# Rational Transformations and Invariant Polynomials

Max Schulz
University of Rostock  Germany
<EMAIL_ADDRESS>
###### Abstract

Rational transformations of polynomials are extensively studied in the context
of finite fields, especially for the construction of irreducible polynomials.
In this paper, we consider the factorization of rational transformations with
(normalized) generators of the field $K(x)^{G}$ of $G$-invariant rational
functions for $G$ a finite subgroup of $\operatorname{PGL}_{2}(K)$, where $K$
is an arbitrary field. Our main theorem shows that the factorization is
related to a well-known group action of $G$ on a subset of monic polynomials.
With this, we are able to extend a result by Lucas Reis for $G$-invariant
irreducible polynomials. Additionally, some new results about the number of
irreducible factors of rational transformations for $Q$ a generator of
$\mathbb{F}_{q}(x)^{G}$ are given when $G$ is non-cyclic.

## Introduction

Let $K$ be an arbitrary field, $K^{\ast}=K\setminus\\{0\\}$ the set of its
units, $K[x]$ the set of polynomials with coefficients in $K$ and
$\mathcal{I}_{K}$ the set of monic irreducible polynomials in $K[x]$, $K(x)$
the rational function field over $K$ and $\mathbb{F}_{q}$ the field with $q$
elements. For a rational function $Q(x)\in K(x)$ we always denote its
numerator and denominator as $g$ and $h$, i.e. $Q(x)=g(x)/h(x)$. Furthermore,
we assume that rational functions are represented as reduced fractions, so
$\gcd(g,h)=1$. Recall that the degree of $Q$ is
$\deg(Q)=\max\\{\deg(g),\deg(h)\\}$. The $Q$-transform of a polynomial
$F(x)=\sum_{i=0}^{k}a_{i}x^{i}\in K[x]$ is defined as

$F^{Q}(x):=h(x)^{\deg(F)}F\left(\frac{g(x)}{h(x)}\right)=\sum\limits_{i=0}^{k}a_{i}g(x)^{i}h(x)^{k-i}.$

This is not yet well-defined since for all $a\in K^{\ast}$ we have

$Q(x)=\frac{a\cdot g(x)}{a\cdot h(x)}$

which leads to

$F^{Q}(x)=\sum\limits_{i=0}^{k}a_{i}(ag(x))^{i}(ah(x))^{k-i}=a^{k}\cdot\sum\limits_{i=0}^{k}a_{i}g(x)^{i}h(x)^{k-i}.$

One might make this transformation unambiguous by normalizing either the
numerator $g$ of $Q$ or the resulting polynomial $F^{Q}$. In our setup we most
often have that $Q$ satisfies $\deg(g)>\deg(h)$ and if $F,g$ are monic so is
$F^{Q}$.

The transformation $F^{Q}$ is often used for constructing irreducible
polynomials of high degree over finite fields starting with an irreducible
polynomial and a rational function. There is a rich literature on this topic,
for example [1], [3], [7], [17], [18] and [19]. The main criterion in use is

######  Lemma ([6, Lemma 1]).

Let $Q(x)=g(x)/h(x)\in K(x)$ and $F\in K[x]$. Then $F^{Q}$ is irreducible if
and only if $F\in K[x]$ is irreducible and $g(x)-\alpha h(x)$ is irreducible
over $K(\alpha)[x]$, where $\alpha$ is a root of $F$.

The original version is only stated for finite fields, but the proof does work
for arbitrary fields as well. The concrete application of this lemma for
arbitrary rational functions and starting polynomials $F$ is very hard, which
is why the best one can do is to focus on specific rational functions or
”small” families of rational functions.

This paper considers two specific $Q$-transformations: The first is $Q$ being
a rational function of degree 1, i.e.

$Q(x)=\frac{ax+b}{cx+d}$

where $ad-bc\neq 0$. The $Q$-transform of $F$ looks like this

$F^{Q}(x)=\lambda_{Q,F}(cx+d)^{\deg(F)}F\left(\frac{ax+b}{cx+d}\right),$

where $\lambda_{Q,F}\in K^{\ast}$ makes the resulting polynomial monic. This
transformation preserves the irreducibility and degree of $F$ if $\deg(F)\geq
2$ by the previous lemma. There is another way to interpret this particular
$Q$-transformation: Let $\operatorname{GL}_{2}(K)$ be the set of invertible
$2\times 2$-matrices over $K$ and let

$A=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in\operatorname{GL}_{2}(K).$ (1)

We make the convention that if we write $A\in\operatorname{GL}_{2}(K)$ then we
assume that $A$ is of the form (1). We consider the projective general linear
group $\operatorname{PGL}_{2}(K)=\operatorname{GL}_{2}(K)/Z$ over $K$, where
$Z=K^{\ast}I_{2}$ is the set of invertible scalar multiples of the identity
matrix $I_{2}$, which is the center of $\operatorname{GL}_{2}(K)$. The group
$\operatorname{PGL}_{2}(K)$ is isomorphic to the set of degree 1 rational
functions in $K(x)$, where the multiplication is composition. We denote by
$[A]$ the coset of $A$ in $\operatorname{PGL}_{2}(K)$, that is,

$[A]:=\\{\alpha\cdot A|\alpha\in K^{\ast}\\}.$

We define $\ast:\operatorname{PGL}_{2}(K)\times K[x]\to K[x]$ by

$[A]\ast
f(x):=\lambda_{A,f}\cdot(cx+d)^{\deg(f)}f\left(\frac{ax+b}{cx+d}\right),$ (2)

where $\lambda_{A,f}\in K^{\ast}$ makes the output-polynomial monic. We call
$f\in K[x]$ $[A]$-invariant for an $[A]\in\operatorname{PGL}_{2}(K)$ if
$[A]\ast f(x)=f(x)$. Moreover $f$ is called $G$-invariant for a subgroup
$G\leq\operatorname{PGL}_{2}(K)$ if it is $[A]$-invariant for all $[A]\in G$.
It can be shown that an $[A]$-invariant polynomial is also
$\langle[A]\rangle$-invariant. There is a substantial amount of literature on
this transformation and its variations in the context of finite fields, for
example [12], [20], [21], [22], [25], [26]. For instance, it is shown that
this transformation induces a (right) group action of
$\operatorname{PGL}_{2}(K)$ on the set of monic polynomials with no roots in
$K$. The following theorem shows that $[A]$-invariant irreducible monic
polynomials over finite fields are always $Q_{A}$-transformations for specific
rational functions $Q_{A}$ depending on
$A\in\operatorname{GL}_{2}(\mathbb{F}_{q})$:

######  Theorem R ([20, Theorem 6.0.7.]).

Let $[A]\in\operatorname{PGL}_{2}(\mathbb{F}_{q})$ be an element of order
$D=\operatorname{ord}([A])$. Then there exists a rational function
$Q_{A}(x)=g_{A}(x)/h_{A}(x)$ of degree $D$ with the property that the
$[A]$-invariant monic irreducible polynomials of degree $Dm>2$ are exactly the
monic irreducible polynomials of the form

$F^{Q_{A}}(x)=h_{A}(x)^{m}\cdot F\left(\frac{g_{A}(x)}{h_{A}(x)}\right),$

where $\deg(F)=m$. In addition, $Q_{A}$ can be explicitly computed from $A$.

This theorem is proved by dividing $\operatorname{PGL}_{2}(\mathbb{F}_{q})$
into four types of conjugacy classes and showing it for a nice representative
of each class.

Let $G\leq\operatorname{PGL}_{2}(K)$ be a finite subgroup and for
$A\in\operatorname{GL}_{2}(K)$ set

$[A]\circ x:=\frac{ax+b}{cx+d}.$ (3)

There exists a rational function $Q_{G}\in K(x)$ of degree $|G|$ so that
$K(x)^{G}=K(Q_{G}(x))$ where

$K(x)^{G}:=\\{Q\in K(x)|~{}Q([A]\circ x)=Q(x)\text{ for all }[A]\in G\\}$

is the fixed field of $G$ (for reference see [4]). Moreover, every rational
function $Q\in K(x)^{G}$ of degree $|G|$ is a generator of $K(x)^{G}$, so we
can normalize $Q_{G}$ in such a way that $Q_{G}(x)=g(x)/h(x)$ with
$0\leq\deg(h)<\deg(g)=|G|$ and $g$ monic. Based on [4], we call these
generators quotient maps for $G$ and this is the second class of rational
functions we consider in this paper. In [20] it is noted that for some
$[A]\in\operatorname{PGL}_{2}(\mathbb{F}_{q})$ the functions $Q_{A}$ in
Theorem R are in fact generators of the fixed field
$K(x)^{\langle[A]\rangle}$.

A natural question to ask is whether the function $Q_{A}$ in Theorem R is
always a generator of $K(x)^{\langle[A]\rangle}$ for all
$[A]\in\operatorname{PGL}_{2}(\mathbb{F}_{q})$. An understanding of this
question is of interest since many constructions of irreducible polynomials
over finite fields via $Q$-transformations that work very well use specific
generators of specific fields of invariant functions, see for example [7],
[18] and [19]. Another natural question is whether the theorem still holds if
we consider $G$-invariant and not necessarily irreducible polynomials for
arbitrary finite subgroups of $\operatorname{PGL}_{2}(K)$. These two questions
led us to study the $Q_{G}$-transformations of irreducible polynomials and
their factorization. We did not want to necessarily restrict ourselves to the
case that $K$ is finite, so we formulate the results for arbitrary fields.
However, the theory is especially beautiful in characteristic $p>0$ because
the finite subgroups of $\operatorname{PGL}_{2}(K)$ are more diverse there
(see [10], [11] and [27]).

The main result and starting point of this paper can be summarized as the
following theorem about the factorization of $F^{Q_{G}}$ for $F$ an
irreducible monic polynomial and $Q_{G}$ a quotient map for $G$:

######  Main Theorem.

Let $F\in K[x]$ be monic and irreducible, $G\leq\operatorname{PGL}_{2}(K)$ a
finite subgroup and $Q_{G}=g/h\in K(x)$ a quotient map for $G$. Then there is
an irreducible monic polynomial $r\in K[x]$ with $\deg(F)|\deg(r)$ and an
integer $k>0$ such that

$F^{Q_{G}}(x)=\left(\prod\limits_{t\in G\ast r}t(x)\right)^{k},$

where $G\ast r:=\\{[A]\ast r|[A]\in G\\}$ is the $G$-orbit of $r$.
Additionally $k=1$ for all but finitely many irreducible monic polynomials
$F\in K[x]$.

The main difficulty of the proof is to show that $k=1$ for all but finitely
many irreducible and monic $F\in K[x]$ in non-perfect fields.

We want to point out that a very similar result is known for the case that
$F\in\mathcal{I}_{K}$ is of degree 1 and $K=\mathbb{F}_{q}$; we state said
theorem for convenience:

######  Theorem ([13, Theorem 26]).

Let $G$ be a subgroup of $\operatorname{PGL}_{2}(\mathbb{F}_{q})$ and
$Q(x)=g(x)/h(x)$ a generator for $\mathbb{F}_{q}(x)^{G}$. Let
$\alpha\in\overline{\mathbb{F}}_{q}$ have the property that
$Q(\alpha)\in\mathbb{F}_{q}$ and assume that $G$ acts regularly on the roots
of $F_{\alpha}(T):=g(T)-Q(\alpha)h(T)\in\mathbb{F}_{q}[T]$ via Möbius-
Transformation, then

  1. 1.

$F_{\alpha}$ will factor into irreducible polynomials of the same degree over
$\mathbb{F}_{q}[T]$

  2. 2.

The minimal polynomial of $\alpha$ is one of the factors of $F_{\alpha}$

  3. 3.

The degree of each factor must be the order of an element of $G$.

To see that both theorems are connected notice that for
$\beta=Q(\alpha)\in\mathbb{F}_{q}$ we have that $F^{Q_{G}}(T)=g(T)-\beta h(T)$
for $F=T-\beta$, which factors into a $G$-orbit of an irreducible polynomial
by our Main Theorem and all elements in a $G$-orbit have the same degree,
which explains item 1. The second item is also true in our setup, that is, if
$\beta\in\overline{K}$ is a root of $F$, then $\alpha\in Q_{G}^{-1}(\beta)$ is
a root of $F^{Q_{G}}$. The third item, however, is a finite field specific
result and generalizes to arbitrary fields and irreducible polynomials $F$ of
arbitrary degree as follows: Every irreducible factor of $F^{Q_{G}}$ has
degree $\deg(F)$ times the size of a subgroup of $G$. The condition that the
set of roots of $F_{\alpha}$ only contains regular $G$-orbits is a crucial one
for the case that $k=1$ in the Main Theorem. All of this will be explained in
depth in this paper. The phenomenon that $F^{Q_{G}}$ factorizes into a
$G$-orbit of an irreducible polynomial was, until now, only noted for some
instances of generators of specific invariant rational function fields over
finite fields.

######  Example 1.

  1. 1.

We start with $Q_{1}=x+1/x\in\mathbb{F}_{q}(x)$ and look at the factorization
of $F^{Q_{1}}$, where $F\in\mathcal{I}_{q}:=\mathcal{I}_{\mathbb{F}_{q}}$. It
is proved in [18, Lemma 4] that $F^{Q_{1}}$ is either irreducible and self-
reciprocal or factorizes into a reciprocal pair. For
$r\in\mathcal{I}_{q}\setminus\\{x\\}$ we set
$r^{\ast}(x):=a_{0}^{-1}x^{\deg(r)}r(1/x)$ as its reciprocal polynomial, where
$a_{0}$ is the constant term of $r$. A polynomial is said to be self-
reciprocal if $r(x)=r^{\ast}(x)$ and a reciprocal pair is a pair $r,r^{\ast}$
such that $r\neq r^{\ast}$. This result can be explained with our Main
Theorem: Let

$G_{1}=\left\langle\left[\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\right]\right\rangle.$

This is a subgroup of order 2 and a generator of $G_{1}$ is
$Q_{G_{1}}(x)=x+1/x=(x^{2}+1)/x\in K(x)$. Then, for all but finitely many
irreducible monic polynomials $F\in\mathcal{I}_{K}$ we obtain that there
exists an irreducible monic polynomial $r\in K[x]$ such that

$F^{Q_{G_{1}}}(x)=\begin{cases}r(x),&\text{ if }F^{Q_{G_{1}}}\text{ is
irreducible}\\\ r(x)\cdot a_{0}^{-1}x^{\deg(r)}r(1/x),&\text{ if
}F^{Q_{G_{1}}}\text{ is not irreducible}\end{cases}.$

  2. 2.

