On self-dual constacyclic codes over finite fields

Abstract

This paper is devoted to the study of self-dual codes arising from constacyclic codes. Necessary and sufficient conditions are given for the existence of Hermitian self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$ of length $n$. As an application of these necessary and sufficient conditions, some conditions under which MDS Hermitian self-orthogonal and self-dual constacyclic codes exist are obtained.

1 Introduction

Let $q$ be a prime power and $\mathbb{F }_{q}$ be the finite field with $q$ elements. An $[n,k]$ linear code $C$ of length $n$ over $\mathbb{F }_{q}$ is a $k$-dimensional subspace of the vector space $\mathbb{F }_{q}^{n}$. We call $\mathbf{c}=(c_{0},c_{1,}\ldots ,c_{n-1})\in C$ a codeword. The Hamming weight $w(\mathbf{c})$ of $\mathbf{c}\in \mathbb{F }_{q}^{n}$ is the number of nonzero coordinates of $\mathbf{c}$. The minimum distance of $C$ is defined to be $d=\min \left\{ w(\mathbf{c})\mid 0\ne \mathbf{c}\in C\right\}$. An $[n,k,d]$ code, which is defined to be an $[n,k]$ code with the minimum distance $d$, is said to be maximum distance separable (MDS) if $d=n-k+1$. The Euclidean dual code of $C$ is defined to be $C^{\perp }=\{\mathbf{x}\in \mathbb{F }_{q}^{n}\mid \sum \nolimits _{i=0}^{n-1}x_{i}y_{i}=0,\forall \mathbf{y}\in C\}$. A code $C$ is Euclidean self-orthogonal provided $C\subseteq C^{\perp }$ and Euclidean self-dual provided $C=C^{\perp }$. Let $\left( \mathbf{x}, \mathbf{y}\right) _{H}=\sum \nolimits _{i=0}^{n-1}x_{i}y_{i}^{q}$ be the Hermitian inner product of $\mathbf{x},\mathbf{y}\in \mathbb{F }_{q}^{n}$, and $C$ be a code of length $n$ over $\mathbb{F }_{q^{2}}$. The Hermitian dual code $C^{\perp H}$ of $C$ is defined by $C^{\perp H}=\{\mathbf{x}\in \mathbb{F }_{q^{2}}^{n}\mid \sum \nolimits _{i=0}^{n-1}x_{i}y_{i}^{q}=0,\forall \mathbf{y}\in C\}$. Hermitian self-orthogonality and Hermitian self-duality are defined as follows: $C$ is Hermitian self-orthogonal if $C\subseteq C^{\perp H}$ and Hermitian self-dual if $C=C^{\perp H}$.

Let $\alpha \in \mathbb{F }_{q}^{*}$. A linear code $C$ is called $\alpha$ -constacyclic [2] provided that for each codeword $(c_{0},c_{1,}\ldots ,c_{n-1})\$in $C, (\alpha c_{n-1},c_{0},\ldots ,c_{n-2})$ is also a codeword in $C$. An $\alpha$-constacyclic code of length $n$ over $\mathbb F _{q}$ corresponds to the principal ideal $\left\langle g(x)\right\rangle$ of the quotient ring $\mathbb{F }_{q}[x]/(x^{n}-\alpha )$, where $g(x)$ is a divisor of $x^{n}-\alpha$. Since the cases when the code length $n$ is divisible by the characteristic of $\mathbb{F }_{q}$ are cases involving repeated root codes, for the remainder of this paper we assume $n$ and $q$ are relatively prime. Because the code length $n$ must be even if there exist Euclidean or Hermitian self-dual codes, we assume $q$ is an odd prime power.

Self-dual codes are an important class of codes which have been extensively studied in coding theory. This paper is mainly concerned with self-dual codes that are constacyclic codes. In recent years, many papers, for example [3, 5, 6, 9], have been written on this subject. Aydin et al. [1] dealt with constacyclic codes and a constacyclic BCH bound was given. In 2008, Gulliver et al. [6] showed that there exists a Euclidean self-dual MDS code of length $q$ over $\mathbb{F }_{q}$ when $q=2^{m}$ by using a Reed-Solomon (RS) code and its extension. They also constructed many new Euclidean and Hermitian self-dual MDS codes over finite fields. In the same year, Blackford [3] studied negacyclic codes over finite fields by using multipliers. He gave conditions on the existence of Euclidean self-dual codes. Recently, Guenda [5] generalized Blackford’s work [3]. She constructed MDS Euclidean and Hermitian self-dual codes from extended cyclic duadic or negacyclic codes and gave necessary and sufficient conditions on the existence of Hermitian self-dual negacyclic codes arising from negacyclic codes. In this paper, we extend Guenda’s work to constacyclic codes and study the existence of Hermitian self-dual codes. We give conditions on the existence of MDS Hermitian self-orthogonal and self-dual codes.

2 Preliminaries

Throughout this paper, let $q$ be an odd prime power and $n$ be a positive integer relatively prime to $q$. Let $C$ be an $[n,k]\, \alpha$-constacyclic code over $\mathbb{F }_{q}$; then the code $C$ is a vector space over $\mathbb{F }_{q}$ and corresponds to an ideal of $\mathbb{F } _{q}[x]/(x^{n}-\alpha )$. By abuse of notation, we let $C$ represent both a set of polynomials and a set of vectors.

