椭圆曲线、除子和直线
拉尔斯·A·瓦伦博恩
2010年6月21日 - 7月19日
目录
1 符号和全局定义
2 椭圆曲线
3 多项式和有理函数
4 零点和极点
5 除子和直线
摘要
该脚本是基于2010年夏季学期由Prof. Nitin Saxena在“计算复杂性中的代数方法”研讨会上所做的报告的基础。它大量借鉴了Leonard S. Charlap和David P. Robbins于1988年的论文[CRD]。我们将给出椭圆曲线在有限域上所有常见的定义、结果以及这些结果的证明。
1 符号和全局定义
对于一个域 K , n ∈ N K,n \in \mathbb{N} K , n ∈ N k ∈ K k \in K k ∈ K
n ⋅ k := k + … + k ⏟ n − times n \cdot k \mathrel{\text{:=}} \underset{n - \text{ times }}{\underbrace{k + \ldots + k}} n ⋅ k := n − times k + … + k 一个域 K K K
char ( K ) := { 0 for C k = ∅ min ( C K ) else \operatorname{char}\left( K\right) \mathrel{\text{:=}} \left\{ \begin{matrix} 0 & \text{ for }{C}_{k} = \varnothing \\ \min \left( {C}_{K}\right) & \text{ else } \end{matrix}\right. char ( K ) := { 0 min ( C K ) for C k = ∅ else 伴随 C k := { p ∈ N > 0 ∣ p ⋅ 1 = 0 , 1 ∈ K additive neutral } {C}_{k} \mathrel{\text{:=}} \left\{ {p \in {\mathbb{N}}_{ > 0} \mid p \cdot 1 = 0,1 \in K\text{additive neutral}}\right\} C k := { p ∈ N > 0 ∣ p ⋅ 1 = 0 , 1 ∈ K additive neutral }
命题 1.1. char(K) 要么是 0 要么是素数。
全局定义/符号 1.2. 从现在起让
K K K char ( K ) ∉ { 2 , 3 } \operatorname{char}\left( K\right) \notin \{ 2,3\} char ( K ) ∈ / { 2 , 3 }
字母 X X X Y Y Y
K [ X ] K\left\lbrack X\right\rbrack K [ X ] K [ X , Y ] K\left\lbrack {X,Y}\right\rbrack K [ X , Y ]
K ( X ) K\left( X\right) K ( X ) K ( X , Y ) K\left( {X,Y}\right) K ( X , Y )
2 椭圆曲线
定义2.1（消失集）。对于 f ∈ K [ X , Y ] f \in K\left\lbrack {X,Y}\right\rbrack f ∈ K [ X , Y ]
V ( f ) := { ( a , b ) ∈ K 2 ∣ f ( a , b ) = 0 } V\left( f\right) \mathrel{\text{:=}} \left\{ {\left( {a,b}\right) \in {K}^{2} \mid f\left( {a,b}\right) = 0}\right\} V ( f ) := { ( a , b ) ∈ K 2 ∣ f ( a , b ) = 0 } 定义2.2（椭圆曲线）。对于 A , B ∈ K A,B \in K A , B ∈ K
E := E A , B := V ( Y 2 − X 3 − A X − B ) ∪ { O } E \mathrel{\text{:=}} {E}_{A,B} \mathrel{\text{:=}} V\left( {{Y}^{2} - {X}^{3} - {AX} - B}\right) \cup \{ \mathcal{O}\} E := E A , B := V ( Y 2 − X 3 − A X − B ) ∪ { O } 称为 K K K s ( x ) := s A , B ( x ) := x 3 + A x + B s\left( x\right) \mathrel{\text{:=}} {s}_{A,B}\left( x\right) \mathrel{\text{:=}} {x}^{3} + {Ax} + B s ( x ) := s A , B ( x ) := x 3 + A x + B O ∈ E \mathcal{O} \in E O ∈ E ∈ E ∖ { O } \in E \smallsetminus \{ \mathcal{O}\} ∈ E ∖ { O } P = ( a , b ) ∈ E P = \left( {a,b}\right) \in E P = ( a , b ) ∈ E − P - P − P
Δ ( E A , B ) := − 4 A 3 − 27 B 2 \Delta \left( {E}_{A,B}\right) \mathrel{\text{:=}} - 4{A}^{3} - {27}{B}^{2} Δ ( E A , B ) := − 4 A 3 − 27 B 2 被称为判别式。
备注2.3。有时人们将 E A , B {E}_{A,B} E A , B s A , B {s}_{A,B} s A , B
定义2.4（二阶点）。设 E A , B {E}_{A,B} E A , B ω 1 , ω 2 , ω 3 {\omega }_{1},{\omega }_{2},{\omega }_{3} ω 1 , ω 2 , ω 3 s A , B ( x ) {s}_{A,B}\left( x\right) s A , B ( x ) Ω i := ( ω i , 0 ) ∈ E A , B {\Omega }_{i} \mathrel{\text{:=}} \left( {{\omega }_{i},0}\right) \in {E}_{A,B} Ω i := ( ω i , 0 ) ∈ E A , B
命题2.5。对于任意的 f ( x ) = x 3 + A x + B f\left( x\right) = {x}^{3} + {Ax} + B f ( x ) = x 3 + A x + B A , B ∈ K A,B \in K A , B ∈ K ω 1 , ω 2 {\omega }_{1},{\omega }_{2} ω 1 , ω 2 ω 3 {\omega }_{3} ω 3
0 = ω 1 + ω 2 + ω 3 0 = {\omega }_{1} + {\omega }_{2} + {\omega }_{3} 0 = ω 1 + ω 2 + ω 3
A = ω 2 ω 3 + ω 1 ω 3 + ω 1 ω 2 A = {\omega }_{2}{\omega }_{3} + {\omega }_{1}{\omega }_{3} + {\omega }_{1}{\omega }_{2} A = ω 2 ω 3 + ω 1 ω 3 + ω 1 ω 2
B = − ω 1 ω 2 ω 3 B = - {\omega }_{1}{\omega }_{2}{\omega }_{3} B = − ω 1 ω 2 ω 3
证明。由于 K K K
