CUBO, A Mathematical Journal

Vol.22, N◦03, (379–393). December 2020
http://dx.doi.org/10.4067/S0719-06462020000300379

Received: 21 May, 2020 | Accepted: 24 November, 2020

Curves in low dimensional projective spaces
with the lowest ranks

Edoardo Ballico

Department of Mathematics,

University of Trento,

38123 Povo (TN), Italy.

ballico@science.unitn.it

ABSTRACT

Let X ⊂ Pr be an integral and non-degenerate curve. For each q ∈ Pr the X-rank

rX(q) of q is the minimal number of points of X spanning q. A general point of P
r has

X-rank ⌈(r + 1)/2⌉. For r = 3 (resp. r = 4) we construct many smooth curves such

that rX(q) ≤ 2 (resp. rX(q) ≤ 3) for all q ∈ P
r (the best possible upper bound). We

also construct nodal curves with the same properties and almost all geometric genera

allowed by Castelnuovo’s upper bound for the arithmetic genus.

RESUMEN

Sea X ⊂ Pr una curva integral y no-degenerada. Para cada q ∈ Pr el X-rango rX(q)

de q es el mı́nimo número de puntos de X que generan q. Un punto general de Pr

tiene X-rango ⌈(r + 1)/2⌉. Para r = 3 (resp. r = 4) construimos muchas curvas

suaves tales que rX(q) ≤ 2 (resp. rX(q) ≤ 3) para todo q ∈ P
r (la mejor cota superior

posible). También construimos curvas nodales con las mismas propiedades y casi todos

los géneros geométricos permitidos por la cota superior de Castelnuovo para el género

aritmético.

Keywords and Phrases: X-rank, projective curve, space curve, curve in projective spaces.

2020 AMS Mathematics Subject Classification: 14H51, 14N05.

c©2020 by the author. This open access article is licensed under a Creative
Commons Attribution-NonCommercial 4.0 International License.

http://dx.doi.org/10.4067/S0719-06462020000300379


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1 Introduction

Let X ⊂ Pr be an integral and non-degenerate variety defined over an algebraically closed field

with characteristic 0. For each q ∈ Pr the X-rank rX(q) of q is the minimal cardinality of a finite

set S ⊂ X such that q ∈ 〈S〉, where 〈 〉 denotes the linear span. An interesting problem is the

maximum of all integers rX(q), q ∈ P
r ([2, 8]). An obvious lower bound for this integer is the generic

X-rank rgen(X), i.e. the only integer such there is a non-empty Zariski open subset U ⊂ P
r such

that rX(q) = rgen(X) for all q ∈ U. For each positive integer t set W
0
t (X) := {q ∈ P

r | rX(q) = t}.

Let Wt(X) denote the closure of W
0
t (X) in P

r. If t ≤ rgen(X) the algebraic set Wt(X) is the

t-secant variety σt(X) of X. Hence if 1 ≤ t ≤ rgen(X) the algebraic set Wt(X) is non-empty,

irreducible and dim Wt(X) ≤ min{r, t(dim X + 1) − 1} with equality if dim X = 1 ([1, Remark

1.6]). Thus rgen(X) = ⌈(r + 1)/2⌉ if dim X = 1. For t > rgen(X) the geometry of Wt(X) is

described in [3, Theorem 3.1], assuming of course Wt(X) 6= ∅, i.e. W
0
t (X) 6= ∅.

We prove the following results.

Theorem 1.1. Fix integers b ≥ a > 0 such that a + b ≥ 5. Set d := a + b and γ := ab − a − b + 1.

Then there exists an integral and non-degenerate nodal curve X ⊂ P3 with geometric genus g,

deg(X) = d, exactly γ − g ordinary nodes and W 03 (X) = ∅.

Theorem 1.2. Fix integers a, b such that a ≥ 2 and b ≥ 2a + 3. Set d := a + b and γ :=

1 + ab− a(a+ 1)/2 − b. Fix an integer g such that 0 ≤ g ≤ γ. Then there is an integral nodal curve

X ⊂ P4 with degree d, geometric genus g, exactly γ − g ordinary nodes and with W 04 (X) = ∅.

Question 1.1. Is there an integral and non-degenerate curve X ⊂ P5 with W 04 (X) = ∅ ? Take an

odd integer r > 5. Is there an integral and non-degenerate curve X ⊂ Pr with W 0
(r+3)/2

(X) = ∅ ?

By [9, Theorem 1] W 03 (X) 6= ∅ for X as in Theorem 1.1, but with (a, b) ∈ {(1, 2), (1, 3), (2, 2)}.

The case (a, b) = (3, 3) of Theorem 1.1 is [9, Theorem 2]. When a ≤ b ≤ a + 1 the integer γ

appearing in Theorem 1.1 is the maximal arithmetic genus of all non-degenerate space curves ([6,

Ch. III]).

Many thanks are due to the referees for useful remarks.

2 Preliminaries

Notation 2.1. For any q ∈ Pr let ℓq : P
r \ {q} −→ Pr−1 denote the linear projection from q.

Let M be a projective scheme. Let D ⊂ M be an effective Cartier divisor of M. For any

zero-dimensional scheme Z ⊂ M the residual scheme ResD(Z) of Z with respect to D is the closed

subscheme of M with IZ : ID as its ideal sheaf. We have ResD(Z) ⊆ Z and hence ResD(Z) is a



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Curves in low dimensional projective spaces with the lowest ranks 381

zero-dimensional scheme. We have deg(Z) = deg(Z ∩ D) + deg(ResD(Z)) and for any line bundle

L on M we have an exact sequence of coherent sheaves on M:

0 −→ IResD(Z) ⊗ L(−D) −→ IZ ⊗ L −→ IZ∩D,D ⊗ L|D −→ 0 (2.1)

We will call (2.1) the residual exact sequence of D or the residual exact sequence of D in M.

