2 - Dept. Math, Hokkaido Univ. EPrints Server

where 11, E M" CIO(W, Co + 6*(Iic + Dm + D,)), E HO(C, Iic +
Dm + D,) and E see (7) and (8)).
Hence by the expression (13) and (14), the curve A = ~ f=,(~i' + hy)
is a part of the base locus of D.
Set xi0) = X2\A the Zariski open set of X2. We prove that the linear
system 1 Dl separates points on Xi0) ; i.e., For any points El, R2 E X2 (0) ,
one can find 4 t H0(X2, D) such that q5(R1) = 0 and +(R2) # 0. Put
R: = a2(Ri) and R: = 6r(Ri) for i = 1,2.
1) When R;' # Ry : IS the condition (a) holds, then Iic + Dm + D, is
a very ample divisor on C. Thus ir*6*(Iic + Dm + D,) + C g, ( separates
R; and R;. IIence I Dl separates R1 and R2. If the condition (b) holds,
then the system hep pa rates R;' and Ry. Hence 1 Dl also separa.tes Rl
and R2.
2) When R; # R; and R;' = R',' : Since Iic + D,, + D, 2 0 and
lic+2Dm+ D, -E-----& 2 0, the system lr*(Co + b*(ICc+ Dm+
D,) + ~ ( g - , hi))l separates R; and R;. Hence ID1 separates R1 and
R2.
3) When R; = Ry : We consider the restriction map
x : fiO(x, D) -+ IIO(T;~(R;), DI,;I(~~)) 2. c2
By (13), the map X is given by
Thus by the same argument as above, X is surjective. Hence ID1 separates
Rl and 122.
By 1) ,2) and 3), I Dl separates points on Xi0). Therefore 1 Kx, + S2 (
(0)
incluce a biratjonal map on X2, and this map is h.olomorphic on X, .
(0)
Since the generic point of S2 is contained in X2 , IwS2
1 also induces a
biration.al. map on S2. Hence QIG is a birational map.
5
Next we prove that the base locns Bslws51 is empty. By the above
argument, we have Bslws, 1 c U;h;. Since ws5 = (p5)*(a o p3 o
p)*ws @ 0s5(- C; hy), the divisor xi hy is not a fixed component of
Iws, 1. For 1 5 i 5 k, let P:, P: be the points on hy' which are the image
of two exceptional curves (E)-l(g) by ps. For a general member