On the third weight of generalized Reed-Muller codes

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by  Elodie Leducq

On the third weight of generalized Reed-Muller
codes
Elodie Leducq
Abstract
4
1 In this paper, we study the third weight of generalized Reed-Muller
0 codes. Using results from [6], we prove under some restrictive condition
2
that the third weight of generalized Reed-Muller codes depends on the

n third weight of generalized Reed-Muller codes of small order with two
a variables. In some cases, we are able to determine the third weight and
J thethird weight codewords of generalized Reed-Muller codes.

1
2
1 Introduction

]
T
In this paper, we study the third weight of generalized Reed-Muller codes.
I
.
s
c We first introduce some notations :
[

1 Let p be a prime number, e a positive integer, q = pe and Fq a finite field
v with q elements.
4
3 If m is a positive integer, we denote by Bq the F -algebra of the functions
2 from Fm to F and by F [X ,...,X ] the F -malgebraqof polynomials in m vari-
5 q q q 1 m q
ables with coefficients in F .
. q
1 We consider the morphism of F -algebras ϕ : F [X ,...,X ] → Bq which
0 q q 1 m m
associates to P ∈F [X ,...,X ] the function f ∈Bq such that
4 q 1 m m
1
∀x=(x ,...,x )∈Fm, f(x)=P(x ,...,x ).
: 1 m q 1 m
v
i Themorphismϕisontoanditskernelistheidealgeneratedbythepolynomials
X
Xq−X ,...,Xq −X . So, for eachf ∈Bq , there exists a unique polynomial
r 1 1 m m m
a P ∈Fq[X1,...,Xm] such that the degree of P in each variable is at most q−1
and ϕ(P) = f. We say that P is the reduced form of f and we define the
degree deg(f) of f as the degree of its reduced form. The support of f is the
set {x ∈ Fm : f(x) 6= 0} and we denote by |f| the cardinal of its support (by
q
identifying canonically Bq and Fqm, |f| is actually the Hamming weight of f).
m q
For 0≤r ≤m(q−1), the rth order generalizedReed-Muller code of length
qm is
R (r,m):={f ∈Bq :deg(f)≤r}.
q m
For 1 ≤ r ≤ m(q −1)−2, the automorphism group of generalized Reed-
Muller codes R (r,m) is the affine group of Fm (see [2]).
q q
1
For more results on generalized Reed-Muller codes, we refer to [5].
In the following of the article, we write r = a(q−1)+b, 0 ≤ a ≤ m−1,
1≤b≤q−1.
In [9], interpreting generalizedReed-Muller codes in terms of BCH codes, it
isprovedthattheminimalweightofthegeneralizedReed-MullercodeR (r,m)
q
is (q−b)qm−a−1. The minimum weight codewords of generalized Reed-Muller
codes are described in [5] (see also [12]).
In his Ph.D thesis [6], Ericksonproves that if we know the second weight of
R (b,2),thenweknowthe secondweightforallgeneralizedReed-Mullercodes.
q
Fromaconjectureonblockingsets,Ericksonconjecturesthatthesecondweight
ofR (b,2)is(q−b)q+b−1. Bruenprovesthe conjectureonblockingsetin[3].
q
Geil also provesthis result in [7] using Groebner basis. An altenative approach
can be found in [14] where the second weight of most R (r,m) is established
q
without using Erickson’s results. Second weight codewords have been studied
in [4], [15] and finally completely described in [13].
For q = 2, small weights and small weight codewords are described in [10],
the third weight for r > (m−1)(q−1)+1 is given in [7], we can find results
on small weight codewords in [1]. In the following, we consider only q ≥3 and
r ≤m(q−1)+1.
