# On groups occurring as absolute centers
of finite groups

Georgiana Fasolă and Marius Tărnăuceanu

(October 19, 2023)

###### Abstract

Given a construction $f$ on groups, we say that a group $G$ is $f$-realisable
if there is a group $H$ such that $G\cong f(H)$, and completely $f$-realisable
if there is a group $H$ such that $G\cong f(H)$ and every subgroup of $G$ is
isomorphic to $f(H_{1})$ for some subgroup $H_{1}$ of $H$ and vice versa.

Denote by $L(G)$ the absolute center of a group $G$, that is the set of
elements of $G$ fixed by all automorphisms of $G$. By using the structure of
the automorphism group of a ZM-group, in this paper we prove that cyclic
groups $C_{N}$, $N\in\mathbb{N}^{*}$, are completely $L$-realisable.

MSC2020 : Primary 20D30; Secondary 20D45, 20D25.

Key words : inverse group theory, (completely) $f$-realisable group,
automorphism group, absolute center of a group, ZM-group.

##  1 Introduction

In group theory, there are many constructions $f$ which start from a group $H$
and produce another group $f(H)$. Examples of such group-theoretical
constructions are: center, central quotient, derived quotient, Frattini
subgroup, Fitting subgroup, Chermak-Delgado subgroup, automorphism group,
Schur multiplier, other cohomology groups, and various constructions from
permutation groups. For each of these constructions, there is an inverse
problem:

$\mbox{Given a group }G,\mbox{ is there a group }H\mbox{ such that }G\cong
f(H)?$ (1)

Several new results related to this problem have been obtained in [1, 2, 10,
11] for $f(H)=D(H)$, the derived subgroup of $H$. Note that in these papers
the group $H$ with the property $G\cong D(H)$ has been called an integral of
$G$ by analogy with calculus. Moreover, we recall Problem 10.19 in [1] that
asks to classify the groups in which all subgroups are integrable.

Other results of the same type are given by [7, 15, 21, 22, 23] for
$f(H)=\Phi(H)$, the Frattini subgroup of $H$. In this case, there is a precise
characterization of finite groups $G$ for which (1) has solutions, namely

$G\cong\Phi(H)\mbox{ for some group }H\mbox{ if and only if }{\rm
Inn}(G)\subseteq\Phi({\rm Aut}(G))$

(see [7]).

The same problem for $f(H)={\rm Aut}(H)$, the automorphism group of $H$, has
been studied in [16, 17]. We also recall the well-known class of capable
groups, i.e. the groups $G$ such that (1) has solutions for $f(H)={\rm
Inn}(H)$, the inner automorphism group of $H$. Their study was initiated by R.
Baer [3] and continued in many other papers (see e.g. [4, 8]).

Inspired by these studies, in [9] we introduced the following two notions.
Given a construction $f$ on groups, we say that a group $G$ is

a)

$f$-realisable if there is a group $H$ such that $G\cong f(H)$

and

b)

completely $f$-realisable if there is a group $H$ such that:

  * i)

$G\cong f(H)$;

  * ii)

$\forall\,G_{1}\leq G,\exists\,H_{1}\leq H$ such that $G_{1}\cong f(H_{1})$;

  * iii)

$\forall\,H_{1}\leq H,\exists\,G_{1}\leq G$ such that $f(H_{1})\cong G_{1}$.

We determined completely ${\rm Aut}$-realisable groups. We also provided some
results about (completely) $f$-realisable groups for $f=Z,F,M,D,\Phi$, where
$Z(H)$, $F(H)$ and $M(H)$ denote the center, the Fitting subgroup and the
Chermak-Delgado subgroup of the group $H$, respectively.

The current paper deals only with finite groups. We will continue the above
study for $f=L$, where $L(H)$ is the absolute center of the group $H$, i.e.

$L(H)=\\{h\in H\mid\alpha(h)=h,\forall\,\alpha\in{\rm Aut}(H)\\}.$

It has been introduced by Hegarty [12], together with the autocommutator
subgroup of the group $H$:

$K(H)=\langle h^{-1}\alpha(h)\mid h\in H,\alpha\in{\rm Aut}(H)\rangle.$

Hegarty proved an analogue of Schur’s theorem for the absolute center and the
autocommutator subgroup, namely that if $H$ is a group such that $H/L(H)$ is
finite, then so are $K(H)$ and ${\rm Aut}(H)$. The converse of this result is
also true, as it has been shown in [13].

Clearly, the absolute center of a group is abelian, since $L(H)\subseteq Z(H)$
for all groups $H$. It was determined for several classes of $p$-groups, such
as for abelian $p$-groups in [6], for minimal non-abelian $p$-groups in [18]
or for $p$-groups of maximal class in [19]. Note that in all these cases we
have

$L(H)\cong C_{p}^{k}\mbox{ for certain }k=0,1,2.$

Also, absolute centers of direct products of such groups of coprime orders
will be direct products of elementary abelian $p$-groups of rank $\leq 2$,
according to Lemma 2.1 of [18]. This leads to the following natural question:

$\mbox{Which (abelian) groups occur as absolute centers of finite groups}?$
(2)

In what follows, we will show that finite cyclic groups satisfy this property.
More precisely, the following more powerful result holds.

######  Theorem 1.1.

All finite cyclic groups are completely $L$-realisable.

The proof of Theorem 1.1 is based on computing the absolute center of a ZM-
group, that is a finite group all of whose Sylow subgroups are cyclic. By
[14], such a group is of type

${\rm ZM}(m,n,r)=\langle a,b\mid a^{m}=b^{n}=1,\hskip
2.84526ptb^{-1}ab=a^{r}\rangle,$

where the triple $(m,n,r)$ satisfies the conditions

${\rm gcd}(m,n)={\rm gcd}(m,r-1)=1\mbox{ and }r^{n}\equiv 1\hskip
2.84526pt({\rm mod}\hskip 2.84526ptm).$

Note that $|{\rm ZM}(m,n,r)|=mn$, ${\rm ZM}(m,n,r)^{\prime}=\langle a\rangle$
and $Z({\rm ZM}(m,n,r))=\langle b^{d}\rangle$, where $d$ is the multiplicative
order of $r$ modulo $m$, i.e.

$d=o_{m}(r)={\rm min}\\{k\in\mathbb{N}^{*}\mid r^{k}\equiv 1\hskip
2.84526pt({\rm mod}\hskip 2.84526ptm)\\}.$

Finally, we mention that we were unable to give a complete answer to question
(2), but, based on the above results, the following conjecture seems
reasonable:

Conjecture. All finite abelian groups are completely $L$-realisable.

Most of our notation is standard and will usually not be repeated here.
Elementary notions and results on groups can be found in [14]. For subgroup
lattice concepts we refer the reader to [20].

##  2 Automorphisms of ZM-groups

The general form of automorphisms of a metacyclic group has been determined in
the main result of [5]. In our case, we get:

######  Theorem 2.1.

Each automorphism of ${\rm ZM}(m,n,r)$ is given by

$b^{u}a^{v}\mapsto b^{yu}a^{x_{1}v+x_{2}[u]_{r}},\,u,v\geq 0,$

for a unique triple of integers $(x_{1},x_{2},y)$ such that

$0\leq x_{1},x_{2}<m,\,(x_{1},m)=1,\,0\leq y<n\mbox{ and }y\equiv 1\hskip
2.84526pt({\rm mod}\hskip 2.84526ptd),$

where

$[u]_{r}=\left\\{\begin{array}[]{lll}\\!\\!1+r+\cdots+r^{u-1},&u>0\\\
\\!\\!0,&u=0.\end{array}\right.$

In particular, we have

$|{\rm Aut}({\rm ZM}(m,n,r))|=m\varphi(m)\frac{n}{d}\,.$

Since ${\rm Inn}({\rm ZM}(m,n,r))$ is of order $md$, Theorem 2.1 leads to the
following corollary.

######  Corollary 2.2.

We have

$|{\rm Out}({\rm ZM}(m,n,r))|=\frac{\varphi(m)n}{d^{2}}$

and

${\rm ZM}(m,n,r)\mbox{ is a complete group
}\Leftrightarrow\,\varphi(m)=n=d.\,$

The form of central automorphisms and of IA-automorphisms of ${\rm ZM}(m,n,r)$
can be also inferred from Theorem 2.1.

######  Corollary 2.3.

Each central automorphism of ${\rm ZM}(m,n,r)$ is given by

$b^{u}a^{v}\mapsto b^{yu}a^{v},\,u,v\geq 0,$

for a unique integer $y$ such that

$0\leq y<n\mbox{ and }y\equiv 1\hskip 2.84526pt({\rm mod}\hskip 2.84526ptd)$

and each IA-automorphism of ${\rm ZM}(m,n,r)$ is given by

$b^{u}a^{v}\mapsto b^{u}a^{x_{1}v+x_{2}[u]_{r}},\,u,v\geq 0,$

for a unique pair of integers $(x_{1},x_{2})$ such that

$0\leq x_{1},x_{2}<m\mbox{ and }(x_{1},m)=1.$

In particular, we have

$|{\rm Aut_{C}}({\rm ZM}(m,n,r))|=\frac{n}{d}\mbox{ and }|{\rm Aut_{IA}}({\rm
ZM}(m,n,r))|=m\varphi(m).$

Next, we determine the absolute center of ${\rm ZM}(m,n,r)$.

