The Gray images of $(1+u)$ constacyclic codes over $F_{2^m}[u]/\langle u^{k} \rangle$

Abstract

Let $R_{k}$ denote the polynomial residue ring $F_{2^m}[u]/\langle u^{k} \rangle$, where $2^{j-1}+1\le k\le 2^{j}$ for some positive integer $j$. Motivated by the work in [1], we introduce a new Gray map from $R_{k}$ to $F_{2^m}^{2^{j}}$. It is proved that the Gray image of a linear $(1+u)$ constacyclic code of an arbitrary length $N$ over $R_{k}$ is a distance invariant linear cyclic code of length $2^{j}N$ over $F_{2^m}$. Moreover, the generator polynomial of the Gray image of such a constacyclic code is determined, and some optimal linear cyclic codes over $F_{2}$ and $F_{4}$ are constructed under this Gray map.

1 Introduction

Linear codes over finite rings have been studied since the early 1970s in [3, 11], that study has regained attention by the works of Nechaev in [8] and Hammons et al in [4] on some efficient nonlinear binary codes. These works deal with nonlinear binary codes such as the Nordstrom-Robinson, Kerdock, Preparata, Goethals and Delsarte–Goethals codes which are considered as binary images under the Gray map of linear codes over the ring $Z_{4}$. Since then, many researchers have paid more and more attentions to study the codes over finite rings. In [15], Wolfmann showed that the Gray image of a linear single-root negacyclic code over $Z_{4}$ is a distance invariant binary cyclic code (not necessarily linear). This result was later generalized to a $(1+2^{k})$ constacyclic code over $Z_{2^{k+1}}$ by Tapia-Recillas and Vega in [13], where the corresponding Gray image was a binary distance invariant quasi-cyclic code. In [7], Ling and Blackford generalized most of the results of [13, 15] to the ring $Z_{p^{k+1}}$, where $p$ is any prime and $k$ is a positive integer.

$(1+u)$ constacyclic codes over $F_{2}+uF_{2}$ were first introduced by Qian et al in [9], where it was proved that the Gray image of a linear single-root $(1+u)$ constacyclic code over $F_{2}+uF_{2}$ is a binary distance invariant linear cyclic code. In [1], Abular and Siap extended the Gray map of [9] to an arbitrary length over $F_{2}+uF_{2}$ and the generator polynomial of the correspoding Gray image was obtained, some optimal binary codes were also constructed via the Gray map. Amarra et al in [2] discussed the Gray image of a single-root $(1-u)$ constacyclic code over $F_{p^k}+uF_{p^k}$, which was a quasi-cyclic code over $F_{p}$. Later, the result of [2] were extended to single-root $(1+u^{t})$ constacyclic codes over $F_{q}[u]/\langle u^{t+1}\rangle$ and single-root $(1-u^{m})$ constacyclic codes over $F_{p^k}[u]/\langle u^{m+1}\rangle$ in [12] and [14] respectively, where the Gray images were quasi-cyclic codes. Kai et al in [5] showed that the Gray image of a linear $(1+\lambda u)$ constacyclic code with an arbitrary length over $F_{p}+uF_{p}$ was a distance invariant linear code over $F_{p}$. The Gray image of a single-root $(1+u+u^{2})$ constacyclic code over the ring $F_{2}[u]/\langle u^{3}\rangle$ were proved to be a binary distance invariant linear cyclic code in [10], but the generator polynomials of the corresponding Gray images in [10] and [5] were not acquired. In this paper, we extend the result of [1] about the Gray map to the polynomial residue ring $R_{k}$, where the generator polynomials of the corresponding Gray images are obtained and some optimal linear cyclic code over $F_{2}$ and $F_{4}$ are constructed via the Gray map.

