#### 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 [1–5]. 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, [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 [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