Codes over an infinite family of algebras

Transcription

1 J Algebra Comb Discrete Appl 4(2) Received: 12 June 2015 Accepted: 17 February 2016 Journal of Algebra Combinatorics Discrete tructures and Applications Codes over an infinite family of algebras Research Article Irwansyah, Intan Muchtadi-Alamsyah, Ahmad Muchlis, Aleams Barra, Djoko uprijanto Abstract: In this paper, we will show some properties of codes over the ring B k = F p[v 1,, v k ]/(v 2 i = v i, i = 1,, k) These rings, form a family of commutative algebras over finite field F p We first discuss about the form of maximal ideals and characterization of automorphisms for the ring B k Then, we define certain Gray map which can be used to give a connection between codes over B k and codes over F p Using the previous connection, we give a characterization for equivalence of codes over B k and Euclidean self-dual codes Furthermore, we give generators for invariant ring of Euclidean self-dual codes over B k through MacWilliams relation of Hamming weight enumerator for such codes 2010 MC: 11T71 Keywords: Gray map, Equivalence of codes, Euclidean self-dual, Hamming weight enumerator, MacWilliams relation, Invariant ring 1 Introduction Codes over finite rings has been an interesting topic in algebraic coding theory since the discovery of codes over Z 4, see [4] An example of finite rings which has interesting properties is the ring A k = F 2 [v 1,, v k ], where v 2 i = v i, for 1 i k, because it has two Gray maps which relate codes over such ring and binary codes, see [2] This ring also has non-trivial automorphisms which can be used to define skew-cyclic codes, for example in [1], skew-cyclic codes over the ring A 1 = F 2 + vf 2, where v 2 = v, which give some optimal Euclidean and Hermitian self-dual codes Furthermore, Abualrub et al show that skew-cyclic codes over A 1 have a connection to left submodules over a skew-polynomial ring and The author is supported by Beasiswa Unggulan BPKLN Direktorat Jenderal Pendidikan Tinggi The authors are supported by Hibah Desentralisasi DIKTI 2016 Irwansyah (Corresponding Author); Algebra Research Group, Institut Teknologi Bandung, Bandung, Indonesia, and Department of Mathematics, Universitas Mataram, Mataram, Indonesia ( irw@unramacid) Intan Muchtadi-Alamsyah, Ahmad Muchlis, Aleams Barra; Algebra Research Group, Institut Teknologi Bandung, Bandung, Indonesia ( ntan@mathitbacid, muchlis@mathitbacid, barra@mathitbacid) Djoko uprijanto; Combinatorial Research Group, Institut Teknologi Bandung, Bandung, Indonesia ( djoko@mathitbacid) IN X 131