######  Theorem 2.4.

We have

$L({\rm ZM}(m,n,r))=\langle b^{de}\rangle,$

where

$e=\min\\{s\in\mathbb{N}^{*}\mid d^{2}s\equiv 0\hskip 2.84526pt({\rm
mod}\hskip 2.84526ptn)\\}.$

Moreover

$|L({\rm ZM}(m,n,r))|=\frac{n}{(de,n)}=\frac{\frac{n}{d}}{(e,\frac{n}{d})}\,.$

###### Proof.

Since $L({\rm ZM}(m,n,r))\subseteq Z({\rm ZM}(m,n,r))$, it follows that

$L({\rm ZM}(m,n,r))=\langle b^{de}\rangle\mbox{ for some }e\in\mathbb{N}^{*}.$

By Theorem 2.1, the condition $b^{de}\in L({\rm ZM}(m,n,r))$ means

$b^{yde}a^{x_{2}[de]_{r}}=b^{de},$

that is

$n\,|\,de(y-1)\mbox{ and }m\,|\,x_{2}[de]_{r}$ (3)

for all pairs of integers $(x_{2},y)$ satisfying $0\leq x_{2}<m$, $0\leq y<n$
and $y\equiv 1\hskip 2.84526pt({\rm mod}\hskip 2.84526ptd)$. Clearly, the
relations (3) hold if and only if they hold for $x_{2}=1$ and $y=d$, that is

$n\,|\,d^{2}e\mbox{ and }m\,|\,[de]_{r}.$

Since $m\,|\,[de]_{r}$ is true for all $e\in\mathbb{N}^{*}$, we must have only
$n\,|\,d^{2}e$ and the proof is completed by the remark that $e$ must be
chosen minimal with respect to this condition. ∎

For example, for ${\rm ZM}(5,16,2)=SmallGroup(80,3)$ we get $d=4$, $e=1$ and
so $L({\rm ZM}(5,16,2))=Z({\rm ZM}(5,16,2))\cong C_{4}$, while for ${\rm
ZM}(5,48,2)=SmallGroup(240,5)$ we get $d=4$, $e=3$ and so $L({\rm
ZM}(5,48,2))\cong C_{4}$. Note that in the second case we have $L({\rm
ZM}(5,48,2))\neq Z({\rm ZM}(5,48,2))\cong C_{12}$.222More precisely, since
${\rm ZM}(5,48,2)\cong C_{3}\times{\rm ZM}(5,16,2)$, we have $L({\rm
ZM}(5,48,2))\cong L(C_{3})\times L({\rm ZM}(5,16,2))\cong 1\times C_{4}\cong
C_{4}$ and $Z({\rm ZM}(5,48,2))\cong Z(C_{3})\times Z({\rm ZM}(5,16,2))\cong
C_{3}\times C_{4}\cong C_{12}$.

##  3 Proof of the main result

First of all, we will show that $L({\rm ZM}(m,n,r))$ can be any cyclic group
$C_{N}$ when $N=q^{\alpha}$ is a prime power. By Theorem 2.4, we must find a
triple of positive integers $(m,n,r)$ such that

$\left\\{\begin{array}[]{lll}\\!\\!{\rm gcd}(m,n)={\rm gcd}(m,r-1)=1\\\
\\!\\!r^{n}\equiv 1\hskip 2.84526pt({\rm mod}\hskip 2.84526ptm)\\\
\\!\\!\frac{n}{(de,n)}=N,\end{array}\right.$ (4)

where

$\left\\{\begin{array}[]{lll}\\!\\!d=o_{m}(r)\\\
\\!\\!e=\min\\{s\in\mathbb{N}^{*}\mid d^{2}s\equiv 0\hskip 2.84526pt({\rm
mod}\hskip 2.84526ptn)\\}.\end{array}\right.$

We start with the following auxiliary result.

######  Lemma 3.1.

There exist a prime $p$ and a positive integer $r$ such that
$o_{p}(r)=q^{\alpha}$.

###### Proof.

By Dirichlet’s Theorem, the arithmetic progression $1+tq^{\alpha}$,
$t\in\mathbb{N}$, contains an infinite number of primes. Let $p$ be one of
them. Then $q^{\alpha}$ divides $p-1$. Since the multiplicative group of
integers modulo $p$ is cyclic of order $p-1$, it contains elements of order
$q^{\alpha}$. Thus there is $r\in\mathbb{N}^{*}$ such that
$o_{p}(r)=q^{\alpha}$, as desired. ∎

Clearly, the triple $(p,q^{2\alpha},r)$, where $p$ and $r$ are given by Lemma
3.1, satisfies the conditions (4), that is we have

$L({\rm ZM}(p,q^{2\alpha},r))\cong C_{q^{\alpha}}.$

For the proof of Theorem 1.1, we also need the following corollary.

######  Corollary 3.2.

Let $P$ be the set of all primes. Then for any $q_{1},...,q_{k}\in P$ and any
$\alpha_{1},...,\alpha_{k}\in\mathbb{N}^{*}$, there exist distinct primes
$p_{1},...,p_{k}\in P\setminus\\{q_{1},...,q_{k}\\}$ and positive integers
$r_{1},...,r_{k}$ such that $o_{p_{i}}(r_{i})=q_{i}^{\alpha_{i}}$,
$i=1,...,k$.

###### Proof.

We apply Lemma 3.1 for all $i=1,...,k$. It suffices to observe that the primes
$p_{1}$, …, $p_{k}$ can be chosen to be distinct, and different from
$q_{1},...,q_{k}$, by Dirichlet’s Theorem. ∎

We are now able to prove our main result.

###### Proof of Theorem 1.1..

Let $N\geq 2$ be an integer and $N=q_{1}^{\alpha_{1}}\cdots
q_{k}^{\alpha_{k}}$ be the decomposition of $N$ as a product of prime factors.
We choose $p_{1}$, …, $p_{k}$ and $r_{1}$, …, $r_{k}$ as in Corollary 3.2, and
let

$H_{i}={\rm ZM}(p_{i},q_{i}^{2\alpha_{i}},r_{i}),\,i=1,...,k.$

Then, for each $i$, we have

$L(H_{i})\cong C_{q_{i}^{\alpha_{i}}}.$

Let $H=H_{1}\times\cdots\times H_{k}$. Since the groups $H_{i}$, $i=1,...,k$,
are of coprime orders, we infer that they are characteristic in $H$ and
therefore

$L(H)\cong L(H_{1})\times\cdots\times L(H_{k})\cong
C_{q_{1}^{\alpha_{1}}}\times\cdots\times C_{q_{k}^{\alpha_{k}}}\cong C_{N},$

i.e. $C_{N}$ is $L$-realisable.

Obviously, every subgroup $G_{1}$ of $C_{N}$ is of type $C_{N_{1}}$, where
$N_{1}=q_{1}^{\beta_{1}}\cdots q_{k}^{\beta_{k}}$ and
$0\leq\beta_{i}\leq\alpha_{i}$, $i=1,...,k$. It is easy to see that each group
$H_{i}$ contains a normal subgroup $S_{i}\cong{\rm
ZM}(p_{i},q_{i}^{\alpha_{i}+\beta_{i}},r_{i})$. Since all normal subgroups of
a ZM-group are characteristic, it follows that $S_{i}$ is characteristic in
$H_{i}$, and so in $H$. Moreover, we have $L(S_{i})\cong
C_{q_{i}^{\beta_{i}}}$. Then the subgroup $S=S_{1}\times\cdots\times S_{k}$ of
$H$ satisfies $L(S)\cong G_{1}$.

Conversely, we observe that every subgroup $T$ of $H$ is a direct product
$T_{1}\times\cdots\times T_{k}$, where $T_{i}\leq H_{i}$, $i=1,...,k$. Then
each $T_{i}$ is either cyclic or of type ${\rm
ZM}(p_{i},q_{i}^{\gamma_{i}},r_{i})$ with $\alpha_{i}\leq\gamma_{i}\leq
2\alpha_{i}$. In the second case, we get

$L(T_{i})\leq Z(T_{i})\cong C_{q_{i}^{\gamma_{i}-\alpha_{i}}}.$

Consequently, $L(T_{i})$ is either trivial or isomorphic to a subgroup of
$C_{q_{i}^{\alpha_{i}}}$. It follows that $L(T)\cong
L(T_{1})\times\cdots\times L(T_{k})$ is isomorphic to subgroup of
$C_{q_{1}^{\alpha_{1}}}\times\cdots\times C_{q_{k}^{\alpha_{k}}}\cong C_{N}$.

This completes the proof. ∎

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Georgiana Fasolă

Faculty of Mathematics

”Al.I. Cuza” University

Iaşi, Romania

e-mail<EMAIL_ADDRESS>
Marius Tărnăuceanu

Faculty of Mathematics

”Al.I. Cuza” University

Iaşi, Romania

e-mail<EMAIL_ADDRESS>


# Rational twisted series

Masood Aryapoor
Division of Mathematics and Physics
Mälardalen University
Hamngatan 15, 632 17, Eskilstuna, Sweden

###### Abstract

Rational twisted power series over a (commutative) filed are studied. We give
several characterizations of such series which are similar to the classical
results concerning rational power series over a commutative field. In
particular, we prove a version of Kronecker’s lemma for the rationality of
twisted power series.
Keywords: Rational Twisted series, $\sigma$-Recognizable series, Syntactic
left ideal, Twisted Hadamard algebra

##  1 Introduction

The study of rational formal power series has led to a rich theory with
applications in areas such as number theory and formal language theory, see
[3] and [1]. The main purpose of this paper is to develop a theory of rational
twisted formal power series over a commutative field, similar to the theory of
rational formal power series as presented in [1].

The paper is organized as follows. Section 2 is devoted to preliminaries. Most
of the results in this section are classical. The only new result is
Kronecker’s lemma for rational twisted formal power series, see Subsection
2.5. In Section 3, the concept of $\sigma$-recognizable formal power series is
introduced and studied. The most important result of this section is that a
twisted formal power series is $\sigma$-recognizable iff it is rational. It is
also shown that the Hadamard product of rational twisted formal power series
is rational. Section 4 deals with the notion of syntactic left ideal which is
used to give a characterization of rational twisted formal power series. More
precisely, it is shown that a twisted formal power series is rational iff its
syntactic left ideal is nonzero. The next section is devoted to reduced
$\sigma$-linear representations. In the final section of the paper, we
investigate the connection between fractions of the twisted polynomial ring
and syntactic left ideals. Also, the notion of regularity is introduced and it
is shown that the ableian group of regular twisted formal power series
equipped with the Hadamard product is a commutative algebra.

Throughout this paper, $K$ is a fixed commutative field, and $\sigma:K\to K$
is a fixed endomorphism of $K$. By ”vector space”, we shall mean ”vector space
over $K$”. The notation $K^{m\times n}$ is used to denote the vector space of
all $m\times n$ matrices with entries in $K$.

##  2 Twisted series

In this section, we review basic facts about twisted polynomial rings and
twisted formal power series, see [2] for more details.

