2017 2nd International Conference on Computer, Network Security and Communication Engineering (CNSCE 2017) ISBN: 978-1-60595-439-4
Generating Idempotents of Sixth Residue Codes over the Binary Field
Xue-dong DONG
1,*and Yan ZHANG
21
College of Information Engineering, Dalian University, Dalian 116622, P.R. China
2
School of Mathematics, Liaoning Normal University, Dalian 116029, P.R. China *Corresponding author
Keywords: Generating idempotent, Residue code, Cyclic code.
Abstract. Higher power residue codes over finite fields are generated by factors of the polynomial 1
n
x − . Unfortunately, to decompose the polynomial xn 1
− over finite fields is difficult. Generating idempotents can also generate higher power residue codes. Thus it is important to get generating idempotents of cyclic codes. We find precise expressions of generating idempotents of some sixth residue codes of length over the binary field, where pis a prime such that p≡1 mod 24
(
)
.Introduction
There are many papers on quadratic residue codes and their generalizations [1-3]. In [4-9] higher power residue codes and forms of generating polynomials of these codes were investigated. Higher power residue codes are generated by factors ofxn−1 . Unfortunately, to decompose xn−1 over finite fields is difficult. We have to get the generating idempotents of higher power residue codes since generating polynomials of cyclic codes can be obtained by computing the GCD (greatest common divisors) of generating idempotents and xn−1without factoring xn−1 over finite fields.
Precise expressions of generating idempotents of cubic and quartic residue codes over the fields F2
and F3 were obtained respectively in [6] and [7]. Precise expressions of generating idempotents of
quintic residue codes over F2were obtained in [9]. This paper gives generating idempotents of some
sixth residue codes of length p over the binary field, wherep is a prime such that p≡1 mod 24
(
)
. The paper is organized as follows. We give preliminary results in Section 2. Generating idempotents of some sixth residue codes over the binary field are given in Section 3. Conclusions are given in Section 4.Preliminaries
Definition 2.1.If x is an integer such that x6≡a
(
modp)
, where a∈Z and ( , ) 1a p = ,then a is called a sixth residue modulo p.Let p be an odd prime and ρ a primitive element of the finite fieldFpin the following.
{
F k Z}
R0= ρtk∈ P| ∈ ,R =
{
tk+ ∈FP k∈Z}
|
1
1 ρ ,
( )
{
1}
1
,Rt ρtk+ −t FP|k Z
− = ∈ ∈
. Assume that mis the
smallest positive integer such that2m ≡1 mod
(
p)
,α
is a primitive p-th root of unity in F2m, and( )
(
0)
( )
(
1)
( )
(
1)
0 0 1 1 1 1
0 , 1 , , 1 .
t
t t
r r r
t
r R r R r R
g x x α g x x α g x x α −
− −
−
∈ ∈ ∈
=
∏
− =∏
− =∏
−Lemma 2.1. 1
(
1) ( )
0 1( )
pt
x − = x− g x g− x and
( )
(
)
2[ ]
j
j j r j
r R
g x x α F x
∈
=
∏
− ∈Definition 2.2.[6] The t -th residue codes C0,,Ct−1,C0,,Ct−1 are cyclic codes of
[ ]
2 / ( 1) p
F x x − with generator polynomials g0
( )
x,,gt−1( ) (
x, x−1) ( )
g0 x ,,(
x−1)
gt−1( )
xrespectively.
Definition 2.3.[10] If
( )
2[ ]
(
1)
pE x ∈F x x − and E x
( )
2 ≡E x( )
(
mod(
xp−1)
)
,then it is called an idempotent.Definition 2.4. [10] The functionµais defined on{0,1,,n−1} by iµa ≡ia
(
modn)
,where a isan integer with ( , )a n =1. µa acts on [ ] /( 1) n p
F x x − by f x
( )
µa f xa( )
(modxn 1)≡ − , where
( )
p[ ] / ( n 1).f x ∈F x x −
Lemma 2.2. [10] If Cis a cyclic code of length n over the finite field Fq with generating
idempotent e x
( )
, then the cyclic code Cµais generated by the generating idempotent e x( )
µa.Lemma 2.3. [6] C0,,Ct−1are pairwise equivalent and C0,,Ct−1 are pairwise equivalent.
