# Postulation of General Unions of Lines and Multiplicity Two Points inℙ3

Authors: E. Ballico
Journal: ISRN Geometry
 (2014)

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## Abstract

We prove that a general unionX⊂ℙ3 of prescribed numbers of lines and double points has maximal rank, except a few well-known
exceptional cases.


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## Body

## 1. Introduction

FixP∈ℙr. The 2-point 2P of ℙr is the closed subscheme of ℙr with (ℐP)2 as its ideal sheaf. The scheme 2P is a zero-dimensional scheme with deg⁡(2P)=r+1 and (2P)red={P}. Now assume that r≥3. For all (t,a)∈ℕ2, let Z(r,t,a) be the set of all disjoint unions X⊂ℙr of t lines and a 2-points. Assume that (t,a)≠(0,0); that is, assume that Z(r,t,a)≠∅. For each X∈Z(r,t,a) and each integer x≥0, we have h0(𝒪X(x))=(x+1)t+a(r+1). The algebraic set Z(r,t,a) is an irreducible subset of the Hilbert scheme. We recall that a closed subscheme W⊂ℙr is said to havemaximal rank if for each integer x≥0, the restriction map H0(𝒪ℙr(x))→H0(𝒪X(x)) is a linear map with maximal rank; that is, it is either injective or surjective. W has maximal rank if and only if for each integer x≥0 either h0(ℐW(x))=0 or h1(𝒪X(x))=0. In this paper we prove the following result.Theorem 1.
Fix(t,a)∈ℕ2∖{(0,0)} and an integer k≥3. Let X be a general element of Z(3,t,a). Then either h0(ℐX(k))=0 or h1(ℐX(k))=0, unless either (t,a,k)=(2,3,3) or (t,a,k)=(0,9,4).The case(t,a,k)=(0,9,4) is one of the exceptional cases in the famous Alexander-Hirschowitz theorem [1–5]. See [6, Example 1] for the case (t,a,k)=(2,3,3). Theorem 1 is obviously false for k=2, but it is false in a controlled way [6, Lemma 1], because a disjoint union of 3 lines of ℙ3 is contained in a unique quadric surface and for each P∈ℙ3, the linear system |ℐ2P(2)| is the set of all quadric cones with vertex containing P. Therefore as an immediate corollary of Theorem 1, we get the following result.Corollary 2.
Fix(t,a)∈ℙ3 such that t+a≥4, (t,a)≠(2,3), and (t,a)≠(0,9). Let X be a general element of Z(3,t,a). Then X has maximal rank.The casek=3,4,5 of Theorem 1 was proved in [6, Propositions 1,2,3] and a weaker form of Theorem 1 was proved in [6, Proposition 5] (we required that (k+1)t+4a+3k≤(k+33)). This weaker version was enough to prove the statement corresponding to Theorem 1 first in ℙ4 and then in ℙ5. Then we proved by induction on r the corresponding statement in ℙr, r≥5, but only if k≥4 [7].For the proof, we use the casea=0 proved by Hartshorne and Hirschowitz [8] and the case t=0, that is, the Alexander-Hirschowitz theorem [1, 2, 4, 5]. We use certain nilpotent structures on reducible conics (called sundials in [9]) and on lines (called +lines in [10]; see Section 2 for them).Our interest in this topic (after, of course, [1–3, 8, 11]) was reborn by Carlini et al. who started a long project about the Hilbert functions of multiple structures on unions of linear subspaces of ℙr [9, 12, 13].


## 2. Preliminary Lemmas

LetQ⊂ℙ3 be a smooth quadric surface. For each finite set S⊂ℙ3, set 2S∶=∪P∈S2P. For each closed subscheme X⊂ℙ3, the residual scheme ResQ(X) of X with respect to Q is the closed subscheme of X with ℐX:ℐQ as its ideal sheaf. For each integer x, we have the following exact sequence (often called Castelnuovo's sequence):
(1)0⟶ℐResQ(X)(x-2)⟶ℐX(x)⟶ℐX∩Q(x)⟶0.
From (1) we get h0(ℐX(x))≤h0(ℐResQ(X)(x-2))+h0(Q,ℐX∩Q(x)) and h1(ℐX(x))≤h1(ℐResQ(X)(x-2))+h1(Q,ℐX∩Q(x)). Let Z⊂Q be a zero-dimensional subscheme, and let E⊂Q be a union of y distinct lines of type (0,1). The residual scheme ResE(Z) of Z is the closed subscheme of Q with ℐZ:ℐE as its ideal sheaf. For each x∈ℤ, we have Castelnuovo's exact sequence of coherent sheaves on Q:
(2)0⟶ℐResE(Z)(x,x-y)⟶ℐZ(x,x)⟶ℐE∩Z,E(x,x)⟶0.
