Repeated-root constacyclic codes of length 3lps and their dual codes

Finite Fields and Their Applications 42 (2016) 269–295

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Finite Fields and Their Applications www.elsevier.com/locate/ffa

Repeated-root constacyclic codes of length 3lps and their dual codes ✩ Li Liu, Lanqiang Li ∗ , Xiaoshan Kai, Shixin Zhu School of Mathematics, Hefei University of Technology, Hefei 230009, Anhui, PR China

a r t i c l e

i n f o

Article history: Received 24 January 2016 Received in revised form 14 June 2016 Accepted 25 August 2016 Available online xxxx Communicated by W. Cary Huffman MSC: 94B05 11T71 Keywords: Repeated-root constacyclic codes Cyclic codes Dual codes Generator polynomial Hamming distance

a b s t r a c t Let p = 3 be any prime and l = 3 be any odd prime with gcd(p, l) = 1. The multiplicative group Fq∗ = ξ can be decomposed into mutually disjoint union of gcd(q − 1, 3lps ) s cosets over the subgroup ξ 3lp , where ξ is a primitive (q − 1)th root of unity. We classify all repeated-root constacyclic codes of length 3lps over the finite field Fq into some equivalence classes by this decomposition, where q = pm , s and m are positive integers. According to these equivalence classes, we explicitly determine the generator polynomials of all repeated-root constacyclic codes of length 3lps over Fq and their dual codes. Self-dual cyclic codes of length 3lps over Fq exist only when p = 2. We give all self-dual cyclic codes of length 3 · 2s l over F2m and their enumeration. We also determine the minimum Hamming distance of these codes when gcd(3, q − 1) = 1 and l = 1. © 2016 Elsevier Inc. All rights reserved.

✩ This research is supported by the National Natural Science Foundation of China (No. 61572168, No. 11401154) and the Anhui Provincial Natural Science Foundation (No. 158085MA13). * Corresponding author. E-mail addresses: [email protected] (L. Liu), [email protected] (L. Li).

http://dx.doi.org/10.1016/j.ffa.2016.08.005 1071-5797/© 2016 Elsevier Inc. All rights reserved.

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1. Introduction Constacyclic codes over finite fields play a very important role in the theory of errorcorrecting codes. More importantly, constacyclic codes have practical applications. As these codes have rich algebraic structures, they can be efficiently encoded and decoded using shift registers. They also have very good error-correcting properties. All of those explain their preferred role in engineering. Repeated-root cyclic codes were first investigated in the 1990s by Castagnoli in [1] and Van Lint in [2], where it was proved that repeated-root cyclic codes have a concatenated construction, and are asymptotically bad. However, it is well known that there still exist a few optimal such codes (see [15–17]), which encourages many scholars to study the class of codes. For example, Dinh determined the generator polynomials of all constacyclic codes and their dual codes over Fq of length 2ps , 3ps and 6ps in [3–5]. Since then, these results have been extended to more general code lengths. In 2012, Bakshi and Raka gave the generator polynomials of all constacyclic codes of length 2t ps over Fq in [6], where q is a power of an odd prime p. In the same year, Chen et al. studied repeated root constacyclic codes of length lt ps over Fq in [9]. In 2014, Chen et al. studied all constacyclic codes of length lps over Fq in [7], where l is a prime different from p. In [7], all constacyclic codes of length lps over Fq and their dual codes were obtained, and all self-dual and all linear complementary dual constacyclic codes were given. In 2015, Raka studied repeated root constacyclic codes of length lt ps over Fq in [10]. Recently, in [8], Sharma explicitly determined the generator polynomials of all repeated-root constacyclic codes of length lt ps over Fpm and their dual codes. Further, they listed all self-dual cyclic and negacyclic codes and also determined all self-orthogonal cyclic and negacyclic codes of length lt ps over Fpm . What’s more, Sharma and Rani gave a method to compute repeated root constacyclic codes of length nps over Fq , for any positive integer n coprime to p, however the generator polynomials are not determined explicitly, in [11]. Chen et al. studied all constacyclic codes of length 2lm ps over Fq of characteristic p in [12], and they gave the characterization and enumeration of all linear complementary dual and self-dual constacyclic codes of length 2lm ps over Fq . In the conclusion of their paper, they proposed to study all constacyclic codes of length klm ps over Fq , where p is the characteristic of Fq , l is an odd prime different from p, and k is a prime different from l and p. However, this is not an easy work. In this paper, we study all constacyclic codes of length 3lps over Fq , where p = 3 is any prime and l = 3 is any odd prime with gcd(p, l) = 1. The article is organized as follows. In Section 3, we decompose the multiplicative cyclic group Fq∗ = ξ into s mutually disjoint union of cosets of ξ 3lp  and show that λ-constacyclic code is equivalent to μ-constacyclic code if λ and μ lie in the same coset. Based on this decomposition, Section 4 explicitly determines the generator polynomials of all λ-constacyclic codes of length 3lps over Fq and their dual codes, where λ is any none-zero element of Fq and q = pm is a power of prime. As an application, we give all self-dual cyclic codes of length 3 · 2s l over F2m and their enumeration in Section 5. In Section 6, we also

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determine the minimum Hamming distance of these codes when gcd(3, q − 1) = 1 and l = 1. 2. Preliminaries Let Fq be the finite field of order q = pm , where p = 3 is a prime and the characteristic of the field, and m is a positive integer. Let Fq∗ = ξ be the multiplicative cyclic group of none-zero elements of Fq , where ξ is a primitive (q − 1)th root of unity. For any element λ ∈ Fq∗ , λ-constacyclic codes of length n over Fq are regarded as the ideals g(x) of the quotient ring Fq [x]/(xn − λ), where g(x)|xn − λ. Further, the definition of the dual code of code C is as follows: C ⊥ = {x ∈ Fqn |x · y = 0, ∀y ∈ C}, where x · y denotes the Euclidean inner product of x and y in Fqn . The code C is called to be a self-orthogonal code if C ⊆ C ⊥ and a self-dual code if C = C ⊥ . Let C be a λ-constacyclic code of length n over Fq generated by a polynomial g(x), i.e., C = g(x). n −λ As g(x)|xn − λ, then there must be a polynomial h(x) ∈ Fq [x] such that h(x) = xg(x) . It is clear that h(x) is also monic if g(x) is monic. The polynomial h(x) is called the parity check polynomial of code C. It is well known that the dual code C ⊥ is generated by h(x)∗ , where h(x)∗ is the reciprocal polynomial of h(x). For any f (x) ∈ Fq [x], the reciprocal polynomial of f (x) is defined as f (x)∗ = f (0)−1 xdeg(f (x)) f ( x1 ). Obviously, (f1 f2 )∗ = f1∗ f2∗ , and (f1∗ )∗ = f1 , for any polynomials f1 (x), f2 (x) ∈ Fq [x]. Let n be any positive integer. For any integer s, 0 ≤ s ≤ n − 1, we have the definition of q-cyclotomic coset of s modulo n is as follows: Cs = {s, sq, ..., sq ns −1 }, where ns is the least positive integer such that sq ns ≡ s(mod n). Then it is easy to n see that ns is equal to the multiplicative order of q modulo gcd(s,n) . If α denotes a primitive nth root of unity in some extension field of Fq , then the polynomial Ms (x) =  i s i∈Cs (x − α ) is the minimal polynomial of α over Fq , and xn − 1 =



Ms (x)

gives the factorization of (xn − 1) into irreducible factors over Fq , where s runs over a complete set of representatives from distinct q-cyclotomic coset modulo n. Obviously, when n = l, where l = 3 is an odd prime with gcd(l, p) = 1, we get that all the distinct q-cyclotomic cosets modulo l are C0 = {0} and Ck = {g k , g k q, ..., g k q nk −1 }, for any integer k, 1 ≤ k ≤ e = φ(l) f , by [18, Theorem 1], where g is a generator of the cyclic group Zl∗ to be suitably chosen, f = ordl (q) is the multiplicative order of q in Zl∗ , and φ is Euler’s phi-function. Therefore, the irreducible factorization of xl − 1 over Fq is given by

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xl − 1 = M0 (x)M1 (x)M2 (x)...Me (x),  where Mi (x) = j∈Ci (x − η i ) with η being a primitive lth root of unity. In addition, we determine all the distinct q 3 -cyclotomic cosets modulo l, which is needed to prove our main results. There exist two subcases. When gcd(f, 3) = 1, it is easy to prove that f = ordl (q) = ordl (q 3 ), then q = q 3  in Zl∗ . According to the definition of q 3 -cyclotomic coset modulo l, we have that C0 and Ck , 1 ≤ k ≤ e = φ(l) f , are also all of the distinct q 3 -cyclotomic cosets modulo l. When gcd(f, 3) = 3, we prove that ordl (q 3 ) = f3 . It is easy to verify that A0 = {0}, f