In the following assume that
( )
∑
∈
=
0 0
0
0
R r
r
x x
e
( )
∑
∈
=
1 1
1
1
R r
r
x x
e ,
( )
∑
− −
−
∈
− =
1 1
1
1
,
t t
t R r
r
t x x
e
Lemma 2.4. [6]
( )
( )
( )
11
0 1
1
0 + + + +
∑
=−
= −
p
i i
t x x
e x
e x
e
Lemma 2.5. [6] Let E x( ) be the generating idempotent of a t -th residue code C .
Then
( )
1
0
( )
t
i i i
E x a a e x
−
=
= +
∑
, wherea a a, 0, ,1,at−1∈F2.Lemma 2.6. [6] If E x0( ) is the generating idempotent of C0, then E0
( )
x =E x0( ) 10 p
i
i
x
−
=
+
∑
is thegenerating idempotent of C0.
Lemma 2.7. [6] If E0
( )
x and E x0( ) are respectively the generating idempotents of C0 and C0,and 1 1
tk t t
d ρ + − R
−
= ∈ , then
1.E0
( )
x µd =E x1( )
,E x1( )
µd =E2( )
x ,,Et−2( )
x µd =Et−1( )
x are respectively the generatingidempotents of C1,,Ct−1 .
2. E x0( )µd =E x E x1( ), 1( )µd =E x2( ),,Et−2
( )
x µd =Et−1( )
x are respectively generatingidempotents of C1,,Ct−1 .
Generating Idempotents of Some Sixth Residue Codes
Let 2 be a sixth residue modulo p.Then there exists an integer x such that x6≡2 mod
(
p)
. Thus, 2 is a quadratic residue modulo p . We have p≡ ±1 mod 8(
)
. If p≡1(
mod8)
and 6 |(
p−1)
, thenp≡1 mod 24(
)
. If p≡ −1 mod 8(
)
and 6 |(
p−1)
, then p≡7 mod 24(
)
.The following theorem will determine the set of the generating idempotents of C C C C C C0, 1, 2, 3, 4, 5.By Lemma 2. 6 andLemma 2.7 one can easily determine the set of the generating idempotents of C0,C1,C2,C3,C4,C5.
Theorem 3.1.Let p≡1 mod 24
(
)
and 2 be a sixth residue modulo p.Then the set of the generating idempotents ofC0,C1,C2,C3,C3,C4,C5over F2 is:( )
( )
( )
( )
( )
{
e x1 +e2 x +e3 x +e4 x +e5 x e, 0( )
x +e2( )
x +e3( )
x +e4( )
x +e5( )
x e, 0( )
x +e x1( )
+e3( )
x ,( )
( )
4 5 ,
e x e x
+ + e0
( )
x +e1( )
x +e2( )
x +e4( )
x +e5( )
x ,e0( )
x +e1( )
x +e2( )
x +e3( )
x +e5( )
x ,e0( )
x +( )
( )
( )
( )
}
1 2 3 4
( )
x e( )
x e( )
xe1 + 2 + 4 , e2
( )
x +e3( )
x +e5( )}
x or{
e x1( )
+e3( )
x +e4( )
x e, 2( )
x +e4( )
x +e5( )
x ,( )
x e( )
x e( )
xe0 + 3 + 5 ,e0
( )
x +e1( )
x +e4( )
x ,e1( )
x +e2( )
x +e5( )
x ,e0( )
x +e2( )
x +e3( )}
x or( ) ( ) ( ) ( ) ( )
{
e0 x,e1 x,e2 x ,e3 x,e4 x,e5( )}
x .Proof: From p≡1 mod 24
(
)
it follows that (p−1) / 6 is even and therefore (p−1) / 6=0over2
F .By Lemma 2.5 assume that
( )
∑
( )
= + = 5 0 0 i i ie x
a a x
E is the generating idempotent of the residue
code C0, where a,ai∈F2,0≤i≤5. Then
( )
( )
(
mod2)