###  2.1 The ring $K[T;\sigma]$

Every nonzero element $P$ of the ring $K[T;\sigma]$ of twisted polynomials
over $K$ can uniquely be written as

$P=a_{0}+a_{1}T+\cdots+a_{n}T^{n}=\sum_{i=0}^{n}a_{i}T^{i},$

where $n\geq 0$ and $a_{0},...,a_{n}\in K$ with $a_{n}\neq 0$. The number $n$,
denoted by $\deg P$, is called the degree of $P$. The notation $P(0)$ will be
used to denote the constant term $a_{0}$. In this ring, we have
$Ta=\sigma(a)T$ for all $a\in K$, and more generally,

$\left(\sum_{i}a_{i}T^{i}\right)\left(\sum_{j}b_{j}T^{j}\right)=\sum_{n}\left(\sum_{i+j=n}a_{i}\sigma^{i}(b_{j})\right)T^{n}.$

###  2.2 The field of fractions of $K[T;\sigma]$

It is known that $K[T;\sigma]$ is an integral domain. Moreover, the Euclidean
algorithm holds for right division, that is, given $P,Q\in K[T;\sigma]$ with
$Q\neq 0$ there exist $S,R\in K[T;\sigma]$ such that $\deg R<\deg Q$ and
$P=SQ+R$. It follows that $K[T;\sigma]$ is a left principal ideal domain.
Since $K[T;\sigma]$ is left Noetherian, we see that $K[T;\sigma]$ is a left
Ore domain, and consequently, has a field of fractions which is denoted by
$K(T;\sigma)$. Any element of $K(T;\sigma)$ can be written as a left fraction
$P^{-1}Q$ where $P,Q\in K[T;\sigma]$ with $P\neq 0$. We remark that if
$\sigma$ is an automorphim, then $K[T;\sigma]$ is also a right principal ideal
domain and a right Ore domain.

###  2.3 The ring of twisted formal power series

Every element $f$ of the ring $K[[T;\sigma]]$ of twisted formal series over
$K$ can uniquely be written as

$f=\sum_{i=0}^{\infty}a_{i}T^{i}=a_{0}+a_{1}T+\cdots+a_{n}T^{n}+\cdots,$

where $a_{0},a_{1},...\in K$. For simplicity, elements of $K[[T;\sigma]]$ will
be called twisted series. As before, $f(0)$ denotes the constant term $a_{0}$.
The ring $K[[T;\sigma]]$ can be considered as a topological ring in which the
left ideals $K[[T;\sigma]]T^{n}$, where $n\geq 0$, form a neighborhood basis
of $0$. Note that an element $f\in K[[T;\sigma]]$ has an inverse $f^{-1}\in
K[[T;\sigma]]$ iff $f(0)\neq 0$. In fact, if $g\in K[[T;\sigma]]$ satisfies
$g(0)=0$, then

$(1-g)^{-1}=\sum_{i=0}^{\infty}g^{i}.$

We also have the ring of twisted formal Laurent series $K((T;\sigma))$ which
contains $K[[T;\sigma]]$ as a subring. Every nonzero element $f$ of the ring
$K((T;\sigma))$ can uniquely be written as

$f=\sum_{i=-m}^{\infty}a_{i}T^{i}=a_{-m}T^{-m}+\cdots+a_{-1}T^{-1}+a_{0}+a_{1}T+\cdots,$

where $m\in\mathbb{Z}$ and $a_{-m},a_{-m+1},...\in K$ with $a_{-m}\neq 0$. It
is known that $K((T;\sigma))$ is a (skew) field and we have a natural
embedding of $K(T;\sigma)$ in $K((T;\sigma))$. In this paper, we consider
$K(T;\sigma)$ as a subfield of $K((T;\sigma))$ under this natural embedding.

###  2.4 Rational twisted series

A twisted series $f\in K[[T;\sigma]]$ is called rational if $f\in
K(T;\sigma)$, that is, there exist twisted polynomials $P,Q\in K[T;\sigma]$
such that $P\neq 0$ and

$f=P^{-1}Q.$

We give a proof of the following known rationality criterion, see Theorem
1.5.6 in [2].

######  Theorem 2.1.

A twisted series

$f=\sum_{i=0}^{\infty}a_{i}T^{i}\in K[[T;\sigma]],$

is rational iff there exist elements $c_{1},...,c_{p}\in K$ such that

$a_{n}+\sum_{i=1}^{p}c_{i}\sigma^{i}(a_{n-i})=0,$

for all $n\geq n_{0}$.

###### Proof.

The proof of the ”if” part is easy since

$(1+\sum_{i=1}^{p}c_{i}T^{i})f\in K[T;\sigma].$

To prove the other direction, let $f=\sum_{i=0}^{\infty}a_{i}T^{i}$ be
rational. Then, there exists a twisted polynomial

$P=b_{0}T^{k}+\cdots+b_{p}T^{p+k},$

where $p,k\geq 0$ and $b_{0},...,b_{p}\in K$, such that

$Pf\in K[T;\sigma].$

Since

$Pf=(b_{0}T^{k}+\cdots+b_{p}T^{p+k})(\sum_{i=0}^{\infty}a_{i}T^{i})=\sum_{n\geq
0}\left(\sum_{i+j=n}b_{i}\sigma^{k+i}(a_{j})\right)T^{k+n},$

we obtain

$\sum_{i=0}^{p}b_{i}\sigma^{k+i}(a_{n-i})=0,$

for all $n\geq n_{0}\geq p$. Therefore, the vectors

$v_{i}=(\sigma^{k+i}(a_{n_{0}-i}),\sigma^{k+i}(a_{n_{0}+1-i}),...,\sigma^{k+i}(a_{n-i}),...),$

where $i=0,...,p$, are linearly dependent over $K$. Without loss of
generality, we may assume that $v_{1},...,v_{p}$ are linearly independent over
$K$. Setting

$w_{i}=(\sigma^{i}(a_{n_{0}-i}),\sigma^{i}(a_{n_{0}+1-i}),...,\sigma^{i}(a_{n-i}),...),$

we have $v_{i}=\sigma^{k}(w_{i})$, for $i=0,...,p$. Since $\sigma$ is a field
homomorphism, it follows that $w_{1},...,w_{p}$ are linearly independent, and
$w_{0},...,w_{p}$ are linearly dependent over $K$. Therefore, there exist
elements $c_{1},...,c_{p}\in K$ such that
$w_{0}+c_{1}w_{1}+\cdots+c_{p}w_{p}=0$, or equivalently,

$a_{n}+\sum_{i=1}^{p}c_{i}\sigma^{i}(a_{n-i})=0,$

for all $n\geq n_{0}$, finishing the proof. ∎

The proof of the theorem gives us the following result:

######  Corollary 2.2.

Any rational twisted series $f\in K[[T;\sigma]]$ can be written as $f=P^{-1}Q$
for some $P,Q\in K[T;\sigma]$ such that $P(0)=1$.

###  2.5 Kronecker’s lemma for twisted series

In this part, we prove a criterion for the rationality of a twisted series
which is similar to a classical result for ordinary power series known as
Kronecker’s lemma, see [3].

######  Lemma 2.3.

A twisted series

$f=\sum_{i=0}^{\infty}a_{i}T^{i}\in K[[T;\sigma]],$

is rational iff

$\begin{vmatrix}\sigma^{m}(a_{0})&\sigma^{m-1}(a_{1})&\cdots&\sigma(a_{m-1})&a_{m}\\\
\sigma^{m}(a_{1})&\sigma^{m-1}(a_{2})&\cdots&\sigma(a_{m})&a_{m+1}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
\sigma^{m}(a_{m})&\sigma^{m-1}(a_{m+1})&\cdots&\sigma(a_{2m-1})&a_{2m}\end{vmatrix}=0$

for all $m>m_{0}$.

###### Proof.

Set

$D_{m}=\begin{pmatrix}\sigma^{m}(a_{0})&\sigma^{m-1}(a_{1})&\cdots&\sigma(a_{m-1})&a_{m}\\\
\sigma^{m}(a_{1})&\sigma^{m-1}(a_{2})&\cdots&\sigma(a_{m})&a_{m+1}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
\sigma^{m}(a_{m})&\sigma^{m-1}(a_{m+1})&\cdots&\sigma(a_{2m-1})&a_{2m}\end{pmatrix},$

where $m\geq 0$. First suppose that $f$ is rational. By Theorem 2.1, there are
$c_{1},...,c_{p}\in K$ such that

$a_{n}+\sum_{i=1}^{p}c_{i}\sigma^{i}(a_{n-i})=0,$

for all $n\geq n_{0}$. It follows that the determinant of $D_{m}$ is zero for
all $m\geq n_{0}$.

Conversely, suppose that $|D_{m}|=0$ for all $m>n_{0}$. If
$\left|D_{m}\right|=0$ for all $m=0,1,2,...$, then it is easy to see that we
must have $f=0$, and we are done. Otherwise, we can find $p\in\mathbb{N}$ such
that $|D_{p-1}|\neq 0$ and $|D_{m}|=0$ for all $m\geq p$. Since
$\left|D_{p}\right|=0$ and $\left|D_{p-1}\right|\neq 0$, we can find
$c_{1},...,c_{p}\in K$ such that

$c_{p}\sigma^{p}(a_{j})+c_{p-1}\sigma^{p-1}(a_{j+1})+\cdots+c_{1}\sigma(a_{j+p-1})+a_{j+p}=0,$

where $0\leq j\leq p$. We set

$L_{j}=a_{j+p}+\sum_{k=1}^{p}c_{k}\sigma^{k}(a_{j+p-k}),$

and use induction to prove that $L_{j}=0$ for all $j\geq 0$. Assume that $m>p$
and $L_{j}=0$ when $j<m$. We note that

$\sigma^{r}(L_{j})=\sigma^{r}(a_{j+p})+\sum_{k=1}^{p}\sigma^{r}(c_{k})\sigma^{k+r}(a_{j+p-k}),$

for all $r\geq 0$, since $\sigma$ is a field homomorphism. Now, let
$A=(c_{ij})$ be the following $(m+1)\times(m+1)$ matrix

