Finite Fields and Their Applications 79 (2022) 101993 Contents lists available at ScienceDirect Finite Fields and Their Applications www.elsevier.com/locate/ffa On LCD, self dual and isodual cyclic codes over finite chain rings ✩ Amel Benyettou a , Aicha Batoul a , Cristina Fernández-Córdoba b,∗ a Faculty of Mathematics USTHB, University of Science and Technology of Algiers, Algeria b Department of Information and Communication Engineering, University Autonomous of Barcelona, Spain a r t i c l e i n f o a b s t r a c t Article history: In this paper, LCD cyclic, self dual and isodual codes over Received 24 May 2021 finite chain rings are investigated. It was proven recently Received in revised form 21 that a non-free LCD cyclic code does not exist over finite December 2021 chain rings. Based on algebraic number theory, we introduce Accepted 28 December 2021 Available online 13 January 2022 necessary and sufficient conditions for which all free cyclic Communicated by W. Cary Huffman codes over a finite chain ring are LCD. We have also obtained conditions on the existence of non trivial self dual cyclic codes MSC: of any length when the nilpotency index of the maximal ideal 94B25 of a finite chain ring is even. Further, several constructions 94B60 of isodual codes are given based on the factorization of the polynomial xn − 1 over a finite chain ring. Keywords: © 2022 The Author(s). Published by Elsevier Inc. This is an LCD open access article under the CC BY-NC-ND license Isodual (http://creativecommons.org/licenses/by-nc-nd/4.0/). Self dual Cyclic codes Finite chain rings ✩ This work has been partially supported by the Spanish MINECO under Grant PID2019-104664GB-I00 (AEI/FEDER, UE). * Corresponding author. E-mail addresses: benyettouamel@yahoo.fr (A. Benyettou), a.batoul@hotmail.fr (A. Batoul), cristina.fernandez@uab.cat (C. Fernández-Córdoba). https://doi.org/10.1016/j.ffa.2022.101993 1071-5797/© 2022 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). 2 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 1. Introduction Codes over finite rings were studied in the early 1970s, and this study has grown enormously since the seminal work of Hammons et al. [11], where it is shown in that some of the best nonlinear codes over F2 can be viewed as linear codes over Z4 . Linear complementary dual or LCD codes are linear codes that intersect with their dual trivially. LCD cyclic codes have applications in data storage. Due to a newly discovered application in cryptography [6,9], interest in LCD codes has increased again. An LCD code defined over a finite field Fq which is also known as reversible code was first introduced by Massey in [17]. Following his first study, Massey also showed the existence of asymptotically good LCD codes. Furthermore, Yang and Massey in [20] pro- vided a necessary and sufficient condition under which a cyclic code has a complementary dual. In [15], Liu and Liu studied LCD codes over finite chain rings and provided a nec- essary and sufficient condition for a free linear code to be LCD. In [14], Lina and Nocon give parameters of some LCD codes using generator matrices and give some methods to construct new LCD from previous ones. Recently, in [5], existence conditions are given for LCD codes over F2 which are images under the Gray map of additive codes over β Zα2 × Z4 . Finally, in [7], the authors proved that there are no non-free LCD codes over finite commutative local Frobenius rings. It was also shown that a free cyclic code C over any finite chain ring is LCD code if and only if C is reversible when the length n of the code is relatively prime to the characteristic of a finite chain ring. Several types of equivalence between codes can be defined, but here we only consider monomial equiva- lence. Two codes C and C  are called monomially equivalent if there exists a monomial permutation which sends C to C  . Isodual codes are codes which are monomially equiv- alent to their duals. The class of isodual codes is important in coding theory because it contains the self dual codes as a subclass. In addition, isodual codes are contained in the larger class of formally self dual codes. In [1,3,4] the authors gave some specific constructions of self dual and isodual codes over finite fields and finite chain rings. The purpose of this paper is to examine linear codes with complementary duals and isodual codes over finite chain rings. We give necessary and sufficient conditions for which all codes are LCD, and generalize the result given in [2] by giving condition on the existence of non trivial self dual codes. Some of the codes obtained are both isodual and LCD, and so are called LCD-isodual codes. The paper is organized as follows, the necessary background material on codes over finite chain rings is given in Section 2. In Section 3, we give the first part of the main results of this paper. Based on algebra number theory properties, we provide conditions under which all free cyclic codes over finite chain rings are LCD. In Section 4, we generalize the result given in [2] of the existence of non trivial self dual codes when the nilpotency index of the maximal ideal of the finite chain ring considered is even. In Section 5, the structure of cyclic codes of length 2a m over finite chain rings is given along with conditions on the existence of isodual cyclic codes. Using these results, several constructions of isodual are given. A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 3 2. Preliminaries We begin with some definitions and properties about finite chain rings. Let R be a finite commutative ring with identity. A commutative ring is called a chain ring if the lattice of all its ideals is a chain. It is well known that if R is a finite chain ring, then R is a principal ideal ring and it has a unique maximal ideal γ. Its chain of ideals is {0} = γ e   γ e−1   . . .  