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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.

Let F q denote a finite field with q elements. An [ n , k , d ] linear code C over F q is a k-dimensional subspace of F q n . These parameters n, k and d satisfy d ≤ 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 ⊥ of C is defined as

C ⊥ : = { x ∈ F q n : ∑ i = 1 n x i y i = 0 , ∀ y ∈ C } . (1)

If q = r 2 , the Hermitian dual code C ⊥ H of C is defined as

C ⊥ H : = { x ∈ F r 2 n : ∑ i = 1 n x i y i r = 0 , ∀ y ∈ C } . (2)

If C satisfies C = C ⊥ or C = C ⊥ H , C is called Euclidean self-dual or Hermitian self-dual, respectively. In [

In this paper, we obtain some new Euclidean self-dual codes by studying the solution of an equation in 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.

A cyclic code C of length n over F q can be considered as an ideal, 〈 g ( x ) 〉 , of the ring R = F q [ x ] x n − 1 , where g ( x ) | x n − 1 and ( n , q ) = 1 . The set T = { 0 ≤ i ≤ n − 1 | g ( α i ) = 0 } is called the defining set of C, where o r d α = n .

Let S 1 and S 2 be unions of cyclotomic classes modulo n, such that S 1 ∩ S 2 = ∅ and S 1 ∪ S 2 = ℤ n \ { 0 } and a S i ( mod n ) = S i + 1 ( mod 2 ) . Then the triple μ a , S 1 and S 2 is called a splitting modulo n. Odd-like codes D 1 and D 2 are cyclic codes over F q with defining sets S 1 and S 2 , respectively. D 1 and D 2 can be denoted by μ a ( D i ) = D i + 1 ( mod 2 ) . Even-like duadic codes C 1 and C 2 are cyclic codes over F q with defining sets { 0 } ∪ S 1 and { 0 } ∪ S 2 , respectively. Obviously, μ a ( C i ) = C i + 1 ( mod 2 ) . In [

Let n | q − 1 and n be an odd integer. D 1 is a cyclic code with defining set T = { 1 , 2 , ⋯ , n − 1 2 } . Then D 1 is an [ n , n + 1 2 , n + 1 2 ] MDS code. Its dual C 1 = D 1 ⊥ is also cyclic with defining set T ∪ { 0 } . There are a pair of odd-like duadic codes D 1 = C 1 ⊥ and D 2 = C 2 ⊥ and a pair of even-like duadic codes C 2 = μ − 1 ( C 1 ) .

Lemma 1 [

MDS codes D 1 and D 2 with parameters [ n , n + 1 2 , n + 1 2 ] , and

μ − 1 ( D i ) = D i + 1 ( m o d 2 ) .

Lemma 2 [

1 + γ 2 n = 0 (*)

has a solution in F q . Let D ˜ i = { c ˜ | c ∈ D i } for 1 ≤ i ≤ 2 and c ˜ = ( c 0 , c 1 , ⋯ , c n − 1 , c ∞ ) with c ∞ = − γ ∑ i = 0 n − 1 c i . Then D ˜ 1 and D ˜ 2 are Euclidean self-dual codes.

In [

Definition 1 (Legendre Symbol) [

( a p ) = { 0, if a ≡ 0 ( m o d p ) , 1, if a ( ≠ 0 ) is a quadratic residue modulo p , − 1, if a is not a quadratic residue modulo p . (3)

Proposition 1 [

( a p ) = ( p 1 p ) ⋯ ( p s p ) ,

where a = p 1 ⋯ p s .

Definition 2 (Jacobi Symbol) [

( m n ) = ( m p 1 ) ⋯ ( m p h ) ,

where n = p 1 ⋯ p h .

We cannot obtain m ( ≠ 0 ) is a quadratic residue modulo n from ( m n ) = 1 . But we have the next proposition.

Proposition 2 Let m ( ≠ 0 ) and n be two integers and ( m , n ) = 1 . If m is a quadratic residue modulo n, then

( m n ) = 1.

