# Unique Generators for Cyclic Codes of Arbitrary Length overFpmu/u3 and Their Applications

Authors: Hongju Li; Ping Yu; Jing Liang; Feng Zhao
Journal: Journal of Mathematics
 (2022)

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## Abstract

Cyclic codes play a very important role in the history of coding theory since they have good algebraic structures that can be widely used in coding and decoding. However, generators for repeated-root cyclic codes of arbitrary length overFpmu/uk are not unique in previous works for k≥3, and hence it is impossible to determine their dual codes. In this work, we propose unique generators for cyclic codes of arbitrary length over Fpmu/u3. As its applications, we derive the numbers of their codewords, as well as generators for their dual codes. Furthermore, we propose necessary and sufficient conditions for their self-dualities.


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## Body

## 1. Introduction

In 1994, Hammons et al. [1] discovered that many efficient nonlinear binary codes are closely related to linear codes over integer residue ring Z4. After that, codes over finite rings have got much attention [2–6]. Among various finite rings, F2+uF2 is interesting which is because that it shares some good properties of Z4 and the finite field F4, and cyclic codes over such a ring can be used to construct optimal binary codes [7]. From then on, much work has been done on this kind of rings, especially on cyclic codes over Fpmu/uk for integers m≥1, k≥2 and positive prime number p. Qian et al. [8] determined the structure of simple-root cyclic codes over Fpu/uk, which are principal ideals over Fpmu/uk. However, structures of repeated-root cyclic codes over Fpmu/uk are more complicated than simple-root cyclic codes since some repeated-root cyclic codes are not principal ideals.Abualrub and Saip [9] studied generators for cyclic codes of arbitrary length over F2+uF2 and F2u/u3 and proposed generators for the dual codes of cyclic codes with arbitrary length over F2+uF2. Later, the results about generators in [9] were generalized to cyclic codes of arbitrary length over F2u/uk, Fpu/uk, and Fpmu/uk in a similar way [10–12]. Recently, Cao et al. [13] determined the number of codewords of a cyclic code with oddly even length over F2mu/uk and investigated the dual code of each cyclic code and self-dual cyclic codes with such a length over F2mu/uk.Whenk≥3, we find that generators for cyclic codes of arbitrary length in [9–12] are not unique. This makes the authors unable to determine generators in [9–12] or give wrong generators in [11] for the dual code of a cyclic code with an arbitrary length over Fpmu/u3. Therefore, the dual code of a cyclic code and self-dual cyclic codes over Fpmu/u3 are still open. In this paper, we give unique generators for cyclic codes of arbitrary length over Fpmu/u3 and then determine their numbers of codewords, dual codes, and self-dualities. The remainder of this paper is organized as follows. After a brief introduction of some notations and existing results in Section 2, we give generators for the dual code of cyclic codes with arbitrary length over Fpmu/u3 in Section 3. We then propose necessary and sufficient conditions for the self-duality of cyclic codes over Fpmu/u3 in Section 4 and conclude in Section 5.