The factorization of $F(x^{n})$ in $\mathbb{F}_{q}[x]$ for $n|q-1$ leads to
another nice example. Let $a\in\mathbb{F}_{q}^{\ast}$ be a primitive $n$-th
root of unity. It can be shown that for all
$F\in\mathcal{I}_{q}\setminus\\{x\\}$ there exists $m\mid n$ and
$r\in\mathcal{I}_{q}$ such that

$F(x^{n})=\prod\limits_{i=0}^{m-1}a^{-i\cdot\deg(r)}\cdot r(a^{i}x).$

For reference see [2] and [8]; for a nice application of this see [15]. The
rational function $Q_{2}(x)=x^{n}$ is a quotient map for the subgroup

$G_{2}:=\left\\{\left[\left(\begin{array}[]{cc}a^{i}&0\\\
0&1\end{array}\right)\right]|i\in\mathbb{N}\right\\}$

and $F^{Q_{2}}(x)=F(x^{n})$. The factors belong to the same $G_{2}$-orbit,
since

$\left[\left(\begin{array}[]{cc}a^{i}&0\\\ 0&1\end{array}\right)\right]\ast
r(x)=a^{-i\deg(r)}\cdot r(a^{i}x).$

  3. 3.

In [5] it is noted that if $F(x^{p}-x)$ is not irreducible for
$F\in\mathcal{I}_{q}$ and $p^{l}=q$, then $F(x^{p}-x)$ factorizes into exactly
$p$ irreducible polynomials of degree $\deg(F)$ and, more precisely, there
exists $r\in\mathcal{I}_{q}$ with $\deg(r)=\deg(F)$ such that

$F(x^{p}-x)=r(x)\cdot r(x+1)\cdot\ldots\cdot r(x+(p-1)).$

The rational function $Q_{3}(x)=x^{p}-x$ is a quotient map for

$G_{3}:=\left\\{\left[\left(\begin{array}[]{cc}1&a\\\
0&1\end{array}\right)\right]|a\in\mathbb{F}_{p}\right\\}$

and for $r\in K[x]$ the transformation with an element of $G_{3}$ looks like
this

$\left[\left(\begin{array}[]{cc}1&a\\\ 0&1\end{array}\right)\right]\ast
r(x)=r(x+a).$

All of these examples still hold in every field $K$ in which the corresponding
subgroups of $\operatorname{PGL}_{2}(K)$ exist. The factorization of
$F^{Q_{G}}$ can be easily obtained by finding just one irreducible factor and
calculating the $G$-orbit of this factor. We also see that in literature,
apart from [13, Theorem 26], only small cyclic subgroups of
$\operatorname{PGL}_{2}(\mathbb{F}_{q})$ of prime order were considered. In
contrast, we want to look at big subgroups of
$\operatorname{PGL}_{2}(\mathbb{F}_{q})$ instead. We can obtain the following
new result over finite fields:

######  Theorem 2.

Let $K=\mathbb{F}_{q}$ and $G\leq\operatorname{PGL}_{2}(\mathbb{F}_{q})$ with
quotient map $Q_{G}\in\mathbb{F}_{q}(x)$. Moreover, set $\mu_{G}\in\mathbb{N}$
as the maximal order of an element in $G$ and let $F\in\mathbb{F}_{q}[x]$ be
an irreducible monic polynomial such that $F^{Q_{G}}$ is separable. Then we
have that $F^{Q_{G}}$ has at least $|G|/\mu_{G}$ irreducible factors and every
such factor has degree at most $\mu_{G}\cdot\deg(F)$.

######  Remark 3.

The polynomial $F^{Q_{G}}$ is separable if $F\in\mathcal{I}_{q}$ and
$\deg(F)\geq 3$, so the only exception polynomials for which the theorem does
not necessarily hold are irreducible polynomials of degree less than 3. For an
explanation see Theorem 17, Theorem 22 and Lemma 28.

For example, take $\\{0\\}\neq V\leq_{p}\mathbb{F}_{q}$ as a
$\mathbb{F}_{p}$-subspace of $\mathbb{F}_{q}$, then define

$\overset{\sim}{V}:=\left\\{\left[\left(\begin{array}[]{cc}1&v\\\
0&1\end{array}\right)\right]|v\in V\right\\}.$

We call $\overset{\sim}{V}$ the to $V$ associated subgroup in
$\operatorname{PGL}_{2}(\mathbb{F}_{q})$. Observe that
$V\cong\overset{\sim}{V}$ as groups. A quotient map for $\overset{\sim}{V}$ is
the to $V$ associated subspace polynomial, that is,

$Q_{V}(x)=\prod\limits_{v\in V}(x-v)\in\mathbb{F}_{q}[x].$

Every non-trivial element in $\overset{\sim}{V}$ has order $p$, so
$\mu_{\overset{\sim}{V}}=p$ and therefore we obtain the following corollary

######  Corollary 4.

Let $\\{0\\}\neq V\leq_{p}\mathbb{F}_{q}$ be an $\mathbb{F}_{p}$-subspace of
$\mathbb{F}_{q}$ and $Q_{V}\in\mathbb{F}_{q}[x]$ the associated subspace
polynomial. For every irreducible (monic) polynomial $F\in K[x]$ we have that
$F(Q_{V}(x))$ has at least $|V|/p$ irreducible factors and every irreducible
factor has the same degree, which is at most $p\cdot\deg(F)$.

In the last part of this paper we consider two further examples of big
subgroups of $\operatorname{PGL}_{2}(\mathbb{F}_{q})$ and show how to apply
Theorem 2 to them.

The Main Theorem shows that the irreducible factors of $F^{Q_{G}}$ belong to
the same $G$-orbit. Together with the fact that for every $G$-orbit $G\ast r$
in $\mathcal{I}_{K}$ there exists an irreducible $F\in\mathcal{I}_{K}$ such
that $F^{Q_{G}}$ has all polynomials in $G\ast r$ as its factors we can prove
a generalization of Theorem R:

######  Theorem 5.

All but finitely many $G$-invariant irreducible monic polynomials $f$ can be
written as a $Q_{G}$-transformation, i.e. there is $F\in\mathcal{I}_{K}$ such
that $f=F^{Q_{G}}$.

This result does not say anything about the existence of $G$-invariant
irreducible polynomials, it just makes a statement about them if they exist in
$K[x]$!

Our proof of a general version of Theorem R avoids the original idea of
dividíng $\operatorname{PGL}_{2}(K)$ into different types of conjugacy classes
and showing the theorem for each type, which should be hard as such a list can
become quite large depending on the field, see [11] or [27].

The last result shows that the $G$-invariant but not-necessarily irreducible
polynomials are a product of a $Q_{G}$-transformation and some exception
polynomials:

######  Theorem 6.

Let $G\leq\operatorname{PGL}_{2}(K)$ be a finite subgroup and $Q_{G}$ a
quotient map. There exists $k\in\mathbb{N}\setminus\\{0\\}$ and irreducible
monic polynomials $r_{1},\ldots r_{k}\in K[x]$ and $n_{1},\ldots
n_{k}\in\mathbb{N}\setminus\\{0\\}$ such that for every $G$-invariant monic
polynomial $f\in K[x]$ there is a unique monic $F\in K[x]$ and integers
$k_{i}<n_{i}$ such that

$f=\left(\prod\limits_{i=1}^{k}(\prod\limits_{t\in G\ast
r_{i}}t)^{k_{i}}\right)\cdot F^{Q_{G}}.$

We give full explanations about what the polynomials $r_{1},\ldots,r_{k}$ and
the integers $n_{i}$ are in section 3.

##  1 Preliminaries

###  1.1 Invariant Polynomials

We denote by
$\circ:\operatorname{PGL}_{2}(K)\times(\overline{K}\cup\\{\infty\\})\to\overline{K}\cup\\{\infty\\}$
the Möbius-Transformation on $\overline{K}\cup\\{\infty\\}$, that is,

$[A]\circ v=\frac{av+b}{cv+d}.$

This equation is self-explanatory if $v\notin\\{\infty,-\frac{d}{c}\\}$. For
$c\neq 0$ we set $[A]\circ\infty=\frac{a}{c}$ and
$[A]\circ(-\frac{d}{c})=\infty$; $[A]\circ\infty=\infty$ if $c=0$. The Möbius-
Transformation is a left group action of $\operatorname{PGL}_{2}(K)$ on
$\overline{K}\cup\\{\infty\\}$ and thus every subgroup of
$\operatorname{PGL}_{2}(K)$ acts on $\overline{K}\cup\\{\infty\\}$ too. For
$G\leq\operatorname{PGL}_{2}(K)$ we denote the $G$-orbit of
$v\in\overline{K}\cup\\{\infty\\}$ as $G\circ v$. Let
$G\leq\operatorname{PGL}_{2}(K)$ then define

$\mathcal{NR}_{K}^{G}:=\\{f\in K[x]|f\text{ monic and }f(\alpha)\neq 0\text{
for all }\alpha\in G\circ\infty\\}.$

This set is closed under multiplication, i.e. is a submonoid of $K[x]$. We
make the convention that $f(\infty)=\infty$ for all polynomials of degree
greater than 0 and $a(\infty)=a$ for $a\in K$. The following basic result
about $\ast$ holds:

######  Lemma 7.

Let $G\leq\operatorname{PGL}_{2}(K)$. For all $f,g\in\mathcal{NR}_{K}^{G}$ and
$[A],[B]\in G$ the following hold:

  1. 1.

$\deg([A]\ast f)=\deg(f)$

  2. 2.

$[AB]\ast f=[B]\ast([A]\ast f)$ and $[I_{2}]\ast f=f$, so $\ast$ is a right
group action of $G$ on $\mathcal{NR}_{K}^{G}$

  3. 3.

$[A]\ast(fg)=([A]\ast f)([A]\ast g)$

  4. 4.

$f$ irreducible if and only if $[A]\ast f$ irreducible

We omit the proof as it can be done almost exactly as in [12] or [26]. Because
of the fourth item of the previous lemma we know that $G$ induces a group
action on

$\mathcal{I}_{K}^{G}:=\mathcal{I}_{K}\cap\mathcal{NR}_{K}^{G}.$

Remember that we write $G\ast f$ for the $G$-orbit of $f$. Note that $G\ast
f\subset\mathcal{NR}_{K}^{G}$ if $f\in\mathcal{NR}_{K}^{G}$ and every
polynomial in the orbit has the same degree as $f$. The following lemma
explains the connection between $G$-invariant polynomials and the Möbius-
Transformation and the proof can be done similarly as in [12] or [26] again:

######  Lemma 8.

Let $G\leq\operatorname{PGL}_{2}(K)$ and $f\in\mathcal{NR}_{K}^{G}$. Further
we denote by

$R_{f}:=\\{v\in\overline{K}|f(v)=0\\}$ (4)

the set of roots of $f$ in $\overline{K}$. Then the following hold:

  1. 1.

If $f$ is $G$-invariant, then $[A]\circ R_{f}=R_{f}$ for all $[A]\in G$. Here
$[A]\circ R_{f}:=\\{[A]\circ v|v\in R_{f}\\}$

  2. 2.

If $f$ is irreducible the converse is also true, more precisely: $[A]\circ
R_{f}=R_{f}$ for all $[A]\in G$ implies that $f$ is $G$-invariant

From now on we use $R_{f}$ as the set of roots of a polynomial $f\in K[x]$ as
defined in (4). In the lemma above we did not make the assumption that $G$ has
to be a finite subgroup of $\operatorname{PGL}_{2}(K)$. So now, we want to
explore what happens if $G$ is infinite. Let
$[A]\in\operatorname{PGL}_{2}(K)$, then it is quite obvious that all fixed
points of $[A]$ in $\overline{K}$ under $\circ$ are in $K\cup\\{\infty\\}$ or
in a quadratic extension of $K$. Let $v\in\overline{K}$ with $[K(v):K]\geq 3$,
then $[A]\circ v\neq v$ for all $[A]\in\operatorname{PGL}_{2}(K)$, so $G\circ
v$ contains infinitely many elements. Therefore there can not exist
$G$-invariant irreducible monic polynomials of degree greater than 2 for $G$
an infinite subgroup of $\operatorname{PGL}_{2}(K)$, since otherwise it would
have infinitely many roots by Lemma 8. This is one of the reasons why we focus
on finite subgroups of $\operatorname{PGL}_{2}(K)$. Thus, from now on, $G$
denotes a finite subgroup of $\operatorname{PGL}_{2}(K)$. The following
corollary helps us to understand the factorization of $G$-invariant
polynomials:

######  Corollary 9.

Let $f,s,t\in\mathcal{NR}_{K}^{G}$, where $f$ is a $G$-invariant polynomial
with irreducible factor $r\in\mathcal{I}_{K}^{G}$. Then the following hold:

  1. 1.

$[A]\ast r$ divides $f$ for all $[A]\in G$

  2. 2.

If $\gcd(s,t)=1$, then $\gcd([A]\ast s,[A]\ast t)=1$ for all $[A]\in G$

  3. 3.

