Table of Contents
ISRN Geometry
Volume 2014, Article ID 120850, 7 pages
http://dx.doi.org/10.1155/2014/120850
Research Article

Postulation of General Unions of Lines and Multiplicity Two Points in

Department of Mathematics, University of Trento, 38123 Povo, Italy

Received 3 January 2014; Accepted 11 February 2014; Published 23 March 2014

Academic Editors: D. Franco and A. Simis

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.

Abstract

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

1. Introduction

Fix . The 2-point of is the closed subscheme of with as its ideal sheaf. The scheme is a zero-dimensional scheme with and . Now assume that . For all , let be the set of all disjoint unions of lines and 2-points. Assume that ; that is, assume that . For each and each integer , we have . The algebraic set is an irreducible subset of the Hilbert scheme. We recall that a closed subscheme is said to have maximal rank if for each integer , the restriction map is a linear map with maximal rank; that is, it is either injective or surjective. has maximal rank if and only if for each integer either or . In this paper we prove the following result.

Theorem 1. Fix and an integer . Let be a general element of . Then either or , unless either or .

The case is one of the exceptional cases in the famous Alexander-Hirschowitz theorem [15]. See [6, Example 1] for the case . Theorem 1 is obviously false for , but it is false in a controlled way [6, Lemma 1], because a disjoint union of 3 lines of is contained in a unique quadric surface and for each , the linear system is the set of all quadric cones with vertex containing . Therefore as an immediate corollary of Theorem 1, we get the following result.

Corollary 2. Fix such that , , and . Let be a general element of . Then has maximal rank.

The case of Theorem 1 was proved in [6, Propositions ] and a weaker form of Theorem 1 was proved in [6, Proposition 5] (we required that ). This weaker version was enough to prove the statement corresponding to Theorem 1 first in and then in . Then we proved by induction on the corresponding statement in , , but only if [7].

For the proof, we use the case proved by Hartshorne and Hirschowitz [8] and the case , 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, [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 [9, 12, 13].

2. Preliminary Lemmas

Let be a smooth quadric surface. For each finite set , set . For each closed subscheme , the residual scheme of with respect to is the closed subscheme of with as its ideal sheaf. For each integer , we have the following exact sequence (often called Castelnuovo's sequence): From (1) we get and . Let be a zero-dimensional subscheme, and let be a union of distinct lines of type . The residual scheme of is the closed subscheme of with as its ideal sheaf. For each , we have Castelnuovo's exact sequence of coherent sheaves on : Hence , . We have if and only if for each connected component of .

Let denote the closure of in the Hilbert scheme of . Fix a line with and a line with and . Set . Set (it is the intersection of with the scheme ). As in [10], we call a +line with as its support and as the support of its nilradical.

For all , let be the set of all disjoint unions of lines and +lines. The algebraic set is irreducible and its general element has maximal rank [10, Theorem 1]. For all integers , let denote the set of all such that and the support of the nilradical of is contained in . The algebraic set is irreducible. For a general , we have ; that is, for each point in the support of the nilradical of , the tangent vector representing the nilradical of is not tangent to .

For all integers , set ,  , and . We have and if and , if . We have , , , , , , and .

For each integer , let and denote the following assertions.

Assertion : , , for a general .

Assertion : , , for a general .

Remark 3. Since general points of are contained in a smooth quadric and a general has maximal rank [10, Theorem 1], is true if , that is, if . For the same reason, is true if , that is, if .

Lemma 4. is true for all .

Proof. Fix a general . We have , , and is a general union of points. Let be a general union of lines of type . Fix with and for each line . Write with the union of the lines of containing a point of . Let be a disjoint union of +lines with , and let be the support of the nilradical of and with containing each tangent vector of a +lines of ; that is, we assume that and that . Set . Deforming to a general union of lines of , we get that is a flat limit of a family of elements of . By the semicontinuity theorem for cohomology [14, III.12.8], it is sufficient to prove that ,  . Since , it is sufficient to prove that , ; that is, . These equalities are true, because is the union of general points and general tangent vectors of and ; 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 integer , let denote the following assertion.

Assertion ,  : for every integer such that , we have for a general .

