ABSTRACT

In this paper, we construct MDS Euclidean self-dual codes which are extended 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.

Keywords:

MDS Euclidean Self-Dual Codes, MDS Hermitian Self-Dual Codes, Constacyclic Codes, Cyclic Duadic Codes, Generalized Reed-Solomon Codes

1. Introduction

Let ${\mathbb{F}}_q$ denote a finite field with q elements. An $\left[n\mathrm{,}k\mathrm{,}d\right]$ 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

$C^{\perp }:=\left\{x\in {\mathbb{F}}_q^n:{\displaystyle \underset{i=1}{\overset{n}{\sum }}}{x}_i{y}_i=0,\forall y\in C\right\}.$ (1)

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

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

If C satisfies $C=C^{\perp }$ or $C=C^{\perp H}$ , C is called Euclidean self-dual or Hermitian self-dual, respectively. In [1] [2] 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 [2] - [9] study the MDS self-dual codes. One of these problems in this topic is to determine existence of MDS self-dual codes. When $\mathrm{2\; <mo>\; |\; </mo>}q$ , Grassl and Gulliver completely solve the existence of MDS Euclidean self-dual codes in [5] . In [6] , Guenda obtain some new MDS Euclidean self-dual codes and MDS Hermitian self-dual codes. In [8] , 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, $\langle g\left(x\right)\rangle$ , of the ring $R=\frac{{\mathbb{F}}_q\left[x\right]}{x^n-1}$ , where $g\left(x\right)\mathrm{<mo>\; |\; </mo>}{x}^n-1$ 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 $ord\alpha =n$ .

Let $S_1$ and $S_2$ be unions of cyclotomic classes modulo n, such that $S_1\cap S_2=\varnothing$ and $S_1\cup S_2={\mathbb{Z}}_n\backslash \left\{0\right\}$ 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 $\mathrm{<mo>\; \{\; </mo>0\; <mo>\; \}\; </mo>}\cup S_2$ , respectively. Obviously, ${\mu }_a\left(C_i\right)=C_{i+1\left(\mathrm{mod}2\right)}$ . In [10] , A duadic code of length n over ${\mathbb{F}}_q$ exists if and only if q is a quadratic residue modulo n.

Let $n\mathrm{<mo>\; |\; </mo>}q-1$ 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 $\left[n\mathrm{,}\frac{n+1}{2}\mathrm{,}\frac{n+1}{2}\right]$ 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 [6] Let $n\mathrm{<mo>\; |\; </mo>}q-1$ and n be an odd integer. There exists a pair of

MDS codes $D_1$ and $D_2$ with parameters $\left[n\mathrm{,}\frac{n+1}{2}\mathrm{,}\frac{n+1}{2}\right]$ , and

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

Lemma 2 [11] 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 ${\tilde{D}}_i=\left\{\tilde{c}|c\in D_i\right\}$ for $1\le i\le 2$ and $\tilde{c}=\left(c_0,c_1,\cdots ,c_{n-1},c_{\infty }\right)$ with $c_{\infty }=-\gamma {\displaystyle {\sum }_{i=0}^{n-1}}{c}_i$ . Then ${\tilde{D}}_1$ and ${\tilde{D}}_2$ are Euclidean self-dual codes.

In [11] , the solution of (*) is discussed when n is an odd prime. In [5] , the solution of (*) is discussed when n is an odd prime power. Next, we discuss the solution of (*) for any odd integer n with $n\mathrm{<mo>\; |\; </mo>}q-1$ .

Definition 1 (Legendre Symbol) [12] Let p be an prime and a be an integer.

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

Proposition 1 [12]

$\left(\frac{a}{p}\right)=\left(\frac{p_1}{p}\right)\cdots \left(\frac{p_s}{p}\right),$

where $a=p_1\cdots p_s$ .

Definition 2 (Jacobi Symbol) [12] 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) [12] 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 $n\mathrm{<mo>\; |\; </mo>}q-1$ 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

$p_1\equiv \cdots \equiv p_s\equiv 3\left(mod4\right)\mathrm{,}{p}_{s+1}\equiv \cdots \equiv p_h\equiv 1\left(mod4\right)\mathrm{.}$

(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 ${\displaystyle {\sum }_{i=1}^s}{e}_i$ 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) $r\equiv \mathrm{1\; <mo\; stretchy="false">\; (\; </mo>}mod\mathrm{4\; <mo\; stretchy="false">\; )\; </mo>}$ and $\mathrm{2\; <mo>\; |\; </mo>}t$ . The proof is similar as (1.1).

(1.3) $r\equiv \mathrm{1\; <mo\; stretchy="false">\; (\; </mo>}mod\mathrm{4\; <mo\; stretchy="false">\; )\; </mo>}$ 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.

