# Isomorphism problems and groups of automorphisms for Ore extensions
$K[x][y;f\frac{d}{dx}]$ (prime characteristic)

V. V. Bavula

###### Abstract

Let $\Lambda(f)=K[x][y;f\frac{d}{dx}]$ be an Ore extension of a polynomial
algebra $K[x]$ over an arbitrary field $K$ of characteristic $p>0$ where $f\in
K[x]$. For each polynomial $f$, the automorphism group of the algebras
$\Lambda(f)$ is explicitly described. The automorphism group ${\rm
Aut}_{K}(\Lambda(f))=\mathbb{S}\rtimes G_{f}$ is a semidirect product of two
explicit groups where $G_{f}$ is the eigengroup of the polynomial $f$ (the set
of all automorphisms of $K[x]$ such that $f$ is their common eigenvector). For
each polynomial $f$, the eigengroup $G_{f}$ is explicitly described. It is
proven that every subgroup of ${\rm Aut}_{K}(K[x])$ is the eigengroup of a
polynomial.

It is proven that the Krull and global dimensions of the algebra $\Lambda(f)$
are 2. The prime, completely prime, primitive and maximal ideals of the
algebra $\Lambda(f)$ are classified.

Key Words: a skew polynomial ring, automorphism, the eigengroup of a
polynomial, a prime ideal, a completely prime ideal, a primitive ideal, a
maximal ideal, simple module, the Krull dimension, the global dimension, the
centre, localization, a left denominator set, an Ore set, a normal element.

Mathematics subject classification 2010: 16D60, 13N10, 16S32, 16P90, 16U20.

${\bf Contents}$

  1. 1.

Introduction.

  2. 2.

Spectra, the centre, the Krull and global dimensions of the algebra $\Lambda$.

  3. 3.

Isomorphism problems and groups of automorphisms for Ore extensions
$K[x][y;f\frac{d}{dx}]$.

  4. 4.

The eigengroup $G_{f}$ of a polynomial $f$.

##  1 Introduction

In this paper, module means a left module, $K$ is a field of characteristic
$p>0$ and $\overline{K}$ is its algebraic closure,
$K^{\times}:=K\backslash\\{0\\}$, $K[x]$ be a polynomial algebra in the
variable $x$ over $K$, ${\rm Der}_{K}(K[x])=K[x]\frac{d}{dx}$ is the set of
all $K$-derivations of the algebra $K[x]$,

$\Lambda:=\Lambda(f):=K[x][y;\delta:=f\frac{d}{dx}]=K\langle x,y\,|\,yx-
xy=f\rangle=\bigoplus_{i\geq 0}K[x]y^{i}$

is an Ore extension of the algebra $K[x]$ where $f=f(x)\in K[x]$. Given an
algebra $D$ and its derivation $\delta$, the Ore extension of $D$, denoted
$D[y;\delta]$, is an algebra generated by the algebra $D$ and $y$ subject to
the defining relations $yd-dy=\delta(d)$ for all $d\in D$. The algebra
$\Lambda$ is a Noetherian domain of Gelfand-Kirillov dimension 2.

The aim of the paper is for each polynomial $f$ to give an explicit
description of the automorphism group ${\rm Aut}_{K}(\Lambda(f))$ of the
algebra $\Lambda(f)$.

By dividing the element $y$ by the leading coefficient of the polynomial $f$
we can assume that the polynomial $f$ is monic, i.e. its leading coefficient
is 1 provided $f\neq 0$. Then the algebras $\\{\Lambda(f)\,|\,f\in K[x]\\}$ as
a class can be partitioned into four subclasses: $f=0$, $f=1$, the polynomial
$f$ has only a single root in $\overline{K}$ and the polynomial $f$ has at
least two distinct roots in $\overline{K}$.

If $f=0$ then the algebra $\Lambda(0)=K[x,y]$ is a polynomial algebra in two
variables and its group of automorphisms is well-known [15]: The group ${\rm
Aut}_{K}(K[x,y])$ is generated by the automorphisms:

$\displaystyle t_{l}$ $\displaystyle:$ $\displaystyle x\mapsto\lambda
x,\;\;\;\;\;\;\;\;\;\;\;y\mapsto y,$ $\displaystyle\Phi_{n,\lambda}$
$\displaystyle:$ $\displaystyle x\mapsto x+\lambda y^{n},\;\;y\mapsto y,$
$\displaystyle\Phi_{n,\lambda}^{\prime}$ $\displaystyle:$ $\displaystyle
x\mapsto x,\;\;\;\;\;\;\;\;\;\;\;\;\;\;y\mapsto y+\lambda x^{n},$

where $n\geq 0$ and $\lambda\in K$.

If $f=1$ then the algebra $\Lambda(1)$ is the (first) Weyl algebra

$A_{1}=K\langle x,\partial\,|\,\partial x-x\partial=1\rangle\simeq
K[x][y;\frac{d}{dx}].$

In characteristic zero Dixmier [10], and in prime characteristic Makar-Limanov
[13], gave an explicit generators for the automorphism group ${\rm
Aut}_{K}(A_{1})$ (see also [4] for more results on ${\rm Aut}_{K}(A_{1})$):
The group ${\rm Aut}_{K}(A_{1})$ is generated by the automorphisms:

$\displaystyle\Phi_{n,\lambda}$ $\displaystyle:$ $\displaystyle x\mapsto
x+\lambda y^{n},\;\;y\mapsto y,$ $\displaystyle\Phi_{n,\lambda}^{\prime}$
$\displaystyle:$ $\displaystyle x\mapsto x,\;\,\;\;\;\;\;\;\;\;\;\;\;y\mapsto
y+\lambda x^{n},$

where $n\geq 0$ and $\lambda\in K$.

Extending results of Dixmier [10] on the automorphisms of the Weyl algebra
$A_{1}$, Bavula and Jordan [3] considered isomorphisms and automorphisms of
generalized Weyl algebras over polynomial algebras of characteristic 0.
Explicit generators of the automorphism group were given (they are more
involved comparing to the case of the Weyl algebra to present them here). Alev
and Dumas [1] initiated the study of automorphisms of Ore extensions
$\Lambda(f)$ in characteristic zero case. Their results were extended also to
prime characteristic by Benkart, Lopes and Ondrus [6]. The algebra
$\Lambda(x^{2})$ (the Jordan plane) was studied by Shirikov [14], Cibils,
Lauve, and [9], and Iyudu [11]. Gadis [8] studied isomorphism problems for
algebras on two generators that satisfy a single quadratic relation.

Isomorphism problems for the algebras $\Lambda(f)$. Theorems 1.1 is an
isomorphism criterion for the algebras $\Lambda$. It also describes the
automorphism group of each algebra $\Lambda(f)$.

###### Theorem 1.1

Let $f,g\in K[x]$ be polynomials. Then $\Lambda(f)\simeq\Lambda(g)$ iff
$g(x)=\lambda f(\alpha x+\beta)$ for some elements $\lambda,\alpha\in
K^{\times}$ and $\beta\in K$.

In characteristic zero, Theorem 1.1 was proven by Alev and Dumas [1,
Proposition 3.6] (1997) and in prime characteristic – by Benkart, Lopes and
Ondrus [6, Theorem 8.2] (2015).

Benkart, Lopes and Ondrus [6] (Theorems 8.3 and 8.6) gave a description of the
set of automorphisms groups of algebras $\Lambda(f)$ over arbitrary fields and
if the automorphism group of $\Lambda(f)$ is given they presented information
on the type of the polynomial $f$, [6, Corollary 8.7] (in general, if one
fixes the type of the polynomial then the automorphism group is larger than
the one which is naively expected). In this paper, we proceed in the opposite
direction: if the polynomial $f$ is given then the automorphism group ${\rm
Aut}_{K}\,\Lambda(f)$ is explicitly described.

The eigengroup $G_{f}(K)$ of a polynomial $f\in K[x]$. Recall that ${\rm
Aut}_{K}(K[x])=\\{\sigma_{\lambda,\mu}\,|\,\lambda\in K^{\times},\mu\in K\\}$
where $\sigma_{\lambda,\mu}(x)=\lambda x+\mu$.

Definition, [5]. For a polynomial $f\in K[x]$,

$G_{f}=G_{f}(K):=\\{\sigma\in{\rm
Aut}_{K}(K[x])\,|\,\sigma(f)=\lambda_{\sigma}f\;\;{\rm
for\;some}\;\;\lambda_{\sigma}\in K^{\times}\\}$ (1)

is called the eigengroup of the polynomial $f$.

Clearly, the set $G_{f}(K)$ is a subgroup of ${\rm Aut}_{K}(K[x])$, it is the
largest subgroup of ${\rm Aut}_{K}(K[x])$ such that the polynomial $f$ is
their common eigenvector. For all scalars $\mu\in K$, $G_{\mu}={\rm
Aut}_{K}(K[x])$. For all scalars $\nu\in K^{\times}$, $G_{f}=G_{\nu f}$. So,
in order to describe the eigengroup $G_{f}(K)$ we can assume that the
polynomial $f$ is a monic polynomial. It is proven that every subgroup of
${\rm Aut}_{K}(K[x])$ is the eigengroup of a polynomial (Theorem 4.26). For
each subgroup $G$ of ${\rm Aut}_{K}(K[x])$ all the polynomials $f\in K[x]$
with $G_{f}=G$ are explicitly described in the case when the field $K$ is
algebraically closed. The most interesting and difficult case is when the
group $G$ is a finite group. There are three types of finite subgroups in
${\rm Aut}_{K}(K[x])$ that are not the identity group. For each such group
$G$, the polynomials $f$ with $G_{f}=G$ has a unique form/presentation the,
so-called. eigenform of $f$.

The eigengroup $G_{f}$ has an isomorphic copy in the automorphism group ${\rm
Aut}_{K}(\Lambda(f))$: The map

$G_{f}(K)\rightarrow{\rm
Aut}_{K}(\Lambda(f)),\sigma_{\lambda\mu}\mapsto\sigma_{\lambda,\mu}:x\mapsto\lambda
x+\mu,\;\;y\mapsto y\mapsto\lambda^{\deg(f)-1}y$ (2)

is a group monomorphism where $\deg(f)$ is the degree of the polynomial $f$.
We identify the group $G_{f}(K)$ with its image in ${\rm
Aut}_{K}(\Lambda(f))$. The group $G_{f}(K)$ is the most important/difficult
part of the group ${\rm Aut}_{K}(\Lambda(f))$. Approximately half of the paper
is about how to find it. If the group $G_{f}(K)$ is a finite group it is a
semidirect product of two subgroups,
$G_{f}(K)=\widetilde{G}_{f}(K)\rtimes\overline{G}_{f}(K)$ (Theorem 4.3).

There are four distinguish cases:

  1. 1.

$\widetilde{G_{f}}\neq\\{e\\}$, $\overline{G}_{f}\neq\\{e\\}$,

  2. 2.

$\widetilde{G_{f}}\neq\\{e\\}$, $\overline{G}_{f}=\\{e\\}$,

  3. 3.

$\widetilde{G_{f}}=\\{e\\}$, $\overline{G}_{f}\neq\\{e\\}$,

  4. 4.

$\widetilde{G_{f}}=\\{e\\}$, $\overline{G}_{f}=\\{e\\}$.

In the case when $K=\overline{K}$, Theorem 4.18, Theorem 4.20, Theorem 4.22
and Theorem 4.23 are criteria for each case to hold, respectively (see also
Proposition 4.8). These four theorems are also explicit descriptions of the
eigengroup $G_{f}(K)$. They also show that in each of four cases the
polynomial $f$ admits a unique presentation – the eigenform of $f$ (introduced
in the paper).

At the end of Section 4, a finite algorithm is given of finding the
eigengroups $G_{f}(\overline{K})$ and $G_{f}(K)$, and the eigenform of $f$.

In the case when $K\neq\overline{K}$, similar results are obtained, see
Theorem 4.24. Proposition 4.25 gives criteria for the groups
$\widetilde{G}_{f}(K)$, $\overline{G}_{f}(K)$ and $G_{f}(K)$ to be $\\{e\\}$.

The group of automorphisms of the algebra $\Lambda(f)$. Given a group $G$, a
normal subgroup $N$ and a subgroup $H$. The group $G$ is called the semidirect
product of $N$ and $H$, written $G=N\rtimes H$, if $G=NH:=\\{nh\,|\,n\in
N,h\in H\\}$ and $N\cap H=\\{e\\}$ where $e$ is the identity of the group $G$.

The automorphism group ${\rm Aut}_{K}(\Lambda(f))$ of the algebra $\Lambda(f)$
contains an obvious subgroup

$\mathbb{S}:=\mathbb{S}(K):=\\{s_{p}\,|\,p\in
K[x]\\}\simeq(K[x],+),\;s_{p}\mapsto p\;\;{\rm where}\;\;s_{p}(x)=x\;{\rm
and}\;s_{p}(y)=y+p.$ (3)

###### Theorem 1.2

Suppose that $f\in K[x]$ is a monic nonscalar polynomial. Then

${\rm Aut}_{K}(\Lambda(f))=\mathbb{S}(K)\rtimes G_{f}(K).$

The Krull and global dimensions of the algebra $\Lambda(f)$. It it proven that
the Krull and global dimensions of the algebra $\Lambda(f)$ are 2 (Theorem
2.5).

Classifications of prime, completely prime, primitive and maximal ideals of
the algebra $\Lambda(f)$. Theorem 2.3 classifies prime, completely prime,
primitive and maximal ideals of the algebra $\Lambda(f)$. The height 1 prime
ideals were classified in [6, Theorem 7.6]. It is proven that every nonzero
ideal of the algebra $\Lambda(f)$ meets the centre of $\Lambda(f)$ (Corollary
2.2).

Classifications of simple $\Lambda(f)$-modules. In [2], simple modules were
classified for all Ore extensions $A=D[x;\sigma,\partial]$ where $D$ is a
Dedekind domain, $\sigma\in$ is an automorphism of $D$ and $\partial$ is a
$\sigma$-derivation of $D$ (for all $a,b\in D$,
$\partial(ab)=\partial(a)b+\sigma(a)\partial(b)$). Recall that the ring $A$ is
generated by $D$ and $x$ subject to the defining relations: For all elements
$d\in D$, $xd=\sigma(d)x+\partial(d)$. The algebras $\Lambda(f)$ is a a very
special case of the rings $A$.

Theorem 2.4 classifies simple left $\Lambda(f)$-modules, see also [7]. For
each simple left $\Lambda(f)$-module an explicit $K$-basis is given and the
actions of the canonical generators $x$ and $y$ of the algebra $\Lambda(f)$ on
the basis is explicitly described.

A classification of simple right $\Lambda(f)$-modules is obtained at once from
the classification of simple left $\Lambda(f)$-modules by using the fact that
the opposite algebra $\Lambda(f)^{op}$ of the algebra $\Lambda(f)$ is
isomorphic to

$\Lambda(f)^{op}\simeq\Lambda(-f).$ (4)

Recall that the opposite algebra $A^{op}$ of an algebra $A$ coincides with the
algebra $A$ as vector space but the multiplication in $A^{op}$ is given by the
rule $a\cdot b=ba$. Every right $A$-module is a left $A^{op}$-module and vice
versa.

##  2 Spectra, the centre, the Krull and global dimensions of the algebra
$\Lambda$

In this section, $K$ is a field of characteristic $p>0$ (not necessarily
algebraically closed) and $f=p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$ is a nonscalar
polynomial of $K[x]$ where $p_{1},\ldots,p_{s}$ are irreducible, co-prime
divisors of $f$ (i.e. $K[x]p_{i}+K[x]p_{j}=K[x]$ for all $i\neq j$). The aim
of this section is to find the centre of the algebra $\Lambda(f)$; to classify
simple $\Lambda(f)$-modules; to classify prime, completely prime, primitive
and maximal ideals of the algebra $\Lambda(f)$; and to prove that the Krull
and global dimension of the algebra $\Lambda(f)$ is 2.

The centre of the algebra $\Lambda(f)$. It follows from the direct sum

$A_{1}=\bigoplus_{i,j=0}^{p-1}K[x^{p},\partial^{p}]x^{i}\partial^{j}$

and the commutation relation $[\partial,x]=1$, that the centre $Z(A_{1})$ of
the Weyl algebra is equal to $K[x^{p},\partial^{p}]$, a polynomial algebra in
two variables. Let $K(x^{p},\partial^{p})$ be the field of fractions of the
polynomial algebra $K[x^{p},\partial^{p}]$. The localization ${\cal A}_{1}$ of
the Weyl algebra $A_{1}$ by the Ore set
$K[x^{p},\partial^{p}]\backslash\\{0\\}$ is a simple $p^{2}$-dimensional
algebra

${\cal A}_{1}=\bigoplus_{i,j=0}^{p-1}K(x^{p},\partial^{p})x^{i}\partial^{j}.$

This follows from the relation $[\partial,x]=1$. So, $Z({\cal
A}_{1})=K(x^{p},\partial^{p})$. Hence, every nonzero ideal of the Weyl algebra
$A_{1}$ meets the centre of $A_{1}$.

The polynomial algebra $K[x]$ is a left $A_{1}$-module where $\partial$ acts
as the derivation $\frac{d}{dx}$. Furthermore, the kernel of the corresponding
algebra homomorphism

$A_{1}\rightarrow{\rm End}_{K}(K[x]),\;\;x\mapsto
x,\;\;\partial\mapsto\frac{d}{dx}$ (5)

is generated by the central element $\partial^{p}$. So,
$A_{1}/(\partial^{p})=\bigoplus_{i=0}^{p-1}K[x]\partial^{i}\subseteq{\rm
End}_{K}(K[x])$. The factor algebra $A_{1}/(\partial^{p})$ is a subalgebra of
the algebra ${\cal D}(K[x])$ of differential operators on the polynomial
algebra $K[x]$ and the Weyl algebra $A_{1}$ is not. This is in sharp contrast
with the characteristic zero case where $A_{1}={\cal D}(K[x])$.

The algebra $\Lambda=\Lambda(f)$ can be identified with a subalgebra of the
Weyl algebra $A_{1}$ by the monomorphism:

$\Lambda\rightarrow A_{1},\;\;x\mapsto x,\;\;y\mapsto f\partial.$ (6)

So, $\Lambda=K\langle x,y=f\partial\rangle\subset A_{1}$. Theorem 2.1
describes the centre of the algebra $\Lambda(f)$. It also gives explicit
expressions for the $p$’th power of various elements of the algebras
$\Lambda(f)$ and $A_{1}$ that are key facts in finding the centre of
$\Lambda(f)$.

The fact that the centre of the algebra $\Lambda(f)$ is equal to
$K[x^{p},y^{p}-(\delta^{p-2}(f))^{\prime}y]$ was proven by Benkart, Lopes and
Ondrus, [6, Theorem 5.3,(2)]. Here we present a short proof of this fact.

###### Theorem 2.1

Let $\delta=f\frac{d}{dx}\in{\rm Der}_{K}(K[x])$ where $f\in
K[x]\backslash\\{0\\}$, and $g^{\prime}:=\frac{dg}{dx}$ where $g\in K[x]$.
Then

  1. 1.

$\delta^{p}=(\delta^{p-2}(f))^{\prime}\delta\in{\rm Der}_{K}(K[x])$.

  2. 2.

In the algebra $\Lambda(f)$,
$y^{p}=f^{p}\partial^{p}+(\delta^{p-2}(f))^{\prime}y$. In particular, in the
Weyl algebra $A_{1}$,
$(f\partial)^{p}=f^{p}\partial^{p}+(\delta^{p-2}(f))^{\prime}f\partial$.

  3. 3.

The centre $Z(\Lambda(f))$ of the algebra $\Lambda(f)$ is the polynomial
algebra

$K[x^{p},y^{p}-(\delta^{p-2}(f))^{\prime}y]=K[x^{p},f^{p}\partial^{p}]$

and $f^{p}\partial^{p}=y^{p}-(\delta^{p-2}(f))^{\prime}y$.

  4. 4.

The algebra $\Lambda(f)=\bigoplus_{i,j=0}^{p-1}Z(\Lambda(f))x^{i}y^{j}$ is a
free $Z(\Lambda(f))$-module of rank $p^{2}$.

  5. 5.

The localization of the algebra $\Lambda(f)$ at the Ore set
$Z(\Lambda(f))\backslash\\{0\\}$ is ${\cal A}_{1}$.

Proof. 1. Since $\delta^{p}\in{\rm Der}_{K}(K[x])$, we have that
$\delta^{p}=g\frac{d}{dx}$ where
$g=\delta^{p}(x)=\delta^{p-1}(f)=(\delta^{p-2}(f))^{\prime}f$. Therefore,

$\delta^{p}=(\delta^{p-2}(f))^{\prime}f\frac{d}{dx}=(\delta^{p-2}(f))^{\prime}\delta.$

2\. Notice that
$y^{p}=(f\partial)^{p}=f^{p}\partial^{p}+\sum_{i=1}^{p-1}a_{i}\partial^{i}$
for some elements $a_{i}\in K[x]$. Recall that
$A_{1}/(\partial^{p})=\bigoplus_{i=0}^{p-1}K[x]\partial^{i}\subseteq{\rm
End}_{K}(K[x])$. By statement 1,

$(f\partial)^{p}\equiv\sum_{i=1}^{p-1}a_{i}\partial^{i}\equiv(\delta^{p-2}(f))^{\prime}\delta\equiv(\delta^{p-2}(f))^{\prime}f\partial\equiv(\delta^{p-2}(f))^{\prime}y\mod(\partial^{p}),$

and so $a_{1}=(\delta^{p-2}(f))^{\prime}f$ and $a_{2}=\cdots=a_{p-1}=0$.

