A class of p -ary cyclic codes and their weight enumerators
aa r X i v : . [ c s . I T ] J u l A class of p -ary cyclic codes and their weight enumerators Long Yu, Hongwei Liu ∗ School of Mathematics and Statistics, Central China Normal University,Wuhan, Hubei 430079, China
Abstract
Let m , k be positive integers such that m gcd( m,k ) ≥ p be an odd prime and π be aprimitive element of F p m . Let h ( x ) and h ( x ) be the minimal polynomials of − π − and π − pk +12 over F p , respectively. In the case of odd m gcd( m,k ) , when k is even, gcd( m, k ) is oddor when k gcd( m,k ) is odd, Zhou et al. in [25] obtained the weight distribution of a class ofcyclic codes C over F p with parity-check polynomial h ( x ) h ( x ). In this paper, we furtherinvestigate this class of cyclic codes C over F p in the rest case of odd m gcd( m,k ) and the caseof even m gcd( m,k ) . Moreover, we determine the weight distribution of cyclic codes C . Key Words cyclic code, exponential sum, quadratic form, weight distribution.
Mathematics Subject Classification · Let p be an odd prime and q be a power of p . An [ n, k, d ] linear code over the finite field F q isa k -dimensional subspace of F nq with minimum Hamming distance d . Let A i denote the numberof codewords with Hamming weight i in a linear code C of length n . The weight enumerator of C is defined by A + A X + A X + · · · + A n X n , where A = 1 . The sequence ( A , A , · · · , A n ) is called the weight distribution of the code C .A linear code C of length n is called cyclic if ( c , c , · · · , c n − ) ∈ C implies ( c n − , c , · · · , c n − ) ∈C . By identifying a codeword ( c , c , · · · , c n − ) ∈ C with the polynomial c + c X + · · · + c n − X n − ∈ F q [ X ] / ( X n − , a cyclic code C of length n over F q corresponds to an ideal of F q [ X ] / ( X n − g ( X ) of this ideal is called the generator polynomial of C , which satisfies that g ( X ) | ( X n − ∗ Corresponding author.Email addresses: [email protected] (Hongwei Liu), [email protected] (Long Yu). h ( X ) = ( X n − /g ( X ) is referred to as the parity-check polynomial of C [13].In general the weight distribution of cyclic codes are difficult to be determined and they areknown only for a few special classes. There are some results on the weight distribution of cycliccodes whose duals have two or more zeros (see [2, 3, 5, 6, 10, 11, 14–26], and the referencestherein).The following notations are fixed throughout this paper. • Let p be an odd prime, m , k be positive integers, d = gcd( m, k ), s = md ≥ q = p d and q ∗ = ( − q − q . • Let F p m be the finite field and F ∗ p m = F q \ { } , π be a primitive element of F p m . For agiven divisor l of m , the trace function from F p m to F p l is defined by Tr ml ( x ) = P ml − i =0 x p li . • Let v ( j ) denote the 2-adic valuation of integer j (i.e., the maximal power of 2 dividing j ). • Let SQ denote the set of square elements in F ∗ p m , SQ p denote the set of square elementsin F ∗ p , u p be a primitive element in F p , and ζ p be a primitive p -th unity root.Let h ( x ) and h ( x ) be the minimal polynomials of − π − and π − pk +12 over F p , respectively.It is easy to check that h ( x ) and h ( x ) are polynomials of degree m and are pairwise distinct.In this paper, we let C be a cyclic code with parity-check polynomial h ( x ) h ( x ), and thendim F p C = 2 m . By the well-known Delsarte’s Theorem [1], cyclic code C can be expressed as C = ( c ( α, β ) = (cid:18) Tr m ( απ pk +12 i + β ( − π ) i ) (cid:19) p m − i =0 | α, β ∈ F p m ) . (1.1)Recently, Zhou et al. in [25] studied this class of cyclic codes C in the case of odd s , when k iseven, d is odd or when kd is odd. They showed cyclic codes C have only three nonzero weights.In this paper, we further investigate this class of cyclic codes C for the rest cases, and determinethe weight distribution of C .The aim of this paper is to determine the weight distribution of a class of cyclic codes C over F p defined by (1.1). To this end, by using the value distribution of the exponentialsum P x ∈ F pm ζ Tr m ( αx pk +1 + βx ) p , α, β ∈ F p m (see Theorem 1[11]) and investigating some overdeter-minate equations over finite fields, we obtain the value distribution of the exponential sum P x ∈ F pm ( ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p ) , α, β ∈ F p m . Applying these results, the weightdistribution of cyclic codes C over F p defined by (1.1) is obtained.This paper is organized as follows. Section gives some preliminary results. In Section ,we give the weight distribution of a class of cyclic codes C defined by (1.1) over F p . In the following, we give a brief introduction to the theory of quadratic forms over finitefields, which is needed to calculate the weight distribution of cyclic codes C in the next section.2 efinition 2.1. Let { ω , ω , · · · , ω s } be a basis for F q s over F q and x = P si =1 x i ω i , where x i ∈ F q . A function f ( x ) from F q s to F q is called a quadratic form if it can be represented as f ( x ) = f ( s X i =1 x i ω i ) = X ≤ i ≤ j ≤ s a ij x i x j , a ij ∈ F q . The rank of the quadratic form f ( x ) is defined as the codimension of the F q -vector space V = { x ∈ F q s | f ( x + z ) − f ( x ) − f ( z ) = 0 , for all z ∈ F q s } , denotes by rank r . Then | V | = q s − r .For a quadratic form f ( x ) with s variables over F q , there exists a symmetric matrix A oforder s over F q such that f ( x ) = XAX ′ , where X = ( x , x , · · · , x s ) ∈ F sq and X ′ denotesthe transpose of X . It is known that there exists a nonsingular matrix T over F q such that T AT ′ is a diagonal matrix [8]. Making a nonsingular linear substitution X = ZT with Z =( z , z , · · · , z s ) ∈ F sq , we have f ( x ) = Z ( T AT ′ ) Z ′ = r X i =1 a i z i , a i ∈ F ∗ q , (2.1)where r is the rank of f ( x ). We have the following result (see Lemma 1 [11]). Lemma 2.2. [11]
Let η d be the quadratic (multiplicative) character of F p d . For the quadraticform f ( x ) defined by (2.1), we have X x ∈ F sq ζ Tr d ( f ( x )) p = η d ( a · · · a r ) q s − r , if q ≡ ; ( √− r η d ( a · · · a r ) q s − r , if q ≡ . By Definition 2 .