If

( m n ) = − 1 ,

then m is not a quadratic residue modulo n.

Proof Obviously.

Lemma 3 (Law of Quadratic Reciprocity) [

( p r ) ( r p ) = ( − 1 ) r − 1 2 ⋅ p − 1 2 . (4)

Corollary 1 Let p and r be odd primes.

(1) When p ≡ 1 ( m o d 4 ) or r ≡ 1 ( m o d 4 ) ,

( p r ) = ( r p ) .

(2) When p ≡ r ≡ 3 ( m o d 4 ) ,

( p r ) = − ( r p ) .

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 ⋯ p s e s p s + 1 e s + 1 ⋯ p h e h ,

where

p 1 ≡ ⋯ ≡ p s ≡ 3 ( m o d 4 ) , p s + 1 ≡ ⋯ ≡ p h ≡ 1 ( m o d 4 ) .

(1) When q ≡ 1 ( m o d 4 ) , there is a solution to (*) in F q .

(2) Let q ≡ 3 ( m o d 4 ) . If ∑ i = 1 s e i is an odd integer, there is a solution to (*) in F q .

Proof (1) q ≡ 1 ( m o d 4 ) .

(1.1) r ≡ 3 ( m o d 4 ) . So we have that t is even. Then every quadratic equation with coefficients in F r , such as Eq. (*), has a solution in F r 2 ⊆ F q .

(1.2) r ≡ 1 ( m o d 4 ) and 2 | t . The proof is similar as (1.1).

(1.3) r ≡ 1 ( m o d 4 ) and 2 ∤ t .

1 = ( q n ) = ( r n ) = ( r p 1 ) e 1 ⋯ ( r p h ) e h = ( p 1 r ) e 1 ⋯ ( p h r ) e h = ( n r ) .

So n is a quadratic residue modulo r. And −1 is a quadratic residue modulo r. So there is a solution to (*) in F q .

(2) q ≡ 3 ( m o d 4 ) . Then r ≡ 3 ( m o d 4 ) and t is odd.

1 = ( q n ) = ( r n ) = ( r p 1 ) e 1 ⋯ ( r p s ) e s ( r p s + 1 ) e s + 1 ⋯ ( r p h ) e h = ( − 1 ) e 1 ( p 1 r ) e 1 ⋯ ( − 1 ) e s ( p s r ) e s ( p s + 1 r ) e s + 1 ⋯ ( p h r ) e h = ( − 1 ) ∑ i = 1 s e i ( p 1 r ) e 1 ⋯ ( p s r ) e s ( p s + 1 r ) e s + 1 ⋯ ( p h r ) e h = ( − 1 ) ∑ i = 1 s e i ( n r ) .

If ∑ 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 F q .

Remark In fact, n | q − 1 , and n is an odd integer and q ≡ 3 ( m o d 4 ) . We can easily prove that there is a solution to (*) in F q if and only if ∑ i = 1 s e i is an odd integer.

Let n | q − 1 , q ≡ 1 ( m o d n ) . q is a quadratic residue modulo n. y 2 ≡ q ( m o d n ) . Let q = r t and q ≡ 3 ( m o d 4 ) , where r is a prime. Then r ≡ 3 ( m o d 4 ) and t is odd. Equation (*) has solutions in F q if and only if Equation (*) has solutions in F r . And r is a quadratic residue modulo n.

( y r − t − 1 2 ) 2 ≡ r ( m o d n ) . Let p be an odd prime divisor of n. r is a quadratic residue modulo p. Then ( r p ) = 1 . By Law of Quadratic Reciprocity, p | n ,

( p r ) = { 1 , p ≡ 1 ( mod 4 ) − 1 , p ≡ 3 ( mod 4 ) .

