写清楚了,会觉得根本没必要写出来;但不写出来,分分钟绊住你.
Step2 第二段 (逐句评论).
Let Kw + Λw be the pullback of Kx + Λ.
---- 做回拉仅是为了得到 Λw.
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Since (X, Λ) is lc near x, Λw ≤ Θw over some neighbourhood of x, hence 0 = a(T, X, Λ) = a(T, W, Λw) ≥ a(T, W, Θw) ≥ 0 which shows that T is a lc place of (W, Θw).
---- 核心结果是 0 ≥ a(T, W, Θw) ≥ 0.
---- 由此得到 a(T, W, Θw) = 0.
---- 由此得到 T 是 (W, Θw) 的 lc place.
疑问: Λw ≤ Θw ...?
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Moreover, if C is the centre of T on W and if ω is its generic point, then Λw = Θw near ω.
---- Λw = Θw near ω? 暂时看不出.
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小结:Step2读写完毕.
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温习:Step2(第一段)
先画出图解:
A ?
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X Λ
1. nΛ 是整系数的;
2. (度式)乘法: degAΛ< Aᵈ <=r.
3. (支撑)加法: Supp(Λ + A).
4. 1, 2,3 ==> (X, Supp(Λ + A)) “入格”.
5. 4 ==> 存在 φ: W --> X 及 Aw (very ample).
参另一图解:
Aw ?
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W E(φ)
其中, E(φ) 是φ的所有exceptional divisor的和.
6. E(φ) = Supp(bir(Λ))
7. 6 ==> (W, E(φ) + Aw) “入格”.
注:E(φ) 对应原作给出的记号Θw.