γ  R, where γ i  (with i ∈ {1, · · · , e}) is the ideal of R generated by γ i . The integer e is called the nilpotency index of γ. The nilradical of R is γ, so all the elements of γ are nilpotent. Therefore, the elements of R\γ are units. We denote this group by R∗ . Since γ is maximal, the residue ring R/γ is a field with q elements which we denote by Fq . Let |R| denote the cardinality of R, then |R| = |Fq | · |γ| = |Fq | · |Fq |e−1 = |Fq |e = q e . Moreover, from [19], |γ i | = |Fq |e−i = q e−i for i = 1, 2, . . . , e − 1 The natural surjective ring morphism is given by − : R −→ Fq (1) a −→ ā = a mod γ The map given in (1) extends naturally to a map from R[x] −→ Fq [x]. A polynomial f (x) of R[x] is called basic irreducible if f (x) is irreducible in Fq [x]. Two polynomials f1 (x) and f2 (x) in R[x] are called coprime if f1 (x) + f2 (x) = R[x] or equivalently there exist λ1 (x) and λ2 (x) in R[x] such that λ1 (x)f1 (x) + λ2 (x)f2 (x) = 1. A polynomial f (x) of R[x] is a unit if and only if f (x) is a unit and also f is a zero divisor if and only if f (x) = 0. Hensel’s lemma is an important tool for studying finite chain rings, which can lift the factorization into a product of pairwise coprime polynomials over Fq to such factorization over R. Lemma 1. [18, Hensel lifting, Theorem XIII.4] Let g(x) in R[x] be monic. Assume that there are monic, pairwise coprime polynomials f1 (x), f2 (x), . . . , fk (x) in Fq [x] i=k such that g(x) = i=1 fi (x), then there are  monic pairwise coprime polynomials i=k g1 (x), g2 (x), . . . , gk (x) in R[x] such that g(x) = i=1 gi (x) and gi (x) = fi (x), for all 0 ≤ i ≤ k. Lemma 2. [19, Theorem 2.7] If f (x) is a monic polynomial over R such that f (x) is square free, then f (x) factors uniquely as product of monic basic irreducible pairwise coprime polynomials. 4 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Let R be a finite chain ring, Fq its residue field, and γ its maximal ideal with nilpotency index e. A linear code C of length n over R is a submodule of Rn . A linear code C of length n over R is called cyclic if (cn−1 , c0 , c1 , . . . , cn−2 ) is in C whenever (c0 , c1 , . . . , cn−1 ) is in C. Each codeword c where c = (c0 , c1 , ..., cn−1 ) is customarily identified with its polynomial representation c(x) where c(x) = c0 + c1 x + ... + cn−1 xn−1 . In this way, any cyclic code of length n over R is identified with exactly one ideal of the ring R[x]/xn − 1 and xc(x) correspond to a cyclic shift of c(x). A monomial transformation over Rn is an R-linear homomorphism τ such that there exist scalars λ1 , λ2 , ..., λn in R∗ and a permutation σ in Sn , where Sn is the group of permutation of {1, 2, ..., n}, such that for all (x1 , x2 , ..., xn ) in Rn , we have τ (x1 , x2 , ..., xn ) = (λ1 xσ(1) , λ2 xσ(2) , ..., λn xσ(n) ). Recall that two codes are called equivalent if there is a monomial permutation which sends one to the other. In this paper, whenever we say that two codes are equivalent, we mean that they are monomially equivalent. Suppose that f (x) = a0 + a1 x + ... + ar xr is a polynomial of R[x] of degree r such that f (0) = a0 is a unit in R. The monic reciprocal polynomial of f (x) is defined by f ∗ (x) = f (0)−1 xr f (x−1 ). If f ∗ (x) = f (x), the polynomial f (x) is called self reciprocal. A code C over R is reversible if for each code word (c0 , c1 , ..., cn−1 ) in C implies that the code word (cn−1 , cn−2 , ..., c0 ) is also in C. It is known that a cyclic code C is reversible if and only if its generator polynomial is self reciprocal. We attach the standard inner product to Rn  n x·y = xi yi , for each x = (x1 , x2 , ..., xn ) and y = (y1 , y2 , ..., yn ) in Rn . i=1 The Euclidean dual code C ⊥ of C is defined as C ⊥ = {x ∈ Rn : ∀y ∈ C; x · y = 0}. A code C is said to be self dual if C = C ⊥ , it is isodual if C = τ (C ⊥ ), where τ is a monomial transformation, and it is called LCD or linear complementary dual if C ∩ C ⊥ = {0}. A code C over a finite chain ring and its dual satisfies the following |C||C ⊥ | = q en = |R|n and (C ⊥ )⊥ = C. The following theorem gives the structure of a cyclic code and its dual over a finite chain ring. A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 5 Lemma 3. [8, Theorem 3.8] Let R be a finite chain ring with maximal ideal γ with nilpotency index e and residue field Fq . Let C be a cyclic code over R of length n such that gcd(n, q) = 1, where p is the characteristic of R̄. Then there exists a unique family of pairwise coprime polynomials Fi (x), 0 ≤ i ≤ e in R[x] satisfying F0 (x)F1 (x)...Fe (x) = xn − 1 such that     C = F̂1 (x), γ F̂2 (x), ..., γ e−1 F̂e (x) , C ⊥ = F̂0∗ (x), γ F̂e∗ (x), ..., γ e−1 F̂2∗ (x) , xn − 1 where F̂i (x) = for 0 ≤ i ≤ e. Moreover, we have that the ring R[x]/ xn − 1 is Fi (x) a principal ideal ring. In particular, when the code is free as a submodule, we have the following statement. Lemma 4. [10, Theorem 4.16] Let C be a cyclic code of length n over a finite chain ring R with residue field Fq such that gcd(n, q) = 1. Then, C is a free cyclic code with rank k if and only if there is a monic polynomial f (x) in R[x] such that f (x) divides xn − 1 and f (x) generates C. In this case, we have   k = n − deg(f ). Further the dual code of C ∗ xn − 1 is also free and it is generated by . f (x) 3. On LCD cyclic codes over finite chain rings The aim of this section is to present some new constructions of LCD cyclic codes, and provide necessary and sufficient conditions for the existence of non trivial LCD cyclic codes over finite chain rings. Let n be a positive integer and q a prime power coprime to n. We denote by ordn (q) the multiplicative order of q modulo n. This is the smallest integer l such that q l ≡ 1 mod n. To process cyclic codes of length n, we have to study the factorization into irreducible polynomials of xn − 1 over Fq . To this end, we need to introduce the q-cyclotomic cosets modulo n. Note that xn − 1 has no repeated factors over Fq if and only if gcd(n, q) = 1. For any s in {0, 1, 2, ..., n − 1}, the q-cyclotomic coset of s modulo n is defined by Cs = s, sq, sq 2 , ..., sq ls −1 , where ls is the smallest positive integer such that s ≡ sq ls ( mod n), and is the size of the q-cyclotomic coset. The smallest integer in Cs is called the coset leader of Cs . Let Pn,q be the set of all the coset leaders. We have then Cs ∩ Ct = ∅ for any two distinct elements s and t in Pn,q , and Cs = {0, 1, 2, ..., n − 1}. s∈Pn,q 6 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Hence, the distinct q-cyclotomic cosets modulo n partition {0, 1, 2, ..., n − 1}. The cyclotomic coset Cs is said to be reversible if and only if Cn−s = Cs if and only if n − s is in Cs . Lemma 5. [2, Lemma 4] If C1 is reversible then Cs is reversible for all s in Pn,q . q r −1 Let r = ordn (q), and let α be a generator of (Fqr )∗ . Put β = α n , then β is a primitive n-th root of unity in Fqr . The minimal polynomial ms (x) of β s for s in Pn,q over Fqr is given by   ms (x) = x − βj , j∈Cs which is irreducible over Fq , and hence the factorization of xn − 1 into irreducible factors over Fq is given by xn − 1 = ms (x). s∈Pn,q The following lemma is well known in literature. Lemma 6. [13, Lemma 5] The minimal polynomial ms (x) is self reciprocal if and only if the cyclotomic coset associated Cs is reversible. 