2 Preliminaries

Let $R_{k}$ denote the ring $F_{2^m}[u]/\langle u^{k} \rangle$, where $2^{j-1}+1\le k\le 2^{j}$ for some positive integer $j$ and $u^{k}=0$. If $x^{n}-1=f_{1}f_{2}\cdots f_{q}$ is the factorization of $(x^{n}-1)$ into a product of monic basic irreducible pairwise coprime polynomials over $R_{k}$ for an odd positive integer $n$, then this factorization is unique and can be directly carried over $R_{k}$ from over $F_{2^m}$. Let $C$ be a code of length $N=2^{e}n$ over $R_{k}$, where $e$ is a non-negative integer. For some fixed unit $\alpha$ of $R_{k}$, the $\alpha$ constacyclic shift $\nu _{\alpha }$ on $R_{k}^{N}$ is the shift $\nu _{\alpha }(c_{0},c_{1},\cdots ,c_{N-1})=(\alpha c_{N-1},c_{0},c_{1},\cdots ,c_{N-2})$. The code C is said to be an $\alpha$ constacyclic code if $\nu (C)=C$. Now, we identify a codeword $c=(c_{0},c_{1},\cdots ,c_{N-1})$ with its polynomial representation $c(x)=c_{0}+c_{1}x+\cdots +c_{N-1}x^{N-1}$, then $xc(x)$ corresponds to an $\alpha$ constacyclic shift of $c(x)$ in the ring $R_{k}[x]/\langle x^{N}-\alpha \rangle$. Thus $\alpha$ constacyclic codes of length $N$ over $R_{k}$ can be identified as ideals in the ring $R_{k}[x]/\langle x^{N}-\alpha \rangle$. In the following, we let $j$ be a positive integer and $2^{j-1}+1\le k\le 2^{j}$.

3 A class of matrices over $F_{2^{m}}$

Definition 3.1

If $j=1$, then $A_{2}=\left( {\begin{array}{ll} {1} &{} {0} \\ {1} &{} {1} \\ \end{array}} \right)$. If $j>1$, then $A_{2^{j}}=\left( {\begin{array}{cc} {A_{2^{j-1}}} &{} {0} \\ {A_{2^{j-1}}} &{} {A_{2^{j-1}}} \\ \end{array}} \right)$.

From the definition of matrix $A_{2^{j}}$, we see that there are $2^{j}$ rows and $2^{j}$ columns in $A_{2^{j}}$. Besides, the first row of $A_{2^{j}}$ is $(1,\underbrace{0,\cdots ,0}_{(2^{j}-1)~zeros})$, the first and $2^{j}$th columns of $A_{2^{j}}$ are $A_{2^{j}}(1)$ and $A_{2^{j}}(2^{j})$ respectively, where $A_{2^{j}}(1)=(\underbrace{1,1,\cdots ,1}_{2^{j}~ones})^ {T}$ and $A_{2^{j}}(2^{j})=(\underbrace{0,\cdots ,0}_{(2^{j}-1)~ones},1)^ {T}$.

Lemma 3.1

$A_{2^{j}}$ is an invertible matrix over $F_{2^m}$.

Proof

From the definition of matrix $A_{2^{j}}$, we see that $A_{2^{j}}$ is a lower triangular matrix and each element of its main diagonal is one. So $A_{2^{j}}$ is invertible over $F_{2^m}$. $\square$

Lemma 3.2

Let ${\small B_{2^{j}}=\left( {\begin{array}{cccc} {0} &{} {0} &{} {\cdots } &{} {0} \\ {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots }\\ {0} &{} {0} &{} {\cdots } &{} {0} \\ {1} &{} {0} &{} {\cdots } &{} {0} \\ \end{array}} \right) }$ with $2^{j}$ rows and $2^{j}$ columns, then $B_{2^{j}}A_{2^{j}}=(A_{2^{j}}(2^{j}),\underbrace{0,\cdots ,0}_{(2^{j}-1)~zero~vectors})$ in $F_{2^{m}}$.

Proof

Since the first row of $A_{2^{j}}$ is $(1,\underbrace{0,\cdots ,0}_{(2^{j}-1)~zeros})$, then $B_{2^{j}}A_{2^{j}}=B_{2^{j}}=(A_{2^{j}}(2^{j}),\underbrace{0,\cdots ,0}_{(2^{j}-1)~zero~vectors})$ in $F_{2^{m}}$. $\square$