2 give skew-polynomial generators for these codes In [6], skew-cyclic codes over the ring A 1 have been characterized using a Gray map This characterization gives a way to construct skew-cyclic codes over the ring A 1 from binary cyclic or quasi-cyclic codes, and also gives decoding algorithm for some codes over such ring Meanwhile, Gao [3] consider skew-cyclic codes over the ring B 1 = F p + vf p, where v 2 = v, and found that these codes are equivalent to either cyclic codes or quasi-cyclic codes Using this connection, Gao is able to give an enumeration for skew-cyclic codes which are constructed using an automorphism with order relatively prime to the length of the codes In this paper, we consider codes over the ring B k = F p [v 1,, v k ], where vi 2 = v i for 1 i k, which is a generalization of the ring A k in [2] and B 1 in [3] We study its maximal ideals, automorphisms, equivalence codes, and Euclidean self-dual codes over these rings, including the generators for its invariant ring This paper is organized as follows: ection 2 describes some properties of the ring B k such as maximal ideals and automorphisms Meanwhile, in ection 3, we describe a Gray map for the ring B k, and we characterize linear codes and equivalent codes over the ring B k Finally, in ection 4, we characterize Euclidean self-dual codes, give the shape of MacWilliams relation and generators of invariant rings for Euclidean self-dual codes 2 The ring B k As we readily see, the ring B k forms a commutative algebra over prime field F p Let Ω = {1, 2,, k} and 2 Ω is the collection of all subsets of Ω Also, let w i be an element in the set {v i, 1 v i }, for 1 i k Then, we will prove the following observation Lemma 21 ω B k is a zero divisor if and only if ω w 1, w 2,, w k Proof ( =) It is clear that, v i (1 v i ) = 0, for all i = 1,, k Therefore, if ω w 1, w 2,, w k, then it is a zero divisor in B k (= ) Consider the equation, (α + βv k )(γ + ɛv k ) = a + bv k given α + βv k, a + bv k B k, for some α, β, a, b B k 1 We have γ = aα 1 and ɛ = (b βa)(α(β + α)) 1 Therefore, if a + bv k = 1, then γ = 1 and ɛ = β(α(β + α)) 1 Which implies, α + βv k is a unit if and only if α and α + β are also units Considering this observation for elements in B k 1, B k 2,, B 1, we have α + βv B 1 is a unit if and only if α, α + β F p are non zero elements ince, every element in finite commutative ring is either a unit or a zero divisor, we can see that the only zero divisors in B 1 are the elements in the ideals generated by βv or α(1 v) By generalizing this result recursively, we have the intended conclusion Also, we can easily show that I = w 1, w 2,, w k is a maximal ideal in B k Lemma 22 Let I = w 1, w 2,, w k Then I is a maximal ideal in B k Proof Consider quotient ring B k /I If v i I, then 1 v i 1 mod I, and if 1 v i I, then v i = 1 (1 v i ) 1 mod I Consequently, B k /I is a field o, I is a maximal ideal Moreover, B k /I = F p The following lemma is needed to prove Proposition 24 Lemma 23 α p = α, for all α B k 132

3 Proof Let α = A {1,,k} α Av A, for some α A F p, where v A = j A v j Then, consider α p = p i=0 ( ) p αa i i 1 v A1 α A v A A A 1 p i = α A1 v A1 + α A v A A A 1 p since F p has characteristic p and β p 1 = 1 for all β F p If we continue this procedure, then we have α p = α The following result shows that the ring B k is a principal ideal ring Proposition 24 Let I = α 1,, α m be an ideal in B k, for some α 1,, α m B k Then, I = α j ) p 1 A {1,,m},A ( 1) A +1 ( j A Proof Consider α i A {1,,m},A ( 1) A +1 ( j A α j) p 1 For any A {1,, m}, if i A, then α i ( 1) A +1 ( α j ) p 1 = ( 1) A +1 α i ( α j ) p 1 j A j A {i} since α p i = α i by Lemma 23 Consequently, there is a unique A = A {i} {1,, m}, such that α i ( 1) A +1 ( j A α j ) p 1 + ( 1) A +1 ( α j ) p 1 = 0 j A Otherwise, if i A, then there is a unique A = A {i} {1,, m} such that α i ( 1) A +1 ( j A α j ) p 1 + ( 1) A +1 ( α j ) p 1 = 0 j A o, every term will be vanish except α i α p 1 i = α i Therefore, I A {1,,m},A ( 1) A +1 ( j A α j ) p 1 It is clear that α j ) p 1 I A {1,,m},A ( 1) A +1 ( j A Thus, I = A {1,,m},A ( 1) A +1 ( j A α j) p 1 The following proposition shows that the ideal in Lemma 22 is the only maximal ideal in B k Proposition 25 An ideal I in B k is maximal if and only if I = w 1, w 2,, w k Proof ( =) It is clear by Lemma 22 (= ) Let J be a maximal ideal in B k By Proposition 24, B k is a principal ideal ring Then, let J = ω, for some ω B k Note that, ω is not a unit in B k, so it is a zero divisor By Lemma 21, ω is an element of some m i = w 1, w 2,, w k, which means J m i Consequently, J = m i, because J is a maximal ideal 133

