The Homogeneous Gray image of linear codes over the Galois ring GR(4, m)

Abstract

Let R be the Galois ring of characteristic 4 and cardinality $4^{m}$, where m is a natural number. Let $\mathcal {C}$ be a linear code of length n over R and $\Phi$ be the Homogeneous Gray map on $R^n$. In this paper, we show that $\Phi (\mathcal {C})$ is linear if and only if for every $\varvec{X}, \varvec{Y}\in \mathcal {C}$, $2(\varvec{X} \odot \varvec{Y})\in \mathcal {C}$. Using the generator polynomial of a cyclic code of odd length over R, a necessary and sufficient condition is given which its Gray image is linear.

1 Introduction

For any positive integer $s\ge 1$, let $\mathbb {Z}_{2^s}$ be the ring of integers modulo $2^s$. A Galois ring is defined to be a finite commutative ring R with identity 1 such that the set of zero divisors of R with 0 added is a principal ideal $\langle p\rangle$ for some prime number p. A polynomial over $\mathbb {Z}_{2^s}$ is called basic irreducible if its reduction modulo 2 is irreducible over $\mathbb {F}_2$, and it is monic if its leading coefficient is 1. Let h(x) be a monic basic irreducible polynomial of degree $m>0$ over $\mathbb {Z}_{2^s}$. It is known that the residue class ring $\frac{\mathbb {Z}_{2^s}[x]}{\langle h(x) \rangle }$ is a Galois ring of characteristic $2^s$ and cardinality $2^{sm}$, and is denoted by $GR(2^s,m)$. Also, it is local ring with maximal ideal $\mathfrak {m}=\langle 2 \rangle$. Furthermore, any ideal of $GR(2^s,m)$ is of the form $\langle 2^i \rangle$ for $1 \le i \le s$, and there is a chain of ideals: $GR(2^s,m)=\langle 1 \rangle \supset \langle 2 \rangle \supset \cdots \supset \langle 2^{s}\rangle =\{0\}$. In this paper, we focus on the special case $s=2$ (i.e., GR(4, m)). Throughout this paper, we denote GR(4, m) by R. The residue classes $a_0+a_1x+\cdots +a_{m-1}x^{m-1}+\langle h(x) \rangle$, where $a_0, a_1,\ldots ,a_{m-1}\in \mathbb {Z}_4$, are all the distinct elements of R. The natural homomorphism $\bar{}: \mathbb {Z}_4\longrightarrow \mathbb {F}_2$ is defined by $c_0+2c_1\longmapsto c_0$, can be extended to a ring epimorphism from $\mathbb {Z}_4[x]$ to $\mathbb {F}_2[x]$ as following:

$$\begin{aligned} \bar{~}: \mathbb {Z}_4[x]&\longrightarrow \mathbb {F}_2[x]\\ a_0+a_1x+\cdots +a_nx^n&\longmapsto \bar{a}_0+\bar{a}_1x+\cdots +\bar{a}_nx^n, \nonumber \end{aligned}$$

(1)

therefore the ring homomorphism (1) induces a ring homomorphism

$$\begin{aligned} R&\longrightarrow \frac{\mathbb {F}_{2}[x]}{\langle \bar{h}(x)\rangle }\\ a_0+a_1x+\cdots +a_nx^n+\langle h(x)\rangle&\longmapsto \bar{a}_0+\bar{a}_1x+\cdots +\bar{a}_nx^n+\langle \bar{h}(x)\rangle , \nonumber \end{aligned}$$

(2)

which will also be denoted by $\bar{~}$. Write $\xi = x + \langle h(x)\rangle$, then $h(\xi )=0$, and $\xi$ is also of order $2^{m}-1$. Every element of R can be written uniquely in the following form

$$\begin{aligned} a_0 + a_1 \xi + \cdots + a_{m-1} \xi ^{m-1},\quad a_{i}\in \mathbb {Z}_{4}\ (0\le i \le m-1), \end{aligned}$$

(3)

and $R = \mathbb {Z}_{4}[\xi ]$. The representation (3) is called the additive representation of the elements of the Galois ring R. Denote the image of $\xi$ under the map (2) by $\bar{\xi }$, then $\bar{\xi }=x+\langle \bar{h}(x)\rangle ,\bar{h}(\bar{\xi })=0$, and $\frac{\mathbb {F}_{2}[x]}{\langle \bar{h}(x)\rangle }=\mathbb {F}_{2}[\bar{\xi }]\simeq \mathbb {F}_{2^m}$. Moreover, $\frac{R}{\langle 2 \rangle }\simeq \mathbb {F}_{2^m}$. Let $\tau = \{0,1,\ldots ,\xi ^{2^{m}-2}\}$, known as Teichmüller set. Then any element $c\in R$ can be written uniquely as

$$\begin{aligned} c = a + 2b, \quad (a,b\in \tau ). \end{aligned}$$

(4)

The representation (4) is called the 2-adic representation of the element $c\in R$, and we denote by $[c]=[a, b]$, see [12, 16,17,18]. Note that, we do not make distinction between the field $\mathbb {F}_{2^m}$ and the $\tau$. A code of length n over $\mathbb {F}_{2^m}$ is a non-empty subset of $\mathbb {F}_{2^m}^n$, and it called linear if it is a subspace of $\mathbb {F}_{2^m}^n$. A code of length n is linear over R if it is a R-submodule of $R^n$. In addition, a linear code is called cyclic, if a cyclic shift of any codeword is a codeword.