As mentioned above, a nonzero $[n,k]\, \alpha$-constacyclic code $C$ has a unique monic generator polynomial $g(x)$ of degree $n-k$, where $g(x)\mid (x^{n}-\alpha )$. The roots of the code $C$ are the roots of $g(x)$. So if $\eta _{1},\ldots ,\eta _{n-k}$ are the zeros of $g(x)$ in the splitting field of $x^{n}-\alpha$, then $\mathbf{c}=\left( c_{0},c_{1},\ldots ,c_{n-1}\right) \in C$ if and only if $c(\eta _{1})=\cdots =c(\eta _{n-k})=0$, where $c(x)=c_{0}+c_{1}x+\cdots +c_{n-1}x^{n-1}$. Let $h(x)=(x^{n}-\alpha )/g(x)=\sum _{i=0}^{k}h_{i}x^{i}$, then $h(x)$ is called the check polynomial of $C$ [7, 10].

Let $C^{(q)}$ denote the code defined by $C^{(q)}=\{\mathbf{c}^{q}\mid \forall \mathbf{c}=(c_{0},c_{1},\ldots ,c_{n-1})\in C\}$, where $\mathbf{c} ^{q}=(c_{0},c_{1},\ldots ,c_{n-1})^{q}=(c_{0}^{q},c_{1}^{q},\ldots ,c_{n-1}^{q})$.

Lemma 2.1

([4, Proposition 2.4]) (i) Let $C$ be an $\alpha$-constacyclic code over $\mathbb{F }_{q}$, then the Euclidean dual code $C^{\perp }$ is an $\alpha ^{-1}$-constacyclic code generated by $g^{\perp }(x)=\sum _{i=0}^{k}h_{i}h_{0}^{-1}x^{k-i}$.

(ii) Let $C$ be an $\alpha$-constacyclic code over $\mathbb{F } _{q^{2}}$, then the Hermitian dual code $C^{\perp H}$ is an $\alpha ^{-q}$ -constacyclic code generated by $g^{\perp (q)}(x)=\sum _{i=0}^{k}h_{i}^{q}h_{0}^{-q}x^{k-i}$.

Proof

(i) The proof can be found in [4, Proposition 2.4].

(ii) $g^{\perp }(x)$ is the generator polynomial of $C^{\perp }$. Let $C^{*}$ denote the code generated by $g^{\perp (q)}(x)=\sum _{i=0}^{k}h_{i}^{q}h_{0}^{-q}x^{k-i}$ and $\xi _{1},\ldots ,\xi _{k}$ be the zeros of $g^{\perp }(x)$, then $\xi _{1}^{q},\ldots ,\xi _{k}^{q}$ are the zeros of $g^{\perp (q)}(x)$. Thus if $\mathbf{c}=\left( c_{0},c_{1},\ldots ,c_{n-1}\right)$ is a codeword in $C$, then we have $c_{0}+c_{1}\xi _{i}+\cdots +c_{n-1}\xi _{i}^{n-1}=0$ ($i=1,\ldots ,k$). It is obvious that $c_{0}^{q}+c_{1}^{q}\xi _{i}^{q}+\cdots +c_{n-1}^{q}\left( \xi _{i}^{q}\right) ^{n-1}=0$ ($i=1,\ldots ,k$). This implies that $\left( c_{0}^{q},c_{1}^{q},\ldots ,c_{n-1}^{q}\right)$ is a codeword in $C^{*}$. So $C^{\perp H}\subset C^{*}$. Because $\dim C^{\perp H}=\dim C^{*}=n-k$, we get $C^{*}=C^{\perp H}$.

Since $C^{\perp }$ is an $\alpha ^{-1}$-constacyclic code generated by $g^{\perp }(x)$, we have

$$\begin{aligned} \xi _{i}^{n}=\alpha ^{-1}\Longrightarrow \left( \xi _{i}^{q}\right) ^{n}=\left( \alpha ^{q}\right) ^{-1}(i=1,\ldots ,k)\text {.} \end{aligned}$$

So $\xi _{1}^{q},\ldots ,\xi _{k}^{q}$ are roots of $x^{n}-\alpha ^{-q}$, which implies $g^{\perp (q)}(x)$ is a divisor of $x^{n}-\alpha ^{-q}$. Therefore, the Hermitian dual code $C^{\perp H}$ is an $\alpha ^{-q}$-constacyclic code. $\square$

