Abstract

Necessary and sufficient conditions for the existence of Hermitian self-orthogonal constacyclic codes of length over a finite field , coprime to , are found. The defining sets and corresponding generator polynomials of these codes are also characterised. A formula for the number of Hermitian self-orthogonal constacyclic codes of length over a finite field is obtained. Conditions for the existence of numerous MDS Hermitian self-orthogonal constacyclic codes are obtained. The defining set and the number of such MDS codes are also found.

1. Introduction

Let denote a finite field with elements. An linear code of length and dimension over is a -dimensional subspace of the vector space . Elements of the subspace are called codewords and are written as row vectors . A linear code over is called -constacyclic if is in for every in . Let be the map given by . One can easily check that is an -module isomorphism. We can therefore identify -constacyclic codes of length over with ideals in . The Hamming weight of is the number of nonzero coordinates of . The minimum distance of is defined to be . An code, that is, a linear code with minimum distance , is said to be maximum distance separable (MDS) if . The Hermitian inner product of elements is defined as , for and . For a linear code of length over , the Hermitian dual code of is defined by . If , then is known as Hermitian self-dual and is Hermitian self-orthogonal if .

Aydin et al. [1] dealt with constacyclic codes and a constacyclic BCH bound was given. Gulliver et al. [2] showed that there exists Euclidean self-dual MDS code of length over when 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. Blackford [3] studied negacyclic codes over finite fields by using multipliers. He gave conditions on the existence of Euclidean self-dual codes. Recently, Guenda [4] 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 [5] the authors gave formulae to enumerate the number of Euclidean self-dual and self-orthogonal negacyclic codes of length over a finite field , where is coprime to . In [6] Yang and Cai gave the necessary and sufficient conditions for the existence of Hermitian self-dual constacyclic codes. They also gave some conditions under which Hermitian self-dual and self-orthogonal MDS codes exist. In this paper, we find necessary and sufficient conditions for the existence of Hermitian self-orthogonal constacyclic codes of length over a finite field , coprime to , and also give a characterization of their defining sets. We obtain a formula to calculate the number of these codes. We give conditions for the existence of some MDS Hermitian self-orthogonal constacyclic codes. We also found their number and defining sets (Table 1).

2. Hermitian Self-Orthogonal Constacyclic Codes

Let be an odd prime power and a positive integer relatively prime to . Let be an   -constacyclic code over with , where denotes the order of in . Let be the generator polynomial of . Then divides . Write . The polynomial is called the check polynomial of . For , let be the -cyclotomic coset modulo containing , where is the least positive integer such that . Let be a primitive th root of unity in some extension field of such that . Then the polynomial is the minimal polynomial of over and where is the set of representatives of all the distinct -cyclotomic cosets modulo . As , one can check that the roots of are precisely , . Define Hence we have where .

Let be a -constacyclic code with defining set is a root of . Clearly is a union of some -cyclotomic cosets mod for . The Hermitian dual of the code is a -constacyclic code over with defining set mod (see Theorem 3.2 of [6]). Write the generator polynomial of the code as , where Then the generator polynomial of the Hermitian dual of is , where It can be easily verified that so that .

Lemma 1. Let be a -constacyclic code over with . If is a Hermitian self-orthogonal code, then .

Proof. The proof is similar to [6, Proposition 2.3].

Theorem 2. Nontrivial Hermitian self-orthogonal -constacyclic codes of length over exist if and only if for some .

Proof. Let be a nontrivial Hermitian self-orthogonal -constacyclic code of length over with defining set . Then there exists such that and . Hence giving us that . Thus, (as and ). Conversely, let be such that . Consider . The code is a nontrivial Hermitian self-orthogonal code since .

Define . The following theorem characterizes the defining set of a Hermitian self-orthogonal constacyclic code of length over .

Theorem 3. Let be a -constacyclic code of length over with the defining set . Then is Hermitian self-orthogonal if and only if (i) and (ii) for each , at least one of and belongs to .

