# “执行定理”的证明(e)

This is an in-mail from TYUST.

(接前: 02 01 31“执行定理” (Th1.6)的证明(e).
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A         L
Tx
X       B|Λ
.

* (X, B) ~ proj., eps-lc. (高配)
*        A ~ very ample, A  r. (有界重)
*  L  0, R-divisor.
* Λ ≥ 0, Q-divisor, nΛ integral.
* A - B, Λ, L ~ ample.
. (X, Λ) ~ lc near x.
. T ~ lc place of (X, Λ), with center x on X.
. a(T, X, B)  1.
--- 结果: 对任何ν: U --> X 带 T (on U), 有 μTνL* ≤ q.
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---- 第2~4段凑齐了5.9的7个条件, 这就“激活”了命题5.9.
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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).
---- 命题5.9 就是为了构造如上 Λ.

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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.
---- 替换即代入: A <~ 2mA, C <~2mC, L <~ 2mL, s <~ s/2m.
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---- ample 有点像比大小(大 - 小 = ample).
1） A - Λ <~ 2mA - Λ.

2） A - B -sL <~ 2mA - B - s/2m · 2mL
= 2mA - B - sL = (mA - B) + (mA -sL)

A - B ample ==> mA - B ample
A - L ample ==>  mA - L ample ==> mA -sL ample (因为 s ≤ 1).
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---- 上面这句是凑 ample 条件, 替换后:
---- A - L 和 A - Λ 都是 ample.
---- 原作特意提出 A - (B + sL) ample, 意味着 B + sL 相当于 5.7 中 B.
---- 恰巧, a(X, T, B + sL) = eps' 1.
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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  μTνL* ≤ q.
---- 命题5.7 作用于 (X, B + sL).
(通常来说, 命题的应用不会照搬, 而是套用).
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Pick such a resolution.
---- 补的这句很必要.
(澄清了上句中 resolution 的来源)
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(5.9 是为 5.7 准备条件).
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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 .

#### Glossary(AG)

*

Introduction
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....
11/9
...
Proposition 5.5 11/5

http://blog.sciencenet.cn/blog-315774-1196501.html

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