Remark 2.1. Let M be a smooth, projective and rational surface. Thus h1(OM ) = 0. Assume

that ω∨M is ample. This will be true in the cases in which we apply this remark, i.e. the case in

which M is the smooth quadric surface and the case in which M is the Hirzebruch surface F1. Fix

an integer e ≥ 2, a very ample line bundle L on M and a nodal curve D = D1 ∪ · · · ∪ De ∈ |L|

with each Di a smooth and connected curve. Note that pa(D) =
∑e

i=1 pa(Di) + ♯(Sing(D)) + 1 − e.

Since L is very ample, D is connected. Since ω∨M is ample, we have Di · ωM < 0 (intersection

number) for all i. A subset A ⊆ Sing(D) is said to be a disconnecting set of nodes if D \ A is not

connected. Fix a set A ⊂ Sing(D) which is not disconnecting and set g := pa(D) − ♯(A). With the

terminology of [10] we will say that A is the set of assigned nodes, while the set Sing(D) \ A is the

set of unassigned nodes. By [10, Corollary 2.14] there are an affine smooth and connected curve

∆, o ∈ ∆, and a flat family {Yt}t∈∆ of elements of |L| such that Yo = D and Yt is integral, nodal

and with geometric genus g for all t ∈ ∆ \ {o}. Moreover, the sets {Sing(Yt)}t∈∆\{o} have A as

a limit. Thus pa(D) = ♯(Sing(D)) + 1 − e. We do not impose (or claim) that all Yt are singular

at the points of A, because it would require very strong restrictions on the integer ♯(A), only that

the nodes of the curves Yt near D are near A and that Yt has only ♯(A) nodes. The quoted result

[10, Corollary 2.14] with movable assigned nodes is optimal, as shown by following particular case,

the only one we will use. Assume that each Di is rational. In this case for each integer g with

0 ≤ g ≤ pa(D) there is a set of assigned nodes A ⊂ Sing(D) such that the corresponding family of

nodal curves has as a general member an integral nodal curve with geometric genus g.

Remark 2.2. Let X be a smooth projective curve, L a line bundle on X and V ⊆ H0(L) a

linear subspace. Set g := pa(X), d := deg(L) and n := dim V − 1. Assume n ≥ 1. For each

p ∈ X and each integer t > 0 set V (−tp) := V ∩ H0(Itp ⊗ L). We get n + 1 integers dim V (−tp),

1 ≤ t ≤ n + 1 ([5, pp. 264–277]). This is also done in details in [9]. The point p is said to be

an osculating point of the pair (L, V ) (or of the linear system PV ) if dim(V (−(n + 1)p)) > 0.

Since we are in characteristic zero, there are only finitely many osculating points of (L, V ), say

p1, . . . , ps, and at each point pi one can associate a positive integer w(pi) (the weight of pi), only

depending on the n + 1 integers dim V (−tp), 1 ≤ t ≤ n + 1. Moreover, there is an integer δ only

depending on g, d and n such that w(p1) + · · · + w(ps) = δ. We have w(pi) = 1 if and only if

dim V (−npi) = dim V (−(n + 1)pi) = 1. Suppose for instance that PV induces an embedding of X

into Pn and see X has a curve of Pn. Since V ⊆ H0(L), X is non-degenerate. The point p ∈ X is

an osculating point if and only if there is a hyperplane H ⊂ Pn such that the connected component

Z of the scheme H ∩ X with p has its reduction has degree ≥ n + 1, i.e. H contains the divisor



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(n + 1)p. The integer deg(Z) is the order of contact of the osculating hyperplane H with X at p.

The integer deg(Z) − n is a lower bound for the weight of p. All non-osculating points have weight

0.

3 Proof of Theorem 1.1

In this section we fix a smooth quadric surface Q ⊂ P3. For any irreducible curve Y ⊂ P3, Y not a

line, let τ(Y ) denote the tangential surface of Y , i.e. the closure in P3 of the union of all tangent

lines of Y at its smooth points. τ(Y ) is a plane if and only if 〈Y 〉 is a plane.

Notation 3.1. For any reduced curve X ⊂ P3 with no irreducible component contained in a plane

let T (X) be the set of all pairs (H, p), where H ⊂ P3 is a plane, p ∈ H ∩ X and the connected

component of the scheme H ∩ X with p as its reduction has degree at least 5.

Remark 3.1. Let ∆ a quasi-projective variety and X ⊂ P3 × ∆ a closed algebraic set such that

the restriction u : X −→ ∆ to X of the projection P3 × ∆ −→ ∆ is proper and flat. Assume that

all fibers of u are reduced curves with no irreducible component contained in a plane. Let T (X)

or T (u) denote the set of all triples (s, H, p), where s ∈ ∆ and (H, p) ∈ T (u−1(s)). The map

uT (X ) : T (X) −→ ∆ is proper. Thus if ∆ is irreducible and if T (u
−1(s0)) = ∅ for some s0 ∈ ∆,

then T (u−1(s)) = ∅ for a general s ∈ ∆.

Let X ⊂ P3 be an integral and non-degenerate curve. Fix p ∈ Xreg. We say that p is a flex

point of X or a flex of X or that the tangent line TpX is a flex tangent of X if the connected

component of the zero-dimensional scheme TpX ∩ X with p as its reduction has degree at least 3.

We say that p is a stall point of X or that TpX is a stall of X if TpX is not a flex tangent, but the

osculating plane Op(X) of X at p has order of contact at least 4 with X at p. Thus a stall point

is an osculating point which is not a flex point.

Remark 3.2. Fix a smooth element Y either of |OQ(1, 1)| or of |OQ(2, 1)| or of |OQ(1, 2)|. Since

Y is a rational normal curve in its linear span, it is easy to check that T (Y ) = ∅ and that each

q ∈ τ(Y ) \ Y is contained in at most 2 tangent lines of Y .

We collect in the next remark some standard tools and ideas which are used in the proofs of

Lemmas 3.1, 3.2 and 3.3 and which may be used in several other cases. In section 4 we will use

this set-up for the Hirzebruch surface F1 and the line bundle OF1(ah + bf).

