the intersection of all hyper- planes of Pr containing A, with the convention hAi =Pr if there is no such a hyperplane

E. BALLICO

Communicated by Vasile Brnzanescu

LetX⊂P^r be an integral and non-degenerate variety. For each linear subspace V ⊂P^rtheX-rankrX(V)ofV is the minimal cardinality of a subset ofX span- ning V. Here, we study theV's with large rX(V) whenV is either a rational normal curve or a linearly normal elliptic curve. WhenX is a Veronese embed- ding of P^m we computerX(V) ifV is spanned either by a degree3subscheme ofZ or by a double fat point ofP³.

AMS 2010 Subject Classication: 14N05, 14M15, 15A69.

Key words: X-rank, Grassmann rank, Veronese variety, symmetric tensor rank, zero-dimensional scheme, Grassmannian, fat point, linearly normal elliptic curve.

1. INTRODUCTION

For all integers r > k ≥ 0 let G(k, r) denote the Grassmannian of all k-dimensional linear subspaces of P^r. For any subset or any closed subscheme A ⊂P^r let hAi denote the linear span of A, i.e. the intersection of all hyper- planes of P^r containing A, with the convention hAi =P^r if there is no such a hyperplane. Fix an integral and non-degenerate variety X ⊂P^r. Let ρ(X) be the maximal integer such thatdim(hAi) = deg(A)−1for every zero-dimensional schemeA⊂X withdeg(A)≤ρ(X). FixV ∈G(k, r). TheX-rankr_X(V)ofV is the minimal cardinality of a nite setS ⊂Xsuch thatV ⊆ hSi. LetS(X, V) denote the set of all S⊂X such that](S) =rX(V)and V ⊆ hSi (i.e. the set of allS⊂Xevincingr_X(V)). In the casek= 0, i.e. whenV ={P}is a point, this is the denition of the X-rank r_X(P) of P. If X is the Segre embedding of a product of projective spaces, then rX(P) is the tensor rank of the tensor P [6, 12, 14]. IfX is the orderdVeronese embedding of a projective spaceP^m (as in Section 3), then P corresponds to a non-zero homogeneous polynomial f of degree d and rX(P) is the minimal number of d-powers of linear forms with f as their sum [4, 5, 11, 13]. The case k ≥ 1 is not new, but most of the papers who studied r_X(V) focused on the case in which V is general in G(k, r), i.e. they studied the X-rank of a general V ∈ G(k, r) [7, 8]. In this

REV. ROUMAINE MATH. PURES APPL. 58 (2013), 4, 437451

paper, we only consider very particular linear spacesV, e.g. the ones with very highr_X(V)or the ones spanned by low degree zero-dimensional subschemes of X. In Section 2, we study the case in which X is a curve. We characterize all V with r_X(V) = r when X is either a rational normal curve or a linearly normal elliptic curve (Propositions 1 and 2). The scheme X-rank z_X(V) of V is the minimal degree of a zero-dimensional schemeZ ⊂X such thatV ⊆ hZi (see [11], Denition 5.66, p. 198, [3, 5] for the case dim(V) = 0). Let Z ⊂X be a zero-dimensional scheme. We say that Z is smoothable inside X if it is a at limit of a family of subsets of X with cardinality deg(Z). The scheme Z is smoothable inside X if and only if each connected component of Z is smoothable inside X. If X is smooth, then Z is smoothable inside X if and only if it is smoothable inside P^r ([5], Proposition 2.1.5). Let ρ⁰(X) be the maximal integertsuch thatdim(hZi) = deg(Z)−1for every zero-dimensional smoothable scheme Z ⊂ X such that deg(Z) ≤ t. The smoothable scheme X-rankz⁰_X(V) is the minimal degree of a smoothable zero-dimensional scheme Z ⊂X such that V ⊆ hZi. In Section 2, we prove the following upper bound for the integer z_X⁰ (V) whenX is a curve andρ⁰(X) is large.

Theorem 1. Let X ⊂ P^r be an integral and non-degenerate curve. Fix integers k, t such that r ≥k≥t≥1 and (r−t+ 1)t≤k+ (k−t)t. Assume ρ⁰(X)≥k. Then z⁰_X(V)≤k for every V ∈G(t−1, r).

For any quasi projective varietyM and anyP ∈Mreg let{2P, M}denote the closed subscheme ofM with(I_P,M)² as its ideal sheaf. The scheme(I_P,M)² is a zero-dimensional subscheme ofM with degree dim(M) + 1 and{P} as its reduction.

In Section 3, we study r_X(V)whenXis the Veronese embedding X_m,d of P^m. See Theorem 3 for the casez_Xm,d(V) = 3. See Theorem 2 and Proposition 4 when V is spanned by a scheme of type {2P, M}.

We work over an algebraically closed eldKsuch that char(K) = 0.

2. CURVES

Proposition 1. Let X ⊂P^r, r ≥2, be a rational normal curve. Fix an integer k such that 1 ≤ k < r. Fix V ∈ G(k, r). We have r_X(V) ≤r if and only if V contains no tangent line of X.

Proof. Since P^r is spanned by r+ 1 points ofX, we have r_X(V)≤r+ 1 for any V ∈G(k, r). Fix V ∈G(k, r) such that r_X(V)> r. Let H ⊂P^r be a general hyperplane containing V. The exact sequence

(1) 0→ I_X → I_X(1)→ IX∩H,H(1)→0

shows that the scheme-theoretic intersectionX∩HspansH. Sincedeg(X) =r, we geth(X∩H)_redi=H if and only ifH is transversal to X, i.e. H contains no tangent line of X. Since H is a general hyperplane containing V, Bertini's theorem gives that H contains no tangent line of X if and only if V contains no tangent line of X. Hence, r_X(V) ≤r if and only if V contains no tangent line ofX.

Proposition 2. Let X ⊂P^r, r ≥2, be a linearly normal elliptic curve.

Fix an integer k ∈ {1, . . . , r−1} and V ∈ G(k, r). Set Z := X∩V (scheme- theoretic intersection) and B := Z_red. We have r_X(V) ≤ r if and only if deg(Z)−](B)≤1.