6 1 1 1 0 5 0 5 0
0 a a
p a e a a E i i i i
i ≡
− + = + = =
∑
∑
= =and therefore
( )
( )
5 0
0 i i i
E x a e x
=
=
∑
.Let αbe a primitive p-th root of unity in F2mas before. It is clearthat each ei
( )
x is an idempotent since 2 is a sixth residue modulo p.Thus we have that ei( )
α =0 or 1 and( )
1( )
2( )
3( )
4( )
5( )
10 +e +e +e +e x +e x =
e α α α α , and the number of 1 amonge0
( )
α ,e1( )
α ,( )
α2
e ,e3
( )
α ,e4( )
α ,e5( )
α is odd. We have to consider three cases in the following.Case 1: One of e0
( )
α ,e1( )
α ,e2( )
α ,e3( )
α ,e4( )
α ,e5( )
x is 1 and the others are 0.Let ei
( )
α =1,ei+1(mod6)( )
α =ei+2(mod6)( )
α =ei+3(mod6)( )
α =ei+4(mod6)( )
α =ei+5(mod6)( )
α =0 0≤i≤5 ,where the subscript is the smallest nonnegative residue modulo 6.Then
( )
aiE R = =
∈ 0,0 0 α
1 ,∀b=ρ6k+1∈R1
,1= 0
( )
= i+5(mod6)b
a
E α ,∀c= ρ6k+2∈R2,
( )
4(mod6)0
1= = i+
c
a
E α ,∀d =ρ6k+3∈R3,1= E0
( )
αd =ai+3(mod6),∀e=ρ6k+4∈R4,( )
2(mod6)0
1=E αe =ai+
, 5
5
6 R
f = k ∈
∀ ρ +
,1= 0
( )
= i+1(mod6) f aE α .Thus
( ) ( ) ( ) ( ) ( ) 1,0 5
,
0 1mod6 = 2mod6 = 3mod6 = 4mod6 = 5mod6 = ≤ ≤
= a+ a+ a+ a+ a+ i
ai i i i i i .
( )
( )( )
( )( )
( )( )
( )( )
( )( )
0 i1 mod 6 i 2 mod 6 i 3 mod 6 i 4 mod 6 i 5 mod 6 , 0 5
E x =e+ x +e+ x +e+ x +e+ x +e+ x ≤ ≤i . By lemma 2.7, the set consisting of generating idempotents of C0,C1,C2,C3,C3,C4,C5is
( )
( )
( )
( )
( )
{
e1 x +e2 x +e3 x +e4 x +e5 x ,e0( )
x +e2( )
x +e3( )
x +e4( )
x +e5( )
x , e0( )
x +e1( )
x +e3( )
x +e4( )
x +e5( )
x ,e0( )
x +e1( )
x +e2( )
x +e4( )
x +e5( )
x ,( )
x e( )
x e( )
x e( )
x e( )
xe0 + 1 + 2 + 3 + 5 ,e0
( )
x +e1( )
x +e2( )
x +e3( )
x +e4( )}
xCase 2: Three of e0
( )
α ,e1( )
α ,e2( )
α ,e3( )
α ,e4( )
α ,e5( )
x are 1 and the other three are 0.1) Letei
( )
α =ei+1(mod6)( )
α =ei+2(mod6)( )
α =1,ei+3(mod6)( )
α =ei+4(mod6)( )
α =ei+5(mod6)( )
α =0,where0≤i≤5.Then
( )
1(mod6) 2(mod6)0
0=E α =ai+ai+ +ai+ (1)
1 1 6
R b= k ∈
∀ ρ +
( )
(mod6) 5(mod6)
1 0
1= b = i+ i+ + i+
a a
a
E α (2)
2 2
6 R
c= k ∈
∀ +
ρ 1= 0
( )
= i+ i+4(mod6)+ i+5(mod6)c
a a
a
E α (3)
3 3 6
R d = k ∈
∀ +
ρ 1= 0
( )
= i+3(mod6)+ i+4(mod6)+ i+5(mod6)d
a a
a
E α (4)
4 4 6
R e= k ∈
∀ ρ + 1= 0
( )
= i+2(mod6)+ i+3(mod6)+ i+4(mod6)e
a a
a
E α (5)
5 5 6
R f = k ∈
∀ +
ρ 1= 0
( )
= i+1(mod6)+ i+2(mod6)+ i+3(mod6)f a a a
From (1) + (6) it follows that ai+ai+3(mod6) =1, from (3)+(4) it follows that ai+ai+3(mod6)=0,a
contradiction.