## 2. Preliminaries

LetR=Fpmu/u2=a|a=r0a+ur1a,r0a,r1a∈Fpm,u2=0 and S=Fpmu/u3=b|b=r0b+ur1b+u2r2b,r0b,r1b,r2b∈Fpm,u3=0. The cyclic shift ν on RN is the shift νc0,c1,…,cN−1=cN−1,c0,c1,…,cN−2, where N=pen and gcdn,p=1 for nonnegative integers n,e and a prime integer p. Let C be a code over R, then C is said to be a cyclic code over R if νC=C. We will identify a codeword c=c0,c1,…,cN−1 with its polynomial representation cx=∑i=0N−1cixi. With this identification, cyclic codes over R can be identified as ideals in the ring Rx/xN−1. Similarly, cyclic codes over S can be identified as ideals in the ring Sx/xN−1. Definitions 1 and 2 are similar to [10–12].Definition 1.
LetC1 be a cyclic code of arbitrary length N over R and let a=r0a+ur1a∈R, then the map φ1:R⟶Fpm is defined by φ1a=r0a, which is a ring homomorphism that can be extended to a homomorphism Φ1:C1⟶Fpmx/xN−1 defined by Φ1cx=∑i=0N−1φ1cixi for cx=∑i=0N−1cixi∈C1.
From Definition1, we know that Φ1C1 is an ideal of Fpmx/xN−1, and hence Φ1C1 is a cyclic code over Fpm. The kernel of Φ1 is(1)kerΦ1=uh1x|uh1x∈C1,h1x∈FpmxxN−1,which is an ideal of Rx/xN−1.Definition 2.
LetC2 be a cyclic code of arbitrary length N over S and let b=r0b+ur1b+u2r2b∈S, then the map φ2:S⟶R is defined by φ2b=r0b+ur1b, which is a ring homomorphism that can be extended to a homomorphism Φ2:C2⟶Rx/xN−1 defined by Φ2cx=∑i=0N−1φ2cixi for cx=∑i=0N−1cixi∈C2.
Definition2 shows that Φ2C2 is an ideal of Rx/xN−1, and hence Φ2C2 is a cyclic code over R. Therefore, we let Φ2C2=C1 in the following. Besides, the kernel of Φ2 is(2)kerΦ2=u2h2x|u2h2x∈C2,h2x∈FpmxxN−1,which is an ideal of Sx/xN−1.Proposition 1 (see [9–12]).
LetC2 be a cyclic code of arbitrary length N over S, then there exist polynomials gx,a1x,a2x,p1x,p2x,q1x∈Fpmx such that(3)C2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2xwhere a2x|a1x|, a1x|p1xxN−1/gx, a2x|q1xxN−1/a1x, a2x|p1xxN−1/gx, a2x|p2xxN−12/a1xgx,dega1x>degp1x, dega2x>degp2x, and dega2x>degq1x.
Next, we give an example to show that the generators forC2 in Proposition 1 are not unique.Example 1.
InF2x, we have x14−1=f12f22f32, where f1=x+1,f2=x3+x+1 and f3=x3+x2+1. Let D1=f12f2f3+u+u2x,uf2f3+u2,u2f3 and D2=f12f2f3+u,uf2f3,u2 be cyclic codes of length N=14 over F2u/u3. Clearly, D1 and D2 are in the form of Proposition 1 but(4)D1=f12f2f3+u+u2x,uf2f3+u2,u2f3+f1uf12f2f3+u+u2x+f12uf2f3+u2=f12f2f3+u+u2x,uf2f3+u2,u2=f12f2f3+u,uf2f3,u2=D2.Definition 3.
For anyc=c0,c1,…,cN−1andb=b0,b1,…,bN−1∈SN, we say c and b are orthogonal if c⋅b=∑i=0N−1cibi=0. The dual code of a cyclic code C2 with length N over S is defined by C2⊥=b∈SN|b⋅c=0forallc∈C2.
It is well known thatC2⊥ is also a cyclic code over S and C2‖C2⊥=p3mN. If fx=arxr+ar−1xr−1+⋯+a1x+a0 is a polynomial of degree r, then the reciprocal of fx is f∗x=a0xr+a1xr−1+⋯+ar−1x+ar=xrf1/x. Now, we define the annihilator of C2 as AnnC2=fx|fxcx=0forallcx∈C2, then AnnC2 is a cyclic code over S and C2⊥=Ann∗C2=f∗x|fx∈AnnC2.