The Legendre symbol

( − n r ) = ( − 1 r ) ( p 1 r ) e 1 ⋯ ( p s r ) e s ( p s + 1 r ) e s + 1 ⋯ ( p h r ) e h = ( − 1 ) 1 + ∑ i = 1 s e i = { 1 , ∑ i = 1 s e i i s odd − 1 , ∑ i = 1 s e i i s even ,

where n = p 1 e 1 ⋯ p s e s p s + 1 e s + 1 ⋯ p h e h , p 1 ≡ ⋯ ≡ p s ≡ 3 ( m o d 4 ) and p s + 1 ≡ ⋯ ≡ p h ≡ 1 ( m o d 4 ) .

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

μ − 1 ( D i ) = D i + 1 ( mod 2 ) , where even-like duadic codes are MDS self-orthogonal, and T 1 = { 1 , ⋯ , n − 1 2 } . Furthermore,

(1) If q = 2 t , then D ˜ i are [ n + 1, n + 1 2 , n + 3 2 ] MDS Euclidean self-dual codes.

(2) If q ≡ 1 ( m o d 4 ) , then D ˜ i are [ n + 1, n + 1 2 , n + 3 2 ] MDS Euclidean self-dual codes.

(3) If q ≡ 3 ( m o d 4 ) and ∑ i = 1 s e i is an odd integer, then D ˜ i are [ n + 1, n + 1 2 , n + 3 2 ] MDS Euclidean self-dual codes, where n = p 1 e 1 ⋯ p s e s p s + 1 e s + 1 ⋯ p t e h and p 1 ≡ ⋯ ≡ p s ≡ 3 ( m o d 4 ) , p s + 1 ≡ ⋯ ≡ p h ≡ 1 ( m o d 4 ) .

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

c ∈ D i and w t ( c ) = n + 1 2 , then w t ( c ˜ ) = n + 1 2 + 1.

This is equivalent to prove that c ∞ ≠ 0 . It can be proved similarly by which proved in [

When q = 2 t , there is a solution to (*) in F 2 t , D ˜ i are [ n + 1, n + 1 2 , n + 3 2 ] 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

Let n ≤ q 2 . We choose n distinct elements { α 1 , ⋯ , α n } from F q 2 and n nonzero elements { v 1 , ⋯ , v n } from F q 2 . The generalized Reed-Solomon code

n | q |
---|---|

4 | 2^{2}, 7 |

6 | 2^{4}, 3^{4} |

8 | 2^{3}, 3^{6} |

10 | 2^{6}, 5^{6} |

12 | 3^{5} |

14 | 2^{12}, 3^{6} |

16 | 31, 31^{2}, 31^{3} |

18 | 3^{16} |

20 | 5^{9} |

22 | 5^{6} |

24 | 3^{11} |

26 | 7^{4} |

28 | 7^{9} |

30 | 59 |

156 | 5^{4} |

G R S k ( α , v ) : = { ( v 1 f ( α 1 ) , ⋯ , v n f ( α n ) ) : f ( x ) ∈ F q 2 [ x ] , deg f ( x ) ≤ k − 1 }

is a q^{2}-ary [ n , k , n − k + 1 ] MDS code, where α = ( α 1 , ⋯ , α n ) and v = ( v 1 , ⋯ , v n ) .

Theorem 3 Let n ≤ q and 2 | n . Let { α 1 , ⋯ , α n } be n distinct elements from F q ( ⊆ F q 2 ) and u i = ∏ 1 ≤ j ≤ n , j ≠ i ( α i − α j ) − 1 , 1 ≤ i ≤ n . Then there exist v i ∈ F q 2 such that u i = v i 2 , for i = 1 , ⋯ , n , and the generalized Reed-Solomon code G R S n 2 ( α , v ) is an [ n , n 2 , n 2 + 1 ] MDS Hermitian self-dual code over F q 2 , where α = ( α 1 , ⋯ , α n ) and v = ( v 1 , ⋯ , v n ) .