f ( x ) = ( x − ω 1 ) ( x − ω 2 ) ( x − ω 3 ) f\left( x\right) = \left( {x - {\omega }_{1}}\right) \left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) f ( x ) = ( x − ω 1 ) ( x − ω 2 ) ( x − ω 3 ) = x 3 + x 2 ( − ω 1 − ω 2 − ω 3 ) + x ( ω 2 ω 3 + ω 1 ω 3 + ω 1 ω 2 ) − ω 1 ω 2 ω 3 = {x}^{3} + {x}^{2}\left( {-{\omega }_{1} - {\omega }_{2} - {\omega }_{3}}\right) + x\left( {{\omega }_{2}{\omega }_{3} + {\omega }_{1}{\omega }_{3} + {\omega }_{1}{\omega }_{2}}\right) - {\omega }_{1}{\omega }_{2}{\omega }_{3} = x 3 + x 2 ( − ω 1 − ω 2 − ω 3 ) + x ( ω 2 ω 3 + ω 1 ω 3 + ω 1 ω 2 ) − ω 1 ω 2 ω 3 将系数与 x 3 + A x + B {x}^{3} + {Ax} + B x 3 + A x + B
命题2.6（椭圆曲线准则）。集合 E A , B {E}_{A,B} E A , B Δ ( E A , B ) ≠ 0 \Delta \left( {E}_{A,B}\right) \neq 0 Δ ( E A , B ) = 0
证明。我们将证明如果 E A , B {E}_{A,B} E A , B s A , B {s}_{A,B} s A , B Δ ( E A , B ) = 0 \Delta \left( {E}_{A,B}\right) = 0 Δ ( E A , B ) = 0
假设 s A , B {s}_{A,B} s A , B ω 1 {\omega }_{1} ω 1
0 = 2 ω 1 + ω 2 0 = 2{\omega }_{1} + {\omega }_{2} 0 = 2 ω 1 + ω 2 A = 2 ω 1 ω 2 + ω 1 2 A = 2{\omega }_{1}{\omega }_{2} + {\omega }_{1}^{2} A = 2 ω 1 ω 2 + ω 1 2 B = − ω 1 2 ω 2 B = - {\omega }_{1}^{2}{\omega }_{2} B = − ω 1 2 ω 2 从第一个关系中我们得到 ω 2 = − 2 ω 1 {\omega }_{2} = - 2{\omega }_{1} ω 2 = − 2 ω 1
A = 2 ω 1 ( − 2 ω 1 ) + ω 1 2 = − 4 ω 1 2 + ω 1 2 = − 3 ω 1 2 \begin{matrix} A & = & 2{\omega }_{1}\left( {-2{\omega }_{1}}\right) + {\omega }_{1}^{2} = - 4{\omega }_{1}^{2} + {\omega }_{1}^{2} = - 3{\omega }_{1}^{2} \end{matrix} A = 2 ω 1 ( − 2 ω 1 ) + ω 1 2 = − 4 ω 1 2 + ω 1 2 = − 3 ω 1 2 B = − ω 1 2 ( − 2 ω 1 ) = 2 ω 1 3 \begin{matrix} B & = & - {\omega }_{1}^{2}\left( {-2{\omega }_{1}}\right) = 2{\omega }_{1}^{3} \end{matrix} B = − ω 1 2 ( − 2 ω 1 ) = 2 ω 1 3 最后我们得到
Δ ( E A , B ) = − 27 B 2 − 4 A 3 = − 27 ( 2 ω 1 3 ) 2 − 4 ( − 3 ω 1 2 ) 3 = − 108 ω 1 6 + 108 ω 1 6 = 0 \Delta \left( {E}_{A,B}\right) = - {27}{B}^{2} - 4{A}^{3} = - {27}{\left( 2{\omega }_{1}^{3}\right) }^{2} - 4{\left( -3{\omega }_{1}^{2}\right) }^{3} = - {108}{\omega }_{1}^{6} + {108}{\omega }_{1}^{6} = 0 Δ ( E A , B ) = − 27 B 2 − 4 A 3 = − 27 ( 2 ω 1 3 ) 2 − 4 ( − 3 ω 1 2 ) 3 = − 108 ω 1 6 + 108 ω 1 6 = 0 假设 s A , B {s}_{A,B} s A , B
假设 Δ ( E A , B ) = 0 \Delta \left( {E}_{A,B}\right) = 0 Δ ( E A , B ) = 0
0 = Δ ( E ) = − 27 B 2 − 4 A 3 0 = \Delta \left( E\right) = - {27}{B}^{2} - 4{A}^{3} 0 = Δ ( E ) = − 27 B 2 − 4 A 3 ⇔ − 27 B 2 8 A 3 = 1 2 \Leftrightarrow \;\frac{-{27}{B}^{2}}{8{A}^{3}} = \frac{1}{2} ⇔ 8 A 3 − 27 B 2 = 2 1 ⇔ 0 = − 27 B 2 8 A 3 − 1 2 \Leftrightarrow \;0 = \frac{-{27}{B}^{2}}{8{A}^{3}} - \frac{1}{2} ⇔ 0 = 8 A 3 − 27 B 2 − 2 1 ⇒ 0 = ( − 27 B 2 8 A 3 − 1 2 ) B = − 27 B 3 8 A 3 − 3 B 2 + B = s A , B ( − 3 B 2 A ) \Rightarrow \;0 = \left( {\frac{-{27}{B}^{2}}{8{A}^{3}} - \frac{1}{2}}\right) B = \frac{-{27}{B}^{3}}{8{A}^{3}} - \frac{3B}{2} + B = {s}_{A,B}\left( \frac{-{3B}}{2A}\right) ⇒ 0 = ( 8 A 3 − 27 B 2 − 2 1 ) B = 8 A 3 − 27 B 3 − 2 3 B + B = s A , B ( 2 A − 3 B ) 所以我们知道 x 1 := − 3 B 2 A {x}_{1} \mathrel{\text{:=}} \frac{-{3B}}{2A} x 1 := 2 A − 3 B s A , B {s}_{A,B} s A , B
( x 3 + A x + B ) : ( x + 3 B 2 A ) = x 2 − 3 B 2 A x + ( A + 9 B 2 4 A 2 ) \left( {{x}^{3} + {Ax} + B}\right) : \left( {x + \frac{3B}{2A}}\right) = {x}^{2} - \frac{3B}{2A}x + \left( {A + \frac{9{B}^{2}}{4{A}^{2}}}\right) ( x 3 + A x + B ) : ( x + 2 A 3 B ) = x 2 − 2 A 3 B x + ( A + 4 A 2 9 B 2 ) p p p
x 2 , 3 = − − 3 B 2 A 2 ± ( 3 B 2 A ) 2 4 − ( A + 9 B 2 4 A 2 ) {x}_{2,3} = - \frac{-\frac{3B}{2A}}{2} \pm \sqrt{\frac{{\left( \frac{3B}{2A}\right) }^{2}}{4} - \left( {A + \frac{9{B}^{2}}{4{A}^{2}}}\right) } x 2 , 3 = − 2 − 2 A 3 B ± 4 ( 2 A 3 B ) 2 − ( A + 4 A 2 9 B 2 ) = − 3 B 4 A ± 9 B 2 16 A 2 − A − 36 B 2 16 A 2 = \; - \frac{3B}{4A} \pm \sqrt{\frac{9{B}^{2}}{{16}{A}^{2}} - A - \frac{{36}{B}^{2}}{{16}{A}^{2}}} = − 4 A 3 B ± 16 A 2 9 B 2 − A − 16 A 2 36 B 2 = − 3 B 4 A ± − 27 B 2 16 A 2 − A = \; - \frac{3B}{4A} \pm \sqrt{\frac{-{27}{B}^{2}}{{16}{A}^{2}} - A} = − 4 A 3 B ± 16 A 2 − 27 B 2 − A 现在假设 s A , B {s}_{A,B} s A , B