3.1. Some properties of positive integers In this section, we give some properties of positive integers which will be needed later. Lemma 7. Let q be a prime power, p an odd prime number coprime to q, then we have (i) If ordp (q) is even then for all k in N ∗ , ordpk (q) is even. (ii) If there is k in N ∗ such that ordpk (q) is even, then ordp (q) is also even. Proof. Since p divides pk , we have q ordpk (q) ≡ 1 mod pk implies that q ordpk (q) ≡ 1 mod p. Hence ordp (q) | ordpk (q), therefore if ordp (q) is even then ordpk (q) is even too. To prove (ii), assume that there is k in N ∗ such that ordpk (q) is even, and by way of contradiction we suppose that ordpk−1 (q) is odd. Therefore, there exist some in- teger i and there exists m in N, such that q 2i+1 = 1 + mpk−1 . Since p is a prime   number, it divides the binomial coefficient pj for all 1 ≤ j ≤ p − 1. Hence we get   p (q 2i+1 )p = 1 + mpk−1 ≡ 1 mod pk . It follows that ordpk (q) | (2i + 1)p. Since (2i + 1)p is odd, this leads to a contradiction. So that ordpk−1 (q) must be even, and by descending recurrence we get that ordp (q) is even.  A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 7 Lemma 8. Let q be a prime power and p an odd prime number such that gcd(p, q) = 1. The three following statements are equivalents. (i) There exits l in N, such that q l ≡ −1 mod p. (ii) For all k in N, there exists lk in N, such that q lk = −1 mod pk . (iii) There is i in N ∗ such that ordpi (q) is even. Further, if q l ≡ −1 mod p, then l = 12 (1 + 2m)ordp (q) for some m in N. Proof. Suppose that (i) is satisfied and we prove (ii) by induction. For k = 1 we have q l ≡ −1 mod p. Assume q lk−1 ≡ −1 mod pk−1 for k ≥ 2. Since p is odd, we can write −1 pk−1 (−q lk−1 )p k−1 −1 q lk−1 p +1 k−1 lk−1 i (−q ) = = . i=0 (−q lk−1 ) − 1 q lk−1 + 1 On the other hand, we have −1 pk−1 −1 pk−1 −1 pk−1 (−q lk−1 i ) = (−1) (q i ) ≡ lk−1 i (−1)i (−1)i mod pk−1 ≡ 0 mod pk−1 , i=0 i=0 i=0 pk−1 −1 which means that pk−1 | i=0 (−q lk−1 )i . Since p | pk−1 | q lk−1 + 1, it follows that −1 pk−1 k−1 p | (q k lk−1 + 1)( (−q lk−1 )i ) = q lk−1 p + 1. i=0 Thus, for lk = lk−1 · pk−1 , we have that q lk ≡ −1 mod pk . Note that when q l ≡ −1 k(k−1) mod p, then lk = lk−1 · pk−1 = lk−2 · pk−2 · pk−1 . We obtain that lk = l · p 2 . Conversely, if the statement (ii) holds, then the statement (i) follows immediately for k = 1. Assume that (iii) is satisfied. Lemma 7 shows that the integer ordp (q) is also even. 1 1 We have, q ordp (q) ≡ 1 mod p if and only if p | (q 2 ordp (q) − 1)(q 2 ordp (q) + 1). Since p is 1 prime it must divide one of the factors and it can not divide (q 2 ordp (q) − 1) because of 1 the definition of the order of q, thus q 2 ordp (q) = −1 mod p. Conversely, if (ii) is satisfied, then by Lemma 8 there exits l in N ∗ such that q l ≡ −1 mod p, which means q 2l ≡ 1 mod p, so that ordp (q) | 2l. If ordp (q) is odd, then ordp (q) | l, which contradicts the fact that q l ≡ −1 mod p. Hence ordp (q) must be even and (iii) holds. It remains to prove that if there exists an integer l such that q l ≡ −1 mod p, then l = 12 (1 + 2m)ordp (q) for some m in N. By definition of order, the integers q ordp (q) and q j are distinct for all 1 ≤ jordp (q). Since p is odd, we obtain that if l is the smallest  integer such that q l ≡ −1 mod p, then on the one hand 1 ≤ l ordp (q) and, on the 8 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 other hand, ordp (q) | 2l . This gives 2l = λordp (q) and λ ≥ 1. Since l ordp (q), then l = 12 ordp (q). Further, if l is an integer that satisfies q l ≡ −1 mod p, then by division algorithm we can write l = sl + r with rl . Hence, we get   q l = q sl +r = (q l )s q r ≡ (−1)s q r mod p ≡ −1 mod p. Which forces that s is odd and r = 0. Thus, 1 l = (2m + 1)l = (2m + 1)ordp (q).  2 Corollary 1. Let q be a prime power and p an odd prime number coprime to q. Let a be a positive integer such that 2a ordp (q). Then for all k ∈ N ∗ we have 2a ordpk (q), where the notation 2a ordp (q) means that 2a |ordp (q) but 2a+1  ordp (q). Proof. Let a be a positive integer such that 2a ordp (q). From Lemma 8, there exists l in N ∗ , such that q l ≡ −1 mod p and l = 12 (1 + 2m)ordp (q) for some m in N. On the other hand, since ordp (q) is even, then ordpk (q) is also even for all k ∈ N ∗ . Hence, from Lemma 8 again, there exists lk in N ∗ , such that q lk ≡ −1 mod pk and lk = 1 2 (1 + 2mk )ordpk (q) for some mk in N. From the proof of Lemma 8, we have that k(k−1) lk = l · p 2 . Therefore 1 1 k(k−1) lk = (1 + 2mk )ordpk (q) = (1 + 2m)ordp (q) · p 2 . 2 2 k(k−1) Since (1 + 2m)p 2 and (1 + 2mk ) are both odd, we conclude that 2a ordpk (q).  3.2. New constructions of LCD cyclic codes over finite chain rings Recall that a cyclic code is an LCD code if it satisfies C ∩ C ⊥ = {0}. It was shown recently in [7] that non-free LCD code don’t exist over finite commutative chain rings. Lemma 9. [7, Theorem 2] Over finite commutative chain rings, any LCD code is free. In [7] again, Bhowmick et al., generalized the characterization of LCD codes on finite chain rings. Lemma 10. [7, Theorem 6] Let C be a cyclic code over a finite chain ring R with residue field Fq of length n such that gcd(n, q) = 1. Let g be a generator polynomial of C. Then C is an LCD code if and only if C is reversible if and only if the polynomial g is self reciprocal. Liu and Wang generalized Massey’s criterion [20] for LCD codes over any finite field of any length to finite chain rings. A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 9 Lemma 11. [16, Theorem 25] A cyclic code C of length n over a finite chain ring R with the residue field Fq is an LCD code if and only if C = g(x), where g(x) is a monic divisor of xn − 1 such that g(x) = g ∗ (x), and g(x) and (xn − 1)/g(x) are coprime. Let q = ps and n = mpr , where gcd(m, p) = 1. Thus the polynomial xm − 1 is a monic square free, hence it factors uniquely as a product of pairwise coprime monic irreducible polynomials f1 (x), ..., fl (x). Hence the factorization of xn − 1 over Fq is given by r r r r xn − 1 = xmp − 1 = (xm − 1)p = f1 (x)p ...fl (x)p (2) Denote the factors fi (x) in the factorization of xm − 1 which are self reciprocal by g1 (x), ...gs (x), and the remaining fj (x) grouped in pairs by h1 (x), h∗1 (x), ..., ht (x), h∗t (x). Hence l = s + 2t, and the factorization given in (2) becomes xn − 1 = g1 (x)p g2 (x)p ...gs (x)p h1 (x)p h∗1 (x)p ...ht (x)p h∗t (x)p r r r r r r r Using Hensel’s Lemma and the properties of the reciprocal polynomial, we get a factor- ization of xn − 1 over R, which is given by xn − 1 = G1 (x)G2 (x)...Gl (x)H1 (x)H1∗ (x)...Ht (x)Ht∗ (x), r where Gi (x), Hj (x) are monic coprime polynomials such that Gi (x) = gip (x), Hj (x) = r hpj (x). By Lemma 11, we obtain a characterization of LCD codes over finite chain rings. Those are codes generated by   C = G1 (x)k1 G2 (x)k2 ...Gl (x)kl H1 (x)r1 H1∗ (x)r1 ...Ht (x)rt Ht∗ (x)rt , where ki , rj ∈ {0, 1} for all 1 ≤ i ≤ l, 1 ≤ j ≤ t. Now, using the algebraic properties of integers given in the section 3.1, and according to the decomposition of n into product of powers of prime numbers, we give some new constructions of LCD codes over R. Theorem 1. Let R be a finite chain ring with residue field Fq , and pk an odd prime power coprime to q. Then, all free cyclic codes of length pk over R are LCD if and only if ordp (q) is even. Proof. Let pk be an odd prime power coprime to q. From Lemma 10 we have that a cyclic code C is an LCD code if it is generated by a self reciprocal polynomial g(x) k which divide xp − 1. On the other hand, Lemma 8 shows that if ordp (q) is even, then there exits l in N ∗ , such that q l ≡ −1 mod pk , which means that −1 is in the cyclotomic coset C1 . Hence, C1 = C−1 mod pk . In other words, C1 is reversible, and so all the other 10 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 k cyclotomic cosets are also reversible by Lemma 5. Therefore, all divisors of xp − 1 are self reciprocal. Conversely, assume that all free cyclic codes of length pk are LCD, then all divisors of pk x − 1 are self reciprocal. Hence all cyclotomic cosets are reversible and, in particular, the cyclotomic coset C1 is reversible. This means that there is an integer l such that q l ≡ −1 mod pk . Finally, Lemma 8 shows that in such case ordp (q) is even.  Example 1. Let R = Z9 with residue field F3 and n = 49. We have ord7 (3) = 6 and the factorization into irreducible polynomials is given by: x49 − 1 = (x + 8)(x6 + x5 + x4 + x3 + x2 + x + 1)(x42 + x35 + x28 + x21 + x14 + x7 + 1) = g1 (x)g2 (x)g3 (x)   3 So, all codes generated by i=1 gi ki (x) , where 0 ≤ ki ≤ 1, are LCD codes over Z9 of length 49. Example 2. Let R = Z4 , n = 17, we have x17 − 1 = (x + 3)(x8 + 2x6 + 3x5 + x4 + 3x3 + 2x2 + 1)(x8 + x7 + 3x6 + 3x4 + 3x2 + x + 1) = g1 (x)g2 (x)g3 (x), and it is easy to see that g1 (x), g2 (x) and g3 (x) are monic basic-irreducibleover Z4 . Since  3 ki ord17 (2) = 8, then all cyclic codes generated by polynomials of the form i=1 gi (x) , with 0 ≤ ki ≤ 1, are LCD codes. Lemma 12. Let q and n be positive integers coprime such that n is odd and the irreducible factorization of n is given by n = pk11 pk22 ...pkt t , with ordpi (q) even for 1 ≤ i ≤ t. Let ai be the positive integers for which 2ai ordpi (q), with 1 ≤ i ≤ t. Then we have a1 = a2 = ... = at = a if and only if there exists l ∈ N ∗ , such that q l ≡ −1 mod n. Further, 2a ordn (q). Proof. Assume that a1 = a2 = ... = at = a. Recall that if 2a ordpi (q) then 2a ordpki (q) i for all ki in N. Thus, we can write ordpki (q) = 2a mi , with mi an odd integer for 1 ≤ i ≤ t. i From Lemma 8 and Corollary 1, we deduce that there exits li ∈ N ∗ , such that q li ≡ −1 mod pki i . The smallest integer li satisfying this congruence is li = 12 ordpki (q) = t a−1 i 2a−1 mi , for 1 ≤ i ≤ t. Let m = i=1 mi . Since m is odd, we get q 2 m ≡ −1 mod pki i . a−1 t a−1 Hence, pki i | q 2 m + 1 for all 1 ≤ i ≤ t. Therefore, n = i=1 pki i | q 2 m + 1. In other a−1 words q 2 m ≡ −1 mod n. Conversely, assume there is an integer l such that q l ≡ −1 mod n. Without loss of generality, we suppose a1 = a2 such that 2a1 ordp1 (q) and 2a2 ordp2 (q). Write ordp1 (q) = 2a1 m1 and ordp2 (q) = 2a2 m2 for odd integers m1 and m2 . We have A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 11 q l ≡ −1 mod n implies q l ≡ −1 mod pi which give q 2l ≡ 1 mod pi , for 1 ≤ i ≤ t. Hence, 2a1 m1 | 2l and 2a2 m2 | 2l. Since both of a1 and a2 are not null, we get 2a1 −1 m1 | l and 2a2 −1 m2 | l. Since a1 = a2 , we can suppose that a1 a2 . Consequently, 2a2 m2 | l. In other words, q l ≡ 1 mod p2 , which is a contradiction. Further, we have n = pk11 pk22 ...pkt t , so ordn (q) = lcm(ordpk1 (q), ordpk2 (q), ..., ordpkt (q)) = 2a (2k + 1), for some k ∈ N 1 2 t Thus, 2a ordn (q).  Theorem 2. Let R be a finite chain ring with residue field Fq and n an odd integer coprime to q such that the factorization of n is given by n = pk11 pk22 ...pkt t with ki ∈ N ∗ for 1 ≤ i ≤ t. Assume that the numbers ordpi (q), 1 ≤ i ≤ t are even and let ai ∈ N ∗ such that 2ai ordpi (q). Then all free cyclic codes of length n over R are LCD if and only if a1 = a2 = ... = at = a. Proof. Assume that there is a positive integer a such that 2a ordpi (q), 1 ≤ i ≤ t. From Lemma 12, there exists an integer l such that q l ≡ −1 mod n. This means that the q cyclotomic coset C1 is reversible. Hence, all the other cyclotomic cosets are reversible by Lemma 5. Thus all divisors of the polynomial xn − 1 are self reciprocal. Therefore, all free cyclic codes of length n over R are LCD. Conversely, suppose that all free cyclic codes are LCD. So that all divisors of xn − 1 are self reciprocal. We deduce that all cyclotomic cosets are reversible. In particular C1 is reversible. Hence −1 is a power of q mod n. The desired result follows immediately from Lemma 12.  As a corollary we construct LCD codes of oddly even length. Corollary 2. Let R be a finite chain ring with residue field Fq such that q is an odd integer. Let n be an oddly even integer coprime to q such that the irreducible factorization of n is given by n = 2pk11 pk22 ...pkt t with ki in N ∗ for 1 ≤ i ≤ t. Assume that for all 1 ≤ i ≤ t the integers ordpi (q) are even. Let ai in N ∗ such that 2ai ordpi (q). Then a1 = a2 = ... = at = a if and only if all free cyclic codes of length n over R are LCD. Proof. On the one hand and according to Lemma 12, we have a1 = a2 = ... = at = a if t and only if there exists l in N ∗ , such that i=1 pki i | q l + 1. On the other hand, since t q is an odd integer then 2 | q l + 1. Hence n = 2 i=1 pki i | q l + 1. This means q l ≡ −1 mod n. Thus the cyclotomic coset C1 is reversible, so according to Lemma 5 all the other cyclotomic cosets are reversibles. Hence, all free cyclic codes of length n are LCD codes. Conversely, assume that all codes of length n are LCD. Then, the cyclotomic coset C1 is reversible. Hence there is an integer l such that q l ≡ −1 mod n. It follows that t q l ≡ −1 mod i=1 pki i . Therefore, from Lemma 12, we get the desired result.  