C OMPOSITIO M ATHEMATICA
X AVIER B ENVENISTE
On the fixed part of certain linear systems on surfaces
Compositio Mathematica, tome 51, n
o2 (1984), p. 237-242
<http://www.numdam.org/item?id=CM_1984__51_2_237_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
237
ON THE FIXED PART OF CERTAIN LINEAR SYSTEMS ON SURFACES
Xavier Benveniste
© 1984 Martinus Nijhoff Publishers, The Hague. Printed in The Netherlands
First we recall what is a
numerically positive
divisor:DEFINITION: Let V be a smooth
projective variety
and D be a divisor onV. We say that D is
numerically positive
if for any curve C on V we haveD.C 0.
Let R be a
numerically positive
divisor on a smoothprojective
surfaceover an
algebraically
closed field of any characteristic such thatR2
>0;
let Wbe the set of irreducible curves C such that R.C = 0
(we
will showthat A is
finite).
The aim of this paper is to prove thefollowing
result:PROPOSITION:
Let t be a
connectedcomponent of UC~AC;
let Z be thefundamental cycle
associated to03BE. If H’(Z, fflz)
=0, then e
is not afixed
component of 1 nR for sufficiently large
n.Let us make a few comments. The first one is that the condition
H1(Z, OZ)
= 0 characterizes rationalsingularities.
The second is that thisproposition generalizes slightly
a well known theorem of Zariski([1]
theorems
6-1, 6-2).
PROOF:
Let e
be as in theproposition
and let(Ci)i~{1,...,m}
the set ofirreducible
components
of03BE.
Then:LEMMA 1: The set A is
finite
and the bilinearsymmetric form defined by
thematrix
(Ç - Cj)i,j~
{1,...,m} is negative definite.
PROOF: We shall show that if
(Ei)i~{1,...,n} is
a finitefamily
of elements ofW,
then the classes[Eil ]
ofthe Ei
areQ-linearly independent
inNS(S)~ZQ.
Assume thecontrary. Any
relation ofdependence
betweenthe
[ Ei in NS(S)~ZQ
can be written:where I and I’ are two non-void
disjoint
subsetsof {1,...,n} and,
for238
any i ~ I ~
l’, ai ~ N - {0}.
Because R -03A3aiEi=0,
the index theoremiEl
on S
implies:
But
Again by
the index theorem on S we have:So
If we observe that
rkz (NS(S))
is finite we have the result. The fact that the bilinear form definedby
the matrix(Ci·Cj)i,j~{1,...,m}
is definitenegative,
is because the intersection form onNS(S)
isnegative
definiteon the
orthogonal
of R. 0We recall now that the fundamental
cycle
Z associatedto t
is definedby
thefollowing
condition:It is the "smallest" effective divisor with
support in e
such that:This is the definition of Artin in
[2].
LEMMA 2: Let D =
03A3ni=1aiCi
and L an invertiblesheaf
on S such thatfor
any i ~ {1,...,m}, degCi(L)0.
ThenPROOF: The
proof
can be found in[2]
Lemma5,
but here wegive
anelementary proof.
First of all we observe that allCi
are rational smoothcurves because we have a
surjective
map:for each
i ~ (1,...,m},
andHl
is aright
exact functor in this case. Nowwe
distinguish
two cases.1 st case: Assume that we have
proved
the result for all divisors of the form nZ with n E 1B1 -{0};
because there exists aninteger n
such thatwe have a
surjective
mapOnZ ~ (9D - 0, we get
asurjective
mapThis
gives
the result becauseH’
isright
exact.2nd case: We can assume D = nZ and shall prove the result
by
inductionon n. If n = 1 it is very easy to see that there exists a sheaf of finite
length
F and an exact sequence
Because
H1(Z, OZ)
=H1(Z, F)
= 0 we have the result. Assume we have the result for n; we shall prove it for n + 1. We have the exact sequence:We observe that:
This
implies
that:LEMMA 3: There exists an
effective
divisor L such that the rational map cP Ldefined by |L| is
a birationalmorphism
on itsimage from
S to asurface
Ysuch that (PL:
S - 03BE
Y -{P},
where P is a closedpoint of
Y and:PROOF: This lemma can also be found in
[2]
minus the lastglobal
condition which could be
easily
obtained. Here weprefer
togive
anotherproof.