Ak = {g k , g k q 3 , ..., g k q 3( 3 −1) }, f

Akq = {g k q, g k qq 3 , ..., g k qq 3( 3 −1) }, f

Akq2 = {g k q 2 , g k q 2 q 3 , ..., g k q 2 q 3( 3 −1) }, consist of all the distinct q 3 -cyclotomic cosets modulo l, where 1 ≤ k ≤ e. Then we have the irreducible factorization of xl − 1 in Fq3 [x] as follows: xl − 1 = A0 (x)A1 (x)Aq (x)Aq2 (x)A2 (x)A2q (x)A2q2 (x)...Ae (x)Aeq (x)Aeq2 (x),   where A0 (x) = (x − 1), Ak (x) = s∈Ak (x − η s ), Akq (x) = t∈Akq (x − η t ) and Akq2 (x) =  j j∈Akq2 (x − η ), 1 ≤ k ≤ e. We also give all the distinct q-cyclotomic cosets modulo 3l. As gcd(q, 3) = 1, we have q φ(3) ≡ 1(mod 3) by Euler’s Theorem, i.e., q 2 ≡ 1(mod 3). Then, it is simple to verify that ⎧ ⎪ q ≡ 1(mod 3); ⎨ f, ord3l (q) = f, q ≡ 2(mod 3) with f even; ⎪ ⎩ 2f , q ≡ 2(mod 3) with f odd. From [19, Chapter 8], there exists a primitive root r modulo l such that gcd( r l −1 , l) = 1. Now taking g = r + (1 − r)l2 , we have g l−1 − 1 ≡ (r + (1 − r)l2 )l−1 − 1 ≡ rl−1 − 1(mod l2 ). l−1 l−1 Therefore, gcd( g l −1 , l) = gcd( r l −1 , l) = 1. It’s clear that g is a primitive root modulo lt , t ≥ 1, such that g ≡ 1(mod 3) as l2 ≡ 1(mod 3). We give all the distinct q-cyclotomic cosets modulo 3l by the following lemma. l−1

Lemma 2.1. (I) If q ≡ 1(mod 3), then all the distinct q-cyclotomic cosets modulo 3l are given by B0 = {0}, Bl = {l}, B−l = {−l}, Bagk = {ag k , ag k q, ..., ag k q f −1 }, for a ∈ R = {1, −1, 3} and 0 ≤ k ≤ e − 1.

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(II) If q ≡ 2(mod 3) and f is even, all the distinct q-cyclotomic cosets modulo 3l are given by B0 = {0}, Bl = {l, lq}, 





Bgk = {g k , g k q, ..., g k q f −1 }, for 0 ≤ k ≤ 2e − 1, B3gk = {3g k , 3g k q, ..., 3g k q f −1 }, for 0 ≤ k ≤ e − 1. (III) If q ≡ 2(mod 3) and f is odd, all the distinct q-cyclotomic cosets modulo 3l are given by B0 = {0}, Bl = {l, lq}, Bgk = {g k , g k q, ..., g k q 2f −1 }, B3gk = {3g k , 3g k q, ..., 3g k q f −1 }, for 0 ≤ k ≤ e − 1. Proof. (I) Firstly, we prove that the cyclotomic cosets Bagk , 0 ≤ k ≤ e − 1, are distinct. If there exist some k1 , k2 , 0 ≤ k1 , k2 ≤ e − 1, such that Bagk1 = Bagk2 , then we have a1 g k1 ≡ a2 g k2 q j (mod 3l), for some integer j, where a1 , a2 ∈ R = {1, −1, 3}. Therefore, we get gcd(a1 g k1 , 3l) = gcd(a2 g k2 q j , 3l) = gcd(a2 g k2 , 3l), as gcd(q, 3l) = 1. From this, we can deduce a1 = a2 or a1 = −a2 = ±1. If a1 = −a2 = ±1, then −g k1 ≡ g k2 q j (mod 3l), i.e., −1 ≡ g k1 −k2 q j (mod 3l), for some integer j. Due to g ≡ 1(mod 3) and q ≡ 1(mod 3), we deduce −1 ≡ 1(mod 3). This is a contradiction. If a1 = a2 , assume that a1 = a2 = a, then we have ag k1 ≡ ag k2 q j (mod 3l), i.e., g k1 −k2 ≡ q j (mod l), for some integer j. Further, we have g (k1 −k2 )f ≡ q jf ≡ 1(mod l). As g is a primitive root modulo l, we get φ(l)|(k1 − k2 )f , i.e., e = φ(l) f |k1 − k2 . Since 0 ≤ k1 , k2 ≤ e − 1, we must have k1 = k2 . Secondly, we get

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|B0 | + |Bl | + |B−l | +

e−1 

|Bagk | = 3 +

a∈R k=0

e−1 

f

a∈R k=0

=3+



ef

a∈R

= 3 + 3φ(l) = 3l. So, the conclusion (I) holds. The conclusions (II) and (III) are also established in a similar way. 2 Assume that Bo (x), Bl (x), B−l (x) and Bagk (x) are the minimal polynomials of the corresponding cosets Bo , Bl , B−l and Bagk . From the above lemma, we get the following theorem immediately. Theorem 2.2. The irreducible factorization of x3l − 1 over Fq is given as follows: (I) If q ≡ 1(mod 3), then x3l − 1 = B0 (x)Bl (x)B−l (x)

  e−1

Bagk (x),

a∈R k=0

where a ∈ R = {1, −1, 3} and 0 ≤ k ≤ e − 1. (II) If q ≡ 2(mod 3) and f is even, then x3l − 1 = B0 (x)Bl (x)

2e−1 

Bgk (x)

k =0

e−1 

B3gk (x),

k=0

where 0 ≤ k ≤ e − 1, 0 ≤ k ≤ 2e − 1. (III) If q ≡ 2(mod 3) and f is odd, then x3l − 1 = B0 (x)Bl (x)

e−1 

Bgk (x)B3gk (x),

k=0

where 0 ≤ k ≤ e − 1. The next two lemmas give the necessary and sufficient conditions for judging the reducibility of binomials and trinomial, which were given by Wan in [20]. Lemma 2.3. Suppose that n ≥ 2. Let k = ord(a) be the multiplicative order of a, for any a ∈ Fq∗ . Then, the binomial xn − a is irreducible over Fq if and only if: (i) Every prime divisor of n divides k, but not (ii) If 4|n, then 4|(q − 1).

(q−1) k ;

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Lemma 2.4. Let t be a positive integer, and H(x) ∈ Fq [x] be irreducible over Fq with deg(H(x)) = n, x does not divide H(x), and e denotes the order of any root of H(x). Then H(xt ) is irreducible over Fq if and only if: (i) Each prime divisor of t divides e; n (ii) gcd(t, q e−1 ) = 1; (iii) If 4|t, then 4|(q n − 1). 3. A classification of constacyclic codes of length 3lps Let ξ be a primitive (q − 1)th root of unity, and Fq∗ = ξ be a cyclic group of s order (q − 1) as before. It is easy to verify that ξ 3lp  = ξ 3l  = ξ d  and the index s |Fq∗ : ξ 3lp | = d, where d = gcd(q − 1, 3lps ). Thus, the multiplicative cyclic group Fq∗ s can be decomposed into mutually disjoint union of cosets over the subgroup ξ 3lp  as follows: Lemma 3.1. Fq∗ = ξ = ξ d 



s

ξ p ξ d 



...



s

ξp

(d−1)

ξ d , where d = gcd(q − 1, 3lps ).

According to the properties of the coset, we obtain the following lemma immediately. Lemma 3.2. For any two none-zero elements λ and μ of Fq , there exists some integer j, s 0 ≤ j ≤ d − 1 such that λ, μ ∈ ξ jp ξ d  if and only if λ−1 μ ∈ ξ d , where d = gcd(q − 1, 3lps ). If λ and μ lie in the same coset, we build a one-to-one correspondence between s s Fq [x]/(x3lp − λ) and Fq [x]/(x3lp − μ) in the following theorem which shows that a λ-constacyclic code is equivalent to μ-constacyclic code. Theorem 3.3. Let λ, μ ∈ Fq∗ . Then there exists some integer a ∈ Fq∗ such that s

s

ϕ : Fq [x]/(x3lp − μ) → Fq [x]/(x3lp − λ), f (x) → f (ax), s

is an isomorphism if and only if λ, μ ∈ ξ jp ξ d , where 0 ≤ j ≤ d − 1. Proof. (⇒) If ϕ is an isomorphism, then we have s

s

s

s

s

s

μ = ϕ(μ) = ϕ(x3lp ) = (ϕ(x))3lp = (ax)3lp = a3lp x3lp = a3lp λ, i.e., λ−1 μ = a3lp . As a = ξ k ∈ Fq∗ , for some positive integer k, then λ−1 μ = ξ k·3lp ∈ s ξ d . By Lemma 3.2, we get that there exists j, 0 ≤ j ≤ d − 1, such that λ, μ ∈ ξ jp ξ d . s

s

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(⇐) If there exists j, 0 ≤ j ≤ d − 1, such that λ, μ ∈ ξ jp ξ d , then we have λ−1 μ ∈ s s ξ d  = ξ 3lp , by Lemma 3.2 again. Thus, λ−1 μ = ξ k·3lp , for some integer k. Set a = ξ k , s then λa3lp = μ. Further, it is easy to prove that the following map is an isomorphism: s

s

s

ϕ : Fq [x]/(x3lp − μ) → Fq [x]/(x3lp − λ), f (x) → f (ax).