1, we have that f α,β ( x ) = Tr md ( αx p k +1 + βx ) is a quadratic form over F q .Let r α,β denote the rank of f α,β ( x ). Therefore, r α,β is determined by the codimension of V = { x ∈ F q s | f α,β ( x + z ) − f α,β ( x ) − f α,β ( z ) = 0 , for all z ∈ F q s } , which is determined by the number of solutions of φ α,β ( x ) = 0 , where φ α,β ( x ) = α p k x p k + 2 β p k x p k + αx. (2.2)Hence, we have the following result (see Lemma 2 [11]). Lemma 2.3. [11]
For ( α, β ) ∈ F p m \ { (0 , } , r α,β is s , s − or s − . Furthermore, let n i bethe number of f α,β ( x ) with r α,β = s − i for ( α, β ) ∈ F p m \ { (0 , } and i = 0 , , , then n = ( p m − p m − − p d − , n = p m − d ( p m − , n = p m − − n − n . We define T ( α, β ) = X x ∈ F pm ζ Tr m ( αx pk +1 + βx ) p , α, β ∈ F p m . (2.3)3 emma 2.4. [11] With the notations given above. The value distribution of the exponentialsum T ( α, β ) defined by (2.3) is shown in the following:i) for the case s is odd,value frequency ±√ q ∗ q s − , p d ( p m − p m − d − p m − d + 1)( p m − / ( p d − p m + d p m − d ( p m − d + 1)( p m − − p m + d p m − d ( p m − d − p m − ±√ q ∗ q s +12 ( p m − d − p m − / ( p d − p m ii) for the case s is even,value frequency p m p d ( p m − p m − d − p m − d + p m − p m − d + 1)( p m − / ( p d − − p m p d ( p m − p m − d − p m − d − p m + p m − d + 1)( p m − / ( p d − ±√ q ∗ q s p m − d ( p m − p m + d ( p m − p m − d + 1)( p m − / ( p d − − p m + d ( p m + 1)( p m − d − p m − / ( p d − p m ε = ± i = 0 , , N i = { ( α, β ) ∈ F q | r α,β = s − i } ,N ε,i = ( { ( α, β ) ∈ F p m \ { (0 , } | T ( α, β ) = εq s + i } , if s + i is even; { ( α, β ) ∈ F p m \ { (0 , } | T ( α, β ) = ε √ q ∗ q s + i − } , if s + i is odd,and n ε,i = | N ε,i | . Then, N i = N ,i ∪ N − ,i and n i = n ,i + n − ,i . By (2.2), we define ψ ( α, x ) = − x − ( αx p k + α p m − k x p m − k ) . (2.4)Hence, β = ψ ( α, x ) if and only if φ α,β ( x ) = 0 with x ∈ F ∗ q . Therefore, it is easy to see that r α,β = s − s − α ∈ F ∗ p m and there exists x ∈ F ∗ p m such that β = ψ ( α, x ).Furthermore, from Lemma 2.3, we obtain the number of solutions of β = ψ ( α, x ) with x ∈ F ∗ p m is { x ∈ F ∗ p m | β = ψ ( α, x ) } = , iff ( α, β ) ∈ N ; p d − , iff ( α, β ) ∈ N ; p d − , iff ( α, β ) ∈ N . (2.5)At the end of this section, we investigate two classes of overdeterminate equations over finitefields. Lemma 2.5.
Let E denote the number of solutions ( x, y ) ∈ F p m of the following systemequations (cid:26) x + y = 0 ,x p k +1 + y p k +1 = 0 . (2.6) i) If ≤ v ( m ) < v ( k ) , then E = 2 p m − .ii) If v ( k ) < v ( m ) , then E = 2 p m − for p k ≡ , E = 1 for p k ≡ . roof. By the first equation of (2.6), we have − x = y . Then x p k +1 + y p k +1 = x p k +1 + ( − x ) pk +12 = (cid:26) , if p k ≡ x p k +1 , if p k ≡ ≤ v ( m ) < v ( k ), then m , k are both even, which implies that p m ≡ p k ≡ E is equal to the numberof solutions of equation − x = y , i.e., x = ± π pm − y . Then E = 2 p m − v ( k ) < v ( m ), we discuss it in two cases.Case I, when p k ≡ E is equal to the number of solutions of equation − x = y by (2.6) and (2.7). Note that m is even, then 4 | p m −
1. So, E = 2 p m − p k ≡ x p k +1 + y p k +1 = 2 x p k +1 = 0, then x = 0and y = 0. Hence, E = 1 . ✷ Lemma 2.6.
Let E denote the number of solutions ( x, y, z ) ∈ F p m of the following systemequations ( x + y − πz = 0 ,x p k +1 + y p k +1 + π pk +12 z p k +1 = 0 . (2.8) If v ( k ) < v ( m ) , then E = 2 p m − for p k ≡ , E = 1 for p k ≡ . Proof.
