Table Of ContentHermitian Self-Dual Cyclic Codes of Length pa over
GR(p2,s)
Somphong Jitman, San Ling, and Ekkasit Sangwisut
4
1
0 Abstract
2 In this paper, we study cyclic codes over the Galois ring GR(p2,s). The main result
n is the characterization and enumeration of Hermitian self-dual cyclic codes of length pa
a over GR(p2,s). Combining with some known results and the standard Discrete Fourier
J
Transform decomposition, we arrive at the characterization and enumeration of Euclidean
6 self-dualcycliccodesofanylengthoverGR(p2,s). SomecorrectionstoresultsonEuclidean
2 self-dual cyclic codes of even length over Z4 in Discrete Appl. Math. 128, (2003), 27 and
Des. Codes Cryptogr. 39, (2006), 127 are provided.
]
A
R 1 Introduction
.
h
t Cyclicandself-dualcodesoverfinitefieldshavebeenextensivelystudiedforboththeoreticaland
a
practical reasons (see [6], [11], and references therein). These concepts have been extended and
m
studiedovertheringZ (see[1],[2],and[4]),afterithasbeenproventhatsomebinarynon-linear
4
[
codes, such as the Kerdock, Preparata, and Goethals codes, are the Gray image of linear cyclic
1 codes over Z in [8]. Later on, the study of cyclic and self-dual codes has been generalized to
4
v codes over Z and Galois rings (see [5], [9], [10], [12], and [13]).
pr
4
In [2], the structure of cyclic and Euclidean self-dual cyclic codes of oddly even length (2m,
3
wheremisodd)overZ hasbeenstudiedviatheDiscreteFourierTransformdecomposition. This
6 4
6 idea has been extended to the case of all even lengths in [4]. However, some results concerning
1. Euclidean self-dual cyclic codes over Z4 in [2] and [4] are not correct. The corrections to these
are provided in Section 4. Using a spectral approach and a generalization of the results in [2]
0
4 and [4], the structure of cyclic codes over Zpe, for any prime p, has been studied in [5]. A nice
1 classification of cyclic and Euclidean self-dual cyclic codes of length pa over GR(p2,s) has been
: given using a different approach in [9], [10], and [12].
v
i Inthis paper, we focus onthe characterizationandenumerationof Hermitianself-dualcyclic
X
codes of length pa over GR(p2,s) and their application to the enumeration of Euclidean self-
r dual cyclic codes of any length over GR(p2,s). Using the standard Discrete Fourier Transform
a
decomposition viewed as an extension of [5], a cyclic code C of any length n =mpa, where p is
a prime and p ∤ m, over GR(p2,s) can be viewed as a product of cyclic codes of length pa over
some Galois extensionsof GR(p2,s). Euclideanself-dualcyclic codes canbe characterizedbased
on this decomposition. Applying some known results concerning cyclic and Euclidean self-dual
cyclic codes of length pa in [9] and [10] and our result on Hermitian self-dual cyclic codes of
S.JitmaniswiththeDepartmentofMathematics,FacultyofScience,SilpakornUniversity,NakhonPathom
73000, Thailand(email: [email protected]).
S.LingiswiththeDivisionofMathematicalSciences,SchoolofPhysicalandMathematicalSciences,Nanyang
TechnologicalUniversity,21NanyangLink,Singapore637371,RepublicofSingapore(email: [email protected]).
E. Sangwisut is with the Department of Mathematics and Computer Science, Faculty of Science, Chula-
longkornUniversity,Bangkok10330, Thailand(email: [email protected]).
This work is partially supported by the National Research Foundation of Singapore under Research Grant
NRF-CRP2-2007-03. E.SangwisutisalsosupportedbytheInstitute forthe PromotionofTeaching Scienceand
TechnologyofThailand.
1
length pa, the number of Euclidean self-dual cyclic codes of arbitrary length over GR(p2,s) can
be determined. Finally, we point out some mistakes on Euclidean self-dual cyclic codes of even
length over Z in [2] and [4]. The corrections to these are given as well.