2

From Theorem 3.3, we get the following obvious corollaries. Corollary 3.4. For any two elements λ and μ of Fq∗ , a λ-constacyclic code is equivalent to s an μ-constacyclic code if and only if there exists j, 0 ≤ j ≤ d−1, such that λ, μ ∈ ξ jp ξ d . s s Further since ξ jp ∈ ξ jp ξ d , λ-constacyclic and μ-constacyclic codes are both equivalent s to ξ jp -constacyclic code. Corollary 3.5. Let λ be any element of Fq∗ , then there exists some integer j, 0 ≤ j ≤ d −1, s such that a λ-constacyclic code is equivalent to a ξ jp -constacyclic code. Obviously, Theorem 3.3 and its two corollaries show that all constacyclic codes of length 3lps over Fq are classified into d = gcd(q − 1, 3lps ) mutually disjoint classes. s It is enough to consider λ-constacyclic codes, where λ = ξ jp , 0 ≤ j ≤ d − 1, and d = gcd(q − 1, 3lps ), which we study in Section 4. 4. All constacyclic codes of length 3lps over Fq Let f (x) be any polynomial of Fq [x] and leading coefficient an = 0, we denote f (x) = Then, f (x) is said to be the monic polynomial of f (x). From the above discussion in the section 3, we know that the number of inequivalent constacyclic classes is equal to d = gcd(q − 1, 3lps ). It is obvious that the value of d has the following four cases:

a−1 n f (x).

(i) (ii) (iii) (iv)

d = gcd(q − 1, 3lps ) = 1. d = gcd(q − 1, 3lps ) = 3. d = gcd(q − 1, 3lps ) = l. d = gcd(q − 1, 3lps ) = 3l.

4.1. All constacyclic codes of length 3lps over Fq when d = 1 From Lemma 3.1, we see that Fq∗ = ξ is the decomposition of coset over the subgroup ξ , when d = gcd(q − 1, 3lps ) = 1 mention that here 3  q − 1 i.e. q = 1(mod 3). In this case, it is clear that all constacyclic codes of length 3lps over Fq are equivalent to the cyclic codes. Therefore, we have the following theorem. d

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Theorem 4.1. Let d = gcd(q − 1, 3lps ) = 1, then λ-constacyclic codes C of length 3lps over Fq are equivalent to the cyclic codes, for any λ ∈ Fq∗ , i.e., there exists a unique s element a ∈ Fq∗ such that a3lp λ = 1. Further, we have the map s

s

ϕa : Fq [x]/(x3lp − 1) → Fq [x]/(x3lp − λ), f (x) → f (ax), s

is an isomorphism, and the irreducible factorization of x3lp − λ in Fq [x] is given by: (I) If f is even, then s

0 (ax)ps B

l (ax)ps x3lp − λ = B

2e−1 

k (ax)ps B g

k =0

e−1 

3gk (ax)ps , B

k=0

where 0 ≤ k ≤ e − 1, 0 ≤ k ≤ 2e − 1. Therefore, we have

0 (ax)ε B

l (ax)ρ C = B

2e−1 

k (ax)τk B g

k =0

0 (a−1 x)ps −ε B

−l (a−1 x)ps −ρ C ⊥ = B

2e−1 

e−1 

3gk (ax)υk , B

k=0

k (a−1 x)ps −τk B −g

k =0

e−1 

−3gk (a−1 x)ps −υk , B

k=0

where 0 ≤ ε, ρ, τk , υk ≤ ps , for any k = 0, 1, 2, ..., e − 1, and k = 0, 1, 2, ..., 2e − 1. (II) If f is odd, then s

0 (ax)ps B

l (ax)ps x3lp − λ = B

e−1 

gk (ax)ps B

3gk (ax)ps , B

k=0

where 0 ≤ k ≤ e − 1. Therefore, we have

0 (ax)ε B

l (ax)ρ C = B

e−1 

gk (ax)τk B

3gk (ax)υk , B

k=0

0 (a−1 x)ps −ε B

−l (a−1 x)ps −ρ C ⊥ = B

e−1 

−gk (a−1 x)ps −τk B

−3gk (a−1 x)ps −υk , B

k=0

where 0 ≤ ε, ρ, τk , υk ≤ ps , for any k = 0, 1, 2, ..., e − 1. Proof. From Theorems 2.2 and 3.3, we see that this theorem is obtained immediately. 2

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4.2. All constacyclic codes of length 3lps over Fq when d = 3 In this subsection, we consider the second case, i.e. when d = gcd(q − 1, 3lps ) = 3, mention that here 3 | q − 1 i.e. q ≡ 1(mod 3).

(1)

s s In this case, we have that Fq∗ = ξ = ξ 3  ξ p ξ 3  ξ 2p ξ 3  is the decomposition of s cosets of Fq∗ over subgroup ξ 3lp  by Lemma 3.1. Therefore, it is enough to consider the s s cyclic codes, ξ p -constacyclic codes and ξ 2p -constacyclic codes if we want to determine all constacyclic codes of length 3lps over Fq , when gcd(q − 1, 3lps ) = 3. From Section 2, we have that the irreducible factorization of xl − 1 is given by xl − 1 = e  j j∈Ci (x − η ) with η is a primitive lth root of unity. i=0 Mi (x), where Mi (x) = According to this, we deduce the following lemma immediately. e Lemma 4.2. Let xl − 1 = i=0 Mi (x) be the factorization of xl − 1 into irreducible factors over Fq where Mi (x) is minimal polynomial of q-cyclotomic coset Ci modulo l. Let ν ∈ Fq∗r , r  1. Suppose there exists a unique element b ∈ Fq∗r such that bl ν = 1, then the factorization of xl − ν into irreducible factors over Fqr is given by xl − ν =

e 

i (bx). M

i=0

Corollary 4.3. Let gcd(q − 1, l) = 1 so that Fq∗ = ξ = ξ l . Then for ξ j ∈ Fq∗ there s j exist unique elements bj ∈ Fq∗ such that blp j ξ = 1 for j = 0, 1, 2, ..., q − 1. Hence the l j factorization of x − ξ into irreducible factors over Fq is given by xl − ξ j =

e 

i (bj x). M

i=0 s

Lemma 4.4. Let gcd(q − 1, 3lps ) = 3. Then the irreducible factorization of x3lp − 1 is given by s

x3lp − 1 =

e 

s i (b q−1 x)ps M i (b 2(q−1) x)ps , Mi (x)p M 3

3

i=0

where Mi (x) is the minimal polynomial of the q-cyclotomic coset Ci modulo l and b q−1 , 3 b 2(q−1) ∈ Fq∗ . 3

Proof. As (1), then it is clear that ξ have s

q−1 3

s

∈ Fq∗ is a primitive 3rd root of unity. Hence, we

x3lp − 1 = (xl − 1)p (xl − ξ

q−1 3

s

)p (xl − ξ

2(q−1) 3

s

)p .

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Further, by Corollary 4.3, we get s

x3lp − 1 =

e 

s i (b q−1 x)ps M i (b 2(q−1) x)ps , Mi (x)p M 3

i=0

3

where Mi (x) is the minimal polynomial of the q-cyclotomic coset Ci modulo l and b q−1 , 3 b 2(q−1) ∈ Fq∗ . 2 3

s

Before determining ξ ip -constacyclic codes, i = 1, 2, we must explicitly decompose s s the polynomial x3lp − ξ ip (i = 1, 2), into the product of monic irreducible factors. Obviously, we only need to determine the irreducible factorization of x3l − ξ i (i = 1, 2). Firstly, we consider the polynomial x3 − ξ i (i = 1, 2). x3 − ξ i , for i = 1, 2, is irreducible in Fq [x] since q ≡ 1(mod 3). We get that Fq3 is a splitting field for x3 − ξ i (i = 1, 2) over Fq . Thus, there exists νi ∈ Fq3 such that νi3 = ξ i (i = 1, 2). Further, it is easy to get that νi , ανi , and α2 νi are all the roots of x3 − ξ i (i = 1, 2), in Fq3 , where α is a primitive 3rd root of unity. In addition, we see that νi ∈ Fq3 but νi ∈ / Fq , and νi is primitive 3(q − 1)th root of unity, i = 1, 2. As gcd(3(q − 1), l) = 1, we can find a bijection θ from the set D to itself such that θ(ν) = ν l , for any ν ∈ D, where D consists of all the primitive 3(q − 1)th roots of unity of Fq∗3 . Therefore, there exists a unique element ωi ∈ D such that νi−1 = θ(ωi ) = ωil , i.e., ωil νi = 1, i = 1, 2. It is easy to deduce that ωi3 ∈ Fq , i = 1, 2. From the above discussion, we have the following two lemmas. Lemma 4.5. The irreducible factorization of x3l − ξ over Fq is given as follows: (I) If gcd(f, 3) = 1,

x3l − ξ =

e 

Ri (x),

i=0

i (ω1 x)M i (αω1 x)M i (α2 ω1 x) for any i = 0, 1, 2, ..., e, with ω1 is a primwhere Ri (x) = M itive 3(q − 1)th root of unity and α is a primitive 3rd root of unity. (II) If gcd(f, 3) = 3,

x3l − ξ = P (x)

e 

Qi (x)Ui (x)Zi (x),

i=1

i (ω1 x)A

iq (αω1 x)A

iq2 (α2 ω1 x), where P (x) = (x −ω1−1 )(x −αω1−1 )(x −α2 ω1−1 ), Qi (x) = A 2 2

i (αω1 x)A

iq (α ω1 x)A

iq2 (ω1 x), and Zi (x) = A

i (α ω1 x)A

iq (ω1 x)A

iq2 (αω1 x) for Ui (x) = A any i = 1, 2, ..., e, with ω1 is a primitive 3(q − 1)th root of unity, α is a primitive 3rd root of unity.