Proof. Riemann-Roch gives ρ(X) = r. Let H ⊂ P^r be a general hy- perplane containing V. By Bertini's theorem, the scheme X∩H is reduced outside (X∩V)_red. Since X is a smooth curve, H is general and B is nite, each O ∈ B appears with the same multiplicity in V ∩X and in H∩X, i.e.

deg(Z)−deg(B) = deg(X∩H)−]((X∩H)_red) =r+ 1−]((X∩H)_red). By (1) the scheme X∩H spansH. First assume k=r−1, i.e. assumeV =H. Hence, deg(Z) = r+ 1 and rX(H) ∈ {r, r+ 1}. We have rX(H) = r if and only if B spans H. Since ρ(X) = r, this is the case if and only if ](B) ≥ r, i.e. if and only if deg(Z)−](B) ≤ 1. Now assume k ≤ r−2. Hence r ≥3. We just proved that rX(H) = r if and only if deg((H ∩X)red) ≥ r. Since r_X(V) ≤ r_X(H), we have r_X(V) ≤ r if deg(Z)−deg(B) ≤ 1. Now assume deg(Z)−deg(B) ≥ 2 and set x := r_X(V) ≤ r. Fix S ∈ S(X, V) and take a general A ⊂ X such that ](A) = r−x. Set M := hS∪Ai. Since ρ(X) = r, M is a hyperplane. Since V ⊆ hSi, we have V ⊆ M. Hence M ⊃ Z. Since deg(M∩X) =r+ 1 =](S∪A) + 1, we getdeg(Z)−deg(B)≤1, a contradic- tion.

Fix integers k, t such that r ≥ k ≥ t ≥ 1. The (t, k)-Grassmann secant variety Gt,k(X) of the n-dimensional variety X ⊂ P^r is the Zariski closure in G(t−1, r) of the set of allV ∈G(t−1, r) such that there isW ∈G(k−1, r) spanned byk points ofX and containing V. A dimensional count gives (2) dim(Gt,k(X))≤min{(r−t+ 1)t, kn+ (k−t)t}

and the dierence, δt,k(X), between the right hand side of (2) and the left hand side of (2) is called the (k, t)-Grassmann defect of X. If n = 1, then δ_k,t(X) = 0 [8]. Hence, if X is a curve we have Gt,k(X) = G(t−1, r) if and only if (r−t+ 1)t≤k+ (k−t)t.

Proof of Theorem 1. Fix a general V ∈ G(t−1, r). Since Gt,k(X) = G(t−1, r) [8], there is a set S_V ⊂X such that](S_V) =k, dim(hS_Vi) =k−1 and V ⊆ hS_Vi. Fix an arbitrary W ∈ G(t−1, r). Since Gt,k(X) is dened

as a closure, there are a smooth and connected ane curve ∆, o ∈ ∆ and a at family {V_λ}_λ∈∆⊂G(k−1, r) such thatV_o=W and for each λ∈∆\ {o}

we have Vλ = hS_Vλi with Sλ ⊂ X and ](Sλ) = k. Since the Hilbert scheme of X is projective, the family {S_λ}_λ∈K\{0} has a at limit, Z, inside X. By the denition of smoothability, Z is a smoothable degree k subscheme of X. Since G(k−1, r) is projective, up to a covering of∆, we may assume that the spaces {V_λ}_λ∈∆\{o} have a at limitV⁰ ∈G(k−1, r). Since S_λ ⊂V_λ for each λ∈∆\ {o}, we haveZ ⊂V⁰. Sincedeg(Z)≤ρ⁰(X)anddeg(Z) = dim(V⁰)−1, we have V⁰ =hZi. HenceW ⊆ hZi.

3. VERONESE VARIETIES

For all positive integersm, dletν_d:P^m →P^N(m,d),N(m, d) := ^m+d_m

−1, denote the orderdVeronese embedding. Set X_m,d :=ν_d(P^m).

LetX be any projective scheme,H any eective Cartier divisor ofX and Z ⊂X any zero-dimensional scheme. Let ResH(Z) denote the residual scheme of Z with respect to H, i.e. the closed subscheme of X with I_Z : I_H as its ideal sheaf. We have ResH(Z)⊆Z anddeg(Z) = deg(ResH(Z)) + deg(Z∩H).

If Z is reduced, i.e. if Z is a nite set, then ResH(Z) =Z \Z∩H. For any line bundleL on X we have an exact sequence

(3) 0→ IRes^H(Z)⊗ L(−H)→ I_Z⊗ L → I_Z∩H,H⊗(L|H)→0

and we call it a residual exact sequence. Hence, for any hyperplane H ⊂P^m and any zero-dimensional schemeF ⊂P^m we have an exact sequence

(4) 0→ IRes^H(F)(d−1)→ I_F(d)→ I_F∩H,H(d)→0 We will use a weak form of the following lemma ([1], Lemma 1):

Lemma 1. Fix integers m, d, z such that m ≥ 2 and 0 < z < 3d. Let Z ⊂ P^m be a zero-dimensional scheme such that deg(Z) =z. If m >2, then assume deg(Z)−deg(Zred)≤ d. We have h¹(I_Z(d))>0 if and only if either there is a line L⊂P^m such thatdeg(L∩Z)≥d+ 2 or there is a conic T ⊂P^m such that deg(T ∩Z)≥2d+ 2.

In the applications, we always have deg(Z)−deg(Z_red)≤4(usually≤3, sometimes≤2). In this case, if there is no lineLsuch thatdeg(Z∩L)≥d+ 2, then any conic T with deg(T ∩Z) ≥ 2d+ 2 must be reduced (and it is also unique).

The proof of [14], Proposition 3.1, i.e. the use of [14], Lemma 2.3, gives the following result.

Proposition 3. Fix a linear subspace M ( P^m. Let V ⊂ hν_d(M)i. Then there is a nite set A⊂M such that ν_d(A) evinces r_Xm,d(V).

Remark 1. Fix an integral variety M ⊆ P^m and P ∈ M_reg. Set y :=

dim(M). LetTPM ⊆P^m be the Zariski tangent space ofM atP. SinceP is a smooth point ofM, we haveTPM ∈G(y, m). We have{2P, M}={2P, T_PM} (as schemes). Hence, it is not restrictive in Theorem 2 and Proposition 4 to take asM a y-dimensional linear subspace ofP^m.

Theorem 2. Fix integers d≥5 andm≥y ≥1, an y-dimensional linear subspace M ⊆ P^m and P ∈ M. Set Z := {2P, M} and V := hν_d(Z)i ∈ G(y, ^m+d_m

−1).

(1)r_Xm,d(V)≤dy+ 1.

(2)Assume y≤3. If y= 3, then assume d≥9. Thenr_Xm,d(V) =dy+ 1. Proof. We have deg(Z) =y+ 1. Sinced > 0, we have dim(V) =y. By Proposition 3 it is sucient to do the case y =m. Indeed, the case y < mis easier (if the casey=mis known) and we do directly it if y= 1.

(a) Assume y = 1. If m = 1, then apply Proposition 1. Now assume m >1. For anyS⊂ν_d(M)with](S) =d+ 1we have V ⊂ hν_d(M)i=hν_d(S)i. Hence, x := rXm,d(V) ≤ d+ 1. Fix any B ∈ S(X_m,d, V) and call A the only subset of P^m such that ν_d(A) = B. If P /∈ A, then set E := A∪ {2P, M}. If P ∈ A, then set E := (A\ {P})∪ {2P, M}. We have deg(E) > x and h⁰(P^m,I_E(d)) =h⁰(P^m,I_A(d)). Hence, h¹(P^m,I_E(d))>0. By [4], Lemma 34, there is a line D ⊂ P^m such that E ⊂ D. Since d > 0, D is the base locus of the linear system |I_E(d)|. Since {2P, M} is contained in the base locus of

|I_A(d)|andM is the only line containing the scheme{2P, M}, we getD=M. Proposition 1 concludes the case y = 1. We also proved the case m = 1 of a conjectural result, i.e. that every S ∈ S(X_m,d, V) is contained in ν_d(M) (see Proposition 4 for the case m= 2).