## 3. Unique Generators for Cyclic Codes of Arbitrary Length overS

In this section, we specify a choice of generator polynomials forC2 such that its generators are unique, and its basic idea is inspired by Ding et al. [3].Lemma 1 (see [11]).
A cyclic codeC1 of arbitrary length N over R can be written uniquely as C1=gx+up1x,ua1x, where p1x and monic gx,a1x are polynomials in Fpmx such that a1xgxxN−1, a1x|p1xxN−1/gx in Fpmx, and dega1x>degp1x.Lemma 2.
LetC1=gx+up1x,ua1x be a cyclic code of arbitrary length N over R, where p1x and monic gx,a1x are polynomials in Fpmx such that a1xgxxN−1 in Fpmx and dega1x>degp1x. Then, kerΦ1=ua1x if and only if a1x|p1xxN−1/gx in Fpmx.Proof.
Assume thatkerΦ1=ua1x. In Rx/xN−1, we have xN−1/gxgx+up1x=up1xxN−1/gx∈kerΦ1, which implies a1x|p1xxN−1/gx. Consequently, there are polynomials l1x,l2x∈Rx such that p1xxN−1/gx=l1xa1x+l2xxN−1.
LetL1x,L2x∈Fpmx, satisfying(5)L1x≡l1xmodu,L2x≡l2xmodu.
Then,L1xa1x=p1xxN−1/gx−L2xxN−1 in Fpmx. Since a1x|xN−1 in Fpmx, then a1x|p1xxN−1/gx in Fpmx. This completes the proof of the necessary part. Next, we will consider the sufficient part.
Suppose thata1x|p1xxN−1/gx in Fpmx. Since ua1x∈C1 and a1x|xN−1 in Fpmx, then ua1x⊆kerΦ1. On the other hand, suppose uax∈kerΦ1 for ax∈Fpmx/xN−1, then we only need to consider the following three cases to construct uax in Rx/xN−1.Case 1:ua1x∈C1Case 2:ugx+up1x=ugxCase 3:gx+up1xxN−1/gx=up1xxN−1/gx
From Cases 1 to 3 and the conditionsa1xgxxN−1 and a1x|p1xxN−1/gx in Fpmx, we derive that a1x|ax if uax∈C1, which implies kerΦ1⊆ua1x. Thus, kerΦ1=ua1x.Theorem 1.
LetC2 be a cyclic code of arbitrary length N over S; then, there exist unique polynomials gx,a1x,a2x,p1x,p2x,q1x∈Fpmx such that(6)C2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2xwhere gx,a1x, and a2x are monic, a2x|a1x| in Fpmx, ua1x=kerΦ1,u2a2x=kerΦ2, dega1x>degp1x, dega2x>degp2x, and dega2x>degq1x.Proof.
According to Lemmas1 and 2 and the note after Definition 2, we know that Φ2C2=C1 is a cyclic code over R, and it can be written uniquely as(7)Φ2C2=gx+up1x,ua1x,where p1x and monic gx,a1x are polynomials in Fpmx such that a1xgxxN−1 in Fpmx, ua1x=kerΦ1, and dega1x>degp1x. Therefore, we only need to prove the uniqueness of p2x,q1x, and a2x.
Sinceu2a2x=kerΦ2, then a2x is an ideal of Fpmx/xN−1, which implies that monic polynomial a2x∈Fpmx with a2x|xN−1 in Fpmx is unique.
Now, we consider the uniqueness ofp2x and q1x. By equation (7), it is easy to check that C2 can be written as(8)C2=gx+up1x+u2p2x,β1xgx+up1x+u2p3x,ua1x+u2q1x,uβ2xa1x+u2q2x,u2a2x,where p2x,p3x,q1x,q2x,β2x∈Fpmx, β1x∈Sx, dega2x>degp2x, dega2x>degp3x, dega2x>degq1x, and dega2x>degq2x. Consequently,(9)β1xgx+up1x+u2p3x−β1xgx+up1x+u2p2x=u2p3x−β1xp2x∈kerΦ2,uβ2xa1x+u2q2x−β2xua1x+u2q1x=u2q2x−β2xq1x∈kerΦ2.
