CUBO A Mathematical Journal
Vol.20, No¯ 3, (31–36). October 2018

http: // dx. doi. org/ 10. 4067/ S0719-06462018000300031

Postulation of general unions of lines and +lines in positive
characteristic

E. Ballico
1

Department of Mathematics,

University of Trento,

38123 Povo (TN), Italy

ballico@science.unitn.it

ABSTRACT

A +line is a scheme R ⊂ Pr with a line as its reduction L = Rred which is the union of

L and a tangent vector v * L with vred ∈ L. Here we prove in arbitrary characteristic

that for r ≥ 4 a general union of lines and +lines has maximal rank. We use the case

r = 3 proved by myself in a previous paper and adapt the characteristic zero proof of

the case r > 3 given in the same paper.

RESUMEN

Una +ĺınea es un esquema R ⊂ Pr con una ĺınea como su reducción L = Rred que es la

unión de L y un vector tangente v * L, con vred ∈ L. Acá demostramos que para r ≥ 4

una unión general de ĺıneas y +ĺıneas tiene rango máximo en caracteŕıstica arbitraria.

Usamos el caso r = 3 demostrado por el autor en un art́ıculo anterior y adaptamos la

demostración en caracteŕıstica cero del caso r > 3 dado en el mismo art́ıculo anterior.

Keywords and Phrases: Hilbert function; decorated line; disjoint unions of lines.

2010 AMS Mathematics Subject Classification: 14N05.

1The author was partially supported by MIUR and GNSAGA of INdAM (Italy).

http://dx.doi.org/10.4067/S0719-06462018000300031


32 E. Ballico CUBO
20, 3 (2018)

1 Introduction

The aim of this note is to extend to the positive characteristic case a results in [1]. This extension

is sufficient to extend [2, 3] to the positive characteristic case.

A scheme X ⊂ Pr is said to have maximal rank if h0(IX(t)) · h
1(IX(t)) = 0 for all t ∈ N. Fix

a line L ⊂ Pr, r ≥ 2, and P ∈ L. A tangent vector of Pr with P as its support is a zero-dimensional

scheme Z ⊂ Pr such that deg(Z) = 2 and Zred = {P}. The tangent vector Z is uniquely determined

by P and the line 〈Z〉 spanned by Z. Conversely, for each line D ⊂ Pr with P ∈ D there is a

unique tangent vector v with vred = P and 〈v〉 = D. A +line M ⊂ P
r supported by L and with

nilradical supported by P is the union v ∪ L of L and a tangent vector v with P as its support

and spanning a line 〈v〉 6= L. The set of all +lines of Pr supported by L and with a nilradical

at P is an irreducible variety of dimension r − 1 (the complement of L in the (r − 1)-dimensional

projective space of all lines of Pr containing P). Hence the set of all +lines of Pr supported by L

is parametrized by an irreducible variety of dimension r. For any +line R and every integer k > 0

we have h0(OR(k)) = k + 2 and h
1(OR(k)) = 0.

For any integers r ≥ 3, t ≥ 0, c ≥ 0 with (t, c) 6= (0, 0) let L(r, t, c) be the set of all schemes

X ⊂ Pr which are the disjoint union of t lines and c +lines. Every element of L(r, c, t) has the map

k 7→ (k + 1)t + (k + 2)c as its Hilbert function.

Consider the following statement.

Theorem 1.1. For all integers r ≥ 3, a ≥ 0 and b ≥ 0, (a, b) 6= (0, 0), a general union X ⊂ Pr of

a lines and b +lines has maximal rank,

This statement was proved in [1] when either r = 3 or r ≥ 4 and the algebraically closed base

field has characteristic zero. The aim of this note is to prove Theorem 1.1 in positive characteristic

(using the case r = 3 proved in [1]). Hence we may assume r ≥ 4. We also use numerical lemmas

and elementary remarks contained in [1]. We only need to change all parts which quote [4, Lemma

1.4] or [6], the only characteristic zero tool used in [1]. We recall that the case c = 0 is due to R.

Hartshorne and A. Hirschowitz ([7]).

2 Proof of Theorem 1.1

For all integers r ≥ 3 and k ≥ 0 let Hr,k denote the following statement:

Assertion Hr,k, r ≥ 3, k ≥ 0: Fix (t, c) ∈ N
2 \ {(0, 0)} and take a general X ∈ L(r, t, c). If

(k+1)t+(k+2)c ≥
(

r+k
k

)

, then h0(IX(k)) = 0. If (k+1)t+(k+2)c ≤
(

r+k
k

)

, then h1(IX(k)) = 0.