If $r^{n}|f$ and $r^{n+1}\nmid f$, then $([A]\ast r)^{n}|f$ and $([A]\ast
r)^{n+1}\nmid f$ for all $[A]\in G$. So all polynomials in $G\ast r$ divide
$f$ with the same multiplicity

###### Proof.

The first statement is immediate, since $[A^{-1}]\circ R_{r}=R_{[A]\ast
r}\subset R_{f}$ by Lemma 8, so $[A]\ast r$ divides $f$ as well.

The condition $\gcd(s,t)=1$ is equivalent to $R_{s}\cap R_{t}=\varnothing$ in
$\overline{K}$ and again $[A^{-1}]\circ R_{s}=R_{[A]\ast s}$ and
$[A^{-1}]\circ R_{t}=R_{[A]\ast t}$. With the fact that every $[A]\in G$
induces a bijection on $\overline{K}\cup\\{\infty\\}$ we obtain $R_{[A]\ast
s}\cap R_{[A]\ast t}=\varnothing$.

For the last item we use the previous statement: Write $f=r^{n}\cdot P$ where
$\gcd(r,P)=1$. Then, with the third item of Lemma 7

$\displaystyle f=[A]\ast f=[A]\ast(r^{n}\cdot P)=([A]\ast r)^{n}\cdot([A]\ast
P)$

and $\gcd([A]\ast r,[A]\ast P)=1$. ∎

We just showed that every $G$-invariant polynomial $f\in\mathcal{NR}_{K}^{G}$
consists of powers of $G$-orbits in $\mathcal{I}_{K}^{G}$ that are glued
together by multiplication. So, in a nutshell, $G$-orbits in
$\mathcal{I}_{K}^{G}$ are the atoms of $G$-invariant polynomials and thus are
quite important for this paper; hence the following definition:

######  Definition 10.

We call $f\in\mathcal{NR}_{K}^{G}$ $G$-orbit polynomial (or simply orbit
polynomial) if there exists an irreducible polynomial
$r\in\mathcal{I}_{K}^{G}$ such that

$f=\prod\limits_{t\in G\ast r}t=:\prod(G\ast r).$

For the sake of completeness we state our observation about the factorization
of $G$-invariant polynomials as a corollary, but we omit the proof since it is
a trivial consequence of Corollary 9.

######  Corollary 11.

Let $f\in\mathcal{NR}_{K}^{G}$ be a $G$-invariant polynomial. Then there are
$r_{1},\ldots,r_{k}\in\mathcal{I}_{K}^{G}$ and
$n_{1},\ldots,n_{k}\in\mathbb{N}\setminus\\{0\\}$ such that

$f=\prod\limits_{i=1}^{k}(\prod(G\ast g_{i}))^{n_{i}}.$

###  1.2 Quotient Maps and Rational Transformations

Throughout the rest of the paper we assume that $Q_{G}\in K(x)$ is a quotient
map for $G$ with monic numerator polynomial $g$. Note that such a rational
function exists for all finite subgroups of $\operatorname{PGL}_{2}(K)$ and if
$Q_{G}^{\prime}$ is another quotient map for $G$, then there are constants
$a,b\in K$ such that $Q_{G}^{\prime}(x)=aQ_{G}(x)+b$ (see [4]). We denote by
$(\overline{K}\cup\\{\infty\\})/G$ the set of $G$-orbits in
$\overline{K}\cup\\{\infty\\}$. The following theorem is an important tool for
proving our Main Theorem and explains the name ”quotient map”:

######  Theorem 12 ([4, Proposition 3.9]).

The quotient map $Q_{G}$ induces a bijection between
$(\overline{K}\cup\\{\infty\\})/G$ and $\overline{K}\cup\\{\infty\\}$, more
precisely

$\displaystyle\psi:\begin{cases}(\overline{K}\cup\\{\infty\\})/G\to\overline{K}\cup\\{\infty\\},\\\
G\circ v\mapsto Q_{G}(G\circ v)=Q_{G}(v)\end{cases}$

is a bijection and $\psi(G\circ\infty)=Q_{G}(\infty)=\infty$.

There are essentially two ways to calculate a quotient map for a given finite
subgroup $G$ that we know of. One of them is explained in [13] and works as
follows: Calculate the polynomial

$F_{G}(y):=\prod\limits_{[A]\in G}(y-([A]\circ x))\in K(x)[y].$

One of the coefficients of $F_{G}(y)$ has to be a generator of $K(x)^{G}$ and
thus can be normalized so that it becomes a quotient map. Another method is
explained in subsection 3.3. in [4].

For an arbitrary rational function $Q(x)=g(x)/h(x)$ with $g,h$ having leading
coefficients $a(g),a(h)\in K^{\ast}$ we set

$Q(\infty):=\begin{cases}\infty,&\deg(g)>\deg(h)\\\
\frac{a(g)}{a(h)},&\deg(g)=\deg(h)\\\ 0,&\deg(h)>\deg(g).\end{cases}$

We collect some known facts about rational transformations in the next lemma:

######  Lemma 13.

Let $Q=\frac{g}{h}$ be such that $g$ is monic and $F\in K[x]$ such that
$F(Q(\infty))\neq 0$, then the following hold:

  1. 1.

$\deg(F^{Q})=\deg(F)\cdot\deg(Q)$

  2. 2.

If $F$ is reducible so is $F^{Q}$, more precisely: $F=rt$ and
$\deg(r),\deg(t)\geq 1$, then $F^{Q}=r^{Q}t^{Q}$

  3. 3.

If $F$ is monic and $\deg(g)>\deg(h)$, then $F^{Q}$ is monic as well

The following lemma shows that rational transformations with $Q_{G}$ yield
$G$-invariant polynomials:

######  Lemma 14.

Let $F\in K[x]$ be a monic polynomial, then $F^{Q_{G}}\in\mathcal{NR}_{K}^{G}$
and $F^{Q_{G}}$ is $G$-invariant.

###### Proof.

By Theorem 3.10 and Proposition 3.4 in [4], the denominator of $Q_{G}$ is of
the form

$h(x)=\prod\limits_{v\in(G\circ\infty)\setminus\\{\infty\\}}(x-v)^{m_{\infty}},$

where
$m_{\infty}:=|\operatorname{Stab}_{G}(\infty)|=\frac{|G|}{|G\circ\infty|}$ is
the cardinality of the stabilizer of $\infty$ in $G$. First of all we want to
show that $F^{Q_{G}}\in\mathcal{NR}_{K}^{G}$. For that we write
$F(x)=\sum\limits_{i=0}^{k}a_{i}x^{i}$ with $a_{k}=1$, then

$F^{Q_{G}}(x)=\sum\limits_{i=0}^{k}a_{i}g(x)^{i}h(x)^{k-i}.$

Since the set of roots of $h$ is $G\circ\infty\setminus\\{\infty\\}$ we obtain
for all $w\in G\circ\infty\setminus\\{\infty\\}$:

$\displaystyle
F^{Q_{G}}(w)=\sum\limits_{i=0}^{k}a_{i}g(w)^{i}h(w)^{k-i}=g(w)^{k}\neq 0.$

The last step in the calculation is a consequence of $\gcd(h,g)=1$ which
implies $g(w)\neq 0$. Thus we got $F^{Q_{G}}\in\mathcal{NR}_{K}^{G}$, since
$\mathcal{NR}_{K}^{G}$ is the set of monic polynomials with no roots in the
orbit of $\infty$ and $f(\infty)=\infty\neq 0$ for all polynomials111Note that
the only polynomial of degree 0 in $\mathcal{NR}_{K}^{G}$ is $1$ with
$\deg(f)\geq 1$.

For the calculation that $F^{Q_{G}}$ is indeed $G$-invariant we verify whether
for all $[A]\in G$ there exists $\alpha_{A}\in K^{\ast}$ such that

$(cx+d)^{\deg(F^{Q_{G}})}F^{Q_{G}}(\frac{ax+b}{cx+d})=\alpha_{A}F^{Q_{G}}(x)$

Writing the left side out gives

$\displaystyle(cx+d)^{\deg(F^{Q_{G}})}F^{Q_{G}}((A\circ
x))=(cx+d)^{\deg(F^{Q_{G}})}h(\frac{ax+b}{cx+d})^{\deg(F)}F(Q_{G}(\frac{ax+b}{cx+d})).$

Since $Q_{G}(\frac{ax+b}{cx+d})=Q_{G}(x)$, we can focus on
$(cx+d)^{\deg(F^{Q_{G}})}h(\frac{ax+b}{cx+d})^{\deg(F)}$. For that we have to
consider 2 separate cases, namely $c\neq 0$ and $c=0$. Here we only consider
the first as the latter can be done similarly. So now $c\neq 0$, then we get:

$\displaystyle(cx+d)^{\deg(F^{Q_{G}})}h(\frac{ax+b}{cx+d})^{\deg(F)}$
$\displaystyle=(cx+d)^{\deg(F)\deg(Q_{G})}\prod\limits_{v\in(G\circ\infty)\setminus\\{\infty\\}}(\frac{ax+b}{cx+d}-v)^{m_{\infty}\deg(F)}$
$\displaystyle=\frac{(cx+d)^{\deg(F)\cdot|G|}}{(cx+d)^{\deg(F)\cdot(|G\circ\infty|-1)\cdot\frac{|G|}{|G\circ\infty|}}}\prod\limits_{v\in(G\circ\infty)\setminus\\{\infty\\}}\left((a-cv)x+(b-dv))\right)^{m_{\infty}\cdot\deg(F)}$
$\displaystyle=\left((cx+d)\left(b-d\cdot\frac{a}{c}\right)\prod\limits_{v\in
G\circ\infty\setminus\\{\infty,\frac{a}{c}\\}}(a-cv)\left(x+\frac{b-dv}{a-cv}\right)\right)^{m_{\infty}\cdot\deg(F)}$
$\displaystyle=\alpha_{A}\left(\left(x+\frac{d}{c}\right)\prod\limits_{v\in
G\circ\infty\setminus\\{\infty,\frac{a}{c}\\}}\left(x-[A^{-1}]\circ
v\right)\right)^{m_{\infty}\cdot\deg(F)}$
$\displaystyle=\alpha_{A}\left(\left(x-[A^{-1}]\circ\infty\right)\prod\limits_{v\in
G\circ\infty\setminus\\{\infty,\frac{a}{c}\\}}\left(x-[A^{-1}]\circ
v\right)\right)^{m_{\infty}\cdot\deg(F)}$
$\displaystyle=\alpha_{A}\left(\prod\limits_{v\in
G\circ\infty\setminus\\{\frac{a}{c}\\}}\left(x-[A^{-1}]\circ
v\right)\right)^{m_{\infty}\cdot\deg(F)}$
$\displaystyle=\alpha_{A}\left(\prod\limits_{u\in
G\circ\infty\setminus\\{\infty\\}}\left(x-u\right)\right)^{m_{\infty}\cdot\deg(F)}=\alpha_{A}h(x)^{\deg(F)}.$

The factor $\alpha_{A}$ has the form

$\alpha_{A}=\left(\underbrace{c(b-d\cdot\frac{a}{c})}_{=-\det(A)}\prod\limits_{v\in
G\circ\infty\setminus\\{\infty,\frac{a}{c}\\}}(a-cv)\right)^{{m_{\infty}\cdot\deg(F)}},$

so it is non-zero, since all its factors are non-zero. Moreover note that the
inverse of $[A]$ in $\operatorname{PGL}_{2}(K)$ is $[B]$ with

$B=\left(\begin{array}[]{cc}d&-b\\\ -c&a\end{array}\right),$

because $A\cdot B=\det(A)I_{2}$. This finishes the proof. ∎

This lemma does not hold in general if we consider a general generator instead
of quotient maps. Consider a quotient map $Q_{G}=g/h$, then
$Q:=Q_{G}^{-1}=h/g$ is a generator of $K(x)^{G}$. However, for $F=x$ we get
$F^{Q}=h(x)$, which is not in $\mathcal{NR}_{K}^{G}$ because $h$ has
$G\circ\infty\setminus\\{\infty\\}$ as its roots; therefore $F^{Q}$ can not be
$G$-invariant. This example suggests that the only exceptions are the monic
polynomials $F\in K[x]$ with root $Q(\infty)$. To show that we need a lemma
first.

######  Lemma 15.

Let $Q_{G}\in K(x)$ be a quotient map for $G$ and $Q\in K(x)$ another
generator of $K(x)^{G}$, then $\deg(Q)=\deg(Q_{G})=|G|$ and there is
$[C]\in\operatorname{PGL}_{2}(K)$ such that $Q=[C]\circ Q_{G}$. More precisely
there is

$C=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in\operatorname{GL}_{2}(K)$

such that

$Q(x)=\frac{aQ_{G}(x)+b}{cQ_{G}(x)+d}.$

###### Proof.

The proof is a combination of Lemma 3.1 and Proposition 3.3 in [4]. ∎

######  Corollary 16.

Let $Q_{G}=\frac{g}{h}$ be a quotient map, $Q\in K(x)$ an arbitrary generator
of $K(x)^{G}$ and $F\in K[x]$ monic. Write $Q=[C]\circ Q_{G}$, which is
possible by the lemma above. If $F([C]\circ\infty)\neq 0$, or equivalently
$F(Q(\infty))\neq 0$, then $a\cdot F^{Q}\in\mathcal{NR}_{K}^{G}$ and $a\cdot
F^{Q}$ is $G$-invariant (the factor $a\in K^{\ast}$ is needed to make
$F^{Q_{G}}$ monic).