Hence hi(Q,ℐZ(x,x))≤hi(Q,ℐResE(Z)(x,x-y))+hi(E,ℐZ∩E(x,x)), i=0,1. We have h1(E,ℐZ∩E(x,x))=0 if and only if deg⁡(Z∩L)≤x+1 for each connected component L of E.LetZ(3,t,a)′ denote the closure of Z(3,t,a) in the Hilbert scheme of ℙ3. Fix a line L⊂ℙ3 with P∈L and a line D⊂ℙ3 with P∈D and D≠L. Set v∶=2P∩D. Set X∶=L∪v (it is the intersection of L∪D with the scheme L∪2P). As in [10], we call X a +line with L as its support and P as the support of its nilradical.For all(x,y)∈ℕ2∖{(0,0)}, let L(3,x,y) be the set of all disjoint unions X⊂ℙ3 of x lines and y +lines. The algebraic set L(3,x,y) is irreducible and its general element has maximal rank [10, Theorem 1]. For all integers (x,y)∈ℕ2, let L(3,x,y)Q denote the set of all X∈L(3,x,y) such that dim⁡(X∩Q)=0 and the support of the nilradical of 𝒪X is contained in Q. The algebraic set L(3,x,y)Q is irreducible. For a general X∈L(3,x,y), we have ResQ(X)=X; that is, for each point P in the support of the nilradical of 𝒪X, the tangent vector v representing the nilradical of 𝒪X is not tangent to Q.For all integersk≥0, set ak∶=⌊(k+3)(k+2)/6⌋,  bk∶=(k+33)-(k+1)ak, and zk∶=⌊(k+3)(k+2)/6⌋-⌊(k+1)k/6⌋. We have ak=(k+3)(k+2)/6 and bk=0 if k≡0,1(mod⁡3) and ak=(k+3)(k+2)/6-1/3, bk=(k+1)/3 if k≡2(mod⁡3). We have z6t=4t+1, z6t+1=4t+2, t>1, z6t+2=4t+2, z6t+3=4t+3, z6t+4=4t+4, and z6t+5=4t+4.For each integerx≥0, let D3x+2′ and D3x+2′′ denote the following assertions.AssertionD3x+2′: hi(ℐX(k))=0, i=0,1, for a general X∈L(3,⌊(k+3)(k+2)/6⌋,(k+1)/3)Q.AssertionD3x+2′′: hi(ℐX(k))=0, i=0,1, for a general X∈L(3,⌊(k+3)(k+2)/6⌋-1,4(k+1)/3)Q.Remark 3.
Since9 general points of ℙ3 are contained in a smooth quadric and a general X∈L(3,⌊(k+3)(k+2)/6⌋,(k+1)/3) has maximal rank [10, Theorem 1], D3x+2′ is true if x+1≤9, that is, if x≤10. For the same reason, D3x+2′′ is true if 4x+3≤9, that is, if x≤1.Lemma 4.
D
3
x
+
2
′ is true for all x≥0.Proof.
Fix a generalY∈L(3,a3x,0). We have hi(ℐY(3x))=0, i=0,1, and Y∩Q is a general union of (x+1)(3x+2) points. Let E⊂Q be a general union of z3x+2=2x+2 lines of type (0,1). Fix S⊂Y∩Q with ♯(S)=x+1 and ♯(S∩L)≤1 for each line L⊂Y. Write Y=Y′⊔Y′′ with Y′′ the union of the lines of Y containing a point of S. Let G⊂ℙ3 be a disjoint union of x+1 +lines with Y′′=Gred, and let S be the support of the nilradical of G and with Q containing each tangent vector of a +lines of G; that is, we assume that deg⁡(G∩Q)=3(x+1) and that ResQ(G)=Y′′. Set X∶=Y′∪E∪G. Deforming E to a general union of z3x+2 lines of ℙ3, we get that X is a flat limit of a family of elements of L(3,a3x+2-x-1,x+1)Q. By the semicontinuity theorem for cohomology [14, III.12.8], it is sufficient to prove that hi(ℐX(k))=0,  i=0,1. Since ResQ(X)=Y, it is sufficient to prove that hi(Q,ℐQ∩X(k))=0, i=0,1; that is, hi(Q,ℐResE(X∩Q)(3x+2,x))=0. These equalities are true, because ResE(X∩Q) is the union of 2a3x-2x-2 general points and x+1 general tangent vectors of Q and 2(x+1)≤(3x+2)+x; in characteristic zero we may also quote two general results, that is, [15] (for general zero-dimensional curvilinear schemes) and [16, Lemma 1.4] (for general tangent vectors).For each integerk≥2, let Dk denote the following assertion.AssertionDk,  k≥2: for every integer y such that 0≤y≤k+1, we have h1(ℐX(k))=0 for a general X∈L(3,ak-1-y,y)Q.Remark 5.
Fix an integerk≥2 and assume the existence of Y∈L(3,ak-k-2,k+1)Q such that h1(ℐY(k))=0. Since h0(𝒪Y(k))=(k+33)-bk, we have h0(ℐY(k))=bk. Fix an integer y such that 0≤y≤k. Take any X∈L(3,ak-1-y,y) such that X⊂Y. Obviously, X∈L(3,ak-1-y,y)Q, h1(ℐX(k))=0, and h0(ℐX(k))=bk+k+1-y. Therefore to prove Dk, it is sufficient to prove its case “y=k+1.”Lemma 6.
D
k is true for all k≥2 and D3x+2′′ is true for all x≥0.Proof.
By Remark5 to prove Dk, we may assume that k≥9 and only check the case “y=k+1” of Dk.
(a) Assume thatk≡2(mod⁡3). Fix a general Y∈L(3,ak-2,0). We have hi(ℐY(k-2))=0, i=0,1 [8], and Y∩Q is a general union of 2ak-2 points of Q. Fix S⊂Y∩Q such that ♯(S)=k+1 and each line of Y contains at most one point of S. Write Y=Y′⊔Y′′ with Y′′ the union of the components of Y containing a point of S. Let G⊂ℙ3 be a disjoint union of k+1 +lines with Y′′=Gred, and let S be the support of the nilradical of G and with Q containing each tangent vector of a +lines of G; that is, we assume that deg⁡(G∩Q)=3(k+1) and that ResQ(G)=Y′′. Let E⊂Q be a general union of zk-1 lines of type (0,1). Set X∶=Y′∪G∪E. Deforming E to a general union of zk-1 lines of ℙ3, we see that X is in the closure of a set of elements of L(3,ak-1,k+1)Q. Hence it is sufficient to prove that h1(ℐX(k))=0; that is, h0(ℐX(k))=(k+1)/3. Since ResQ(X)=Y and h1(ℐY(k-2))=0, it is sufficient to prove that h1(Q,ℐX∩Q(k))=0; that is, h1(Q,ℐResE(Q∩(Y′∪G)(k,k+1-zk))=0. Since E∩X=∅, we have ResE(Q∩(Y′∪G))=Q∩(Y′∪G). The scheme Q∩(Y′∪G) is a general union of 2deg⁡(Y′) points of Q and k+1 tangent vectors of Q. Hence it is sufficient to prove that h1(Q,ℐZ(k,k+1-zk))=0 for a general union Z of k+1 tangent vectors of Q; this is very easy; in characteristic zero we may also quote two general results: [15] and [16, Lemma 1.4].
