Self-Dual Abelian Codes in some Non-Principal Ideal Group Algebras

Transcription

1 Self-Dual Abelian Codes in some Non-Principal Ideal Group Algebras Parinyawat Choosuwan, Somphong Jitman, and Patanee Udomkavanich arxiv: v2 [mathra] 26 Sep 2016 Abstract The main focus of this paper is the complete enumeration of self-dual abelian codes in non-principal ideal group algebras F 2 k[a Z 2 Z 2 s] 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 2 s over some Galois extensions of the ring F 2 k +uf 2 k, where u 2 = 0 Subsequently, general results on the characterization and enumeration of cyclic codes and self-dual codes of length p s over F p k+uf p k are given Combining these results, the complete enumeration of self-dual abelian codes in F 2 k[a Z 2 Z 2 s] is therefore obtained Keywords: self-dual codes, abelian codes, finite chain rings, group algebras 2010 Mathematics Subject Classification: 94B15, 94B05, 16A26 1 Introduction Information media, such as communication systems and storage devices of data, are not 100 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 1960s Linear codes with additional algebraic structures P Choosuwan is with the Department of Mathematics and Computer Science, Faculty of Science, Chulalongkorn University, Bangkok 10330, Thailand parinyawatch@gmailcom S Jitman is with the Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom 73000, Thailand sjitman@gmailcom P Udomkavanich is with the Department of Mathematics and Computer Science, Faculty of Science, Chulalongkorn University, Bangkok 10330, Thailand pattaneeu@chulaacth 1

2 and self-dual codes are important families of codes that have been extensively studied for both theoretical and practical reasons see [2 6,14,16,21 23] and references therein Some major results on Euclidean self-dual cyclic codes have been established in [18] In [13], 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 [14] 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 [16] To the best of our knowledge, the characterization and enumeration of self-dual abelian codes in non-principal ideal group algebras non-pigas have not been well studied It is therefore of natural interest to focus on this open problem In [14] and [16], it has been shown that there exists a Euclidean resp, Hermitian self-dual abelian code in F p k[g] if and only if p = 2 and G is even In order to study self-dual abeliancodes, it is thereforerestricted to the groupalgebra F 2 k[a B], where A is an abelian group of odd order and B is a non-trivial abelian group of two power order In this case, F 2 k[a B] is a PIGA if and only if B = Z 2 s is a cyclic group see [12] Equivalently, F 2 k[a B] is a non-piga if and only if B is non-cyclic To avoid tedious computations, we focus on the simplest case where B = Z 2 Z 2 s, 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 F 2 k[a Z 2 Z 2 s] It turns out that every Euclidean resp, Hermitian self-dual abelian code in F 2 k[a Z 2 Z 2 s] is a suitable Cartesian product of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length 2 s over some Galois extension of the ring F 2 k +uf 2 k, where u 2 = 0 see Section 2 Hence, the number of self-dual abelian codes in F 2 k[a Z 2 Z 2 s] 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 p s over F p k + uf p k are given for all primes p Combining these results, the characterizations and enumerations of Euclidean and Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s] are rewarded The paper is organized as follows In Section 2, some basic results on abelian codes are recalled together with a link between abelian codes in F 2 k[a Z 2 Z 2 s] and cyclic codes of length 2 s over Galois extensions of F 2 k + uf 2 k General results on the characterization and enumeration of cyclic codes of length p s over F p k +uf p k are provided in Section 3 In Section 4, the characterizations and enumerations of Euclidean and Hermitian self-dual cyclic codes of length p s over F p k +uf p k are established Summary and remarks are given in Section 5 2

3 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 non-principal ideal algebras and a product of cyclic codes over rings is given This link plays an important role in determining the algebraic structures an the numbers of Euclidean and Hermitian self-dual abelian codes in non-principal ideal algebras 21 Rings and Abelian Codes in Group Rings For a prime p and a positive integer k, denote by F p k the finite field of order p k Let F p k + uf p k := {a + ub a,b F p k} be a ring, where the addition and multiplication are defined as in the usual polynomial ring over F p k with indeterminate u together with the condition u 2 = 0 We note that F p k + uf p k is isomorphic to F p k[u]/ u 2 as rings The Galois extension of F p k +uf p k of degree m is defined to be the quotient ring F p k +uf p k[x]/ fx, where fx is an irreducible polynomial of degree m over F p k It is not difficult to see that the Galois extension of F p k+uf p k of degree m is isomorphic to F p km +uf p km as rings In the case where k is even, the mapping a+ub a pk/2 +ub pk/2 is a ring automorphism of order 2 on F p k + uf p k The readers may refer to [7,8] for properties of the ring F p k +uf p k For a commutative ring R with identity 1 and a finite abelian group G, written additively, let R[G] denote the group ring of G over R The elements in R[G] will be written as g Gα g Y g, where α g 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 Y 0 := 1 is the multiplicative identity of R[G] resp, R, where 0 is the identity of G We define a conjugation on R[G] to be the map that fixes R and sends Y g to Y g for all g G, ie, for u = g Gα g Y g R[G], we set u := α g Y g = g Y g G g Gα g In the case where, there exists a ring automorphism ρ on R of order 2, we define ũ := g Gρα g Y g for all u = α g Y g R[G] In the g G case where R is a finite field F p k, then F p k[g] can be viewed as an F p k-algebra and it is called a group algebra The group algebra F p k[g] is called a principal ideal group algebra PIGA if its ideals are generated by a single element An abelian code in R[G] is defined to be an ideal in R[G] If G is cyclic, this code is called a cyclic 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]/ x n 1 = R[Z n ] 3