###### Proof.

Let $H=[C]\ast F$, then $\deg(H)=\deg(F)$ by the assumption that
$F([C]\circ\infty)\neq 0$ (see Lemma 7). Write

$C=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in\operatorname{GL}_{2}(K),$

then $H(x)=\lambda_{C,H}(cx+d)^{\deg(F)}F(\frac{ax+b}{cx+d})$ and
$Q(x)=\frac{ag(x)+bh(x)}{cg(x)+dh(x)}$. Moreover

$\displaystyle H^{Q_{G}}(x)$
$\displaystyle=h(x)^{\deg(H)}H(\frac{g(x)}{h(x)})$
$\displaystyle=\lambda_{C,H}h(x)^{\deg(H)}(c\frac{g(x)}{h(x)}+d)^{\deg(F)}F\left(\frac{a\frac{g(x)}{h(x)}+b}{c\frac{g(x)}{h(x)}+d}\right)$
$\displaystyle=\lambda_{C,H}(cg(x)+dh(x))^{\deg(F)}F\left(\frac{ag(x)+bh(x)}{cg(x)+dh(x)}\right)=\lambda_{C,H}F^{Q}(x).$

With Lemma 14 we have that $H^{Q_{G}}$ is $G$-invariant and an element of
$\mathcal{NR}_{K}^{G}$, thus both facts are also true for
$\lambda_{C,H}F^{Q}$. ∎

##  2 Proof of the Main Theorem

Here $\overline{K}$ denotes a fixed algebraic closure of $K$ and for a
polynomial $P\in K[x]$ we denote its splitting field in $\overline{K}$ as
$L_{P}$. Further we define for an extension $K\subset L\subset\overline{K}$
the set $\operatorname{hom}_{K}(L)$ as the $K$-automorphisms
$\sigma:L\to\overline{K}$. If $L$ is normal over $K$ then
$\operatorname{hom}_{K}(L)=\operatorname{Aut}_{K}(L)$. We are going to prove
the first part of the Main Theorem:

######  Theorem 17.

Let $F\in\mathcal{I}_{K}$, then there is $k\in\mathbb{N}\setminus\\{0\\}$ and
$r\in\mathcal{I}_{K}^{G}$ with $\deg(F)|\deg(r)$ such that

$F^{Q_{G}}=(\prod G\ast r)^{k}.$

In words: The $Q_{G}$-transform of an irreducible monic polynomial is a power
of an orbit polynomial.

###### Proof.

We know that $F^{Q_{G}}\in\mathcal{NR}_{K}^{G}$ is $G$-invariant by Lemma 14
and therefore all its irreducible factors are contained in
$\mathcal{I}_{K}^{G}$. First we prove the degree condition for the irreducible
factors of $F^{Q_{G}}$. For that let $r\in\mathcal{I}_{K}^{G}$ be an arbitrary
irreducible factor of $F^{Q_{G}}$ and $v\in\overline{K}$ a root of $r$, then

$0=F^{Q_{G}}(v)=h(v)^{\deg(F)}F(Q_{G}(v))$

and $h(v)\neq 0$ because $r\in\mathcal{NR}_{K}^{G}$. Thus $F(Q_{G}(v))=0$
which shows that $Q_{G}(v)=\alpha\in R_{F}$ and with that
$K(Q_{G}(v))\subseteq K(v)$. We conclude

$\deg(r)=[K(v):K]=[K(v):K(\alpha)]\cdot[K(\alpha):K]=[K(v):K(\alpha)]\cdot\deg(F).$

For the rest note that $G\ast r$ divides $F^{Q_{G}}$ by Corollary 9. So our
goal now is to show that every irreducible factor of $F^{Q_{G}}$ belongs to
$G\ast r$. With Lemma 8 we know that the set of roots of $F^{Q_{G}}$ can be
partitioned into $G$-orbits under the Möbius-Transformation, i.e. there exist
$v_{1},\ldots,v_{l}\in R_{F^{Q_{G}}}$ such that

$R_{F^{Q_{G}}}=\bigcup\limits_{i=1}^{l}(G\circ v_{i})$

and $(G\circ v_{i})\cap(G\circ v_{j})=\varnothing$ for $i\neq j$. We set
w.l.o.g. $v=v_{1}$. By Theorem 12 we know that there are
$\alpha_{1},\ldots,\alpha_{l}\in\overline{K}$ such that $Q_{G}(G\circ
v_{i})=\alpha_{i}$. Note that $R_{F}=\\{\alpha_{1},\ldots,\alpha_{l}\\}$. Now
consider the splitting fields $L_{F},L_{F^{Q_{G}}}$ of $F$ and $F^{Q_{G}}$
over $K$. The extensions $L_{F}/K,L_{F^{Q_{G}}}/K$ and $L_{F^{Q_{G}}}/L_{F}$
are normal and finite. It can be shown that for all $\alpha_{i},\alpha_{j}\in
R_{F}$ there is
$\sigma_{i,j}\in\operatorname{hom}_{K}(L_{F})=\operatorname{Aut_{K}}(L_{F})$
such that $\sigma_{i,j}(\alpha_{i})=\alpha_{j}$ because $F$ is irreducible
(for reference see [24, Theorem 2.8.3] for example). Now let $\beta\in R_{F}$
be arbitrary, then there is $\sigma_{\beta}\in\operatorname{Aut}_{K}(L_{F})$
such that $\sigma_{\beta}(\alpha_{1})=\beta$. The automorphism
$\sigma_{\beta}$ can be extended to an automorphism in
$\operatorname{Aut}_{K}(L_{F^{Q_{G}}})$, we denote it by
$\overline{\sigma}_{\beta}$ (for reference see [24, Theorem 2.8.4]). Finally,
we put everything together: Let $w\in R_{F^{Q_{G}}}$ and $Q_{G}(w)=\gamma\in
R_{F}$, then

$\displaystyle
Q_{G}(w)=\gamma=\overline{\sigma}_{\gamma}(\alpha)=\overline{\sigma}_{\gamma}(Q_{G}(v))=Q_{G}(\overline{\sigma}_{\gamma}(v)),$

so $w$ and $\overline{\sigma}_{\gamma}(v)$ are contained in the same
$G$-orbit. We just showed that every $G$-orbit in $R_{F^{Q_{G}}}$ contains at
least one root of $r$, since $\sigma(v)$ is always a root of $r$ for all
$\sigma\in\operatorname{Aut}_{K}(L_{F^{Q_{G}}})$. To finish the proof let
$t\in\mathcal{I}_{K}^{G}$ be an arbitrary irreducible factor of $F^{Q_{G}}$
and $w$ a root of $t$, then there is $[A]\in G$ and $v\in R_{r}$ such that
$[A]^{-1}\circ v=w$, thus $t=[A]\ast r$. ∎

######  Remark 18.

This theorem still holds for arbitrary generators $Q=\frac{g}{h}$ of
$K(x)^{G}$ if $F\in\mathcal{I}_{K}$ satisfies $F(Q(\infty))\neq 0$ because of
Corollary 16 with proof. Notice that $Q(\infty)\in K\cup\\{\infty\\}$, thus
for $\deg(F)\geq 2$ it always holds. But $F^{Q}$ is not guaranteed to be
monic, so we have to normalize it on occasion.

What is left to show is that $k=1$ for all but finitely many
$F\in\mathcal{I}_{K}$. The next corollary is very helpful:

######  Corollary 19.

Let $F\in\mathcal{I}_{K}$, then every $G$-orbit in $R_{F^{Q_{G}}}$ is of the
same size. So for $v\in R_{F^{Q_{G}}}$ we obtain:

$|R_{F^{Q_{G}}}|=|R_{F}|\cdot|G\circ v|$

###### Proof.

Let $v\in R_{F^{Q_{G}}}$ be such that $Q_{G}(G\circ v)=\alpha\in R_{F}$.
Additionally, for $\beta\in R_{F}$ let
$\overline{\sigma}_{\beta}:L_{F^{Q_{G}}}\to L_{F^{Q_{G}}}$ be an automorphism
of $L_{F^{Q_{G}}}$ such that $\overline{\sigma}_{\beta}(\alpha)=\beta$ as in
the proof of Theorem 17. Moreover let $w_{\beta}$ be a root of $F^{Q_{G}}$
such that $Q_{G}(w_{\beta})=\beta$. We have

$\overline{\sigma}_{\beta}(G\circ v)\subseteq G\circ w_{\beta}$

and since
$\overline{\sigma}_{\beta}^{-1}\in\operatorname{Aut}_{K}(L_{F^{Q_{G}}})$ with
$\overline{\sigma}_{\beta}^{-1}(\beta)=\alpha$ also $G\circ
w_{\beta}\subseteq\overline{\sigma}_{\beta}(G\circ v)$. Hence $G\circ
w_{\beta}=\overline{\sigma}_{\beta}(G\circ v)$ and $|G\circ
v|=|\overline{\sigma}_{\beta}(G\circ v)|$ since $\overline{\sigma}_{\beta}$ is
bijective on $L_{F^{Q_{G}}}$. Thus we obtain

$\displaystyle|R_{F^{Q_{G}}}|=|\bigcup\limits_{\beta\in
R_{F}}Q_{G}^{-1}(\beta)|=\sum\limits_{\beta\in
R_{F}}|Q_{G}^{-1}(\beta)|=|R_{F}|\cdot|Q_{G}^{-1}(\alpha)|=|R_{F}|\cdot|G\circ
v|.$

∎

It follows that if $K$ is perfect, then $F^{Q_{G}}$ is separable if and only
if $G\circ v$ is regular, because then

$|R_{F^{Q_{G}}}|=|R_{F}|\cdot|G\circ v|=\deg(F)\cdot|G|=\deg(F^{Q_{G}}).$

Later we will see that there are only finitely many non-regular $G$-orbits in
$\overline{K}\cup\\{\infty\\}$ and consequentially there are only finitely
many $F\in\mathcal{I}_{K}$ such that $F^{Q_{G}}$ is a proper power of an orbit
polynomial. But before we do that we want to show that $G\circ v$ is regular
for a root of $F^{Q_{G}}$ implies that $F^{Q_{G}}$ is a $G$-orbit polynomial
and not a proper power thereof holds over every field.

###  2.1 Proof of the Second Part of the Main Theorem, Theorem R and Theorem 6

Let $L/K$ be a finite field extension. The separable degree of $L$ over $K$ is
defined as

$[L:K]_{s}:=|\operatorname{hom}_{K}(L)|.$

Recall that it behaves in the same way as the degree of field extensions, that
is, for $M/L/K$ we have

$[M:K]_{s}=[M:L]_{s}\cdot[L:K]_{s}.$

If $K$ is perfect, then $[L:K]_{s}=[L:K]$ for every finite field extension $L$
of $K$. Now let $\operatorname{char}(K)=p>0$ and $r\in\mathcal{I}_{K}$ with
root $v\in R_{r}$. Then there is a natural number $d$ such that

$[K(v):K]=p^{d}\cdot[K(v):K]_{s},$

this $d$ is called the radical exponent of $r$ or $v$ over $K$. It can be
shown that the radical exponent of $r$ is the smallest positive integer such
that there is an irreducible and separable polynomial $s\in K[x]$ such that
$r(x)=s(x^{p^{d}}).$ For a nice reference on this topic see [24]. For the sake
of convenience we use the notation

$\operatorname{rad}(r):=p^{d}=\frac{[K(v):K]}{[K(v):K)]_{s}}$

for $r\in\mathcal{I}_{K}$ with radical exponent $d$ and $v\in R_{r}$.

The essential part of the proof is to show that
$\operatorname{rad}(F)=\operatorname{rad}(r)$ for all irreducible factors
$r\in\mathcal{I}_{K}^{G}$ of $F^{Q_{G}}$. So let $F\in\mathcal{I}_{K}$,
$r\in\mathcal{I}_{K}^{G}$ an irreducible factor of $F^{Q_{G}}$ and $v\in
R_{r}$ a root of $r$ with $Q_{G}(v)=\alpha\in R_{F}$, then

$\displaystyle\operatorname{rad}(r)$
$\displaystyle=\frac{[K(v):K]}{[K(v):K]_{s}}=\frac{[K(v):K(\alpha)]}{[K(v):K(\alpha)]_{s}}\cdot\frac{[K(\alpha):K]}{[K(\alpha):K]_{s}}$
$\displaystyle=\frac{[K(v):K(\alpha)]}{[K(v):K(\alpha)]_{s}}\cdot\operatorname{rad}(F),$

so $\operatorname{rad}(F)\leq\operatorname{rad}(r)$. For
$\operatorname{rad}(F)\geq\operatorname{rad}(r)$ we need to work a bit.