$c_{ij}=\begin{cases}\ \sigma^{m+1-j}(c_{j-i})&j>p\text{ and }j-p\leq i\leq
j-1,\\\ 1&i=j,\\\ 0&\text{otherwise.}\\\ \end{cases}$

Then the $ij$-th entry of $D_{m}A$ is equal to $\sigma^{m-j+1}(a_{i+j-1})$ if
$j\leq p$, and equal to

$\sigma^{m-j+1}(a_{i+j-1})+\sum_{k=j-p}^{j-1}\sigma^{m-k+1}(a_{i+k-1})\sigma^{m+1-j}(c_{j-k})=$
$\displaystyle\sigma^{m-j+1}\left(a_{i+j-1}+\sum_{k=j-p}^{j-1}c_{j-k}\sigma^{j-k}(a_{i+k-1})\right)$
$\displaystyle=\sigma^{m-j+1}\left(a_{i+j-1}+\sum_{k=1}^{p}c_{k}\sigma^{k}(a_{i+j-1-k})\right)$
$\displaystyle=\sigma^{m-j+1}\left(L_{i+j-p-1}\right),$

if $j>p$. Since, $|A|=1$, and $L_{j}=0$ for $j<m$, we see that

$|D_{m}|=(-1)^{m-p}|D_{p-1}|\prod_{j=p+1}^{m}\sigma^{m-j+1}\left(L_{m}\right).$

It follows that $L_{m}=0$ since $|D_{m}|=0$, $|D_{p-1}|\neq 0$, and $\sigma$
is injective. The proof of the induction step is finishes. Since $L_{j}=0$ for
all $j\geq 0$, by Lemma 2.1, $f$ must be rational, and we are done.

∎

##  3 $\sigma$-Recognizable twisted series

A twisted series $f=\sum_{n\geq 0}a_{n}T^{n}\in K[[T;\sigma]]$ is called
$\sigma$-recognizable if there exist a matrix $A\in K^{m\times m}$, a row
vector $X\in K^{1\times m}$ and a column vector $Y\in K^{m\times 1}$ such that

$a_{n}=X\left(A\sigma(A)\cdots\sigma^{n-1}(A)\right)\sigma^{n}(Y),$ (3.1)

for all $n\geq 0$ (note that for $n=0$, the formula reads $a_{0}=XY$). The
triple $(X,A,Y)$ is called a $\sigma$-linear representation of $f$, and the
number $m$ is called the dimension of $(X,A,Y)$. We note that Equation 3.1 can
be written as

$f=X(I_{m}-AT)^{-1}Y,$

where

$(I_{m}-AT)^{-1}=\sum_{n\geq 0}(AT)^{n}=\sum_{n\geq
0}A\sigma(A)\cdots\sigma^{n-1}(A)T^{n}\in K[[T;\sigma]]^{m\times m}.$

In order to study $\sigma$-recognizable twisted series, we use the shift map
$s:K[[T;\sigma]]\to K[[T;\sigma]]$ which is defined as follows

$s(\sum_{n\geq 0}a_{n}T^{n})=\sum_{n\geq 0}a_{n+1}T^{n}.$

The proof of the following lemma is left to the reader.

######  Lemma 3.1.

(1) $s(f+g)=s(f)+s(g)$ for all $f,g\in K[[T;\sigma]]$.
(2) $s(af)=as(f)$ for all $f\in K[[T;\sigma]]$ and $a\in K$.
(3) $s(fa)=s(f)\sigma(a)$ for all $f\in K[[T;\sigma]]$ and $a\in K$.
(4) If $f=\sum_{n\geq 0}a_{n}T^{n}\in K[[T;\sigma]]$, then $a_{n}=s^{n}(f)(0)$
for all $n\geq 0$.

###  3.1 Stability

A subset $P$ of $K[[T;\sigma]]$ is called stable if $s(P)\subset P$.

######  Proposition 3.2.

A twisted series is $\sigma$-recognizable iff it belongs to a finite-
dimensional stable right vector subspace of $K[[T;\sigma]]$. More precisely,
the following hold:
(i) Let $f_{1},...,f_{m}$ be a basis of a stable right vector subspace $V$ of
$K[[T;\sigma]]$ containing $f$. Setting
$F=\begin{pmatrix}f_{1}&f_{2}&\cdots&f_{m}\end{pmatrix}\in
K[[T;\sigma]]^{1\times m},$ we have the following $\sigma$-linear
representation of $f$

$(F(0),A,Y),$

where $Y\in K^{m\times 1}$ and $A\in K^{m\times m}$ satisfy $f=FY$ and
$s(F)=FA$.
(ii) Let $(X,A,Y)$ be a $\sigma$-linear representation of $f$. The right
vector subspace of $K[[T;\sigma]]$ generated by the components of the vector

$X(I_{m}-AT)^{-1}\in K[[T;\sigma]]^{1\times m},$

is a finite-dimensional stable right vector subspace of $K[[T;\sigma]]$
containing $f$.

###### Proof.

(i) We can find a vector $Y\in K^{m\times 1}$ such that $f=FY$ since $f\in V$.
The fact that $V$ is stable, implies that there is a matrix $A\in K^{m\times
m}$ such that $s(F)=FA$. Using Part (3) of Lemma 3.1 and induction on $n$, one
can show that

$s^{n}(f)=\left(FA\sigma(A)\cdots\sigma^{n-1}(A)\right)\sigma^{n}(Y),$

for all $n\geq 0$. It follows that

$a_{n}=s^{n}(f)(0)=F(0)\left(A\sigma(A)\cdots\sigma^{n-1}(A)\right)\sigma^{n}(Y),$

for all $n\geq 0$, proving that $(F(0),A,Y)$ is a $\sigma$-linear
representation of $f$.

(ii) Clearly, $V$ is finite-dimensional, and $f\in V$ because
$f=X(I_{m}-AT)^{-1}Y$. We have

$\displaystyle s\left(X(I_{m}-AT)^{-1}\right)$
$\displaystyle=s\left(\sum_{n\geq
0}\left(XA\sigma(A)\cdots\sigma^{n-1}(A)\right)T^{n}\right)$
$\displaystyle=\sum_{n\geq 0}\left(XA\sigma(A)\cdots\sigma^{n}(A)\right)T^{n}$
$\displaystyle=\sum_{n\geq
0}\left(XA\sigma(A)\cdots\sigma^{n-1}(A)\right)T^{n}A=FA,$

showing that $V$ is stable, and we are done. ∎

We note that if $(X,A,Y)$ is a $\sigma$-linear representation of $f\in
K[T;\sigma]$, then $(aX,A,Y)$ is a $\sigma$-linear representation of $af$, and
$(X,A,aY)$ is a $\sigma$-linear representation of $fa$ for all $a\in K$. Since
the sum of stable sets is stable, we obtain the following result:

######  Corollary 3.3.

Any (left or right) $K$-linear combination of $\sigma$-recognizable twisted
series is $\sigma$-recognizable.

Clearly, the intersection of any family of stable right vector subspaces of
$K[[T;\sigma]]$ is a stable right vector subspace of $K[[T;\sigma]]$.
Therefore, we can define the concept of the smallest stable right vector
subspace $V(S)$ of $K[[T;\sigma]]$ generated by a set $S\subset
K[[T;\sigma]]$. It is easy to see that the vector space

$V(f)=\sum_{i\geq 0}s^{i}(f)K,$

is the smallest stable right vector subspace of $K[[T;\sigma]]$ which contains
$f\in K[[T;\sigma]]$. Therefore, $f$ is $\sigma$-recognizable iff $V(f)$ is
finite-dimensional. Furthermore, we have the following result:

######  Proposition 3.4.

A twisted series $f\in K[T;\sigma]$ is $\sigma$-recognizable iff there is some
$n>0$ such that $V(f)=\sum_{i=0}^{n-1}s^{i}(f)K$, or equivalently

$s^{n}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K.$

###### Proof.

If $s^{n}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K$, then an application of Lemma 3.1
shows that

$s^{m}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K,$

for all $m\geq n$. Therefore, $V(f)$ is finite-dimensional, implying that $f$
is $\sigma$-recognizable, by Proposition 3.2. Conversely, if $f$ is
$\sigma$-recognizable, $V(f)$ must be finite-dimensional. Since
$f,s(f),s^{2}(f),...$ generate this vector space, there exists $n>0$ such that
$f,s(f),s^{2}(f),...,s^{n-1}(f)$ generate $V(f)$ as a right vector space. In
particular, we have $s^{n}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K$. ∎

The condition $s^{n}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K$ can be represented in
terms of the coefficients of $f$. More precisely, let $f=\sum_{j\geq
0}a_{j}T^{j}$. Then the condition $s^{n}(f)\in\sum_{i=0}^{n-1}s^{i}(f)K$
implies that there exist $c_{0},...,c_{n-1}\in K$ such that

$s^{n}(f)=\sum_{i=0}^{n-1}s^{i}(f)c_{i}.$

Since $s^{n}(f)=\sum_{j\geq 0}a_{n+j}T^{j}$, we obtain

$\sum_{j\geq 0}a_{n+j}T^{j}=\sum_{i=0}^{n-1}\left(\sum_{j\geq
0}a_{i+j}T^{j}\right)c_{i}=\sum_{j\geq
0}\left(\sum_{i=0}^{n-1}a_{i+j}\sigma^{j}(c_{i})\right)T^{j}.$

Therefore, we see that

$a_{n+j}=\sum_{i=0}^{n-1}a_{i+j}\sigma^{j}(c_{i}),$

for all $j\geq 0$, proving the following result:

######  Corollary 3.5.

A twisted series $f\in K[T;\sigma]$ is $\sigma$-recognizable iff there exist
$c_{0},...,c_{n-1}\in K$ such that

$a_{n+j}=\sum_{i=0}^{n-1}a_{i+j}\sigma^{j}(c_{i}),$

for all $j\geq 0$.

###  3.2 The main theorem of this section

In this section, we prove that every $\sigma$-recognizable twisted series is
rational, and vice versa. We begin by some lemmas.