(b) Assume y ≥ 2. In this step, we prove that r_Xm,d(V) ≤yd+ 1. Fix y lines T_i ⊂ M, 1 ≤ i ≤ y, such that P ∈ T_i for all i and M is spanned by T1∪· · ·∪T_y. TakeFi ⊂Ti\{P}such that](Fi) =dand setF :={P}∪Sy

i=1Fi. SetF[i] :=F \Fi. SetMi :=h∪_j6=iTji and take any hyperplaneH ⊂P^m such thatM∩H_i=M_i. By induction on ywe get hν_d({2P, M_i})i ⊆ hν_d(F[i])i. Let B be the base locus of |I_F(d)| and B_i the base locus of |I_F∩Hi,Hi(d)| on Hi. By induction on y we get B_i ⊇ {2P, M_i}. Hence, B ∩Hi ⊇ {2P, M_i} for all i. Hence, B ⊇ {2P, M}, i.e. V ⊆ hν_d(F)i.

(c) Assume y = 2. By step (b) we have rX_m,d(V) ≤ 2d+ 1. Assume x:=r_Xm,d(V)≤2d. By Proposition 3 there isA⊂M such thatν_d(A) evinces r_Xm,d(V). Hence, it is sucient to get a contradiction to the assumptionx≤2d

when m= 2. Assumem = 2. Set := 3 if P /∈Aand := 2 if P ∈A. Notice thatdeg(Z∪A) =x+. SinceZis in the base locus of the linear system|I_A(d)|, we haveh¹(I_Z∪A(d))≥. TakeB⊂Asuch that](A\B) =−1. Notice that h¹(I_Z∪B(d))≥−(−1) = 1. Since deg(Z∪B) ≤x+ 1≤2d+ 1, there is a lineD⊂P² such thatdeg(D∩(Z∪B))≥d+ 2. Sinceν_d(A)evinces a rank, it is linearly independent, i.e. h¹(I_A(d)) = 0. Hence](A∩L)≤d+ 1for any line L⊂P^m. HenceP ∈D. Hence deg(Z∩D) = 2 and ResD(Z) ={P}. We have deg(ResD(Z∪A))≤x+−d−2. First assume h¹(IRes^D(Z∪A)(d−1)) = 0. We haveh¹(I_Z∪A(d)) = deg(Z∪A)−dim(hZ∪Ai)and a similar formula holds takingZ,AorZ∩Ainstead ofZ∪A. Sinceh¹(I_A(d)) = 0, the residual exact sequence (4) with F := Z ∪A and H := D gives that hν_d(Z)i ∩ hν_d(A)i is spanned by the union of hν_d(Z∩D)i ∩ hν_d(A∩D)i and ResD(Z)∩ResD(A). Recall that ResD(Z) = {P}. Since P ∈ D and A is a nite set, we have ResD(A) =A\A∩D. Hence ResD(Z)∩ResD(A) =∅. Sincehν_d(Z)i ⊂ hν_d(A)i, we gethν_d(Z)i=hν_d(Z∩D)i, a contradiction.

Now, assumeh¹(IRes^D(Z∪A)(d−1))>0. Hencedeg(ResD(Z∪A))≥d+1 ([4], Lemma 34). We havedeg(ResD(Z∪A)) = deg(Z∪A)−deg(D∩(Z∪A))≤ x+−d−2. We get x = 2d, = 3 (i.e. P /∈ A) and ](A∩D) = d. Since h¹(IRes^D(Z∪A)(d−1))>0anddeg(ResD(Z∪A)) =d+ 1≤2(d−1) + 1, there is a line L containing ResD(Z∪A). Since deg(ResD(A)) =d, we have P ∈L andA⊂L∪D. FixQ∈D∩AandQ⁰ ∈L∩Aand setB⁰ :=A\ {Q, Q⁰}. The rst part gives the existence of a lineD⁰ such thatdeg(D⁰∩(Z∪B⁰))≥d+ 2.

Since deg(Z∩D⁰)≤2and d≥3, the choice of Q, Q⁰ gives a contradiction.

(d) From now on, we assumem=y= 3. By step (b) we haver_X3,d(V)≤ 3d+ 1. Assumex :=rX_3,d(V)≤3dand take A⊂P³ such that ν_d(A) evinces r_X3,d(V). Set := 4 if P /∈ A and := 3 if P ∈A. Set W₀ := Z∪A. Notice that deg(F) = x+. Since A evinces r_Xm,d(V), we have h¹(I_A(d)) = 0 and h⁰(I_W0(d)) =h⁰(I_A(d)). Henceh¹(I_W0(d)) = >0.

We havedeg(W₀)≤x+≤3d+. LetM₁ ⊂P³be a plane such thatb₁ :=

deg(M1∩W0)is maximal. SetW1 :=ResM1(W0). For each integeri≥2dene recursively the plane Mi ⊂P³, the integerbi and the zero-dimensional scheme W_i in the following way. Let M_i be a plane such that b_i := deg(M_i ∩Wi−1) is maximal. Set Wi := ResMi(Wi−1). Notice that bi ≥ bi+1 for all i ≥ 1 and that ifbi ≤2, thenbi+1= 0 and Wi =∅. Sincedeg(Z∩H) = 3 for each plane H ⊂M containing P, we haveb₁ ≥5. Hence b_d+2 = 0 and b_d+1 ≤−2. For each integer i≥1 we have an exact sequence

(5) 0→ I_Wi(d−i)→ I_Wi−1(d−i+ 1)→ I_{Wi−1∩Mi,Mi}(d−i+ 1)→0.

Since h¹(I_W0(d)) > 0, there is an integer i ≥ 1 such that h¹(M_i, I_Wi−1∩M_i,Mi(d−i + 1)) > 0. Call e the minimal such an integer. Since

h¹(I_Wd) = max{0, ](W_d)−1} ≤ −3 and h¹(I_W0(d)) = , we have e ≤ d. Since e ≤d, [4], Lemma 34, gives b_e ≥ d−e+ 3. Since b_i ≥b_e for all i < e, we get x+ ≤ e(d−e+ 3). Set φ(t) := t(d+ 3−t). The function φ(t) in increasing ift≤(d+ 3)/2and decreasing ift >(d+ 3)/2. Sinced≥9, we have φ(4) =φ(d−1) = 4(d−1)>3d+ 4. Hencee∈ {1,2,3, d}.