12 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Example 3. Let R = Z25 , n = 2646 = 2 · 72 · 33 . We have ord7 (5) = 6 and ord3 (5) = 2. Since 2 ord7 (5) and 2 ord3 (5), so all free cyclic codes of length 2646 are LCD codes. In the remainder of this section, we provide necessary and sufficient conditions for cyclic codes to be LCD when the lengths are divisible by 4. The following lemmas are needed. Lemma 13. [12, Theorem 2’] The integer 2k has primitive roots for k = 1 or 2 but not for k ≥ 3. If k ≥ 3, then {(−1)a 5b ; a = 0, 1 and 0 ≤ b ≤ 2k−2 } constitutes a reduced residue system mod 2k . It follows that for k ≥ 3, the group (Z/2k Z)∗ is not cyclic; it is the direct product of two cyclic groups, one of order 2, the other of order 2k−2 Lemma 14. Let q be an odd prime power. Assume that there is an integer l in N ∗ such that q l ≡ −1 mod 2k with k ≥ 2. Then q ≡ −1 mod 2k . Further, the integer l is odd and ord2k (q) = 2. Proof. Assume that there is an integer l such that q l ≡ −1 mod 2k . If k2, then from Lemma 13, q can be written as q = (−1)i · 5j , with (i, j) in N 2 . Hence q l = (−1)il · 5jl ≡ −1 mod 2k , which requires that the integer il must be odd and that the order ord2k (5) of the integer 5 which equal to 2k−2 must divide jl. Thus l is odd and then 2k−2 divides j. Write j = 2k−2 · j  , we get k−2  q = (−1)i · 5j = (−1)i · 52 j ≡ (−1)i mod 2k ≡ −1 mod 2k For k = 2 and since q is odd we have clearly that q l ≡ −1 mod 4 leads to q ≡ −1 mod 4. Hence l must be odd. Further, q ≡ −1 mod 2k implies ord2k (q) = 2.  Theorem 3. Let R be a finite chain ring with residue field Fq , and let n be a doubly even integer coprime to q such that the factorization of n is given by n = 2k0 pk11 pk22 ...pkt t with ki ∈ N for all 1 ≤ i ≤ t and k0 ≥ 2. Then the following statements are equivalent: (i) 2 ordpi (q) for 1 ≤ i ≤ t and 2k0 | q + 1. (ii) All free cyclic codes over R of length n are LCD codes. Proof. Suppose that (i) is satisfied. Proving (ii) is equivalent to proving the existence of an integer l such that q l ≡ −1 mod n. The assumption 2 ordpi (q) and Corollary 1 give that ordpki (q) = 2mi , with mi odd . Using Lemma 8, we get q mi ≡ −1 mod pki i . ti t Therefore, q i=1 mi ≡ −1 mod pki i . Hence, there exists l = i=1 mi an odd integer such t that q l ≡ −1 mod i=1 pki i . On the other hand, q ≡ −1 mod 2k0 implies q l ≡ −1 t mod 2k0 . Consequently n = 2k0 i=1 pki i divide q l + 1. Thus q l ≡ −1 mod n. Conversely, assume that all free cyclic codes over R are LCD. This means that all cyclotomic cosets are reversible and, in particular, the cyclotomic coset C1 . Hence there exists l in N ∗ such that q l ≡ −1 mod n. Therefore A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 13 q l ≡ −1 mod pki i , for 1 ≤ i ≤ t (3) and q l ≡ −1 mod 2k0 (4) 1 Equation (3) and Lemma 8 give that l = (1 + 2mi )ordpi (q), for some integers mi . 2 Equation (4) and Lemma 14 give q ≡ −1 mod 2k0 and that the integer l must be odd. It follows that 2l = (1 + 2mi )ordpi (q). Which means that 2 ordpi (q). This completes the proof.  Example 4. Let R = Z9 , n = 3724 = 22 · 72 · 19. We have ord7 (3) = 6, ord19 (3) = 18. Note that 2 6, 2 18 and 22 | 4. So all free cyclic codes of length n = 3724 are LCD codes. 4. Self dual cyclic codes over finite chain rings Let R be a finite chain ring with maximal ideal γ. If the nilpotency index e of γ e is even, the cyclic code γ 2  is self dual and is called the trivial self dual code. The following result gives a necessary and sufficient conditions for the existence of non trivial self dual cyclic codes of length n over R. Lemma 15. [8, Theorem 4.4] Let R be a finite chain ring with maximal ideal γ, even index of nilpotency e, and residue field Fq . Then non trivial cyclic self dual codes of length n over R exist if and only if there is no positive integer i, such that q i ≡ −1 mod n. Cyclic codes of length n which is not divisible by the characteristic of R are called simple root cyclic codes. It was proven that there are no simple root self dual cyclic codes over finite chain rings when the nilpotency index of the generator of the maximal ideal is odd. Theorem 4. [2, Theorem 12] Let R be a finite chain ring where γ is the maximal ideal with nilpotency index e and Fq is the residue field. If e is odd, then there are no non trivial self dual cyclic codes of length n over R when gcd(n, q) = 1. In [2], authors introduce a simple criterion for the existence of non trivial self dual codes over R when the length is an odd prime power and the nilpotency index of the maximal ideal of the ring is even. Lemma 16. [2, Theorem 6] Let R be a finite chain ring with maximal ideal γ, even index of nilpotency e and residue field Fq . If n is an odd prime power coprime with q, 14 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 then there exists a non trivial cyclic self dual code of length n over R if and only if ordn (q) is odd. Using Lemmas 12 and 15, we will generalize this result and provide conditions on the existence of non trivial self dual codes of arbitrary length over R. Theorem 5. Let R be a finite chain ring with maximal ideal of even index of nilpotency e, and residue field Fq . Let n be an odd integer coprime to q such that the factorization of n is given by n = pk11 pk22 ...pkt t , with ki in N ∗ for all 1 ≤ i ≤ t and t ≥ 2. Denote by ai the integers of N such that 2ai ordpi (q), for all 1 ≤ i ≤ t. Then a non trivial self dual cyclic codes of length n exist if and only if one of the following statements holds: (i) There exists at least i0 , 1 ≤ i0 ≤ t such that ai0 = 0. (ii) For all 1 ≤ i ≤ t, ai = 0, and there exist two distinct integers i1 , i2 with 1 ≤ i1 , i2 ≤ t such that ai1 = ai2 . Proof. Assume that there exits i0 , 1 ≤ i0 ≤ t such that ai0 = 0. This means that ordpi0 (q) is odd. Lemma 8 guarantees that there is no integer l such that q l ≡ −1 mod pi0 . Hence, for all i in N we can’t have q i ≡ −1 mod n. Thus, from Lemma 15, a non trivial self dual cyclic code over R exists. Assume now that (ii) is satisfied. Therefore, all ordpi (q), for 1 ≤ i ≤ t, are even. Lemma 12 shows that if there exist two distinct integers i1 , i2 , 1 ≤ i1 , i2 ≤ t such that ai1 = ai2 , then there is no integer l such that q l ≡ −1 mod n. Hence, a non trivial self dual codes over R exist. Conversely, assume that a non trivial self dual codes exist. So there is no integer l such that q l ≡ −1 mod n. We need to prove that either there exists i0 such that ai0 = 0 or every ai is different to zero and at least two of them are distinct. Suppose that for all 1 ≤ i ≤ t, ai = 0. This implies that ordpi (q) is even for all 1 ≤ i ≤ t. Since there is no integer l such that q l ≡ −1 mod n, by Lemma 12, we have that there exists i1 , i2 with 1 ≤ i1 , i2 ≤ t such that ai1 = ai2 .  