4 Using the above result, we have the following lemmas Lemma 26 The ring B k can be viewed as an F p -vector space with dimension 2 k whose basis consists of elements of the form w = i w i, where 2 Ω Proof As we can see, every element a B k can be written as a = 2 α v Ω, for some α F p, where v = i v i and v = 1 o, B k is a vector space over F p whose basis consists of elements of the form v = i v i, where v = 1 and there are k ( k j=0 j) = 2 k elements of basis Now, we will show that the set {1, w 2,, w 2 } is also a basis Consider, k for some α i F p, for all i = 1,, 2 k, which gives, α 1 + α 2 w α 2 kw 2 k = 0 α 1 = α 2 w α 2 kw 2 k If α 1 0, then ξ 1 = ( α 2 w α 2 kw 2 k ) is a unit, a contradiction to the fact that ξ1 w 1,, w k o, α 1 = 0, which means, ( α 2 w α k+1 w k+1 ) = αk+2 w k α 2 kw 2 k If ( ) α 2 w α k+1 w k+1 0, then it is a contradiction to the fact that j 2, for all j = k + 2,, 2 k Consequently, ( ) α 2 w α k+1 w k+1 = 0 We have to note that, the set with elements of the w, where 2 Ω, is also linearly independent over F p, because k is a vector space over F p with element of basis are of the form v, where Ω Therefore, ( ) α 2 w α k+1 w k+1 = 0 gives α 2 = = α k+1 = 0 By continuing this process, we have α 1 = = α 2 k = 0, which means they are linearly independent over F p Lemma 27 The ring B k has characteristic p and cardinality p 2k Proof It is immediate since characteristic of F p is p, and B k can be viewed as a F p -vector space with dimension k ( k i=0 i) = 2 k o, B k = p 2k The following theorem characterizes the shape of automorphisms in the ring B k Theorem 28 Let θ be an endomorphism in B k Then, θ is an automorphism if and only if θ(v i ) = w j, for every i Ω, and θ, when restricted to F p, is an identity map Proof (= ) Let J = v 1,, v k and J θ = θ(v 1 ),, θ(v k ) Consider the map λ : B k B k J J θ a + J θ(a) + J θ We can see that the map λ is a ring homomorphism For any a, b B k /J where λ(a) = λ(b), let a = a 1 + J and b = b 1 + J for some a 1, b 1 B k As we can see, θ(a 1 b 1 ) J θ, so a 1 b 1 J Consequently, a b = 0 + J, which means a = b, in other words, λ is a monomorphism Moreover, for any a B k /J θ, let a = a 2 + J θ for some a 2 B k, then there exists a = θ 1 (a 2 ) + J such that λ(a) = a Therefore, F p B k /J B k /J θ, which implies J θ is also a maximal ideal By Proposition 25, J θ = w 1,, w k, where w i {v i, 1 v i } for 1 i k By Proposition 24, J θ = w j ) p 1 = θ(v j )) p 1 A Ω,A ( 1) A +1 ( j A A Ω,A ( 1) A +1 ( j A which means, A Ω,A ( 1) A +1 ( j A w j) p 1 and A Ω,A ( 1) A +1 ( j A θ(v j)) p 1 are associate Therefore, θ(v i ) = βw j for some unit β which satisfies ( β A ) p 1 = β, for all A Consequently, we 134