3–5. By statement 2, $f^{p}\partial^{p}=y^{p}-(\delta^{p-2}(f))^{\prime}y\in
Z(\Lambda(f))$. Hence,

$Z^{\prime}:=K[x^{p},y^{p}-(\delta^{p-2}(f))^{\prime}y]=K[x^{p},f^{p}\partial^{p}]\subseteq
Z(\Lambda(f))$

and $\Lambda(f)=\bigoplus_{i,j=0}^{p-1}Z^{\prime}x^{i}y^{j}$. Now,

$(Z^{\prime}\backslash\\{0\\})^{-1}\Lambda(f)=\bigoplus_{i,j=0}^{p-1}K(x^{p},\partial^{p})x^{i}y^{j}=\bigoplus_{i,j=0}^{p-1}K(x^{p},\partial^{p})x^{i}\partial^{j}={\cal
A}_{1},$

and so statement 5 is obvious and statements 3–4 follow. $\Box$

Let $A$ be an algebra and $a\in A$. The map ${\rm ad}_{a}=[a,-]:A\rightarrow
A$, $b\mapsto[a,b]:=ab-ba$ is a derivation of the algebra $A$ which is called
the inner derivation of $A$ associated with the element $a$.

###### Corollary 2.2

Every nonzero ideal of the algebra $\Lambda(f)$ meets the centre of
$\Lambda(f)$.

Proof. Let $I$ be a nonzero ideal of the algebra $\Lambda(f)$. Fix a nonzero
element of $I$, say $a=\sum_{i,j=0}^{p-1}z_{ij}x^{i}\partial^{j}$ for some
elements $z_{ij}\in Z(\Lambda(f))$, by Theorem 2.1.(4). Then applying several
times the inner derivation ${\rm ad}_{x}:=[x,-]$ of the algebra $\Lambda(f)$,
we obtain a nonzero element, say $b\in I\cap Z(\Lambda(f)[x]$. Then $0\neq
b^{p}\in I\cap Z(\Lambda(f))$. $\Box$

The prime, completely prime, primitive and maximal spectra of the algebra
$\Lambda$. An ideal $\mathfrak{p}$ of a ring $R$ is called a completely prime
ideal if the factor ring $R/\mathfrak{p}$ is a domain. A completely prime
ideal is a prime ideal. The sets of prime and completely prime ideals of the
ring $R$ are denoted by ${\rm Spec}(R)$ and ${\rm Spec}_{c}(R)$, respectively.
The annihilator of a simple $R$-module is called a primitive ideal of $R$.
Every primitive ideal is a prime ideal of $R$. The set of all primitive ideals
is denoted by ${\rm Prim}(R)$. The set of all maximal ideals of $R$ is denoted
by ${\rm Max}(R)$. Clearly, ${\rm Max}(R)\subseteq{\rm Prim}(R)\subseteq{\rm
Spec}(R)$.

An element $a$ of an algebra $A$ is called a normal element of $A$ if $Aa=aA$.
An element $a$ of an algebra $A$ is called a regular element if it is not a
zero divisor. The set of all regular elements of the algebra $A$ is denoted by
${\cal C}_{A}$. Each regular normal element $a$ of the algebra $A$ determines
an automorphism of the algebra $A$ given by the rule:

$\omega_{a}:A\rightarrow A,\;\;b\mapsto\omega_{a}(b)\;\;{\rm
where}\;\;ab=\omega_{a}(b)a.$ (7)

The elements $p_{1},\ldots,p_{s}$ are regular normal elements of the algebra
$\Lambda=\Lambda(f)$ (recall that $f=\prod_{i=1}^{s}p_{i}^{n_{i}}$) since

$yp_{i}=p_{i}(y-p_{i}^{-1}f)\;\;{\rm and}\;\;xp_{i}=p_{i}x.$

Therefore, $\omega_{p_{i}}(x)=x$ and $\omega_{p_{i}}(y)=y+p_{i}^{-1}f$.

For an ideal $\mathfrak{a}$ of an algebra $A$, we denote by $V(\mathfrak{a})$
the set of all prime ideals of $A$ that contain the ideal $\mathfrak{a}$. Let
$\min\mathfrak{a}$ be the set of minimal primes of $\mathfrak{a}$. These are
the minimal elements of the set $V(\mathfrak{a})$ with respect to inclusion.
Suppose that the set $S_{a}:=\\{a^{i}\,|\,i\geq 0\\}$ is a left Ore set of a
domain $A$. The algebra $A_{a}:=S_{a}^{-1}A=\\{a^{-i}b\,|\,b\in A\\}$ is
called the localization of $A$ at the powers of the element $a$.

For a commutative algebra $C$ and its non-nilpotent element $s$, the map

${\rm Spec}(C)\backslash V(s)\rightarrow{\rm
Spec}(C_{s}),\;\;\mathfrak{p}\mapsto S_{s}^{-1}\mathfrak{p}$

is a bijection with the inverse map $\mathfrak{q}\mapsto C\cap\mathfrak{q}$.
We identify the sets ${\rm Spec}(C)\backslash V(s)$ and ${\rm Spec}(C_{s})$
via the bijection above, i.e. ${\rm Spec}(C)\backslash V(s)={\rm
Spec}(C_{s})$.

By Theorem 2.1.(4), the centre of the Weyl algebra
$A_{1}=K[x][\partial;\frac{d}{dx}]$ is the polynomial algebra
$K[x^{p},\partial^{p}]$ (the result is well-known). Then

$f^{p}\in K[x^{p}]\subseteq Z(\Lambda(f))=K[x^{p},f^{p}\partial^{p}]\subseteq
K[x^{p},\partial^{p}]=Z(A_{1})$

and

$L=\Lambda(f)\subseteq
L_{f}=A_{1,f}=L_{f^{p}}=A_{1,f^{p}}=\bigoplus_{i,j=0}^{p-1}Z(L)_{f^{p}}x^{i}y^{i}=\bigoplus_{i,j=0}^{p-1}Z(A_{1})_{f^{p}}x^{i}\partial^{i}=\bigoplus_{i,j=0}^{p-1}K[x^{p},\partial^{p}]_{f^{p}}x^{i}\partial^{i}.$
(8)

In particular, $Z(L)_{f^{p}}=Z(A_{1})_{f^{p}}=K[x^{p},\partial^{p}]_{f^{p}}$,
and so we can write

$\displaystyle{\rm Spec}(Z(L))\backslash V(f^{p})$ $\displaystyle=$
$\displaystyle{\rm Spec}(Z(L)_{f^{p}})={\rm Spec}(Z(A_{1})_{f^{p}})={\rm
Spec}(K[x^{p},\partial^{p}]_{f^{p}})$ $\displaystyle=$ $\displaystyle{\rm
Spec}(Z(A_{1}))\backslash V(f^{p})={\rm Spec}(K[x^{p},\partial^{p}])\backslash
V(f^{p}).$

Theorem 2.3 gives explicit descriptions of the sets of prime, completely
prime, primitive and maximal ideals of the algebra $\Lambda$. Let ${\rm
Spec}_{c}(\Lambda,{\rm ht}=1)$ be the set of completely prime ideals of height
1 of the algebra $\Lambda(f)$.

###### Theorem 2.3

Let $K$ be a field of characteristic $p>0$,
$\Lambda=K[x][y;\delta:=f\frac{d}{dx}]$ where $f\in K[x]\backslash K$. Let
$f=p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$ be a unique (up to permutation) product
of irreducible polynomials of $K[x]$. Then

  1. 1.

The elements $p_{1},\ldots,p_{s}$ are regular normal elements of the algebra
$\Lambda$ (i.e. $p_{i}$ is a non-zero-divisor of $\Lambda$ and
$p_{i}\Lambda=\Lambda p_{i}$).

  2. 2.

$\min(f)=\\{(p_{1}),\ldots,(p_{s})\\}$.

  3. 3.

${\rm Spec}_{c}(\Lambda)=\\{0,\Lambda
p_{i},(p_{i},q_{i})\,|\,i=1,\ldots,s;\;q_{i}\in{\rm Irr}_{m}(F_{i}[y])\\}$
where $F_{i}:=K[x]/(p_{i})$ is a field and ${\rm Irr}_{m}(F_{i}[y])$ is the
set of monic irreducible polynomials of the polynomial algebra $F_{i}[y]$ over
the field $F_{i}$ in the variable $y$. If, in addition, the field $K$ is an
algebraically closed and $\lambda_{1},\ldots,\lambda_{s}$ are the roots of the
polynomial $f$ then ${\rm
Spec}_{c}(\Lambda)=\\{0,\Lambda(x-\lambda_{i}),(x-\lambda_{i},y-\mu)\,|\,i=1,\ldots,s;\;\mu\in
K\\}$.

  4. 4.

${\rm Spec}(\Lambda)={\rm
Spec}_{c}(\Lambda)\coprod\\{\Lambda\mathfrak{p}\,|\,\mathfrak{p}\in{\rm
Spec}(Z(\Lambda))\backslash\\{(0),V(f^{p})\\}$.

  5. 5.

For all $\mathfrak{p}\in{\rm Spec}(Z(\Lambda))\backslash V(f^{p})$,

$k(\mathfrak{p})\otimes_{Z(\Lambda)}\Lambda/(\mathfrak{p})\simeq\bigoplus_{i,j=0}^{p-1}{\bf
k}(\mathfrak{p})x^{i}y^{i}=\bigoplus_{i,j=0}^{p-1}{\bf
k}(\mathfrak{p})x^{i}\partial^{i}\simeq M_{p}(k(\mathfrak{p})),$

the algebra of $p\times p$ matrices over the field of fractions ${\bf
k}(\mathfrak{p})$ of the domain $Z(\Lambda)/\mathfrak{p}$.

  6. 6.

${\rm Max}(\Lambda)={\rm
Prim}(\Lambda)=\\{(p_{i},q_{i}),\Lambda\mathfrak{m}\,|\,i=1,\ldots,s;\;q_{i}\in{\rm
Irr}_{m}(F_{i}[y]),\mathfrak{m}\in{\rm Max}(Z(\Lambda))\backslash
V(f^{p})\\}$.

  7. 7.

${\rm Spec}_{c}(\Lambda,{\rm ht}=1)=\\{(p_{1}),\ldots,(p_{s})\\}$. If, in
addition $K=\overline{K}$, then ${\rm Spec}_{c}(\Lambda,{\rm
ht}=1)=\\{(x-\lambda_{1}),\ldots,(x-\lambda_{t})\\}$ where
$\\{\lambda_{1},\ldots,\lambda_{t}\\}$ is the set of roots of the polynomial
$f$.

Proof. 1\. Statements 1 is proven above.

2\. Since

$\Lambda/\Lambda p_{i}\simeq F_{i}[y]$ (9)

is a polynomial algebra with coefficients in the field $F_{i}$ (since $yx-
xy=f\in\Lambda p_{i}$), the ideal $\Lambda p_{i}$ is a completely prime ideal
of $\Lambda$. Now, statement 2 follows from the equality of ideals
$(f)=(p_{1})^{n_{1}}\cdots(p_{s})^{n_{s}}$.

5\. Since $\mathfrak{p}\in{\rm Spec}(Z(\Lambda))\backslash V(f^{p})$, the
element $f^{p}$ is a unit in the field $k(\mathfrak{p})$. Now, the first
isomorphism and the equality in statement 5 follows from (8). Then using the
equalities,

$[y,x^{i}]=ix^{i-1}\;\;{\rm and}\;\;[y^{i},x]=iy^{i-1},$

and the fact that $k(\mathfrak{p})$ is a field, we see that the algebra
$\bigoplus_{i,j=0}^{p-1}{\bf k}(\mathfrak{p})x^{i}\partial^{i}$ is a simple,
central $k(\mathfrak{p})$-algebra of dimension $p^{2}$ over the field
$k(\mathfrak{p})$. Therefore, it is isomorphic to the matrix algebra
$M_{p}(k(\mathfrak{p}))$.

3-4. The algebra $\Lambda$ is a domain, hence $0\in{\rm Spec}_{c}(\Lambda)$.
We have seen in the proof of statement 2 that the ideals
$(p_{1}),\ldots,(p_{s})$ of the algebra $\Lambda$ are completely prime ideals.
By (9),

$V(f)=\\{\Lambda p_{i},(p_{i},q_{i})\,|\,i=1,\ldots,s;\;q_{i}\in{\rm
Irr}_{m}(F_{i}[y])\\}\subseteq{\rm Spec}_{c}(\Lambda).$

Given a nonzero prime ideal $P$ of $\Lambda$ such that $P\not\in
V(f)=V(f^{p})$. Then $P_{f}$ is a nonzero prime ideal of the algebra
$\Lambda_{f^{p}}=A_{1,f^{p}}$. By Corollary 2.2, the intersection
$\mathfrak{p}:=P\cap Z(L)$ is a nonzero prime ideal of the centre $Z(\Lambda)$
of the algebra $\Lambda$. By Theorem 2.1.(4),
$\Lambda=\bigoplus_{i,j=0}^{p-1}Z(\Lambda)x^{i}y^{i}$. Now, by statement 5,
$\Lambda\mathfrak{p}\in{\rm Spec}(L)\backslash V(f)$ and the prime ideal
$\Lambda\mathfrak{p}$ is not completely prime. Now, statements 3 and 4 follows
from statement 5.

6\. Statement 6 follows from statement 4.

7\. Statement 7 follows from statement 3. $\Box$

Classification of simple $\Lambda(f)$-modules. For an $\Lambda$-module $M$, we
denote by ${\rm ann}_{\Lambda}(M)$ the annihilator of the $\Lambda$-module
$M$. For an algebra $A$, we denote by $\widehat{A}$, the set of isomorphism
classes of (left) simple $A$-modules. An isomorphism class of a simple
$A$-modules $M$ is denoted by $[M]$. Let elements $a_{1},\ldots,a_{n}\in A$ be
generators for a left ideal $I$ of the algebra $A$. Then we write
$I=A(a_{1},\ldots,a_{n})$. Theorem 2.4 is a classification of simple
$\Lambda(f)$-modules.

###### Theorem 2.4

Let $K$ be a field of characteristic $p>0$,
$\Lambda=K[x][y;\delta:=f\frac{d}{dx}]$ where $f\in K[x]\backslash K$. Let
$f=p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$ be a unique (up to permutation) product
of irreducible polynomials of $K[x]$. Then

  1. 1.

the map

${\rm Max}(L)\rightarrow\widehat{\Lambda},\;\;\mathfrak{M}\mapsto
L(\mathfrak{M})$

is a bijection with inverse $[M]\mapsto{\rm ann}_{\Lambda}(M)$ where
$L(\mathfrak{M})$ is a unique (up to isomorphism) simple direct
summand/submodule/factor module of the (simple) matrix algebra
$\Lambda/\mathfrak{M}$. In particular, for all $\mathfrak{m}\in{\rm
Max}(\Lambda)\backslash V(f^{p})$, ${\rm
dim}_{K}(L(\Lambda\mathfrak{m}))=p\cdot{\rm
dim}_{K}(Z(\Lambda)/\mathfrak{m})<\infty$.

  2. 2.

For each maximal ideal $(p_{i},q_{i})$ of $\Lambda(f)$, where $i=1,\ldots,s$
and $q_{i}\in{\rm Irr}_{m}(F_{i}[y])\\}$,

$L(p_{i},q_{i})=\Lambda/(p_{i},q_{i})\simeq K[y]/(q_{i})$

and ${\rm dim}_{K}(L(p_{i},q_{i}))={\rm
dim}_{K}(K[y]/(q_{i}))=\deg_{y}(q_{i})<\infty$.

  3. 3.

Suppose that the field $K=\overline{K}$ is an algebraically closed. For each
maximal ideal $\Lambda\mathfrak{m}$ of $\Lambda(f)$, where
$\mathfrak{m}\in{\rm Max}(Z(\Lambda))\backslash V(f^{p})$,

$\displaystyle L(\Lambda\mathfrak{m})$ $\displaystyle\simeq$
$\displaystyle\Lambda/\Lambda(\mathfrak{m},x-\sqrt[p]{\xi})=\bigoplus_{i,j=0}^{p-1}Ky^{i}\bar{1}$
$\displaystyle\simeq$ $\displaystyle
A_{1,f^{p}}/A_{1,f^{p}}(\mathfrak{m},x-\sqrt[p]{\xi})=\bigoplus_{i,j=0}^{p-1}K\partial^{i}\hat{1},$

where $x^{p}-\xi\in\mathfrak{m}$ for a unique element $\xi\in K$,
$\bar{1}=1+\Lambda(\mathfrak{m},x-\sqrt[p]{\xi})$ and
$\hat{1}=1+A_{1,f^{p}}(\mathfrak{m},x-\sqrt[p]{\xi})$, ${\rm
dim}_{K}L(\Lambda\mathfrak{m})=p<\infty$.

Proof. 1\. Statements 1 follows at once from Theorem 2.3.(5,6).

2\. Statement 2 is obvious.

3\. Notice that $(x-\sqrt[p]{\xi}))^{p}=x^{p}-\xi\in\mathfrak{m}$.

By Theorem 2.1.(4) and the choice of the ideal $\mathfrak{m}$, we have that

$\Lambda/\Lambda\mathfrak{m}=\bigoplus_{i=0}^{p-1}Ky^{i}\otimes
K[x]/(x^{p}-\xi)=\bigoplus_{i=0}^{p-1}Ky^{i}\otimes
K[x]/((x-\sqrt[p]{\xi})^{p}),$

direct sums of tensor products of vector spaces. Hence,

$\Lambda/\Lambda(\mathfrak{m},x-\sqrt[p]{\xi})\simeq\bigoplus_{i=0}^{p-1}Ky^{i}\otimes
K[x]/(x-\sqrt[p]{\xi})\simeq\bigoplus_{i,j=0}^{p-1}Ky^{i}\bar{1}$

is a $p$-dimensional $\Lambda$-module that is annihilated by the maximal ideal
$\mathfrak{m}$. By statement 1, it must be $L(\Lambda\mathfrak{m})$. Since
$\mathfrak{m}\not\in V(f^{p})$, the central element $f^{p}$ acts as a
bijection on the module $L(\Lambda\mathfrak{m})$. Therefore,

$L(\Lambda\mathfrak{m})=L(\Lambda\mathfrak{m})_{f^{p}}=\Lambda_{f^{p}}/\Lambda_{f^{p}}(\mathfrak{m},x-\sqrt[p]{\xi})\simeq
A_{1,f^{p}}/A_{1,f^{p}}(\mathfrak{m},x-\sqrt[p]{\xi})=\bigoplus_{i,j=0}^{p-1}K\partial^{i}\hat{1}.\;\;\Box$

The action of the elements $x$ and $y$ on the $K$-basis
$\\{y\bar{1}\,|\,i=0,\ldots,p-1\\}$ of the $\Lambda(f)$-module
$L(\Lambda\mathfrak{m})$ of Theorem 2.4.(3) is given below:

$\displaystyle x\cdot\bar{1}$ $\displaystyle=$
$\displaystyle\xi^{\frac{1}{p}}\bar{1},$ $\displaystyle x\cdot y^{i}\bar{1}$
$\displaystyle=$
$\displaystyle\xi^{\frac{1}{p}}y^{i}\bar{1}+\sum_{j=0}^{i-1}{i\choose
j}\xi_{ij}y^{j}\bar{1}\;\;{\rm
where}\;\;\xi_{ij}=(-1)^{i-j}\phi_{ij}(\xi^{\frac{1}{p}}),\;\;\phi_{ij}=\delta^{i-j-1}(f)\in
K[x],$ $\displaystyle y\cdot y^{i}\bar{1}$ $\displaystyle=$ $\displaystyle
y^{i+1}\bar{1}\;\;{\rm where}\;\;0\leq i\leq p-2,$ $\displaystyle y\cdot
y^{p-1}\bar{1}$ $\displaystyle=$ $\displaystyle\rho\bar{1}\;\;{\rm
where}\;\;y^{p}-\rho\in\mathfrak{m}\;\;{\rm
for\;a\;unique\;element}\;\;\rho\in K.$

The Krull and global dimensions of the algebra $\Lambda(f)$.

###### Theorem 2.5

Let $K$ be a field of characteristic $p>0$,
$\Lambda=K[x][y;\delta:=f\frac{d}{dx}]$ where $f\in K[x]\backslash K$. Let
$f=p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$ be a unique (up to permutation) product
of irreducible polynomials of $K[x]$. Then

  1. 1.

The Krull dimension of $\Lambda$ is ${\rm Kdim}(\Lambda)=2$.

  2. 2.

The global dimension of $\Lambda$ is ${\rm gldim}(\Lambda)=2$.

Proof. Let $\Lambda=\Lambda(f)$.

1\. By Theorem 2.1.(4), the algebra $\Lambda$ is a finitely generated
$Z(\Lambda)$-module. Therefore, the Krull dimension of the algebra $\Lambda$
is equal to the Krull dimension of the polynomial algebra $Z(\Lambda)$ in two
variables (Theorem 2.1.(3)), and statement 1 follows.

2\. By [12, Theorem 7.5.3.(i)], ${\rm gldim}(\Lambda)\leq{\rm
gldim}(K[X])+1=1+1=2$.