4
The paper is organized as follows. Some preliminary concepts and results are recalled in
Section 2. In Section 3, we prove the main result concerning the number of Hermitian self-dual
cycliccodesoflengthpa overGR(p2,s). AnapplicationtotheenumerationofEuclideanself-dual
cyclic codes of any length over GR(p2,s) and corrections to [2] and [4] are discussed in Section
4. A conclusion is provided in Section 5.
2 Preliminaries
Inthissection,werecallsomedefinitionsandbasicpropertiesofcycliccodesovertheGaloisring
GR(p2,s).
2.1 Cyclic Codes over GR(p2,s)
For a prime p and a positive integer s, the Galois ring GR(p2,s) is the Galois extension of
the integer residue ring Z of degree s. Let ξ be an element in GR(p2,s) that generates the
p2
Teichmu¨llersetT ofGR(p2,s). Inotherwords,T ={0,1,ξ,ξ2,...,ξps−2}. Theneveryelement
s s
in GR(p2,s) has a unique p-adic expansion of the form
α=a+bp,
where a,b∈T . If s is even, let¯denote the automorphism on GR(p2,s) defined by
s
α=aps/2 +bps/2p. (2.1)
For more details concerning Galois rings, we refer the readers to [14].
A cyclic code of length n over GR(p2,s) is a GR(p2,s)-submodule of the GR(p2,s)-module
(GR(p2,s))n which is invariant under the cyclic shift. It is well known that every cyclic code
C of length n over GR(p2,s) can be regarded as an ideal in the quotient polynomial ring
GR(p2,s)[X]/hXn−1i and represented by its polynomial representation
n−1
c Xi (c ,c ,...,c )∈C .
i 0 1 n−1
( (cid:12) )
Xi=0 (cid:12)
(cid:12)
For a givencyclic code C of length n(cid:12)overGR(p2,s), denote by C⊥E the Euclidean dual of C
(cid:12)
defined with respect to the form
n−1
hu,vi := u v
E i i
i=0
X
where u = n−1u Xi and v = n−1v Xi. The code C is said to be Euclidean self-dual if
i=0 i i=0 i
C =C⊥E.
In additiPon, if s is even, we caPn also consider the Hermitian dual C⊥H of C defined with
respect to the form
n−1
hu,vi := u v .
H i i
i=0
X
The code C is said to be Hermitian self-dual if C =C⊥H.
The goal of this paper is to characterize and enumerate self-dual codes over GR(p2,s). For
convenience,let N(GR(p2,s),n), N (GR(p2,s),n), andN (GR(p2,s),n) denotethe numbersof
E H
cyclic codes, Euclidean self-dual cyclic codes, and Hermitian self-dual cyclic codes of length n
over GR(p2,s), respectively.
2
2.2 Some Results on Cyclic Codes of Length pa over GR(p2,s)
In this subsection, we recall some results concerning cyclic and Euclidean self-dual cyclic codes
of length pa over GR(p2,s) in terms of ideals in GR(p2,s)[X]/hXpa −1i.
In [9], it has been shown that all the ideals in GR(p2,s)[X]/hXpa −1i have a unique repre-
sentation.
Proposition 2.1 ([9, Theorem 3.8]). Every ideal in GR(p2,s)[X]/hXpa −1i can be uniquely
represented in the form of
i1−1
C =h(X −1)i0 +p h (X −1)j,p(X −1)i1i, (2.2)
j
j=0
X
where i ,i are integers such that 0 ≤ i < pa, 0 ≤ i ≤ min{i ,pa−1}, i +i ≤ pa, and h is
0 1 0 1 0 0 1 j
an element in the Teichmu¨ller set T for all j.
s
The integer i in Proposition 2.1 is called the first torsion index of C.
