ISRN.GEOMETRY ISRN Geometry 2090-6315 Hindawi Publishing Corporation 120850 10.1155/2014/120850 120850 Research Article Postulation of General Unions of Lines and Multiplicity Two Points in 3 Ballico E. Franco D. Simis A. 1 Department of Mathematics University of Trento 38123 Povo Italy unitn.it 2014 2332014 2014 03 01 2014 11 02 2014 23 3 2014 2014 Copyright © 2014 E. Ballico. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove that a general union X3 of prescribed numbers of lines and double points has maximal rank, except a few well-known exceptional cases.

1. Introduction

Fix Pr. The 2-point 2P of r is the closed subscheme of r with (P)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 r3. For all (t,a)2, let Z(r,t,a) be the set of all disjoint unions Xr of t lines and a 2-points. Assume that (t,a)(0,0); that is, assume that Z(r,t,a). For each XZ(r,t,a) and each integer x0, 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 Wr is said to have maximal rank if for each integer x0, the restriction map H0(𝒪r(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 x0 either h0(W(x))=0 or h1(𝒪X(x))=0. In this paper we prove the following result.

Theorem 1.

Fix (t,a)2{(0,0)} and an integer k3. Let X be a general element of Z(3,t,a). Then either h0(X(k))=0 or h1(X(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 . 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 3 is contained in a unique quadric surface and for each P3, the linear system |2P(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)3 such that t+a4, (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 case k=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 4 and then in 5. Then we proved by induction on r the corresponding statement in r, r5, but only if k4 .

For the proof, we use the case a=0 proved by Hartshorne and Hirschowitz  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 ) and on lines (called +lines in ; see Section 2 for them).

Our interest in this topic (after, of course, [13, 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 r [9, 12, 13].

2. Preliminary Lemmas

Let Q3 be a smooth quadric surface. For each finite set S3, set 2S=PS2P. For each closed subscheme X3, the residual scheme ResQ(X) of X with respect to Q is the closed subscheme of X with X:Q as its ideal sheaf. For each integer x, we have the following exact sequence (often called Castelnuovo's sequence): (1)0ResQ(X)(x-2)X(x)XQ(x)0. From (1) we get h0(X(x))h0(ResQ(X)(x-2))+h0(Q,XQ(x)) and h1(X(x))h1(ResQ(X)(x-2))+h1(Q,XQ(x)). Let ZQ be a zero-dimensional subscheme, and let EQ 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 Z:E as its ideal sheaf. For each x, we have Castelnuovo's exact sequence of coherent sheaves on Q: (2)0ResE(Z)(x,x-y)Z(x,x)EZ,E(x,x)0. Hence hi(Q,Z(x,x))hi(Q,ResE(Z)(x,x-y))+hi(E,ZE(x,x)), i=0,1. We have h1(E,ZE(x,x))=0 if and only if deg(ZL)x+1 for each connected component L of E.

Let Z(3,t,a) denote the closure of Z(3,t,a) in the Hilbert scheme of 3. Fix a line L3 with PL and a line D3 with PD and DL. Set v=2PD. Set X=Lv (it is the intersection of LD with the scheme L2P). As in , we call X a +line with L as its support and P as the support of its nilradical.

For all (x,y)2{(0,0)}, let L(3,x,y) be the set of all disjoint unions X3 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)2, let L(3,x,y)Q denote the set of all XL(3,x,y) such that dim(XQ)=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 XL(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 integers k0, 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 k0,1  (mod3) and ak=(k+3)(k+2)/6-1/3, bk=(k+1)/3 if k2  (mod3). 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 integer x0, let D3x+2 and D3x+2′′ denote the following assertions.

Assertion D3x+2: hi(X(k))=0, i=0,1, for a general XL(3,(k+3)(k+2)/6,(k+1)/3)Q.

Assertion D3x+2′′: hi(X(k))=0, i=0,1, for a general XL(3,(k+3)(k+2)/6-1,4(k+1)/3)Q.

Remark 3.

