#  $(\sigma,\delta)$-polycyclic codes in Ore extensions over rings

Maryam Bajalan<EMAIL_ADDRESS>Ivan Landjev<EMAIL_ADDRESS>Edgar Martínez-Moro<EMAIL_ADDRESS>Steve Szabo<EMAIL_ADDRESS>Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Bl.
8, Acad. G. Bonchev Str., 1113, Sofia, Bulgaria  Department of Mathematics,
New Bulgarian University, Sofia, Bulgaria  Institute of Mathematics,
University of Valladolid, Castilla, Spain  Institute of Mathematics and
Statistics, Eastern Kentucky University, Kentucky, USA

###### Abstract

In this paper, we study the algebraic structure of
$(\sigma,\delta)$-polycyclic codes as submodules in the quotient module
$S/Sf$, where $S=R[x,\sigma,\delta]$ is the Ore extension, $f\in S$, and $R$
is a finite but not necessarily commutative ring. We establish that the
Euclidean duals of $(\sigma,\delta)$-polycyclic codes are
$(\sigma,\delta)$-sequential codes. By using $(\sigma,\delta)$-Pseudo Linear
Transformation (PLT), we define the annihilator dual of
$(\sigma,\delta)$-polycyclic codes. Then, we demonstrate that the annihilator
duals of $(\sigma,\delta)$-polycyclic codes maintain their
$(\sigma,\delta)$-polycyclic nature. Furthermore, we classify when two
$(\sigma,\delta)$-polycyclic codes are Hamming isometrical equivalent. By
employing Wedderburn polynomials, we introduce simple-root
$(\sigma,\delta)$-polycyclic codes. Subsequently, we define the
$(\sigma,\delta)$-Mattson-Solomon transform for this class of codes and we
address the problem of decomposing these codes by using the properties of
Wedderburn polynomials.

###### keywords:

Ore extensions, Wedderburn polynomials, Non-commutative rings,
$(\sigma,\delta)$-polycyclic codes, $(\sigma,\delta)$-Mattson-Solomon maps,
Simple-root $(\sigma,\delta)$-polycyclic codes _AMS Subject Classification
2010: 17A01, 94B05_

††journal: FFAmmfootnotetext: Maryam Bajalan is supported by the Bulgarian
Ministry of Education and Science, Scientific Programme "Enhancing the
Research Capacity in Mathematical Sciences (PIKOM)", No. DO1-67/05.05.2022.

##  1 Introduction

Suppose that $R$ is a finite ring with unit, $\sigma\in\operatorname{End}(R)$
and $\delta$ is an additive endomorphism of $R$ satisfying
$\delta(ab)=\sigma(a)\delta(b)+\delta(a)b$ for all $a,b\in R.$ An Ore
extension is a non-commutative polynomial ring $R[x,\sigma,\delta]$
constructed by multiplications $xa=\sigma(a)x+\delta(a)$ and
$x^{r}x^{s}=x^{r+s}$. In coding theory, $(\sigma,\delta)$-polycyclic codes
over the ring $R$ is a novel study, where these codes are defined as ideals
(submodules) in the quotient ring (quotient module) $S/Sf$ where $S$ is the
Ore extension $R[x,\sigma,\delta]$ and $Sf$ as a two-sided (one-sided) ideal.
This class of codes constitutes a powerful generalization that encompasses
various generalizations of cyclic codes, including skew cyclic codes, skew
negacyclic codes, skew constacyclic codes, polycyclic codes, etc, see [2, 3,
5, 6, 7]. Therefore, by studying $(\sigma,\delta)$-polycyclic codes from
several points of view, we achieve a unification of different approaches,
providing a comprehensive survey of all these various generalizations.

In $2012$, Boucher and Ulmer have introduced the concept of
$(\sigma,\delta)$-polycyclic codes as submodules in $S/Sf$, where $R$ is the
finite field $\mathbb{F}_{q}$, see [8]. It is well-known that if
$R=\mathbb{F}_{q}$, then all $\sigma$-derivations $\delta$ are uniquely
determined by $\delta(a)=\beta(\sigma(a)-a)$, where $\beta\in\mathbb{F}_{q}$.
They have proved that if $z=x+\beta$, then there exists a bijective ring
homomorphism between the Ore extension $\mathbb{F}_{q}[x,\sigma,\delta]$ and
the skew polynomial ring $\mathbb{F}_{q}[z,\sigma]$. However, this
homomorphism preserves neither the Hamming distance nor the rank distance. As
a result, $(\sigma,\delta)$-polycyclic codes in
$\mathbb{F}_{q}[x,\sigma,\delta]$ produce new codes that are not isometrically
equivalent to skew cyclic codes in the skew polynomial ring
$\mathbb{F}_{q}[z,\sigma]$. Additionally, they have demonstrated that
$(\sigma,\delta)$-polycyclic codes can provide improved distance bounds
compared to skew cyclic codes in $\mathbb{F}_{q}[x,\sigma]$.

There is a limited account on $(\sigma,\delta)$-polycyclic codes. In [24], the
authors determined the generator and parity-check matrices of
$(\sigma,\delta)$-polycyclic codes over rings. In [41], it is proven that
$(\sigma,\delta)$-polycyclic codes over $\mathbb{F}_{q}$ are
$T_{C_{f}}$-invariant linear codes in $\mathbb{F}_{q}^{n}$, where $T_{C_{f}}$
represents a $(\sigma,\delta)$-Pseudo Linear Transfromation, and $C_{f}$ is
the companion matrix associated with $f$. Under a special assumption for $f$,
they considered $(\sigma,\delta)$-polycyclic codes as invariant subspaces and
generalized some of the main results in [3, 37]. In [31], the authors took a
step further and introduced a generalization of the $(\sigma,\delta)$-Pseudo
Linear Transformation $T_{C_{f}}$ into the transformation $T_{M}$, with $M$
being an arbitrary matrix. Then, they generalized the
$(\sigma,\delta)$-polycyclic codes over fields into $T_{M}$-invariant linear
codes. As applications, in [33, 34], $(\sigma,\delta)$-polycyclic codes were
used to establish good quantum codes, and in [36], these codes were applied to
construct $\mathbb{Z}$-lattices in $\mathbb{R}^{N}$. It is remarkable that
$(\sigma,\delta)$-polycyclic codes have been described by different
terminologies in the above references, such as ``$(\sigma,\delta)$-codes'',
``$(f,\sigma,\delta)$-codes'', ``$(M,\sigma,\delta)$-codes'', and ``skew
generalized cyclic codes''. All these terms represent one concept: the
generalization of polycyclic codes (see [3]) into the non-commutative
polynomial ring $R[x,\sigma,\delta]$. In this paper, we use the term
``$(\sigma,\delta)$-polycyclic codes'' for this concept.

Motivated by these seminal works, this paper aims to investigate
$(\sigma,\delta)$-polycyclic codes as submodules in $S/Sf$, where $R$ is a
finite, but not necessarily commutative, ring. This is the first attempt to
study this class of codes in the non-commutative ring $R$. A challenging
aspect of $(\sigma,\delta)$-polycyclic codes is that their Euclidean duals do
not have $(\sigma,\delta)$-polycyclic code structure. To address this, we
employ $(\sigma,\delta)$-Pseudo Linear Transformations to introduce the non-
degenerate sesquilinear form $\langle\,,\,\rangle_{0}$. We then establish that
the left and right duals related to this form preserve their
$(\sigma,\delta)$-polycyclic code nature.

In coding theory, a well-known problem is the classification of isometrically
equivalent codes. We establish that if there exists a monomial matrix $B$ such
that $C_{f_{1}}B=\sigma(B)C_{f_{2}}+\delta(B)$, then
$(\sigma,\delta)$-polycyclic codes in $S/Sf_{1}$ and
$(\sigma,\delta)$-polycyclic codes in $S/Sf_{2}$ are Hamming isometrically
equivalent.

The Mattson-Solomon transform is a well-known mathematical technique with
diverse applications, including the determination of minimal generator
polynomials and Reed-Solomon codes, see [3, 25, 29]. Our next challenge is
defining the Mattson-Solomon transform in the Ore extension setting. To
achieve this, we employ the theory of Wedderburn polynomials, which is
extensively explored by Lam and Leroy in [20, 21] when $R$ is a division ring.
We classify a $(\sigma,\delta)$-polycyclic code as simple-root if $f$ is a
Wedderburn polynomial. Note that in the case where $R$ is a field, the
Wedderburn polynomials have the form $(x-a_{1})\cdots(x-a_{n})$, where
$a_{i}$s are distinct elements, see [20]. Therefore, the simple-root
$(\sigma,\delta)$-polycyclic codes are the generalization of simple-root
polycyclic codes, as defined in [3]. Our motivation in defining the
$(\sigma,\delta)$-Mattson-Solomon transform for these specific codes lies in
the invertibility of the $(\sigma,\delta)$-Vandermond matrix, see [21] Theorem
5.8.

The next challenge is the decomposition of $(\sigma,\delta)$-polycyclic codes.
Utilizing the Wedderburn polynomial, we define the simple-root
$(\sigma,\delta)$-polycyclic code $C$ and decompose them into the direct sum
of the simple-root $(\sigma,\delta)$-polycyclic codes $C_{i}$ such that each
$C_{i}$ is related to an eigenvalue of $T_{C_{f}}$.

The structure of this paper can be outlined as follows. In Section 2, the
foundation of the Ore extension rings for the subsequent sections is
introduced. In Sections 3 and 4, the concept of right and left
$(\sigma,\delta)$-polycyclic and sequential codes and their relationships are
presented. In Section 5, an alternate concept of left and right dualities is
offered, ensuring the dual of any $(\sigma,\delta)$-polycyclic code remains a
$(\sigma,\delta)$-polycyclic code. In Section 6, the Hamming isometry
equivalence of $(\sigma,\delta)$-polycyclic codes is explored. In Section 7,
the $(\sigma,\delta)$-Mattson-Solomon maps for simple-root
$(\sigma,\delta)$-polycyclic codes are presented. Finally, in Section 8, a
decomposition of a simple-root $(\sigma,\delta)$-polycyclic code is provided.

##  2 Preliminary

###  2.1 Ore extension

Suppose $R$ is a finite but not necessarily commutative ring with unit $1\neq
0$, $\sigma\in\operatorname{End}(R)$ and $\delta$ a left $\sigma$-derivation,
meaning that $\delta$ is an additive endomorphism of $R$ satisfying
$\delta(ab)=\sigma(a)\delta(b)+\delta(a)b$ for all $a,b\in R.$ The set of all
polynomials with coefficients on the left forms a non-commutative ring under
the usual polynomial addition and the multiplications given by

$xa=\sigma(a)x+\delta(a)\quad\text{for all}\,a\in R\qquad\text{and}\qquad
x^{r}x^{s}=x^{r+s}\quad\text{for all}\,r,s\in\mathbb{N}.$ (1)

It is well-known that for every $k\in\mathbb{N}$

$x^{k}a=\sum_{i=0}^{k}\binom{k}{i}\sigma^{i}\delta^{k-i}(a)x^{i}\quad\text{for
all}\,a\in R.$ (2)

This ring is known as the Ore extension and we denote it by
$S=R[x,\sigma,\delta]$. Note that $\sigma(1)=1$ and $\delta(1)=0$, which imply
that $1$ is an identity for $S$ as well. If
$(\sigma,\delta)=(\operatorname{Id},0)$, then the Ore extension is the usual
polynomial ring $R[x]$. In the case where $\delta=0$, the Ore extension is
called skew polynomial ring and is denoted by $R[x,\sigma]$. Moreover, when
$\sigma=0,$ it is called differential polynomial ring and is denoted by
$R[x,\delta].$ In coding theory literature, the term ``skew polynomial ring''
is often used for both $R[x,\sigma]$ and $R[x,\sigma,\delta]$. However, to
avoid confusion, we distinguish between the two cases and refer to the term
``skew polynomial ring'' only when $\delta=0$, while for the broader context
of $R[x,\sigma,\delta]$, we employ the term ``Ore extension''. A comprehensive
overview of the Ore extensions over rings can be found in [10, 15, 16, 17, 20,
21, 28, 30, 38]. This section aims to highlight fundamental characteristics
for use in the next sections.