2) Let ei
( )
α =ei+1(mod6)( )
α =ei+3(mod6)( )
α =1,ei+2(mod6)( )
α =ei+4(mod6)( )
α =ei+5(mod6)( )
α =0 , where5
0≤i≤ .Then 0=E0
( )
α =ai+ai+1(mod6)+ai+3(mod6) ,1 1 6
R b= k ∈
∀ ρ + 1= 0
( )
= i+ i+2(mod6)+ i+5(mod6)b a a a E α 2 2 6 R
c= k ∈
∀ +
ρ 1= 0
( )
= i+1(mod6)+ i+4(mod6)+ i+5(mod6)c a a a
E α
3 3 6
R d = k ∈
∀ ρ +
( )
(mod6) 4(mod6)
3 0
1= d = i + i+ + i+
a a a E α 4 4 6 R e= k ∈
∀ ρ + 1= 0
( )
= i+2(mod6)+ i+3(mod6)+ i+5(mod6) e a a a E α 5 5 6 Rf = k ∈
∀ +
ρ 1= 0
( )
= i+1(mod6)+ i+2(mod6)+ i+4(mod6) fa a
a
E α
By solving system of linear equations in 6 unknownsai(mod 5),ai+1 mod 5( ),ai+2 mod 5( ),ai+3 mod 5( ),ai+4 mod 5( ), ( )
5 mod 6 i
a+ we get ai =1,ai+1(mod6)=0,ai+2(mod6) =0,ai+3(mod6) =1,ai+4(mod6)=1,ai+5(mod6)=0,0≤i≤5, and therefore E0
( )
α =e xi( )
+ei+3 mod 6( )( )
x +ei+4 mod 6( )( )
x . By lemma 2.7, the set consisting of generatingidempotents of C0,C1,C2,C3,C3,C4,C5is
( )
( )
( )
{
e0 x +e3 x +e4 x ,e1( )
x +e4( )
x +e5( )
x ,e0( )
x +e2( )
x +e5( )
x ,( )
x e( )
x e( )
xe0 + 1 + 5 ,e1
( )
x +e2( )
x +e4( )
x ,e2( )
x +e3( )
x +e5( )}
x3) Let
( )
α = i+1(mod6)( )
α = i+4(mod6)( )
α =1, i+2(mod6)( )
α = i+3(mod6)( )
α = i+5(mod6)( )
α =0i e e e e e
e , 0≤i≤5 ,Then
( )
1(mod6) 4(mod6)0
0=E α =ai+ai+ +ai+ ,
1 1 6
R b= k ∈
∀ ρ +
( )
(mod6) 5(mod6)
3 0
1= b = i+ i+ + i+
a a a E α 2 2 6 R c= k ∈
∀ ρ + 1= 0
( )
= i+2(mod6)+ i+4(mod6)+ i+5(mod6)c a a a E α 3 3 6 R
d = k ∈
∀ +
ρ 1= 0
( )
= i+1(mod6)+ i+3(mod6)+ i+4(mod6)d a a a
E α
4 4 6
R e= k ∈
∀ ρ + 1= 0
( )
= i+ i+2(mod6)+ i+3(mod6)e a a a E α 5 5 6 R f = k ∈
∀ ρ + 1=E0
( )
αf =ai+1(mod6)+ai+2(mod6)+ai+5(mod6)By solving system of linear equations in 6 unknowns ai(mod 5),ai+1 mod 5( ),ai+2 mod 5( ),ai+3 mod 5( ),
( ) ( )
4 mod 5 , 5 mod 6
i i
a+ a+ we get that
( ) 1, ( ) 0, ( ) 1, ( ) 1, ( ) 0,0 5
,
0 1mod6 = 2mod6 = 3mod6 = 4mod6 = 5mod6 = ≤ ≤
= a+ a+ a+ a+ a+ i
ai i i i i i .