Example 5. Let R = Z4 and n = 3 · 5. We have ord3 (2) = 2 and ord5 (2) = 4, and hence 21 ord3 (2) and 22 ord5 (2). So there exist non trivial self dual codes over Z4 of length 15. The factorization of x15 − 1 over Z4 is given by x15 − 1 = f1 (x)f2 (x)f3 (x)f4 (x)f4∗ (x), where f1 (x) = x+3, f2 (x) = x2 +x+1, f3 (x) = x4 +x3 +x2 +x+1, and f4 (x) = x4 +2x2 +3x+1. Let g(x) = f1 (x)f2 (x)f3 (x) and h(x) = f4 (x). Then the following codes A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 15 g(x)h(x), 2h(x)h∗ (x) and g(x)h∗ (x), 2h(x)h∗ (x) are non trivial self dual cyclic codes of length 15. Example 6. Let R = Z16 and n = 21. We have ord3 (2) = 2 and ord7 (2) = 3. Then, there exist non trivial self dual cyclic codes over R of length 21. The factorization of x21 − 1 over R is equal to x21 − 1 = f1 (x)f2 (x)f3 (x)f3∗ (x)f4 (x)f4∗ (x), where f1 (x) = x−1, f2 (x) = x2 +x+1, f3 (x) = x3 +6x2 +5x−1, f4 (x) = x6 −6x5 −x4 −x2 +5x+1. Let x21 − 1 = g(x)h(x)h∗ (x), where g(x) = f1 (x)f2 (x) is a self reciprocal polynomial and h(x) = f3 (x)f4 (x). Thus, for example, the code g(x)h(x), 2h(x)h∗ (x) is self dual. We give now the necessary and sufficient conditions for the existence of non trivial self dual codes when the length is an oddly even integer. Theorem 6. Let R be a finite chain ring with maximal ideal of even index of nilpotency e, and residue field Fq . Let n be an oddly even integer coprime to q such that the irreducible factorization of n is given by n = 2.pk11 pk22 ...pkt t , where t ≥ 2, and ki in N ∗ for all 1 ≤ i ≤ t. Let ai ∈ N such that 2ai ordpi (q). Then a non trivial self dual cyclic code of length n exists if and only if one of the following statements holds: (i) There exists at least i0 , 1 ≤ i0 ≤ t such that ai0 = 0. (ii) For all 1 ≤ i ≤ t, ai = 0, there exist two distinct integers i1 , i2 with 1 ≤ i1 , i2 ≤ t such that ai1 = ai2 . Proof. Since gcd(n, q) = 1, then q must be an odd integer. Hence for all l in N ∗ we have 2 | q l + 1. If a1 = a2 = ... = at = a, and a = 0, by Lemma 12, we know that there exists t l in N ∗ such that q l ≡ −1 mod i=1 pki i . Therefore, q l ≡ −1 mod n. Hence, there do not exist non trivial self dual codes on R by Lemma 15. Conversely, assume (i) holds. Thus, from Lemma 8, there is no integer l such that q ≡ −1 mod pi0 . Hence, there does not exist integer l in N, such that q l ≡ −1 mod n. l This proves by Lemma 15 that non trivial self dual cyclic codes over R exist. Assume now that (ii) is satisfied. By Lemma 12, if there exist two distinct integers i1 , i2 , 1 ≤ i1 , i2 ≤ t such that ai1 = ai2 , then there is no integer l such that q l ≡ −1 t mod i=1 pki i . Even if we have 2 divides q l + 1 for all l ∈ N ∗ , we cannot find any integer l such that q l ≡ −1 mod n. Hence by Lemma 15, non trivial self dual codes over R exist.  16 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Example 7. Let R = Z9 and n = 70 = 2 · 5 · 7. We have ord5 (3) = 4, ord7 (3) = 6, and hence 2 ord7 (3) and 22 ord5 (3). So there exist non trivial self dual cyclic codes of length 70 over Z9 . The factorization of x70 − 1 over Z9 is given by x70 − 1 = f1 (x)f2 (x)f3 (x)f4 (x)f5 (x)f6 (x)f7 (x)f7∗ (x)f8 (x)f8∗ (x), where f1 (x) = x + 1, f2 (x) = x + 8, f3 (x) = x4 + x3 + x2 + x + 1, f4 (x) = x4 + 8x3 + x2 + 8x + 1, f5 (x) = x6 + x5 + x4 + x3 + x2 + x + 1, f6 (x) = x6 + 8x5 + x4 + 8x3 + x2 + 8x + 1, f7 (x) = x12 + 4x10 + 6x9 + 8x8 + x7 + 3x6 + 4x5 + 5x4 + 7x3 + 5x2 + 8x + 1, f8 (x) = x12 + 4x10 + 3x9 + 8x8 + 8x7 + 3x6 + 5x5 + 5x4 + 2x3 + 5x2 + x + 1. Let x70 − 1 = g(x)h(x)h∗ (x), where g(x) = f1 (x)f2 (x)f3 (x)f4 (x)f5 (x)f6 (x) is a self reciprocal polynomial and h(x) = f7 (x)f8 (x). Thus, for example, the code g(x)h(x), 3h(x)h∗ (x) is self dual. Example 8. Let R = Z49 and n = 30 = 2 · 3 · 5. We have ord3 (7) = 1, ord5 (7) = 4. We have that ord3 (7) is odd, so there exist non trivial self dual cyclic codes of length 30 over Z49 . The factorization of x30 − 1 over Z49 is given by x30 − 1 = f1 (x)f2 (x)f2∗ (x)f3 (x)f3∗ (x)f4 (x)f5 (x)f6 (x)f7 (x)f7∗ (x)f8 (x)f8∗ (x), where f1 (x) = x + 1, f2 (x) = x − 19 f3 (x) = x − 18, f4 (x) = x − 1, f5 (x) = x4 + x3 + x2 + x + 1, f6 (x) = x4 − x3 + x2 − x + 1, f7 (x) = x4 − 19x3 + 18x2 + x − 19, f8 (x) = x4 − 18x3 − 19x2 − x + 18, Let x30 − 1 = g(x)h(x)h∗ (x), where g(x) = f1 (x)f4 (x)f5 (x)f6 (x) and h(x) = f2 (x)f3 (x)f7 (x)f8 (x). Thus, for example, the code g(x)h(x), 7h(x)h∗ (x) is self dual. We determine now, necessary and sufficient conditions for the existence of non trivial self dual cyclic codes over R for doubly even lengths. Theorem 7. Let R be a finite chain ring with maximal ideal of even index of nilpo- tency e, and residue field Fq . Let n be a doubly even integer coprime to q, such that the irreducible factorization of n is given by n = 2k0 pk11 pk22 ...pkt t where ki in A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 17 N ∗ for all 1 ≤ i ≤ t and k0 ≥ 2. Let ai in N such that 2ai ordpi (q), for 1 ≤ i ≤ t. The following statements are equivalent: (i) Non trivial self dual cyclic codes over R exist. (ii) There exits i, 1 ≤ i ≤ t, such that ai = 1 or 2k0  (q + 1). Proof. Assume that (ii) is not satisfied. This implies that a1 = a2 = ... = at = 1 and 2k0 | (q + 1). Then from Lemma 12, there is an odd integer l such that q l ≡ −1 t mod i=1 pki i . Since q ≡ −1 mod 2k0 and l is odd, it follows that q l ≡ −1 mod 2k0 . Thus q l ≡ −1 mod n. Lemma 15 shows that non trivial self dual cyclic code over R does not exist. Conversely, suppose that it does not exist any non trivial self dual cyclic codes of length n over R. Then, by Lemma 15, there exist some positive integer l such that q l ≡ −1 mod n. Thus, q l ≡ −1 mod 2k0 . By Lemma 14, we have that q ≡ −1 mod 2k0 and that the integer l is odd. On the other hand q l ≡ −1 mod n implies again that q l ≡ −1 mod pi . Lemma 8 gives that 2l = (1 + 2mi )ordpi (q), for some integers mi . Since l is odd, it follows that 2 ordpi (q). This means that ai = 1 for each 1 ≤ i ≤ t.  