For all integers r ≥ 3 and k ≥ 0 define the integers mr,k and nr,k by the relations

(k + 1)mr,k + nr,k =

(

r + k

r

)

, 0 ≤ nr,k ≤ k (2.1)



CUBO
20, 3 (2018)

Postulation of general unions of lines and +lines in positive . . . 33

From (2.1) for the pairs (r, k) and (r, k − 1) we get

mr,k−1 + (k + 1)(mr,k − mr,k−1) + nr,k − nr,k−1 =

(

r + k − 1

r − 1

)

(2.2)

for all k > 0.

For all integers r ≥ 3 and k ≥ 0 set ur,k := ⌈
(

r+k
r

)

/(k + 2)⌉ and vr,k := (k + 2)ur,k −
(

r+k
r

)

.

We have

(k + 2)(ur,k − vr,k) + (k + 1)vr,k =

(

r + k

r

)

, 0 ≤ vr,k ≤ k + 1 (2.3)

As in [1] we need the following assumption Br,k:

Assumption Br,k, r ≥ 4, k > 0. Fix a hyperplane H ⊂ P
r. There is X ∈ L(r, mr,k −

nr,k, nr,k) such that the support of the nilradical sheaf of X is contained in H and h
0(IX(k)) = 0.

For all X ∈ L(r, mr,k − nr,k, nr,k) we have h
0(OX(k)) =

(

r+k
r

)

and so h1(IX(k)) = h
0(IX(k)).

Lemma 2.1. We have mr,k − mr,k−1 ≥ nr,k−1 + nr,k if r ≥ 4 and k ≥ 2.

Proof. Assume mr,k − mr,k−1 ≤ nr,k−1 + nr,k − 1. From (2.1) we get

mr,k−1 + knr,k−1 + (k + 2)nr,k − k − 1 ≥

(

r + k − 1

r − 1

)

Since nr,k−1 ≤ k − 1 and nr,k ≤ k, we get mr,k−1 ≥
(

r+k−1
r−1

)

− 2k2 + 1. Since kmr,k−1 ≤
(

r+k−1
r

)

and k
(

r+k−1
r−1

)

−
(

r+k−1
r

)

= (r − 1)
(

r+k−1
r

)

, we get

2k3 − k ≥ (r − 1)

(

r + k − 1

r

)

(2.4)

This inequality is false if r = 4 and k ≥ 2, because it is equivalent to the inequality k(2k2 − 1) ≥

(k+3)(k+2)(k+1)k/8. Since the right hand side of (2.4) is an increasing function of r, we conclude

for all r ≥ 5 and k ≥ 2.

Lemma 2.2. Fix an integer r ≥ 4 and assume that Theorem 1.1 is true in Pr−1. Then Br,k is

true for all k > 0.

Proof. Since the case k = 1 is true ([1, Remark 3]), we may assume k ≥ 2 and use induction on

k. By Lemma 2.1 we have mr,k − mr,k−1 ≥ nr,k−1 + nr,k. Fix a solution X ∈ L(r, mr,k−1 −

nr,k−1, nr,k−1) of Br,k−1, say X = A⊔B with A ∈ L(r, mr,k−1 −nr,k−1, 0), B ∈ L(r, 0, nr,k−1) and

the tangent vectors of B have support S ⊂ H. By semicontinuity we may assume that no irreducible

component of Xred is contained in H, that no tangent vector associated to the nilradical of B is

contained in H and that S is a general subset of H with cardinality nr,k−1. Let C1 ⊂ H be a general

union of mr,k − mr,k−1 − nr,k − nr,k−1 lines. Let C2 ⊂ H be a general union of nr,k−1 lines, each

of them containing a different point of S. Let E ⊂ H be a general union of nr,k +lines. Since S is



34 E. Ballico CUBO
20, 3 (2018)

general, C1 ∪C2 ∪E is a general element of L(r−1, mr,k −mr,k−1 −nr,k, rn,k). Since Theorem 1.1 is

true in Pr−1, by (2.2) we get h1(H, IC1∪C2∪F(k)) = 0 and h
0(H, IC1∪C2∪E(k)) = mr,k−1 −nr,k−1.

Deforming A with B ∪ C1 ∪ C2 ∪ E fixed, we may assume A ∩ (B ∪ C1 ∪ C2 ∪ E) = ∅ and that

hi(H, IC1∪C2∪E∪(A∩H)(k)) = 0, i = 0, 1. Since A∩(B∪C1 ∪C2 ∪E) = ∅, Y := A∪B∪C1 ∪C2 ∪E is

a disjoint union of nr,k +lines with support in H (even contained in H), mr,k −2nr,k−1 −nr,k lines

and nr,k−1 sundials in the sense of [5]. Hence Y is a flat limit of a family of elements L(r, mr,k−1 −

nr,k−1, nr,k−1) whose nilpotent sheaf is contained in H ([7], [5]). By the semicontinuity theorem

to prove Br,k it is sufficient to prove that h
0(IY(k)) = 0. Since no tangent vector of B is contained

in H, then ResH(Y) = X and Y ∩ H = C1 ∪ C2 ∪ E ∪ (A ∩ H). Since h
0(IX(k − 1)) = 0 and

h0(H, IC1∪C2∪E∪(A∩H)(k)) = 0, a residual exact sequence gives h
0(IY(k)) = 0.