- x 2 = x 3 or \text{-}{x}_{2} = {x}_{3}\text{or} - x 2 = x 3 or - x 1 = x 2 or x 1 = x 3 \text{-}{x}_{1} = {x}_{2}\text{or}{x}_{1} = {x}_{3} - x 1 = x 2 or x 1 = x 3 第一种情况意味着根号下的项为零：
− 27 B 2 16 A 2 − A = 0 ⇔ − 27 B 2 − 16 A 3 = 0 \frac{-{27}{B}^{2}}{{16}{A}^{2}} - A = 0 \Leftrightarrow - {27}{B}^{2} - {16}{A}^{3} = 0 16 A 2 − 27 B 2 − A = 0 ⇔ − 27 B 2 − 16 A 3 = 0 结合 − 27 B 2 − 4 A 3 = 0 - {27}{B}^{2} - 4{A}^{3} = 0 − 27 B 2 − 4 A 3 = 0 A = B = 0 A = B = 0 A = B = 0
对于后两种情况，我们计算：
− 3 B 2 A = − 3 B 4 A ± − 27 B 2 16 A 2 − A \begin{matrix} - \frac{3B}{2A} & = & - \frac{3B}{4A} \pm \sqrt{\frac{-{27}{B}^{2}}{{16}{A}^{2}} - A} \end{matrix} − 2 A 3 B = − 4 A 3 B ± 16 A 2 − 27 B 2 − A − 3 B 4 A = ± − 27 B 2 16 A 2 − A \begin{matrix} - \frac{3B}{4A} & = & \pm \sqrt{\frac{-{27}{B}^{2}}{{16}{A}^{2}} - A} \end{matrix} − 4 A 3 B = ± 16 A 2 − 27 B 2 − A 9 B 2 16 A 2 = − 27 B 2 16 A 2 − A \begin{matrix} \frac{9{B}^{2}}{{16}{A}^{2}} & = & \frac{-{27}{B}^{2}}{{16}{A}^{2}} - A \end{matrix} 16 A 2 9 B 2 = 16 A 2 − 27 B 2 − A 9 B 2 = − 27 B 2 − 16 A 3 9{B}^{2} = - {27}{B}^{2} - {16}{A}^{3} 9 B 2 = − 27 B 2 − 16 A 3 0 = − 36 B 2 − 16 A 3 0 = - {36}{B}^{2} - {16}{A}^{3} 0 = − 36 B 2 − 16 A 3 0 = − 9 B 2 − 4 A 3 0 = - 9{B}^{2} - 4{A}^{3} 0 = − 9 B 2 − 4 A 3 与 − 27 B 2 − 4 A 3 = 0 - {27}{B}^{2} - 4{A}^{3} = 0 − 27 B 2 − 4 A 3 = 0 A = B = 0 A = B = 0 A = B = 0
命题 2.7。椭圆曲线是无限的。
证明。假设 E A , B {E}_{A,B} E A , B K K K a ∈ K a \in K a ∈ K ∀ b ∈ K : ( a , b ) ∉ E A , B \forall b \in K : \left( {a,b}\right) \notin {E}_{A,B} ∀ b ∈ K : ( a , b ) ∈ / E A , B ∀ b ∈ K : ( a , b ) ∉ E A , B \forall b \in K : \left( {a,b}\right) \notin {E}_{A,B} ∀ b ∈ K : ( a , b ) ∈ / E A , B ∄ b ∈ K : b 2 = c \nexists b \in K : {b}^{2} = c ∄ b ∈ K : b 2 = c c = a 3 + A a + B c = {a}^{3} + {Aa} + B c = a 3 + A a + B K K K X 2 − c {X}^{2} - c X 2 − c
定义 2.8。对于一个子域 k ⊆ K k \subseteq K k ⊆ K A , B ∈ k A,B \in k A , B ∈ k
E ( k ) := { ( a , b ) ∈ E A , B ∣ a , b ∈ k } ∪ { O } E\left( k\right) \mathrel{\text{:=}} \left\{ {\left( {a,b}\right) \in {E}_{A,B} \mid a,b \in k}\right\} \cup \{ \mathcal{O}\} E ( k ) := { ( a , b ) ∈ E A , B ∣ a , b ∈ k } ∪ { O } 被称为 k k k
注释 2.9。当 char ( K ) ∈ { 2 , 3 } \operatorname{char}\left( K\right) \in \{ 2,3\} char ( K ) ∈ { 2 , 3 }
k 2 + a 1 h k + a 3 k = h 3 + a 2 h 2 + a 4 h + a 6 {k}^{2} + {a}_{1}{hk} + {a}_{3}k = {h}^{3} + {a}_{2}{h}^{2} + {a}_{4}h + {a}_{6} k 2 + a 1 hk + a 3 k = h 3 + a 2 h 2 + a 4 h + a 6 但在我们的情况下 ( char ( K ) ∉ { 2 , 3 } ) \left( {\operatorname{char}\left( K\right) \notin \{ 2,3\} }\right) ( char ( K ) ∈ / { 2 , 3 } )
3 多项式和有理函数
定义 3.1（椭圆曲线上的多项式）。对于一个椭圆曲线 E = E A , B E = {E}_{A,B} E = E A , B E E E E = E A , B E = {E}_{A,B} E = E A , B
K [ E ] := K [ X , Y ] / ⟨ Y 2 − X 3 − A X − B ⟩ K\left\lbrack E\right\rbrack \mathrel{\text{:=}} K\left\lbrack {X,Y}\right\rbrack /\left\langle {{Y}^{2} - {X}^{3} - {AX} - B}\right\rangle K [ E ] := K [ X , Y ] / ⟨ Y 2 − X 3 − A X − B ⟩ 全局定义/符号 3.2。现在起让小写字母 x x x y y y x ( a , b ) := a x\left( {a,b}\right) \mathrel{\text{:=}} a x ( a , b ) := a y ( a , b ) := b y\left( {a,b}\right) \mathrel{\text{:=}} b y ( a , b ) := b E E E y 2 = s ( x ) {y}^{2} = s\left( x\right) y 2 = s ( x ) K [ E ] = K [ x , y ] K\left\lbrack E\right\rbrack = K\left\lbrack {x,y}\right\rbrack K [ E ] = K [ x , y ]
注释 3.3。通过取商意味着我们可以用一个项 X 3 + A X + B {X}^{3} + {AX} + B X 3 + A X + B f ∈ K [ X , Y ] f \in K\left\lbrack {X,Y}\right\rbrack f ∈ K [ X , Y ] Y 2 {Y}^{2} Y 2 f f f f f f f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) v , w ∈ K [ X ] v,w \in K\left\lbrack X\right\rbrack v , w ∈ K [ X ]
