New MDS Euclidean and Hermitian Self-Dual Codes over Finite Fields

DOI: 10.4236/apm.2017.75019   PDF   HTML     1,110 Downloads   1,512 Views   Citations
In this paper, we construct MDS Euclidean self-dual codes which are ex-tended cyclic duadic codes. And we obtain many new MDS Euclidean self-dual codes. We also construct MDS Hermitian self-dual codes from generalized Reed-Solomon codes and constacyclic codes.

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Tong, H. and Wang, X. (2017) New MDS Euclidean and Hermitian Self-Dual Codes over Finite Fields. Advances in Pure Mathematics, 7, 325-333. doi: 10.4236/apm.2017.75019.

1. Introduction

Let ${\mathbb{F}}_{q}$ denote a finite field with q elements. An $\left[n,k,d\right]$ linear code C over ${\mathbb{F}}_{q}$ is a k-dimensional subspace of ${\mathbb{F}}_{q}^{n}$ . These parameters n, k and d satisfy $d\le n-k+1$ . If $d=n-k+1$ , C is called a maximum distance separable (MDS) code. MDS codes are of practical and theoretical importance. For examples, MDS codes are related to geometric objects called n-arcs.

The Euclidean dual code ${C}^{\perp }$ of $C$ is defined as

${C}^{\perp }:=\left\{x\in {\mathbb{F}}_{q}^{n}:\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{x}_{i}{y}_{i}=0,\text{\hspace{0.17em}}\forall y\in C\right\}.$ (1)

If $q={r}^{2}$ , the Hermitian dual code ${C}^{\perp H}$ of $C$ is defined as

${C}^{\perp H}:=\left\{x\in {\mathbb{F}}_{{r}^{2}}^{n}:\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{x}_{i}{y}_{i}^{r}=0,\text{\hspace{0.17em}}\forall y\in C\right\}.$ (2)

If C satisfies $C={C}^{\perp }$ or $C={C}^{\perp H}$ , C is called Euclidean self-dual or Hermitian self-dual, respectively. In   discussing Euclidean self-dual codes or Hermitian self-dual codes. If C is MDS and Euclidean self-dual or Hermitian self-dual, C is called an MDS Euclidean self-dual code or an MDS Hermitian self-dual code, respectively. In recent years, In  -  study the MDS self-dual codes. One of these problems in this topic is to determine existence of MDS self-dual codes. When $2|q$ , Grassl and Gulliver completely solve the existence of MDS Euclidean self-dual codes in  . In  , Guenda obtain some new MDS Euclidean self-dual codes and MDS Hermitian self-dual codes. In  , Jin and Xing obtain some new MDS Euclidean self-dual codes from generalized Reed- Solomon codes.

In this paper, we obtain some new Euclidean self-dual codes by studying the solution of an equation in ${\mathbb{F}}_{q}$ . And we generalize Jin and Xing’s results to MDS Hermitian self-dual codes. We also construct MDS Hermitian self-dual codes from constacyclic codes. We discuss MDS Hermitian self-dual codes obtained from extended cyclic duadic codes and obtain some new MDS Hermitian self-dual codes.

2. MDS Euclidean Self-Dual Codes

A cyclic code C of length n over ${\mathbb{F}}_{q}$ can be considered as an ideal, $〈g\left(x\right)〉$ , of the ring $R=\frac{{\mathbb{F}}_{q}\left[x\right]}{{x}^{n}-1}$ , where $g\left(x\right)|{x}^{n}-1$ and $\left(n,q\right)=1$ . The set $T=\left\{0\le i\le n-1|g\left({\alpha }^{i}\right)=0\right\}$ is called the defining set of C, where $\text{ }ord\text{ }\alpha =n$ .