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Proof. (I) If gcd(f, 3) = 1, we get that C0 and Ck , 1 ≤ k ≤ e = φ(l) f , are all the 3 3 q -cyclotomic cosets modulo l. Let ν1 be a root of x − ξ, and α be a primitive 3rd root of unity. Then we have ν1 , αν1 , and α2 ν1 are all the roots of x3 − ξ over Fq3 , i.e., x3l − ξ = (xl − ν1 )(xl − αν1 )(xl − α2 ν1 ). From the above discussion, we know that there exists ω1 such that ω1l ν1 = 1, where ω1 is a primitive 3(q − 1)th root of unity. Hence, we have ω1q = αω1 or α2 ω1 . Here, we just consider ω1q = αω1 . However, for the case ω1q = α2 ω1 , we see that the result is still right in the same way. As gcd(3, l) = 1, we know l ≡ 1(mod 3) or l ≡ 2(mod 3). When l ≡ 2(mod 3), we have (αω1 )l αν1 = αl+1 = 1 and (α2 ω1 )l α2 ν1 = α2(l+1) = 1. When l ≡ 1(mod 3), we have (α2 ω1 )l αν1 = α2l+1 = 1 and (αω1 )l α2 ν1 = αl+2 = 1. Hence, there always exist ω1 , αω1 and α2 ω1 such that ω1l ν1 = 1, (αω1 )l αν1 = 1 and (α2 ω1 )l α2 ν1 = 1 or ω1l ν1 = 1, (α2 ω1 )l αν1 = 1 and (αω1 )l α2 ν1 = 1. Further, by Corollary 4.3, we get x3l − ξ =

e 

i (ω1 x)M i (αω1 x)M i (α2 ω1 x), M

i=0

i (ω1 x) = which is the monic irreducible factorization of x3l − ξ over Fq3 . We have M    −1 k 2 −1 k 2   k∈Ci (x − ω1 η ), Mi (αω1 x) = k∈Ci (x − α ω1 η ) and Mi (α ω1 x) = k∈Ci (x − −1 k 2    αω1 η ). Now, we take Ri (x) = Mi (ω1 x)Mi (αω1 x)Mi (α ω1 x). Then we have Ri (x) =   −1 k −1 k −3 3k 2 −1 k 3 ). Therefore, it is k∈Ci (x − ω1 η )(x − α ω1 η )(x − αω1 η ) = k∈Ci (x − ω1 η  3 easy to deduce that Ri (x) is a polynomial over Fq , as ω1 ∈ Fq and k∈Ci (x − η k ) is an irreducible polynomial over Fq . Next, we assume that Ri (x) is reducible over Fq . Then the monic irreducible fact torization of Ri (x) over Fq can be given by Ri (x) = i=0 ai (x), t = 1 or 2, as 2    Ri (x) = Mi (ω1 x)Mi (αω1 x)Mi (α ω1 x) is the monic irreducible factorization of Ri (x) over Fq3 . If t = 1, then Ri (x) = a0 (x)a1 (x) is a monic irreducible factorization of Ri (x) over Fq . There exist three subcases: (i) a0 (x) and a1 (x) are irreducible over Fq3 . (ii) a0 (x) and a1 (x) are reducible over Fq3 . (iii) a0 (x) or a1 (x) is irreducible over Fq3 . i (ω1 x)M i (αω1 x) × If a0 (x) and a1 (x) are irreducible over Fq3 . According to Ri (x) = M 2 i (α ω1 x) is the monic irreducible factorization of Ri (x) over Fq3 , we know this is M impossible. If a0 (x) and a1 (x) are reducible over Fq3 . Then we see that there are at least four irreducible factors of Ri (X) over Fq3 . This is impossible. If a0 (x) or a1 (x) are irreducible over Fq3 . We suppose that a0 (x) is irreducible i (ω1 x) or M i (αω1 x) or M i (α2 ω1 x), according to Ri (x) = over Fq3 . Then a0 (x) = M

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281

i (ω1 x)M i (αω1 x)M i (α2 ω1 x) is the monic irreducible factorization of Ri (x) over Fq3 . M i (ω1 x), M i (αω1 x) and M i (α2 ω1 x) are not polynomials over Fq . This is impossible, as M Therefore, we deduce that Ri (x) is irreducible over Fq . For t = 2, we can also deduce that Ri (x) is irreducible over Fq , using the same way. (II) When gcd(f, 3) = 3, we have that A0 , Ak , Akq , Akq2 consist of all the distinct q -cyclotomic cosets modulo l, where 1 ≤ k ≤ e. Then, the irreducible factorization of xl − 1 over Fq3 is given by 3

xl − 1 = A0 (x)A1 (x)Aq (x)Aq2 (x)A2 (x)A2q (x)A2q2 (x)...Ae (x)Aeq (x)Aeq2 (x). Next, working in the same way as in (I), we can prove that the conclusion (II) holds. 2 Using arguments similar to the proof of Lemma 4.5, we have the following lemma, and we omit its proof here. Lemma 4.6. The irreducible factorization of x3l − ξ 2 over Fq is given as follows: (I) If gcd(f, 3) = 1, x −ξ = 3l

2

e 

Ri (x),

i=0

i (ω2 x)M i (αω2 x)M i (α2 ω2 x) for any i = 0, 1, 2, ..., e, with ω2 is a primwhere Ri (x) = M itive 3(q − 1)th root of unity and α is a primitive 3rd root of unity. (II) If gcd(f, 3) = 3, x3l − ξ 2 = P  (x)

e 

Qi (x)Ui (x)Zi (x),

i=1

i (ω2 x)A

iq (αω2 x)A

iq2 (α2 ω2 x), where P  (x) = (x −ω2−1 )(x −αω2−1 )(x −α2 ω2−1 ), Qi (x) = A

i (αω2 x)A

iq (α2 ω2 x)A

iq2 (ω2 x), and Z  (x) = A

i (α2 ω2 x)A

iq (ω2 x)A

iq2 (αω2 x) for Ui (x) = A i any i = 1, 2, ..., e, with ω2 is a primitive 3(q − 1)th root of unity and α is a primitive 3rd root of unity. Combining Lemmas 4.4, 4.5 and 4.6, we easily obtain the following result. q−1

Theorem 4.7. Let gcd(q − 1, 3lps ) = 3 and α = ξ 3 . For any element λ of Fq∗ and λ-constacyclic code C of length 3lps over Fq , one of the following cases holds: s (I) If λ ∈ ξ 3 , then there exists some element c ∈ Fq∗ such that c3lp λ = 1, and we have C=

e 

i=0

i (cx)εi M i (cb q−1 x)σi M i (cb 2(q−1) x)τi , M 3

3

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L. Liu et al. / Finite Fields and Their Applications 42 (2016) 269–295

C⊥ = 

e 

ps −σi  ps −τi −i (c−1 x)ps −εi M −i (c−1 b−1 M M−i (c−1 b−1 , q−1 x) 2(q−1) x) 3

i=0

3

where 0 ≤ εi , σi , τi ≤ ps for any i = 0, 1, 2, ..., e. s s s (II) If λ ∈ ξ p ξ 3 , then there exists some element c1 ∈ Fq∗ such that c3lp λ = ξp , 1 and one of the following holds: (i) If gcd(f, 3) = 1, C=

e 

i (c1 x)ε , R

i=0

C⊥ = 

e 

−i (c−1 x)ps −ε , R 1

i=0

−i (ω −1 x)M −i (α2 ω −1 x)M −i (αω −1 x) for any i = where 0 ≤ ε ≤ ps , R−i (x) = M 1 1 1 0, 1, 2, ..., e. (ii) If gcd(f, 3) = 3, C = P (c1 x)ε

e 

i (c1 x)εi U

i (c1 x)σi Z i (c1 x)τi , Q

i=1 ps −ε C ⊥ = P ∗ (c−1 1 x)

e 

−i (c−1 x)ps −εi U

−i (c−1 x)ps −σi Z −i (c−1 x)ps −τi , Q 1 1 1

i=1

−i (ω −1 x) · where 0 ≤ εi , σi , τi ≤ ps , P ∗ (x) = (x −ω1 )(x −α2 ω1 )(x −αω1 ), Q−i (x) = A 1 −1 −1 −1 −1 2 2

−iq (α ω x)A

−iq2 (αω x), U−i (x) = A

−i (α ω x)A

−iq (αω x)A

−iq2 (ω −1 x), A 1 1 1 1 1

−i (αω −1 x)A

−iq (ω −1 x)A

−iq2 (α2 ω −1 x), for any i = 1, 2, ..., e. and Z−i (x) = A 1 1 1 s

(III) If λ ∈ ξ 2p ξ 3 , then there exists some element c2 ∈ Fq∗ such that c3lp λ = ξ 2p , 2 and one of the following holds: s

s

(i) If gcd(f, 3) = 1, C=

e 

 i (c2 x)εi , R

i=0

C⊥ = 

e 

 −i (c−1 x)ps −εi , R 2

i=0  −i (ω −1 x)M −i (α2 ω −1 x)M −i (αω −1 x) for any i = where 0 ≤ ε ≤ ps , R−i (x) = M 2 2 2 0, 1, 2, ..., e.