This meansC2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x. Suppose there are p4x,q3x∈Fpmx such that dega2x>degp4x, dega2x>degq3x, and(10)C2=gx+up1x+u2p4x,ua1x+u2q3x,u2a2x,then(11)gx+up1x+u2p2x−gx+up1x+u2p4x=u2p2x−p4x∈kerΦ2,which implies a2x|p2x−p4x. Since dega2x>degp2x and dega2x>degp4x, then we have p2x=p4x, which means p2x is unique. Similarly, we can get q1x=q3x and hence give the proof of the uniqueness of q1x.
Next, we will consider the necessary and sufficient conditions for the constraintsua1x=kerΦ1 and u2a2x=kerΦ2 in Theorem 1, which will be proposed in Lemmas 3 and 4, respectively.Lemma 3.
LetC2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x be a cyclic code of arbitrary length N over S, and the constraints on corresponding polynomials are the same as that in Theorem 1. Then, we have the following results (1) to (5) in Fpmx:(1)
a1x|p1xxN−1/gx(2)
a2x|q1xxN−1/a1x(3)
a2x|p2xxN−1/gcdgx,p1x(4)
a2x|p1x−gxq1x/a1x(5)
a2x|p2xxN−1/gx−p1xq1xxN−1/a1xgxProof.
SinceΦ2C2=gx+up1x,ua1x and ua1x=kerΦ1, then a1x|p1xxN−1/gx in Fpmx from Lemma 2. On the other hand, we have the following in Sx/xN−1.Case 1:xN−1/a1xua1x+u2q1x=u2q1xxN−1/a1x∈kerΦ2Case 2:xN−1/gcdgx,p1xgx+up1x+u2p2x=u2p2xxN−1/gcdgx,p1x∈kerΦ2Case 3:ugx+up1x+u2p2x−gx/a1xua1x+u2q1x=u2p1x−gxq1x/a1x∈kerΦ2Case 4:xN−1/gxgx+up1x+u2p2x−p1xxN−1/a1xgxua1x+u2q1x=u2p2xxN−1/gx−p1xq1xxN−1/a1xgx∈kerΦ2
These cases mean that the results (2) to (5) are true inSx/xN−1, which are also true in Fpmx through a similar discussion to Lemma 2.Lemma 4.
LetC2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x be a cyclic code of arbitrary length N over S, where p1x,p2x,q1x and monic gx,a1x,a2x are polynomials in Fpmx such that a2x|a1x| in Fpmx, dega1x>degp1x, dega2x>degp2x, and dega2x>degq1x. Then, ua1x=kerΦ1 and u2a2x=kerΦ2 if the corresponding polynomials satisfy all conditions (1) to (5) in Lemma 3.Proof.
It is straight forward thatua1x=kerΦ1 from Lemmas 2 and 3 (1). Next, we will construct u2bx∈C2 for bx∈Fpmx. On the one hand, we use one generator of C2.Case 1: sinceu2gx+up1x+u2p2x=u2gx∈C2, then we have u2gx∈u2a2x for a2xgxxN−1 in Fpmx.Case 2: sinceuxN−1/gxgx+up1x+u2p2x=u2p1xxN−1/gx∈C2, then u2p1xxN−1/gx∈u2a2x from Lemma 3 (1) and a2xa1xxN−1 in Fpmx.Case 3: sincexN−1/gcdgx,p1xgx+up1x+u2p2x=u2p2xxN−1/gcdgx,p1x∈C2, then u2p2xxN−1/gcdgx,p1x∈u2a2x from Lemma 3 (3).Case 4: sinceuua1x+u2q1x=u2a1x∈C2, then we have u2a1x∈u2a2x from a2xa1xxN−1 in Fpmx.Case 5: sincexN−1/a1xua1x+u2q1x=u2q1xxN−1/a1x∈C2, then we have u2q1xN−1/a1x∈u2a2x from Lemma 3 (2).Case 6:u2a2x∈u2a2x.On the other hand, we use two generators ofC2. Indeed, we only need to consider gx+up1x+u2p2x and ua1x+u2q1x, since u2a2x∈u2a2x.Case 7: from Lemma3 (4), we have(12)ugx+up1x+u2p2x−gxa1xua1x+u2q1x=u2p1x−gxa1xq1x∈C2,and hencep1x−gx/a1xq1x∈u2a2x.Case 8: from Lemma3 (5), we have(13)xN−1gxgx+up1x+u2p2x−xN−1a1xgxp1xua1x+u2q1x=u2xN−1gxp2x−xN−1a1xgxp1xq1x∈u2a2x.
