Generalized Hamming Weights of Duals of Algebraic-geometric Codes

Here we extends a work of A. Couvreur on the Hamming distance of the dual of an evaluation code to its generalized Hamming weights. We prove the following result. Fix integers r ≥ 2, m > 0 and e ≥ 1. Let Z ⊂ P r be a zero-dimensional scheme such that deg(Z) ≤ 3m + r − 3. If r > 2 assume that Z spans P r and that the sum of the degrees of the non-reduced connected components of Z is at most 2m + 1. We have h 1 (I Z (m)) ≥ e if and only if there is W ⊆ Z as one of the schemes in the following list: (a) deg(W) = m + 1 + e and W is contained in a line; (b) deg(W) = 2m + 1 + e and W is contained in a reduced plane conic;


Introduction
Fix a prime p and a p-power q.We recall that an affine [n, k]-code C over F q is often given in the following way.Fix positive integers n, r, distinct points P 1 , . . ., P n of the affine space F m q and a k-dimensional linear subspace W of F q [t 1 , . . ., t m ] such that no f ∈ W \ {0} vanishes at all points P 1 , . . ., P n .Fix a basis f 1 , . . ., f k of W .The k × n matrix (f i (P j )) gives an injective linear map F k → F n q , i.e., this matrix is the generator matrix of an [n, k]-code C ( [11]). he dual code C ∨ is the [n, (n − k)]-code whose words are the elements of F n q orthogonal to the words of C with respect to the canonical inner product, i.e., (a 1 , . . ., a n ) ∈ F n q is a word of C ∨ if and only if a 1 b 1 + • • • + a n b n = 0 for all words (b 1 , . . ., b n ) of C (i.e. the generator matrix of C ∨ is the parity check matrix of C, and conversely).We usually consider the following projective set-up, leaving to the interested reader the task to translate the projective language into the affine language.For any field K let K[x 0 , . . ., x r ] m denote the set of homogeneous polynomials over K with degree m.The K-vector space K[x 0 , . . ., x r ] has dimension m+r r .Fix a finite set S ⊆ P r (F q ) and a k-dimensional linear subspace of H 0 (P r , O P r (m)) defined over F q .We fix homogeneous coordinates x 0 , . . ., x r of P r , so that we may identify V with a k-dimensional linear subspace of F q [x 0 , . . ., x r ].For any P ∈ S we fix P ′ ∈ F r+1 q \ {0} whose equivalence class modulo scalars induces P .We use these choices to identify S with a subset of F r+1 q \ {0} (also called S).With these choices every f ∈ V gives a map S → F q .Therefore V induces a linear map V → F S q .If this map is injective, then we get an [n, k]-code C over F q .If we identify V with a linear subspace of F S q , then we denote with V ⊥ its orthogonal linear subspace, i.e., the linear subspace of F S q inducing C ∨ .A. Couvreur proved that quite often it is easier to compute the minimum distance of C ∨ and classify all the codewords of C ∨ with small weights (not only the ones with minimum distance) than to solve the corresponding problems for C ( [7], [5]).As clear from [7] in evaluation codes arising from H 0 (P n , O P n (m)) very simple objects (lines, small degree plane curves, finite sets which are complete intersections) are often useful.We extend [7] to the higher distances of a code (a.k.a.higher support weights ( [10], [12]) (see Lemma 1).In [2] we translated the issue directly on plane curve giving the associated Goppa code.Lemma 1 is more flexible.By Lemma 1 to get the extension it is sufficient to prove the case e ≥ 2 of the following result.
Theorem 1. Fix integers r ≥ 2, m > 0 and e ≥ 1.Let Z ⊂ P r be a zerodimensional scheme such that deg(Z) ≤ max{3m + ρ − 3, 3m + e − 2} where ρ is the dimension of the linear span of Z.If ρ > 2 assume that the sum of the degrees of the non-reduced connected components of Z is at most 2m + 1.We have h 1 (I Z (m)) ≥ e if and only if there is W ⊆ Z as one of the schemes in the following list: Case (c) is the one arising only in the set-up of generalized Hamming weight, because if W is as in case (c), then there is W ′ W with h 1 (I W ′ (m)) > 0, and it is W ′ , not W , that count for the minimum distance.