If ${\delta}(ab)={\delta}(a){\sigma}(b)+a{\delta}(b)$, it is called right
${\sigma}$-derivation. Note that a right ${\sigma}$-derivation over $R$ is a
left ${\sigma}$-derivation over the opposite ring $R^{op}$ (the ring
constructed by $R$ with the same underlying Abelian group but with the
multiplication operation reversed). Throughout this paper, our focus is on
left ${\sigma}$-derivations, and we simply refer to them as
${\sigma}$-derivations, unless stated otherwise. According to our definition
of Ore extension, $S=R[x,\sigma,\delta]$ is a left $R$-module with the basis
$\mathcal{B}=\\{x^{n}:n\geqslant 0\\}$. Now, if we want to consider $S$ as the
set of all polynomials with coefficients on the right and under the usual
polynomial addition and the multiplications given by $x^{r}x^{s}=x^{r+s}$ for
$r,s\in\mathbb{N}$ and $ax=x{\sigma}(a)+{\delta}(a)$ for $a\in R,$ then the
ring $S$ is referred to as the opposite Ore extension. If
$\sigma\in\operatorname{Aut}(R)$, then $-\delta\sigma^{-1}$ is a left
$\sigma^{-1}$-derivation of $R^{op}$ and
$S^{op}=(R[x,\sigma,\delta])^{op}\cong
R^{op}[x,\sigma^{-1},-\delta\sigma^{-1}]$, see [11, 22].

The ring $S$ has a division algorithm that works for left and right divisors.
More precisely, given any two non-zero polynomials $g,k\in S$, there exist
polynomials $q,r\in S$ such that $g=qk+r$ and $\deg(r)<\deg(k)$, see [10]
Theorem 4.34. Consider a sequence of mappings $N_{i}:R\to R$ such that
$N_{0}(a)=1$ and $N_{i+1}(a)=\sigma(N_{i}(a))a+\delta(N_{i}(a))$ for $a\in R$.
If $g(x)=\sum_{i=0}^{n}g_{i}x^{i}\in S$ and $g(x)=q(x-a)+r$ then,
$r=\sum_{i=0}^{n}g_{i}N_{i}(a)=g(a)$, see [18] Lemma 2.4. Therefore, $g(a)=0$
if and only if $x-a$ is a right factor of $g(x)$. In the case $g(a)=0$, $a$ is
called the right root of $g$. Furthermore, when $\sigma$ is injective and $R$
is a division ring, then $S$ is a left PID, and thereby it is a UFD, see [10]
Corollary 4.36 and Theorem 4.25.

Two elements $a$ and $b$ in $R$ are called $(\sigma,\delta)$-conjugate if
there exists an invertible element $c$ in $R^{\times}$ such that
$b=\sigma(c)ac^{-1}+\delta(c)c^{-1}$. It is straightforward to verify that
$(a^{c})^{d}=a^{dc}$, which implies that the relation is an equivalence
relation on $R$, see [18]. We will denote $\sigma(c)ac^{-1}+\delta(c)c^{-1}$
by $a^{c}$ and the $(\sigma,\delta)$-conjugacy class of $a$ by
$\Delta(a)=\\{a^{c}:c\in R^{\times}\\}.$

Throughout the paper, we assume that $f\in S$ is the monic polynomial
$f(x)=-f_{0}-f_{1}x-f_{n-1}x^{n-1}+x^{n}$ and we use $Sf$ to represent the
left $S$-submodule generated by $f$. If we consider the set $\\{g\in
S:\deg{g}<n\\}$ with the usual addition and the multiplication
$gh\,(\operatorname{mod}f)$, then this set has the structure of a left free
$R$-module with the $R$-basis $\mathcal{B}=\\{1,x,\ldots,x^{n-1}\\}$. We
denote this module by $S/Sf$. If $Sf$ is a two-sided ideal in $S$, then $S/Sf$
also possesses a ring structure.

###  2.2 $(\sigma,\delta)$-Pseudo Linear Transformation

Let $V$ be a left $R$-module. An additive map $T:V\rightarrow V$ is called
$(\sigma,\delta)$-Pseudo Linear Transformation (or $(\sigma,\delta)$-PLT for
short) if

$T(av)=\sigma(a)T(v)+\delta(a)v\quad\text{ for all }a\in R.$

For more details on $(\sigma,\delta)$-PLT, the reader is referred to [9, 22,
23]. Suppose $V$ is a free left $R$-module with a basis
$\\{e_{1},e_{2},\ldots,e_{n}\\}$. For any matrix $C\in M_{n}(R)$, the map
$T_{C}:V\to V$ defined as $T_{C}(v)=\sigma(v)C+\delta(v),$ where $\sigma(v)$
and $\delta(v)$ are the extensions of $\sigma$ and $\delta$ from $R$ to
$R^{n}$, is a $(\sigma,\delta)$-PLT. In particular, for $n=1$ and $a\in R$,
the map $T_{a}:R\to R$ given by $T_{a}(x)=\sigma(x)a+\delta(x)$ is a
$(\sigma,\delta)$-PLT. Clearly, $T_{0}=\delta$.

######  Proposition 1 ([22] Lemma 2 and [23] Proposition 1.2).

The following conditions are equivalent for an additive group $(V,+)$.

  1. 1.

$V$ is a left $R$-module and there exists a $(\sigma,\delta)$-Pseudo Linear
Transformation $T:V\rightarrow V.$

  2. 2.

$V$ is a left $S=R[x,\sigma,\delta]$-module induced by the product $g(x)\cdot
v:=g(T)(v).$

  3. 3.

There exists a ring homomorphism $\Lambda:S\to End(V,+)$ such that for
$g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}\in S$,
$\Lambda(g):=g(T)=\sum_{i=0}^{n-1}g_{i}T^{i}\in\operatorname{End}(V,+).$

######  Corollary 1 ([23] Corollary 1.3).

For any $g,h\in S$ and any $(\sigma,\delta)$-Pseudo Linear Transformation $T$,
$(gh)(T)=g(T)h(T)$.

In the classical polynomial ring, evaluating a polynomial at a point is
obtained by substituting the variable x with the point. However, the process
is different in the Ore extension.

######  Lemma 1 ([22] Corollary 2.10).

For $g,h\in S$ and $a\in R$, $gh(a)=g(T_{a})(h(a)).$ In particularly, if
$h(a)$ is invertible then $(gh)(a)=g(a^{h(a)})h(a)$, and if $h(a)=0$ then
$(gh)(a)=0.$

##  3 $(\sigma,\delta)$-polycyclic codes

A code $C$ of length $n$ over the ring $R$ is a subset of $R^{n}$. A left
(right) linear code $C$ of length $n$ over $R$ is a left (right) $R$-submodule
of $R^{n}$. Let $f(x)=-f_{0}-f_{1}x-\cdots-f_{n-1}x^{n-1}+x^{n}\in S$ be a
monic polynomial.

######  Definition 1.

Let $C$ be a left linear code over $R$.

  1. 1.

The code $C$ is called right $(\sigma,\delta)$-polycyclic if for every
$g=(g_{0},g_{1},\dots,g_{n-1})\in C$ we have

$\bigg{(}0,\,\,\sigma(g_{0}),\,\,\sigma(g_{1}),\,\,\ldots,\,\,\sigma(g_{n-2})\bigg{)}+\sigma(g_{n-1})\bigg{(}f_{0},f_{1},\ldots,f_{n-1}\bigg{)}+\delta(g)\in
C.$ (3)

  2. 2.

The code $C$ is called left $(\sigma,\delta)$-polycyclic if for every
$g=(g_{0},g_{1},\dots,g_{n-1})\in C$ we have

$\bigg{(}\sigma(g_{1}),\,\,\sigma(g_{2}),\,\,\dots,\,\,\sigma(g_{n-1}),\,\,0\bigg{)}+\sigma(g_{0})\bigg{(}f_{0},f_{1},\dots,f_{n-1}\bigg{)}+\delta(g)\in
C.$ (4)

######  Remark 1.

In the case $(\sigma,\delta)=(\operatorname{Id},0),$ right and left
$(\sigma,\delta)$-polycyclic codes coincide with right and left polycyclic
codes, respectively, see for example [26].

Let us denote the companion matrix related to $f$ as

$C_{f}=\begin{pmatrix}0&1&0&\cdots&0\\\ 0&0&1&\cdots&0\\\
\vdots&\vdots&\ddots&\cdots&\vdots\\\ 0&0&0&\ddots&1\\\
f_{0}&f_{1}&f_{2}&\cdots&f_{n-1}\end{pmatrix}.$

######  Lemma 2.

The code $C$ is a right $(\sigma,\delta)$-polycyclic code if and only if it is
invariant under $T_{C_{f}}=\sigma(v)C_{f}+\delta(v)$.

###### Proof.

For each $g=(g_{0},g_{1},\ldots,g_{n-1})\in R^{n},$ it is easy to see that

$\displaystyle T_{C_{f}}(g)$ $\displaystyle=\sigma(g)C_{f}+\delta(g)$
$\displaystyle=\bigg{(}\sigma(g_{n-1})f_{0},\sigma(g_{0})+\sigma(g_{n-1})f_{1},\sigma(g_{1})+\sigma(g_{n-1})f_{2},\ldots,\sigma(g_{n-2})+\sigma(g_{n-1})f_{n}\bigg{)}+\delta(g)$
$\displaystyle=\bigg{(}0,\sigma(g_{0}),\sigma(g_{1}),\ldots,\sigma(g_{n-2})\bigg{)}+\sigma(g_{n-1})\bigg{(}f_{0},f_{1},\ldots,f_{n-1}\bigg{)}+\delta(g).$

Therefore, $C$ is a right $(\sigma,\delta)$-polycyclic code if and only if
$T_{C_{f}}(C)\subseteq C.$ ∎

The left $R$-module isomorphism $\Omega_{f}:R^{n}\rightarrow S/Sf$ associates
each element $g=(g_{0},g_{1},\ldots,g_{n-1})$ to its corresponding polynomial
$g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}.$

######  Lemma 3.

The $(\sigma,\delta)$-Pseudo Linear Transformation $T_{C_{f}}$ corresponds to
the left multiplication by $x$ on $S/Sf.$

###### Proof.

Let $g(x)$ be the image of $g=(g_{0},g_{1},\ldots,g_{n-1})\in R^{n}$ under
$\Omega_{f}.$ Then

$\displaystyle xg(x)$
$\displaystyle=\sum_{i=0}^{n-1}xg_{i}x^{i}=\sum_{i=0}^{n-1}\bigg{(}\sigma(g_{i})x+\delta(g_{i})\bigg{)}x^{i}$
$\displaystyle=\sigma(g_{0})x+\delta(g_{0})+\sigma(g_{1})x^{2}+\delta(g_{1})x+\sigma(g_{2})x^{3}+\delta(g_{2})x^{2}+\cdots+\sigma(g_{n-1})x^{n}+\delta(g_{n-1})x^{n-1}$
$\displaystyle=\bigg{(}\sigma(g_{n-1})f_{0}\bigg{)}+\bigg{(}\sigma(g_{n-1})f_{1}+\sigma(g_{0})\bigg{)}x+\cdots+\bigg{(}\sigma(g_{n-1})f_{n-1}+\sigma(g_{n-2})\bigg{)}x^{n-1}+\sum_{i=0}^{n-1}\delta(g_{i})x^{i}.$

It is easy to see that
$\bigg{(}\sigma(g_{n-1})f_{0},\,\,\sigma(g_{n-1})f_{1}+\sigma(g_{0}),\,\,\ldots,\,\,\sigma(g_{n-1})f_{n-1}+\sigma(g_{n-2})\bigg{)}=\sigma(g)C_{f}$.
Consequently, $xg(x)$ corresponds to the image of
$\sigma(g)C_{f}+\delta(g)=T_{C_{f}}(g)$ under the mapping $\Omega_{f},$ which
completes the proof. ∎

######  Lemma 4.

The code $C$ is a right $(\sigma,\delta)$-polycyclic code if and only if
$\Omega_{f}(C)$ is a left $S$-submodule of $S/Sf$.