4 The Euclidean inner product in R[G] is defined as follows For u = α g Y g and v = β g Y g g G g G in R[G], we set u,v E := g Gα g β g 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 u,v ρ := g Gα g ρβ g If R = F q 2 resp, R = F q 2 + uf q 2 and ρa = a q resp, ρa + ub = a q + ub q for all a F q 2 resp, a+ub F q 2 +uf q 2, the ρ-inner product is called the Hermitian inner product and denoted by u,v H The Euclidean dual and Hermitian dual of C in R[G] are defined to be the sets C E := {u R[G] u,v E = 0 for all v C} and C H := {u R[G] u,v H = 0 for all v C}, respectively An abelian code C is said to be Euclidean self-dual resp, Hermitian self-dual if C = C E resp, C = C H For convenience, denote by Np k,n, NEp k,n, and NHp k,n the number of cyclic codes, the number of Euclidean self-dual cyclic codes, and the number of Hermitian selfdual cyclic codes of length n over F p k +uf p k, respectively 22 Decomposition of Abelian Codes in F 2 k[a Z 2 Z 2 s] In [14] and [16], it has been shown that there exists a Euclidean resp, Hermitian selfdual abelian code in F p k[g] if and only if p = 2 and G is even To study self-dual abelian codes, it is sufficient to focus on F 2 k[a B], where A is an abelian group of odd order and B is a non-trivial abelian group of two power order In this case, F 2 k[a B] is a PIGA if and only if B = Z 2 s is a cyclic group for some positive integer s see [12] The complete characterization and enumeration of self-dual abelian codes in PIGAs have been given in [14] and [16] 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 4

5 the simplest case where B = Z 2 Z 2 s Useful decompositions of F 2 k[a Z 2 Z 2 s] are given in this section First, we consider the decomposition of R := F 2 k[a] In this case, R is semi-simple [2] which can be decomposed using the Discrete Fourier Transform in [22] see [16] and [14] for more details For completeness, the decompositions used in this paper are summarized as follows For an odd positive integer i and a positive integer k, let ord i 2 k denote the multiplicative order of 2 k modulo i For each a A, denote by orda the additive order of a in A A 2 k -cyclotomic class of A containing a A, denoted by S 2 ka, is defined to be the set where 2 ki a := 2ki S 2 ka :={2 ki a i = 0,1,} = {2 ki a 0 i < ord orda 2 k }, j=1 a in A An idempotent in R is a non-zero element e such that e 2 = e It is called primitive if for every other idempotent f, either ef = e or ef = 0 The primitive idempotents in R are induced by the 2 k -cyclotomic classes of A see [6, Proposition II4] Let {a 1,a 2,,a t } be a complete set of representatives of 2 k -cyclotomic classes of A and let e i be the primitive idempotent induced by S 2 ka i for all 1 i t From [22], R can be decomposed as and hence, R = t Re i, 21 i=1 F 2 k[a Z 2 ] = R[Z 2 ] t = Re i [Z 2 ] 22 It is well known see [14,16] that Re i := F 2 k i, where k i is a multiple of k Precisely, k i = k S 2 ka i = k ord ordai 2 k provided that e i is induced by S 2 ka i It follows that Re i [Z 2 ] = F 2 k i [Z 2 ] Under the ring isomorphism that fixes the elements in F 2 k i and Y 1 u+1, F 2 k i [Z 2 ] is isomorphic to the ring F 2 k i +uf 2 k i, where u 2 = 0 We note that this ring plays an important role in coding theory and codes over rings in this family have extensively been studied [7, 8, 10, 17] and references therein From 22 and the ring isomorphism discussed above, we have i=1 F 2 k[a Z 2 Z 2 s] = t R i [Z 2 s], 23 i=1 5

6 where R i := F 2 k i +uf 2 k i for all 1 i t In order to study the algebraic structures of Euclidean and Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s], the two rearrangements of R i s in the decomposition 23 are needed The details are given in the following two subsections 221 Euclidean Case A 2 k -cyclotomic class S 2 ka is said to be of type I if a = a in this case, S 2 ka = S 2 k a, type II if S 2 ka = S 2 k a and a a, or type III if S 2 k a S 2 ka The primitive idempotent e induced by S 2 ka is said to be of type λ {I,II,III} if S 2 ka is a 2 k -cyclotomic class of type λ Without loss of generality, the representatives a 1,a 2,,a t of 2 k -cyclotomic classes of A can be chosen such that {a i i = 1,2,,r I }, {a ri +j j = 1,2,,r II } and {a ri +r I +l,a ri +r I +r I +l = a ri +r I +l l = 1,2,,r III } are sets of representatives of 2 k - cyclotomic classes of types I,II, and III, respectively, where t = r I +r II +2r III Rearranging the terms in the decomposition of R in 21 based on the 3 types primitive idempotents, we have F 2 k[a Z 2 ] = t Re i [Z 2 ] = i=1 ri R i i=1 ri S j j=1 r I T l T l, 24 where R i := F 2 k + uf 2 k for all i = 1,2,,r I, S j := F 2 k ri +j + uf 2 k ri +j for all j = 1,2,,r II, and T l := F 2 k ri +r I +l +uf 2 k ri +r I +l for all l = 1,2,,r III From 24, we have F 2 k[a Z 2 Z 2 s] ri = R i [Z 2 s] i=1 S j [Z 2 s] ri j=1 r I l=1 T l [Z 2 s] T l [Z 2 s] 25 It follows that, an abelian code C in F 2 k[a Z 2 Z 2 s] can be viewed as C ri ri r I = B i C j D l D l, 26 i=1 j=1 where B i, C j, D s and D s are cyclic codes in R i [Z 2 s], S j [Z 2 s], T l [Z 2 s] and T l [Z 2 s], respectively, for all i = 1,2,,r I, j = 1,2,,r II and l = 1,2,,r III Using the analysis similar to those in [14, Section IID], the Euclidean dual of C in 26 is of the form ri C E = i=1 B E i ri j=1 C H j l=1 r I l=1 D l E D E l l=1 6