Example 9. Let R = Z25 and n = 84 = 22 · 3 · 7. We have that 22  (5 + 1). Then there are non trivial self dual cyclic codes of length 84 over Z25 . The factorization of x84 − 1 over Z25 is given by 10 15 x84 − 1 = fi (x) fi (x)fi∗ (x), i=1 i=11 where f1 (x) = x + 1, f2 (x) = x + 24, f3 (x) = x2 + x + 1, f4 (x) = x2 + 24x + 1, f5 (x) = x6 + 5x5 + 22x4 + 2x3 + 22x2 + 5x + 1, f6 (x) = x6 + 20x5 + 22x4 + 23x3 + 22x2 + 20x + 1, f7 (x) = x6 + x5 + x4 + x3 + x2 + x + 1, f8 (x) = x6 + 6x5 + 8x4 + 9x3 + 8x2 + 6x + 1, f9 (x) = x6 + 24x5 + x4 + 24x3 + x2 + 24x + 1, f10 (x) = x6 + 19x5 + 8x4 + 16x3 + 8x2 + 19x + 1, f11 (x) = x + 7, f12 (x) = x2 + 7x + 24, f13 (x) = x6 + 10x5 + 3x4 + 11x3 + 22x2 + 10x + 24, f14 (x) = x6 + 17x5 + 17x4 + 12x3 + 8x2 + 17x + 24, f15 (x) = x6 + 7x5 + 24x4 + 18x3 + x2 + 7x + 24. 18 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 10 15 Let x84 − 1 = g(x)h(x)h∗ (x) where g(x) = i=1 fi (x) and h(x) = i=11 fi (x). Thus for example the code g(x)h(x), 5h(x)h∗ (x) is self dual. Example 10. Let R = Z81 and n = 140 = 22 · 5 · 7. We have that 22 | (3 + 1), ord7 (3) = 6 and ord5 (3) = 4. Thus 21 ord7 (3) and 22 ord5 (3). Therefore, there exist non trivial self dual cyclic codes of length 140 over Z81 . The factorization of x140 − 1 over Z81 is given by x140 − 1 = f1 (x)f2 (x)f3 (x)f4 (x)f4∗ (x)f5 (x)f6 (x)f7 (x)f8 (x)f9 (x) ∗ ∗ ∗ ∗ f10 (x)f11 (x)f11 (x)f12 (x)f12 (x)f13 (x)f13 (x)f14 (x)f14 (x), where f1 (x) = x + 1, f2 (x) = x − 1, f3 (x) = x2 + 1, f4 (x) = x4 − 20x3 − 3x2 + 20x + 1, f5 (x) = x4 + x3 + x2 + x + 1, f6 (x) = x4 − x3 + x2 − x + 1, f7 (x) = x6 + 13x5 + 3x4 + 13x3 + 3x2 + 13x + 1, f8 (x) = x6 + x5 + x4 + x3 + x2 + x + 1, f9 (x) = x6 − 13x5 + 3x4 − 13x3 + 3x2 − 13x + 1, f10 (x) = x6 − x5 + x4 − x3 + x2 − x + 1, f11 (x) = x12 − 24x11 + 9x10 − 17x9 + 5x8 + 31x7 − 4x6 + 11x5 − 12x4 + 4x3 − 8x2 + 37x + 1, f12 (x) = x12 + 24x11 + 9x10 + 17x9 + 5x8 + 50x7 − 4x6 − 11x5 − 12x4 − 4x3 − 8x2 + 44x + 1, f13 (x) = x12 − 9x11 + 4x10 + 15x9 + 53x8 + 19x7 + 12x6 + 49x5 + 23x4 − 2x3 − 13x2 + 8x + 1, f14 (x) = x12 + 9x11 + 4x10 − 15x9 + 53x8 − 19x7 + 12x6 + 32x5 + 23x4 + 2x3 − 13x2 − 8x + 1. 10 14 Let x140 − 1 = g(x)h(x)h∗ (x) where g(x) = i=1 fi (x)f4∗ (x) and h(x) = i=11 fi (x). Thus the code g(x)h(x), 9h(x)h∗ (x) is self dual. 5. Construction of new isodual cyclic codes over finite chain rings In this section and according to different factorizations of the polynomial xn − 1, we give some new constructions of isodual cyclic codes over finite chain rings as a general- ization of those obtained in [1]. First, we recall the structure of cyclic codes of length 2a m given in [1]. A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 19 Lemma 17. [1, Lemma 4] Let R be a finite chain ring with residue field Fq such that q is an odd prime power, and let m an odd integer coprime to q. Then there is a primitive 2a -th root of unity ξ in R∗ if and only if q ≡ 1 mod 2a . Further, a 2a 1. x2 − 1 = k=1 (x − ξ k ) in R[x], 2. ξ m is also a primitive 2a -th root of unity, 2a 3. k=1 ξ k = 1, if a  2. l Proposition 1. If R∗ contains a primitive 2a -th root of unity ξ, and xm − 1 = i=1 fi (x), where fi (x), 1 ≤ i ≤ l, are monic basic irreducible pairwise coprime factors in R[x], then 2a l a x2 m −1= fi (ξ k x). k=1 i=1 l Proof. Assume that xm − 1 = i=1 fi (x). Let ξ be a primitive 2a -th root of unity. Then l l (ξ k x)m − 1 = i=1 fi (ξ k x). Thus xm − ξ −km = ξ −km i=1 fi (ξ k x). Since ξ is a primitive  a 2a -th root of unity and k=1 ξ −km = 1, we get 2 2a 2a − ξ −km ) a a x2 m − 1 = (xm )2 − 1 = k=1 (x m − ξk ) = k=1 (x m 2a −km l 2a l = k=1 (ξ i=1 fi (ξ k x)) = k=1 i=1 fi (ξ k x))  For the construction of isodual codes we need the following result given in [1] Lemma 18. [1, Theorem 3] Let R be a finite chain ring, C a free cyclic code of length n over R generated by a polynomial g(x) and δ a unit in R such that δ n = 1. Then the following holds: (i) C is equivalent to the cyclic code generated by g ∗(x). (ii) C is equivalent to the cyclic code generated by g(δx). (iii) C is equivalent to the cyclic code generated by g ∗(δx) or (g(δx))∗ . (iv) If n is even, then C is equivalent to the cyclic code generated by g(−x). Proposition 2. Let n be a positive integer. If f (x) and g(x) are polynomials in R[x] such that xn − 1 = g(x)f (x), then the cyclic code generated by g(x) is equivalent to the dual of the code generated by f (x). Proof. Let C be a cyclic code generated by g(x) and C  a cyclic code generated by f (x). We have that the dual of C  is generated by g ∗ (x). By Lemma 18, C is equivalent to C  ⊥.  The following theorem gives a natural construction for free cyclic isodual codes. 20 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Proposition 3. Let n be a positive integer such that there exists δ in R∗ verifying δ n = 1. If xn −1 = αg(x)g(δx) or xn −1 = αg(x)g(δx)∗ for some α in R∗ , then the code generated by g(x) is isodual. Proof. Assume that xn − 1 = αg(x)g(δx)∗ . Let C be a code generated by the polynomial g(x). From Theorem 4, C is free code. From Proposition 2, C is equivalent to the dual of the cyclic code C  generated by αg(δx)∗ . Up to normalization and since δ n = 1 and α in R∗ , we obtain that the code C  is equivalent to the code generated by g(x) which is C itself. Therefore C is isodual. With the same argument, we get the result for the second part.  