(b) Assume thatk≡0(mod⁡3). Take Y,S,G,X as in step (a). In this case we easily get hi(ℐX(k))=0,  i=0,1.
(c) By Remark3 to prove D3x+2′′ for all x≥0, it is sufficient to prove D3x+2′′ for all x≥2. As in step (a), we get D3x+2′′.
(d) Assume thatk≡1(mod⁡3). Take Y∈L(3,ak-2-(k-1)/3,(k-1)/3)Q satisfying Dk-2′. For a general Y, we may assume that ResQ(Y)=Y and that Y∩Q is a general union of 2ak-2 points of Q. Write Y=Y′⊔Y′′ with Y′′ union of the +lines. Fix a set S⊂Y′∩Q such that ♯(S)=k+1-(k-1)/3 and each line of Y′ contains at most one point of S. Write Y′=Y1⊔Y2 with Y2 the union of the lines of Y′′ containing a point of S. Let G⊂ℙ3 be a disjoint union of k+1-(k-1)/3 +lines with Y2 as its support and S as the support of its nilradical and with Q containing each tangent vector of a +lines of G. Let E⊂Q be a general union of zk-1 lines of type (0,1). Set X∶=Y1∪G∪Y′′∪E. As in step (a), we get hi(ℐX(k))=0, i=0,1, and X is a flat limit of a family of elements of L(3,ak-1,k+1)Q.AssertionCk,  k≥3. Set ϵ∶=0 if k≡0,1(mod⁡3), ϵ∶=0 if k=5,  ϵ∶=(k+1)/3-1 if k≡2(mod⁡6), and ϵ:(k+1)/3-2 if k≡5(mod⁡6) and k≥11. There are a union F⊂Q of ⌈y/2⌉ different lines of type (1,0), B⊂F, ♯(B)=y, and ♯(B∩L)≤2 for each component L of F, Y∈L(3,ak-1-y,y), B is the support of the nilpotent sheaf of Y, and h1(ℐY(k))=0.Remark 7.
Take a generalB⊂ℙ3 with ♯(B)≤6. B is a contained in the union T of 3 disjoint lines and T is contained in a smooth quadric with, say, T of type (3,0). Since a general element of L(3,ak-1-y,y) has maximal rank [10], to prove Ck, it is sufficient to check the existence of Y for all y>7. In particular Ck is true if k≤6.Lemma 8.
C
k is true for all k≥3.Proof.
By Remark7, we may assume that k≥y≥7. In steps (a), (b), and (c), we assume that y≥zk, while in step (d), we handle the easier case y≤zk-1.
(a) Assume thatk≡2(mod⁡3) and y≥zk. Take a general Y∈L(3,ak-4,0). We have hi(ℐY(k-4))=0,  i=0,1. Let E⊂Q be a general union of zk-2 lines of type (0,1). Fix the union F⊂Q of ⌊zk-2/2⌋ general lines of type (1,0). As in [8] (or as in the proof of Lemma 4), we get hi(ℐY∪E(k-2))=0,  i=0,1. For each line L⊂F, we fix two of the lines of E (and call them EL′ and EL′′) with the condition that ∪L⊆FEL′∪EL′′ is the union of 2⌊zk-2/2⌋ distinct lines. Set S1∶=∪L⊆FL∩(EL′∪EL′′). Write E=E1⊔E0 with E0=∅ if zk-2 is even and E0 the only line of E not containing a point of S1 if zk-2 is odd. Equivalently, take S1⊂F∩E such that ♯(S1∩L)=2 for every line L⊂F and ♯(S1∩L)≤1 for every line L⊂E. There are a smooth and connected affine curve Δ, o∈Δ, a flat family {Eλ}λ∈Δ of subschemes of ℙ3 with Eλ∈L(3,zk-2,0) for all λ∈Δ, Eo=E, Eλ transversal to Q for all λ≠o and such that for each L∈F and each λ∈Δ∖{o} exactly two of the lines of Eλ meet L; that is, we deform the lines of EL′ and EL′′ to lines EλL′, EλL′′ transversal to Q, but meeting L. Notice that for a general λ∈Δ, the point of EλL′∩(Q∖L) and the point of EλL′′∩(Q∖L) may be general points of Q. Hence we may find this flat family with the additional condition that for a general λ∈Δ the set (Y∪Eλ)∩(Q∖F) is a general union of 2ak-2+2zk-2-2⌊zk-2/2⌋ points of Q. Fix a general λ∈Δ and write Eλ=E1λ⊔E0λ with S1⊂E1λ and E0λ either empty (case zk-2 even) or E0λ a line (case zk-2 odd). Let G⊂Q be a general union of zk-1 lines of type (0,1). Fix a general union F′⊂Q of ⌈y/2⌉-⌊zk-2/2⌋ lines of type (1,0) and any union G2⊂G of y-2⌊zk-2/2⌋ components of G. Write G=G2⊔G0. We take S2⊂F′∩G2 so that ♯(S2)=y-2⌊zk-2/2⌋, ♯(S2∩L)≤2 for every line L⊂F′ and ♯(S2∩L)≤1 for every line L⊂E′; to check that this is possible we need to check