We first give some tools that we will use through all the paper. Then we
give an upper bound on the third weight of generalized Reed-Muller codes. In
Section 4 is the main result of this article : we describe the third weight of
generalized Reed-Muller codes with some restrictive conditions. In section 5,
we study more particularly the case of two variables which is quite essential in
thedeterminationofthethirdweight. InSection6,wedescribedthecodewords
reachingthe third weight. In section7, we summarize the results obtain in this
article. This article ends with an Appendix which gives more precisions on the
results in Section 3.
2 Preliminaries
2.1 Notation and preliminary remark
Let f ∈Bq , λ∈F . We define f ∈Bq by
m q λ m
∀x=(x ,...,x )∈Fm, f (x)=f(λ,x ...,x ).
2 m q λ 2 m
Let 0≤r ≤(m−1)(q−1) and f ∈R (r,m). We denote by S the support
q
of f. Consider H an affine hyperplane in Fm, by an affine transformation,
q
we can assume x = 0 is an equation of H. Then S ∩H is the support of
1
f ∈R (r,m−1) or the support of (1−xq−1)f ∈R (r+(q−1),m).
0 q 1 q
2.2 Useful lemmas
Lemma 2.1 Let q ≥ 3, m ≥ 3 and S a set of points of Fm such that #S =
q
uqn < qm, u 6≡ 0 mod q. If for all hyperplane H #(S ∩H) = 0, #(S ∩H) =
wqn−1, #(S ∩H) = vqn−1 or #(S ∩H) ≥ uqn−1, with w < v < u, then there
2
exists H an affine hyperplane such that #(S ∩H) = 0, #(S ∩H) = wqn−1 or
#(S∩H)=vqn−1.
Proof : Assume for all H hyperplane, #(S ∩H) ≥ uqn−1. Consider an
affine hyperplane H; then for all H′ hyperplane parallel to H, #(S ∩H′) ≥
u.qn−1. Since u.qn =#S = #(S∩H′), we get that for all H hyperplane,
HX′//H
#(S∩H)=u.qn−1.
NowconsiderAanaffinesubspaceofcodimension2andthe(q+1)hyperplanes
through A. These hyperplanes intersect only in A and their union is equal to
Fm. So
q
uqn =#S =(q+1)u.qn−1−q#(S∩A).
Finally we get a contradiction if n = 1 since u 6≡ 0 mod q. Otherwise, #(S ∩
A) = u.qn−2. Iterating this argument, we get that for all A affine subspace of
codimension k ≤n, #(S∩A)=u.qn−k.
Let A be an affine subspace of codimension n+1 and A′ an affine subspace of
codimension n−1 containing A. We consider the (q +1) affine subspaces of
codimension n containing A and included in A′, then
u.q =#(S∩A′)=(q+1)u−q#(S∩A)
which is absurd since #(S∩A) is an integer and u6≡0 mod q. So there exists
H anhyperplanesuchthat#(S∩H )=vqn−1, #(S∩H )=wqn−1 orS does
0 0 0
not meet H .
0
✷
The following lemma is proved in [5].
Lemma 2.2 Let m ≥1, q ≥2, f ∈Bq and w ∈F . If for all (x ,...,x ) in
m q 2 m
Fm−1, f(w,x ,...,x )=0 then for all (x ,...,x )∈Fm,
q 2 m 1 m q
f(x ,...,x )=(x −w)g(x ,...,x )
1 m 1 1 m
with deg (g)≤deg (f)−1 and deg(g)≤deg(f)−1.
x1 x1
The following lemmas are proved in [6]
Lemma 2.3 Let m ≥ 2, q ≥ 3, 0 ≤ r ≤ m(q −1). If f ∈ R (r,m), f 6= 0
q
and there exists y ∈ R (1,m) and (λ ) n elements in F such that the
q i 1≤i≤n q
hyperplanes of equation y =λ do not meet the support of f, then
i
n(b−n)qm−a−2 if n<b
|f|≥(q−b)qm−a−1+
(n−b)(q−1−n)qm−a−1 if n≥b
(cid:26)
where r =a(q−1)+b, 1≤b≤q−1.