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(ii) If gcd(f, 3) = 3,  (c2 x)ε C = P

e 

 i (c2 x)εi U  i (c2 x)σi Z  i (c2 x)τi ), Q

i=1 ∗

 (c−1 x)ps −ε C ⊥ = P 2

e 

 −i (c−1 x)ps −εi U  −i (c−1 x)ps −σi Z  −i (c−1 x)ps −τi ), Q 2 2 2

i=1

where 0 ≤ εi , σi , τi ≤ ps , P  ∗ (x) = (x − ω2 )(x − α2 ω2 )(x − αω2 ), Q−i (x) =

−i (ω −1 x) · A

−iq (α2 ω −1 x)A

−iq2 (αω −1 x), U  (x) = A

−i (α2 ω −1 x)A

−iq (αω −1 x) × A −i 2 2 2 2 2

−iq2 (ω −1 x), and Z  (x) = A

−i (αω −1 x)A

−iq (ω −1 x)A

−iq2 (α2 ω −1 x) for any i = A −i 2 2 2 2 1, 2, ..., e. 4.3. All constacyclic codes of length 3lps over Fq when d = l Let d = gcd(q − 1, 3lps ) = l, i.e., l|(q − 1) and gcd(q − 1, 3) = 1. Then η = ξ is a primitive lth root of unity. Therefore, we have the following lemma. Lemma 4.8. Assume that gcd(q − 1, 3lps ) = l, and let η = ξ s factorization of x3lp − 1 over Fq is given by:

x

3lps

−1=

l−1 

s

q−1 l

q−1 l

∈ Fq∗

. Then, the irreducible

s

(x − η k )p (x2 + η k x + η 2k )p .

k=0

Proof. As gcd(q−1, 3) = 1, there exists no primitive 3rd root of unity in Fq , which implies x2 + x + 1 is irreducible. By this, we can deduce that x2 + η k x + η 2k is irreducible, for any k = 1, 2, ..., l. Otherwise if x2 +η k x +η 2k is reducible, then we have η −2k (x2 +η k x +η 2k ) = (η −k x)2 + η −k x + 1 is reducible. By substituting x for η −k x in the above polynomial, we q−1 have that x2 + x + 1 is reducible, which is a contradiction. Since η = ξ l is a primitive lth root of unity, then η 3 is also a primitive lth root of unity. Hence, the irreducible s factorization of x3lp − 1 over Fq is given by

x

3lps

ps

− 1 = (x − 1) 3l

=

l−1 

3k ps

(x − η ) 3

=

k=0

l−1 

s

s

(x − η k )p (x2 + η k x + η 2k )p .

2

k=0 s

Lemma 4.9. Assume that gcd(q − 1, 3lps ) = l, then the irreducible factorization of x3lp − s ξ jp , 1 ≤ j ≤ l − 1, over Fq is given by: (I) When gcd(3, j) = 3, let j = 3k, for some integer k. Then we have s

s

s

s

x3lp − ξ jp = (xl − ξ k )p (x2l + ξ k xl + ξ 2k )p .

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(II) When gcd(3, j) = 1, there must exist some integer i, 1 ≤ i ≤ q − 1, such that 3i = j + q − 1 or 3i = j + 2(q − 1). Then we have s

s

s

s

x3lp − ξ jp = (xl − ξ i )p (x2l + ξ i xl + ξ 2i )p . Proof. (I) Obviously, gcd(l, k) = 1. From Lemma 2.3, it is very easy to verify that xl −ξ k is irreducible. By the proof of Lemma 4.8, we see that x2 +ξ k x +ξ 2k is irreducible over Fq . Now, we suppose that δ is any root of x2 + ξ k x + ξ 2k in some extended field of Fq , and e q−1 e is the order of δ. Then, we have δ 3 = ξ 3k . Further, we deduce that (e,3) = (q−1,3k) , i.e., 2l k l 2k e = (q−1)(e,3) (q−1,k) , as gcd(q − 1, 3) = 1. From Lemma 2.4, we can verify that x + ξ x + ξ is irreducible, where l is an odd prime. (II) When (3, j) = 1, we get that x3 − ξ j , 1 ≤ j ≤ l − 1, are all reducible, from Lemma 2.3. Therefore, there must exist some ξ i ∈ Fq∗ , which is a root of x3 − ξ j , for any j = 1, 2, ..., l − 1. Then, ξ 3i − ξ j = 0, i.e., ξ 3i = ξ j . As 1 ≤ i ≤ q − 1 and 1 ≤ j ≤ l − 1, we deduce that 3i = j + q − 1 or 3i = j + 2(q − 1) and gcd(l, i) = 1. Next, working similarly to the proof of (I), we get that the result holds. 2

According to Lemma 4.8 and Lemma 4.9, we get the following theorem immediately. q−1

Theorem 4.10. Assume that gcd(q − 1, 3lps ) = l, and let η = ξ l . For any element λ of Fq∗ and λ-constacyclic code C of length 3lps over Fq , one of the following cases holds: s (I) If λ ∈ ξ l , then there exists c1 ∈ Fq∗ such that c3lp λ = 1, and we have 1 C=

l−1 

−1 k −2 2k τk k εk 2 (x − c−1 1 η ) (x + c1 η x + c1 η ) ,

k=0

C⊥ = 

l−1 

(x − c1 η −k )p

s

−εk

(x2 + c1 η −k x + c21 η −2k )p

s

−τk

,

k=0

where 0 ≤ εk , τk ≤ ps , for any k = 0, 1, 2, ..., l − 1. s s s (II) If λ ∈ ξ jp ξ l , 1 ≤ j ≤ l − 1, then there exists c2 ∈ Fq∗ such that c3lp λ = ξ jp , 2 and one of the following holds: (i) When (3, j) = 3, let j = 3k, for some integer k. Then we have −l k l −2l 2k τk k εk 2l C = (xl − c−l 2 ξ ) (x + c2 ξ x + c2 ξ ) ,

C ⊥ = (xl − cl2 ξ −k )p

s

−εk

−2k p (x2l + cl2 ξ −k xl + c2l ) 2 ξ

s

−τk

,

where 0 ≤ εk , τk ≤ ps . (ii) When gcd(3, j) = 1, there must exist some integer i, 1 ≤ i ≤ q − 1, such that 3i = j + q − 1 or 3i = j + 2(q − 1). Then we have

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285

−l i l −2l 2i τi i εi 2l C = (xl − c−l 2 ξ ) (x + c2 ξ x + c2 ξ ) ,

C ⊥ = (xl − cl2 ξ −i )p

s

−εi

−2i p (x2l + cl2 ξ −i xl + c2l ) 2 ξ

s

−τi

,

where 0 ≤ εi , τi ≤ ps . 4.4. All constacyclic codes of length 3lps over Fq when d = 3l In this subsection, we assume that d = gcd(3lps , q −1) = 3l, namely 3l|(q −1). Clearly, q−1 there exists an element γ = ξ 3l ∈ Fq∗ , which is a primitive 3lth root of unity. Further, q−1 q−1 due to l|(q −1), and 3|(q −1), it is easy to know that η = ξ l and β = ξ 3 are primitive lth and 3rd roots of unity respectively. From Lemma 3.1, we get that the Fq∗ = ξ = ξ 3l  ∪ ξ p ξ 3l  ∪ ξ 2p ξ 3l  ∪ ... ∪ (3l−1)ps 3l ξ ξ . Therefore, any element λ of Fq∗ belongs to exactly one of the cosets, s i.e., there is a unique integer j, 0 ≤ j ≤ 3l − 1, such that λ ∈ ξ jp ξ 3l , namely s λ-constacyclic codes are equivalent to ξ jp -constacyclic codes. Hence, we just need to s determine ξ jp -constacyclic codes, where 0 ≤ j ≤ 3l − 1. s

Lemma 4.11. Let d = gcd(3lps , q − 1) = 3l and γ = ξ s of x3lp − 1 over Fq is given as follows: s

x3lp − 1 =

3l−1 

q−1 3l

s

. Then irreducible factorization

s

(x − γ i )p .

i=0 q−1

Lemma 4.12. Let η = ξ l and β = ξ over Fq is given as follows: (I) When gcd(3l, j) = l, we have s

q−1 3

s

s

s

x3lp − ξ jp = (x3l − ξ tl )p =

l−1 

s

(x3 − ξ t η i )p ,

i=0

where t = 1 or 2. (II) When gcd(3l, j) = 3, we have s

s

s

x3lp − ξ jp = (x3l − ξ 3k )p =

2 

s

(xl − ξ k β i )p ,

i=0

where k is some integer such that j = 3k. (III) Otherwise, we can see that gcd(3l, j) = 1. Then we have s

s

. Then the irreducible factorization of x3lp −ξ jp

s

s

x3lp − ξ jp = (x3l − ξ j )p .

