GENERICALLY AMPLE DIVISORS ON NORMAL GORENSTEIN SURFACES 17
By the hypothesis that
g(L)
=
h1•0(S) we conclude that r(wc) ;;
H1(w8). Since C can be chosen to be a general element of ILl, we
conclude that the map r(w8
®
L) ... f(wc) can be the 0 map only if
r(w8
®
L)
=
0. Therefore we conclude that S has only rational sin-
gularities by Corollary (0.2.2) of [So2]. From this it is easy to see
that
-
is an isomorphism where
p :
S ... S is a desingularization of S.
Therefore we can assume that S is smooth without loss of
generality.
If
g(L)
=
0, then (w8
+
L) ·L
0 and therefore h0((w8
®
L)n)
=
0
for all
n
0 and hence ln(S,L)
= -.
Therefore we can assume that h1•0(S) 0. We know that the
Albanese mapping a :
S ... R
has a one dimensional image since
h2(w8) h0(w8
®
L)
=
0. The general fibre,
I,
if a is smooth and
connected. Being a general fibre of a map between smooth
varieties we also know that w8,f
;; Wr
If we can show that the
L
·I
=
1 then we will be done. This will follow because
L,
being
spanned and deg(L,)
=
1 for a irreducible curve
I,
imply that
I
is a
smooth Pc1· From this we conclude that
deg(w8
®
L),
=
deg(w8) + deg(L,)
=
deg(w,) + deg(L,)
=
-2 +
I
0.
To see that
L
·I
=
I
it suffices to show that ac gives a
biholomorphism between C and R. Since the genus of C equals
h1•0(S), the genus of R, this is obvious by the Hurwitz formula if
h1•0(S)
I.
If h1•0(S)
=
I, then the Hurwitz formula tells us that ac
is an unbranched cover. We also know from the vanishing theorem
(0.6) that elements of
ILl
are connected. Using these two facts the
argument of the third paragraph of the proof of Theorem (3.6) of
[So2] applies without change to our situation here. 0
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