Let $r=ord_{q}(\alpha )$ (i.e., the smallest integer $r$ such that $\alpha ^{r}=1$) and the multiplicative order of $q$ modulo $rn$ be $m$ [i.e., the smallest integer $m$ such that $q^{m}\equiv 1\pmod {rn}$]. There exists $\delta \in \mathbb{F }_{q^{m}}^{*}$, called a primitive $rn$th root of unity, such that $\delta ^{n}=\alpha$. Let $\zeta =\delta ^{r}$, then $\zeta$ is a primitive $n$th root of unity. Therefore, the roots of $x^{n}-\alpha$ are $\{\delta ,\delta ^{1+r},\ldots ,\delta ^{1+(n-1)r}\}$ and the roots of $x^{n}-\alpha ^{-1}$ are $\{\delta ^{-1},\delta ^{-1+r},\ldots ,\delta ^{-1+(n-1)r}\}$. Define $O_{r,n}(1)$ and $O_{r,n}(-1)$ as follows:

$$\begin{aligned} O_{r,n}(1)&= \left\{ ir+1\mid 0\le i\le n-1\right\} \pmod {rn}\subseteq \mathbb Z _{rn}\text {;} \\ O_{r,n}(-1)&= \left\{ ir-1\mid 0\le i\le n-1\right\} \pmod {rn}\subseteq \mathbb Z _{rn}\text {.} \end{aligned}$$

The defining set of the $\alpha$-constacyclic code $C$ is defined as $T=\{ir+1\in O_{r,n}(1)\mid \delta ^{ir+1}$ is a root of $C\}$. It is clear that $T\subset O_{r,n}(1)$ and the dimension of $C$ is $n-\left| T\right|$. Let $Cl_{q}(s)$ be the $q$ -cyclotomic coset modulo $rn$ which contains $s$, i.e. $Cl_{q}(s)=\{sq^{j}\pmod {rn}\mid j\in \mathbb Z \}$. Assume the generator polynomial of $C$ is $g(x)= \sum _{i=0}^{k}g_{i}x^{i}$, where $g_{i}\in \mathbb{F }_{q}$. If $g(\nu )=0$ for some $\nu \in \mathbb{F }_{q^{m}}$, then

$$\begin{aligned} g(\nu ^{q})=\sum _{i=0}^{k}g_{i}\big (\nu ^{q}\big )^{i}=\sum _{i=0}^{k}g_{i}^{q}\big (\nu ^{i}\big )^{q}=\left( \sum _{i=0}^{k}g_{i}\nu ^{i}\right) ^{q}=(g(\nu ))^{q}=0. \end{aligned}$$

Therefore, the defining set $T$ is a union of some $q$-cyclotomic cosets modulo $rn$ and a union of some $q$-cyclotomic cosets modulo $rn$ is also the defining set of some $\alpha$-constacyclic code.

Proposition 2.2

There exists a Euclidean self-dual $\alpha$-constacyclic code over $\mathbb{F }_{q}$ if and only if $r=2$.

Proof

By Lemma 2.1, the Euclidean dual code of an $\alpha$-constacyclic code is an $\alpha ^{-1}$-constacyclic code. To prove that if there is a Euclidean self-dual $\alpha$-constacyclic code, we need to verify $\alpha ^{2}=1$. This indicates that either $r=1$ or $r=2$. If $r=1$, then $\alpha =1$. It has been proved by Jian et al. [8, Theorem 1] that there exists at least one self-dual cyclic code if and only if $q$ is a power of 2. Since $q$ is odd, this leads to the unique solution $r=2$.

If $r=2$, then $\alpha =-1$. Guenda [5] has proved that there exist Euclidean self-dual negacyclic codes over $\mathbb{F }_{q}$ (i.e. for $r=2$ there exists a Euclidean self-dual $\alpha$-constacyclic code over $\mathbb{F }_{q}$). $\square$

Proposition 2.3

Let $\alpha \in \mathbb{F }_{q^{2}}^{*}, r=ord_{q^{2}}(\alpha )$, and $C$ be an $\alpha$-constacyclic code over $\mathbb{F }_{q^{2}}$. If $C$ is a Hermitian self-dual code, then $r\mid q+1$.

Proof

If the $\alpha$-constacyclic code $C$ is a Hermitian self-dual code, then $C=C^{\perp H}$. By Lemma 2.1, the Hermitian dual code $C^{\perp H}$ is an $\alpha ^{-q}$-constacyclic code. Hence, we have

$$\begin{aligned} C=C^{\perp H}\Longrightarrow \alpha =\alpha ^{-q}\Longrightarrow \alpha ^{q+1}=1. \end{aligned}$$

Since $r=ord_{q}(\alpha )$, we obtain $r\mid q+1$. $\square$

3 Hermitian self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$

This section is devoted to the Hermitian self-dual $\alpha$-constacyclic codes over $\mathbb{F }_{q^{2}}$, where $\alpha \in \mathbb{F }_{q^{2}}^{*}$. Let $r=ord_{q^{2}}(\alpha )$, then $r\mid q^{2}-1$. By Proposition 2.3, we can further assume $r\mid q+1$ and $rs=q+1$ for some integer $s$. Note that if $T\subset O_{r,n}(1)$ is a union of some $q^{2}$-cyclotomic cosets, $C_T$ is an $\alpha$-constacyclic code over $\mathbb{F }_{q^{2}}$ with the defining set $T$.

Lemma 3.1

$-qO_{r,n}(1)=O_{r,n}(1)\pmod {rn}$.

