$ such that $i\in E_{l}$ and $n_{O,l}\neq 1$,
$\frac{n_{O,l}}{\operatorname{denom}(\beta)}|f_{i,\operatorname{rel}}$.

  3. (3)

If $\operatorname{denom}(\beta)\neq 1$, then there is at most one $l_{0}$ so
that $n_{O,l_{0}}=1$.

  4. (4)

If $I\neq[m]$, then there is some $l_{0}$, such that $E_{l_{0}}$ contains all
of $[m]\setminus I$.

If all of the above conditions are satisfied, let $S$ be the set of $k$ for
which there is some $l$ such that $n_{O,l}=k$. Also, let $F_{k}=\\{i:i\in
E_{l},n_{O,l}=k\\}$ be the set of all indices $i$ such that $i$ is in a
partition $E_{l}$ with $n_{O,l}=k$. Then,

$P(\sigma,\\{E_{l}\\},\\{b_{i}\\},\\{n_{O,l}\\},p)=\prod_{k\in S}N_{k},$

where for $k\neq 1$,

$N_{k}=\left(\frac{k}{\operatorname{denom}(\beta)}\right)^{\\#F_{k}}\frac{P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)!}{(P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)-\\#\\{l,n_{O,l}=k\\})!},$

and

$N_{1}=\begin{cases}1&\operatorname{denom}(\beta)\neq 1\\\
\frac{p^{f_{\operatorname{base}}}!}{(p^{f_{\operatorname{base}}}-\\{l,n_{O,l}=1\\})!}&\operatorname{denom}(\beta)=1,I=[m]\\\
\frac{(p^{f_{\operatorname{base}}}-1)!}{((p^{f_{\operatorname{base}}}-1)-(\\{l,n_{O,l}=1\\}-1))!}&\operatorname{denom}(\beta)=1,I\neq[m]\end{cases}.$

Note that it seems that we are missing one condition, which was that for all
$k$, $\\#\\{l:n_{O,l}=k\\}\leq
P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)$. But this
condition is subsumed by the fact that if this condition is not satisfied,
then

$\frac{P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)!}{(P(f_{\operatorname{base}},k/\operatorname{denom}(\beta),p)-\\#\\{l,n_{O,l}=k\\})!},$

as a polynomial, are $0$ when evaluated at $p$.

We end this section with the proof of Lemmas 7.1 and 7.2. First we have the
following short lemma

######  Lemma 7.4.

Fix $f$ and $e$ such that $p\nmid e$. Then,
$x\frac{e}{\gcd(b,e)}+y\frac{b}{\gcd(b,e)}\mod(p^{f}-1)$ as $x$ ranges in
$[p^{f}-1]$ and as $y$ ranges in $[\gcd(p^{f}-1,e)]$ is equidistributed among
residue classes of $p^{f}-1$, with each class having $\gcd(p^{f}-1,e)$
elements.

###### Proof.

This comes down to seeing that
$\gcd(p^{f}-1,\frac{e}{\gcd(b,e)})|\gcd(p^{f}-1,e)$.

∎

###### Proof of Lemma 7.1.

Recall
$\pi_{L_{j_{1}}}=(\zeta_{p^{f_{1}}-1}^{j_{1}}\pi_{K})^{1/e_{1,\operatorname{rel}}}$.
Also, write $a_{1,b_{1}}=\zeta_{p^{f_{1}}-1}^{s}$. Then,
$n_{O}(a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}})$ is the number of Galois conjugates
of
$\zeta_{p^{f_{1}}-1}^{s}(\zeta_{p^{f_{1}}-1}^{jb_{1}/e_{1,\operatorname{rel}}}\pi_{K})^{b_{1}/e_{1,\operatorname{rel}}}$.
The set of these conjugates is precisely

$\\{(\zeta_{p^{f_{1}}-1}^{s}\zeta_{p^{f_{1}}-1}^{jb_{1}/e_{1,\operatorname{rel}}})^{p^{f_{\operatorname{base}}r_{1}}}\zeta_{e_{1,\operatorname{rel}}}^{b_{1}r_{2}}\pi_{K}^{b_{1}/e_{1,\operatorname{rel}}}:r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/(e_{1,\operatorname{rel}}){\mathbb{Z}}\\}.$

