# “执行定理”的证明(d)

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(接前: 01 31 30) “执行定理” (Th1.6) 的证明(d).
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* |A - M|R ≠ Ø ~> 只须证 lct(X, B, |A|) 有正下界.
* A ~  有界重 ~> X' --> X 系 X 的 Q-因式化.
* X 非Q-factorial ~> X', B', H' 替换 X, B, H.
* 取 L∈|C| ~> 设最大 s 使 B + sL eps'-lc.
(则只须证 s 有正下界).
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There is a prime divisor T on birational models of X such that a(T, X, B +sL) = eps'.
---- 这句话凑出了5.9的第6个条件, 但相当突兀..
(第5个条件由上一段构造, 见前贴).
---- T 的出现和句尾的式子, 都很突兀...
---- 换句话说, 看不出其中的逻辑起点在哪里.
---- 又, 此处还涉及到 birational models (注意复数形式).
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Let x be the generic point of the centre of T on X.
---- 从 T 的中心取 “generic point” 设为 x.
(这里准备做出5.9的最后一个条件)
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Assume x is not a closed point.
---- 这里做个反向假设.
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Then cutting by general elements of |A| and applying induction...
---- cutting by 可能是做个降维(?), 以便运用归纳假设.
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...there is a positive number v bounded from below away from zero such that (X, B + vL) is lc near x.
---- 在 x 附近做归纳, 得到 v 有正下界.
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Then (X, B + (1 - eps'/eps)vL) is eps'-lc near x, by Lemma 2.3, because B + (1 - eps'/eps)vL = eps'/eps B + (1 - eps'/eps)(B + vL).
---- 句尾的式子, 配对类型的凸组合为:
eps'/eps · eps-lc + (1 - eps'/eps) ·0-lc
=(eps'/eps · eps +  (1 - eps'/eps) ·0 )-lc
= eps'-lc

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In particular, s ≥ (1 - eps'/eps)v.
---- 此不等式源于 s 的最大设定.
(奇怪的是, 到此 s 还不算有正下界?)
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Thus we can assume x is a closed point.
---- 看不出逻辑关系.
(暂不晓得概念 “closed point” ?)
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