Remark 5. Fix an integer and assume the existence of such that . Since , we have . Fix an integer such that . Take any such that . Obviously, , , and . Therefore to prove , it is sufficient to prove its case “.”

Lemma 6. is true for all and is true for all .

Proof. By Remark 5 to prove , we may assume that and only check the case “” of .
(a) Assume that . Fix a general . We have , [8], and is a general union of points of . Fix such that and each line of contains at most one point of . Write with the union of the components of containing a point of . Let be a disjoint union of +lines with , and let be the support of the nilradical of and with containing each tangent vector of a +lines of ; that is, we assume that and that . Let be a general union of lines of type . Set . Deforming to a general union of lines of , we see that is in the closure of a set of elements of . Hence it is sufficient to prove that ; that is, . Since and , it is sufficient to prove that ; that is, . Since , we have . The scheme is a general union of points of and tangent vectors of . Hence it is sufficient to prove that for a general union of tangent vectors of ; this is very easy; in characteristic zero we may also quote two general results: [15] and [16, Lemma 1.4].
(b) Assume that . Take as in step (a). In this case we easily get ,  .
(c) By Remark 3 to prove for all , it is sufficient to prove for all . As in step (a), we get .
(d) Assume that . Take satisfying . For a general , we may assume that and that is a general union of points of . Write with union of the +lines. Fix a set such that and each line of contains at most one point of . Write with the union of the lines of containing a point of . Let be a disjoint union of +lines with as its support and as the support of its nilradical and with containing each tangent vector of a +lines of . Let be a general union of lines of type . Set . As in step (a), we get , , and is a flat limit of a family of elements of .

Assertion ,  . Set if , if ,   if , and if and . There are a union of different lines of type , , , and for each component of , , is the support of the nilpotent sheaf of , and .

Remark 7. Take a general with . is a contained in the union of disjoint lines and is contained in a smooth quadric with, say, of type . Since a general element of has maximal rank [10], to prove , it is sufficient to check the existence of for all . In particular is true if .

Lemma 8. is true for all .