记号 3.4（规范形式）。一个多项式 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ] f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x )
备注 3.5。规范形式是唯一的。
证明。设 f ( x , y ) = v ~ ~ ( x ) + y w ~ ~ ( x ) = v ~ ( x ) + y w ~ ( x ) f\left( {x,y}\right) = \widetilde{\widetilde{v}}\left( x\right) + y\widetilde{\widetilde{w}}\left( x\right) = \widetilde{v}\left( x\right) + y\widetilde{w}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) = v ( x ) + y w ( x ) v ~ ‾ ( x ) − v ~ ( x ) ) + y ( w ~ ~ ( x ) − w ~ ( x ) ) = 0 \overline{\widetilde{v}}\left( x\right) - \widetilde{v}\left( x\right) ) + y\left( {\widetilde{\widetilde{w}}\left( x\right) - \widetilde{w}\left( x\right) }\right) = 0 v ( x ) − v ( x ) ) + y ( w ( x ) − w ( x ) ) = 0 v ( x ) = v ~ ~ ( x ) − v ~ ( x ) v\left( x\right) = \widetilde{\widetilde{v}}\left( x\right) - \widetilde{v}\left( x\right) v ( x ) = v ( x ) − v ( x ) w ( x ) = w ~ ~ ( x ) − w ~ ( x ) w\left( x\right) = \widetilde{\widetilde{w}}\left( x\right) - \widetilde{w}\left( x\right) w ( x ) = w ( x ) − w ( x ) v ( x ) + y w ( x ) = 0 v\left( x\right) + {yw}\left( x\right) = 0 v ( x ) + y w ( x ) = 0 v ( x ) = w ( x ) = 0 v\left( x\right) = w\left( x\right) = 0 v ( x ) = w ( x ) = 0
0 = 0 ⋅ ( v ( x ) − y w ( x ) ) 0 = 0 \cdot \left( {v\left( x\right) - {yw}\left( x\right) }\right) 0 = 0 ⋅ ( v ( x ) − y w ( x ) ) = ( v ( x ) + y w ( x ) ) ⋅ ( v ( x ) − y w ( x ) ) = \left( {v\left( x\right) + {yw}\left( x\right) }\right) \cdot \left( {v\left( x\right) - {yw}\left( x\right) }\right) = ( v ( x ) + y w ( x ) ) ⋅ ( v ( x ) − y w ( x ) ) = v 2 ( x ) − y 2 w 2 ( x ) = {v}^{2}\left( x\right) - {y}^{2}{w}^{2}\left( x\right) = v 2 ( x ) − y 2 w 2 ( x ) = v 2 ( x ) + ( − s ( x ) ) w 2 ( x ) = {v}^{2}\left( x\right) + \left( {-s\left( x\right) }\right) {w}^{2}\left( x\right) = v 2 ( x ) + ( − s ( x ) ) w 2 ( x ) 由于 deg x ( s ) {\deg }_{x}\left( s\right) deg x ( s ) deg x ( v 2 ) {\deg }_{x}\left( {v}^{2}\right) deg x ( v 2 ) deg x ( w 2 ) {\deg }_{x}\left( {w}^{2}\right) deg x ( w 2 ) w w w v v v
定义 3.6（共轭和范数）。将 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ] f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) f ‾ \underline{f} f f ˉ ( x , y ) := v ( x ) − y w ( x ) \bar{f}\left( {x,y}\right) \mathrel{\text{:=}} v\left( x\right) - {yw}\left( x\right) f ˉ ( x , y ) := v ( x ) − y w ( x ) f ‾ \underline{f} f N f := f ⋅ f ˉ {N}_{f} \mathrel{\text{:=}} f \cdot \bar{f} N f := f ⋅ f ˉ
备注 3.7。
可以计算 N f = v 2 ( x ) − s ( x ) w 2 ( x ) {N}_{f} = {v}^{2}\left( x\right) - s\left( x\right) {w}^{2}\left( x\right) N f = v 2 ( x ) − s ( x ) w 2 ( x ) N f ∈ K [ X ] {N}_{f} \in K\left\lbrack X\right\rbrack N f ∈ K [ X ]
因为我们可以轻易看出 f g ‾ = f ˉ g ˉ \overline{fg} = \bar{f}\bar{g} f g = f ˉ g ˉ N f g = N f N g {N}_{fg} = {N}_{f}{N}_{g} N f g = N f N g
定义 3.8（椭圆曲线上的有理函数）。对于椭圆曲线 E E E E E E
K ( E ) := K [ E ] 2 / ∼ K\left( E\right) \mathrel{\text{:=}} K{\left\lbrack E\right\rbrack }^{2}/ \sim K ( E ) := K [ E ] 2 / ∼ 满足以下等价关系：对于 ( f , g ) , ( h , k ) ∈ K [ E ] 2 \left( {f,g}\right) ,\left( {h,k}\right) \in K{\left\lbrack E\right\rbrack }^{2} ( f , g ) , ( h , k ) ∈ K [ E ] 2