Lemma 2.4 Let m ≥ 2, q ≥ 3, 1 ≤ b ≤ q−1. Assume f ∈ R (b,m) is such
q
that f depends only on x and g ∈R (b−k,m), 1≤k ≤b. Then either f +g
1 q
depends only on x or |f +g|≥(q−b+k)qm−1.
1
3
Lemma 2.5 Let m ≥ 2, q ≥ 3, 1 ≤ a ≤ m −1, 1 ≤ b ≤ q −2. Assume
f ∈R (a(q−1)+b,m) is such that ∀x=(x ,...,x )∈Fm,
q 1 m q
f(x)=(1−xq−1)f(x ,...,x )
1 2 m
and g ∈ R (a(q−1)+b−k), 1 ≤ k ≤ q−1, is such that (1−xq−1) does not
q e 1
divide g. Then, either |f +g|≥(q−b+k)qm−a−1 or k =1.
Lemma 2.6 Let m≥2, q ≥3, 1≤a≤m−2, 1≤b≤q−2 and f ∈R (a(q−
q
1)+b,m). We set an order on the elements of F such that |f |≤...≤|f |.
q λ1 λq
If f has no linear factor and there exists k ≥1 such that (1−xq−1) divides
2
f for i≤k but (1−xq−1) does not divide f then,
λi 2 λk+1
|f|≥(q−b)qm−a−1+k(q−k)qm−a−2.
Lemma 2.7 Let m ≥ 2, q ≥ 3, 1 ≤ a ≤ m and f ∈ R (a(q−1),m) such that
q
|f|=qm−a and g ∈R (a(q−1)−k,m), 1≤k ≤q−1, such that g 6=0. Then,
q
either |f +g|=kqm−a or |f +g|≥(k+1)qm−a.
Lemma 2.8 Let m ≥ 2, q ≥ 3, 1 ≤ a ≤ m− 1 and f ∈ R (a(q − 1),m).
q
If for some u, v ∈ F , |f | = |f | = qm−a−1, then there exists T an affine
q u v
transformation fixing x such that
1
(f ◦T) =(f ◦T) .
u v
3 An upper bound on the third weight
Theorem 3.1 Let q ≥3, m ≥2, 0≤ a ≤ m−1, 1 ≤ b ≤q−1, then if W is
3
the third weight of R (a(q−1)+b,m), we have
q
• If b=1 then,
– For q =3, m≥3, 1≤a≤m−2,
W ≤3m−a.
3
– For q =4, m≥3 and 1≤a≤m−2,
W ≤18.4m−a−2.
3
– For q =3 and a=m−1 or q =4 and a=m−1,
W ≤2(q−1).
3
– For q ≥5 and 1≤a≤m−1,
W ≤2(q−2)qm−a−1.
3
• If 2≤b≤q−1
– For q ≥5, 0≤a≤m−2 and 4≤b≤⌊q +2⌋,
2
W ≤(q−2)(q−b+2)qm−a−2.
3
4
– For q ≥ 7, 0 ≤ a ≤ m− 2 and ⌈q +2⌉ ≤ b ≤ q −1 or q ≥ 4,
2
0 ≤ a ≤ m−2 and b = 2 or q ≥ 4, a = m−2 and b = 3 or q = 3,
a∈{0,m−2} and b=2
W ≤(q−b+1)qm−a−1.
3
– For q ≥4, m≥3, 0≤a≤m−3 and b=3,
W ≤(q−1)3qm−a−3.
3
– For q =3, m≥4, 1≤a≤m−3 and b=2,
W ≤16.3m−a−3.
3
Proof :
• If b=1 then,
– For q =3, m≥3, 1≤a≤m−2, define for x=(x ,...,x )∈Fm,
1 m q
a
f(x)= (1−x2).
i
i=1
Y
Then, f ∈R (2a+1,m) and |f|=3m−a >8.3m−a−2.