######  Lemma 3.6.

Let $f,g\in K[[T;\sigma]]$ be arbitrary twisted series. Then,

$s(fg)=fs(g)+s(f)\sigma(g(0)).$

In particular, if $f(0)\neq 0$, then

$s(f^{-1})=-f^{-1}s(f)\sigma(f^{-1}(0)).$

###### Proof.

Given $f=\sum_{i\geq 0}a_{i}T^{i},g=\sum_{i\geq 0}b_{i}T^{i},$ we write

$fg=\sum_{n\geq 0}\left(\sum_{i+j=n}a_{i}\sigma^{i}(b_{j})\right)T^{n}\implies
s(fg)=\sum_{n\geq 0}\left(\sum_{i+j=n+1}a_{i}\sigma^{i}(b_{j})\right)T^{n}.$

We also have

$fs(g)=\left(\sum_{i\geq 0}a_{i}T^{i}\right)\left(\sum_{j\geq
0}b_{j+1}T^{j}\right)=\sum_{n\geq
0}\left(\sum_{i+j=n}a_{i}\sigma^{i}(b_{j+1})\right)T^{n}.$

It follows that

$s(fg)-fs(g)=\sum_{n\geq
0}a_{n+1}\sigma^{n+1}(b_{0})T^{n}=s(f)\sigma(b_{0})=s(f)\sigma(g(0)).$

The second formula can be derived by setting $g=f^{-1}$ in the first formula
and using the fact that $s(1)=0$. ∎

In what follows, we identify the ring $K[[T;\sigma]]^{m\times m}$ of $m\times
m$ matrices over $K[[T;\sigma]]$ with the ring $K^{m\times m}[[T;\sigma]]$ of
twisted polynomials over the ring $K^{m\times m}$.

######  Lemma 3.7.

Let $M\in K[[T;\sigma]]^{m\times m}$ be a matrix with $M(0)=0$. If the entries
of $M$ are rational twisted series, then so are the entries of the matrix
$(I_{m}-M)^{-1}=\sum_{n\geq 0}M^{n}$, where $I_{m}$ denotes the identity
matrix of order $m$.

###### Proof.

We prove the lemma by induction on $n$. The assertion is trivial when $n=1$.
To prove the induction step, we write $M$ as

$M=\begin{pmatrix}N&X\\\ Y&f\end{pmatrix}.$

where $N$ is a square matrix of order $m-1$, $X$ is a column vector of
appropriate size, $Y$ is a row vector of appropriate size, and $f\in
K[[T;\sigma]]$. We write $(I_{m}-M)^{-1}$ in a similar way:

$(I_{m}-M)^{-1}=\begin{pmatrix}N_{1}&X_{1}\\\ Y_{1}&f_{1}\end{pmatrix}.$

The relation $(I_{m}-M)^{-1}(I_{m}-M)=1$ implies that

$\begin{cases}N_{1}(I_{m-1}-N)-X_{1}Y=I_{m-1}\\\ -N_{1}X+X_{1}(1-f)=0\\\
Y_{1}(I_{m-1}-N)-f_{1}Y=0\\\ -Y_{1}X+f_{1}(1-f)=1\end{cases}.$

We can solve this system of equations: The second and third equations give

$X_{1}=N_{1}X(1-f)^{-1},\,\,Y_{1}=f_{1}Y(I_{m-1}-N)^{-1},$

using which we obtain the equations

$\begin{cases}N_{1}(I_{m-1}-N)-N_{1}X(1-f)^{-1}Y=I_{m-1}\\\
-f_{1}Y(I_{m-1}-N)^{-1}X+f_{1}(1-f)=1\end{cases}.$

It follows that

$\begin{cases}N_{1}=\left(I_{m-1}-N-X(1-f)^{-1}Y\right)^{-1}\\\
X_{1}=\left(I_{m-1}-N-X(1-f)^{-1}Y\right)^{-1}X(1-f)^{-1}\\\
f_{1}=\left(1-f-Y(I_{m-1}-N)^{-1}X\right)^{-1}\\\
Y_{1}=\left(1-f-Y(I_{m-1}-N)^{-1}X\right)^{-1}Y(I_{m-1}-N)^{-1}\end{cases}.$

By induction and using the fact that sums and products of rational twisted
series are rational, we see that $N_{1},X_{1},f_{1},Y_{1}$ are rational, and
the proof of the induction step is finished. ∎

Next, we show that the product of $\sigma$-recognizable twisted series is
$\sigma$-recognizable, and furthermore, the inverse of an invertible
$\sigma$-recognizable twisted series is $\sigma$-recognizable.

######  Lemma 3.8.

Let $f,g\in K[[T;\sigma]]$. If $f,g$ are $\sigma$-recognizable, then $fg$ is
$\sigma$-recognizable too. If $f$ is $\sigma$-recognizable and $f(0)\neq 0$,
then $f^{-1}\in K[[T;\sigma]]$ is $\sigma$-recognizable.

###### Proof.

Let $f,g\in K[[T;\sigma]]$ be $\sigma$-recognizable. By Proposition 3.2, there
are stable finite-dimensional right vector subspaces $V$ and $W$ of
$K[[T;\sigma]]$ such that $f\in V$ and $g\in W$. Then, the right vector space
$fW+V$ contains $fg$, is finite-dimensional, and is stable since, by Lemma
3.6,

$s(fh)=fs(h)+s(f)\sigma(h(0))\in fW+V,$

for all $h\in W$. Now, an application of Proposition 3.2 proves that $fg$ is
$\sigma$-recognizable.

To prove the second statement, let $f\in K[[T;\sigma]]$ be
$\sigma$-recognizable and satisfy $f(0)\neq 0$. Without loss of generality, we
may assume that $f(0)=1$. By Corollary 3.3, $1-f$ is $\sigma$-recognizable.
Therefore, there exists a stable finite-dimensional right vector subspace $V$
of $K[[T;\sigma]]$ such that $1-f\in V$. The right vector space $K+f^{-1}V$ is
finite-dimensional and contains $f^{-1}$ since $f^{-1}=1+f^{-1}(1-f)$.
Furthermore, using Lemma 3.6, we see that

$s(f^{-1}h)=f^{-1}s(h)+s(f^{-1})\sigma(h(0))=$
$f^{-1}s(h)+f^{-1}s(1-f)\sigma(f^{-1}(0))\sigma(h(0))\in K+f^{-1}V,$

for all $h\in V$. Therefore, $V$ is stable. By Proposition 3.2, $f^{-1}$ must
be $\sigma$-recognizable, and we are done. ∎

Now, we can prove the following result:

######  Theorem 3.9.

A twisted series is rational iff it is $\sigma$-recognizable.

###### Proof.

First, we prove that every rational twisted series is $\sigma$-recognizable.
We note that that any twisted polynomial is $\sigma$-recognizable. In fact, if
$P$ is a twisted polynomial, then $s^{n}(P)=0$ for all $n>\deg P$. By
Proposition 3.4, $P$ must be $\sigma$-recognizable. Since any rational twisted
series is of the form $P^{-1}Q$, where $P,Q$ are twisted polynomials, the
result follows from Lemma 3.8.

Conversely, let $f=\sum_{n\geq 0}a_{n}T^{n}$ be $\sigma$-recognizable. Suppose
that $(X,A,Y)$ is a $\sigma$-linear representation of $f$. Consider the matrix
$AT\in K[[T;\sigma]]^{m\times m}$. By Lemma 3.7, all the entries of the matrix
$(I_{m}-AT)^{-1}$ are rational. Since

$(I_{m}-AT)^{-1}=\sum_{n\geq 0}(AT)^{n}=\sum_{n\geq
0}\left(A\sigma(A)\cdots\sigma^{n-1}(A)\right)T^{n},$

and $f=X(I_{m}-AT)^{-1}Y$, we conclude that $f$ is rational.

∎

###  3.3 Hadamard product

In this part, we prove that the Hadamard product of rational twisted series is
rational. The Hadamard product of twisted series $f=\sum_{i\geq 0}a_{i}T^{i}$
and $g(T)=\sum_{i\geq 0}b_{i}T^{i}$ is defined to be the twisted series

$f\odot g=\sum_{i\geq 0}a_{i}b_{i}T^{i}.$

The proof of the following lemma is left to the reader.

######  Lemma 3.10.

If $f,f_{1},f_{2},g,g_{1},g_{2}\in K[[T;\sigma]]$ and $a,b,c,d\in K$, then

$(f_{1}+f_{2})\odot g=f_{1}\odot g+f_{2}\odot g,$ $f\odot(g_{1}+g_{2})=f\odot
g_{1}+f\odot g_{2},$ $(afb)\odot(cgd)=ac(f\odot g)bd,$ $s(f\odot g)=s(f)\odot
s(g).$

Using this lemma, we prove the following result:

######  Theorem 3.11.

The Hadamard product of rational twisted series is rational.

###### Proof.

Let $f,g\in K[[T;\sigma]]$ be rational. By Theorem 3.9, $f$ and $g$ are
$\sigma$-recognizable. Therefore, by Proposition 3.2, there are stable finite-
dimensional right vector subspaces $V$ and $W$ of $K[[T;\sigma]]$ such that
$f\in V$ and $g\in W$. Let $f_{1},...,f_{m}$ be a basis of $V$ and
$g_{1},...,g_{n}$ be a basis of $W$. Let $V\odot W$ be the right vector
subspace of $K[[T;\sigma]]$ generated by twisted series of the form $h\odot l$
where $h\in V$ and $l\in W$. Clearly, $f\odot g\in V\odot W$. By Lemma 3.10,
we have

$h\odot l=\sum_{i}\sum_{j}(f_{i}\odot g_{j})a_{i}b_{j}\in V\odot W,$

where

$h=\sum_{i}f_{i}a_{i}\in V\text{ and }l=\sum_{j}g_{j}b_{j}\in W.$

Therefore, the twisted series $f_{i}\odot g_{j}$, where $i=1,...,m$ and
$j=1,...,n$, generate $V\odot W$ as a right vector space. In particular, we
see that $V\odot W$ is finite-dimensional as a right vector space over $K$.
The identity (see Lemma 3.10)

$s(h\odot l)=s(h)\odot s(l),$

shows that $V\odot W$ is stable. Therefore, by Proposition 3.2, $f\odot g$
must be $\sigma$-recognizable. The result follows by applying Theorem 3.9. ∎

##  4 Syntactic left ideals

In this section, we assign a left ideal $I_{f}$ of $K[T;\sigma]$ to a given
twisted series $I_{f}$. It is shown that a twsited series $f$ is rational iff
$I_{f}$ is nonzero.