So, we wish to find the number of distinct elements of this set. We do this by
turning to modular arithmetic by looking at the exponents. That is, we have a
bijection from

$\\{p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})+b_{1}r_{2}(p^{f_{1}}-1)\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}:r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/(e_{1,\operatorname{rel}}){\mathbb{Z}}\\}$

to

$\\{(\zeta_{p^{f_{1}}-1}^{s}\zeta_{p^{f_{1}}-1}^{jb_{1}/e_{1,\operatorname{rel}}})^{p^{f_{\operatorname{base}}r_{1}}}\zeta_{e_{1,\operatorname{rel}}}^{b_{1}r_{2}}\pi_{K}^{b_{1}/e_{1,\operatorname{rel}}}:r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/(e_{1,\operatorname{rel}}){\mathbb{Z}}\\}$

sending $x\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}$ to
$\zeta_{(p^{f_{1}}-1)e_{1,\operatorname{rel}}}^{x}\pi_{K}^{b_{1}/e_{1,\operatorname{rel}}}$.
So, equivalently, we wish to find the number of distinct elements of

$\\{p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})+b_{1}r_{2}(p^{f_{1}}-1)\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}:r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/(e_{1,\operatorname{rel}}){\mathbb{Z}}\\}.$

Taking these elements modulo $b_{1}(p^{f_{1}}-1)$, we see that the number of
such distinct elements is then equal to
$\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}$ times
the number of distinct elements of

$S_{s,j_{1}}=\\{p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}}):r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}}\\}.$

Now the number of distinct elements to the above equation is the smallest
$r_{1}$ such that

(7.1) $(p^{r_{1}}-1)(se_{1,\operatorname{rel}}+j_{1}b_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}}).$

We will denote the LHS of this congruence to be $E(r_{1},s,j_{1})$. One can
see by minimality of such $r_{1}$ that $r_{1}|f_{1}/f_{\operatorname{base}}$.
Thus, so far, what we have is that

$n_{O}(\zeta_{p^{f_{1}}-1}^{s}(\zeta_{p^{f_{1}}-1}^{jb_{1}/e_{1,\operatorname{rel}}}\pi_{K})^{b_{1}/e_{1,\operatorname{rel}}})=\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}\min(\\{r_{1}^{\prime}:r_{1}^{\prime},E(r_{1}^{\prime},s,j_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})\\}).$

However, what we want to find is the number of choices $(s,j_{1})$ that give a
certain number of Galois conjugates. So, we want to fix $r_{1}$, and find the
number of choices of $(s,j_{1})$ such that

$r_{1}=\min(\\{r_{1}^{\prime}:r_{1}^{\prime},E(r_{1}^{\prime},s,j_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})\\}).$

These choices will all give

$\frac{r_{1}e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}=r_{1}\operatorname{denom}(\beta)$

conjugates.

First, we will find the number of solutions $s,j_{1}$ to

$E(r_{1},s,j_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}}),$

after fixing any $r_{1}$ such that $r_{1}|f_{1}/f_{\operatorname{base}}$.
These solutions $(s,j_{1})$ are precisely the ones such that

$\min(\\{r_{1}^{\prime}:r_{1}^{\prime},E(r_{1}^{\prime},s,j_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})\\})$

divide $r_{1}$, which is not exactly what we want, but it will be after
applying inclusion exclusion, as we will see. So, for now, we will fix an
$r_{1}|f_{1}/f_{\operatorname{base}}$ and find the number of solutions
$s,j_{1}$ to Equation 7.1. Dividing both sides by
$(p^{r_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})$, our equation turns into

$s\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}}))}+j_{1}\frac{b_{1}}{\gcd(b_{1},e_{1,\operatorname{rel}})}\equiv
0\mod\frac{p^{f_{1}}-1}{p^{r_{1}}-1}.$

Now we apply Lemma 7.4, obtain that there are

$(p^{r_{1}}-1)\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})$

solutions. With this, we may apply inclusion exclusion principle, and obtain
that the number of choices of $(s,j_{1})$ such that

