# Self-Dual Abelian Codes in Some Nonprincipal Ideal Group Algebras

Authors: Parinyawat Choosuwan; Somphong Jitman; Patanee Udomkavanich
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

The main focus of this paper is the complete enumeration of self-dual abelian codes in nonprincipal ideal group algebrasF2k[A×Z2×Z2s] with respect to both the Euclidean and Hermitian inner products, where k and s are positive integers and A is an abelian group of odd order. Based on the well-known characterization of Euclidean and Hermitian self-dual abelian codes, we show that such enumeration can be obtained in terms of a suitable product of the number of cyclic codes, the number of Euclidean self-dual cyclic codes, and the number of Hermitian self-dual cyclic codes of length 2s over some Galois extensions of the ring F2k+uF2k, where u2=0. Subsequently, general results on the characterization and enumeration of cyclic codes and self-dual codes of length ps over Fpk+uFpk are given. Combining these results, the complete enumeration of self-dual abelian codes in F2k[A×Z2×Z2s] is therefore obtained.


---

## Body

## 1. Introduction

Information media, such as communication systems and storage devices of data, are not100 percent reliable in practice because of noise or other forms of introduced interference. The art of error correcting codes is a branch of Mathematics that has been introduced to deal with this problem since the 1960s. Linear codes with additional algebraic structures and self-dual codes are important families of codes that have been extensively studied for both theoretical and practical reasons (see [1–10] and references therein). Some major results on Euclidean self-dual cyclic codes have been established in [11]. In [12], the complete characterization and enumeration of such self-dual codes have been given. These results on Euclidean self-dual cyclic codes have been generalized to abelian codes in group algebras [6] and the complete characterization and enumeration of Euclidean self-dual abelian codes in principal ideal group algebras (PIGAs) have been established. Extensively, the characterization and enumeration of Hermitian self-dual abelian codes in PIGAs have been studied in [9]. To the best of our knowledge, the characterization and enumeration of self-dual abelian codes in nonprincipal ideal group algebras (non-PIGAs) have not been well studied. It is therefore of natural interest to focus on this open problem.In [6, 9], it has been shown that there exists a Euclidean (resp., Hermitian) self-dual abelian code in FpkG if and only if p=2 and G is even. In order to study self-dual abelian codes, it is therefore restricted to the group algebra F2kA×B, where A is an abelian group of odd order and B is a nontrivial abelian group of two power order. In this case, F2kA×B is a PIGA if and only if B=Z2s is a cyclic group (see [13]). Equivalently, F2kA×B is a non-PIGA if and only if B is noncyclic. To avoid tedious computations, we focus on the simplest case  where B=Z2×Z2s, where s is a positive integer. Precisely, the goal of this paper is to determine the algebraic structures and the numbers of Euclidean and Hermitian self-dual abelian codes in F2kA×Z2×Z2s.It turns out that every Euclidean (resp., Hermitian) self-dual abelian code inF2k[A×Z2×Z2s] is isomorphic to a suitable Cartesian product of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length 2s over some Galois extension of the ring F2k+uF2k, where u2=0 (see Section 2). Hence, the number of self-dual abelian codes in F2k[A×Z2×Z2s] can be determined in terms of the cyclic codes mentioned earlier. Subsequently, useful properties of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length ps over Fpk+uFpk are given for all primes p. Combining these results, the characterizations and enumerations of Euclidean and Hermitian self-dual abelian codes in F2kA×Z2×Z2s are rewarded.The paper is organized as follows. In Section2, some basic results on abelian codes are recalled together with a link between abelian codes in F2kA×Z2×Z2s and cyclic codes of length 2s over Galois extensions of F2k+uF2k. General results on the characterization and enumeration of cyclic codes of length ps over Fpk+uFpk are provided in Section 3. In Section 4, the characterizations and enumerations of Euclidean and Hermitian self-dual cyclic codes of length ps over Fpk+uFpk are established. Summary and remarks are given in Section 5.


## 2. Preliminaries

In this section, we recall some definitions and basic properties of rings and abelian codes. Subsequently, a link between an abelian code in nonprincipal ideal algebras and a product of cyclic codes over rings is given. This link plays an important role in determining the algebraic structures and the numbers of Euclidean and Hermitian self-dual abelian codes in nonprincipal ideal algebras.


### 2.1. Rings and Abelian Codes in Group Rings

For a primep and a positive integer k, denote by Fpk the finite field of order pk. Let Fpk+uFpk≔a+ub∣a,b∈Fpk be a ring, where the addition and multiplication are defined as in the usual polynomial ring over Fpk with indeterminate u together with the condition u2=0. We note that Fpk+uFpk is isomorphic to Fpku/u2 as rings. The Galois extension of Fpk+uFpk of degree m is defined to be the quotient ring Fpk+uFpkx/fx, where f(x) is an irreducible polynomial of degree m over Fpk. It is not difficult to see that the Galois extension of Fpk+uFpk of degree m is isomorphic to Fpkm+uFpkm as rings. In the case  where k is even, the mapping a+ub↦apk/2+ubpk/2 is a ring automorphism of order 2 on Fpk+uFpk. The readers may refer to [14, 15] for properties of the ring Fpk+uFpk.For a commutative ringR with identity 1 and a finite abelian group G, written additively, let RG denote thegroup ring of G over R. The elements in R[G] will be written as ∑g∈G‍αgYg, where αg∈R. The addition and the multiplication in R[G] are given as in the usual polynomial ring over R with indeterminate Y, where the indices are computed additively in G. Note that Y0≔1 is the multiplicative identity of R[G] (resp., R), where 0 is the identity of G. We define aconjugation— on R[G] to be the map that fixes R and sends Yg to Y-g for all g∈G; that is, for u=∑g∈G‍αgYg∈R[G], we set u¯≔∑g∈G‍αgY-g=∑g∈G‍α-gYg. In the case  where there exists a ring automorphism ρ on R of order 2, we define u~≔∑g∈G‍ρ(αg)Y-g for all u=∑g∈G‍αgYg∈R[G]. In the case  where R is a finite field Fpk, then Fpk[G] can be viewed as an Fpk-algebra and it is called agroup algebra. The group algebra Fpk[G] is called aprincipal ideal group algebra (PIGA) if its ideals are generated by a single element.Anabelian code in R[G] is defined to be an ideal in R[G]. If G is cyclic, this code is called acyclic code, a code which is invariant under the right cyclic shift. It is well known that cyclic codes of length n over R can be regarded as ideals in the quotient polynomial ring R[x]/xn-1≅RZn.TheEuclidean inner product in R[G] is defined as follows. For(1)u=∑g∈GαgYg,v=∑g∈GβgYg in R[G], we set (2)u,vE≔∑g∈Gαgβg. In addition, if there exists a ring automorphism ρ of order 2 on R, the ρ-inner product of u and v is defined by (3)u,vρ≔∑g∈Gαgρβg. If R=Fq2 (resp., R=Fq2+uFq2) and ρ(a)=aq (resp., ρ(a+ub)=aq+ubq) for all a∈Fq2 (resp., a+ub∈Fq2+uFq2), the ρ-inner product is called theHermitian inner product and denoted by u,vH.TheEuclidean dual andHermitian dual of C in R[G] are defined to be the sets(4)C⊥E≔u∈RG∣u,vE=0∀v∈C,(5)C⊥H≔u∈RG∣u,vH=0∀v∈C,respectively.An abelian codeC is said to be Euclidean self-dual (resp., Hermitian self-dual) if C=C⊥E (resp., C=C⊥H).For convenience, denote byN(pk,n), NE(pk,n), and NH(pk,n) the number of cyclic codes, the number of Euclidean self-dual cyclic codes, and the number of Hermitian self-dual cyclic codes of length n over Fpk+uFpk, respectively.


### 2.2. Decomposition of Abelian Codes inF2kA×Z2×Z2s

In [6, 9], it has been shown that there exists a Euclidean (resp., Hermitian) self-dual abelian code in FpkG if and only if p=2 and G is even. To study self-dual abelian codes, it is sufficient to focus on F2kA×B, where A is an abelian group of odd order and B is a nontrivial abelian group of two power order. In this case, F2kA×B is a PIGA if and only if B=Z2s is a cyclic group for some positive integer s (see [13]). The complete characterization and enumeration of self-dual abelian codes in PIGAs have been given in [6, 9]. Here, we focus on self-dual abelian codes in non-PIGAs, or equivalently, B is non-cyclic. To avoid tedious computations, we establish results for the simplest case  where B=Z2×Z2s. Useful decompositions of F2kA×Z2×Z2s are given in this section.First, we consider the decomposition ofR≔F2k[A]. In this case, R is semisimple [2] which can be decomposed using the Discrete Fourier Transform in [7] (see [6, 9] for more details). For completeness, the decompositions used in this paper are summarized as follows.For an odd positive integeri and a positive integer k, let ordi(2k) denote the multiplicative order of 2k modulo i. For each a∈A, denote by ord(a) the additive order of a in A. A 2k-cyclotomic class of A containing a∈A, denoted by S2k(a), is defined to be the set (6)S2ka≔2ki·a∣i=0,1,…=2ki·a∣0≤i<ordorda2k,where 2ki·a≔∑j=12kia in A.Anidempotent in R is a nonzero element e such that e2=e. It is calledprimitive if, for every other idempotent f, either ef=e or ef=0. The primitive idempotents in R are induced by the 2k-cyclotomic classes of A (see [5, Proposition  II.4]). Let {a1,a2,…,at} be a complete set of representatives of 2k-cyclotomic classes of A and let ei be the primitive idempotent induced by S2k(ai) for all 1≤i≤t. From [7], R can be decomposed as(7)R=⨁i=1tRei,and hence,(8)F2kA×Z2≅RZ2≅⨁i=1tReiZ2.It is well known (see [6, 9]) that Rei≅F2ki, where ki is a multiple of k. Precisely, ki=kS2kai=k·ordord(ai)(2k) provided that ei is induced by S2k(ai). It follows that Rei[Z2]≅F2ki[Z2]. Under the ring isomorphism that fixes the elements in F2ki and Y1↦u+1, F2ki[Z2] is isomorphic to the ring F2ki+uF2ki, where u2=0. We note that this ring plays an important role in coding theory and codes over rings in this family have extensively been studied in [14–17] and references therein.From (8) and the ring isomorphism discussed above, we have(9)F2kA×Z2×Z2s≅∏i=1tRiZ2s,where Ri≔F2ki+uF2ki for all 1≤i≤t.In order to study the algebraic structures of Euclidean and Hermitian self-dual abelian codes inF2k[A×Z2×Z2s], the two rearrangements of Ri’s in the decomposition (9) are needed. The details are given in the following two subsections.