######  Lemma 20.

Let $F\in\mathcal{I}_{K}$ be such that $|G\circ v|=|G|$ for $v\in
R_{F^{Q_{G}}}$ and $\operatorname{char}(K)=p>0$. Further $F(x)=H(x^{q})$ for
$q$ a power of $p$ and $H\in\mathcal{I}_{K}$ also separable, thus
$\operatorname{rad}(F)=q$. Then there is a separable polynomial $S\in K[x]$
such that

$F^{Q_{G}}(x)=S(x^{q}).$

###### Proof.

This proof has two parts. At first we show that we can write $F^{Q_{G}}$ as
$S(x^{q})$ and afterwards show that the polynomial $S$ is separable. The first
part is a calculation exercise. For that let $Q_{G}=\frac{g}{h}$ and
$H=\sum_{i=0}^{n}a_{i}x^{i}$. Additionally, for an arbitrary polynomial
$P:=\sum_{i=0}^{m}c_{i}x^{i}$ we define

$P^{(q)}:=\sum\limits_{i=0}^{m}c_{i}^{q}x^{i}.$

Observe that $P(x)^{q}=P^{(q)}(x^{q})$ for $q$ a power of
$\operatorname{char}(K)=p$. With that we obtain the following:

$\displaystyle F^{Q_{G}}(x)$
$\displaystyle=h(x)^{\deg(F)}F(\frac{g(x)}{h(x)})=h(x)^{q\deg(H)}H(\frac{g(x)^{q}}{h(x)^{q}})$
$\displaystyle=\sum\limits_{i=0}^{n}a_{i}g(x)^{iq}h(x)^{(n-i)q}=\sum\limits_{i=0}^{n}a_{i}g^{(q)}(x^{q})^{i}h^{(q)}(x^{q})^{(n-i)}$

Since $K[x^{q}]\subseteq K[x]$ is a subring there exists a polynomial $S$ such
that $F^{Q_{G}}(x)=S(x^{q})$, which is exactly what we wanted. The polynomial
$S$ is of degree $\deg(S)=\deg(F^{Q_{G}})/q=|G|\cdot\deg(H)$. For every $v\in
R_{F^{Q_{G}}}$ holds $v^{q}\in R_{S}$. The map $y\mapsto y^{q}$ is bijective
on $\overline{K}$, thus $\rho:R_{F^{Q_{G}}}\to R_{S}$ with $v\mapsto v^{q}$ is
injective and therefore $|R_{F^{Q_{G}}}|\leq|R_{S}|$. Conversely, for
$\alpha\in R_{S}$ the $q$-th root $\alpha^{1/q}$ is a root of $F^{Q_{G}}$,
because $F(\alpha^{1/q})=S((\alpha^{1/q})^{q})=S(\alpha)=0$. This shows that
$\rho$ is actually a bijection and $|R_{S}|=|R_{F^{Q_{G}}}|$. We finish this
proof by applying Corollary 19 and using $\deg(H)=|R_{F}|$ as well as our
assumption that $|G|=|G\circ v|$:

$|R_{S}|=|R_{F^{Q_{G}}}|=|R_{F}|\cdot|G\circ v|=\deg(H)\cdot|G|=\deg(S)$

So $S$ is separable because it has $\deg(S)$ many roots in $\overline{K}$. ∎

If $S\in K[x]$ is the separable polynomial such that $F^{Q_{G}}=S(x^{q})$ as
in the lemma above, then $S$ factorizes into separable irreducible factors
$S=s_{1}\cdot\ldots\cdot s_{l}$. Hence $S(x^{q})=s_{1}(x^{q})\cdot\ldots\cdot
s_{l}(x^{q})$, so it should be beneficial to study the factorization of
polynomials of the form $s(x^{q})$ for $s\in\mathcal{I}_{K}$ irreducible and
separable. The next lemma shows that such polynomials consist of only one
irreducible factor. We give a proof of this lemma as it is a crucial tool for
the following theorem. In the proof we employ a similar method as in the proof
of Theorem 17. We want to point out that there is no finite subgroup
$G\leq\operatorname{PGL}_{2}(K)$ with $x^{q}\in K(x)$ as a quotient map for
$q$ a power of $\operatorname{char}(K)$, so this is not a particular case of
Theorem 17.

######  Lemma 21.

Let $s\in K[x]$ be an irreducible, separable and monic polynomial. Furthermore
let $\operatorname{char}(K)=p>0$ and $q=p^{d}$ for $d>0$. Then there is an
irreducible,separable and monic polynomial $f\in K[x]$ with $\deg(f)=\deg(s)$
and $a,b\in\mathbb{N}$ with $a+b=d$ such that

$s(x^{q})=(f(x^{p^{a}}))^{p^{b}}$

and $f(x^{p^{a}})$ is irreducible.

###### Proof.

At first we show that $s(x^{q})$ only has one irreducible factor. Since $s$ is
both irreducible and separable, $L_{s}/K$ is a Galois extension. If we set
$P(x):=s(x^{q})$ and consider the splitting field $L_{P}/K$ of $P$, then, with
similar arguments as in the proof of Theorem 17, we can extend every
$\sigma\in\operatorname{Gal}(L_{s}/K)$ to a $K$-homomorphism
$\overline{\sigma}\in\operatorname{hom}_{K}(L_{P})$. Since splitting fields
are normal, every such $K$-homomorphism is actually an automorphism on
$L_{P}$. Let $F\in\mathcal{I}_{K}$ be an irreducible factor of $P$ and $v\in
R_{F}$ one of its roots. Then $v^{q}=:\alpha$ is a root of $s$ and similarly
$w^{q}=:\beta\in R_{s}$ for $w\in R_{P}$. Let
$\overline{\sigma}\in\operatorname{hom}(L_{P})$ be the extension of the
homomorphism $\sigma\in\operatorname{Gal}(L_{s}/K)$ with
$\sigma(\alpha)=\beta$, then

$w^{q}=\beta=\overline{\sigma}(\alpha)=\overline{\sigma}(v^{q})=(\overline{\sigma}(v))^{q}.$

As $y\mapsto y^{q}$ is injective on $\overline{K}$ we obtain that
$w=\overline{\sigma}(v)$. Therefore $w$ has to be a root of $F\in K[x]$ as
well, since $\overline{\sigma}$ is an automorphism of $L_{P}$ that fixes $K$.
So we just showed that all roots of $P$ are also roots of the irreducible
factor $F$, thus $P=F^{k}$ for $k\in\mathbb{N}$. With the degree formula for
field extensions we get

$\deg(F)=[K(v):K]=[K(v):K(\alpha)]\cdot[K(\alpha):K]=[K(v):K(\alpha)]\cdot\deg(s),$

which shows $\deg(s)|\deg(F)$. Together with the fact that
$k\cdot\deg(F)=\deg(P)=q\cdot\deg(s)$ we get $\deg(F)=p^{a}\cdot\deg(s)$ for
an $a\in\\{0,\ldots,d\\}$ and $k=p^{b}$ such that $a+b=d$. Moreover
$p^{a}=[K(v):K(\alpha)]$, which is the degree of the minimal polynomial
$m_{v}\in K(\alpha)[x]$ of $v$ in $K(\alpha)$. Since $v^{q}=\alpha$, it is
also a root of $x^{q}-\alpha$ and therefore $m_{v}|x^{q}-\alpha$. A simple
calculation shows that $m_{v}=x^{p^{a}}-\alpha^{\frac{1}{p^{b}}}$, so
$\alpha^{\frac{1}{p^{b}}}\in K(\alpha)$. Conversely $\alpha\in
K(\alpha^{\frac{1}{p^{b}}})$, since $\alpha$ is the $p^{b}$-th power of
$\alpha^{\frac{1}{p^{b}}}$, hence $K(\alpha)=K(\alpha^{\frac{1}{p^{b}}})$. Let
$f\in\mathcal{I}_{K}$ be the minimal polynomial of $\alpha^{\frac{1}{p^{b}}}$
in $K[x]$. We just showed that $\deg(f)=[K(\alpha):K]=\deg(s)$. All that shows
that $v$ is also root of the monic polynomial $F^{\prime}:=f(x^{p^{a}})\in
K[x]$. Since $F$ is the minimal polynomial of $v$ over $K$ it has to divide
$F^{\prime}$. Moreover, $F$ and $F^{\prime}$ have the same degree and are
monic, thus have to be equal. ∎

This is enough to prove that $k=1$ if $G\circ v$ is regular for $v\in
R_{F^{Q_{G}}}$:

######  Theorem 22.

Let $F\in\mathcal{I}_{K}$ be such that $|G\circ v|=|G|$ for $v\in
R_{F^{Q_{G}}}$. Then $F^{Q_{G}}=\prod(G\ast r)$, i.e. it is an orbit
polynomial.

###### Proof.

If $F$ is separable and $|G\circ v|=|G|$ for a root of $F^{Q_{G}}$, then all
$G$-orbits in $R_{F^{Q_{G}}}$ are of the same size by Corollary 19 and as a
consequence $F^{Q_{G}}$ is also separable since
$\deg(F^{Q_{G}})=|R_{F^{Q_{G}}}|$. If $\operatorname{rad}(F)=q>1$ then by
Lemma 20

$F^{Q_{G}}(x)=S(x^{q})=\prod\limits_{i=1}^{l}s_{i}(x^{q})=\prod\limits_{i=1}^{l}(f_{i}(x^{p^{a_{i}}}))^{p^{b_{i}}},$

where $s_{i}\in\mathcal{I}_{K}$ are the irreducible and separable factors of
$S$ and $f_{i}\in\mathcal{I}_{K}$ the separable and irreducible polynomials as
in the lemma above, so $a_{i}+b_{i}=d$ for $q=p^{d}$. Observe that
$\gcd(f_{i}(x^{p^{a_{i}}}),f_{j}(x^{p^{a_{j}}}))=1$ for $i\neq j$. The reason
for that is that since $S$ is separable, $s_{i}$ and $s_{j}$ have to be
different irreducible polynomials, i.e. $\gcd(s_{i},s_{j})=1$ which is
equivalent to $R_{s_{i}}\cap R_{s_{j}}=\varnothing$. Now, the roots of
$s_{i}(x^{q})$ and $s_{j}(x^{q})$ are the preimages of $R_{s_{i}}$ and
$R_{s_{j}}$ under the map $y\mapsto y^{q}$ on $\overline{K}$. This map is
bijective and therefore also injective, so the preimages are also different
and thus $s_{i}(x^{q})$ and $s_{j}(x^{q})$ have no roots in common. This small
observation is very important because now, with the help of Theorem 17, we can
deduce that

$\displaystyle S(x^{q})$
$\displaystyle=\prod\limits_{i=1}^{l}(f_{i}(x^{p^{a_{i}}}))^{p^{b_{i}}}$
$\displaystyle=(\prod\limits_{t\in G\ast r}t)^{k}.$

The irreducible factors $f_{i}(x^{p^{a_{i}}})$ and $t\in G\ast r$ have to
coincide with each other because $K[x]$ is a factorial ring. So we obtain with
the remarks above (and the fact that all $t_{1},t_{2}\in G\ast r$ with
$t_{1}\neq t_{2}$ also satisfy $\gcd(t_{1},t_{2})=1$ since they are
irreducible) that for all $t\in G\ast r$ there is exactly one $i\in[l]$ such
that $t(x)=f_{i}(x^{p^{a_{i}}})$. This also implies $k=p^{b_{i}}$ for all
$i\in[l]$ and thus $p^{a_{i}}=p^{a_{j}}$ for all $i,j\in[l]$. To summarize
what we could obtain:

$\displaystyle
S(x^{q})=\prod\limits_{i=1}^{l}(f_{i}(x^{p^{a}}))^{k}=\prod\limits_{t\in G\ast
r}t^{k}=F^{Q_{G}}(x).$

This shows that $\operatorname{rad}(t)=p^{a}\leq q=\operatorname{rad}(F)$ for
all $t\in G\ast r$, thus $\operatorname{rad}(t)=\operatorname{rad}(F)$, since
we already explained that $\operatorname{rad}(r)\geq\operatorname{rad}(F)$.
Hence $p^{a}=q$ and $b_{i}=0$, which shows $k=1$. ∎

Before we finish the proof of the Main Theorem we want to state an immediate
consequence of this result

######  Corollary 23.

Let $F\in\mathcal{I}_{K}$ be such that $|G\circ v|=|G|$ for a root $v$ of
$F^{Q_{G}}$, then

  1. 1.

The degree of every irreducible factor $r$ of $F^{Q_{G}}$ satisfies

$\deg(r)=\frac{|G|}{|G\ast r|}\cdot\deg(F)$

  2. 2.

If $\deg(F)<\deg(r)$ for an irreducible factor $r$ of $F^{Q_{G}}$, then
$\\{[I_{2}]\\}\neq\operatorname{Stab}_{G}(r)\leq G$ is non-trivial

###### Proof.

For the first we use Theorem 22 (so $k=1$) and calculate:

$|G|\cdot\deg(F)=\deg(F^{Q_{G}})=|G\ast r|\cdot\deg(r).$

The last is obvious because $\deg(r)=\frac{|G|}{|G\ast
r|}\cdot\deg(F)>\deg(F)$, so $|\operatorname{Stab}_{G}(r)|=\frac{|G|}{|G\ast
r|}>1$. ∎

######  Remark 24.

Observe that for $Q$ an arbitrary generator of $K(x)^{G}$ Theorem 22 and
Corollary 23 still hold if $F\in\mathcal{I}_{K}$ and $F(Q(\infty))\neq 0$. The
reason for this is again the proof of Corollary 16: Write

$Q(x)=[C]\circ Q_{G}(x)=\frac{aQ_{G}(x)+b}{cQ_{G}(x)+d}$

as in Corollary 16, where $Q_{G}$ is a quotient map, then there is
$H\in\mathcal{I}_{K}$ such that $H^{Q_{G}}=a\cdot F^{Q}$. Since we always
assume something about the roots of the resulting polynomial, i.e. about
$F^{Q}$ and thus also about $H^{Q_{G}}$, both Theorem 22 and Corollary 23 hold
because if $F^{Q}$ only contains roots in regular $G$-orbits so does
$H^{Q_{G}}$.

We state an analogue of Theorem 12 for polynomials:

######  Corollary 25.

The map $\delta_{Q_{G}}:\mathcal{I}_{K}\to\mathcal{I}_{K}^{G}/G$ with
$F\mapsto G\ast r$ such that $F^{Q_{G}}=\prod(G\ast r)^{k}$ is a bijection.