Proof

Since $q+1=rs$, for $ir+1\in O_{r,n}(1)$, we have

$$\begin{aligned} -q(ir+1)\!=\!-qir-(q\!+\!1)\!+\!1\!=\!-qir-rs\!+\!1\!=\!(-qi-s)r\!+\!1\pmod {rn}\in O_{r,n}(1)\text {. } \end{aligned}$$

By this, we have $-qO_{r,n}(1)=O_{r,n}(1)\pmod {rn}$. $\square$

Let $T^{\bot }=-\left[ O_{r,n}(1)\backslash T\right] \subset O_{r,n}(-1)$ be the defining set of code $C_{T^{\bot }}$. Then

$$\begin{aligned} x^{n}-\alpha =\prod \limits _{i\in O_{r,n}(1)}(x-\delta ^{i})=\prod \limits _{i\in T}(x-\delta ^{i})\cdot \prod \limits _{i\in T^{\bot }}(x-\delta ^{-i})=g(x)h(x), \end{aligned}$$

where $g(x)$ is the generator polynomial of $C_{T}$. By Lemma 2.1, $g^{\bot }(x)=\prod \limits _{i\in T^{\bot }}(x-\delta ^{i})$. Therefore, $T^{\bot }$ is the defining set of the $\alpha ^{-1}$-constacyclic code $C_{T}^{\bot }$ (i.e. the Euclidean dual code of $C_{T}$). Thus we have $C_{T^{\bot }}=C_{T}^{\bot }$.

Let $\bar{T}=-q\left[ O_{r,n}(1)\backslash T\right] =qT^{\bot }$. According to Lemma 3.1, $\bar{T}\subset O_{r,n}(1)$. It is clear that $\bar{T}$ is a union of some $q^{2}$-cyclotomic cosets and $\left| T\right| +\left| \bar{T}\right| =n$. Similarly, $g^{\bot (q)}(x)=\prod \limits _{i\in T^{\bot }}(x-\delta ^{iq})$. Therefore, $\bar{T}$ is the defining set of the $\alpha ^{-q}$-constacyclic code $C_{T}^{\bot H}$. Thus we have the following theorem.

Theorem 3.2

$C_{\bar{T}}$ is the Hermitian dual code of $C_{T}$.

Based on Theorem 3.2, two necessary and sufficient conditions are given as follows:

Corollary 3.3

Let $T\subset O_{r,n}(1)$ be the defining set of code $C_{T}$ and let $\bar{T }=-q\left[ O_{r,n}(1)\backslash T\right]$. Then

• (i) $C_{T}$ is a Hermitian self-orthogonal constacyclic code if and only if $\bar{T}\subset T$;

• (ii) $C_{T}$ is a Hermitian self-dual constacyclic code if and only if $\bar{T}=T$.

Example 3.4

Let $q=5, n=4$, and $r=2$, then $q^{2}=25$. Consider the $\alpha$-constacyclic code of length 4 over $\mathbb{F }_{25}$ with $\alpha =-1$.

We notice that $r\mid q+1$ and $O_{2,4}(1)=\left\{ 1,3,5,7\right\}$. Let $T=\{3,5\}$, then $\bar{T}=-5\left[ O_{2,4}(1)\backslash T\right] =\left\{ 3,5\right\} \pmod {8}$. By Corollary 3.3, the code $C_{T}$ with defining set $T=\{3,5\}$ is a Hermitian self-dual negacyclic code.

Example 3.5

Let $q^{2}=31^{2}, n=16$, and $r=4$. Now we consider the $\alpha$-constacyclic code of length 16 over $\mathbb{F }_{31^{2}}$ with $\alpha$ a primitive 4th root of unity.

Clearly, $O_{4,16}(1)=\{1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61 \}$ and $r\mid q+1$. Let $T=\{33,37,41,45,49,53,57,61\}$, then

$$\begin{aligned} \bar{T}=-31\left[ O_{4,16}(1)\backslash T\right] =\left\{ 33,37,41,45,49,53,57,61\right\} \pmod {64}. \end{aligned}$$

Hence, $\bar{T}=T$. By Corollary 3.3, $C_{T}$ is a Hermitian self-dual $\alpha$-constacyclic code.

Lemma 3.6

Let $n$ be an odd integer with $n\mid q+1$, then there exists an integer $m$ such that $n\mid \frac{q^{2m+1}+1}{q+1}$ and $n\mid 2m+1$.

Proof

$$\begin{aligned} \frac{q^{2m+1}+1}{q+1}&= \sum _{j=0}^{2m}(-1)^{j}q^{j}=\sum _{i=0}^{2m} \sum _{j=i}^{2m}\genfrac(){0.0pt}1{j}{i}(q+1)^{i}(-1)^{i} \\&= \sum _{i=0}^{2m}(q+1)^{i}(-1)^{i}\sum _{j=i}^{2m}\genfrac(){0.0pt}1{j}{i} =\sum _{i=0}^{2m}(q+1)^{i}(-1)^{i}\genfrac(){0.0pt}1{2m+1}{i+1} \\&= \left( \sum _{i=1}^{2m}(q+1)^{i-1}(-1)^{i}\genfrac(){0.0pt}1{2m+1}{i+1}\right) (q+1)+(2m+1). \end{aligned}$$

We can choose an integer $m$ such that $n\mid 2m+1$, which further implies that $n\mid \frac{q^{2m+1}+1}{q+1}$.