###### Proof.

Assume that $C$ is a $(\sigma,\delta)$-polycyclic code. Choose an arbitrary
element $g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}\in\Omega_{f}(C)$ and denote its
corresponding element by $g=(g_{0},g_{1},\ldots,g_{n-1})\in C$. Applying Lemma
3 and the fact that $T_{C_{f}}(g)\in C$, we have $xg(x)\in\Omega_{f}(C).$ By
induction, $x^{i}g(x)$ also belongs to $\Omega_{f}(C)$ for all $i$, which
implies $h(x)g(x)\in\Omega_{f}(C)$ for any $h(x)\in S$. Hence, $\Omega_{f}(C)$
is a left $S$-submodule of $S/Sf$. Using the same argument, the converse can
be proved. ∎

From now on, let us denote $\Omega_{f}(C)$ simply as $C$. Moreover, elements
of any $(\sigma,\delta)$-polycyclic code will be denoted interchangeably by
both $g$ and $g(x)$.

The matrix $A\in M_{n}(R)$ is called $(\sigma,\delta)$-similar to the matrix
$B\in M_{n}(R)$ if there exists an invertible matrix $P$ such that
$B=\sigma(P)AP^{-1}+\delta(P)P^{-1}$, where $\sigma(P)$ and $\delta(P)$ are
the extensions of $\sigma$ and $\delta$ from the ring $R$ to $M_{n}(R)$ by
applying them on each entry. Furthermore, the matrix $A$ is called
$(\sigma,\delta)$-diagonalizable if it is $(\sigma,\delta)$-similar to a
diagonal matrix, see [22].

######  Lemma 5.

Let $M$ be a $(\sigma,\delta)$-similar matrix to $C_{f}$ and $C\subseteq
R^{n}.$ Then $C$ is invariant under $T_{M}$ if and only if $CP^{-1}$ is
invariant under $T_{C_{f}},$ or equivalently, $CP$ is invariant under $T_{M}$
if and only if $C$ is invariant under $T_{C_{f}}$.

###### Proof.

Let $C$ be invariant under $T_{M}$. Then

$\displaystyle T_{C_{f}}(CP^{-1})$
$\displaystyle=\sigma(CP^{-1})C_{f}+\delta(CP^{-1})$
$\displaystyle=\sigma(CP^{-1})\bigg{(}\sigma(P)MP^{-1}+\delta(P)P^{-1}\bigg{)}+\delta(CP^{-1})$
$\displaystyle=\sigma(CP^{-1})\sigma(P)MP^{-1}+\sigma(CP^{-1})\delta(P)P^{-1}+\delta(CP^{-1})$
$\displaystyle=\sigma(C)MP^{-1}+\delta(CP^{-1}P)P^{-1}$
$\displaystyle=\sigma(C)MP^{-1}+\delta(C)P^{-1}$
$\displaystyle=\bigg{(}\sigma(C)M+\delta(C)\bigg{)}P^{-1}\subseteq CP^{-1}.$

Moreover,

$\displaystyle T_{M}(C)$
$\displaystyle=T_{M}(CP^{-1}P)=\sigma(CP^{-1}P)M+\delta(CP^{-1}P)$
$\displaystyle=\sigma(CP^{-1})\sigma(P)M+\sigma(CP^{-1})\delta(P)+\delta(CP^{-1})P$
$\displaystyle=\sigma(CP^{-1})\bigg{(}\sigma(P)M+\delta(P)\bigg{)}+\delta(CP^{-1})P$
$\displaystyle=\sigma(CP^{-1})\bigg{(}\sigma(P)MP^{-1}+\delta(P)P^{-1}\bigg{)}P+\delta(CP^{-1})P$
$\displaystyle=\bigg{(}\sigma(CP^{-1})(C_{f})+\delta(CP^{-1})\bigg{)}P$
$\displaystyle\subseteq(CP^{-1})P=C.$

∎

Associated with $f(x)=-f_{0}-f_{1}x-f_{n-1}x^{n-1}+x^{n}$, define the matrix

$E_{f}=\begin{pmatrix}f_{0}&f_{1}&f_{2}&\cdots&f_{n-1}\\\ 1&0&0&\cdots&0\\\
0&1&0&\cdots&0\\\ \vdots&\vdots&\ddots&\ddots&\vdots\\\
0&0&\cdots&1&0\end{pmatrix}.$

Similar to the Lemma 2, we have the following lemma.

######  Lemma 6.

A left linear code $C$ is a left $(\sigma,\delta)$-polycyclic code if and only
if it is invariant under $T_{E_{f}}$.

######  Lemma 7.

The matrix $E_{d}$ is $(\sigma,\delta)$-similar to $C_{f}$, where
$d=-f_{n-1}-f_{n-2}x\cdots-f_{0}x^{n-1}+x^{n}$.

###### Proof.

It is easy to see that $C_{f}=PE_{d}P,$ where

$P=\begin{pmatrix}0&0&\cdots&1\\\ 0&0&1&0\\\ \vdots&\iddots&\vdots&\vdots\\\
1&\cdots&0&0\\\ \end{pmatrix}.$ (5)

Note that $P$ is invertible, $P^{-1}=P$, $\sigma(P)=P$, and $\delta(P)=0,$
which imply $C_{f}P=\sigma(P)E_{d}+\delta(P).$ ∎

######  Corollary 2.

If $C$ is a right $(\sigma,\delta)$-polycyclic code which is invariant under
$T_{C_{f}}$, then $CP$ is a left $(\sigma,\delta)$-polycyclic code which is
invariant under $T_{E_{d}}$.

We define the left $R$-module isomorphism
${\Omega}^{*}_{d}:R^{n}\rightarrow{S}/{S}{d}$ which corresponds each element
$g=(g_{0},g_{1},\ldots,g_{n-1})\in R^{n}$ to its reciprocal polynomial
$g^{*}(x)=g_{n-1}+g_{n-2}x+\cdots+g_{1}x^{n-2}+g_{0}x^{n-1}.$

######  Lemma 8.

The $(\sigma,\delta)$-Pseudo Linear Transformation
$T_{E_{d}}(g)=\sigma(g)E_{d}+\delta(g)$ corresponds to the left multiplication
$g^{*}(x)$ by $x$ in $S/Sd$, but under the reciprocal correspondence.

###### Proof.

Note that in the factor ring $S/S{d}$, we have
$x^{n}=f_{n-1}+f_{n-2}x+\cdots+f_{1}x^{n-2}+f_{0}x^{n-1}.$ We get

$\displaystyle xg^{*}(x)$
$\displaystyle=\sum_{i=0}^{n-1}xg_{n-1-i}x^{i}=\sum_{i=0}^{n-1}\bigg{(}\sigma(g_{n-1-i})x+\delta(g_{n-1-i})\bigg{)}x^{i}$
$\displaystyle=\sigma(g_{n-1})x+\delta(g_{n-1})+\sigma(g_{n-2})x^{2}+\delta(g_{n-2})x+\sigma(g_{n-3})x^{3}+\delta(g_{n-3})x^{2}+\cdots+\sigma(g_{0})x^{n}+\delta(g_{0})x^{n-1}$
$\displaystyle=\bigg{(}f_{n-1}\sigma(g_{0})\bigg{)}+\bigg{(}f_{n-2}\sigma(g_{0})+\sigma(g_{n-1})\bigg{)}x+\cdots+\bigg{(}f_{1}\sigma(g_{0})+\sigma(g_{2})\bigg{)}x^{n-2}+\bigg{(}f_{0}\sigma(g_{0})+\sigma(g_{1})\bigg{)}x^{n-1}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{i=0}^{n-1}\delta(g_{n-1-i})x^{i}.$

It is easy to see that
$\bigg{(}f_{0}\sigma(g_{0})+\sigma(g_{1}),\,f_{1}\sigma(g_{0})+\sigma(g_{2}),\,\ldots,\,f_{n-2}\sigma(g_{0})+\sigma(g_{n-1}),\,f_{n-1}\sigma(g_{0})\bigg{)}={\sigma(g)}E_{d}.$
Therefore, the reciprocal of $xg^{*}(x)$ is associated with the image of
$\sigma(g)E_{d}+\delta(g)=T_{E_{d}}(g)$. ∎

######  Corollary 3.

A code $C$ is a left $(\sigma,\delta)$-polycyclic code if and only if
$\Omega^{*}_{d}(C)$ is a left $S$-submodule of $S/Sd$. Moreover, if
$\sigma\in\operatorname{Aut}(R)$, then $C$ is a left
$(\sigma,\delta)$-polycyclic code if and only if it is a right
$R^{op}[x,\sigma^{-1},-\delta\sigma^{-1}]$-submodule of
$R^{op}[x,\sigma^{-1},-\delta\sigma^{-1}]/R^{op}[x,\sigma^{-1},-\delta\sigma^{-1}]d^{op},$
where $d^{op}$ is obtained by considering $d$ with coefficients on the right.

###### Proof.

The first statement is an immediate consequence of Lemma 8. The second
statement follows from the facts that a left $S$-submodule of $S/Sd$ is
equivalent to a right $S^{op}$-submodules $S^{op}/S^{op}d$, and the
isomorphism $(R[x,\sigma,\delta])^{op}\cong
R^{op}[x,\sigma^{-1},-\delta\sigma^{-1}].$ ∎

##  4 $(\sigma,\delta)$-sequential codes

######  Definition 2.

Let $C$ be a left linear code of length $n$ over $R$.

  1. 1.

The code $C$ is called right $(\sigma,\delta)$-sequential if for every
$(g_{0},g_{1},\dots,g_{n-1})\in C$ we have

$\bigg{(}\sigma(g_{1}),\,\,\sigma(g_{2}),\,\,\dots,\,\,\sigma(g_{n-2}),\,\,\sigma(g_{0})f_{0}+\sigma(g_{1})f_{1}+\cdots+\sigma(g_{n-1})f_{n-1}\bigg{)}+\delta(g)\in
C.$ (6)

  2. 2.

The code $C$ is called left $(\sigma,\delta)$-sequential if for every
$(g_{0},g_{1},\dots,g_{n-1})\in C$ we have

$\bigg{(}\sigma(g_{0})f_{0}+\sigma(g_{1})f_{1}+\cdots+\sigma(g_{n-1})f_{n-1},\,\,\sigma(g_{0}),\,\,\sigma(g_{2}),\,\,\dots,\,\,\sigma(g_{n-2})\bigg{)}+\delta(g)\in
C.$ (7)

Applying the above definitions, the proof of the next lemma is immediate.

######  Lemma 9.

The code $C$ is right $(\sigma,\delta)$-sequential if and only if it is
invariant under $T_{C_{f}^{t}}=\sigma(g)C_{f}^{t}+\delta(g)$. Moreover, the
code $C$ is left $(\sigma,\delta)$-sequential if and only if it is invariant
under $T_{E_{d}^{t}}=\sigma(g)E_{d}^{t}+\delta(g).$ Here, $t$ denotes the
transpose matrix.

Recall that an anti-automorphism $\zeta:R\to R^{op}$ is a map which sends $ab$
to $\zeta(b)\zeta(a),$ where $R^{op}$ denotes the opposite ring of $R.$ A
$\zeta$-sesquilinear form on an $R$-module $M$ associated with an anti-
automorphism $\zeta$ is given by $\langle\cdot,\cdot\rangle:M\times M\to R$
that satisfies the following conditions: for all $x,y,z$ in $M$ and all
$\alpha,\beta\in R$, $\langle x+y,z\rangle=\langle x,z\rangle+\langle
y,z\rangle,$ $\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle,$
$\langle\alpha x,y\rangle=\alpha\langle x,y\rangle$ and $\langle x,\beta
y\rangle=\langle x,y\rangle\zeta(\beta).$ For an anti-automorphism $\zeta$ on
the ring $R$, define the map $\langle\cdot,\cdot\rangle:R^{n}\times
R^{n}\rightarrow R$ by

$\langle g,h\rangle=\sum_{i=0}^{n-1}g_{i}\zeta(h_{i}),$ (8)

where $g=(g_{0},g_{1},\ldots,g_{n-1})$ and $h=(h_{0},h_{1},\ldots,h_{n-1}).$
This map is a $\zeta$-sesquilinear form, see [40] Proposition 3.1. For any
code $C\subseteq R^{n}$, either linear or not, its left dual code (denoted by
$l(C)$) and its right dual code (denoted by $r(C)$) are defined as

$l(C)=\big{\\{}g\in R^{n}:\langle g,h\rangle=0\,\,\text{for all}\,\,h\in
C\big{\\}}\quad\text{and}\quad r(C)=\big{\\{}g\in R^{n}:\langle
h,g\rangle=0\,\,\text{for all}\,\,h\in C\big{\\}}.$ (9)

In the case $\zeta(\langle x,y\rangle)=\langle y,x\rangle$, it follows that
$l(C)=r(C)$. In the following theorem, Szabo and Wood have presented the
double-dual and cardinality complementary properties of codes over finite
Frobenius rings for non-degenerate sesquilinear forms.