$\begin{array}{c}1=\left(\frac{q}{n}\right)=\left(\frac{r}{n}\right)={\left(\frac{r}{p_1}\right)}^{e_1}\cdots {\left(\frac{r}{p_s}\right)}^{e_s}{\left(\frac{r}{p_{s+1}}\right)}^{e_{s+1}}\cdots {\left(\frac{r}{p_h}\right)}^{e_h}\\ ={(}^{-}{\left(\frac{p_1}{r}\right)}^{e_1}\cdots {\left(-1\right)}^{e_s}{\left(\frac{p_s}{r}\right)}^{e_s}{\left(\frac{p_{s+1}}{r}\right)}^{e_{s+1}}\cdots {\left(\frac{p_h}{r}\right)}^{e_h}\\ ={\left(-1\right)}^{{\displaystyle \underset{i=1}{\overset{s}{\sum }}}{e}_i}{\left(\frac{p_1}{r}\right)}^{e_1}\cdots {\left(\frac{p_s}{r}\right)}^{e_s}{\left(\frac{p_{s+1}}{r}\right)}^{e_{s+1}}\cdots {\left(\frac{p_h}{r}\right)}^{e_h}={(}^{-}\left(\frac{n}{r}\right).\end{array}$

If ${\displaystyle {\sum }_{i=1}^s}{e}_i$ 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, $n\mathrm{<mo>\; |\; </mo>}q-1$ , 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 ${\displaystyle {\sum }_{i=1}^s}{e}_i$ is an odd integer.

Let $n\mathrm{<mo>\; |\; </mo>}q-1$ , $q\equiv 1\left(modn\right)$ . 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 $r\equiv \mathrm{3\; <mo\; stretchy="false">\; (\; </mo>}mod\mathrm{4\; <mo\; stretchy="false">\; )\; </mo>}$ 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, $p\mathrm{<mo>\; |\; </mo>}n$ ,

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

The Legendre symbol

$\begin{array}{c}\left(\frac{-n}{r}\right)=\left(\frac{-1}{r}\right){\left(\frac{p_1}{r}\right)}^{e_1}\cdots {\left(\frac{p_s}{r}\right)}^{e_s}{\left(\frac{p_{s+1}}{r}\right)}^{e_{s+1}}\cdots {\left(\frac{p_h}{r}\right)}^{e_h}\\ ={\left(-1\right)}^{1+{\displaystyle \underset{i=1}{\overset{s}{\sum }}}{e}_i}=\{\begin{array}{ll}1,\hfill & {\displaystyle \underset{i=1}{\overset{s}{\sum }}}{e}_{i}is\text{odd}\hfill \\ -1,\hfill & {\displaystyle \underset{i=1}{\overset{s}{\sum }}}{e}_{i}is\text{even}\hfill \end{array},\end{array}$

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, $n\mathrm{<mo>\; |\; </mo>}q-1$ 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 ${\tilde{D}}_i$ are $\left[n+\mathrm{1,}\frac{n+1}{2}\mathrm{,}\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes.

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

(3) If $q\equiv 3\left(mod4\right)$ and ${\displaystyle {\sum }_{i=1}^s}{e}_i$ is an odd integer, then ${\tilde{D}}_i$ are $\left[n+\mathrm{1,}\frac{n+1}{2}\mathrm{,}\frac{n+3}{2}\right]$ 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 $\left[n\mathrm{,}\frac{n+1}{2}\mathrm{,}\frac{n+1}{2}\right]$ MDS odd-like duadic codes. If there is a solution to (*), we want to prove ${\tilde{D}}_i$ are $\left[n+\mathrm{1,}\frac{n+1}{2}\mathrm{,}\frac{n+3}{2}\right]$ MDS Euclidean self-dual codes, and we only need to prove that

$c\in D_i\text{and}wt\left(c\right)=\frac{n+1}{2},\text{then}wt\left(\tilde{c}\right)=\frac{n+1}{2}+1.$

This is equivalent to prove that $c_{\infty }\ne 0$ . It can be proved similarly by which proved in [5] .

When $q=2^t$ , there is a solution to (*) in ${\mathbb{F}}_{2^t}$ , ${\tilde{D}}_i$ are $\left[n+\mathrm{1,}\frac{n+1}{2}\mathrm{,}\frac{n+3}{2}\right]$ 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 $\left\{{\alpha }_1\mathrm{,}\cdots \mathrm{,}{\alpha }_n\right\}$ from ${\mathbb{F}}_{q^2}$ and n nonzero elements $\left\{v_1\mathrm{,}\cdots \mathrm{,}{v}_n\right\}$ from ${\mathbb{F}}_{q^2}$ . The generalized Reed-Solomon code

$GR{S}_k\left(\alpha ,v\right):=\left\{\left(v_{1}f\left({\alpha }_1\right),\cdots ,v_{n}f\left({\alpha }_n\right)\right):f\left(x\right)\in {\mathbb{F}}_{q^2}\left[x\right],\mathrm{deg}f\left(x\right)\le k-1\right\}$

is a q²-ary $\left[n\mathrm{,}k\mathrm{,}n-k+1\right]$ MDS code, where $\alpha =\left({\alpha }_1\mathrm{,}\cdots \mathrm{,}{\alpha }_n\right)$ and $v=\left(v_1,\cdots ,v_n\right)$ .