###  4.1 $K[[T;\sigma]]$ as a left $K[T;\sigma]$-module

We consider $K[T;\sigma]$ as a left vector space (over $K$). Clearly, the
elements $1,T,T^{2},...$ form a basis for this vector space. Given a twisted
series $f=\sum_{n\geq 0}a_{n}T^{n}\in K[[T;\sigma]]$, we define a map
$L_{f}:K[T;\sigma]\to K[T;\sigma]$ by

$L_{f}(\sum_{n}b_{n}T^{n})=\sum_{n}b_{n}a_{n}.$

It is easy to see that $L_{f}$ is a (left) $K$-linear map, and moreover, the
assignment

$f\mapsto L_{f}$

establishes an isomorphism between $K[[T;\sigma]]$ and the dual of
$K[T;\sigma]$ as (left) vector spaces. In fact, $L_{f}(T^{n})=a_{n}$ for all
$n\geq 0$. Using this assignment, one can consider $K[[T;\sigma]]$ as a left
$K[T;\sigma]$-module. More precisely, given $f\in K[[T;\sigma]]$ and $P\in
K[T;\sigma]$, we define a map $K[T;\sigma]\to K$ as follows

$Q\mapsto L_{f}(QP).$

It is clear that this map is (left) $K$-linear. Thus, there is a unique
twisted series $P\circ f\in K[[T;\sigma]]$ such that $L_{P\circ
f}(Q)=L_{f}(QP)$.

######  Lemma 4.1.

The operation $(P,f)\mapsto P\circ f$ turns $K[[T;\sigma]]$ into a left
$K[T;\sigma]$-module.

###### Proof.

It is easy to check that

$(P_{1}+P_{2})\circ f=P_{1}\circ f+P_{2}\circ f,$ $P\circ(f_{1}+f_{2})=P\circ
f_{1}+P\circ f_{2},$ $1\circ f=f,$

for all $P,P_{1},P_{2}\in K[T;\sigma]$ and $f,f_{1},f_{2}\in K[[T;\sigma]]$.
We also have

$L_{(P_{1}P_{2})\circ f}(Q)=L_{f}(QP_{1}P_{2})=L_{P_{2}\circ
f}(QP_{1})=L_{P_{1}\circ(P_{2}\circ f)}(Q),$

showing that $(P_{1}P_{2})\circ f=P_{1}\circ(P_{2}\circ f)$. Therefore,
$(P,f)\mapsto P\circ f$ turns $K[[T;\sigma]]$ into a left
$K[T;\sigma]$-module. ∎

Note that if $f=\sum_{n}a_{n}T^{n}\in K[[T;\sigma]]$ and
$P=\sum_{n}b_{n}T^{n}\in K[T;\sigma]$, then

$P\circ f=\sum_{n\geq 0}\left(\sum_{i\geq
0}a_{n+i}\sigma^{n}(b_{i})\right)T^{n}.$ (4.1)

In particular, we have $(P\circ f)(0)=\sum_{i}b_{i}a_{i}=L_{f}(P)$, $T\circ
f=s(f)$ and $(a\circ f)=fa$ for all $a\in K$.

###  4.2  The syntactic left ideal of a twisted series

Considering $K[[T;\sigma]]$ as a left $K[T;\sigma]$-module, we define the
syntactic left ideal $I_{f}$ of $f\in K[[T;\sigma]]$ to be the annihilator of
$f$, that is,

$I_{f}=\\{P\in K[T;\sigma]\,|\,P\circ f=0\\}.$

Clearly, $I_{f}$ is a left ideal of $K[T;\sigma]$. Note that the map $P\mapsto
P\circ f$ establishes an isomorphism

$K[T;\sigma]/I_{f}\to K[T;\sigma]\circ f$

of left $K[T;\sigma]$-modules.

######  Lemma 4.2.

The syntactic left ideal of $f\in K[[T;\sigma]]$ is the largest left ideal of
$K[T;\sigma]$ contained in $\ker L_{f}$.

###### Proof.

If $P\in I_{f}$, then $P\circ f=0$. It follows that $L_{f}(P)=(P\circ
f)(0)=0$, that is, $P\in\ker L_{f}$. This proves that $I_{f}\subset\ker
L_{f}$. To finish the proof of the lemma, we need to show that any left ideal
contained in $\ker L_{f}$ is a subset of $I_{f}$. Suppose that $I\subset\ker
L_{f}$ is a left ideal of $K[T;\sigma]$. Let $P\in I$. We have
$K[T;\sigma]P\subset I_{f}$, that is, $L_{f}(QP)=0$ for all $Q\in
K[T;\sigma]$. Since

$L_{P\circ f}(Q)=L_{f}(QP),$

we see that $P\circ f=0$, hence $P\in I_{f}$. ∎

###  4.3 The rank of a twisted series

The rank of a twisted series $f\in K[[T;\sigma]]$ is defined to be the
dimension of $K[T;\sigma]/I_{f}$ as a (left) vector space. Note that the rank
of $f\in K[[T;\sigma]]$ is equal to the co-dimension of $I_{f}$ in
$K[T;\sigma]$.

######  Theorem 4.3.

(i) The rank of a rational twisted series is less than or equal to the
dimension of any of its $\sigma$-linear representations.
(ii) The rank of a twisted series $f$ is equal to the dimension of the right
vector space $V(f)$ defined in Subsection 3.1. In particular, a twisted series
is rational iff it has a finite rank.

###### Proof.

Note that by Theorem 3.9, any rational twisted series is
$\sigma$-recognizable.
(i) Let $(X,A,Y)$ be a $\sigma$-linear representation of a rational twisted
series $f$. We consider the map $L:K[T;\sigma]\to K^{m\times 1}$ defined by

$L\left(\sum_{n}b_{n}T^{n}\right)=\sum_{n}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y).$

Clearly, $L$ is a linear map of left vector spaces. I claim that $\ker L$ is a
left ideal of $K[T;\sigma]$ contained in $\ker L_{f}$. If
$P=\sum_{n}b_{n}T^{n}\in\ker L$, then

$L_{f}(P)=\sum_{n}b_{n}a_{n}=\sum_{n}b_{n}XA\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)=$
$X\left(\sum_{n}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)\right)=0,$

proving that $\ker L\subset\ker L_{f}$. It remains to show that $T^{m}P\in\ker
L$ for all $m$ and $P=\sum_{n}b_{n}T^{n}\in\ker L$: We have

$T^{m}P=T^{m}\left(\sum_{n}b_{n}T^{n}\right)=\sum_{n}\sigma^{m}(b_{n})T^{m+n}=\sum_{n\geq
m}\sigma^{m}(b_{n-m})T^{n}.$

Since

$\sum_{n\geq
m}\sigma^{m}(b_{n-m})A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)$
$=A\sigma(A)\cdots\sigma^{m-1}(A)\sigma^{m}\left(\sum_{n\geq
m}b_{n-m}A\sigma(A)\cdots\sigma^{n-m-1}(A)\sigma^{n-m}(Y)\right)$
$=A\sigma(A)\cdots\sigma^{m-1}(A)\sigma^{m}\left(\sum_{n\geq
0}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)\right)=0,$

we see that $T^{m}P\in\ker L$. This finishes the proof of the claim. By Lemma
4.2, we must have $\ker L\subset I_{f}$. Since the co-dimension of $\ker L$ in
$K[T;\sigma]$ is less than or equal to $m$, we see that the co-dimension of
$I_{f}$, which is equal to the rank of $f$, is at most $m$.
(ii) Let $f=\sum_{n\geq 0}a_{n}T^{n}\in K[[T;\sigma]]$. The
$K[T;\sigma]$-module isomorphism

$K[T;\sigma]/I_{f}\to K[T;\sigma]\circ f,\,\,P\mapsto P\circ f,$

induces a structure of a vector space over $K$ on $K[T;\sigma]\circ f$. This
structure is given by $(a,g)\mapsto a\circ g=ga$, that is, $K[T;\sigma]\circ
f$ is a right vector subspace of $K[[T;\sigma]]$. Since $s(g)=T\circ g$, for
all $g\in K[[T;\sigma]]$, we see that

$K[T;\sigma]\circ f=\sum_{n\geq 0}s^{n}(f)K=V(f).$

Since $K[T;\sigma]/I_{f}$ is isomorphic to $K[T;\sigma]\circ f$ (as vector
spaces), we conclude that the rank of $f$ is equal to the dimension of $V(f)$
(as a right vector subspace of $K[[T;\sigma]]$). ∎

As an application of this theorem, we give the following result:

######  Corollary 4.4.

A twisted series is rational iff its syntactic left ideal is nonzero.

###### Proof.

It is enough to note that since $K[T;\sigma]$ is a left principal ideal domain
and satisfies the Euclidean algorithm for right division, any nonzero ideal of
$K[T;\sigma]$ is finite-co-dimensional in $K[T;\sigma]$. ∎

##  5 Reduced $\sigma$-linear representations

A $\sigma$-linear representation of a twisted series $f$ is called reduced if
it has the smallest dimension among all $\sigma$-linear representations of
$f$. In this section, we study some properties of reduced $\sigma$-linear
representations.

###  5.1 Ranks and reduced $\sigma$-linear representations

We begin with the following lemma.

######  Lemma 5.1.

The rank of a rational twisted series is equal to the dimension of any reduced
$\sigma$-linear representation of $f$.

###### Proof.

According to Part (i) of Theorem 4.3, the dimension of any $\sigma$-linear
representation of $f$ is at least the rank of $f$. On the other hand, by Part
(i) of Proposition 3.2, the vector space $V(f)$ gives rise to a
$\sigma$-linear representation of dimension $\dim V(f)$. Since, by Theorem
4.3, the rank of $f$ is equal to the dimension of $V(f)$, the result follows.
∎

In the following proposition, we give more properties of reduced
$\sigma$-linear representations.