From Cases 1 to 8, we haveu2bx∈u2a2x if u2bx∈C2, which implies kerΦ2⊆u2a2x. Since u2a2x∈C2, then u2a2x⊆kerΦ2. Thus, we have kerΦ2=u2a2x.
Finally, we consider the number of codewords of a given cyclic code overS. Let cx=∑i=0N−1cixi∈Sx. If we denote rjcx=∑i=0N−1rjci)xi∈Fpmx for 0≤j≤2, then cx=∑j=02ujrjcx. Consequently, the η-th torsion code of C2 is defined by TorηC2=r0cx|uηcx∈C2, which is a cyclic code of length N over Fpm.Lemma 5.
LetC2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x be a cyclic code of arbitrary length N over S, where p1x,p2x,q1x and monic gx,a1x,a2x are polynomials in Fpmx such that a2x|a1x| in Fpmx, dega1x>degp1x, dega2x>degp2x, and dega2x>degq1x. Then, ua1x=kerΦ1 and u2a2x=kerΦ2 if and only if a1x=Tor1C2 and a2x=Tor2C2.Proof.
Letcx=∑i=0N−1cixi=∑j=02ujrjcx∈Sx, then(14)Tor1C=r0cx|ur0cx+u2r1cx∈C2=r0cx|ur0cx∈Φ2C2.
Therefore,Tor1C=a1x if and only if(15)ua1x=uh1x|uh1x∈Φ2C2=C1,h1x∈FpmxxN−1=kerΦ1.
On the other hand,(16)Tor2C=r0cx|u2cx∈C2=r0cx|u2r0cx∈C2
Hence,Tor2C=a2x if and only if(17)u2a2x=u2h2x|u2h2x∈C2,h2x∈FpmxxN−1=kerΦ2.Theorem 2.
LetC2 be a cyclic code of arbitrary length N over S, then there exist unique polynomials gx,a1x,a2x,p1x,p2x,q1x∈Fpmx, such that(18)C2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x,where gx,a1x, and a2x are monic, a2x|a1x|, a1x|p1xxN−1/gx,a2x|q1xxN−1/a1x, a2x|p1x−gxq1x/a1x, a2x|p2xxN−1/gx−p1xq1xxN−1/a1xgx, a2x|p2xxN−1/p1x in Fpmx,dega1x>degp1x, dega2x>degp2x, and dega2x>degq1x. Besides, the number of codewords of C2 is(19)C2=pm3N−deggx−dega1x−dega2x.Proof.
From Theorem1 and Lemmas 3 and 4, the uniqueness of generators for C2 is clear. Consequently, we only need to calculate the number of codewords of C2. It is easy to check that Tor0C2=gx from the definition of torsion code. On the other hand, a1x=Tor1C2 and a2x=Tor2C2 from Lemma 5. Therefore, by Han et al. [14], we have(20)C2=∏η=02TorηC=pm3N−deggx−dega1x−dega2x.
IfC2=gx+up1x+u2p2x,ua1x+u2q1x,u2a2x is a cyclic code of arbitrary length N over S and the corresponding polynomials satisfy the conditions in Theorem 2, then we denote C2 by(21)C2=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫.
From Theorem2, the following corollary is straightforward.Corollary 1.
LetC2 be a cyclic code of length N=pe over S, then there exit unique polynomials p1x,p2x,andq1x and unique integers t0,t1,t2 such that(22)C2=x−1t0+up1x+u2p2x,ux−1t1+u2q1x,u2x−1t2where 0≤t2≤t1≤t0≤pe, t1>degp1x, t2>degp2x, t2>degq1x, and x−1t0+t1|x−1pep1x, x−1t1+t2|x−1peq1x, x−1t2|p1x−x−1t0−t1q1x, x−1t2|x−1pe−t0p2x−x−1pe−t0−t1p1xq1x, and x−1t2|p2xx−1pe/gcdx−1t0,p1x) in Fpmx. Besides, the number of codewords of C2 is C2=pm3pe−t0−t1−t2.