Let ${S}_{1}$ and ${S}_{2}$ be unions of cyclotomic classes modulo n, such that ${S}_{1}\cap {S}_{2}=\varnothing$ and ${S}_{1}\cup {S}_{2}={ℤ}_{n}\\left\{0\right\}$ and $a{S}_{i}\left(\mathrm{mod}n\right)={S}_{i+1\left(\mathrm{mod}2\right)}$ . Then the triple ${\mu }_{a}$ , ${S}_{1}$ and ${S}_{2}$ is called a splitting modulo n. Odd-like codes ${D}_{1}$ and ${D}_{2}$ are cyclic codes over ${\mathbb{F}}_{q}$ with defining sets ${S}_{1}$ and ${S}_{2}$ , respectively. ${D}_{1}$ and ${D}_{2}$ can be denoted by ${\mu }_{a}\left({D}_{i}\right)={D}_{i+1\left(\mathrm{mod}2\right)}$ . Even-like duadic codes ${C}_{1}$ and ${C}_{2}$ are cyclic codes over ${\mathbb{F}}_{q}$ with defining sets $\left\{0\right\}\cup {S}_{1}$ and $\left\{0\right\}\cup {S}_{2}$ , respectively. Obviously, ${\mu }_{a}\left({C}_{i}\right)={C}_{i+1\left(\mathrm{mod}2\right)}$ . In  , A duadic code of length n over ${\mathbb{F}}_{q}$ exists if and only if q is a quadratic residue modulo n.

Let $n|q-1$ and n be an odd integer. ${D}_{1}$ is a cyclic code with defining set $T=\left\{1,2,\cdots ,\frac{n-1}{2}\right\}$ . Then ${D}_{1}$ is an $\left[n,\frac{n+1}{2},\frac{n+1}{2}\right]$ MDS code. Its dual ${C}_{1}={D}_{1}^{\perp }$ is also cyclic with defining set $T\cup \left\{0\right\}$ . There are a pair of odd-like duadic codes ${D}_{1}={C}_{1}^{\perp }$ and ${D}_{2}={C}_{2}^{\perp }$ and a pair of even-like duadic codes ${C}_{2}={\mu }_{-1}\left({C}_{1}\right)$ .

Lemma 1  Let $n|q-1$ and n be an odd integer. There exists a pair of

MDS codes ${D}_{1}$ and ${D}_{2}$ with parameters $\left[n,\frac{n+1}{2},\frac{n+1}{2}\right]$ , and

${\mu }_{-1}\left({D}_{i}\right)={D}_{i+1\left(mod2\right)}$ .

Lemma 2  Let ${D}_{1}$ and ${D}_{2}$ be a pair of odd-like duadic codes of length n over ${\mathbb{F}}_{q}$ , ${\mu }_{-1}\left({D}_{i}\right)={D}_{i+1\left(\mathrm{mod}2\right)}$ . Assume that

$1+{\gamma }^{2}n=0$ (*)

has a solution in ${\mathbb{F}}_{q}$ . Let ${\stackrel{˜}{D}}_{i}=\left\{\stackrel{˜}{c}|c\in {D}_{i}\right\}$ for $1\le i\le 2$ and $\stackrel{˜}{c}=\left({c}_{0},{c}_{1},\cdots ,{c}_{n-1},{c}_{\infty }\right)$ with ${c}_{\infty }=-\gamma {\sum }_{i=0}^{n-1}\text{ }\text{ }{c}_{i}$ . Then ${\stackrel{˜}{D}}_{1}$ and ${\stackrel{˜}{D}}_{2}$ are Euclidean self-dual codes.

In  , the solution of (*) is discussed when n is an odd prime. In  , the solution of (*) is discussed when n is an odd prime power. Next, we discuss the solution of (*) for any odd integer n with $n|q-1$ .

Definition 1 (Legendre Symbol)  Let p be an prime and a be an integer.

$\left(\frac{a}{p}\right)=\left\{\begin{array}{ll}0,\hfill & \text{ }\text{if}\text{ }\text{\hspace{0.17em}}a\equiv 0\left(modp\right),\hfill \\ 1,\hfill & \text{ }\text{if}\text{ }\text{\hspace{0.17em}}a\left(\ne 0\right)\text{\hspace{0.17em}}\text{ }\text{is}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{quadratic}\text{\hspace{0.17em}}\text{residue}\text{\hspace{0.17em}}\text{modulo}\text{ }\text{\hspace{0.17em}}p,\hfill \\ -1,\hfill & \text{ }\text{if}\text{ }\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\text{ }\text{is}\text{\hspace{0.17em}}\text{not}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{quadratic}\text{\hspace{0.17em}}\text{residue}\text{\hspace{0.17em}}\text{modulo}\text{ }\text{\hspace{0.17em}}p.\hfill \end{array}$ (3)

Proposition 1 

$\left(\frac{a}{p}\right)=\left(\frac{{p}_{1}}{p}\right)\cdots \left(\frac{{p}_{s}}{p}\right),$

where $a={p}_{1}\cdots {p}_{s}$ .