Theorem 3.3

Let $H_{2^{j}}=\left( {\begin{array}{cccccc} {1} &{} {1} &{} {0} &{} {\cdots } &{} {0} &{} {0} \\ {0} &{} {1} &{} {1} &{} {\cdots } &{} {0} &{} {0} \\ {0} &{} {0} &{} {1} &{} {\cdots } &{} {0} &{} {0} \\ {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots }\\ {0} &{} {0} &{} {0} &{} {\cdots } &{} {1} &{} {1} \\ {0} &{} {0} &{} {0} &{} {\cdots } &{} {0} &{} {1} \\ \end{array}} \right)$ and $D_{2{j}}=\left( {\begin{array}{ccccc} {0} &{} {1} &{} {0}&{} {\cdots } &{} {0} \\ {0} &{} {0} &{} {1}&{} {\cdots } &{} {0}\\ {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots } &{} {\vdots }\\ {0} &{} {0} &{} {0} &{} {\cdots } &{} {1}\\ {1} &{} {0} &{} {0}&{} {\cdots } &{} {0}\\ \end{array}} \right)$, where $H_{2^{j}}$ and $D_{2^{j}}$ are both square matrices with $2^{j}$ rows and $2^{j}$ columns, then $H_{2^{j}}A_{2^{j}}=A_{2^{j}}D_{2^{j}}$ in $F_{2^{m}}$.

Proof

We prove the result by induction on $j$. In $F_{2^{m}}$, if $j=1$, then

$$\begin{aligned} H_{2}A_{2}=\left( {\begin{array}{ll} {1} &{} {1} \\ {0} &{} {1} \\ \end{array}} \right) \left( {\begin{array}{ll} {1} &{} {0} \\ {1} &{} {1} \\ \end{array}} \right) =\left( {\begin{array}{ll} {0} &{} {1} \\ {1} &{} {1} \\ \end{array}} \right) =\left( {\begin{array}{ll} {1} &{} {0} \\ {1} &{} {1} \\ \end{array}} \right) \left( {\begin{array}{ll} {0} &{} {1} \\ {1} &{} {0} \\ \end{array}} \right) =A_{2}D_{2}. \end{aligned}$$

Suppose $H_{2^{j_{1}}}A_{2^{j_{1}}}=A_{2^{j_{1}}}D_{2^{j_{1}}}$ for some positive integer $j_{1}$, then

$$\begin{aligned} H_{2^{j_{1}}}A_{2^{j_{1}}}=(A_{2^{j_{1}}}(2^{j_{1}}), A_{2^{j_{1}}}(1),\cdots ,A_{2^{j_{1}}}(2^{j_{1}}-1)). \end{aligned}$$

From lemma 3.2, we can get

$$\begin{aligned}&H_{2^{j_{1}+1}}A_{2^{j_{1}+1}} = \left( {\begin{array}{cc} {H_{2^{j_{1}}}} &{} {B_{2^{j_{1}}}} \\ {0} &{} {H_{2^{j_{1}}}} \\ \end{array}} \right) \left( {\begin{array}{cc} {A_{2^{j_{1}}}} &{} {0} \\ {A_{2^{j_{1}}}} &{} {A_{2^{j_{1}}}} \\ \end{array}} \right) =\left( {\begin{array}{cc} {H_{2^{j_{1}}}A_{2^{j_{1}}}+B_{2^{j_{1}}}A_{2^{j_{1}}}} &{} {B_{2^{j_{1}}}A_{2^{j_{1}}}} \\ {H_{2^{j_{1}}}A_{2^{j_{1}}}} &{} {H_{2^{j_{1}}}A_{2^{j_{1}}}} \\ \end{array}} \right) \\&\quad \!=\! \left( {\begin{array}{cc} {(0,A_{2^{j_{1}}}(1),\cdots ,A_{2^{j_{1}}}(2^{j_{1}}-1))} &{} {(A_{2^{j_{1}}}(2^{j_{1}}),\underbrace{0,\cdots ,0}_{(2^{j_{1}}-1)~ zero~vectors})} \\ {(A_{2^{j_{1}}}(2^{j_{1}}),A_{2^{j_{1}}}(1),\cdots ,A_{2^{j_{1}}}(2^{j_{1}}-1))} &{} {(A_{2^{j_{1}}}(2^{j_{1}}),A_{2^{j_{1}}}(1),\cdots ,A_{2^{j_{1}}}(2^{j_{1}}-1))} \\ \end{array}} \right) . \end{aligned}$$

so $H_{2^{j_{1}+1}}A_{2^{j_{1}+1}}=A_{2^{j_{1}+1}}D_{2^{j_{1}+1}}$, which gives the proof. $\square$