## 4. The Dual Codes of Cyclic Codes of Arbitrary Length overS

Lemma 6.
LetC2=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫ be a cyclic code of arbitrary length N over S, then(23)xN−1a2x−uq1xxN−1a1xa2x+u2θx∈AnnC2,uxN−1a1x−u2p1xxN−1a1xgx∈AnnC2,u2xN−1gx∈AnnC2,where θx=p1xq1xxN−1/gxa1x−p2xxN−1/gx/a2x.Proof.
From Theorem2, we know that xN−1/a2x,q1xxN−1/a1xa2x,θx,xN−1/a1x,p1xxN−1/a1xgx, and xN−1/gx are polynomials in Fpmx and all their degrees are not more than N. In Sx/xN−1, we have(24)xN−1a2x−uq1xxN−1a1xa2x+u2θxgx+up1x+u2p2x=−ugxq1xxN−1a1xa2x+up1xxN−1a2x=uxN−1p1x−gxq1x/a1xa2x=0.
Besides, it is easy to check that(25)xN−1a2x−uq1xxN−1a1xa2x+u2θxua1x+u2q1x=0,xN−1a2x−uq1xxN−1a1xa2x+u2θxu2a2x=0,which implies(26)xN−1a2x−uq1xxN−1a1xa2x+u2θx∈AnnC2.
Similarly, we can proveuxN−1/a1x−u2p1xxN−1/a1xgx∈AnnC2 and u2xN−1/gx∈AnnC2.Lemma 7.
LetC2=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫ be a cyclic code of arbitrary length N over S, then(27)AnnC2=≪xN−1a2x−uq1xxN−1a1xa2x+u2θx,uxN−1a1x−u2p1xxN−1a1xgx,u2xN−1gx≫,where θx=p1xq1xxN−1/gxa1x−p2xxN−1/gx/a2x.Proof.
Denote by(28)J=gx˜+up1x˜+u2p2x˜,ua1x˜+u2q1x˜,u2a2x˜,where gx˜=xN−1/a2x, p1x˜=−q1xxN−1/a1xa2x,p2x˜=θx, a1x˜=xN−1/a1x,q1x˜=−p1xxN−1/a1xgx, and a2x˜=xN−1/gx. It is straightforward that J⊆AnnC2 from Lemma 6. Next, we will prove(29)J=≪gx˜+up1x˜+u2p2x˜,ua1x˜+u2q1x˜,u2a2x˜≫.
From Theorem2, we have p1x˜,p2x˜, q1x˜∈Fpmx. Besides, the polynomials gx˜, a1x˜, a2x˜∈Fpmx are monic, dega1x˜>degp1x˜, dega2x˜>degp2x˜, and dega2x˜>degq1x˜.Case 1: sincea2x|a1x| in Fpmx, then a2x˜|a1x˜| in Fpmx.Case 2: sincep1x˜xN−1/gx˜=−q1xxN−1/a1x, then a1x˜|p1x˜xN−1/gx˜ in Fpmx.Case 3: sinceq1x˜xN−1/a1x˜=−p1xxN−1/gx, then a2x˜|q1x˜xN−1/a1x˜ in Fpmx.Case 4: sincep1x˜−gx˜q1x˜/a1x˜=xN−1/gxp1x−gxq1x/a1x/a2x, then a2x˜|p1x˜−gx˜q1x˜/a1x˜ in Fpmx.Case 5: sincep2x˜xN−1/gx˜−p1x˜q1x˜xN−1/a1x˜gx˜=−p2xxN−1/gx, then in Fpmx, we have(30)a2x˜|xN−1gx˜p2x˜−xN−1a1x˜gx˜p1x˜q1x˜.Case 6: since(31)xN−1gcdgx˜,p1x˜p2x˜xN−1gxp1xq1xxN−1/a2xa1x−p2xxN−1/a2xgcdgx˜,p1x˜,then in Fpmx, we have a2x˜|[p2x˜xN−1/gcdgx˜,p1x˜, which is because(32)gcdgx˜,p1x˜|p2xxN−1a2x,gcdgx˜,p1x˜|p1xq1xxN−1a2xa1x.
From the above discussion and Theorem2, we have(33)J=≪gx˜+up1x˜+u2p2x˜,ua1x˜+u2q1x˜,u2a2x˜≫and hence J=pmdeggx+dega1x+dega2x=C2⊥=AnnC2. Since J⊆AnnC2, then J=AnnC2. This gives the proof.Theorem 3.
LetC2=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫ be a cyclic code of arbitrary length N over S. Degrees of gx,a1x,anda2x are denoted by 0≤t0,t1,t2≤N, respectively. Degrees of q1xxN−1/a1xa2x,θx,p1xxN−1/a1xgx are denoted by 0≤t3,t4,t5<N, respectively. Constant terms of xN−1/gx,xN−1/a1x,andxN−1/a2x are written as γ0,γ1,andγ2, respectively. Then,(34)C2⊥=≪γ2−1xN−1a2x∗+uP1x+u2P2x,uγ1−1xN−1a1x∗+u2Q1x,u2γ0−1xN−1gx∗≫,where(35)Q1x≡−γ1−1xN−t1−t5p1xxN−1a1xgx∗modγ0−1xN−1gx∗,uP1x+u2P2x≡−uγ2−1xN−t2−t3q1xxN−1a1xa2x∗+u2γ2−1xN−t2−t4θx∗moduγ1−1xN−1a1x∗+u2Q1x,u2γ0−1xN−1gx∗,for(36)θx=p1xq1xxN−1/gxa1x−p2xxN−1/gxa2x,degP1x<degγ1−1xN−1a1x∗,degQ1x<degγ0−1xN−1gx∗,degP2x<degγ0−1xN−1gx∗.Proof.