We distinguish between the following two cases to calculate the number of solutions( x, y, z ) ∈ F p m of (2.8).Case I, when z = 0: If z = 0, by Lemma 2.5 ii), we get that (2.8) has 2 p m − p k ≡ p k ≡ z = 0: It is easy to check that x = 0 and y = 0. In this case, one has that(2.8) has ( p m − M solutions, where M is the number of solutions of the following systemequations ( x + y = π,x p k +1 + y p k +1 = − π pk +12 . (2.9)By (2.9), we have ( x + y ) p k +1 = π p k +1 and ( x p k +1 + y p k +1 ) = π p k +1 . Combining these twoequations (the first one minus the second one) leads to( xy p k − x p k y ) = 0 , which shows that x = λy , where λ ∈ F ∗ p d (since gcd( p k − , p m −
1) = p d − x = λy into (2.9), we get ( ( λ + 1) y = π, ( λ + 1) y p k +1 = − π pk +12 . (2.10)Eliminating λ + 1 of (2.10), one has y p k − = − π pk − = π pk − + pm − , which induces that p d − | p k − + p m − . Since v ( k ) < v ( m ), then s = m/d is even. Hence, p d − | p m − . From v ( k ) < v ( m ), we have k/d is odd and then p d − ∤ p k − . It is acontradiction with p d − | p k − + p m − . So the number of solutions of (2.10) is 0, i.e., M = 0.Therefore, E = 2 p m − p k ≡ E = 1 if p k ≡ ✷ The weight distribution of C In this section, we first calculate the weight of the codeword c ( α, β ) ∈ C defined by (1.1), wt ( c ( α, β )) = { ≤ i ≤ p m − c i = 0 } = p m − − p p m − X i =0 X u ∈ F p ζ u Tr m ( απ pk +12 i + β ( − π ) i ) p = p m − − p X u ∈ F p pm − X i =0 (cid:18) ζ u Tr m ( απ pk +12 2 i + βπ i ) p + ζ u Tr m ( απ pk +12 (2 i +1) − βπ (2 i +1) ) p (cid:19) = p m − − p X u ∈ F p X x ∈ SQ (cid:18) ζ u Tr m ( αx pk +12 + βx ) p + ζ u Tr m ( απ pk +12 x pk +12 − βπx ) p (cid:19) = p m − − p X u ∈ F p X x ∈ F ∗ pm (cid:18) ζ u Tr m ( αx pk +1 + βx ) p + ζ u Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) = p m − p m − − p X u ∈ F ∗ p S ( uα, uβ ) , (3.1)where S ( α, β ) = X x ∈ F pm (cid:18) ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) = T ( α, β ) + T ( π pk +12 α, − πβ ) , (3.2)where T ( α, β ) is defined by (2.3).From (3.1), the weight distribution of C is completely determined by the value distributionof S ( α, β ). To obtain the value distribution of S ( α, β ), we need a series of lemmas. Beforeintroducing them, we define f ( x ) = Tr md ( αx p k +1 + βx ) and g ( x ) = Tr md ( π pk +12 αx p k +1 − πβx ).Let r f and r g be the rank of f ( x ) and g ( x ), respectively. Then, we have the following result. Lemma 3.1.
For ( α, β ) ∈ F p m \ { (0 , } , when ≤ v ( m ) < v ( k ) or v ( k ) < v ( m ) , we havethat at least one of r f and f g is s . Proof.
Assume that r f and f g are both less than s . By (2.2), we have that there exists x , x ∈ F ∗ p m such that φ α,β ( x ) = α p k x p k + 2 β p k x p k + αx = 0 , (3.3)and φ π ( pk +1) / α, − πβ ( x ) = ( π pk +12 α ) p k x p k − πβ ) p k x p k + π pk +12 αx = 0 . (3.4)If α = 0, we have x = x = 0, which is a contradiction with x , x ∈ F ∗ p m . In the following, weassume that α = 0. Simplifying (3.3) × ( πx ) p k +(3.4) × x p k , one has − π pk − (cid:18) αx x ( x p k − + π pk − x p k − ) (cid:19) p k = αx x ( x p k − + π pk − x p k − ) , (3.5)6hich implies that x p k − + π pk − x p k − = 0or (cid:18) αx x ( x p k − + π pk − x p k − ) (cid:19) p k − = − π − pk − . If one of the above two equations holds, we have that p d − | p m − − p k − .Case I, when 1 ≤ v ( m ) < v ( k ): In this case, s = m/d is odd and we get that p d − ∤ p m − .From 1 ≤ v ( m ) < v ( k ), we have that k/d is even. This implies that p d − | p k − and then itis a contradiction with p d − | p m − − p k − .Case II, v ( k ) < v ( m ): Similarly, we can prove that p d − ∤ p m − − p k − , which is also acontradiction with p d − | p m − − p k − .Therefore, we have that at least one of r f and f g is s . ✷ C for ≤ v ( m ) < v ( k ) In this subsection, we always assume that 1 ≤ v ( m ) < v ( k ). To determine the valuedistribution of S ( α, β ), we need the following lemma. Lemma 3.2.
With the notations given above, we have the following result:1. P α,β ∈ F pm S ( α, β ) = 4 p m . ( p d − P ( α,β ) ∈ N S ( α, β ) +( p d − P ( α,β ) ∈ N S ( α, β ) = p m ( p m − p m + d − p m +2 p d − p d − . Proof.
1. By (3.2), we get that X α,β ∈ F pm S ( α, β ) = X α,β ∈ F pm (cid:18) T ( α, β ) + 2 T ( α, β ) T ( π pk +12 α, − πβ ) + T ( π pk +12 α, − πβ ) (cid:19) . (3.6)On one hand, by (2.3), we have X α,β ∈ F pm T ( α, β ) = X x,y ∈ F pm X α ∈ F pm ζ Tr m ( α ( x pk +1 + y pk +1 )) p X β ∈ F pm ζ Tr m ( α ( x + y )) p = M p m , (3.7)where M = { ( x, y ) ∈ F p m | x + y = 0 , x p k +1 + y p k +1 = 0 } . By Lemma 2.5 i) and (3.7), X α,β ∈ F pm T ( α, β ) = p m (2 p m − . Similarly, X α,β ∈ F pm T ( π pk +12 α, − πβ ) = p m (2 p m − .