that zk-1≥y-2⌊zk-2/2⌋ (call ⋄ this inequality). First assume that k=6t+2. We have z6t+2=4t+2 and z6t=4t+1; we have z6t+1-1+2⌊z6t/2⌋=4t+1+4t=k+(k+1)/3-1, and hence ⋄ is true by our definition of the integer ϵ. Now assume that k=6t+5; we have z6t+5=4t+4, z6t+3=4t+3, and hence z6t+5-1+2⌊z6t+3/2⌋=4t+3+4t+2=k+(k+1)/3-2; hence ⋄ is true in this case. Write E′=E1′⊔E0′ with E0′ the union of all lines of E′ containing no point of S2. Let Y1 be a general union of +lines with E1 as its support and S1 as the support of the nilradical and the nilradical contained in Q, so that ResQ(Y1)=E1λ and deg⁡(Y1∩Q)=3deg⁡(Y1). Let Y2⊂Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X∶=Y∪Y1∪Y2∪E0λ∪G0. It is sufficient to prove that h1(ℐX(k))=0. Since ResQ(X)=Y∪Eλ, it is sufficient to prove that h1(Q,ℐQ∩X(k))=0. Since ResG(Q∩S)=Y∩Q∪(Y1∩Q)∪S2 and y≤k+1+(k+1)/3, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1∩Q satisfies h1(Q,ℐW(k,k+1-zk))=0. Write W=W′⊔S2. Since F′ is general, ♯(S2∩L)≤2 for each line L⊆F′, and S2∩L is general in L for each line L⊆F′ and k+1-zk≥2, it is sufficient to prove that h1(Q,ℐW′(k,k+1-zk))=0. In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only ♯(S2) tangent vectors and k+1+(k+1-zk)~y-⌊zk-2/2⌋=♯(S2).
(b) Assume thatk≡0(mod⁡3) and y≥zk. Fix a general Y∈L(3,ak-4-(k-3)/2,(k-3)/2)Q. By Dk-4′, we have hi(ℐY(k-4))=0,  i=0,1 (Lemma 4). Let E⊂Q be a general union of zk-2 lines of type (0,1), with the restriction that (k-3)/2 of them contain one of the points of the support of the nilradical of Y. As in step (d) of the proof of Lemma 6, we get hi(ℐY(k-2))=0,  i=0,1. Let F⊂Q be a general union of ⌊zk-2/2⌋ lines of type (1,0). Fix S1⊂F∩E such that ♯(S1∩L)=2 for every line L⊂F (and hence ♯(S1)=2⌊zk-2/2⌋) and ♯(S1∩L)≤1 for every line L⊂E. There are a smooth and connected affine curve Δ,  o∈Δ, a flat family {Eλ}λ∈Δ with Eλ∈L(3,zk-2,0) for all λ∈Δ, Eo=E, Eλ transversal to Q for all λ≠o and that for each L∈F and each λ∈Δ∖{o} exactly two of the lines of Eλ meet L. As in step (a) for a general λ, the set (Y∪Eλ)∩(Y∖F) is a general union of 2ak-2+2zk-2-2⌊zk-2/2⌋ points of Q. Fix a general λ∈Δ and write Eλ=E1λ⊔E0λ with S1⊂E1λ and E0λ either empty (case zk-2 even) or a line (case zk-2 odd). Let G⊂Q be a general union of zk-1 lines of type (0,1). Fix a general union F′⊂Q of ⌈y/2⌉-⌊zk-2/2⌋ lines of type (1,0) and any union G2⊂G or y-2⌊zk-2/2⌋ components of G. Write G=G2⊔G0 with G0 the union of all lines of G not intersecting S2. We take S2⊂F′∩G′ so that ♯(S2)=y-2⌊zk-2/2⌋, ♯(S2∩L)≤2 for every line L⊂F′, and ♯(S2∩L)≤1 for every line L⊂E′; to check that this is possible we need to check that zk-1≥y-2⌊zk-2/2⌋ and this is true and easier than in step (a), because ϵ=0. Write E′=E1′⊔E0′ with E0′ the union of all lines of E′ containing no point of S1. Let Y1 be a general union of +lines with E1 as its support and S1 as the support of the nilradical and the nilradical contained in Q, so that ResQ(Y1)=E1λ and deg⁡(Y1∩Q)=3deg⁡(Y1). Let Y2⊂Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X∶=Y∪Y1∪Y2∪E0λ∪G0. It is sufficient to prove that h1(ℐX(k))=0. Since ResQ(X)=Y∪Eλ, it is sufficient to prove that h1(Q,ℐQ∩X(k))=0. Since ResG(Q∩S)=Y∩Q∪(Y1∩Q)∪S2 and y≤k+1, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1∩Q satisfies h1(Q,ℐW(k,k+1-zk))=0. Write W=W′⊔S2. Since F′ is general, ♯(S2∩L)≤2 for each line L⊆F′, and S2∩L is general in L for each line L⊆F′ and k+1-zk≥2, it is sufficient to prove that h1(Q,ℐW′(k,k+1-zk))=0. In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only 2⌊zk-2/2⌋ tangent vectors and k+1+k+1-zk~2⌊zk-2/2⌋.