1
Corollary 2.2 ([9, Corollary 3.9]). In GR(p2,s)[X]/hXpa −1i, the number of distinct ideals
with i +i =d, where 0≤d≤pa, is
0 1
ps(ℓ+1)−1
,
ps−1
where ℓ=min{⌊d⌋,pa−1}.
2
The next corollary follows immediately from Corollary 2.2 and [9, Theorem 3.6].
Corollary 2.3. The number of all cyclic codes of length pa over GR(p2,s) is
N(GR(p2,s),pa)=2 pa−1ps(min{⌊d2⌋,pa−1}+1)−1 + ps(pa−1+1)−1. (2.3)
ps−1 ps−1
!
d=0
X
Combining the results in [9], [10], and [12], the complete enumeration of Euclidean self-dual
cyclic codes of length pa over GR(p2,s) can be summarized as follows.
Proposition 2.4 ([10, Corollary3.5]). The number of Euclidean self-dual cyclic codes of length
2a over GR(22,s) is
1 if a=1,
N (GR(22,s),2a)=1+2s if a=2,
E
1+2s+22s+1 (2s)(2a−2−1)−1 if a≥3.
2s−1
(cid:18) (cid:19)
If p is an odd prime, then the number of Euclidean self-dual cyclic codes of length pa over
GR(p2,s) is
pa−1+1
(ps) 2 −1
N (GR(p2,s),pa)=2 .
E ps−1
3 Hermitian Self-Dual Cyclic Codes of Length pa over GR(p2,s)
Inthis section,weassumethats is evenandfocus oncharacterizingandenumeratingHermitian
self-dual cyclic codes of length pa over GR(p2,s).
3
Ithas been provenin[12, Theorem2] thatthe Euclideandual ofthe cyclic code C in (2.2) is
of the form
i1−1 t i −j
C⊥E = (X −1)pa−i1 −p(X −1)pa−i0−i1 (−1)i0+j 0 h (X −1)t
j
t−j
* t=0 j=0 (cid:18) (cid:19)
X X
K min{t,p−1}
+ (−1)j+1 p−j p (X −1)tpa−1−i1,p(X −1)pa−i0 ,
t−j j
t=1 j=1 (cid:18) (cid:19)(cid:18) (cid:19) +
X X
where K=⌊pa−i0+i1−1⌋ and the empty sum is regarded as zero.
pa−1
For a subset A of GR(p2,s)[X]/hXpa −1i, let A denote the set
pa−1 pa−1
a Xi a Xi ∈A ,
i i
( (cid:12) )
Xi=0 (cid:12) Xi=0
(cid:12)
where¯is the automorphism defined in (2.1).(cid:12)(cid:12) Since it is well know that C⊥H =C⊥E, we have
i1−1 t i −j
C⊥H = (X −1)pa−i1 −p(X −1)pa−i0−i1 (−1)i0+j 0 h ps/2 (X −1)t
j
r−j
* t=0 j=0 (cid:18) (cid:19)
X X
K min{t,p−1}
+ (−1)j+1 p−j p (X −1)tpa−1−i1,p(X −1)pa−i0 .
t−j j
t=1 j=1 (cid:18) (cid:19)(cid:18) (cid:19) +
X X
If C =C⊥H, then |C|=(ps)pa which implies that i0+i1 =pa. Then we can write
(X −1)i0 −p ti1=−01 tj=0(−1)i0+j it0−−jj hjps/2 (X −1)t,p(X −1)i1
C⊥H =D(X −1)i0 −pPti1=−01(cid:16)Ptj=0(−1)i0+j(cid:0)ir0−−jj(cid:1)hjps/2(cid:17)(X −1)tif i1 <⌊pa−21+E1⌋, (3.1)
If i =0D+, tph(eXn−pG1R)p(ap−21,−Psi)1[,Xp(]/X(cid:16)hXP−p1a)−i1E1i is the(cid:0)only(cid:1)Herm(cid:17)itian self-difuia1l≥cy⌊cpliac−21c+od1⌋e.of length
1
pa over GR(p2,s). Next, we assume that i ≥1.