Since 9 general points of 3 are contained in a smooth quadric and a general XL(3,(k+3)(k+2)/6,(k+1)/3) has maximal rank [10, Theorem 1], D3x+2 is true if x+19, that is, if x10. For the same reason, D3x+2′′ is true if 4x+39, that is, if x1.

Lemma 4.

D 3 x + 2 is true for all x0.

Proof.

Fix a general YL(3,a3x,0). We have hi(Y(3x))=0, i=0,1, and YQ is a general union of (x+1)(3x+2) points. Let EQ be a general union of z3x+2=2x+2 lines of type (0,1). Fix SYQ with (S)=x+1 and (SL)1 for each line LY. Write Y=YY′′ with Y′′ the union of the lines of Y containing a point of S. Let G3 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(GQ)=3(x+1) and that ResQ(G)=Y′′. Set X=YEG. Deforming E to a general union of z3x+2 lines of 3, 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(X(k))=0,  i=0,1. Since ResQ(X)=Y, it is sufficient to prove that hi(Q,QX(k))=0, i=0,1; that is, hi(Q,ResE(XQ)(3x+2,x))=0. These equalities are true, because ResE(XQ) 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,  (for general zero-dimensional curvilinear schemes) and [16, Lemma 1.4] (for general tangent vectors).

For each integer k2, let Dk denote the following assertion.

Assertion Dk,  k2: for every integer y such that 0yk+1, we have h1(X(k))=0 for a general XL(3,ak-1-y,y)Q.

Remark 5.

Fix an integer k2 and assume the existence of YL(3,ak-k-2,k+1)Q such that h1(Y(k))=0. Since h0(𝒪Y(k))=(k+33)-bk, we have h0(Y(k))=bk. Fix an integer y such that 0yk. Take any XL(3,ak-1-y,y) such that XY. Obviously, XL(3,ak-1-y,y)Q, h1(X(k))=0, and h0(X(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 k2 and D3x+2′′ is true for all x0.

Proof.

By Remark 5 to prove Dk, we may assume that k9 and only check the case “y=k+1” of Dk.

(a) Assume that k2  (mod3). Fix a general YL(3,ak-2,0). We have hi(Y(k-2))=0, i=0,1 , and YQ is a general union of 2ak-2 points of Q. Fix SYQ such that (S)=k+1 and each line of Y contains at most one point of S. Write Y=YY′′ with Y′′ the union of the components of Y containing a point of S. Let G3 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(GQ)=3(k+1) and that ResQ(G)=Y′′. Let EQ be a general union of zk-1 lines of type (0,1). Set X=YGE. Deforming E to a general union of zk-1 lines of 3, 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(X(k))=0; that is, h0(X(k))=(k+1)/3. Since ResQ(X)=Y and h1(Y(k-2))=0, it is sufficient to prove that h1(Q,XQ(k))=0; that is, h1(Q,ResE(Q(YG)(k,k+1-zk))=0. Since EX=, we have ResE(Q(YG))=Q(YG). The scheme Q(YG) 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,Z(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:  and [16, Lemma 1.4].

(b) Assume that k0  (mod3). Take Y,S,G,X as in step (a). In this case we easily get hi(X(k))=0,  i=0,1.

(c) By Remark 3 to prove D3x+2′′ for all x0, it is sufficient to prove D3x+2′′ for all x2. As in step (a), we get D3x+2′′.

(d) Assume that k1  (mod3). Take YL(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 YQ is a general union of 2ak-2 points of Q. Write Y=YY′′ with Y′′ union of the +lines. Fix a set SYQ such that (S)=k+1-(k-1)/3 and each line of Y contains at most one point of S. Write Y=Y1Y2 with Y2 the union of the lines of Y′′ containing a point of S. Let G3 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 EQ be a general union of zk-1 lines of type (0,1). Set X=Y1GY′′E. As in step (a), we get hi(X(k))=0, i=0,1, and X is a flat limit of a family of elements of L(3,ak-1,k+1)Q.

