Non-trivial Linear Systems on Smooth Plane Curves

16 Pages·0.16 MB·English

by  Marc Coppens

Non-trivial Linear Systems on Smooth Plane
3 Curves
9
9
1
Marc Coppens∗ Takao Kato†

n
a
J

6

1 0 Introduction
v
3
0 Let C be a smooth plane curve of degree d defined over an algebraicallyclosed field k. In
0 [10],whilestudyingspacecurves,MaxNoetherconsideredthefollowingquestion. Forn∈
1 Z findℓ(n)∈Z suchthatthereexistsalinearsystemgℓ(n) onC butnolinearsystem
0 ≥1 ≥0 n
3 gℓ(n)+1 and classify those linear systems gℓ(n) on C. The arguments given by Noether in
n n
9 the answer to this question contained a gap. In [1] C. Ciliberto gave a complete proof
/
m for Noether’s claim using different arguments. In [6] R. Hartshorne completed Noether’s
originalargumentsbysolvingthe problemalsoforintegral(notnecessarilysmooth)plane
o
curves (see Remark 1.3).
e
g The linear systems gℓ(n) are either non-special,or specialbut not veryspecial, or very
n
-
specialbuttrivial. Bya veryspecial(resp. trivial)linearsystemonasmoothplane curve
g
l C we mean:
a
:
v Definition. A linear system gr on C is very special if r ≥ 1 and dim|K −gr| ≥ 1.
i n C n
X (Here K is a canonical divisor on C). A base point free complete very special linear
C
r system gr on C is trivial if there exists an integer m≥0 and an effective divisor E on C
a n
∗This authoris relatedto the UniversityatLeuven(Celestijnenlaan 200BB-3001Leu-
ven Belgium) as a Research Fellow.
†This author represents his thanks to Katholieke Universiteit Leuven, Departement
Wiskunde for their kind hospitality during his stayand to the JSPSfor financialsupport.
1
of degree md−n such that gr = |mg2−E| and r = m2+3m −(md−n). A complete
n d 2
very special linear system gr on C is trivial if its associated base point free linear system
n
is trivial.
In this paper, we consider the following question. For n ∈ Z find r(n) such that
≥1
thereexistsaveryspecialnon-trivialcompletelinearsystemgr(n) onC butnosuchlinear
n
system gr(n)+1. Our main result is the following:
n
Theorem. Let gr be a base point free very special non-trivial complete linear system on
n
(x+1)(x+2)
C. Write r = −β with x,β integers satisfying x≥1,0≤β ≤x. Then
2
n≥n(r):=(d−3)(x+3)−β.
This theorem only concerns linear systems of dimension r ≥ 2. But 1-dimensional
linear systems are studied in [3]. From these results one finds that C has no non-trivial
very special linear system of dimension 1 if d ≤ 5 and for d ≥ 6, C has non-trivial very
special complete linear systems g1 but no such linear system g1 with n<3d−9. The
3d−9 n
proof of this theorem is also effective for case r =1 if one modifies it little bit.
Concerningtheoriginalproblemonecanmakethefollowingobservation. Forx≥d−2
one has r > g(C) and of course C has no non-trivial very special linear systems gr. For
n
x ≤ d−3 one has (d−3)(x+2) ≤ (d−3)(x+3)−x. So, if the bound n(r) is sharp,
then also the bound r(n) can be found. Concerning the sharpness of the bound n(r), we
prove it in case char(k)=0 for x≤d−6. In case char(k)6=0 we prove that there exists
smooth plane curves of degree d with a very special non-trivial gr in case x ≤ d−6.
n(r)
Finally for the case x >d−6 we prove that there exist no base point free complete very
specialnon-triviallinear systemsof dimensionr onC. Hence, at leastin casechar(k)=0
the numbers r(n) are determined.
Some Notations
We write P to denote the space of effective divisors of degree a on P2. If P is
a
a linear subspace of some P then we write P.C for the linear system on C obtained
a
by intersection with divisors Γ ∈ P not containing C. We write F(P.C) for the fixed
point divisor of P.C and f(P.C) for the associated base point free linear system on C,
2
so f(P.C) = {D−F(P.C) : D ∈ P.C}. If Z is a 0-dimensional subscheme of P2 then
P (−Z) is the subspace of divisors D ∈P with Z ⊂D.
a a
1 Bound on the degree of non-trivial linear
systems
A complete linear system gr on a smooth curve C is called very special if r ≥ 1 and
n
dim|K −gr|≥1. From now on, C is a smooth plane curve of degree d and gr is a very
C n n
special base point free linear system on C with r ≥2.