(d1) In this step we assumee=d. Since7(d−1)>3d+4, we haveb_d≤6. Hence, bd ≥4 and there is a line L containing at least 4points of Wd−1∩Md

([4], Lemma 34). Taking a plane containing Land a further point ofWd−2 we getbd−1≥5. Hencedeg(W₀)≥5(d−1) + 4, a contradiction.

(d2) Assume e = 1. Since h¹(M₁,I_W0∩M1,M1(d)) > 0 and h¹(I_A(d)) = 0, we have P ∈ M1. Hence Z ∩ M1 = {2P, M₁} and ResM1(Z) = {P}. First assume h¹(I_W1(d−1)) = 0. The residual exact sequence (5) gives that V = hν_d(Z)i is spanned by the union of hν_d(M₁ ∩Z)i ∩ hν_d(M₁ ∩A)i and of νd(ResM1(Z)∩ResM1(A)). Since P ∈ M1 and A is a nite set, we have P /∈ A\A∩M1 =ResM1(A). Hence V ⊆ hν_d(Z ∩M1)i. Hence dim(V) ≤2, a contradiction. Hence h¹(I_W1(d−1)) > 0. Hence deg(W₁) ≥ d+ 1. Hence b1 ≤x+−d−1≤2d−1+. Hence](A\A∩M₁)≥d. Hence](A∩M₁)≤2d. By Lemma 1 either there is a lineD⊂M1such thatdeg(D∩W₀)≥d+2or there is a reduced conicT ⊂M₁ such thatdeg(T∩W₀)≥2d+ 2. Sinceh¹(I_A(d)) = 0, we have deg(A∩D) ≤ d+ 1 in the rst case and deg(A∩T) ≤ 2d+ 1 in the second case. Hence, P ∈ D in the rst case and P ∈ T in the second case. In both cases, we havedeg(W₁)<3(d−1). Sinceh¹(I_W1(d−1))>0and deg(W1)<3(d−1), either there is a lineD1 ⊂P³such thatdeg(D1∩W₁)≥d+1 or there is a conic T1 such that deg(T1∩W1)≥2d; in the latter case we would have b₂ ≥ 2d and hence, 3d+ 4 ≥ b₁ +b₂ ≥ 4d, a contradiction. Hence, there is a line D₁ such that deg(W₁ ∩D₁) ≥d+ 1. If P /∈D₁, then we have ](A∩D1) =d+ 1, because h¹(I_A(d)) = 0. SinceD1∩W1∩A 6=∅, D1 is not contained inM₁. Since b₂ ≥d+ 1 and deg(A∪Z) =x+≤3d+, we have b₁ ≤2d+ 3.

(d2.1) Assume the non-existence of the line D. Hence, there is a re- duced conic T ⊂ M1 such that deg(T ∩W0) ≥ 2d+ 2. See T as a divisor of M₁. Since b₁ −deg(T ∩W₀) ≤ 1, we have h¹(M₁,IRes^{T(W0∩M1),M1}(d− 2)) = 0. Hence a residual exact sequence gives h¹(M1,I_M1∩W0,M1(d)) = h¹(M₁,I_{W0∩T ,M1}(d)). Since D does not exist and h⁰(T,O_T(d)) = 2d+ 1, we have h¹(M₁,I_W0∩T,M₁(d)) = deg(T ∩W₀)−2d−1. We also haveh¹(I_W1(d− 1))≤deg(W1)−d. Sinceh¹(I_W0(d))≥, we havex≥3d+ 1, a contradiction.

(d2.2) By step (d2.1) there is a lineD⊂M1such thatdeg(D∩W₀)≥d+2. We saw that P ∈D.

(d2.3) Assume P /∈D₁. First assumeD₁∩D6=∅. SetΠ₂ :=hD∪D₁i. Since ](D₁∩(A∪ {P})) ≥ d+ 1 and P /∈ D₁, we have deg(ResΠ2(W₀)) =

1 + deg(ResΠ2(A))≤1 +d−1. Henceh¹(IResΠ2(W0)(d−1)) = 0. As in the rst part of step (d2) we getV ⊆ hν_d(Z∩Π₂)i, a contradiction. HenceD∩D₁=∅. FixO⁰∈A\A∩(D₁∪D). LetE ⊂P³ be a general quadric surface containing D1∪D∪ {O⁰}; it exists, becauseh⁰(I_D1∪D∪{O⁰_}(2)) = 3>0. The quadricE is smooth, because no irreducible quadric cone contains the two disjoint lines D andD₁and the set of all reducible quadric surfaces containingD₁∪D∪{O⁰}has dimension one. Call (1,0)the ruling ofE containingDandD1. LetR1 be the line of type(0,1)of E passing through O⁰. The linear system |I_D1∪D∪{O0}(2)|

hasD₁∪D∪R₁ as its base locus. Since E is general and A is a nite set, we haveA∩E =A∩(D1∪D∪R1). SinceP ∈D2 ⊂E andE is a smooth surface, we have ResE(Z) ={P}. Since](ResE(Z∪A))≤1+(3d−d−d−1−1) =d−1, we have h¹(IRes^E(Z∪A)(d−2)) = 0. Hence, the residual exact sequence

0→ IResE(Z∪A)(d−2)→ I_Z∪A(d)→ I_{E∩(Z∪A),E}(d)→0

on E gives h¹(E,I_{(Z∪A)∩E,E}(d)) ≥ . Since deg(D1 ∩(Z ∪A)) ≤ d+ 1 and deg(D∩(Z∪A))≤d+ 2, the residual exact sequence

0→ IRes^D1∪D(E∩(A∪Z)),E(d−2, d)→ I_{E∩(A∪Z),E}(d)

→ I_(D1∪D)∩(A∪Z),D1∪D(d)→0 on E gives h¹(E,IRes^D1∪D(Z∪A)∩E,E(d−2, d)) ≥−1. Since ResD1∪D((Z∪ A)∩E) ={O⁰, P}, we get a contradiction.

(d2.4) Assume P ∈ D₁. Since D₁∩W₁ ∩A 6= ∅ and D ⊂M₁, we have D 6= D₁. Hence ](D∩D₁) ≤ 1. The plane Π₁ := hD₁∪Di intersects W₀ in a scheme of degree 3 supported by P and in at least 2d other points. Hence deg(ResΠ1(W₀))≤d+−3. Hence,h¹(IResΠ1(W0)(d−1))≤−3. SinceV * hν_d(Π1∩Z)i, as in the rst part of step (d2) we geth¹(IResΠ1(W0)(d−1))>0.