$r_{1}=\min(\\{r_{1}^{\prime}:r_{1}^{\prime},E(r_{1}^{\prime},s,j_{1})\equiv
0\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})\\})$

is

$\displaystyle=\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\big{(}p^{r_{1}}-1-\sum_{q_{1}|r_{1}}(p^{f_{\operatorname{base}}\frac{r_{1}}{q_{1}}}-1)$
$\displaystyle+\sum_{q_{1}\neq
q_{2},q_{i}|r_{1}}(p^{f_{\operatorname{base}}\frac{r_{1}}{q_{1}q_{2}}}-1)-\dots\big{)}$
$\displaystyle=\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})r_{1}P(f_{\operatorname{base}},r_{1},p).$

In other words, the number of choices of $(s,j_{1})$ such that
$\zeta_{p^{f_{1}}-1}^{s}(\zeta_{p^{f_{1}}-1}^{jb_{1}/e_{1,\operatorname{rel}}}\pi_{K})^{b_{1}/e_{1,\operatorname{rel}}}$
has $k$ Galois conjugates is
$\gcd(p^{f_{1}}-1,e_{1,\operatorname{rel}})\frac{k}{\operatorname{denom}(\beta)}P(f_{\operatorname{base}},\frac{k}{\operatorname{denom}(\beta)},p)$.

∎

###### proof of Lemma 7.2.

We again convert the problem into modular arithmetic. Write
$a_{i,b_{i}}=\zeta_{p^{f_{i}}-1}^{s_{i}}$ and
$a_{2,b_{2}}=\zeta_{p^{f_{2}}-1}^{s_{2}}$, where
$s_{i}\in{\mathbb{Z}}/(p^{f_{i}}-1){\mathbb{Z}}$. Then, as in the proof of
Lemma 7.1, we had that

$\\{p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})+b_{1}r_{2}(p^{f_{1}}-1)\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}:r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/(e_{1,\operatorname{rel}}){\mathbb{Z}}\\}$

corresponds to the Galois conjugates of $a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}}$,
via the map that sends $x\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}$ to
$\zeta_{(p^{f_{1}}-1)e_{1,\operatorname{rel}}}^{x}\pi_{K}^{b_{1}/e_{1,\operatorname{rel}}}$.
Since we need to consider $a_{2,b_{2}}\pi_{L_{j_{2}}}^{b_{2}}$, we need to
actually express our conjugates in terms of roots of unity with order
$(p^{f_{1}}-1)e_{1,\operatorname{rel}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}$.
Thus, for the sake of this proof, the Galois conjugates of
$a_{1,b_{1}}\pi_{L_{j_{1}}}^{b_{1}}$ will correspond to

$\\{(p^{f_{2}}-1)e_{2,\operatorname{rel}}(p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})+b_{1}r_{2}(p^{f_{1}}-1))\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}:$
$r_{1}\in{\mathbb{Z}}/(f_{1}/f_{\operatorname{base}}){\mathbb{Z}},r_{2}\in{\mathbb{Z}}/e_{1,\operatorname{rel}}{\mathbb{Z}}\\}.$

Similarly, elements $a_{2,b_{2}}\pi_{L_{j_{2}}}^{b_{2}}$ will correspond to

$(p^{f_{1}}-1)e_{1,\operatorname{rel}}(s_{2}e_{2,\operatorname{rel}}+j_{2}b_{2})\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}.$

Thus, in terms of modular arithmetic, we must find

$(s_{2},j_{2})\in{\mathbb{Z}}/(p^{f_{2}}-1){\mathbb{Z}}\times[\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})]$

so that there is $r_{1},r_{2}$ such that

$(p^{f_{2}}-1)e_{2,\operatorname{rel}}(p^{f_{\operatorname{base}}r_{1}}(se_{1,\operatorname{rel}}+jb_{1})+b_{1}r_{2}(p^{f_{1}}-1))\equiv(p^{f_{1}}-1)e_{1,\operatorname{rel}}(s_{2}e_{2,\operatorname{rel}}+j_{2}b_{2})$
$\mod(p^{f_{1}}-1)e_{1,\operatorname{rel}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}.$

