The Generalized Reed-Muller codes and the radical powers of a modular algebra

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by  Harinaivo Andriatahiny

The Generalized Reed-Muller codes and the
radical powers of a modular algebra
6
1
Harinaivo ANDRIATAHINY
0
2 Mention: Mathématiques et informatique,
n Domaine: Sciences et Technologies,
a Université d’Antananarivo, Madagascar
J
e-mail: [email protected]
8
2
January 29, 2016

]
T
I
. Abstract
s
c First, a new proof of Berman and Charpin’s characterization of the
[
Reed-Muller codes over the binary field or over an arbitrary prime field

1 ispresented. Thesecodesareconsidered asthepowers oftheradical ofa
v modular algebra. Secondly,thesame method isused for thestudy of the
3 Generalized Reed-Muller codes overa non primefield.
3
6 Keywords: Reed-Muller codes, modular algebra,nilpotent radical
7 MSC 2010: 16N40, 94B05, 12E05
0
.
1
1 Introduction
0
6
1 S.D. Berman [4] showed that the binary Reed-Muller codes may be identified
: with the powers of the radical of the group algebra over the two elements field
v
i F2 of an elementary abelian 2-group. P. Charpin [6] gave a generalization of
X
Berman’s result for Reed-Muller codes over a prime field.
r ManyauthorsexploredBerman’sideaandgaveanotherproofsofBerman’sthe-
a
orem (see [2],[9]).
Recently, I.N. Tumaikin [11][12] studied the connections between Basic Reed-
Muller codes and the radicalpowersof the modular groupalgebraF [H] where
q
H is a multiplicative group isomorphic to the additive group of the field F of
q
order q = pr where p is a prime number and r is an integer. The index of
nilpotency of the radical of F [H] is r(p−1)+1.
q
This paper presents an elementary proof of Berman and Charpin’s characteri-
zationof the Reed-Muller codes by using a polynomial approach. The modular
algebra F [X ,...,X ]/(Xp−1,...,Xp −1) where m ≥ 1 is an integer is
p 1 m 1 m
used to representthe ambient space of the codes. It is isomorphic to the group
algebra Fp[Fpm]. We utilize some properties of a linear basis of the modular
1
algebra.
WestudyalsothecaseoftheGeneralizedReed-Muller(GRM)codesoveranon
prime field F (with r >1). We consider the modular algebra
q
A=F [X ,...,X ]/(Xq−1,...,Xq −1).
q 1 m 1 m
The index of nilpotency of the radical M of A is m(q −1)+1, so there are
m(q−1)+1 non-zeropowersof M (with M0 =A). It is well-knownthat there
are m(q−1)+1 non-zero Reed-Muller codes of length qm over F . The main
q
result is Theorem 6.1 which gives the GRM codes over F which are radical
q
powersofA. We showthatexceptforM0,M andMm(q−1), noneofthe radical
powers of A is a GRM code over the non prime field F .
q
The outline of the paper is as follows. In section 2, we give the definition and
some properties of the GRM codes of length qm over a finite field F (see [5]).
q
This section is also devoted to the modular algebra A. In section 3, we give a
proofforthelinks (see[1],[4]and[6])betweenGRMcodesoveraprimefieldF
p
and the radical powers of F [X ,...,X ]/(Xp−1,...,Xp −1). In section
p 1 m 1 m
4,westudythefunctions(x−1)i,0≤i≤q−1,overafinitefieldF . Insection
q
5, we examine the case of the GRM codes over a non prime field. In section 6,
an exemple is given.
2 The modular algebra A = F [X ,...,X ]/(Xq −
q 1 m 1
1,...,Xq − 1) and the GRM codes
m
Let q =pr with p a prime number and r ≥1 an integer. We consider the finite
field F of order q.
q
Let P(m,q) be the vectorspace of the reducedpolynomials in m variables over
F :
q
q−1 q−1
P(m,q):= P(Y ,...,Y )= ··· u Yi1...Yim |u ∈F .
1 m i1...im 1 m i1...im q
( )
iX1=0 iXm=0
(1)
The polynomial functions from (F )m to F are given by the polynomials of
q q
P(m,q).