7 Similar to [14, Corollary 29], necessary and sufficient conditions for an abelian code in F 2 k[a Z 2 Z 2 s] to be Euclidean self-dual can be given using the notions of cyclic codes of length 2 s over R i, S j, and T l in the following corollary Corollary 21 An abelian code C in F 2 k[a Z 2 Z 2 s] is Euclidean self-dual if and only if in the decomposition 26, i B i is a Euclidean self-dual cyclic code of length 2 s over R i for all all i = 1,2,,r I, ii C j is a Hermitian self-dual cyclic code of length 2 s over S j for all j = 1,2,,r II, and iii D l = D E l is a cyclic code of length 2 s over T l for all l = 1,2,,r III Given a positive integer k and an odd positive integer j, the pair j,2 k is said to be good if j divides 2 kt + 1 for some integer t 1, and bad otherwise This notions have been introduced in [13, 14] for the enumeration of self-dual cyclic codes and self-dual abelian codes over finite fields Let χ be a function defined on the pair j,2 k, where j is an odd positive integer, as follows 0 if j,2 k is good, χj,2 k = 1 otherwise 27 The number of Euclidean self-dual abelian codes in F 2 k[a Z 2 Z 2 s] can be determined as follows Theorem 22 Let k 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 F 2 k[a Z 2 Z 2 s] is NE2 k,2 s 1 χd,2 k N A d d M,ord d 2 k =1 d M ord d 2 k 1 d M NH2 k ord d2 k 1 χd,2 k,2 s N A d ord d 2 k N2 k ord d2 k,2 s χd,2 k N A d 2ord d 2 k, where N A d denotes the number of elements in A of order d determined in [1] Proof From 26 and Corollary 21, it suffices to determine the numbers of cyclic codes B i s, C j s, and D l s such that B i and C j are Euclidean and Hermitian self-dual, respectively 7

8 From [16, Remark 25], the elements in A of the same order are partitioned into 2 k -cyclotomic classes of the same type For each divisor d of M, a 2 k -cyclotomic class containing an element of order d has cardinality ord d 2 k and the number of such 2 k - cyclotomic classes is N Ad ord d We consider the following 3 cases 2 k Case 1: χd,2 k = 0 and ord d 2 k = 1 By [14, Remark 26], every 2 k -cyclotomic class of A containing an element of order d is of type I Since there are N Ad such ord d 2 k 2 k -cyclotomic classes, the number of Euclidean self-dual cyclic codes B i s of length 2 s corresponding to d is NE2 k,2 s N A d ord d 2 k = NE2 k,2 s 1 χd,2 k N A d Case 2: χd,2 k = 0 and ord d 2 k 1 By [14, Remark 26], every 2 k -cyclotomic class of A containing an element of order d is of type II Hence, the number of Hermitian self-dual cyclic codes C j s of length 2 s corresponding to d is NH2 k ord d2 k N A d,2 s ord d 2 k = NH2 k ord d2 k,2 s 1 χd,2 k N A d ord d 2 k Case 3: χd,2 k = 1 By [14, Lemma 45], every 2 k -cyclotomic class of A containing an element of order d is of type III Then the number of cyclic codes D l s of length 2 s corresponding to d is N2 k ord d2 k N A d,2 s 2ord d 2 k = N2 k ord d2 k,2 s The desired result follows since d runs over all divisors of M χd,2 k N A d 2ord d 2 k This enumeration will be completed by counting the above numbers NE, NH, and N in Corollaries 45, 48, and 313, respectively 222 Hermitian Case We focus on the case where k is even A 2 k -cyclotomic class S 2 ka is said to be of type I if S 2 ka = S 2 k 2 k 2a or type II if S 2 ka S 2 k 2 k 2a The primitive idempotent e induced by S 2 ka is said to be of type λ {I,II } if S 2 ka is a 2 k -cyclotomic class of type λ Without loss of generality, the representatives a 1,a 2,,a t of 2 k -cyclotomic classes can be chosen such that {a i i = 1,2,,r I } and {a ri +j,a ri +r I +j = 2 k 2a ri +j j = 1,2,,r II } are sets of representatives of 2 k -cyclotomic classes of types I and II, respectively, where t = r I +2r II Rearranging the terms in the decomposition of R in 21 based on the above 2 types primitive idempotents, we have F 2 k[a Z 2 ] t = Re i [Z 2 ] = i=1 ri S j j=1 ri T l T l, 28 l=1 8

9 where S j := F 2 k j + uf 2 k j for all j = 1,2,,r I and T l := F 2 k ri +l + uf 2 k ri +l for all l = 1,2,,r II From 28, we have F 2 k[a Z 2 Z 2 s] = ri S j [Z 2 s] j=1 T l [Z 2 s] T l [Z 2 s] 29 Hence, an abelian code C in F 2 k[a Z 2 Z 2 s] can be viewed as ri C ri = C j D l D l, 210 j=1 where C j, D l and D l are cyclic codes in S j[z 2 s], T l [Z 2 s] and T l [Z 2 s], respectively, for all j = 1,2,,r I and l = 1,2,,r II l=1 ri Using the analysis similar to those in [16, Section IID], the Hermitian dual of C in 210 is of the form ri C H = j=1 C H j ri l=1 l=1 D l E D E l Similar to [16, Corollary 28], necessary and sufficient conditions for an abelian code in F 2 k[a Z 2 Z 2 s] to be Hermitian self-dual are now given using the notions of cyclic codes of length 2 s over S j and T l in the following corollary Corollary 23 An abelian code C in F 2 k[a Z 2 Z 2 s] is Hermitian self-dual if and only if in the decomposition 210, i C j is a Hermitian self-dual cyclic code of length 2 s over S j for all j = 1,2,,r I, and ii D l = D E l is a cyclic code of length 2 s over T l for all l = 1,2,,r II Given a positive integer k and an odd positive integer j, the pair j,2 k is said to be oddly good if j divides 2 kt + 1 for some odd integer t 1, and evenly good if j divides 2 kt + 1 for some even integer t 2 These notions have been introduced in [16] for characterizing the Hermitian self-dual abelian codes in PIGAs Let λ be a function defined on the pair j,2 k, where j is an odd positive integer, as 0 if j,2 k is oddly good, λj,2 k = otherwise The number of Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s] can be determined as follows 9