Remark 3.3. Fix positive integers a, b and an integral quasi-projective family F of zero-dimensional

subschemes of the smooth quadric Q. Suppose you want to compute the dimension of the fam-

ily Ψ of all C ∈ |OQ(a, b)| containing at least one Z ∈ F or of the family Φ of all smooth

C ∈ |OQ(a, b)| containing at least one Z ∈ F. In most lemmas we will need to check that



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Curves in low dimensional projective spaces with the lowest ranks 383

dim Φ < dim |OQ(a, b)|, i.e. that a general C ∈ |OQ(a, b)| contains no Z ∈ F. Consider the

incidence variety I := {(Z, C) ∈ F × |OQ(a, b)| : Z ∈ C}. Let π1 : I −→ F and π2 : I −→ |OQ(a, b)|

denote the restriction to I of the projections of F ×|OQ(a, b)| onto its factors. Note that Ψ = π2(I).

The algebraic set I is a closed subset of F × |OQ(a, b)|. Thus by Chevalley’s theorem Ψ is a con-

structible set ([7, ex. II.3.18 and II.3.19]). If I is irreducible, then Ψ is irreducible. Obviously

Φ = ∅, unless at least some Z ∈ F is curvilinear. Call U the set of all smooth C ∈ |OQ(a, b)|.

Assume that at least some Z ∈ F is curvilinear and let G denote the set of all curvilinear Z ∈ F.

The set G is an open subset of F. Since F is assumed to be irreducible, G is irreducible. Set

J := I ∩ G × U. Usually, if we are only interested in smooth curves C ∈ |OQ(a, b)| it is better to

start with G, i.e. take an irreducible family of curvilinear schemes. Thus from now on we assume

F = G, but we use I, i.e. we also consider singular curves, to quote below [7, III.9.3, III.9.6,

III.9.7]. Suppose there is an integer z > 0 such that h0(Q, IZ(a, b)) = z for all Z ∈ G. With this

assumption all fibers of π1 are projective spaces of dimension z − 1. Hence π1 is a proper flat map.

Since G is assumed to be irreducible, I is irreducible and dim I = dim G + z ([7, III.9.3, III.9.6,

III.9.7]). Since J is a non-empty open subset of I, J is irreducible and dim J = dim I = dim G + z.

Thus Φ is irreducible and dim Φ ≤ dim G + z. If this inequality is not sufficient to conclude, one

should look at a general C ∈ Φ and try to compute dim(J∩π−12 (C)). Suppose dim(J∩π
−1
2 (C)) = x

for a general C ∈ |OQ(a, b)|. Then dim Φ = dim G + z − x. Since C is smooth and dim C = 1,

dim(J ∩ π−12 (C)) ≤ x if ♯(Zred) ≤ x for all Z ∈ J ∩ π
−1
2 (C). Moreover, dim(J ∩ π

−1
2 (C)) = x if

varying Z ∈ J ∩ π−12 (C) the sets Zred form an x-dimensional family of x distinct points of C. This

set-up is classically summarized by the words “ A dimensional count shows that Φ has dimension

dim G + z − x ”. If our family G is not irreducible, we try to study separately each of its irreducible

components. Now we drop the assumption that all integers h0(Q, IZ(a, b)) are the same. There

are a non-empty open subset G′ of G and an integer z such that h0(Q, IZ(a, b)) = z for all Z ∈ G
′.

Moreover, there are a positive integer s and integers zi ≥ z, 1 ≤ i ≤ s, such that G \ G
′ is the

union of finitely many irreducible quasi-projective varieties, say G \ G′ = G1 ∪ · · · ∪ Gs, such that

h0(Q, IZ(a, b)) = zi for all Z ∈ Gi. Then we use the irreducible families G
′, G1, . . . , Gs of curvilinear

schemes.

We will need only the case a = 1 of the next lemma, but its proof when a ≥ 2 requires no

modification.

Lemma 3.1. Fix integers a > 0, b > 0 such that a + b ≥ 4. Let D be a general element of

|OQ(a, b)|. Then D has no flex and T (D) = ∅.

Proof. We follow the classical approach outlined in Remark 3.3. The key step in the proof of the

lemma is the computation of the integer h0(Q, IZ(a, b)) for two types of zero-dimensional schemes

Z.



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With no loss of generality we may assume b ≥ a and hence b ≥ 2. By Bertini’s theorem D is

smooth. Since D ⊂ Q, Bezout theorem implies that each flex tangent line of D is contained in Q

and hence it is either an element of |OQ(1, 0)| or an element of |OQ(0, 1)|.

(a) Take L ∈ |OQ(1, 0)| and any connected zero-dimensional scheme F ⊂ L such that

deg(F) = 3. Since deg(OL(a, b)) = b ≥ 2, we have h
1(L, IF,L(a, b)) = 0. Since h

1(OQ(0, b)) = 0,

the residual exact sequence of L gives h1(IF (a, b)) = 0, i.e. h
0(IF (a, b)) = h

0(OQ(a, b)) − 3. Since

dim |OQ(1, 0)| = 1 and each L ∈ |OQ(1, 0)| contains ∞
1 connected degree 3 subschemes, a general

D ∈ |OQ(a, b)| contains no F (for any L), i.e. no L ∈ |OQ(1, 0)| is a flex tangent of D.

(b) If a ≥ 2 step (a) shows that no R ∈ |OQ(0, 1)| is a flex tangent of D. Now assume

a = 1. Since D ∈ |OQ(a, b)|, we have deg(R ∩ D) = 1 for all R ∈ |OQ(0, 1)|. Thus no element of

|OQ(0, 1)| is a flex tangent line of D.

By steps (a) and (b) D has no flex. Thus it is sufficient to prove that each osculating plane

of D has order of contact 4 with D at the osculating point. Fix a smooth element A ∈ |OQ(1, 1)|

and p ∈ A. Let E be the connected zero-dimensional subscheme of A such that Ered = {p} and

deg(E) = 5.