#### 2.2.1. Euclidean Case

A2k-cyclotomic class S2k(a) is said to be oftypeI if a=-a (in this case, S2k(a)=S2k(-a)),typeII if S2k(a)=S2k(-a) and a≠-a, ortypeIII if S2k(-a)≠S2k(a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈{I,II,III} if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes of A can be chosen such that ai∣i=1,2,…,rI, arI+j∣j=1,2,…,rII, and arI+rII+l,arI+rII+rIII+l=-arI+rII+l∣l=1,2,…,rIII are sets of representatives of 2k-cyclotomic classes of types I, II, and III, respectively, where t=rI+rII+2rIII.Rearranging the terms in the decomposition ofR in (7) based on the 3 types of primitive idempotents, we have(10)F2kA×Z2≅⨁i=1tReiZ2≅∏i=1rIRi∏j=1rIISj∏l=1rIIITl×Tl,where Ri≔F2k+uF2k for all i=1,2,…,rI, Sj≔F2krI+j+uF2krI+j for all j=1,2,…,rII, and Tl≔F2krI+rII+l+uF2krI+rII+l for all l=1,2,…,rIII.From (10), we have(11)F2kA×Z2×Z2s≅∏i=1rIRiZ2s∏j=1rIISjZ2s∏l=1rIIITlZ2s×TlZ2s.It follows that an abelian code C in F2k[A×Z2×Z2s] can be viewed as(12)C≅∏i=1rIBi∏j=1rIICj∏l=1rIIIDl×Dl′,where Bi, Cj, Dl, and Dl′ are cyclic codes in RiZ2s, SjZ2s, TlZ2s, and TlZ2s, respectively, for all i=1,2,…,rI, j=1,2,…,rII, and l=1,2,…,rIII.Using the analysis similar to those in [6, Section  II.D], the Euclidean dual of C in (12) is of the form(13)C⊥E≅∏i=1rIBi⊥E∏j=1rIICj⊥H∏l=1rIIIDl′⊥E×Dl⊥E.Similar to [6, Corollary  2.9], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Euclidean self-dual can be given using the notions of cyclic codes of length 2s over Ri, Sj, and Tl in the following corollary.Corollary 1.
An abelian codeC in F2k[A×Z2×Z2s] is Euclidean self-dual if and only if in the decomposition (12)(i)
Bi is a Euclidean self-dual cyclic code of length 2s over Ri for all i=1,2,…,rI,(ii)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rII,(iii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rIII.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to begood if j divides 2ki+1 for some integer i≥1 andbad otherwise. These notions have been introduced in [6, 12] for the enumeration of self-dual cyclic codes and self-dual abelian codes over finite fields.Letχ be a function defined on the pair (j,2k), where j is an odd positive integer, as follows:(14)χj,2k=0ifj,2kisgood,1otherwise.The number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 2.
Letk and s be positive integers and let A be a finite abelian group of odd order and exponent M. Then the number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] is (15)NE2k,2s∑d∣M,ordd2k=11-χd,2kNAd∏d∣Mordd2k≠1NH2k·ordd2k,2s1-χd,2kNAd/ordd2k∏d∣MN2k·ordd2k,2sχd,2kNAd/2ordd2k,where NA(d) denotes the number of elements in A of order d determined in [18].Proof.
From (12) and Corollary 1, it suffices to determine the numbers of cyclic codes Bi’s, Cj’s, and Dl’s such that Bi and Cj are Euclidean and Hermitian self-dual, respectively.
From [9, Remark  2.5], the elements in A of the same order are partitioned into 2k-cyclotomic classes of the same type. For each divisor d of M, a 2k-cyclotomic class containing an element of order d has cardinality ordd(2k) and the number of such 2k-cyclotomic classes is NA(d)/ordd(2k). We consider the following 3 cases.
Case  1 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) = 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type I. Since there are NAd/ordd(2k) such 2k-cyclotomic classes, the number of Euclidean self-dual cyclic codes Bi’s of length 2s corresponding to d is(16)NE2k,2sNAd/ordd2k=NE2k,2s1-χd,2kNAd.
Case  2 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) ≠ 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type II. Hence, the number of Hermitian self-dual cyclic codes Cj’s of length 2s corresponding to d is (17)NH2k·ordd2k,2sNAd/ordd2k=NH2k·ordd2k,2s1-χd,2kNAd/ordd2k.
Case  3 ( χ ( d , 2 k ) = 1 ). By [6, Lemma  4.5], every 2k-cyclotomic class of A containing an element of order d is of type III. Then the number of cyclic codes Dl’s of length 2s corresponding to d is(18)N2k·ordd2k,2sNAd/2ordd2k=N2k·ordd2k,2sχd,2kNAd/2ordd2k.
The desired result follows sinced runs over all divisors of M.This enumeration will be completed by counting the above numbersNE, NH, and N in Corollaries 22, 25, and 17, respectively.


#### 2.2.2. Hermitian Case

We focus on the case  wherek is even. A 2k-cyclotomic class S2k(a) is said to be oftypeI′ if S2k(a)=S2k(-2k/2a) ortypeII′ if S2k(a)≠S2k(-2k/2a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈I′,II′ if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes can be chosen such that ai∣i=1,2,…,rI′ and arI′+j,arI′+rII′+j=-2k/2arI′+j∣j=1,2,…,rII′ are sets of representatives of 2k-cyclotomic classes of types I′ and II′, respectively, where t=rI′+2rII′.Rearranging the terms in the decomposition ofR in (7) based on the above 2 types of primitive idempotents, we have(19)F2kA×Z2≅⨁i=1tReiZ2≅∏j=1rI′Sj∏l=1rII′Tl×Tl,where Sj≔F2kj+uF2kj for all j=1,2,…,rI′ and Tl≔F2krI′+l+uF2krI′+l for all l=1,2,…,rII′.From (19), we have(20)F2kA×Z2×Z2s≅∏j=1rI′SjZ2s∏l=1rII′TlZ2s×TlZ2s.Hence, an abelian code C in F2k[A×Z2×Z2s] can be viewed as(21)C≅∏j=1rI′Cj∏l=1rII′Dl×Dl′,where Cj, Dl, and Dl′ are cyclic codes in Sj[Z2s], Tl[Z2s], and Tl[Z2s], respectively, for all j=1,2,…,rI′ and l=1,2,…,rII′.Using the analysis similar to those in [9, Section  II.D], the Hermitian dual of C in (21) is of the form(22)C⊥H≅∏j=1rI′Cj⊥H∏l=1rII′Dl′⊥E×Dl⊥E.Similar to [9, Corollary  2.8], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Hermitian self-dual are now given using the notions of cyclic codes of length 2s over Sj and Tl in the following corollary.Corollary 3.
An abelian codeC in F2k[A×Z2×Z2s] is Hermitian self-dual if and only if in the decomposition (21)(i)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rI′,(ii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rII′.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to beoddly good if j divides 2ki+1 for some odd integer i≥1 andevenly good if j divides 2ki+1 for some even integer i≥2. These notions have been introduced in [9] for characterizing the Hermitian self-dual abelian codes in PIGAs.Letλ be a function defined on the pair (j,2k), where j is an odd positive integer, as(23)λj,2k=0ifj,2kisoddlygood,1otherwise.The number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 4.
Letk be an even positive integer and let s be a positive integer. Let A be a finite abelian group of odd order and exponent M. Then the number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] is (24)∏d∣MNH2k·ordd2k,2s1-λd,2k/2NAd/ordd2k∏d∣MN2k·ordd2k,2sλd,2k/2NAd/2ordd2k,where NA(d) denotes the number of elements of order d in A determined in [18].Proof.
By Corollary3 and (21), it is enough to determine the numbers of cyclic codes Cj’s and Dl’s of length 2s in (21) such that Cj is Hermitian self-dual.
The desired result can be obtained using arguments similar to those in the proof of Theorem2, where [9, Lemma  3.5] is applied instead of [6, Lemma  4.5].This enumeration will be completed by counting the above numbersNH and N in Corollaries 25 and 17, respectively.