5 have β p 1 = β, but by Lemma 23, β p = β ince β is a unit, we have that β p 1 = 1 Therefore, β must be equal to 1 Moreover, since θ is an automorphism, θ(v i ) θ(v j ) whenever i j Also, since the only automorphism in F p is identity map, we have the conclusion ( =) uppose that θ(v i ) = w j, and θ(v i ) θ(v j ) whenever i j By Lemma 26, we can see that θ is also an automorphism Now, we have to note that every element a in B k can be written as a = 2 Ω α w for some α F p, where w = i w i Define a map ϕ as follows ϕ : B k F 2k p a = 2 k i=1 α i w i ( 1 α, 2 α,, α 2 ) k We can show that this map ϕ is a bijection map Furthermore, this map can be extended n tuples of B k as follows ϕ : Bk n F n2k p (a 1,, a n ) (ϕ(a 1 ),, ϕ(a n )) ince ϕ is a bijection map, we also have ϕ is a bijection map We have to note that, the map ϕ is a permutation, based on the choice of subsets i 2 Ω, of Gray maps in [2] 3 Codes over the ring B k A subset C Bk n is called code over B k of length n If C is a B k -submodule of Bk n, then C called linear code The following proposition gives a characterization of B k -linear codes using the map ϕ Proposition 31 C is a linear code over B k if and only if there exist linear codes C 1,, C 2 k such that C = ϕ 1 (C 1,, C 2 k) over F p Proof (= ) ince ϕ is a bijection, there exist C 1,, C 2 k such that C = ϕ 1 (C 1,, C 2 k) Now, we only need to show that C i is a linear code over F p for all i = 1,, 2 k For any C i, let c 1 and c 2 be two codewords in C i For l = 1, 2, let c l = (α (l) 1,, α(l) n ), for some α (l) j in F p Consider c l = ϕ( 1 (0,, 0, λ l c l, 0, 0) ) = ϕ 1 (0,, 0, λ l α (l) 1, 0,, 0),, ϕ 1 (0,, 0, λ l α n (l), 0,, 0) ( ( = λ l α (l) 1 w l ) j {1,,k} l w l {j},, λ l α n (l) ( w l j {1,,k} l w l {j} )), for any λ l in F p for all l = 1, 2 The last equality holds since ϕ α (l) t w l j {1,,k} l w l {j} = (0,, 0, α (l) t, 0,, 0) for all 1 t n ince C = ϕ 1 (C 1,, C 2 k), we have c l is in C for all l = 1, 2, and c 1 + c 2 is also in C Then, consider 135

6 ϕ(c 1 + c 2) = λ 1 α (1) 1 + λ 2 α (2) 1 λ 1 α (1) l + λ 2 α (2) l λ 1 α n (1) + λ 2 α n (2) Hence, λ 1 c 1 + λ 2 c 2 is also in C i ( =) Take any two codewords c 3 and c 4 in C Let ( c 3 = α (1) w,, ) α (n) w 2 Ω 2 Ω and ( c 4 = β (1) w,, ) β (n) w, 2 Ω 2 Ω for some α i, β i in F p, where i = 1,, 2 k For any λ 3 and λ 4 in F p we have ϕ(λ 3 c 3 + λ 4 c 4 ) = λ 3 α (1) 1 + λ 4 β (1) 1 λ 3 α (n) 1 + λ 4 β (n) 1 + λ 4 2 β (1) λ 3 2 α (n) + λ 4 λ 3 2 α (1) λ 3 2 k α(1) + λ 4 2k β(1) λ 3 2 k α(n) + λ 4 2 β (n) 2k β(n) is also in (C 1,, C 2 k), since C i is a linear code for every i = 1,, 2 k Therefore, λ 3 c 3 + λ 4 c 4 is also in C Now, following [5], we define permutation equivalence of codes as follows Definition 32 Two codes are permutation equivalent if one can be obtained from the other by permuting the coordinates Using Definition 32, we can define the following notion of equivalence between two codes Definition 33 Two codes C and C over B k are equivalent if either they are permutation-equivalent or C is permutation equivalent to the code θ(c ) for some automorphism θ in B k, ie the code θ(c ) obtained from C by changing α with θ(α) in all coordinates Note that, the above definition is similar to the one in [5] Now, let Π θ be a permutation on 2 k tuples of F p induced by automorphism θ Then we have for any c Bk n Then, we have the following characterization (Π θ ϕ) (c) = ϕ(θ(c)) (1) Theorem 34 Let C and C be two codes over B k Then, C and C are equivalent if and only if there exists a permutation which sends (C 1,, C 2 k) to (C 1,, C 2 k ) or to (Π θ (C 1),, Π θ (C 2 k )) 136