( f , g ) ∼ ( h , k ) : ⇔ f ⋅ k = g ⋅ h \left( {f,g}\right) \sim \left( {h,k}\right) : \Leftrightarrow f \cdot k = g \cdot h ( f , g ) ∼ ( h , k ) :⇔ f ⋅ k = g ⋅ h (为了检查相等性，可以将 f ⋅ k f \cdot k f ⋅ k g ⋅ h g \cdot h g ⋅ h f g \frac{f}{g} g f ( f , g ) ∈ K ( E ) \left( {f,g}\right) \in K\left( E\right) ( f , g ) ∈ K ( E ) r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E ) P ∈ E P \in E P ∈ E r = f g r = \frac{f}{g} r = g f f , g ∈ K [ E ] f,g \in K\left\lbrack E\right\rbrack f , g ∈ K [ E ] g ( P ) ≠ 0 g\left( P\right) \neq 0 g ( P ) = 0 r r r P P P r ( P ) := f ( P ) g ( P ) r\left( P\right) \mathrel{\text{:=}} \frac{f\left( P\right) }{g\left( P\right) } r ( P ) := g ( P ) f ( P ) r r r P P P r ( P ) = ∞ r\left( P\right) = \infty r ( P ) = ∞
注释 3.9（有理函数的规范形式）。可以计算
r = f g ∈ K ( E ) : r = \frac{f}{g} \in K\left( E\right) : r = g f ∈ K ( E ) :
f g = f g ˉ g g ˉ = f g ˉ N g \frac{f}{g} = \frac{f\bar{g}}{g\bar{g}} = \frac{f\bar{g}}{{N}_{g}} g f = g g ˉ f g ˉ = N g f g ˉ 将 ( f g ˉ ) ( x , y ) = v ( x ) + y w ( x ) \left( {f\bar{g}}\right) \left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) ( f g ˉ ) ( x , y ) = v ( x ) + y w ( x )
f g = v ( x ) + y w ( x ) N g = v ( x ) N g + y w ( x ) N g \frac{f}{g} = \frac{v\left( x\right) + {yw}\left( x\right) }{{N}_{g}} = \frac{v\left( x\right) }{{N}_{g}} + y\frac{w\left( x\right) }{{N}_{g}} g f = N g v ( x ) + y w ( x ) = N g v ( x ) + y N g w ( x ) 因此每个有理函数也可以写成规范形式。
命题 3.10。在 P ∈ E P \in E P ∈ E
证明。我们想要证明
R P := { r ∈ K ( E ) ∣ r is finite at P } {R}_{P} \mathrel{\text{:=}} \{ r \in K\left( E\right) \mid r\text{ is finite at }P\} R P := { r ∈ K ( E ) ∣ r is finite at P } 以及逐点加法和乘法构成一个环。加法和乘法的结合性和交换性以及分配性继承自基础字段。元素 0 1 , 1 1 ∈ R P \frac{0}{1},\frac{1}{1} \in {R}_{P} 1 0 , 1 1 ∈ R P P P P
− f g = − f g . - \frac{f}{g} = \frac{-f}{g}. − g f = g − f .
在下面的内容中，我们想要定义有理函数在 O \mathcal{O} O f ∈ K ( X ) f \in K\left( X\right) f ∈ K ( X ) ∞ \infty ∞ y 2 = x 3 + A x + B {y}^{2} = {x}^{3} + {Ax} + B y 2 = x 3 + A x + B y y y x x x 2 3 \frac{2}{3} 3 2 y y y x x x f ∈ K [ X ] f \in K\left\lbrack X\right\rbrack f ∈ K [ X ] deg x ( f ) {\deg }_{x}\left( f\right) deg x ( f )
定义 3.11（多项式的次数）。设 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ] f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) f f f
deg ( f ) := max { 2 ⋅ deg x ( v ) , 3 + 2 ⋅ deg x ( w ) } \deg \left( f\right) \mathrel{\text{:=}} \max \left\{ {2 \cdot {\deg }_{x}\left( v\right) ,3 + 2 \cdot {\deg }_{x}\left( w\right) }\right\} deg ( f ) := max { 2 ⋅ deg x ( v ) , 3 + 2 ⋅ deg x ( w ) } 备注 3.12。回顾 deg x ( 0 ) = − ∞ {\deg }_{x}\left( 0\right) = - \infty deg x ( 0 ) = − ∞ deg x ( c ) = 0 ∀ c ∈ K ∖ { 0 } {\deg }_{x}\left( c\right) = 0\forall c \in K \smallsetminus \{ 0\} deg x ( c ) = 0∀ c ∈ K ∖ { 0 }
多项式的经典次数与多项式在 E E E
定理3.13（度数与经典度数的连接）。对于 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ]
deg ( f ) = deg x ( N f ) \deg \left( f\right) = {\deg }_{x}\left( {N}_{f}\right) deg ( f ) = deg x ( N f ) 证明。将 f f f f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) N f = v 2 ( x ) − {N}_{f} = {v}^{2}\left( x\right) - N f = v 2 ( x ) − s ( x ) w 2 ( x ) ‾ s\left( x\right) \overline{{w}^{2}\left( x\right) } s ( x ) w 2 ( x ) deg x ( v 2 ) {\deg }_{x}\left( {v}^{2}\right) deg x ( v 2 ) deg x ( w 2 ) {\deg }_{x}\left( {w}^{2}\right) deg x ( w 2 ) deg x ( s ) {\deg }_{x}\left( s\right) deg x ( s )
那就是
deg x ( N f ) = deg x ( v 2 ( x ) − s ( x ) w 2 ( x ) ) {\deg }_{x}\left( {N}_{f}\right) = {\deg }_{x}\left( {{v}^{2}\left( x\right) - s\left( x\right) {w}^{2}\left( x\right) }\right) deg x ( N f ) = deg x ( v 2 ( x ) − s ( x ) w 2 ( x ) ) = max { deg x ( v 2 ) , deg x ( s ) + deg x ( w 2 ) } = \max \left\{ {{\deg }_{x}\left( {v}^{2}\right) ,{\deg }_{x}\left( s\right) + {\deg }_{x}\left( {w}^{2}\right) }\right\} = max { deg x ( v 2 ) , deg x ( s ) + deg x ( w 2 ) } = max { 2 ⋅ deg x ( v ) , 3 + 2 ⋅ deg x ( w ) } = \max \left\{ {2 \cdot {\deg }_{x}\left( v\right) ,3 + 2 \cdot {\deg }_{x}\left( w\right) }\right\} = max { 2 ⋅ deg x ( v ) , 3 + 2 ⋅ deg x ( w ) } = deg ( f ) = \deg \left( f\right) = deg ( f ) 此外，在3.11中定义的度数具有我们期望度数应具有的基本性质：