Hence = 4, i.e. P /∈A. We also get that A∩Π1 =A∩(D∪D1) and that deg(W₀ ∩Π₁) = 2d+ 3. Since P ∈ D₁ ∩D, we have {2P,Π₁} ⊂ D∪D₁. Hence W₀ ∩Π₁ = W₀ ∩(D₁ ∪D). Hence h¹(M₁,I_W0∩M1,M1(d)) = h¹(D∪ D1,I_{W0∩(D∪D1),D∪D1}(d)). Sincedeg(W0∩D)≥d+1anddeg(W0∩D₁)≥d+1, we havehν_d(W₀∩(D∪D₁))i=hν_d(D∪D₁)i. Sincedim(hν_d(D∪D₁)i) = 2d, we get h¹(D∪D₁,I_{W0∩(D∪D1),D∪D1}(d)) = 2. Since h¹(IResΠ1(W0)(d−1))≤1, a residual exact sequence givesh¹(I_W0(d))≤3, contradicting the equality= 4. (d3) In this step, we assume e = 2. Since b₁ ≥ b₂, b₁ +b₂ ≤ 3d+ 4 and d ≥ 5, we have b₂ ≤ 2(d−1) + 1. Hence there is a line R ⊂ M₂ such that deg(R∩W1) ≥d+ 1. Assume for the moment P /∈ R. Since deg(W0∩ h{P} ∪Ri) ≥ deg(W₁ ∩R) + 3, we get b₁ ≥ deg(R∩W₁) + 3 ≥ d+ 4. Now assumeP ∈R. Let U be a plane containingR and a point ofA\A∩R. Since

deg(U ∩Z) = 3, we havedeg(U∩W₀)≥deg(R∩W₁) + 2≥d+ 3. Hence, in both cases we haveb₂ ≥deg(R∩W₁)≥d+ 1andb₁ ≥deg(R∩W₁) + 2≥d+ 3. (d3.1) Assume h¹(I_W2(d−2)) > 0. We get deg(W2) ≥ d and that if deg(W₂) =d, then there is a line L₁ such that W₃ ⊂L₁. We saw in the rst part of step (d3) that b₁+b₂ ≥2d+ 4. We get deg(W₂) =d, = 4, x = 3d and W3 = ∅. Since deg(W2) = d, we have h¹(I_W2(d−2)) = 1. The exact sequences (5) for i = 1,2 give h¹(M₂,I_M2∩W1,M2(d−1)) ≥−1 = 3. Hence deg(R∩W₁)≥d+2. Henceb₁+b₂+deg(W₃)≥d+5+d+2+d, a contradiction.

(d3.2) Now assume h¹(I_W2(d−2)) = 0. From the exact sequence (5) for i= 1,2we geth¹(M2,I_W1∩M2,M2(d−1))≥. Sincedeg(W1)−deg(W1∩R)≤ d−2, taking a general plane throughRand using a residual exact sequence we geth¹(M2,I_W1∩M2,M2(d−1)) = deg(W1∩R)−d. Hencedeg(W1∩R)≥d+. Since h¹(I_A(d)) = 0, we have deg(A∩R) ≤d+ 1. Hence P ∈ R, = 3(i.e.

P ∈A) anddeg(A∩R) =d+1. Sincedeg(W₁∩R)≥d+,](A∩R) =d+1and deg(Z∩R) = 2, we getdeg(W1∩A∩R) = deg(A∩R). SinceP ∈A∩W1, we getP /∈M1. SinceM2 is a smooth surface andP /∈M1, we haveZ*M1∪M2. Since h¹(I_W2(d−2)) = 0, the residual exact sequence

0→ I_W2(d−2)→ I_W0(d)→ I_{W0∩(M1∪M2),M1∪M2}(d)→0 gives the surjectivity of the restriction map

H⁰(P³,I_W0(d))→H⁰(M1∪M2,I_{W0∩(M1∪M2),M1∪M2}(d)).

As in step (d2), we get thatV is the linear span of the union ofνd(Z∩(M₁∪ M2))and ofν_d(ResM1∪M2(Z)∩ResM1∪M2(A)). SinceP ∈M2 andP /∈M1, we have ResM1∪M₂(Z) = {P}. SinceP ∈M₂, we haveP /∈ResM1∪M₂(A). Hence ResM1∪M₂(Z)∩ResM1∪M₂(A) = ∅. Hence V ⊆ hν_d(Z ∩(M1 ∪M2))i. Since Z *M1∪M2, we get a contradiction.

(d4) Assumee= 3. Sinceh¹(M₃,I_W2∩M3,M3(d−2))>0, we haveb₃ ≥d. First assume h¹(I_W3(d−3)) = 0. From the exact sequences (5) for i= 1,2,3 we get h¹(M3,I_W2∩M3,M3(d−2)) ≥ ≥ 3. Hence b3 ≥ d+ 2. Hence 3d+ 4

≤ b₁+b₂+b₃ ≤ 3d+ 6, a contradiction. Now assume h¹(I_W3(d−3)) > 0. Hence deg(W3) ≥ d−1. Since b1 ≥ b2 ≥ b3 ≥ d, we get 3d+ 4 ≤ 4d−1, a contradiction.

Proposition 4. Assume d≥5. Fix a plane M ⊆P^m and P ∈M. Set Z :={2P, M} and V :=hν_d(Z)i ∈G(2, ^m+d_m

−1). Then rXm,d(V) = 2d+ 1 andν_d(A)evincesr_Xm,d(V) if and only if there are linesD1, D2⊂M such that {P}=D₁∩D₂, A⊂D₁∪D₂, and either P ∈A,](D_i∩(A\ {P})) =dfor all i or P /∈A and ](D2∩A) =](D1∩A)−1 =d.

Proof. The case y = 2 of Theorem 2 gives r_Xm,d(V) = 2d+ 1. Take lines D₁, D₂ ⊂ M such that {P} = D₁ ∩D₂ and A ⊂ D₁∪D₂ ⊂ M with

P =D₁∩D₂ and either P ∈A,](D_i∩(A\ {P})) =d for all ior P /∈A and ](D₂∩A) =](D₁∩A)−1 =d. In both cases, we havehν_d(A)i=hν_d(D₁∪D₂)i. SinceP is a singular point ofD1∪D₂ andD1∪D2 ⊂M, we haveZ⊂D1∪D2. HenceV ⊂ hν_d(D₁∪D₂)i. Sincer_Xm,d(V) =](A),ν_d(A) evinces r_Xm,d(V).

Now, we prove the only if part.

Fix A ⊂ P^m such that ν_d(A) evinces r_Xm,d(V). Since ν_d(A) evinces rXm,d(V), we haveh¹(I_A(d)) = 0. Take A⁰ ⊂Asuch that](A⁰) =](A)−2and P /∈A⁰. Sinceh¹(I_A(d)) = 0, we haveh¹(I_A⁰(d)) = 0. Since h¹(I_A⁰_∪Z(d))>0 and deg(A⁰ ∪Z) ≤ 2d+ 2 < 3d, either there is a line D_A⁰ ⊂ P^m such that deg(DA⁰ ∩(A⁰∪Z)) ≥ d+ 2 or there is a reduced conic TA⁰ such that deg(T_A⁰ ∩(Z ∩A⁰)) ≥ 2d+ 2. Since h¹(I_A(d)) = 0, in the rst (resp. sec- ond) case, we have deg(A∩D_A⁰) ≤ d+ 1 (resp. deg(T_A⁰ ∩A) ≤ 2d+ 1).