For any $u\in \mathbb {Z}_{4}$, the 2-adic expansion of $u=u_{0}+2u_{1}$ is denoted by $[u]=[u_{0},u_{1}]$, where $u_{0},u_{1}\in \mathbb {F}_2$. As stated in [15], three operations “$\odot$”, “$\oplus$”and “$\circledast$”are introduced as follows; if $u, v\in {\mathbb {Z}_{4}}$, $[u]=[u_{0}, u_{1}]$ and $[v]=[v_{0}, v_{1}]$, then

$$\begin{aligned} u\odot v&=u_{0}v_{0}+2u_{1}v_{1},\\ u\oplus v&=u_{0}\oplus v_{0}+ 2(u_{1}\oplus v_{1}),\\ u\circledast v&=2u_{0}v_{0}. \end{aligned}$$

Note that $u\circledast v$ has been denoted by $(0,\zeta _1)$ in [15]. It is easy to see that $u+v=u\oplus v \oplus (u\circledast v)$, $u\odot v=v\odot u$, $u\oplus u=0$ and $u\odot u=u$. The vectors in $\mathbb {Z}_{4}^{n}$ are written in the form $(u^{1}, u^{2},\ldots , u^{n})$, where $u^{i}\in \mathbb {Z}_{4}$ and $[u^{i}]=[u_{0}^{i}, u_{1}^{i}]$, for all $i=1,\ldots ,n$. The application of the operation “$\odot$”on $\mathbb {Z}_{4}$ to $\mathbb {Z}_{4}^n$ is extended in the natural way; that is, if $\varvec{u}=(u^{1}, u^{2},\ldots , u^{n}), \varvec{v}=(v^{1}, v^{2},\ldots , v^{n})\in \mathbb {Z}_{4}^n$, then $\varvec{u} \odot \varvec{v}=(u^{1} \odot v^{1}, u^{2} \odot v^{2}, \ldots , u^{n}\odot v^{n})$. Recall that for each $\varvec{u}=(u^1, \ldots , u^n)$ and $\varvec{v}=(v^1, \ldots , v^n)$ belong to $\mathbb {Z}_{4}^n$, $\varvec{u} *\varvec{v}=(u^{1}v^{1},\ldots ,u^{n}v^{n})$, and $2(\varvec{u}*\varvec{v})=2(\varvec{u}\odot \varvec{v})$. Also, if U and V are two subsets of $\mathbb {Z}_{4}^n$, then

$$\begin{aligned} U *V =\{\varvec{u} *\varvec{v} :\ \varvec{u} \in U, \varvec{v} \in V \}. \end{aligned}$$

(5)

A vector $\varvec{c}\in R^{n}$ is written in the form $(c^1, c^2, \ldots , c^n)$, where $c^{i}\in R$, and $[c^{i}]=[a^i, b^i]$ $(a^i, b^i\in \mathbb {F}_{2^m})$, for all $i=1,2,\ldots ,n$.

Unlike linear codes over finite fields, linear codes over rings do not have a basis, but there exists a generator matrix for these codes. A matrix $\mathcal {G}$ is called a generator matrix for $\mathcal {C}$ if the rows of $\mathcal {G}$ span $\mathcal {C}$ and none of them can be written as a linear combination of the other rows of $\mathcal {G}$. A linear code of length n over R is permutation equivalent to a code whose generator matrix $\mathcal {G}$ has the form

$$\begin{aligned} \mathcal {G}=\left( \begin{matrix} I_{k_0}&{}A&{}B\\ 0&{}2I_{k_1}&{}2C \end{matrix}\right) , \end{aligned}$$

where A and C have entries only from $\tau$, and B has entries from R, see [7].

There are different Gray maps from $\mathbb {Z}_{2^s}$ to $\mathbb {Z}_2^{2^{s-1}}$, see for instance [1, 2, 10]. One of them is Homogeneous Gray map which has been generalized for finite chain rings [8, 14]. Let $\mathcal {C}$ be a linear code of length n over R and let $\Phi$ be Homogeneous Gray map on $R^n$. In general, $\Phi (\mathcal {C})$ does not need to be linear. One of the interesting topics in algebraic coding theory is to see under what conditions the image of this code is linear or non-linear. In [6, 18], it has been given some necessary and sufficient conditions for which the binary Gray image of a quaternary linear code is linear. Specifically, if $\mathcal {G}$ is a generator matrix of a linear code $\mathcal {C}$ over $\mathbb {Z}_{4}$ and $\{\varvec{v}_{j}\}_{j=1}^{k_0}$ are the rows of order four in $\mathcal {G}$, then, $\Phi (\mathcal {C})$ is a binary linear code if and only if $2(\varvec{v}_{t}*\varvec{v}_{l}) \in \mathcal {C}$, for all t, l satisfying $1\le t<l\le k_0$. In [15] by applying the operation “$\odot$”on $\mathbb {Z}_{2^s}$, a necessary and sufficient condition for linearity $\Phi (\mathcal {C})$ has been stated. The linearity problem has been studied for the family of Hadamard linear codes over $\mathbb {Z}_{8}$ and $\mathbb {Z}_{2^s}$, see [4, 5]. Moreover, Wolfmann [19] determined all cyclic codes of odd length over $\mathbb {Z}_4$ whose Gray images are linear. In [3], we investigated the linearity of the Gray image of a linear code and a linear cyclic code of odd length over $\mathbb {Z}_8$ using their generator matrix and generator polynomial, respectively. Also, the necessary and sufficient conditions are given for which the Gray image of a linear code defined over the Galois ring $GR(p^2,m)$ to be linear [11]. The paper is organized as follows. In Section 2, the Homogeneous Gray image over finite chain rings is introduced. The linearity of Gray image of a linear code of length n over R is investigated in Section 3. In Section 4, we give a condition on the generator polynomial of a linear cyclic code of odd length over R which its Gray image is linear.

2 Preliminaries

A finite commutative ring with identity $1\ne 0$ is called a finite chain ring if its ideals are linearly ordered by inclusion. It is well known that it is a local ring. Let S be a finite chain ring and $\mathfrak {m}=\langle \gamma \rangle$ be the unique maximal ideal of S. Since S is finite, so, there exist the number i such that $\langle \gamma ^i\rangle =\{0\}$. Let s be the minimal number such that $\langle \gamma ^s\rangle =\{0\}$. Therefore, s is called the nilpotency index of $\gamma$, see [8, 13, 14]. Suppose that the residue field $\frac{S}{\langle \gamma \rangle }$ is $\mathbb {F}_{p^m}$, where p is a prime. For every $r\in S$, there are unique $a_0(r),a_1(r),\ldots ,a_{s-1}(r)\in \mathbb {F}_{p^m}$ such that $r=a_0(r)+a_1(r)\gamma +\cdots +a_{s-1}(r)\gamma ^{s-1}$. Thus, each element $\varvec{r} \in S^n$ can be written uniquely as

$$ \varvec{r}=a_0(\varvec{r})+a_1(\varvec{r})\gamma +\cdots +a_{s-1}(\varvec{r})\gamma ^{s-1}, $$

where $a_i(\varvec{r})=(r_{i,0},r_{i,1},\ldots ,r_{i,n-1})\in \mathbb {F}_{p^m}^n$, for every $0\le i\le {s-1}$. In [8], for a finite chain ring S, the Homogeneous Gray map from $S^n$ to $\mathbb {F}_{p^m}^{p^{m(s-1)}n}$ is defined as follows.