Definition 1.1 gr is called a trivial linear system on C if there exists an integer m ≥ 0
n
and an effective divisor E on C of degree md−n such that gr = |mg2 −E| and r =
n d
m2+3m
−(md−n).
2
(x+1)(x+2)
Theorem 1.2 Writer = −β withx,β integerssatisfyingx≥1,0≤β ≤x.
2
If gr is not trivial, then
n
n≥n(r):=(d−3)(x+3)−β.
Remark 1.3 In the proof of this theoremwe are going to make use of the main result of
Hartshorne[6]whichdescribesthelinearsystemsonC ofmaximaldimensionwithrespect
to their degrees. The result is as follows:
(d−1)(d−2)
Letgr bealinearsystemonC (notnecessarilyveryspecial). Writeg(C)= .
n 2
i) If n>d(d−3) then r =n−g (the non-special case)
ii) If n≤d(d−3) then write n=kd−e with 0≤k ≤d−3,0≤e<d, one has
(k−1)(k+2)
r ≤ if e>k+1
2

k(k+3)
 r ≤ −e if e≤k+1.
2

3
Hartshorne also gives a description for the case one has equality. This theorem (a
claim originally stated by M. Noether with an incomplete proof) is also proved in [1].
However, Hartshorne also proves the theorem for integral plane curves using the concept
of generalized linear systems on Gorenstein curves. We need this more general result in
the proof of Theorem 1.2
Proof of Theorem 1.2. Assume gr = rg1 and n < (x + 3)(d − 3) − β. Noting
n n/r
(x+3)(d−3)−β
2r = (x+1)(x+2)−2β ≥ x2+x+2 ≥ x+3, we have < 2(d−2).
r
Hence, g1 = |g2−P| for some P ∈ C. Since dim|rg1 | = r, certainly dim|2g1 | = 2.
n/r d n/r n/r
But dim|2g2−2P|=3. A contradiction.
d
Since gr is special, there exist an integer 1≤m≤d−3 and a linear system P⊂P
n m
such that gr = f(P.C) and P has no fixed components. In Lemma 1.4 we are going to
n
provethat,becausegr isnotamultipleofapencil,ageneralelementΓ ofPisirreducible.
n
Now, for each element Γ′ of P we have F(P.C)⊂Γ′ (inclusion of subschemes of P2).
InparticularF(P.C)⊂Γ∩Γ′. ThisremarkisknownintheliteratureasNamba’slemma.
As a subscheme ofΓ,F(P.C) is aneffective generalizeddivisoronΓ (terminologyof[6]).
We find that for each Γ′ ∈ P with Γ′ 6= Γ the residual of F(P.C) in Γ ∩Γ′ (we denote
it by Γ ∩Γ′ −F(P.C)) is an element of the generalized complete linear system on Γ
associatedto O (m−F(P.C)). Hence,weobtainageneralizedlinearsystemgr−1
Γ m2−(md−n)
on Γ.
NowwearegoingtoapplyHartshorne’stheorem(Remark1.3)tothisgr−1 onΓ.
m2−(md−n)
Since gr is non-trivialon C, we know that r> m2+3m −(md−n). If m2−(md−n)>
n 2
(m−1)(m−2)
m(m − 3), then i) in Remark 1.3 implies r −1 ≤ m2 + n −md − so
2
m2+3m
r ≤ −(md−n), a contradiction.
2
So m2−(md−n)≤m(m−3)andwe applyii)in Remark1.3. We findx≤m−3and
m2+n−md≥mx−β, so n≥ϕ(m):=−m2+m(d+x)−β. Since x+3≤m≤d−3,
we find n≥ϕ(x+3)=ϕ(d−3)=(d−3)(x+3)−β =n(r). This completes the proof of
the theorem except for the proof of Lemma 1.4.
Lemma 1.4 Let C be a smooth plane curve of degree d and let gr be a base point free
n
complete linear system on C which is not a multiple of a one-dimensional linear system.