Definition 2 (Jacobi Symbol)  Let $m$ and $n\left(\ne 0\right)$ be two integers.

$\left(\frac{m}{n}\right)=\left(\frac{m}{{p}_{1}}\right)\cdots \left(\frac{m}{{p}_{h}}\right),$

where $n={p}_{1}\cdots {p}_{h}$ .

We cannot obtain $m\left(\ne 0\right)$ is a quadratic residue modulo n from $\left(\frac{m}{n}\right)=1$ . But we have the next proposition.

Proposition 2 Let $m\left(\ne 0\right)$ and n be two integers and $\left(m,n\right)=1$ . If m is a quadratic residue modulo n, then

$\left(\frac{m}{n}\right)=1.$

If

$\left(\frac{m}{n}\right)=-1,$

then m is not a quadratic residue modulo n.

Proof Obviously.

Lemma 3 (Law of Quadratic Reciprocity)  Let p and r be odd primes, $\left(p,r\right)=1$ .

$\left(\frac{p}{r}\right)\left(\frac{r}{p}\right)={\left(-1\right)}^{\frac{r-1}{2}\cdot \frac{p-1}{2}}.$ (4)

Corollary 1 Let p and r be odd primes.

(1) When $p\equiv 1\left(mod4\right)$ or $r\equiv 1\left(mod4\right)$ ,

$\left(\frac{p}{r}\right)=\left(\frac{r}{p}\right).$

(2) When $p\equiv r\equiv 3\left(mod4\right)$ ,

$\left(\frac{p}{r}\right)=-\left(\frac{r}{p}\right).$

Theorem 1 Let $q={r}^{t}$ and r be an odd prime. Let $n|q-1$ and n be an odd integer. And

$n={p}_{1}^{{e}_{1}}\cdots {p}_{s}^{{e}_{s}}{p}_{s+1}^{{e}_{s+1}}\cdots {p}_{h}^{{e}_{h}},$

where

${p}_{1}\equiv \cdots \equiv {p}_{s}\equiv 3\left(mod4\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}_{s+1}\equiv \cdots \equiv {p}_{h}\equiv 1\left(mod4\right).$

(1) When $q\equiv 1\left(mod4\right)$ , there is a solution to (*) in ${\mathbb{F}}_{q}$ .

(2) Let $q\equiv 3\left(mod4\right)$ . If ${\sum }_{i=1}^{s}\text{ }{e}_{i}$ is an odd integer, there is a solution to (*) in ${\mathbb{F}}_{q}$ .

Proof (1) $q\equiv 1\left(mod4\right)$ .

(1.1) $r\equiv 3\left(mod4\right)$ . So we have that t is even. Then every quadratic equation with coefficients in ${\mathbb{F}}_{r}$ , such as Eq. (*), has a solution in ${\mathbb{F}}_{{r}^{2}}\subseteq {\mathbb{F}}_{q}$ .

(1.2) $r\equiv 1\left(mod4\right)$ and $2|t$ . The proof is similar as (1.1).

(1.3) $r\equiv 1\left(mod4\right)$ and $2\nmid t$ .

$1=\left(\frac{q}{n}\right)=\left(\frac{r}{n}\right)={\left(\frac{r}{{p}_{1}}\right)}^{{e}_{1}}\cdots {\left(\frac{r}{{p}_{h}}\right)}^{{e}_{h}}={\left(\frac{{p}_{1}}{r}\right)}^{{e}_{1}}\cdots {\left(\frac{{p}_{h}}{r}\right)}^{{e}_{h}}=\left(\frac{n}{r}\right).$

So n is a quadratic residue modulo r. And −1 is a quadratic residue modulo r. So there is a solution to (*) in ${\mathbb{F}}_{q}$ .

(2) $q\equiv 3\left(mod4\right)$ . Then $r\equiv 3\left(mod4\right)$ and t is odd.