Let $\alpha$ be a fixed primitive element of $\mathbb {F}_{p^m}$. An element $\alpha _{\epsilon }\in \mathbb {F}_{p^m}$ corresponding to $\epsilon \in \mathbb {Z}_{p^m}$ is given by

$$ \alpha _\epsilon :=\xi _0(\epsilon )+\xi _1(\epsilon )\alpha +\cdots +\xi _{m-1}(\epsilon )\alpha ^{m-1}, $$

if the p-adic representation of $\epsilon$ is

$$ \epsilon =\xi _0(\epsilon )+\xi _1(\epsilon )p+\cdots +\xi _{m-1}(\epsilon )p^{m-1}, $$

where $\xi _i(\epsilon )\in \{0,1, \ldots ,p-1\}$, for every $0 \le i \le m-1$.

Likewise, each $\omega \in \mathbb {Z}_{p^{m(s-1)}}$ is viewed uniquely as the $p^m$-adic representation

$$ \omega =\bar{\xi }_0(\omega )+\bar{\xi }_1(\omega )p^m +\cdots +\bar{\xi }_{s-2}(\omega )p^{m(s-2)}, $$

where $\bar{\xi }_i(\omega )\in \{0,1,\ldots ,p^{m}-1\}$, for every $0\le i\le s-2$. Define the Homogeneous Gray map $\Phi : S^n\longrightarrow \mathbb {F}_{p^m}^{p^{m(s-1)}n}$ by $\Phi (\varvec{r})=(\varvec{b}_0,\varvec{b}_1,\ldots ,\varvec{b}_{p^{m(s-1)}-1}),$ for all $\varvec{r}=a_0(\varvec{r})+a_1(\varvec{r})\gamma +\cdots +a_{s-1}(\varvec{r})\gamma ^{s-1}\in S^n$, where

$$ \varvec{b}_{\omega p^{m}+\epsilon }=\alpha _\epsilon \widetilde{a_0(\varvec{r})}+\sum _{l=1}^{s-2}\alpha _{\bar{\xi }_{l-1}(\omega )}\widetilde{a_{l}(\varvec{r})}+\widetilde{a_{s-1}(\varvec{r})}, $$

for all $0\le \omega \le p^{m(s-2)}-1$ and $0\le \epsilon \le p^{m}-1$, and $\widetilde{a_i(\varvec{r})}=(\widetilde{r_{i,0}},\widetilde{r_{i,1}},\ldots ,\widetilde{r_{i,n-1}})$, where $\widetilde{~} :S\longrightarrow \mathbb {F}_{p^m}^n$ is the canonical map. It is easy to see that $\Phi$ is injective. The Homogeneous Gray map $\Phi : S^n\longrightarrow \mathbb {F}_{p^m}^{p^{m(s-1)}n}$ can be written in the following form

$$\begin{aligned} \Phi (\varvec{r})=[\varvec{r}]\mathcal {\mathcal {A}}=[\widetilde{a_0(\varvec{r})}, \widetilde{a_1(\varvec{r})}, \ldots , \widetilde{a_{s-2}(\varvec{r})}, \widetilde{a_{s-1}(\varvec{r})}]\mathcal {\mathcal {A}}, \end{aligned}$$

(6)

where

$$\begin{aligned} \mathcal {A}\!=\!\left( \begin{matrix} \alpha _0&{}\cdots &{}\alpha _{p^{m}-1}&{}\alpha _0&{}\cdots &{}\alpha _{p^{m}-1}&{}\cdots &{}\cdots &{}\alpha _0&{}\cdots \\ \alpha _{\bar{\xi }_{0}(0)}&{}\cdots &{}\alpha _{\bar{\xi }_{0}(0)}&{}\alpha _{\bar{\xi }_{0}(1)}&{}\cdots &{}\alpha _{\bar{\xi }_{0}(1)}&{}\cdots &{}\cdots &{}\alpha _{\bar{\xi }_{0}(p^{m(s-2)}-1)}&{}\cdots \\ \alpha _{\bar{\xi }_{1}(0)}&{}\cdots &{}\alpha _{\bar{\xi }_{1}(0)}&{}\alpha _{\bar{\xi }_{1}(1)}&{}\cdots &{}\alpha _{\bar{\xi }_{1}(1)}&{}\cdots &{}\cdots &{}\alpha _{\bar{\xi }_{1}(p^{m(s-2)}-1)}&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots \\ \alpha _{\bar{\xi }_{s-3}(0)}&{}\cdots &{}\alpha _{\bar{\xi }_{s-3}(0)}&{}\alpha _{\bar{\xi }_{s-3}(1)}&{}\cdots &{}\alpha _{\bar{\xi }_{s-3}(1)}&{}\cdots &{}\cdots &{}\alpha _{\bar{\xi }_{s-3}(p^{m(s-2)}-1)}&{}\cdots \\ 1&{}\cdots &{}1&{}1&{}\cdots &{}1&{}\cdots &{}\cdots &{}1&{}\cdots \end{matrix}\quad \begin{matrix} \alpha _{{p^m}-1}\\ \alpha _{\bar{\xi }_{0}(p^{m(s-2)}-1)}\\ \alpha _{\bar{\xi }_{1}(p^{m(s-2)}-1)}\\ \vdots \\ \alpha _{\bar{\xi }_{s-3}(p^{m(s-2)}-1)}\\ 1 \end{matrix} \right) ._{s\times p^{m(s-1)}} \end{aligned}$$