3
– For q = 4, m ≥ 3 and 1 ≤ a ≤ m−2, define for x = (x ,...,x ) ∈
1 m
Fm,
q
a−1
f(x ,...,x )= (1−x3)(x −u)(x −v)(x −w)(x −z)
1 m i a a a+1 a+2
i=1
Y
with u, v, w, z ∈ F and u 6= v. Then, f ∈ R (3a + 1,m) and
q 4
|f|=18.4m−a−2>4m−a
– For q = 3 and a = m − 1 or q = 4 and a = m − 1, define for
x=(x ,...,x )∈Fm,
1 m q
m−2 q−2
f(x)= (1−xq−1) (x −b )(x −c)
i m−1 j m
i=1 j=1
Y Y
with b ∈ F , b 6= b for j 6= k and c ∈ F . Then, f ∈ R ((m−
j q j k q q
1)(q−1)+1,m) and |f|=2(q−1)>q
– For q ≥5 and 1≤a≤m−1, define for x=(x ,...,x )∈Fm,
1 m q
a−1 q−2
f(x)= (1−xq−1) (x −b )(x −u)(x −v)
i a j a+1 a+1
i=1 j=1
Y Y
with b ∈ F , b 6= b for j 6= k and u, v ∈ F , u 6= v. Then,
j q j k q
f ∈R (a(q−1)+1,m) and |f|=2(q−2)qm−a−1 >qm−a
q
• If 2≤b≤q−1
5
– For q ≥ 5, 0 ≤ a ≤ m−2 and 4 ≤ b ≤ ⌊q +2⌋, define for x =
2
(x ,...,x )∈Fm,
1 m q
a b−2
f(x)= (1−xq−1) (x −b )(x −u)(x −v)
i a+1 j a+2 a+2
i=1 j=1
Y Y
with b ∈ F , b 6= b for j 6= k and u, v ∈ F , u 6= v. Then,
j q j k q
f ∈ R (a(q − 1) + b,m) and |f| = (q − 2)(q − b + 2)qm−a−2 >
q
(q−b+1)(q−1)qm−a−2
– For q ≥ 7, 0 ≤ a ≤ m − 2 and ⌈q + 2⌉ ≤ b ≤ q − 1 or q ≥ 4,
2
0 ≤ a ≤ m−2 and b = 2 or q ≥ 4, a = m−2 and b = 3 or q = 3,
a∈{0,m−2} and b=2, define for x=(x ,...,x )∈Fm,
1 m q
a b−1
f(x)= (1−xq−1) (x −b )
i a+1 j
i=1 j=1
Y Y
with b ∈ F , b 6= b for j 6= k. Then, f ∈ R (a(q−1)+b,m) and
j q j k q
|f|=(q−b+1)qm−a−1 >(q−b+1)(q−1)qm−a−2.
– For q ≥ 4, m ≥ 3, 0 ≤ a ≤ m − 3 and b = 3, define for x =
(x ,...,x )∈Fm,
1 m q
a
f(x)= (1−xq−1)(x −u)(x −v)(x −w)
i a+1 a+2 a+3
i=1
Y
with u, v, w ∈ F . Then, f ∈ R (a(q −1)+3,m) and |f| = (q −
q q
1)3qm−a−3 >(q−2)(q−1)qm−a−2
– For q = 3, m ≥ 4, 1 ≤ a ≤ m − 3 and b = 2, define for x =
(x ,...,x )∈Fm,
1 m q
a−1 4
f(x)= (1−x2) (x −u )
i a−1+j j
i=1 j=1
Y Y
with u ∈ F . Then, f ∈ R (2(a+1),m) and |f| = 16.3m−a−3 >
j q 3
4.3m−a−2
✷
Remark 3.2 We say that B is an hyperplane arrangement in L if there exist
d
k ∈ N∗, (d ,...,d ) ∈ (N∗)k such that k d ≤ d and f ,...,f are k inde-
1 k i=1 i 1 k
pendent linear forms over Fm such that B is composed of k blocks of d parallel
hyperplanes of equation f (xq)=u ,j whePre 1≤i≤k, 1≤j ≤d , u ∈i F and
i i i i,j q
if k 6=j, u 6=u . The upper bound given in the Theorem above are the third
i,j i,k
weight among hyperplane arrangements in L . The proof of this result is given
d
in Appendix.