######  Proposition 5.2.

Let $(X,A,Y)$ be a reduced $\sigma$-linear representation of a twisted series
$f\in K[[T;\sigma]]$. Then, we have

$I_{f}=\\{\sum_{n}b_{n}T^{n}\in
K[T;\sigma]\,|\,\sum_{n}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)=0\\}.$

Moreover, we have

$\sum_{n}KA\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)=K^{r\times 1},$

where $r$ is the dimension of $(X,A,Y)$.

###### Proof.

We consider the linear map

$L:K[T;\sigma]\to K^{r\times
1},\,\,L(\sum_{n}b_{n}T^{n})=\sum_{n}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y),$

introduced in the proof of Theorem 4.3. As seen before, $\ker L\subset I_{f}$.
The co-dimension of $\ker L$ in $K[T;\sigma]$ is not larger than the dimension
of $(X,A,Y)$. On the other hand, the dimension of $(X,A,Y)$ is equal to the
rank of $f$, see Lemma 5.1. Since the rank of $f$ is the co-dimension of
$I_{f}$ in $K[T;\sigma]$ and $\ker L\subset I_{f}$, we see that $\ker
L=I_{f}$. It follows that $L$ is onto because the dimension of
$K[T;\sigma]/\ker L=K[T;\sigma]/I_{f}$ is $r$. Therefore, we have

$\sum_{n\geq 0}KA\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)=K^{r\times 1}.$

∎

###  5.2 $\sigma$-Similarity

Two $\sigma$-linear representations $(X_{1},A_{1},Y_{1})$ and
$(X_{2},A_{2},Y_{2})$ are said to be $\sigma$-similar if there is an
invertible matrix $B$ over $K$ such that

$X_{2}=X_{1}B,\,A_{2}=B^{-1}A_{1}\sigma(B),\,Y_{2}=B^{-1}Y_{1}.$

It is easy to see that any two $\sigma$-similar $\sigma$-linear
representations give rise to the same twisted series.

######  Theorem 5.3.

Let $f\in K[[T;\sigma]]$ be a rational series. Then, any two reduced
$\sigma$-linear representations of $f$ are $\sigma$-similar.

###### Proof.

Let $(X_{1},A_{1},Y_{1})$ and $(X_{2},A_{2},Y_{2})$ be arbitrary reduced
$\sigma$-linear representations of $f$. Let $r$ be the rank of $f$. By Lemma
5.1, both $(X_{1},A_{1},Y_{1})$ and $(X_{2},A_{2},Y_{2})$ have dimension $r$.
By Part (ii) of Proposition 3.2, the right vector spaces

$\\{X_{1}(I_{r}-A_{1}T)^{-1}Z|Z\in K^{r\times 1}\\},$

and

$\\{X_{2}(I_{r}-A_{2}T)^{-1}Z|Z\in K^{r\times 1}\\},$

are stable right vector subspaces of $K[[T;\sigma]]$. It follows that these
vector subspaces are equal to $V(f)$ since $(X_{1},A_{1},Y_{1})$ and
$(X_{2},A_{2},Y_{2})$ are reduced. As the components of each of the vectors

$F_{1}=X_{1}(I_{r}-A_{1}T)^{-1}\,\text{ and
}\,F_{2}=X_{2}(I_{r}-A_{2}T)^{-1},$

form a basis for $V(f)$, we see that there exists an invertible matrix $B$ of
order $r$ over $K$ such that $F_{2}=F_{1}B$. Now, we have

$F_{2}=F_{1}B\implies F_{2}(0)=F_{1}(0)B\implies X_{2}=X_{1}B,$
$f=F_{1}Y=F_{2}Y_{2}\implies F_{1}Y_{1}=F_{1}BY_{2}\implies
Y_{2}=B^{-1}Y_{1},$

and

$s(F_{1})=F_{1}A_{1}\,\text{ and }\,s(F_{2})=F_{2}A_{2}\implies
A_{2}=B^{-1}A_{1}\sigma(B),$

finishing the proof.

∎

##  6 Minimal fractions

In this section, we introduce the concepts of minimal fraction and minimal
twisted polynomial. It turns out that these two concepts are quite related. We
also define the class of regular twisted series and show that they form a
commutative algebra over $K$ under the Hadamard product. Throughout this
section, we assume that $\sigma$ is a filed automorphism of $K$.

###  6.1 Minimal fractions

A fraction $P^{-1}Q\in K(T;\sigma)$ is called reduced if $P$ and $Q$ are left
co-prime, that is, there does not exist a nonconstant polynomial $R$ such that

$P\in RK[T;\sigma]\text{ and }Q\in RK[T;\sigma].$

Note that every fraction $P^{-1}Q\in K(T;\sigma)$ is equal to a reduced
fraction since $(RP)^{-1}(RQ)=P^{-1}Q$ for all (nonzero) $P,Q,R\in
K[T;\sigma]$. A reduced fraction $P^{-1}Q\in K(T;\sigma)$ which also satisfies
$P(0)=1$ will be called a minimal left fraction. The notion of a minimal right
fraction is defined in a similar way.

Using the Euclidean algorithm in $K[T;\sigma]$, one can show that skew
polynomials $P,Q\in K[T;\sigma]$ are left co-prime iff they are right co-
maximal, that is,

$PK[T;\sigma]+QK[T;\sigma]=K[T;\sigma].$

######  Lemma 6.1.

Every rational twisted series has a unique minimal left fraction. Moreover, if
$P^{-1}Q$ is a minimal left fraction and $P^{-1}Q=P^{-1}_{1}Q_{1}$, where
$P_{1},Q_{1}\in K[T;\sigma]$, then $P_{1}=SP$ and $Q_{1}=SQ$ for some $S\in
K[T;\sigma]$. A similar result holds for right fractions.

###### Proof.

Let $f\in K[[T;\sigma]]$ be rational. First, we prove the existence part.
Consider a fraction $f=P^{-1}Q$ with $\deg P$ as small as possible. If
$P(0)=0$, then $Pf=Q$ implies that $Q(0)=0$ and $f=(T^{-1}P)^{-1}(T^{-1}Q)$
would give a fraction with $\deg(T^{-1}P)<\deg P$, a contradiction. Therefore,
$P(0)\neq 0$. We may assume that $P(0)=1$ since $f=(aP)^{-1}(aQ)$ for all
nonzero $a\in K$. Clearly, the fraction $f=P^{-1}Q$ is minimal.

To prove the uniqueness part, let $f=P_{1}^{-1}Q_{1}$ where $P_{1},Q_{1}\in
K[T;\sigma]$. Using the Euclidean division algorithm, we write

$P_{1}=SP+R,\text{ where }R,S\in K[T;\sigma]\text{ and }\deg R<\deg P.$

It follows that $P_{1}f=SPf+Rf$. Since $P_{1}f,Pf\in K[T;\sigma]$, we see that
$Rf\in K[T;\sigma]$. By the minimality of $\deg P$, we must have $R=0$. It
follows that $P_{1}=SP$ and $Q_{1}=SQ$. If $P_{1}^{-1}Q_{1}$ is minimal, $S$
must be a constant. It follows $P_{1}=P$ and $Q_{1}=Q$ since
$P(0)=P_{1}(0)=1$. This proves the uniqueness of minimal left fraction. The
existence and uniqueness of minimal right fractions can be proved in a similar
way. ∎

The unique fraction introduced in this lemma is called the minimal left
fraction of $f$.

###  6.2 Twisted reciprocal polynomials

Given $Q=\sum_{i=0}^{m}a_{i}T^{i}\in K[T;\sigma]$ with $a_{m}\neq 0$, the left
and right twisted reciprocal polynomials of $Q$, denoted by $Q_{*}$ and
$Q^{*}$ respectively, are defined as

$Q_{*}=\sum_{i=0}^{m}\sigma^{i}(a_{m-i})T^{i}=\sum_{i=0}^{m}T^{i}a_{m-i}=T^{m}(\sum_{i=0}^{m}T^{-i}a_{i})\in
K[T;\sigma],$
$Q^{*}=\sum_{i=0}^{m}\sigma^{i-m}(a_{m-i})T^{i}=(\sum_{i=0}^{m}T^{-i}a_{i})T^{m}\in
K[T;\sigma].$

###  6.3 Minimal fractions and rational twisted series

Let $f\in K[[T;\sigma]]$ be rational. Then, the syntactic left ideal $I_{f}$
of $f$ is nonzero, see Corollary 4.4. Since $K[T;\sigma]$ is a (left and
right) principal ideal domain, there must exist a unique twisted polynomial
$R\in K[T;\sigma]$ with the leading coefficient equal to $1$ such that

$I_{f}=K[T;\sigma]R.$

The twisted polynomial $R$ is called the minimal twisted polynomial of $f$.
Note that the rank of $f$ is equal to the degree of $R$ since the rank of $f$
is equal to the dimension of the right vector space $K[T;\sigma]/I_{f}$.

In the following proposition, we examine the relation between minimal right
fractions and minimal twisted polynomials.

######  Proposition 6.2.

Let $f\in K[[T;\sigma]]$ be rational. Suppose that $PQ^{-1}$ is the minimal
right fraction of $f$ and $R$ is the minimal twisted polynomial of $f$. Then,
$T^{k}Q_{*}=R$ and $R^{*}=Q$, where $k=\max(0,\deg P-\deg Q+1)$.