4 A new Gray map and the structure of the corresponding Gray image

Let $a,b$ be two elements in $R_{k}$, then $a,b$ can be written as $a=\sum \limits _{i = 0}^{k-1}u^{i}r_{i}(a)$ and $b=\sum \limits _{i = 0}^{k-1}u^{i}r_{i}(b)$ respectively, where $r_{i}(a),r_{i}(b)\in F_{2^{m}}$ for $0\le i\le k-1$. It is easy to check that $r_{i}(a+b)=r_{i}(a)+r_{i}(b)$ for $0\le i\le k-1$.

Definition 4.1

For an arbitrary element $a$ in $R_{k}$, a new Gray map $\Phi _{k}$ from $R_{k}$ to $F_{2^{m}}^{2^{j}}$ is defined as follows:

$$\begin{aligned} \Phi _{k}(a)=(\underbrace{0,\cdots ,0}_{(2^{j}-k)~zeros},r_{0}(a),r_{1}(a),\cdots ,r_{k-1}(a))A_{2^{j}}. \end{aligned}$$

From the definition of $\Phi _{k}$, we see that this Gray map is linear that because

$$\begin{aligned} \Phi _{k}(a+b)&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(a+b),r_{1}(a+b),\cdots ,r_{k-1}(a+b))A_{2^{j}}\\&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(a),r_{1}(a),\cdots ,r_{k-1}(a))A_{2^{j}}\\&+\,(\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(b),r_{1}(b),\cdots ,r_{k-1}(b))A_{2^{j}}\\&= \Phi _{k}(a)+\Phi _{k}(b). \end{aligned}$$

According to lemma 3.1, $A_{2^{j}}$ is an invertible matrix over $F_{2^{m}}$, so $\Phi _{k}$ is a bijection from $R_{k}$ to $F_{2^{m}}^{2^{j}}$. we identify a codeword $c=(c_{0},c_{1},\cdots ,c_{N-1})\in R_{k}^{N}$ with its polynomial representation $c(x)=c_{0}+c_{1}x+\cdots +c_{N-1}x^{N-1}$ and denote $P_{i}[c(x)]=\sum \limits _{l = 0}^{N-1}r_{i}(c_{l})x^{l}$ for $0\le i\le k-1$, then $c(x)=\sum \limits _{i = 0}^{k-1}u^{i}P_{i}[c(x)]$. Thus, the Gray map $\Phi _{k}$ can be extended to $R_{k}[x]$ in an obvious way.

Definition 4.2

For an arbitrary codeword $c=(c_{0},c_{1},\cdots ,c_{N-1})\in R_{k}^{N}$, its polynomial representation is $c(x)=c_{0}+c_{1}x+\cdots +c_{N-1}x^{N-1}$. The polynomial Gray map $\Phi _{k}$ from $R_{k}[x]$ to $F_{2^{m}}[x]$ is defined as follows:

$$\begin{aligned} \Phi _{k}[c(x)]=(\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},P_{0}[c(x)],P_{1}[c(x)],\cdots ,P_{k-1}[c(x)])A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) . \end{aligned}$$

Obviously, $\Phi _{k}$ is not only linear, but also a bijection from $R_{k}[x]$ to $F_{2^{m}}[x]$.

Definition 4.3

Let $W_{L}$ be the Lee weight of the element of $R_{k}$ and $W_{H}$ be the Hamming weight of the element of $F_{2^{m}}^{2^{j}}$. We define that $W_{L}(a)=W_{H}[\Phi _{k}(a)]$ for an arbitrary element $a$ in $R_{k}$. The Lee weight of a codeword in $R_{k}[x]$ is the rational integer sum of the Lee weight of its coefficients. The Lee distance between two codewords $c$ and $\acute{c}$ is defined as the Lee weight of $(c-\acute{c})$.

The following lemma 4.1 is straightforward from the definitions of the polynomial Gray map and Lee distance.