By (9), ${\rm gldim}(\Lambda/\Lambda p_{i})={\rm gldim}(F_{i}[Y])=1<\infty$.
Now, by [12, Theorem 7.3.5.(i)],

${\rm gldim}(\Lambda)\geq{\rm gldim}(\Lambda/\Lambda
p_{i})+1\stackrel{{\scriptstyle(\ref{LLpi})}}{{=}}{\rm
gldim}(F_{i}[Y])+1=1+1=2.$

Therefore, ${\rm gldim}(\Lambda)=2$. $\Box$

##  3 Isomorphism problems and groups of automorphisms for Ore extensions
$K[x][y;f\frac{d}{dx}]$

In this section, a proof Theorem 1.2 is given. It can be deduced from Theorem
1.1 but we give a different proof.

Let $K(x)$ be the field of rational functions in the variable $x$. Then the
Ore extension $B_{1}:=K(x)[\partial;\frac{d}{dx}]$ is the localization
$B_{1}=(K[x]\backslash\\{0\\})^{-1}A_{1}$ of the Weyl algebra $A_{1}$ at the
Ore set $K[x]\backslash\\{0\\}$.

Notice that

$\mathbb{S}(K)\rtimes G_{f}(K)=\\{\sigma_{\lambda,\mu,p}\,|\,\lambda\in
K^{\times},\mu\in K,p\in K[x]\\}$ (10)

where

$\sigma_{\lambda,\mu,p}(x)=\lambda x+\mu\;\;{\rm
and}\;\;\sigma_{\lambda,\mu,p}(y)=\lambda^{d-1}y+p$

since $\sigma_{\lambda,\mu,p}=s_{\lambda^{-d+1}p}\sigma_{\lambda,\mu}$ where
$d=\deg(f)$.

Proof of Theorem 1.2. Let $\sigma$ be an automorphism of the $K$-algebra
$\Lambda=\Lambda(f)$. It can be uniquely extended to a
$\overline{K}$-automorphism, say $\sigma$, of the algebra
$\overline{K}\otimes_{K}\Lambda$. Let $\lambda_{1},\ldots,\lambda_{t}$ be the
roots of the polynomial $f$ in $\overline{K}$. By Theorem 2.3.(7), the
automorphism $\sigma$ permutes the set

${\rm Spec}(\overline{K}\otimes_{K}\Lambda,{\rm
ht}=1)=\\{(x-\lambda_{1}),\ldots,(x-\lambda_{t})\\}$

of height 1 prime ideals of the algebra $\overline{K}\otimes_{K}\Lambda$ that
are generated by regular normal elements $x-\lambda_{1},\ldots,x-\lambda_{t}$
of the domain $\overline{K}\otimes_{K}\Lambda$ and the set
$\overline{K}^{\times}$ is the group of units of the algebra
$\overline{K}\otimes_{K}\Lambda$. So, we must have that

$\sigma(x)=\lambda x+\mu$

for some elements $\lambda\in\overline{K}^{\times}$ and $\mu\in\overline{K}$.
Since $K[x]=\Lambda\cap\overline{K}[x]$, we must have that

$\sigma(x)\in\sigma(\Lambda)\cap\sigma(\overline{K}[x])=\Lambda\cap\overline{K}[x]=K[x],$

and so $\lambda\in K^{\times}$ and $\mu\in K$. So, the automorphism $\sigma$
respects the polynomial algebra $K[x]$. In particular, it respects the Ore set
$S=K[x]\backslash\\{0\\}$ of the algebra $\Lambda$. The automorphism $\sigma$
can be uniquely extended to an automorphism of the algebra
$B_{1}=S^{-1}\Lambda$. Then $\sigma(\partial)=\lambda^{-1}\partial+q$ for some
element $q\in K(x)$. In particular,

$\sigma(y)=\sigma(f\partial)=\sigma(f)(\lambda^{-1}\partial+q)=\lambda^{-1}\frac{\sigma(f)}{f}y+p\;\;{\rm
where}\;\;p:=\sigma(f)q\in K[x]$

and $\sigma(f)=\gamma f$ for some element $\gamma\in K^{\times}$. Clearly,
$\gamma=\lambda^{d}$ where $d=\deg(f)$ is the degree of the polynomial $f$
(since $\sigma(x)=\lambda x+\mu$). So,

$\sigma(x)=\lambda x+\mu\;\;{\rm and}\;\;\sigma(y)=\lambda^{d-1}y+p,$

i.e. $\sigma\in\mathbb{S}(K)\rtimes G_{f}(K)$, as required. $\Box$.

By Theorem 1.2,

$G_{f}(K)=\mathbb{S}(K)\rtimes
G_{f}(K)=\\{\sigma_{\lambda,\mu,p}\,|\,\lambda\in K^{\times},\mu\in K,p\in
K[x]\\}$ (11)

where the multiplication and the inversion in the group $G_{f}(K)$ are given
by the rule (where $d=\deg(f)$):

$\displaystyle\sigma_{\lambda_{1},\mu_{1},p_{1}}\sigma_{\lambda_{2},\mu_{2},p_{2}}$
$\displaystyle=$
$\displaystyle\sigma_{\lambda_{1}\lambda_{2},\lambda_{2}\mu_{1}+\mu_{2},\lambda_{2}^{d-1}p_{1}+p_{2}},$
$\displaystyle\sigma_{\lambda,\mu,p}^{-1}$ $\displaystyle=$
$\displaystyle\sigma_{\lambda^{-1},-\lambda^{-1}\mu,-\lambda^{-d+1}p}.$

The algebra $B_{1}$ and its automorphism group. The element $f$ is a regular
normal element of $\Lambda$ (i.e. $\Lambda f=f\Lambda)$ since

$fy=yf-f^{\prime}f=(y-f^{\prime})f\;\;{\rm
where}\;\;f^{\prime}=\frac{df}{dx}.$

It determines the $K$-automorphism $\omega_{f}$ of the algebra $\Lambda$:

$fu=\omega_{f}(u)f,\;\;u\in\Lambda,$ $\omega_{f}:x\mapsto x,\;\;y\mapsto
y-f^{\prime}.$

We denote by $\Lambda_{f}$ and $A_{1,f}$ the localizations of the algebras
$\Lambda$ and $A_{1}$ at the powers of the element $f$, i.e.

$\Lambda_{f}=S^{-1}_{f}\Lambda\;\;{\rm and}\;\;A_{1,f}=S^{-1}_{f}A_{1}\;\;{\rm
where}\;\;S_{f}=\\{f^{i}\\}_{i\geq 0}.$

By (6),

$\Lambda\subset
A_{1}\subset\Lambda_{f}=A_{1,f}=K[x,f^{-1}][\partial;\frac{d}{dx}]\subset
B_{1}.$ (12)

Recall that ${\rm Aut}_{K}(K[x])=\\{\sigma_{\lambda,\mu}\,|\,\lambda\in
K^{\times},\mu\in K\\}$, $\sigma_{\lambda,\mu}(x)=\lambda x+\mu$ and

$\displaystyle{\rm Aut}_{K}(K(x))$ $\displaystyle=$
$\displaystyle\\{\sigma_{M}\,|\,M\in{\rm PGL}_{2}(K)\\}\simeq{\rm
PGL}_{2}(K),\;\sigma_{M}\rightarrow M\;{\rm
where}\;\sigma_{M}(x)=\frac{ax+b}{cx+d},$ $\displaystyle M$ $\displaystyle=$
$\displaystyle\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in{\rm PGL}_{2}(K):={\rm
GL}_{2}(K)/K^{\times}E\;{\rm and}\;E=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}.$

The maps

$\displaystyle{\rm Aut}_{K}(K[x])$ $\displaystyle\rightarrow$
$\displaystyle{\rm
Aut}_{K}(A_{1}),\;\;\sigma_{\lambda,\mu}\mapsto\sigma_{\lambda,\mu}:x\mapsto\lambda
x+\mu,\;\;\partial\mapsto\lambda^{-1}\partial,$ $\displaystyle{\rm
Aut}_{K}(K(x))$ $\displaystyle\rightarrow$ $\displaystyle{\rm
Aut}_{K}(B_{1}),\;\;\sigma_{M}\mapsto\sigma_{M}:x\mapsto\frac{ax+b}{cx+d},\;\;\partial\mapsto\frac{\sigma_{M}(f)}{f\sigma_{M}(x)^{\prime}}\partial,$

are group monomorphisms where $g^{\prime}=\frac{dg}{dx}$ for $g\in K(x)$. We
identify these groups with their images, i.e.

${\rm Aut}_{K}(K[x])\subseteq{\rm Aut}_{K}(A_{1})\;\;{\rm and}\;\;{\rm
Aut}_{K}(K(x))\subseteq{\rm Aut}_{K}(B_{1}).$

The automorphism group ${\rm Aut}_{K}(B_{1})$ acts in the obvious way on the
algebra $B_{1}$. Let

$\mathbb{S}_{1}:={\rm St}_{{\rm Aut}_{K}(B_{1})}(x):=\\{\sigma\in{\rm
Aut}_{K}(B_{1})\,|\,\sigma(x)=x\\},$

the stabilizer of the element $x\in B_{1}$ in ${\rm Aut}_{K}(B_{1})$. Clearly,

$\mathbb{S}_{1}=\\{s_{q}:\,|\,q\in K(x)\\}\simeq(K(x),+),\;\;s_{q}\mapsto q$

where $s_{q}(x)=x$ and $s_{q}(\partial)=\partial+q$.

###### Lemma 3.1

  1. 1.

${\rm Aut}_{K}(B_{1})=\mathbb{S}_{1}\rtimes{\rm
Aut}_{K}(K(x))=\\{\sigma_{M,q}:=s_{q}\sigma_{M}\,|\,M\in{\rm PGL}_{2}(K),q\in
K(x)\\}$.

  2. 2.

${\rm Aut}_{K}(B_{1},K[x]):=\\{\sigma\in{\rm
Aut}_{K}(B_{1})\,|\,\sigma(K[x])=K[x]\\}=\mathbb{S}_{1}\rtimes{\rm
Aut}_{K}(K[x])=\\{\sigma_{l,\mu,q}\,|\,\lambda\in K^{\times},\mu\in K,q\in
K(x)\\}$ where $\sigma_{l,\mu,q}(x)=\lambda x+\mu$ and
$\sigma_{\lambda,\mu,q}(\partial)=\lambda^{-1}\partial+q$.

Proof. 1. Since $\mathbb{S}_{1}:={\rm St}_{{\rm Aut}_{K}(B_{1})}(x)$, we must
have $\mathbb{S}_{1}\cap{\rm Aut}_{K}(B_{1})=\\{e\\}$ and
$\sigma\mathbb{S}_{1}\sigma^{-1}\subseteq\mathbb{S}_{1}$ for all automorphisms
$\sigma\in{\rm Aut}_{K}(B_{1})$. Hence, ${\rm
Aut}_{K}(B_{1})\supseteq\mathbb{S}_{1}\rtimes{\rm Aut}_{K}(K(x))$.

To prove that the reverse inclusion holds we have to show that every element
$\sigma\in{\rm Aut}_{K}(K(x))$ belongs to the group $\mathbb{S}_{1}\rtimes{\rm
Aut}_{K}(K(x))$. The group of units $K(x)^{\times}:=K(x)\backslash\\{0\\}$ of
the algebra $B_{1}$ is a $\sigma$-invariant set, i.e.
$\sigma(K(x)^{\times})=K(x)^{\times}$. Hence so is the field $K(x)$. Let
$\tau$ be the restriction of the automorphism $\sigma$ to the field $K(x)$.
Then $\sigma_{1}:=\tau^{-1}\sigma\in\mathbb{S}_{1}$, and so
$\sigma=\tau\sigma_{1}\in\mathbb{S}_{1}\rtimes{\rm Aut}_{K}(K(x))$, as
required.

2\. Statement 2 follows from statement 1. $\Box$

Below is a different proof of Theorem 1.1 is given.

Proof of Theorem 1.1. Let $\sigma:\Lambda(f)\rightarrow\Lambda(g)$ be an
isomorphism of the $K$-algebras. It can be uniquely extended to a
$\overline{K}$-isomorphism
$\sigma:\overline{K}\otimes_{K}\Lambda(f)\rightarrow\overline{K}\otimes_{K}\Lambda(g)$.
Let $\lambda_{1},\ldots,\lambda_{s^{\prime}}$ (resp.,
$\lambda_{1}^{\prime},\ldots,\lambda_{t}^{\prime}$) be the roots of the
polynomial $f$ (resp., $g$) in $\overline{K}$. By Theorem 2.3.(7), the
automorphism $\sigma$ maps bijectively the set
$\\{(x-\lambda_{1}),\ldots,(x-\lambda_{s^{\prime}})\\}$ of height 1 completely
prime ideals of the algebra $\overline{K}\otimes_{K}\Lambda(f)$ to the set
$\\{(x-\lambda_{1}^{\prime}),\ldots,(x-\lambda_{t}^{\prime})\\}$ of height 1
completely prime ideals of the algebra $\overline{K}\otimes_{K}\Lambda(g)$.
Therefore, $s^{\prime}=t$. Since the elements
$x-\lambda_{1}^{\prime},\ldots,x-\lambda_{t}^{\prime}$ are regular normal
elements of the domain $\overline{K}\otimes_{K}\Lambda(g)$ and the set
$\overline{K}^{\times}$ is the group of units of the algebra $\Lambda(g)$, we
must have that

$\sigma(x)=\lambda x+\mu$

for some elements $\lambda\in\overline{K}^{\times}$ and $\mu\in\overline{K}$.
Since $K[x]=\Lambda(g)\cap\overline{K}[x]$, we must have that
$\sigma(x)\in\sigma(\Lambda(f))\cap\sigma(\overline{K}[x])=\Lambda(g)\cap\overline{K}[x]=K[x]$,
and so $\lambda\in K^{\times}$ and $\mu\in K$. So, the isomorphism $\sigma$
respects the polynomial algebra $K[x]$ of the algebras $\Lambda(f)$ and
$\Lambda(g)$. In particular, it respects the Ore sets
$S=K[x]\backslash\\{0\\}$ of the algebras $\Lambda(f)$ and $\Lambda(g)$, i.e.
$\sigma(S)=S$. The isomorphism $\sigma$ can be uniquely extended to an
isomorphism

$\sigma:B_{1}=S^{-1}\Lambda(f)\rightarrow B_{1}=S^{-1}\Lambda(g).$

Then $\sigma(\partial)=\lambda^{-1}\partial+q$ for some element $q\in K[x]$.
In particular,

$\sigma(y)=\sigma(f\partial)=\sigma(f)(\lambda^{-1}\partial+q)=\lambda^{-1}\frac{\sigma(f)}{g}y+p\;\;{\rm
where}\;\;p:=\sigma(f)q\in K[x]$

and $\sigma(f)=\gamma g$ for some element $0\neq\gamma\in K[x]$. Applying the
same argument for the isomorphism
$\sigma^{-1}:\Lambda(g)\rightarrow\Lambda(f)$, we have that
$\sigma^{-1}(g)=\gamma_{1}f$ for some element $0\neq\gamma_{1}\in K[x]$.
Therefore,

$f=\sigma^{-1}\sigma(f)=\sigma^{-1}(\gamma g)=\sigma^{-1}(\gamma)\gamma_{1}f,$

and so $\gamma,\gamma_{1}\in K^{\times}$, and $\gamma_{1}=\gamma^{-1}$.
Clearly, $\gamma=\lambda^{d}$ where $d=\deg(f)$ is the degree of the
polynomial $f$ (since $\sigma(x)=\lambda x+\mu$), and the theorem follows.
Furthermore,

$\sigma(x)=\lambda x+\mu\;\;{\rm and}\;\;\sigma(y)=\lambda^{d-1}y+p.\;\;\Box$

##  4 The eigengroup $G_{f}$ of a polynomial $f$

For each nonscalar monic polynomial $f(x)$, Proposition 4.8, Theorem 4.18,
Theorem 4.20, Theorem 4.22 and Theorem 4.23 are explicit descriptions of the
eigengroup $G_{f}(K)$ in the case when the field $K$ is algebraically closed.
The case of an arbitrary field is obtained from these results, Theorem 4.24.
The aim of this section is to prove these results.

The eigengroup $G_{U}(K)$.

Definition, [5]. Given a group, a $G$-module $V$ over a field $K$ and a non-
empty subset $U$ of $V$. The eigengroup of the set $U$ in $G$, denoted by
$G_{U}(K)$, is the set of all elements of the group $G$ such that the elements
of the set $U$ are eigenvectors of them with eigenvalues in the field $K$.
Clearly, the eigengroup is a subgroup of $G$.

Clearly,

$G_{U}=\bigcap_{u\in U}G_{u}$

where $G_{u}:=G_{\\{u\\}}=\\{g\in G\,|\,gu=\lambda(g)u$ for some
$\lambda(g)\in K\\}$. If $K$ is a subfield of a field $K^{\prime}$ then
$G_{U}(K)\subseteq G_{U}(K^{\prime})$ where $U$ is a subset of the $G$-module
$K^{\prime}\otimes_{K}V$ over the field $K^{\prime}$.

Finite subgroups of ${\rm Aut}_{K}(K[x])$. Let $K$ be a field of prime
characteristic $p>0$, $\mathbb{F}_{p}=\mathbb{Z}/\mathbb{Z}p$ is the field
that contains $p$ elements, for each $n\geq 1$, $\mathbb{F}_{p^{n}}$ is the
finite field that contains $p^{n}$ elements,
$\overline{\mathbb{F}}_{p}=\bigcup_{n\geq 1}\mathbb{F}_{p^{n}}$ is the algebra
is closure of the field $\mathbb{F}_{p}$. Clearly,
$\overline{\mathbb{F}}_{p}\subseteq\overline{K}$ and group of roots of 1 in
the field $\overline{K}$ is
$\overline{\mathbb{F}}_{p}^{\times}:=\overline{\mathbb{F}}_{p}\backslash\\{0\\}$.
The group ${\rm Aut}_{K}(K[x])=\\{\sigma_{\lambda,\mu}\,|\,\lambda\in
K^{\times},\mu\in K\\}$ where $\sigma_{\lambda,\mu}(x)=\lambda x+\mu$ and

${\rm Aut}_{K}(K[x])\simeq{\rm Sh}(K)\rtimes\mathbb{T}\simeq K\rtimes
K^{\times}$

where ${\rm Sh}(K):=\\{\sigma_{1,\mu}\,|\,\mu\in K\\}\simeq(K,+)$,
$\sigma_{1,\mu}\mapsto\mu$ and
$\mathbb{T}:=\\{\sigma_{\lambda},0\\}\,|\,\lambda\in K^{\times}\\}\simeq
T^{\times}$, $\sigma_{\lambda,0}\mapsto\lambda$ is the the algebraic
1-dimensional torus.

The set $\mathbb{U}=\mathbb{U}(K)=K\cap\overline{\mathbb{F}}_{p}^{\times}$ is
the group of roots of 1 of the field $K$. The map
$\mathbb{U}\rightarrow\mathbb{T}$, $u\mapsto\sigma_{u,0}$ is a group
monomorphism and we identify the group $\mathbb{U}$ with its image, i.e.
$\mathbb{U}=\\{\sigma_{u,0}\,|\,u\in\mathbb{U}\\}$. Let ${\rm or}(g)$ be the
order of an element of a group $G$.

###### Lemma 4.1

The group ${\rm
Sh}(K)\rtimes\mathbb{U}(K)=\\{\sigma_{u,\mu}\,|\,u\in\mathbb{U}(K),\mu\in
K\\}$ is the set of all finite order automorphisms of the group ${\rm
Aut}_{K}(K[x])$. The order of the element
$\sigma_{u,\mu}=\sigma_{u,0}\sigma_{1,\mu}$ is

${\rm or}(\sigma_{\lambda,\mu})=\begin{cases}{\rm or}(\lambda)&\text{if
}\lambda\neq 1,\\\ p&\text{if }\lambda=1,\mu\neq 0,\\\ 1&\text{if
}\lambda=1,\mu=0.\\\ \end{cases}$

Proof. For all $\lambda\in K^{\times}\backslash\\{1\\}$ and $\mu\in K$,
$\sigma_{1,\mu}^{i}=\sigma_{1,i\mu}$ and
$\sigma_{\lambda,\mu}^{i}=\sigma_{\lambda^{i},\frac{1-\lambda^{i}}{1-\lambda}\mu}$,
and statement 1 follows. $\Box$

By Lemma 4.1,

$1\rightarrow{\rm Sh}(K)\rightarrow{\rm
Sh}(K)\rtimes\mathbb{U}(K)\stackrel{{\scriptstyle\varphi}}{{\rightarrow}}\mathbb{U}(K)\rightarrow
1,\;\;{\rm where}\;\;\varphi(\sigma_{\lambda,\mu})=\lambda,$ (13)

is a short exact sequence of group homomorphisms.

###### Lemma 4.2

If an element $\lambda\in\mathbb{U}(\overline{K})$ is a primitive $n$’th root
of unity then $\mathbb{F}_{p}(\lambda)=\mathbb{F}_{p^{m}}$ where
$m=\min\\{k\geq 1\,|\;n|(p^{k}-1)\\}=$ the degree of the minimal polynomial of
$\lambda$ over the field $\mathbb{F}_{p}$.