$\square$

Lemma 3.7

Let $n$ be an odd integer with prime decomposition $n=p_{1}^{t_{1}}p_{2}^{t_{2}}\cdots p_{s}^{t_{s}}$, where $p_{i}$ are such that $p_{i}\mid q+1, p_{i}\ne p_{j}, t_{i}>0\, (1\le i\le s)$. Then there exists an integer $m$ such that $n\mid \frac{q^{2m+1}+1}{q+1}$.

Proof

First, let $n_{1}=p_{_{1}}^{t_{1}}$. We use induction to prove that there exists $m_{t}$ such that $n_{1}\mid \frac{q^{2m_{t}+1}+1 }{q+1}$ and $p_{1}\mid 2m_{t}+1$.

When $t_{1}=1$, by Lemma 3.6, there exists $m_{1}$ such that $p_{1}\mid \frac{q^{2m_{1}+1}+1}{q+1}$ and $p_{1}\mid 2m_{1}+1$. When $t_{1}\ge 2$, assume there exists $m_{t-1}$ such that $p_{1}^{t_{1}-1}\mid \frac{q^{2m_{t-1}+1}+1}{q+1}$ and $p_{1}\mid 2m_{t-1}+1$. Then by the proof of Lemma 3.6, we know

$$\begin{aligned} \frac{q^{(2m_{t-1}+1)(2m_{t-1}+1)}+1}{q+1}&= \frac{q^{2m_{t-1}+1}+1}{q+1}\frac{\left( q^{2m_{t-1}+1}\right) ^{2m_{t-1}+1}+1}{q^{2m_{t-1}+1}+1} \\&= \frac{q^{2m_{t-1}+1}+1}{q+1}\left[ \left( \sum _{i=1}^{2m_{t-1}}(q^{2m_{t-1}+1}+1)^{i-1}(-1)^{i}\genfrac(){0.0pt}1{2m_{t-1}+1}{i+1}\right) \right. \\&\left. \qquad \qquad \qquad \qquad (q^{2m_{t-1}+1}+1)+(2m_{t-1}+1) \right] . \end{aligned}$$

According to the assumption, we have

$$\begin{aligned} p_{1}\mid \left[ \left( \sum _{i=1}^{2m_{t-1}}(q^{2m_{t-1}+1}+1)^{i-1}(-1)^{i}\genfrac(){0.0pt}1{2m_{t-1}+1}{i+1}\right) (q^{2m_{t-1}+1}+1)+(2m_{t-1}+1)\right] \end{aligned}$$

and $p_{1}^{t_{1}-1}\mid \frac{q^{2m_{t-1}+1}+1}{q+1}$, thus $p_{_{1}}^{t_{1}}\mid \frac{ q^{(2m_{t-1}+1)(2m_{t-1}+1)}+1}{q+1}$. Let $2m_{t}+1=(2m_{t-1}+1)^2$. It follows that $p_{_{1}}^{t_{1}}\mid \frac{q^{2m_{t}+1}+1}{q+1}$ and $p_{1}\mid 2m_{t}+1$.

Next, we prove there exists some $m$ such that $n\mid \frac{ q^{2m+1}+1}{q+1}$.

Let $n_{i}=p_{i}^{t_{i}}$ for $1\le i\le s$. The case $s=1$ has been proven above. Similarly, there exists $m_{s}^{^{\prime }}$ such that $n_{s}\mid \frac{q^{2m_{s}^{^{\prime }}+1}+1}{q+1}$. Let $n^{\prime }=n_{1}n_{2}\cdots n_{s-1}$. We assume that there exists $m^{\prime }$ such that $n^{\prime }\mid \frac{ q^{2m^{\prime }+1}+1}{q+1}$. Let $2m+1=(2m^{\prime }+1)(2m_{s}^{^{\prime }}+1)$. Then

$$\begin{aligned} \frac{q^{2m+1}+1}{q+1}&= \frac{q^{(2m^{\prime }+1)(2m_{s}^{^{\prime }}+1)}+1}{q+1} \\&= \frac{q^{2m^{^{\prime }}+1}+1}{q+1}\left[ \left( \sum _{i=1}^{2m_{s}^{^{ \prime }}}(q^{2m_{s}^{^{\prime }}+1}+1)^{i-1}(-1)^{i}\genfrac(){0.0pt}1{2m_{s}^{^{\prime }}+1 }{i+1}\right) \right. \\&\left. \qquad \qquad \qquad \qquad (q^{2m_{s}^{^{\prime }}+1}+1)+(2m_{s}^{^{\prime }}+1)\right] \!. \end{aligned}$$

Because $n^{^{\prime }}\mid \frac{q^{2m^{^{\prime }}+1}+1}{q+1}$, we have $n^{^{\prime }}\mid \frac{q^{2m+1}+1}{q+1}$. Similarly, we have $n_{s}\mid \frac{q^{2m+1}+1}{q+1}$. Since $(n^{\prime },n_{s})=1$, we obtain $n^{\prime }n_{s}\mid \frac{q^{2m+1}+1}{q+1}$, i.e., $n\mid \frac{q^{2m+1}+1}{q+1}$. $\square$

Proposition 3.8

Hermitian self-dual $\alpha$-constacyclic codes over $\mathbb{F }_{q^{2}}$ of length $n$ exist if and only if $Cl_{q^{2}}(j)\ne Cl_{q^{2}}(-qj)$ for any $j\in O_{r,n}(1)$.