For example,

• Consider $S=\mathbb {Z}_{27}$. So, the Gray map $\Phi$ for every $\varvec{r}\in \mathbb {Z}_{27}^n$ is defined from $\mathbb {Z}_{27}^{n}$ to $\mathbb {F}_{3}^{9n}$ by $\Phi (\varvec{r})=[\varvec{r}]\mathcal {A}=[\widetilde{a_0(\varvec{r})}, \widetilde{a_1(\varvec{r})}, \widetilde{a_2(\varvec{r})}]\mathcal {A}$. It is easy to see that

  $$ \mathcal {A}={\left( \begin{matrix} 0&{}1&{}2&{}0&{}1&{}2&{}0&{}1&{}2\\ 0&{}0&{}0&{}1&{}1&{}1&{}2&{}2&{}2\\ 1&{}1&{}1&{}1&{}1&{}1&{}1&{}1&{}1 \end{matrix}\right) .}_{3\times 9} $$

• Consider $S=GR(4,2)$. Thus, the Gray map $\Phi$ for every $\varvec{r}\in GR(4,2)^n$ is defined from $GR(4,2)^{n}$ to $\mathbb {F}_{4}^{4n}$ by $\Phi (\varvec{r})=[\varvec{r}]\mathcal {A}=[\widetilde{a_0(\varvec{r})}, \widetilde{a_1(\varvec{r})}]\mathcal {A}$. It is easy to check that

  $$ \mathcal {A}={\left( \begin{matrix} 0&{}1&{}\xi &{}1+\xi \\ 1&{}1&{}1&{}1 \end{matrix}\right) .}_{2\times 4} $$

3 Gray images of linear codes over R

In this section, we will present a method that can be used to check the linearity or non-linearity of $\Phi (\mathcal {C})$. The following two definitions will be useful for the proof of the main result of this section.

Definition 1

Let $a=a_{0}+a_{1}\bar{\xi }+\cdots +a_{m-1}\bar{\xi }^{m-1}$ and $b= b_{0}+b_{1}\bar{\xi }+\cdots +b_{m-1}\bar{\xi }^{m-1}$ belong to $\mathbb {F}_{2^{m}}$, where $a_0,a_1,\ldots ,a_{m-1}$, $b_0,b_1,\ldots ,b_{m-1}\in \mathbb {F}_2$. We define

$$\begin{aligned} a\odot b&=a_{0}b_{0}+a_{1}b_{1}\bar{\xi }+\cdots +a_{m-1}b_{m-1}\bar{\xi }^{m-1},\\ a\oplus b&=a_{0}\oplus b_{0}+(a_{1}\oplus b_{1})\bar{\xi }+\cdots +(a_{m-1}\oplus b_{m-1})\bar{\xi }^{m-1}. \end{aligned}$$

One can see that $a\odot b=b\odot a$, $a\odot a=a$, and $a\oplus a=0$. We extend the application of the operations “$\odot$”and “$\oplus$”on $\mathbb {F}_{2^{m}}$ to $\mathbb {F}_{2^{m}}^{n}$ in the natural way; that is, if $\varvec{a}=(u^{1}, u^{2},\ldots , u^{n}), \varvec{b}=(v^{1}, v^{2},\ldots , v^{n})\in \mathbb {F}_{2^{m}}^{n}$, then $\varvec{a} \odot \varvec{b}=(u^{1} \odot v^{1}, u^{2} \odot v^{2}, \ldots , u^{n}\odot v^{n})$, $\varvec{a} \oplus \varvec{b}=(u^1\oplus v^1, u^2\oplus v^2,\ldots ,u^n\oplus v^n)$.

Definition 2

Let $c=a_0+a_1\xi +\cdots +a_{m-1}{\xi }^{m-1}$ and $c'=a'_0+a'_1\xi +\cdots +a'_{m-1}{\xi }^{m-1}$ belong to R, where $a_i=a_{i0}+2a_{i1},a'_i=a'_{i0}+2a'_{i1}\in \mathbb {Z}_4$ for $i=0,\ldots ,m-1$. We define

$$\begin{aligned} c\odot c'&=a_0\odot a'_0+(a_1\odot a'_1)\xi +\cdots +(a_{m-1}\odot a'_{m-1}){\xi }^{m-1},\\ c\oplus c'&=a_0\oplus a'_0+(a_1\oplus a'_1)\xi +\cdots +(a_{m-1}\oplus a'_{m-1}){\xi }^{m-1},\\ c\circledast c'&=2(a_{00}a'_{00}+a_{10}a'_{10}\xi +\cdots +a_{m-10}a'_{m-10}\xi ^{m-1}). \end{aligned}$$

It is easy to see that $c\odot c'=c'\odot c$ and $c\odot c=c$. The operations “$\odot$”, “$\oplus$”and “$\circledast$”can be extended to $R^n$ in an obvious way; in other words, for $\varvec{c}=(u^{1}, u^{2},\ldots , u^{n}), \varvec{c'}=(v^{1}, v^{2},\ldots , v^{n})\in R^n$, we have $\varvec{c} \odot \varvec{c'}=(u^{1} \odot v^{1}, u^{2} \odot v^{2}, \ldots , u^{n}\odot v^{n})$, $\varvec{c} \oplus \varvec{c'}=(u^1\oplus v^1, u^2\oplus v^2,\ldots ,u^n\oplus v^n)$ and $\varvec{c}\circledast \varvec{c'}=(u^1\circledast v^1,\ldots ,u^n\circledast v^n)$. Furthermore, if U and V are two subsets of $R^n$, we define $U *V$ as (5).

Lemma 3.1

For all $\varvec{X}, \varvec{Y}\in R^n$, we have $\Phi (\varvec{X}+\varvec{Y})=\Phi (\varvec{X})\oplus \Phi (\varvec{Y})\oplus \Phi (\varvec{X}\circledast \varvec{Y})$.