Assertion Ck,  k3. Set ϵ=0 if k0,1  (mod3), ϵ=0 if k=5,  ϵ=(k+1)/3-1 if k2  (mod6), and ϵ:(k+1)/3-2 if k5  (mod6) and k11. There are a union FQ of y/2 different lines of type (1,0), BF, (B)=y, and (BL)2 for each component L of F, YL(3,ak-1-y,y), B is the support of the nilpotent sheaf of Y, and h1(Y(k))=0.

Remark 7.

Take a general B3 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 , to prove Ck, it is sufficient to check the existence of Y for all y>7. In particular Ck is true if k6.

Lemma 8.

C k is true for all k3.

Proof.

By Remark 7, we may assume that ky7. In steps (a), (b), and (c), we assume that yzk, while in step (d), we handle the easier case yzk-1.

(a) Assume that k2  (mod3) and yzk. Take a general YL(3,ak-4,0). We have hi(Y(k-4))=0,  i=0,1. Let EQ be a general union of zk-2 lines of type (0,1). Fix the union FQ of zk-2/2 general lines of type (1,0). As in  (or as in the proof of Lemma 4), we get hi(YE(k-2))=0,  i=0,1. For each line LF, we fix two of the lines of E (and call them EL and EL′′) with the condition that LFELEL′′ is the union of 2zk-2/2 distinct lines. Set S1=LFL(ELEL′′). Write E=E1E0 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 S1FE such that (S1L)=2 for every line LF and (S1L)1 for every line LE. There are a smooth and connected affine curve Δ, oΔ, a flat family {Eλ}λΔ of subschemes of 3 with EλL(3,zk-2,0) for all λΔ, Eo=E, Eλ transversal to Q for all λo and such that for each LF and each λΔ{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 λΔ, the point of EλL(QL) and the point of EλL′′(QL) may be general points of Q. Hence we may find this flat family with the additional condition that for a general λΔ the set (YEλ)(QF) is a general union of 2ak-2+2zk-2-2zk-2/2 points of Q. Fix a general λΔ and write Eλ=E1λE0λ with S1E1λ and E0λ either empty (case zk-2 even) or E0λ a line (case zk-2 odd). Let GQ be a general union of zk-1 lines of type (0,1). Fix a general union FQ of y/2-zk-2/2 lines of type (1,0) and any union G2G of y-2zk-2/2 components of G. Write G=G2G0. We take S2FG2 so that (S2)=y-2zk-2/2, (S2L)2 for every line LF and (S2L)1 for every line LE; to check that this is possible we need to check that zk-1y-2zk-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+2z6t/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+2z6t+3/2=4t+3+4t+2=k+(k+1)/3-2; hence is true in this case. Write E=E1E0 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(Y1Q)=3deg(Y1). Let Y2Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X=YY1Y2E0λG0. It is sufficient to prove that h1(X(k))=0. Since ResQ(X)=YEλ, it is sufficient to prove that h1(Q,QX(k))=0. Since ResG(QS)=YQ(Y1Q)S2 and yk+1+(k+1)/3, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1Q satisfies h1(Q,W(k,k+1-zk))=0. Write W=WS2. Since F is general, (S2L)2 for each line LF, and S2L is general in L for each line LF and k+1-zk2, it is sufficient to prove that h1(Q,W(k,k+1-zk))=0. In characteristic zero this is true by either  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 that k0  (mod3) and yzk. Fix a general YL(3,ak-4-(k-3)/2,(k-3)/2)Q. By Dk-4, we have hi(Y(k-4))=0,  i=0,1 (Lemma 4). Let EQ 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(Y(k-2))=0,  i=0,1. Let FQ be a general union of zk-2/2 lines of type (1,0). Fix S1FE such that (S1L)=2 for every line LF (and hence (S1)=2zk-2/2) and (S1L)1 for every line LE. There are a smooth and connected affine curve Δ,  oΔ, a flat family {Eλ}λΔ with EλL(3,zk-2,0) for all λΔ, Eo=E, Eλ transversal to Q for all λo and that for each LF and each λΔ{o} exactly two of the lines of Eλ meet L. As in step (a) for a general λ, the set (YEλ)(YF) is a general union of 2ak-2+2zk-2-2zk-2/2 points of Q. Fix a general λΔ and write Eλ=E1λE0λ with S1E1λ and E0λ either empty (case zk-2 even) or a line (case zk-2 odd). Let GQ be a general union of zk-1 lines of type (0,1). Fix a general union FQ of y/2-zk-2/2 lines of type (1,0) and any union G2G or y-2zk-2/2 components of G. Write G=G2G0 with G0 the union of all lines of G not intersecting S2. We take S2FG so that (S2)=y-2zk-2/2, (S2L)2 for every line LF, and (S2L)1 for every line LE; to check that this is possible we need to check that zk-1y-2zk-2/2 and this is true and easier than in step (a), because ϵ=0. Write E=E1E0 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(Y1Q)=3deg(Y1). Let Y2Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X=YY1Y2E0λG0. It is sufficient to prove that h1(X(k))=0. Since ResQ(X)=YEλ, it is sufficient to prove that h1(Q,QX(k))=0. Since ResG(QS)=YQ(Y1Q)S2 and yk+1, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1Q satisfies h1(Q,W(k,k+1-zk))=0. Write W=WS2. Since F is general, (S2L)2 for each line LF, and S2L is general in L for each line LF and k+1-zk2, it is sufficient to prove that h1(Q,W(k,k+1-zk))=0. In characteristic zero this is true by either  or [16, Lemma 1.4]; in positive characteristic one could see that there are only 2zk-2/2 tangent vectors and k+1+k+1-zk~2zk-2/2.

