平面曲線的線性系統及曲線上的有理點(1/2)

行政院國家科學委員會專題研究計畫 期中進度報告

平面曲線的線性系統及曲線上的有理點(1/2)

中 華 民 國 93 年 5 月 28 日

P.-Y. HUANG AND K.S. TAN

Abstract. We study linear systems of plan curves.

1. Introduction

In the present note, we study the regularity (see Definition 3.1) of linear systems of plane curves passing through given base points with prescribed multiplicities. Research on this subject could be dated back to the previous two centuries ([Cas91, Sev26, Seg47, Seg61, Na601, Na602]) and could be seen flourishing in recent years (for example, see [AbC81, AlH00, Bru97, Cat89, ClM98, Eva98, Gim89, Hbn89, Hsz89, Mig01]).

Our approach is elementary and different form most of the previous works which take use of the Hilbert function of the corresponding zero dimensional sub-scheme of P2_{. Instead, we analyze how a generic curve factorizes into irreducible}

com-ponents. From this, we are able to prove the equivalence between a conjecture of Segre ([Seg61], or see Conjecture 3.2) and that of Harbourne and Hirschowitz ([Hbn89, Hsz89], or see Conjecture 3.1). Also, we show that these conjectures imply a more detailed conjecture of Gimigliano ([Gim87]). It has been announced ([Cil04]) that same kind of result have been proved through a different method by Ciliberto, Clemens and Miranda.

Assuming the above mentioned conjectures, we show that if a generic curve in a linear system is irreducible of genus g ≥ 1, then the dimension of the system is at most 3g + 6.

Finally, we consider the case where all the base points are located on a smooth cubic curve and deduce certain conditions for the regularity. In particular, this gives

This work was supported in part by the National Science Council of Taiwan, NSC92-2115-M-002-024.

a new proof for the above conjectures in the case where the number of base points are at most 9.

2. Factorizations

Let K be an algebraically closed field. Over the projective plane P2_{/K, consider a}

0-cycle a = m1P1+ m2P2+ ... + mnPn where P1, ..., Pnare distinct K-rational points

on P2 _{and m}

1, ..., mn are non-negative integers. For a positive integer d, We denote

by V(d; a) the linear system of plane curves of degree d with multiplicity at least mi on each Pi. Thus, a curve ξ in V(d; a) is determined by a degree d homogeneous

equation

Fξ(X, Y, Z) =

X

ξi,j,kXiYjZk= 0,

with the coefficients ξi,j,k satisfying a system of

f (d; a) := (d + 1)(d + 2)/2 − 1 − n X i=1 mi(mi+ 1)/2 = d(d + 3)/2 − n X i=1 mi(mi+ 1)/2

linear equations. Here we view Fξ(X, Y, Z) = 0 and c · Fξ(X, Y, Z) = 0, c ∈ K∗

as the same equation and the assignment ξ 7→ [ξ] := [ξi,j,k]i,j,k gives an embedding

from V(d; a) onto a linear subspace of Pd(d+1)/2 _{= V(d; 0).}

In this note the phrase ”linear system” will be used only to indicate linear systems of the form V(d; a).

Suppose that d = d1+ d2 and a = a1+ a2. Then there is a product morphism

V(d1; a1) × V(d2; a2) −→ V(d; a),

which sends ([ξ0_{], [ξ}00_{]) to [ξ}0_{· ξ}00_{] with}

Fξ0_·ξ00(X, Y, Z) = F_ξ0(X, Y, Z) · F_ξ00(X, Y, Z).

The theory of polynomial factorization says that this morphism is quasi-finite. We use V(d1; a1) · V(d2; a2) to denote its image and have

Lemma 2.1. If

V(d; a) = [

b+c=d

V(b; 0) · V(c; 0),

then d = d1+ d2 and a= a1+ a2 for some d1, d2, a1, a2, such that

V(d; a) = V(d1; a1) · V(d2; a2).

Proof. Suppose that ξ = ξ0 _{· ξ}00 _{∈ V(d; a) with ξ}0 _{∈ V(d}

1; 0) and ξ00 ∈ V(d2; 0). Let

m0

i and m00i be respectively the multiplicities of ξ0 and ξ00at Pi. Then m0i+ m00i ≥ mi.