Proof

It is enough to prove for the case $n=1$. Let X, Y be two arbitrary elements in R. There are $a_0,\ldots ,a_{m-1},a'_0,\ldots ,a'_{m-1}\in \mathbb {Z}_4$ such that $X=a_0+a_1\xi +\cdots +a_{m-1}{\xi }^{m-1}$ and $Y=a'_0+a'_1\xi +\cdots +a'_{m-1}{\xi }^{m-1}$, where $a_i=a_{i0}+2a_{i1}$ and $a'_i=a'_{i0}+2a'_{i1}$, for $i=0,\ldots ,m-1$. Note that, for every $u,v\in \mathbb {Z}_4$, $u+ v=u\oplus v\oplus (u\circledast v)$. So,

$$\begin{aligned} X+Y&=(a_0+a_1\xi +\cdots +a_{m-1}{\xi }^{m-1})+(a'_0+a'_1\xi +\cdots +a'_{m-1}{\xi }^{m-1})\\&=((a_{00}+2a_{01})+(a_{10}+2a_{11})\xi +\cdots +(a_{m-10}+2a_{m-11})\xi ^{m-1})\\&\quad +((a'_{00}+2a'_{01})+(a'_{10}+2a'_{11})\xi +\cdots +(a'_{m-10}+2a'_{m-11})\xi ^{m-1})\\&= ((a_{00}+2a_{01})+(a'_{00}+2a'_{01}))+((a_{10}+2a_{11})+(a'_{10}+2a'_{11}))\xi +\\&\quad \cdots +((a_{m-10}+2a_{m-11})+(a'_{m-10}+2a'_{m-11}))\xi ^{m-1}\\&=((a_{00}+a'_{00})+2(a_{01}+a'_{01}))+((a_{10}+a'_{10})+2(a_{11}+a'_{11}))\xi +\\&\quad \cdots +((a_{m-10}+a'_{m-10})+2(a_{m-11}+a'_{m-11}))\xi ^{m-1}\\&\quad +2(a_{00}a'_{00}+a_{10}a'_{10}\xi +\cdots +a_{m-10}a'_{m-10}\xi ^{m-1})\\&=X\oplus Y\oplus (X\circledast Y). \end{aligned}$$

Therefore, $[X+Y]=[X]\oplus [Y]\oplus [X\circledast Y]$, but then $[X+Y]\mathcal {A}=[X]\mathcal {A}\oplus [Y]\mathcal {A}\oplus [X\circledast Y]\mathcal {A}$. Consequently, $\Phi (X+Y)=\Phi (X)\oplus \Phi (Y)\oplus \Phi (X\circledast Y)$.$\square$

Note that for $\varvec{X}, \varvec{Y}$ belong to $R^n$, $\varvec{X}\circledast \varvec{Y}=2(\varvec{X}\odot \varvec{Y})$. Equivalently, we have

$$\begin{aligned} \Phi (\varvec{X}+\varvec{Y})=\Phi (\varvec{X})\oplus \Phi (\varvec{Y})\oplus \Phi (2(\varvec{X}\odot \varvec{Y})). \end{aligned}$$

(7)

Proposition 3.1

Let $\varvec{X}, \varvec{Y} \in R^n$. Then $\Phi (\varvec{X}+\varvec{Y}+2(\varvec{X}\odot \varvec{Y}))=\Phi (\varvec{X})\oplus \Phi (\varvec{Y})$.

Proof

The proof follows from (7).$\square$

Let $\mathcal {C}$ be a linear code of length n over R, in the following theorem, we state the necessary and sufficient condition for the linearity of $\Phi (\mathcal {C})$.

Theorem 3.1

Let $\mathcal {C}$ be a linear code of length n over R. Then, $\Phi (\mathcal {C})$ is a linear code if and only if $2(\varvec{X} \odot \varvec{Y})\in \mathcal {C}$, for all $\varvec{X},\varvec{Y}\in \mathcal {C}$.

Proof

By Proposition 3.1, we have $\Phi (\varvec{X}+\varvec{Y}+2(\varvec{X}\odot \varvec{Y}))=\Phi (\varvec{X})\oplus \Phi (\varvec{Y})$. Therefore, we deduce that $\Phi (\varvec{X})\oplus \Phi (\varvec{Y})\in \Phi (\mathcal {C})$ if and only if $\Phi (\varvec{X}+\varvec{Y}+2(\varvec{X}\odot \varvec{Y}))\in \Phi (\mathcal {C})$ for all $\varvec{X},\varvec{Y}\in \mathcal {C}$. So, the results follows.$\square$

Theorems 5 of [6] and $4.13 (s=2)$ of [15] can be concluded from the above theorem.

Corollary 3.1

Let $\mathcal {C}$ be a code of length n over R, and $\varvec{X}_1,\ldots ,\varvec{X}_r$ be a set of generators of $\mathcal {C}$. Then, $\Phi (\mathcal {C})$ is linear if and only if $2(\varvec{X}_{t}\odot \varvec{X}_{l})\in \mathcal {C}$ for all t, l satisfying $1\le t< l\le r$. In particular, if $\mathcal {G}$ is a generator matrix of a linear code $\mathcal {C}$ and $\{\varvec{v}_j\}_{j=1}^{k_0}$ are the row vectors of order four in $\mathcal {G}$. Thus, $\Phi (\mathcal {C})$ is linear if and only if $2(\varvec{v}_{t}\odot \varvec{v}_{l})\in \mathcal {C}$ for all t, l satisfying $1\le t< l\le k_0$.

Example 3.1

Let $R=GR(4,2)=\mathbb {Z}_4[\xi ]$, where $h(\xi )=0$ and $h(x)=x^2+x+1$ is a basic irreducible polynomial over $\mathbb {Z}_{4}$. Suppose that $\mathcal {C}$ be a linear code of length 7 over R with generator matrix

$$ \mathcal {G}=\left( \begin{array}{c} \varvec{v}_1\\ \varvec{v}_2\\ \varvec{v}_3\\ \varvec{v}_4\\ \varvec{u} \end{array}\right) =\left( \begin{array}{ccccccc} 1&{}0&{}0&{}0&{}\xi &{}0&{}3\\ 0&{}1&{}0&{}0&{}0&{}2&{}2\xi \\ 0&{}0&{}1&{}0&{}1&{}\xi &{}0\\ 0&{}0&{}0&{}1&{}\xi &{}0&{}1\\ 0&{}0&{}0&{}0&{}2&{}2\xi &{}2 \end{array}\right) . $$

Since $2(\varvec{v}_{1}\odot \varvec{v}_{4})\notin \mathcal {C}$, thus Theorem 3.1 implies that $\Phi (\mathcal {C})$ is nonlinear.