命题3.14（多项式次数的性质）。对于 f , g ∈ K [ E ] f,g \in K\left\lbrack E\right\rbrack f , g ∈ K [ E ]
deg ( f ⋅ g ) = deg ( f ) + deg ( g ) \deg \left( {f \cdot g}\right) = \deg \left( f\right) + \deg \left( g\right) deg ( f ⋅ g ) = deg ( f ) + deg ( g ) 证明。我们可以轻松计算：
deg x ( N f ) + deg x ( N g ) {\deg }_{x}\left( {N}_{f}\right) + {\deg }_{x}\left( {N}_{g}\right) deg x ( N f ) + deg x ( N g ) 谈及“有理函数在 E u {E}^{u} E u
x + 1 x y − 2 = x 2 + x x 2 y − 2 x \frac{x + 1}{{xy} - 2} = \frac{{x}^{2} + x}{{x}^{2}y - {2x}} x y − 2 x + 1 = x 2 y − 2 x x 2 + x 但根据命题3.14，我们得到对于 r = f g = h k ∈ K ( E ) r = \frac{f}{g} = \frac{h}{k} \in K\left( E\right) r = g f = k h ∈ K ( E ) deg ( f ) − deg ( g ) = deg ( h ) − deg ( k ) \deg \left( f\right) - \deg \left( g\right) = \deg \left( h\right) - \deg \left( k\right) deg ( f ) − deg ( g ) = deg ( h ) − deg ( k ) f k = g h {fk} = {gh} f k = g h O \mathcal{O} O
定义 3.15（在 O \mathcal{O} O r = f g ∈ K ( E ) r = \frac{f}{g} \in K\left( E\right) r = g f ∈ K ( E )
deg ( f ) < deg ( g ) : set r ( O ) = 0 \deg \left( f\right) < \deg \left( g\right) : \operatorname{set}r\left( \mathcal{O}\right) = 0 deg ( f ) < deg ( g ) : set r ( O ) = 0 deg ( f ) > deg ( g ) \deg \left( f\right) > \deg \left( g\right) deg ( f ) > deg ( g ) r r r O \mathcal{O} O
deg ( f ) = deg ( g ) \deg \left( f\right) = \deg \left( g\right) deg ( f ) = deg ( g ) deg ( f ) \deg \left( f\right) deg ( f ) f f f g g g a x d a{x}^{d} a x d b x d b{x}^{d} b x d a , b ∈ K a,b \in K a , b ∈ K d = d = d = deg ( f ) 2 ) \left. \frac{\deg \left( f\right) }{2}\right) 2 d e g ( f ) ) r ( O ) = a b r\left( \mathcal{O}\right) = \frac{a}{b} r ( O ) = b a deg ( f ) = deg ( g ) \deg \left( f\right) = \deg \left( g\right) deg ( f ) = deg ( g ) deg ( f ) \deg \left( f\right) deg ( f ) f f f g g g x d {x}^{d} x d x d {x}^{d} x d a , b ∈ K a,b \in K a , b ∈ K d = deg ( f ) − 3 2 d = \frac{\deg \left( f\right) - 3}{2} d = 2 d e g ( f ) − 3 r ( O ) = a b r\left( \mathcal{O}\right) = \frac{a}{b} r ( O ) = b a
注释 3.16。似乎很自然地定义有理函数 r = f g r = \frac{f}{g} r = g f deg ( f ) − deg ( g ) \deg \left( f\right) - \deg \left( g\right) deg ( f ) − deg ( g ) O \mathcal{O} O
示例 3.17。对于
r ( x , y ) = x 3 + 2 x + y + 2 x 4 y x + x 2 + 5 x y 3 r\left( {x,y}\right) = \frac{{x}^{3} + {2x} + y + 2{x}^{4}y}{x + {x}^{2} + {5x}{y}^{3}} r ( x , y ) = x + x 2 + 5 x y 3 x 3 + 2 x + y + 2 x 4 y 可以写成
r ( x , y ) = x 3 + 2 x + y + 2 x 4 y x + x 2 + 5 x y ( x 3 + A x + B ) = ( x 3 + 2 x ) + y ( 1 + 2 x 4 ) ( x + x 2 ) + y ( 5 x 4 + 5 A x 2 + 5 B x ) r\left( {x,y}\right) = \frac{{x}^{3} + {2x} + y + 2{x}^{4}y}{x + {x}^{2} + {5xy}\left( {{x}^{3} + {Ax} + B}\right) } = \frac{\left( {{x}^{3} + {2x}}\right) + y\left( {1 + 2{x}^{4}}\right) }{\left( {x + {x}^{2}}\right) + y\left( {5{x}^{4} + {5A}{x}^{2} + {5Bx}}\right) } r ( x , y ) = x + x 2 + 5 x y ( x 3 + A x + B ) x 3 + 2 x + y + 2 x 4 y = ( x + x 2 ) + y ( 5 x 4 + 5 A x 2 + 5 B x ) ( x 3 + 2 x ) + y ( 1 + 2 x 4 ) 这个代表具有分子次数 max { 2 ⋅ 3 , 3 + 2 ⋅ 4 } = 11 \max \{ 2 \cdot 3,3 + 2 \cdot 4\} = {11} max { 2 ⋅ 3 , 3 + 2 ⋅ 4 } = 11 max { 2 ⋅ 2 , 3 + 2 ⋅ 4 } = 11 \max \{ 2 \cdot 2,3 + 2 \cdot 4\} = {11} max { 2 ⋅ 2 , 3 + 2 ⋅ 4 } = 11 r ( O ) = 2 5 r\left( \mathcal{O}\right) = \frac{2}{5} r ( O ) = 5 2
命题 3.18。对于 r , s ∈ K ( E ) r,s \in K\left( E\right) r , s ∈ K ( E ) r ( O ) r\left( \mathcal{O}\right) r ( O ) s ( O ) s\left( \mathcal{O}\right) s ( O )