Lemma 4.1

The polynomial Gray map $\Phi _{k}$ is not only a linear bijection from $R_{k}[x]$ to $F_{2^{m}}[x]$, but also a distance-preserving map from ($R_{k}[x]$, Lee distance) to ($F_{2^{m}}[x]$, Hamming distance).

Theorem 4.2

If $C$ is a $(1+u)$ constacyclic code of length $N$ over $R_{k}$, then $\Phi _{k}(C)$ is a linear cyclic code of length $2^{j}N$ over $F_{2^m}$.

Proof

It only needs to prove $\Phi _{k}[xc(x)]=x\Phi _{k}[c(x)]$. In fact, $x^{N}=1+u$ in $R_{k}[x]/\langle x^{N}-(1+u)\rangle$, so $x^{2^{j}N}=1$. For an arbitrary codeword $c(x)=c_{0}+c_{1}x+\cdots +c_{N-1}x^{N-1}\in C$, it can be written in the form $c(x)=\sum \limits _{i = 0}^{k-1}u^{i}P_{i}[c(x)]$. Then, we have $xc(x)=(1+u)c_{N-1}+c_{0}x+c_{1}x^{2}+\cdots +c_{N-2}x^{N-1}=\sum \limits _{i = 0}^{k-1}u^{i}P_{i}[xc(x)]$, where $P_{0}[xc(x)]=r_{0}[(c_{N-1})]+xP_{0}[c(x)]+x^{N}r_{0}(c_{N-1})$ and $P_{i}[xc(x)]=r_{i}[(1+u)c_{N-1}]+\sum \limits _{l = 0}^{k-2}r_{i}(c_{l})x^{l+1}=[r_{i-1}(c_{N-1})+r_{i}(c_{N-1})]+xP_{i}[c(x)]+x^{N}r_{i}(c_{N-1})$ for $1\le i\le k-1$. Therefore

$$\begin{aligned} \Phi _{k}[xc(x)]&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},P_{0}[xc(x)],P_{1}[xc(x)],\cdots ,P_{k-1}[xc(x)])A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) \\&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(c_{N-1}),\sum \limits _{i = 0}^{1}r_{i}(c_{N-1}),\cdots ,\sum \limits _{i = k-2}^{k-1}r_{i}(c_{N-1}))A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) \\&+x^{N}(\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(c_{N-1}),r_{1}(c_{N-1}),\cdots ,r_{k-1}(c_{N-1}))A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) \\&+x(\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},P_{0}[c(x)],P_{1}[c(x)],\cdots ,P_{k-1}[c(x)])A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) \\&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},r_{0}(c_{N-1}),r_{1}(c_{N\!-\!1}),\cdots ,r_{k\!-\!1}(c_{N-1}))(H_{2^{j}}A_{2^{j}}\!+\!A_{2^{j}}D_{2^{j}})\\&\times \left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) +x\Phi _{k}[c(x)]. \end{aligned}$$

By theorem 3.3, $\Phi _{k}[xc(x)]=x\Phi _{k}[c(x)]$. $\square$

5 The generator polynomial of the Gray image

Lemma 5.1

In $F_{2^m}[x]$, $A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) =\left( {\begin{array}{c} {1} \\ {1+x^{N}}\\ {\vdots } \\ {(1+x^{N})^{2^{j}-1}} \\ \end{array}} \right)$.

Proof

We prove the result by induction on $j$. In $F_{2^m}[x]$, if $j=1$, then $A_{2}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ \end{array}} \right) =\left( {\begin{array}{ll} {1} &{} {0} \\ {1} &{} {1}\\ \end{array}} \right) \left( {\begin{array}{c} {1} \\ {x^{N}} \\ \end{array}} \right) =\left( {\begin{array}{c} {1} \\ {1+x^{N}} \\ \end{array}} \right)$.