Fixing $r_{1}$, we have a solution $r_{2}$ to the above equation if and only
if

$p^{r_{1}f_{\operatorname{base}}}s_{1}e_{1,\operatorname{rel}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}+p^{r_{1}f_{\operatorname{base}}}j_{1}b_{1}(p^{f_{2}}-1)e_{2,\operatorname{rel}}$
$\equiv
s_{2}e_{2,\operatorname{rel}}(p^{f_{1}}-1)e_{1,\operatorname{rel}}+j_{2}b_{2}(p^{f_{1}}-1)e_{1,\operatorname{rel}}\mod(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})(p^{f_{2}}-1)e_{2,\operatorname{rel}}.$

Thus, this is the equation that we need to solve. Now by Equation 7.1, we have

$\frac{(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})}{p^{f_{\operatorname{base}}k}-1}|(s_{1}e_{1,\operatorname{rel}}+j_{1}b_{1}).$

Thus, we may write
$s_{1}e_{1,\operatorname{rel}}+j_{1}b_{1}=y\frac{(p^{f_{1}}-1)\gcd(b_{1},e_{1,\operatorname{rel}})}{p^{f_{\operatorname{base}}k}-1}$
for some integer $y$. Thus, our congruence becomes

$\frac{yp^{r_{1}f_{\operatorname{base}}}(p^{f_{2}}-1)e_{2,\operatorname{rel}}}{p^{f_{\operatorname{base}}k}-1}\equiv
s_{2}e_{2,\operatorname{rel}}\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}+j_{2}b_{2}\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}\mod(p^{f_{2}}-1)e_{2,\operatorname{rel}}.$

Now as we had in the statement of the lemma,
$\frac{e_{1,\operatorname{rel}}}{\gcd(b_{1},e_{1,\operatorname{rel}})}=\frac{e_{2,\operatorname{rel}}}{\gcd(b_{2},e_{2,\operatorname{rel}})}$.
Thus, we may divide by $e_{2,\operatorname{rel}}$ on both sides and obtain

$\frac{yp^{r_{1}f_{\operatorname{base}}}(p^{f_{2}}-1)}{p^{f_{\operatorname{base}}k}-1}\equiv
s_{2}\frac{e_{2,\operatorname{rel}}}{\gcd(b_{2},e_{2,\operatorname{rel}})}+j_{2}\frac{b_{2}}{\gcd(b_{2},e_{2,\operatorname{rel}})}\mod(p^{f_{2}}-1).$

Now, by Lemma 7.4, for each choice of $r_{1}$, there are
$\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})$ choices of $s_{2},j_{2}$
satisfying the above congruence. Thus, these choices $s_{2},j_{2}$ are
precisely the ones such that
$\zeta_{p^{f_{2}}-1}^{s_{2}}\pi_{L_{j_{2}}}^{b_{2}}$ is a Galois conjugate of
$\zeta_{p^{f_{1}}-1}^{s_{1}}\pi_{L_{j_{1}}}^{b_{1}}$. Furthermore, for
$r_{1}\in[\frac{k}{\operatorname{denom}(\beta)}]$, we get different elements
on the LHS. Otherwise, we would have that

$(p^{r_{1}^{\prime}f_{\operatorname{base}}}-1)(s_{2}\frac{e_{2,\operatorname{rel}}}{\gcd(b_{2},e_{2,\operatorname{rel}})}+j_{2}\frac{b_{2}}{\gcd(b_{2},e_{2,\operatorname{rel}})})\equiv
0\mod p^{f_{2}}-1,$

for some $r_{1}^{\prime}<r_{1}$, which would imply that our choice of
$s_{2},j_{2}$ gives us that
$\zeta_{p^{f_{2}}-1}^{s_{2}}\pi_{L_{j_{2}}}^{b_{2}}$ has less than
$k=r_{1}\operatorname{denom}(\beta)$ Galois conjugates, which would be a
contradiction to the fact that these are Galois conjugates of
$\zeta_{p^{f_{1}}-1}^{s_{1}}\pi_{L_{j_{1}}}^{b_{1}}$. Thus, we get a total of

$\frac{k}{\operatorname{denom}(\beta)}\gcd(p^{f_{2}}-1,e_{2,\operatorname{rel}})$

choices of $s_{2},j_{2}$ ∎

##  8\. Large $q$ limit of $\rho$

In this section, we study the large $q$ limit of $\rho(\sigma,e,f;p)$ and
prove Theorem 3.2. By large $q$ limit, we mean asymptotics as $p$ or $f$ (or
both) approaches infinity. For the rest of this section, we will use absolute
notation, as we did for section 4 and so in particular we have that
$e_{\operatorname{base}}|e_{i}$ and $f_{\operatorname{base}}|f_{i}$ for all
$e_{i},f_{i}$ in $\sigma$.