Thus, we haveE0
( )
α =ei+1 mod 6( )( )
x +ei+3 mod 6( )( )
x +ei+4 mod 6( )( )
x . By lemma 2.7, the set consisting ofgenerating idempotents of C0,C1,C2,C3,C3,C4,C5 is
( )
( )
( )
{
e1 x +e3 x +e4 x ,e2( )
x +e4( )
x +e5( )
x ,e0( )
x +e3( )
x +e5( )
x ,( )
x e( )
x e( )
xe0 + 1 + 4 ,e1
( )
x +e2( )
x +e5( )
x ,e0( )
x +e2( )
x +e3( )}
x4) Let ei
( )
α =ei+2(mod6)( )
α =ei+4(mod6)( )
α =1,ei+1(mod6)( )
α =ei+3(mod6)( )
α =ei+5(mod6)( )
α =0,0≤i≤5.( )
2(mod6) 4(mod6)0
0=E α =ai+ai+ +ai+ (7)
1 1 6
R b= k ∈
∀ ρ +
( )
(mod6) 3(mod6) 5(mod6)
1 0
1= b = i+ + i+ + i+
a a
a
E α (8)
2 2
6 R
c= k ∈
∀ +
ρ 1= 0
( )
= i+ i+2(mod6)+ i+4(mod6)c
a a
a
Case 3: Five of e0
( )
α ,e1( )
α ,e2( )
α ,e3( )
α ,e4( )
α ,e5( )
x are 1 and the other is 0.Let ei
( )
α =0,ei+2(mod6)( )
α =ei+1(mod6)( )
α =ei+3(mod6)( )
α =ei+4(mod6)( )
α =ei+5(mod6)( )
α =1, 0≤i≤5 .Then0=E0
( )
α =ai+1(mod6)+ai+2(mod6)+ai+3(mod6)+ai+4(mod6)+ai+5(mod6)1 1 6
R b= k ∈
∀ ρ + 1= 0
( )
= i+ i+1(mod6)+ i+2(mod6)+ i+3(mod6)+ i+4(mod6) ba a
a a
a E α
2 2
6 R
c= k ∈
∀ +
ρ 1= 0
( )
= i+ i+1(mod6)+ i+2(mod6)+ i+3(mod6)+ i+5(mod6)c a a a a a
E α
3 3 6
R d = k ∈
∀ ρ +
( )
(mod6) 2(mod6) 4(mod6) 5(mod6)
1
0 ,
1= d = i+ i+ + i+ + i+ + i+
a a
a a
a E α
4 4 6
R e= k ∈
∀ ρ + 1= 0
( )
= i+ i+1(mod6)+ i+3(mod6)+ i+4(mod6)+ i+5(mod6) ea a
a a
a E α
5 5
6 R
f = k ∈
∀ +
ρ 1= 0
( )
= i+ i+2(mod6)+ i+3(mod6)+ i+4(mod6)+ i+5(mod6)f
a a
a a
a E α
By solving system of linear equations in 6 unknowns ai(mod 5),ai+1 mod 5( ),ai+2 mod 5( ),ai+3 mod 5( ), ( ) ( )
4 mod 5 , 5 mod 6
i i
a+ a+ we get that ai=1,ai+1 mod 6( )=ai+2 mod 6( ) =ai+3 mod 6( ) =ai+4 mod 6( ) =ai+5 mod 6( ) =0,
( )
( )
0
0≤ ≤i 5,E α =e xi . By lemma 2.7, the set consisting of generating idempotents of
5 4 3 3 2 1
0,C ,C ,C ,C ,C ,C
C is
{
e0( ) ( ) ( ) ( ) ( )
x ,e1 x ,e2 x ,e3 x ,e4 x ,e5( )}
x .Summary
Using coding theory, we have given precise expressions of generating idempotents of some sixth residue codes of length p over the binary field, wherep is a prime such that p≡1 mod 24
(
)
. Thus, the generating polynomials can be found by computing GCD of these generating idempotents and1
p
x − without factoringxp−1. Whenp is a prime, wherep≡7 mod 24
(
)
, precise expressions of generating idempotents of some sixth residue codes of length pover the binary field will be discussed in the other paper.Acknowledgement
This research was financially supported by the Research Project of Liaoning Education Bureau under Project Code L2014490.
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