(c) Assume that k1  (mod3) and yzk. Fix a general YL(3,ak-4,0). We have hi(Y(k-4))=0,  i=0,1. Let EQ be a general union of zk-2 lines of type (0,1). Let FQ be a general union of zk-2/2 lines. Fix S1FE such that (S1L)=2 for every line LF and (S1L)1 for every line LE. There are a smooth and connected affine curve Δ, oΔ, a flat family {Eλ}λΔ with EλL(3,zk-2,0) for all λΔ, Eo=E, Eλ transversal to Q for all λo and that for each LF and each λΔ{o} exactly two of the lines of Eλ meet L. As in step (a) for a general λ, the set (YEλ)(YF) is a general union of 2ak-2+2zk-2-2zk-2/2 points of Q. In this flat family each Eλ, λ0, is a disjoint union of lines, and only Eo has sundials as some of its connected components. Fix a general λΔ and write Eλ=E1λE0λ with S1E1λ and E0λ either empty (case zk-2 even) or a line (case zk-2 odd). Fix γE1E1F such that (γ)=bk-2=(k-1)/3. For each line R of E and each μΔ call Rμ, the line of Eμ such that {Rμ} is an algebraic family of lines. Let γλ be the intersection with QF of the subcurve Eλ(γ) of Eλ corresponding to the lines Rλ with Rγ. The set γλ is a general subset of Q with cardinality bk-2. Let E~λ be the union of EλEλ(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 JQ has degree two and it is the disjoint union of two points; with these restrictions we have ResQ(E~λ)=E~λ. Let GQ 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 FQ of y/2-zk-2/2 lines of type (1,0) and any union G2G or y-2(ak-2-ak-4)/2 components of G. Write G=G2G0 with G0 the union of all lines of G not intersecting S2. We take S2FG so that (S2)=y-zk-2/2, (S2L)2 for every line LF and (S2L)1 for every line LE; to check that this is possible we need to check that zk-1y-2zk-2/2 and this is true and easier than in step (a), because ϵ=0. Write E=E1E0 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(Y1Q)=3deg(Y1). Let Y2Q be a general union of +lines contained in Q, supported by G2 and with S2 as the support of the nilradical. Set X=YY1Y2E~0λG0. It is sufficient to prove that h1(X(k))=0. Since ResQ(X)=YEλ, it is sufficient to prove that h1(Q,QX(k))=0. Since ResG(QS)=YQ(Y1Q)S2 and yk+1+(k+1)/3, it is sufficient to prove that the union W of S2 and the degree two connected components of Y1Q satisfies h1(Q,W(k,k+1-zk))=0. Write W=WS2. Since F is general, (S2L)2 for each line LG, and S2L is general in L for each line LG and k+1-zk2, it is sufficient to prove that h1(Q,W(k,k+1-zk))=0. In characteristic zero this is true by either  or [16, Lemma 1.4]; in positive characteristic one could see that there are only 2zk-2/2 tangent vectors and k+1+k+1-zk~2zk-2/2.