From Theorem2, we know that gx,a1x,anda2x are monic polynomials in Fpmx and a2x|a1x| in Fpmx, which implies that γ0,γ1,γ2≠0. From Lemma 7, we have (37)C2⊥=xN−1a2x−uq1xxN−1a1xa2x+u2θx∗,uxN−1a1x−u2p1xxN−1a1xgx∗,u2xN−1gx∗=xN−1a2x∗−uxN−t2−t3q1xxN−1a1xa2x∗+u2xN−t2−t4θx∗,uxN−1a1x∗−u2xN−t1−t5p1xxN−1a1xgx∗,u2xN−1gx∗=γ2−1xN−1a2x∗−uγ2−1xN−t2−t3q1xxN−1a1xa2x∗+u2γ2−1xN−t2−t4θx∗,uγ1−1xN−1a1x∗−u2γ1−1xN−t1−t5p1xxN−1a1xgx∗,u2γ0−1xN−1gx∗=γ2−1xN−1a2x∗+uP1x+u2P2x,uγ1−1xN−1a1x∗+u2Q1x,u2γ0−1xN−1gx∗.
From the definitions of torsion code, we have(38)γ2−1xN−1a2x∗⊆Tor0C2⊥,γ1−1xN−1a1x∗⊆Tor1C2⊥,γ0−1xN−1gx∗⊆Tor2C2⊥.
Since(39)γ2−1xN−1a2x∗⋅γ1−1xN−1a1x∗⋅γ0−1xN−1gx∗=pmdeggx+dega1x+dega2x=C2⊥=Tor0C2⊥⋅Tor1C2⊥⋅Tor2C2⊥,then(40)γ2−1xN−1a2x∗=Tor0C2⊥,γ1−1xN−1a1x∗=Tor1C2⊥,γ0−1xN−1gx∗=Tor2C2⊥,which implies that(41)C2⊥=≪γ2−1xN−1a2x∗+uP1x+u2P2x,uγ1−1xN−1a1x∗+u2Q1x,u2γ0−1xN−1gx∗≫from Theorem 1 and Lemma 5.Lemma 8.
Leta1x be a monic divisor of xN−1 in Fpmx and γ1 be the constant term of xN−1/a1x. If a1xa1∗x=γ1−1xN−1 in Fpmx, then N is a positive even number and p=2.Proof.
SupposeN=pen for nonnegative e and gcdp,n=1. Let xn−1=x−1∏i=1jfi be the factorization of xn−1 into a product of monic basic irreducible pairwise coprime polynomials in Fpmx, then xN−1=xn−1pe=x−1pe∏i=1jfipe and gcd[x−1pe,∏i=1jfipe]=gcd[x−1pe,∏i=1jfipe∗=1. Since a1x is a monic divisor of xN−1 in Fpmx, then we let a1x=x−1k0∏i=1jfiki for 0≤k0≤pe and 0≤ki≤pe. Thus, a1∗x=−1k0x−1k0∏i=1jfiki∗. Since a1xa1∗x=γ1−1xN−1 and gcdx−1pe,∏i=1jfipe]=gcd[x−1pe,∏i=1jfipe∗=1 in Fpmx, then we have 2k0=pe, which implies N=pen is a positive even number and p=2.Theorem 4.
LetC2=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫ be a cyclic code of arbitrary length N over S. Degrees of gx,a1x,anda2x are denoted by 0≤t0,t1,t2≤N, respectively. Degrees of q1xxN−1/a1xa2x,θx,p1xxN−1/a1xgx are denoted by 0≤t3,t4,t5<N, respectively. Constant terms of xN−1/gx,xN−1/a1x,xN−1/a2x are written as γ0,γ1,andγ2, respectively. Then, C2 is a self-dual cyclic code of length N over S if and only if N is a positive even number, p=2, a2∗xgx=γ0−1xN−1, a1xa1∗x=γ1−1xN−1, γ0+γ2=0, q1x≡−γ1−1xN−t1−t5p1xxN−1/a1xgx∗moda2x,and(42)uP1x+u2P2x≡−uγ2−1xN−t2−t3q1xxN−1a1xa2x∗+u2γ2−1xN−t2−t4θx∗modua1x+u2q1x,u2a2x,for θx=p1xq1xxN−1/gxa1x−p2xxN−1/gx/a2x.Proof.
According to Theorem3, we have(43)C2isaself−dualcycliccodeoflengthNoverS⇔gx=γ2−1xN−1a2x∗,a1x=γ1−1xN−1a1x∗,a2x=γ0−1xN−1gx∗,q1x=Q1x,up1x+u2p2x=P1x+u2P2x.⇔gx=γ2−1xN−1a2x∗,a1x=γ1−1xN−1a1x∗,a2x=γ0−1xN−1gx∗,q1x≡−γ1−1xN−t1−t5p1xxN−1a1xgx∗moda2x,uP1x+u2P2x≡−uγ2−1xN−t2−t3q1xxN−1a1xa2x∗+u2γ2−1xN−t2−t4θx∗modua1x+u2q1x,u2a2x.
Besides,(44)gx=γ2−1xN−1a2x∗,a1x=γ1−1xN−1a1x∗,a2x=γ0−1xN−1gx∗.⇔a2xgx∗=γ2−1xN−1,a1xa1∗x=γ1−1xN−1,a2∗xgx=γ0−1xN−1.⇔a2∗xgx=γ0−1xN−1,a1xa1∗x=γ1−1xN−1,γ0+γ2=0.
Sincea1x is a monic divisor of xN−1 and a1xa1∗x=γ1−1xN−1, then N is a positive even number and p=2 from Lemma 8. This gives the proof.Corollary 2.