6
4 Third weight
4.1 The case where a = 0
We denote by c the third weight of R (b,2), for 2≤b≤q−1. From Theorem
b q
3.1, we get that
(q−2)(q−b+2) for q ≥5, 4≤b≤ q+3
2
c ≤
b for q ≥7 and q +2≤b≤q−1 or q ≥3 and b=2
 (q−b+1)q or q ≥4 and b2=3
Lemma 4.1 Let m ≥ 2, q ≥ 3, 4 ≤ b ≤ q −1 and f ∈ Rq(b,m). Assume
c <(q−b+1)q. If |f|>(q−b+1)(q−1)qm−2 then |f|≥c qm−2.
b b
Proof : We prove this result by induction on m. For m=2, it is the defini-
tion of c .
b
Letm≥3. Assumeiff ∈R (b,m−1)issuchthat|f|>(q−b+1)(q−1)qm−3
q
then |f|≥c qm−3.
b
Letf ∈R (b,m)suchthat|f|>(q−b+1)(q−1)qm−2. Assume|f|<c qm−2.
q b
We denote by S the support of f.
Assume S meets all affine hyperplanes. Then, for all H hyperplane, #(S ∩
H)≥(q−b)qm−2. Assume there existsH suchthat #(S∩H )=(q−b)qm−2.
1 1
By applying an affine transformation, we can assume x = λ, λ ∈ F is an
1 q
equationofH . We setanorderonthe elementsofF suchthat|f |≤|f |≤
1 q λ1 λ2
... ≤ |f |. Then, f is a minimal weight codeword of R (b,m−1). So, by
λq λ1 q
applying an affine transformation, we can assume f depends only on x . Let
λ1 2
k ≥ 1 be such that for all i ≤ k, f depends only on x and f does not
λi 2 λk+1
depend only on x .
2
If k>b, we can write for all x=(x ,...,x )∈Fm),
1 m q
b
f(x)= f(i) (x ,...,x ) (x −λ ) (see [6]).
λi+1 2 m 1 j
i=0 1≤j≤i
X Y
Since for i ≤ b+1, f depends only on x then f depends only on x and x
λi 2 1 2
which is a contradiction by the case m = 2. So k ≤ b and we can write for all
x=(x ,...,x )∈Fm,
1 m q
k
f(x)=g(x ,x )+ (x −λ )h(x)
1 2 1 i
i=1
Y
where deg(h)≤b−k. Then, for all x ∈F and all x∈Fm−2,
2 q q
f (x ,x)=g (x )+α.h(x ,x)
λk+1 2 λk+1 2 2
where α ∈ F∗. So, by Lemma 2.4, since f does not depend only on x ,
q λk+1 2
|f |≥(q−b+k)qm−2. We get
λk+1
|f|≥k(q−b)qm−2+(q−k)(q−b+k)qm−2
=(q−b)qm−1+(q−k)kqm−2
|f|≥(q−b)qm−1+(q−1)qm−2
7
Since c <(q−b+1)q and |f|<c qm−2, we get a contradiction.
b b
Then,forallH hyperplane,#(S∩H)≥(q−1)(q−b+1)qm−3. Byinduction
hypothesis, since |f| < c qm−2 there exists an affine hyperplane H such that
b 2
#(S ∩H ) = (q −1)(q −b+1)qm−3. So, there exists A an affine subspace
2
of codimension 2 included in H which does not meet S (see [13]). Then,
2
considering all affine hyperplanes through A, we must have
(q+1)(q−1)(q−b+1)qm−3 <c qm−2
b
which gives, since c ≤ (q−b+1)q−1, q < q−b+1. We get a contradiction
b
since b≥4.