This implies

V(d; a) =[V(d1; a1) · V(d2; a2),

where the union runs through all the possible d1, d2, a1, a2 with d = d1+ d2 and

a= a1+ a2. The lemma is proved by the observation that V(d; a) is irreducible and

every V(d1; a1) · V(d2; a2) is a closed set. 

Definition 2.1. If for some d1, d2, a1, a2,

V(d; a) = V(d1; a1) · V(d2; a2),

then the set V(d; a) is decomposable and V(di; ai), i = 1, 2, are its factors.

Other-wise, it is indecomposable.

A linear system V(d; a) is decomposable if and only if every curve in it is reducible, and in an indecomposable linear system irreducible curves form a dense open set. It is easy to see that the operation ”·” is commutative and associative. Using the theory of polynomial factorization, we can prove the following.

Lemma 2.2. We have a decomposition

V(d; a) = V(d1; a1) · · · V(dl; al), (2)

such that each V(di; ai) is indecomposable and d = d1+ · · · + dl, a= a1+ · · · + al.

The factors, V(d1; a1),..., V(dl; al) are unique up to permutations.

Lemma 2.3. If

V(d; a) = V(d1; a1) · · · V(dl; al),

then

Proof. We have V(d2; a2) · · · V(dl; al) ⊂ V(d − d1; a − a1). From (1), we have

dim V(d; a) ≥ dim V(d1, a1) + dim V(d − d1; a − a1)

≥ dim V(d1, a1) + dim V(d2; a2) · · · V(dl; al)

≥ dim V(d1, a1) + dim V(d2; a2) + · · · + dim V(dl; al)

= dim V(d; a).

Equality must hold everywhere, and therefore, the lemma is proved.

 Definition 2.2. A zero dimensional linear system is called isolated.

By abuse of language, we will identify an isolated linear system with the unique curve contained in it.

Suppose that V(d; a) = V(d1; a1) · · · V(dl; al) and each V(di; ai) is

indecom-posable. Assume that the first n of them are isolated and the others have positive dimensions. For i > n, the number of base points in V(di; ai) is finite, and we can

find a point Q ∈ P2_\S

1≤j≤nV(dj; aj), which is not a base point of any V(di; ai), for

n < i ≤ l. Let b = a + Q, bi = ai+ Q. Then

dim V(d; b) = dim V(d; a) − 1,

and, for n < i ≤ l,

dim V(di; bi) = dim V(di; ai) − 1.

Consider, for n < i ≤ l,

Wi := V(d1; ai) · · · V(di−1; ai−1) · V(di; bi) · V(di+1; ai+1) · · · V(dl; al),

which is a sub-set of V(d; b). But by (1), dim Wi = dim V(d; b) and hence Wi =

V(d; b). If i0

is another integer such that n < i0

≤ l, then Wi0 = V(d; b), too. By

Lemma 2.2, we must have

V(di; ai) = V(di0; a_i0).

In fact, the following is true.

Proposition 2.1. If a linear system has at least two positive dimensional indecom-posable factors, then all of them are the same and of dimension one.

Proof. In view of the above discussion, we only need to check the dimension. Let i and i0 _{be as above. If dim V(d}

i; ai) > 1, then V(di; bi), which is a factor of

V(d; b), will have an indecomposable factor of dimension at least 1 but at most dim V(di; bi) = dim V(di; ai) − 1. On the other hand, V(di; ai) is also a positive

dimensional indecomposable factor of V(d; b), and this will contradict the first part

of the proposition. 

3. The Regularity

In this section, we fix the points P1,..., Pr, and for each vector u = (d, m1, ..., mr)

where the entries are all non-negative integers, denote

a(u) = m1P1+ · · · + mrPr, V(u) = V(d; a), f (u) = f (d; a), g(u) = g(d; a) := d(d − 3)/2 − n X i=1 mi(mi− 1)/2 + 1. If v = (d0_{, m}0

1, ..., m0r) is another vector, denote

u · v = d · d0

− m1· m01− · · · − mr· m0r.

Then

u · u = f (u) + g(u) − 1. (3) Also, B´ezout Theorem implies the following.

Lemma 3.1. If V(u1) and V(u2) are both indecomposable with either V(u1) 6= V(u2)

or dim V(u1) > 0 and dim V(u2) > 0, then

u1· u2 ≥ 0.

We have

Definition 3.1. If a linear system V(u) is an empty set, then we put

dim V(u) = −1.