Lemma 2.3. Assume r ≥ 4 and that Theorem 1.1 is true in H = Pr−1. Fix an integer k ≥ 2

and assume that Hr,k−1 is true. Fix integers a ≥ 0, b ≥ 0, e ≥ 0 such that e ≤ 2⌊(k + 2)/2⌋,

(k + 2)a + (k + 1)b + 4⌊(k + 2)/2⌋ ≤
(

r+k−1
r−1

)

. Let X ⊂ H be a general union of a +lines, b lines

and e tangent vectors. Then h1(H, IX(k)) = 0.

Proof. It is sufficient to do the case e = ⌊(k + 2)/2⌋. Let A ⊂ H be a general union of a lines and

b 2-lines.

First assume that k is even. Let L1, L2 ⊂ H be general lines. Fix a general Si ⊂ Li with

♯(Si) = k/2 and a general Pi ∈ Li, i = 1, 2. Let vi ⊂ H be a general tangent vector of H

with Pi as its support; in particular we assume vi * Li. Let Ei ⊂ Li be the union of the k/2

tangent vectors of Li with (Ei)red = Si. Set Y := A ∪ E1 ∪ v1 ∪ E2 ∪ v2. Let Ri the +lines with

Li as their supports and with vi as the tangent vectors associated to their nilpotent sheaf. We

have h0(OA∪E1∪E2∪v1∪v2(k)) = h
0(OA∪R1∪R2(k)), h

1(OA∪E1∪E2∪v1∪v2(k)) = h
1(OA∪R1∪R2(k))

and h0(IA∪E1∪E2∪v1∪v2(k)) = h
0(IA∪R1∪R2(k)). Therefore we have h

1(IA∪E1∪E2∪v1∪v2(k)) =

h1(IA∪R1∪R2(k)). Since (k + 2)a + (k + 1)b + 2(k + 2) ≤
(

r+k−1
r−1

)

and Theorem 1.1 is true in Pr−1,

we have h1(IA∪R1∪R2(k)) = 0. Hence h
1(IA∪E1∪E2∪v1∪v2(k)) = 0. The semicontinuity theorem

gives h1(H, IX(k)) = 0.

Now assume that k is even. Let Fi ⊂ Li be any disjoint union of (k + 1)/2 tangent vec-

tors. We have h0(OA∪F1∪F2(k)) = h
0(OA∪L1∪L2(k)), h

1(OA∪F1∪F2(k)) = h
1(OA∪L1∪L2(k)) and

h0(IA∪F1∪F2(k)) = h
0(IA∪L1∪L2(k)). Therefore we obtain h

1(IA∪F1∪F2(k)) = h
1(IA∪L1∪L2(k)).

Since (k + 2)a + (k + 1)b + 2(k + 1) ≤
(

r+k−1
r−1

)

and Theorem 1.1 is true in Pr−1, we have

h1(IA∪L1∪L2(k)) = 0. Therefore h
1(IA∪F1∪F2(k)) = 0. The semicontinuity theorem gives

h1(H, IX(k)) = 0.

Proof of Theorem 1.1: By [1] we may assume r ≥ 4. By induction on r we may also assume

that Theorem 1.1 is true in Pr−1. By [1, Remark 3] it is sufficient to prove Hr,k for all integers

k ≥ 1. Hr,1 is true ([1, Lemma 3]). Hence we may assume k ≥ 2 and that Hr,k−1 is true.

By [1, Remark 4] it is sufficient to prove Hr,k for the pairs (t, c) such that either t = 0 and
(

r+k
r

)

− k − 1 ≤ c(k + 2) ≤
(

r+k
r

)

or t(k + 1) + (k + 2)c =
(

r+k
r

)

and c > 0; in the former case either



CUBO
20, 3 (2018)

Postulation of general unions of lines and +lines in positive . . . 35

vr,k = 0 and c = ur,k or vr,k > 0 and c = ur,k − 1; in the latter case we have t + c ≥ ur,k. If

c < nr,k−1, then we use step (b) of the proof of Theorem 1 in [1], because we gave a characteristic

free proof of Br,k (Lemma 2.2). The case c ≥ nr,k−1 and t ≥ mr,k−1 − nr,k−1 was proved as

step (a1) without using the characteristic zero assumption. Hence we may assume c ≥ nr,k−1 and

t < mr,k−1 − nr,k−1, i.e. the case of step (a2) of the proof in [1].