(d) From now on we assume that yzk-1.

(d1) Assume that k2  (mod3). Fix a general YL(3,ak-2,0). We have hi(Y(k-2))=0, i=0,1 . Fix a general union FQ of y/2 lines of type (1,0) and a general union EQ of zk-1 lines of type (0,1). Write E=E1E0 with deg(E1)=y. Fix SE1F such that (S)=y, (SL)2 for all lines LF and (SR)=1 for each line RE1. Let GQ be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X=YGE0; it works, because h1(Q,XQ(k))=0 since yk+1+(k+1)/3, deg(E)=zk-1, and h0(𝒪QX(k))(k+1)2.

(d2) Assume that k0  (mod3). Fix a general YL(3,ak-2,0). We have hi(Y(k-2))=0,  i=0,1. Fix a general union FQ of y/2 lines of type (1,0) and a general union EQ of zk-1 lines of type (0,1). Write E=E1E0 with deg(E1)=y. Fix SE1F such that (S)=y, (SL)2 for all lines LF and (SR)=1 for each line RE1. Let GQ be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X=YGE0; it works, because h1(Q,XQ(k))=0 since yk+1, deg(E)=zk-1, and h0(𝒪QX(k))(k+1)2.

(d3) Assume that k1  (mod3). We have bk-2=(k-1)/3. Fix a general YL(3,ak-2-(k-1)/3,(k-1)/3)Q. We have hi(Y(k-2))=0, i=0,1 by Dk-2 (Lemma 6). Fix a general union FQ of y/2 lines of type (1,0) and a general union EQ 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=E1E0 with deg(E1)=y. Fix SE1F such that (S)=y, (SL)2 for all lines LF and (SR)=1 for each line RE1. Let GQ be a general union of y +lines with E1 as their support and S as the support of their nilradical. Set X=YGE0; it works, because h1(Q,XQ(k))=0 since yk+1, deg(E)=zk-1, and h0(𝒪QX(k))(k+1)2.

We also need the following elementary and well-known lemma (see, e.g., [6, Lemma 2]).

Lemma 9.

Fix integer k2, y>0, a closed scheme X3 such that h1(X(k))=0 and a general set SQ such that (S)=y. If yh0(X(k))-h0(ResQ(X)(k-2)), then h1(XS(k))=0. If h0(ResQ(k-2))=0 and yh0(X(k)), then h0(XS(k))=0.

3. Proof of Theorem <xref ref-type="statement" rid="thm1">1</xref>

In this section we prove Theorem 1. We may assume that a>0 (by ) 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 -3δ3. 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(X(k))=1 for a general XZ(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 -3δ2. In characteristic zero we may avoid all cases with δ=2 quoting either  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 k9 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 that a(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 have gk-1.

Proof of Claim <xref ref-type="statement" rid="claim1">1</xref>.

Assume that gk. We get 3a+2t+(k-1)2(k+1)2+3; that is, 3a+2t4k+3. Since t>0, we get 3a4k+1, a contradiction.

Claim 2.

We have g<t.

Proof of Claim <xref ref-type="statement" rid="claim2">2</xref>.

Assume that gt. The minimality of the integer g implies 3a+2t+(t-1)(k-1)(k+1)2+4; that is, 3a+(k+1)tk2+3k+4, contradicting the inequality δ-3.