7 Proof (= ) Let C = ϕ 1 (C 1,, C 2 k) and C = ϕ 1 (C 1,, C 2 ), where C k i and C i are codes over F p, for all 1 i 2 k If there exists an automorphism θ such that C is permutation equivalent to θ(c ), then by equation 1, we have C = ϕ 1 (C 1,, C 2 k) is permutation equivalent to ( Π θ (C 1),, Π θ (C 2 ) ) k ( =) If there exists a permutation which sends (C 1,, C 2 k) to (Π θ (C 1),, Π θ (C 2 k)), for some bijective map Π θ, then we can have the automorphism θ using the equation 1 4 Invariant ring In this section, we describe some aspect of Euclidean self-dual codes as well as MacWiliams identity and invariant ring Related to Euclidean self-dual codes over the ring B k, we have the following result Proposition 41 Let C = ϕ 1 (C 1, C 2,, C 2 k), for some p-ary codes C 1,, C 2 k Then, C is Eulidean self-dual codes over B k if and only if C i is also Euclidean self-dual codes, for 1 i 2 k Proof (= ) For any c i C i, let c i = (α (0) i,, α (n 1) i ), for some α (j) i F p, where 0 j n 1 Let c = ϕ 1 (0,, 0, c i, 0,, 0) C, then we have c, c = 0 for every c C To make the representation for any element in the ring B k easier, we will use the basis whose elements are of the form v, for all {1, 2,, k} Now, let c = β (0) i v i +,, β (n 1) i v i + v Consider, β (0) 2 Ω, i β (n 1) 2 Ω, i c = ϕ( 1 (0,, 0, c i, 0,, 0) = α (0) i (v i j {1,,k} i v i {j}),, α (n 1) (v i ) j {1,,k} i v i {j}) ince c, c = 0 for every c C and v 2 = v for every 2 Ω, we have n 1 α (j) i β (j) i v i α (j) i β (j) v i {j} i {j} = 0 j=0 j {1,,k} i i Consequently, n 1 j=0 α(j) i β (j) i = 0 Take any c i C i Let c i = (γ(0) i,, γ (n 1) i ), for some γ (j) i c = ϕ 1 (0,, 0, c i, 0,, 0) C, we have c, c = 0 o F p, where 0 j n 1 ince Therefore C i C i n 1 c i, c i = α (j) i γ (j) i = 0 j=0 For any c 1 Ci, let c 1 = (ζ 0,, ζ n 1 ) for some ζ j F p, where 0 j n 1 ince c 1, c i = 0, we have n 1 j=0 ζ jα (j) i = 0 We can see that c 1 = ϕ( 1 (0,, 0, c 1, 0,, 0) = ζ 0 (v i j {1,,k} i v i {j}),, ζ n 1 (v i ) j {1,,k} i v i {j}) 137

8 Now, since n 1 j=0 ζ jα (j) i = 0, we also have c 1, c 2 = 0 for every c 2 C Remember that C = C, which gives c 1 C o, c 1 C i, or in other words Ci C i Thus, C i is a Euclidean self-dual code, for all i = 1,, 2 k ( =) Take any c 1, c 2 C For every i = 1, 2, let c i = c (i,0),, for some c (i,j) {1,,k} c (i,n 1) {1,,k} F p, where i = 1, 2, and j = 0,, n 1 Consider,, ϕ(c i ) = ( c(i,0) 1,, 1 c (i,n 1),, l c (i,0),, l c (i,n 1),, c(i,0) 2 k,, ) c(i,n 1) 2 k, where i = 1, 2 ince C l is a Euclidean self-dual code, for all l = 1,, 2 k, we have o, C C c 1, c 2 = n 1 j=0 = 0 l 2 Ω l c (1,j) Now, take any c 3 C ince c 3, c = 0 for all c C, we have n 1 j=0 c (1,j) l c (2,j) v = 0, c (2,j) v for all 2 Ω Remember that C l is a Euclidean self-dual code, for all l = 1, 2,, 2 k, which give n 1 j=0 c (1,j) l c (2,j) v = 0, for all 2 Ω, and moreover c 3 C o, C C Therefore, C is a Euclidean self-dual code The following lemma gives MacWilliams identity for codes over the ring B k Lemma 42 The MacWilliams identity for Hamming weight enumerators for codes over B k is : W C (X, Y ) = 1 C W C(X + (p 2 k 1)Y, X Y ) (2) Proof The identity follows from [7, Theorem 83] and Proposition 41 As we can see from Lemma 42, MacWilliams identity gives a transformation between polynomial representing a code and polynomial representing its corresponding dual code We have to note that if C is an Euclidean self-dual code, then the weight enumerator of C is invariant under this transformation The above transformation can be formulated as an action by a matrix group G generated by matrices T = 1 p 2 k 1 ( ) ( ) p 2k 1 p 2k a1 a and D = The action of any g = 2 G to a polynomial 0 1 a 3 a 4 1 p 2k 1 1 p 2k 1 f(x, Y ) is written as g f(x, Y ) = f(a 1 X + a 2 Y, a 3 X + a 4 Y ) 138