LetC2=≪x−1t0+up1x+u2p2x,ux−1t1+u2q1x,u2x−1t2≫ be a cyclic code of length N=pe over S. Degrees of q1xx−12e−1−t2,θx,x−12e−1−t0p1x are denoted by 0≤t3,t4,t5<N, respectively. Then, C2 is a self-dual cyclic code of N=pe over S if and only if p=2, t0+t2=2e, t1=2e−1, q1x≡x2e−1−t5x−12e−1−t0p1∗xmodx−1t2, and up1x+u2p2x≡ux2e−t2−t3q1xx−12e−1−t2∗+u2x2e−t2−t4θx∗modux−12e−1+u2q1x,u2x−1t2 for θx=x−1−2e−1p1xq1x+p2x.Proof.
LetN=pe,gx=x−1t0,a1x=x−1t1,a2x=x−1t2 in Theorem 4, then γ0=γ1=γ2=1 and p=2.
Observe thata2∗xgx=γ0−1xN−1, a1xa1∗x=γ1−1xN−1, and γ0+γ2=0 if and only if t0+t2=2e and t1=2e−1. So, q1xxN−1/a1xa2x=q1xx−12e−1−t2,θx=x−1−2e−1p1xq1x+p2x, and p1xxN−1/a1xgx=x−12e−1−t0p1x. Thus,(45)q1x≡−γ1−1xN−t1−t5p1xxN−1a1xgx∗≡x2e−1−t5x−12e−1−t0p1∗xmodx−1t2,up1x+u2p2x≡−uγ2−1xN−t2−t3q1xxN−1a1xa2x∗+u2γ2−1xN−t2−t4θx∗≡ux2e−t2−t3q1xx−12e−1−t2∗+u2x2e−t2−t4θx∗modux−12e−1+u2q1x,u2x−1t2,where θx=x−1−2e−1p1xq1x+p2x.Example 2.
InF2x, we have x8−1=x−18. Let(46)D3=x−15+ux−13+u2,ux−14+u2x−12,u2x−13,be a cyclic code of length N=23 over F2u/u3. It is easy to check that D3 is in the unique form in Corollary 1, and hence D3=224−5−4−3=212. From the unique form in Corollary 1, we get that D3 is a new cyclic code of length N=23 over F2u/u3 comparing with codes in [10]. Furthermore,(47)N=2e=23,t0=5,t1=4,t2=3,p1x=x−13,p2x=1,q1x=x−12.
Thus,p=2, t0+t2=23, and t1=22. Besides,(48)q1xx−12e−1−t2=x−13,θx=x,x−12e−1−t0p1x=x−12,which implies that t3=3,t4=1, and t5=2 from Corollary 2.
Sinceq1x−x2e−1−t5x−12e−1−t0p1∗x=x−12−x2x−1−1−x+13=x−14, then q1x−x2e−1−t5x−12e−1−t0p1∗x≡0modx−1t2, which implies that q1x≡x2e−1−t5x−12e−1−t0p1∗xmodx−1t2. Since(49)up1x+u2p2x−ux2e−t2−t3q1xx−12e−1−t2∗−u2x2e−t2−t4θx∗=ux−13+u2−ux2x−13−u2x4=ux−15−u2x−14=x−1ux−14+u2x−12−u2xx−13≡0modux−12e−1+u2q1x,u2x−1t2,then(50)up1x+u2p2x≡ux2e−t2−t3q1xx−12e−1−t2∗+u2x2e−t2−t4θx∗modux−12e−1+u2q1x,u2x−1t2.
Thus,D3 is a self-dual cyclic code of length N=23 over F2u/u3.Example 3.
InF2x, we have x12−1=x+14x2+x+14. Suppose gx=x+12x2+x+14, p1x=0, p2x=x, a1x=x+12x2+x+12, q1x=0, and a2x=x+12. From Theorem 2, we get D4=≪gx+up1x+u2p2x,ua1x+u2q1x,u2a2x≫ is a cyclic code of length N=22⋅3 over F2u/u3 and D4=218.
Besides, degrees ofgx,a1x,anda2x are t0=10,t1=6,andt2=2, respectively. Constant terms of xN−1/gx,xN−1/a1x,andxN−1/a2x are γ0=γ1=γ2=1, and hence γ0+γ2=0. Since q1xxN−1/a1xa2x=0,θx=x,andp1xxN−1/a1xgx=0, then t3=0,t4=1,andt5=0. It is easy to check that a2∗xgx=γ0−1xN−1,a1xa1∗x=γ1−1xN−1, and q1x≡−γ1−1xN−t1−t5p1xxN−1/a1xgx∗moda2x.
SinceuP1x+u2P2x≡−uγ2−1xN−t2−t3q1xxN−1/a1xa2x∗+u2γ2−1xN−t2−t4θx∗=u2x−u2x9=u2xx+18≡0modua1x+u2q1x,u2a2x, then D4 is a self-dual cyclic code of length N=22⋅3 over F2u/u3 from Theorem 4.


## 5. Conclusion

In this paper, we obtain unique generators for cyclic codes with arbitrary length overFpmu/u3, as well as their numbers of codewords. Based on these, we determine the dual of cyclic codes of arbitrary length and propose necessary and sufficient conditions for the self-dualities of cyclic codes. It is interesting to consider the number of self-dual cyclic codes of even length over F2mu/u3.



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