ISRN Geometry

Volume 2013 (2013), Article ID 391721, 6 pages

http://dx.doi.org/10.1155/2013/391721

## Postulation of a Union , , of a Given Zero-Dimensional Scheme and Several General Lines

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

Received 10 October 2013; Accepted 5 November 2013

Academic Editors: E. Previato and F. Shi

Copyright © 2013 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 study the Hilbert function of the union of a given zero-dimensional scheme , , and general lines of , proving, for example, that if , then has maximal rank.

#### 1. Introduction

A schemeis said to have *maximal rank* if for all integers , the restriction mapis either injective or surjective, that is, if eitheror, that is, ifimposes the “expected” number of conditions to the vector space of the homogeneous degreepolynomials invariables. Hartshorne and Hirschowitz proved that for alland, a general unionofgeneral lines has maximal rank [1]. Carlini et al. considered several cases in which we allow unions of linear spaces with certain multiplicities (for zero-dimensional subspaces these are general unions of fat points) [2–4]. Letdenote the set of all unionsofdisjoint lines. In this paper (following [4]) we consider the case in whichis a disjoint union of a given zero-dimensional schemeand a general.

Fix a zero-dimensional scheme,, and integersand. Set. In this note we study the Hilbert function of a general union ofanddisjoint lines. We assume almost nothing on. We only prescribe the integerand an integer. Take a general. Sinceis fixed, whileis general, we have, and hencefor all. To study the Hilbert function of a general union ofanddisjoint lines, we need the following integersand,, defined by the following relations: The schemehas maximal rank if and only iffor all integerswithandfor the minimal integersuch that eitherorand.

We first prove the following result, which says that if we may handle general unions ofand general lines with respect to homogeneous polynomials in two consecutive degrees (not small with respect to the integer), then these information propagates to higher degree polynomials.

Theorem 1. *Fix a zero-dimensional scheme,, and an integer. Set. Assume. If, assume. Letbe any nonnegative integer such thatfor a general. Letbe any nonnegative integer such thatfor a general. Set.**Fix an integer, and letbe a general element of.*(a)*For all integers, eitheror.*(b)*Assume. Then for all integers, eitheror.*

Then we go to a case in which we only assume something about the integerand prove the following result.

Theorem 2. *Fix positive integers , , and such thatand a zero-dimensional schemesuch thatand. Set. If eitherorand, then set. Ifand, then set. Fix any integer. Letbe a general union ofandlines. Thenhas maximal rank.*

If, then we have the following result.

Proposition 3. *Fix integersand. Letbe a zero-dimensional scheme such that and. Then for each integer, a general union ofandlines has maximal rank.*

In the case, we get the caseof [4]. We omit the proof of Proposition 3, because the proof of [4] works verbatim (e.g., quote [4, Lemma] instead of Lemmas 10, 12, and 13).

#### 2. The Proofs

For all integers,,,, and, define the integersandby the relations

Notice thatand. For any,, the *critical value* of the triple(resp.,) is the minimal integer(resp.,) such that, that is, the minimal integer(resp.,) such that.

Taking the equation in (2) minus the same equation for the integer, we get

Take , , , , and as in Theorem 1. For each integerconsider the following assertion::we havefor a general.

Take anysuch that. Since, we have. Henceis true if and only iffor a general.

Call, the same assertion withinstead of.

*Remark 4. *By the definition of the integersand, the assertions,, andare true.

For any hyperplane, and any closed subscheme, letbe the closed subscheme ofwithas its ideal sheaf. Ifwithand is a reduced scheme, then, whereis the union of the irreducible components ofnot contained in.

Lemma 5. *Fix a hyperplane,, an integerand a closed subscheme. If, thenfor a general.*

*Proof. *The assumption implies thatis not contained in the base scheme of. The long cohomology exact sequence associated to the following exact sequence
gives the following inequalities: (1);(2).

We haveand, because a generalis not in the base locus of. Apply (4) first toand then to.