9 Note that the matrix T is derived from the identity in Lemma 42 and the matrix D is derived from the condition that n is always even Also, it is easy to see that G = {I, D, T, T } Formally, we have the following result Lemma 43 If W C (X, Y ) is a Hamming weight enumerator for an Euclidean self-dual code C over B k, then W C (X, Y ) is invariant under the action of G Let R G be a set of all polynomials in two variables which are invariant under the action of G We can easily prove that R G is a ring, and by the above Lemma we can see that every Hamming weight enumerator of Euclidean self-dual codes must be inside R G This ring R G called invariant ring for Euclidean self-dual codes over B k The following theorem gives generators for R G Theorem 44 Invariant ring of G is generated by W C0 (x, y) = x 2 + (p 2k 1)y 2 and Proof f(x, y) = 1 4 2p2k 1 Consider the Molien series, Φ(λ) = 1 G ( = x 2 + p 2k A G 1 det(i λa) ( ) 4 p 2k 1 1 (1+λ) (1 λ) (1 λ 2 ) xy + 2(p2 k 1) 2 y 2 p 2k 1 p 2k 1 = 1 (1 λ 2 ) 2 = 1 + 2λ 2 + 3λ 4 + 4λ 6 + 5λ nλ 2(n 1) + ) we can see that, the invariant ring generated by two invariants of degree 2 Consider the weight enumerator for self-dual code C 0 = {cc c A k } ie W C0 (x, y) = x 2 + (p 2k 1)y 2 This weight enumerator is of degree 2 and invariant under the action of G o, this weight enumerator is one of the generator We use averaging method to find the other one Let f(x) = x 2, then by averaging method, we have ( ) f(x, y) = 1 2p2k p 2k 1 x 2 + xy + 2(p2 k 1) 2 y 2 4 p 2k p 2k 1 p 2k 1 f(x, y) are algebraically independent References [1] T Abualrub, N Aydin, P eneviratne, On θ cyclic codes over F 2 + vf 2, Australas J Combin 54 (2012) [2] Y Cengellenmis, A Dertli, T Dougherty, Codes over an infinite family of rings with a Gray map, Des Codes Cryptogr 72(3) (2014) [3] J Gao, kew cyclic codes over F p + vf p, J Appl Math Inform 31(3 4) (2013)

10 [4] AR Hammons, P V Kumar, A R Calderbank, N J A loane, P ole, The Z 4 linearity of Kerdock, Preparata, Goethals and related codes, IEEE Trans Inform Theory 40(2) (1994) [5] W Huffman, V Pless, Fundamentals of Error Correcting Codes, Cambridge University Press, 2003 [6] Irwansyah, I Muchtadi Alamsyah, A Muchlis, A Barra, D uprijanto, Construction of θ cyclic codes over an algebra of order 4, Proceeding of the Third International Conference on Computation for cience and Technology (ICCT 3), Atlantis Press, 2015 [7] J Wood, Duality for modules over finite rings and applications to coding theory, Amer J Math 121(3) (1999)