(i) Assume t = 0 and hence either vr,k = 0 and c = ur,k or vr,k > 0 and c = ur,k − 1. Fix a

general U ∈ L(r, 0, vr,k−1, ur,k−1 −vr,k−1), say U = A⊔B with A the union of the vr,k−1 lines. By

Hr,k−1 we have h
i(IU(k−1)) = 0, i = 0, 1. It is easy to check using (2.3) that ur,k > ur,k−1. Hence

c ≥ ur,k−1. Let E ⊂ H be a general union of c − ur,k−1 +lines. We may assume E ∩ (H ∩ U)) = ∅.

Let G ⊂ H be a general union of vr,k−1 tangent vectors of H with the only restriction that

Gred = A ∩ H. For general A (and hence a general A ∩ H)) the scheme E ∪ G is a general union

inside H of ur,k − uk−1 +lines and vr,k−1 tangent vectors. We have vr,k−1 ≤ k. Using (2.3) for

the integer k − 1 is easy to check that if vr,k−1 > 0, then ur,k−1 − vr,k−1 ≥ 2(k + 2) − 2vr,k−1.

Hence Lemma 2.3 gives h1(H, IE∪G(k)) = 0. Since B ∩ H is a general union of

(ii) Assume t > 0, c > 0, t(k+1)+(k+2)c =
(

r+k
r

)

and t < mr,k−1 −nr,k−1. First assume

t ≤ 2⌊(k + 2)/2⌋. In this case we may use the proof given in [1] (step (a2)) quoting Lemma 2.3

instead of [4, Lemma 1.4] for the postulation of the t tangent vectors, because mr,k−1 − t ≥ 2k + 2

in this case. Therefore we may assume t ≥ k + 1. Since t < mr,k−1 − nr,k−1, we have k ≥ 3

and kt <
(

r+k−1
r

)

. Set d := ⌊(
(

r+k−1
r

)

− kt)/(k + 1)⌋ and z := (k + 1)d + kt −
(

r+k
r

)

. We have

0 ≤ z ≤ k + 1. Fix a general W ∈ L(r, t, d). Since Hr,x holds for x = k − 1, k − 2, we have

h0(IW(k − 2)) = 0 and h
1(IW(k)) = 0 and h

0(IW(k)) = z. Since S is general in H and ♯(S) = z,

we get hi(IW∪S(k−1)) = 0, i = 0, 1. Since kt+(k+1)t+z =
(

r+k−1
r

)

and t(k+1)+(k+2)c =
(

r+k
r

)

,

we get

t + d + (k + 2)(c − d − z) + (k + 1)z =

(

r + k − 1

r − 1

)

(2.5)

Claim 1: We have c ≥ d + z.

Proof of Claim 1: Assume c ≤ d+z−1. From (2.5) we get t+d+(k+1)z−(k+1) ≥
(

r+k−1
r−1

)

and hence k(t + d) + (k + 1)kz − (k + 1)k ≥ k
(

r+k−1
r−1

)

. Since kt + (k + 1)d + z =
(

r+k−1
r

)

and

z ≤ k, we get (k + 1)k2 − k(k + 1) − k ≥ k
(

r+k−1
r−1

)

−
(

r+k−1
r

)

, i.e. k3 − 2k ≥ (r − 1)
(

r+k−1
r

)

. Call

φ(r, k) the difference between the right hand side and the left hand side of this inequality. We

have φ(r, k) = (r − 1)
(

r+k−1
r

)

− k3 + 2k, which is positive if r ≥ 4 and k ≥ 2.

Let M ⊂ H be a general union of c − d − z +lines of H. Let N ⊂ H be z general lines of H,

each of them containing a different point of Z. Since S is general, M ∪ N has the Hilbert function

of a general element of L(r − 1, z, c − d − z) and hence it has maximal rank. By (2.5) we have

h1(H, IM∪N(k)) = 0 and h
0(IM∪N(k)) = t + d. Let Z ⊂ P

r be a general union of z +lines of Pr

with N as their support. We have G ∩ H = N and ResH(Z) = S. Since W ∪ M ∪ Z ∈ L(r, t, c),

it is sufficient to prove that hi(IW∪M∪Z(k)) = 0, i = 0, 1. Since ResH(W ∪ M ∪ Z) = W ∪ S, we

have hi(IResH(W∪M∪Z)(k − 1)) = 0. Since W ∩ H is a general union of d + c points of H and

(W ∪ M ∪ Z) = (W ∩ H) ∪ M ∪ N as schemes, (2.5) gives hi(H, IH∩(W∪M∪Z(k)) = 0. Apply the



36 E. Ballico CUBO
20, 3 (2018)

Castelnuovo’s lemma.

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	Introduction
	Proof of Theorem 1.1