## 2.1. Rings and Abelian Codes in Group Rings

For a primep and a positive integer k, denote by Fpk the finite field of order pk. Let Fpk+uFpk≔a+ub∣a,b∈Fpk be a ring, where the addition and multiplication are defined as in the usual polynomial ring over Fpk with indeterminate u together with the condition u2=0. We note that Fpk+uFpk is isomorphic to Fpku/u2 as rings. The Galois extension of Fpk+uFpk of degree m is defined to be the quotient ring Fpk+uFpkx/fx, where f(x) is an irreducible polynomial of degree m over Fpk. It is not difficult to see that the Galois extension of Fpk+uFpk of degree m is isomorphic to Fpkm+uFpkm as rings. In the case  where k is even, the mapping a+ub↦apk/2+ubpk/2 is a ring automorphism of order 2 on Fpk+uFpk. The readers may refer to [14, 15] for properties of the ring Fpk+uFpk.For a commutative ringR with identity 1 and a finite abelian group G, written additively, let RG denote thegroup ring of G over R. The elements in R[G] will be written as ∑g∈G‍αgYg, where αg∈R. The addition and the multiplication in R[G] are given as in the usual polynomial ring over R with indeterminate Y, where the indices are computed additively in G. Note that Y0≔1 is the multiplicative identity of R[G] (resp., R), where 0 is the identity of G. We define aconjugation— on R[G] to be the map that fixes R and sends Yg to Y-g for all g∈G; that is, for u=∑g∈G‍αgYg∈R[G], we set u¯≔∑g∈G‍αgY-g=∑g∈G‍α-gYg. In the case  where there exists a ring automorphism ρ on R of order 2, we define u~≔∑g∈G‍ρ(αg)Y-g for all u=∑g∈G‍αgYg∈R[G]. In the case  where R is a finite field Fpk, then Fpk[G] can be viewed as an Fpk-algebra and it is called agroup algebra. The group algebra Fpk[G] is called aprincipal ideal group algebra (PIGA) if its ideals are generated by a single element.Anabelian code in R[G] is defined to be an ideal in R[G]. If G is cyclic, this code is called acyclic code, a code which is invariant under the right cyclic shift. It is well known that cyclic codes of length n over R can be regarded as ideals in the quotient polynomial ring R[x]/xn-1≅RZn.TheEuclidean inner product in R[G] is defined as follows. For(1)u=∑g∈GαgYg,v=∑g∈GβgYg in R[G], we set (2)u,vE≔∑g∈Gαgβg. In addition, if there exists a ring automorphism ρ of order 2 on R, the ρ-inner product of u and v is defined by (3)u,vρ≔∑g∈Gαgρβg. If R=Fq2 (resp., R=Fq2+uFq2) and ρ(a)=aq (resp., ρ(a+ub)=aq+ubq) for all a∈Fq2 (resp., a+ub∈Fq2+uFq2), the ρ-inner product is called theHermitian inner product and denoted by u,vH.TheEuclidean dual andHermitian dual of C in R[G] are defined to be the sets(4)C⊥E≔u∈RG∣u,vE=0∀v∈C,(5)C⊥H≔u∈RG∣u,vH=0∀v∈C,respectively.An abelian codeC is said to be Euclidean self-dual (resp., Hermitian self-dual) if C=C⊥E (resp., C=C⊥H).For convenience, denote byN(pk,n), NE(pk,n), and NH(pk,n) the number of cyclic codes, the number of Euclidean self-dual cyclic codes, and the number of Hermitian self-dual cyclic codes of length n over Fpk+uFpk, respectively.


## 2.2. Decomposition of Abelian Codes inF2kA×Z2×Z2s

In [6, 9], it has been shown that there exists a Euclidean (resp., Hermitian) self-dual abelian code in FpkG if and only if p=2 and G is even. To study self-dual abelian codes, it is sufficient to focus on F2kA×B, where A is an abelian group of odd order and B is a nontrivial abelian group of two power order. In this case, F2kA×B is a PIGA if and only if B=Z2s is a cyclic group for some positive integer s (see [13]). The complete characterization and enumeration of self-dual abelian codes in PIGAs have been given in [6, 9]. Here, we focus on self-dual abelian codes in non-PIGAs, or equivalently, B is non-cyclic. To avoid tedious computations, we establish results for the simplest case  where B=Z2×Z2s. Useful decompositions of F2kA×Z2×Z2s are given in this section.First, we consider the decomposition ofR≔F2k[A]. In this case, R is semisimple [2] which can be decomposed using the Discrete Fourier Transform in [7] (see [6, 9] for more details). For completeness, the decompositions used in this paper are summarized as follows.For an odd positive integeri and a positive integer k, let ordi(2k) denote the multiplicative order of 2k modulo i. For each a∈A, denote by ord(a) the additive order of a in A. A 2k-cyclotomic class of A containing a∈A, denoted by S2k(a), is defined to be the set (6)S2ka≔2ki·a∣i=0,1,…=2ki·a∣0≤i<ordorda2k,where 2ki·a≔∑j=12kia in A.Anidempotent in R is a nonzero element e such that e2=e. It is calledprimitive if, for every other idempotent f, either ef=e or ef=0. The primitive idempotents in R are induced by the 2k-cyclotomic classes of A (see [5, Proposition  II.4]). Let {a1,a2,…,at} be a complete set of representatives of 2k-cyclotomic classes of A and let ei be the primitive idempotent induced by S2k(ai) for all 1≤i≤t. From [7], R can be decomposed as(7)R=⨁i=1tRei,and hence,(8)F2kA×Z2≅RZ2≅⨁i=1tReiZ2.It is well known (see [6, 9]) that Rei≅F2ki, where ki is a multiple of k. Precisely, ki=kS2kai=k·ordord(ai)(2k) provided that ei is induced by S2k(ai). It follows that Rei[Z2]≅F2ki[Z2]. Under the ring isomorphism that fixes the elements in F2ki and Y1↦u+1, F2ki[Z2] is isomorphic to the ring F2ki+uF2ki, where u2=0. We note that this ring plays an important role in coding theory and codes over rings in this family have extensively been studied in [14–17] and references therein.From (8) and the ring isomorphism discussed above, we have(9)F2kA×Z2×Z2s≅∏i=1tRiZ2s,where Ri≔F2ki+uF2ki for all 1≤i≤t.In order to study the algebraic structures of Euclidean and Hermitian self-dual abelian codes inF2k[A×Z2×Z2s], the two rearrangements of Ri’s in the decomposition (9) are needed. The details are given in the following two subsections.


### 2.2.1. Euclidean Case

A2k-cyclotomic class S2k(a) is said to be oftypeI if a=-a (in this case, S2k(a)=S2k(-a)),typeII if S2k(a)=S2k(-a) and a≠-a, ortypeIII if S2k(-a)≠S2k(a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈{I,II,III} if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes of A can be chosen such that ai∣i=1,2,…,rI, arI+j∣j=1,2,…,rII, and arI+rII+l,arI+rII+rIII+l=-arI+rII+l∣l=1,2,…,rIII are sets of representatives of 2k-cyclotomic classes of types I, II, and III, respectively, where t=rI+rII+2rIII.Rearranging the terms in the decomposition ofR in (7) based on the 3 types of primitive idempotents, we have(10)F2kA×Z2≅⨁i=1tReiZ2≅∏i=1rIRi∏j=1rIISj∏l=1rIIITl×Tl,where Ri≔F2k+uF2k for all i=1,2,…,rI, Sj≔F2krI+j+uF2krI+j for all j=1,2,…,rII, and Tl≔F2krI+rII+l+uF2krI+rII+l for all l=1,2,…,rIII.From (10), we have(11)F2kA×Z2×Z2s≅∏i=1rIRiZ2s∏j=1rIISjZ2s∏l=1rIIITlZ2s×TlZ2s.It follows that an abelian code C in F2k[A×Z2×Z2s] can be viewed as(12)C≅∏i=1rIBi∏j=1rIICj∏l=1rIIIDl×Dl′,where Bi, Cj, Dl, and Dl′ are cyclic codes in RiZ2s, SjZ2s, TlZ2s, and TlZ2s, respectively, for all i=1,2,…,rI, j=1,2,…,rII, and l=1,2,…,rIII.Using the analysis similar to those in [6, Section  II.D], the Euclidean dual of C in (12) is of the form(13)C⊥E≅∏i=1rIBi⊥E∏j=1rIICj⊥H∏l=1rIIIDl′⊥E×Dl⊥E.Similar to [6, Corollary  2.9], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Euclidean self-dual can be given using the notions of cyclic codes of length 2s over Ri, Sj, and Tl in the following corollary.Corollary 1.
An abelian codeC in F2k[A×Z2×Z2s] is Euclidean self-dual if and only if in the decomposition (12)(i)
Bi is a Euclidean self-dual cyclic code of length 2s over Ri for all i=1,2,…,rI,(ii)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rII,(iii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rIII.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to begood if j divides 2ki+1 for some integer i≥1 andbad otherwise. These notions have been introduced in [6, 12] for the enumeration of self-dual cyclic codes and self-dual abelian codes over finite fields.Letχ be a function defined on the pair (j,2k), where j is an odd positive integer, as follows:(14)χj,2k=0ifj,2kisgood,1otherwise.The number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 2.
Letk and s be positive integers and let A be a finite abelian group of odd order and exponent M. Then the number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] is (15)NE2k,2s∑d∣M,ordd2k=11-χd,2kNAd∏d∣Mordd2k≠1NH2k·ordd2k,2s1-χd,2kNAd/ordd2k∏d∣MN2k·ordd2k,2sχd,2kNAd/2ordd2k,where NA(d) denotes the number of elements in A of order d determined in [18].Proof.
From (12) and Corollary 1, it suffices to determine the numbers of cyclic codes Bi’s, Cj’s, and Dl’s such that Bi and Cj are Euclidean and Hermitian self-dual, respectively.
From [9, Remark  2.5], the elements in A of the same order are partitioned into 2k-cyclotomic classes of the same type. For each divisor d of M, a 2k-cyclotomic class containing an element of order d has cardinality ordd(2k) and the number of such 2k-cyclotomic classes is NA(d)/ordd(2k). We consider the following 3 cases.
Case  1 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) = 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type I. Since there are NAd/ordd(2k) such 2k-cyclotomic classes, the number of Euclidean self-dual cyclic codes Bi’s of length 2s corresponding to d is(16)NE2k,2sNAd/ordd2k=NE2k,2s1-χd,2kNAd.
Case  2 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) ≠ 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type II. Hence, the number of Hermitian self-dual cyclic codes Cj’s of length 2s corresponding to d is (17)NH2k·ordd2k,2sNAd/ordd2k=NH2k·ordd2k,2s1-χd,2kNAd/ordd2k.
Case  3 ( χ ( d , 2 k ) = 1 ). By [6, Lemma  4.5], every 2k-cyclotomic class of A containing an element of order d is of type III. Then the number of cyclic codes Dl’s of length 2s corresponding to d is(18)N2k·ordd2k,2sNAd/2ordd2k=N2k·ordd2k,2sχd,2kNAd/2ordd2k.
The desired result follows sinced runs over all divisors of M.This enumeration will be completed by counting the above numbersNE, NH, and N in Corollaries 22, 25, and 17, respectively.