1
By the unique representation of C =C⊥H, (2.2), and (3.1), we have
t
i −j
ph ps/2 =p b + (−1)i0+j−1 0 h (3.2)
t t j
r−j
j=0 (cid:18) (cid:19)
X
for all t=0,1,...,i −1, where b =1 if t=pa−1−i and b =0 otherwise.
1 t 1 t
Let M(pa,i ) be an i ×i matrix defined by
1 1 1
(−1)i0 +1 0 0 ... 0
(−1)i0 i0 (−1)i0+1+1 0 ... 0
M(pa,i1)= (−1)i0(cid:0)i210(cid:1) (−1)i0+1 i01−1 (−1)i0+2+1 ... 0 . (3.3)
.. .. .. .. ..
. (cid:0) (cid:1) . (cid:0) (cid:1) . . .
(−1)i0 i0 (−1)i0+1 i0−1 (−1)i0+2 i0−2 ... (−1)i0+i1−1+1
i1−1 i1−2 i1−3
For a matrix V, denote(cid:0)by V(cid:1)T the trans(cid:0)pose(cid:1)of V. Then(cid:0)(3.2)(cid:1)forms a matrix equation
M(pa,i )x+(xps/2 −x)=b, (3.4)
1
4
where x := (x ,x ,...,x )T, xps/2 := (xps/2,xps/2,...,xps/2)T, and b := (b ,b ,...,b )T is a
1 2 i1 1 2 i1 1 2 i1
zero vector except for the case i ≥ pa−1−1 where, for each 1≤i≤i , b is defined by
1 2 1 i
1 if i=pa−1−i +1,
1
b = (3.5)
i
(0 otherwise.
Therefore,the cyclic code C in (2.2) is Hermitian self-dual if and only if the matrix equation
(3.4) has a solution in Fi1. Moreover, the number of Hermitian self-dual cyclic codes of length
ps
pa over GR(p2,s) with first torsion degree i equals the number of solutions of (3.4) in Fi1.
1 ps
From (3.3), we observe that M(pa,i ) has the following properties. For 1≤j,i≤i , let m
1 1 ij
denote the entry in the ith row and jth column of M(pa,i ). Then, for integers 1≤k≤j ≤i≤
1
i , we have
1
i −j+1
m =(−1)i0+j−1 0 ,
ij
i−j
(cid:18) (cid:19)
and hence,
i−k
m m =(−1)i0+j−1 m . (3.6)
ij jk ik
j−k
(cid:18) (cid:19)
Lemma 3.1. Let i be an integer such that 1≤i<i . Then
1
i
m b =0 in F .
i+1,j j ps
j=1
X
Proof. If i < pa−1−1, then b =0 for all 1≤i≤i , and hence, the result follows.
1 2 i 1
Assume that i ≥ pa−1−1. If i < pa−1+1, then b = 0 for all 1 ≤ j ≤ i, and hence,
1 2 2 j
i m b =0 in F . Assume that i≥ pa−1+1. Then, by (3.5), we have
j=1 i+1,j j ps 2
P
i
mi+1,jbj =mi+1,pa−1−i1+1
j=1
X
=(−1)i0+pa−1−i1 pa−pa−1
i +1−i
(cid:18) 0 (cid:19)
=0 in F
ps
by Lucas’s Theorem (see [3, Theorem 26]) and the fact that 1≤i −i<pa−1.
0
In order to determine the number of solutions of (3.4) in Fi1, we recall two maps which are
ps
important tools.
1. The trace map Tr:F →F defined by α7→α+αps/2 for all α∈F .
ps ps/2 ps
2. The map Ψ:F →F defined by Ψ(α)=αps/2 −α for all α∈F .
ps ps ps
It is well know that Tr is F -linear and it is not difficult to see that Ψ is also F -linear.
ps/2 ps/2
If p=2, then Tr=Ψ. Moreover,we have the following properties.
