Hartshornes Algebraic Geometry1판 [Hartshornespringer저]
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작성일 20-11-07 20:12
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have unwanted cancellation. Suppose (g) has zeroes of order nj at some (or
none, or all) Pj . Then for > maxfnj ; 1g, (f) has poles of higher order at
n > maxf2g 2; gg. Then similar to (Ex. 1.1), we nd an eective divisor
순서
there is a rational function f 2 K(X) having poles (of some order) at each
zeros of (g) cannot cancel the poles at the remaining Pj of (f), so this is
D = n(2P Q) of degree n, l(K D) = 0 (1.3.4), so Riemann-Roch gives
l(D) = n + 1 g > 1. Thus there is an eective divisor D0 such that
D with D + n(r 1)Q nP1 nP2 : : : nPr = (f). Again since (f) has
the diviso
Hartshornes Algebraic Geometry1판 [Hartshornespringer저] 입니다. 다운받기전에 여러 참고하시고 일치하시면 다운받아주세요.그리고 잘못된 자료일시 환불요청 하시면 환불을 받을수 있습니다.
레포트 > 공학,기술계열
nonconstant rational function f 2 K(X), which is regular everywhere except
regular everywhere except at P. Note we cannot control the zeros of f with
D0 D = (f). Since (f) is degree 0 (II 6.10), D0 has degree n, so D0 cannot
of the Pi, and regular elsewhere.
설명
Hartshornes Algebraic Geometry1판 [Hartshornespringer저] 입니다.
Pj 6= Pi. Then the principal divisor (fg) must have a pole at Pi, and the
the pole, or D = nPi for some xed i. In the former case were done, in the
latter use (Ex. 1.1) to get a principal divisor (g) with a pole at Pi. (fg) may
1.2. Again let X be a curve, and let P1; P2; : : : ; Pr 2 X be points. Then
this proof.
솔루션
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have a zero of order large enough to kill the pole of D of order 2n. f is
outside of D, either D cannot have a zero at any Pi large enough to cancel
Proof. We have to be careful. Multiplying functions from (Ex. 1.1) may
from the Pi, and consider the divisor n(P1 + P2 + : : : + Pr (r 1)Q), with
at P.
다운받기전에 여러 참고하시고 일치하시면 다운받아주세요.그리고 잘못된 자료(data)일시 환불요청 하시면 환불을 받을수 있습니다.
Proof. Let X have genus g. Since X is dimension 1, there exists a point
degree 0 (II 6.10), degree of D is n. Since each Pi occurs with order n
4.1 Section 1
1.1. Let X be a curve, and let P 2 X be a point. Then there exists a
Hartshornes Algebraic Geometry1판 [Hartshornespringer저]
Q 2 X;Q 6= P. Pick an n > maxfg; 2g 2; 1g. Then for the divisor
result in zeroes cancelling poles. So proceed as follows: Fix a point Q distinct
다.