A linear system V(u) is regular, if

dim V(u) = max{−1, f (u)},

otherwise, it is called special.

Thus, an empty linear system is always regular.

Lemma 3.2. (1) Suppose both V(u1) and V(u2) are indecomposable and regular

with u1· u2 = 0. If V(u1) is isolated and V(u2) is non-isolated, then

u1· u1 = −1.

(2) Let

V(u1+ · · · + un) = V(u1) · · · V(un)

be the factorization into indecomposable factors. Suppose that bothV(ui) and

V(uj) are regular. Then we have

ui· uj ≤ 0.

Furthermore, if either V(ui) 6= V(uj), or V(ui) = V(uj) but i 6= j with

dim V(ui) > 0, then

ui· uj = 0.

Proof. Put u1 = (d1, m1, ..., mr), u2 = (d2, n1, ..., nr), a1 = m1P1+ · · · + mrPr and

a₂ = n1P1 + · · · + nrPr. Then we choose a point Q ∈ V(u1) = V(d1; a1), which is

not a base point of V(u2) = V(d2; a2). Since u1· u2 = 0, Lemma 3.1 implies that

V(d1; a1) = V(d1; a1+ Q) is a factor of V(d2; a2+ Q). If we put u2 = u1+ u3, then

Now,

dim V(u2) = f (u2)

= f (u1) + u1· u3+ f (u3)

≤ 0 + u1· u2− u1· u1+ dim V(u3)

= −u1· u1− 1 + dim V(u2).

This implies

u1· u1 ≤ −1.

Since g(u1) is at least the genus of V(u1) which is non-negative, by (3), we also have

u1· u1 ≥ −1. This proves (1).

By Lemma 2.3, we have

V(ui+ uj) = V(ui) · V(uj).

Then

f (ui+ uj) ≤ dim V(ui+ uj)

= dim V(ui) + dim V(uj)

= f (ui) + f (uj).

This together with (4) and Lemma 3.1 proves (2).

 Lemma 3.2 (1) leads us to recall the following definition.

Definition 3.2. A linear system V(u) or the vector u is of (−1) type, if V(u) is regular, isolated, indecomposable with

u · u = −1.

A linear systemV(v) or the vector v is (−1)-special if there is a u of (−1) type such that

v · u < −1.

Conjecture 3.1. (Harbourne-Hirschowitz, [Hbn89, Hsz89]) Suppose that P1,...,Pr

are in general position. Then a linear system is special if and only if it is (−1)-special. Conjecture 3.2. ( Segre, [Seg61]) Suppose that P1,...,Pr are in general position. If

Conjecture 3.3. Suppose that P1,...,Pr are in general position. Then every

inde-composable linear system is regular.

Theorem 3.1. Conjectures 3.1, 3.2, 3.3 are equivalent.

Proof. If V(v) is (−1)-special with a u of (−1) type such that v · u < −1, then V(u) is a multiple factor of V(v). Therefore, Conjecture 3.1 implies Conjecture 3.2. Also, Conjecture 3.2 obviously implies Conjecture 3.3.

Now we show that Conjecture 3.3 implies Conjecture 3.1. Suppose u is not (−1)-special. Let

V(u) = V(u1) · · · V(un)

be the factorization into indecomposable factors. Then

dim V(u) = dim V(u1) + · · · + dim V(n),

and dim V(ui) = f (ui) for i = 1, ..., n. By Lemma 3.2, for every pair i, j, ui· uj ≤ 0.

If ui · uj < 0 for some i 6= j, then by Lemma 3.1, ui = uj and f (ui) = 0. This

implies that ui is of (−1) type and u · ui < −1. Since u u is not (−1)-special, this

cannot happen. Therefore, for i 6= j,

ui· uj = 0,

and by (4), we have

f (u) = f (u1) + · · · + f (un) = dim V(u).

 Lemma 3.3. If u1· u1 ≥ 0, u2· u2 ≥ 0, and u1· u2 = 0, then u1 = u2, and u1· u1 = 0.

Proof. Let u1 = (d1, m1, ..., mr), u2 = (d2, n1, ..., nr). Then,

d2₁ ≥ m2₁+ · · · + m2_r,

d22 ≥ n21+ · · · + n2r,

and

d1d2 = m1n1 + · · · + mrnr.

The Lemma is a consequence of Cauchy-Schwartz Inequality.  Most of the following theorem is proved and for the time being we omit the rest of the proof.