7n the other hand, by (2.3), X α,β ∈ F pm T ( α, β ) T ( π pk +12 α, − πβ )= X x,y ∈ F pm X α ∈ F pm ζ Tr m ( α ( x pk +1 + π pk +12 y pk +1 )) p X β ∈ F pm ζ Tr m ( β ( x − πy )) p = { ( x, y ) ∈ F p m | x − πy = 0 , x p k +1 + π pk +12 y p k +1 = 0 } · p m = { ( x, y ) ∈ F p m | x − πy = 0 , π pk +12 y p k +1 = 0 } · p m = p m . Hence, from (3.6), X α,β ∈ F pm S ( α, β ) = 4 p m .
2. By (2.5) and (3.2), we have that( p d − X ( α,β ) ∈ N S ( α, β ) + ( p d − X ( α,β ) ∈ N S ( α, β ) = X x,α ∈ F ∗ pm S ( α, ψ ( α, x )) = X x,α ∈ F ∗ pm (cid:18) T ( α, ψ ( α, x )) + T ( π pk +12 α, − πψ ( α, x )) (cid:19) . (3.8)By (2.3) and (2.4), X x,α ∈ F ∗ pm T ( α, ψ ( α, x )) = X x,α ∈ F ∗ pm X y,z ∈ F pm ζ Tr m ( α ( y pk +1 + y pk +1 ) − ( αx pk − + α pm − k x pm − k − )( y + z )) p = X x,α ∈ F ∗ pm X y,z ∈ F pm ζ Tr m ( α ( y pk +1 + z pk +1 ) − αx pk − ( y + z ) − x − pk α ( y pk + z pk )) p = X x ∈ F ∗ pm ,y,z ∈ F pm X α ∈ F ∗ pm ζ Tr m ( − αx − pk (( x pk − y − y pk ) +( x pk − z − z pk ) )) p = ( p m − M − ( p m ( p m − − M )= p m M − p m + p m , (3.9)where M = { x ∈ F ∗ p m , y, z ∈ F p m | ( x p k − y − y p k ) + ( x p k − z − z p k ) = 0 } . For a fixed x ∈ F ∗ p m , we investigate the following equation( x p k − y − y p k ) = − ( x p k − z − z p k ) . (3.10)8ince m is even, then 4 | p m −
1. This shows that − π pm − is a square element in F p m .In the following, we discuss it case by case.Case I, when y = 0: In this case, we have x p k − z − z p k = 0, i.e., z = λx , where λ ∈ F p d .Hence, (3.10) has p d solutions.Case II, when z = 0 and y = 0: We get that y = tx , where t ∈ F ∗ p d , which implies that(3.10) has p d − yz = 0: Since k is even, then ( π pm − ) p k = π pm − . Together with − π pm − and by (3.10), we have y p k − yx p k − = ± π pm − ( z p k − zx p k − ) , which is equivalent to ( y − π pm − z ) p k = ( y − π pm − z ) x p k − (3.11)or ( y + π pm − z ) p k = ( y + π pm − z ) x p k − . (3.12)Since gcd( p k − , p m −
1) = p d − y − π pm − z = rx and y + π pm − z = r ′ x , respectively, where r, r ′ ∈ F p d . Hence, (3.11) and (3.12) have p m + d solutions,respectively. Note that the common solutions of (3.11) and (3.12) are p d . Therefore, the totalnumber of pairs ( y, z ) ∈ F p d satisfying (3.11) and (3.12) is 2 p m + d − p d . In which the numbersatisfying yz = 0 is 2 p d −
1. Then, when yz = 0, the total number of pairs ( y, z ) ∈ F p d satisfying(3.11) and (3.12) is 2 p m + d − p d − p d + 1.As a result, we have M = ( p d + p d − p m + d − p d − p d + 1)( p m −
1) = p d (2 p m − p d )( p m − . Similarly, by (2.3), X x,α ∈ F ∗ pm T ( α, ψ ( α, x )) T ( π pk +12 α, − πψ ( α, x ))= X x,α ∈ F ∗ pm X y,z ∈ F pm ζ Tr m ( α ( y pk +1 + π pk +12 z pk +1 ) − ( αx pk − + α pm − k x pm − k − )( y − πz )) p = X x ∈ F ∗ pm ,y,z ∈ F pm X α ∈ F ∗ pm ζ Tr m ( αx − pk ( − ( x pk − y − y pk ) + π ( x pk − z + π pk − z pk ) )) p = ( p m − M − ( p m ( p m − − M )= p m M − p m + p m , (3.13)where M = { x ∈ F ∗ p m , y, z ∈ F p m | ( x p k − y − y p k ) = π ( x p k − z + π pk − z p k ) } . For a given x ∈ F ∗ p m , we study the following equation( x p k − y − y p k ) = π ( x p k − z + π pk − z p k ) . (3.14)9ase I, when y = 0: In this case, we get that z = 0 or x p k − = − π pk − z p k − . Note that s is odd, then p d − ∤ p m − . Since kd is even, then p d − | p k − . Hence, there is no solution of x p k − = − π pk − z p k − for z ∈ F ∗ p d . In this case, (3.14) has only one solution.Case II, when z = 0 and y = 0: We have that x p k − y − y p k = 0, which is equivalent to x p k − = y p k − , i.e., y = tx , where t ∈ F ∗ p d . Hence, (3.14) has p d − yz = 0: From case I, we get that x p k − + π pk − z p k − = 0 for z = 0, whichimplies that x p k − y − y p k = 0 by (3.14). Then there is no solution of (3.14).Therefore, M = p d ( p m − X x,α ∈ F ∗ pm T ( π pk +12 α, − πψ ( α, x )) = X x,α ∈ F ∗ pm X y,z ∈ F pm ζ Tr m ( απ pk +12 ( y pk +1 + z pk +1 )+ π ( αx pk − + α pm − k x pm − k − )( y + z )) p = X x ∈ F ∗ pm ,y,z ∈ F pm X α ∈ F ∗ pm ζ Tr m ( παx − pk (( x pk − y + π pk − y pk ) +( x pk − z + π pk − z pk ) )) p = ( p m − M − ( p m ( p m − − M )= p m M − p m + p m , (3.15)where M = { x ∈ F ∗ p m , y, z ∈ F p m | ( x p k − y + π pk − y p k ) + ( x p k − z + π pk − z p k ) = 0 } . For a given x ∈ F ∗ p m , we investigate the following equation( x p k − y + π pk − y p k ) = − ( x p k − z + π pk − z p k ) , which is equivalent to − π pk − ( y − π pm − z ) p k = ( y − π pm − z ) x p k − (3.16)or − π pk − ( y + π pm − z ) p k = ( y + π pm − z ) x p k − . (3.17)Case I, when y − π pm − z = 0 and y + π pm − z = 0: In this case, y = z = 0. (3.16) has onesolutions and (3.17) has one solution, respectively.Case II, when y − π pm − z = 0 and y + π pm − z = 0: (3.17) is equivalent to − π pk − ( y + π pm − z ) p k − = x p k − . Note that m/d is odd and k/d is even, then p d − ∤ p m − and p d − | p k − .This implies that there is no solution of − π pk − ( y + π pm − z ) p k − = x p k − . Hence, (3.16) has p m − y − π pm − z = 0 and y + π pm − z = 0: We can discuss it by a similar way inthe case II. Hence, (3.16) has no solution and (3.17) has p m − y − π pm − z )( y + π pm − z ) = 0: Similarly, we have that (3.16) and (3.17)have no solution, respectively.So, we get that M = (2 p m − p m − ✷ In the following, we define N ε,µ,i,j = { ( α, β ) | ( α, β ) ∈ N ε,i , ( π pk +12 α, − πβ ) ∈ N µ,j } and n ε,µ,i,j = | N ε,µ,i,j | for ε, µ = ±
1, 0 ≤ i, j ≤ ij = 0, then n ε,µ,i,j = 0. Moreover, it is easy to checkthat n ε,µ,i,j = n µ,ε,j,i by the symmetry of T ( α, β ) and T ( π pk +12 α, − πβ ). Hence, we need onlyto calculate n ε,µ,i, . For convenience, we let n ε,µ,i, = n ε,µ,i . With above preparation we candetermine the value distribution of the exponential sum S ( α, β ). Theorem 3.3.
With the notations given above. Then the value distribution of S ( α, β ) definedby (3.2) is as follows:value frequency p m ( p m + d − p m + p d + 1)( p m − / ( p d − ± ( p d − p m ( p m − d − p m − / ( p d − ± − p d ) p m p m − d ( p m − d − p m − ± p d ) p m p m − d ( p m − d + 1)( p m − ± p m ( p d − p m − / ( p d + 1) Proof.
From Lemma 3.1 and Lemma 2.4, we get that (cid:26) n , = n , , + n − , , + n , , + n − , , + n , , + n − , , ,n − , = n , − , + n − , − , + n , − , + n − , − , + n , − , + n − , − , . (3.18)By (3.2) and Lemma 3.1, we have that n , = n , , + n , − , ,n − , = n − , , + n − , − , ,n , = n , , + n , − , ,n − , = n − , , + n − , − , . (3.19)Note that if the rank of Tr md ( αx p k +1 + βx ) is odd and we compute the value of T ( α, β ) and T ( aα, aβ ), where a is a nonsquare element in F q . By Lemma 2.2, we have T ( aα, aβ ) = − T ( α, β ) , s is odd, we have n , , = n − , − , ,n , − , = n − , , ,n , , = n , − , ,n − , , = n − , − , ,n , , = n − , − , ,n , − , = n − , , . (3.20)By Lemma 3.2 1), we get that4 p m = 4 p m + 4 p m ( n , , + n − , − , ) + 2( p d + 1) p m ( n , , + n − , − , )+2( p d − p m ( n , − , + n − , , ) + 2( p d + 1) p m ( n , , + n − , − , )+2( p d − p m ( n , − , + n − , , ) . (3.21)By Lemma 3.2 2), we have that p m ( p m − p m + d − p m + 2 p d − p d − p d − p m (( p d + 1) ( n , , + n − , − , ) + ( p d − ( n , − , + n − , , )+( p d + 1) ( n , , + n − , − , ) + ( p d + 1)( p d − ( n , − , + n − , , )) . (3.22)Solving the system of equations consisting of (3.18)-(3.22), we have n , , = n − , − , = 14 ( p d − p m − / ( p d + 1) ,n , − , = n − , , = 14 ( p m + d − p m + p d + 1)( p m − / ( p d − ,n − , , = n − , − , = 14 p m − d ( p m − d − p m − ,n , , = n , − , = 14 p m − d ( p m − d + 1)( p m − ,n , − , = n − , , = 12 ( p m − d − p m − / ( p d − ,n , , = n − , − , = 0 . We complete the proof. ✷ Therefore, we can give the weight distribution of cyclic code C . Theorem 3.4.
With the notations given above. Then C defined by (1.1) is a cyclic code over F p with length p m − and dimension m . Moreover, the weight distribution of cyclic code C isgiven in Table . Proof.