We begin with a formula for $\rho$ in terms of $G$. Combining Equation 3.1 and
Lemma 3.8, we may rewrite $\rho$ as

$\rho(\sigma,e_{\operatorname{base}},f_{\operatorname{base}};p)=\frac{w(d)\sum_{A\subset[m]}N_{1,A}N_{2,A}}{\operatorname{perm}(\sigma)q^{\sum_{i=1}^{m}(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}/2}\prod_{i=1}^{m}f_{i,\operatorname{rel}}},$

where

$N_{1,A}=\frac{\sum_{\begin{subarray}{c}L_{A}=\prod_{i\in A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i,\operatorname{rel}}}-1,e_{i,\operatorname{rel}})]\end{subarray}}\int_{\alpha\in{\mathcal{O}}_{L_{A}}}q^{-\frac{1}{2}v_{K}(\Delta_{P_{\alpha/K}})}d\lambda_{L_{A}}}{\prod_{i\in
A}\gcd(q^{f_{i,\operatorname{rel}}}-1,e_{i,\operatorname{rel}})}$

and

$N_{2,A}=\frac{\sum_{\begin{subarray}{c}L_{A^{c}}=\prod_{i\notin
A}L_{j_{i}}\\\
j_{i}\in[\gcd(q^{f_{i,\operatorname{rel}}}-1,e_{i,\operatorname{rel}})]\end{subarray}}\int_{\alpha\in
m_{L_{A^{c}}}}q^{-\frac{1}{2}v_{K}(\Delta_{P_{\alpha/K}})}d\lambda_{L_{A^{c}}}}{\prod_{i\notin
A}\gcd(q^{f_{i,\operatorname{rel}}}-1,e_{i,\operatorname{rel}})}.$

Here $q=p^{f_{\operatorname{base}}}$ is the size of the residue field of $K$.
We recognize that $N_{1,A}$ is just
$G(\sigma_{A},e_{\operatorname{base}},f_{\operatorname{base}},\\{0\\}_{i\in
A},p)(q^{-e_{\operatorname{base}}/2})$, where
$\sigma_{A}=(e_{i}^{f_{i}})_{i\in A}$, and similarly for $N_{2,A}$. Writing

$G(A,0)(q,t)=G(\sigma_{A},e_{\operatorname{base}},f_{\operatorname{base}},\\{0\\}_{i\in
A},p)(t)$

and

$G(A,1)(q,t)=G(\sigma_{A},e_{\operatorname{base}},f_{\operatorname{base}},\\{1\\}_{i\in
A},p)(t),$

where $q=p^{f_{\operatorname{base}}}$, we obtain the formula

(8.1)
$\rho(\sigma,e_{\operatorname{base}},f_{\operatorname{base}};p)=\frac{w(d)\sum_{A\subset[m]}G(A,0)(q^{-e_{\operatorname{base}}/2})G(A^{c},1)(q^{-e_{\operatorname{base}}/2})}{\operatorname{perm}(\sigma)q^{\sum_{i=1}^{m}(e_{i,\operatorname{rel}}-1)f_{i,\operatorname{rel}}/2}\prod_{i=1}^{m}f_{i,\operatorname{rel}}},$

which we may use to compute $\rho$ from $G$ (here we define $G(A,1)=1$ if $A$
is empty.

Thus, if suffices to compute asymptotics of $G$. Let $L/K$ be an étale
extension with the appropriate splitting type. Our approach revolves around
the following observation: Most Galois conjugates of elements in
${\mathcal{O}}_{L}$ try to be as far as possible from one another. In other
words, most elements $\alpha$ in ${\mathcal{O}}_{L}$ have the least possible
$v_{p}(\Delta_{P_{\alpha}/K})$. Thus, what we will show is:

  1. (1)

Find the smallest possible $v_{p}(\Delta_{P_{\alpha}/K})$ in
${\mathcal{O}}_{L}$.