Claim 1: We have h1(Q, IE(a, b)) = 0.

Proof of Claim 1: We have h1(A, IE,A(a, b)) = 0, because A ∼= P
1 and deg(OA(1, b)) =

b + 1 ≥ 4. Since E ⊂ A, ResA(E) = ∅. Thus it is sufficient to use the residual exact sequence of A

in Q and that h1(OQ(0, b − 1)) = 0.

By Claim 1 we have h0(IE(a, b)) = (a + 1)(b + 1) − 5 for all E. Since dim |OQ(1, 1)| = 3 and

each smooth A ∈ |OQ(1, 1)| has ∞
1 points and hence ∞1 schemes E’s. Use Claim 1.

Notation 3.2. Let D ⊂ Q be a reduced curve with no irreducible component of D being an element

of |OQ(1, 0)| or |OQ(1, 0)| or |OQ(1, 1)|. Let F(D) be denote the set of all C ∈ |OQ(1, 1)| such that

the scheme C ∩ D contains at least two connected components, both of them of degree at least 4.

Lemma 3.2. Fix integers a > 0 and b > 0 such that a + b ≥ 3. Take a general D ∈ |OQ(a, b)|.

Then F(D) = ∅.

Proof. The curve D is smooth and for each line L ⊂ Q every connected component of the scheme

L ∩ D has connected components of degree 1 or 2, with at most one having degree 2. Thus

it is sufficient to test the smooth C ∈ |OQ(1, 1)|. Since deg(D ∩ C) = a + b, we may assume

a + b ≥ 8. Call G the set of all zero-dimensional schemes Z with 2 connected components, both

of degree 4 and with Z contained in some smooth C ∈ |OQ(1, 1)|. Each C contains ∞
2 elements

of G. Fix Z ∈ G and take C containing it. As in the proof of Lemma 3.1 it is sufficient to

observe that h1(IZ(a, b)) = 0, because C ∼= P
1 and deg(OC(a, b)) = a + b ≥ deg(Z) − 1 and hence



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h1(C, OC(a, b)(−Z)) = 0. Since ResC(Z) = ∅ and h
1(Q, OQ(a − 1, b − 1)) = 0, the residual exact

sequence of C gives h1(Q, IZ,Q(a, b)) = 0.

Lemma 3.3. Fix positive integers a, b, a′, b′ such that (a, b) 6= (1, 1) and (a′, b′) 6= (1, 1). Take a

general (D, D′) ∈ |OQ(a, b)| × |OQ(a
′, b′)|. Set Y := D ∪ D′. Then there is no C ∈ |OQ(1, 1)| such

that the scheme C ∩ Y has two connected components of degree at least 4.

Proof. By Bertini’s theorem D and D′ are smooth and Y is nodal. For a general pair (D, D′) for

each line L ⊂ Q the scheme L ∩ D has connected components of degree 1 or 2, with at most one

being of degree 2. Thus it is sufficient to test all smooth C ∈ |OQ(1, 1)|. Since (D, D
′) is general,

each C contains at most 2 points of D ∩ D′. Thus every smooth C ∈ |OQ(1, 1)| containing some

p ∈ D ∩ D′ satisfies the property that the connected component of C ∩ Y with p as its reduction

has degree ≤ 3. Thus we only need to consider the schemes C ∩ (Y \ D ∩ D′) with C smooth. By

Lemma 3.2 it is sufficient to exclude the smooth C such that C ∩ D has a connected component

Z1 of degree at least 4 and C ∩ (D
′ \ D ∩ D′) has a connected component Z2 of degree at least

4. We may assume a + b ≥ 4 and a′ + b′ ≥ 4. As in the proof of Lemma 3.1 we find only finitely

many smooth Ci ∈ |OQ(1, 1)|, say Ci, 1 ≤ i ≤ t, such that C ∩ D has a connected component of

degree at least 4. For a general D′, the curve D′ is transversal to all Ci, 1 ≤ i ≤ t.

Lemma 3.4. Fix positive integers e ≥ 2, ai, bi, 1 ≤ i ≤ e, such that for each i ∈ {1, . . . , e} exactly

one among ai and bi is 1. Let D = D1 ∪ · · · ∪ Ds ⊂ Q be a general union with each Di general in

|OQ(ai, bi)|. Then D is nodal, no two of the nodes of D are contained in the same line of Q, each

line of Q passing through a singular point of D is transversal to each Di, T (D) = ∅ and there is

no line J ⊂ Q such that J ∩ D has a connected component of degree at least 3.

Proof. D is nodal by Bertini’s theorem. Lemma 3.1 gives T (D) ⊆ Sing(D). Fix p ∈ Sing(D). Call

Di and Dj the irreducible components of D containing p. Since D is general, neither Di nor Dj

have a osculating plane at p with weight ≥ 2 and the tangent plane to one component, does not

contain the tangent line to the other component. Thus p /∈ T (D).

For a general (D1, . . . , De) no two of the nodes of D are on the same line of Q, because aibi 6= 0

for all i. We also see by induction on e that each line of Q passing through a singular point of D

is transversal to each Di.

Fix any line J ⊂ P3. Since D ⊂ Q, we have deg(D ∩ J) ≤ 2 if J * Q. Now assume

L ∈ |OQ(1, 0)| (resp. R ∈ |OQ(1, 0)|). We have deg(L ∩ D) = b (resp. deg(R ∩ D) = a). By

Lemma 3.1 each connected component of the zero-dimensional schemes L ∩ D and R ∩ D has

degree ≤ 2.

Lemma 3.5. Fix positive integers a, b and q ∈ P3 \Q. Then q /∈ τ(Y ) for a general Y ∈ |OQ(a, b)|.