### 2.2.2. Hermitian Case

We focus on the case  wherek is even. A 2k-cyclotomic class S2k(a) is said to be oftypeI′ if S2k(a)=S2k(-2k/2a) ortypeII′ if S2k(a)≠S2k(-2k/2a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈I′,II′ if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes can be chosen such that ai∣i=1,2,…,rI′ and arI′+j,arI′+rII′+j=-2k/2arI′+j∣j=1,2,…,rII′ are sets of representatives of 2k-cyclotomic classes of types I′ and II′, respectively, where t=rI′+2rII′.Rearranging the terms in the decomposition ofR in (7) based on the above 2 types of primitive idempotents, we have(19)F2kA×Z2≅⨁i=1tReiZ2≅∏j=1rI′Sj∏l=1rII′Tl×Tl,where Sj≔F2kj+uF2kj for all j=1,2,…,rI′ and Tl≔F2krI′+l+uF2krI′+l for all l=1,2,…,rII′.From (19), we have(20)F2kA×Z2×Z2s≅∏j=1rI′SjZ2s∏l=1rII′TlZ2s×TlZ2s.Hence, an abelian code C in F2k[A×Z2×Z2s] can be viewed as(21)C≅∏j=1rI′Cj∏l=1rII′Dl×Dl′,where Cj, Dl, and Dl′ are cyclic codes in Sj[Z2s], Tl[Z2s], and Tl[Z2s], respectively, for all j=1,2,…,rI′ and l=1,2,…,rII′.Using the analysis similar to those in [9, Section  II.D], the Hermitian dual of C in (21) is of the form(22)C⊥H≅∏j=1rI′Cj⊥H∏l=1rII′Dl′⊥E×Dl⊥E.Similar to [9, Corollary  2.8], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Hermitian self-dual are now given using the notions of cyclic codes of length 2s over Sj and Tl in the following corollary.Corollary 3.
An abelian codeC in F2k[A×Z2×Z2s] is Hermitian self-dual if and only if in the decomposition (21)(i)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rI′,(ii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rII′.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to beoddly good if j divides 2ki+1 for some odd integer i≥1 andevenly good if j divides 2ki+1 for some even integer i≥2. These notions have been introduced in [9] for characterizing the Hermitian self-dual abelian codes in PIGAs.Letλ be a function defined on the pair (j,2k), where j is an odd positive integer, as(23)λj,2k=0ifj,2kisoddlygood,1otherwise.The number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 4.
Letk be an even positive integer and let s be a positive integer. Let A be a finite abelian group of odd order and exponent M. Then the number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] is (24)∏d∣MNH2k·ordd2k,2s1-λd,2k/2NAd/ordd2k∏d∣MN2k·ordd2k,2sλd,2k/2NAd/2ordd2k,where NA(d) denotes the number of elements of order d in A determined in [18].Proof.
By Corollary3 and (21), it is enough to determine the numbers of cyclic codes Cj’s and Dl’s of length 2s in (21) such that Cj is Hermitian self-dual.
The desired result can be obtained using arguments similar to those in the proof of Theorem2, where [9, Lemma  3.5] is applied instead of [6, Lemma  4.5].This enumeration will be completed by counting the above numbersNH and N in Corollaries 25 and 17, respectively.


## 2.2.1. Euclidean Case

A2k-cyclotomic class S2k(a) is said to be oftypeI if a=-a (in this case, S2k(a)=S2k(-a)),typeII if S2k(a)=S2k(-a) and a≠-a, ortypeIII if S2k(-a)≠S2k(a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈{I,II,III} if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes of A can be chosen such that ai∣i=1,2,…,rI, arI+j∣j=1,2,…,rII, and arI+rII+l,arI+rII+rIII+l=-arI+rII+l∣l=1,2,…,rIII are sets of representatives of 2k-cyclotomic classes of types I, II, and III, respectively, where t=rI+rII+2rIII.Rearranging the terms in the decomposition ofR in (7) based on the 3 types of primitive idempotents, we have(10)F2kA×Z2≅⨁i=1tReiZ2≅∏i=1rIRi∏j=1rIISj∏l=1rIIITl×Tl,where Ri≔F2k+uF2k for all i=1,2,…,rI, Sj≔F2krI+j+uF2krI+j for all j=1,2,…,rII, and Tl≔F2krI+rII+l+uF2krI+rII+l for all l=1,2,…,rIII.From (10), we have(11)F2kA×Z2×Z2s≅∏i=1rIRiZ2s∏j=1rIISjZ2s∏l=1rIIITlZ2s×TlZ2s.It follows that an abelian code C in F2k[A×Z2×Z2s] can be viewed as(12)C≅∏i=1rIBi∏j=1rIICj∏l=1rIIIDl×Dl′,where Bi, Cj, Dl, and Dl′ are cyclic codes in RiZ2s, SjZ2s, TlZ2s, and TlZ2s, respectively, for all i=1,2,…,rI, j=1,2,…,rII, and l=1,2,…,rIII.Using the analysis similar to those in [6, Section  II.D], the Euclidean dual of C in (12) is of the form(13)C⊥E≅∏i=1rIBi⊥E∏j=1rIICj⊥H∏l=1rIIIDl′⊥E×Dl⊥E.Similar to [6, Corollary  2.9], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Euclidean self-dual can be given using the notions of cyclic codes of length 2s over Ri, Sj, and Tl in the following corollary.Corollary 1.
An abelian codeC in F2k[A×Z2×Z2s] is Euclidean self-dual if and only if in the decomposition (12)(i)
Bi is a Euclidean self-dual cyclic code of length 2s over Ri for all i=1,2,…,rI,(ii)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rII,(iii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rIII.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to begood if j divides 2ki+1 for some integer i≥1 andbad otherwise. These notions have been introduced in [6, 12] for the enumeration of self-dual cyclic codes and self-dual abelian codes over finite fields.Letχ be a function defined on the pair (j,2k), where j is an odd positive integer, as follows:(14)χj,2k=0ifj,2kisgood,1otherwise.The number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 2.
Letk and s be positive integers and let A be a finite abelian group of odd order and exponent M. Then the number of Euclidean self-dual abelian codes in F2k[A×Z2×Z2s] is (15)NE2k,2s∑d∣M,ordd2k=11-χd,2kNAd∏d∣Mordd2k≠1NH2k·ordd2k,2s1-χd,2kNAd/ordd2k∏d∣MN2k·ordd2k,2sχd,2kNAd/2ordd2k,where NA(d) denotes the number of elements in A of order d determined in [18].Proof.
From (12) and Corollary 1, it suffices to determine the numbers of cyclic codes Bi’s, Cj’s, and Dl’s such that Bi and Cj are Euclidean and Hermitian self-dual, respectively.
From [9, Remark  2.5], the elements in A of the same order are partitioned into 2k-cyclotomic classes of the same type. For each divisor d of M, a 2k-cyclotomic class containing an element of order d has cardinality ordd(2k) and the number of such 2k-cyclotomic classes is NA(d)/ordd(2k). We consider the following 3 cases.
Case  1 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) = 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type I. Since there are NAd/ordd(2k) such 2k-cyclotomic classes, the number of Euclidean self-dual cyclic codes Bi’s of length 2s corresponding to d is(16)NE2k,2sNAd/ordd2k=NE2k,2s1-χd,2kNAd.
Case  2 ( χ ( d , 2 k ) = 0 and o r d d ( 2 k ) ≠ 1 ). By [6, Remark  2.6], every 2k-cyclotomic class of A containing an element of order d is of type II. Hence, the number of Hermitian self-dual cyclic codes Cj’s of length 2s corresponding to d is (17)NH2k·ordd2k,2sNAd/ordd2k=NH2k·ordd2k,2s1-χd,2kNAd/ordd2k.
Case  3 ( χ ( d , 2 k ) = 1 ). By [6, Lemma  4.5], every 2k-cyclotomic class of A containing an element of order d is of type III. Then the number of cyclic codes Dl’s of length 2s corresponding to d is(18)N2k·ordd2k,2sNAd/2ordd2k=N2k·ordd2k,2sχd,2kNAd/2ordd2k.
The desired result follows sinced runs over all divisors of M.This enumeration will be completed by counting the above numbersNE, NH, and N in Corollaries 22, 25, and 17, respectively.