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286

Proof. (I) As gcd(3l, j) = l and 1 ≤ j ≤ 3l − 1, we have j = tl, where t = 1, 2. Obviously, q−1 η = ξ l is a primitive l-th root of unity in Fq . Therefore, we get

s

s

s

x3lp − ξ jp = (x3l − ξ tl )p =

l−1 

s

(x3 − ξ t η i )p .

i=0

Next, we prove that the polynomial x3 − ξ t η i , for any i = 0, 1, 2, ..., l − 1, is irreducible in i(q−1) Fq [x]. Firstly, we know that the multiplicative order of ξ t η i = ξ t+ l , t = 1, 2, is ei = i(q−1) q−1 ) = 1. i(q−1) . As 3|(q − 1), gcd(3, t) = 1 and gcd(3, l) = 1, we get that (3, t + l (q−1,t+

l

)

Thus, 3 divides ei but not q−1 = (q − 1, t + i(q−1) ). From Lemma 2.3, we get the ei l 3 t i polynomial x − ξ η , for any i = 0, 1, 2, ..., l − 1, is irreducible in Fq [x]. In the same way, we have (II) and (III) hold. 2 In the following theorem, we determine all constacyclic codes of length 3lps over Fq and their dual codes, when d = gcd(3lps , q − 1) = 3l. q−1

q−1

q−1

Theorem 4.13. Assume that gcd(3lps , q − 1) = 3l, let γ = ξ 3l , η = ξ l and β = ξ 3 be primitive 3lth, lth and 3rd root of unity in Fq respectively. For any element λ of Fq∗ and λ-constacyclic codes C of length 3lps over Fq , one of the following holds: s (I) If λ ∈ ξ 3l , then there exists d1 ∈ Fq∗ such that d3lp λ = 1, and we have 1 3l−1 

C=

i εi (x − d−1 1 γ ) ,

i=0 3l−1 

C⊥ = 

(x − d1 γ −i )p

s

−εi

,

i=0

where 0 ≤ εi ≤ ps , for any i = 0, 1, 2, ..., 3l − 1. s s s (II) If λ ∈ ξ jp ξ 3l , 1 ≤ j ≤ 3l − 1, then there exists d2 ∈ Fq∗ such that d3lp λ = ξ jp , 2 and one of the following holds: (i) When gcd(3l, j) = l, we have

C=

l−1 

t i εi (x3 − d−3 2 ξ η ) ,

i=0

C⊥ = 

l−1 

(x3 − d32 ξ −t η −i )p

i=0

where 0 ≤ εi ≤ ps , for any i = 0, 1, 2, ..., l − 1.

s

−εi

,

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287

(ii) When gcd(3l, j) = 3, we have C=

2 

k i εi (xl − d−l 2 ξ β ) ,

i=0

C⊥ = 

2 

(xl − dl2 ξ −k β −i )p

s

−εi

,

i=0

where 0 ≤ εi ≤ ps for i = 0, 1, 2, and j = 3k. (iii) When gcd(3l, j) = 1, we have C = (x3l − d2−3l ξ j )ε , −j p ) C ⊥ = (x3l − d3l 2 ξ

s

−ε

,

where 0 ≤ ε ≤ ps . 5. All self-dual cyclic codes of length 3lps over Fq In Section 4, we gave the generator polynomials of all the constacyclic codes and their dual codes of length 3lps over Fq . We will determine all the self-dual cyclic codes of length 3lps over Fq in detail, in this section. It is well known that there exist self-dual cyclic codes of length N over Fq if and only if N is even and the characteristic of Fq is p = 2 [13,14]. Therefore, we get that self-dual cyclic codes of length 3lps over Fq exist only when p = 2. s s s s s s s s s Let x3lp −1 = (x3l −1)p = f1 (x)p f2 (x)p ·· ·fa (x)p h1 (x)p h∗1 (x)p ·· ·hb (x)p h∗b (x)p s be the irreducible factorization of x3lp − 1, where f1 (x), f2 (x), ..., fa (x) are monic irreducible self-reciprocal polynomials over Fq , hj (x) and its reciprocal polynomial h∗j (x), 1 ≤ j ≤ b, are also monic irreducible polynomials over Fq . Hence, for any cyclic code C = g(x) of length 3lps over Fq , we suppose that g(x) = f1 (x)τ1 f2 (x)τ2 · · · fa (x)τa h1 (x)δ1 h∗1 (x)σ1 · · · hb (x)δb h∗b (x)σb , where 0 ≤ τi , δj , σj ≤ ps , for any i = 1, 2, ..., a, and j = 1, 2, ..., b. Then, we have s

h(x) = f1 (x)p

−τ1

s

f2 (x)p

−τ2

s

· · · fa (x)p

−τa

s

h1 (x)p

s −δ1 ∗ h1 (x)p −σ1

s

· · · hb (x)p

s −δb ∗ hb (x)p −σb .

Therefore, h∗ (x) = f1 (x)p

s

−τ1

s

f2 (x)p

−τ2

s

···fa (x)p

−τa

s

h1 (x)p

s s s −σ1 ∗ h1 (x)p −δ1 ···hb (x)p −σb h∗b (x)p −δb .

If C is a self-dual cyclic code, we get the following theorem. Theorem 5.1. With the above notations, we have that C is a self-dual cyclic code if and only if 2τi = ps , 1 ≤ i ≤ a, and δj + σj = ps , 1 ≤ j ≤ b.

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288

Proof. C is a self-dual cyclic code if and only if g(x) = h∗ (x), i.e., 2τi = ps , 1 ≤ i ≤ a, and δj + σj = ps , 1 ≤ j ≤ b. 2 According to this theorem, we see that it is enough to determine the irreducible s factorization of x3lp − 1 as above. And if we do this, we can give all self-dual cyclic codes immediately. Similar to the definition of reciprocal polynomial, we give the following definition. Definition 5.2. Let Cs = {s, sq, ..., sq f −1 } be any q-cyclotomic coset modulo l, then Cs∗ = {−s, −sq, ..., −sq f −1 } is said to be the reciprocal coset of Cs . The coset Cs is called self-reciprocal if Cs = Cs∗ . Obviously, Cs∗ is still a q-cyclotomic coset modulo l. And the reciprocal polynomial of the minimal polynomial of Cs is the minimal polynomial of Cs∗ . Hence, the minimal polynomial of Cs is also self-reciprocal if Cs is self-reciprocal. Lemma 5.3. Assume that q ≡ 1(mod 3). For the q-cyclotomic cosets, which have been described in Lemma 2.1, one of the following holds: (I) If f = ordl (q) is even, we have ∗ B0∗ = B0 , Bl∗ = B−l , Bg∗k = B−gk , B3g k = B3g k ,

where 0 ≤ k ≤ e − 1. (II) If f = ordl (q) is odd, we have ∗ B0∗ = B0 , Bl∗ = B−l , Bg∗k = B−gk , B3g k = B−3g k

where

e−1 k=0

B3gk =

e2 −1

k =0 (B3g k





B−3gk ).

∗ Proof. (I) Obviously, we only need to prove B3g k = B3g k . If f = ordl (q) is even, we f

deduce that q 2 ≡ −1(mod l). According to this, we get there exist i, j, 0 ≤ i, j ≤ f − 1, and |j − i| = f2 , such that 3g k q i ≡ −3g k q j (mod 3l), for any 3g k q i ∈ B3gk , 0 ≤ k ≤ e − 1. ∗ Therefore, we have B3g k = B3g k , for any k = 0, 1, ..., e − 1. (II) In the same way as Lemma 2.1, we get that B0 , Bl , B−l , Bgk , B−gk , B3gk and B−3gk , 0 ≤ k ≤ e − 1, 0 ≤ k ≤ 2e − 1, are all the distinct q-cyclotomic cosets modulo 3l. The next result is obvious. 2 Lemma 5.4. Let q ≡ 2(mod 3) and f be even. For the q-cyclotomic cosets, which have been described in Lemma 2.1, one of the following holds:

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(I) When f = 2t and t is even, we have ∗ B0∗ = B0 , Bl∗ = Bl , Bg∗k = B−gk , B3g k = B3g k ,

e−1 2e−1 where k =0 Bgk = k=0 (Bgk B−gk ). (II) When f = 2t and t is odd, we have ∗ B0∗ = B0 , Bl∗ = Bl , Bg∗k = Bgk , B3g k = B3g k ,

where 0 ≤ k ≤ e − 1 and 0 ≤ k ≤ 2e − 1. Proof. (I) In the same way as Lemma 2.1, we get B0 , Bl , Bgk , B−gk and B3gk , 0 ≤ k ≤ e −1, are all the distinct q-cyclotomic cosets modulo 3l. Next, we first prove that Bl∗ = Bl , i.e., {l, lq}∗ = {l, lq}. As q ≡ 2(mod 3), i.e., q ≡ −1(mod 3), then lq ≡ −l(mod 3). Since gcd(3, l) = 1, we have lq ≡ −l(mod 3l), which implies Bl∗ = Bl . Otherwise, by the conclusion (I) of Lemma 5.3, we get the other results immediately. (II) According to (I), it is obvious that we only need to prove Bg∗k = Bgk . As t is odd, we have q t ≡ −1(mod 3). Since q t ≡ −1(mod l) and gcd(3, l) = 1, we get q t ≡ −1(mod 3l). Then, we deduce that there exist i, j, 0 ≤ i, j ≤ f − 1, and |j − i| = t, such    that g k q i ≡ −g k q j (mod 3l), for any g k q i ∈ Bk , 0 ≤ k ≤ 2e − 1, i.e., Bg∗k = Bgk . 2 Lemma 5.5. Let q ≡ 2(mod 3) and f be odd. For the q-cyclotomic cosets, which have been described in Lemma 2.1, we have ∗ B0∗ = B0 , Bl∗ = Bl , Bg∗k = B−gk , B3g k = B−3g k ,

where

e−1 k=0

Bgk =

e2 −1

k =0 (Bg k





B−gk ) and

e−1 k=0

B3gk =

e2 −1

k =0 (B3g k





B−3gk ).