Proof. The element $\lambda$ is algebraic over the field $\mathbb{F}_{p}$. Let
$\varphi$ be its minimal polynomial over $\mathbb{F}_{p}$. Then the field
$\mathbb{F}_{p}(\lambda)\simeq\mathbb{F}_{p}[x]/(\varphi)$ is a finite field,
and so $\mathbb{F}_{p}(\lambda)=\mathbb{F}_{p^{m}}$ for some $m\geq 1$. Now,

$\deg(\varphi)=[\mathbb{F}_{p}(\lambda):\mathbb{F}_{p}]=[\mathbb{F}_{p^{m}}:\mathbb{F}_{p}]=m.$

Notice that the order of the group $\langle\lambda\rangle$, which is $n$
(since $\lambda$ is a primitive $n$’th root of unity) divides the order of the
group $\mathbb{F}_{p^{m}}^{\times}$, which is $p^{m}-1$. Clearly, $m\geq
m^{\prime}:=\min\\{k,|\,n|(p^{k}-1)\\}$. We claim that $m=m^{\prime}$. Suppose
that $m>m^{\prime}$, wee seek a contradiction. Then
$\lambda^{p^{m^{\prime}}}=\lambda$, and so
$\lambda\in\mathbb{F}_{p^{m^{\prime}}}$, hence
$F_{p^{m}}=\mathbb{F}_{p}(\lambda)\subseteq\mathbb{F}_{p^{m^{\prime}}}$.
Therefore, $m|m^{\prime}$, a contradiction. $\Box$

The next theorem is a classification of all the finite subgroups of the
automorphism group ${\rm Aut}_{K}(K[x])$.

###### Theorem 4.3

Let $G$ be a finite subgroup of ${\rm Aut}_{K}(K[x])$. Then
$G=\widetilde{G}\rtimes\overline{G}$ where $\widetilde{G}=G\cap{\rm
Sh}(K)=\\{\sigma_{1,\mu}\,|\,\mu\in V\\}$, $V\subseteq K$ is a finite
dimensional $\mathbb{F}_{p}(\lambda_{n})$-subspace of $K$ and
$\overline{G}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ is a
cyclic group of order $n$ where $\lambda_{n}$ is a primitive $n$’th root of 1
and $\nu\in K$. In particular, the order of the group $G$ is $np^{l}$ where
$l={\rm dim}_{\mathbb{F}_{p}}(V)$ such that $m|l$ where
$\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for some $m\geq 1$ (Lemma
4.2). Conversely, given a finite dimensional
$\mathbb{F}_{p}(\lambda_{n})$-subspace $V$ of $K$, an automorphism
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}$ where $\lambda_{n}\in K$ is a
primitive $n$’th root of unity and $\nu\in K$. Let
$\widetilde{G}:=\\{\sigma_{1,\mu}\,|\,\mu\in V\\}$ and
$\overline{G}:=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$. Then
the semidirect product $\widetilde{G}\rtimes\overline{G}$ is a finite subgroup
of ${\rm Aut}_{K}(K[x])$ of order $np^{l}$ where $l={\rm
dim}_{\mathbb{F}_{p}}(V)$ such that $m|l$. The element $\nu\in K$ is unique up
to adding an arbitrary element of $V$, i.e.
$\widetilde{G}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\simeq\widetilde{G}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu^{\prime}}\rangle$
iff $\nu^{\prime}-\nu\in V$. Furthermore,
$G=\\{\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{i}\sigma_{1,v}\,|\,0\leq i\leq
n-1,v\in V\\}$ and
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{i}\sigma_{1,v}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu}\sigma_{1,v}:x\mapsto\lambda_{n}^{i}x+(1-\lambda_{n}^{i})\nu+v$.

Proof. Let $G$ be a finite subgroup of ${\rm Aut}_{K}(K[x])$. Then, by (13),
the group $\varphi(G)$ is a finite subgroup of $\mathbb{U}(K)$ of order $n$,
hence $\varphi(G)=\langle\lambda_{n}\rangle$ where $\lambda_{n}$ is a
primitive $n$’th root of 1. Fix an element, say
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\in G$ where $\nu\in K$, such that
$\varphi(\sigma_{\lambda_{n},(1-\lambda_{n})\nu})=\lambda_{n}$. Then
$\overline{G}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ is a
cyclic group of order $n=|\overline{G}|$ since
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{i}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu}$
for all $i\geq 1$. Therefore,

$G=\widetilde{G}\rtimes\overline{G}\;\;{\rm where}\;\;\widetilde{G}:=G\cap{\rm
Sh}(K)=\\{\sigma_{1,\mu}\,|\,\mu\in V\\},$

$V\subseteq K$ is a finite dimensional $\mathbb{F}_{p}$-subspace of $K$ since
$\sigma_{1,\mu}^{i}=\sigma_{1,i\mu}$ for all $i\geq 0$. Furthermore,
$\lambda_{n}V\subseteq V$, i.e. the $\mathbb{F}_{p}$-vector space $V$ is a
$\mathbb{F}_{p}(\lambda)$-module since

$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{-1}\sigma_{1,\mu}\sigma_{\lambda_{n},(1-\lambda_{n})\nu}=\sigma_{1,\lambda\mu}.$

Clearly, $|G|=|\widetilde{G}||\overline{G}|=p^{l}n$ and $m|l$ since $V$ is a
$\mathbb{F}_{p^{m}}$-module and ${\rm
dim}_{\mathbb{F}_{p^{m}}}(V)=\frac{l}{m}$.

The converse, is obvious.

Clearly,
$\widetilde{G}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\simeq\widetilde{G}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu^{\prime}}\rangle$
iff there is natural number $i$ such that $1\leq i\leq n-1$ and a vector $v\in
V$ such that

$\sigma_{\lambda_{n},(1-\lambda_{n})\nu^{\prime}}=\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{i}\sigma_{1,v}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})(\nu+(1-\lambda_{n}^{i})^{-1}v)}$

iff $i=1$ and $\nu^{\prime}=\nu+(1-\lambda_{n}^{i})^{-1}v\in V$ iff
$\nu^{\prime}-\nu\in V$ since $(1-\lambda_{n}^{i})^{-1}V=V$. $\Box$

The automorphism group ${\rm Aut}_{K}(K[x])$ acts on the set ${\rm Max}(K[x])$
of maximal ideals of $K[x]$ in the obvious way. If $K=\overline{K}$ then ${\rm
Max}(K[x])=\\{(x-\gamma)\,|\,\gamma\in K\\}$ and the action takes the form:
For all $\sigma\in{\rm Aut}_{K}(K[x])$ and $\gamma\in K$,

$\sigma((x_{\gamma}))=(x-\sigma^{-1}(\gamma))\;\;{\rm
where}\;\;\sigma^{-1}(\gamma):=\sigma^{-1}(x)|_{x=\gamma}.$ (14)

Let us identify the set ${\rm Max}(K[x])$ with $K$ via
$(x-\gamma)\mapsto\gamma$. Then the action of the group ${\rm Aut}_{K}(K[x])$
on ${\rm Max}(K[x])=K$ is given below:

${\rm Aut}_{K}(K[x])\times K\rightarrow
K,\;\;(\sigma,\gamma)\mapsto\sigma*\gamma:=\sigma^{-1}(\gamma)=\sigma^{-1}(x)|_{x=\gamma}.$
(15)

If $\sigma=\sigma_{\lambda,\mu}$ then
$\sigma_{\lambda,\mu}^{-1}=\sigma_{\lambda^{-1},-\lambda^{-1}\mu}$ and
$\sigma_{\lambda,\mu}*\gamma=\sigma_{\lambda,\mu}^{-1}(\gamma)=\sigma_{\lambda^{-1},-\lambda^{-1}\mu}(\gamma)=\lambda^{-1}\gamma-\lambda^{-1}\mu$.

Every automorphism $\sigma_{\lambda\mu}\in{\rm Aut}_{K}(K[x])$ with
$\lambda\neq 1$ can be uniquely written in the form
$\sigma_{\lambda,(1-\lambda)\nu}$ where $\nu=(1-\lambda)^{-1}\mu$. Notice that

$\sigma_{\lambda,(1-\lambda)\nu}*(\nu)=\nu.$

Furthermore, the set $\\{\nu\\}$ is the only 1-element orbit in $K$ of the
cyclic group $\langle\sigma_{\lambda,(1-\lambda)\nu}\rangle$ generated by the
automorphism $\sigma_{\lambda,(1-\lambda)\nu}$. The number of elements in any
other orbit is equal to the order of the group
$\langle\sigma_{\lambda,(1-\lambda)\nu}\rangle$ which is the order of the
element $\lambda$ in the group $(K^{\times},\cdot)$.

Suppose that $K=\overline{K}$. Let $f\in K[x]$ be a nonscalar monic polynomial
that has at least two distinct roots in $K=\overline{K}$. Recall that ${\cal
R}(f)$ is the set of all roots of the polynomial $f$ counted with multiplicity
and ${\cal R}_{d}(f)$ be the set of all distinct roots of $f$ (i.e. each root
has multiplicity 1). Example. For $f=(x-1)^{2}(x-2)^{3}$, ${\cal
R}(f)=\\{1,1,2,2,2\\}$ and ${\cal R}(f)=\\{1,2\\}$.

The group $G_{f}$ permutes the roots in ${\cal R}(f)$ and ${\cal R}_{d}(f)$
via the action (14). Let us stress that the action of $G_{f}$ on ${\cal R}(f)$
respects the multiplicity. If the group $G_{f}$ is finite then
$G_{f}=\widetilde{G}_{f}\rtimes\overline{G}_{f}$ and $\widetilde{G}_{f}={\rm
Sh}_{V}$ is a normal subgroup of $G_{f}$. For a set ${\cal R}={\cal
R}(f),{\cal R}_{d}(f)$ and a group
$G=G_{f},\widetilde{G}_{f},\overline{G}_{f}$, we denote by ${\cal R}/G$ is the
set of $G$-orbits in ${\cal R}$.

Invariants and eigenalgebras of finite subgroups of ${\rm Aut}_{K}(K[x])$.
Notice that

$x^{p}-x=\prod_{i\in\mathbb{F}_{p}}(x-i).$ (16)

For each element $\mu\in K^{\times}$, let

$f_{\mu}(x):=\prod_{i=0}^{p-1}\sigma_{1,\mu}^{i}(x)=\prod_{i=0}^{p-1}(x-i\mu)=\prod_{i\in\mathbb{F}_{p}}(x-i\mu)=x^{p}-\mu^{p-1}x.$
(17)

The equality above follows at once from (16):
$f_{\mu}(x)=\prod_{i=0}^{p-1}(x-i\mu)=\mu^{p}\prod_{i=0}^{p-1}(\mu^{-1}x-i)=\mu^{p}((\mu^{-1}x)^{p}-\mu^{-1}x)=x^{p}-\mu^{p-1}x$.
For all $\alpha,\beta\in\mathbb{F}_{p}$:

$f_{\mu}(\alpha x+\beta x^{\prime})=\alpha f_{\mu}(x)+\beta
f_{\mu}(x^{\prime})$

since $\gamma^{p}=\gamma$ for all $\gamma\in\mathbb{F}_{p}$. In particular,
the map $K\rightarrow K$, $\lambda\mapsto f(\lambda)$ is a
$\mathbb{F}_{p}$-linear map. Hence for all elements $\lambda\in K$,

$x^{p}-\mu^{p-1}x-(\lambda^{p}-\mu^{p-1}\lambda)=f_{\mu}(x)-f(\lambda)=f_{\mu}(x-\lambda)=\prod_{i=0}^{p-1}(x-\lambda-i\mu)=\prod_{i\in\mathbb{F}_{p}}(x-\lambda-i\mu).$
(18)

Given a $K$-algebra $A$, and a subgroup $G$ of the automorphism group ${\rm
Aut}_{K}(A)$. A group homomorphism $\chi:G\rightarrow K^{\times}$ is called a
character of the group $G$ in $K$. Let $\widehat{G}(K)$ be the
(multiplicative) group of characters of the group $G$ in $K^{\times}$ . The
multiplication in the group $\widehat{G}(K)$ is given by the rule: For all
$\chi,\psi\in\widehat{G}(K)$, $(\chi\psi)(g)=\chi(g)\psi(g)$ for all elements
$g\in G$.

Definition. The direct sum of $G$-eigenspaces,

$\mathbb{E}(A)=\mathbb{E}(A,G):=\bigoplus_{\chi\in\widehat{G}(K)}A^{\chi},\;\;{\rm
where}\;\;A^{\chi}:=\\{a\in A\,|\,g(a)=\chi(g)a\;\;{\rm for\;all}\;\;g\in
G\\},$

is called the $G$-eigenalgebra of $A$. The direct sum is a
$\widehat{G}(K)$-graded algebra since $A^{\chi}A^{\psi}\subseteq A^{\chi\psi}$
for all $\chi,\psi\in\widehat{G}(K)$.

If $e$ is the identity element of the character group $\widehat{G}(K)$ then
$A^{e}=A^{G}$ is the algebra of $G$-invariants. In particular
$A^{G}\subseteq\mathbb{E}(A,G)$.

The set ${\rm Supp}(A,G):=\\{\chi\in\widehat{G}(K)\,|\,A^{\chi}\neq 0\\}$ is
called the support of $G$ in $A$. For each character $\chi\in{\rm Supp}(A,G)$,
the vector space $A^{\chi}$ is called the $\chi$-weight/eigenvalue subspace of
the algebra $A$. If the algebra $\mathbb{E}(A,G)$ is a domain (eg, the algebra
$A$ is a domain) then the support ${\rm Supp}(A,G)$ is a submonoid of
$\widehat{G}(K)$ and the algebra $\mathbb{E}(A,G)$ is a ${\rm
Supp}(A,G)$-graded algebra. If the algebra $A$ is a commutative algebra then
the Frobenius endomorphism ${\rm Fr}:A\rightarrow A$, $a\mapsto a^{p}$ is a
$\mathbb{F}_{p}$-algebra endomorphism of $A$. It is a monomorphism if the
algebra $A$ is a domain. By the very definition, the Frobenius endomorphism
commute with all endomorphisms of the ring $A$.

###### Lemma 4.4

Let $A$ be a commutative $K$-algebra and $G$ be a subgroup of the automorphism
group ${\rm Aut}_{K}(A)$.

  1. 1.

The algebras $\mathbb{E}(A,G)$ and $A^{G}$ are ${\rm Fr}$-stable (that is
${\rm Fr}(\mathbb{E}(A,G))\subseteq\mathbb{E}(A,G)$ and ${\rm
Fr}(A^{G})\subseteq A^{G}$).

  2. 2.

Suppose that the algebra $A$ is reduced and ${\rm Fr}(K)=K$. If $g({\rm
Fr}(a))=\chi(g){\rm Fr}(a)$ for all $g\in G$ then
$g(a)=(\chi(g))^{\frac{1}{p}}a$. In particular, ${\rm Fr}\in{\rm
Aut}_{\mathbb{F}_{p}}(\mathbb{E}(A,G))$ and ${\rm Fr}\in{\rm
Aut}_{\mathbb{F}_{p}}(A^{G})$.

Proof. 1. The Frobenius endomorphism commutes with all endomorphisms of the
ring $A$, and statement 1 follows.

2\. The equality ${\rm Fr}(K)=K$ implies that ${\rm Fr}\in{\rm
Aut}_{\mathbb{F}_{p}}(K)$. Since the algebra $A$ is reduced the Frobenius
endomorphism $A$ is a monomorphism. If $g({\rm Fr}(a))=\chi(g){\rm Fr}(a)$ for
all $g\in G$ then $(g(a)-(\chi(g))^{\frac{1}{p}}a)^{p}=0$, and so
$g(a)=(\chi(g))^{\frac{1}{p}}a$ for all $g\in G$, and statement 2 follows.
$\Box$

Let $V\subseteq K$ be a $\mathbb{F}_{p^{m}}$-subspace of $K$. The subgroup
${\rm Sh}_{V}:=\\{\sigma_{1,v}\,|\,v\in V\\}$ of ${\rm Aut}_{K}(K[x])$ is
called the shift group that is determined by the $\mathbb{F}_{p^{m}}$-subspace
$V$. Proposition 4.5 describes the algebra of invariants and the eigenalgebra
of the shift group ${\rm Sh}_{V}$.

###### Proposition 4.5

Let $V\subseteq K$ be a nonzero $\mathbb{F}_{p^{m}}$-subspace of $K$. Then
$\mathbb{E}(K[x],{\rm Sh}_{V})=K[x]^{{\rm Sh}_{V}}$.

  1. 1.

If ${\rm dim}_{\mathbb{F}_{p^{m}}}(V)=\infty$ then $K[x]^{{\rm Sh}_{V}}=K$.

  2. 2.

If $l={\rm dim}_{\mathbb{F}_{p^{m}}}(V)<\infty$ and
$\\{\mu_{1},\ldots,\mu_{l}\\}$ is a basis of the vector space $V$ over
$\mathbb{F}_{p^{m}}$ then

    1. (a)

the fixed algebra $K[x]^{{\rm Sh}_{V}}=K[f_{V}]$ is a polynomial algebra in
$f_{V}:=\prod_{v\in V}(x-v)$, the polynomial $f_{V}$ is divisible by the
polynomial $\prod_{i=1}^{l}f_{\mu_{i}}$,

    2. (b)

for all elements $\alpha,\beta\in\mathbb{F}_{p^{m}}$ and $\lambda\in K$,
$f_{V}(\alpha x+\beta\lambda)=\alpha f_{V}(x)+\beta f_{V}(\lambda)$. In
particular, the map $K\rightarrow K$, $\lambda\mapsto f_{V}(\lambda)$ is a
$\mathbb{F}_{p^{m}}$-linear map,

    3. (c)

If $V\subset V^{\prime}$ are distinct $\mathbb{F}_{p^{m}}$-subspaces of $K$
then $f_{V}|f_{V^{\prime}}$ and $f_{V}\neq f_{V^{\prime}}$, and

    4. (d)

$\frac{df_{V}}{dx}\neq 0$.

  3. 3.

In particular, for all elements $\mu\in K^{\times}$,
$K[x]^{\sigma_{1,\mu}}=K[f_{\mu}]$.

Proof. For all elements $\mu\in K$, the map

$\sigma_{1,\mu}-1:K[x]\rightarrow K[x],\;\;\psi(x)\mapsto\psi(x+\mu)-\psi(x)$

is locally nilpotent map. Therefore, the element 1 is the only eigenvalue for
the map $\sigma_{1,\mu}$, and so $\mathbb{E}(K[x],{\rm Sh}_{V})=K[x]^{{\rm
Sh}_{V}}$.

1\. Statement 1 follows from statement 2. Let $\\{\mu_{i}\\}_{i\in\mathbb{N}}$
be an $\mathbb{F}_{p^{m}}$-linearly independent elements of the vector space
$V$ and $V_{i}=\bigoplus_{j=1}^{i}\mathbb{F}_{p^{m}}\mu_{i}$. Then
$V_{1}\subset V_{2}\subset\cdots\subset V_{\infty}:=\bigoplus_{i\geq
1}\mathbb{F}_{p^{m}}\mu_{i}\subseteq V$. Hence,

$K[x]^{{\rm Sh}_{V_{1}}}\supseteq K[x]^{{\rm
Sh}_{V_{2}}}\supseteq\cdots\supseteq K[x]^{{\rm
Sh}_{V_{\infty}}}=\bigcap_{i\geq 1}K[x]^{{\rm Sh}_{V_{i}}}=\bigcap_{i\geq
1}K[f_{V_{i}}]=K\supseteq K[x]^{{\rm Sh}_{V}}\supseteq K,$

and so $K[x]^{{\rm Sh}_{V}}=K$.

2.(a,b). For all elements $v^{\prime}\in V$,

$\sigma_{1,v^{\prime}}(f_{V})=\prod_{v\in V}(x-v^{\prime}-v)=f_{V}.$

Therefore, $K[x]^{{\rm Sh}_{V}}\supseteq K[f_{V}]$. By (17), the polynomial
$\prod_{i=1}^{l}\prod_{u\in\mathbb{F}_{p}}(x-u\mu_{i})=\prod_{i=1}^{l}f_{\mu_{i}}$
is a divisor of the polynomial $f_{V}$. In particular, $f_{V}(0)=0$.

First, we prove that the statements (a) and (b) hold in the case when
$K=\overline{K}$ and then we deduce that the statements (a) and (b) hold for
an arbitrary field $K$.

So, suppose that $K=\overline{K}$. Let $g(x)\in K[x]^{{\rm Sh}_{V}}$ be a
noscalar monic polynomial. Let $\gamma$ be a root of $g(x)$. Then for all
elements $v\in V$, the element $\gamma+v$ is also a root of the polynomial
$g(x)$. So, the set of all roots of the polynomial $g(x)$ is a disjoin union
of the sets $\coprod_{i=1}^{s}\\{\gamma_{i}+V\\}$ for some roots $\gamma_{i}$
of $g(x)$.