Proof. Let be a Hermitian self-orthogonal constacyclic code. Let . Then . Suppose that . Then so that and , which contradicts the hypothesis that . Thus, . Now, let . Then either or , as required.
Conversely, let the defining set be such that and for each , at least one of and is in . Then , by condition (ii) so that the code having as a defining set is a Hermitian self-orthogonal code.

Corollary 4. A -constacyclic code of length over generated by is a Hermitian self-orthogonal if and only if for all .

Define and . Observe that .

Example 5. Let , , and ; then . We consider the -constacyclic code of length 15 over , where with order 7. Clearly . Let ; then . Hence, the code with defining set is a Hermitian self-orthogonal -constacyclic code.

Theorem 6. The number of Hermitian self-orthogonal -constacyclic codes of length over is .

Proof. Let be a Hermitian self-orthogonal -constacyclic code of length over generated by . Then . For , . However for , the pairs have three choices , , and . Hence, the number of Hermitian self-orthogonal -constacyclic codes of length over is .

In order to find the number of Hermitian self-orthogonal -constacyclic codes of length over , we need to compute the value of . Our aim is to prove the following.

Theorem 7. Let , , and be positive integers such that . Then the number of solutions for the linear congruence in the set and is exactly .

Since , the linear congruence has a unique solution modulo . Let it be . Then . We write . The solutions of (7) will be amongst . Clearly, the elements of are relatively prime to . We need to count the number of elements of which are coprime to . Also, . Therefore, the required number .

Lemma 8. Let be a prime divisor of such that . The number of multiples of in is .

Proof. Write , where for each , . Since each contains elements which are pairwise incongruent mod , each forms a complete residue system mod . Hence exactly one element in each is divisible by . Consequently, there exist elements in which are divisible by .

Lemma 9. Let and be two distinct prime divisors of with . Then the number of multiples of or in is .

Proof. The number of multiples of in equals , while the number of multiples of in is . By a similar argument as in Lemma 8, the number of multiples of both and equals . Therefore the required number is .

Theorem 10. Let be all the distinct prime divisors of which are relatively prime to . The number of elements in which are not coprime to is

Proof. The proof follows by induction on Lemmas 8 and 9.

In order to prove Theorem 7, it is enough to show that .

Now,

Let be all the distinct prime divisors of . Also, let be all the distinct prime divisors of . Then , which completes the proof of Theorem 7.

Pick . Let . Define .

Theorem 11. Consider where .

Proof. As , , so that . Since is a divisor of , we have that . Thus , whenever is not coprime to .
As , we write . Also . As , holds giving us that holds with and (). By Theorem 7, there is number of elements in satisfying , , and . However, we have to calculate the number of such in . Now, for , , Consequently, , whenever .

Define Observe that, for , holds if and only if , where , as defined earlier. Recall that, for , .

Theorem 12. One has

Proof. The proof follows from Theorem 11 and the above definition.

Example 13 (let , , and ). Then , , , , and . Take . Thus (as and so that . By Theorem 12, are possible values of on right hand side. Now, as . However, as there does not exist any odd integer such that , so that

We will now investigate the behavior of the function .

Lemma 14. Let and be two integers coprime to such that and for some odd integers and . If , where and , then there exists an odd integer such that .

Proof. Write , being odd distinct primes, . Let be an odd prime divisor of , so that there exists an odd integer such that . Therefore, . Consequently, but , giving us that . For we have , since for some positive integer . Hence , where is odd. Thus, where is odd.

Lemma 15. Let . Then holds for some integer if and only if . In fact, such a is odd.

Proof. Proof is trivial.

Theorem 16. if and only if and for all odd prime divisors of , where and .

Proof. If , then there exists an odd integer such that . Thus so that, by Lemma 15, . Also for every odd prime divisor of . Thus, showing that and . Therefore for all odd prime divisors of .
Conversely, let and for all odd prime divisors of . To prove , we need to find an odd integer such that . For any odd prime divisor of , as , , where is odd. As in the proof of Lemma 14, there exists an odd integer such that , where . Also, . Therefore, with odd. By Lemma 1, . Using Lemma 14, we get that , for some odd integer .