Remark 1. Assume that Z is defined over a perfect field K and call K the algebraic closure of K.It is easy to check that we may find W with the additional condition that W and the curve (a line, or a conic or a disjoint union of two lines in cases (a), (b) or (c)) are defined over a finite extension K ′ of K with deg([K ′ : K]) ≤ 2. To get K ′ = K in cases (a) and (b) it is sufficient that one of the points of W red is defined over K.In case 2 if the conic E is either smooth or a double line, then it is defined over K.If E is a reduced, but a reducible conic to get that W and E are defined over K it is sufficient that two points of W red are defined over K.For the application to coding theory we have K = F q and, as in [7] and [5], many of the connected components of Z are reduced and defined over K.
If Z is just a finite set, then the proofs are easier and (at least if e = 1) (see [1]).Even weaker assumption on Z may be used (as in [3]).Basically in the proof of Theorem 1 we need in some way to handle the case D = L (or in the case of Lemma 3 the case D ′ = D).

The Proof and a Result in P 2
Lemma 1. Fix S ⊆ P r (F q ) and a linear subspace V ⊆ F q [x 0 , . . ., x r ] m and call C the evaluation code induced by (V, S).Set n := ♯(S) and k := dim(V ).Assume V (−S) = {0}, i.e., assume that C is an [n, k]-code.
(a) Fix B ⊆ S and an integer e > 0. There is an e-dimensional linear subspace of C ∨ = (V ⊥ , S) with support contained in B if and only if i(V, B) ≥ e.

E. Ballico (b) Fix an integer
. The linear projection of F S q onto its factor F B q and the inclusion V ֒→ F S q induces an inclusion V /V (−B) ֒→ F B q .Fix f ∈ F S q with support on B. Since f has support on B, we have P ∈S f (P )g(P ) = P ∈B f (P )g(P ) for all g ∈ F S q .We have i(V, B) = 0 if and only if the evaluations of V at the points of B are linearly independent.In general i(V, B) is the number of independent linear relations among the evaluations of V at the points of B. Hence i(V, B) is the dimension of the linear subspace of C ∨ formed by the words with support on B. Hence we get part (a).Part (b) follows from part (a).
Remark 2. Let X be any projective scheme and D any effective Cartier divisor of X.For any closed subscheme Z of X let Res D (Z) denote the residual scheme of Z with respect to D, i.e. the closed subscheme of X with I Z : I D as its ideal sheaf.We have deg From (1) we get for every integer i ≥ 0.
Lemma 2. Let A, B be zero-dimensional subschemes of P r such that In P 2 by far the strongest tool is the following result due to Ph. Ellia and Ch.Peskine ([8], Corollaire 2) (see [8], Remarques (i) at page 116 for a use of it).
Proposition 1. Fix an integer m ≥ 3 and a zero-dimensional scheme Z ⊂ P 2 such that either deg(Z) ≤ 10 (case m = 3) or deg(Z) ≤ 4m − 4 (case m ≥ 4).We have h 1 (I Z (m)) > 0 if and only if there is a scheme W ⊆ Z such that one of the following cases occurs: Lemma 3. Fix integers r ≥ 2 and m > 1.Let Z ⊂ P r be a zerodimensional scheme such that deg(Z) ≤ max{3m + 1, 3m + r − 2}.If r > 2 assume that Z spans P r and that the union of the non-reduced connected components of Z has degree ≤ 2m + 1.We have h 1 (I Z (m)) > 0 if and only if there is a scheme W ⊆ Z as one of the schemes in the following list: Proof.The " if " part follows from Lemma 2 and the cohomology of low degree plane curves.Now assume h 1 (I Z (m)) > 0. If m = 1, then deg(Z) ≤ r + 1.Since Z spans P r and any degree r subscheme of P r is contained in a hyperplane, we have deg(Z) = r + 1.Since h 0 (I Z (1)) = 0 and deg(Z) = r + 1, we have h 1 (I Z (1)) = 0. Hence Proposition 3 is true if m = 1.We use induction on r.For fixed r we use induction on m, starting the induction with the case m = 1 just done.In the case r = 2 use Proposition 1.Now assume r > 2. Let H ⊂ P r be a hyperplane such that deg(Z ∩ H) is maximal.Since Z spans P r , we have deg Hence if h 1 (H, I Z∩H (m)) > 0, then we may take W ⊆ Z ∩ H by the inductive assumption on r.Hence we may assume h 1 (H, I Z∩H (m)) = 0. Remark 2 implies h 1 (I Res H (Z) (m − 1)) > 0. By the inductive assumption on m there is W ′ ⊆ Res H (Z) satisfying one of the assumptions (a), (b), (c) or (d) of Proposition 3 with respect to the integer m ′ := m − 1.In particular there is a plane M ⊂ P r such that W ′ ⊂ M and deg(W ′ ) ≥ m + 1, with strict inequality, unless W ′ is contained in a line.Every zero-dimensional scheme with degree at most r is contained in a hyperplane.Since Z is not contained in a hyperplane, there is a hyperplane U ⊂ P r such that W ′ ⊂ U and either First assume r = 3.Take a smooth quadric Q containing the two disjoint lines D and D ′ , say as lines of type (1, 0).Since deg(Res Now assume r > 3. Let H ′ be any hyperplane containing D ∪ D ′ and (among these hyperplanes) with maximal deg(Z ∩ H ′ ).Since Z spans P r , we have deg (ii) deg(W ) = 2m + e and W is contained in a plane conic; (iii) e ≥ 3, deg(W ) = 2m+e+1, and there are an integer f ∈ {1, . . ., e−2} and lines (a) First assume that W is as in cases (i) or (ii) or that r ≥ 4 and that we are as in case (iii).Let H ⊂ P r be any hyperplane containing W , where denote the linear span, and with maximal u := deg(Z ∩ H).Since Z spans P r , we have deg(Z ∩ H) < deg(Z) ≤ 3m + r − 2. Since Z spans P r , the maximality property of the integer u gives (a3) Now assume D = L. Let E be the union of all non-reduced connected components A of Z with A red ∈ L, but with A L. Let F be the union of the connected components of Z contained in L (among them there are all the reduced connected components of Z contained in F .We have Res H (a) deg(W ) = m + 1 + e and W is contained in a line; (b) deg(W ) = 2m + 1 + e and W is contained in a reduced plane conic; (c) r ≥ 3, e ≥ 2, and there are an integer f ∈ {1, . . ., e − 1} and lines (a) deg(W ) = m + 2 and W is contained in a line; (b) deg(W ) = 2m + 2 and W is contained in a conic; (c) deg(W ) = 3m and W is the complete intersection of a curve of degree 3 and a curve of degree m; (d) deg(W ) = 3m + 1 and W is contained in a degree 3 curve; (e) deg(W ) = 4m − 4 and W is the complete intersection of a degree 4 curve and a degree m − 1 curve.Proof.Let T be the degree ≤ 4 curve listed in cases (a), (b), (c), (d), (e).In cases (a), (b), (c), (d), (e) we have h 1 (I W (m)) > 0 either because deg(W ) > h 0 (T, O T (m)) (cases (a), (b), (d)) or because W is a complete intersection, say W = T ∩ T 1 , and we may use the Koszul complex induced by the equations of the curves T and T 1 .Hence the " if " part follows from Lemma 2.Now assume h 1 (I Z (m)) > 0. Fix any subscheme W ⊆ Z such that h 1 (I W (m)) = 1 and h 1 (I Z ′ (m)) = 0 for every Z ′ W (to prove the existence of W it is essential to work over an algebraically closed base field).Therefore h 1 (I W (t)) = 0 for every t > m.Set d := deg(W ).First assume d ≤ 3m and d ≥ 9.In the set-up of[8].Corollaire 2, take τ := m and s := 3. We have τ ≥ (3 − 3) + d/3.By[8], Corollaire 2, we get that either W is the complete intersection of a cubic curve and a degree m curve (and hence W = Z and deg(Z) = 3m) or there is t ∈ {1, 2} and a degree t curve F ⊂ P 2 such that deg(F ∩ W ) ≥ tm + t(3 − t).Now assume d ≤ 8. Adapt the case of a reduced set done, for instance, in[9], p. 715.Now assume d ≥ 3m + 1 and d ≥ 16.Since d ≥ 16, we may apply the case s = 4, τ = m of[8], Corollaire 2 (we may apply it, because d ≤ 4m − 4, i.e., τ = m ≥ 1+d/4 = s−3+d/s).We get that either W is the complete intersection of a degree 4 curve and a degree m − 1 curve (case (e)) or there is t ∈ {1, 2, 3} and W t ⊆ W with W t contained in a degree t curve and either deg(W t ) = tm+2 (case t ∈ {1, 2}) or t = 3 and 3m ≤ deg(W 3 ) ≤ 3m + 1.If t = 1 (resp.t = 2), then we are in case (a) (resp.(b)).If t = 3 and deg(W 3 ) = 3m + 1, then we are in case (d).Hence we may assume deg(W 3 ) = 3m.Let E be a degree 3 curve containing W 3 .To prove that we are in case (c) (taking W 3 as the subscheme of Z) it is sufficient to prove that W 3 is the complete intersection of E and a degree m plane curve.Since W ∩ E ⊇ W 3 either we are in case (d) or W ∩ E = W 3 .Hence we may assume W ∩ E = W 3 (as schemes).Since deg(Res E (W )) = deg(W )−deg(W 3 ) ≤ m−4, we have h 1 (I Res E (W ) (m−1)) = 0. Hence Remark 2 gives h 1 (I W 3 (m)) > 0. Apply the case d ≤ 3m just done to W 3 .Now assume d ≤ 15, and 3m + 1 ≤ d ≤ 4m − 4. Hence m = 3 and 10 ≤ d ≤ 11.Since deg(Z) ≤ 10 if m = 3, it is sufficient to do the case m = 3 and d = 10.Since deg(W ) = 10 = h 0 (O P 2 (3)), we have h 0 (I W (3)) = h 1 (I W (3)). Hence we are in case(d).