Therefore, the polynomial

$g(x)=\prod_{i=1}^{s}\prod_{v\in
V}(x-\gamma_{i}-v)=\prod_{i=1}^{s}f_{V}(x-\gamma_{i})$

is a product of ${\rm Sh}_{V}$-invariant polynomials $f_{V}(x-\gamma_{i})$ of
the same degree $p^{lm}$. In particular, every nonscalar ${\rm
Sh}_{V}$-invariant polynomials has degree at least $p^{lm}$. Therefore, for
all elements $\lambda,\mu\in K$, the difference

$c_{\lambda,\mu}:=f_{V}(x+\lambda)-f_{V}(x+\mu)$

of two monic ${\rm Sh}_{V}$-invariant polynomials of degree $p^{lm}$ must be a
constant which is equal to $f_{V}(\lambda)-f_{V}(\mu)$. Therefore,

$f_{V}(x+\lambda)-f_{V}(\lambda)=f_{V}(x+\mu)-f_{V}(\mu).$

When $\mu=0$, we have that

$f_{V}(x+\lambda)=f_{V}(x)+f_{V}(\lambda)-f_{V}(0)=f_{V}(x)+f_{V}(\lambda)$

since $f_{V}(0)=0$. Since for all elements $u\in\mathbb{F}_{p^{m}}^{\times}$,

$f_{V}(ux)=u^{p^{lm}}\prod_{v\in V}(x-u^{-1}v)=u\prod_{v\in
V}(x-v)=uf_{V}(x),$

and $f_{V}(0x)=0=0f_{V}(x)$, we see that for all elements
$\xi\in\mathbb{F}_{p^{m}}$, $f_{V}(\xi x)=\xi f_{V}(x)$. Now, the statement
(b) follows.

Now, the polynomial

$g(x)=\prod_{i=1}^{s}f_{V}(x-\gamma_{i})=\prod_{i=1}^{s}(f_{V}(x)-f_{V}(\gamma_{i}))\in
K[f_{V}(x)]$

and the statement (a) follows.

Suppose that $K$ is not necessarily algebraically closed field and $g(x)\in
K[x]^{{\rm Sh}_{V}}$ be a noscalar monic polynomial. Then
$g(x)\in\overline{K}[f_{V}(x)]$. Since the $\overline{K}=K\oplus W$ for some
$K$-subspace of $K$ and $f_{V}(x)\in K[x]$, we must have that $g(x)\in
K[f_{V}(x)]$ since

$\overline{K}[f_{V}(x)]=K[f_{V}(x)]\oplus\bigoplus_{i\geq
0}Wf_{V}(x)^{i}\subseteq K[x]\oplus\bigoplus_{i\geq 0}Wf_{V}(x)^{i}.$

Now, the statements (a) and (b) hold for the field $K$.

(c) The statement (c) follows from the statement (a).

(d) WLOG we may assume that $K=\overline{K}$. Suppose that
$\frac{df_{V}}{dx}=0$. Then $f_{V}=g^{p}$ for some polynomial $g$. This is not
possible as every root of $f$ has multiplicity 1.

3\. Statement 3 is a particular case of statement 2. $\Box$

Notice that for all natural numbers $m\geq 1$,

$x^{p^{m}}-x=\prod_{i\in\mathbb{F}_{p^{m}}}(x-i).$ (19)

By (19), for each element $\mu\in K^{\times}$, let

$f_{p^{m},\mu}(x):=f_{\mathbb{F}_{p^{m}}\mu}(x)=\prod_{i\in\mathbb{F}_{p^{m}}}(x-i\mu)=x^{p^{m}}-\mu^{p^{m}-1}x.$
(20)

For all $\alpha,\beta\in\mathbb{F}_{p^{m}}$:

$f_{p^{m},\mu}(\alpha x+\beta x^{\prime})=\alpha f_{p^{m},\mu}(x)+\beta
f_{p^{m},\mu}(x^{\prime})$ (21)

since $\gamma^{p^{m}}=\gamma$ for all $\gamma\in\mathbb{F}_{p^{m}}$. In
particular, the map $K\rightarrow K$, $\lambda\mapsto f_{p^{m},\mu}(\lambda)$
is a $\mathbb{F}_{p^{m}}$-linear map. Hence, for all elements $\lambda\in K$,

$x^{p^{m}}-\mu^{p^{m}-1}x-(\lambda^{p^{m}}-\mu^{p^{m}-1}\lambda)=f_{p^{m},\mu}(x)-f_{p^{m},\mu}(\lambda)=f_{p^{m},\mu}(x-\lambda)=\prod_{i\in\mathbb{F}_{p^{m}}}(x-\lambda-i\mu).$
(22)

Theorem 4.6 describes the algebra of invariants and the eigenalgebra of a
‘generic’ finite subgroup of ${\rm Aut}_{K}(K[x])$.

###### Theorem 4.6

Let $G=\widetilde{G}\rtimes\overline{G}$ be a finite subgroup of ${\rm
Aut}_{K}(K[x])$ (Theorem 4.3) where
$\overline{G}:=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ and
$\widetilde{G}:=\\{\sigma_{1,\mu}\,|\,\mu\in V\\}$, $\lambda_{n}\in K$ is a
primitive $n$’th root of unity, $n\geq 2$ and $\nu\in K$, $V$ is a nonzero
finite dimensional $\mathbb{F}_{p}(\lambda_{n})$-subspace of $K$, and
$\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for some $m\geq 1$ (Lemma
4.2). Then

  1. 1.

$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(f_{V}(x-\nu))=\lambda_{n}f_{V}(x-\nu)$.

  2. 2.

$K[x]^{G}=K[f_{V}^{n}(x-\nu)]$ is a polynomial algebra in
$f_{V}^{n}(x-\nu):=\Big{(}f_{V}(x-\nu)\Big{)}^{n}$.

  3. 3.

The $G$-eigenvalue subalgebra of $K[x]$ is
$\mathbb{E}(K[x],G)=\bigoplus_{i=0}^{n-1}f_{V}^{i}(x-\nu)K[x]^{G}$, a direct
sum of distinct $G$-eigenspaces.

Proof. 1. Let $\sigma=\sigma_{\lambda_{n},(1-\lambda_{n})\nu}$. Suppose that
$l={\rm dim}_{\mathbb{F}_{p^{m}}}(V)$. Then $\deg(f_{V}(x))=p^{lm}$. Now, by
Proposition 4.5.(2b),

$\sigma(f_{V}(x-\nu))=f_{V}(\sigma(x-\nu))=f_{V}(\lambda_{n}(x-\nu))=\lambda_{n}f_{V}(x-\nu)$

since $\lambda_{n}\in K(\lambda_{n})=\mathbb{F}_{p^{m}}$.

2 and 3. By Proposition 4.5.(2b), the $\widetilde{G}$-eigenvalue subalgebra of
$K[x]$ is the fixed algebra $K[x]^{\widetilde{G}}=K[f_{V}(x)]=K[f_{V}(x-\nu)]$
since

$f_{V}(x)=f_{V}(x-\nu+\nu)=f_{V}(x-\nu)+f_{V}(\nu).$

By statement 1, $\sigma(f_{V}(x-\nu))=\lambda_{n}f_{V}(x-\nu)$, hence the
$\widetilde{G}$-eigenvalue subalgebra of $K[x]$, $K[f_{V}(x-\nu)]$, is
$\sigma$-invariant, and statements 2 and 3 follow. $\Box$

Proposition 4.7 describes the algebra of invariants and the eigenalgebra of
the subgroup $G=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ of
${\rm Aut}_{K}(K[x])$ where $\lambda_{n}\in K$ is a primitive $n$’th root of
unity, $n\geq 2$ and $\nu\in K$.

###### Proposition 4.7

Let $G=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ be a finite
subgroup of ${\rm Aut}_{K}(K[x])$ where $\lambda_{n}\in K$ is a primitive
$n$’th root of unity, $n\geq 2$ and $\nu\in K$. Then

  1. 1.

$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(x-\nu)=\lambda_{n}(x-\nu)$.

  2. 2.

$K[x]^{G}=K[(x-\nu)^{n}]$ is a polynomial algebra in $(x-\nu)^{n}$.

  3. 3.

$\mathbb{E}(K[x],G)=K[x]=\bigoplus_{i=0}^{n-1}(x-\nu)^{i}K[x]^{G}$ is a direct
sum of distinct $G$-eigenspaces.

Proof. 1. Statement 1 is obvious.

2\. Statement 2 follows from statement 1 and the fact that $\lambda_{n}$ is a
primitive $n$’th root of unity.

3\. Statement 3 follows from statement 3. $\Box$

The eigengroup $G_{f}(K)$ of a polynomial $f\in K[x]$ that has single root in
$\overline{K}$. For an element $\nu\in K$, the subset

$\mathbb{T}_{\nu}(K):=\\{\sigma_{\lambda,(1-\lambda)\nu}\,|\,\lambda\in
K^{\times}\\}$ (23)

of ${\rm Aut}_{K}(K[x])$ is a subgroup which is isomorphic to the algebraic
torus $\mathbb{T}=(K^{\times},\cdot)$ via

$\mathbb{T}_{\nu}(K)\rightarrow\mathbb{T},\;\;\sigma_{\lambda,(1-\lambda)\nu}\mapsto\lambda$

since
$\sigma_{\lambda,(1-\lambda)\nu}\sigma_{\lambda^{\prime},(1-\lambda^{\prime})\nu}=\sigma_{\lambda\lambda^{\prime},(1-\lambda)\nu}$
for all $\lambda,\lambda^{\prime}\in K^{\times}$.

Suppose that a monic nonscalar polynomial $f\in\overline{K}[x]$ of degree $d$
has single root, say $\nu\in\overline{K}$. Then $d=p^{r}d_{1}$ where for
unique natural numbers $r$ and $d_{1}$ such that $p\nmid d_{1}$. Then

$f=(x-\nu)^{d}=(x-\nu)^{p^{r}d_{1}}=(x^{p^{r}}-\nu^{p^{r}})^{d_{1}}=x^{p^{r}d_{1}}-d_{1}\nu^{p^{r}}x^{p^{r}(d_{1}-1)}+\cdots.$
(24)

Therefore, $f=(x-\nu)^{d}\in K[x]$ iff $\nu^{p^{r}}\in K$ (since $p\nmid
d_{1}$).

The next proposition describes all the monic polynomial $f\in K[x]$ such that
$G_{f}(K)=\mathbb{T}_{\nu}(K)$ for some $\nu\in K$. Furthermore, it describes
the group $G_{f}$ for all polynomial $f\in K[x]$ that has a single root in
$\overline{K}$.

###### Proposition 4.8

  1. 1.

Let $f(x)\in\overline{K}[x]$ be a monic nonscalar polynomial of degree $d$.
Then $f(x)=(x-\nu)^{d}$ for some $\nu\in\overline{K}$ iff
$G_{f}(\overline{K})=\mathbb{T}_{\nu}(\overline{K})$.

  2. 2.

Let $f(x)\in K[x]$ be a monic nonscalar polynomial of degree $d$ that has a
single root $\nu\in\overline{K}$. Then
$G_{f}(K)=\begin{cases}\mathbb{T}_{\nu}(K)&\text{if }\nu\in K,\\\
\\{e\\}&\text{if }\nu\not\in K.\\\ \end{cases}$

  3. 3.

Let $f(x)\in K[x]$ be a monic nonscalar polynomial of degree $d$ that has a
single root $\nu\in\overline{K}$. Then $f(x)=(x-\nu)^{d}$ for some $\nu\in K$
iff $G_{f}(K)=\mathbb{T}_{\nu}(K)$.

Proof. 1. $(\Rightarrow)$ Suppose that $f=(x-\nu)^{d}$. An automorphism
$\sigma\in{\rm Aut}_{K}(K[x])$ belongs to the group $G_{f}(\overline{K})$ iff
$\sigma(x-\nu)=\lambda(x-\nu)$ for some element
$\lambda\in\overline{K}^{\times}$ iff $\sigma=\sigma_{\lambda,(1-\lambda)\nu}$
iff $G_{f}=\mathbb{T}_{\nu}(\overline{K})$.

$(\Leftarrow)$ Suppose that $G_{f}=\mathbb{T}_{\nu}(\overline{K})$. Then
$K[x]=\bigoplus_{i\geq 0}K(x-\nu)^{i}$ is a direct sum of the eigenspaces of
the group $\mathbb{T}_{\nu}(\overline{K})$. Therefore, $f(x)=(x-\nu)^{d}$ for
some $d\geq 1$.

2\. Clearly, $G_{f}(K)=G_{f}(\overline{K})\cap{\rm
Aut}_{K}(K[x])=\mathbb{T}_{\nu}(\overline{K})\cap{\rm Aut}_{K}(K[x])$. By
statement 1, $G_{f}(K)\neq\\{e\\}$ iff
$e\neq\sigma_{\lambda,(1-\lambda)\nu}\in\mathbb{T}_{\nu}(\overline{K})\cap{\rm
Aut}_{K}(K[x])$ where $1\neq\lambda\in K^{\times}$ and (necessarily) $\nu\in
K^{\times}$ iff $G_{f}(K)=\mathbb{T}_{\nu}(K)$.

3\. Statement 3 follows from statement 2. $\Box$

The eigengroup $G_{f}(K)$ of a polynomial $f\in K[x]$ that has at least two
distinct roots in $\overline{K}$. For each nonscalar monic polynomial $f(x)$,
Theorem 4.18, Theorem 4.20, Theorem 4.22 and Theorem 4.23 (resp., Theorem
4.24) are explicit descriptions of the eigengroup $G_{f}(K)$ in the case when
the field $K$ is algebraically closed (resp., in general case).

Lemma 4.9 is an explicit description of the roots of the polynomials of the
type $g(f^{n}_{V}(x-\nu))$.

###### Lemma 4.9

Suppose that $\lambda_{n}\in K$ is a primitive $n$’th root of unity, $n\geq 2$
and $\nu\in K$, $V$ is a nonzero finite dimensional
$\mathbb{F}_{p}(\lambda_{n})$-subspace of $K$, and
$K(\lambda_{n})=\mathbb{F}_{p^{m}}$ for some $m\geq 1$ (Lemma 4.2). Then

  1. 1.

For all elements $\rho\in K$,

$f^{n}_{V}(x-\nu)-f_{V}^{n}(\rho)=\prod_{i=0}^{n-1}\prod_{v\in
V}(x-\nu-\lambda_{n}^{i}\rho-v).$

  2. 2.

Let $g(x)=\prod_{j=1}^{k}(x-\xi_{j})\in\overline{K}[x]$ where ${\cal
R}(g)=\\{\xi_{1},\ldots,\xi_{k}\\}$ is the set of roots of the polynomial
$g(x)$ counted with multiplicity. Then $\xi_{j}=f_{V}^{n}(\rho_{j})$ for some
element $\rho_{j}\in\overline{K}$ and

$g(f^{n}_{V}(x-\nu))=\prod_{j=1}^{k}\prod_{i=0}^{n-1}\prod_{v\in
V}(x-\nu-\lambda_{n}^{i}\rho_{j}-v).$

Proof. 1. By Proposition 4.5.(2b), the map $f_{V}$ is a
$K(\lambda_{n})$-linear map (since $K(\lambda_{n})=\mathbb{F}_{p^{m}}$), and
the result follows:

$\displaystyle f^{n}_{V}(x-\nu)-f_{V}^{n}(\rho)$ $\displaystyle=$
$\displaystyle\prod_{i=0}^{n-1}(f_{V}(x-\nu)-\lambda_{n}^{i}f_{V}(\rho))=\prod_{i=0}^{n-1}f_{V}(x-\nu-\lambda_{n}^{i}\rho)$
$\displaystyle=$ $\displaystyle\prod_{i=0}^{n-1}\prod_{v\in
V}(x-\nu-\lambda_{n}^{i}\rho-v).$

2\. Notice that $g(x)=\prod_{j=1}^{k}(f^{n}_{V}(x-\nu)-f_{V}^{n}(\rho_{j}))$,
and statement 2 follows from statement 1. $\Box$

###### Theorem 4.10

Suppose that a monic polynomial $f(x)\in K[x]$ has at least two distinct roots
in $\overline{K}$. Then the group $G_{f}(K)$ is a finite group,
$G_{f}(K)=\widetilde{G}_{f}(K)\rtimes\overline{G}_{f}(K)$ where
$\overline{G}_{f}(K)=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
and $\widetilde{G}_{f}(K)={\rm Sh}_{V}(K)$, $\lambda_{n}\in K$ is a primitive
$n$’th root of unity, $\nu\in K$, $V$ is a finite dimensional
$\mathbb{F}_{p}(\lambda_{n})$-subspace of $K$, and
$\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for some $m\geq 1$ (Lemma
4.2).

Proof. Since $G_{f}(K)=G_{f}(\overline{K})\cap{\rm Aut}_{K}(K[x])$, it
suffices to prove the theorem in the case when the field $K$ is an
algebraically closed field. So, we assume that $K=\overline{K}$. Let ${\cal
R}^{\prime}$ be the set of distinct roots of the polynomial $f$. The subgroup
$\widetilde{G_{f}}=G_{f}\cap{\rm Sh}(K)$ is equal to a group ${\rm Sh}_{V}$
where $V$ is a finite dimensional vector space over the field $\mathbb{F}_{p}$
since $|V|\leq|{\cal R}^{\prime}|$ (as ${\cal R}^{\prime}+V\subseteq{\cal
R}^{\prime}$).

If $G$ is a finite subgroup of $G_{f}$ that contains the group ${\rm Sh}_{V}$
then $G={\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ where
$\lambda_{n}$ is a primitive $n$’th root of unity and $\nu\in K$. The cyclic
group $\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ of order $n$
acts on the field $K$ and on the set ${\cal R}^{\prime}$, see (14). The point
$\nu$ is the only fixed point of the action and the orbit of every element
$\lambda\neq\nu$ contains precisely $n$ elements. The polynomial $f$ contains
at least two distinct roots. Therefore $n\leq|{\cal R}^{\prime}|$. Then, by
(13), the group $\varphi(G_{f})$ is equal to
$\langle\sigma_{\lambda_{n^{\prime}},(1-\lambda_{n^{\prime}})\nu^{\prime}}\rangle$
where $\lambda_{n^{\prime}}$ is a primitive $n^{\prime}$’th root of unity and
$\nu^{\prime}\in K$. By Theorem 4.3, $G_{f}={\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{n^{\prime}},(1-\lambda_{n^{\prime}})\nu^{\prime}}\rangle$
is a finite group. $\Box$

Corollary 4.11 is a criterion for the group $G_{f}$ to be an infinite group.

###### Corollary 4.11

Let $f(x)\in K[x]$ be a nonscalar monic polynomial. Then the group $G_{f}$ is
an infinite group iff $f(x)=(x-\nu)^{d}$ for some $\nu\in K$ iff
$G_{f}(K)=\mathbb{T}_{\nu}(K)$ (see (23) for the definition of the group
$\mathbb{T}_{\nu}(K)$).

Proof. The corollary follows from Proposition 4.8.(3) and Theorem 4.10. $\Box$

Recall that the field $K$ is an algebraically closed field, $f(x)\in K[x]$ is
a monic polynomial that has at least two distinct roots, and ${\cal R}(f)$ is
the set of all roots of $f$ counted with multiplicity. Recall that (Theorem
4.10 and Theorem 4.3) the group
$G_{f}=\widetilde{G_{f}}\rtimes\overline{G}_{f}$ is a finite subgroup of ${\rm
Aut}_{K}(K[x])$ where
$\overline{G}_{f}:=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
provided $\overline{G}_{f}\neq\\{e\\}$ and
$\widetilde{G_{f}}:=\\{\sigma_{1,\mu}\,|\,\mu\in V\\}$, $\lambda_{n}\in K$ is
a primitive $n$’th root of unity, $\nu\in K$, $V$ is a finite dimensional
$\mathbb{F}_{p}(\lambda_{n})$-subspace of $K$, and
$\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for some $m\geq 1$ (Lemma
4.2).

There are four distinguish cases:

  1. 1.

$\widetilde{G_{f}}\neq\\{e\\}$, $\overline{G}_{f}\neq\\{e\\}$,

  2. 2.

$\widetilde{G_{f}}\neq\\{e\\}$, $\overline{G}_{f}=\\{e\\}$,

  3. 3.

$\widetilde{G_{f}}=\\{e\\}$, $\overline{G}_{f}\neq\\{e\\}$,

  4. 4.

$\widetilde{G_{f}}=\\{e\\}$, $\overline{G}_{f}=\\{e\\}$.

Below, in the case of $K=\overline{K}$, for the polynomial $f$ criteria are
given in terms of its roots for each case to hold.

A description of the group $\widetilde{G_{f}}(K)$ and a criterion for
$\widetilde{G_{f}}(K)\neq\\{e\\}$.

Definition. Let $f(x)\in K[x]$ be a nonscalar polynomial. Two distinct roots
$\lambda,\lambda^{\prime}\in\overline{K}$ of the polynomial $f(x)$ are called
a $K$-shift pair of $f(x)$ if

$\lambda-\lambda^{\prime}\in K\;\;{\rm
and}\;\;\mathbb{F}_{p}(\lambda-\lambda^{\prime})+{\cal R}(f)\subseteq{\cal
R}(f)$ (25)

where ${\cal R}(f)$ is the set of all roots in $\overline{K}$ of the
polynomial $F(x)$ counted with multiplicity.

The set of all $K$-shift pairs of the polynomial $f(x)$ is denoted by ${\rm
SP}(f,K)$. The vector space over $K$,

$V(f,K)=\begin{cases}\sum_{\\{\lambda,\lambda^{\prime}\\}\in{\rm
SP}(f,K)}\mathbb{F}_{p}(\lambda-\lambda^{\prime})&\text{if }{\rm
SP}(f,K)\neq\emptyset,\\\ 0&\text{if }{\rm SP}(f,K)=\emptyset.\\\ \end{cases}$
(26)

is called the $K$-shift vector space of the polynomial $f(x)$.