Theorem 3.2. Assume that Conjecture 3.1 is true and P1,...,Pr are in general

position. Suppose that V(u) is non-empty. Then one of the following holds. (1) The system is regular and indecomposable.

(2) The system is regular, of positive dimension, all its isolated factors are dis-tinct and of (−1)-type, all its non-isolated factors are identical and, if they are more than one, are of one dimensional and composed curves of(−1)-type in a pencil.

(3) The system is regular, of zero dimension, all its factors are isolated and all but at most one of these factors are of (−1)-type.

(4) The system is regular, of zero dimension, all its factors are identical and they are isolated consist of an elliptic curve.

(5) The system is special and is (−1)-special.

4. Degenerated Cases

Let t = (3, 1, ..., 1) and let E ∈ V(t) be a smooth cubic curve. If r ≥ 10, then E = V(t). In this section, we consider the case where the base points P1,..., Pr are

in genral position on the smooth cubic curve E.

Lemma 4.1. If V(u) is non-empty and t · u ≤ 0, then V(u) = V(t) · V(u − t). Proof. In view of Lemma 3.1, we only need to consider the case where t · u = 0. Let u = (d, m1, ..., mr). We fix P1,...,Pr−1 and let Pr move along E. If in

general V(t) is not a factor of V(u), then we can, for general Pr(1), Pr(2) on E, find

homogeneous polynomials F1(X, Y, Z), F2(X, Y, Z) which respectively defines curves

in V(d; m1P1+ · · · + mr−1Pr−1 + mrPr(1)), V(d; m1P1 + · · · + mr−1Pr−1 + mrPr(2)),

containing no component equal E. Then φ = F1/F2 is a non-trivial rational function

of E with divisor

Div(φ) = mr[Pr(1)] − mr[Pr(2)].

This means that the divisor [Pr(1)] − [Pr(2)] is in the mr-torsions of the Jacobian

variety of E. But since Pr(2) is in general position, this cannot happen. 

Lemma 4.2. Suppose that P1, ..., Pr ∈ E are in general position. If V(u) is isolated,

indecomposable and u 6= t, then u is regular and is of (−1)-type.

Proof. We have f (u) ≤ dim V(u) = 0, g(u) ≥ 0 and also, by Lemma 4.1,

f (u) − g(u) + 1 = t · u > 0.



Theorem 4.1. Suppose that P1, ..., Pr ∈ E are in general position. If u satisfies

t · u ≥ 0,

and, for every z of (−1)-type,

u · z ≥ 0, then u is regular.

Proof. Let u = (d, m1, ..., mr) and a = m1P1+ · · · + mrPr. We prove the theorem

by Induction on d. If u0 _{= (d}0_{, m}0

1, ..., m0r) is not of (−1)-type, with d0 < d, V(u0) 6=

V(t) and V(u0_{) indecomposable, then by Lemma 4.1, t · u}0 _{> 0 and the induction}

hypothesis implies that V(u0_{) is regular. In particular, a proper indecomposable}

factor, which is not V(t), of V(u) is regular.

First, consider the case where V(u) is indecomposable and dim V(u) = 0. By Lemma 4.1, we have

and hence

f (u) ≥ g(u). Then from

0 = dim V(u) ≥ f (u) ≥ g(u) ≥ 0, we get f (u) = g(u) = 0.

Next, we consider the case where V(u) is decomposable. By Lemma 4.2, the isolated factors of V(u) are either V(t) = E (this happen only when r < 9) or those of (−1)-type. Let V(z1),..., V(zn) be all the factors of (−1)-type and a1,...,

an be their multiplicities, if r ≥ 9 let b be the multiplicity of V(t), and let V(w)

be the product of all the non-isolated factors. Thus, if V(w) is non-empty, then it is either indecomposable or a multiple of an indecomposable factor which is one dimensional. If it is indecomposable and not equal V(u), then we already see that it is regular. If it is a multiple of an indecomposable factor V(x), then V(x) is regular and from Lemma3.2, we see that x · x = 0. This implies that if w = mx, then dim V(w) = m = mf (x) = f (w) and V(w) is also regular. Again, from Lemma3.2, we have

zi· zj = 0, (5)

and

zi· w = 0. (6)

If b = 0, Then the condition that u · z ≥ 0 for every z of (−1)-type implies n = 0 and V(u) = V(w). Since V(u) is decomposable, it must be a multiple of V(x) as described above with m > 1, and hence V(u) is regular. If b > 0, then Lemma 2.3 says that V(t), V(z1),..., V(zn) and V(w) are factors of V(t + z1+ · · · + zn+ w). Now

V(z1)..., V(zn) and V(w) are all regular. Using (4), (5) and (6), we get

f (w) = dim V(w)

= dim V(t + z1+ · · · + zn+ w)

≥ f (t + z1 + · · · + zn+ w)

= f (t) + f (w) + n + t · w = 9 − r + f (w) + n + t · w.