Suppose there exists a linear system P⊂ P without fixed component for some e≤ d−1
e
such that gr =f(P.C). Then the general element of P is irreducible.
n
4
Proof. Let F = F(P.C) = s n P with n ≥ 1 and P 6= P for i 6= j. For
j=1 j j j i
t∈Z ,e=(e ,...,e )∈(Z )tPwith t e =e and m=[m ] , let
≥1 1 t ≥1 i=1 i ij 1≤i≤t,1≤j≤s
P
V(t,e,m)={Γ +···+Γ :Γ ∈P is irreducible and i(Γ ,C;P )=m }.
1 t i ei i j ij
It is not so difficult to prove that this defines a stratification of P by means of locally
e
closed subspaces.
SincePisirreduciblethereisauniquetriple(t ,e ,m )suchthatP∩V(t ,e ,m )isan
0 0 0 0 0 0
opennon-emptysubsetofP. Inparticular,P⊂{Γ +···+Γ :Γ ∈P and i(Γ ,C;P )≥
1 t0 i e0i i j
m }. We need to prove that t = 1, so assume that t > 1. Let forget the subscript 0
0ij 0 0
from now on.
Let F = s m P ⊂ C. For each D ∈ gr there exists Γ = Γ +...+Γ with
i j=1 ij j n 1 t
Γ ∈P (−F )PandD =Γ.C−(F +...+F )= t (Γ .C−F ). WritingD =Γ .C−F ∈
i i i 1 s j=1 i i i i i
|e g2−F | we find D = t D . Suppose forPsome 1≤i≤t we havedim|e g2−F |=0.
i d i i=1 i i d i
If Γ′ and Γ′′ are in Pi(P−Fi) then Γ′.C = Γ′′.C, but ei < d so Γ′ = Γ′′. This implies
that P (−F ) = {Γ }, but then Γ is a fixed component of P, a contradiction. Hence,
i i 0 0
for 1 ≤ i ≤ t, we have dim|e g2 −F | ≥ 1. Now, let L be the irreducible sheaf on C
i d i i
associated to |e g2−F | and let L be the irreducible sheaf on C associated to gr. Then
i d i n
L=L ⊗···⊗L and we find that the natural map
1 t
H0(C,L )⊗···⊗H0(C,L )→H0(C,L)
1 t
is surjective, while dimH0(C,L ) ≥ 2 for 1 ≤ i ≤ t. From [4, Corollary 5.2], it follows
i
that gr is a multiple of a pencil. But this is a contradiction.
n
Remark 1.5 In [9] one makes a classification of linear systems on smooth plane curves
for which r is one less than the maximal dimension with respect to the degree. In that
paper one uses arguments like in [1]. That classification is completely contained in our
Theorem 1.2
(d−1)(d−2)
Remark 1.6 If r ≥ n− +d− 1 then dim|K −gr| ≥ d− 2. Hence,
2 C n
|K −g2−gr| = |(d−4)g2−gr| =6 ∅. So in this case we can assume m ≤ d−4 in the
C d n d n
proofofTheorem1.2. Then,in the proofofTheorem1.2,using Bertini’s theorem,wecan
5
prove that, for D ∈ gr general there exists an irreducible curve Γ of degree d−3 with
n
Γ.C ≥ D (see argument in [6, p.384]). So we don’t need the proof of Lemma 1.4 under
that assumption on r.
Remark 1.7 Inthatcasen=n(r),wefindm=x+3andm2+n−md=mx−β. Sothe
generalized linear system gr−1 =gr−1 on Γ is of maximal dimension with respect
m2+n−md mx−β
to its degree. The description of those linear systems in [6] implies that it is induced by a
family of plane curves of degree x containing some subspace E ⊂ Γ of length β. Writing
Z = F(P.C) ⊂ Γ we find |P .Γ −Z|= |P .Γ −E| and so Z ∈|P .Γ +E|. In order to
m x 3
find non-trivial gr ’s it is interesting to find for a smooth curve C of degree d, a curve
n(r)
Γ of degree m and a curve Γ′ of degree 3 such that Γ ∩Γ′ ⊂ C. We discuss this in §3.