By (3.1), we have wt ( c ( α, β )) = p m − p m − − p X u ∈ F ∗ p S ( uα, uβ ) . C for 1 ≤ v ( m ) < v ( k )Weight Frequency0 1 p m − ( p −
1) + ( p − p d − p m − ( p m − d − p m − / ( p d − p m − ( p − − ( p − p d − p m − ( p m − d − p m − / ( p d − p m − ( p −
1) + ( p − p d − p m − p m − d ( p m − d − p m − p m − ( p −
1) + ( p − p d + 1) p m − p m − d ( p m − d − p m − p m − ( p − − ( p − p d + 1) p m − p m − d ( p m − d + 1)( p m − p m − ( p − − ( p − p d − p m − p m − d ( p m − d + 1)( p m − p m − ( p −
1) + p m − ( p − ( p d − p m − / ( p d + 1) p m − ( p − − p m − ( p − ( p d − p m − / ( p d + 1) p m − ( p − ( p m + d − p m + p d + 1)( p m − / ( p d − m is even, then u p = π pm − p − = π p m − + ··· +1 is a square element in F p m . For a given u ∈ F ∗ p , we have that u ∈ SQ . Since k is even, then u pk +12 = u . So we have S ( uα, uβ ) = X x ∈ F pm (cid:18) ζ Tr m ( uαx pk +1 + uβx ) p + ζ Tr m ( uαπ pk +12 x pk +1 − uβπx ) p (cid:19) = X x ∈ F pm (cid:18) ζ Tr m ( α ( u x ) pk +1 + β ( u x ) ) p + ζ Tr m ( απ pk +12 ( u x ) pk +1 − βπ ( u x ) ) p (cid:19) = X x ∈ F pm (cid:18) ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) = S ( α, β ) . Therefore, we obtain that wt ( c ( α, β )) = p m − p m − − p − p S ( α, β ) . By Theorem 3.3, we get the result. ✷ In the following, we give an example to verify our result in Theorem 3.4.
Example 3.5.
Let p = 3 , m = 6 , k = 4 , the code C is a [728 , , cyclic code over F withweight enumerator X + 32760 X + 139048 X + 199472 X + 132496 X + 26208 X + 728 X , which confirms the weight distribution in Table . C for v ( k ) < v ( m ) In this subsection, we always assume that v ( k ) < v ( m ). To determine the value distribu-tion of S ( α, β ), we need some identities on S ( α, β ).13 emma 3.6. With the notations given above, we have the following result:1. P α,β ∈ F pm S ( α, β ) = 2 p m . P α,β ∈ F pm S ( α, β ) = (cid:26) p m , if p k ≡ ; p m , if p k ≡ .3. P α,β ∈ F pm S ( α, β ) = (cid:26) p m (7 p m −
3) + 2 p m + d ( p m − , if p k ≡ ; p m ( p m + 3) + 2 p m + d ( p m − , if p k ≡ .4. ( p d − P ( α,β ) ∈ N S ( α, β ) + ( p d − P ( α,β ) ∈ N S ( α, β ) = p m ( p d − p m − . Proof.
1. We compute X α,β ∈ F pm S ( α, β ) = X α,β ∈ F pm (cid:18) T ( α, β ) + T ( π pk +12 α, − πβ ) (cid:19) = X α,β ∈ F pm T ( α, β ) + X α ′ ,β ′ ∈ F pm T ( α ′ , β ′ )= 2 X x ∈ F pm X α ∈ F pm ζ Tr m ( αx pk +1 ) p X β ∈ F pm ζ Tr m ( βx ) p = 2 p m .
2. By (3.2), we get that X α,β ∈ F pm S ( α, β ) = X α,β ∈ F pm (cid:18) T ( α, β ) + 2 T ( α, β ) T ( π pk +12 α, − πβ ) + T ( π pk +12 α, − πβ ) (cid:19) . (3.23)On one hand, X α,β ∈ F pm T ( α, β ) = X x,y ∈ F pm X α ∈ F pm ζ Tr m ( α ( x pk +1 + y pk +1 )) p X β ∈ F pm ζ Tr m ( α ( x + y )) p = M p m , (3.24)where M = { ( x, y ) ∈ F p m | x + y = 0 , x p k +1 + y p k +1 = 0 } . Together with Lemma 2.5 ii) and(3.24), we obtain that X α,β ∈ F pm T ( α, β ) = (cid:26) p m (2 p m − , if p k ≡ p m , if p k ≡ X α,β ∈ F pm T ( π pk +12 α, − πβ ) = (cid:26) p m (2 p m − , if p k ≡ p m , if p k ≡ X α,β ∈ F pm T ( α, β ) T ( π pk +12 α, − πβ )= X x,y ∈ F pm X α ∈ F pm ζ Tr m ( α ( x pk +1 + π pk +12 y pk +1 )) p X β ∈ F pm ζ Tr m ( β ( x − πy )) p = { ( x, y ) ∈ F p m | x − πy = 0 , x p k +1 + π pk +12 y p k +1 = 0 } · p m = { ( x, y ) ∈ F p m | x − πy = 0 , π pk +12 y p k +1 = 0 } · p m = p m . Hence, from (3.23), we get X α,β ∈ F pm S ( α, β ) = (cid:26) p m , if p k ≡ p m , if p k ≡ X α,β ∈ F pm S ( α, β ) = X α,β ∈ F pm ( T ( α, β ) + 3 T ( α, β ) T ( π pk +12 α, − πβ )+3 T ( α, β ) T ( π pk +12 α, − πβ ) + T ( π pk +12 α, − πβ ) ) . (3.25)It is easy to check that X α,β ∈ F pm T ( α, β ) = X α,β ∈ F pm T ( π pk +12 α, − πβ ) (3.26)and X α,β ∈ F pm T ( α, β ) T ( π pk +12 α, − πβ ) = X α,β ∈ F pm T ( α, β ) T ( π pk +12 α, − πβ ) . (3.27)By Lemma 2.4 ii), X α,β ∈ F pm T ( α, β ) = p m + p m + d ( p m − . (3.28)By Lemma 2.6, we have X α,β ∈ F pm T ( α, β ) T ( π pk +12 α, − πβ )= X x,y,z ∈ F pm X α ∈ F pm ζ Tr m ( α ( x pk +1 + y pk +1 + π pk +12 z pk +1 )) p X β ∈ F pm ζ Tr m ( β ( x + y − πz )) p = { ( x, y, z ) ∈ F p m | x + y − πz = 0 , x p k +1 + y p k +1 + π pk +12 z p k +1 ) = 0 } · p m = (cid:26) (2 p m − p m , if p k ≡ p m , if p k ≡ X α,β ∈ F pm S ( α, β ) = (cid:26) p m (7 p m −