Let ν be an integer such that 0 ≤ ν ≤ m(q −1). Consider the subspace of
P(m,q) defined by
P (m,q):={P(Y ,...,Y )∈P(m,q)|deg(P(Y ,...,Y ))≤ν}.
ν 1 m 1 m
Consider the ideal I =(Xq−1,...,Xq −1) of the ring F [X ,...,X ].
1 m q 1 m
Set x =X +I,...,x =X +I. Then
1 1 m m
q−1 q−1
A= ··· a xi1...xim |a ∈F (2)
i1...im 1 m i1...im q
( )
iX1=0 iXm=0
2
A is a local ring with maximal ideal M which is the radical of A.
Letdbe anintegersuchthat0≤d≤m(q−1). Consider the powersMd ofM.
A linear basis of Md over F is
q
B := (x −1)i1...(x −1)im |0≤i ,...,i ≤q−1,i +...+i ≥d
d 1 m 1 m 1 m
(3)
(cid:8) (cid:9)
We have the following ascending sequence of ideals:
{0}=Mm(q−1)+1 ⊂Mm(q−1) ⊂···⊂M2 ⊂M ⊂A (4)
Let us fix an order on the set of monomials
xi1...xim |0≤i ,...,i ≤q−1 .
1 m 1 m
Then we have the fol(cid:8)lowing remark: (cid:9)
2.1 Remark. Each element iq1−=10··· iqm−=10ai1...imxi11...ximm of A can be
identifiedwiththevector(a ) of(F )qm andvice-versa. Hence
i1...iPm 0≤i1,...,iPm≤q−1 q
the modular algebra A is identified with (F )qm.
q
Let α be a primitive element of the finite field F . It is clear that F =
q q
0,1,α,α2,...,αq−2 .
Set
(cid:8) β =(cid:9)0 and β =αi−1 for 1≤i≤q−1. (5)
0 i
WhenconsideringP(m,q)andAasvectorspacesoverF ,wehavethefollowing
q
isomorphism:
φ: P(m,q)−→A
q−1 q−1 (6)
P(Y ,...,Y )7−→ ··· P(β ,...,β )xi1...xim
1 m i1 im 1 m
iX1=0 iXm=0
The GeneralizedReed-Muller code of length qm andof order ν (0≤ν ≤m(q−
1)) over F is defined by
q
C (m,q):={(P(β ,...,β )) |P(Y ,...,Y )∈P (m,q)}. (7)
ν i1 im 0≤i1,...,im≤q−1 1 m ν
It is a subspace of (F )qm and we have the following ascending sequence:
q
{0}⊂C (m,q)⊂C (m,q)⊂···⊂C (m,q)⊂C (m,q)=(F )qm
0 1 m(q−1)−1 m(q−1) q
(8)
We need the following notations:
2.2 Notations. Set [0,q−1]={0,1,2,...,q−1},
i:=(i ,...,i )∈([0,q−1])m,
1 m
|i|:=i +...+i ,
1 m
j ≤i if j ≤i for all l =1,2,...,m where j :=(j ,...,j )∈([0,q−1])m,
l l 1 m
x:=(x ,...,x ),
1 m
xi :=xi1...xim.
1 m
3
Consider the polynomial
B (x):=(x −1)i1...(x −1)im. (9)
i 1 m
It is clear that
il i
(x −1)il = (−1)il−j l xj (10)
l j l
j=0 (cid:18) (cid:19)
X
for all l =1,2,...,m.
2.3 Proposition. Considering the sequences (8)and (4), we have
dimFq(Md)=dimFq(Cm(q−1)−d(m,q))
for 0≤d≤m(q−1).
Proof. Consider the set E :={i∈([0,q−1])m ||i|≥d}.
Since Bd = Bi(x)||i|≥d is a basis of Md, then dimFq(Md) = Card(E)
where Card(E) denotes the number of elements in the set E.
(cid:8) (cid:9)
Consider the set F :={i∈([0,q−1])m ||i|≤m(q−1)−d}.
We have dimFq(Cm(q−1)−d(m,q))=dimFq(Pm(q−1)−d(m,q))=Card(F).