First we solve the following postulation problem: let Γ′ be the union of 3 distinct lines
L ,L ,L and let D be an effective divisor of degree a on L with D ∩(L ∪L )=∅ for
1 2 3 i i i j k
{i,j,k}={1,2,3}. Give necessary and sufficient conditions for the existence of a smooth
curve Γ of degree a such that Γ.L =D for i=1,2,3.
i i
2 Carnot’s theorem
We begin with pointing out the following elementary fact:
ℓ
Lemma 2.1 Let m(≥4) and m (j =1,...,ℓ) be positive integers satisfying m =m
j j
Xj=1
m
andletΦ(X)= a Xj beanon-zeropolynomialofdegreeatmostm. IfΦ(X)isdivisible
j
Xj=1
by (X −Xj)mj for ℓ distinct values X1,...,Xℓ, then the ratio a0 : ··· : am is uniquely
ℓ ℓ
determined. In particular, a 6=0,a =(−1)ma Xmj and a =−a m X .
m 0 m j m−1 m j j
jY=1 Xj=1
Usingthis,wehavethefollowingCarnot’stheoremandinfinitesimalCarnot’stheorems.
Another generalization of this theorem is given by Thas et al. [11] (see also [12]). They
call this B. Segre’s generalization of Menelaus’ theorem.
6
Lemma 2.2 (Carnot, cf. [12, Proposition 1.8], [8, p.219]) Let L ,L ,L be lines in P2
1 2 3
ℓi ℓi
so that L ∩L ∩L =∅ and let D = m P , ( m =m,i=1,2,3) be an effective
1 2 3 i ij ij ij
Xj=1 Xj=1
divisoronL suchthatD ∩L =∅ifi 6=i . Assume(x:y :z)isacoordinatesystemon
i i1 i2 1 2
P2 such that L ,L ,L correspond to the coordinate axes x=0,y =0,z =0, respectively.
1 2 3
Let the coordinate of P be given by (x :y :z ) (of course x =y =z =0). Then,
ij ij ij ij 1j 2j 3j
there exists a curve Γ not containing any one of the lines L ,L ,L of degree m satisfying
1 2 3
(Γ.L )=D (i=1,2,3) if and only if
i i
ℓ1 y m1j ℓ2 z m2j ℓ3 x m3j
(2.1) 1j 2j 3j =(−1)m.
(cid:18)z (cid:19) (cid:18)x (cid:19) (cid:18)y (cid:19)
jY=1 1j jY=1 2j jY=1 3j
Proof. Assume there exists a curve Γ of degree m not containing any one of the
lines L ,L ,L satisfying (Γ.L ) = D (i = 1,2,3). Such a curve is given by Φ(x,y,z) =
1 2 3 i i
a xiyjzk = 0. In this description, if i(Γ,L ;P ) = m , then Φ(0,y,z) is
ijk 1 1j 1j
i+jX+k=m
ℓ1 y m1j
divisible by (y1jz −z1jy)m1j. This implies a00m = (−1)ma0m0 1j . Similarly,
(cid:18)z (cid:19)
jY=1 1j
ℓ2 z m2j ℓ3 x m3j
we have a = (−1)ma 2j and a = (−1)ma 3j . Since
m00 00m 0m0 m00
(cid:18)x (cid:19) (cid:18)y (cid:19)
jY=1 2j jY=1 3j
Γ doesnot containanintersectionpointL ∩L fori 6=i ,we havea a a 6=0.
i1 i2 1 2 m00 0m0 00m
Hence, we have (2.1).
For the converse, take a = 1. Then, by (2.1) we can determine a so that Γ has
m00 ijk
the desired property. This completes the proof.
Next, we see two infinitesimal casesi. e. case D ∩L 6=∅ andcase L ∩L ∩L 6=∅.
i1 i2 1 2 3
Lemma 2.3 Let L ,L ,L be lines in P2 so that L ∩ L ∩ L = ∅ and let D =
1 2 3 1 2 3 i
ℓi ℓi
m P ,( m = m) be an effective divisor on L such that m = m = 1,P =
ij ij ij i 11 21 11
Xj=1 Xj=1
7
P = L ∩L and D ∩L = D ∩L = ∅ (i = 1,2). Let (x : y : z) and (x : y : z )
21 1 2 i 3 3 i ij ij ij
be as in Lemma 2.2. Then, there exists a curve Γ, not containing any one of the lines
L ,L ,L , of degree m such that (Γ.L ) = D (i = 1,2,3) and whose tangent line T at
1 2 3 i i
P =(0:0:1) is given by αx+y =0 (α6=0) if and only if
11
ℓ1 y m1j ℓ2 z m2j ℓ3 x m3j
(2.2) α 1j 2j 3j =(−1)m.