######  Theorem 1 ([40] Theorem 6.2).

Consider a left $R$-linear code $C\subseteq R^{n}$ over a finite Frobenius
ring $R$. Assume $R^{n}$ is equipped with a non-degenerate sesquilinear form.
Then

  1. 1.

$l(C)$ and $r(C)$ are left $R$-linear codes.

  2. 2.

$l(r(C))=C$ and $r(l(C))=C$.

  3. 3.

$|C|\cdot|l(C)|=|R^{n}|$ and $|C|\cdot|r(C)|=|R^{n}|$.

From now one, assume that the anti-automorphism $\zeta$ is the identity
operator $\operatorname{Id}:R\to R^{op},$ which maps $ab$ to $ba$. Therefore,
$\zeta$-sesquilinear form $\langle
g,h\rangle=\sum_{i=0}^{n-1}g_{i}\zeta(h_{i})$ is equal to
$\sum_{i=0}^{n-1}g_{i}h_{i}=gh^{t}$, where $t$ denotes the transpose. We have,
$\langle
h,g\rangle=\sum_{i=1}^{n}h_{i}g_{i}=\sum_{i=1}^{n}\zeta(g_{i}h_{i})=\sum_{i=1}^{n}g_{i}h_{i}=\langle
g,h\rangle$. Moreover, $\langle g,\beta h\rangle=\langle
g,h\rangle\zeta(\beta)=\zeta(\langle
g,h\rangle)\zeta(\beta)=\zeta(\beta(\langle g,h\rangle))=\beta\langle
g,h\rangle.$ Therefore, in the case $\zeta=\operatorname{Id}$, the
$\zeta$-sesquilinear form $\langle\cdot\,,\cdot\rangle$ is exactly the
standard inner product and $l(C)=r(C)=C^{\perp}$.

Before establishing the relationship between $(\sigma,\delta)$-polycyclic
codes and their duals, we need the following easy lemma.

######  Lemma 10.

If $\sigma\in\operatorname{Aut}(R)$, then $\delta_{1}:=-\sigma^{-1}\delta$ is
a left $\sigma^{-1}$-derivation on $R.$ In other words, it is a right
$\sigma^{-1}$-derivation on the opposite ring $R^{op}$.

###### Proof.

For any $a,b\in R$, we have

$\delta_{1}(ab)=-\sigma^{-1}\big{(}\sigma(a)\delta(b)+\delta(a)b\big{)}=-a\sigma^{-1}\delta(b)-\sigma^{-1}\delta(a)\sigma^{-1}(b)=a\delta_{1}(b)+\delta_{1}(a)\sigma^{-1}(b).$

∎

######  Theorem 2.

Assume $\sigma\in\operatorname{Aut}(R)$. If $C$ is a right
$(\sigma,\delta)$-polycyclic code in $S/Sf,$ then $C^{\perp}$ is invariant
under
$T_{C_{\sigma^{\prime}(f)}^{t}}(v)=\sigma^{\prime}(v)C_{\sigma^{\prime}(f)}^{t}+\delta^{\prime}(v)$,
where $\sigma^{\prime}=\sigma^{-1}$, $\delta^{\prime}=-\sigma^{-1}\delta$. In
fact, $C^{\perp}$ is a right $(\sigma^{\prime},\delta^{\prime})$-sequential
code in $S^{\prime}/S^{\prime}f^{\prime}$, where $S^{\prime}$ is the Ore
extension $R^{op}[x,\sigma^{\prime},\delta^{\prime}]$ and $f^{\prime}$ is
obtained by applying $\sigma^{\prime}$ to the coefficients of $f$.

###### Proof.

Assume $C$ is a right $(\sigma,\delta)$-polycyclic code and $g\in C$ is
arbitrary. Since $T_{C_{f}}(g)\in C,$ for $h\in C^{\perp}$ we have

$0=\langle\sigma(g)C_{f}+\delta(g),\,h\rangle=\langle\sigma(g)C_{f},\,h\rangle+\langle\delta(g),\,h\rangle.$
(10)

On the other side, since $0=\langle g,h\rangle=gh^{t}$, we obtain
$0=\delta(gh^{t})=\sigma(g)\delta(h^{t})+\delta(g)h^{t}=\sigma(g)\delta(h)^{t}+\delta(g)h^{t}$.
Hence $-\sigma(g)\delta(h)^{t}=\delta(g)h^{t},$ that is
$\langle\sigma(g),-\delta(h)\rangle=\langle\delta(g),h\rangle.$ By
substituting the last equality in Equation (10) and using the fact that

$\langle\sigma(g)C_{f},\,h\rangle=\sigma(g_{n-1})f_{0}h_{0}+\sigma(g_{n-1})f_{1}h_{1}+\sigma(g_{0})h_{1}+\cdots+\sigma(g_{n-1})f_{n-1}h_{n-1}+\sigma(g_{n-2})h_{n-1}=\langle\sigma(g),\,hC_{f}^{t}\rangle,$

we get

$\displaystyle 0$
$\displaystyle=\langle\sigma(g),\,hC_{f}^{t}\rangle+\langle\sigma(g),\,-\delta(h)\rangle$
$\displaystyle=\langle\sigma(g),\,hC_{f}^{t}-\delta(h)\rangle$
$\displaystyle=\langle hC_{f}^{t}-\delta(h),\,\sigma(g)\rangle$
$\displaystyle=\big{(}hC_{f}^{t}-\delta(h)\big{)}\sigma(g^{t}).$

So,
$\bigg{(}\sigma^{-1}(h)\sigma^{-1}(C_{f}^{t}))-\sigma^{-1}\delta(h)\bigg{)}g^{t}=0$.
It can easily be seen that $\sigma^{-1}(C_{f}^{t}))=C_{\sigma^{-1}(f)}^{t},$
where $\sigma^{-1}(f)$ is obtained by applying $\sigma^{-1}$ to the
coefficients of $f$. Thus, $\langle
g,\,\,\sigma^{-1}(h)C_{\sigma^{-1}(f)}^{t}-\sigma^{-1}\delta(h)\rangle=0.$ Let
us define

$T_{C_{\sigma^{-1}(f)}^{t}}(v):=\sigma^{-1}(v)C_{\sigma^{-1}(f)}^{t}-\sigma^{-1}\delta(v).$

For any given $g$, we have proved that $\langle
g,T_{C_{\sigma^{-1}(f)}}(h)\rangle=0$, that is $T_{C_{\sigma^{-1}(f)}}(h)\in
C^{\perp}.$ To sum up, for any given $h\in C^{\perp},$ we get
$T_{C_{\sigma^{-1}(f)}^{t}}(h)\in C^{\perp},$ which implies that $C^{\perp}$
is a $(\sigma^{-1},-\sigma^{-1}\delta)$-sequential code in
$R^{op}[x,\sigma^{-1},-\sigma^{-1}\delta]/R^{op}[x,\sigma^{-1},-\sigma^{-1}\delta]{\sigma}^{-1}(f)$.
∎

##  5 Duality

In Section 4, we observed that the Euclidean dual of a
$(\sigma,\delta)$-polycyclic code may not necessarily be a
$(\sigma,\delta)$-polycyclic code. In the current section, we aim to introduce
an alternative duality that ensures the dual of any
$(\sigma,\delta)$-polycyclic code remains a $(\sigma,\delta)$-polycyclic code.

######  Lemma 11.

Suppose that $f_{0}$ is invertible and $\sigma(f_{i})=f_{i}$ and
$\delta(f_{i})=0$ for all $i$ satisfying $0\leqslant i\leqslant n-1.$ Then
$x\in S/Sf$ is a right invertible element.

###### Proof.

Define $k(x)=\sum_{i=0}^{n-2}-f_{0}^{-1}f_{i+1}x^{i}+f_{0}^{-1}x^{n-1}.$ We
get
$0=\delta(f_{0}^{-1}f_{0})=\sigma(f_{0}^{-1})\delta(f_{0})+\delta(f_{0}^{-1})f_{0},$
and thus $\delta(f_{0}^{-1})f_{0}=0.$ This implies that
$\delta(f_{0}^{-1})=0.$ Moreover, we obtain
$\delta(f_{0}^{-1}f_{i})=\sigma(f_{0}^{-1})\delta(f_{i})+\delta(f_{0}^{-1})f_{i}=0.$
Therefore,

$\displaystyle xk(x)$
$\displaystyle=\sum_{i=0}^{n-2}x\bigg{(}-f_{0}^{-1}f_{i+1}\bigg{)}x^{i}+x\bigg{(}f_{0}^{-1}\bigg{)}x^{n-1}$
(11)
$\displaystyle=-\sum_{i=0}^{n-2}\bigg{(}\sigma(f_{0}^{-1}f_{i+1})x+\delta(f_{0}^{-1}f_{i+1})\bigg{)}x^{i}+\bigg{(}\sigma(f_{0}^{-1})x+\delta(f_{0}^{-1})\bigg{)}x^{n-1}$
$\displaystyle=-\sigma(f_{0}^{-1})\bigg{(}\sum_{i=0}^{n-2}f_{i+1}x^{i+1}+x^{n}\bigg{)}$
$\displaystyle=-\sigma(f_{0}^{-1})\bigg{(}\sum_{i=0}^{n-2}f_{i+1}x^{i+1}+x^{n}\bigg{)}-\sigma(f_{0}^{-1}f_{0})+\sigma(f_{0}^{-1}f_{0})$
$\displaystyle=\sigma(f_{0}^{-1})f(x)+1=1\,(\operatorname{mod}f).$

∎

######  Definition 3.

Define the map $\langle\cdot\,,\cdot\rangle_{0}:S/Sf\times S/Sf\to R$ by

$\langle g,h\rangle_{0}=gh(0).$

Recall that $T_{0}=\delta$ is a $(\sigma,\delta)$-PLT and $\delta(1)=0.$ Let
$g,\,h\in S/Sf$ and $gh\,(\operatorname{mod}f)$ be denoted by the polynomial
$r(x)=\sum_{i=0}^{n-1}r_{i}x^{i}$. Note that
$r(0)=\sum_{i=0}^{n-1}r_{i}N_{i}(0)=r_{0}$, so $gh(T_{0})(1)=r_{0}=gh(0)$.
Therefore, we have the equivalent definition $\langle
g,h\rangle_{0}=gh(T_{0})(1)$, which provides technical assistance in our
proofs.

######  Lemma 12.

Suppose that $f_{0}$ is invertible and $\sigma(f_{i})=f_{i}$ and
$\delta(f_{i})=0$ for all $i$ satisfying $0\leqslant i\leqslant n-1.$ Then
$\langle\cdot,\cdot\rangle_{0}$ is a non-degenerate $\zeta$-sesquilinear form,
where $\zeta:=\operatorname{Id}$.