From the above lemmas, we can give all self-dual cyclic codes of length 3 · 2s l over F2m and their enumeration in the following theorem. Theorem 5.6. Let l = 3 be an odd prime, p = 2, f = ordl (2m ), and e = cyclic self-dual codes of length 3 · 2s l over F2m , we have: (I) When q ≡ 1(mod 3), one of the following holds:

l−1 f .

Then, for

(i) If f = ordl (q) is even, then there exist (2s + 1)e+1 cyclic self-dual codes of length 3 · 2s l over F2m given by s−1

(x − 1)2

s

Bl (x)δ B−l (x)2

−δ

e−1 

s

Bgk (x)δk B−gk (x)2

k=0

where 0 ≤ δ, δk ≤ 2s , for any 0 ≤ k ≤ e − 1.

−δk

s−1

B3gk (x)2

,

290

L. Liu et al. / Finite Fields and Their Applications 42 (2016) 269–295

3e

(ii) If f = ordl (q) is odd, then there exist (2s + 1) 2 +1 cyclic self-dual codes of length 3 · 2s l over F2m given by s−1

(x − 1)2

s

Bl (x)δ B−l (x)2

−δ

e−1 

s

Bgk (x)δk B−gk (x)2

−δk

k=0 e 2 −1

×



s

B3gk (x)σk B−3gk (x)2

−σk

,

k =0

where 0 ≤ δ, δk , σk ≤ 2s , for any 0 ≤ k ≤ e and 0 ≤ k ≤

e 2

− 1.

(II) When q ≡ 2(mod 3), we have: (i) If f = 2t and t is even, then there exist (2s + 1)e cyclic self-dual codes of length 3 · 2s l over F2m given by s−1

e−1 

s−1

(x − 1)2

Bl (x)2

s

Bgk (x)δk B−gk (x)2

−δk

s−1

B3gk (x)2

,

k=0

where 0 ≤ δk ≤ 2s , for any 0 ≤ k ≤ e − 1. (ii) If f = 2t and t is odd, then there exists only one cyclic self-dual code of length 3 · 2s l over F2m given by s−1

(x − 1)2

s−1

Bl (x)2

2e−1 

s−1

Bgk (x)2

k =0

e−1 

s−1

B3gk (x)2

.

k=0

(iii) If f is odd, then there exist (2s + 1)e cyclic self-dual codes of length 3 · 2s l over F2m given by (x − 1)

2s−1

Bl (x)

2s−1

e 2 −1



s

Bgk (x)δk B−gk (x)2

−δk

s

B3gk (x)σk B−3gk (x)2

−σk

,

k =0

where 0 ≤ δk , σk ≤ 2s , for any 0 ≤ k ≤

e 2

− 1.

6. A discussion of the minimum Hamming distance In this section, we will give the minimum Hamming distance of these codes when l = 1 and gcd(q − 1, 3) = 1. From Section 4, we have that all constacyclic codes of length 3ps over Fq are equivalent to the cyclic codes when gcd(q − 1, 3) = 1. Therefore, we only need to determine the minimum Hamming distance of the cyclic codes, in this case. It is clear that all cyclic codes of length 3ps have the form C = (x − 1)i (x2 + x + 1)j  with 0 ≤ i, j ≤ ps , when l = 1 and gcd(q − 1, 3) = 1. If i = 0 and j = 0, then

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291

s

C = Fq /x3p − 1. If i = ps and j = ps , then C = {0}. These are the trivial cyclic codes of length 3ps over Fq . Next, we determine the minimum Hamming distance of all non-trivial cyclic codes of length 3ps over Fq . We denote the minimum Hamming distance of a code C by dH (C). Lemma 6.1. Let C be a non-trivial cyclic code of length 3ps over Fq . Then dH (C) ≥ 2. Proof. If there exists a codeword c ∈ C with weight 1, then c must be a unite. This is impossible, as C is a non-trivial cyclic code of length 3ps over Fq . 2 Lemma 6.2. Let C = (x − 1)i (x2 + x + 1)j , for some 0 ≤ i, j ≤ ps−1 and (i, j) = (0, 0), be a non-trivial cyclic code of length 3ps over Fq . Then dH (C) = 2. s−1

− 1) with Proof. We could see that dH (C) ≤ 2, as (x − 1)i (x2 + x + 1)j | (x3p 0 ≤ i, j ≤ ps−1 and (i, j) = (0, 0). From Lemma 6.1, the result is obvious. 2 Let g(x) = (x − 1)i (x2 + x + 1)j , where i > ps−1 or j > ps−1 but not i = j = ps . Similarly to [21], we can compute the Hamming distance of code C = g(x). Let t be an integer such that 0 ≤ t ≤ ps − 1. We define the simple-root cyclic code Ct = gt (x) ⊂ Fq [x]/x3 − 1 depending on C and t, where gt (x) = (x − 1)ei,t (x2 + x + 1)ej,t with ei,t =

1, if i > t, and ej,t = 0, otherwise,

1, if j > t, 0, otherwise.

Then, it is clear that ⎧ ⎪ ⎪ 1, if gt = 1, ⎪ ⎨ 2, if g = x − 1, t dH (Ct ) = ⎪ 3, if gt = x2 + x + 1, ⎪ ⎪ ⎩ ∝ if g = x3 − 1. t

(2)

For 0 ≤ t ≤ ps − 1, let t = t0 + t1 p + ... + ts−1 ps−1 be the p-adic representation of t, s−1 and Pt = m=0 (tm + 1). We define the set T = {t : t = (p − 1)ps−1 + (p − 1)ps−2 + ... + (p − 1)ps−(j−1) + rps−j , 1 ≤ j ≤ s, 1 ≤ r ≤ p − 1} {0}. Let β, τ , k be integers such that 0 ≤ β ≤ p − 2, 1 ≤ τ ≤ p − 1 and 1 ≤ k ≤ s − 1. From [21, Lemma 6], we have min{Pt : t ≥ l and t ∈ T } = β + 2,

(3)

where l is an integer such that βps−1 ≤ l ≤ (β + 1)ps−1 . From [21, Lemma 7], we have min{Pt : t ≥ l and t ∈ T } = (τ + 1)pk ,

(4)

where l is an integer such that ps −ps−k +(τ −1)ps−k−1 +1 ≤ l ≤ ps −ps−k +τ ps−k−1 +1. C, Ct and Pt are defined as above. By Ref. [21], we know that

292

L. Liu et al. / Finite Fields and Their Applications 42 (2016) 269–295

dH (C) = min{Pt dH (Ct ) : t ∈ T }.

(5)

Next, we will begin to discuss the minimum Hamming distance of all non-trivial cyclic codes when i ≥ j. For j > i, the results of the minimum Hamming distance can be computed by the similar technique, and the statements are given in Table 2. Lemma 6.3. Let C = (x − 1)i (x2 + x + 1)j , where 2ps−1 ≥ i > ps−1 ≥ j > 0. Then dH (C) = 3. Proof. According to (3), we have min{Pt : t ∈ T and i > t ≥ j} = 2. If i > t ≥ j, we have dH (Ct ) = 2 by (2), thus min{Pt dH (Ct ) : t ∈ T and i > t ≥ j} = 4. If t ≥ i > j and 2ps−1 ≥ i > ps−1 , we get min{Pt : t ∈ T and t ≥ i > j} = 3 by (3), and dH (Ct ) = 1 by (2). Thus min{Pt dH (Ct ) : t ∈ T and t ≥ i > j} = 3. From (5), we get dH (C) = 3. 2 Lemma 6.4. Let C = (x − 1)i (x2 + x + 1)j , where ps ≥ i > 2ps−1 and ps−1 ≥ j > 0. Then dH (C) = 4. Proof. Similarly to the proof of Lemma 6.3, we get dH (C) = 4. 2 Lemma 6.5. Let i ≥ j be integers with βps−1 + 1 ≤ i ≤ (β + 1)ps−1 and β  ps−1 + 1 ≤ j ≤ (β  + 1)ps−1 , where 1 ≤ β  ≤ β ≤ p − 2 are integers. Let C = (x − 1)i (x2 + x + 1)j , then dH (C) = min{β + 2, 2(β  + 2)}. Proof. First we assume that β = β  , then there must exist no t such that i > t > j. So, we only need to consider t ≥ i ≥ j. We have min{Pt : t ∈ T and t ≥ i ≥ j} = β + 2 by (3), and dH (C) = 1 by (2). Thus, we have min{Pt dH (Ct ) : t ∈ T and t ≥ i ≥ j} = β + 2. Next, suppose that β > β  , then we will get min{Pt : t ∈ T and i > t ≥ j} = β  + 2 by (3), and dH (C) = 2 by (2). Therefore, we have min{Pt dH (Ct ) : t ∈ T and i > t ≥ j} = 2(β  + 2). If t ≥ i ≥ j, we will get min{Pt : t ∈ T and t ≥ i ≥ j} = β + 2 by (3), and dH (C) = 1 by (2). Therefore, we have min{Pt dH (Ct ) : t ∈ T and t ≥ i ≥ j} = β + 2. According to (5), we could get dH (C) = min{β + 2, 2(β  + 2)}.