(c) Assume thatk≡1(mod⁡3) and y≥zk. Fix a general Y∈L(3,ak-4,0). We have hi(ℐY(k-4))=0,  i=0,1. Let E⊂Q be a general union of zk-2 lines of type (0,1). Let F⊂Q be a general union of ⌊zk-2/2⌋ lines. Fix S1⊂F∩E such that ♯(S1∩L)=2 for every line L⊂F and ♯(S1∩L)≤1 for every line L⊂E. There are a smooth and connected affine curve Δ, o∈Δ, a flat family {Eλ}λ∈Δ with Eλ∈L(3,zk-2,0) for all λ∈Δ, Eo=E, Eλ transversal to Q for all λ≠o and that for each L∈F and each λ∈Δ∖{o} exactly two of the lines of Eλ meet L. As in step (a) for a general λ, the set (Y∪Eλ)∩(Y∖F) is a general union of 2ak-2+2zk-2-2⌊zk-2/2⌋ points of Q. In this flat family each Eλ, λ≠0, is a disjoint union of lines, and only Eo has sundials as some of its connected components. Fix a general λ∈Δ and write Eλ=E1λ⊔E0λ with S1⊂E1λ and E0λ either empty (case zk-2 even) or a line (case zk-2 odd). Fix γ⊂E1∖E1∩F such that ♯(γ)=bk-2=(k-1)/3. For each line R of E and each μ∈Δ call Rμ, the line of Eμ such that {Rμ} is an algebraic family of lines. Let γλ be the intersection with Q∖F of the subcurve Eλ(γ) of Eλ corresponding to the lines Rλ with R∩γ≠∅. The set γλ is a general subset of Q with cardinality bk-2. Let E~λ be the union of Eλ∖Eλ(γ) and bk-2 general +lines contained in Q, with Eλ(γ) as their support and with γλ as the support of their nilradicals; for each +line J of E~, we assume that the tangent vector v of J corresponding to the nilpotent sheaf of J is not tangent to Q; that is, we assume that the scheme J∩Q has degree two and it is the disjoint union of two points; with these restrictions we have ResQ(E~λ)=E~λ. Let G⊂Q be a general union of zk-1 lines of type (0,1) with the only restriction that bk-2 of them contain a point of γλ. Fix a general union F′⊂Q of ⌈y/2⌉-⌊zk-2/2⌋ lines of type (1,0) and any union G2⊂G or y-2⌊(ak-2-ak-4)/2⌋ components of G. Write G=G2⊔G0 with G0 the union of all lines of G not intersecting S2. We take S2⊂F′∩G′ so that ♯(S2)=y-⌊zk-2/2⌋, ♯(S2∩L)≤2 for every line L⊂F′ and ♯(S2∩L)≤1 for every line L⊂E′; to check that this is possible we need to check that zk-1≥y-2⌊zk-2/2⌋ and this is true and easier than in step (a), because ϵ=0. Write E′=E1′⊔E0′ with E0′ the union of all lines of E′ not containing a point of S2. Let Y1 be a general union of +lines with E1 as its support and S1 as the support of the nilradical and the nilradical contained in Q, so that ResQ(Y1)=E1λ and deg⁡(Y1∩Q)=3deg⁡(Y1). Let Y2⊂Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X∶=Y∪Y1∪Y2∪E~0λ∪G0. It is sufficient to prove that h1(ℐX(k))=0. Since ResQ(X)=Y∪Eλ, it is sufficient to prove that h1(Q,ℐQ∩X(k))=0. Since ResG(Q∩S)=Y∩Q∪(Y1∩Q)∪S2 and y≤k+1+(k+1)/3, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1∩Q satisfies h1(Q,ℐW(k,k+1-zk))=0. Write W=W′⊔S2. Since F′ is general, ♯(S2∩L)≤2 for each line L⊆G, and S2∩L is general in L for each line L⊆G and k+1-zk≥2, it is sufficient to prove that h1(Q,ℐW′(k,k+1-zk))=0. In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only 2⌊zk-2/2⌋ tangent vectors and k+1+k+1-zk~2⌊zk-2/2⌋.
(d) From now on we assume thaty≤zk-1.
(d1) Assume thatk≡2(mod⁡3). Fix a general Y∈L(3,ak-2,0). We have hi(ℐY(k-2))=0, i=0,1 [8]. Fix a general union F⊂Q of ⌈y/2⌉ lines of type (1,0) and a general union E⊂Q of zk-1 lines of type (0,1). Write E=E1⊔E0 with deg⁡(E1)=y. Fix S⊂E1∩F such that ♯(S)=y, ♯(S∩L)≤2 for all lines L⊆F and ♯(S∩R)=1 for each line R⊆E1. Let G⊂Q be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X∶=Y∪G∪E0; it works, because h1(Q,ℐX∩Q(k))=0 since y≤k+1+(k+1)/3, deg⁡(E)=zk-1, and h0(𝒪Q∩X(k))≤(k+1)2.
(d2) Assume thatk≡0(mod⁡3). Fix a general Y∈L(3,ak-2,0). We have hi(ℐY(k-2))=0,  i=0,1. Fix a general union F⊂Q of ⌈y/2⌉ lines of type (1,0) and a general union E⊂Q of zk-1 lines of type (0,1). Write E=E1⊔E0 with deg⁡(E1)=y. Fix S⊂E1∩F such that ♯(S)=y, ♯(S∩L)≤2 for all lines L⊆F and ♯(S∩R)=1 for each line R⊆E1. Let G⊂Q be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X∶=Y∪G∪E0; it works, because h1(Q,ℐX∩Q(k))=0 since y≤k+1, deg⁡(E)=zk-1, and h0(𝒪Q∩X(k))≤(k+1)2.