The mapping
θ :([0,q−1])m −→([0,q−1])m
(i ,...,i )7−→(q−1−i ,...,q−1−i )
1 m 1 m
is a bijection and the inverse mapping is θ−1 =θ.
Let i ∈ E. Then |i| ≥ d, and |θ(i)| = m(q −1)−|i| ≤ m(q−1)−d. Hence
θ(i)∈F. And it follows that θ(E)⊆F.
Conversely, let i∈F. Then |i|≤m(q−1)−d, and |θ(i)|=m(q−1)−|i|≥d.
So θ(i)∈E. Note that θ(θ(i))=i. Thus F ⊆θ(E).
Therefore, F =θ(E) and Card(E)=Card(F).
Let β ∈F . Consider the indicator function
k q
F (Y )=1−(Y −β )q−1 (11)
βk l l k
with 1≤l≤m.
Then F (Y )∈P(m,q) and
βk l
1 if j =k,
F (β )=
βk j (0 otherwise.
Consider the interpolation function
il i
Hil(Yl):= (−1)il−k kl Fβk(Yl). (12)
k=0 (cid:18) (cid:19)
X
4
Then, we have H (Y )∈P(m,q),
il l
H (β )= (−1)il−j ijl if 0≤j ≤il,
il j (0 (cid:0) (cid:1) if il <j ≤q−1
and
il
(xl−1)il = Hil(βj)xjl. (13)
j=0
X
Set
m
H (Y):= H (Y ). (14)
i il l
l=1
Y
Thus
m
deg(H (Y))= deg(H (Y )). (15)
i il l
l=1
X
2.4 Proposition. We have H (Y) = φ−1(B (x)), where φ is the isomorphism
i i
defined in (6), i.e.
B (x)= H (β ,...,β )xj
i i j1 jm
j≤i
X
Proof.
m
Bi(x)= (xl−1)il
l=1
Y
m il
= ( H (β )xjl)
il jl l
Yl=1 jXl=0
m
= ( H (β ))xj
il jl
j≤i l=1
X Y
= H (β ,...,β )xj.
i j1 jm
j≤i
X
3 The GRM codes over the prime field F and the
p
radical powers of F [X ,...,X ] / ( Xp−1,...,Xp −
p 1 m 1 m
1 )
In this section, we consider the case r = 1, i.e. q = p a prime number and
F =F a prime field.
q p
Let F ={0,1,2,...,p−1} and set β =k for all k =0,1,...,p−1.
p k
5
Let us study (x−1)i for 0≤i≤p−1 over F .
p
It is clear that
i
i
(x−1)i = (−1)i−j xj.
j
j=0 (cid:18) (cid:19)
X
For k ∈F , according to (1), we have
p
p−1
F (Y)=1−(Y −k)p−1 =− (Y −j)∈P(1,p)
k
j=0Y(j6=k)
and
1 if j =k,
F (j)=
k
(0 otherwise.
Let us consider the interpolation function
i
i
H (Y):= (−1)i−k F (Y) (16)
i k
k
k=0 (cid:18) (cid:19)
X
which is in P(1,p).
Thus, H (j)=(−1)i−j i for 0≤j ≤i
i j
and
(cid:0) (cid:1) i
(x−1)i = H (j)xj.
i
j=0
X
Let a be the elements of F defined by
i,d p
i
i
a := (−1)i−j jd (17)
i,d
j
j=0 (cid:18) (cid:19)
X
where 0≤i≤p−1 and 0≤d≤p−1.
3.1 Proposition. The interpolation function (16) satisfies the relation
p−1
H (Y)=a − a Yp−1−d
i i,0 i,d
d=0
X
Proof.