Proof

Assume the $q^2$-cyclotomic cosets of $O_{r,n}(1)$ are

$$\begin{aligned} Cl_{q^{2}}(j_{1}),Cl_{q^{2}}(j_{2}),\ldots ,Cl_{q^{2}}(j_{t}) \end{aligned}$$

denoted simply by $Cl_{1},Cl_{2},\ldots ,Cl_{t}$ for convenience. By Lemma 3.1, $-qCl_{i}, i\in \{1,2,\ldots ,t\}$, is also a $q^2$-cyclotomic coset of $O_{r,n}(1)$. Let $Cl_{\bar{i}}=-qCl_{i}$ and $\sigma$ be a permutation of $\{1,2,\ldots ,t\}$ which satisfies $\sigma (i)=\bar{i}$ for any $i\in \{1,2,\ldots ,t\}$. Because $q^{2}Cl_{i}=Cl_{i}$ for any $i\in \{1,2,\ldots ,t\}$, we obtain $Cl_{\sigma (\bar{i} )}=Cl_{\sigma ^{2}(i)}=Cl_{i}$. This implies $\sigma ^{2}(i)=i$, i.e., $\sigma ^{2}$ is the identity permutation of $\{1,2,\ldots ,t\}$.

Now we prove necessity. Assume there exists a Hermitian self-dual code $C_{T}$ with defining set $T\subset O_{r,n}(1)$. Then by Corollary 3.3, $\bar{T}=-q[O_{r,n}(1)\backslash T] =T$. Therefore, if there exists $j$ such that $Cl_{q^{2}}(j)=Cl_{q^{2}}(-qj)$, we will have the following two cases.

Case 1: If $j\in T$, then $-qj\in T$. By the fact that $\bar{T} =-q[O_{r,n}(1)\backslash T] =T$, there exists some $i\notin T$ such that $-qi=j$. Thus $q^{2}i=-qj\notin T$. This is a contradiction.

Case 2: If $j\notin T$, by the fact that $\bar{T}=-q[O_{r,n}(1)\backslash T] =T$, we have $-qj\in T$. Because $Cl_{q^{2}}(j)=Cl_{q^{2}}(-qj)\subset T$, we have $j\in T$ which contradicts the assumption.

Next, we prove the sufficiency. We assume $Cl_{q^{2}}(j)\ne Cl_{q^{2}}(-qj)$ for any $j\in O_{r,n}(1)$. This implies $\sigma (i)\ne i$ for any $i\in \{1,2,\ldots ,t\}$. Since $\sigma ^{2}(i)=i, \sigma$ must be a product of mutually disjoint transpositions like $(a_{1}\, b_{1})(a_{2}\, b_{2})\cdots (a_{k}\, b_{k})$. We might assume $t=2k$ and let $\sigma (i)=k+i$ and $\sigma (k+i)=i$ for $1\le i\le k$. If we let $T=Cl_{1}\cup Cl_{2}\cup \cdots \cup Cl_{k}$, then the code $C_{T}$ with defining set $T$ is a Hermitian self-dual code. Therefore, if $Cl_{q^{2}}(j)\ne Cl_{q^{2}}(-qj)$ for $\forall j\in O_{r,n}(1)$, there exist Hermitian self-dual codes. $\square$

Based on this proposition, we have the following theorem. This theorem is an extension of Theorem 3 in [3] (the case of $b=1$ and $r^{\prime }=1$).

Theorem 3.9

Let $n=2^{a}n^{\prime }\left( a>0\right)$ and $r=2^{b}r^{\prime }$ be integers such that $2\not \mid n^{\prime }$ and $2\not \mid r^{\prime }$. Let $q$ be an odd prime power such that $(n,q)=1$ and $r\mid q+1$, and let $\alpha \in \mathbb{F }_{q^{2}}^{*}$ has order $r$. Then Hermitian self-dual $\alpha$ -constacyclic codes over $\mathbb{F }_{q^{2}}$ of length $n$ exist if and only if $b>0$ and $q+1\not \equiv 0\pmod {2^{a+b}}$.

Proof

$n^{\prime }$ can be written as $n^{\prime }=r_{1}^{t_{1}}\cdots r_{j}^{t_{j}}r_{j+1}^{t_{j+1}}\cdots r_{s}^{t_{s}}$, where $r_{1},\ldots ,r_{s}$ are distinct primes, $r_{1},\ldots ,r_{j}\mid r$, and $r_{j+1},\ldots ,r_{s}\not \mid r$. Assume $n_{1}=r_{1}^{t_{1}} \cdots r_{j}^{t_{j}}, n_{2}=r_{j+1}^{t_{j+1}}\cdots r_{s}^{t_{s}}$, and $n^{\prime }=n_{1}n_{2}$. Since $r_{j+1},\ldots ,r_{s}\not \mid r$, it follows $(n_{2},r)=1$. Because $r_{1},\ldots ,r_{j}\mid r$, we know $r_{1},\ldots ,r_{j}\mid q+1$. By Lemma 3.7, there exists $m$ such that $n_{1}\mid \frac{ q^{2m+1}+1}{q+1}$.