(d3) Assume thatk≡1(mod⁡3). We have bk-2=(k-1)/3. Fix a general Y∈L(3,ak-2-(k-1)/3,(k-1)/3)Q. We have hi(ℐY(k-2))=0, i=0,1 by Dk-2′ (Lemma 6). Fix a general union F⊂Q of ⌈y/2⌉ lines of type (1,0) and a general union E⊂Q of zk-1 lines of type (0,1) with the only restriction that (k-1)/3 of them contain a point of the support of the nilradical of Y. Write E=E1⊔E0 with deg⁡(E1)=y. Fix S⊂E1∩F such that ♯(S)=y, ♯(S∩L)≤2 for all lines L⊆F and ♯(S∩R)=1 for each line R⊆E1. Let G⊂Q be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X∶=Y∪G∪E0; it works, because h1(Q,ℐX∩Q(k))=0 since y≤k+1, deg⁡(E)=zk-1, and h0(𝒪Q∩X(k))≤(k+1)2.We also need the following elementary and well-known lemma (see, e.g., [6, Lemma 2]).Lemma 9.
Fix integerk≥2, y>0, a closed scheme X⊂ℙ3 such that h1(ℐX(k))=0 and a general set S⊂Q such that ♯(S)=y. If y≤h0(ℐX(k))-h0(ℐResQ(X)(k-2)), then h1(ℐX∪S(k))=0. If h0(ℐResQ(k-2))=0 and y≥h0(ℐX(k)), then h0(ℐX∪S(k))=0.


## 3. Proof of Theorem1

In this section we prove Theorem1. We may assume that a>0 (by [8]) and t>0 (by the Alexander-Hirschowitz theorem). Set δ∶=(k+33)-(k+1)t-4a. By increasing or decreasing a, we reduce to the case -3≤δ≤3. Since any 2-point contains a point if a>0 and Theorem 1 is true for the triple (t,a-1,k) with δ=-1 (i.e., if h0(ℐX(k))=1 for a general X∈Z(3,t,a-1), then it is true for the triple (t,a,k) with δ=4). Hence it is sufficient to check all triples (t,a,k) with -3≤δ≤2. In characteristic zero we may avoid all cases with δ=2 quoting either [15] or [16, Lemma 1.4]. The cases k=3,4,5 are true by [6, Propositions 1,2,3]. In steps (a) and (b), we assume that k≥9 and that Theorem 1 is true in degree k-2 and k-4. See steps (c), (d), and (e) for the cases k=6,7,8.(a) Assume thata≥(4k+2)/3. If 3a+2t≥(k+1)2+4, then set g∶=0. If 3a+2t≤(k+1)2+3, then let g be the minimal positive integer such that 3a+2t+g(k-1)≥(k+1)2+4.Claim 1.
We haveg≤k-1.Proof of Claim1.
Assume thatg≥k. We get 3a+2t+(k-1)2≤(k+1)2+3; that is, 3a+2t≤4k+3. Since t>0, we get 3a≤4k+1, a contradiction.Claim 2.
We haveg<t.Proof of Claim2.
Assume thatg≤t. The minimality of the integer g implies 3a+2t+(t-1)(k-1)≤(k+1)2+4; that is, 3a+(k+1)t≤k2+3k+4, contradicting the inequality δ≥-3.Setw∶=⌊(k+1)2-2t-(k-1)g)/3⌋ and z∶=(k+1)2-2t-(k-1)g-3w. We have 0≤z≤2 and 3w+z+2(t-g)=(k+1)(k+1-g). Since z≤2 and 3a+2t+g(k-1)≥(k+1)2+4, we get a≥w+z. Let Y be a general element of Z(3,t-g,a-w-z). Fix a general S∪S′⊂Q such that ♯(S)=w, ♯(S′)=z, and S∩S′=∅. Let E⊂Q be a general union of g lines of type (0,1). Since g<t, we have 3w<(k+1)(k+1-g). Since k-g>0, we have h1(Q,ℐ2S∩Q(k,k-g))=0 [17, 18] and [19, Corollary 4.6] (the latter one only if k-g≥2, but the case k-g=1 is elementary, because 3w<k+1). Since Q∩Y is a general subset with cardinality 2t-2g and 3w+z+2t=(k+1)2, we get hi(ℐ(Y∪2S)∩Q∪S′∪E(k))=0,  i=0,1. By the Differential Horace Lemma for double points [1] and [5, Lemma 5] to prove that h1(ℐX′(k))=max⁡{0,δ} for a general union X′ of Y∪2S∪E and z general 2-points (and hence to prove Theorem 1 in this case), it is sufficient to prove that h1(ℐY∪S∪(S′∩Q)(k-2))=max⁡{0,δ}. We have h0(𝒪Y∪S∪(S′∩Q)(k-2))=(k+13)+δ+2z-2w. Assume that for the moment δ≤0. Since z≤w, we have h0(𝒪Y∪S∪(S′∩Q)(k-2))≤(k+13). Since w≥z and deg⁡(2S′)-deg⁡(2S′∩Q)=z, we have h0(𝒪Y∪2S′(k-2))≤(k+13). Since ♯(S′)=z≤2 and any two points of ℙ3 are contained in a smooth quadric surface, Y∪2S′ may be considered as a general element of Z(3,t-g,a-u). Since h0(𝒪Y∪2S′(k-2))≤(k+13), the inductive assumption gives h1(ℐY∪2S′(k-2))=0. Hence h1(ℐY∪(S′∩Q)(k-2))=0. By Lemma 9 to prove that h1(ℐY∪S∪(S′∩Q)(k-2)), it is sufficient to prove that h0(ℐY(k-4))≤h0(ℐY(k-2))-w-2z. Since h1(ℐY(k-2))≤h1(ℐY∪2S′(k-2))=0, we have h0(ℐY(k-2))=w+2z-δ≥w+2z. Hence it is sufficient to prove that h0(ℐY(k-4)). The inductive assumption in degree k-4 shows that it is sufficient to prove that h0(𝒪Y(k-4))≥(k-13). We have h0(𝒪Y(k-4))=h0(𝒪Y(k-2))-2(t-g)=(k+13)+δ-w-2z-2(t-g). Since 3w+z+2(t-g)=(k+1)2 and (k+13)-(k-13)=(k-1)2, it is sufficient to check that 2w≥z-δ+4k. Since z≤2 and -δ≤3, it is sufficient to prove that w≥2k+3. Assume that w≤2k+2. Hence 3w+z≤6k+8. Since h0(ℐY(t-2))>0, we have (k-1)(t-g)≤(k+13); that is, t-g≤k(k+1)/6. Hence 3w+z+2(t-g)≤6k+8+k(k+1)/3, contradicting the inequality δ≥-3 for all k≥6.Now assume thatδ=1,2. If δ≥z, then we just saw that h0(ℐY∪S(k-2))=0. Hence we may assume z=2 and δ=1. Take a general P∈Q and set S1∶=S∪{P}. We have h0(Q,ℐ2S1∩Q(k))=0. Hence it is sufficient to prove that h0(ℐY∪S1(k-2))=0. Since we checked that h0(ℐA(k-4))=0 (easier for δ>0) and that h1(ℐA(k-2))=0 (i.e., h0(ℐA(k-2))=♯(S1)), Lemma 9 gives hi(ℐA∪S1(k-2))=0,  i=0,1.(b) In this step we assume thata<(4k+2)/3. Define the integers f and f′ by the relations
(3)3a+2t+(k-1)f-f′=(k+1)2,0≤f′≤k-2.