10 Theorem 24 Let k 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 F 2 k[a Z 2 Z 2 s] is d M k NH2 k ord d2 k 1 λd,2 2 N A d,2 s ord d 2 k k N2 k ord d2 k λd,2 2 N A d,2 s 2ord d 2 k, d M where N A d denotes the number of elements of order d in A determined in [1] Proof By Corollary 23 and 210, it is enough to determine the numbers cyclic codes C j s and D l s of length 2 s in 210 such that C j is Hermitian self-dual The desired result can be obtained using arguments similar to those in the proof of Theorem 22, where [16, Lemma 35] is applied instead of [14, Lemma 45] This enumeration will be completed by counting the above numbers NH and N in Corollaries 48 and 313, respectively 3 Cyclic Codes of Length p s over F p k +uf p k The enumeration of self-dual abelian codes in non-pigas in the previous section requires properties of cyclic codes of length 2 s over F 2 k + uf 2 k In this section, a more general situation is discussed Precisely, properties cyclic of length p s over F p k+uf p k are studied for all primes p We note that algebraic structures of cyclic codes of length p s over F p k + uf p k was studied in [7] and [8] Here, based on [20], we give an alternative characterization of such codes which is useful in studying self-dual cyclic codes of length p s over F p k +uf p k First, we note that there exists a one-to-one correspondence between the cyclic codes of length p s over F p k+uf p k and the ideals in the quotient ring F p k+uf p k[x]/ x ps 1 Precisely, a cyclic code C of length p s can be represented by the ideal { p } s 1 v i x i v 0,v 1,,v p s 1 C in F p k +uf p k[x]/ x ps 1 i=0 Formnowon,acycliccodeC willbereferredtoastheabovepolynomialpresentation Note that the map µ : F p k +uf p k[x]/ x n 1 F p k[x]/ x n 1 defined by µfx = fx mod u is a surjective ring homomorphism For each cyclic code C in F p k +uf p k[x]/ x ps 1 and i {0,1}, let Tor i C = {µvx vx F p k +uf p k[x]/ x n 1 and u i vx C} 10

11 For each i {0,1}, Tor i C is called the ith torsion code of C The codes Tor 0 C = µc and Tor 1 C are some time called the residue and torsion codes of C, respectively It is not difficult to see that for each i {0,1}, cx Tor i C if and only if u i cx + uzx C for some zx F p k[x]/ x ps 1 Consequently, we have that Tor 0 C Tor 1 C are ideals in F p k[x]/ x ps 1 cyclic codes of length p s over F p k We note that every ideal C in F p k[x]/ x ps 1 is of the form x 1 i for some 0 i p s and the cardinality of C is p s i From the structures of cyclic codes of length p s over F p k discussed above and [8, Proposition 25], we have the following properties of the torsion and residue codes Proposition 31 Let C be an ideal in F p k + uf p k[x]/ x ps 1 and let i {0,1} Then the following statements hold i Tor i C is an ideal of F p k[x]/ x ps 1 and Tor i C = x 1 T i for some 0 T i p s ii If Tor i C = x 1 T i, then Tor i C = p k ps T i iii C = Tor 0 C Tor 1 C = p k 2ps T 0 +T 1 With the notations given in Proposition 31, for each i {0,1}, T i C := T i is called the ith-torsional degree of C Remark 32 From Proposition 31 and the definition above, we have the following facts i Since Tor 0 C Tor 1 C, we have 0 T 1 C T 0 C p s ii If ux 1 t C, then t T 1 C Next, we determine a generator set of an ideal in F p k +uf p k[x]/ x ps 1 Theorem 33 Let C be an ideal of F p k +uf p k[x]/ x ps 1 Then C = s 0 x,us 1 x, where, for each i {0,1}, i either s j x = 0 or s j x = x 1 t j + uz j x for some z j x F p k[x]/ x ps 1 and 0 t j < p s, ii s j x 0 if and only if Tor j C {0} and Tor 0 C Tor 1 C, and iii if s j x 0, then Tor j C = x 1 t j 11