2

Lemma 6.6. Let i, j be integers with ps −ps−k +(τ −1)ps−k−1 +1 ≤ i ≤ ps −ps−k +τ ps−k−1 and βps−1 + 1 ≤ j ≤ (β + 1)ps−1 , where 1 ≤ τ ≤ p − 1, 1 ≤ β ≤ p − 2 and 1 ≤ k ≤ s − 1. Let C = (x − 1)i (x2 + x + 1)j , then dH (C) = 2(β + 2).

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293

Proof. Combining (2), (3) and (4), we deduce that min{Pt dH (Ct ) : t ∈ T and i > t ≥ j} = 2(β + 2), min{Pt dH (Ct ) : t ∈ T and t ≥ i > j} = (τ + 1)pk . It is clear that 2(β + 2) ≤ 2p ≤ (τ + 1)pk , as β ≤ p − 2, 1 ≤ τ and 1 ≤ k. Therefore, we could obtain dH (C) = 2(β + 2). 2 By the similar arguments as above, we can obtain the following lemmas immediately. Lemma 6.7. Let i, j be integers with ps −ps−k +(τ −1)ps−k−1 +1 ≤ i ≤ ps −ps−k +τ ps−k−1 and ps −ps−k +(τ  −1)ps−k−1 +1 ≤ j ≤ ps −ps−k +τ  ps−k−1 , where 1 ≤ τ  ≤ τ ≤ p −1 and 1 ≤ k ≤ s − 1. Let C = (x − 1)i (x2 + x + 1)j , then dH (C) = min{2(τ  + 1)pk , (τ + 1)pk }. Lemma 6.8. Let i, j be integers with ps −ps−k +(τ −1)ps−k−1 +1 ≤ i ≤ ps −ps−k +τ ps−k−1     and ps − ps−k + (τ  − 1)ps−k −1 + 1 ≤ j ≤ ps − ps−k + τ  ps−k −1 , where 1 ≤ τ  , τ ≤ p − 1  and 1 ≤ k < k ≤ s − 1. Let C = (x − 1)i (x2 + x + 1)j , then dH (C) = 2(τ  + 1)pk . s

Lemma 6.9. Let C = (x − 1)p (x2 + x + 1)j , then ⎧ ⎪ ⎨ 4, dH (C) = 2(β + 2), ⎪ ⎩ 2(τ + 1)pk ,

if 0 < j ≤ ps−1 , if βps−1 + 1 ≤ j ≤ (β + 1)ps−1 , if ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ ps−k−1 ,

where 1 ≤ τ ≤ p − 1, 1 ≤ β ≤ p − 2 and 1 ≤ k ≤ s − 1. According to the lemmas as above, we have the following theorem. Theorem 6.10. The minimum Hamming distance of all non-trivial cyclic codes is given in Table 1 with i ≥ j, when gcd(3, q − 1) = 1 and l = 1. For i < j, the minimum Hamming distance of all non-trivial cyclic codes is also shown in Table 2. Remark. It is no doubt that results on the minimum Hamming distance of codes are very important in error-correcting coding theory. So, it would be a worthwhile problem to discuss the minimum Hamming distance of these codes. As is known to all, the minimum Hamming distance of constacyclic codes has been an interesting subject for many years. However, determining explicitly the minimum Hamming distance of them, in general, is a very hard problem. In this paper, we only determine the minimum Hamming distance of these codes when gcd(3, q − 1) = 1 and l = 1. For other cases, the discussions on the minimum Hamming distance of all repeated-root constacyclic codes are not given explicitly. A new or improved technique is needed to discuss the more general cases. We

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Table 1 The minimum Hamming distance of all non-trivial cyclic codes, of the form (x −1)i (x2 +x +1)j  with i ≥ j, of length 3ps over Fq . The parameters 1 ≤ β  ≤ β ≤ p − 2, 1 ≤ τ (2) ≤ τ (1) ≤ p − 1, 1 ≤ τ, τ (3) , τ (4) ≤ p − 1, 1 ≤ k ≤ s − 1, 1 ≤ k < k ≤ s − 1 are integers. Case

i

j

dH (C)

1 2 3 4 5

0 < i ≤ ps 0 ≤ i ≤ ps−1 ps−1 < i ≤ 2ps−1 2ps−1 < i ≤ ps βps−1 + 1 ≤ i ≤ (β + 1)ps−1

j=0 0 ≤ j ≤ ps−1 0 < j ≤ ps−1 0 < j ≤ ps−1 β  ps−1 + 1 ≤ j ≤ (β  + 1)ps−1

6

ps − ps−k + (τ − 1)ps−k−1 + 1 i ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ − 1)ps−k−1 + 1 i ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ 1 − 1)ps−k−1 + 1 i ≤ ps − ps−k + τ 1 ps−k−1 + 1   ps −ps−k +(τ 3 −1)ps−k −1 +1 s s−k 3 s−k 1 i≤p −p +τ p +1 i = ps s i=p

2 2 3 4 min{β + 2, 2(β  + 2)} 2(β + 2)

7 8 9 10 11

≤

βps−1 + 1 ≤ j ≤ (β + 1)ps−1

≤

ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ 2 − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ 2 ps−k−1 + 1   ps −ps−k +(τ 4 −1)ps−k −1 +1 ≤ s s−k 4 s−k −1 j ≤p −p +τ p +1 βps−1 + 1 ≤ j ≤ (β + 1)ps−1 s s−k s−k−1 p −p + (τ − 1)p +1 ≤ j ≤ ps − ps−k + τ ps−k−1 + 1

≤ ≤

(τ + 1)pk min{2(τ 2 + 1)pk , (τ 1 + 1)pk }  2(τ 4 + 1)pk 2(β + 2) 2(τ + 1)pk

Table 2 With respect to the above notations, the minimum Hamming distance of all non-trivial cyclic codes, of the form (x − 1)i (x2 + x + 1)j  with j > i, of length 3ps over Fq . Case

i

j

dH (C)

1 2 3 4 5 6

0 ≤ i ≤ ps−1 i=0 0 < i ≤ ps−1 0 < i ≤ ps−1 0 < i ≤ ps−1 β  ps−1 + 1 ≤ i ≤ (β  + 1)ps−1

0 < j ≤ ps−1 2ps−1 < j ≤ ps 2ps−1 < j ≤ 3ps−1 3ps−1 < j ≤ 4ps−1 4ps−1 + 1 ≤ j ≤ ps βps−1 + 1 ≤ j ≤ (β + 1)ps−1

7

βps−1 + 1 ≤ i ≤ (β + 1)ps−1

8

ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ i ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ 2 − 1)ps−k−1 + 1 ≤ i ≤ ps − ps−k + τ 2 ps−k−1 + 1   ps −ps−k +(τ 4 −1)ps−k −1 +1 ≤ s s−k 4 s−k −1 i≤p −p +τ p +1 βps−1 + 1 ≤ i ≤ (β + 1)ps−1 ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ i ≤ ps − ps−k + τ ps−k−1 + 1

ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ ps−k−1 + 1 ps − ps−k + (τ 1 − 1)ps−k−1 + 1 ≤ j ≤ ps − ps−k + τ 1 ps−k−1 + 1   ps −ps−k +(τ 3 −1)ps−k −1 +1 ≤ s s−k 3 s−k −1 j ≤p −p +τ p +1 j = ps j = ps

2 3 4 5 6 min{β + 2, 3(β  + 2)} min{(τ + 1)pk , 3(β + 2)} (τ + 1)pk

9 10 11 12

min{3(τ 2 + 1)pk , (τ 1 + 1)pk }  3(τ 4 + 1)pk 3(β + 2) 3(τ + 1)pk

will discuss the minimum Hamming distance of these codes of more general lengths in further studies. Acknowledgments The authors would like to thank the referees for their helpful comments and meticulous reading of this manuscript. Their suggestions are all valuable and very helpful for revising and improving our paper, as well as an important guide to our researches.

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