Hence P ∈D_A⁰ (resp. P ∈T_A⁰) and if P ∈ A⁰ then deg(Z ∩D_A⁰) ≥2 (resp.

deg(Z∩T_A⁰)≥2), i.e. D_A⁰ ⊂M (resp. P ∈T_A⁰ and M intersects the tangent space to T_A⁰ at P at least in a line).

(i) In this step, we assume the existence of D_A⁰. Hence P ∈ D_A⁰ and d≤](A∩D_A⁰)≤d+ 1. Moreover, eitherD_A⁰ ⊂M or ](D_A⁰∩A) =d+ 1. Fix Q∈A⁰∩D_A⁰. Since](A∩D_A⁰)≤](A⁰∩D_A⁰)+1, there isQ⁰ ∈A\(A⁰∪(A∩D)). Set B := (A⁰\ {Q})∪ {Q⁰}. We have ](A⁰ ∪B) = 2d. Either there is a line D_B with deg(D_B ∩(A∪Z)) ≥ d+ 2 or there is a reduced conic T_B with deg(T_B∩(A∪Z)) ≥ 2d+ 2. Assume that T_B exists, but that D_B does not exist. Hence if TB is reducible, say TB = R1 ∪ R2 with R1 and R2 lines, then deg((Z ∪B)∩R_i) ≤ d+ 1 for all i. Since ](T_B ∩B) ≥ 2d−1 and ](B∩D_A⁰) ≥ d−1, T_B must be reducible with D_A⁰ as one of its irreducible components. Write TB = DA⁰ ∪L with L a line and L∩DA⁰ 6= ∅. Since ](B) = 2d−1 and deg((Z ∪B)∩T_B) ≥ 2d+ 2, we have Z ∪B ⊂ T_B and Z ∩B = ∅. Since Z ⊂ T_B, we have T_B ⊂ M and {P} = D_A⁰ ∩L. We get B ⊂ M. Since A⁰ ⊂ DA⁰ ⊂ M, we see that M contains all points of A, except at most one; call α this hypothetical point; take a general hyperplane H containingM; since h¹(I_α(d−1)) = 0, a residual exact sequence as in step (d2) of the proof of Theorem 2 givesV ⊆ hν_d(A\ {α})i, a contradiction. Hence A ⊂M. Assume for the moment m = 2, so that T_B is a degree 2 divisor of M. SinceZ ⊂T_B and](A\A∩T_B)≤1, we haveh¹(IResTB(Z∪A)(d−2)) = 0. Hence, the residual exact sequence

0→ IResTB(Z∪A)(d−2)→ IZ∪A(d)→ I_{TB∩(A∪Z),TB}(d)→0

gives V ⊆ hν_d(T_B∩A)i. Hence A ⊂ T_B. If m > 2, then we work as above inside M and get a contradiction, unless A ⊂ TB. We have ](A) = 2d+ 1, A ⊂ T_B and ](A∩ D) ≤ d+ 1 for each line. Hence, either P ∈ A and ]((D \ {P}) ∩A) = d for each irreducible component D of T_B or P /∈ A

and there is a numbering, say R₁ and R₂, of the irreducible components of T_B such that ](A∩R₁) = d+ 1 and ](A∩R₂) = d. In both cases, we are done. Now, assume that DB exists. First assume DB 6= D_A⁰. We saw that P ∈ D_B ⊂ M. Hence Z ⊂ D_A⁰ ∪D_B. Since at most one point of A is not contained in D_B∪D_A⁰, a residual exact sequence with D_B ∪D_A⁰ instead of TB gives A ⊂ DA⁰ ∪DB. We nd that A is as claimed in the statement of Proposition 4. Now assume D_B =D_A⁰. Since ](B∩D_A⁰) =](A⁰∩D_A⁰)−1, we get ](A⁰ ∩D_A⁰) =d+ 1. Let H ⊂P^m be a general hyperplane containing DA⁰. We have ResH(Z) = {P}. Since A is a nite set, we have A∩H = A∩D. First assumeh¹(IResH(Z∪A)(d−1)) = 0. We get thatV is spanned by hν_d(Z∩D_A⁰)i ∩ hν_d(A∩D_A⁰)iand by ResH(Z)∩(A\A∩H). SinceAis a nite set andP ∈A, we get ResH(Z)∩(A\A∩H) =∅. HenceV ⊆ hν_d(A∩D_A⁰)i. Hence r_Xm,d(V) ≤ d+ 1, a contradiction. Henceh¹(IResH(Z∪A)(d−1)) >0. Since ResH(Z∪A) ={P} ∪(A\A∩D_A⁰)has cardinalityd+ 1, there is a line L such that{P} ∪(A\A∩D_A⁰)⊂L([4], Lemma 34). Notice that P ∈Land that A ⊂ D_A⁰ ∪L. If L ⊂ M, then A and the conic D_A⁰ ∪L are as claimed by the statement of Proposition 4. Now assume L*M. Hence Z *D_A⁰ ∪L. Let H⁰ be a general hyperplane containing D_A⁰ ∪L. Since A ⊂D_A⁰∪L and L * M, we have ResH⁰(Z ∪A) = ResH⁰(Z) = {P}. Since V ⊂ hν_d(A)i, h¹(IResH0(Z∪A)(d−1)) = 0 and ResH⁰(Z)∩ResH⁰(A) = ∅, a residual exact sequence gives V =hν_d(Z∩H)i. SinceZ∩H =Z∩D6=Z and dim(V) = 2, we get a contradiction.