3) + 2 p m + d ( p m − , if p k ≡ p m ( p m + 3) + 2 p m + d ( p m − , if p k ≡ p d − X ( α,β ) ∈ N S ( α, β ) + ( p d − X ( α,β ) ∈ N S ( α, β )= X x,α ∈ F ∗ pm (cid:18) T ( α, ψ ( α, x )) + T ( π pk +12 α, − πψ ( α, x )) (cid:19) . (3.30)By (2.4), X x,α ∈ F ∗ pm T ( α, ψ ( α, x )) = X x,α ∈ F ∗ pm X y ∈ F pm ζ Tr m ( αy pk +1 − x − ( αx pk + α pm − k x pm − k ) y ) p = X x,α ∈ F ∗ pm X y ∈ F pm ζ Tr m ( αy pk +1 − x − αx pk y − x − pk αxy pk ) p = X x ∈ F ∗ pm ,y ∈ F pm X α ∈ F ∗ pm ζ Tr m ( − α ( y x − pk ( x pk − − y pk − ) )) p = ( p m − M − ( p m ( p m − − M )= p m M − p m + p m , (3.31)where M = { x ∈ F ∗ p m , y ∈ F p m | y ( x p k − − y p k − ) = 0 } = { x ∈ F ∗ p m , y ∈ F p m | y = λx, where λ ∈ F p d } = ( p m − p d . (3.32)On the other hand, by (2.5), X x,α ∈ F ∗ pm T ( π pk +12 α, − πψ ( α, x )) = X x,α ∈ F ∗ pm X y ∈ F pm ζ Tr m ( π pk +12 αy pk +1 + π x − ( αx pk + α pm − k x pm − k ) y ) p = X x ∈ F ∗ pm ,y ∈ F pm X α ∈ F ∗ pm ζ Tr m ( πα ( y x − pk ( x pk − + π pk − y pk − ) )) p = ( p m − I − ( p m ( p m − − I )= p m I − p m + p m , (3.33)where I = { x ∈ F ∗ p m , y ∈ F p m | y ( x p k − + π pk − y p k − ) = 0 }
16 { x ∈ F ∗ p m , y ∈ F p m | y = 0 or ( xy ) p k − = − π pk − } . Since kd is odd and m is even, then p d − ∤ p k − + p m − , which implies that there is no solutionof ( xy ) p k − = − π pk − . Hence, I = p m −
1. Together with (3.30), (3.31) and (3.33), we have that( p d − X ( α,β ) ∈ N S ( α, β ) + ( p d − X ( α,β ) ∈ N S ( α, β ) = p m ( p d − p m − . ✷ Theorem 3.7.
With the notations given above. Then the value distribution of S ( α, β ) definedby (3.2) is as follows:value frequency p m ( p m + d − p m + p d + 1)( p m − / ( p d − ± ( p d − p m ( p m − d ± p m ∓ p m − / ( p d − ±√ q ∗ − p m p m − d ( p m − p m − ±√ q ∗ + 1) p m p m − d ( p m + 1)( p m − ± p m ( p m ± ( p d − p m − / ( p d + 1) Proof.
From Lemma 3.1 and Lemma 2.4, we get that (cid:26) n , = n , , + n − , , + n , , + n − , , + n , , + n − , , ,n − , = n , − , + n − , − , + n , − , + n − , − , + n , − , + n − , − , . (3.34)By (3.2) and Lemma 3.1, n , = n , , + n , − , ,n − , = n − , , + n − , − , ,n , = n , , + n , − , ,n − , = n − , , + n − , − , . (3.35)Note that if the rank of Tr md ( αx p k +1 + βx ) is odd and we compute the value of T ( α, β ) and T ( aα, aβ ), where a is a nonsquare element in F q . Then, by Lemma 2.2, we have T ( aα, aβ ) = − T ( α, β ) , and since s is even , we get (cid:26) n , , = n − , , ,n , − , = n − , − , . (3.36)On the other hand, we have P α,β ∈ F pm S ( α, β ) = 2 p m + 2 p m ( n , , − n − , − , + 2( √ q ∗ + 1)( n , , − n − , − , )+2( √ q ∗ − n , − , − n − , , ) + ( p d + 1)( n , , − n − , − , ) + ( p d − n , − , − n − , , )) , (3.37)17 α,β ∈ F pm S ( α, β ) = 4 p m + 2 p m (2( n , , + n − , − , ) + ( √ q ∗ + 1) ( n , , + n − , − , )+( √ q ∗ − ( n , − , − n − , , ) + ( p d + 1) ( n , , + n − , − , ) + ( p d − ( n , − , + n − , , )) , (3.38) P α,β ∈ F pm S ( α, β ) = 8 p m + 2 p m (4( n , , − n − , − , ) + ( √ q ∗ + 1) ( n , , − n − , − , )+( √ q ∗ − ( n , − , − n − , , ) + ( p d + 1) ( n , , − n − , − , ) + ( p d − ( n , − , − n − , , )) , (3.39)( p d − P ( α,β ) ∈ N S ( α, β ) + ( p d − P ( α,β ) ∈ N S ( α, β ) = ( p d − p m (( √ q ∗ + 1)( n , , − n − , − , )+( √ q ∗ − n , − , − n − , , ) + ( p d + 1) ( n , , − n − , − , ) + ( p d − n , − , − n − , , )) . (3.40)Applying Lemma 3.6 1) −
4) and solving the system equations consisting of (3.34)-(3.40), weget n , , = 14 ( p m + 1) ( p d − p m − / ( p d + 1) ,n − , − , = 14 ( p m − ( p d − p m − / ( p d + 1) ,n , − , = n − , , = 14 ( p m + d − p m + p d + 1)( p m − / ( p d − ,n , , = n − , , = 14 p m − d ( p m + 1)( p m − ,n , − , = n − , − , = 14 p m − d ( p m − p m − ,n , − , = 12 ( p m − d + 1)( p m − p m − / ( p d − ,n − , , = 12 ( p m − d − p m + 1)( p m − / ( p d − ,n , , = n − , − , = 0 . This finishes the proof. ✷ Therefore, we can determine the weight distribution of cyclic code C . Theorem 3.8.