$\begin{array}{c}1=\left(\frac{q}{n}\right)=\left(\frac{r}{n}\right)={\left(\frac{r}{{p}_{1}}\right)}^{{e}_{1}}\cdots {\left(\frac{r}{{p}_{s}}\right)}^{{e}_{s}}{\left(\frac{r}{{p}_{s+1}}\right)}^{{e}_{s+1}}\cdots {\left(\frac{r}{{p}_{h}}\right)}^{{e}_{h}}\\ ={\left(}^{-}{\left(\frac{{p}_{1}}{r}\right)}^{{e}_{1}}\cdots {\left(-1\right)}^{{e}_{s}}{\left(\frac{{p}_{s}}{r}\right)}^{{e}_{s}}{\left(\frac{{p}_{s+1}}{r}\right)}^{{e}_{s+1}}\cdots {\left(\frac{{p}_{h}}{r}\right)}^{{e}_{h}}\\ ={\left(-1\right)}^{\underset{i=1}{\overset{s}{\sum }}{e}_{i}}{\left(\frac{{p}_{1}}{r}\right)}^{{e}_{1}}\cdots {\left(\frac{{p}_{s}}{r}\right)}^{{e}_{s}}{\left(\frac{{p}_{s+1}}{r}\right)}^{{e}_{s+1}}\cdots {\left(\frac{{p}_{h}}{r}\right)}^{{e}_{h}}={\left(}^{-}\left(\frac{n}{r}\right).\end{array}$

If ${\sum }_{i=1}^{s}{e}_{i}$ is odd, n is not a quadratic residue modulo r. And −1 is not a quadratic residue modulo r. So $-n$ is a quadratic residue modulo r. There is a solution to (*) in ${\mathbb{F}}_{q}$ .

Remark In fact, $n|q-1$ , and n is an odd integer and $q\equiv 3\left(mod4\right)$ . We can easily prove that there is a solution to (*) in ${\mathbb{F}}_{q}$ if and only if ${\sum }_{i=1}^{s}{e}_{i}$ is an odd integer.

Let $n|q-1$ , $q\equiv 1\left(modn\right)$ . q is a quadratic residue modulo n. ${y}^{2}\equiv q\left(modn\right)$ . Let $q={r}^{t}$ and $q\equiv 3\left(mod4\right)$ , where r is a prime. Then $r\equiv 3\left(mod4\right)$ and t is odd. Equation (*) has solutions in ${\mathbb{F}}_{q}$ if and only if Equation (*) has solutions in ${\mathbb{F}}_{r}$ . And r is a quadratic residue modulo n.

${\left(y{r}^{-\frac{t-1}{2}}\right)}^{2}\equiv r\left(modn\right)$ . Let p be an odd prime divisor of n. r is a quadratic residue modulo p. Then $\left(\frac{r}{p}\right)=1$ . By Law of Quadratic Reciprocity, $p|n$ ,

$\left(\frac{p}{r}\right)=\left\{\begin{array}{ll}1,\hfill & p\equiv 1\left(\mathrm{mod}4\right)\hfill \\ -1,\hfill & p\equiv 3\left(\mathrm{mod}4\right)\hfill \end{array}.$

The Legendre symbol

$\begin{array}{c}\left(\frac{-n}{r}\right)=\left(\frac{-1}{r}\right){\left(\frac{{p}_{1}}{r}\right)}^{{e}_{1}}\cdots {\left(\frac{{p}_{s}}{r}\right)}^{{e}_{s}}{\left(\frac{{p}_{s+1}}{r}\right)}^{{e}_{s+1}}\cdots {\left(\frac{{p}_{h}}{r}\right)}^{{e}_{h}}\\ ={\left(-1\right)}^{1+\underset{i=1}{\overset{s}{\sum }}{e}_{i}}=\left\{\begin{array}{ll}1,\hfill & \underset{i=1}{\overset{s}{\sum }}\text{ }{e}_{i}\text{\hspace{0.17em}}\text{ }is\text{\hspace{0.17em}}\text{odd}\text{ }\hfill \\ -1,\hfill & \underset{i=1}{\overset{s}{\sum }}\text{ }{e}_{i}\text{\hspace{0.17em}}\text{ }is\text{\hspace{0.17em}}\text{even}\text{ }\hfill \end{array},\end{array}$

where $n={p}_{1}^{{e}_{1}}\cdots {p}_{s}^{{e}_{s}}{p}_{s+1}^{{e}_{s+1}}\cdots {p}_{h}^{{e}_{h}}$ , ${p}_{1}\equiv \cdots \equiv {p}_{s}\equiv 3\left(mod4\right)$ and ${p}_{s+1}\equiv \cdots \equiv {p}_{h}\equiv 1\left(mod4\right)$ .