###### Proof.

Let $g,h\in S/Sf$. For a fixed $\beta\in R$ and for any $r\in R,$ we have

$\beta h(T_{0})(r)=\operatorname{Id}\bigg{(}\beta
h(T_{0})(r)\bigg{)}=h(T_{0})(r)\beta=h(T_{0})(r)\operatorname{Id}(\beta)=h(T_{0})(r)\zeta(\beta).$

We thus have $g\beta h(T_{0})(r)=gh(T_{0})(r)\zeta(\beta),$ which implies
$\langle g,\beta h\rangle_{0}=\langle g,h\rangle_{0}\zeta(\beta).$ Verifying
the other conditions of the $\zeta$-sesquilinear form definition is easy.
Therefore, $\langle\cdot,\cdot\rangle_{0}$ is a $\zeta$-sesquilinear form. To
prove the non-degeneracy of the form, suppose that
$g=g_{0}+g_{1}x+\cdots+g_{n-1}x^{n-1}\in S/Sf$ satisfies $\langle
g,h\rangle_{0}=0$ for all $h\in S/Sf.$ Let us assume $h=1.$ This gives
$0=g(T_{0})(1)=g(\delta)(1)=g_{0}.$ Consequently, we can rewrite
$g=(g_{1}+g_{2}x+\cdots+g_{n-1}x^{n-2})x$. By Lemma 11, $x$ is right
invertible. Let us denote its inverse by $x^{-1}$. If we assume $h=x^{-1}$, we
get $0=\langle g,h\rangle_{0}=gh(\delta)(1)=g_{1}$. By applying the same
discussion, we deduce that $g_{i}=0$ for all $i$. Consequently, we have $g=0$,
implying that the form is non-degenerate. ∎

######  Definition 4.

For the right $(\sigma,\delta)$-polycyclic code $C$, define the left
annihilator dual by

$l_{0}(C)=\\{g\in S/Sf:\langle g,h\rangle_{0}=0,\,\,\text{for all}\,\,h\in
C\\}=\\{g\in S/Sf:\langle g,C\rangle_{0}=0\\}.$

Furthermore, for the left $(\sigma,\delta)$-polycyclic code $C$, define the
right annihilator dual by

$r_{0}(C)=\\{g\in S/Sf:\langle h,g\rangle_{0}=0,\,\,\text{for all}\,\,h\in
C\\}=\\{g\in S/Sf:\langle C,g\rangle_{0}=0\\}.$

######  Remark 2.

By Lemma 1, we have $\langle g,h\rangle_{0}=gh(0)=g(T_{0})(h(0)),$ which is
not equal to $h(T_{0})(g(0))=\langle h,g\rangle_{0}$ in generally. Therefore,
$l_{0}(C)$ and $r_{0}(C)$ are not equal in generally. Assume that $R$ is
commutative and $(\sigma,\delta)=(\operatorname{Id},0).$ In this case,
$T_{0}=0$, and $\langle g,h\rangle_{0}=\langle h,g\rangle_{0}$. Therefore
$l_{0}(C)=r_{0}(C)$ and both are equal to the annihilator dual $C^{\perp_{0}}$
as defined in [1, 3].

Referring to [13], for a subset $X$ in a non-commutative ring $A$, the left
annihilator $\operatorname{Ann}_{l}(X)$ and the right annihilator
$\operatorname{Ann}_{r}(X)$ of $X$ are defined by

$\operatorname{Ann}_{l}(X)=\\{a\in A:ax=0,\,\,\text{for all}\,\,x\in
X\\}\quad\text{and}\quad\operatorname{Ann}_{r}(X)=\\{a\in A:xa=0,\,\,\text{for
all}\,\,x\in X\\}.$

######  Theorem 3.

Suppose that $f_{0}$ is invertible and $\sigma(f_{i})=f_{i}$ and
$\delta(f_{i})=0$ for all $i$ satisfying $0\leqslant i\leqslant n-1.$ If $C$
is a right $(\sigma,\delta)$-polycyclic code, then
$l_{0}(C)=\operatorname{Ann}_{l}(C)$. In addition, if $C$ is a left
$(\sigma,\delta)$-polycyclic code, then $r_{0}(C)=\operatorname{Ann}_{r}(C).$

###### Proof.

We have $\operatorname{Ann}_{l}(C)\subseteq l_{0}(C).$ Conversely, let $g\in
l_{0}(C)$, $h$ be an arbitrary element of $C$, and $gh\,(\operatorname{mod}f)$
be denoted by the polynomial $r_{0}+r_{1}x+\cdots+r_{n-1}x^{n-1}$. We have
$gh(0)=0,$ which leads to $r_{0}=0$. So,
$gh(x)=(r_{1}+r_{2}x+\cdots+r_{n-1}x^{n-2})x.$ Using Lemma 11,

$ghk(x)=(r_{1}+r_{2}x+\cdots+r_{n-1}x^{n-2})\,\,(\operatorname{mod}f).$ (12)

On the other side, we have
$\delta(k(0))=\delta(-f_{0}^{-1}f_{1})=-\sigma(f_{0}^{-1})\delta(f_{1})-\delta(f_{0}^{-1})f_{1}$.
Since $\delta(f_{1})=\delta(f_{0}^{-1})=0$, we get $\delta(k(0))=0.$ By Lemma
1, $ghk(0)=gh(T_{0})(k(0))=gh(\delta)(k(0))=gh(0)=0$. Therefore $r_{1}=0$ by
Equation (12). Applying the same discussion, we can deduce that $r_{i}=0$ for
all $i$. Consequently, $gh=0$, which implies to
$g\in\operatorname{Ann}_{l}(C).$ Therefore,
$l_{0}(C)=\operatorname{Ann}_{l}(C)$. The proof of the second equality is
similar.

∎

######  Corollary 4.

Suppose that $f_{0}$ is invertible and $\sigma(f_{i})=f_{i}$ and
$\delta(f_{i})=0$ for all $i$ satisfying $0\leqslant i\leqslant n-1.$

  1. 1.

If $C$ is a right $(\sigma,\delta)$-polycyclic code then $l_{0}(C)$ is a right
$(\sigma,\delta)$-polycyclic code.

  2. 2.

If $C$ is a left $(\sigma,\delta)$-polycyclic code then $r_{0}(C)$ is a left
$(\sigma,\delta)$-polycyclic code.

###### Proof.

Assume that $C$ is a right $(\sigma,\delta)$-polycyclic code. Let
$a\in\operatorname{Ann}_{l}(C)$ and $s\in S$. For any given arbitrary element
$c\in C$, we know that $sc$ also belongs to $C$. Hence, $(as)c=a(sc)=0,$ which
implies $as\in\operatorname{Ann}_{l}(C).$ Thus $\operatorname{Ann}_{l}(C)$ is
a left $S$-submodule of $S/Sf.$ Using Theorem 3, $l_{0}(C)$ is a left
$S$-submodule of $S/Sf,$ which means $l_{0}(C)$ is a right
$(\sigma,\delta)$-polycyclic code. The proof of the second statement is
similar. ∎

##  6 Equivalent $(\sigma,\delta)$-polycyclic codes

The Hamming weight $\omega_{H}(g)$ of a vector $g\in R^{n}$ is the number of
nonzero coordinates in $g$. A linear map $\psi:C_{1}\rightarrow C_{2}$ is
called Hamming isometry if it is an isomorphism that preserves the Hamming
weight, meaning that for every codeword $c\in C_{1}$, we have
$\omega_{H}(\psi(c))=\omega_{H}(c)$. Two codes $C_{1}$ and $C_{2}$ are called
Hamming isometrically equivalent if there exists a Hamming isometry
$\psi:C_{1}\rightarrow C_{2}$.

######  Proposition 2 ([23] Proposition 1.6).

Let $T_{1}$ and $T_{2}$ be $(\sigma,\delta)$-PLTs defined on free $R$-modules
$V_{1}$ and $V_{2}$ with basis $\mathcal{B}_{1}$ and $\mathcal{B}_{2},$
respectively. Suppose $\varphi\in\operatorname{Hom}_{R}(V_{1},V_{2})$ is an
$R$-module homomorphism, and let $B$, $M_{1}$ and $M_{2}$ denote matrix
representations of $\varphi$, $T_{1}$ and $T_{2},$ respectively. The following
conditions are equivalent.

  1. 1.

$\varphi\in\operatorname{Hom}_{R}(V_{1},V_{2})$.

  2. 2.

$\varphi T_{1}=T_{2}\varphi$.

  3. 3.

$M_{1}B=\sigma(B)M_{2}+\delta(B)$.

Leroy's paper [23] includes the following corollary without proof. In order to
provide context for the next theorem, we will present its proof.

######  Corollary 5 ([23] Corollary 1.8).

Consider the same notations as the previous proposition. Let $f_{1},f_{2}\in
S$ be two monic polynomials of degree $n$ with companion matrices
$C_{f_{1}},C_{f_{2}}$. The map $\varphi:S/Sf_{1}\to S/Sf_{2}$ is an $R$-module
isomorphism if and only if $B$ is an invertible matrix and
$C_{f_{1}}B=\sigma(B)C_{f_{2}}+\delta(B)$.

###### Proof.

Define the linear transformation $\psi_{B}:R^{n}\to R^{n}$ by
$\psi_{B}(v)=vB.$ Assume that $B$ is an invertible matrix and
$C_{f_{1}}B=\sigma(B)C_{f_{2}}+\delta(B)$. Proposition 2 implies that
$\varphi:S/Sf_{1}\to S/Sf_{2}$ is an $R$-module homomorphism. Now, consider
the following commutative diagram.

${S/Sf_{1}}$${S/Sf_{2}}$${R^{n}}$${R^{n}}$$\scriptstyle{\varphi}$$\scriptstyle{\Omega_{f_{1}}^{-1}}$$\scriptstyle{\Omega_{f_{2}}^{-1}}$$\scriptstyle{\psi_{B}}$

Figure 1:

The map $\psi_{B}$ is injective, and due to the equal cardinality of its
domain and codomain, it is an $R$-module isomorphism. Additionally, since
$\Omega_{f_{1}}$ and $\Omega_{f_{2}}$ are both isomorphisms, it follows that
$\varphi$ is also an isomorphism. The converse can be proved using the same
discussion. ∎

######  Corollary 6.

$\psi_{B}$ is an $R$-module isomorphism if and only if $B$ is an invertible
and $C_{f_{1}}B=\sigma(B)C_{f_{2}}+\delta(B)$.

######  Theorem 4.

Suppose that $B$ is a monomial matrix and
$C_{f_{1}}B=\sigma(B)C_{f_{2}}+\delta(B)$. Then, there is a one-to-one Hamming
isometrically equivalent correspondence between the set of right
$(\sigma,\delta)$-polycyclic codes in $S/Sf_{1}$ and right
$(\sigma,\delta)$-polycyclic codes in $S/Sf_{2}.$

###### Proof.

Claim 1: $\varphi$ is a Hamming isometry if and only if $\psi_{B}$ is a
Hamming isometry. To prove the claim, note that for every $g_{1}(x)\in
S/Sf_{1}$ (which associated with $g_{1}\in R^{n}$), and for every $g_{2}(x)\in
S/Sf_{2}$ (which associated with $g_{2}\in R^{n}$), we have
$\omega_{H}(g_{1}(x))=\omega_{H}(g_{1})$ and
$\omega_{H}(g_{2}(x))=\omega_{H}(g_{2}).$ Let $\varphi$ be a Hamming isometry.
From the commutativity of the diagram in Figure 1, we have
$\Omega_{f_{2}}^{-1}\varphi=\psi_{B}\Omega_{f_{1}}^{-1}.$ Therefore,
$\omega_{H}(g_{1})=\omega_{H}(g_{1}(x))=\omega_{H}(\varphi(g_{1}(x)))=\omega_{H}(\Omega^{-1}_{f_{2}}\varphi(g_{1}(x)))=\omega_{H}(\psi_{B}\Omega^{-1}_{f_{1}}(g_{1}(x)))=\omega_{H}(\psi_{B}(g_{1})).$
Thus, $\psi_{B}$ is a Hamming isometry. The proof of the converse is
straightforward.

Claim 2: $\psi_{B}$ is a Hamming isometry if and only if $B$ is a monomial
matrix. The proof from right to left is clear and the proof of another
direction derives from Proposition 6.1 in [42].

To prove the theorem, let $C_{1}$ be a right $(\sigma,\delta)$-polycyclic code
in $S/Sf_{1}$. Corollaries 5, 6 and Claims 1, 2 imply that the restriction of
$\psi_{B}$ to $C_{1}$, denoted as $\psi_{B}:C_{1}\to C_{2}:=\psi_{B}(C_{1})$,
is a Hamming isometry. ∎

As an application of the above theorem, we have the next theorem, which says
that both left and right $(\sigma,\delta)$-polycyclic codes lead to equivalent
theories. For this reason, we have only focused on the right
$(\sigma,\delta)$-polycyclic codes.