12 Proof The statement can be obtained using a slight modification of the proof of [11, Theorem 65] For completeness, the details are given as follows Foreach ideal I inf p k[x]/ x ps 1, it canberepresented asi = {0}orC = x 1 t 0 where 0 t 0 < p s If I = {0}, then we are done by choosing s 0 x = 0 For 0 t 0 < p s, let s 0 x = x 1 t 0 By abuse of notation, Tor 0 C = C = x 1 t 0 Hence, s 0 x satisfies the conditions i, ii and iii Since C is an ideal of the ring F p k + uf p k[x]/ x ps 1 and µ is a surjective ring homomorphism, µc is an ideal of F p k[x]/ x ps 1 which implies that µc = s 0 x where s 0 x satisfies the conditions i, ii and iii If s 0 x = 0, then take s 0x = 0 Assume that s 0 x 0, then s 0 x = x 1t 0 where 0 t 0 < p s Then there exists z 0 x F p k[x]/ x ps 1 such that s 0 x = x 1 t 0 +uz 0 x C and µs 0 x = s 0 x, ie, s 0 x = s 0x + uz 0 x Since Tor 1 C is an ideal of F p k[x]/ x ps 1, it follows that Tor 1 C = x 1 t 1 for some 0 t 1 p s Let s 1 x = x 1 t 1 Claim that C = s 0 x,us 1 x Since C is an ideal of F p k+uf p k[x]/ x ps 1, we have ux 1 t 1 C Thus, s 0 x,us 1 x C To show that C s 0 x,us 1 x, let cx C Then µcx = wxs 0x for some wx F p k[x]/ x ps 1 Thus, µcx = wxs 0x which implies that cx = wxs 0 x + uw x = wxs 0 x uwxz 0 x + uw x = wxs 0 x+u z 0 xwx+w x for some w x Tor 1 C Since w x Tor 1 C, it follows that cx s 0 x,us 1 x Therefore, C = s 0 x,us 1 x as desired Note that s 1 x = x 1 pk = 0 implies C = {0} Assume that C {0} Then s 1 x = x 1 t 1 where 0 t 1 < p s If s 0 x = 0, then we are done Assume that s 0 x 0 Then s 0 x = x 1 t where 0 t < p s Hence, Tor 0 C = x 1 t Since Tor 0 C Tor 1 C, we have t 1 t If t 1 < t, then we are done Assume that t 1 = t Then s 0 x = x 1 t + uz 0 x for some z 0 x F p k[x]/ x ps 1 It follows that us 1 x = ux 1 t 1 = ux 1 t = us 0 x s 0 x, a contradiction Therefore, C = s 0 x = s 0 x,us 0 x = s 0 x,us 1 x However, the generator set given in Theorem 33 does not need to be unique The unique presentation is given in the following theorem Theorem 34 Let C be an ideal in F p k + uf p k[x]/ x ps 1, T 0 := T 0 C and T 1 := T 1 C Then C = f 0 x,f 1 x, where and x 1 T 0 +ux 1 t hx if T 0 < p s, f 0 x = 0 if T 0 = p s, 12

13 ux 1 T 1 if T 1 < p s, f 1 x = 0 if T 1 = p s, with hx F p k[x]/ x ps 1 is either zero or a unit with t+deghx < T 0 Moreover, f 0 x,f 1 x is unique in the sense that if there exists a pair g 0 x,g 1 x of polynomials satisfying the conditions in the theorem, then f 0 x = g 0 x and f 1 x = g 1 x Proof If C = {0}, then Tor 1 C = {0} and Tor 0 C = {0} which imply that T 0 = p s and T 1 = p s The polynomials f 0 x = 0 and f 1 x = 0 have the desired properties Next, assume that C {0} Then there exists the smallest nonnegative integer r {0,1} such that T r < p s From Theorem 33, it can be concluded that where and C = s 0 x,us 1 x, x 1 T 0 +ugx for some gx F p k[x]/ x ps 1 if r = 0, s 0 x = 0 if r = 1, s 1 x = x 1 T 1 Case 1: r = 0 Then s 0 x = x 1 T 0 + ugx for some gx F p k[x]/ x ps 1 and s 1 x = x 1 T 1 It follows that C = x 1 T 0 + ugx,x 1 T 1, T 1 C = T 1 and T 0 C = T 0 Let f 1 x = us 1 x Since gx F p k[x]/ x ps 1, we have p s 1 s 0 x = x 1 T 0 +u a j x 1 j, where a j F p k for all j = 0,1,,p k 1 Since ux 1 T 1 = us 1 x C, we have p s 1 ua j x 1 j C for all j = T 1,T 1 +1,,p s 1 It follows that u a j x 1 j C j=t 1 T 1 1 p s 1 Let f 0 x = x 1 T 0 +u a j x 1 j Then f 0 x = s 0 x u a j x 1 j C j=t 1 WeshowthatC = f 0 x,f 1 x Fromthediscussionabove, wehave f 0 x,f 1 x C Since us 1 x = ux 1 T 1 = f 1 x, it follows that ua j x 1 j f 0 x,f 1 x for all p s 1 j = T 1,T 1 +1,,p s 1 Hence, u a j x 1 j f 0 x,f 1 x which implies that j=t 1 p s 1 s 0 x = f 0 x+u a j x 1 j f 0 x,f 1 x j=t 1 13

14 Therefore, C = s 0 x,us 1 x f 0 x,f 1 x As desired, C = f 0 x,f 1 x Case 2: r = 1 Then s 0 x = 0 and s 1 x = ux 1 T 1 which implies that Tor 1 C = x 1 T 1 and Tor 0 C = {0} By choosing f 0 x = 0 and f 1 x = ux 1 T 1, the result follows Toprovetheuniqueness, letc = g 0 x,g 1 x besuchthatg 0 xandg 1 xsatisfying the conditions in the theorem Then g 1 x = ux 1 T 1 = f 1 x T 1 1 Write g 0 x = x 1 T 0 +u c j x 1 j where c j F p k Then T 1 1 f 0 x g 0 x = u a j c j x 1 j It can be seen that f 0 x g 0 x = ux 1 l hx, where hx = 0 or hx is a unit with l T 1 1 < T 1 If hx is a unit, then ux 1 l C which implies that l T 1, a contradiction Hence, hx = 0 which means that f 0 x = g 0 x as desired Definition 35 ForeachidealC inf p k+uf p k[x]/ x ps 1,denotebyC = f 0 x,f 1 x the unique representation of the ideal C obtained in Theorem 34 Illustrative examples of the representations in Theorem 33 and Theorem 34 are given as follows Example 36 Consider the ideal C = x 1 2 in F 2 + uf 2 [x]/ x 4 1 Using Theorem 34, we obtain that C has the unique representation x 1 2,ux 1 2 Based on Theorem 33, C can be represented as x 1 2,0, x 1 2 +ux 1 2,0, and x 1 2 +ux 1 3,0 TheannihilatorofanidealC inf p k+uf p k[x]/ x ps 1 iskeytodetermineproperties C as well as the number of ideals in F p k +uf p k[x]/ x ps 1 Definition 37 Let C be an ideal in F p k + uf p k[x]/ x ps 1 The annihilator of C, denoted by AnnC, is defined to be the set {fx F p k + uf p k[x]/ x ps 1 fxgx = 0 for all gx C} The following properties of the annihilator can be obtained using arguments similar to those in the case of Galois rings in [20] Theorem 38 Let C be an ideal of F p k + uf p k[x]/ x ps 1 Then the following statements hold i AnnC is an ideal of F p k +uf p k[x]/ x ps 1 ii If C = p k d, then AnnC = p k 2 ps d 14