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Proof. The polar surface of Q with respect to Q is a plane, H, intersecting transversally Q and

q ∈ TpQ if and only if p ∈ H ∩ Q. Take Y intersecting transversally H ∩ Q and not containing the

degree 2 subscheme of 〈{p, q}〉 with p as its reduction at all p ∈ H ∩ Q ∩ Y .

Lemma 3.6. Fix positive integers s ≥ 4, ai, bi, 1 ≤ i ≤ s. Take a general (D1, . . . , Ds) ∈
∏s

i=1 |OQ(ai, bi)|. Then for every q ∈ P
3 \ Q there is Sq ⊂ {1, . . . , s} such that ♯(Sq) ≤ 3 and

q /∈ τ(Di) for all i ∈ {1, . . . , s} \ Sq.

Proof. By Lemma 3.5 and the generality of (D1, . . . , Ds) we have dim((P
3 \ Q) ∩ τ(D1)) = 2,

dim((P3 \ Q) ∩ τ(D1) ∩ τ(D2)) ≤ 1, dim((P
3 \ Q) ∩ τ(D1) ∩ τ(D2) ∩ τ(D3)) ≤ 0 and (P

3 \ Q) ∩

τ(D1) ∩ τ(D2) ∩ τ(D3) ∩ τ(D4) = ∅. Using all subsets of {1, . . . , s} with cardinality 4 we get the

lemma.

Lemma 3.7. Fix positive integers a, b. Take a general Y ∈ |OQ(a, b)|. Then for every q ∈ P
3 \ Q

there are at most 3 points p ∈ Y such that q ∈ TpY .

Proof. With no loss of generality we may assume b ≥ a. Y is smooth. If a + b ≤ 3, then Y is a

rational normal curve in its linear span and the lemma is trivial in this case. Thus we may assume

a + b ≥ 4. The lemma is also easy to check using the linear projection ℓq and the genus formula

for plane curves if (a, b) ∈ {(2, 2), (1, 3), (2, 3)} (all these cases are discussed in [9]).

For any q ∈ P3 \ Q the polar plane Hq of Q with respect to q has the following properties.

The curve Cq := Hq ∩ Q is a smooth conic and q ∈ T pQ, p ∈ Q, if and only if p ∈ Cq. For

any p ∈ Cq let zp denote the degree 2 connected zero-dimensional subscheme of the line 〈{p, q}〉

with p as its reduction. For any curve E ⊂ Q such that p ∈ Ereg we have q ∈ TpE if and only

if zp ⊂ E. Let U denote the set of quadruples (Z1, Z2, Z3, Z4) with each Zi a connected degree 2

zero-dimensional subscheme of Q such that there is q ∈ P3 \ Q and (p1, p2, p3, p4) ∈ C
4
q such that

pi 6= pj for all i 6= j and Zi = zpi. The lemma is equivalent to proving that a general Y contains

no scheme Z1 ∪ Z2 ∪ Z3 ∪ Z4 with (Z1, Z2, Z3, Z4) ∈ U. For each smooth C ∈ |OQ(1, 1)| there is a

unique q ∈ P3 \ Q such that C = Cq. Each smooth C ∈ |OQ(1, 1)| has ∞
4 quadruples of distinct

points. Since dim |OQ(1, 1)| = 3, we get dim U = 7. Thus to prove the lemma it is sufficient to

prove that dim |IZ1∪Z2∪Z3∪Z4(a, b)| = dim |OQ(a, b)| − 8. Fix (Z1, Z2, Z3, Z4) ∈ U, say Zi = zpi

with p1, p2, p3, p4 distinct points of a smooth C ∈ |OQ(1, 1)|. Set Z := Z1 ∪ Z2 ∪ Z3 ∪ Z4. Since

deg(Z) = 8, it is sufficient to prove that h1(IZ(a, b)) = 0. We have C∩Z = {p1, p2, p3, p4} (scheme-

theoretically), because each tangent line of C is contained in the plane 〈C〉 and if C = Cq, then

q /∈ 〈C〉. Hence ResC(Z) = {p1, p2, p3, p4}. We have h
1(C, IZ∩C(a, b)) = 0, because C ∼= P

1 and

deg(OC(a, b)) = a + b. We have h
1(C, IResC(Z)(a − 1, b − 1)) = 0, because deg(OC(a − 1, b − 1)) =

a + b − 2 ≥ 3. We have h1(OQ(a − 2, b − 2)) = 0. Use twice the residual exact sequence of C, first

with IZ(a, b) as its middle term and then with IResC(Z)(a − 1, b − 1) as its middle term.



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Lemma 3.8. Fix positive integers a, b such that (a, b) 6= (1, 1). Take a general Y ∈ |OQ(a, b)|.

The set of all q ∈ P3 \ Q such that there are 2 (resp. 3) points p ∈ Y with q ∈ TpY has dimension

≤ 1 (resp. ≤ 0).

Proof. Adapt the proof of Lemma 3.7 using Z1 ∪ Z2 ∪ Z3 (resp. Z1 ∪ Z2) instead of Z1 ∪ Z2 ∪ Z3 ∪

Z4.

Lemma 3.9. Fix positive integers a1, b1, a2, b2. Take a general pair (D1, D2) ∈ |OQ(a1, b1)| ×

|OQ(a2, b2)|. For each q ∈ P
3 \ Q the following properties are true:.

(a) there is no (p1, p2, p3, p4) ∈ D1 × D1 × D2 × D2 such that p1 6= p2, p3 6= p4 and

q ∈ Tp1D1 ∩ Tp2D1 ∩ Tp3D2 ∩ Tp4D2;

(b) there is no (p1, p2, p3, p4) ∈ D1 × D1 × D1 × D2 such that ♯({p1, p2, p3}) = 3 and

q ∈ Tp1D1 ∩ Tp2D1 ∩ Tp3D1 ∩ Tp4D2.

Proof. Part (b) follows from Lemmas 3.5 and 3.8.