## 2.2.2. Hermitian Case

We focus on the case  wherek is even. A 2k-cyclotomic class S2k(a) is said to be oftypeI′ if S2k(a)=S2k(-2k/2a) ortypeII′ if S2k(a)≠S2k(-2k/2a). The primitive idempotent e induced by S2k(a) is said to be of type λ∈I′,II′ if S2k(a) is a 2k-cyclotomic class of type λ.Without loss of generality, the representativesa1,a2,…,at of 2k-cyclotomic classes can be chosen such that ai∣i=1,2,…,rI′ and arI′+j,arI′+rII′+j=-2k/2arI′+j∣j=1,2,…,rII′ are sets of representatives of 2k-cyclotomic classes of types I′ and II′, respectively, where t=rI′+2rII′.Rearranging the terms in the decomposition ofR in (7) based on the above 2 types of primitive idempotents, we have(19)F2kA×Z2≅⨁i=1tReiZ2≅∏j=1rI′Sj∏l=1rII′Tl×Tl,where Sj≔F2kj+uF2kj for all j=1,2,…,rI′ and Tl≔F2krI′+l+uF2krI′+l for all l=1,2,…,rII′.From (19), we have(20)F2kA×Z2×Z2s≅∏j=1rI′SjZ2s∏l=1rII′TlZ2s×TlZ2s.Hence, an abelian code C in F2k[A×Z2×Z2s] can be viewed as(21)C≅∏j=1rI′Cj∏l=1rII′Dl×Dl′,where Cj, Dl, and Dl′ are cyclic codes in Sj[Z2s], Tl[Z2s], and Tl[Z2s], respectively, for all j=1,2,…,rI′ and l=1,2,…,rII′.Using the analysis similar to those in [9, Section  II.D], the Hermitian dual of C in (21) is of the form(22)C⊥H≅∏j=1rI′Cj⊥H∏l=1rII′Dl′⊥E×Dl⊥E.Similar to [9, Corollary  2.8], necessary and sufficient conditions for an abelian code in F2k[A×Z2×Z2s] to be Hermitian self-dual are now given using the notions of cyclic codes of length 2s over Sj and Tl in the following corollary.Corollary 3.
An abelian codeC in F2k[A×Z2×Z2s] is Hermitian self-dual if and only if in the decomposition (21)(i)
Cj is a Hermitian self-dual cyclic code of length 2s over Sj for all j=1,2,…,rI′,(ii)
Dl′=Dl⊥E is a cyclic code of length 2s over Tl for all l=1,2,…,rII′.Given a positive integerk and an odd positive integer j, the pair (j,2k) is said to beoddly good if j divides 2ki+1 for some odd integer i≥1 andevenly good if j divides 2ki+1 for some even integer i≥2. These notions have been introduced in [9] for characterizing the Hermitian self-dual abelian codes in PIGAs.Letλ be a function defined on the pair (j,2k), where j is an odd positive integer, as(23)λj,2k=0ifj,2kisoddlygood,1otherwise.The number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] can be determined as follows.Theorem 4.
Letk be an even positive integer and let s be a positive integer. Let A be a finite abelian group of odd order and exponent M. Then the number of Hermitian self-dual abelian codes in F2k[A×Z2×Z2s] is (24)∏d∣MNH2k·ordd2k,2s1-λd,2k/2NAd/ordd2k∏d∣MN2k·ordd2k,2sλd,2k/2NAd/2ordd2k,where NA(d) denotes the number of elements of order d in A determined in [18].Proof.
By Corollary3 and (21), it is enough to determine the numbers of cyclic codes Cj’s and Dl’s of length 2s in (21) such that Cj is Hermitian self-dual.
The desired result can be obtained using arguments similar to those in the proof of Theorem2, where [9, Lemma  3.5] is applied instead of [6, Lemma  4.5].This enumeration will be completed by counting the above numbersNH and N in Corollaries 25 and 17, respectively.