(cid:18)z (cid:19) (cid:18)x (cid:19) (cid:18)y (cid:19)
jY=2 1j jY=2 2j jY=1 3j
Proof. We use the same notationasin the proofofLemma 2.2. Assume there exists a
curve Γ having the desired property. The condition that the tangent line at P is given
11
by αx+y = 0 implies that Φ(0,0,1) = 0 and the linear term of Φ(x,y,1) is divisible by
αx+y. Hence, a = 0 and a −αa = 0. As in the proof of the previous
00m 10,m−1 01,m−1
ℓ3 x m3j
lemma,we havea a 6=0,a 6=0, a =(−1)ma 3j ,a =
m00 0m0 01,m−1 0m0 m00 01,m−1
(cid:18)y (cid:19)
jY=1 3j
ℓ1 y m1j ℓ2 x m2j
(−1)m−1a 1j andαa =a =(−1)m−1a 2j . Hence,
0m0 01,m−1 10,m−1 m00
(cid:18)z (cid:19) (cid:18)z (cid:19)
jY=2 1j jY=2 2j
we have (2.2).
Similar to the proof of the previous lemma, we have the converse.
Lemma 2.4 Let L ,L ,L be lines in P2 so that L ∩ L ∩ L 6= ∅ and let D =
1 2 3 1 2 3 i
ℓi ℓi
m P ,( m = m) be an effective divisor on L such that D ∩L = ∅ if i 6= j.
ij ij ij i i j
Xj=1 Xj=1
Let (x : y : z) be a coordinate system on P2 such that L ,L ,L correspond to the line
1 2 3
y−z =0,y =0,z =0, respectively. Letthe coordinate of P begiven by (x :y :z ) (of
ij ij ij ij
course y =z ,y =0,z =0). Then, there exists a curve Γ of degree d not containing
1j 1j 2j 3j
any one of the lines L ,L ,L such that (Γ.L )=D (i=1,2,3) if and only if
1 2 3 i i
ℓ1 x ℓ2 x ℓ3 x
1j 2j 3j
(2.3) m − m − m =0.
1j 2j 3j
y z y
Xj=1 1j Xj=1 2j Xj=1 3j
8
Proof. Again, we use the same notation as in the proof of Lemma 2.2. Assume there
exists a curve Γ having the desired property. Since Γ does not contain L ∩L , we have
2 3
a 6=0. By Lemma 2.1,
m00
ℓ3 x ℓ2 x
3j 2j
a =− m a and a =− m a .
m−1,10 3j m00 m−1,01 2j m00
y z
Xj=1 3j Xj=1 2j
To consider the condition on L , we take the coordinate system (ξ :η :ζ) with ξ =x,η =
1
y,ζ =z−y. Put
Ψ(ξ,η,ζ)=Φ(ξ,η,η+ζ)= a ξiηj(ζ+η)k = b ξiηjζk.
ijk ijk
i+jX+k=m i+jX+k=m
Then,b =a andb =a +a . Inthisdescription,ifi(Γ,L ;P )=
m00 m00 m−1,10 m−1,10 m−1,01 1 1j
ℓ1 x
m1j,thenΨ(ξ,η,0)isdivisibleby(ξ1jη−η1jξ)m1j. Thisimpliesthatbm−1,10 =− m1j 1jbm00.
y
Xj=1 1j
Then, we have (2.3).
For the converse, noting that if a 6= 0 then Γ does not contains L (i = 1,2,3) as
m00 i
a component, we can find a Γ having the desired property.