(a) deg(W ) = m + 2 and W is contained in a line; (b) deg(W ) = 2m + 2 and W is contained in a plane conic; (c) deg(W ) = 3m and W is the complete intersection of a degree 3 plane curve and a degree m hypersurface; (d) deg(W ) = 3m + 1 and W is contained in a plane cubic.
part of Remark 2).Since deg(Z) < 2m + 2 + 2(r − 2), we get that W ′ is collinear.Call D the line spanned by W ′ and set W ′′ := D ∩ Z.If W ′′ = W ′ , then deg(W ′′ ) ≥ m + 2 and hence we are in case (a).Therefore we may assume W = W ′ .If h 1 (I Z∩U (m)) > 0, then we may take as W a subscheme of Z ∩ U .Therefore we may assume h 1 (I Z∩U (m)) = 0. Therefore h 1 (I Res U (Z) (m − 1)) > 0. Hence there is A ⊆ Res U (Z) as in one of the cases (a), (b), (c), (d) for the integer m − 1.As in the case with H we get that A is collinear, deg(A) = m + 1 and A = Z ∩ D ′ with D ′ a line.First assume D = D ′ .Let E be the union of all non-reduced connected components A of Z with A red ∈ D, but with A D.Let F be the union of the connected components of Z contained in D (among them there are all the reduced connected components of Z contained in D).We have Res U Apply the inductive assumption on r to the scheme Z ∩ H ′ .Proof of Theorem 1.First assume deg(Z) ≤ 3m + ρ − 2. With no loss of generality we may assume r = ρ.Since the case r = 2 is true by Proposition 1, we may assume r > 2 and use induction on r.Since the case e = 1 is true by Proposition 1 we may assume e ≥ 2 and use induction on e.Since h 1 (I Z (m)) ≥ e ≥ e − 1, there is W ⊆ Z with W one of the schemes in the following list: (i) deg(W ) = m + e and W is contained in a line L; which is not reduced, then the connected component B of Z with B red = A red contains A and hence deg(A) ≤ deg(B).Therefore we may use induction on r (for an arbirary integer e) for the schemeZ ∩ H ⊂ H. Since Z ∩ H ⊇ W , we have h 1 (H, I Z∩H (m)) ≥ e − 1.If h 1 (H, I Z∩H (m)) ≥ e,then we may use the inductive assumption on r.Hence we may assume that h 1 (H, I Z∩H (m)) = e − 1.By the residual exact sequence (1) we get h 1 (I Res H (Z) (m − 1)) > 0. Therefore deg(Res H (Z)) ≥ m + 1 ([6, Lemma 34]).Since Z ∩ H spans H, we have u ≥ r − 1− dim( W )+ deg(W ).Hence deg(Z) ≥ m + r − dim( W ) + deg(W ).In case (ii) we get deg(Z) ≥ m + r − 2 + 2m + e = 3m+r−2+e, a contradiction.In case (iii) we get deg(Z) ≥ m+r−3+2m+e+1, a contradiction.Now assume that we are in case (i).We may assume W = Z ∩L (as schemes), because otherwise deg(Z ∩L) ≥ m+1+e and we may take as new W a subscheme of Z ∩L with degree m+1+e.Since deg(Z ∩H) ≥ m+e+r −2, we have deg(Res H (Z)) ≤ 2m − e − 1.By [6, Lemma 34] there is a line D ⊂ P r such that deg(D ∩ Res H (Z)) ≥ m + 1. (a1) First assume D ∩ L = ∅ and D = L. Therefore D ∪ L is a plane.We are in case (b).