Proof Obviously, u i ( ≠ 0 ) ∈ F q ( ⊆ F q 2 ) for 1 ≤ i ≤ n . So there exist v i ( ≠ 0 ) ∈ F q 2 such that u i = v i 2 for 1 ≤ i ≤ n . The generalized Reed-Solomon

code G R S n 2 ( α , v ) is an [ n , n 2 , n 2 + 1 ] MDS code over F q 2 . For proving the generalized Reed-Solomon code G R S n 2 ( α , v ) is Hermitian self-dual over F q 2 , we only prove

( v 1 α 1 l , ⋯ , v n α n l ) ⋅ ( v 1 q α 1 k q , ⋯ , v n q α n k q ) = 0, 0 ≤ l , k ≤ n 2 − 1.

From the choose of α i , v i and [8, Corollary 2.3],

( v 1 α 1 l , ⋯ , v n α n l ) ⋅ ( v 1 q α 1 k q , ⋯ , v n q α n k q ) = ( v 1 α 1 l , ⋯ , v n α n l ) ⋅ ( v 1 α 1 k , ⋯ , v n α n k ) = 0, 0 ≤ l , k ≤ n 2 − 1.

So the generalized Reed-Solomon code G R S n 2 ( α , v ) is an [ n , n 2 , n 2 + 1 ] MDS Hermitian self-dual code over F q 2 .

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

Let C be an [ n , k ] l-constacyclic code over F q 2 and ( n , q ) = 1 . C is considered as an ideal, 〈 g ( x ) 〉 , of F q 2 [ x ] x n − λ , where g ( x ) | ( x n − λ ) . Simply, C = 〈 g ( x ) 〉 .

Lemma 4 [

Lemma 5 [

Let r = ord q 2 ( λ ) and r | q + 1 .

O r , n = { 1 + r j | j = 0 , 1 , ⋯ , n − 1 } .

Then α i ( i ∈ O r , n ) are all solutions of x n − λ = 0 in some extension field of F q 2 , where ord α = r n . C is called a l-constacyclic code with defining set T ⊆ O r , n , if

C = 〈 g ( x ) 〉 and g ( α i ) = 0 , ∀ i ∈ T .

Theorem 4 Let n = 2 a n ′ ( a > 0 ) and r = 2 b r ′ ( b > 0 ) . r n | q 2 − 1 . λ ∈ F q 2 * with ord λ = r . q ≡ − 1 ( m o d 2 a + b ) . If r n | 2 ( q + 1 ) , there exists an MDS Hermitian self-dual code C over F q 2 with length n, C is a l-constacyclic code with defining set

T = { 1 + r j | 0 ≤ j ≤ n 2 − 1 } .

Proof If r n | q 2 − 1 , C q 2 ( i ) = { i } , for i ∈ O r , n , where C q 2 ( i ) denote the q^{2}-cyclotomic coset of i mod r n . And | T | = n 2 , C is an [ n , n 2 , n 2 + 1 ] MDS l-constacyclic code by the BCH bound of constacyclic code.

When r n | 2 ( q + 1 ) , q = r n l 2 − 1 . Because q ≡ − 1 ( m o d 2 a + b ) , l is odd.

( − q ) ( 1 + r j ) = − q − q r j ≡ 1 − r n l 2 + r j ≡ 1 + r ( n 2 + j ) ( mod r n ) .

So

( − q ) T ∩ T = ∅ .

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 [

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

Corollary 2 Let n = 2 a n ′ ( a ≥ 1 ) and n ′ is odd. Let

q ≡ − 1 ( mod 2 a n ″ ) and q ≡ 2 a − 1 ( mod 2 a + 1 ) ,

where n ′ | 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 = { 1 + 2 j | j = 0 , 1 , ⋯ , n 2 − 1 } .

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 ≤ q + 1 and n be even. There exists an MDS Hermitian self-dual code with length n over F q 2 .

In this paper, we obtain many new MDS Euclidean self-dual codes by solving the Equation (*) in F q . We generalize the work of [

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