Now we prove part (a). This is trivial if (a2, b2) ∈ {(2, 1), (1, 2)}, i.e. if D2 is a rational normal

curve. Thus we may assume a2 + b2 ≥ 4. As in the proof of Lemma 3.7 let Hq be the polar

hyperplane of Q with respect to q and Cq := Hq ∩ Q. For any p ∈ Cq let zp denote the degree 2

connected zero-dimensional subscheme of the line 〈{p, q}〉 with p as its reduction. Let U denote the

set of all quadruples Z1, Z2, Z3, Z4 such that there is a smooth C ∈ |OQ(1, 1)| and 4 distinct points

pi ∈ C, 1 ≤ i ≤ 4, such that zpi = Zi for all i. For a fixed D1 Lemma 3.8 shows that we have at

most ∞1 pairs (p1, p2) which may be prolonged to be the reduction of some (Z1, Z2, Z3, Z4). For a

fixed p1, p2 we have h
0(Q, Ip1,p2(1, 1)) = 2 and hence there are only ∞

1 C ∈ |OQ(1, 1)| containing

{p1, p2}. For a fixed C we have ∞
2 pairs (p3, p4) ∈ C × C. We fix the general D1. To prove that a

general D2 satisfies part (a) of the lemma it is sufficient to prove that h
1(IZ1∪Z2∪Z3∪Z4(a2, b2)) ≤ 2.

We prove this inequality in the following way. Recall that C ∩ (Z∪Z2 ∪ Z3 ∪ Z4) = {p1, p2, p3, p4}

(scheme-theoretically), because each tangent line of C is contained in the plane 〈C〉 and if C =

Cq, then q /∈ 〈C〉. Thus ResC(Z1 ∪ Z2 ∪ Z3 ∪ Z4) = {p1, p2, p3, p4}. Since a2 + b2 ≥ 4, we

have h1(C, I{p1,p2,p3,p4}(a2, b2)) = 0 and h
1(C, I{p1,p2,p3,p4}(a2 − 1, b2 − 1)) ≤ 1. Use twice the

residual exact sequence of C, first with IZ1∪Z2∪Z3∪Z4(a2, b2) as its middle term and then with

I{p1,p2,p3,p4}(a2 − 1, b2 − 1) as its middle term.

Lemma 3.10. Fix positive integers a1, b1, a2, b2, a3, b3 such that (ai, bi) 6= (1, 1), 1 ≤ i ≤ 3. Take

a general (D1, D2, D3) ∈ |OQ(a1, b1)|× |OQ(a2, b2)|× |OQ(a3, b3)|. Take any q ∈ P
3 \ Q. There are

no (p1, p2, p3, p4) ∈ D1 × D1 × D2 × D3 such that p1 6= p2 and q ∈ Tp1D1 ∩Tp2D1 ∩Tp3D2 ∩Tp4D3.

Proof. The proof of part (a) of Lemma 3.9 shows that there are only finitely many triples (p1, p2, p3) ∈

D1 × D1 × D2 such that p1 6= p2 and Tp1D1 ∩ Tp2D1 ∩ Tp3D2 is a point of P
3 \ Q. Apply Lemma



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3.5 to D3.

Proof of Theorem 1.1: Any Y ∈ |OQ(a, b)| has arithmetic genus γ.

Claim 1: There are integers e ≥ 2, ai, bi, 1 ≤ i ≤ e, such that for each i ∈ {1, . . . , e} exactly

one among ai and bi is 1, a1 + · · · + ae = a and b1 + · · · + be = b.

Proof of Claim 1: If d ≡ 0 (mod 6) we take e = d/3, (ai, bi) = (1, 2) for odd i and

(ai, bi) = (2, 1) for even i. If d ≡ i (mod 6), 1 ≤ i ≤ 5, we take e = (d − i)/3, (a1, b1) = (1, 2 + i),

(ai, bi) = (1, 2) for odd i ≥ 3 and (ai, bi) = (2, 1) for even i.

Take a nodal curve D = D1 ∪ · · · ∪ De ⊂ Q satisfying the thesis of Lemma 3.4. Since each

Di is smooth and rational and pa(D) = γ, we have ♯(Sing(D)) = γ + e − 1. Since 0 ≤ g ≤ γ and

each Di is irreducible, there is a set A ⊂ Sing(D) such that ♯(A) = γ − g and D \ A is connected.

We fix one such set A and call it the set of all assigned nodes. The set Sing(D) is called the set

of all unassigned nodes (we are using the terminology of A. Tannenbaum ([10]) who extended to

other rational surfaces the classical theory of nodal plane curves due to Severi). Since D \ A is

connected, [10, Lemma 2.2 and Theorem 2.13] gives the existence of a flat family {Dt}t∈∆, ∆

an integral affine curve, and o ∈ ∆ such that Dt ∈ |OQ(a, b)| for all t ∈ ∆, Do = D, each Dt,

t ∈ ∆ \ {o}, is integral, nodal and with geometric genus g, and the nodes of Dt, t ∈ ∆ \ {o}, go to

the set of assigned nodes. By Remark 3.1 we have T (Dt) = ∅ for a general t ∈ ∆. Fix c ∈ ∆ \ {o}

such that T (Dc) = ∅ and set X := Dc. X is an integral and nodal curve with geometric genus g.

To conclude the proof of the theorem it is sufficient to prove that rX(q) = 2 for all q ∈ P
3 \ X.

(a) Fix q ∈ Q. Let L be the element of |OQ(1, 0)| containing q. We have deg(L ∩ X) = b.

By Lemma 3.1 each connected component of L ∩ X has degree ≤ 2. Thus ♯((L ∩ X)red) ≥ ⌈ba/2⌉.

Since b ≥ 3, we get rX(q) = 2.