15 iii AnnAnnC = C Theorem 39 Let I denote the set of ideals of F p k +uf p k[x]/ x ps 1 and let A = {C I T 0 C+T 1 C p s } and A = {C I T 0 C+T 1 C p s } Then the map φ : A A defined by C AnnC is a bijection The rest of this section is devoted to the determination of all ideals in F p k + uf p k[x]/ x ps 1 In view of Theorem 39, it suffices to focus on the ideals in A For each C = f 0 x,f 1 x in A, if f 0 x = 0, then T 0 C = p s and T 1 C = 0 Hence, the only ideal in A with f 0 x = 0 is of the form 0,u In the following two theorems, we assume that f 0 x 0 Theorem 310 Let x 1 i 0 + ux 1 t hx,ux 1 i 1 be the representation of an ideal in F p k + uf p k[x]/ x ps 1 Then it is a representation of an ideal in A if and only if i 0,i 1,t are integers and hx F p k[x]/ x ps 1 such that 0 i 0 < p s, 0 i 1 min{i 0,p s i 0 }, t 0, t+deghx < i 1 and hx is either zero or a unit in F p k[x]/ x ps 1 Proof Form Theorem 34, we have that i 0,i 1,t are integers and hx F p k[x]/ x ps 1 such that 0 i 0 < p s, 0 i 1 i 0, t 0, t+deghx < i 1 and hx is either zero or a unit in F p k[x]/ x ps 1 Assume that x 1 i 0 +ux 1 t hx,ux 1 i 1 is a representation of an ideal in A Then i 0 +i 1 p s which implies that i 1 p s i 0 Hence, we have i 1 min{i 0,p s i 0 } Conversely, assume that C = x 1 i 0 +ux 1 t hx,ux 1 i 1, where i 0,i 1,t are integers and hx F p k[x]/ x ps 1 such that 0 i 0 < p s, 0 i 1 min{i 0,p s i 0 }, t 0, t+deghx < i 1 and hx is either zero or a unit in F p k[x]/ x ps 1 Clearly, i 0 + i 1 p s To show that C = x 1 i 0 + ux 1 t hx,ux 1 i 1 in A and it remains to proof that T 0 C = i 0 and T 1 C = i 1 that Since Let D = x 1 ps i 1 ux 1 ps i 0 i 1 +t hx,ux 1 ps i 0 It is not difficult to see D = x 1 ps i 1 +ux 1 ps i 0 ux 1 ps i 0 i 1 +t hx,ux 1 ps i 0 x 1 i 0 +ux 1 i 1 hx x 1 ps i 1 +ux 1 ps i 0 ux 1 ps i 0 i 1 +t hx = 0, x 1 i 0 +ux 1 i 1 hx ux 1 ps i 0 = 0, and ux 1 i 1 x 1 p s i 1 +ux 1 ps i 0 ux 1 ps i 0 i 1 +t hx = 0, 15

16 wehaved AnnCByProposition31, weobtainthatt j C i j andt j D p s i 1 j for all j {0,1} Hence, C p k 2 ps i 0 i 1 and AnnC D p k i 0+i 1 Since C AnnC = p k 2 ps, we have AnnC = D Therefore, T 0 C = i 0, T 1 C = i 1 as desired Since every polynomial m i=0 a ix 1 i in F p k[x] is either 0 or x 1 t hx, where hx is a unit in F p k[x] and 0 t m deghx, Theorem 310 is rewritten as follows: i 1 1 Theorem 311 The expression x 1 i 0 + u h j x 1 j,ux 1 i 1 represents an ideal in A if and only if i 0 and i 1 are integers such that 0 i 0 < p s, 0 i 1 min{i 0,p s i 0 }, i 0 +i 1 p s, and h j F p k for all 0 j < i 1 The number of distinct ideals of F p k +uf p k[x]/ x ps 1 of a fixed d = T 0 +T 1 is given in the following proposition Proposition 312 Let 0 d p s Then the number of distinct ideals in F p k + uf p k[x]/ x ps 1 with T 0 +T 1 = d is where K = min{ d 2,ps d 2 } p kk+1 1, p k 1 Proof Let T 1 = i 1 and i 0 := T 0 = d T 1 be fixed Case 1 : d < p s Then i 0 i 0 +i 1 = T 0 +T 1 = d < p s By Theorem 311, it follows that i 1 1 i 1 1 C = x 1 i 0 + u h j x 1 j,ux 1 i 1 Then the choices for h j x 1 j is p k i 1 By Theorem 311 again, we also have T 1 min{t 0,p s T 0 } Since T 0 +T 1 = d, we obtain that T 1 d 2 T 0, an hence, T 1 min{ d 2,ps T 0 } min{ d 2,ps d 2 } Now, vary T 1 from 0 to K, we obtain that there are 1 + p k + + p k K = pkk+1 1 p k 1 ideals with T 0 +T 1 = d Case 2 : d = p s If i 0 = p s, then the only ideal with T 0 +T 1 = p s is the ideal represented by 0,u If i 0 < p s, then we have p k + p k 2 +p k K ideals by arguments similar to those in Case 1 For a cyclic code C in A, we have C AnnC whenever T 0 C + T 1 C < p s In the case where T 0 C + T 1 C = p s, by the proof of Theorem 310, the annihilator of the cyclic code C = x 1 i 0 + ux 1 t hx,ux 1 i 1 is of the form AnnC = x 1 i 0 ux 1 t hx,ux 1 i 1 If p is odd, then C = AnnC occurs only the case hx = 0 In the case where p = 2, C = AnnC is always true By Proposition 312 and the bijection given in Theorem 39, the number of cyclic codes of length p s over F p k +uf p k can be summarized as follows 16