  2. (2)

Show that as $q$ tends to infinity, the probability that an element
$\alpha\in{\mathcal{O}}_{L}$ attains this minimal valuation approaches $1$.

Now recall our definition of
$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{b_{i}\\})$ in
terms of coefficients
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\}_{i=1}^{m},p)$.
The above list translates to

  1. (1)

Find the smallest $c$ such that
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{b_{i}\\}_{i=1}^{m},p)$
is nonzero. Call this $c_{0}$.

  2. (2)

Show
$\lim_{q\to\infty}a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c_{0},\\{b_{i}\\}_{i=1}^{m},p)=1$.

The combination of these two facts then gives us large $q$ limit asymptotics
for

$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{0\\}_{i=1}^{m},p)(q^{-e_{\operatorname{base}}/2}).$

We will then give a bound on the size of

$G(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},\\{1\\}_{i=1}^{m},p)(q^{-e_{\operatorname{base}}/2}).$

It turns out that these facts will suffice to calculate the asymptotics of
$\rho$.

Now we start with our first step. We need to find the minimal $c$ so that
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{0\\}_{i=1}^{m},p)$
is nonzero. Fix a base field $K$ with
$e(K/{\mathbb{Q}}_{p})=e_{\operatorname{base}}$ and
$f(K/{\mathbb{Q}}_{p})=f_{\operatorname{base}}$. Then,
$a(\sigma,e_{\operatorname{base}},f_{\operatorname{base}},c,\\{0\\}_{i=1}^{m},p)$
is nonzero if and only if there is some element $(\alpha_{i})$ generating an
extension of $K$ with splitting type $\sigma$ such that
$v_{p}(\Delta_{P_{(\alpha_{i})}/K})=c$.

######  Lemma 8.1.

The smallest $c$ such that there is some element $(\alpha_{i})$ generating an
extension of $K$ with splitting type $\sigma$ such that
$v_{p}(\Delta_{P_{(\alpha_{i})}/K})=c$ is

$c_{0}:=\sum_{i=1}^{m}\frac{1}{e_{\operatorname{base}}}f_{i,\operatorname{rel}}(e_{i,\operatorname{rel}}-1).$

Moreover, the elements $(\alpha_{i})$ that achieve this are precisely the ones
that satisfy

  1. (1)

For all $i$, the first term in the Teichmüller expansion of $\alpha_{i}$ are
primitive $p^{f_{i}}-1$ roots of unity, and these roots of unity belong in
different Galois orbits over $K$ as $i$ varies.

  2. (2)

For all $i$, the second term in the Teichmüller expansion of $\alpha_{i}$ is
nonzero.

###### Proof.

We approach this problem by essentially following the recursive steps as in
Proposition 4.2. We will prove lower bounds on the valuation of discriminant
of a minimal polynomial generating an extension with appropriate splitting
type, and see along the way that these bounds are achieved.

We have that (here we use the same notation as Section 4)

$\Delta_{P_{(\alpha_{i})}/K}=\prod_{(i,k)\neq(i^{\prime},k^{\prime}),k,k^{\prime}\in[d_{i}]}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}}),$

where $\alpha_{i,k}$ are Galois conjugates of $\alpha_{i}$ over $K$. So, we
wish to maximize distances between Galois conjugates. Now we use equation 4.1
(and its associated notations) and try to minimize distances in each of the
terms. So, we have two cases: $(i,k)\sim(i^{\prime},k^{\prime})$ and
$(i,k)\not\sim(i^{\prime},k^{\prime})$. In the first case, where
$(i,k)\sim(i^{\prime},k^{\prime})$, we have the trivial bound that

$v_{p}(\alpha_{i,k}-\alpha_{i^{\prime},k^{\prime}})\geq 0.$

In the second case, where $(i,k)\sim(i^{\prime},k^{\prime})$, we have that
$\alpha_{i,k}$ and $\alpha_{i^{\prime},k^{\prime}}$ share a common lowest
order term $\gamma$ (in terms of the Teichmüller expansion). Thus,