(ii) Now assume that D_A⁰ does not exist for any A⁰. Hence T_A⁰ exists for all A⁰ and 2d−1≤](A∩T_A⁰) ≤2d+ 1 for all A⁰. We get that the conicT_A⁰ does not depend from the choice of A⁰. Call it T. Since A⁰ ⊂ T, varying A⁰ among the subsets of A we get A ⊂ T. First assume T ⊂ M. Since A ⊂ T, we have ResT(A) = ∅, where ResT is taken seeing T as a divisor of M. Since deg(ResT(Z ∪A)) ≤deg(Z), we have h¹(IRes^T(Z∪A)(d−2)) = 0. Hence the residual exact sequence

0→ IRes^T(Z∪A)(d−2)→ I_Z∪A(d)→ I_{(Z∪A)∩T,T}(d)→0

gives thathZi ∩ hAi is in the linear span of ofhν_d(T∩Z)iand ofν_d(ResT(Z)∩ ResT(A)) = ∅. Since hZi = hZi ∩ hAi, we get Z ⊂ T. Since Z = {2P, M} and T is a reduced conic, we get T =D₁∪D₂ with D₁ and D₂ distinct lines throughP. Hencedeg(Z∩Di) = 2for alli. Since there is no lineD_A⁰, we have deg(D_i∩(A∪Z))≤d+ 1 for all i. Since P is an ordinary node of D₁∪D₂, we havedeg((A∪Z)∩(D₁∪D₂))≤deg(D₁∩(A∪Z)) + deg(D₂∩(A∪Z))≤ 2d+ 2. Hence P ∈ A and ](Di ∩(A \ {P})) = d for all i as claimed by the statement of Proposition 4. Now assume T * M. Hence, m ≥ 3. Let H ⊂ P^m be a general hyperplane containing M. Since T is a reduced conic,

P ∈ T and M intersects the tangent space to T at P at least in a line, we get T ∩H = T ∩M as schemes. Since A ⊂ M ⊆ H, we have ResH(A) = ∅. Hence, deg(ResH(Z ∪A)) ≤ 3 ≤ d. Hence, h¹(IRes^H(Z∪A)(d−1)) = 0. A residual exact sequence giveshν_d(Z)i ∩ hν_d(A)i ⊆ hν_d(A∩H)i. SinceAevinces r_Xm,d(hν_d(Z)i), we get A=A∩H. Since A⊂T and T∩H is either a line or at most two points, we get](A)≤d+ 1, a contradiction.

The case dim(V) = 0 of the next theorem is [4], Theorem 37.

Theorem 3. Fix integers m≥2, d≥5 and a degree3 scheme Z ⊂P^m. Let V be a linear subspace of hν_d(Z)i such thatdim(V)>0 and V *hν_d(Z⁰)i for any Z⁰(Z. Then exactly one of the following cases occurs:

1.Z is reduced; in this case r_Xm,d(V) = 3.

2.Z is not reduced and contained in a line; in this case rXm,d(V) = d+ dim(V)−1.

3.Z has two connected components, say Z ={Q} ∪v with deg(v) = 2 and v connected, and it is not contained in a line; we have rXm,d(V) =d+ 1 if dim(V) = 1 andν_d(Q)∈V, while r_Xm,d(V) =d+ 2 in all other cases.

4.Z is connected, curvilinear and not contained in a line; in this case rXm,d(V) = 2d+ dim(V)−1.

5.There is a plane M ⊆P^m and P ∈ M such that Z = {2P, M}; in this case we have r_Xm,d(V) = 2d+ dim(V)−1.

In every case, Z is the only scheme evincing z_Xd,m(V).

Proof. Let Z1⊂P^m be any zero-dimensional scheme such thatdeg(Z1)≤ 3 and Z1 6= Z. Since d ≥ 5, we have dim(hν_d(Z1∪Z)i) = deg(Z₁∪Z)−1, dim(hν_d(Z)i) = 2, dim(hν_d(Z₁)i) = deg(Z₁) −1 and dim(hν_d(Z ∩Z₁)i) = deg(Z1 ∩Z) −1 (with the convention dim(∅) = −1). Since νd(Z ∩Z1) ⊆ hν_d(Z ∩Z1)i, Grassmann's formula gives hν_d(Z ∩Z1)i = hν_d(Z)i ∩ hν_d(Z1)i. Hence, if V ⊆ hν_d(Z₁)i ∩ hν_d(Z)i, then V ⊆ hν_d(Z₁∩Z)i. Since Z₁∩Z (Z, we get a contradiction.

The listed cases cover all the degree 3 subschemes ofP^m,m ≥2. Hence, it is sucient to show that in each case the value ofrXm,d(V)is the listed one.

First assume that Z is reduced. Since we just proved that zX_m,d(V) = 3, we have r_Xm,d(V) = 3. Hence, from now on, we assume thatZ is not reduced.

(a) Take Z as in case (5). Proposition 4 gives rX_m,d(V) = 2d+ 1 if dim(V) = 2. Assume dim(V) = 1. The point P and the intersections of Z with a line of M trough P are the only proper subschemes of Z. Set G :=

{g ∈ Aut(M) : g(P) = P}. The group G acts transitively on the set of all lines of M not containing P. Hence the group G acts transitively on the set of all V ⊂ hν_d(Z)i such that dim(V) = 1 and V * hν_d(Z⁰)i for any Z⁰ ( Z. Hence, r_Xm,d(V) is the same for all V ( hν_d(Z)i with dim(V) = 1 and V 6=

hν_d(Z⁰)ifor some degree2 subschemeZ⁰ ofZ. Fix linesL₁,L₂ ofM such that L1∩L2 ={P}. FixSi ⊂Li\ {P}such that](Si) =dand setS:=S1∪S2. As in the proof of Proposition 4 we see thatW :=hν_d(Z)i ∩ hν_d(S)iis a line. Since P /∈S, we also see thatW ∩ν_d(Z) =∅. Hence W *hν_d(Z⁰)i for any Z⁰ ⊆Z. Hence rXm,d(W) ≤2d. Assume x :=rXm,d(W) <2d. Repeating (with several simplications) the proof of Proposition 4 we get a contradiction.

(b) Assume the existence of a line D ⊂ P^m such that Z ⊂ D. Since V ⊆ hν_d(Z)i ⊂ hν_d(D)i, Proposition 3 gives the existence of A ⊂ D evincing r_Xm,d(V). Ifdim(V) = 1, then the schemeV∩ν_d(Z)has degree≤1, becauseV is not spanned by a degree2subscheme ofZ. Proposition 1 givesr_Xm,d(V) =d if dim(V) = 1. Let Z2 ⊂Z the connected degree two subscheme. Proposition 1 gives r_Xm,d(hν_d(Z2)i) = d+ 1. If dim(V) = 2, then V ⊃ hν_d(Z2)i. Hence r_Xm,d(V) =d+ 1if dim(V) = 2.

(c) TakeZ ={Q}tvwithva degree2connected zero-dimensional scheme and Z not contained in a line. Set {O}:= v_red. Since rX_m,d(hvi) ≤d+ 1, we have r_Xm,d(hν_d(Z)i)≤d+ 2.

(c1) Assume dim(V) = 1 and νd(Q) ∈ V. Since V is not spanned by a proper subscheme of ν_d(Z), the set V ∩ hν_d(v)i is a unique point, O⁰, and O⁰ 6=ν_d(O). Sincer_Xm,d(O⁰) =dby a theorem of Sylvester ([4, 9], Theorem 23, [13], Theorem 5.1, [12]) andν_d(Q)∈V, we haved≤r_Xm,d(V)≤d+ 1. Assume rX_m,d(V) =dand takeS ⊂X_m,devincingrX_m,d(V). SinceSevincesrX_m,d(O⁰), we have S ⊂ hν_d(hvi)i ([12], Ex. 3.2.2.2). Hence, V is in the linear span of a proper subscheme of ν_d(Z), a contradiction.