###### Proof.

By Theorem 17 $\delta_{Q_{G}}$ defines a mapping between $\mathcal{I}_{K}$ and
$\mathcal{I}_{K}^{G}/G$. First we show that $\delta_{Q_{G}}$ is surjective.
Let $r\in\mathcal{I}_{K}^{G}$ and $v\in R_{r}$ a root of $r$. Then $r$ is the
minimal polynomial of $v$ over $K$. Moreover, let $\alpha\in\overline{K}$ be
such that $Q_{G}(v)=\alpha$ and denote by $F\in\mathcal{I}_{K}$ the minimal
polynomial of $\alpha$. Then $F^{Q_{G}}$ has $v$ as a root, thus $r|F^{Q_{G}}$
and $F^{Q_{G}}=(\prod(G\ast r))^{k}$ by Theorem 17, so
$\delta_{Q_{G}}(F)=G\ast r$. Now onto the injectivity: Let
$F,H\in\mathcal{I}_{K}$ be such that
$\delta_{Q_{G}}(F)=\delta_{Q_{G}}(H)=G\ast r$ for $r\in\mathcal{I}_{K}^{G}$,
so $F^{Q_{G}}=(\prod(G\ast r))^{k}$ and $H^{Q_{G}}=(\prod(G\ast r))^{l}$ and
both $F^{Q_{G}}$ and $H^{Q_{G}}$ have the same roots. With the help of what we
observed in the proof of Theorem 17 we obtain:

$\bigcup\limits_{\alpha\in
R_{F}}Q_{G}^{-1}(\alpha)=R_{F^{Q_{G}}}=R_{H^{Q_{G}}}=\bigcup\limits_{\beta\in
R_{H}}Q_{G}^{-1}(\beta).$

Therefore $R_{F}=R_{H}$ since $Q_{G}$ induces a bijection between
$\overline{K}\cup\\{\infty\\}$ and $(\overline{K}\cup\\{\infty\\})/G$ by
Theorem 12. As $F$ and $H$ are irreducible, monic and share the same roots
they have to be equal. ∎

As an immediate consequence of this corollary we obtain the main part of the
general version of Theorem R:

######  Theorem 26.

If $f\in\mathcal{I}_{K}^{G}$ is a $G$-invariant monic irreducible polynomial
with root $v\in R_{f}$ that is contained in a regular $G$-orbit, then there is
$F\in\mathcal{I}_{K}$ such that $f=F^{Q_{G}}$.

###### Proof.

If $f\in\mathcal{I}_{K}^{G}$ is $G$-invariant, then $G\ast f=\\{f\\}$. By
Corollary 25 we get that there is $F\in\mathcal{I}_{K}$ such that
$\delta_{Q_{G}}(F)=\\{f\\}$, which translates to $F^{Q_{G}}=f^{k}$. Further
$k=1$ because of Theorem 22 and the assumption that $v\in R_{f}$ is contained
in a regular $G$-orbit. ∎

To complete the proofs of the Main Theorem and the general Theorem R we need
to show that the number of irreducible polynomials for which $F^{Q_{G}}=(\prod
G\ast r)^{k}$ with $k>1$ is finite. By Theorem 22 this is equivalent to
showing that the number of irreducible polynomials $F\in\mathcal{I}_{K}$ for
which $F^{Q_{G}}$ has roots in non-regular $G$-orbits is finite. We give these
polynomials the following name:

######  Definition 27.

We call $F\in\mathcal{I}_{K}$ $Q_{G}$-non-conformal if $F^{Q_{G}}=(\prod(G\ast
r))^{k}$ for $r\in\mathcal{I}_{K}^{G}$, $k\in\mathbb{N}$ and $k>1$. For the
set of $Q_{G}$-non-conformal polynomials we write $\mathcal{NC}^{Q_{G}}$.

Further we set

$P_{G}:=\\{u\in\overline{K}\cup\\{\infty\\}|~{}|G\circ u|<|G|\\}$

as the set of elements in $\overline{K}\cup\\{\infty\\}$ contained in non-
regular $G$-orbits. We have the following:

######  Lemma 28 ([4, Lemma 2.1]).

Let $G\leq\operatorname{PGL}_{2}(K)$ be finite and
$v\in\overline{K}\cup\\{\infty\\}$. Then $G\circ v$ is non-regular if and only
if there is $[A]\in G\setminus\\{[I_{2}]\\}$ such that $[A]\circ v=v$, thus

$P_{G}=\\{u\in\overline{K}\cup\\{\infty\\}|~{}\exists[A]\in
G\setminus\\{[I_{2}]\\}:~{}[A]\circ u=u\\}$

and this set is finite; more precisely $|P_{G}|\leq 2(|G|-1)$. Furthermore
$[K(u):K]\leq 2$ for all $u\in P_{G}\setminus\\{\infty\\}$.

We denote by $\mathcal{P}_{G}$ the set of minimal polynomials of elements in
$P_{G}$, that is,

$\mathcal{P}_{G}:=\\{r\in\mathcal{I}_{K}^{G}|\exists\alpha\in
P_{G}:~{}r(\alpha)=0\\}.$ (5)

Notice that $\mathcal{P}_{G}\subseteq\mathcal{I}_{K}^{G}$ by definition, thus
$\mathcal{P}_{G}$ does not contain polynomials with roots in
$(G\circ\infty)\setminus\\{\infty\\}$, even if this orbit is non-regular. We
obtain

######  Lemma 29.

We have222The set $\mathcal{NC}^{Q_{G}}$ can be empty. In that case we define
the left side of the subset equation to be empty as well.

$\bigcup\limits_{F\in\mathcal{NC}^{Q_{G}}}\delta_{Q_{G}}(F)\subseteq\mathcal{P}_{G}.$

In particular

$|\mathcal{NC}^{Q_{G}}|\leq|\mathcal{P}_{G}|\leq|P_{G}|\leq 2(|G|-1),$

so there are only finitely many $Q_{G}$-non-conformal polynomials.

###### Proof.

The map $\mathcal{NC}^{Q_{G}}\to\mathcal{P}_{G}/G$ with
$F\mapsto\delta_{Q_{G}}(F)$ defines an injective mapping by Theorem 25 and
Theorem 22, thus the subset equation

$\bigcup\limits_{F\in\mathcal{NC}^{Q_{G}}}\delta_{Q_{G}}(F)\subseteq\mathcal{P}_{G}$

follows. Since $\mathcal{P}_{G}$ only contains minimal-polynomials of elements
in $P_{G}$ and the degree of $r\in\mathcal{P}_{G}$ is either 1 or 2 by Lemma
28 we get that $|\mathcal{P}_{G}|\leq|P_{G}|$, so $\mathcal{P}_{G}$ is finite
because $|P_{G}|\leq 2(|G|-1)$ by Lemma 28. Hence $\mathcal{NC}^{Q_{G}}$ is
finite as well and

$|\mathcal{P}_{G}|\geq\sum\limits_{F\in\mathcal{NC}^{Q_{G}}}|\delta_{Q_{G}}(F)|\geq\sum\limits_{F\in\mathcal{NC}^{Q_{G}}}1=|\mathcal{NC}^{Q_{G}}|.$

∎

We close this section by proving Theorem 6:

######  Theorem 30.

Let $F_{1},\ldots F_{l}\in\mathcal{NC}^{Q_{G}}$ be all $Q_{G}$-non-conformal
polynomials. Further let $r_{1},\ldots,r_{l}\in\mathcal{I}_{K}^{G}$ be such
that $\delta_{Q_{G}}(F_{i})=G\ast r_{i}$ and $n_{i}\in\mathbb{N}$ such that
$F_{i}^{Q_{G}}=(\prod G\ast r_{i})^{n_{i}}$. Then for every $G$-invariant
polynomial $f\in\mathcal{NR}_{K}^{G}$ there is a unique monic polynomial $F\in
K[x]$ and unique natural numbers $0\leq k_{i}<n_{i}$ such that

$f=\left(\prod\limits_{i=1}^{l}(\prod\limits G\ast r_{i})^{k_{i}}\right)\cdot
F^{Q_{G}}.$

###### Proof.

By Corollary 11 we have

$f=\prod\limits_{i=1}^{e}(\prod(G\ast g_{i}))^{m_{i}}$

with $g_{i}\in\mathcal{I}_{K}^{G}$ and
$m_{1},\ldots,m_{e}\in\mathbb{N}\setminus\\{0\\}$. We refine this
factorization by grouping the orbit polynomials into either belonging to
$\mathcal{P}_{G}$ or not, which gives

$f=\prod\limits_{i=1}^{l}(\prod(G\ast
r_{i}))^{l_{i}}\cdot\prod\limits_{j=1}^{c}(\prod(G\ast h_{j}))^{d_{j}};$

here we allow $l_{i}=0$. Since
$h_{j}\in\mathcal{I}_{K}^{G}\setminus\mathcal{P}_{G}$ for all $j\in[c]$ there
exists a unique monic polynomial $H_{1}\in K[x]$ such that

$H_{1}^{Q_{G}}=\prod\limits_{j=1}^{c}(\prod(G\ast h_{j}))^{d_{j}}$

by Theorem 22 together with Lemma 13 item 2. For the remaining factor we
divide $l_{i}$ by $n_{i}$ and write $k_{i}$ for the remainder, so
$l_{i}=a_{i}\cdot n_{i}+k_{i}$ and $0\leq k_{i}<n_{i}$. We have

$(F_{i}^{a_{i}})^{Q_{G}}=(\prod(G\ast r_{i}))^{a_{i}\cdot n_{i}}.$

Hence we define

$H_{2}:=\prod_{i=1}^{l}F_{i}^{a_{i}}$

and get

$f=\left(\prod\limits_{i=1}^{l}(\prod\limits G\ast
r_{i})^{k_{i}}\right)\cdot(H_{2}\cdot H_{1})^{Q_{G}}.$

Set $F=H_{2}\cdot H_{1}$, which is unique since both $H_{1}$ and $H_{2}$ are
unique. ∎

##  3 Some Notes on the Galois Theory of Invariant Polynomials

In this section we want to explain some statements about the Galois theory of
$G$-invariant polynomials and their implications for finite fields. In
particular, we give an alternative proof of the fact that all irreducible
monic polynomials $f\in\mathbb{F}_{q}[x]$ of degree $\deg(f)\geq 3$ have
cyclic stabilizers in $\operatorname{PGL}_{2}(\mathbb{F}_{q})$ ([22, Theorem
1.3]). This means that if $f\in\mathcal{I}_{\mathbb{F}_{q}}=:\mathcal{I}_{q}$
is of degree $\deg(f)\geq 3$ and $G\leq\operatorname{PGL}_{2}(\mathbb{F}_{q})$
is non-cyclic, then $|G\ast f|>1$. We will exploit this in the proof of
Theorem 2.

Consider the $G$-invariant separable and monic polynomial
$f\in\mathcal{I}_{K}^{G}$ with roots belonging to regular $G$-orbits and its
splitting field $L_{f}$ in a fixed algebraic closure of $K$. As seen before we
can partition the set of roots of $f$ into $G$-orbits

$R_{f}=\bigcup_{i=1}^{k}(G\circ v_{i})$

where $v_{1},\ldots v_{k}\in R_{f}$ is a set of representatives of $R_{f}/G$.
Thus $\deg(f)=|G|\cdot k$, since all $G$-orbits in $R_{f}$ are regular. Let
$\sigma\in\operatorname{Gal}(f):=\operatorname{Gal}(L_{f}/K)$ and

$A=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$

such that $[A]\in G$, then

$\sigma([A]\circ
v)=\sigma(\frac{av+b}{cv+d})=\frac{a\sigma(v)+b}{c\sigma(v)+d}=[A]\circ\sigma(v)$

for all $v\in R_{f}$. This shows that the actions of $G$ and
$\operatorname{Gal}(f)$ on $R_{f}$ commute and that $G\circ
v_{1},\ldots,G\circ v_{k}$ is a non-trivial block system for
$\operatorname{Gal}(f)$:

######  Definition 31 (See [9]).

Let $G$ be a finite group acting transitively on a non-empty finite set $X$.
We say that a subset $Y\subseteq X$ is a block for $G$ if $g\cdot Y=Y$ or
$g\cdot Y\cap Y=\varnothing$. Moreover, $Y$ is a non-trivial block if
$1<|Y|<|X|$. If $Y$ is a block then $\\{g\cdot Y|g\in G\\}$ is a partition of
$X$ and is called a block system of $X$ for $G$.

We define the point-wise stabilizer subgroup of a block $G\circ v$ as

$\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ
v):=\\{\sigma\in\operatorname{Gal}(f)|\sigma(w)=w\text{ for all }w\in G\circ
v\\}.$

Similarly, the set-wise stabilizer is

$\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v):=\\{\sigma\in\operatorname{Gal}(f)|\sigma(w)\in G\circ v\text{ for all
}w\in G\circ v\\}.$

Notice that $\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ
v)\trianglelefteq\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ v)$. Our
first goal is to show that $G$ is isomorphic to the quotient of these
stabilizers. For that, we need a nice lemma about commuting group actions
stated in [13] and [14]:

######  Lemma 32 ([13] & [14]).

Let $X$ be a finite non-empty set and $G,H$ groups acting transitively and
faithful333$G$ acts faithful on $X$ if $g\cdot x=x$ for all $x\in X$ implies
$g=1$ on $X$. Moreover, $G$ acts regularly on $X$, that is,
$\operatorname{Stab}_{G}(x)=\\{1\\}$ for all $x\in X$ and the actions of $G$
and $H$ commute, i.e.

$g(h(x))=h(g(x))$

for all $h\in H$, $g\in G$ and $x\in X$. Then $H$ acts regularly on $X$ and is
isomorphic to $G$.

Additionally we prove the following

######  Lemma 33.

Let $f\in\mathcal{I}_{K}^{G}$ be $G$-invariant, separable and $R_{f}$ only
contains regular $G$-orbits. Moreover let $v\in R_{f}$ be a root of $f$. Then
we have:

  1. 1.

$G$ acts transitively and regularly on $G\circ v$

  2. 2.

$U:=\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)/\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ v)$ acts faithful and
transitively on $G\circ v$

  3. 3.