###### Proof.

Let

$f=\sum_{i\geq
0}a_{i}T^{i},\,R=\sum_{i=0}^{r}b_{i}T^{i},\,Q=\sum_{i=0}^{q}c_{i}T^{i},$

where $r=\deg R$ and $q=\deg Q$. Since $R\circ f=0$, Equation 4.1 implies that

$\sum_{i=0}^{r}a_{n+i}\sigma^{n}(b_{i})=0,$

for all $n\geq 0$. We have

$fR^{*}=\sum_{n\geq
0}\left(\sum_{i+j=n}a_{i}\sigma^{i}(\sigma^{j-r}(b_{r-j}))\right)T^{n}.$

Since

$\displaystyle\sum_{i+j=n+r}a_{i}\sigma^{i}(\sigma^{r-j}(b_{r-j})$
$\displaystyle=\sum_{j=0}^{r}a_{n+r-j}\sigma^{n+r-j}(\sigma^{j-r}(b_{r-j})$
$\displaystyle=\sum_{j=0}^{r}a_{n+r-j}\sigma^{n}(b_{r-j})$
$\displaystyle=\sum_{j=0}^{r}a_{n+j}\sigma^{n}(b_{j})=0,$

for all $n\geq 0$, we see that $fR^{*}\in K[T;\sigma]$, and moreover,
$\deg(fR^{*})<r$. By Lemma 6.1, $R^{*}=QS$ and $fR^{*}=PS$ for some $S\in
K[T;\sigma]$. In particular, we have

$p\leq\deg P+\deg S=\deg(fR^{*})<r,$ (6.1)

and

$q\leq\deg Q+\deg S=\deg R^{*}\leq r,$ (6.2)

where $p=\deg P$. Since

$P=fQ=\sum_{n\geq 0}\left(\sum_{i+j=n}a_{i}\sigma^{i}(c_{j})\right)T^{n},$

we see that

$\sum_{i+j=n}a_{i}\sigma^{i}(c_{j})=0,$

for all $n>p=\deg P$. We can rewrite this equation as

$\sum_{i=\max(0,n-q)}^{n}a_{i}\sigma^{i}(c_{n-i})=0,$

for all $n>p=\deg P$. It follows that

$\displaystyle\left(\sum_{i=k}^{q+k}\sigma^{i}(c_{q+k-i})T^{i}\right)\circ f$
$\displaystyle=\sum_{n\geq
0}\left(\sum_{i=k}^{q+k}a_{n+i}\sigma^{n}(\sigma^{i}(c_{q+k-i}))\right)T^{n}$
$\displaystyle=\sum_{n\geq
0}\left(\sum_{i=k}^{q+k}a_{n+i}\sigma^{n+i}(c_{q+k-i})\right)T^{n}$
$\displaystyle=\sum_{n\geq
0}\left(\sum_{i=n+k}^{n+q+k}a_{i}\sigma^{i}(c_{n+q+k-i})\right)T^{n}$
$\displaystyle=\sum_{n\geq
0}\left(\sum_{i=max(0,(n+q+k)-q)}^{n+q+k}a_{i}\sigma^{i}(c_{n+q+k-i})\right)T^{n}=0,$

since $n+q+k>p$ for all $n\geq 0$. Therefore, we have

$\sum_{i=k}^{q+k}\sigma^{i}(c_{q+k-i})T^{i}\in I_{f},$

which implies that

$T^{k}Q_{*}=T^{k}\sum_{i=0}^{q}T^{i}c_{q-i}=\sum_{i=k}^{q+k}\sigma^{i}(c_{q+k-i})T^{i}=S_{1}R,$

for some $S_{1}\in K[T;\sigma]$. In particular, we have $r\leq q+k$. If $k=0$,
since $q\leq r$ by Inequality 6.2, $r$ must be equal to $q$, which implies
that

$T^{k}Q_{*}=\sum_{i=0}^{q}T^{i}c_{q-i}=R.$

If $k>0$, then $r\leq q+k=p+1$ and $p<r$ (see Inequality 6.1) imply that
$r=q+k$. Since $c_{0}=Q(0)=R(0)=1$, we obtain

$T^{k}Q_{*}=R.$

To prove $R^{*}=Q$, we write

$R^{*}=(T^{k}Q_{*})^{*}=\left(\sum_{i=k}^{q+k}\sigma^{i}(c_{q+k-i})T^{i}\right)^{*}=\sum_{i=0}^{q}\sigma^{i-q-k}(\sigma^{i-q-k}(c_{i}))T^{i}=Q.$

∎

Here is an application of this proposition:

######  Corollary 6.3.

Let $f\in K[[T;\sigma]]$ be rational. Suppose that $PQ^{-1}$ is the minimal
right fraction of $f$ and $R$ is the minimal twisted polynomial of $f$. Then,
the rank of $f$ is equal to $\max(\deg Q,\deg P+1)$.

###### Proof.

By Proposition 6.2, we have

$\deg Q=\deg R-\max(0,\deg P-\deg Q+1)\implies\deg R=\max(\deg Q,\deg P+1).$

Since $\deg R$ is equal to the rank of $f$, the result follows.

∎

###  6.4 Regular twisted series

By a regular twisted series, we mean a rational twisted series in
$K[[T;\sigma]]$ which has a negative degree as an element of $K(T;\sigma)$. It
follows from the properties of the degree function that the set $H(\sigma)$ of
all regular twisted series in $K[[T;\sigma]]$ is a subring of $K[[T;\sigma]]$.
In what follows, we give a number of characterizations of regular twisted
series.

######  Lemma 6.4.

Let $f\in K[[T;\sigma]]$. Then, the following are equivalent:
(i) $f\in H(\sigma)$.
(ii) The minimal twisted polynomial of $f$ has a nonzero constant term.
(iii) For any reduced $\sigma$-linear representation $(X,A,Y)$ of $f$, the
matrix $A$ is invertible.
(iv) There exists a $\sigma$-linear representation $(X,A,Y)$ of $f$ such that
$A$ is invertible.

###### Proof.

The equivalence of (i) and (ii) follows from Proposition 6.2 since the minimal
twisted polynomial of $f$ has a nonzero constant term iff the number $k$ in
Proposition 6.2 is zero, which in turn is equivalent to $f$ belonging to
$H(\sigma)$.

(ii)$\implies$(iii): Let $(X,A,Y)$ be an $r$-dimensional reduced
$\sigma$-linear representation of $f$. Let

$R=b_{0}+b_{1}T+\cdots+b_{r-1}T^{r-1}+b_{r}T^{r},$

be the minimal twisted polynomial of $f$. Note that $b_{r}=1$. By Proposition
5.2, we have

$b_{0}Y+b_{1}A\sigma(Y)+\cdots+b_{r}A\sigma(A)\cdots\sigma^{r-1}(A)\sigma^{r}(Y)=0.$

Since $b_{0}\neq 0$, we have

$ZA=0\implies ZY=0,\text{ for all }Z\in K^{1\times r}.$

This fact and the equality (see Proposition 5.2)

$\sum_{n\geq 0}KA\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y)=K^{r\times 1},$

imply that $A$ is invertible.

Since (iii)$\implies$(iv) is trivial, it only remains to show that (iv)
implies (ii). Let $(X,A,Y)$ be an $m$-dimensional $\sigma$-linear
representation of $f$ such that $A$ is invertible. We consider the linear map

$L:K[T;\sigma]\to K^{m\times
1},\,\,L(\sum_{n}b_{n}T^{n})=\sum_{n}b_{n}A\sigma(A)\cdots\sigma^{n-1}(A)\sigma^{n}(Y),$

introduced in the proof of Theorem 4.3. Clearly, the co-dimension of $\ker L$
in $K[T;\sigma]$ (as a left vector space) is at most $m$. Arguing as in the
proof of Theorem 4.3, we see that $\ker L$ is a left ideal of $K[T;\sigma]$.
It follows that there exists a (nonzero) polynomial $P\in K[T;\sigma]$ such
that

$\ker L=K[T;\sigma]P.$

Let

$P=c_{0}+c_{1}T+\cdots+c_{p-1}T^{p-1}+c_{p}T^{p}.$

Assume that $c_{0}=0$. Therefore, we obtain

$c_{1}A\sigma(Y)+\cdots+c_{p}A\sigma(A)\cdots\sigma^{p-1}(A)\sigma^{p}(Y)=0.$

Since $A$ is invertible, it follows that

$c_{1}\sigma(Y)+c_{2}\sigma(A)\sigma^{2}(Y)+\cdots+c_{p}\sigma(A)\cdots\sigma^{p-1}(A)\sigma^{p}(Y)=0,$

or equivalently,

$\sigma\left(\sigma^{-1}(c_{1})Y+\sigma^{-1}(c_{2})A\sigma(Y)+\cdots+\sigma^{-1}(c_{p})A\sigma(A)\cdots\sigma^{p-2}(A)\sigma^{p-1}(Y)\right)=0.$

Since $\sigma$ is injective, we see that

$\sigma^{-1}(c_{1})Y+\sigma^{-1}(c_{2})A\sigma^{1}(Y)+\cdots+\sigma^{-1}(c_{p})A\cdots\sigma^{p-2}(A)\sigma^{p-1}(Y)=0.$

It follows that

$\sigma^{-1}(c_{1})+\sigma^{-1}(c_{2})T+\cdots+\sigma^{-1}(c_{p})T^{p-1}\in\ker
L,$

contradicting the choice of $P$ since $P$ has the least possible degree among
the elements of $\ker L$. Therefore, $P$ has a nonzero constant term. Since

$P\in\ker L\subset I_{f}=K[T;\sigma]R,$

$R$ must have a nonzero constant term as well. ∎

###  6.5 The twisted Hadamard algebra $H(\sigma)$

We begin by proving that the Hadamarr product of regular twisted series is
regular.

######  Proposition 6.5.

If $f,g\in H(\sigma)$, then $f\odot g\in H(\sigma)$.

###### Proof.

Let $(X,A,Y)$ and $(X_{1},A_{1},Y_{1})$ be $\sigma$-linear representations of
$f$ and $g$, respectively. Since we have $\sigma(C\otimes
D)=\sigma(C)\otimes\sigma(D)$ for all matrices $C$ and $D$ over $K$, one can
see that

$(X\otimes X_{1},A\otimes A_{1},Y\otimes Y_{1})$

is a $\sigma$-linear representations of $f\odot g$. Here, $C\otimes D$ denotes
the Hadamard product of matrices $C$ and $D$. Therefore, we see that $f\odot
g$ has a $\sigma$-linear representation with an invertible matrix. By Lemma
6.4, we conclude that $f\odot g$ is regular. ∎

Since sums of regular twisted series are regular, this proposition implies
that the set $H(\sigma)$ equipped with the Hadamard product is an algebra over
$K$ which we will call the twisted Hadamard algebra. The unit element of the
twisted Hadamard algebra is the series

$(1-T)^{-1}=\sum_{n\geq 0}T^{n}.$

Clearly, the twisted Hadamard algebra is commutative.

## References

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  * [2] P.M. Cohn.  Free ideal rings and localisation in general rings, volume 3.  Cambridge university press, 2006.
  * [3] R. Salem.  Algebraic numbers and Fourier analysis.  Wadsworth Publiching Company, 1983.