For fields $K\subseteq L\subseteq\overline{K}$, we have ${\rm
SP}(f,K)\subseteq{\rm SP}(f,L)\subseteq{\rm SP}(f,\overline{K})$ and
$V(f,K)\subseteq V(f,L)\subseteq V(f,\overline{K})$.

Proposition 4.12 gives an explicit description of the group
$\widetilde{G}_{f}(K)$ and a criterion for $\widetilde{G}_{f}(K)\neq\\{e\\}$.

###### Proposition 4.12

Let $f(x)\in K[x]$ be a nonscalar monic polynomial. Then

  1. 1.

$\widetilde{G}_{f}(K)={\rm Sh}_{V}$ where $V=V(f,K)$ is the $K$-shift vector
space of $f$.

  2. 2.

$\widetilde{G}_{f}(K)=\\{e\\}$ iff ${\rm SP}(f,K)=\emptyset$ iff for all
distinct roots $\lambda,\lambda^{\prime}\in\overline{K}$ of the polynomial $f$
such that $\lambda-\lambda^{\prime}\in K$ (if they exist),
$\mathbb{F}_{p}(\lambda-\lambda^{\prime})+{\cal R}(f)\not\subseteq{\cal
R}(f)$.

  3. 3.

$\widetilde{G}_{f}(\overline{K})=\\{e\\}$ iff ${\rm
SP}(f,\overline{K})=\emptyset$ iff for all distinct roots
$\lambda,\lambda^{\prime}\in\overline{K}$ of the polynomial $f$,
$\mathbb{F}_{p}(\lambda-\lambda^{\prime})+{\cal R}(f)\not\subseteq{\cal
R}(f)$.

Proof. 1. It follows from the description of the group $\widetilde{G}_{f}(K)$
as a shift group, $\widetilde{G}_{f}(K)={\rm Sh}_{V}$, that $V=V(f,K)$.

2 and 3. Statement 2 follows from statement 1 and statement 3 is a particular
case of statement 2. $\Box$

Definition. Let $V$ be a nonzero finite dimensional $\mathbb{F}_{p}$-subspace
of $K$. The largest finite field, denoted $\mathbb{F}_{p^{e}}$, where
$e=e(V)\geq 1$, such that $\mathbb{F}_{p^{e}}V\subseteq V$ is called the
multiplier field of $V$. The natural number $e=e(V)$ is called the
$p$-exponent of $V$.

The multiplier field $\mathbb{F}_{p^{e}}$ is the composite of all finite
fields $\mathbb{F}_{p^{m}}$ such that $\mathbb{F}_{p^{m}}V\subseteq V$. If
${\rm dim}_{\mathbb{F}_{p}}(V)=n$ then $p^{m}\leq|V|=p^{n}$, and so $m\leq n$.

Let $\lambda_{p^{e}-1}$ be a primitive $p^{e}-1$’st root of unity (a generator
of the cyclic group $\mathbb{F}_{p^{e}}^{\times}$). Since
$\lambda_{p^{e}-1}\in\mathbb{F}_{p^{e}}$ and $\mathbb{F}_{p^{e}}V\subseteq V$,
we have that $\lambda_{p^{e}-1}V\subseteq V$.

For each finite dimensional $\mathbb{F}_{p}$-subspace $V$ of the field $K$,
Lemma 4.13 describes all the roots of unity $\lambda_{n}$ such that
$\lambda_{n}V\subseteq V$.

###### Lemma 4.13

Let $V$ be a nonzero finite dimensional $\mathbb{F}_{p}$-subspace of the field
$K$, $\mathbb{F}_{p^{e}}$ be its multiplier field.

  1. 1.

Suppose that $\lambda_{n}$ is a primitive $n$’th root of unity. Then
$\lambda_{n}V\subseteq V$ iff $n|p^{e}-1$.

  2. 2.

$|\mathbb{F}_{p^{e}}^{\times}|=1$ iff $(p,e)=(2,1)$ iff $\lambda_{n}V\subseteq
V$ (where $\lambda_{n}$ is a primitive $n$’th root of unity) implies
$\lambda_{n}=1$.

Proof. 1. $\lambda_{n}V\subseteq V$ iff
$\lambda_{n}\in\mathbb{F}_{p^{e}}^{\times}$ iff
$n||\mathbb{F}_{p^{e}}^{\times}|$ iff $n|p^{e}-1$.

2\. Statement 2 follows from statement 1. $\Box$

Classification of subgroups $G$ of ${\rm Aut}_{K}(K[x])$ maximal satisfying
the property $G\cap{\rm Sh}(K)={\rm Sh}_{V}$.

###### Corollary 4.14

Let $V$ be a nonzero finite dimensional $\mathbb{F}_{p}$-subspace of the field
$K$, $\mathbb{F}_{p^{e}}$ be its multiplier field and $\lambda_{p^{e}-1}$ be a
primitive $p^{e}-1$’st root of unity. Then the finite groups $G_{V,\nu}:={\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{p^{e}-1},(1-\lambda_{p^{e}-1})\nu}\rangle$,
where $\nu\in K/V$, are the maximal subgroups $G$ of the group ${\rm
Aut}_{K}(K[x])$ that satisfy the property that $G\cap{\rm Sh}(K)={\rm
Sh}_{V}$.

Proof. Recall that for all elements $\lambda\in K^{\times}$ and $\mu,v\in K$,
$\sigma_{\lambda,\mu}\sigma_{1,v}\sigma^{-1}_{\lambda,\mu}=\sigma_{1,\lambda^{-1}v}$.
Hence the groups $G_{V,\nu}$ are well defined and every subgroup $H$ of ${\rm
Aut}_{K}(K[x])$ such that $H\cap{\rm Sh}(K)={\rm Sh}_{V}$ is a finite group.

Given a finite subgroup $G^{\prime}$ of ${\rm Aut}_{K}(K[x])$ such that
$G^{\prime}\cap{\rm Sh}(K)={\rm Sh}_{V}$. By Theorem 4.3, $G^{\prime}={\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ for some
$\nu\in K$ and a primitive $n$’th root of unity $\lambda_{n}$ such that
$n|p^{e}-1$, by Lemma 4.13.(1), and so $G^{\prime}\subseteq G_{V,\nu}$. Since
the groups $\\{G_{V,\nu}\\}_{\nu\in K}$ are distinct, the corollary follows
($G_{V,\nu}=G_{V,\nu^{\prime}}$ iff
$\sigma_{\lambda,(1-\lambda)\nu^{\prime}}=\sigma^{i}_{\lambda,(1-\lambda)\nu}\sigma_{1,v}$
for some natural number $i$ such that $1\leq i<p$ and ${\rm gcd}(i,p)=1$ and
an element $v\in V$ where $\lambda=\lambda_{p^{e}-1}$ iff
$\nu^{\prime}=\nu+(1-\lambda^{i})^{-1}v$ since
$\sigma^{i}_{\lambda,(1-\lambda)\nu}\sigma_{1,v}=\sigma_{\lambda^{i},(1-\lambda^{i})(\nu+(1-\lambda^{i})^{-1}v)}$
iff $\nu^{\prime}\equiv\nu\mod V$ since
$\mathbb{F}_{p}(\lambda)V=\mathbb{F}_{p^{e}}V=V$). $\Box$

Criterion for $\widetilde{G}_{f}\neq\\{e\\}$ and
$\overline{G}_{f}\neq\\{e\\}$.

###### Lemma 4.15

Suppose that $K$ is an algebraically closed field, $f(x)\in K[x]$ is monic
nonscalar polynomial that has at least two distinct roots,
$\widetilde{G}_{f}={\rm Sh}_{V}\neq\\{e\\}$ and
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq\\{e\\}$
where $V$ is a nonzero $\mathbb{F}_{p}(\lambda_{n})$-subspace of the field $K$
and $\lambda_{n}$ is primitive $n$’th root of unity. Then

  1. 1.

$\lambda_{n}\in\mathbb{F}_{p^{e}}$ where $\mathbb{F}_{p^{e}}$ is the
multiplier field of the $\mathbb{F}_{p}$-subspace $V$ of $K$, or,
equivalently, $n|p^{e}-1$.

  2. 2.

The group $G_{f}={\rm Sh}_{V}\rtimes\overline{G}_{f}$ is a subgroup of ${\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda,(1-\lambda)\nu}\rangle$ where $\lambda$
is a cyclic generator of the group $\mathbb{F}_{p^{e}}^{\times}$, i.e.
$\lambda=\lambda_{p^{e}-1}$ is primitive $p^{e}-1$’st root of unity.

Proof. 1. Since $\lambda_{n}V\subseteq V$, $\lambda_{n}\in\mathbb{F}_{p^{e}}$
and the lemma follows from Corollary 4.14. $\Box$

Definition. Let
$f(x)=x^{d}+a_{d-1}x^{d-1}+\cdots+a_{1}x+a_{0}\in\overline{K}[x]$ be a monic
polynomial of degree $d\geq 1$ where $a_{i}\in\overline{K}$ are the
coefficients of the polynomial $f(x)$. Then the natural number

${\rm gcd}(f(x)):={\rm gcd}\\{i\geq 1\,|\,a_{i}\neq 0\\}$

is called the exponent of $f(x)$.

Clearly, the exponent of $f(x)$ is the largest natural number $m\geq 0$ such
that $f(x)=g(x^{m})$ for some polynomial $g(x)\in K[x]$.

Definition. For a nonscalar polynomial $f\in K[x]$, we have the unique product

${\rm gcd}(f)=p^{s}{\rm gcd}_{p}(f)\;\;{\rm where}\;\;s\geq 0,\;\;{\rm
gcd}_{p}(f)\in\mathbb{N}\;\;{\rm and}\;\;p\nmid{\rm gcd}_{p}(f).$ (27)

###### Proposition 4.16

Suppose that $f(x)=f_{V}^{i}(x-\nu)$ for some nonzero finite dimensional
$\mathbb{F}_{p}$-subspace $V$ of $K$, $\nu\in K$ and a natural number $i\geq
1$ (i.e. ${\cal R}_{d}(f)=\nu+V$ and each root of $f(x)$ has multiplicity
$i$). Let $\mathbb{F}_{p^{e}}$ be the multiplier field of $V$,
$\mathbb{F}_{p^{e}}^{\times}=\langle\lambda_{n}\rangle$ where $n=p^{e}-1$.
Then

$G_{f}=\begin{cases}{\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq{\rm
Sh}_{V}&\text{if }(p,e)\neq(2,1),\\\ {\rm Sh}_{V}&\text{if }(p,e)=(2,1).\\\
\end{cases}$

Proof. Since $|{\cal R}_{d}(f)|=|\nu+V|=|V|\geq 2$, the group
$G_{f}=\widetilde{G}_{f}\rtimes\overline{G}_{f}$ is a finite group. Clearly,
${\rm Sh}_{V}\subseteq\widetilde{G}_{f}$. In fact, ${\rm
Sh}_{V}=\widetilde{G}_{f}$ since ${\cal R}_{d}(f)=\nu+V$.

Suppose that $(p,e)\neq(2,1)$. Recall that $n=p^{e}-1>1$ and
$\mathbb{F}_{p^{e}}^{\times}=\langle\lambda_{n}\rangle$, the multiplier field
of $V$. In particular, $\lambda_{n}V\subseteq V$. Therefore,
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$, by
Lemma 4.13.(1).

The case $(p,e)=(2,1)$ is obvious, see Lemma 4.13.(2). $\Box$

###### Lemma 4.17

Suppose that ${\rm Fr}(K)=K$ (where ${\rm Fr}(a)=a^{p}$, the Frobenius
endomorphism). Then $G_{f^{p^{n}}}(K)=G_{f}(K)$ for all polynomials $f\in
K[x]$ and all natural numbers $n\geq 1$.

Proof. (i) $G_{f^{p^{n}}}(K)\supseteq G_{f}(K)$: If $\sigma\in G_{f}(K)$ then
$\sigma(f)=\lambda f$ for some $\lambda\in K$, and so
$\sigma(f^{p^{n}})=\lambda^{p^{n}}f^{p^{n}}$. This means that $\sigma\in
G_{f^{p^{n}}}(K)$.

(ii) $G_{f^{p^{n}}}(K)\subseteq G_{f}(K)$: If $\tau\in G_{f^{p^{n}}}(K)$ then
$\tau(f^{p^{n}})=\mu f^{p^{n}}$ for some $\mu\in K$, and so
$(\tau(f)-\mu^{\frac{1}{p^{n}}}f)^{p^{n}}=0$, i.e.
$\tau(f)=\mu^{\frac{1}{p^{n}}}f$. This means that $\tau\in G_{f}(K)$. $\Box$

Let $f(x)=\sum a_{i}x^{i}\in K[x]$ be a monic nonscalar polynomial and ${\rm
gcd}(f)=p^{s}{\rm gcd}_{p}(f)$. Suppose that $K=\overline{K}$. Then there is a
unique monic nonscalar polynomial $f_{1}(x)\in K[x]$ such that

$f(x)=f_{1}^{p^{s}}(x).$ (28)

Clearly, ${\rm gcd}(f_{1})={\rm gcd}_{p}(f)$, $\deg(f)=p^{s}\deg(f_{1})$ and
$f_{1}^{\prime}\neq 0$ (the derivative of $f_{1}$).

Theorem 4.18 is a criterion for the group
$G_{f}=\widetilde{G_{f}}\rtimes\overline{G}_{f}$ to have nontrivial subgroups
$\widetilde{G_{f}}$ and $\overline{G}_{f}$, it also gives an explicit
description of the group $G_{f}$.

###### Theorem 4.18

Suppose the field $K$ is an algebraically closed field and $f(x)\in K[x]$ is a
monic polynomial that has at least two distinct roots, $gcd(f)=p^{s}{\rm
gcd}_{p}(f)$ and $f(x)=f_{1}^{p^{s}}(x)$ for a unique monic nonscalar
polynomial $f_{1}(x)\in K[x]$, see (28). Suppose that $V\neq 0$ is a finite
dimensional $\mathbb{F}_{p}$-subspace of $K$ and $\mathbb{F}_{p^{e}}$ is the
multiplier field of $V$. Then the following statements are equivalent:

  1. 1.

$\widetilde{G_{f}}={\rm Sh}_{V}\neq\\{e\\}$ and
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq\\{e\\}$.

  2. 2.

There is a primitive $n$’th root of unity $\lambda_{n}\neq 1$ (in particular,
$n\geq 2$) such that $\lambda_{n}V\subseteq V$, and an element $\nu\in K$ such
that either $f(x)=f_{V}^{i}(x-\nu)$ for some natural number $i\geq 1$ and
$n=p^{e}-1$ (in this case, $(p,e)\neq(2,1)$, $\widetilde{G}_{f}={\rm Sh}_{V}$
and
$\overline{G}_{f}=\langle\sigma_{\lambda_{p^{e}-1},(1-\lambda_{p^{e}-1})\nu}\rangle$
where $\mathbb{F}_{p^{e}}^{\times}=\langle\lambda_{p^{e}-1}\rangle$) or
otherwise $f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$ for some natural number
$i\geq 0$ and a monic nonscalar polynomial $g(x)\in K[x]$ such that $g(0)\neq
0$, and the following two conditions hold:

    1. (a)

$n\geq 2$ and ${\rm gcd}(\frac{p^{e}-1}{n},{\rm gcd}_{p}(g))=1$, and

    2. (b)

${\cal
R}(f)+\mathbb{F}_{p}(\lambda_{n})(\lambda-\lambda^{\prime})\not\subseteq{\cal
R}(f)$ for all distinct roots $\lambda$ and $\lambda^{\prime}$ of the
polynomial $f$ such that $\lambda-\lambda^{\prime}\not\in V$.

Suppose that statement 1 holds. Then the natural number $i$ and the polynomial
$g(x)$ in statement 2 are unique, the element $\nu$ is unique up to adding an
arbitrary element of $V$ (i.e. $\nu$ can be replaced by $\nu+v$ for any
element $v\in V$), and the equality $f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$
is unique (since $f_{V}(x-\nu-v)=f_{V}(x-\nu)$ for all $v\in V$). Furthermore,
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(f)=\lambda_{n}^{i}f$, and $f\in
K[x]^{G_{f}}$ iff $n|i$.

In the second case, i.e. $f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$,

$n={\rm gcd}(p^{e}-1,{\rm gcd}_{p}(h)$

where $h(x)\in K[x]$ is a unique polynomial such that
$f(x)=f_{V}^{i}(x-\nu)h(f_{V}(x-\nu))$ (i.e. $h(x)=g(x^{n})$), and either
$\nu$ is a root of $f(x)$ (i.e. $i\geq 1$) or otherwise (i.e. $i=0$) $\nu$ is
a root of $f_{1}^{\prime}(x)$ (the derivative of $f_{1}(x)$).

Proof. $(1\Rightarrow 2)$ Suppose that statement 1 holds. By Theorem 4.10,
$\widetilde{G}_{f}={\rm Sh}_{V}$ and
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ for a
nonzero $\mathbb{F}_{p}$-subspace $V$ of $K$, a primitive $n$’th root of unity
$\lambda_{n}\neq 1$ (in particular, $n\geq 2$) such that
$\lambda_{n}V\subseteq V$ and an element $\nu\in K$. Then, by Theorem 4.6.(3),
either $f(x)=f^{i}_{V}(x-\nu)$ for some natural number $i\geq 1$ or otherwise

$f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$

for some natural number $i\geq 0$, $n\geq 2$, and a monic nonscalar polynomial
$g(x)\in K[x]$ such that $g(0)\neq 0$.

In the first case, by Proposition 4.16, $\widetilde{G}_{f}={\rm
Sh}_{V}\neq\\{e\\}$ and
$\overline{G}_{f}=\langle\sigma_{\lambda_{p^{e}-1},(1-\lambda_{p^{e}-1})\nu}\rangle\neq\\{e\\}$.

Now let us consider the second case. By Lemma 4.13, $n|p^{e}-1$ (since
$\lambda_{n}\in\mathbb{F}_{p^{e}}$). Suppose that $l:={\rm
gcd}(\frac{p^{e}-1}{n},{\rm gcd}_{p}(g))>1$. Then
$\sigma_{\lambda_{ln},(1-\lambda_{ln})\nu}\in\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
where $\lambda_{ln}$ is a primitive $ln$’th root of unity, a contradiction
(since the order of the element $\sigma_{\lambda_{ln},(1-\lambda_{ln})\nu}$ is
$ln>n=|\overline{G}_{f}|$). Therefore, the statement (a) holds.

Suppose that the condition (b) does not holds, i.e. there are two distinct
roots $\lambda$ and $\lambda^{\prime}$ of the polynomial $f(x)$ such that
$v^{\prime}:=\lambda-\lambda^{\prime}\not\in V$ and ${\cal
R}(f)+\mathbb{F}_{p}(\lambda_{n})(\lambda-\lambda^{\prime})\subseteq{\cal
R}(f)$. Then

$V^{\prime}:=V+\mathbb{F}_{p}(\lambda_{n})v^{\prime}$

is a $\mathbb{F}_{p}(\lambda_{n})$-submodule of $K$ that properly contains the
$\mathbb{F}_{p}(\lambda_{n})$-module $V$. Then ${\rm
Sh}_{V^{\prime}}\subseteq{\rm Sh}_{V}$, a contradiction.

$(2\Rightarrow 1)$ In the first case, i.e. $f(x)=f_{V}^{i}(x-\nu)$, the
implication follows from Proposition 4.16. In the second case, i.e.
$f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$,

$\overline{G}_{f}\supseteq\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq\\{e\\}\;\;{\rm
and}\;\;\widetilde{G_{f}}\supseteq\\{\sigma_{1,\mu}\,|\,\mu\in
V\\}\neq\\{e\\}.$

The conditions (a) and (b) imply that the inclusions above are equalities
hold, see the proof of the implication $(1\Rightarrow 2)$.

Suppose that statement 1 holds. Then statement 2 holds and vice versa. So, we
have the equality

$f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$

in statement 2 (the case $g=1$ corresponds to the first case). The polynomial
$f_{V}(x-\nu)$ is $\widetilde{G}_{f}$-invariant, i.e. for all elements $v\in
V$, $f_{V}(x-\nu)=\sigma_{1,-v}(f_{V}(x-\nu))=f_{V}(x-(\nu+v))$. Therefore,
for all elements $v\in V$, $f(x)=f_{V}^{i}(x-(\nu+v))g(f_{V}^{n}(x-(\nu+v)))$,
i.e. the element $\nu$ can be replaced by the element $\nu+v$ for any element
$v\in V$. This is the only freedom for the choice of the element $\nu$.
Indeed,
$G_{f}=\\{\sigma^{j}_{\lambda_{n},(1-\lambda_{n})\nu}\sigma_{1,v}\,|\,0\leq
j\leq n-1,v\in V\\}$. Since

$\sigma^{j}_{\lambda_{n},(1-\lambda_{n})\nu}\sigma_{1,v}=\sigma_{\lambda_{n}^{j},(1-\lambda_{n}^{j})\nu}\sigma_{1,v}=\sigma_{\lambda_{n}^{j},(1-\lambda_{n}^{j})(\nu+(1-\lambda_{n}^{j})^{-1}v)}\;\;{\rm
for}\;\;1\leq j\leq n-1,$

it follows that the only freedom in choosing the generator
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}$ in Theorem 4.6 is an element of the
type
$\sigma_{\lambda_{n}^{j},(1-\lambda_{n}^{j})(\nu+(1-\lambda_{n}^{j})^{-1}v)}$
where $j$ is a natural number such that $1\leq j\leq n-1$, ${\rm gcd}(j,n)=1$
and $v$ is an arbitrary element of $V$. Now, by Theorem 4.6, $\nu$ is unique
up to addition an element of $V$, and the elements $i$ and $g(x)$ are unique.