The proof consists of two parts. First we prove the necessity. If $r$ is odd, which is equivalent to $b=0$, clearly, we have $(r,2^{a}n_{2})=1$. There exists $i\in \mathbb Z$ such that $2^{a}n_{2}\mid ir+1$. Thus by $n_{1}\mid \frac{q^{2m+1}+1}{q+1}$, we have

$$\begin{aligned}&(q+1)2^{a}n_{1}n_{2}\mid (q^{2m+1}+1)(ir+1)\Longrightarrow (q+1)n\mid (q^{2m+1}+1)(ir+1)\\&\quad \Longrightarrow rn\mid (q^{2m+1}+1)(ir+1). \end{aligned}$$

Therefore, $ir+1=q^{2m}(-q(ir+1))\pmod {rn}$. This implies $Cl_{q^{2}}(ir+1)=Cl_{q^{2}}(-q(ir+1))$. Since $ir+1\in O_{r,n}(1)$, by Proposition 3.8, there is no Hermitian self-dual $\alpha$ -constacyclic code over $\mathbb{F }_{q^{2}}$, which contradicts the assumption. Therefore, $r$ must be even, i.e., $b>0$.

Let $q+1=2^{c}rt$ with $c\ge 0$ and $(t,2)=1$. If $q+1\equiv 0\pmod {2^{a+b}}$, then $c\ge a$. Because $(n_{2},r)=1$, there exists $i^{\prime }\in \mathbb Z$ such that $n_{2}\mid i^{\prime }r+1$. Since $n_{1}\mid \frac{q^{2m+1}+1}{ q+1}$, we have

$$\begin{aligned} (q+1)n_{1}n_{2}\mid (q^{2m+1}+1)(i^{\prime }r+1)&\Longrightarrow&2^{c}rtn_{1}n_{2}\mid (q^{2m+1}+1)(i^{\prime }r+1)\\&\Longrightarrow&rn\mid (q^{2m+1}+1)(i^{\prime }r+1). \end{aligned}$$

Similarly, we have $Cl_{q^{2}}(i^{\prime }r+1)=Cl_{q^{2}}(-q(i^{\prime }r+1))$. By Proposition 3.8, we get a contradiction. So it is necessary to have $q+1\not \equiv 0\pmod {2^{a+b}}$.

Now we prove the sufficiency. Assume $b>0$ and $q+1\not \equiv 0\pmod {2^{a+b}}$. If there is no Hermitian self-dual code, by Proposition 3.8, there exists $ir+1\in O_{r,n}(1)$ such that $Cl_{q^{2}}(ir+1)=Cl_{q^{2}}(-q(ir+1))$. Therefore, for some $m\in \mathbb Z ^{+}$,

$$\begin{aligned} rn\mid (q^{2m+1}+1)(ir+1)\Longrightarrow 2^{a+b}r^{\prime }n^{\prime }\mid \frac{q^{2m+1}+1}{q+1}(q+1)(ir+1). \end{aligned}$$

Since $b>0, ir+1$ must be odd. Together with the fact that $\frac{ q^{2m+1}+1}{q+1}$ is odd, we get $2^{a+b}\mid q+1$, which contradicts the assumption that $q+1\not \equiv 0\pmod {2^{a+b}}$. $\square$

4 MDS hermitian self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$

We study MDS Hermitian self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$ in this section. The following theorem will give the BCH bound for constacyclic codes (cf. [1, Theorem 2.2]).

Theorem 4.1

Let $C$ be an $\alpha$-constacyclic code of length $n$ over $\mathbb{F }_{q^2}$. Let $r=ord_{q^{2}}(\alpha )$. Let $\delta$ be a primitive $rn$th root of unity in an extension field of $\mathbb{F }_{q^2}$ such that $\delta ^{n}=\alpha$, and let $\zeta =\delta ^r$. Assume the generator polynomial of $C$ has roots that include the set $\{\delta \zeta ^{^{i}}\mid i_{1}\le i\le i_{1}+d-1\}$. Then the minimum distance of $C\ge d$.

Example 4.2

Let $q^{2}=17^{2}, n=8$ and $r=18$. We consider the $\alpha$-constacyclic code of length 8 over $\mathbb{F }_{17^{2}}$ with $\alpha$ a primitive 18th root of unity.

Obviously, we have $O_{18,8}(1)=\{1,19,37,55,73,91,109,127\}$ and $r\mid q+1$. Let $T=\{73,91,109,127\}$, then $\bar{T}=-17\left[ O_{18,8}(1)\backslash T\right] =\left\{ 73,91,109,127\right\} \pmod {144}$. Thus $\bar{T}=T$. By Corollary 3.3, $C_{T}$ is a Hermitian self-dual $\alpha$-constacyclic code.

Furthermore, we notice that the generator polynomial of $C_{T}$ has roots:

$$\begin{aligned} \delta ^{1+4r}, \delta ^{1+5r}, \delta ^{1+6r}, \delta ^{1+7r}. \end{aligned}$$

By Theorem 4.1, the minimum distance $d$ is at least 5. Since $n-k+1=8-4+1=5, C_{T}$ is an [3, 4, 7] MDS Hermitian self-dual $\alpha$-constacyclic code.