Since a>0, we have f≤k+2.Claim 3.
We havef≤k-1 for all k≥7 and f≤k-2 for all k≥9.Proof of Claim3.
Assume thatf≥k. We get 3a+2t+(k-1)2<(k+1)2; that is, 3a+2t≤4k+1. For all k≥7, we get 4a+(k+1)t≤(k+1)(4k+1)/2<-3+(k+33), a contradiction. Now assume that f=k-1. We get 3a+2t≤5k-2. For all k≥9, we get 4a+(k+1)t≤(k+1)(5k-2)/2<-3+(k+33), a contradiction.Claim 4.
We havet-f≥k-2 for all k≥7.Proof of Claim4.
Assume thatt≤f+k-1. Since f≤k-1 by Claim 3, we get t≤2k. Hence 4a+(k+1)t≤4(4k+1)/3+2k(k+1)<-3+(k+33), a contradiction for all k≥7.Claim 5.
We have2f≥f′ for all k≥4.Proof of Claim5.
Assume thatf≤(f′-1)/2. We get 3a+(k-1)(f′-1)/2-f′≥(k+1)2; that is, 3a+(k-3)f′/2≥k2+3k/2+3/2. Since f′≤k-2 and 3a≤4k+1, we get a contradiction.Claim 6.
Eithert-f<ak-2 or t-f=ak-2, k≡1(mod⁡3) and a<(k-1)/3, or Theorem 1 holds for the pair (t,a).Proof of Claim6.
First assume thatt-f≥ak-2+1. Fix a general U∈L(3,t-f,0). By [8], we have h0(ℐU(k-2))=0 and h1(ℐU(k-2))=k-1-bk-2≥2(k-1)/3. Let E⊂Q be a general union of f lines. Fix a general S⊂Q such that ♯(S)=a. Since h0(Q,ℐQ∩(Y∪2S)∪E(k))=0, Castelnuovo's sequence gives h0(ℐE∪2S∪Y(k))=0 and hence δ≥0 and Theorem 1 is true for the pairs (t,a). The same proof works if t-f=ak-2 and either bk-2=0 (i.e., k-2≡0,1(mod⁡3)) or k-2≡2(mod⁡3) and a≥(k-1)/3 (in the latter case we quote Lemma 9 to get h0(ℐY∪S(k-2))=0.(b1) Assume for the moment that eitherk≡0,2(mod⁡3), t-f≠ak-2, or a≥(k-1)/3. Let S⊂Q be a general set with ♯(S)=a. Let E⊂Q be a general union of f lines of type (0,1). Claim 4 implies t≥f+f′. By Claim 6, we may assume that f′≤2f. Since f′≤2f, f′≤k-2, t-f-f′≥0, and t-f<ak-2, Lemma 6 shows the existence Y∈L(3,t-f-f′,f′) such that h1(ℐY(k-2))=0, the nilradical of 𝒪Y is supported by points of E, each scheme Y∩Q is reduced, and it is the union of 2(t-f) points, f′ on E, and the remaining ones forming a general subset of Q with cardinality 2(t-f)-f′>0. Since k-f≥1 and 2(t-f)-f′>0, we have h1(Q,ℐ2S∩Q(k,k-f))=0 [17]. Hence hi(Q,ℐ(Y∪2S)∩Q∪E(k))=0, i=0,1.(b2) Now assume thatk≡1(mod⁡3), ak-2=t-f, and a<(k-1)/3. We modify step (b1) in the following way. Let W be a general element of L(3,ak-2-(k-3)/3,(k-3)/3)Q. Since Dk-2′ is true (Lemma 4), we have hi(ℐW(k-2))=0,  i=0,1. Take any Y⊆W with Y∈L(3,t-f-f′,f′)Q. We have h1(ℐY(k-2))=0, because for each +line T and each integer t≥1, the restriction map H0(T,𝒪T(t))→H0(Tred,𝒪Tred(t)) is surjective. Since (t-3)(ak-2-(k-1)/3)≥(k+13), we have h0(ℐY(k-4))=0. Hence Lemma 9 gives that either h0(ℐY∪S(k-2))=0 or h1(ℐY∪S(k-2))=0. Since 3a+2(t-f)+(k+1)f-f′=(k+1)2, we have f≤zk≤k-2. Hence we may apply [17, 18] or [19, Corollary 4.6].(c) Assume thatk=6. Since (93)=84, it is sufficient to consider the following triples (t,a,δ): (1,19,-1), (1,20,3), (2,17,-2), (2,18,2), (3,15,-3), (3,16,1), (4,14,0), (5,12,-1), (5,13,3), (6,10,-2), (6,11,2), (7,8,-3), (7,9,1), (8,7,0), (9,5,-1), (9,6,3), (10,3,-2), (10,4,2), (11,1,-3), and (11,2,1). In all cases we take inside Q a general union of f lines of type (0,1) and S∪S′ with ♯(S)=u, ♯(S′)=v, 3u+v+2(t-f)=8(8-f), and 0≤v≤2. Outside Q, we have a general Y∈Z(3,t-f,a-u-v).