In the sequel of this paper, assume that $R$ is the field $\mathbb{F}_{q}$ and
$\sigma\in\operatorname{Aut}(R)$.

##  7 $(\sigma,\delta)$-Mattson-Solomon transform for simple-root
$(\sigma,\delta)$-polycyclic codes

The theory of Wedderburn polynomials in Ore extensions, extensively explored
by Lam and Leroy in [20, 21], forms a foundational basis for two next
sections. In this section, our first objective is to define simple-root
$(\sigma,\delta)$-polycyclic codes, using the concept of Wedderburn
polynomials in the Ore extension $S=\mathbb{F}_{q}[x,\sigma,\delta]$. Our
second objective is to present a definition for $(\sigma,\delta)$-Mattson-
Solomon transform for simple-root $(\sigma,\delta)$-polycyclic codes.

###  7.1 Algebraic sets and minimal polynomials

For any polynomial $g\in S$, define $V(g):=\\{a\in\mathbb{F}_{q}:g(a)=0\\},$
and for any subset $X\subseteq\mathbb{F}_{q}$, define $I(X):=\\{g\in
S:g(X)=0\\}.$ The set $X$ is called $(\sigma,\delta)$-algebraic if $I(X)\neq
0$. The set $V(g)$ is $(\sigma,\delta)$-algebraic because $g(V(g))=0$.
Moreover, any finite set is $(\sigma,\delta)$-algebraic, see [19] Section $6$.
If $X$ is $(\sigma,\delta)$-algebraic, then the monic generator of $I(X)$ is
called the minimal polynomial of $X$ and is denoted by $g_{X}.$ Clearly
$g_{X}(X)=0$. By the Remainder Theorem, the minimal polynomial $g_{X}$ is the
monic least right common multiple of the linear polynomials $\\{x-a:a\in
X\\}$. So it has the form $(x-a_{0})(x-a_{1})\cdots(x-a_{n-1}),$ where each
$a_{i}$ is $(\sigma,\delta)$-conjugate to some element of $X$ and
$n=\operatorname{deg}g_{X}$. Furthermore, any right root of $g_{X}$ is also
$(\sigma,\delta)$-conjugate to some element of $X$, see [17] Section 4 and
[20] Section 2. An element $a\in\mathbb{F}_{q}$ is called P-dependent on an
algebraic set $X$ if $g(a)=0$ for every $g\in I(X)$. A
$(\sigma,\delta)$-algebraic set $X$ is P-independent if no element $b\in X$ is
P-dependent on $X\setminus{b}$, see [20].

###  7.2 Wedderburn polynomials

A monic polynomial $g\in S$ is called Wedderburn polynomial (or W-polynomial
for short) if the minimal polynomial of $V(g)$ is $g.$ If
$(\sigma,\delta)=(\operatorname{Id},0)$, then W-polynomials are in the form of
$(x-a_{0})\cdots(x-a_{n-1})$, where $a_{i}$s are distinct elements in
$\mathbb{F}_{q}$, see [20] Example 3.5.

Recall from Section 2 that if $g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}\in S$, then
$g(a)=\sum_{i=0}^{n-1}g_{i}N_{i}(a).$

The Vandermonde matrix is a valuable tool for polynomial evaluation. Hence, in
[18], the Vandermonde matrix in the $(\sigma,\delta)$-setting is naturally
defined as follows. A $(\sigma,\delta)$-Vandermonde matrix related to the
elements $a_{0},\ldots,a_{n-1}\in\mathbb{F}_{q},$ denoted by
$V(a_{0},a_{1},\ldots,a_{n-1})$ or simply by $V$, is

$V=V(a_{0},a_{1},\ldots,a_{n-1})=\begin{pmatrix}1&1&\cdots&1\\\
N_{0}(a_{0})&N_{0}(a_{1})&\cdots&N_{0}(a_{n-1})\\\
\vdots&\vdots&\cdots&\vdots\\\
N_{n-1}(a_{0})&N_{n-1}(a_{1})&\cdots&N_{n-1}(a_{n-1})\end{pmatrix}.$

The following theorem, which plays a crucial role in both this section and the
subsequent one, presents some equivalent conditions for W-polynomials.

######  Theorem 5 ([21] Theorem 5.2, Lemma 5.7 and Theorem 5.8).

Let $f\in S$ be a monic polynomial of degree $n$, then the following
statements are equivalent.

  1. 1.

$f$ is a W-polynomial.

  2. 2.

There is a P-independent set
$A=\\{a_{0},a_{1},\ldots,a_{n-1}\\}\subseteq\mathbb{F}_{q}$ such that
$f=f_{A}$.

  3. 3.

There is a set $\\{a_{0},a_{1},\ldots,a_{n-1}\\}\subseteq\mathbb{F}_{q}$ such
that the $(\sigma,\delta)$-Vandermond matrix $V=V(a_{0},a_{1},\ldots,a_{n-1})$
is invertible and

$C_{f}V=\sigma(V)\operatorname{diag}(a_{0},a_{1},\ldots,a_{n-1})+\delta(V).$

  4. 4.

The companion matrix $C_{f}$ is $(\sigma,\delta)$-diagonalizable.

  5. 5.

$Sf=\bigcap_{i=0}^{n-1}S(x-a_{i})$.

  6. 6.

The left $S$-module $S/Sf$ is decomposed as
$\bigoplus_{i=0}^{n-1}S/S{(x-a_{i})}$ .

###  7.3 $(\sigma,\delta)$-Mattson-Solomon transform

######  Definition 5.

A $(\sigma,\delta)$-polycyclic code in $S/Sf$ is called simple-root
$(\sigma,\delta)$-polycyclic code if $f$ is a W-polynomial.

A polycyclic code in the quotient ring $\mathbb{F}_{q}[x]/\langle f(x)\rangle$
is called simple-root if $f(x)$ is a simple-root polynomial in
$\mathbb{F}_{q}[x]$, see [3]. In the case where
$(\sigma,\delta)=(\operatorname{Id},0)$, W-polynomials are precisely simple-
root polynomials. Therefore, simple-root $(\sigma,\delta)$-polycyclic codes
are the generalization of simple-root polycyclic codes.

Let $f$ be a W-polynomial with a P-independent set
$\\{a_{0},a_{1},\ldots,a_{n-1}\\}$ . Define $T:S/Sf\to S/Sf$ by
$T(\sum_{i=0}^{n-1}b_{i}x^{i})=\sum_{i=0}^{n-1}T_{a_{i}}(b_{i})x^{i}$, where
$T_{a_{i}}:R\to R$ is the $(\sigma,\delta)$-PLT given by
$T_{a_{i}}(x)=\sigma(x)a_{i}+\delta(x)$.

######  Lemma 13.

The additive map $T$ is a $(\sigma,\delta)$-PLT.

###### Proof.

For all $a\in\mathbb{F}_{q}$ and $v=\sum_{i=0}^{n-1}b_{i}x^{i}\in S/Sf$ we get

$\displaystyle T(av)$
$\displaystyle=T\bigg{(}\sum_{i=0}^{n-1}ab_{i}x^{i}\bigg{)}=\sum_{i=0}^{n-1}T_{a_{i}}(ab_{i})x^{i}$
$\displaystyle=\sum_{i=0}^{n-1}\bigg{(}\sigma(ab_{i})a_{i}+\delta(ab_{i})\bigg{)}x^{i}$
$\displaystyle=\sum_{i=0}^{n-1}\bigg{(}\sigma(a)\sigma(b_{i})a_{i}+\sigma(a)\delta(b_{i})+\delta(a)b_{i}\bigg{)}x^{i}$
$\displaystyle=\sigma(a)\sum_{i=0}^{n-1}\bigg{(}\sigma(b_{i})a_{i}+\delta(b_{i})\bigg{)}x^{i}+\delta(a)\sum_{i=0}^{n-1}b_{i}x^{i}=\sigma(a)T(v)+\delta(a)V.$

∎

For all $v=\sum_{i=0}^{n-1}b_{i}x^{i}\in S/Sf$ and for all $k\in\mathbb{N},$
we have
$T^{k}(\sum_{i=0}^{n-1}b_{i}x^{i})=\sum_{i=0}^{n-1}T_{a_{i}}^{k}(b_{i})x^{i}$.
Therefore

$g(T)v=g(T)\bigg{(}\sum_{i=0}^{n-1}b_{i}x^{i}\bigg{)}=\sum_{i=0}^{n-1}g(T_{a_{i}})(b_{i})x^{i},\quad\text{for
all}\,\,g\in S/Sf.$ (13)

Using Lemma 13 and Proposition 1, the $\mathbb{F}_{q}$-module $S/Sf$ has the
$S$-module structure induced by the product $g(x)\cdot v:=g(T)v$. Let us
denote this structure by $(S/Sf,\cdot)$. Furthermore, we know $S/Sf$ also
possesses the $S$-module structure induced by the product $g(x)\bullet
h(x)=g(x)h(x)\,(\operatorname{mod}f)$. This structure will be denoted by
$(S/Sf,\bullet)$.

######  Definition 6.

For a W-polynomial $f$ with an associated P-independent set
$A=\\{a_{0},a_{1},\ldots,a_{n-1}\\}$, define a $(\sigma,\delta)$-Mattson-
Solomon transform as

$\displaystyle\operatorname{MS}_{(\sigma,\delta)}:(S/Sf,\bullet)$
$\displaystyle\longrightarrow(S/Sf,\cdot)$ $\displaystyle g(x)$
$\displaystyle\mapsto\sum_{i=0}^{n-1}g(a_{i})x^{i}.$

######  Theorem 6.

The $(\sigma,\delta)$-Mattson-Solomon transform
$\operatorname{MS}_{(\sigma,\delta)}$ is an $S$-module isomorphism.

###### Proof.

Recall that if $g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}\in S$, then
$g(a)=\sum_{i=0}^{n-1}g_{i}N_{i}(a)$. If we associate the polynomial $g(x)$
with the vector $g=(g_{0},g_{1},\ldots,g_{n-1}),$ then
$gV=(g(a_{0}),g(a_{1}),\ldots,g(a_{n-1}))$, where $V$ is the
$(\sigma,\delta)$-Vandermond matrix related to $a_{0},a_{1},\cdots,a_{n-1}.$
Hence, as depicted in the Figure 2, the vector form of
$\operatorname{MS}_{(\sigma,\delta)}$ is given by
$\operatorname{MS}_{(\sigma,\delta)}(g)=gV$.

${S/Sf}$${S/Sf}$${R^{n}}$${R^{n}}$$\scriptstyle{\operatorname{MS}_{(\sigma,\delta)}}$$\scriptstyle{\Omega_{f}^{-1}}$$\scriptstyle{\Omega_{f}^{-1}}$$\scriptstyle{\operatorname{MS}_{(\sigma,\delta)}(g)=gV}$

Figure 2:

Using Theorem 5, $V$ is an invertible matrix, and hence
$\operatorname{MS}_{(\sigma,\delta)}$ is a bijection. Furthermore,
$\operatorname{MS}_{(\sigma,\delta)}$ is an $S$-module homomorphism because

$\displaystyle\operatorname{MS}_{(\sigma,\delta)}(g(x)h(x))$
$\displaystyle=\sum_{i=0}^{n-1}gh(a_{i})x^{i}$
$\displaystyle=\sum_{i=0}^{n-1}g(T_{a_{i}})h(a_{i})x^{i}\qquad\text{By Lemma
\ref{20} }$
$\displaystyle=g(T)\bigg{(}\sum_{i=0}^{n-1}h(a_{i})x^{i}\bigg{)}\qquad\text{By
Equation \eqref{60003}}$
$\displaystyle=g(T)\bigg{(}\operatorname{MS_{(\sigma,\delta)}(h(x))}\bigg{)}=g(x)\cdot\operatorname{MS}_{(\sigma,\delta)}(h(x)).$

∎

######  Remark 3.

Let us equip $S/Sf$ with the component-wise product
$(\sum_{i=0}^{n-1}b_{i}x^{i})\star(\sum_{i=0}^{n-1}c_{i}x^{i})=\sum_{i=0}^{n-1}b_{i}c_{i}x^{i}$.
Now, suppose that $(\sigma,\delta)=(\operatorname{Id},0)$. Then

$\displaystyle g(x)\cdot\sum_{i=0}^{n-1}h(a_{i})x^{i}$
$\displaystyle=g(T)\bigg{(}\sum_{i=0}^{n-1}h(a_{i})x^{i}\bigg{)}$
$\displaystyle=\sum_{i=0}^{n-1}g(T_{a_{i}})(h(a_{i}))x^{i}\qquad\text{Equation
\eqref{60003}}$ $\displaystyle=\sum_{i=0}^{n-1}gh(a_{i})x^{i}\qquad\text{Lemma
\ref{20} }$
$\displaystyle=\sum_{i=0}^{n-1}g(a_{i})h(a_{i})x^{i}=\sum_{i=0}^{n-1}g(a_{i})x^{i}\star\sum_{i=0}^{{n-1}}h(a_{i})x^{i}.$

So, due to the bijectivity of $\operatorname{MS}_{(\sigma,\delta)}$, the two
operations $\cdot$ and $\star$ are equivalent. Furthermore,

$\operatorname{MS}_{(\sigma,\delta)}(g(x)h(x))=\operatorname{MS}_{(\sigma,\delta)}(g(x))\star\operatorname{MS}_{(\sigma,\delta)}(h(x)).$

Thus, $\operatorname{MS}_{(\sigma,\delta)}$ is a ring homomorphism. Therefore,
in the case where $(\sigma,\delta)=(\operatorname{Id},0)$, the
$(\sigma,\delta)$-Mattson Solomon map
$\operatorname{MS}_{(\sigma,\delta)}:(S/Sf,\bullet)\to(S/Sf,\star)$
corresponds to the Mattson Solomon map of simple-root polycyclic codes defined
in [3].