## 3. Cyclic Codes of Lengthps over Fpk+uFpk

The enumeration of self-dual abelian codes in non-PIGAs in the previous section requires properties of cyclic codes of length2s over F2k+uF2k. In this section, a more general situation is discussed. Precisely, properties cyclic of length ps over Fpk+uFpk are studied for all primes p. We note that algebraic structures of cyclic codes of length ps over Fpk+uFpk were studied in [14, 15]. Here, based on [19], we give an alternative characterization of such codes which is useful in studying self-dual cyclic codes of length ps over Fpk+uFpk.First, we note that there exists a one-to-one correspondence between the cyclic codes of lengthps over Fpk+uFpk and the ideals in the quotient ring Fpk+uFpkx/xps-1. Precisely, a cyclic code C of length ps can be represented by the ideal(25)∑i=0ps-1vixi∣v0,v1,…,vps-1∈C in Fpk+uFpkx/xps-1.From now on, a cyclic codeC will be referred to as the above polynomial presentation. Note that the map μ:Fpk+uFpkx/xps-1→Fpkx/xps-1 defined by(26)μfx=fxmod⁡uis a surjective ring homomorphism. For each cyclic code C in Fpk+uFpkx/xps-1 and i∈0,1, let(27)ToriC=μvx∣vx∈Fpk+uFpkxxps-1,uivx∈C.For each i∈0,1, Tori(C) is called the ithtorsion code of C. The codes Tor0(C)=μ(C) and Tor1(C) are sometimes called theresidue andtorsion codes of C, respectively.It is not difficult to see that, for eachi∈0,1, cx∈ToriC if and only if ui(cx+uz(x))∈C for some z(x)∈Fpkx/xps-1. Consequently, we have that Tor0(C)⊆Tor1(C) are ideals in Fpk[x]/〈xps-1〉 (cyclic codes of length ps over Fpk). We note that every ideal C in Fpk[x]/xps-1 is of the form x-1i for some 0≤i≤ps and the cardinality of C is ps-i.From the structures of cyclic codes of lengthps over Fpk discussed above and [14, Proposition  2.5], we have the following properties of the torsion and residue codes.Proposition 5.
LetC be an ideal in Fpk+uFpkx/xps-1 and let i∈0,1. Then the following statements hold:(i)
Tori(C) is an ideal of Fpkx/xps-1 and Tori(C)=x-1Ti for some 0≤Ti≤ps.(ii)
IfTori(C)=〈x-1)Ti, then ToriC=pkps-Ti.(iii)
C=Tor0C·Tor1C=pk2ps-(T0+T1).With the notations given in Proposition5, for each i∈{0,1}, Ti(C)≔Ti is called the ith-torsional degree of C.Remark 6.
From Proposition5 and the definition above, we have the following facts:(i)
SinceTor0(C)⊆Tor1(C), we have 0≤T1(C)≤T0(C)≤ps.(ii)
Ifu(x-1)t∈C, then t≥T1(C).Next, we determine a generator set of an ideal inFpk+uFpkx/xps-1.Theorem 7.
LetC be an ideal of Fpk+uFpkx/xps-1. Then(28)C=s0x,us1x,where, for each i∈{0,1},(i)
eithersj(x)=0 or sj(x)=x-1tj+uzj(x) for some zj(x)∈Fpkx/xps-1 and 0≤tj<ps,(ii)
sj(x)≠0 if and only if Torj(C)≠{0} and Tor0(C)≠Tor1(C),(iii)
ifsj(x)≠0, then Torj(C)=x-1tj.Proof.
The statement can be obtained using a slight modification of the proof of [20, Theorem  6.5]. For completeness, the details are given as follows.
For each idealC in Fpkx/xps-1, it can be represented as C=0 or C=x-1t0, where 0≤t0<ps. If C=0, then we are done by choosing s0x=0. For 0≤t0<ps, let s0(x)=x-1t0. By abuse of notation, Tor0(C)=C=x-1t0. Hence, s0(x) satisfies conditions (i), (ii), and (iii).
SinceC is an ideal of the ring Fpk+uFpkx/xps-1 and μ is a surjective ring homomorphism, μ(C) is an ideal of Fpkx/xps-1 which implies that μ(C)=〈s0′(x)〉, where s0′x satisfies conditions (i), (ii), and (iii). If s0′(x)=0, then take s0(x)=0. Assume that s0′(x)≠0; then s0′(x)=x-1t0, where 0≤t0<ps. Then there exists z0(x)∈Fpkx/xps-1 such that s0(x)=x-1t0+uz0(x)∈C and μ(s0(x))=s0′(x); that is, s0(x)=s0′(x)+uz0(x). Since Tor1(C) is an ideal of Fpkx/xps-1, it follows that Tor1(C)=x-1t1 for some 0≤t1≤ps. Let s1(x)=x-1t1. Claim that C=s0x,us1x. Since C is an ideal of Fpk+uFpkx/xps-1, we have ux-1t1∈C. Thus, s0x,us1x⊆C. To show that C⊆s0x,us1x, let c(x)∈C. Then μ(c(x))=w(x)s0′(x) for some w(x)∈Fpkx/xps-1. Thus, μ(c(x))=w(x)s0′(x) which implies that c(x)=w(x)s0′(x)+uw′(x)=w(x)s0(x)-uw(x)z0(x)+uw′(x)=w(x)s0(x)+u-z0(x)w(x)+w′(x) for some w′(x)∈Tor1(C). Since w′(x)∈Tor1(C), it follows that c(x)∈s0x,us1x. Therefore, C=s0x,us1x as desired.
Note thats1(x)=x-1ps=0 implies C={0}. Assume that C≠{0}. Then s1(x)=x-1t1, where 0≤t1<ps. If s0(x)=0, then we are done. Assume that s0(x)≠0. Then s0(x)=(x-1)t, where 0≤t<ps. Hence, Tor0(C)=x-1t. Since Tor0(C)⊆Tor1(C), we have t1≤t. If t1<t, then we are done. Assume that t1=t. Then s0(x)=x-1t+uz0(x) for some z0(x)∈Fpkx/xps-1. It follows that us1(x)=ux-1t1=ux-1t=us0(x)∈s0x, a contradiction. Therefore, C=s0x=s0x,us0x=s0x,us1x.However, the generator set given in Theorem7 does not need to be unique. The unique presentation is given in the following theorem.Theorem 8.
LetC be an ideal in Fpk+uFpkx/xps-1, T0≔T0(C), and T1≔T1(C). Then(29)C=f0x,f1x,where(30)f0x=x-1T0+ux-1thxifT0<ps,0ifT0=ps,f1x=ux-1T1ifT1<ps,0ifT1=ps,with hx∈Fpkx/xps-1 being either zero or a unit with t+deg⁡(h(x))<T0.
Moreover,f0x,f1x is unique in the sense that if there exists a pair (g0(x),g1(x)) of polynomials satisfying the conditions in the theorem, then f0(x)=g0(x) and f1(x)=g1(x).Proof.
IfC=0, then Tor1(C)={0} and Tor0(C)={0} which imply that T0=ps and T1=ps. The polynomials f0(x)=0 and f1(x)=0 have the desired properties.
Next, assume thatC≠0. Then there exists the smallest nonnegative integer r∈0,1 such that Tr<ps. From Theorem 7, it can be concluded that(31)C=s0x,us1x,where(32)s0x=x-1T0+ugxifr=0,0ifr=1,s1x=x-1T1for some gx∈Fpkx/xps-1.
Case  1 ( r = 0 ). Then s0(x)=(x-1)T0+ug(x)forsomeg(x)∈Fpkx/xps-1 and s1(x)=x-1T1. It follows that C=〈x-1T0+ug(x),x-1T1〉, T1(C)=T1, and T0(C)=T0. Let f1(x)=us1(x). Since g(x)∈Fpkx/xps-1, we have(33)s0x=x-1T0+u∑j=0ps-1ajx-1j,where aj∈Fpk for all j=0,1,…,pk-1. Since ux-1T1=us1(x)∈C, we have uajx-1j∈C for all j=T1,T1+1,…,ps-1. It follows that u∑j=T1ps-1ajx-1j∈C. Let f0(x)=x-1T0+u∑j=0T1-1ajx-1j. Then f0(x)=s0(x)-u∑j=T1ps-1ajx-1j∈C.
We show thatC=f0x,f1x. From the discussion above, we have f0x,f1x⊆C. Since us1(x)=ux-1T1=f1(x), it follows that uajx-1j∈f0x,f1x for all j=T1,T1+1,…,ps-1. Hence, u∑j=T1ps-1ajx-1j∈f0x,f1x which implies that(34)s0x=f0x+u∑j=T1ps-1ajx-1j∈f0x,f1x.Therefore, C=s0x,us1x⊆f0x,f1x. As desired, C=f0x,f1x.
Case  2 ( r = 1 ). Then s0(x)=0 and s1(x)=ux-1T1 which implies that Tor1(C)=x-1T1 and Tor0(C)={0}. By choosing f0(x)=0 and f1(x)=ux-1T1, the result follows.
To prove the uniqueness, letC=g0x,g1x be such that g0(x) and g1(x) satisfy the conditions in the theorem. Then g1(x)=ux-1T1=f1(x).
Writeg0(x)=x-1T0+u∑j=0T1-1cjx-1j, where cj∈Fpk. Then(35)f0x-g0x=u∑j=0T1-1aj-cjx-1j.
It can be seen thatf0(x)-g0(x)=ux-1lh(x), where h(x)=0 or h(x) is a unit with l≤T1-1<T1. If h(x) is a unit, then ux-1l∈C which implies that l≥T1, a contradiction. Hence, h(x)=0 which means that f0(x)=g0(x) as desired.Definition 9.
For each idealC in Fpk+uFpkx/xps-1, denote by C=f0x,f1x the unique representation of the ideal C obtained in Theorem 8.Illustrative examples of the representations in Theorems7 and 8 are given as follows.Example 10.
Consider the idealC=x-12 in F2+uF2x/x4-1. Using Theorem 8, we obtain that C has the unique representation x-12,ux-12. Based on Theorem 7, C can be represented as 〈x-12,0〉, 〈(x-1)2+ux-12,0〉, and x-12+ux-13,0.The annihilator of an idealC in Fpk+uFpkx/xps-1 is key to determine properties C as well as the number of ideals in Fpk+uFpkx/xps-1.Definition 11.
LetC be an ideal in Fpk+uFpkx/xps-1. Theannihilator of C, denoted by Ann(C), is defined to be the set {f(x)∈Fpk+uFpkx/xps-1∣f(x)g(x)=0forallg(x)∈C}.The following properties of the annihilator can be obtained using arguments similar to those in the case  of Galois rings in [19].Theorem 12.
LetC be an ideal of Fpk+uFpkx/xps-1. Then the following statements hold:(i)
Ann(C) is an ideal of Fpk+uFpkx/xps-1.(ii)
If|C|=pkd, then AnnC=pk(2·ps-d).(iii)
Ann(Ann(C))=C.Theorem 13.
LetI denote the set of ideals of Fpk+uFpkx/xps-1 and let A=C∈I∣T0C+T1C≤ps and A′=C∈I∣T0C+T1C≥ps. Then the map ϕ:A→A′ defined by C↦Ann(C) is a bijection.The rest of this section is devoted to the determination of all ideals inFpk+uFpkx/xps-1. In view of Theorem 13, it suffices to focus on the ideals in A.For eachC=f0x,f1x in A, if f0(x)=0, then T0(C)=ps and T1(C)=0. Hence, the only ideal in A with f0(x)=0 is of the form 0,u. In the following two theorems, we assume that f0(x)≠0.Theorem 14.
Letx-1i0+ux-1thx,ux-1i1 be the representation of an ideal in Fpk+uFpkx/xps-1. Then it is a representation of an ideal in A if and only if i0, i1, and t are integers and h(x)∈Fpkx/xps-1 such that 0≤i0<ps, 0≤i1≤min⁡{i0,ps-i0}, t≥0, t+deg⁡(h(x))<i1, and h(x) is either zero or a unit in Fpkx/xps-1.Proof.
Form Theorem8, we have that i0, i1, and t are integers and h(x)∈Fpkx/xps-1 such that 0≤i0<ps, 0≤i1≤i0, t≥0, t+deg⁡(h(x))<i1, and h(x) is either zero or a unit in Fpkx/xps-1.
Assume thatx-1i0+ux-1thx,ux-1i1 is a representation of an ideal in A. Then i0+i1≤ps which implies that i1≤ps-i0. Hence, we have i1≤min⁡i0,ps-i0.
Conversely, assume thatC=x-1i0+ux-1thx,ux-1i1, where i0, i1, and t are integers and h(x)∈Fpkx/xps-1 such that 0≤i0<ps, 0≤i1≤min⁡i0,ps-i0, t≥0, t+deg⁡(h(x))<i1, and h(x) is either zero or a unit in Fpkx/xps-1. Clearly, i0+i1≤ps. To show that C=x-1i0+ux-1thx,ux-1i1 in A, it remains to prove that T0(C)=i0 and T1(C)=i1.
LetD=x-1ps-i1-ux-1ps-i0-i1+thx,ux-1ps-i0. It is not difficult to see that (36)D=x-1ps-i1+ux-1ps-i0-ux-1ps-i0-i1+thx,ux-1ps-i0.Since(37)x-1i0+ux-1i1hxx-1ps-i1+ux-1ps-i0-ux-1ps-i0-i1+thx=0,x-1i0+ux-1i1hxux-1ps-i0=0,ux-1i1x-1ps-i1+ux-1ps-i0-ux-1ps-i0-i1+thx=0,we have D⊆Ann(C). By Proposition 5, we obtain that Tj(C)≤ij and Tj(D)≤ps-i1-j for all j∈{0,1}. Hence, C≥pk2·ps-i0-i1 and AnnC≥|D|≥pki0+i1. Since |C|·|Ann(C)|=pk2·ps, we have AnnC=D. Therefore, T0(C)=i0 and T1(C)=i1 as desired.Since every polynomial∑i=0maix-1i in Fpkx is either 0 or x-1th(x), where h(x) is a unit in Fpkx and 0≤t≤m-deg⁡(h(x)), Theorem 14 is rewritten as follows.Theorem 15.
The expressionx-1i0+u∑j=0i1-1hjx-1j,ux-1i1 represents an ideal in A if and only if i0 and i1 are integers such that 0≤i0<ps, 0≤i1≤min⁡i0,ps-i0, i0+i1≤ps, and hj∈Fpk for all 0≤j<i1.The number of distinct ideals ofFpk+uFpkx/xps-1 of a fixed d=T0+T1 is given in the following proposition.Proposition 16.
Let0≤d≤ps. Then the number of distinct ideals in (Fpk+uFpk)[x]/〈xps-1〉 with T0+T1=d is(38)pkK+1-1pk-1,where K=min⁡d/2,ps-d/2.Proof.
LetT1=i1 and i0≔T0=d-T1 be fixed.
Case  1 ( d < p s ). Then i0≤i0+i1=T0+T1=d<ps. By Theorem 15, it follows that C=〈〈x-1i0+u∑j=0i1-1hjx-1j,ux-1i1〉〉. Then the choice for ∑j=0i1-1hjx-1j is pki1. By Theorem 15 again, we also have T1≤min⁡T0,ps-T0. Since T0+T1=d, we obtain that T1≤⌊d/2⌋≤T0, and hence, T1≤min⁡{⌊d/2⌋,ps-T0}≤min⁡{⌊d/2⌋,ps-⌊d/2⌋}. Now, vary T1 from 0 to K; we obtain that there are 1+pk+⋯+pkK=pk(K+1)-1/pk-1 ideals with T0+T1=d.
Case  2 ( d = p s ). If i0=ps, then the only ideal with T0+T1=ps is the ideal represented by 0,u. If i0<ps, then we have pk+pk2+⋯+pkK ideals by arguments similar to those in Case  1.For a cyclic codeC in A, we have C≠Ann(C) whenever T0(C)+T1(C)<ps. In the case  where T0(C)+T1(C)=ps, by the proof of Theorem 14, the annihilator of the cyclic code C=x-1i0+ux-1thx,ux-1i1 is of the form Ann(C)=x-1i0-ux-1thx,ux-1i1. If p is odd, then C=Ann(C) occurs only in the case  h(x)=0. In the case  where p=2, C=Ann(C) is always true. By Proposition 16 and the bijection given in Theorem 13, the number of cyclic codes of length ps over Fpk+uFpk can be summarized as follows.Corollary 17.
The number of cyclic codes of lengthps over Fpk+uFpk is(39)Npk,ps=2∑d=0pspkmin⁡d/2,ps-d/2+1-1pk-1-pkps-1+1-1pk-1ifp=2,2∑d=0pspkmin⁡d/2,ps-d/2+1-1pk-1-1ifpisodd.Proof.
From Theorem13, the number of cyclic codes of length ps over Fpk+uFpk is A∪A′=A+A′-A∩A′. The desired results follow immediately form the discussion above.