$\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ
v)=\operatorname{Stab}_{\operatorname{Gal}(f)}(v)$, thus $U$ acts regularly on
$G\circ v$

###### Proof.

For the first item note that the action of $G$ restricted to any of its orbits
in $R_{f}$ is always transitive. Additionally, $G$ acts regularly on $G\circ
v$ since all orbits are regular. That the induced action of $U$ on $G\circ v$
is faithful and transitive follows from standard facts about group actions, so
onto the last item: The inclusion
$\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ
v)\subseteq\operatorname{Stab}_{\operatorname{Gal}(f)}(v)$ is obvious. For the
other let $\sigma\in\operatorname{Stab}_{\operatorname{Gal}(f)}(v)$, so
$\sigma(v)=v$. Moreover, by the first item, $G$ acts transitively on $G\circ
v$, so for all $w\in G\circ v$ there is $[A]\in G$ such that $[A]\circ v=w$.
As a consequence

$\sigma(w)=\sigma([A]\circ v)=[A]\circ\sigma(v)=[A]\circ v=w,$

so both sets are equal. Moreover, all stabilizers of elements in $G\circ v$
are equal to $\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ v)$. So for
all $w\in G\circ v$ we get

$U=\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)/\operatorname{Stab}_{\operatorname{Gal}(f)}(w),$

thus $U$ acts regularly on $G\circ v$. ∎

We apply Lemma 32 to our setup. We set $G$ as $G$ in Lemma 32 and $H=U$ as in
the previous lemma, then

$G\cong U=\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)/\operatorname{Stab}_{\operatorname{Gal}(f)}(v).$ (6)

With that we can obtain

######  Corollary 34.

Let $f\in\mathcal{I}_{K}^{G}$ be $G$-invariant, separable and all $G$-orbits
in $R_{f}$ are regular. Moreover let $v\in R_{f}$ be a root of $f$.

  1. 1.

If $\operatorname{Gal}(f)$ is abelian, then $G\cong
U\leq\operatorname{Gal}(f)$

  2. 2.

If $\deg(f)=|G|$, then $G\cong\operatorname{Gal}(f)$

###### Proof.

We want to show that $\operatorname{Gal}(f)$ is abelian implies
$\operatorname{Stab}_{\operatorname{Gal}(f)}(v)=\\{\operatorname{id}\\}$. To
see this let $\sigma\in\operatorname{Stab}_{\operatorname{Gal}(f)}(v)$ and
$w\in R_{f}$. Additionally set $\tau\in\operatorname{Gal}(f)$ such that
$\tau(v)=w$ (exists since $f$ is irreducible and thus $\operatorname{Gal}(f)$
acts transitively on $R_{f}$). Then we obtain

$\sigma(w)=\sigma(\tau(v))=\tau(\sigma(v))=\tau(v)=w,$

so $\sigma(w)=w$ for all $w\in R_{f}$ and thus $\sigma=\operatorname{id}$
because $\operatorname{Gal}(f)$ acts faithful on $R_{f}$. Consequentially
$G\cong\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)\leq\operatorname{Gal}(f)$, which finishes the first part.

Now let $f$ be of degree $|G|$, so also $|R_{f}|=|G|$ and $R_{f}=G\circ v$ for
all $v\in R_{f}$. Therefore
$\operatorname{Gal}(f)=\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)$ by definition and $\operatorname{p-Stab}_{\operatorname{Gal}(f)}(G\circ
v)=\\{\operatorname{id}\\}$ because $\operatorname{Gal}(f)$ acts faithful on
$R_{f}$. This shows

$\operatorname{Gal}(f)=\operatorname{s-Stab}_{\operatorname{Gal}(f)}(G\circ
v)\cong G.$

∎

Further, we get:

######  Corollary 35.

Let $f\in\mathcal{I}_{q}$. If $f$ is $G$-invariant for a subgroup
$G\leq\operatorname{PGL}_{2}(\mathbb{F}_{q})$ and $R_{f}$ only contains
regular $G$-orbits then $G$ has to be cyclic and $|G|\mid\deg(f)$.

###### Proof.

It is well-known that $\operatorname{Gal}(f)\cong C_{\deg(f)}$, where $C_{n}$
is the cyclic group of order $n$, so $\operatorname{Gal}(f)$ is also abelian.
If $f$ is $G$-invariant for $G\leq\operatorname{PGL}_{2}(\mathbb{F}_{q})$,
then $G$ has to be isomorphic to a subgroup of $C_{\deg(f)}$ by Corollary 34
(1) and therefore has to be cyclic as well. By Lagrange’s theorem $|G|$
divides $|C_{\deg(f)}|=\deg(f)$. ∎

######  Remark 36.

This result does not hold for quadratic irreducible polynomials over finite
fields. Define $\mathcal{I}_{q}^{n}$ as the set of monic irreducible
polynomials over $\mathbb{F}_{q}$ of degree $n$. It can be shown that for
$g\in\mathcal{I}_{q}^{2}$ we have

$G\ast g=\mathcal{I}_{q}^{2}$

and thus

$|\operatorname{Stab}_{\operatorname{PGL}_{2}(\mathbb{F}_{q})}(g)|=\frac{|\operatorname{PGL}_{2}(\mathbb{F}_{q})|}{|\mathcal{I}_{q}^{2}|}=2(q+1).$

Since the biggest order of an element in
$\operatorname{PGL}_{2}(\mathbb{F}_{q})$ is $q+1$ the stabilizer can not be
cyclic. In fact, it is dihedral. The reason why Corollary 35 fails is that the
set of roots of a quadratic irreducible polynomial $g\in\mathcal{I}_{q}^{2}$
does not contain regular
$\operatorname{Stab}_{\operatorname{PGL}_{2}(\mathbb{F}_{q})}(g)$-orbits.

We have enough to prove Theorem 2

###### Proof. (Theorem 2).

Since $F^{Q_{G}}$ is separable it has degree many roots and together with
Corollary 19 we have that all $G$-orbits in $R_{F^{Q_{G}}}$ are regular. Note
that $K=\mathbb{F}_{q}$ is perfect so $|R_{r}|=\deg(r)$ for all irreducible
polynomials in $\mathbb{F}_{q}[x]$. With Theorem 17 and 22 we know that there
exists an irreducible polynomial $r\in\mathcal{I}_{K}^{G}$ such that
$F^{Q_{G}}=\prod G\ast r$. Observe that $r$ is $\operatorname{Stab}_{G}(r)\leq
G$ invariant and the $\operatorname{Stab}_{G}(r)$-orbits in $R_{r}$ are
regular because the $G$-orbits in $R_{F^{Q_{G}}}$ are regular. Consequentially
$\operatorname{Stab}_{G}(r)$ has to be cyclic by Corollary 35 and with
Corollary 23 we obtain

$\deg(r)=|\operatorname{Stab}_{G}(r)|\cdot\deg(F)\leq\mu_{G}\cdot\deg(F)$

and

$|G\ast r|=\frac{|G|}{|\operatorname{Stab}_{G}(r)|}\geq\frac{|G|}{\mu_{G}}.$

∎

######  Corollary 37.

If $G\leq\operatorname{PGL}_{2}(\mathbb{F}_{q})$ is non-cyclic and
$Q_{G}\in\mathbb{F}_{q}(x)$ is a quotient map for $G$. Then $F^{Q_{G}}$ is
reducible for all $F\in\mathbb{F}_{q}[x]$.

###### Proof.

Note that ”$F^{Q_{G}}$ is irreducible” implies ”$F$ is irreducible”, so we can
focus on $F$ being irreducible. If
$F\in\mathcal{I}_{K}\setminus\mathcal{NC}^{Q_{G}}$, then $F^{Q_{G}}$ has at
least $|G|/\mu_{G}$ irreducible factors by Theorem 2 and $|G|/\mu_{G}>1$ if
$G$ is non-cyclic. If $F\in\mathcal{NC}^{Q_{G}}$, then

$F^{Q_{G}}=\prod(G\ast r)^{k}$

for $r\in\mathcal{I}_{K}^{G}$ and $k>1$. ∎

##  4 Examples of Invariant Polynomials

In this section we show how our result apply to specific subgroups of
$\operatorname{PGL}_{2}(K)$ where $\operatorname{char}(K)>0$.

###  4.1 Unipotent Subgroups

Consider a field $K$ with $\operatorname{char}(K)=p>0$ and $q=p^{l}$ for
$l>0$. Moreover assume $\mathbb{F}_{q}\subseteq K$ and let $V\leq_{q}K$ be a
$\mathbb{F}_{q}$-subspace of $K$ of dimension
$n\in\mathbb{N}\setminus\\{0\\}$. For the subspace $V$ we define

$\overset{\sim}{V}:=\left\\{\left[\left(\begin{array}[]{cc}1&v\\\
0&1\end{array}\right)\right]:v\in V\right\\}\leq\operatorname{PGL}_{2}(K).$

Observe that $\overset{\sim}{V}\cong\mathbb{F}_{q}^{n}$ as groups, so
$\overset{\sim}{V}$ is abelian and every non-trivial element
$[A]\in\overset{\sim}{V}$ has order $p$. Additionally,
$\overset{\sim}{V}\subseteq\operatorname{Stab}_{\operatorname{PGL}_{2}(K)}(\infty)$,
so $\mathcal{NR}_{K}^{\overset{\sim}{V}}$ is just the set of monic polynomials
over $K$. A quotient map is the to $V$ associated subspace polynomial (see [4,
§10])

$Q_{\overset{\sim}{V}}(x)=\prod_{v\in V}(x-v).$ (7)

The set $P_{\overset{\sim}{V}}$ only contains $\infty$, so
$\mathcal{P}_{\overset{\sim}{V}}=\varnothing=\mathcal{NC}^{Q_{\overset{\sim}{V}}}$.
This makes the classification of $\overset{\sim}{V}$-invariant polynomials
especially nice:

######  Corollary 38.

For every monic $\overset{\sim}{V}$-invariant polynomial
$f\in\mathcal{NR}_{K}^{\overset{\sim}{V}}$ exists a unique monic polynomial
$F\in K[x]$ such that

$f(x)=F\left(\prod_{v\in V}(x-v)\right).$

######  Remark 39.

This result is already known, see [23, Theorem 2.5.]. There,
$\overset{\sim}{V}$-invariant polynomials are called $V$-translation invariant
polynomials since

$[A]\ast f(x)=f(x+v)$

for

$A=\left(\begin{array}[]{cc}1&v\\\ 0&1\end{array}\right).$

Even though the proof of Theorem 2.5. is only stated for finite fields, the
assumption that $K$ is finite is not used at all so it also holds if $K$ is
infinite. We gave an alternative proof of this result.

Next we look at the factorization of $F(Q_{\overset{\sim}{V}}(x))$.

######  Lemma 40.

Let $F\in\mathcal{I}_{K}$, then we obtain:

  1. 1.

All irreducible factors of $F(Q_{\overset{\sim}{V}}(x))$ have the same
stabilizer in $\overset{\sim}{V}$, i.e. there is $W\leq V$ such that all
irreducible factors of $F(Q_{\overset{\sim}{V}}(x))$ are
$\overset{\sim}{W}$-invariant.

  2. 2.

Let $F(Q_{\overset{\sim}{V}}(x))=\prod(\overset{\sim}{V}\ast r)$ for an
$r\in\mathcal{I}_{K}$ and
$\operatorname{Stab}_{\overset{\sim}{V}}(r)=\overset{\sim}{W}$ for $W\leq V$.
Moreover let $v_{1},\ldots,v_{k}\in V$ be a complete set of representatives
for $V/W$, then

$F(Q_{\overset{\sim}{V}}(x))=\prod\limits_{i=1}^{k}r(x+v_{i}).$ (8)

###### Proof.

By Theorem 17 and 22 we know that
$F(Q_{\overset{\sim}{V}}(x))=\prod(\overset{\sim}{V}\ast r)$ for an
$r\in\mathcal{I}_{K}^{\overset{\sim}{V}}=\mathcal{I}_{K}$. All elements in the
same orbit have conjugated stabilizers. Since $\overset{\sim}{V}$ is abelian
every subgroup is normal, thus
$\operatorname{Stab}_{\overset{\sim}{V}}(t)=\operatorname{Stab}_{\overset{\sim}{V}}(r)$
for all $t\in\overset{\sim}{V}\ast r$, so the first part is proved.

For the second item notice that for $v,u\in V$ we have

$r(x+v)=r(x+u)\Leftrightarrow r(x)=r(x+(u-v))\Leftrightarrow u-v\in W,$

hence all $r(x+v_{i})$ are different irreducible polynomials for
$v_{1},\ldots,v_{k}$ a complete set of representatives of $V/W$. Since
$F(Q_{\overset{\sim}{V}}(x))$ has exactly $|V|/|W|$ irreducible factors by
Corollary 23 equation (8) follows. ∎

Note that Corollary 4 is an immediate consequence of Theorem 2.

###  4.2 Borel-Subgroup

Here we consider the Borel-subgroup of $\operatorname{PGL}_{2}(q)$ in fields
$K$ with $\mathbb{F}_{q}\subseteq K$. This groups is defined as

$B(q)=\left\\{\left[\left(\begin{array}[]{cc}a&b\\\
0&1\end{array}\right)\right]:a\in\mathbb{F}_{q}^{\ast},b\in\mathbb{F}_{q}\right\\}.$

The transformation with $[A]\in B(q)$ looks like

$[A]\ast f(x)=a^{-\deg(f)}\cdot f(ax+b)$

for

$A=\left(\begin{array}[]{cc}a&b\\\ 0&1\end{array}\right).$ (9)

The group can be seen as

$B(q)\cong\mathbb{F}_{q}\rtimes\mathbb{F}_{q}^{\ast}$

where the multiplication is defined as

$(b,a)\cdot(d,c)=(b+ad,ac)$

and for $A\in\operatorname{GL}_{2}(K)$ as in (9) we have

$\operatorname{ord}([A])=\begin{cases}\operatorname{ord}_{\mathbb{F}_{q}^{\ast}}(a),&\text{
if }a\neq 1\\\ p,&\text{ if }a=1\text{ and }b\neq 0\end{cases}$

Moreover $B(q)=\operatorname{Stab}_{\operatorname{PGL}_{2}(q)}(\infty)$, so
$\mathcal{NR}_{K}^{B(q)}$ is the set of monic polynomials in
$\mathbb{F}_{q}[x]$. We calculate a quotient map for $B(q)$ using

######  Lemma 41 ([4, Theorem 3.10]).

Let $G\leq\operatorname{PGL}_{2}(K)$ be a finite subgroup and for
$v\in\overline{K}$ let

$g_{v}(x):=\prod\limits_{u\in G\circ v}(x-u)^{m_{v}}\text{ and
}h_{\infty}(x)=\prod\limits_{u\in
G\circ\infty\setminus\\{\infty\\}}(x-u)^{m_{\infty}},$

where $m_{u}:=|\operatorname{Stab}_{G}(u)|$ for
$u\in\overline{K}\cup\\{\infty\\}$. Then there is $w\in\overline{K}$ such that

$Q_{G}(x):=\frac{g_{v}(x)}{h_{\infty}(x)}+w\in K(x)$

is a quotient map for $G$. Conversely, if there is $w\in\overline{K}$ with
$\frac{g_{v}(x)}{h_{\infty}(x)}+w\in K(x)$, then
$Q(x):=\frac{g_{v}(x)}{h_{\infty}(x)}+w$ is a quotient map for $G$.