So there exists H an hyperplane which does not meet S. We denote by n
0
the numberofhyperplanesparalleltoH whichdo notmeetS. ByLemma2.3,
0
since c ≤(q−b+2)(q−2),we getthatn=b,n=b−1orn=1. By applying
b
an affine transformation,we canassume x =λ , λ ∈F is an equationof H .
1 1 1 q 0
If n=b, then for all x=(x ,...,x )∈Fm, we have
1 m q
b
f(x)= (x −λ )
1 i
i=1
Y
with λ ∈ F and for i 6= j, λ 6= λ . In this case, f is a minimum weight
i q i j
codeword of R (b,m) which is absurd.
q
If n=b−1, then for all x=(x ,...,x )∈Fm, we have
1 m q
b−1
f(x)= (x −λ )g(x)
1 i
i=1
Y
with λ ∈ F , for i 6= j, λ 6= λ and g ∈ R (1,m). If deg(g) = 0 then f is a
i q i j q
minimum weight codeword of R (b−1,m). If deg(g) = 1 then f is a second
q
weight codeword of R (b,m). Both cases give a contradiction.
q
If n=1 then for all x=(x ,...,x )∈Fm, we have
1 m q
f(x)=(x −λ )g(x)
1 1
with g ∈R (b−1,m). Then, for i ≥2, deg(f )≤ (b−1), so, |f |≥ (q−b+
q λi λi
1)qm−2. We denote by N =#{i:|f |=(q−b+1)qm−2}. Then,
λi
N(q−b+1)qm−2+(q−1−N)(q−b+2)(q−1)qm−3 <c qm−2
b
which gives N > (q−1)2(q−b+2)−cbq ≥0 so N ≥1.
b−2
Denote by H anhyperplanesuchthat #(S∩H )=(q−b+1)qm−2. Then,
1 1
S ∩H is the support of a minimal weight codeword of R (b−1,m−1) so it
1 q
is the union of (q−b+1) parallelaffine subspaces of codimension2 included in
H .
1
Now, consider P an affine subspace of codimension 2 included in H such
1
that #(S ∩P) = (q − b + 1)qm−3. Then, for all H hyperplane through P,
8
#(S ∩ H) ≥ (q − b + 1)(q − 1)qm−3. Indeed, by definition of P, S meets
all hyperplanes through P, so, for all H hyperplane through P, #(S ∩H) ≥
(q−b)qm−2. If #(S ∩H) = (q −b)qm−2, then S ∩H is the union of (q−b)
parallel affine subspaces of codimension 2 which is absurd since it intersects
P in (q −b+1) affine subspaces of codimension 3. We can apply the same
argument to all affine subspaces of codimension 2 included in H parallel to P.
1
Now consider an hyperplane through P and the q hyperplanes parallel to this
hyperplane, since |f| < c qm−2, one of these hyperplanes, say H , meets S in
b 2
(q−b+1)(q−1)qm−3 points.
We denote by (A ) the b affine subspaces of codimension 2 included in
i 1≤i≤b
H which do not meet S. Let 1≤ i≤ b, suppose that S meets all hyperplanes
2
throughA andletH be one hyperplanethroughA . Ifallhyperplanesparallel
i i
to H meet S then as in the beginning ofthe proofof this lemma, we get #(S∩
H)≥(q−1)(q−b+1)qm−3. If there exists an hyperplane parallel to H which
does not meet S then #(S ∩H) ≥ (q −b+1)qm−2. In both cases we get a
contradictionsince (q+1)(q−b+1)(q−1)qm−3 ≥c qm−2. So,for all1≤i≤b
b
there exists an hyperplane through A which does not meet S.
i
Then at least b−1 of the hyperplanes through the (A ) which do not meet
i
S must intersect H , we get that |f| = (q−1)(q−b+1)qm−2 (see [13]) which
1
is absurd.