(b) Fix q ∈ P3 \ Q. Assume rX(q) > 2, i.e. assume ℓq|X is injective. Since ℓq(X) has

degree d = a + b, it has arithmetic genus (a + b − 1)(a + b − 2)/2, while X has arithmetic genus

γ = ab − a − b + 1. We silently use a small modification of Remark 3.1 to get F(X) = ∅ (for

a general partial smoothing X) knowing that F(D) = ∅. We use Lemmas 3.1, 3.2, 3.3 to get

T (D) = ∅ and hence (Remark 3.1) we get T (X) = ∅.

(b1) Assume for the moment that q is not in the tangent space of one of the nodes of X.

Call oi, 1 ≤ i ≤ s, the points of Xreg such that q ∈ ToiX.

The following observation summarize lemmas 3.1, 3.2, 3.3, 3.4, 3.6, 3.7, 3.9, 3.10 first on D

and then on X.

Observation 1: X has no flex, its osculating planes have weight 1 and each point of P3 \ Q

is contained in at most 3 tangent lines to smooth points of X.

A dimensional count similar to the one needed to prove Lemmas 3.2 and 3.7 gives the following



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observation.

Observation 2: At each q ∈ P3 \ Q such that there are 3 different smooth points p1, p2, p3 of

Xreg with q ∈ TpiX, no Tpi(X) is a stall. At each point of X at which there are 2 different smooth

points p1, p2 of Xreg with q ∈ TpiX at most one among Tp1X and Tp2X is a stall.

By Observations 1 and 2 we have pa(ℓq(X)) ≤ pa(X) + 3. Since ℓq(X) is a plane curve of

degree a + b and pa(X) = ab − a − b + 1, we get (a + b − 1)(a + b − 2)/2 ≤ ab − a − b + 4, i.e.

a2 + b2 ≤ a + b + 6, which is false if a = 1 and b ≥ 4 or a ≥ 2 and b ≥ 3.

(b2) Assume g < γ and that q is contained in at least one tangent plane at X at one of its

points.

First assume that q is contained in the tangent cone at one of the nodes, o, of X. For a general

D (and hence a general partial smoothing X) no line in the tangent cones of X at its singular points

are stalls and tangent cones at different singular points are disjoints. At most another singular

point o′ of X has tangent plane containing q.

Now assume that q is not contained in any tangent cone at singular points. It is contained in

at most 3 tangent spaces of X at its singular points and if at 3 it is not contained in any tangent

line at a smooth point of X. We get a contradiction if (a + b − 1)(a + b − 2)/2 ≥ γ + 4, i.e. if

a2 + b2 ≥ a + b + 8, which is true (for positive a, b) if and only if a + b ≥ 5.

4 Curves in P4

Let F1 ⊂ P
4 be a smooth and non-degenerate surface such that deg(F1) = 3. All such surfaces are

projectively equivalent. The smooth or nodal curves we use to prove Theorem 1.2 are contained in

F1. The surface F1 is an embedding of the Hirzebruch surface with the same name ([7, §V.2]). We

have Pic(F1) ∼= Z
2 and we take as free generators of it the class f, of a fiber of the ruling of F1 and

the section h of its ruling with negative self-intersection. We have h2 = −1, f2 = 0 and h · f = 1.

We have OF1(1)
∼= OF1(h+2f) and h and the elements of the ruling |f| are the only lines contained

in F1. Each OF1(ah + bf), b ≥ a ≥ 0, is globally generated; it is ample (and very ample, too) if

and only if b > a > 0. Fix D ∈ |ah + bf|, b ≥ a > 0. Since ωF1
∼= OF1(−2h − 3f), the adjunction

formula gives ωD ∼= OD((a−2)h+(b−3)f). Thus pa(D) = 1+ab−a(a+1)/2−b. For all b ≥ a−1

we have h1(OF1(ah + bf)) = 0 and h
0(OF1(ah + bf)) =

∑a
i=0(b + 1 − i) = (2b + 2 − a)(a + 1)/2.

Remark 4.1. Take any curve D ⊂ F1 and any line L ⊂ P
4 such that deg(D ∩ L) ≥ 3. Since F1

is scheme-theoretically cut out by quadrics and D ⊂ F1, Bezout theorem gives L ⊂ F1.

Lemma 4.1. Fix an integer q ∈ P4 \ F1. Then there is C ∈ |h + f| such that q ∈ 〈C〉.

Proof. Since 3 is a prime integer and q /∈ F1, ℓq(F1) is an irreducible degree 3 ruled surface. This



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surface has a double line L meeting all lines of the ruling of ℓq(F1) ([4, Theorem 9.2.1]). Thus

there is a plane conic C ⊂ F1 (a priori even a double line) mapped by ℓq onto L. All conics C ⊂ F1

are elements of |h + f|.

Up to projective transformations there are exactly two degree 3 surfaces ℓq(F1), q ∈ P
4 \ F1,

distinguished by the fact that the unique conic C ∈ |h + f| given by Lemma 4.1 is smooth or not

([4, Theorem 9.2.1]).

Proposition 4.1. Let X ⊂ F1 ⊂ P
4 be a reduced and non-degenerate curve whose irreducible

component have degrees at least 3. Assume the following conditions:

(1) ♯((L ∩ X)red) ≥ 2 for all L ∈ |f|;

(2) ♯((h ∩ X)red) ≥ 2;

(3) ♯((C ∩ X)red) ≥ 3 for all smooth C ∈ |h + f|.

Then rX(q) ≤ 3 for all q ∈ P
4.

Proof. The assumptions on the irreducible components of X is equivalent to assuming that X ∩ C

contains no curve for all C ∈ |h + f|. First assume q ∈ F1. Let L be the only element of |f|

containing q. Since L is a line and ♯((L ∩ X)red) ≥ 2, we have rX(q) ≤ 2.

Now assume q /∈ F1. Take C ∈ |h + f| such that q ∈ 〈C〉. Note that 〈C〉 is a plane. If C is

smooth (and hence it is a smooth conic), we have rX(q) ≤ 3, because ♯((C ∩ X)red) ≥ 3 and hence

(C ∩X)red spans 〈C〉. Now assume that C is singular, i.e. C = h+L for some L ∈ |f|. Both h and

L are lines and h ∩ L is a single point. By assumption there are p1, p2 ∈ (L ∩ X)red with p1 6= p2

and hence L = 〈{p1, p2}〉. Since ♯((h ∩ X)red) ≥ 2, there is p3 ∈ (h ∩ X)red such that p3 6= h ∩ L.