As in [2–4], we will call *Casteluonovo’s sequence* for any of the inequalities in the proof of Lemma 5.

Lemma 6. *Fix integers,,, andsuch that. Letbe a general union oflines andreducible conics. Then.*

*Proof. *For each, let be the connected component ofcontaining. Write. For each, letbe a general-dimensional linear space containing. Letbe the sundial withas its support. Set. Sinceand eachare general,is a general union oflines andsundials. Hencehas maximal rank [3]. Since, we get. Fix. Letbe the nilpotent sheaf of. We have an exact sequence
Sinceis supported by a single point,, we have. Hence (5) gives the surjectivity of the restriction map. Looking at all the connected components ofand, we get the surjectivity of the restriction map. Hence.

Lemma 7. *is true for all integers.*

*Proof. *By Remark 4 we may assume. We use induction on. Sinceis true, we may assume thatis true. Fix a hyperplanesuch that. Fix a general. We have,is a general union ofpoints of, and.

(i) First assume. Letbe a general union oflines; this is possible, because(Lemma 10). By (3), we have. Sincehas maximal rank in[1] andis general in, we have. Castelnuovo’s sequence gives, provingin this case.

(ii) Now assume. In this case we have(Lemma 13). Letbe a general union ofsundials ofandlines. By [2],has maximal rank in. Lemma 10 for the integergives. Hence. Hence (3) gives. Setand. Letbe the union of the degree one connected components of. Letbe a general union of disjoint lines and sundials withhas its reduction. Sinceandis a degeneration of a family of disjoint unions oflines, the semicontinuity theorem for cohomology ([5, Theorem III.12.8]) implies that it is sufficient to prove that. We have. The setis a general union ofpoints of. Ifwere general points of, then obviously. To applytimes Lemma 5 and get the inequality, we need to check the restriction on the residual scheme in its statement. Assume that this lemma cannot be applied and thatfails in this case; that is,. We would get. If, we get for a general(contrary to the definition of), becauseby Lemma 10. Now assume, and leta union oflines. By, we have. Any two points ofare contained in a line. Hence. Lemma 10 gives. Hence = .

We have. Letbe the nilpotent sheaf of. Sinceis supported by the finite set, we havefor each line bundleon. The exact sequence
gives the surjectivity of the restriction map . Hence . Hence + . Sinceis formed by general points of, (3) gives . Castelnuovo’s sequence gives , concluding the proof of.

Lemma 8. *Assume. Thenis true for all.*

*Proof. * is true (Remark 4). Now assume. To copy step (i) of the proof of Lemma 7, it is sufficient to have. Use Lemma 12. The caseis done as in the proof of Lemma 7.

Lemma 9. *Fixas in Theorem 1 and integersand. Letbe a general element of. Then.*

*Proof. *It is sufficient to do the case. Fix a hyperplane. First assume. Take a general. We have.gives. Letbe a general union oflines ofandsundials whose support in contained in. We conclude as in step (ii) of the proof of Lemma 7.

*Proof of Theorem 1. *First assume. Sinceandfor a generaland, we may assume. In this range the assumption is satisfied if and only if eitherorand, becauseis a strictly increasing function of(Lemma 10). Henceeither by Lemma 9 (case) or by(case).

Now assume (we allow the equality). By the semicontinuity theorem for cohomology ([5, Theorem III.12.8]), it is sufficient to findsuch thatand,. By (2) and the assumption, we haveand. Fix a general. Letbe the union ofof the lines of. By, we have. Sinceis a union of some of the connected components of, we have.

Now we check part (b). Since, Lemma 8 givesfor all. Then we get Lemma 9 withinstead of. Then we continue as in the proof of part (a).

*Proof of Theorem 2. *Fix a hyperplanesuch that. If eitherorandfor each integer, set. Ifand, set. Hence . Ifand, we havefor all. Consider the following statement ,::we havefor a general.