( r ⋅ s ) ( O ) = r ( O ) s ( O ) \left( {r \cdot s}\right) \left( \mathcal{O}\right) = r\left( \mathcal{O}\right) s\left( \mathcal{O}\right) ( r ⋅ s ) ( O ) = r ( O ) s ( O ) 且
( r + s ) ( O ) = r ( O ) + s ( O ) \left( {r + s}\right) \left( \mathcal{O}\right) = r\left( \mathcal{O}\right) + s\left( \mathcal{O}\right) ( r + s ) ( O ) = r ( O ) + s ( O ) 4 零点和极点
定义 4.1（零点和极点）。设 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E ) r r r P ∈ E P \in E P ∈ E r ( P ) = 0 r\left( P\right) = 0 r ( P ) = 0 r ( P ) r\left( P\right) r ( P ) P P P
在以下内容中，我们将定义零点和极点的重数。这是由单变量函数分析中零点的重数所启发的：考虑椭圆曲线 E = E 1 , 0 E = {E}_{1,0} E = E 1 , 0
Y 2 = X 3 + X {Y}^{2} = {X}^{3} + X Y 2 = X 3 + X 然后 P = ( 0 , 0 ) ∈ E P = \left( {0,0}\right) \in E P = ( 0 , 0 ) ∈ E P P P x x x y y y x = y 2 − x 3 x = {y}^{2} - {x}^{3} x = y 2 − x 3 x → 0 x \rightarrow 0 x → 0 x 3 {x}^{3} x 3 x x x P P P y y y P ′ ′ {P}^{\prime \prime } P ′′
保持不变
" f f f : ⇔ : \Leftrightarrow :⇔ f f f
定义 4.2（统一化子）。对于椭圆曲线 E E E P ∈ E P \in E P ∈ E u ∈ K ( E ) u \in K\left( E\right) u ∈ K ( E ) u ( P ) = 0 u\left( P\right) = 0 u ( P ) = 0 P P P ∀ r ∈ K ( E ) ∖ { 0 } : ∃ d ∈ Z , s ∈ K ( E ) ‾ \forall r \in K\left( E\right) \smallsetminus \{ 0\} : \exists d \in \mathbb{Z},s \in \overline{K\left( E\right) } ∀ r ∈ K ( E ) ∖ { 0 } : ∃ d ∈ Z , s ∈ K ( E ) P ˉ \bar{P} P ˉ s ( P ) ≠ 0 s\left( P\right) \neq 0 s ( P ) = 0
r = u d ⋅ s r = {u}^{d} \cdot s r = u d ⋅ s 引理 4.3（一般情况下的统一化子）。设 E E E P ∈ E P \in E P ∈ E P = ( a , b ) P = \left( {a,b}\right) P = ( a , b ) u ( x , y ) := x − a u\left( {x,y}\right) \mathrel{\text{:=}} x - a u ( x , y ) := x − a P P P
证明。首先注意到 u ( a , b ) = 0 u\left( {a,b}\right) = 0 u ( a , b ) = 0 r ∈ K ( E ) ∖ { 0 } r \in K\left( E\right) \smallsetminus \{ 0\} r ∈ K ( E ) ∖ { 0 } r r r P P P d = 0 d = 0 d = 0 s = r s = r s = r u u u r ( P ) = 0 r\left( P\right) = 0 r ( P ) = 0 r = f g r = \frac{f}{g} r = g f f ( P ) = 0 f\left( P\right) = 0 f ( P ) = 0 g ( P ) ≠ 0 g\left( P\right) \neq 0 g ( P ) = 0 f = u d s f = {u}^{d}s f = u d s
r = f g = u d s g = u d s g = u d s ~ ( x ) r = \frac{f}{g} = \frac{{u}^{d}s}{g} = {u}^{d}\frac{s}{g} = {u}^{d}\widetilde{s}\left( x\right) r = g f = g u d s = u d g s = u d s ( x ) 并且我们找到了所需的 s ~ := s q ∈ K ( E ) \widetilde{s} \mathrel{\text{:=}} \frac{s}{q} \in K\left( E\right) s := q s ∈ K ( E )
将 s 0 ( x , y ) := f ( x , y ) {s}_{0}\left( {x,y}\right) \mathrel{\text{:=}} f\left( {x,y}\right) s 0 ( x , y ) := f ( x , y ) i = 0 i = 0 i = 0 s i ( P ) = 0 {s}_{i}\left( P\right) = 0 s i ( P ) = 0 s i ( x , y ) = v i ( x ) + y w i ( x ) {s}_{i}\left( {x,y}\right) = {v}_{i}\left( x\right) + y{w}_{i}\left( x\right) s i ( x , y ) = v i ( x ) + y w i ( x ) s i ‾ ( P ) = 0 \overline{{s}_{i}}\left( P\right) = 0 s i ( P ) = 0 s i ‾ ( P ) ≠ 0 \overline{{s}_{i}}\left( P\right) \neq 0 s i ( P ) = 0
情况 s i ‾ ( P ) = 0 \overline{{s}_{i}}\left( P\right) = 0 s i ( P ) = 0 y ( P ) = b ≠ 0 y\left( P\right) = b \neq 0 y ( P ) = b = 0
v i ( a ) + b w i ( a ) = 0 {v}_{i}\left( a\right) + b{w}_{i}\left( a\right) = 0 v i ( a ) + b w i ( a ) = 0 v i ( a ) − b w i ( a ) = 0 {v}_{i}\left( a\right) - b{w}_{i}\left( a\right) = 0 v i ( a ) − b w i ( a ) = 0 的秩为2（小于特征值），因此产生 v i ( a ) = w i ( a ) = 0 {v}_{i}\left( a\right) = {w}_{i}\left( a\right) = 0 v i ( a ) = w i ( a ) = 0