Example 4.2 shows that there exist MDS Hermitian self-dual constacyclic codes. The following theorem is a generalization of Example 4.2.

Theorem 4.3

Let $\alpha \in \mathbb{F }_{q^{2}}^{*}$ have order $r$ with $rs=q+1$ for some positive integer $s$. Let $n$ be even and $n\mid q-1$. Let

$$\begin{aligned} T=O_{r,n}(1)\backslash \left\{ ir+1\mid -[\frac{s-1}{2}]\le i\le [\frac{n-1-s}{2}]\right\} \pmod {rn}. \end{aligned}$$

Then the following holds.

• (i) If $s$ is odd, then $C_{T}$ is a Hermitian self-dual $\alpha$ -constacyclic MDS code with parameters $[n,\frac{n}{2},\frac{n}{2}+1]$;

• (ii) If $s$ is even, then $C_{T}$ is a Hermitian self-orthogonal $\alpha$-constacyclic MDS code with parameters $[n,\frac{n}{2}-1,\frac{n}{2} +2]$.

Proof

Let $T_{1}=\left\{ ir+1\mid -[\frac{s-1}{2}]\le i\le [\frac{ n-1-s}{2}]\right\} \pmod {rn}$. If $s$ is odd, then $T_{1}$ has $\frac{n}{2}$ elements, and therefore, the dimension of $C_{T}$ is $\frac{n}{2}$; if $s$ is even, then $T_{1}$ has $\frac{n}{2}-1$ elements, and therefore, the dimension of $C_{T}$ is $\frac{n}{2}-1$. Let $I_{1}=\left\{ i\mid -[\frac{s-1}{2} ]\le i\le [\frac{n-1-s}{2}]\right\} \pmod {n}$. The set $\{0,1,\ldots ,n-1\}\backslash I_{1}$ has $\frac{n}{2}$ consecutive elements (modulo $n$) when $s$ is odd and $\frac{ n}{2}+1$ consecutive elements when $s$ is even. Using the Singleton Bound and Theorem 4.1, the minimum distance of $C_{T}$ is $\frac{n}{2}+1$ when $s$ is odd and $\frac{n}{2}+2$ when $s$ is even, making $C_{T}$ MDS.

The proof is complete if we show that $C_{T}$ is Hermitian self-orthogonal. By Corollary 3.3, this can be verified if we show $T_{1}\cap (-qT_{1})=\varnothing$ where we reduce the entries in $T_{1}$ and $-qT_{1}$ modulo $rn$ before taking the intersection. Since $n\mid q-1$, we know $-q\equiv -1\pmod {n}$, which implies that $-qr\equiv -r\pmod {rn}$. So $-q(ir+1)\equiv -ir-q\equiv -ir-(q+1)+1\equiv (-i-s)r+1\equiv (n-i-s)r+1\pmod {rn}$. Therefore, showing that $T_{1}\cap (-qT_{1})=\varnothing$ is equivalent to showing that $I_{1}\cap I_{2}=\varnothing$, where $I_{2}=\left\{ n-i-s\mid i\in I_{1}\right\} \pmod {n}$ and the intersection $I_{1}\cap I_{2}$ is taken after reducing modulo $n$. Consider the case that $s$ is odd. The elements of $I_{1}$ are the $\frac{n}{2}$ consecutive integers $-(\frac{s-1}{2}),\ldots , \frac{n-1-s}{2}$. Using this, the elements of $I_{2}$ are the $\frac{n}{2}$ consecutive integers $\frac{n+1-s}{2},\ldots ,n-(\frac{ s+1}{2})$. These two lists together make up $n$ consecutive integers, and hence, when reducing modulo $n, I_{1}\cap I_{2}=\varnothing$. Consider the case that $s$ is even. The elements of $I_{1}$ are the $\frac{n}{2}-1$ consecutive integers $-\frac{s}{2}+1,\ldots ,\frac{n-s}{2}-1$. Using this, the elements of $I_{2}$ are the $\frac{n}{2}-1$ consecutive integers $\frac{n-s}{2}+1,\ldots ,n-\frac{s}{2}-1$. These two lists together make up $n-1$ consecutive integers, excluding the single integer $\frac{n-s}{2}$. Therefore, when reducing modulo $n, I_{1}\cap I_{2}=\varnothing$. $\square$

Table 1 gives some Hermitian self-dual codes over $\mathbb{F }_{q^{2}}$ for $q\le 19$ with lower bounds on the minimum distance $d$.

Table 1 $[n,k,d]$ Hermitian self-dual codes over $\mathbb{F }_{q^{2}}$ (where $q\le 19$)

5 Conclusion

We have studied Hermitian self-dual codes arising from constacyclic codes in this paper. In Sect. 3, necessary and sufficient conditions have been given for the existence of Hermitian self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$ of length $n$. In Sect. 4, we have given conditions for the existence of MDS Hermitian self-orthogonal and self-dual constacyclic codes over $\mathbb{F }_{q^{2}}$.