Theorem 2 Let $q={r}^{t}$ be a prime power, $n|q-1$ and n be an odd integer. Then there exists a pair ${D}_{1}$ , ${D}_{2}$ of MDS odd-like duadic codes of length n and

${\mu }_{-1}\left({D}_{i}\right)={D}_{i+1\left(\mathrm{mod}2\right)}$ , where even-like duadic codes are MDS self-orthogonal, and ${T}_{1}=\left\{1,\cdots ,\frac{n-1}{2}\right\}$ . Furthermore,

(1) If $q={2}^{t}$ , then ${\stackrel{˜}{D}}_{i}$ are $\left[n+1,\frac{n+1}{2},\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes.

(2) If $q\equiv 1\left(mod4\right)$ , then ${\stackrel{˜}{D}}_{i}$ are $\left[n+1,\frac{n+1}{2},\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes.

(3) If $q\equiv 3\left(mod4\right)$ and ${\sum }_{i=1}^{s}{e}_{i}$ is an odd integer, then ${\stackrel{˜}{D}}_{i}$ are $\left[n+1,\frac{n+1}{2},\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes, where $n={p}_{1}^{{e}_{1}}\cdots {p}_{s}^{{e}_{s}}{p}_{s+1}^{{e}_{s+1}}\cdots {p}_{t}^{{e}_{h}}$ and ${p}_{1}\equiv \cdots \equiv {p}_{s}\equiv 3\left(mod4\right)$ , ${p}_{s+1}\equiv \cdots \equiv {p}_{h}\equiv 1\left(mod4\right)$ .

Proof Obviously, ${D}_{i}$ are $\left[n,\frac{n+1}{2},\frac{n+1}{2}\right]$ MDS odd-like duadic codes. If there is a solution to (*), we want to prove ${\stackrel{˜}{D}}_{i}$ are $\left[n+1,\frac{n+1}{2},\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes, and we only need to prove that

$c\in {D}_{i}\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}wt\left(c\right)=\frac{n+1}{2},\text{\hspace{0.17em}}\text{ }\text{then}\text{ }\text{\hspace{0.17em}}wt\left(\stackrel{˜}{c}\right)=\frac{n+1}{2}+1.$

This is equivalent to prove that ${c}_{\infty }\ne 0$ . It can be proved similarly by which proved in  .

When $q={2}^{t}$ , there is a solution to (*) in ${\mathbb{F}}_{{2}^{t}}$ , ${\stackrel{˜}{D}}_{i}$ are $\left[n+1,\frac{n+1}{2},\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes by Lemma 2.

We can obtain (2) and (3) from Theorem 1 and Lemma 2. Theorem 2 is proved.

We list some new MDS Euclidean self-dual codes in the next Table 1.

3. MDS Hermitian Self-Dual Codes

Let $n\le {q}^{2}$ . We choose n distinct elements $\left\{{\alpha }_{1},\cdots ,{\alpha }_{n}\right\}$ from ${\mathbb{F}}_{{q}^{2}}$ and n nonzero elements $\left\{{v}_{1},\cdots ,{v}_{n}\right\}$ from ${\mathbb{F}}_{{q}^{2}}$ . The generalized Reed-Solomon code

Table 1. Some new MDS Euclidean self-dual codes.

$GR{S}_{k}\left(\alpha ,v\right):=\left\{\left({v}_{1}f\left({\alpha }_{1}\right),\cdots ,{v}_{n}f\left({\alpha }_{n}\right)\right):f\left(x\right)\in {\mathbb{F}}_{{q}^{2}}\left[x\right],\mathrm{deg}f\left(x\right)\le k-1\right\}$

is a q2-ary $\left[n,k,n-k+1\right]$ MDS code, where $\alpha =\left({\alpha }_{1},\cdots ,{\alpha }_{n}\right)$ and $v=\left({v}_{1},\cdots ,{v}_{n}\right)$ .

Theorem 3 Let $n\le q$ and $2|n$ . Let $\left\{{\alpha }_{1},\cdots ,{\alpha }_{n}\right\}$ be n distinct elements from ${\mathbb{F}}_{q}\left(\subseteq {\mathbb{F}}_{{q}^{2}}\right)$ and ${u}_{i}={\prod }_{1\le j\le n,j\ne i}{\left({\alpha }_{i}-{\alpha }_{j}\right)}^{-1}$ , $1\le i\le n$ . Then there exist ${v}_{i}\in {\mathbb{F}}_{{q}^{2}}$ such that ${u}_{i}={v}_{i}^{2}$ , for $i=1,\cdots ,n$ , and the generalized Reed-Solomon code $GR{S}_{\frac{n}{2}}\left(\alpha ,v\right)$ is an $\left[n,\frac{n}{2},\frac{n}{2}+1\right]$ MDS Hermitian self-dual code over ${\mathbb{F}}_{{q}^{2}}$ , where $\alpha =\left({\alpha }_{1},\cdots ,{\alpha }_{n}\right)$ and $v=\left({v}_{1},\cdots ,{v}_{n}\right)$ .