Set w=(k+1)2-2t-(k-1)g)/3 and z=(k+1)2-2t-(k-1)g-3w. We have 0z2 and 3w+z+2(t-g)=(k+1)(k+1-g). Since z2 and 3a+2t+g(k-1)(k+1)2+4, we get aw+z. Let Y be a general element of Z(3,t-g,a-w-z). Fix a general SSQ such that (S)=w, (S)=z, and SS=. Let EQ 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,2SQ(k,k-g))=0 [17, 18] and [19, Corollary 4.6] (the latter one only if k-g2, but the case k-g=1 is elementary, because 3w<k+1). Since QY is a general subset with cardinality 2t-2g and 3w+z+2t=(k+1)2, we get hi((Y2S)QSE(k))=0,  i=0,1. By the Differential Horace Lemma for double points  and [5, Lemma 5] to prove that h1(X(k))=max{0,δ} for a general union X of Y2SE and z general 2-points (and hence to prove Theorem 1 in this case), it is sufficient to prove that h1(YS(SQ)(k-2))=max{0,δ}. We have h0(𝒪YS(SQ)(k-2))=(k+13)+δ+2z-2w. Assume that for the moment δ0. Since zw, we have h0(𝒪YS(SQ)(k-2))(k+13). Since wz and deg(2S)-deg(2SQ)=z, we have h0(𝒪Y2S(k-2))(k+13). Since (S)=z2 and any two points of 3 are contained in a smooth quadric surface, Y2S may be considered as a general element of Z(3,t-g,a-u). Since h0(𝒪Y2S(k-2))(k+13), the inductive assumption gives h1(Y2S(k-2))=0. Hence h1(Y(SQ)(k-2))=0. By Lemma 9 to prove that h1(YS(SQ)(k-2)), it is sufficient to prove that h0(Y(k-4))h0(Y(k-2))-w-2z. Since h1(Y(k-2))h1(Y2S(k-2))=0, we have h0(Y(k-2))=w+2z-δw+2z. Hence it is sufficient to prove that h0(Y(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 2wz-δ+4k. Since z2 and -δ3, it is sufficient to prove that w2k+3. Assume that w2k+2. Hence 3w+z6k+8. Since h0(Y(t-2))>0, we have (k-1)(t-g)(k+13); that is, t-gk(k+1)/6. Hence 3w+z+2(t-g)6k+8+k(k+1)/3, contradicting the inequality δ-3 for all k6.

Now assume that δ=1,2. If δz, then we just saw that h0(YS(k-2))=0. Hence we may assume z=2 and δ=1. Take a general PQ and set S1=S{P}. We have h0(Q,2S1Q(k))=0. Hence it is sufficient to prove that h0(YS1(k-2))=0. Since we checked that h0(A(k-4))=0 (easier for δ>0) and that h1(A(k-2))=0 (i.e., h0(A(k-2))=(S1)), Lemma 9 gives hi(AS1(k-2))=0,  i=0,1.

(b) In this step we assume that a<(4k+2)/3. Define the integers f and f by the relations (3)3a+2t+(k-1)f-f=(k+1)2,0fk-2. Since a>0, we have fk+2.

Claim 3.

We have fk-1 for all k7 and fk-2 for all k9.

Proof of Claim <xref ref-type="statement" rid="claim3">3</xref>.

Assume that fk. We get 3a+2t+(k-1)2<(k+1)2; that is, 3a+2t4k+1. For all k7, 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+2t5k-2. For all k9, we get 4a+(k+1)t(k+1)(5k-2)/2<-3+(k+33), a contradiction.

Claim 4.

We have t-fk-2 for all k7.

Proof of Claim <xref ref-type="statement" rid="claim4">4</xref>.

Assume that tf+k-1. Since fk-1 by Claim 3, we get t2k. Hence 4a+(k+1)t4(4k+1)/3+2k(k+1)<-3+(k+33), a contradiction for all k7.

Claim 5.

We have 2ff for all k4.

Proof of Claim <xref ref-type="statement" rid="claim5">5</xref>.