Clearly, $\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(f)=\lambda_{n}^{i}f$
(Theorem 4.6.(1)), and so $f\in K[x]^{G_{f}}$ iff $n|i$ (Theorem 4.6.(2,3)).

Suppose that $g(x)\neq 1$. Clearly, $\nu$ is a root of $f(x)$ iff $i\geq 1$.
Suppose that $\nu$ is not a root of $f(x)$, i.e. $i=0$ and
$f(x)=g(f_{V}^{n}(x-\nu))$. Recall that $f(x)=f_{1}^{p^{s}}(x)$ and
$G_{f}=G_{f_{1}}$, by Lemma 4.17. Then $\nu$ is not a root of $f_{1}(x)$, i.e.
$f_{1}(x)=g_{1}(f_{V}^{n}(x-\nu))$ for a unique polynomial $g_{1}(x)\in K[x]$
such that $g=g_{1}^{p^{s}}$. Hence, $\nu$ is a root of the polynomial
$f_{1}(x)$ since

$0\neq
f_{1}^{\prime}(x)=nf_{V}^{n-1}(x-\nu)f_{V}^{\prime}(x-\nu)g_{1}^{\prime}(f_{V}^{n}(x-\nu)),$

$n\geq 2$ and $f_{V}(0)=0$.

In the second case, i.e. $f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu))$, $n={\rm
gcd}(p^{e}-1,{\rm gcd}_{p}(h)$, by the statement (a). $\Box$

Definition. The unique presentation of the polynomial $f(x)$,

$f(x)=f_{V}^{i}(x-\nu)\;\;{\rm
or}\;\;f(x)=f_{V}^{i}(x-\nu)g(f_{V}^{n}(x-\nu)),$

in Theorem 4.18.(2) is called the eigenform or the eigenpresentation of the
polynomial $f(x)$. The scalars $\nu+V$ and the natural number $i\geq 0$ are
called the eigenroots of $f(x)$ and their multiplicity, respectively. The
natural number $n\geq 2$ and the monic polynomial $g(x)$ are called the
eigenorder and the eigenfactor of $f(x)$. In the second case, the eigenroots
may not be roots of the polynomial $f(x)$. They are iff $i\neq 0$.

###### Corollary 4.19

Suppose the field $K$ is an algebraically closed field and $f(x)\in K[x]$ is a
monic polynomial that has at least two distinct roots, $gcd(f)=p^{s}{\rm
gcd}_{p}(f)$ and $f(x)=f_{1}^{p^{s}}(x)$ for a unique monic nonscalar
polynomial $f_{1}(x)\in K[x]$. Suppose that the polynomial $f$ satisfies the
assumption of Theorem 4.18, and $f_{1}(x)=f_{V}^{j}(x-\nu)$ or
$f_{1}(x)=f_{V}^{j}(x-\nu)g_{1}(f_{V}^{n}(x-\nu))$, is the eigenform of the
polynomial $f_{1}(x)$. Then $f(x)=f_{V}^{p^{s}j}(x-\nu)$ or
$f(x)=f_{V}^{p^{s}j}(x-\nu)g_{1}^{p^{s}}(f_{V}^{n}(x-\nu))$, is the eigenform
of the polynomial $f_{1}(x)$.

Proof. The statement follows from the facts that $f(x)=f_{1}^{p^{s}}(x)$,
$G_{f}=G_{f_{1}}$ (Lemma 4.17) and the uniqueness of the eigenform (Theorem
4.18). $\Box$

Criterion for $\widetilde{G}_{f}=\\{e\\}$ and $\overline{G}_{f}\neq\\{e\\}$.
Theorem 4.20 is a criterion for the group
$G_{f}=\widetilde{G_{f}}\rtimes\overline{G}_{f}$ to be equal to
$\overline{G}_{f}\neq\\{e\\}$.

###### Theorem 4.20

Suppose the field $K$ is an algebraically closed field, $f(x)\in K[x]$ is a
monic polynomial that has at least two distinct roots, $gcd(f)=p^{s}{\rm
gcd}_{p}(f)$ and $f(x)=f_{1}^{p^{s}}(x)$ for a unique monic nonscalar
polynomial $f_{1}(x)\in K[x]$, see (28). Then the following statements are
equivalent:

  1. 1.

$\widetilde{G_{f}}=\\{e\\}$ and
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq\\{e\\}$
where $\lambda_{n}$ is a primitive $n$’th root of unity and $\nu\in K$.

  2. 2.

$f(x)=(x-\nu)^{i}g((x-\nu)^{n})$ for some natural number $i\geq 0$ and a monic
nonscalar polynomial $g(x)\in K[x]$ such that $g(0)\neq 0$,

    1. (a)

$n\geq 2$, $p\nmid n$ and ${\rm gcd}_{p}(g(x))=1$, and

    2. (b)

${\cal R}(f)+\mathbb{F}_{p}(\lambda-\lambda^{\prime})\not\subseteq{\cal R}(f)$
for all distinct roots $\lambda$ and $\lambda^{\prime}$ of the polynomial $f$.

Suppose that statement 1 holds. Then the presentation
$f(x)=(x-\nu)^{i}g((x-\nu)^{n})$ is unique, i.e. the triple $(\nu,i,g(x))$ is
unique. Either $\nu$ is a root of $f(x)$ (i.e. $i\geq 1$) or otherwise (i.e.
$i=0$) $\nu$ is a root of $f_{1}^{\prime}(x)$ (the derivative of $f_{1}(x)$).
If $\nu$ is a root of $f(x)$ then $n={\rm gcd}_{p}(x^{-i}f(x+\nu))$. If $\nu$
is not a root of $f(x)$ then $n={\rm gcd}_{p}(f(x+\nu))$. Furthermore,
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(f)=\lambda_{n}^{i}f$, and $f\in
K[x]^{G_{f}}$ iff $n|i$.

Proof. By Proposition 4.12.(3), $\widetilde{G_{f}}=\\{e\\}$ iff the condition
(b) holds.

$(1\Rightarrow 2)$ Suppose that statement 1 holds. Then, by Theorem 4.7.(2,3),

$f(x)=(x-\nu)^{i}g((x-\nu)^{n})$

for some natural number $i\geq 0$ and a monic nonscalar polynomial $g(x)\in
K[x]$ such that $g(0)\neq 0$ (since $|{\cal R}_{d}(f)|\geq 2$). Clearly,
$n\geq 2$ and $p\nmid n$.

Suppose that $l:={\rm gcd}_{p}(g(x))>1$. Then
$\sigma_{\lambda_{ln},(1-\lambda_{ln})\nu}\in\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
where $\lambda_{ln}$ is a primitive $ln$’th root of unity, a contradiction
(since the order of the element $\sigma_{\lambda_{ln},(1-\lambda_{ln})\nu}$ is
$ln>n=|\overline{G}_{f}|$). Therefore, the statement (a) holds.

$(2\Rightarrow 1)$ By the statement (b), $\widetilde{G_{f}}=\\{e\\}$. Since
$f(x)=(x-\nu)^{i}g((x-\nu)^{n})$ for some natural number $i\geq 0$, $n\geq 2$,
$p\nmid n$ and a monic nonscalar polynomial $g(x)\in K[x]$ such that $g(0)\neq
0$,

$\overline{G}_{f}\supseteq\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle\neq\\{e\\}.$

The condition (a) implies that the inclusion above is the equality, see the
proof of the implication $(1\Rightarrow 2)$.

Suppose that statement 1 holds. Then statement 2 holds and vice versa. So, we
have the equality $f(x)=(x-\nu)^{i}g((x-\nu)^{n})$ as in statement 2. To prove
uniqueness of this presentation it suffices to show that the element $\nu$ is
unique. The set of cyclic generators for the group
$G_{f}=\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
is equal to $\\{\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{j}\,|\,1\leq j\leq
n-1,{\rm gcd}(j,n)=1\\}$. Since
$\sigma_{\lambda_{n},(1-\lambda_{n})\nu}^{j}=\sigma_{\lambda_{n}^{j},(1-\lambda_{n}^{j})\nu}$,
the element $\nu$ is unique.

Clearly, $\nu$ is a root of $f(x)$ iff $i\geq 1$. Suppose that $\nu$ is not a
root of $f(x)$, i.e. $i=0$ and $f(x)=g((x-\nu)^{n})$, then
$f_{1}(x)=h((x-\nu)^{n})$ for a unique monic nonscalar polynomial
$h(x):=g^{\frac{1}{p^{s}}}(x)\in K[x]$ (since $p\nmid n$). Hence, $\nu$ is a
root of the polynomial $f_{1}(x)$ since

$0\neq f_{1}^{\prime}(x)=n(x-\nu)^{n-1}h^{\prime}((x-\nu)^{n})$

and $n\geq 2$.

If $\nu$ is a root of $f(x)$ then $n={\rm gcd}_{p}(x^{-i}f(x+\nu))$ (since
$f(x+\nu)=x^{i}g(x^{n})$). If $\nu$ is not a root of $f(x)$ then $n={\rm
gcd}_{p}(f(x+\nu))$ (since $f(x+\nu)=g(x^{n})$).

Clearly, $\sigma_{\lambda_{n},(1-\lambda_{n})\nu}(f)=\lambda_{n}^{i}f$, and so
$f\in K[x]^{G_{f}}$ iff $n|i$. $\Box$

Definition. The unique presentation of the polynomial $f(x)$,

$f(x)=(x-\nu)^{i}g((x-\nu)^{n}),$

in Theorem 4.20.(2) is called the eigenform or the eigenpresentation of the
polynomial $f(x)$. The scalar $\nu$ and the natural number $i\geq 0$ are
called the eigenroot of $f(x)$ and its multiplicity, respectively. The natural
number $n\geq 2$ and the monic polynomial $g(x)$ are called the eigenorder and
the eigenfactor of $f(x)$. In general, the eigenroot may not be a root of the
polynomial $f(x)$. It is iff $i\neq 1$.

Criterion for $\widetilde{G}_{f}\neq\\{e\\}$ and $\overline{G}_{f}=\\{e\\}$.

###### Lemma 4.21

Let $g(x)\in K[x]$ be a monic nonscalar polynomial such that $g^{\prime}\neq
0$ (the derivative of $g$), and $V$ be an $\mathbb{F}_{p}$-subspace of $K$ and
$\mathbb{F}_{p^{e}}$ be its multiplier field (for $V=0$, $f_{V}(x)=x$). Then

  1. 1.

$g(f_{V}(x))^{\prime}\neq 0$.

  2. 2.

For each $\nu\in K$, $g(f_{V}(x))=f_{V}^{i(\nu)}(x-\nu)g_{\nu}(f_{V}(x-\nu))$
for a unique natural number $i(\nu)\geq 0$ and a unique monic polynomial
$g_{\nu}(x)\in K[x]$ such that $g_{\nu}(0)\neq 0$; $i(\nu)\neq 0$ iff $\nu$ is
a root of the polynomial $g(f_{V}(x))$. If $n={\rm gcd}(p^{e}-1,{\rm
gcd}_{p}(g_{\nu}(x)))\geq 2$ then
$e\neq\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\in\overline{G}_{g(f_{V}(x))}(\overline{K})$
where $\lambda_{n}\in\overline{K}$ is a primitive $n$’th root of unity. If, in
addition, $\lambda_{n}\in K$ then
$e\neq\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\in\overline{G}_{g(f_{V}(x))}(K)$.

  3. 3.

Suppose that $\nu$ is not a root of the polynomial $g(f_{V}(x))$, i.e.
$i(\nu)=0$ and $g(f_{V}(x))=g_{\nu}(f_{V}(x-\nu))$, and ${\rm
gcd}_{p}(g_{\nu}(x))\neq 1$ then $\nu$ is a root of the derivative
$g(f_{V}(x))^{\prime}$ of the polynomial $g(f_{V}(x))$.

Proof. 1. $g(f_{V}(x))^{\prime}=g^{\prime}(f_{V}(x))f_{V}^{\prime}(x)\neq 0$,
by Proposition 4.5.(2d).

2\.
$g(f_{V}(x))=g(f_{V}(x-\nu+\nu))=g(f_{V}(x-\nu)+f_{V}(\nu))=f_{V}^{i(\nu)}(x-\nu)g_{\nu}(f_{V}(x-\nu))$
for a unique natural number $i(\nu)\geq 0$ and a unique monic polynomial
$g_{\nu}(x)\in K[x]$ such that $g_{\nu}(0)\neq 0$. Since $f_{V}(0)=0$ and
$g_{\nu}(0)\neq 0$, we see that $i(\nu)\neq 0$ iff $\nu$ is a root of the
polynomial $g(f_{V}(x))$.

If $n\geq 2$ then
$e\neq\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\in\overline{G}_{g(f_{V}(x))}(\overline{K})$,
by Theorem 4.6.(1). If, in addition, $\lambda_{n}\in K$ then
$e\neq\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\in\overline{G}_{g(f_{V}(x))}(K)$.

3\. Suppose that $\nu$ is not a root of the polynomial $g(f_{V}(x))$ and
$m={\rm gcd}_{p}(g_{\nu}(x))\neq 1$, i.e.
$g(f_{V}(x))=g_{\nu}(f_{V}(x-\nu))=h_{\nu}(f_{V}^{m}(x-\nu))$ for some monic
nonscalar polynomial $h_{\nu}(x)\in K[x]$. Then

$g(f_{V}(x))^{\prime}=h_{\nu}(f_{V}^{m}(x-\nu))^{\prime}=mf_{V}^{m-1}(x-\nu)f^{\prime}_{V}(x-\nu)h_{\nu}^{\prime}(f_{V}^{m}(x-\nu)),$

and so $\nu$ is a root of the polynomial $g(f_{V}(x))^{\prime}$. $\Box$

Given monic nonscalar polynomials $f(x),h(x)\in K[x]$. If $f(x)=g(h(x))$ for
some polynomial $g(x)\in K[x]$ then the polynomial $g(x)$ is unique and
necessarily monic. (Proof. If $f(x)=g(h(x))$ then the polynomial $g(x)$ is
monic, $\deg(f)=\deg(g)\deg(h)$, $K[h]\ni f_{1}:=f-h^{\deg(g)}$ and
$\deg(f_{1})<\deg(f)$. Now, the induction on $\deg(f)$ completes the proof).

Theorem 4.22 is a criterion for the group
$G_{f}=\widetilde{G_{f}}\rtimes\overline{G}_{f}$ to be equal to
$\widetilde{G_{f}}\neq\\{e\\}$.

###### Theorem 4.22

Suppose the field $K$ is an algebraically closed field, $f(x)\in K[x]$ is a
monic nonscalar polynomial that has at least two distinct roots,
$gcd(f)=p^{s}{\rm gcd}_{p}(f)$ and $f(x)=f_{1}^{p^{s}}(x)$ for a unique monic
nonscalar polynomial $f_{1}(x)\in K[x]$. Suppose that $V\neq 0$ is a finite
dimensional $\mathbb{F}_{p}$-subspace of $K$ and $\mathbb{F}_{p^{e}}$ is the
multiplier field of $V$. Then the following statements are equivalent:

  1. 1.

$\widetilde{G_{f}}={\rm Sh}_{V}\neq\\{e\\}$ and $\overline{G}_{f}=\\{e\\}$.

  2. 2.

$f_{1}(x)=g(f_{V}(x))$ for a (unique) monic polynomial $g(x)\in K[x]$ such
that

    1. (a)

either $|{\cal R}_{d}(g)|=1$ and $(p,e)=(2,1)$ or otherwise $|{\cal
R}_{d}(g)|\geq 2$ and ${\rm gcd}(p^{e}-1,{\rm gcd}_{p}(g_{\nu}(x)))=1$ for all
roots $\nu\in{\cal R}_{d}(f_{1}(x))\cup{\cal R}_{d}(f_{1}(x)^{\prime})$ where
$g_{\nu}$ is as in Lemma 4.21.(2) (i.e.
$f_{1}(x)=f_{V}^{i(\nu)}(x-\nu)g_{\nu}(f_{V}(x-\nu))$), and

    2. (b)

${\cal R}(f_{1})+\mathbb{F}_{p}(\lambda-\lambda^{\prime})\not\subseteq{\cal
R}(f_{1})$ for all distinct roots $\lambda$ and $\lambda^{\prime}$ of the
polynomial $f_{1}$ such that $\lambda-\lambda^{\prime}\not\in V$
($\Leftrightarrow$ ${\cal
R}(f)+\mathbb{F}_{p}(\lambda-\lambda^{\prime})\not\subseteq{\cal R}(f)$ for
all distinct roots $\lambda$ and $\lambda^{\prime}$ of the polynomial $f$ such
that $\lambda-\lambda^{\prime}\not\in V$).

Suppose that statement 1 holds. Then $f,f_{1}\in
K[x]^{G_{f}}=K[x]^{G_{f_{1}}}$.

Remark. By Lemma 4.17, $G_{f}=G_{f_{1}^{p^{s}}}=G_{f_{1}}$. This explains why
statement 2 is given via properties of the polynomial $f_{1}$ rather than $f$.

Proof. By Proposition 4.5 and Proposition 4.12.(1),
$\widetilde{G}_{f}=\widetilde{G}_{f_{1}}={\rm Sh}_{V}(\neq\\{e\\})$ iff
$f_{1}(x)=g(f_{V}(x))$ for a (unique) monic nonscalar polynomial $g(x)\in
K[x]$ such that the condition 2(b) holds.

Suppose that $|{\cal R}_{d}(g)|=1$, i.e. ${\cal R}_{d}(g)=\\{\rho\\}$ and let
$i(\rho)$ be the multiplicity of the root $\rho$. Fix an element
$\nu^{\prime}\in K=\overline{K}$ such that $f_{V}(\nu^{\prime})=\rho$. Then
$f_{1}(x)=(f_{V}(x)-f_{V}(\nu^{\prime}))^{i(\rho)}=f_{V}^{i(\rho)}(x-\nu^{\prime})$,
by Proposition 4.5.(2b). By Proposition 4.16,
$G_{f_{1}}=\widetilde{G}_{f_{1}}={\rm Sh}_{V}$ iff $(p,e)=(2,1)$.

Suppose that $|{\cal R}_{d}(g)|\geq 2$. By Lemma 4.21.(2),

$f_{1}(x)=g(f_{V}(x))=f_{V}^{i(\nu)}(x-\nu)g_{\nu}(f_{V}(x-\nu))$

for a unique monic nonscalar polynomial $g_{\nu}(x)\in K[x]$ such that
$g_{\nu}(0)\neq 0$ where $i(\nu)\geq 0$ is the multiplicity of the root $\nu$
(if $g_{\nu}(x)=1$ then
$f_{1}(x)=g(f_{V}(x))=f_{V}^{i(\nu)}(x-\nu)=(f_{V}(x)-f_{V}(\nu))^{i(\nu)}$,
and so $|{\cal R}_{d}(g)|=1$, a contradiction).

By Theorem 4.6 and Lemma 4.21.(2), $\overline{G}_{f_{1}}=\\{e\\}$ iff ${\rm
gcd}(p^{e}-1,{\rm gcd}_{p}(g_{\nu}(x)))=1$ for all $\nu\in K$ iff ${\rm
gcd}(p^{e}-1,{\rm gcd}_{p}(g_{\nu}(x)))=1$ for all $\nu\in{\cal
R}_{d}(f_{1}(x))\cup{\cal R}_{d}(f_{1}(x)^{\prime})$, Lemma 4.21.(2,3).

Clearly, $f,f_{1}\in K[x]^{G_{f}}=K[x]^{G_{f_{1}}}$ (Proposition 4.5 and Lemma
4.17). $\Box$

Definition. The unique presentation $f(x)=g^{p^{s}}(f_{V}(x))$ in Theorem 4.22
(where ${\rm gcd}(f)=p^{s}{\rm gcd}_{p}(f)$) is called the eigenform or
eigenpresentation of the polynomial $f(x)$ and the polynomial $g(x)$ is called
the eigenfactor of $f(x)$.

Criterion for $G_{f}=\\{e\\}$. Given a monic nonscalar polynomial $g(x)\in
K[x]$ with $g^{\prime}(x)\neq 0$. By Lemma 4.21.(2) (where $V=0$), for each
$\nu\in K$,

$g(x)=(x-\nu)^{i(\nu)}g_{\nu}(x-\nu)$ (29)

for a natural number $i(\nu)\geq 0$ and a unique monic polynomial
$g_{\nu}(x)\in K[x]$ such that $g_{\nu}(0)\neq 0$. Clearly, $i(\nu)\neq 0$ iff
$\nu$ is a root of the polynomial $g(x)$.

Theorem 4.23 is a criterion for $G_{f}=\\{e\\}$.

###### Theorem 4.23

Suppose the field $K$ is an algebraically closed field, $f(x)\in K[x]$ is a
monic polynomial that has at least two distinct roots, ${\rm gcd}(f)=p^{s}{\rm
gcd}_{p}(f)$ and $f(x)=g^{p^{s}}(x)$ for a unique monic nonscalar polynomial
$g(x)\in K[x]$, see (28). The following statements are equivalent:

  1. 1.

$G_{f}=\\{e\\}$.

  2. 2.