i
i
H (Y)= (−1)i−k F (Y)
i k
k
k=0 (cid:18) (cid:19)
X
i
i
= (−1)i−k 1−(Y −k)p−1
k
k=0 (cid:18) (cid:19)
X (cid:2) (cid:3)
i p−1
i p−1
= (−1)i−k 1− (−1)dkd Yp−1−d
k d
k=0 (cid:18) (cid:19)" d=0 (cid:18) (cid:19) #
X X
6
Since we consider a field of characteristic p, then p−1 = (−1)d mod p. It
d
follows that
(cid:0) (cid:1)
i p−1
i
H (Y)= (−1)i−k 1− kdYp−1−d
i
k
k=0 (cid:18) (cid:19)" d=0 #
X X
i p−1 i
i i
= (−1)i−k − (−1)i−k kd Yp−1−d
k k
k=0 (cid:18) (cid:19) d=0"k=0 (cid:18) (cid:19) #
X X X
3.2 Proposition. The coefficients a satisfy the following properties
i,d
d−1
d−1
(i)- a =i a for 1≤d≤p−1 and 1≤i≤p−1.
i,d i−1,k
k
k=0(cid:18) (cid:19)
X
(ii)- a =0 for 1≤i≤p−1 and 0≤d≤i−1.
i,d
(iii)- a 6=0 for all i=1,2,...,p−1.
i,i
Proof. (i)- For d≥1,
i i
i i
a = (−1)i−j jd = (−1)i−j jd
i,d
j j
j=0 (cid:18) (cid:19) j=1 (cid:18) (cid:19)
X X
i
i
= (−1)i−j ·j· jd−1.
j
j=1 (cid:18) (cid:19)
X
But
i i i−1
= ,
j j j−1
(cid:18) (cid:19) (cid:18) (cid:19)
thus
i i
i i−1 i−1
a = (−1)i−j ·j· jd−1 =i (−1)i−j jd−1
i,d
j j−1 j−1
j=1 (cid:18) (cid:19) j=1 (cid:18) (cid:19)
X X
i
i−1
=i (−1)i−1−(j−1) ((j−1)+1)d−1.
j−1
j=1 (cid:18) (cid:19)
X
By using the relation
d−1
d−1
((j−1)+1)d−1 = (j−1)k,
k
k=0(cid:18) (cid:19)
X
and setting J =j−1, we have
d−1 i
d−1 i−1
a =i ( (−1)i−1−(j−1) (j−1)k)
i,d
k j−1
k=0(cid:18) (cid:19) j=1 (cid:18) (cid:19)
X X
d−1 i−1
d−1 i−1
=i ( (−1)i−1−J Jk).
k J
k=0(cid:18) (cid:19) J=0 (cid:18) (cid:19)
X X
7
(ii) and (iii) can be easily proved by induction on i and by using (i).
3.3 Proposition.
deg(H (Y))=p−1−i.
i
Proof. It is obvious by Proposition 3.1 and Proposition 3.2.
3.4Remark. AnexplicitexpressionofH (Y)asapolynomialofdegreep−1−i
i
is
p−1−i
H (Y)=α (Y +j),
i i
j=1
Y
where α =−i! mod p.
i
3.5 Remark. Polynomials H (Y), 0 ≤ i ≤ p−1, satisfy the backward recur-
i
rence relation
H (Y)=1,
p−1
1
H (Y)= (Y −i−1)H (Y) , 0≤i≤p−2.
i i+1
i+1
It is clear that in this section, the Proposition 2.4 become
B (x)= H (j)xj (18)
i i
j≤i
X
The following Theorem is well-known (see [4],[6]).
3.6 Theorem (Berman-Charpin). Let C (m,p) be the Reed-Muller code of
ν
length pm and of order ν (0≤ν ≤m(p−1)) over the prime field F and M the
p
radical of F [X ,...,X ]/(Xp−1,...,Xp −1). Then
p 1 m 1 m
C (m,p)=Mm(p−1)−ν.
ν
Proof. Setd:=m(p−1)−ν. The setB = B (x)||i|≥d is a linear basisof
d i
Md over F . Consider B (x) = H (j)xj ∈ Md. By (15) and Proposition
p i j≤i i (cid:8) (cid:9)
m
3.3, we have deg(H (Y)) = p−1−i = m(p−1)−|i| ≤ m(p−1)−d =
i l=P1 l
ν. It follows from Remark 2.1 and (7) that B (x) ∈ C (m,p). Thus Md ⊆
i ν
Cν(m,p). Moreover,if we taPke r =1 in Proposition 2.3, we have dimFp(Md)=
dimFp(Cν(m,p)).