Remark 2.5 In each of the lemmas 2.2, 2.3 and 2.4, if (2.1) (resp. (2.2), (2.3)) holds,
we can find a smooth curve Γ of degree m with Γ.L = D for i = 1,2,3. Indeed, let
i i
P ⊂ P be the linear system of divisors Γ of degree m on P2 satisfying, as schemes,
m
D ⊂ Γ ∩L . Clearly L +L +L +P ⊂ P and we proved that U = {Γ ∈ P :
i i 1 2 3 m−3
Γ does not contain any of the lines L ,L ,L }isanon-emptyopensubsetofP. Because
1 2 3
of Bertini’s theorem, for Γ ∈ U we have L ∩Γ = {P ,...,P }. Consider the linear
i i1 iℓi
systemP′ ={Γ∩P2\(L ∪L ∪L ):Γ ∈P}onM =P2\(L ∪L ∪L ). SinceΓ∩M ∈P′
1 2 3 1 2 3
for anyΓ ∈P , P′ separatestangentdirections andpoints onM. BecauseofBertini’s
m−3
theorem in arbitrary characteristics (see [7]), we find that a general element Γ of P′ is
smooth. So,a generalelement Γ ofP satisfies Sing(Γ)⊂{P :i=1,2,3 and 1≤j ≤ℓ}.
ij
But, using Γ′ ∈P suited we find Γ =Γ′+L +L +L is smooth at P , except for
m−3 1 2 3 ij
the case P = P in Lemma 2.3. In that case, however, if Γ ∈ U then Γ is smooth at
11 21
P because of Bezout’s theorem. This completes the proof of the remark. (For Bertini’s
11
theorem in arbitrary characteristics, one can also use [5].)
9
3 Sharpness of the bound
Thenextpropositionimpliesthatitisenoughtosolvethepostulationproblemmentioned
in Remark 1.7 in order to prove sharpness for the bound n(r) in Theorem 1.2.
(x+1)(x+2)
Proposition 3.1 LetC beasmoothplanecurveofdegreedandletr = −β
2
with x,β ∈Z satisfying 4≤x+3≤d−3,0≤β ≤x. Let n=n(r)=(d−3)(x+3)−β.
Suppose there exists a smooth curve Γ of degree m=x+3, a curve Γ′ of degree 3 and an
effective divisor E of degree β on Γ such that Z ⊂C, where the divisor Z =(Γ ∩Γ′)+E
on Γ is considered as a closed subscheme of P2. Then |mg2−Z| is a non-trivial gr on
d n
C.
Proof. WewriteE =P +···+P . LetL ,...,L begenerallinesthroughP ,...,P ,
1 β 1 β 1 β
resp., and let L ...,L be general lines in P2. If P ∈C, then we write µ (Z) for the
β+1 x P
multiplicity of Z at P.
i) |mg2−Z| is base point free.
d
Suppose P isabasepointfor|mg2−Z|. Since Γ.C−Z ∈|mg2−Z|onefinds P+Z ≤
d d
Γ.C, hence i(Γ,C;P) > µ (Z) ≥ i(Γ,Γ′;P). Also (Γ′ + x L ).C −Z ∈ |mg2 −Z|,
P i=1 i d
hence P ∈ (Γ′ + xi=1Li).C −Z = (Γ′.C −Γ′ ∩Γ)+(Pxi=1Li.C −E) (sum of two
effective divisors).PSince P 6∈ xi=1Li.C −E, we find P ∈PΓ′.C −Γ′∩Γ. This implies
i(Γ′,C;P)>i(Γ,Γ′;P). ButiP(Γ,Γ′;P)≥min(i(Γ,C;P),i(Γ′,C;P))(socalledNamba’s
lemma), hence we have a contradiction.
ii) dim(|mg2−Z|)≥r.
d
(x+1)(x+2)
Indeed, (Γ′+P (−E)).C−Z ⊂|mg2−Z|anddim(Γ′+P (−E))= −
x d x 2
β −1. But also Γ.C −Z ⊂ |mg2−Z| while Γ.C 6∈ (Γ′ +P (−E)).C. This proves the
d x
claim.
iii) dim(|mg2−Z|)=r.
d
If dim(|mg2−Z|) > r then on Γ it induces a linear system gr′ with r′ ≥ r. But
d mx−β
Hartshorne’s theorem (see 1.3) implies that this is impossible.
iv) |mg2−Z| is not trivial.
d
First of all, |mg2−Z| is very special. Indeed (d−3−m)g2+Z ⊂|K −(mg2−Z)|.
d d C d
If d−3=m then from the Riemann-Roch theorem, one finds dim|Z|=1.
10