Since h = 〈{p3, h ∩ L}〉, we have 〈C〉 = 〈{p1, p2, p3}〉 and hence rX(q) ≤ 3.

Lemma 4.2. Let ∆ a quasi-projective variety and X ⊂ F1 × ∆ a closed algebraic set such that

the restriction u : X −→ ∆ to X of the projection F1 × ∆ −→ ∆ is proper and flat. For each t ∈ ∆

set Xt := u
−1(t). Assume that all fibers of u are reduced curves with no irreducible component of

degree ≤ 2. Fix o ∈ ∆ and assume ♯((C ∩ Xo)red) ≥ 3 for all C ∈ |h + f|. Then for a general

t ∈ ∆ we have ♯((C ∩ Xt)red) ≥ 3 for all C ∈ |h + f|.

Proof. Assume that the lemma is false. Taking a neighborhood Ω of o in ∆ and then a branch

covering of Ω we may assume that for each t ∈ Ω\{o} there is Ct ∈ |h+f| with ♯((Ct ∩Xt)red) ≤ 2.

Since |h + f| is a projective set, the family {Ct}t∈Ω\{o} has at least one limit point, C
′, and

♯((C′ ∩ Xo)red) ≤ 2.



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Lemma 4.3. Fix integers a, b such that either a = 1 and b ≥ 5 or a ≥ 2 and b ≥ max{4, a}. Let

X be a general element of |ah + bf|. Then ♯((C ∩ X)red) ≥ 3 for all smooth C ∈ |h + f|.

Proof. For each C ∈ |h + f| we have deg(X ∩ C) = b. For e ∈ {1, 2} let U(e) denote the set

of all degree zero-dimensional schemes Z ⊂ F1 such that deg(Z) = b, Z has exactly e connected

components and there is a smooth C ∈ |h+f| containing Z. Since each smooth C ∈ |h+f| has ∞e

elements of U(e), we have dim U(e) = 2 + e. Thus (since e ≤ 2) to prove the lemma it is sufficient

to prove that dim |IZ(ah + bf)| = dim |ah + bf| − 5 for all Z ∈ U(e), i = 1, 2. Fix Z ∈ U(e)

and take a smooth C ∈ |h + f| containing it. Since deg(Z) = b ≥ 5, it is sufficient to prove that

h1(IZ(ah + bf)) = 0. Since h
1(OF1((a − 1)h + (b − 1)f)) = 0, the residual exact sequence of C

shows that it is sufficient to prove that h1(C, IZ,C(ah + bf)) = 0. This is true, because C ∼= P
1

and deg(OC(ah + bf)) = b.

Lemma 4.4. Fix q ∈ F1. There is a smooth C ∈ |h + f| such that q ∈ C if and only if q ∈ F1 \ h.

Proof. Since h · (h + f) = 0, no irreducible C ∈ |h + f| (i.e. no smooth C ∈ |h + f|) meets h. Now

assume q ∈ |h + f|. Since dim |Iq(h + f)| = dim |h + f| − 1 = 1 and there is a unique singular

element of |h + f| containing q, there is a smooth C ∈ |h + f| such that q ∈ C.

Proposition 4.2. Fix integer a, b such that a ≥ 1 and b ≥ 2a + 3.

(1) There is a nodal D ∈ |ah+bf| with exactly a smooth irreducible components, all of them

rational and neither lines nor conics, such that ♯((D ∩ C)red) ≥ 3 for all C ∈ |h + f|.

(2) If a ≥ 2 we have rD(q) ≤ 3 for all q ∈ P
4.

Proof. Set bi := 2 for 2 ≤ i ≤ a and b1 := b − 2a + 2. Take a general (D1, . . . , Da) ∈
∏a

i=1 |h + bif|

and set D := D1 ∪ · · · ∪ Da. By Bertini’s theorem each Di is smooth and connected and D is

nodal. Set S := Sing(D). Each Di is rational and pa(D) = 1 + ab − a(a + 1)/2 − b. Thus

♯(S) = pa(D) + a − 1 = ab − a(a − 1)/2 − b. In the case a = 1 we have D = D1 with D1 a general

element of |h + bf|. For a general (D1, . . . , Da) the nodal curve D is transversal to h and hence

♯((h ∩ D)red) = b − a ≥ 4. Hence part (1) is true for all singular C ∈ |h + f|.

Now we check part (1) for all smooth C ∈ |h + f|.

If a = 1 it is sufficient to quote Lemma 4.3.

Now assume a ≥ 2. Since ♯((D1 ∩ C)red) ≥ 3 by Lemma 4.3, we get part (1) for all smooth

C ∈ |h + f|.

Now we prove part (2). By Lemma 4.4 we have rD(q) ≤ 3 for all q ∈ F1\h. Since ♯((h∩D)red) =

b − a ≥ 2, we have rD(q) ≤ 2 for all q ∈ h.



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Take q ∈ P4 \ F1. Take C ∈ |h + f| such that q ∈ 〈C〉 (Lemma 4.1). If C is smooth we get

rD(q) ≤ 3 by Proposition 4.1. Now assume C singular, say C = h ∪ L with L ∈ |f|. Since D

contains b − a points of D, it is sufficient to prove that L contains a point of D \ D ∩ h. This is

true, because a ≥ 2 and D is transversal to h.

Proof of Theorem 1.2: Take the curve D given by Proposition 4.2. Use Remark 2.1 to get X as in

the proof of Theorem 1.1. Apply part (1) of Proposition 4.2 and Lemma 4.2.



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	Introduction
	Preliminaries
	Proof of Theorem 1.1
	Curves in P4