Let $v\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$, then $B(q)\circ
v=\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ because $\\{1,v\\}$ is a
$\mathbb{F}_{q}$ basis of $\mathbb{F}_{q^{2}}$, thus

$\\{a\cdot
v+b|a\in\mathbb{F}_{q}^{\ast},b\in\mathbb{F}_{q}\\}=\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}.$

Hence, a quotient map is given by

$\displaystyle Q_{B(q)}(x)$ $\displaystyle=\prod\limits_{w\in B(q)\circ
v}(x-w)^{|\operatorname{Stab}_{B(q)}(v)|}=\prod\limits_{w\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}}(x-w)^{|\operatorname{Stab}_{B(q)}(v)|}$
$\displaystyle=\prod\limits_{g\in\mathcal{I}_{q}^{2}}g(x).$

Recall that $\mathcal{I}_{q}^{2}$ is the set of monic irreducible polynomials
of degree $2$ over $\mathbb{F}_{q}$. That $|\operatorname{Stab}_{B(q)}(v)|=1$
holds is a consequence of $av+b=v$ only having solutions
$v\in\mathbb{F}_{q}\cup\\{\infty\\}$ if $(a,b)\neq(1,0)$.

If $q\neq 2$ then $P_{B(q)}$ is equal to $\mathbb{F}_{q}\cup\\{\infty\\}$
because $A$ as in (9) fixes $-b/(a-1)$. For $q=2$ we have
$B(2)\cong\mathbb{F}_{2}$ and $P_{B(2)}=\\{\infty\\}$; so this case belongs to
the previous example. The group $B(q)$ acts transitively on
$P_{G}\setminus\\{\infty\\}=\mathbb{F}_{q}$ because for fixed
$c\in\mathbb{F}_{q}$ and $b\in\mathbb{F}_{q}$ arbitrarily take the matrix

$B=\left(\begin{array}[]{cc}1&b-c\\\ 0&1\end{array}\right)$

and we see $[B]\circ c=c+(b-c)=b$. Consequentially, $B(q)$ also acts
transitively on $\mathcal{P}_{G}$, which consists of polynomials of the form
$x-c$ for $c\in\mathbb{F}_{q}$. Hence there is a monic irreducible polynomial
$F$ of degree $1$ (so $F=x-\alpha$ for $\alpha\in K$) and an exponent $k>1$
such that

$\left(\prod\limits_{g\in\mathcal{I}_{q}^{2}}g(x)\right)-\alpha=F^{Q_{B(q)}}(x)=\left(\prod\limits_{v\in\mathbb{F}_{q}}(x-v)\right)^{k}=(x^{q}-x)^{k}.$

We can deduce the exponent $k$ from comparing the degree of the polynomials on
both sides of the equality. The left side has degree $2\cdot
N_{q}(2)=2\cdot(\frac{1}{2}(q^{2}-q))=q^{2}-q=q(q-1)$, so $k=q-1$. To obtain
$\alpha$ we calculate

$\displaystyle(x^{q}-x)^{q-1}=\frac{x^{q^{2}}-x^{q}}{x^{q}-x}=\frac{x^{q^{2}}-x}{x^{q}-x}-\frac{x^{q}-x}{x^{q}-x}=\prod\limits_{g\in\mathcal{I}_{q}^{2}}g(x)-1$

so $\alpha=1$ and thus

$\left(\prod\limits_{g\in\mathcal{I}_{q}^{2}}g(x)\right)-1=(x^{q}-x)^{q-1}.$

Hence $(x-1)\in\mathcal{NC}^{Q_{b(q)}}$ and
$\delta_{Q_{B(q)}}(x-1)=\mathcal{P}_{B(q)}$, so
$\mathcal{NC}^{Q_{b(q)}}=\\{x-1\\}$ by Lemma 29. With this we can characterize
all $B(q)$-invariant polynomials as follows:

######  Corollary 42.

For every monic $B(q)$-invariant polynomial $f\in\mathcal{NR}_{K}^{B(q)}$
exists a unique monic polynomial $F\in K[x]$ and $m\in\mathbb{N}$ with $0\leq
m<q-1$ such that

$f(x)=(x^{q}-x)^{m}\cdot F\left(\prod_{g\in\mathcal{I}_{q}^{2}}g(x)\right).$

######  Remark 43.

The polynomial

$Q(x)=(x^{q}-x)^{q-1}$

is another quotient map for $B(q)$. For $Q$ the sets $P_{B(q)}$ and
$\mathcal{P}_{B(q)}$ remain the same (notice that these sets are always the
same regardless which quotient map we choose), just $\mathcal{NC}^{Q}=\\{x\\}$
is different. Therefore we can reformulate the previous Corollary in the
following way:

> For every monic $B(q)$-invariant polynomial $f\in\mathcal{NR}_{K}^{B(q)}$
> exists a unique monic polynomial $F\in K[x]$ and $m\in\mathbb{N}$ with
> $0\leq m<q-1$ such that
>
> $f(x)=(x^{q}-x)^{m}\cdot F\left((x^{q}-x)^{q-1}\right)$

Changing the quotient maps for $B(q)$ in the representation of
$B(q)$-invariant polynomials is like changing the basis of a vector space. The
polynomials $F$ are, in this analogy, like the coefficients of the vectors
written as the linear combination of the basis elements.

The factorization over finite $K$ can be explained with Theorem 2 again:

######  Corollary 44.

Let $K=\mathbb{F}_{q^{s}}$ and $F\in\mathcal{I}_{q^{s}}$ for an
$s\in\mathbb{N}\setminus\\{0\\}$ and $q=p^{n}$. If $F\neq x-1$ then
$F^{Q_{B(q)}}$ has at least

  1. 1.

$q$ irreducible factors if $q$ is not prime, i.e. $n>1$, or

  2. 2.

$q-1$ irreducible factors if $q$ is prime, i.e. $n=1$

Every such factor has a cyclic stabilizer in $B(q)$ and thus has degree at
most $(q-1)\cdot\deg(F)$ if $q$ is not prime and $q\cdot\deg(F)$ if $q$ is
prime.

###  4.3 Projective General Linear Groups

Let $\operatorname{char}(K)=p>0$ and assume that $\mathbb{F}_{q}\subseteq K$
for $q$ a power of $p$, then
$G:=\operatorname{PGL}_{2}(\mathbb{F}_{q})\leq\operatorname{PGL}_{2}(K)$.
First of all we need to calculate a quotient map for $G$. As shown in [4,
Example 3.12]

$Q_{G}(x)=\frac{\prod_{r\in\mathcal{I}_{q}^{3}}r(x)}{(\prod_{h\in\mathcal{I}_{q}^{1}}h(x))^{(q-1)q}}=\frac{\prod_{r\in\mathcal{I}_{q}^{3}}r(x)}{(x^{q}-x)^{(q-1)q}}$
(10)

is a quotient map for $G$ over every field that contains $\mathbb{F}_{q}$, so
also over $K$. The sets $\mathcal{I}_{q}^{1}$ and $\mathcal{I}_{q}^{3}$ are
the sets of monic irreducible polynomials in $\mathbb{F}_{q}[x]$ of degree 1
and 3 respectively. We want to determine $P_{G},\mathcal{P}_{G}$ and
$\mathcal{NC}^{Q_{G}}$. By Lemma 28 we know that
$P_{G}\subseteq\mathbb{F}_{q^{2}}\cup\\{\infty\\}$ since the equation

$[A]\circ v=v$

for $[A]\in\operatorname{PGL}_{2}(q)$ is, in essence, a polynomial equation
over $\mathbb{F}_{q}$, hence all solutions are algebraic over $\mathbb{F}_{q}$
(except $\infty$) and thus $[\mathbb{F}_{q}(v):\mathbb{F}_{q}]\leq 2$. Indeed,
$P_{G}=\mathbb{F}_{q^{2}}\cup\\{\infty\\}$ since for $a\in\mathbb{F}_{q}$ and
$v\in\mathbb{F}_{q^{2}}$ we have

$\displaystyle G\circ a$ $\displaystyle=\mathbb{F}_{q}\cup\\{\infty\\}$
$\displaystyle G\circ v$
$\displaystyle=\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}.$

Therefore $\mathcal{NR}_{K}^{G}$ consists of monic polynomials in $K[x]$ with
no roots in $\mathbb{F}_{q}$. The set $\mathcal{P}_{G}$ contains minimal
polynomials of elements in
$P_{G}\setminus(G\circ\infty)=\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$, thus
we have two cases:

  1. 1.

If $\mathbb{F}_{q^{2}}\not\subseteq K$, then
$\mathcal{P}_{G}=\mathcal{I}_{q}^{2}$ and every $g\in\mathcal{I}_{q}^{2}$ is
also irreducible over $K[x]$

  2. 2.

If $\mathbb{F}_{q^{2}}\subseteq K$, then

$\mathcal{P}_{G}=\\{x-v|v\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}\\}.$

Since $B(q)\subseteq G$ and $B(q)$ acts transitively on $\mathcal{P}_{G}$ so
does $G$ and thus $G\ast g=\mathcal{P}_{G}$ for all $g\in\mathcal{P}_{G}$.
Since $\delta_{Q_{G}}(\mathcal{NC}^{Q_{G}})\subseteq\mathcal{P}_{G}$ we are
looking for an irreducible polynomial $F\in\mathcal{I}_{K}$ such that

$F^{Q_{G}}=(\prod G\ast
h)^{k}=(\prod\mathcal{P}_{G})^{k}=(\prod\limits_{g\in\mathcal{I}_{q}^{2}}g)^{k}$

for an $h\in\mathcal{P}_{G}$ and $k>1$. Looking back at Example 3.12 in [4]
gives $F=x+1$ and $k=q+1$, thus $\mathcal{NC}^{Q_{G}}=\\{x+1\\}$. With Theorem
30 we obtain

######  Corollary 45.

For every monic $\operatorname{PGL}_{2}(q)$-invariant polynomial
$f\in\mathcal{NR}_{K}^{G}$ exists a unique monic polynomial $F\in K[x]$ and
$m\in\mathbb{N}$ with $0\leq m<q+1$ such that

$f(x)=\left(\prod\limits_{g\in\mathcal{I}_{q}^{2}}g(x)\right)^{m}\cdot\left((x^{q}-x)^{(q-1)q\cdot\deg(F)}F\left(\frac{\prod_{r\in\mathcal{I}_{q}^{3}}r(x)}{(x^{q}-x)^{(q-1)q}}\right)\right).$

For the factorization over finite fields we shortly recall the 3 types of
conjugacy classes of cyclic subgroups of
$\operatorname{PGL}_{2}(\mathbb{F}_{q})$ (for reference see [4, Proposition
11.1] or [16, §8]).

Every $[A]\in\operatorname{PGL}_{2}(q)$ is contained in one of the following
three types of conjugacy classes:

  1. 1.

$[A]$ fixes a unique element in $\mathbb{F}_{q}\cup\\{\infty\\}$, i.e.
$[A]\circ v=v$ for a unique $v\in\mathbb{F}_{q}\cup\\{\infty\\}$

  2. 2.

$[A]$ fixes two different elements in $\mathbb{F}_{q}\cup\\{\infty\\}$ under
Möbius-transformation

  3. 3.

$[A]$ fixes $\lambda,\lambda^{q}\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$
under Möbius-transformation

We then say that $[A]$ is of type $1,2$ or $3$ respectively. If $[A]$ is of
type 1 then $\operatorname{ord}([A])=p$ and $p$ is the prime dividing $q$.
Every element $[B]$ of type 2 has an order dividing $q-1$ and if $[C]$ is of
type 3 then $\operatorname{ord}([C])|q+1$. So
$\mu_{\operatorname{PGL}_{2}(\mathbb{F}_{q})}=q+1$ and we obtain

######  Corollary 46.

Let $K=\mathbb{F}_{q^{s}}$ for $s>0$ and $F\in\mathcal{I}_{q^{s}}$ and
$G=\operatorname{PGL}_{2}(\mathbb{F}_{q})$. If $F\neq x+1$ then $F^{Q_{G}}$
has at least $q^{2}-q$ irreducible factors and every such factor has degree at
most $(q+1)\cdot\deg(F)$.

###### Proof.

Follows immediately from Theorem 2 together with $\mu_{G}=q+1$ and
$|G|=q^{3}-q$. ∎

## Acknowledgements

I want to thank Alev Topuzoğlu and Henning Stichtenoth for their helpful
remarks and the advice they have given me. I am especially grateful to Henning
Stichenoth helping me with some technicalities of section 2.1. and making me
aware of the paper [27].

I am also very grateful for all of the invaluable help my supervisor Gohar
Kyureghyan has given me. Without her, this paper would probably not exist.

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