17 Corollary 313 The number of cyclic codes of length p s over F p k +uf p k is 2 Np k,p s = 2 p s d=0 p s d=0 p kmin{ d 2,ps d 2 }+1 1 p k 1 p kmin{ d 2,ps d 2 }+1 1 p k 1 pkps p k 1 if p = 2, 1 if p is odd Proof From Theorem 39, the number of cyclic codes of length p s over F p k + uf p k is A A = A + A A A The desired results follow immediately form the discussion above 4 Self-Dual Cyclic Codes of Length p s over F p k+uf p k In this section, characterization and enumeration self-dual cyclic codes of length p s over F p k +uf p k are given under the Euclidean and Hermitian inner products 41 Euclidean Self-Dual Cyclic Codes of Length p s over F p k + uf p k Characterization and enumeration of Euclidean self-dual cyclic codes of length p s over F p k +uf p k are given in this subsection Foreachsubset AofF p k+uf p k[x]/ x ps 1, denotebyathesetofpolynomialsfx for all fx ina, where is viewed as the conjugation onthe group ring F p k+uf p k[z p s] defined in Section 2 From the definition of the annihilator, the next theorem can be derived similar to [9, Proposition 212] Theorem 41 Let C be an ideal in F p k +uf p k[x]/ x ps 1 Then C E = AnnC Using the unique generators of an ideal C in F p k +uf p k[x]/ x ps 1 determined in Theorem 311, the Euclidean dual of C can be given in the following theorem i 1 1 Theorem 42 Let C = x 1 i 0 +u h j x 1 j,ux 1 i 1 be an ideal in A, where h j F p k for all 0 j i i 1 Then C E is in the form of i1 1 C E = x 1 ps i 1 ux 1 ps i 0 i 1 r=0 r 1 i 0+j i0 j h j x 1 r, r j ux 1 ps i

18 Proof From the proof of Theorem 310, we have i 1 1 AnnC = x 1 ps i 1 ux 1 ps i 0 i 1 h j x 1 j,ux 1 ps i 0 By Theorem 41, it follows that C E = AnnC Hence, C E contains the elements ux 1 ps i 0 and i 1 1 x 1 ps i 1 ux 1 ps i 0 i 1 1 i0+j h j x 1 j x i0 j By writing x = x 1+1 and using the Binomial Theorem, it follows that C E contains the element i1 1 x 1 ps i 1 ux 1 ps i 0 i 1 Hence, i1 1 x 1 ps i 1 ux 1 ps i 0 i 1 x 1 ps i 1 C E i0 j 1 i 0+j l=0 i0 j 1 i 0+j l=0 i0 j l i0 j l h j x 1 i 0 l h j x 1 i 0 l, By counting the number of elements, the two sets are equal as desired Updating the indices, it can be concluded that i1 1 r C E = x 1 ps i 1 ux 1 ps i 0 i 1 x 1 ps i 1 r=0 1 i 0+j i0 j h j x 1 r, r j i 1 1 Assume that an ideal C = x 1 i 0 +u h j x 1 j,ux 1 i 1 in F p k + uf p k[x]/ x ps 1 is Euclidean self-dual, ie, C = C E By Theorem 41, it is equivalent to p s = i 0 +i 1 and in F p k for all 0 t i i 1 h t = t 1 i 0+l i0 j h j 42 r j We note that, if i 1 = 0, then it is not difficult to see that only the ideal generated by u is Euclidean self-dual 18

19 For the case i 1 1, the situation is more complicated First, we recall an i 1 i 1 matrix Mp s,i 1 over F p k defined in [19] as 1 i i i0 1 i Mp s,i 1 = 1 0 i i0 1 i 0+1 i i i i0 i i 0 +1 i 0 1 i i 0 +2 i 0 2 i i 0 +i It is not difficult to see that i 1 equations from 42 are equivalent to the matrix equation where h = h 0,h 1,,h i1 1 T and 0 = 0,0,,0 T Mp s,i 1 h = 0 44 Moreover, it can be concluded that the ideal C is Euclidean self-dual if and only if p s = i 0 + i 1 and h 0,h 1,,h i1 1 satisfy 44 Since h 0 = h 1 = = h i1 1 = 0 is a solution of 44, the corresponding idea x 1 ps i 1,ux 1 i 1 is Euclidean self-dual in F p k + uf p k[x]/ x ps 1 Hence, for a fixed first torsion degree 1 i 1 p s, a Euclidean self-dual ideal in F p k +uf p k[x]/ x ps 1 always exists By solving 44, all Euclidean self-dual ideals in F p k + uf p k[x]/ x ps 1 can be constructed Therefore, for a fixed first torsion degree 1 i 1 p s, the number of Euclidean self-dual ideals in F p k +uf p k[x]/ x ps 1 equals the number of solutions of 44 which is p kκ, where κ is the nullity of Mp s,i 1 determined in [19] Proposition 43 [19, Proposition 33] Let κ be the nullity of Mp s,i 1 Then i 1 κ = 2 if p is odd; if p = 2 i The number of Euclidean self-dual cyclic codes in F p k+uf p k[x]/ x ps 1 with first torsional degree i 1 is given in terms of the nullity of Mp s,i 1 as follows Proposition 44 Let i 1 > 0 and let κ be the nullity of Mp s,i 1 over F p k Then the number of Euclidean self-dual cyclic codes of length p s over F p k with first torsional degree i 1 is p k κ From Theorem 310, we have 0 i 1 ps 2 since i 0 + i 1 = p s Hence, the number of Euclidean self-dual cyclic codes of length p s over F p k +uf p k is given by the following corollary 19

