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 n8.3m−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 definition 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|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| (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

On the third weight of generalized Reed-Muller codes

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