With the notations given above.i) If k is odd, then C defined by (1.1) is a cyclic code over F p with length p m − anddimension m . Moreover, the weight distribution of cyclic code C is given in Table .ii) If k is even, then C defined by (1.1) is a cyclic code over F p with length p m − anddimension m . Moreover, the weight distribution of cyclic code C is given in Table . Proof.
By (3.1), we have wt ( c ( α, β )) = p m − p m − − p X u ∈ F ∗ p S ( uα, uβ ) . C for odd k value frequency0 1 p m − p m − ( p m + d − p m + p d + 1)( p m − / ( p d − p m − ( p − − ( p − p d − p m − ( p m − d + 1)( p m − p m − / ( p d − p m − ( p −
1) + ( p − p d − p m − ( p m − d − p m + 1)( p m − / ( p d − p m − ( p −
1) + ( p − p m − p m − d ( p m − p m − p m − ( p − − ( p − p m − p m − d ( p m + 1)( p m − p m − ( p −
1) + ( p − p m − ( p m − ( p d − p m − / ( p d + 1) p m − ( p − − ( p − p m − ( p m + 1) ( p d − p m − / ( p d + 1)Table 3: weight distribution of C for even k Weight Frequency0 1 p m − p m − ( p m + d − p m + p d + 1)( p m − / ( p d − p m − ( p −
1) + ( p − p d − p m − ( p m − d − p m + 1)( p m − / ( p d − p m − ( p − − ( p − p d − p m − ( p m − d + 1)( p m − p m − / ( p d − p m − ( p − − ( p − p d − p m − p m − d ( p m − p m − p m − ( p −
1) + ( p − p d + 1) p m − p m − d ( p m − p m − p m − ( p − − ( p − p d + 1) p m − p m − d ( p m + 1)( p m − p m − ( p −
1) + ( p − p d − p m − p m − d ( p m + 1)( p m − p m − ( p −
1) + ( p − p m − ( p m − ( p d − p m − / ( p d + 1) p m − ( p − − ( p − p m − ( p m + 1) ( p d − p m − / ( p d + 1)If k is odd, then d is odd and u p = π pm − p − = π pm − pd − · pd − p − is a nonsquare element in F p d . Sowe have η d ( u p ) = −
1. If u ∈ SQ p , then u pk +12 = u . Hence, by Lemma 2.2, we have X u ∈ F ∗ p S ( uα, uβ )= X u ∈ F ∗ p X x ∈ F pm (cid:18) ζ Tr m ( uαx pk +1 + uβx ) p + ζ Tr m ( uαπ pk +12 x pk +1 − uβπx ) p (cid:19) = X u ∈ SQ p X x ∈ F pm (cid:18) ζ Tr m ( α ( u x ) pk +1 + β ( u x ) ) p + ζ Tr m ( απ pk +12 ( u x ) pk +1 − βπ ( u x ) ) p (cid:19) + X u ∈ SQ p X x ∈ F pm (cid:18) ζ u p Tr m ( α ( u x ) pk +1 + β ( u x ) ) p + ζ u p Tr m ( απ pk +12 ( u x ) pk +1 − βπ ( u x ) ) p (cid:19) X u ∈ SQ p X x ∈ F pm (cid:18) ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) + X u ∈ SQ p X x ∈ F pm (cid:18) ζ u p Tr m ( αx pk +1 + βx ) p + ζ u p Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) = p − (1 + η d ( u p ) r X x ∈ F pm ζ Tr m ( αx pk +1 + βx ) p + (1 + η d ( u p ) r ′ X x ∈ F pm ζ Tr m ( απ pk +12 x pk +1 − βπx ) p , where r and r ′ are the rank of Tr md ( αx p k +1 + βx ) and Tr md ( απ pk +12 x p k +1 − βπx ), respectively.Note that η d ( u p ) = − s is even. By Lemma 3.1, we have if r = s −
1, then r ′ = s and X u ∈ F ∗ p S ( uα, uβ ) = ( p − X x ∈ F pm ζ Tr m ( απ pk +12 x pk +1 − βπx ) p . If r ′ = s −
1, then r = s and X u ∈ F ∗ p S ( uα, uβ ) = ( p − X x ∈ F pm ζ Tr m ( αx pk +1 + βx ) p . Otherwise, X u ∈ F ∗ p S ( uα, uβ ) = ( p − X x ∈ F pm (cid:18) ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) . Therefore, by Theorem 3.7, we obtain the weight distribution of cyclic code C for odd k .If k is even, then u pk +12 = u for all u ∈ F p . Since m is even, then u is a square element in F p m . Hence, we have S ( uα, uβ ) = X x ∈ F pm (cid:18) ζ Tr m ( uαx pk +1 + uβx ) p + ζ Tr m ( uαπ pk +12 x pk +1 − uβπx ) p (cid:19) = X x ∈ F pm (cid:18) ζ Tr m ( α ( u x ) pk +1 + β ( u x ) ) p + ζ Tr m ( απ pk +12 ( u x ) pk +1 − βπ ( u x ) ) p (cid:19) = X x ∈ F pm (cid:18) ζ Tr m ( αx pk +1 + βx ) p + ζ Tr m ( απ pk +12 x pk +1 − βπx ) p (cid:19) = S ( α, β ) . Therefore, wt ( c ( α, β )) = p m − p m − − p − p S ( α, β ) . By Theorem 3.7, we get the weight distribution of cyclic code C for even k . ✷ In the following, we give an example to verify the result in Theorem 3.8 for the case of odd k . For the case of even k , we are not able to give an example to verify the result in Theorem 3.8because of our limited computation ability. 20 xample 3.9. Let p = 3 , m = 6 , k = 1 , the code C is a [728 , , cyclic code over F withweight enumerator X + 183456 X + 728 X + 170352 X + 81900 X , which confirms the weight distribution in Table . References [1] P. Delsarte, “On subfield subcodes of modified Reed-Solomon codes,”
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