## 4. Self-Dual Cyclic Codes of Lengthps over Fpk+uFpk

In this section, characterization and enumeration self-dual cyclic codes of lengthps over Fpk+uFpk are given under the Euclidean and Hermitian inner products.


### 4.1. Euclidean Self-Dual Cyclic Codes of Lengthps over Fpk+uFpk

Characterization and enumeration of Euclidean self-dual cyclic codes of lengthps over Fpk+uFpk are given in this subsection.For each subsetA of Fpk+uFpkx/xps-1, denote by A¯ the set of polynomials f(x)¯ for all f(x) in A, where — is viewed as the conjugation on the group ring (Fpk+uFpk)[Zps] defined in Section 2. From the definition of the annihilator, the next theorem can be derived similar to [21, Proposition  2.12].Theorem 18.
LetC be an ideal in Fpk+uFpkx/xps-1. Then C⊥E=Ann(C)¯.Using the unique generators of an idealC in Fpk+uFpkx/xps-1 determined in Theorem 15, the Euclidean dual of C can be given in the following theorem.Theorem 19.
LetC=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1 be an ideal in A, where hj∈Fpk for all 0≤j≤ii-1. Then C⊥E is in the form of(40)C⊥E=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjx-1r,ux-1ps-i0.Proof.
From the proof of Theorem14, we have(41)AnnC=x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1hjx-1j,ux-1ps-i0.By Theorem 18, it follows that C⊥E=AnnC¯. Hence, C⊥E contains the elements u(x-1)ps-i0 and(42)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1-1i0+jhjx-1jxi0-j. By writing x=(x-1)+1 and using the Binomial Theorem, it follows that C⊥E contains the element(43)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1∑l=0i0-j-1i0+ji0-jlhjx-1i0-l.Hence, (44)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1∑l=0i0-j-1i0+ji0-jlhjx-1i0-l,x-1ps-i1⊆C⊥E.By counting the number of elements, the two sets are equal as desired. Updating the indices, it can be concluded that (45)C⊥E=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjx-1r,x-1ps-i1.Assume that an idealC=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1 in Fpk+uFpkx/xps-1 is Euclidean self-dual; that is, C=C⊥E. By Theorem 18, it is equivalent to ps=i0+i1 and(46)-ht=∑j=0t-1i0+li0-jr-jhjin Fpk for all 0≤t≤ii-1.We note that ifi1=0, then it is not difficult to see that only the ideal generated by u is Euclidean self-dual.For the casei1≥1, the situation is more complicated. First, we recall an i1×i1 matrix M(ps,i1) over Fpk defined in [22] as(47)Mps,i1=-1i0+100⋯0-1i0i01-1i0+1+10⋯0-1i0i02-1i0+1i0-11-1i0+2+1⋯0⋮⋮⋮⋱⋮-1i0i0i1-1-1i0+1i0-1i1-2-1i0+2i0-2i1-3⋯-1i0+i1-1+1.It is not difficult to see thati1 equations from (46) are equivalent to the matrix equation(48)Mps,i1h=0,where h=h0,h1,…,hi1-1T and 0=0,0,…,0T.Moreover, it can be concluded that the idealC is Euclidean self-dual if and only if ps=i0+i1 and h0,h1,…,hi1-1 satisfy (48). Since h0=h1=⋯=hi1-1=0 is a solution of (48), the corresponding idea x-1ps-i1,ux-1i1 is Euclidean self-dual in Fpk+uFpkx/xps-1. Hence, for a fixed first torsion degree 1≤i1≤ps, a Euclidean self-dual ideal in Fpk+uFpkx/xps-1 always exists. By solving (48), all Euclidean self-dual ideals in Fpk+uFpkx/xps-1 can be constructed. Therefore, for a fixed first torsion degree 1≤i1≤ps, the number of Euclidean self-dual ideals in Fpk+uFpkx/xps-1 equals the number of solutions of (48) which is pkκ, where κ is the nullity of M(ps,i1) determined in [22].Proposition 20 (see [22, Proposition  3.3]).
Letκ be the nullity of M(ps,i1). Then(49)κ=i12ifpisodd;i1+12ifp=2.The number of Euclidean self-dual cyclic codes inFpk+uFpkx/xps-1 with first torsional degree i1 is given in terms of the nullity of M(ps,i1) as follows.Proposition 21.
Leti1>0 and let κ be the nullity of M(ps,i1) over Fpk. Then the number of Euclidean self-dual cyclic codes of length ps over Fpk with first torsional degree i1 is pkκ.From Theorem14, we have 0≤i1≤⌊ps/2⌋ since i0+i1=ps. Hence, the number of Euclidean self-dual cyclic codes of length ps over Fpk+uFpk is given by the following corollary.Corollary 22.
Letp be a prime and let s and k be positive integers. Then the following statements hold: (i)
Ifp is odd, then the number of Euclidean self-dual cyclic codes of length ps over Fpk+uFpk is (50)NEpk,ps=2pkps+1/4-1pk-1ifps≡3mod⁡4,2pkps-1/4-1pk-1+pkps-1/4ifps≡1mod⁡4.(ii)
Ifp=2, then the number of Euclidean self-dual cyclic codes of length 2s over F2k+uF2k is (51)NE2k,2s=1+2kifs=1,1+2k+2k2ifs=2,1+2k+22k22k2s-2-1-12k-1ifs≥3.Proof.
From Propositions20 and 21, the number of Euclidean self-dual cyclic codes of length 2s over F2k+uF2k is ∑i1=0ps/2pki1/2. Apply a suitable geometric sum; the results follow.


### 4.2. Hermitian Self-Dual Cyclic Codes of Lengthps over Fpk+uFpk

Under the assumption thatk is even, characterization and enumeration of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk are given in this section.For a subsetA of Fpk+uFpkx/xps-1, let (52)ρA≔∑i=0ps-1ρaixi∣∑i=0ps-1aixi∈A, where ρ(a+ub)=apk/2+ubpk/2.Based on the structural characterization ofC given in Theorem 15, the Hermitian dual of C is determined as follows.Theorem 23.
LetC be an ideal in A and(53)C=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1,
wherehj∈Fpk. Then C⊥H has the representation (54)C⊥H=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjpk/2x-1r,ux-1ps-i0.Proof.
From Theorem19 and the fact that C⊥H=ρ(C⊥E), the result follows.Assume thatC is Hermitian self-dual. Then C=C⊥H which implies that C=pkps and i0+i1=ps.Ifi1=0, then it is not difficult to see that the ideal generated by u is only Hermitian self-dual cyclic code of length ps over Fpk+uFpk.Assume thati1≥1. Then(55)-uhtpk/2=u∑j=0t-1i0+ji0-jt-jhjfor all 0≤t≤i1-1.Fromii equations above and the definition of M(ps,i1), we have(56)Mps,i1h+hpk/2-h=0,where h=(h1,h2,…,hi1), hpk/2=(h1pk/2,h2pk/2,…,hi1pk/2), and 0=(0,0,…,0).From Theorem14, we have 0≤i1≤ps/2 since i0+i1=ps.Proposition 24.
Letk be an even positive integer and let i1 be a positive integer such that i1≤pk/2. Then the number of solutions of (56) in Fpki1 is pki1/2.Proof.
LetΨ:Fpk→Fpk defined by α↦αpk/2-α for all α∈Fpk. Using the fact that Ψ(1)=0=Ψ(0) and arguments similar to those in [23, Proposition  3.3], the result follows.For a prime numberp, a positive integer s, and an even positive integer k, the number of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk can be determined in the following corollary.Corollary 25.
Letp be a prime and let s and k be positive integers such that k is even. Then the number of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk is(57)NHpk,ps=∑i1=0ps/2pki1/2=pk/2ps/2+1-1pk/2-1.