(c2) Assume dim(V) = 1 and νd(Q) ∈/ V. We claim that rX_m,d(V) = d+ 2. Assume x := r_Xm,d(V) ≤ d+ 1 and take A ⊂ P^m evincing r_Xm,d(V). Since h¹(I_A∪Z(d)) > 0 and deg(A∪Z) ≤ d+ 4 ≤ 2d+ 1, there is a line D ⊂ P^m such that deg(D∩(A ∪Z)) ≥ d+ 2 ([4], Lemma 34). Let H ⊂ P^m be a general hyperplane containing D. Since deg(ResH(A∪Z)) ≤ 2, we have h¹(IRes^H(Z∪A)(d−1)) = 0. Since V * hν_d(Z⁰)i for any Z⁰ ( Z and V *hν_d(A⁰)i for any A⁰ (A, a residual exact sequence gives that ResH(Z) = ResH(A)and thathν_d(A)i ∩ hν_d(Z)iis the linear span of ResH(A)andhν_d(Z∩ H)i∩hν_d(A∩H)i. SinceZhas only nitely many subschemes andHis a general hyperplane containingD, we haveZ∩H=Z∩DandA∩H=A∩D. SinceZis not contained in a line, we have ResH(Z)6=∅and henceA\A∩D=ResH(A)6=

∅. Since deg(D∩(A∪Z))≥d+ 2and deg(Z) + deg(A)≤d+ 4, we get that ResH(Z) is a single point, that](A) =d+ 1, that A\ResH(Z)⊂D and that

D∩A∩Z =∅. Sincedeg(ResH(Z)) = 1and Z∩H =Z∩D, we haveO∈D. SinceD∩A∩Z =∅, we haveO /∈A. Hence,{Q}=ResH(Z) =A\A∩Dand D=hvi. SinceV ⊂ hν_d(A)iand V *hν_d(A\ {Q})i, the set V ∩ hν_d(A\ {Q})i is a single point,O₁, andV =h{O₁, ν_d(Q)i. Hence ν_d(Q)∈V, a contradiction.

(c3) Assumedim(V) = 2. By step (c2) there is a one-dimensionalV⁰ ⊂V withr_Xm,d(V⁰) =d+ 2. Sincer_Xm,d(hZi)≤d+ 2, we haver_Xm,d(V) =d+ 2. (d) LetZ be a degree3connected and curvilinear scheme not contained in any line. Set{Q}:=Zred. Hence Z is contained in a smooth conic,C. SinceZ is connected and curvilinear, it has a unique subscheme of degree2. Call itZ2. Since dim(hν_d(C)i) = 2d, we have r_Xm,d(V) ≤ 2d+ 1. Proposition 1 implies thatrXm,d(V)≤2difdim(V) = 1, becauseV 6=hν_d(Z2)i. SetM :=hZi ⊆P^m. M is a plane. By Proposition 3 we may nd A ⊂M such that ν_d(A) evinces r_Xm,d(V). Set x := ](A). Set := 0 if Q /∈A and := 1 if Q ∈A. We have deg(A∪Z) =x+ 3−.

(d1) Take dim(V) = 2, i.e. take V = hν_d(Z)i. Assume x ≤ 2d. Since ν_d(A) evincesrX_m,d(V), we haveh¹(I_A(d)) = 0and h⁰(I_Z∪A(d)) =h⁰(I_A(d)). Hence, h¹(I_A∪Z(d)) = 3−. Fix B ⊂ A such that ](B) = x−2 +. Since h¹(IA∪Z(d)) = 3−, we have h¹(IZ∪B(d))>0. Since deg(Z∪B) ≤x+ 1≤ 2d+ 1, there is a line D ⊂ M such that deg(D∩(Z ∪B)) ≥ d+ 2. Since h¹(I_A(d)) = 0, we have ](A∩D) ≤d+ 1. Hence, Q∈ D and ](B∩D) ≥d. Moreover if Q ∈ B, then D = hZ₂i. In each case, we see that D does not depend from the choice ofB. TakingBnot containingQwe see thatD=hZ₂i. VaryingB we getdeg(D∩(Z∪A))≥d+ 4−. LetH ⊂P^m be any hyperplane such that H ∩M = D. Since deg(ResH(Z ∪ A)) ≤ x −d ≤ d, we have h¹(IResH(Z∪A)(d−1)) = 0. A residual exact sequence gives thatV is spanned by the union of hν_d(Z∩D)i ∩ hν_d(A∩D)iand ResH(Z)∩ResH(A). We have D∩A 6= A, because hν_d(C)i ∩ hν_d(D)i = hν_d(Z2)i. We have Q /∈ ResH(A), becauseQ∈D⊆H. Hence, ResH(Z)∩ResH(A) =∅. Hence,V ⊂ hν_d(A∩D)i, a contradiction.

(d2) Now, we take dim(V) = 1. Assumex ≤2d−1. Since V ⊂ hν_d(A)i andV has codimension1inhν_d(Z)i, we haveh¹(IZ∪A(d))≥2−. Fix B ⊆A such that](A\B) = 1−. We haveh¹(I_B∪Z(d))≥(2−)−(1−) = 1. Since deg(B∪Z) ≤2d, there is a lineD⊂M such thatdeg(D∩(B∪Z))≥d+ 2. Varying B we get the existence of a line D ⊂ M such that deg(D∩(Z ∪ A)) ≥ d+ 3−. Since h¹(I_A(d)) = 0, we have ](A∩D) ≤ d+ 1. Hence deg(Z ∩D) ≥ 2 (even in the case = 1, because Z ∪A = Z∪(A\ {Q}) if = 1). HenceD=hZ₂i. LetH ⊂P^mbe any hyperplane such thatH∩M =D. Since deg(ResH(Z ∪A)) ≤ x−d, we have h¹(IRes^H(Z∪A)(d−1)) = 0. A residual exact sequence gives thathν_d(Z)i ∩ hν_d(A)iis spanned by the union of hν_d(Z∩D)i∩hν_d(A∩D)iand ResH(Z)∩ResH(A). We haveD∩A6=A, because

hν_d(C)i ∩ hν_d(D)i = hν_d(Z₂)i. We have Q /∈ ResH(A), because Q ∈ D ⊆ H. Hence, ResH(Z)∩ResH(A) =∅. HenceV ⊂ hν_d(A∩D)i, a contradiction.

Question 1. Fix positive integers m, d and k. Which is the maximal value of the integer r_Xm,d(V) among allV ∈G(k, ^m+d_m

−1)? Proposition 1 answers Question 1 in the case m= 1.

Acknowledgments. The author was partially supported by MIUR and GNSAGA of INdAM (Italy).

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Received 14 January 2013 University of Trento,

Department of Mathematics 38123 Povo (TN), Italy

ballico@science.unitn.it