Take anysuch that. We have. Henceis true if and only iffor a general.(a)In this step we proveand.is true, because we assumed that. To check, we first notice that, and hence we may apply (3) for the integer. Since, we have. Since, Lemma 15 gives. Hence the construction in step (i) of the proof of Lemma 7 works verbatim.(b)In this step we provefor all. Since the casesare true by step (a), we may assume thatand thatis true. Taking the difference of (2) withandand integers, we get
We apply (7) withand. Hence either(case) or. In both cases, we have. If, then we may apply the numerical Lemmas 10, 12, and 13 used in the proof of Theorem 1. We need different numerical lemmas if, that is, ifand. Fix a general. We have,,, andis a general union ofpoints of. Since,, we have for a general(and even a generalby Lemma 5 if. This is always the case if, that is, if.(b.1)First assume. Since, Lemma 15 gives. Letbe a general union oflines. Take a general reducible conic. Letbe a general sundial of, and letbe a general sundial ofwithas its support. Set. Sinceis a specialization of a union ofandlines, it is sufficient to prove that . Castelnuovo’s sequence gives . Lemma 14 gives . By (7) to apply Lemma 5, we need that has as least one point; this is the reason why this proof does not work if .(b.2)Now assume. Since, Lemma 16 gives. We repeat step (ii) of the proof of Lemma 7 taking insideinstead ofa general unionoflines,reducible conics. Then we take the general sundials (inand in) with these conics as their supports. To apply Lemma 6, we need. This inequality is true by Lemma 15.(c)Fix an integer . Let be the critical value for the triple , i.e. the minimal positive integer such that . Since , either or and . To prove Theorem 2 for the integer it is sufficient to prove and for a general . Since is irreducible, the semicontinuity theorem for cohomology says that to prove Theorem 2 for the integer it is sufficient to prove the existence of and such that , and . Since , we have . Hence Since the existence of is proved as.

#### 3. Numerical Lemmas

Lemma 10. *Assume. Fix an integer. One hasfor all.*

*Proof. *Assume. From (3), we get
From (2) for the integer, we get
Sinceand, from (11) and (12), we get
Set. Sinceand, to get a contradiction, it is sufficient to prove that. First assume. We have. Obviouslyif. We have,, and .

We have. By induction on, we get the lemma for all.

Lemma 11. *Fix integers,,, and. Thenfor all.*

*Proof. *Assume. From (3), we get
From (2) for the integer, we get
Since,,,, and, from (11) and (12), we get
Set. Since, to get a contradiction it is sufficient to prove that. First assume. We have. Hencefor all. Now assume. We have. By induction on, we get the lemma.

Lemma 12. *Assume. Fix integerssuch thatand. One hasfor all.*

*Proof. *By Lemma 10, it would be sufficient to do the case. Assume. From (3) and (12), we get
which is obviously false.

Lemma 13. *Assume,, and. If, assume. Then.*

*Proof. *Assume. From (3), we get
Since,, and, from (15), we get
Since, we get
Set. Since, to get a contradiction and hence to prove the lemma, it is sufficient to prove that. First assume. We have . Hence if and only if.

Now assume. We havefor all. We have, andis an increasing function of. Henceif eitheror.

Lemma 14. *Assume. Fix integers , , , , and such that, , , and. One has.*

*Proof. *Assume. By (7), we get
We have. Since, we have. Hence
Since,, and, we get a contradiction.

Lemma 15. *Take the setup of Theorem 2. For each integerwith, one has.*

*Proof. *By Lemma 14, it is sufficient to do the case. In this case, we have. Assume. Since, we get. The right hand side of this inequality is an increasing function of. For, this inequality fails for all, becausefor all.

Lemma 16. *Take the setup of Theorem 2. Assume,,, and. Then.*

*Proof. *Assume. Since, we have.

From (7) with, we get
Since, we get
This inequality is false if eitherandorand.

#### Acknowledgment

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

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