Example 3.2

Suppose that $R=GR(4,2)=\mathbb {Z}_4[\xi ]$, where $h(\xi )=0$ and $h(x)=x^2+x+1$ is a basic irreducible polynomial over $\mathbb {Z}_{4}$. Let $\mathcal {C}$ be a linear code of length 6 over R with generator matrix

$$ \mathcal {G}=\left( \begin{array}{c} \varvec{v}_1\\ \varvec{v}_2\\ \varvec{v}_3 \end{array}\right) =\left( \begin{array}{cccccc} 1&{}0&{}0&{}\xi &{}0&{}2\\ 0&{}1&{}0&{}0&{}1&{}2\xi \\ 0&{}0&{}1&{}1&{}\xi &{}1 \end{array}\right) . $$

It is to check that $2(\varvec{v}_{1}\odot \varvec{v}_{2})$, $2(\varvec{v}_{1}\odot \varvec{v}_{3})$ and $2(\varvec{v}_{2}\odot \varvec{v}_{3})$ belong to $\mathcal {C}$, so Theorem 3.1 implies that $\Phi (\mathcal {C})$ is linear.

4 Gray images of linear cyclic codes over R

Generators for cyclic $\mathbb {Z}_4$-codes have been studied by Kanwar and López in [9]. Also, all cyclic codes of odd length over $\mathbb {Z}_{4}$ whose Gray images are linear were determined by Wolfmann, see [19]. In this section, we generalize Wolfmann’s idea for these codes of odd length n over the Galois ring R. So, throughout this section, we let n be odd. Recall that a linear code $\mathcal {C}$ of length n is called cyclic, if $c=(c_0, c_1, \ldots , c_{n-1}) \in \mathcal {C}$ implies that $(c_{n-1}, c_0, \ldots , c_{n-2}) \in \mathcal {C}$. As is customary, $\mathcal {R}$ denote the quotient ring $\frac{R[x]}{\langle x^{n}-1\rangle }$ and the elements of $\mathcal {R}$ will be identified with polynomials over R of degree $\leqslant n-1$. For simplicity, we write a polynomial f(x) as f, where $f(x) \in \mathcal {R}$. We represent any n-tuple $(c_0,c_1, \ldots , c_{n-1}) \in R^n$ by the residue class of the polynomial $c_0+c_{1}x+ \dots +c_{n-1}x^{n-1}$ over R mod $x^n-1$. It is known that a cyclic code $\mathcal {C}$ corresponds to an ideal of $\mathcal {R}$. We start this section with a lemma where we discuss the structure of the polynomial generator of the cyclic code of odd length n over R. This lemma will play a crucial role in the linearity of $\Phi (\mathcal {C})$. Before we go on, we need to state the following well-known proposition.

Proposition 4.1

[16] Let $\mathcal {C}$ be a cyclic code of odd length n over the Galois ring R. Then there is a unique set of pairwaise coprime monic polynomials $g_{0},g_{1}, g_{2}$ over R (possibly, some of them are equal to 1) such that $g_{0}g_{2} g_{1}=x^{n}-1$ in R[x] and $\mathcal {C}=<\hat{g}_2, 2\hat{g}_1>$, where $\hat{g}_{i}=\prod _{i\ne j}g_{j}$.

Lemma 4.1

Under the conditions stated in Proposition 4.1, we have:

• (i) $\mathcal {C}=\langle g_0g_1,2g_0 \rangle =\langle g_0g_1+2g_0 \rangle$.

• (ii) $\mathcal {C} \cap 2\mathcal {R} = \langle 2g_0 \rangle$.

Proof

(i) By Proposition 4.1, $\mathcal {C}= \langle g_0g_1, 2g_0g_2 \rangle$ and $x^n-1=g_0g_2g_1$. Set $\mathcal {C}'=\langle g_0g_1, 2g_0 \rangle$. Clearly $\mathcal {C} \subseteq \mathcal {C}'$. Inasmuch as gcd$(g_1, g_2)=1$, thus there exist $s_1$ and $s_2 \in R[x]$ such that

$$ s_1g_1+s_2g_2=1. $$

Multiplying the above equation by $2g_0$, we obtain

$$ 2g_0=2s_1g_0g_1+2s_2g_0g_2 \in \mathcal {C}. $$

This means that $\mathcal {C}' \subseteq \mathcal {C}$; i.e., $\mathcal {C}' = \mathcal {C}$.

For the last equality, it is enough to show that $\mathcal {C} \subseteq \langle g \rangle$, where $g=g_0g_1+2g_0$. Inasmuch as gcd$(g_1, g_2)=1$, thus there exist $r_1$ and $r_2 \in R[x]$ such that

$$ r_1g_1+r_2g_2=1. $$

Multiplying the above equation by $2g_0$, we conclude

$$ 2g_0=2r_1g_0g_1+2r_2g_0g_2=g(2r_1)+g(r_2g_2) \in \langle g \rangle . $$

Furthermore, $g_0g_1= g-2g_0 \in \langle g \rangle$. Consequently, $\mathcal {C}=\langle g \rangle$.

(ii) It is easy to see that $(\supseteq )$ holds. Now suppose that $b \in \mathcal {C} \cap 2\mathcal {R}$. So, there exist $r \in \mathcal {R}$ such that

$$ b= rg_0g_1+2rg_0. $$

Therefore, there exists $r' \in \mathcal {R}$ such that $r=2r'$. Hence, we have

$$ b=2r'g_0g_1 \in \langle 2g_0 \rangle . $$

Consequently, $(\subseteq )$ also holds.$\square$

Remark 4.1

The elements of Galois ring R can be expressed in the form $a+2b$, where $a,b \in \tau$. The natural epimorphism $\bar{~}: R \longrightarrow \mathbb {F}_{2^m}$ is defined by $a+2b \longmapsto a$ can be extended to a ring epimorphism from R[x] to $\mathbb {F}_{2^m}[x]$ as following:

$$\begin{aligned} \bar{~}: R[x]&\longrightarrow \mathbb {F}_{2^m}[x]\\ a(x)+2b(x)&\longmapsto a(x). \end{aligned}$$

One can see that if $a(x) \in R[x]$, then $2a(x)=2 \bar{a}(x)$.

Lemma 4.2

Let $\mathcal {C}$ be a cyclic code of odd length n over the Galois ring R. Then $\Phi (\mathcal {C})$ is a linear code if and only if $\langle \bar{g}_0\bar{g}_1 \rangle *\langle \bar{g}_0\bar{g}_1 \rangle \subset \langle \bar{g}_0 \rangle$.