    1. (a)

For each root $\nu$ of the polynomial $g(x)$, ${\rm gcd}_{p}(g_{\nu}(x))=1$
where the polynomial $g_{\nu}(x)$ is defined in (29),

    2. (b)

for each root $\nu^{\prime}$ of the derivative $g^{\prime}(x)$ of the
polynomial $g(x)$ such that $g(\nu^{\prime})\neq 0$, ${\rm
gcd}_{p}(g_{\nu}(x))=1$, and

    3. (c)

${\cal R}(g)+\mathbb{F}_{p}(\lambda-\lambda^{\prime})\not\subseteq{\cal R}(g)$
for all distinct roots $\lambda$ and $\lambda^{\prime}$ of the polynomial $g$
($\Leftrightarrow$ ${\cal
R}(f)+\mathbb{F}_{p}(\lambda-\lambda^{\prime})\not\subseteq{\cal R}(f)$ for
all distinct roots $\lambda$ and $\lambda^{\prime}$ of the polynomial $f$).

Proof. Notice that $G_{f}=G_{g^{p^{s}}}=G_{g}$. The condition (c) is
equivalent to the condition that $\widetilde{G}_{g}=\\{e\\}$ (Proposition
4.12.(3)). It remains to show that provided $\widetilde{G}_{g}=\\{e\\}$ the
condition $\overline{G}_{g}=\\{e\\}$ is equivalent to the conditions (a) and
(b). Equivalently, $\widetilde{G}_{g}=\\{e\\}$ and
$\overline{G}_{g}\neq\\{e\\}$ iff one of the conditions (a) or (b) does not
hold and the condition (c) holds. This follows from Theorem 4.20. Indeed, by
Theorem 4.20, $\widetilde{G}_{g}=\\{e\\}$ and $\overline{G}_{g}\neq\\{e\\}$
iff the condition (c) holds and $g(x)=(x-\nu)^{i}g_{\nu}(x-\nu)$ for a unique
$\nu\in K$, a natural number $i\geq 0$ and a monic nonscalar polynomial
$g_{\nu}(x)$ such that $g_{\nu}(0)\neq 0$ (since $|{\cal R}_{d}(f)|\geq 2$)
and ${\rm gcd}_{p}(g_{\nu}(x))\geq 2$. We have two options either $\nu$ is a
root of the polynomial $g(x)$ or not. If $\nu$ is not a root of the polynomial
$g(x)$, i.e. $i=0$, then $g(x)=g_{\nu}(x-\nu)$, and so $\nu$ is a root of
$g^{\prime}(x)$ since ${\rm gcd}_{p}(g_{\nu}(x))\geq 2$. Now, it follows that
statements 1 and 2 are equivalent. $\Box$.

Theorem 4.24 describes the group $G_{f}(K)$ in terms of the group
$G_{f}(\overline{K})$.

###### Theorem 4.24

Suppose the field $K$ is not necessarily algebraically closed and $f(x)\in
K[x]$ is a monic polynomial that has at least two distinct roots in
$\overline{K}$. Recall that (Theorem 4.10 and Theorem 4.3) the group
$G_{f}(\overline{K})=\widetilde{G_{f}}(\overline{K})\rtimes\overline{G}_{f}(\overline{K})$
where
$\overline{G}_{f}(\overline{K}):=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
and
$\widetilde{G_{f}}(\overline{K}):=\\{\sigma_{1,\mu}\,|\,\mu\in\overline{V}\\}$,
$\lambda_{n}\in\overline{K}$ is a primitive $n$’th root of unity provided
$\overline{G}_{f}(\overline{K})\neq\\{e\\}$, $\nu\in\overline{K}$,
$\overline{V}$ is a finite dimensional $\mathbb{F}_{p}(\lambda_{n})$-subspace
of $\overline{K}$, and $\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for
some $m\geq 1$ (Lemma 4.2). Then
$G_{f}(K)=\widetilde{G_{f}}(K)\rtimes\overline{G}_{f}(K)={\rm
Aut}_{K}(K[x])\cap G_{f}(\overline{K})$, $\widetilde{G_{f}}(K)={\rm Sh}_{V}$
where $V:=K\cap\overline{V}$ and if $\overline{G}_{f}(K)\neq\\{e\\}$ then
$\overline{G}_{f}(K)=\langle\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu+\overline{v}}\rangle$
where
$i=\min\\{i^{\prime}=1,\ldots,n-1\,|\,i^{\prime}|n,\lambda_{n}^{i^{\prime}}\in
K,(1-\lambda_{n}^{i^{\prime}})\nu\in\overline{V}+K\\}$ and
$\overline{v}\in\overline{V}$ is any (fixed) element such that
$(1-\lambda_{n}^{i})\nu+\overline{v}\in K$.

Proof. It is obvious that $G_{f}(K)={\rm Aut}_{K}(K[x])\cap
G_{f}(\overline{K})$. By Theorem 4.3,
$G_{f}(K)=\widetilde{G_{f}}(K)\rtimes\overline{G}_{f}(K)$. It is obvious that
$\widetilde{G_{f}}(K)={\rm Sh}_{V}$ where $V:=K\cap\overline{V}$. By Theorem
4.3,
$\overline{G}_{f}(K)=\langle\sigma_{\lambda_{n^{\prime}},(1-\lambda_{n^{\prime}})\nu^{\prime}}\rangle$
where $\lambda_{n^{\prime}}\in K$ is a primitive $n^{\prime}$’th root of unity
and $\nu^{\prime}\in K$ provided $\overline{G}_{f}(K)\neq\\{e\\}$. Notice that

$\sigma_{\lambda_{n^{\prime}},(1-\lambda_{n^{\prime}})\nu^{\prime}}=\sigma^{i}_{\lambda_{n},(1-\lambda_{n})\nu}\sigma_{1,\overline{v}}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu}\sigma_{1,\overline{v}}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu+\overline{v}}$

for unique elements $i$ and $\overline{v}\in\overline{V}$ such that $0\leq
i\leq n-1$. So, the elements $i$ can be chosen such that

$\displaystyle i$ $\displaystyle=$
$\displaystyle\min\\{i^{\prime}=1,\ldots,n-1\,|\,i^{\prime}|n,\;\lambda_{n}^{i^{\prime}}\in
K,\;(1-\lambda_{n}^{i^{\prime}})\nu+\overline{v}\in K\;\;{\rm
for\;some\;element}\;\;\overline{v}\in\overline{V}\\}$ $\displaystyle=$
$\displaystyle\min\\{i^{\prime}=1,\ldots,n-1\,|\,i^{\prime}|n,\;\lambda_{n}^{i^{\prime}}\in
K,\;(1-\lambda_{n}^{i^{\prime}})\nu\in\overline{V}+K\\}$

and $\overline{v}\in\overline{V}$ is any (fixed) element such that
$(1-\lambda_{n}^{i})\nu+\overline{v}\in K$. $\Box$

Proposition 4.25 gives criteria for the groups $\widetilde{G}_{f}(K)$,
$\overline{G}_{f}(K)$ and $G_{f}(K)$ to be $\\{e\\}$.

###### Proposition 4.25

Suppose the field $K$ is not necessarily algebraically closed and $f(x)\in
K[x]$ is a monic polynomial that has at least two distinct roots in
$\overline{K}$. Recall that (Theorem 4.10 and Theorem 4.3) the group
$G_{f}(\overline{K})=\widetilde{G_{f}}(\overline{K})\rtimes\overline{G}_{f}(\overline{K})$
where
$\overline{G}_{f}(\overline{K}):=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
and
$\widetilde{G_{f}}(\overline{K}):=\\{\sigma_{1,\mu}\,|\,\mu\in\overline{V}\\}$,
$\lambda_{n}\in\overline{K}$ is a primitive $n$’th root of unity provided
$\overline{G}_{f}(\overline{K})\neq\\{e\\}$, $\nu\in\overline{K}$,
$\overline{V}$ is a finite dimensional $\mathbb{F}_{p}(\lambda_{n})$-subspace
of $\overline{K}$, and $\mathbb{F}_{p}(\lambda_{n})=\mathbb{F}_{p^{m}}$ for
some $m\geq 1$ (Lemma 4.2). Then

  1. 1.

$\widetilde{G}_{f}(K)=\\{e\\}$ iff $\overline{V}\cap K=0$.

  2. 2.

$\overline{G}_{f}(K)=\\{e\\}$ iff $\overline{G}_{f}(\overline{K})=\\{e\\}$ or
otherwise
$\overline{G}_{f}(\overline{K}):=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
and there is no a natural number $i^{\prime}$ such that $1\leq i^{\prime}\leq
n-1$ such that $i^{\prime}|n$, $\lambda_{n}^{i^{\prime}}\in K$ and
$(1-\lambda_{n}^{i^{\prime}})\nu\in\overline{V}+K$.

  3. 3.

$G_{f}(K)=\\{e\\}$ iff $G_{f}(\overline{K})=\\{e\\}$ or otherwise
$\overline{V}\cap K=0$,
$\overline{G}_{f}(\overline{K}):=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
and there is no a natural number $i^{\prime}$ such that $1\leq i^{\prime}\leq
n-1$ such that $i^{\prime}|n$, $\lambda_{n}^{i^{\prime}}\in K$ and
$(1-\lambda_{n}^{i^{\prime}})\nu\in\overline{V}+K$.

Proof. Statements 1 and 2 follow at once from Theorem 4.24. Then statement 3
follows from statements 1 and 2. $\Box$

Every subgroup of ${\rm Aut}_{K}(K[x])$ is of type $G_{f}$. Theorem 4.26 shows
that all subgroups of ${\rm Aut}_{K}(K[x])$ are eigengroups of polynomials.

###### Theorem 4.26

Let $K$ be an arbitrary field of characteristic $p>0$. Then for each subgroup
$H$ of ${\rm Aut}_{K}(K[x])$ there is a monic polynomial $f_{H}$ such that
$G_{f_{H}}=H$:

  1. 1.

For $H=\\{e\\}$, $f_{H}=x(x+1)^{2}$.

  2. 2.

For $H=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ where
$\lambda_{n}\in K$ is a primitive $n$’th root of unity and $\nu\in K$,
$f_{H}=(x-\nu)^{n}-1$.

  3. 3.

For $H={\rm Sh}_{V}$ where $V$ is a nonzero $\mathbb{F}_{p}$-subspace of $K$,

    1. (a)

if $K=\mathbb{F}_{p^{n}}$ then $f_{H}(x)=f_{V}(x-\nu)-\rho$ where $\rho$ is
any element of $\mathbb{F}_{p^{n}}$ that does not belong to the image of the
map $f_{V}(x-\nu):\mathbb{F}_{p^{n}}\rightarrow\mathbb{F}_{p^{n}}$, $x\mapsto
f_{V}(x-\nu)$ (the map $f_{V}(x-\nu)$ is not a surjection since the set
$\nu+V$ is mapped to $0$).

    2. (b)

If $|K|=\infty$ then $f_{H}(x)=f_{V}(x)f_{V}^{2}(x-\nu)$ where $\nu\in
K\backslash V$.

  4. 4.

For $H={\rm
Sh}_{V}\rtimes\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ where $V$
is a nonzero $\mathbb{F}_{p}$-subspace of $K$, $\mathbb{F}_{p^{e}}$ is its
multiplier field, and $\lambda_{n}$ is primitive $n$’th root of unity such
that $\lambda_{n}V\subseteq V$, $f_{H}(x)=\begin{cases}f_{V}(x-\nu)&\text{if
}n=p^{e}-1,\\\ f_{V}^{n}(x-\nu)+1&\text{if }n<p^{e}-1.\\\ \end{cases}$

  5. 5.

For $H=\mathbb{T}_{\nu}(K)=\\{\sigma_{\lambda,(1-\lambda)\nu}\,|\,\lambda\in
K^{\times}\\}$, $f_{H}=x-\nu$.

  6. 6.

For $H={\rm Aut}_{K}(K[x])$, $f_{H}=\begin{cases}1&\text{if }|K|=\infty,\\\
x^{p^{n}}-x&\text{if }K=\mathbb{F}_{p^{n}}.\\\ \end{cases}$

Proof. 1. If $\sigma\in G_{f_{H}}$ then the maximal ideals $(x)$ and $(x+1)$
of $K[x]$ are $\sigma$-stable, hence $\sigma=e$.

2\. Clearly, $G_{f_{H}}\supseteq H$.

(i) $\widetilde{G}_{f_{H}}=\\{e\\}$: The polynomial $f_{H}$ has $n$ distinct
roots, namely, $\\{\nu+\lambda_{n}^{i}\,|\,i=0,1,\ldots,n-1\\}$, and $p\nmid
n$. Suppose that $\widetilde{G}_{f_{H}}={\rm Sh}_{V}\neq\\{e\\}$ for some
nonzero $\mathbb{F}_{p}$-subspace $V$ of $K$. Then $p||V|$. Since $V+{\cal
R}(f)\subseteq{\cal R}(f)$, we must have $|V|||{\cal R}(f)|$, i.e. $|V||n$,
and so $p|n$, a contradiction.

(ii) $G_{f_{H}}=H$: Let $\sigma=\sigma_{\lambda_{n},(1-\lambda_{n})\nu}$. By
the statement (i),
$G_{f_{H}}=\overline{G}_{f_{H}}=\langle\sigma^{\prime}\rangle$ where
$\sigma=\sigma_{\lambda_{m},(1-\lambda_{m})\nu^{\prime}}$ for some primitive
$m$’th root of unity $\lambda_{m}$ and $\nu^{\prime}\in K$. Since
$H\subseteq\overline{G}_{f_{H}}$, we must have $\nu^{\prime}=\nu$ and $n|m$
(since $(x-\nu^{\prime})$ is the only
$\langle\sigma^{\prime}\rangle$-invariant maximal ideal of $K[x]$, $(x-\nu)$
is the only $\langle\sigma\rangle$-invariant maximal ideal of $K[x]$ and
$\langle\sigma\rangle\subseteq\langle\sigma^{\prime}\rangle$).

The polynomial $f_{H}$ is an eigenvector for the automorphism
$\sigma^{\prime}$ with (necessarily) eigenvalue $\lambda_{m}^{n}$ since
$n=\deg(f_{H})$. Now,
$\lambda_{m}^{n}f_{H}=\lambda_{m}^{n}((x-\nu)^{n}-1)=\sigma^{\prime}(f_{H})=\lambda_{m}^{n}(x-\nu)^{n}-1$.
Therefore, $\lambda_{m}^{n}=1$, and so
$\langle\sigma\rangle=\langle\sigma^{\prime}\rangle$. This means that
$\overline{G}_{f_{H}}=H$, and the statement (ii) follows from the statement
(i).

3(a). Clearly, $G_{f_{H}}\supseteq H={\rm Sh}_{V}$.

(i) $\widetilde{G}_{f_{H}}=H$: The statement follows at once from the fact
that the polynomial $f_{H}$ has $|V|$ distinct roots in $\overline{K}$ (if
$\widetilde{G}_{f_{H}}={\rm Sh}_{V^{\prime}}$ for some
$\mathbb{F}_{p}$-subspace $V^{\prime}$ of $K$ that properly contains $V$ then
the polynomial $f_{H}$ contains at least $|V^{\prime}|$ distinct roots in
$\overline{K}$, a contradiction).

(ii) $\overline{G}_{f_{H}}=\\{e\\}$: Suppose that
$\overline{G}_{f_{H}}\neq\\{e\\}$. Then
$\overline{G}_{f_{H}}=\langle\sigma\rangle$ where
$\sigma=\sigma_{\lambda_{n},(1-\lambda_{n})\nu^{\prime}}$,
$1\neq\lambda_{n}\in K$ is a primitive $n$’th root of unity such that
$\lambda_{n}V\subseteq V$ and $\nu^{\prime}\in K$. Notice that
$\sigma(f_{H})=\lambda_{n}^{|V|}f_{H}$ and

$f_{H}(x)=f_{V}(x-\nu^{\prime}-(\nu-\nu^{\prime}))-\rho=f_{V}(x-\nu^{\prime})-f_{V}(\nu-\nu^{\prime})-\rho=f_{V}(x-\nu^{\prime})+f_{V}(\nu^{\prime}-\nu)-\rho.$

By Theorem 4.6.(1),
$\sigma(f_{V}(x-\nu^{\prime}))=\lambda_{n}f_{V}(x-\nu^{\prime})$. Let
$a=f_{V}(\nu^{\prime}-\nu)-\rho$. Now,

$\lambda_{n}^{|V|}(f_{V}(x-\nu^{\prime})+a)=\lambda_{n}^{|V|}f_{H}=\sigma(f_{H})=\sigma(f_{V}(x-\nu^{\prime})+a)=\lambda_{n}f_{V}(x-\nu^{\prime})+a.$

Hence, $\lambda_{n}^{|V|}=\lambda_{n}\neq 1$ and $(\lambda_{n}-1)a=0$, i.e.
$a=0$. The last equality implies that $\rho\in{\rm im}\,f_{V}(x-\nu)$, a
contradiction, and the statement (ii) follows.

(b). Clearly, $G_{f_{H}}\supseteq H={\rm Sh}_{V}$.

(i) $\widetilde{G}_{f_{H}}=H$: The statement follows at once from the fact
that ${\cal R}(f)=V\coprod(\nu+V)^{2}$ where the upper index $`2^{\prime}$
means that the multiplicity of each root in $\nu+V$ is $2$.

(ii) $\overline{G}_{f_{H}}=\\{e\\}$: Suppose that
$e\neq\sigma\in\overline{G}_{f_{H}}$. Then
$\sigma\in\overline{G}_{f_{V}(x)}\cap\overline{G}_{f_{V}(x-\nu)}$. Notice that
$\overline{G}_{f_{V}(x-\nu)}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$
for some primitive $n$’th root of unity $\lambda_{n}$ such that
$\mathbb{F}_{p}(\lambda_{n})V\subseteq V$. Then there is a natural number $i$
such that
$\sigma=\sigma^{i}_{\lambda_{n},(1-\lambda_{n})\nu}=\sigma_{\lambda_{n}^{i},(1-\lambda_{n}^{i})\nu}\in\overline{G}_{f_{V}(x)}$.
In particular, $V\ni\sigma^{-1}*(0)=(1-\lambda_{n}^{i})\nu$, and so
$\nu\in(1-\lambda_{n}^{i})^{-1}V=V$, a contradiction ($1-\lambda_{n}^{i}\neq
0$ since $\sigma\neq e$).

4\. Statement 4 follows from Theorem 4.18.

5 and 6. Statements 5 and 6 are obvious. $\Box$

Algorithm of finding the eigengroup $G_{f}(\overline{K})$ and the eigenform of
$f$. The algorithm consists of finitely many steps and is based on
Prpoposition 4.8, Theorem 4.18, Theorem 4.20, Theorem 4.22 and Theorem 4.23.
We assume that $K=\overline{K}$.

Step 1. If $|{\cal R}_{d}(f)|=1$ then apply Proposition 4.8 to find $G_{f}$.

From this moment on we assume that $|{\cal R}_{d}(f)|\geq 2$.

Step 2. Use Theorem 4.23 to check whether $G_{f}=\\{e\\}$ or
$G_{f}\neq\\{e\\}$.

From this moment on we assume that $G_{f}\neq\\{e\\}$.

Step 3. By Proposition 4.12.(1), the group $\widetilde{G}_{f}={\rm Sh}_{V}$
can be found.

Step 4. Suppose that $\widetilde{G}_{f}=\\{e\\}$. Then necessarily
$\overline{G}_{f}\neq\\{e\\}$, and using Theorem 4.20 the group
$\overline{G}_{f}$ is found. In more detail, we know that
$\overline{G}_{f}=\langle\sigma_{\lambda_{n},(1-\lambda_{n})\nu}\rangle$ and
that $f(x)=(x-\nu)^{i}g((x-\nu)^{n})$ for a unique $\nu\in{\cal
R}_{d}(f)\cup{\cal R}_{d}(f_{1}^{\prime})$ and $n\geq 2$ such that if $\nu$ is
a root of $f(x)$ then $n={\rm gcd}_{p}(x^{-i}f(x+\nu))$, and if $\nu$ is not a
root of $f(x)$ then $n={\rm gcd}_{p}(f(x+\nu))$.

From this moment on we assume that $\widetilde{G}_{f}={\rm
Sh}_{V}\neq\\{e\\}$, the $\mathbb{F}_{p}$-subspace $V$ of the field $K$ is
non-zero. Let $\mathbb{F}_{p^{e}}$ be the multiplier field of $V$. It can be
easily found since the multiplier field $\mathbb{F}_{p^{e}}$ is the largest
among finite fields $\mathbb{F}_{p^{m}}$ such that
$\mathbb{F}_{p^{m}}V\subseteq V$ and $m\leq{\rm dim}_{\mathbb{F}_{p}}(V)$.

Step 5. Now, we check whether the conditions of Theorem 4.22 hold or not. If
they do then $\overline{G}_{f}=\\{e\\}$.

If they do not then necessarily $\overline{G}_{f}\neq\\{e\\}$ and hence the
conditions of Theorem Theorem 4.18 hold. Using Theorem 4.18 the group
$\overline{G}_{f}$ and the eigenform of $f$ are found in finitely many steps.
$\Box$

Algorithm of finding the eigengroup $G_{f}(K)$ where $K\neq\overline{K}$.

Step 1. Using the algorithm above the group $G_{f}(\overline{K})$ is found.

Step 2. The group $G_{f}(K)$ is found by using Theorem 4.24. $\Box$

${\bf Acnowledgements}$

The author would like to thank the Royal Society for support.

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