20 Corollary 45 Let p be a prime and let s and k be positive integers Then the following statements hold i If p is odd, then the number of Euclidean self-dual cyclic codes of length p s over F p k +uf p k is 2 NEp k,p s = 2 p k ps p k 1 p k ps p k 1 if p s 3mod4, +p k ps 1 4 if p s 1mod4 ii If p = 2, then the number of Euclidean self-dual cyclic codes of length 2 s over F 2 k +uf 2 k is 1+2 k if s = 1, NEp k,p s = 1+2 k +2 k 2 if s = 2, 1+2 k +22 k 2 2 k 2s if s 3 Proof From Propositions 43 and 44, the number of Euclidean self-dual cyclic codes of length 2 s over F 2 k + uf 2 k is follow ps 2 2 k 1 p k i 1 2 Apply a suitable geometric sum, the results i 1 =0 42 Hermitian Self-Dual Cyclic Codes of Length p s over F p k + uf p k Under the assumption that k is even, characterization and enumeration Hermitian selfdual cyclic codes of length p s over F p k +uf p k are given in this section For a subset A of F p k +uf p k[x]/ x ps 1, let ρa := where ρa+ub = a pk 2 +ub pk 2 { p s 1 i=0 ps 1 ρa i x i i=0 a i x i A Based on the structural characterization of C given in Theorem 311, the Hermitian dual of C is determined as follows Theorem 46 Let C be an ideal in A and i 1 1 C = x 1 i 0 +u h j x 1 j,ux 1 i 1, }, 20

21 where h j F p k Then C H has the representation i1 1 r C H = x 1 ps i 1 ux 1 ps i 0 i 1 ux 1 ps i 0 r=0 1 i 0+j i0 j h pk/2 j x 1 r, r j Proof From Theorem 42 and the fact that C H = ρc E, the result follows AssumethatC ishermitianself-dual ThenC = C H whichimpliesthat C = p k ps and i 0 +i 1 = p s If i 1 = 0, then it is not difficult to see that the ideal generated by u is only Hermitian self-dual cyclic code of length p s over F p k +uf p k Assume that i 1 1 Then for all 0 t i 1 1 uh pk/2 t = u t 1 i 0+j i0 j h j t j From i i equations above and the definition of Mp s,i 1, we have Mp s,i 1 h+h pk/2 h = 0 45 where h = h 1,h 2,,h i1, h pk/2 = h pk/2 1,h pk/2 2,,h pk/2 i 1 and 0 = 0,0,,0 From Theorem 310, we have 0 i 1 ps 2 since i 0 +i 1 = p s Proposition 47 Let k be an even positive integer and let i 1 be a positive integer such that i 1 pk 2 Then the number of solution of 45 in Fi 1 p k is p ki 1/2 Proof Let Ψ : F p k F p k defined by α α pk/2 α for all α F p k Using the fact that Ψ1 = 0 = Ψ0 and arguments similar to those in [15, Proposition 33], the result follows For a prime number p, a positive integer s and an even positive integer k, the number of Hermitian self-dual cyclic codes of length p s over F p k+uf p k can be determined in the following corollary Corollary 48 Let p 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 p s over F p k +uf p k is ps 2 NHp k,p s = i 1 =0 p ki1/2 = pk/2 p s p k/2 1 21

22 5 Conclusions and Remarks Euclidean and Hermitian self-dual abelian codes in non-pigas F 2 k[a Z 2 Z 2 s] are studied The complete characterization and enumeration of such abelian codes are given and summarized as follows In Corollaries 21 and 23, self-dual abelian code in F 2 k[a Z 2 Z 2 s] are shown to be a suitable Cartesian product of cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length 2 s over some Galois extension of the ring F 2 k+uf 2 k Subsequently, the characterizations and enumerations of cyclic and self-dual cyclic codes of length p s over F p k +uf p k are studied for all primes p Combining these results, the following enumerations of Euclidean and Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s] are rewarded For each abelian group A of odd order and positive integers s and k, the number of Euclidean self-dual abelian codes in F 2 k[a Z 2 Z 2 s] is given in Theorem 22 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 2 s over a Galois extension of F 2 k +uf 2 k, respectively In addition, if k is even, the number of Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s] is given in Theorem 24 in terms of the numbers N and NH of cyclic codes and Hermitian self-dual cyclic codes of length 2 s over a Galois extension of F 2 k +uf 2 k, respectively We note that all numbers N, NE, and NH are determined in Corollaries 313, 45, and 48, respectively Therefore, the complete enumerations of Euclidean and Hermitian self-dual abelian codes in F 2 k[a Z 2 Z 2 s] are established One of the interesting problems concerning the enumeration of self-dual abelian codes in F k 2 [A B], where A is an abelian group of odd order, is the case where B is a 2-group of other types Acknowledgements This work is supported by the Thailand Research Fund under Research Grant TRG and the DPST Research Grant 005/2557 References [1] S Benson, Students ask the darnedest things: A result in elementary group theory, Math Mag, 70, , 1997 [2] S D Berman, Semi-simple cyclic and abelian codes, Kibernetika, 3, 21 30,

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