(a2) Now assume D ∩ L = ∅.If D ∩ Z D ∩ Res H (L) (or if deg(Z ∩ D) ≥ m + 2), then we are as in case (c) with f = e − 1, L 1 = L and L 2 = D. Now assume deg(D ∩ Z) = m + 1.Let M := D ∪ L be the 3-dimensional linear space.First assume r = 3.Let Q ⊂ M be any smooth quadric containing D∪L.Since deg(Q ∩ Z) ≥ 2m + 1 + e, we have deg(Res Q (Z)) ≤ m − 1 − e. Hence h 1 (I Res Q (Z) (m − 2)) = 0. Remark 2 gives h 1 (Q, I Z∩Q (m)) ≥ e.Call (1, 0) the type of the ruling of Q containing D and L. Since deg(Res D∪L contradicting the inequality h 1 (Q, I Z∩Q (m)) ≥ e just proved.Now assume r ≥ 4. Let N ⊂ P r be any hyperplane containing M and with maximal deg(Z ∩ N ) among all hyperplanes containing M .Since Z is non-degenerate, we have deg(Z ∩ N ) ≥ deg(Z ∩ M ) + r − 4 ≥ 2m + e + r − 3.By induction on r we get h 1 (N, I N ∩Z (m)) = e − 1.Since deg(Res N we get a contradiction.(b)Now assume r = 3, e ≥ 3, and that W is as in case (iii).If either W ∩ L 1 = Z ∩L 1 or W ∩L 2 = Z ∩L 2 , then we are in case ( c) with as integer f either f or f + 1. Therefore we may assumeZ ∩ (L 1 ∪ L 2 ) = W .Let Q be any smooth quadric containing L 1 ∪ L 2 .We have deg(Res Q (Z)) = deg(Z) − deg(Z ∩ Q) ≤ deg(Z) − deg(Z ∩ L 1 ∪ L 2 ) ≤ m − 1 − e. Therefore h 1 (I Res Z (Q) (m − 2)) = 0 ([6, Lemma 34]).Remark 2 gives h 1 (Q, I Q∩Z (m)) = 0. Set T := L 1 ∪ L 2 seeing as a divisors of Q, say of type (2, 0).Since T ∩ Z = W , we have h 1 (T, I T ∩Z (m)) = e − 1.We have deg(Res T (Z ∩ Q)) = deg(Z ∩ Q) − deg(Z ∩ T ) ≤ m − 1 − e and hence h 1 (Q, I Res T (Z∩Q) (m − 2, m)) ≤ h 1 (Q, I Res T (Z∩Q) (m − 2, m − 2)) = 0.The residual exact sequence of the inclusion T ⊂ Q gives a contradiction.(c)Now assume deg(Z)≥ 3m + r − 1, but deg(Z) ≤ 3m + e − 2.Taking the linear span Z of Z instead of P r , we may assume Z = P r .Since the case r = 2 is true by Proposition 1 we may assume r > 2 and use induction on r.Since the case e = 1 is true by Lemma 3 we may assume e ≥ 2 and use induction on e.Take H, W as in the first part of the proof.In case (ii) (resp.case (iii) with r > 3) of step (a) we use that deg(Res H (W )) ≤ m − 2 (resp.m − 3).Now assume that we are in case (i).We have deg(Z) − deg(Z ∩ H) ≤ 3m + e − 2 − (m + 1 + e) − (r − 2) ≤ 2(m − 1) + 1.As in the proof of Theorem 1 quoting [6, Lemma 34] we get a line D with deg(D ∩ Res H (Z)) ≥ m + 1.The assumption on the non-reduced connected components of Z gives D = L.If D ∩ L = ∅, then we get deg(Z ∩ (D ∪ L)) ≥ 2m + 1 + e and hence we are in case (b).If D ∩ L = ∅, we are in case ( c) with f = e − 1, L 1 = L and L 2 = D.If r = 3 and we are in case (iii), then step (b) gives a contradiction, unless