Lemma 3.2. For each α∈F , Ψ(α)=0 if and only if α∈F . Moreover,
ps ps/2
Ψ◦Tr≡0≡Tr◦Ψ. (3.7)
Proof. It follows immediately from the definitions.
5
Lemma 3.3. The following statements hold.
i) For each a∈Ψ(F ), |Ψ−1(a)|=ps/2.
ps
ii) For each a∈Tr(F ), |Tr−1(a)|=ps/2.
ps
Proof. We note that, for each a ∈ Ψ(F ) and b ∈ F , b ∈ Ψ−1(a) if and only if b+ker(Ψ) =
ps ps
Ψ−1(a). Since ker(Ψ)=F , the statement i) follows.
ps/2
Similarly, for each a∈Tr(F ) and b∈F , b∈Tr−1(a) if and only if b+ker(Tr)=Tr−1(a).
ps ps
Since Tr is a surjective F -linear map from F to F , we have
ps/2 ps ps/2
|F |
|ker(Tr)|= ps =ps/2,
|F |
ps/2
and hence, ii) follows.
Proposition 3.4. Let s be an even positive integer and let i be a positive integer such that
1
i ≤pa−1. Then the number of solutions of (3.4) in Fi1 is
1 ps
psi1/2.
Proof. From[10],M(pa,i )has4presentationsdependingontheparityofpandi . Wetherefore
1 1
separate the proof into 4 cases.
Case 1. p is odd and i =2µ +1 is odd.
1 1
From [10], the matrix M(pa,i ) can be written as
1
2 0 0 0 ··· 0
∗ 0 0 0 ··· 0
∗ ∗ 2 0 ··· 0
M(pa,i1)= ∗ ∗ ∗ 0 ··· 0 , (3.8)
... ... ... ... ... ...
∗ ∗ ∗ ∗ ··· 2
where ∗’s denote entries of M(pa,i ) defined in (3.3).
1
From (3.4) and (3.8), we have
Tr(x )=b , (3.9)
1 1
2i−1
Ψ(x )=b − m x , (3.10)
2i 2i 2i,j j
j=1
X
and
2i
Tr(x )=b − m x (3.11)
2i+1 2i+1 2i+1,j j
j=1
X
for all integers 1≤i≤µ .
1
This implies that (3.4) has a solution if and only if the right hand sides of (3.9) and (3.11)
are in F and the right hand side of (3.10) is in Ψ(F ). In this case, we have
ps/2 ps
x ∈Tr−1(b ),
1 1
2i−1
x ∈Ψ−1 b − m x ,
2i 2i 2i,j j
j=1
X
6
and
2i
x ∈Tr−1 b − m x
2i+1 2i+1 2i+1,j j
j=1
X
for all 1≤i≤µ1. Hence, by Lemma 3.3, the number of solutions of (3.4) is psi1/2.
By Lemma 3.2, it suffices to show that the images under Ψ of the right hand sides of (3.9)
and (3.11) are 0 and the image under the trace map of the right hand side of (3.10) is 0.