Proof

By Theorem 3.1$\Phi (\mathcal {C})$ is a linear code if and only if

$$ 2(\bar{\varvec{X}}*\bar{\varvec{Y}})=2(\varvec{X} *\varvec{Y})=2(\varvec{X} \odot \varvec{Y}) \in \mathcal {C}, $$

for all $\varvec{X},\varvec{Y}\in \mathcal {C}$. The translation in terms of polynomial representations gives that $\Phi (\mathcal {C})$ is a linear code if and only if $2(\bar{m}_1\bar{g} *\bar{m}_2\bar{g}) \in \mathcal {C}$, for all $m_1, m_2 \in \mathcal {R}$, where $g=g_0g_1+2g_0$ is the generator of $\mathcal {C}$. Since $\bar{g}= \bar{g}_0\bar{g}_1$, the condition from Lemma 4.1 becomes $2(\bar{m}_1\bar{g}_0\bar{g}_1 *\bar{m}_2\bar{g}_0\bar{g}_1) \in \mathcal {C} \cap 2\mathcal {R} = \langle 2g_0 \rangle$, for all $m_1, m_2 \in \mathcal {R}$. Therefore the above condition is equivalent to $\bar{m}_1\bar{g}_0\bar{g}_1 *\bar{m}_2\bar{g}_0\bar{g}_1 \in \langle \bar{g}_0 \rangle$, for all $m_1, m_2 \in \mathcal {R}$, which is the expected result.$\square$

Remark 4.2

Let n be odd, $\mathbb {F}_{2^m}$ be the finite field of order $2^m$, $\mathbb {F}_{2^{mt}}$ be the splitting field of $x^{n}-1$ over $\mathbb {F}_{2^m}$. Let $\beta$ be a primitive nth root in $\mathbb {F}_{{2^m}^t}$. We define the generalized Mattson-Solomon transform as the function $\mathcal {T} : \mathbb {F}_{{2^m}^t}[x]/\langle x^{n}-1 \rangle \longrightarrow \mathbb {F}_{{2^m}^t}[x]/\langle x^{n}-1 \rangle$ such that

$$ A(x)=\mathcal {T}({a}(x))=\sum _{k=0}^{n-1}{a}(\beta ^{-k})x^k. $$

It is easy to see that $\mathcal {T}$ is a bijection. The inverse transformation is given by

$$ a(x)=\mathcal {T}^{-1}(A(x))=\sum _{k=0}^{n-1}A(\beta ^{k})x^k. $$

Lemma 4.3

Let $a(x), b(x) \in \mathbb {F}_{{2^m}^t}[x]/\langle x^{n}-1 \rangle$, then $\mathcal {T}(a(x) b(x))=\mathcal {T}(a(x))*\mathcal {T}(b(x))$.

In the following, $g=g_0g_1+2g_0$ and $\bar{g}_2\otimes \bar{g}_2$ is the divisor of $x^n-1$ in $\mathbb {F}_{2^m}[x]$ whose roots are the $\beta ^i\beta ^j$ such that $\beta ^i$ and $\beta ^j$ are roots of $\bar{g}_2$.

Lemma 4.4

Let $\bar{d}$ and $\bar{g}_2$ be in $\mathbb {F}_{2^m}[x]$ such that $x^{n}-1=\bar{d}\bar{g}_2$, with n odd. Let $\bar{e}$ be such that $x^{n}-1=(\bar{g}_2\otimes \bar{g}_2)\bar{e}$. Then $\langle \bar{d} \rangle *\langle \bar{d} \rangle \subset \langle \bar{e} \rangle$.

Proof

Let $\bar{s}=\bar{m}_1\bar{d} *\bar{m}_2\bar{d}$ be in $\langle \bar{d} \rangle *\langle \bar{d} \rangle$. Using the inverse Mattson-Solomon transform we obtain $\sum _{k=0}^{n-1}\bar{s}(\beta ^{k})x^{k}=\sum _{i=0}^{n-1}\bar{m}_1(\beta ^{i})\bar{d}(\beta ^{i})x^{i}\sum _{j=0}^{n-1}\bar{m}_2(\beta ^{j})\bar{d}(\beta ^{j})x^{j}$. Since $\bar{d}(\beta ^t) \ne 0$ if and only if $\bar{g}_2(\beta ^t)= 0$ then the right side of the previous equality is $\sum _{\bar{g}_2(\beta ^i)=0}\bar{m}_1(\beta ^{i})\bar{d}(\beta ^{i})x^{i}\sum _{\bar{g}_2(\beta ^j)=0}\bar{m}_2(\beta ^{j})\bar{d}(\beta ^{j})x^{j}$. Consequently, $\sum _{k=0}^{n-1}\bar{s}(\beta ^{k})x^{k}= \sum _{(\bar{g}_2\otimes \bar{g}_2)(\beta ^k)=0}\sum _{i+j=k}\bar{m}_1(\beta ^{i})\bar{d}(\beta ^{i})\bar{m}_2(\beta ^{j})\bar{d}(\beta ^{j})x^{k}$. It follows that if $\bar{s}(\beta ^{k}) \ne 0$ then $(\bar{g}_2\otimes \bar{g}_2)(\beta ^k)=0$ or, equivalently, if $\bar{e}(\beta ^k)=0$ then $\bar{s}(\beta ^{k})= 0$. This means that $\bar{e}$ divides $\bar{s}$ and, therefore, $\bar{s}$ belongs to $\langle \bar{e} \rangle$.

Theorem 4.1

Let $\mathcal {C}=\langle g \rangle$ be a cyclic code of odd length n over the Galois ring R, where $g=g_0g_1+2g_0$ with $x^{n}-1=g_{0}g_{2} g_{1}$. Let $\bar{e}$ be such that

$$\begin{aligned} x^n-1=(\bar{g}_2\otimes \bar{g}_2)\bar{e} \end{aligned}$$

in $\mathbb {F}_{2^m}[x]$. The following statements are equivalent.

• (i) $\Phi (\mathcal {C})$ is linear.

• (ii) $\bar{g}_0$ divides $\bar{e}$ in $\mathbb {F}_{2^m}[x]$.