Theorem 3 Let $n\le q$ and $\mathrm{2\; <mo>\; |\; </mo>}n$ . Let $\left\{{\alpha }_1\mathrm{,}\cdots \mathrm{,}{\alpha }_n\right\}$ be n distinct elements from ${\mathbb{F}}_q\left(\subseteq {\mathbb{F}}_{q^2}\right)$ and $u_i={\displaystyle {\prod }_{1\le j\le n,j\ne i}}{\left({\alpha }_i-{\alpha }_j\right)}^{-1}$ , $1\le i\le n$ . 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 $GR{S}_{\frac{n}{2}}\left(\alpha \mathrm{,}v\right)$ is an $\left[n\mathrm{,}\frac{n}{2}\mathrm{,}\frac{n}{2}+1\right]$ 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 $GR{S}_{\frac{n}{2}}\left(\alpha \mathrm{,}v\right)$ is an $\left[n\mathrm{,}\frac{n}{2}\mathrm{,}\frac{n}{2}+1\right]$ MDS code over ${\mathbb{F}}_{q^2}$ . For proving the generalized Reed-Solomon code $GR{S}_{\frac{n}{2}}\left(\alpha \mathrm{,}v\right)$ is Hermitian self-dual over ${\mathbb{F}}_{q^2}$ , we only prove

$\left(v_1{\alpha }_1^l\mathrm{,}\cdots \mathrm{,}{v}_n{\alpha }_n^l\right)\cdot \left(v_1^q{\alpha }_1^{kq}\mathrm{,}\cdots \mathrm{,}{v}_n^q{\alpha }_n^{kq}\right)=\mathrm{0,}0\le l\mathrm{,}k\le \frac{n}{2}-1.$

From the choose of ${\alpha }_i$ , $v_i$ and [8, Corollary 2.3],

$\begin{array}{l}\left(v_1{\alpha }_1^l\mathrm{,}\cdots \mathrm{,}{v}_n{\alpha }_n^l\right)\cdot \left(v_1^q{\alpha }_1^{kq}\mathrm{,}\cdots \mathrm{,}{v}_n^q{\alpha }_n^{kq}\right)\\ =\left(v_1{\alpha }_1^l\mathrm{,}\cdots \mathrm{,}{v}_n{\alpha }_n^l\right)\cdot \left(v_1{\alpha }_1^k\mathrm{,}\cdots \mathrm{,}{v}_n{\alpha }_n^k\right)=\mathrm{0,}0\le l\mathrm{,}k\le \frac{n}{2}-1.\end{array}$

So the generalized Reed-Solomon code $GR{S}_{\frac{n}{2}}\left(\alpha \mathrm{,}v\right)$ is an $\left[n\mathrm{,}\frac{n}{2}\mathrm{,}\frac{n}{2}+1\right]$ 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 $\left[n\mathrm{,}k\right]$ l-constacyclic code over ${\mathbb{F}}_{q^2}$ and $\left(n,q\right)=1$ . C is considered as an ideal, $\langle g\left(x\right)\rangle$ , of $\frac{F_{q^2}\left[x\right]}{x^n-\lambda }$ , where $g\left(x\right)\mathrm{<mo>\; |\; </mo>}\left(x^n-\lambda \right)$ . Simply, $C=\langle g\left(x\right)\rangle$ .

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

Lemma 5 [2] 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 $r\mathrm{<mo>\; |\; </mo>}q+1$ , 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 $q\cancel{\equiv }-1\left(mod{2}^{a+b}\right)$ .

Let $r={\text{ord}}_{q^2}\left(\lambda \right)$ and $r\mathrm{<mo>\; |\; </mo>}q+1$ .

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

Then ${\alpha }^i\left(i\in O_{r\mathrm{,}n}\right)$ 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 $T\subseteq O_{r\mathrm{,}n}$ , if

$C=\langle g\left(x\right)\rangle \text{and}g\left({\alpha }^i\right)=0,\forall i\in T.$

Theorem 4 Let $n=2^a{n}^{\prime }\left(a>0\right)$ and $r=2^b{r}^{\prime }\left(b>0\right)$ . $rn\mathrm{<mo>\; |\; </mo>}{q}^2-1$ . $\lambda \in {\mathbb{F}}_{q^2}^{\mathrm{<mo>\; *\; </mo>}}$ with $\text{ord}\lambda =r$ . $q\cancel{\equiv }-1\left(mod{2}^{a+b}\right)$ . If $rn\mathrm{<mo>\; |\; </mo>2}\left(q+1\right)$ , 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

$T\mathrm{<mo>\; =\; </mo>}\left\{1+rj\mathrm{<mo>\; |\; </mo>0}\le j\le \frac{n}{2}-1\right\}\mathrm{.}$

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

When $rn\mathrm{<mo>\; |\; </mo>2}\left(q+1\right)$ , $q=\frac{rnl}{2}-1$ . Because $q\cancel{\equiv }-1\left(mod{2}^{a+b}\right)$ , 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

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

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 [12] , 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

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

where $n^{\prime }\mathrm{<mo>\; |\; </mo>}{n}^{″}$ 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 [8] 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.

Cite this paper

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

References