## 4.1. Euclidean Self-Dual Cyclic Codes of Lengthps over Fpk+uFpk

Characterization and enumeration of Euclidean self-dual cyclic codes of lengthps over Fpk+uFpk are given in this subsection.For each subsetA of Fpk+uFpkx/xps-1, denote by A¯ the set of polynomials f(x)¯ for all f(x) in A, where — is viewed as the conjugation on the group ring (Fpk+uFpk)[Zps] defined in Section 2. From the definition of the annihilator, the next theorem can be derived similar to [21, Proposition  2.12].Theorem 18.
LetC be an ideal in Fpk+uFpkx/xps-1. Then C⊥E=Ann(C)¯.Using the unique generators of an idealC in Fpk+uFpkx/xps-1 determined in Theorem 15, the Euclidean dual of C can be given in the following theorem.Theorem 19.
LetC=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1 be an ideal in A, where hj∈Fpk for all 0≤j≤ii-1. Then C⊥E is in the form of(40)C⊥E=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjx-1r,ux-1ps-i0.Proof.
From the proof of Theorem14, we have(41)AnnC=x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1hjx-1j,ux-1ps-i0.By Theorem 18, it follows that C⊥E=AnnC¯. Hence, C⊥E contains the elements u(x-1)ps-i0 and(42)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1-1i0+jhjx-1jxi0-j. By writing x=(x-1)+1 and using the Binomial Theorem, it follows that C⊥E contains the element(43)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1∑l=0i0-j-1i0+ji0-jlhjx-1i0-l.Hence, (44)x-1ps-i1-ux-1ps-i0-i1∑j=0i1-1∑l=0i0-j-1i0+ji0-jlhjx-1i0-l,x-1ps-i1⊆C⊥E.By counting the number of elements, the two sets are equal as desired. Updating the indices, it can be concluded that (45)C⊥E=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjx-1r,x-1ps-i1.Assume that an idealC=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1 in Fpk+uFpkx/xps-1 is Euclidean self-dual; that is, C=C⊥E. By Theorem 18, it is equivalent to ps=i0+i1 and(46)-ht=∑j=0t-1i0+li0-jr-jhjin Fpk for all 0≤t≤ii-1.We note that ifi1=0, then it is not difficult to see that only the ideal generated by u is Euclidean self-dual.For the casei1≥1, the situation is more complicated. First, we recall an i1×i1 matrix M(ps,i1) over Fpk defined in [22] as(47)Mps,i1=-1i0+100⋯0-1i0i01-1i0+1+10⋯0-1i0i02-1i0+1i0-11-1i0+2+1⋯0⋮⋮⋮⋱⋮-1i0i0i1-1-1i0+1i0-1i1-2-1i0+2i0-2i1-3⋯-1i0+i1-1+1.It is not difficult to see thati1 equations from (46) are equivalent to the matrix equation(48)Mps,i1h=0,where h=h0,h1,…,hi1-1T and 0=0,0,…,0T.Moreover, it can be concluded that the idealC is Euclidean self-dual if and only if ps=i0+i1 and h0,h1,…,hi1-1 satisfy (48). Since h0=h1=⋯=hi1-1=0 is a solution of (48), the corresponding idea x-1ps-i1,ux-1i1 is Euclidean self-dual in Fpk+uFpkx/xps-1. Hence, for a fixed first torsion degree 1≤i1≤ps, a Euclidean self-dual ideal in Fpk+uFpkx/xps-1 always exists. By solving (48), all Euclidean self-dual ideals in Fpk+uFpkx/xps-1 can be constructed. Therefore, for a fixed first torsion degree 1≤i1≤ps, the number of Euclidean self-dual ideals in Fpk+uFpkx/xps-1 equals the number of solutions of (48) which is pkκ, where κ is the nullity of M(ps,i1) determined in [22].Proposition 20 (see [22, Proposition  3.3]).
Letκ be the nullity of M(ps,i1). Then(49)κ=i12ifpisodd;i1+12ifp=2.The number of Euclidean self-dual cyclic codes inFpk+uFpkx/xps-1 with first torsional degree i1 is given in terms of the nullity of M(ps,i1) as follows.Proposition 21.
Leti1>0 and let κ be the nullity of M(ps,i1) over Fpk. Then the number of Euclidean self-dual cyclic codes of length ps over Fpk with first torsional degree i1 is pkκ.From Theorem14, we have 0≤i1≤⌊ps/2⌋ since i0+i1=ps. Hence, the number of Euclidean self-dual cyclic codes of length ps over Fpk+uFpk is given by the following corollary.Corollary 22.
Letp be a prime and let s and k be positive integers. Then the following statements hold: (i)
Ifp is odd, then the number of Euclidean self-dual cyclic codes of length ps over Fpk+uFpk is (50)NEpk,ps=2pkps+1/4-1pk-1ifps≡3mod⁡4,2pkps-1/4-1pk-1+pkps-1/4ifps≡1mod⁡4.(ii)
Ifp=2, then the number of Euclidean self-dual cyclic codes of length 2s over F2k+uF2k is (51)NE2k,2s=1+2kifs=1,1+2k+2k2ifs=2,1+2k+22k22k2s-2-1-12k-1ifs≥3.Proof.
From Propositions20 and 21, the number of Euclidean self-dual cyclic codes of length 2s over F2k+uF2k is ∑i1=0ps/2pki1/2. Apply a suitable geometric sum; the results follow.


## 4.2. Hermitian Self-Dual Cyclic Codes of Lengthps over Fpk+uFpk

Under the assumption thatk is even, characterization and enumeration of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk are given in this section.For a subsetA of Fpk+uFpkx/xps-1, let (52)ρA≔∑i=0ps-1ρaixi∣∑i=0ps-1aixi∈A, where ρ(a+ub)=apk/2+ubpk/2.Based on the structural characterization ofC given in Theorem 15, the Hermitian dual of C is determined as follows.Theorem 23.
LetC be an ideal in A and(53)C=x-1i0+u∑j=0i1-1hjx-1j,ux-1i1,
wherehj∈Fpk. Then C⊥H has the representation (54)C⊥H=x-1ps-i1-ux-1ps-i0-i1∑r=0i1-1∑j=0r-1i0+ji0-jr-jhjpk/2x-1r,ux-1ps-i0.Proof.
From Theorem19 and the fact that C⊥H=ρ(C⊥E), the result follows.Assume thatC is Hermitian self-dual. Then C=C⊥H which implies that C=pkps and i0+i1=ps.Ifi1=0, then it is not difficult to see that the ideal generated by u is only Hermitian self-dual cyclic code of length ps over Fpk+uFpk.Assume thati1≥1. Then(55)-uhtpk/2=u∑j=0t-1i0+ji0-jt-jhjfor all 0≤t≤i1-1.Fromii equations above and the definition of M(ps,i1), we have(56)Mps,i1h+hpk/2-h=0,where h=(h1,h2,…,hi1), hpk/2=(h1pk/2,h2pk/2,…,hi1pk/2), and 0=(0,0,…,0).From Theorem14, we have 0≤i1≤ps/2 since i0+i1=ps.Proposition 24.
Letk be an even positive integer and let i1 be a positive integer such that i1≤pk/2. Then the number of solutions of (56) in Fpki1 is pki1/2.Proof.
LetΨ:Fpk→Fpk defined by α↦αpk/2-α for all α∈Fpk. Using the fact that Ψ(1)=0=Ψ(0) and arguments similar to those in [23, Proposition  3.3], the result follows.For a prime numberp, a positive integer s, and an even positive integer k, the number of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk can be determined in the following corollary.Corollary 25.
Letp be a prime and let s and k be positive integers such that k is even. Then the number of Hermitian self-dual cyclic codes of length ps over Fpk+uFpk is(57)NHpk,ps=∑i1=0ps/2pki1/2=pk/2ps/2+1-1pk/2-1.


## 5. Conclusions and Remarks

Euclidean and Hermitian self-dual abelian codes in non-PIGAsF2kA×Z2×Z2s are studied. The complete characterization and enumeration of such abelian codes are given and summarized as follows.In Corollaries1 and 3, self-dual abelian codes in F2kA×Z2×Z2s are shown to be a suitable Cartesian product of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length 2s over some Galois extension of the ring F2k+uF2k. Subsequently, the characterizations and enumerations of cyclic and self-dual cyclic codes of length ps over Fpk+uFpk are studied for all primes p. Combining these results, the following enumerations of Euclidean and Hermitian self-dual abelian codes in F2kA×Z2×Z2s are rewarded.For each abelian groupA of odd order and positive integers s and k, the number of Euclidean self-dual abelian codes in F2kA×Z2×Z2s is given in Theorem 2 in terms of the numbers N, NE, and NH of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length 2s over a Galois extension of F2k+uF2k, respectively.In addition, ifk is even, the number of Hermitian self-dual abelian codes in F2kA×Z2×Z2s is given in Theorem 4 in terms of the numbers N and NH of cyclic codes and Hermitian self-dual cyclic codes of length 2s over a Galois extension of F2k+uF2k, respectively.We note that all numbersN, NE, and NH are determined in Corollaries 17, 22, and 25, respectively. Therefore, the complete enumerations of Euclidean and Hermitian self-dual abelian codes in F2kA×Z2×Z2s are established.One of the interesting problems concerning the enumeration of self-dual abelian codes inF2kA×B, where A is an abelian group of odd order, is the case  where B is a 2-group of other types.



---

*Source: 9020173-2016-10-27.xml*