Figure 1
(a) (b)
H H
1 1
H H
2 2
P P
✷
Lemma 4.2 Let m ≥ 2, q ≥ 3 and f ∈ R (2,m). If |f| > (q−1)2qm−2 then
q
|f|≥(q2−q−1)qm−2.
Proof : Let f ∈R (2,m) such that |f| > (q−1)2qm−2. If deg(f)≤ 1 then
q
|f|≥(q−1)qm−1. From now, assume that deg(f)=2.
First we recall some results on quadratic forms. These results can be found in
[8] for example. If Q is a quadratic form of rank R on Fm then, there exists a
q
linear transformation such that for all x=(x ,...,x )∈Fm,
1 m q
if R=2r+1
r
Q(x)= x x +ax2 (1)
2i−1 2i 2r+1
i=1
X
or if R=2r
r
Q(x)= x x (2)
2i−1 2i
i=1
X
9
or
r−1
Q(x)= x x +ax2 +bx x +cx2 (3)
2i−1 2i 2r−1 2r−1 2r 2r
i=1
X
with ax2+bx+c is irreducible over F .
q
Then N(Q) the number of zeros of Q is
N(Q)=qm−1+(w−1)(q−1)qm−R2−1
where
1 if R is odd
w = 2 if R is even and Q is of type (2) .

0 if R is even and Q is of type (3)

We write for all x=(x ,...,x )∈Fm, f(x)=q (x)+α x +...+α x +β
1 m q 0 1 1 m m
where q is a quadratic form of rank r and w is defined as above. Then the
0 0
number of zeros of f is the number of affine zeros of the homogeneized form
Q(x)=q (x)+α x z+...+α x z+βz2. We denote by R the rankofQ and
0 1 1 m m
w is defined as above. Then, using the formula above,
|f|=(q−1)qm−1+(w0−1)qm−r2−1−(w−1)qm−R2.
. By applying affine transformation (see [11]), we get that :
• If r is odd then, either R=r, w =1 and |f|=(q−1)qm−1 or R=r+1,
w=2 and |f|=(q−1)qm−1−qm−r+21.
• If r is even and w = 2 then, R = r+2, w = 2 and |f| = (q−1)qm−1,
0
R=r, w=2 and |f|=(q−1)(qm−1−qm−1−r2) or R=r+1, w =1 and
|f|=(q−1)qm−1+qm−1−r2.
• If r is even and w = 0 then, R = r+2, w = 0 and |f| = (q−1)qm−1,
0
R = r+1, w = 1 and |f| = (q−1)qm−1−qm−1−r2 or R = r, w = 0 and
|f|=(q−1)(qm−1+qm−1−r2).
Finally, the third weight of R (2,m) is (q2−q−1)qm−2.
q
✷
Lemma 4.3 For q ≥4, c =q2−3q+3.
3
Furthermore, for q ≥ 7, if f ∈ R (3,2) is such that |f| = q2−3q+3 then
q
up to affine transformation for all (x,y)∈F2,
q
f(x,y)=(a x+b y)(a x+b y)(a x+b y+c)
1 1 2 2 3 3
where (a ,b )∈F2\{(0,0)} such that for i6=j, a b −a b 6=0 and c∈F∗.
i i q i j j i q
Proof : The second weight in this case is (q−2)(q−1) = q2−3q+2. So
we only have to find a codeword of R (3,2) such that its weight is q2−3q+3
q
to provethe firstpart of this proposition. Consider 3 lines whichmeet pairwise
but do not intersect in one point. Then the union of this 3 lines has 3q −3
points. Let a x+b y+c = 0, a x+b y+c = 0 and a x+b y+c = 0 be
1 1 1 2 2 2 3 3 3
10