Suppose $A_{2^{j_{1}}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j_{1}}-1)N}} \\ \end{array}} \right) =\left( {\begin{array}{c} {1} \\ {1+x^{N}}\\ {\vdots } \\ {(1+x^{N})^{2^{j_{1}}-1}} \\ \end{array}} \right)$ for some positive integer $j_{1}$ in $F_{2^m}[x]$, then

$$\begin{aligned} A_{2^{j_{1}+1}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j_{1}+1}-1)N}} \\ \end{array}} \right)&= \left( {\begin{array}{ll} {A_{2^{j_{1}}}} &{} {0} \\ {A_{2^{j_{1}}}} &{} {A_{2^{j_{1}}}} \\ \end{array}} \right) \left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j_{1}+1}-1)N}} \\ \end{array}} \right) \\ \!&= \!\left( \!{\begin{array}{c} {A_{2^{j_{1}}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j_{1}}-1)N}} \\ \end{array}} \right) } \\ {(1+x^{2^{j_{1}}N})A_{2^{j_{1}}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j_{1}}-1)N}} \\ \end{array}} \right) } \\ \end{array}} \!\right) \!=\!\left( \!{\begin{array}{c} {1} \\ {1+x^{N}}\\ {\vdots } \\ {(1+x^{N})^{2^{j_{1}+1}-1}}\\ \end{array}} \!\right) . \end{aligned}$$

This gives the proof. $\square$

Let $n$ be an odd positive integer and let $x^{n}-1=f_{1}f_{2}\cdots f_{q}$ be the factorization of $(x^{n}-1)$ into a product of monic basic irreducible pairwise coprime polynomials in $F_{2^{m}}[x]$, then the following lemma is straightforward from the theorem 4 and lemma 3 of [7].

Lemma 5.2

Let $C$ be a $(1+u)$ constacyclic code of length $N=2^{e}n$ over $R_{k}$, then $C=\langle f_{1}^{k_{1}}f_{2}^{k_{2}}\cdots f_{q}^{k_{q}}\rangle$, where $0\le k_{i}\le 2^{e}k$ for $i=0,1,\cdots ,q$. Furthermore, $|C|=2^{m(kN-\eta )}$, where $\eta =\sum \limits _{i = 1}^{q}{k_{i}deg(f_{i})}$.

Theorem 5.3

Let $C=\langle f_{1}^{k_{1}}f_{2}^{k_{2}}\cdots f_{q}^{k_{q}}\rangle$ be a $(1+u)$ constacyclic code of length $N=2^{e}n$ over $R_{k}$, where $e$ is a non-negative integer and $0\le k_{i}\le 2^{e}k$ for $i=0,1,\cdots ,q$. Then the Gray image $\Phi _{k}(C)$ is a linear cyclic code of length $2^{j}N$ over $F_{2^{m}}$ and $\Phi _{k}(C)=\langle f_{1}^{k_{1}+2^{e}(2^{j}-k)}f_{2}^{k_{2}+2^{e}(2^{j}-k)}\cdots f_{q}^{k_{q}+2^{e}(2^{j}-k)}\rangle$.

Proof

By theorem 4.2, $\Phi _{k}(C)$ is a linear cyclic code of length $2^{j}N$ over $F_{2^{m}}$. So, we only need to prove $\Phi _{k}(C)=\langle f_{1}^{k_{1}+2^{e}(2^{j}-k)}f_{2}^{k_{2}+2^{e}(2^{j}-k)}\cdots f_{q}^{k_{q}+2^{e}(2^{j}-k)}\rangle$. In fact, we denote $c(x)=f_{1}^{k_{1}}f_{2}^{k_{2}}\cdots f_{q}^{k_{q}}\in C$, then $c(x)=\sum \limits _{i = 0}^{k-1}u^{i}P_{i}[c(x)]=\sum \limits _{i = 0}^{k-1}(1+x^{N})^{i}P_{i}[c(x)]$. By lemma 5.1,

$$\begin{aligned} \Phi _{k}[c(x)]&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},P_{0}[c(x)],P_{1}[c(x)],\cdots ,P_{k-1}[c(x)])A_{2^{j}}\left( {\begin{array}{c} {1} \\ {x^{N}} \\ {\vdots } \\ {x^{(2^{j}-1)N}} \\ \end{array}} \right) \\&= (\underbrace{0,\cdots ,0}_{(2^{j}-k)~ zeros},P_{0}[c(x)],P_{1}[c(x)],\cdots ,P_{k-1}[c(x)])\left( {\begin{array}{c} {1} \\ {1+x^{N}} \\ {\vdots } \\ {(1+x^N)^{2^{j}-1}} \\ \end{array}} \right) \\&= (1+x^{N})^{2^{j}-k}\sum \limits _{i = 0}^{k-1}(1+x^{N})^{i}P_{i}[c(x)]=(1+x^{N})^{2^{j}-k}c(x)\in \Phi _{k}(C). \end{aligned}$$