Proof Obviously, ${u}_{i}\left(\ne 0\right)\in {\mathbb{F}}_{q}\left(\subseteq {\mathbb{F}}_{{q}^{2}}\right)$ for $1\le i\le n$ . So there exist ${v}_{i}\left(\ne 0\right)\in {\mathbb{F}}_{{q}^{2}}$ such that ${u}_{i}={v}_{i}^{2}$ for $1\le i\le n$ . The generalized Reed-Solomon

code $GR{S}_{\frac{n}{2}}\left(\alpha ,v\right)$ is an $\left[n,\frac{n}{2},\frac{n}{2}+1\right]$ MDS code over ${\mathbb{F}}_{{q}^{2}}$ . For proving the generalized Reed-Solomon code $GR{S}_{\frac{n}{2}}\left(\alpha ,v\right)$ is Hermitian self-dual over ${\mathbb{F}}_{{q}^{2}}$ , we only prove

$\left({v}_{1}{\alpha }_{1}^{l},\cdots ,{v}_{n}{\alpha }_{n}^{l}\right)\cdot \left({v}_{1}^{q}{\alpha }_{1}^{kq},\cdots ,{v}_{n}^{q}{\alpha }_{n}^{kq}\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le l,k\le \frac{n}{2}-1.$

From the choose of ${\alpha }_{i}$ , ${v}_{i}$ and [8, Corollary 2.3],

$\begin{array}{l}\left({v}_{1}{\alpha }_{1}^{l},\cdots ,{v}_{n}{\alpha }_{n}^{l}\right)\cdot \left({v}_{1}^{q}{\alpha }_{1}^{kq},\cdots ,{v}_{n}^{q}{\alpha }_{n}^{kq}\right)\\ =\left({v}_{1}{\alpha }_{1}^{l},\cdots ,{v}_{n}{\alpha }_{n}^{l}\right)\cdot \left({v}_{1}{\alpha }_{1}^{k},\cdots ,{v}_{n}{\alpha }_{n}^{k}\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le l,k\le \frac{n}{2}-1.\end{array}$

So the generalized Reed-Solomon code $GR{S}_{\frac{n}{2}}\left(\alpha ,v\right)$ is an $\left[n,\frac{n}{2},\frac{n}{2}+1\right]$ MDS Hermitian self-dual code over ${\mathbb{F}}_{{q}^{2}}$ .

Next we construct MDS Hermitian self-dual codes from constacyclic codes.

Let C be an $\left[n,k\right]$ l-constacyclic code over ${\mathbb{F}}_{{q}^{2}}$ and $\left(n,q\right)=1$ . C is considered as an ideal, $〈g\left(x\right)〉$ , of $\frac{{F}_{{q}^{2}}\left[x\right]}{{x}^{n}-\lambda }$ , where $g\left(x\right)|\left({x}^{n}-\lambda \right)$ . Simply, $C=〈g\left(x\right)〉$ .

Lemma 4  Let $\lambda \in {\mathbb{F}}_{{q}^{2}}^{*}$ , $r={\text{ord}}_{{q}^{2}}\left(\lambda \right)$ , and C be a l-constacyclic code over ${\mathbb{F}}_{{q}^{2}}$ . If C is Hermitian self-dual, then $r|q+1$ .

Lemma 5  Let $n={2}^{a}{n}^{\prime }$ $\left(a>0\right)$ and $r={2}^{b}{r}^{\prime }$ be integers such that $2\nmid {n}^{\prime }$ and $2\nmid {r}^{\prime }$ . Let q be an odd prime power such that $\left(n,q\right)=1$ and $r|q+1$ , and let $\lambda \in {\mathbb{F}}_{{q}^{2}}$ has order r. Then Hermitian self-dual l-constacyclic codes over ${\mathbb{F}}_{{q}^{2}}$ of length n exist if and only if $b>0$ and $q\overline{)\equiv }-1\left(mod{2}^{a+b}\right)$ .