4 Some properties of (x − 1)i,0 ≤ i ≤ q − 1, over
an arbitrary finite field F
q
In this section, we consider the finite field F where q = pr with p a prime
q
number and r ≥1 an integer.
We have already seen from (10), (11) and (12) of Section 2 that
i
i
(x−1)i = (−1)i−j xj , 0≤i≤q−1,
j
j=0 (cid:18) (cid:19)
X
8
and the interpolation function is
i
i
H (Y):= (−1)i−k F (Y) , 0≤i≤q−1. (19)
i k βk
k=0 (cid:18) (cid:19)
X
And we have
i
(x−1)i = H (β )xj , 0≤i≤q−1.
i j
j=0
X
4.1 Lemma.
pr−1
=(−1)d mod p
d
(cid:18) (cid:19)
where p is a prime number, r ≥1 an integer and 0≤d≤pr−1.
Proof. It can be proved easily by induction on d.
The following proposition is fundamental.
4.2 Proposition. The interpolation function (19) satisfies the relation
q−1
H (Y)= α−d (−1)i−(αd−1)i Yq−1−d
i
d=1
X (cid:2) (cid:3)
where α is a primitive element of F and 1≤i≤q−1.
q
Proof. We have
i
i
H (Y)= (−1)i−k F (Y)
i k βk
k=0 (cid:18) (cid:19)
X
i
i
= (−1)i−k 1−(Y −β )q−1
k
k
k=0 (cid:18) (cid:19)
X (cid:2) (cid:3)
i
i
=(−1)i(1−Yq−1)+ (−1)i−k 1−(Y −β )q−1
k
k
k=1 (cid:18) (cid:19)
X (cid:2) (cid:3)
i i
i i
=(−1)i(1−Yq−1)+ (−1)i−k − (−1)i−k (Y −β )q−1.
k
k k
k=1 (cid:18) (cid:19) k=1 (cid:18) (cid:19)
X X
Since
q−1
q−1
(Y −β )q−1 = (−1)d(β )d Yq−1−d
k k
d
d=0 (cid:18) (cid:19)
X
and by Lemma 4.1, we have
q−1
(Y −β )q−1 = (β )dYq−1−d.
k k
d=0
X
9
Thus, by (5),
i
i
H (Y)=(−1)i(1−Yq−1)+ (−1)i−k
i
k
k=1 (cid:18) (cid:19)
X
i q−1
i
− (−1)i−k (αk−1)dYq−1−d
k
k=1 (cid:18) (cid:19)"d=0 #
X X
i
i
=(−1)i(1−Yq−1)+ (−1)i−k
k
k=1 (cid:18) (cid:19)
X
q−1 i
i
− ( (−1)i−k (αk−1)d)Yq−1−d.
k
d=0 k=1 (cid:18) (cid:19)
X X
Since
i
i
(−1)i−k =(−1)i+1
k
k=1 (cid:18) (cid:19)
X
then
i
i
H (Y)=(−1)i−(−1)iYq−1−(−1)i−( (−1)i−k )Yq−1
i
k
k=1 (cid:18) (cid:19)
X
q−1 i
i
− ( (−1)i−k (αk−1)d)Yq−1−d
k
d=1 k=1 (cid:18) (cid:19)
X X
q−1 i
i
=− ( (−1)i−k (αk−1)d)Yq−1−d
k
d=1 k=1 (cid:18) (cid:19)
X X
q−1 i
i
=− α−d( (−1)i−k (αk)d)Yq−1−d
k
d=1 k=1 (cid:18) (cid:19)
X X
q−1 i
i
=− α−d (−1)i−k (αk)d−(−1)i Yq−1−d
k
d=1 "k=0 (cid:18) (cid:19) #
X X
q−1
=− α−d (αd−1)i−(−1)i Yq−1−d
d=1
X (cid:2) (cid:3)
q−1
= α−d (−1)i−(αd−1)i Yq−1−d.
d=1
X (cid:2) (cid:3)
4.3 Corollary. 1. H (Y)=− q−2Yd.
1 d=0
2. H (Y)=− q−2(2−α−k)YPk.
2 k=0
3. H (Y)=P1.
q−1
10