In the following, we take q is an odd prime power such that q ≡ 1 mod 2a with a ≥ 1 and m an odd integer with gcd(m, q) = 1. We next give new construction of isodual cyclic codes of length 2a m over R. Theorem 8. Assume that xm − 1 = f1 (x)f2 (x). Then the free cyclic codes of length 2a m generated by 2a−1 2a−1 −1 2k fi (ξ x) fj (ξ 2k+1 x), i,j ∈ {1, 2}, i = j, k=1 k=0 or 2a−1 f1 (ξ 2k x)f2 (ξ 2k x), k=1 or 2a−1 −1 f1 (ξ 2k+1 x)f2 (ξ 2k+1 x) k=0 are isodual, where ξ is a primitive 2a -th root of unity. Proof. Let xm − 1 = f1 (x)f2 (x). From Proposition 1, we have 2a 2a−1 2a−1 −1 a x2 m −1= f1 (ξ k x)f2 (ξ k x) = f1 (ξ 2k x)f2 (ξ 2k x) f1 (ξ 2k+1 x)f2 (ξ 2k+1 x). k=1 k=1 k=0 Let 2a−1 2a−1 −1 g(x) = f1 (ξ 2k x) f2 (ξ 2k+1 x). k=1 k=0 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 21 Table 1 List of isodual codes obtained of length 36 over Z25 . Polynomial generator g(x) of the isodual code C g(x) = x9 + (ξ − ξ 2 )x6 + (ξ 2 − ξ 3 )x3 − 1 g(x) = ξ 3 x9 + (ξ − 1)x6 + (ξ 3 − ξ 2 )x3 − 1 g(x) = ξ 3 x9 − (ξ 2 − ξ)x8 + (ξ 3 − 1)x7 − (ξ 2 − 1)x6 + (ξ 2 − ξ 3 )x5 − (ξ − 1)x4 + (ξ − ξ 3 )x3 + (ξ 3 − 1)x2 + (ξ − ξ 2 )x − 1 g(x) = x9 − (ξ 2 − ξ 3 )x8 − (ξ − ξ 2 )x7 + (ξ 2 − 1)x6 + (ξ − 1)x5 − (ξ 3 − 1)x4 + (ξ 2 − 1)x3 − (ξ 2 − ξ 3 )x2 − (ξ − ξ 2 )x − 1 g(x) = ξ 2 x9 − (ξ − 1)x8 − (ξ 3 − 1)x7 − (ξ 2 − ξ 3 )x6 − (ξ − ξ 2 )x5 + (ξ − 1)x4 − (ξ 3 − 1)x3 − (ξ 2 − ξ 3 )x2 − (ξ − ξ 2 )x − 1 g(x) = ξx9 + (ξ 3 − 1)x8 + (ξ − ξ 2 )x7 + (ξ 3 − 1)x6 + (ξ − ξ 2 )x5 + (ξ 3 − 1)x4 + (ξ − ξ 2 )x3 + (ξ 3 − 1)x2 + (ξ − ξ 2 )x − 1 g(x) = ξ 2 x9 − 1 g(x) = ξx9 − 1 a Knowing that ξ 2 = 1, we get 2a−1 2a−1 −1 2a−1 −1 2a−1 2k+1 2k+2 2k+1 g(ξx) = f1 (ξ x) f2 (ξ x) = f1 (ξ x) f2 (ξ 2k x). k=1 k=0 k=0 k=1 a a In other words, we have that x2 m −1 = g(x)g(ξx). Since ξ 2 m = 1, then from Theorem 3 the code generated by g(x) is isodual. A similar argument is employed to prove that codes generated by 2a−1 f1 (ξ 2k x)f2 (ξ 2k x) k=1 or 2a−1 −1 f1 (ξ 2k+1 x)f2 (ξ 2k+1 x) k=0 are also isodual.  Example 11. Let R = Z25 and n = 36 = 22 · 32 , q = 5 ≡ 1 mod 22 . We have x9 − 1 = (x − 1)(x2 + x + 1)(x6 + x3 + 1). Thus, we get the isodual codes given in Table 1, where ξ is a primitive 4-th root of unity. k Corollary 3. Let pk be a prime power. Assume that xp − 1 = f1 (x)f2 (x). Then the cyclic codes generated by k k f1 (x)f2 (−x) or f1 (−x)f2 (x) or xp − 1 or xp + 1 22 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 Table 2 List of LCD-isodual codes obtained of length 50 over Z9 . Polynomial generator g(x) of LCD-isodual code C g(x) = x25 − 2x20 + 2x15 − 2x10 + 2x5 − 1 g(x) = x25 + 2x20 + 2x15 + 2x10 + 2x5 + 1 g(x) = −x25 − 2x24 − 2x23 − 2x22 − 2x21 + 2x19 + 2x18 + 2x17 + 2x16 − 2x14 − 2x13 − 2x12 − 2x11 + 2x9 + 2x8 + 2x7 + 2x6 − 2x4 − 2x3 − 2x2 − 2x − 1 g(x) = x25 − 2x24 + 2x23 − 2x22 + 2x21 − 2x19 + 2x18 − 2x17 + 2x16 − 2x14 + 2x13 − 2x12 + 2x11 − 2x9 + 2x8 − 2x7 + 2x6 − 2x4 + 2x3 − 2x2 + 2x − 1 g(x) = x25 − 2x24 + 2x23 − 2x22 + 2x21 − 2x20 + 2x19 − 2x18 + 2x17 − 2x16 + 2x15 − 2x14 + 2x13 − 2x12 + 2x11 − 2x10 + 2x9 − 2x8 + 2x7 − 2x6 + 2x5 − 2x4 + 2x3 − 2x2 + 2x − 1 g(x) = x25 + 2x24 + 2x23 + 2x22 + 2x21 + 2x20 + 2x19 + 2x18 + 2x17 + 2x16 + 2x15 + 2x14 + 2x13 + 2x12 + 2x11 + 2x10 + 2x9 + 2x8 + 2x7 + 2x6 + 2x5 − 2x4 + 2x3 + 2x2 + 2x + 1 g(x) = x25 − 1 g(x) = x25 + 1 are isodual codes of length 2pk . Further, if ordp (q) is even, then these codes are LCD- isodual codes. Proof. The result follows from Theorem 1 and Theorem 8 for a = 1, m = pk and ξ = −1.  Example 12. Let R = Z9 and n = 50 = 2 · 52 . We have x25 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)(x20 + x15 + x10 + x5 + 1). Further ord5 (3) = 4. So we get the LCD isodual codes shown in Table 2. Corollary 4. Let m be an odd integer such that the irreducible factorization of m is given by m = pk11 pk22 ...pkt t and xm − 1 = f1 (x)f2 (x). Assume that there exists a in N ∗ such that 2a ordpi (q), for all 1 ≤ i ≤ t. Then, the cyclic codes generated by f1 (x)f2 (−x) or f1 (−x)f2 (x) or xm − 1 or xm + 1 are LCD-isodual codes of length 2m. Proof. The result follows immediately from Theorem 2 and Theorem 8 with ξ = −1.  Another construction of isodual cyclic codes is given by the following theorem. Theorem 9. Assume that we have the factorization xm − 1 = f1 (x)f2 (x)f2∗ (x), such that the polynomial f1 is self reciprocal. Then the cyclic codes of length 2a m over R generated by A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993 23 2a−1 2a 2k f1 (ξ x) f2 (ξ k x) k=1 k=1 or 2a−1 2a 2k f1 (ξ x) f2∗ (ξ k x) k=1 k=1 are isodual, where ξ is a primitive 2a -th root of unity. Proof. Let xm − 1 = f1 (x)f2 (x)f2∗ (x) then 2a 2a−1 2a−1 −1 2a f1 (ξ k x)f2 (ξ k x)f2∗ (ξ k x) = f2 (ξ k x)f2∗ (ξ k x) a x2 m −1 = f1 (ξ 2k ) f1 (ξ 2k+1 ) k=1 k=1 k=0 k=1 2a−1 2a a Let g(x) = k=1 f1 (ξ 2k x) k=1 f2 (ξ k x). Since ξ 2 = 1, we get 2a−1 2a 2a−1 −1 2a g(ξx) = f1 (ξ 2k+1 x) f2 (ξ k+1 x) = f1 (ξ 2k+1 x) f2 (ξ k x). k=1 k=1 k=0 k=1 a Since the polynomial f1 is self reciprocal, we obtain the factorization x2 m − 1 = g(x)g(ξx)∗ . The desired result follows from Theorem 3. The same result is obtained for the codes generated by 2a−1 2a 2k f1 (ξ x) f2∗ (ξ k x).  k=1 k=1 Example 13. Let R = Z25 and n = 132 = 22 · 33. We have 5 ≡ 1 mod 22 and x33 − 1 = (x − 1)(x2 + x + 1)(x5 − 8x4 − x3 + x2 − 9x − 1) (x5 + 9x4 − x3 + x2 + 8x − 1) (x10 − 9x9 + 7x8 + 11x7 + 9x6 − 4x5 − 7x4 − 6x3 − 10x2 + 8x + 1) (x10 + 8x9 − 10x8 − 6x7 − 7x6 − 4x5 + 9x4 + 11x3 + 7x2 − 9x + 1) Let x33 − 1 = f1 (x)f2 (x)f2∗ (x), where f1 (x) = (x − 1)(x2 + x + 1) and f2 (x) = (x5 −8x4 −x3 +x2 −9x −1)(x10 −9x9 +7x8 +11x7 +9x6 −4x5 −7x4 −6x3 −10x2 +8x +1). Thus, for example, the codes generated by   f1 (x)f1 (ξ 2 x)f2 (x)f2 (ξ 2 x)f2 (ξ 3 x) or 24 A. Benyettou et al. / Finite Fields and Their Applications 79 (2022) 101993   f1 (x)f1 (ξ 2 x)f2∗ (x)f2∗ (ξx)f2∗ (ξ 2 x)f2∗ (ξ 3 x) are isodual, where ξ is a primitive 4-th root of unity. k Corollary 5. Let p be an odd prime number and k in N such that xp − 1 = f1 (x)f2 (x)f2∗ (x). Assume that ordp (q) is even. Then the cyclic codes of length 2pk gen- erated by f1 (x)f2 (x)f2 (−x) or f1 (x)f2∗ (x)f2∗ (−x) are LCD-isodual codes. Proof. The result follows from Theorem 1 and Theorem 9 with ξ = −1.  Corollary 6. Let m be an odd integer such that the irreducible factorization of m is given by m = pk11 pk22 ...pkt t and xm − 1 = f1 (x)f2 (x)f2∗ (x). Assume that there exists a in N ∗ such that 2a ordpi (q), for all 1 ≤ i ≤ t. Then the cyclic codes of length 2m generated by f1 (x)f2 (x)f2 (−x) or f1 (x)f2∗ (x)f2∗ (−x) are LCD-isodual codes. Proof. The result follows immediately from Theorem 2 and Theorem 9 with ξ = −1.  Example 14. Let R = Z27 and n = 70 = 2 · 5 · 7. We have x35 − 1 = f1 (x)f2 (x)f2∗ (x), where f1 (x) = (x − 1)(x4 + x3 + x2 + x + 1)(x6 + x5 + x4 + x3 + x2 + x + 1) f2 (x) = (x12 − 9x11 + 4x10 − 12x9 − 3x8 − 7x7 + 12x6 − 5x5 − 4x4 − 2x3 + 14x2 + 8x + 1). So the cyclic codes of length 70 over Z27 generated by g(x) = f1 (x)f2 (x)f2 (−x) or h(x) = f1 (x)f2∗ (x)f2∗ (−x) are isodual. 6. Conclusion In this work, cyclic codes over finite chain rings were studied. 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