Assume that f(f-1)/2. We get 3a+(k-1)(f-1)/2-f(k+1)2; that is, 3a+(k-3)f/2k2+3k/2+3/2. Since fk-2 and 3a4k+1, we get a contradiction.

Claim 6.

Either t-f<ak-2 or t-f=ak-2, k1  (mod3) and a<(k-1)/3, or Theorem 1 holds for the pair (t,a).

Proof of Claim <xref ref-type="statement" rid="claim6">6</xref>.

First assume that t-fak-2+1. Fix a general UL(3,t-f,0). By , we have h0(U(k-2))=0 and h1(U(k-2))=k-1-bk-22(k-1)/3. Let EQ be a general union of f lines. Fix a general SQ such that (S)=a. Since h0(Q,Q(Y2S)E(k))=0, Castelnuovo's sequence gives h0(E2SY(k))=0 and hence δ0 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-20,1  (mod3)) or k-22  (mod3) and a(k-1)/3 (in the latter case we quote Lemma 9 to get h0(YS(k-2))=0.

(b1) Assume for the moment that either k0,2  (mod3), t-fak-2, or a(k-1)/3. Let SQ be a general set with (S)=a. Let EQ be a general union of f lines of type (0,1). Claim 4 implies tf+f. By Claim 6, we may assume that f2f. Since f2f, fk-2, t-f-f0, and t-f<ak-2, Lemma 6 shows the existence YL(3,t-f-f,f) such that h1(Y(k-2))=0, the nilradical of 𝒪Y is supported by points of E, each scheme YQ 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-f1 and 2(t-f)-f>0, we have h1(Q,2SQ(k,k-f))=0 . Hence hi(Q,(Y2S)QE(k))=0, i=0,1.

(b2) Now assume that k1  (mod3), 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(W(k-2))=0,  i=0,1. Take any YW with YL(3,t-f-f,f)Q. We have h1(Y(k-2))=0, because for each +line T and each integer t1, 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(Y(k-4))=0. Hence Lemma 9 gives that either h0(YS(k-2))=0 or h1(YS(k-2))=0. Since 3a+2(t-f)+(k+1)f-f=(k+1)2, we have fzkk-2. Hence we may apply [17, 18] or [19, Corollary 4.6].

(c) Assume that k=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 SS with (S)=u, (S)=v, 3u+v+2(t-f)=8(8-f), and 0v2. Outside Q, we have a general YZ(3,t-f,a-u-v).

(c1) Assume that 1t3. 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 au+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(Y(2))h0(Y(4))-u-2v. In all cases we have a-u-v2, and hence h1(Y(1))1. Hence h0(Y(4))-h0(Y(2))52-1.

(c2) Assume that t=4. Take f=3, and hence (u,v)=(12,2). Hence Y is a line, h1(Y2S(4))=0 and h0(Y(4))-h0(Y(2))=25u+2v.

(c3) Assume that 5t10. Take f=4. Since t-f>0 and YQ is a general union of 2(t-f) points, we have h1(Q,2SQ(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 uv, uv+δ, au+v, h1(Y2S(4))=0 [6, Proposition 2], and h0(Y(2))=0.

(c4) Assume that t=11. If δ=-3, then we take f=5 and S=S=. It is sufficient to use that h1(W(4))=0 for a general WZ(3,6,1) [6, Proposition 2].

(d) Assume that k=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 1t14. In all cases we take inside Q a general union of f lines of type (0,1) and SS 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(Y(3))·h1(Y(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(Y(3))1 and even h0(Y(3))3 would be enough.

(d1) Assume that 1t4. Take t=f. We have uv, au+v.

(d2) Assume that t=5. Take f=4, u=10, and v=0. Since v=0, we do not need to check the value of h0(Y(3)).

(d3) Assume that 6t14. Take f=5. Since t-f>0, we have h1(Q,2SQ(7,2))=0. Obviously au+v and uv, 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 that k=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 t17, 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 0g7, t>g (and hence in all cases we may apply [17, 18] or [19, Corollary 4.6]), au+v, and uv+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.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

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

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