Consequently,

r ≥ 9 + n + t · w. But, we also have

0 ≤ t · u = bt · t + a1+ · · · + an+ t · w,

and

0 ≤ zi· u = b − ai.

The above three inequalities imply that either w is null, r = 9 + n and ai = b for

every i or b = ai = 1 for every i and r = 9 + n + t · w. In both cases, we can deduce,

by using (4), dim V(u) = f (u).

Finally, we treat the case where V(u) is indecomposable and dim V(u) > 0. In this case, we should have t · u > 0. Choose a point Q1 ∈ E, which is not a base

point of V(u) and is not on any curve of (−1)-type with degree less than d. We then replace the number r by r + 1, t by t + e, u by u + e, where e is the vector in Zr+2_,

with the first r + 1 entries equal 0 and the last entry equal 1. Then the conditions t · u ≥ 0 and z · u ≥) for every z of (−1)-type still holds. After this replacement, the dimension of V(u) becomes one less and so does f (u). If now V(u) becomes either isolated or decomposable, then we reduce the proof to the previous two cases. Otherwise, we repeat this procedure. 

5. The Dimensions of Indecomposable Systems

In this section, we put P1,..., Pr in general position and assume Conjecture 3.1.

Let t = (3, 1, ..., 1). Then for a z of (−1)-type,

t · z = 1. (7)

Lemma 5.1. If z1,...,zn are of (−1)-type and

zi · zj = 0, for every i 6= j,

then

Proof. The condition for (−1)-type and (7), imply that the vectors t, z1,...,zn are

linearly independent. 

Theorem 5.1. Assume that Conjecture 3.1 is true and P1,..., Pr are in general

position. If V(u) is indecomposable and a generic curve in its is of genus g ≥ 1, then

dim V(u) ≤ 3g + 6.

Proof. By a suitable Cremona transformation, we can assume that g = g(u). Then f (u − t) = g − 1 ≥ 0. If z is of (−1)-type, then u · z ≥ 0 and therefore, by (7), (u − t) · z ≥ −1. This implies that u − t is not (−1)-special and hence regular with dim V(u − t) = g − 1. We shall analyze the factorization of V(u − t).

Suppose that g(u) = 1. Then dim V(u − t) = 0 and by Theorem 3.2, we have

u − t = z1+ · · · + zn+ kw,

where each V(zi) is of (−1)-type, w is an elliptic curve and either n = 0 or k = 0, 1.

From this, we get t · w ≤ 0 and

f (u) − g + 1 = t · u

= t · (u − t) + t · t = n + kt · w + 9 − r ≤ 9 + n − r.

The by Lemma 5.1, n ≤ r and f (u) ≤ 9.

Suppose that g(u) > 1. Then dim V(u − t) > 0 and by Theorem 3.2, we also have

u − t = z1+ · · · + zn+ kw,

and

V(u − t) = V(z1) · · · V(zn) · V(kw),

where each V(zi) is of (−1)-type, V(kw) is indecomposable, of positive dimension,

and t · kw = f (kw) − g(kw) + 1 ≤ dim V(kw) + k = dim V(u − t) + k = g + k − 1. Consequently, f (u) − g + 1 = t · u = t · t + t · (u − t) = 9 − r + n + t · kw ≤ 9 + g + k − 1.

If k = 1, then f (u) ≤ 2g + 8 ≤ 3g + 6. Otherwise, then k = dim V(u − t) = g − 1, and f (u) ≤ 3g + 6.



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Department of Mathematics, National Taiwan University, Taipei,Taiwan E-mail address: pyhuang@math.ntu.edu.tw

Department of Mathematics, National Taiwan University, Taipei,Taiwan E-mail address: tan@math.ntu.edu.tw