Proof. By Remark 7, we may assume that . In steps (a), (b), and (c), we assume that , while in step (d), we handle the easier case .
(a) Assume that and . Take a general . We have ,  . Let be a general union of lines of type . Fix the union of general lines of type . As in [8] (or as in the proof of Lemma 4), we get ,  . For each line , we fix two of the lines of (and call them and ) with the condition that is the union of distinct lines. Set . Write with if is even and the only line of not containing a point of if is odd. Equivalently, take such that for every line and for every line . There are a smooth and connected affine curve , , a flat family of subschemes of with for all , , transversal to for all and such that for each and each exactly two of the lines of meet ; that is, we deform the lines of and to lines , transversal to , but meeting . Notice that for a general , the point of and the point of may be general points of . Hence we may find this flat family with the additional condition that for a general the set is a general union of points of . Fix a general and write with and either empty (case even) or a line (case odd). Let be a general union of lines of type . Fix a general union of lines of type and any union of components of . Write . We take so that , for every line and for every line ; to check that this is possible we need to check that (call this inequality). First assume that . We have and ; we have , and hence is true by our definition of the integer . Now assume that ; we have , , and hence ; hence is true in this case. Write with the union of all lines of containing no point of . Let be a general union of +lines with as its support and as the support of the nilradical and the nilradical contained in , so that and . Let be a general union of +lines contained in , supported by and with as the support of the nilradical. Set . It is sufficient to prove that . Since , it is sufficient to prove that . Since and , it is sufficient to prove that the union of and the degree two connected components of satisfies . Write . Since is general, for each line , and is general in for each line and , it is sufficient to prove that . In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only tangent vectors and .
(b) Assume that and . Fix a general . By , we have ,   (Lemma 4). Let be a general union of lines of type , with the restriction that of them contain one of the points of the support of the nilradical of . As in step (d) of the proof of Lemma 6, we get ,  . Let be a general union of lines of type . Fix such that for every line (and hence ) and for every line . There are a smooth and connected affine curve ,  , a flat family with for all , , transversal to for all and that for each and each exactly two of the lines of meet . As in step (a) for a general , the set is a general union of points of . Fix a general and write with and either empty (case even) or a line (case odd). Let be a general union of lines of type . Fix a general union of lines of type and any union or components of . Write with the union of all lines of not intersecting . We take so that , for every line , and for every line ; to check that this is possible we need to check that and this is true and easier than in step (a), because . Write with the union of all lines of containing no point of . Let be a general union of +lines with as its support and as the support of the nilradical and the nilradical contained in , so that and . Let be a general union of +lines contained in , supported by and with as the support of the nilradical. Set . It is sufficient to prove that . Since , it is sufficient to prove that . Since and , it is sufficient to prove that the union of and the degree two connected components of satisfies . Write . Since is general, for each line , and is general in for each line and , it is sufficient to prove that . In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only tangent vectors and .
(c) Assume that and . Fix a general . We have ,  . Let be a general union of lines of type . Let be a general union of lines. Fix such that for every line and for every line . There are a smooth and connected affine curve , , a flat family with for all , , transversal to for all and that for each and each exactly two of the lines of meet . As in step (a) for a general , the set is a general union of points of . In this flat family each , , is a disjoint union of lines, and only has sundials as some of its connected components. Fix a general and write with and either empty (case even) or a line (case odd). Fix such that . For each line of and each call , the line of such that is an algebraic family of lines. Let be the intersection with of the subcurve of corresponding to the lines with . The set is a general subset of with cardinality . Let be the union of and general +lines contained in , with as their support and with as the support of their nilradicals; for each +line of , we assume that the tangent vector of corresponding to the nilpotent sheaf of is not tangent to ; that is, we assume that the scheme has degree two and it is the disjoint union of two points; with these restrictions we have . Let be a general union of lines of type with the only restriction that of them contain a point of . Fix a general union of lines of type and any union or components of . Write with the union of all lines of not intersecting . We take so that , for every line and for every line ; to check that this is possible we need to check that and this is true and easier than in step (a), because . Write with the union of all lines of not containing a point of . Let be a general union of +lines with as its support and as the support of the nilradical and the nilradical contained in , so that and . Let be a general union of +lines contained in , supported by and with as the support of the nilradical. Set . It is sufficient to prove that . Since , it is sufficient to prove that . Since and , it is sufficient to prove that the union of and the degree two connected components of satisfies . Write . Since is general, for each line , and is general in for each line and , it is sufficient to prove that . In characteristic zero this is true by either [15] or [16, Lemma 1.4]; in positive characteristic one could see that there are only tangent vectors and .
(d) From now on we assume that .
(d1) Assume that . Fix a general . We have , [8]. Fix a general union of lines of type and a general union of lines of type . Write with . Fix such that , for all lines and for each line . Let be a general union of +lines with as their support and as the support of their nilradical. Set ; it works, because since , , and .
(d2) Assume that . Fix a general . We have ,  . Fix a general union of lines of type and a general union of lines of type . Write with . Fix such that , for all lines and for each line . Let be a general union of +lines with as their support and as the support of their nilradical. Set ; it works, because since , , and .
(d3) Assume that . We have . Fix a general . We have , by (Lemma 6). Fix a general union of lines of type and a general union of lines of type with the only restriction that of them contain a point of the support of the nilradical of . Write with . Fix such that , for all lines and for each line . Let be a general union of +lines with as their support and as the support of their nilradical. Set ; it works, because since , , and .

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

Lemma 9. Fix integer , , a closed scheme such that and a general set such that . If , then . If and , then .

3. Proof of Theorem 1

In this section we prove Theorem 1. We may assume that (by [8]) and (by the Alexander-Hirschowitz theorem). Set . By increasing or decreasing , we reduce to the case . Since any 2-point contains a point if and Theorem 1 is true for the triple with (i.e., if for a general , then it is true for the triple with ). Hence it is sufficient to check all triples with . In characteristic zero we may avoid all cases with quoting either [15] or [16, Lemma 1.4]. The cases are true by [6, Propositions ]. In steps (a) and (b), we assume that and that Theorem 1 is true in degree and . See steps (c), (d), and (e) for the cases .

(a) Assume that . If , then set . If , then let be the minimal positive integer such that .

Claim 1. We have .

Proof of Claim 1. Assume that . We get ; that is, . Since , we get , a contradiction.

Claim 2. We have .