Let $r={\text{ord}}_{{q}^{2}}\left(\lambda \right)$ and $r|q+1$ .

${O}_{r,n}=\left\{1+rj|j=0,1,\cdots ,n-1\right\}.$

Then ${\alpha }^{i}\left(i\in {O}_{r,n}\right)$ are all solutions of ${x}^{n}-\lambda =0$ in some extension field of ${\mathbb{F}}_{{q}^{2}}$ , where $\text{ord}\alpha =rn$ . C is called a l-constacyclic code with defining set $T\subseteq {O}_{r,n}$ , if

$C=〈g\left(x\right)〉\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left({\alpha }^{i}\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall i\in T.$

Theorem 4 Let $n={2}^{a}{n}^{\prime }\left(a>0\right)$ and $r={2}^{b}{r}^{\prime }\left(b>0\right)$ . $rn|{q}^{2}-1$ . $\lambda \in {\mathbb{F}}_{{q}^{2}}^{*}$ with $\text{ord}\lambda =r$ . $q\overline{)\equiv }-1\left(mod{2}^{a+b}\right)$ . If $rn|2\left(q+1\right)$ , there exists an MDS Hermitian self-dual code C over ${\mathbb{F}}_{{q}^{2}}$ with length n, C is a l-constacyclic code with defining set

$T=\left\{1+rj|0\le j\le \frac{n}{2}-1\right\}.$

Proof If $rn|{q}^{2}-1$ , ${C}_{{q}^{2}}\left(i\right)=\left\{i\right\}$ , for $i\in {O}_{r,n}$ , where ${C}_{{q}^{2}}\left(i\right)$ denote the q2-cyclotomic coset of $i\mathrm{mod}rn$ . And $|T|=\frac{n}{2}$ , C is an $\left[n,\frac{n}{2},\frac{n}{2}+1\right]$ MDS l-constacyclic code by the BCH bound of constacyclic code.

When $rn|2\left(q+1\right)$ , $q=\frac{rnl}{2}-1$ . Because $q\overline{)\equiv }-1\left(mod{2}^{a+b}\right)$ , l is odd.

$\left(-q\right)\left(1+rj\right)=-q-qrj\equiv 1-\frac{rnl}{2}+rj\equiv 1+r\left(\frac{n}{2}+j\right)\left(\mathrm{mod}rn\right).$

So

$\left(-q\right)T\cap T=\varnothing .$

C is MDS Hermitian self-dual by the relationship of roots of a constacyclic code and its Hermitian dual code’s roots.

Remark The MDS Hermitian self-dual constacyclic code obtained from Theorem 4 is different with the MDS Hermitian self-dual constacyclic code in  , because $\left(q+1,q-1\right)=2$ for an odd prime power q.

If $r=2$ , C is negacyclic. Theorem 4 can be stated as follow.

Corollary 2 Let $n={2}^{a}{n}^{\prime }\left(a\ge 1\right)$ and ${n}^{\prime }$ is odd. Let

$q\equiv -1\left(\mathrm{mod}{2}^{a}{n}^{″}\right)\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}q\equiv {2}^{a}-1\left(\mathrm{mod}{2}^{a+1}\right),$

where ${n}^{\prime }|{n}^{″}$ and ${n}^{″}$ is odd. Then there exists an MDS Hermitian self-dual code C of length n which is negacyclic with defining set

$T=\left\{1+2j|j=0,1,\cdots ,\frac{n}{2}-1\right\}.$

Especially, when $a=1$ , Corollary 2 is similar as [5, Theorem 11].

From Theorem 3 and Theorem 4, we obtain the next theorem.

Theorem 5 Let $n\le q+1$ and n be even. There exists an MDS Hermitian self-dual code with length n over ${\mathbb{F}}_{{q}^{2}}$ .

4. Conclusion

In this paper, we obtain many new MDS Euclidean self-dual codes by solving the Equation (*) in ${\mathbb{F}}_{q}$ . We generalize the work of  to MDS Hermitian self-dual codes, and we construct new MDS Hermitian self-dual codes from constacyclic codes. We obtain that there exists an MDS Hermitian self-dual code with length n over ${\mathbb{F}}_{{q}^{2}}$ , where $n\le q+1$ and n is even. And we also discuss these MDS Hermitian self-dual codes, which are extended cyclic duadic codes. Some new MDS Hermitian self-dual codes are obtained.

Conflicts of Interest

The authors declare no conflicts of interest.

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