##  8 Decomposition of simple-root $(\sigma,\delta)$-polycyclic codes

Let us recall that $S=\mathbb{F}_{q}[x,\sigma,\delta]$, and consider $C$ as a
$(\sigma,\delta)$-polycyclic code. In this section, we aim to find the direct
sum decomposition $C=C_{0}\oplus C_{1}\oplus\cdots\oplus C_{n-1},$ where each
$C_{i}$ is a $(\sigma,\delta)$-polycyclic code. Additionally, recall the
definition of the idealizer ring of $Sf$ as $\operatorname{Idl}(Sf)=\\{g\in
S:fg\in Sf\\}$. A polynomial $g\in S$ is called right invariant if
$gS\subseteq Sg$ and is called invariant if $Sg=gS.$ It is easy to see that
$x\in\operatorname{Idl}(Sf)$ if and only if $fS\subseteq Sf$. Therefore,
Proposition 4.1 in [22] gives the following lemma.

######  Lemma 14.

If $x\in\operatorname{Idl}(Sf)$, then the following statements are true.

  1. 1.

$f$ is right invariant.

  2. 2.

$f$ is invariant.

  3. 3.

If $a\in\mathbb{F}_{q}$ and $f(a)=0$, then $f(\Delta(a))=0.$

######  Lemma 15 ([23] Corollary 1.12).

Let $f\in S$ be a polynomial of degree $n>1$. Then the following statements
are equivalent.

  1. 1.

$x\in\operatorname{Idl}(Sf)$.

  2. 2.

For any $g\in S$, $g\in Sf$ if and only if $g(C_{f})=0$.

  3. 3.

$f(C_{f})=0$.

######  Lemma 16 ([20] Proposition 2.6).

Let $f\in S$ be a polynomial of degree $n.$ Then the following statements
hold.

  1. 1.

The right roots of $f$ can exist in at most $n$ of $(\sigma,\delta)$-conjugacy
classes of $\mathbb{F}_{q}$.

  2. 2.

If $f(x)=(x-a_{0})\cdots(x-a_{n-1})$, then each right root of $f$ in
$\mathbb{F}_{q}$ is $(\sigma,\delta)$-conjugate to some $a_{i}$.

######  Lemma 17.

Let $x\in\operatorname{Idl}(Sf)$ and $f$ be a W-polynomial with a
P-independent set $A=\\{a_{0},a_{1},\ldots,a_{n-1}\\}.$ If $\Gamma$ denotes
the set of all eigenvalues of $T_{C_{f}}$, then
$\Gamma=\bigcup_{i=0}^{n-1}\Delta(a_{i}).$

###### Proof.

Applying Theorem 5, $f=f_{A}.$ According to Subsection 7.1,
$f(x)=(x-b_{0})\cdots(x-b_{n-1}),$ where each $b_{i}$ is
$(\sigma,\delta)$-conjugate to some element of $A$. For a fixed $i$, there is
a non-zero polynomial $g_{i}(x)\in S$ such that $f(x)=g_{i}(x)(x-b_{i}).$
Using Lemma 14, we have $Sf=fS,$ which implies the existence of a non-zero
polynomial $h_{i}(x)\in S$ such that $f(x)=(x-b_{i})h_{i}(x).$ We claim that
there is an element $v\in\mathbb{F}_{q}^{n}$ such that
$w_{i}:=h_{i}(T_{C_{f}})(v)\neq 0$. Otherwise, we have $h_{i}(T_{C_{f}})=0$.
However, Lemma 15 and the fact $\deg(h_{i})<\deg(f)$ lead to $h_{i}(x)=0$,
which is a contradiction. Consequently, there is a non-zero $w_{i}$ such that
$0=f(T_{C_{f}})(v)=(T_{C_{f}}-b_{i}I)h_{i}(T_{C_{f}})(v)=(T_{C_{f}}-b_{i}I)w_{i}=T_{C_{f}}(w_{i})-b_{i}w_{i}I$,
showing that $b_{i}$ is an eigenvalue of $T_{C_{f}}$. Thus, we have
$T_{C_{f}}(\beta
v)=\sigma(\beta)T_{C_{f}}(v)+\delta(\beta)v=(\sigma(\beta)b_{i}+\delta(\beta))v$
for $\beta\in\mathbb{F}_{q}^{*}$ and $v\in\mathbb{F}_{q}^{n}.$ Thus
$T_{C_{f}}(\beta
v)=(\sigma(\beta)b_{i}\beta^{-1}+\delta(\beta)\beta^{-1})\beta
v=b_{i}^{\beta}\beta v.$ Consequently, $b_{i}^{\beta}$ is an eigenvalue of
$T_{C_{f}}$ for all $\beta\in\mathbb{F}_{q}^{*}$, showing
$\bigcup_{i=0}^{n-1}\Delta(b_{i})\subseteq\Gamma$. Conversely, let
$\alpha\in\Gamma$. Then, there exists a non-zero vector $v\in\mathbb{F}_{q}$
such that $T_{C_{f}}(v)=\alpha v$. By Proposition 4.3 in [22], we have
$0=f(T_{C_{f}})(v)=f(\alpha)v$, which implies $f(\alpha)=0.$ Utilizing Lemma
16, we conclude that $\alpha\in\bigcup_{i=0}^{n-1}\Delta(b_{i}).$ Therefore,
$\Gamma=\bigcup_{i=0}^{n-1}\Delta(b_{i}).$ Since each $b_{i}$ is
$(\sigma,\delta)$-conjugate to some element of $A$, we have
$\Gamma=\bigcup_{i=0}^{n-1}\Delta(a_{i}).$

∎

A $(\sigma,\delta)$-Pseudo Linear Transformation $T:V\to V$ is called
algebraic if there is a polynomial $g(x)=\sum_{i=0}^{n-1}g_{i}x^{i}$ such that
$g(T)=\sum_{i=0}^{n-1}g_{i}T^{i}=0.$ The minimal polynomial possessing this
property is called the minimal polynomial of $T$ and is denoted by $g_{T}$,
see [22]. Let $\mathbb{F}_{q}^{n}$ be equipped with the basis
$\mathcal{D}=\\{e_{1},e_{2},\cdots,e_{n}\\}$, where each $e_{i}$ is a vector
with a $1$ in position $i$ and $0$ in all other positions. With the facts that
$\delta(1)=0=\delta(0)$ and $\sigma(1)=1$, it is easily to see that
$[T_{C_{f}}]_{\mathcal{D}}=C_{f}$. The main theorem of this section follows,
providing a decomposition of $\mathbb{F}_{q}^{n}$ into $T_{C_{f}}$-invariant
submodules associated with distinct eigenvalues of $T_{C_{f}}$.

######  Theorem 7.

Let $x\in\operatorname{Idl}(Sf)$ and $f$ be a W-polynomial with a
P-independent set $A=\\{a_{0},a_{1},\ldots,a_{n-1}\\}.$ Additionally, let
$T_{C_{f}}:\mathbb{F}_{q}^{n}\to\mathbb{F}_{q}^{n}$ be a
$(\sigma,\delta)$-PLT. Then, we have the following statements.

  1. 1.

$\mathbb{F}_{q}^{n}=\oplus_{i=0}^{n-1}V_{\Gamma_{i}},$ where $V_{\Gamma_{i}}$
is the vector space spanned by eigenvectors of $T_{C_{f}}$ associated with an
eigenvalue in $\Gamma_{i}:=\Delta(a_{i})$.

  2. 2.

Each $V_{\Gamma_{i}}$ is a $T_{C_{f}}$-invariant left $S$-submodule of
$\mathbb{F}_{q}^{n}$.

  3. 3.

There are linear transformations $E_{i}$ such that
$\sum_{i=0}^{n-1}E_{i}=\operatorname{Id},$ $E_{i}E_{j}=0$ for $i\neq j$, and
the range of each $E_{i}$ is $V_{\Gamma_{i}}.$

###### Proof.

  1. 1.

Applying Theorem 5, $C_{f}$ is $(\sigma,\delta)$-similar to the diagonal
matrix $\operatorname{diag}(a_{0},a_{1},\cdots,a_{n-1})$. Let us define
$\mathscr{D}({T_{C_{f}}}):=\big{\\{}[T_{C_{f}}]_{\mathcal{B}}\,:\,\mathcal{B}\,\,\text{is
a basis of }\,\,\mathbb{F}_{q}^{n}\big{\\}}$. According to Proposition 2.4 in
[22], $\mathscr{D}({T_{C_{f}}})$ is a $(\sigma,\delta)$-similar class of
matrices in $M_{n}(\mathbb{F}_{q})$. Therefore, since
$C_{f}=[T_{C_{f}}]_{\mathcal{D}}\in\mathscr{D}({T_{C_{f}}})$ we have
$\operatorname{diag}(a_{0},a_{1},\cdots,a_{n-1})\in\mathscr{D}({T_{C_{f}}})$.
Now, applying Theorem 4.9 in [22], we conclude that $T_{C_{f}}$ is
diagonalizable, meaning that
$\mathbb{F}_{q}^{n}=\oplus_{i=0}^{n-1}V_{\Gamma_{i}}$, where $V_{\Gamma_{i}}$
is the set of eigenvectors of $T_{C_{f}}$ associated with an eigenvalue in
$\Gamma_{i}:=\Delta(a_{i}).$

  2. 2.

