本期开始分组发送邮件,搭载数学类学院等链接。 (接前: 16 15 10) “执行定理” 的证明(e+). .
By Proposition 5.9, there is a Q-divisor Λ ≥ 0 such that nΛ is integral, mA - Λ is ample, (X, Λ) is lc near x, and T is a lc place of (X, Λ).
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18 19 20 15 21 24 25 27 28 35
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para 5.1 (Pro.5.9)
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36 37 38 39 40
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注: 第2~4段凑齐了条件, 从而 invoke 命题5.9.
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18 ~ 35: (已定义).
---- 18,19 联立相当于a.即(X, B) eps-lc, proj.Qf.
---- 20: A.
(18~20是替换来的, 满足原有条件, 附带修正).
---- 15, 21: A - B, A - C ample.
(21系C, 减半的A, 由它产生 L, 则 A - L ample)
---- 24: L.
---- 25, 27: (X, B +sL) eps'-lc, s (≤1) ≤ r.
---- 28: T, 即 a(T, X, B + sL) = eps'.
---- 35: x, centre of T, is a closed point.
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36 ~ 40: 调用命题5.9得到的输出.
---- 36: 存在 Q-divisor Λ ≥ 0.
---- 37: nΛ integral.
---- 38: mA - Λ ample.
---- 39: (X, Λ) lc near x.
---- 40: T is lc place of (X, Λ).
(37~40系约束条件).
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评论: 这句话只是表述5.9的输出, 而输入来自前几段(2~4).
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Replacing A, C, L, s with 2mA, 2mC, 2mL, s/(2m), respectively, and replacing r accordingly, we can assume A - B - sL and A - Λ are ample.
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38 20' 21' 24' 25' r'
41 42
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注: 输入端带撇数字表示替换操作.
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---- 20', 21', 24', 25': A, C, L, s 替换为 2mA, 2mC, 2mL, s/(2m).
(38 指示 m 的来源, r' 指示 r 是替换的).
---- 41: A - (B + sL) ample.
---- 42: A - Λ ample.
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Applying Proposition 5.7 to (X, B + sL), there is a natural number q depending only on d, r, n, eps', such that if ν: U --> X is a resolution so that T is a divisor on U, then μTν*L ≤ q.
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25' 20' 24' 36 37 41 42 21' 39 40 28
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para 5.3 (Pro.5.7)
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43 44
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注: 凑齐了条件, 调用命题5.7.
---- 输入: (已定义)
---- 43: ν: U --> X.
---- 44: q, 即 μTν*L ≤ q.
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小结: 完成第5段的逻辑模块图.
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温习: 命题 5.9 与 命题 5.7.(图解)
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1) 5.9.
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A L
Tx
X B|t
↓
(m)A T
. x
X (n)Λ
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2) 5.7.
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符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈ ∉ ∪ ∩ ⊆ ⊇ ⊂ ⊃ ≤ ≥ ⌊ ⌋ ⌈ ⌉ ≠ ≡ ⁻⁰ ¹ ² ³ ᵈ ⁺ ₊ ₀ ₁ ₂ ₃ ᵢ .