Abstract

Let be a finite field with elements and a positive integer. In this paper, we use matrix method to give all primitive idempotents of irreducible cyclic codes of length , whose prime divisors divide .

1. Introduction

Let be a finite field with elements, where and is a prime. Let be a linear code over , i.e., it is a -dimensional subspace of with minimum Hamming distance . If for each codeword , is also in , then we call a cyclic code. In fact, each cyclic code of length over can be viewed as an ideal in the ring and each irreducible cyclic code of length over is an ideal of generated by a primitive idempotent.

A lot of papers investigate primitive idempotents of . We list some results about the length .(1)In [1, 2], , and , where is an odd prime and is a primitive root modulo .(2)In [3, 4], .(3)In [5], , where are distinct odd primes with and is a common primitive root modulo and .(4)In [6], , where , and are three distinct odd primes, , , and .(5)In [7, 8], , where is an odd prime different from the characteristic of , and ; , where is an odd prime and .(6)In [9, 10], , where are two distinct primes with ; and , where is an odd prime with .(7)In [11], , where are two distinct primes with .(8)In [12], , where are distinct odd primes with .

In this paper, suppose that . We shall use matrix method to give all primitive idempotents of the ring . The rest of paper is organized as follows: in Section 2, we give some basic results, in Section 3, we obtain all primitive idempotents in under the condition: , and in Section 4, we conclude this paper.

2. Preliminaries

If a positive integer has a prime factorization, , where are distinct primes and positive integers for , we denote and , , and is the order of . Through this paper, we always assume that .

Every cyclic code of length over a finite field is identified with exactly one ideal of the quotient algebra . Some explicit factorizations of can be found in [7–11, 13–16]. We need the following results about the irreducible factorization of over .

Lemma 1 ([14, Corollary 1]). Let be a finite field and a positive integer such that both and either or . Let , , and be a generator of . Then one has the following:
(1) The factorization of into irreducible factors in is (2) For each , the number of irreducible factors of degree is , where denotes the Euler Totient function, and the number of irreducible factors is

Lemma 2 ([14, Corollary 2]). Let be a finite field and a positive integer such that , , and . Let , , , , and be a generator of satisfying . Then one has the following:
(1) The factorization of into irreducible factors in is where is the set and denotes the remainder of the division of by .
(2) For each odd with , the number of irreducible polynomials of degree is , and the number irreducible polynomials of degree isThe total number of irreducible factors is

Lemma 3 (see [17]). Let be positive integers. For a set of integers , the system of congruences , has solutions if and only ifIf (7) is satisfied, the solution is unique modulo .

3. Primitive Idempotents in

In this section, we shall give all primitive idempotents in if .

First, we consider the case or .

In Lemma 1, let be all positive factors of . For each with , there are positive integers satisfying and , . Since and , is of order . Then the irreducible factorization of over can be rewritten as

Note that the number of primitive idempotents in coincides with the number of irreducible factors of over .

Theorem 4. Let and either or . Then there are primitive idempotents in as follows: corresponding to the irreducible polynomials over ,

Proof. For each , , let be a ring with direct summands; for , , , and . By (8) and Chinese Remainder Theorem, there is an -algebra isomorphism: where each is an -algebraic epimorphism and each Note that . Hence there is a -linear space isomorphism: where each is a -linear space epimorphism. Hence there is a -linear space isomorphism: where is a invertible matrix over . Now we shall determine and .
In (14), let be a matrix, where each , is a matrix and each , , is a matrix: In fact, each is determined by these rows, where , .
We know that , , , and are an irreducible polynomial of , so . Fix and , . If , , . Then and . Let be a matrix over . Then, i.e., where and are the identity matrices of order and , respectively.
Let be a matrix. Next, we shall prove that In fact, Hence we only need to show thatWe consider the following congruence equations:Suppose that . Then it has no solution in (22) by Lemma 3, so it holds in (21).
Suppose that . Then this is unique solution in (22) with . Let . Then are all solutions in (22) with . Let be a matrix over . Then for , the entry is where , , and Since is an irreducible divisor of over , ; similarly, . Hence On the other hand, by we assume that there is a prime such that . Then and by , so and . Hence . Therefore, , and it holds in (21).
In conclusion, , , and In the following, we present all primitive idempotents in by lifting some primitive idempotents in through the isomorphism .
By Lemma 1, the number of irreducible factors of , which coincides with the number of primitive idempotents in , is . Let denote the standard basis of . Hence the vectors of , , , correspond to all primitive idempotents in . Hence for , , let be a primitive idempotent in , which is corresponding to . By (14),and So we have proved the theorem.