We know $V_{\Gamma_{i}}$ is the vector space spanned by eigenvectors of
$T_{C_{f}}$ associated with an eigenvalue in $\Gamma_{i}:=\Delta(a_{i})$. Let
$x\in\mathbb{F}_{q}^{*}$ and $T_{C_{f}}(v)=a_{i}^{x}v$, then
$T_{C_{f}}(x^{-1}v)=\sigma(x^{-1})T_{C_{f}}(v)+\sigma(x^{-1})v=\big{(}\sigma(x^{-1})a_{i}^{x}+\sigma(x^{-1})\big{)}v=\big{(}\sigma(x^{-1})a_{i}^{x}+\sigma(x^{-1})\big{)}xx^{-1}v=(a_{i}^{x})^{x^{-1}}x^{-1}v=a_{i}x^{-1}v.$
Therefore, without loss of generality, we can take
$V_{\Gamma_{i}}=\operatorname{span}\\{v\in\mathbb{F}_{q}^{n}:T_{C_{f}}(v)=a_{i}v\\}$.
To prove $V_{\Gamma_{i}}$ is an $S$-submodule of $\mathbb{F}_{q}^{n}$,
consider two cases. In the case $a_{i}=0$, then
$V_{\Gamma_{i}}=\operatorname{ker}(T_{C_{f}}),$ which is an $S$-submodule of
$\mathbb{F}_{q}^{n}$ (notice that by Proposition 1, $\mathbb{F}_{q}^{n}$
itself is an $S$-module). In the case $a_{i}\in\mathbb{F}_{q}^{*},$ we have
$T_{C_{f}}(a_{i}v)=\sigma(a_{i})T_{C_{f}}(v)+\sigma(a_{i})v=\big{(}\sigma(a_{i})a_{i}+\delta(a_{i})\big{)}a_{i}^{-1}a_{i}v=a_{i}^{a_{i}}a_{i}v$,
and hence $a_{i}v\in V_{\Gamma_{i}}$. Applying Proposition 1, $x\cdot
v=T_{C_{f}}(v)=a_{i}v$ for all $v\in V_{\Gamma_{i}}$, which follows $x\cdot
v\in V_{\Gamma_{i}}$. So, the linearity of $V_{\Gamma_{i}}$ implies that it is
an $S$-submodule of $\mathbb{F}_{q}^{n}$. To complete the proof, let $v\in
V_{\Gamma_{i}}$. Then $T_{C_{f}}(v)=a_{i}v\in V_{\Gamma_{i}}.$ So
$V_{\Gamma_{i}}$ is $T_{C_{f}}$-invariant.

  3. 3.

By Lemma 1.11 in [23] and Lemma 15, we have
$0=f(C_{f})=f([T_{C_{f}}]_{\mathcal{D}})=[f(T_{C_{f}})]_{\mathcal{D}}$ and $f$
is the minimal polynomial with this property. Therefore, $f$ is the minimal
polynomial of $T_{C_{f}}$. Now, since $T$ is diagonalizable (as proved in Part
1), Theorem 4.9 in [22] implies that $f=\prod_{i=0}^{n-1}f_{\Gamma_{i}},$
where $f_{\Gamma_{i}}$ is a minimal polynomial of $\Gamma_{i}=\Delta(a_{i}).$

Put $h_{i}=\prod_{j\neq i}f_{\Gamma_{j}},$ $i=0,\ldots,n-1.$ We claim that
$\sum_{i=0}^{n-1}h_{i}S=S.$ By contradiction, assume that
$\sum_{i=0}^{n-1}h_{i}S=gS$ or $\sum_{i=0}^{n-1}h_{i}S=Sg$ for some $g\in S$
with $\operatorname{deg}(g)\geqslant 1.$ If $\sum_{i=0}^{n-1}h_{i}S=gS$, since
$h_{1}\in\sum_{i=0}^{n-1}h_{i}S=gS$, then there is $g_{1}\in S$ such that
$h_{1}=gg_{1}.$ We know that $h_{1}$ is the minimal polynomial of the set
$\cup_{j\neq 1}\Gamma_{j}=\cup_{j\neq 1}\Delta(a_{j}).$ So $h_{1}$ vanished on
$\cup_{j\neq 1}\Gamma_{j}.$ Thus, there exists an element $d\in\cup_{j\neq
1}\Gamma_{j}$ such that $h_{1}(d)=0$ and $g_{1}(d)\neq 0.$ Otherwise,
$g_{1}(d)=0$ for all $d\in\cup_{j\neq 1}\Gamma_{j},$ that is $h_{1}$ can
divide $g_{1},$ which contradicts with $h_{1}=gg_{1}$ and degree of $g$. Take
$x:=g_{1}(d)\neq 0.$ Using Lemma 1, we obtain
$0=h_{1}(d)=gg_{1}(d)=g(d^{x})x$, and hence $g(d^{x})=0$. Now assume that
$d\in\Gamma_{l},$ where $l\neq 1.$ With the same argument, we can say there is
a $g_{l}\in S$ such that $h_{l}=gg_{l}.$ The polynomial $h_{l}$ is a minimal
polynomial of $\cup_{j\neq l}\Gamma_{j}=\cup_{j\neq l}\Delta(a_{j})$. Applying
Lemma 5.2 in [17], $h_{l}$ is right invariant, and applying Lemma 14, it is
invariant. Thus, there is a polynomial $k_{l}$ such that $h_{l}=k_{l}g$. So,
$g(d^{x})=0$ leads to $h_{l}(d^{x})=0.$ Using Lemma 14 and considering
$h_{l}(d^{x})=0$, we obtain $h_{l}(\Gamma_{l})=0,$ which contradicts with the
definition of $h_{1}$. With the same argument, we can deduct
$\sum_{i=0}^{n-1}h_{i}S=Sg$ also leads to a contradiction. So, the claim is
proved.

Using this claim, there are polynomials $g_{0}(x),\ldots,g_{n-1}(x)$ such that
$\sum_{i=0}^{n-1}h_{i}g_{i}(x)=1.$ Let us define
$E_{i}:=h_{i}g_{i}(T_{C_{f}})$. Therefore,
$\sum_{i=0}^{n-1}E_{i}=\operatorname{Id}.$ Moreover, applying the same proof
used for $h_{l}$, we can show that $h_{i}$ is the right invariant for all $i$.
So, there are polynomials $k_{i}(x)\in S$ such that
$h_{i}g_{i}(x)=k_{i}h_{i}(x)$. By Corollary 1, we have
$E_{i}E_{j}=h_{i}g_{i}h_{j}g_{j}(T_{C_{f}})=k_{i}h_{i}h_{j}g_{j}(T_{C_{f}})=k_{i}(T_{C_{f}})h_{i}h_{j}(T_{C_{f}})g_{j}(T_{C_{f}})$
for $i\neq j$. Since $h_{i}h_{j}$ is divided by $f$ and $f(T_{C_{f}})=0,$ we
get $E_{i}E_{j}=0.$

To complete the proof, we need to determine the range of $E_{i}$. As proved in
Part 2,
$V_{\Gamma_{i}}=\operatorname{span}\\{v\in\mathbb{F}_{q}^{n}:T_{C_{f}}(v)=a_{i}v\\}=\operatorname{span}\\{v\in\mathbb{F}_{q}^{n}:(T_{C_{f}}-a_{i}I)v=0\\}.$
We have $f=\prod_{i=0}^{n-1}f_{\Gamma_{i}}$ and $n=\deg f$. So, each
$f_{\Gamma_{i}}$ is a linear polynomial in the form $(x-b_{i})$. On the other
side, $f_{\Gamma_{i}}$ is the minimal polynomial of
$\Gamma_{i}=\Delta_{(}a_{i}).$ So $f_{\Gamma_{i}}(a_{i})=0$, which follows
that $b_{i}=a_{i}$, that is $f_{\Gamma_{i}}(x)=x-a_{i}$. Thus,
$V_{\Gamma_{i}}=\operatorname{ker}f_{\Gamma_{i}}(T_{C_{f}}).$ Now let $\alpha$
be in the range of $E_{i}$. Since $\sum_{i=0}^{n-1}E_{i}=\operatorname{Id}$,
$E_{j}E_{i}=0$, and $\alpha$ is in the range of $E_{i}$, we have
$E_{i}(\alpha)=\alpha.$ Thus
$f_{\Gamma_{i}}(T_{C_{f}})(\alpha)=f_{\Gamma_{i}}(T_{C_{f}})E_{i}(\alpha)=f_{\Gamma_{i}}(T_{C_{f}})h_{i}(T_{C_{f}})g_{i}(T_{C_{f}})(\alpha)=0,$
because $f=\prod_{i=0}^{n-1}f_{\Gamma_{i}}$ divides $f_{\Gamma_{i}}h_{i}g_{i}$
and $f(T_{C_{f}})=0.$ Hence $V_{\Gamma_{i}}$ contains the range of $E_{i}$.
Conversely, suppose $\alpha\in V_{\Gamma_{i}}.$ Then
$f_{\Gamma_{i}}(T_{C_{f}})(\alpha)=0.$ Note that for all $j\neq i$,
$h_{j}g_{j}$ is divided by $f_{\Gamma_{i}}$. So $E_{j}(\alpha)=0.$ Since
$\sum_{i=0}^{n-1}E_{i}=\operatorname{Id},$ we can immediately conclude that
$E_{i}(\alpha)=\alpha$, which means $\alpha$ is in the range of $E_{i}$.

∎

######  Corollary 7.

Let $x\in\operatorname{Idl}(Sf)$ and $f$ be a W-polynomial with a
P-independent set $A=\\{a_{0},a_{1},\ldots,a_{n-1}\\}.$ If $C$ is a
$T_{C_{f}}$-invariant $S$-submodule of $\mathbb{F}_{q}^{n}$, then

$C=\big{(}C\cap V_{\Gamma_{0}}\big{)}\oplus\big{(}C\cap
V_{\Gamma_{1}}\big{)}\oplus\cdots\oplus\big{(}C\cap V_{\Gamma_{n-1}}\big{)}$
(14)

and each $S$-submodule $C\cap V_{\Gamma_{i}}$ is $T_{C_{f}}$-invariant.

###### Proof.

According to Theorem 7 and its proof,
$\mathbb{F}_{q}^{n}=\oplus_{i=0}^{n-1}V_{\Gamma_{i}},$ and each projection
$E_{i}$ is associated with a polynomial $p_{i}$ such that
$E_{i}=p_{i}(T_{C_{f}})$. Let $\alpha\in C$, then
$\alpha=\alpha_{0}+\alpha_{1}+\cdots+\alpha_{n-1}$, where each $\alpha_{i}\in
V_{\Gamma_{i}}$. Applying Statement (3) in Theorem 7, we deduce that
$\alpha_{i}=E_{i}(\alpha)=p_{i}(T_{C_{f}})(\alpha)$. Since $C$ is invariant
under $T_{C_{f}}$, it follows that $\alpha_{i}\in C.$ In other words,
$\alpha_{i}\in C\cap V_{\Gamma_{i}},$ and hence $C\subseteq\big{(}C\cap
V_{\Gamma_{0}}\big{)}\oplus\big{(}C\cap
V_{\Gamma_{1}}\big{)}\oplus\cdots\oplus\big{(}C\cap V_{\Gamma_{n-1}}\big{)},$
which completes the proof of Equation (14). Applying Theorem 7, each
$V_{\Gamma_{i}}$ is a $T_{C_{f}}$-invariant left $S$-submodule of
$\mathbb{F}_{q}^{n}$. Thus each $C\cap V_{\Gamma_{i}}$ is
$T_{C_{f}}$-invariant. ∎

######  Corollary 8.

If $C$ is a simple-root $(\sigma,\delta)$-polycyclic code, then there are
simple-root $(\sigma,\delta)$-polycyclic codes $C_{i}$ such that

$C=C_{0}\oplus C_{1}\oplus\cdots\oplus C_{n-1}.$

###### Proof.

Take $C_{i}=C\cap V_{\Gamma_{i}}$ and use Corollary 7. ∎

##  9 Conclusions and future work

In this paper, we have explored $(\sigma,\delta)$-polycyclic codes in the Ore
extension $S=R[x,\sigma,\delta]$, where $R$ is a finite, not necessarily
commutative, ring. We have defined these codes as S-submodules in the quotient
module $S/Sf$. We have established that the Euclidean duals of
$(\sigma,\delta)$-polycyclic codes are $(\sigma,\delta)$-sequential codes.
Using $(\sigma,\delta)$-Pseudo Linear Transformation, we have introduced the
annihilator dual of $(\sigma,\delta)$-polycyclic codes. Moreover, we have
demonstrated that the annihilator dual of $(\sigma,\delta)$-polycyclic codes
preserves their $(\sigma,\delta)$-polycyclic nature. Furthermore, we have
classified the Hamming isometrical equivalence $(\sigma,\delta)$-polycyclic
codes, a crucial aspect in the study of each class of codes. Employing
Wedderburn polynomials, we have defined simple-root
$(\sigma,\delta)$-polycyclic codes and have constructed the
$(\sigma,\delta)$-Mattson-Solomon transform for this class of codes. Finally,
we have provided a decomposition of the simple-root
$(\sigma,\delta)$-polycyclic codes associated with distinct eigenvalues of
$T_{C_{f}}$. To further our research we plan to explore
$(\sigma,\delta)$-multivariable codes and derive optimal codes from
$(\sigma,\delta)$-polycyclic codes.

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