Let H be a veryample
divisor on S such that:where b1 E
N and d =|det(Ci· Cj)|.
Because the bilinear form definedby
the matrix
(Ci·Cj)i,j~{1,...,m}
is definitenegative by
Lemma1,
there240
exists an effective divisor D =
03A3mi=1aiCi with a; E
N such that:We consider the exact sequence
Observe that
r2D(H
+D) = (9D.
Hence we have the exact sequences:This
implies by
Lemma 2 thatOn the other hand the constant function
equal
to 1belongs
toHO( r2D)
sothe linear
system |H + D| (
does not have any fixedcomponent
ine.
Because H is very
ample
it is clear that L = H + D satisfies the conditions of Lemma 3. 0LEMMA 4: There exists no E N such that
for
n n o there existsr(n)
E 1B1such that
for
rr(n), 1 rL
+nR doesn’t have t as fixed
component.PROOF: Because we have
R2
>0,
if wereplace R by
a suitablemultiple,
we can assume R to be an effective divisor. So we can write R = F + G with F and G effective divisors such that all the irreducible
components
of F are containedin t
and none of the irreduciblecomponents
of G are ine.
But we have the exact sequences for r E 1B1 -{0},
The exact sequence of
cohomology gives:
If r is
big enough,
because of thevanishing
theorem ofSerre,
we have:so we have a
surjective
map:But the constant function
equal
to 1belongs
toH0(OF),
so weget
theresult. D
Now we show the
proposition;
we assume that there existinfinitely
many n such that
InRI has 03BE
as a fixedcomponent.
Forany n ~ N - {0}
we write
where
M,
is themoving part
of|nR|, Fn
its fixedpart.
By
Theorem 9-1 of[1]
we have that forsufficiently large n,
R · C = 0for any irreducible
component
C ofFn,
because R isnumerically positive.
Replacing
Rby
a suitablemultiple
we can assume thatwhere A is the
moving part of |R|,
F + G its fixedpart
such thatR - F = R · G =
0,
and all the irreduciblecomponents
of F are containedin e,
while none of the irreduciblecomponents
of G are contained int.
By
thehypothesis
for any i ~{1,...,m}
we haveCi.
G = 0.Because nR --- n0394 + nF +
nG,
we can assumethat t
is a fixed compo- nentof |n(0394
+F)| for infinitely
many n. But if R’ = à + F we haveBy
the definitionof e
andapplying
theproof
of Lemma4,
thereexists ro
such that for r
/o, rL
+R’l
doesn’thave e
as a fixedcomponent.
Let
{P1,..., P, 1
be the basepoints of |0394|.
Then there exists a smooth irreducible curveE ~ L such
that:For any
i, j E 1B1
we denoteby Hij
the trace on E of the linearsystems
|iL + jR’1
on S. Then it follows from the choice of E(recall
thatIR’I
hasonly
fixedcomponents
int)
that:(B) Hlo
andHo,
are free from basepoints.
By
the Theorem 4-2(Relation 24)
of[1]
we have for a suitableinteger N,
It follows that
|iL + jR’|
isspanned by
thefollowing
twosubsystems
242
where the second
system
has E as a fixedcomponent.
Forfixed j
thelinear
system |iL + jR’1
has no basepoint
if i issufficiently large by
theproof
of Lemma 4. Letthen j
a fixedinteger >
N and let i be such thatliL + jR’|
has no basepoints.
If P is any basepoint
ofIR’I
then P is also abase
point
of the first of the two linearsystems (1).
Therefore P is not a base
point of j(1
-1)L + jR/I. Applying
the sameargument
to this lastsystem (if
i - 1 >0),
we find that P is not a basepoint of j( 1 - 2) L + jR’I. Ultimately
we reach the conclusion that P is not a basepoint of IjR’1 (if j > N).
DReferences
[1] O. ZARISKI: The theorem of Riemann Roch for high multiples of an effective divisor on an algebraic surface. Ann. Math. 76 (3) (1962) 560-615.
[2] M. ARTIN: On isolated rational singularities of surfaces. Amer. J. Math. 88 (1966)
129-136.
(Oblatum 20-11-1982 & 21-111-1983) Mathematics Department
Centro de
Investigaciôn
y de Estudios Avanzados del IPN.México, D.F. 07000