From (3.9), we have Ψ(b )=0 since b ∈{0,1}⊆F . Let 1≤i≤µ be an integer. From
1 1 ps/2 1
(3.10), we have
2i−1 2i−1
Tr b − m x =Tr(b )− m Tr(x )
2i 2i,j j 2i 2i,j j
j=1 j=1
X X
i i−1
=0− m Tr(x )+ m Tr(x ) ,
2i,2j−1 2j−1 2i,2j 2j
j=1 j=1
X X
since pa−1−i +1 is odd and then b =0 forall i=1,2,...,µ ,
1 2i 1
i i−1
=− m Tr(x )+ m (Φ(x )+2x ) ,
2i,2j−1 2j−1 2i,2j 2j 2j
j=1 j=1
X X
since Tr(α)=Φ(α)+2α for all α∈F ,
ps
2i−1 2i−1j−1 i−1
=− m b + m m x − 2m x
2i,j j 2i,j j,k k 2i,2j 2j
j=1 j=1 k=1 j=1
X X X X
2i−2 2i−1 i−1
=0+ m m x − 2m x , by Lemma 3.1,
2i,j j,k k 2i,2j 2j
k=1 j=k+1 j=1
X X X
2i−2 2i−1 i−1
2i−k
= (−1)i0+j−1 m x − 2m x , by (3.6),
2i,k k 2i,2j 2j
j−k
k=1 j=k+1 (cid:18) (cid:19) j=1
X X X
2i−2 2i−k−1 i−1
2i−k
= m (−1)j+k−1 x − 2m x
2i,k k 2i,2j 2j
j
k=1 j=1 (cid:18) (cid:19) j=1
X X X
i 2i−2k
2i−2k+1
= m (−1)j x
2i,2k−1 2k−1
j
k=1 j=1 (cid:18) (cid:19)
X X
i−1 2i−2k−1 i−1
2i−2k
+ m −(−1)j x − 2m x
2i,2k 2k 2i,2j 2j
j
k=1 j=1 (cid:18) (cid:19) j=1
X X X
i
= m (−1−(−1)2i−2k+1) x
2i,2k−1 2k−1
k=1
X(cid:0) (cid:1)
i−1 i−1
+ −(−1−(−1)2i−2k)m x − 2m x
2i,2k 2k 2i,2j 2j
k=1 j=1
X(cid:0) (cid:1) X
=0 in F . (3.12)
ps/2
7
From (3.11), we have
2i 2i
Ψ b − m x =Ψ(b )− m Ψ(x )
2i+1 2i+1,j j 2i+1 2i+1,j j
j=1 j=1
X X
i i
=0− m Ψ(x )+ m Ψ(x ) ,
2i+1,2j−1 2j−1 2i+1,2j 2j
j=1 j=1
X X
since b ∈F for all i=1,2,...,µ ,
2i+1 pν 1
i i
=− m (Tr(x )−2x )+ m Ψ(x ) ,
2i+1,2j−1 2j−1 2j−1 2i+1,2j 2j
j=1 j=1
X X
since Ψ(α)=Tr(α)−2α for all α∈F ,
ps
2i 2i j−1 i
=− m b + m m x + 2m x
2i+1,j j 2i+1,j j,k k 2i+1,2j−1 2j−1
j=1 j=1k=1 j=1
X XX X
2i−1 2i i
=0+ m m x + 2m x ,
2i+1,j j,k k 2i+1,2j−1 2j−1
k=1 j=k+1 j=1
X X X
by Lemma 3.1,
2i−1 2i
2i−k+1
= (−1)i0+j−1 m x
2i+1,k k
j−k
k=1 j=k+1 (cid:18) (cid:19)
X X
i
+ 2m x , by (3.6),
2i+1,2j−1 2j−1
j=1
X
i 2i−2k−1
2i−2k+2
= m (−1)j x
2i+1,2k−1 2k−1
j
k=1 j=1 (cid:18) (cid:19)
X X
i 2i−2k i
2i−2k+1
+ m −(−1)j x + 2m x
2i,2k 2k 2i+1,2j−1 2j−1
j
k=1 j=1 (cid:18) (cid:19) j=1
X X X
i
= (−1−(−1)2i−2k+2)m x
2i+1,2k−1 2k−1
k=1
X(cid:0) (cid:1)
i i
+ −(−1−(−1)2i−2k+1)m x + 2m x
2i,2k 2k 2i+1,2j−1 2j−1
k=1 j=1
X(cid:0) (cid:1) X
=0 in F . (3.13)
ps/2
This case is completed.
Case 2. p is odd and i =2µ is even.
1 1
8
From [10], we have
0 0 0 0 ··· 0
∗ 2 0 0 ··· 0
∗ ∗ 0 0 ··· 0
M(pa,i1)= ∗ ∗ ∗ 2 ··· 0 , (3.14)
... ... ... ... ... ...