(c1) Assume that1≤t≤3. Take f=t. The triples (t,u,v) are the following ones: (1,18,2), (2,16,0), and (3,13,1). Hence we need to modify the proof in the case (t,a,δ)=(1,19,-1). In that case we take S′ with ♯(S′)=1; that is, take v=1. With this modification, we get a≥u+v in all cases. To use the Alexander-Hirschowitz theorem in degree 4, we need to check that a-u<9. We need to check that h0(ℐY(2))≤h0(ℐY(4))-u-2v. In all cases we have a-u-v≤2, and hence h1(ℐY(1))≤1. Hence h0(ℐY(4))-h0(ℐY(2))≥52-1.(c2) Assume thatt=4. Take f=3, and hence (u,v)=(12,2). Hence Y is a line, h1(ℐY∪2S′(4))=0 and h0(ℐY(4))-h0(ℐY(2))=25≥u+2v.(c3) Assume that5≤t≤10. Take f=4. Since t-f>0 and Y∩Q is a general union of 2(t-f) points, we have h1(Q,ℐ2S∩Q(6,2))=0. We have u=⌊(29-2t)/3⌋ and v=29-2t-3u. Hence the triples (t,u,v) are the following ones: (5,6,1), (6,5,2), (7,5,0), (8,4,1), (9,3,2), and (10,3,0). We have u≥v, u≥v+δ, a≥u+v, h1(ℐY∪2S′(4))=0 [6, Proposition 2], and h0(ℐY(2))=0.(c4) Assume thatt=11. If δ=-3, then we take f=5 and S=S′=∅. It is sufficient to use that h1(ℐW(4))=0 for a general W∈Z(3,6,1) [6, Proposition 2].(d) Assume thatk=7. Since (103)=120 and 120 is divided both by 8 and by 4, it is sufficient to check all cases with δ=0, a>0, t>0, that is, all pairs (t,a) with a=30-2t and 1≤t≤14. In all cases we take inside Q a general union of f lines of type (0,1) and S∪S′ with ♯(S)=u, ♯(S′)=v, u=⌊(8(8-f)-2(t-f))/3⌋, and 3u+v+2(t-f)=8(8-f). To apply [6, Proposition 1] and get h0(ℐY(3))·h1(ℐY(3))=0, it is sufficient to have (t-f,a-u-v)≠(2,3) [6, Proposition 1]; in the exceptional case we would have h0(ℐY(3))≤1 and even h0(ℐY(3))≤3 would be enough.(d1) Assume that1≤t≤4. Take t=f. We have u≥v, a≥u+v.(d2) Assume thatt=5. Take f=4, u=10, and v=0. Since v=0, we do not need to check the value of h0(ℐY(3)).(d3) Assume that6≤t≤14. Take f=5. Since t-f>0, we have h1(Q,ℐ2S∩Q(7,2))=0. Obviously a≥u+v and u≥v, but in a few cases one of these inequality is an equality (if t=12, then (a,u,v)=(6,3,1); if t=13, then (a,u,v)=(4,2,2); if t=14, then (a,u,v)=(2,2,0)).(e) Assume thatk=8. Since (113)=165, we need to control the following triples (t,a,δ): (1,39,0), (2,36,-3), (2,37,1), (3,34,-2), (3,35,2), (4,32,-1), (5,30,0), (6,27,-3), (6,28,1), (7,25,-2), (7,26,2), (8,23,-1), (9,21,0), (10,18,3), (10,19,1), (11,16,-2), (11,17,2), (12,14,-1), (13,12,0), (14,9,-3), (14,10,1), (15,7,-2), (15,8,2), (16,5,-1), and (17,3,0). For all t≠17, we use the proof of step (a) with the following quintuples (t,a,δ,g,u,v) (to show that the proof of step (a) works, we only need to check that in all cases 0≤g≤7, t>g (and hence in all cases we may apply [17, 18] or [19, Corollary 4.6]), a≥u+v, and u≥v+max⁡{0,δ}): (1,39,0,0,26,1), (2,36,-3,25,2), (2,37,1,25,2), (3,34,-2,25,0), (3,35,2,25,0), (4,32,-1,24,1), (5,30,0,23,2), (6,27,-3,23,0), (6,28,1,23,0), (7,25,-2,1,22,1), (7,26,2,1,22,1), (8,23,-1,3,18,0), (9,21,0,3,17,1), (10,18,-3,3,13,1), (10,19,1,3,13,1), (11,16,-2,5,8,0), (11,17,25,8,0), (12,14,-1,5,7,1), (13,12,0,5,6,2), (14,10,-1,5,6,0), (15,7,-2,5,5,1), (15,8,2,5,5,1), and (16,5,-1,6,2,1). If (t,a,δ)=(17,3,0), then we apply the proof of step (b) with f=6 and f′=4.



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*Source: 120850-2014-03-23.xml*