So $\langle (1+x^{N})^{2^{j}-k}f_{1}^{k_{1}}f_{2}^{k_{2}}\cdots f_{q}^{k_{q}}\rangle \subseteq \Phi _{k}(C)$. Comparing the number of codewords, we have $\Phi _{k}(C)=\langle (1+x^{N})^{2^{j}-k}f_{1}^{k_{1}}f_{2}^{k_{2}}\cdots f_{q}^{k_{q}}\rangle$. Since $1+x^{N}=(f_{1}f_{2}\cdots f_{q})^{2^{e}}$ in $F_{2^{m}}[x]$, then we get the result. $\square$

6 Examples

Let $\omega$ be a primitive element of $F_{4}$. In $F_{2}[x]$, $x^{3}-1=Q_{1}Q_{2}$, where $Q_{1}=x+1$ and $Q_{2}=x^{2}+x+1$. In $F_{4}[x]$, $x^{3}-1=g_{1}g_{2}g_{3}$, where $g_{1}=x+1$, $g_{2}=x+\omega$ and $g_{3}=x+\omega ^{2}$. In $F_{2}[x]$ and $F_{4}[x]$, $x^{7}-1=f_{1}f_{2}f_{3}$, where $f_{1}=x+1$, $f_{2}=x^{3}+x+1$ and $f_{3}=x^{3}+x^{2}+1$.

Example 1

Let $k=2$ and $m=1$ in definition 4.2 and theorem 5.3, we get the result of Abularub et al about the Gray map in [1]. Now, we let $k=2$ and $m=2$, we can get some other optimal codes. For example, $C_{1}=\langle g_{1}g_{2}g_{3}^{3}\rangle$ is a $(1+u)$ constacyclic code of length $N=6$ over $F_{4}+uF_{4}$. According to theorem 5.3, $\Phi _{2}(C_{1})=\langle g_{1}g_{2}g_{3}^{3}\rangle$, which is a [4, 7, 12] linear cyclic code over $F_{4}$ and an optimal code. Table 1 presents several optimal linear codes over $F_{4}$ obtained from $(1+u)$ constacyclic codes over $F_{4}+uF_{4}$ of some lengths.

Table 1 Optimal linear codes over $F_{4}$ obtained from $(1+u)$ constacyclic codes over $F_{4}+uF_{4}$

Tables 2, 3, 4, 5, 6 and 7 present some optimal codes obtained from $(1+u)$ constacyclic codes of some lengths over $R_{k}$ under the Gray map.

Table 2 Optimal binary linear codes obtained from $(1+u)$ constacyclic codes over $F_{2}[u]/\langle u^{3}\rangle$

Table 3 Optimal linear codes over $F_{4}$ obtained from $(1+u)$ constacyclic codes over $F_{4}[u]/\langle u^{3}\rangle$

Table 4 Optimal binary linear codes obtained from $(1+u)$ constacyclic codes over $F_{2}[u]/\langle u^{4}\rangle$

Table 5 Optimal linear codes over $F_{4}$ obtained from $(1+u)$ constacyclic codes over $F_{4}[u]/\langle u^{4}\rangle$

Table 6 Optimal binary linear codes obtained from $(1+u)$ constacyclic codes over $F_{2}[u]/\langle u^{8}\rangle$

Table 7 Optimal binary linear codes obtained from $(1+u)$ constacyclic codes over $F_{2}[u]/\langle u^{16}\rangle$

7 Conclusion

In this paper, we extend the result of [1] about the Gray map to the polynomial residue ring $R_{k}=F_{2^m}[u]/\langle u^{k} \rangle$, where $2^{j-1}+1\le k\le 2^{j}$ for some positive integer $j$. Some optimal linear cyclic code over $F_{2}$ and $F_{4}$ have been constructed via the Gray map. A nature problem is to extend the results to the ring $F_{q}[u]/\langle u^{k} \rangle$.