∗ ∗ ∗ ∗ ··· 2
where ∗’s denote entries of M(pa,i ) defined in (3.3).
1
From (3.4) and (3.14), we have
2i−2
Ψ(x )=b − m x (3.15)
2i−1 2i−1 2i−1,j j
j=1
X
and
2i−1
Tr(x )=b − m x (3.16)
2i 2i 2i,j j
j=1
X
for all integers 1≤i≤µ .
1
This implies that (3.4) has a solution if and only if the right hand side of (3.15) is in Ψ(F )
ps
andthe righthandside of(3.16) is inF . Inthis case,by Lemma 3.3, the number ofsolutions
ps/2
of (3.4) is psi1/2.
By Lemma 3.2, it is sufficient to show that the image under the trace map of the right hand
side of (3.15) and the image under Ψ of the right hand side of (3.16) are 0. Using computations
similar to those in (3.12) and (3.13), the desired properties can be concluded.
Case 3. p=2 and i =2µ +1 is odd.
1 1
From [10], we have
0 0 0 0 ··· 0
1 0 0 0 ··· 0
∗ 0 0 0 ··· 0
M(pa,i1)= ∗ ∗ 1 0 ··· 0 , (3.17)
... ... ... ... ... ...
∗ ∗ ∗ ∗ ··· 0
where ∗’s denote entries of M(pa,i ) defined in (3.3).
1
From (3.4) and (3.17), we conclude that
Tr(x )=b , (3.18)
1 1
2i−1
Tr(x )=b + m x , (3.19)
2i 2i 2i,j j
j=1
X
and
2i−1
Tr(x )=b + m x (3.20)
2i+1 2i+1 2i+1,j j
j=1
X
for all integers 1≤i≤µ .
1
9
Similar to Cases 1 and 2, we need to show that the right hand sides of (3.18), (3.19), and
(3.20) are in F , or equivalently, the images under the trace map of the right hand sides of
2s/2
(3.18), (3.19), and (3.20) are 0. Clearly, the right hand side of (3.18) is b ∈{0,1}⊆ F and
1 2s/2
Tr(b )=0. Let 1≤i≤µ be an integer.
1 1
From (3.19), we have
2i−1 2i−1
Tr b + m x = m Tr(x )
2i 2i,j j 2i,j j
j=1 j=1
X X
2i−1 j−1
= m b + m x
2i,j j j,k k
!
j=1 k=1
X X
2i−1 2i−1j−1
= m b + m m x
2i,j j 2i,j j,k k
j=1 j=1 k=1
X X X
2i−2 2i−1
=0+ m m x , by Lemma 3.1,
2i,j j,k k
k=1 j=k+1
X X
2i−2 2i−1
2i−k
= m x , by (3.6),
2i,k k
j−k
k=1 j=k+1(cid:18) (cid:19)
X X
2i−2 2i−k−1
2i−k
= m x
2i,k k
j
k=1 j=1 (cid:18) (cid:19)
X X
2i−2
= m (22i−k−2) x
2i,k k
k=1
X (cid:0) (cid:1)
=0 in F . (3.21)
2s/2
Applying a similar computation to (3.20) yields
2i
Tr b + m x =0 in F .
2i+1 2i+1,j j 2s/2
j=1
X
Case 4. p=2 and i =2µ +2 is even.
1 1
Form [10], we have
0 0 0 0 0 ··· 0
0 0 0 0 0 ··· 0
∗ 1 0 0 0 ··· 0
M(pa,i )= ∗ ∗ 0 0 0 ··· 0 , (3.22)
1
∗ ∗ ∗ 1 0 ··· 0
.. .. .. .. .. .. ..
. . . . . . .
∗ ∗ ∗ ∗ ∗ ··· 0
where ∗’s denote entries of M(pa,i ) defined in (3.3).
1
From (3.4) and (3.22), it follows that
Tr(x )=b , (3.23)
1 1
Tr(x )=b , (3.24)
2 2
10