Proof of Claim 2. Assume that . The minimality of the integer implies ; that is, , contradicting the inequality .

Set and . We have and . Since and , we get . Let be a general element of . Fix a general such that , , and . Let be a general union of lines of type . Since , we have . Since , we have [17, 18] and [19, Corollary 4.6] (the latter one only if , but the case is elementary, because ). Since is a general subset with cardinality and , we get ,  . By the Differential Horace Lemma for double points [1] and [5, Lemma 5] to prove that for a general union of and general 2-points (and hence to prove Theorem 1 in this case), it is sufficient to prove that . We have . Assume that for the moment . Since , we have . Since and , we have . Since and any two points of are contained in a smooth quadric surface, may be considered as a general element of . Since , the inductive assumption gives . Hence . By Lemma 9 to prove that , it is sufficient to prove that . Since , we have . Hence it is sufficient to prove that . The inductive assumption in degree shows that it is sufficient to prove that . We have . Since and , it is sufficient to check that . Since and , it is sufficient to prove that . Assume that . Hence . Since , we have ; that is, . Hence , contradicting the inequality for all .

Now assume that . If , then we just saw that . Hence we may assume and . Take a general and set . We have . Hence it is sufficient to prove that . Since we checked that (easier for ) and that (i.e., ), Lemma 9 gives ,  .

(b) In this step we assume that . Define the integers and by the relations Since , we have .

Claim 3. We have for all and for all .

Proof of Claim 3. Assume that . We get ; that is, . For all , we get , a contradiction. Now assume that . We get . For all , we get , a contradiction.

Claim 4. We have for all .

Proof of Claim 4. Assume that . Since by Claim 3, we get . Hence , a contradiction for all .

Claim 5. We have for all .

Proof of Claim 5. Assume that . We get ; that is, . Since and , we get a contradiction.

Claim 6. Either or , and , or Theorem 1 holds for the pair .

Proof of Claim 6. First assume that . Fix a general . By [8], we have and . Let be a general union of lines. Fix a general such that . Since , Castelnuovo's sequence gives and hence and Theorem 1 is true for the pairs . The same proof works if and either (i.e., ) or and (in the latter case we quote Lemma 9 to get .

(b1) Assume for the moment that either , , or . Let be a general set with . Let be a general union of lines of type . Claim 4 implies . By Claim 6, we may assume that . Since , , , and , Lemma 6 shows the existence such that , the nilradical of is supported by points of , each scheme is reduced, and it is the union of points, on , and the remaining ones forming a general subset of with cardinality . Since and , we have [17]. Hence , .

(b2) Now assume that , , and . We modify step (b1) in the following way. Let be a general element of . Since is true (Lemma 4), we have ,  . Take any with . We have , because for each +line and each integer , the restriction map is surjective. Since , we have . Hence Lemma 9 gives that either or . Since , we have . Hence we may apply [17, 18] or [19, Corollary 4.6].

(c) Assume that . Since , it is sufficient to consider the following triples : , , , , , , , , , , , , , , , , , , , and . In all cases we take inside a general union of lines of type and with , , , and . Outside , we have a general .

(c1) Assume that . Take . The triples are the following ones: , , and . Hence we need to modify the proof in the case . In that case we take with ; that is, take . With this modification, we get in all cases. To use the Alexander-Hirschowitz theorem in degree 4, we need to check that . We need to check that . In all cases we have , and hence . Hence .

(c2) Assume that . Take , and hence . Hence is a line, and .

(c3) Assume that . Take . Since and is a general union of points, we have . We have and . Hence the triples are the following ones: , , , , , and . We have , , , [6, Proposition 2], and .

(c4) Assume that . If , then we take and . It is sufficient to use that for a general [6, Proposition 2].

(d) Assume that . Since and is divided both by and by , it is sufficient to check all cases with , , , that is, all pairs with and . In all cases we take inside a general union of lines of type and with , , , and . To apply [6, Proposition 1] and get , it is sufficient to have [6, Proposition 1]; in the exceptional case we would have and even would be enough.

(d1) Assume that . Take . We have , .

(d2) Assume that . Take , , and . Since , we do not need to check the value of .

(d3) Assume that . Take . Since , we have . Obviously and , but in a few cases one of these inequality is an equality (if