Proof

$(i)\Rightarrow (ii)$ If $\bar{g}_0=1$ or $\bar{g}_2=1$, then obviously $\bar{g}_0$ divides $\bar{e}$. From now on $\bar{g}_0 \ne 1$ and $\bar{g}_2 \ne 1$. Define

$$\begin{aligned} K=\{k:\ (\bar{g}_2\otimes \bar{g}_2)(\beta ^{k})=0, \ 0 \le k \le n-1\ \} \end{aligned}$$

and

$$\begin{aligned} J(k)=\{j:\ \bar{g}_{2}(\beta ^{j})= \bar{g}_{2}(\beta ^{k-j})=0, \ 0 \le j \le n-1\ \}. \end{aligned}$$

Since $\bar{g}_2 \ne 1$ then $K \ne \emptyset$ and if $k \in K$ then $J(k) \ne \emptyset$. Following Lemma 4.2, for every integer m, $0 \le m \le n-1$, there exists $\bar{\lambda }_m$ such that $\bar{g}_0\bar{g}_1 *x^{m} \bar{g}_0\bar{g}_1= \bar{\lambda }_m \bar{g}_0$. Using the Mattson-Solomon transform with $\bar{d}= \bar{g}_0\bar{g}_1$ we find $\sum _{k \in K}(\sum _{j \in J(k)}A_{j}(k)(\beta ^m)^j)x^k=\sum _{k \in K}\bar{\lambda }_m(\beta ^k)\bar{g}_0(\beta ^k)x^k$, with $A_{j}(k)=\bar{d}(\beta ^j)\bar{d}(\beta ^{k-j})$. Let k be in K and let us consider $S_{k}(X)=\sum _{j \in J(k)}A_{j}(k)X^{j}$. From the definitions of K and J(k), we have $A_j(k) \ne 0$ for every $k \in K$ and every $j \in J(k)$ and, therefore, $S_k(x)$ is not zero. The degree of $S_k(x)$ is at most $n-1$ which also is the maximum number of roots of $S_k(x)$. Consequently, there exists m, $0 \le m \le n-1$ such that $S_k(\beta ^m) \ne 0$. It follows that $\bar{\lambda }_m(\beta ^k)\bar{g}_0(\beta ^k) \ne 0$ and, therefore, $\bar{g}_0(\beta ^k) \ne 0$. We deduce that $(\bar{g}_2\otimes \bar{g}_2)(\beta ^k) = 0$ implies $\bar{g}_0(\beta ^k) \ne 0$ or, equivalently, $\bar{g}_0(\beta ^k)=0$ implies $(\bar{g}_2\otimes \bar{g}_2)(\beta ^k) \ne 0$. In other word, if $\bar{g}_0(\beta ^k)=0$ then $\bar{e}(\beta ^k)=0$ which means that $\bar{g}_0$ divides $\bar{e}$.

$(ii)\Rightarrow (i)$ If $\bar{g}_0$ divides $\bar{e}$ then $\langle \bar{e} \rangle \subset \langle \bar{g}_0 \rangle$. Applying Lemma 4.4 with $\bar{d}=\bar{g}_0\bar{g}_1$ we show that the condition of Lemma 4.2 is satisfied and, therefore, $\Phi (\mathcal {C})$ ia a linear code.$\square$

Corollary 4.1

Let $\mathcal {C}=\langle g \rangle$ be a cyclic code of odd length n over the Galois ring R, where $x^n-1=g_0g_2g_1$.

• (i) If $g_0=1$, then $\Phi (\mathcal {C})$ is linear.

• (ii) If $g_2=1$, then $\Phi (\mathcal {C})$ is linear.

• (iii) If $g_2=x-1$, then $\Phi (\mathcal {C})$ is linear.

Example 4.1

Let $R=GR(4,2) = \mathbb {Z}_4[\xi ]$, where $h(\xi )=0$ and $h=x^2+x+1$ is a basic irreducible polynomial over $\mathbb {Z}_4$. In R[x], $x^{3}-1=g_{0}g_{2} g_{1}$, where $g_0=x-1$, $g_1=x-\xi$ and $g_2=x-\xi ^2$. Let $\mathcal {C}$ be a cyclic code of length 3 over R generated by polynomial $g=g_0g_1+2g_0$. Obviously, $\bar{g}_0$ divides $\bar{e}= (x-1)(x-\bar{\xi }^2)$ in $\mathbb {F}_{4}[x]$. Hence, Theorem 4.1 implies that $\Phi (\mathcal {C})$ is linear.

Example 4.2

Let $R=GR(4,2) = \mathbb {Z}_4[\xi ]$, where $h(\xi )=0$ and $h(x)=x^2+x+1$ is a basic irreducible polynomial over $\mathbb {Z}_4$. In R[x], $x^{7}-1=g_{0}g_{2} g_{1}$, where $g_0 = x^3 - x^2 + 2x - 1$, $g_1 = x^3 + 2x^2 + x - 1$ and $g_2 = x - 1$. Let $\mathcal {C}$ be a cyclic code of length 7 over R generated by polynomial $g=g_0g_1+2g_0$. It is easy to see that $\bar{g}_0$ divides $\bar{e}=(x^3-x^2-1)(x^3+x-1)$ in $\mathbb {F}_{4}[x]$. Therefore, Theorem 4.1 implies that $\Phi (\mathcal {C})$ is linear.

5 Conclusion

For a linear code $\mathcal {C}$ of arbitrary length n over R, we showed that Gray image of $\mathcal {C}$ is linear if and only if for every $\varvec{X},\varvec{Y}\in \mathcal {C}$, $2(\varvec{X} \odot \varvec{Y})\in \mathcal {C}$. In addition, we gave a necessary and sufficient condition whose Gray image is linear for a cyclic code of odd length over R by using the generator polynomial of this code. As a future research, it would be interesting to check the linearity of the Gray image of a linear code and a cyclic code over $GR(2^s,m)$ (for $s > 2$) using their generator matrix and generator polynomial, respectively. As mentioned in [19], an open problem is check the linearity of the Gray image of a cyclic code over $\mathbb {Z}_4$ of even length by using its generator polynomials.This problem can be studied for the Galois ring GR(4, m) too.