s i ( x , y ) = v i ( x ) + y w i ( x ) = ( x − a ) v i + 1 ( x ) + ( x − a ) y w i + 1 ( x ) = ( x − a ) s i + 1 ( x , y ) {s}_{i}\left( {x,y}\right) = {v}_{i}\left( x\right) + y{w}_{i}\left( x\right) = \left( {x - a}\right) {v}_{i + 1}\left( x\right) + \left( {x - a}\right) y{w}_{i + 1}\left( x\right) = \left( {x - a}\right) {s}_{i + 1}\left( {x,y}\right) s i ( x , y ) = v i ( x ) + y w i ( x ) = ( x − a ) v i + 1 ( x ) + ( x − a ) y w i + 1 ( x ) = ( x − a ) s i + 1 ( x , y ) 对于 s i + 1 ( x , y ) = v i + 1 ( x ) + y w i + 1 ( x ) {s}_{i + 1}\left( {x,y}\right) = {v}_{i + 1}\left( x\right) + y{w}_{i + 1}\left( x\right) s i + 1 ( x , y ) = v i + 1 ( x ) + y w i + 1 ( x ) v i + 1 , w i + 1 ∈ {v}_{i + 1},{w}_{i + 1} \in v i + 1 , w i + 1 ∈ K [ E ] K\left\lbrack E\right\rbrack K [ E ]
情况 s i ‾ ( P ) ≠ 0 : ‾ \underline{\overline{{s}_{i}}\left( P\right) \neq 0 : } s i ( P ) = 0 : s i {s}_{i} s i 1 = s i ‾ s i ‾ 1 = \frac{\overline{{s}_{i}}}{\overline{{s}_{i}}} 1 = s i s i
s i ( x , y ) = N s i ( x ) s i ‾ ( x , y ) {s}_{i}\left( {x,y}\right) = \frac{{N}_{{s}_{i}}\left( x\right) }{\overline{{s}_{i}}\left( {x,y}\right) } s i ( x , y ) = s i ( x , y ) N s i ( x ) 现在 s i ( P ) = 0 {s}_{i}\left( P\right) = 0 s i ( P ) = 0 s i ‾ ( P ) ≠ 0 \overline{{s}_{i}}\left( P\right) \neq 0 s i ( P ) = 0 N s i ( a ) = 0 {N}_{{s}_{i}}\left( a\right) = 0 N s i ( a ) = 0 N s i ( x ) = ( x − a ) n ( x ) {N}_{{s}_{i}}\left( x\right) = \left( {x - a}\right) n\left( x\right) N s i ( x ) = ( x − a ) n ( x ) s i + 1 ( x , y ) := n ( x ) s i ‾ ( x , y ) {s}_{i + 1}\left( {x,y}\right) \mathrel{\text{:=}} \frac{n\left( x\right) }{\overline{{s}_{i}}\left( {x,y}\right) } s i + 1 ( x , y ) := s i ( x , y ) n ( x ) P P P
s i ( x , y ) = N s i ( x ) s i ‾ ( x , y ) = ( x − a ) n i + 1 ( x ) s i ‾ ( x , y ) = ( x − a ) s i + 1 ( x , y ) {s}_{i}\left( {x,y}\right) = \frac{{N}_{{s}_{i}}\left( x\right) }{\overline{{s}_{i}}\left( {x,y}\right) } = \frac{\left( {x - a}\right) {n}_{i + 1}\left( x\right) }{\overline{{s}_{i}}\left( {x,y}\right) } = \left( {x - a}\right) {s}_{i + 1}\left( {x,y}\right) s i ( x , y ) = s i ( x , y ) N s i ( x ) = s i ( x , y ) ( x − a ) n i + 1 ( x ) = ( x − a ) s i + 1 ( x , y ) 如果这个过程终止，我们最终得到
f ( x , y ) = ( x − a ) i s i ( x , y ) f\left( {x,y}\right) = {\left( x - a\right) }^{i}{s}_{i}\left( {x,y}\right) f ( x , y ) = ( x − a ) i s i ( x , y ) 其中 s := s i s \mathrel{\text{:=}} {s}_{i} s := s i x − a = u ( x , y ) x - a = u\left( {x,y}\right) x − a = u ( x , y ) d := i d \mathrel{\text{:=}} i d := i f = u d s f = {u}^{d}s f = u d s
由于 s i {s}_{i} s i
N f ( x ) = N u i s i ( x ) {N}_{f}\left( x\right) = {N}_{{u}^{i}{s}_{i}}\left( x\right) N f ( x ) = N u i s i ( x ) = ( ( x − a ) i v i ( x ) ) 2 − s ( x ) ( ( x − a ) i w i ( x ) ) 2 = {\left( {\left( x - a\right) }^{i}{v}_{i}\left( x\right) \right) }^{2} - s\left( x\right) {\left( {\left( x - a\right) }^{i}{w}_{i}\left( x\right) \right) }^{2} = ( ( x − a ) i v i ( x ) ) 2 − s ( x ) ( ( x − a ) i w i ( x ) ) 2 = ( x − a ) 2 i ( v i 2 ( x ) − s ( x ) w i 2 ( x ) ) \begin{matrix} & = & {\left( x - a\right) }^{2i}\left( {{v}_{i}^{2}\left( x\right) - s\left( x\right) {w}_{i}^{2}\left( x\right) }\right) \end{matrix} = ( x − a ) 2 i ( v i 2 ( x ) − s ( x ) w i 2 ( x ) ) = ( x − a ) 2 i N s i ( x ) = {\left( x - a\right) }^{2i}{N}_{{s}_{i}}\left( x\right) = ( x − a ) 2 i N s i ( x ) 因此我们得到 deg x ( N f ) = 2 i + deg x ( N s i ) {\deg }_{x}\left( {N}_{f}\right) = {2i} + {\deg }_{x}\left( {N}_{{s}_{i}}\right) deg x ( N f ) = 2 i + deg x ( N s i ) deg x ( N s i ) > 0 {\deg }_{x}\left( {N}_{{s}_{i}}\right) > 0 deg x ( N s i ) > 0 deg x ( N f ) > 2 i {\deg }_{x}\left( {N}_{f}\right) > {2i} deg x ( N f ) > 2 i 2 i {2i} 2 i
因此如果 r r r P P P r r r r r r P , 1 r P,\frac{1}{r} P , r 1 u u u d d d u u u
引理 4.4（二阶点处的统一化子）。设 E E E P := Ω i ∈ E P \mathrel{\text{:=}} {\Omega }_{i} \in E P := Ω i ∈ E u ( x , y ) := y u\left( {x,y}\right) \mathrel{\text{:=}} y u ( x , y ) := y Ω i {\Omega }_{i} Ω i
证明。不失一般性，我们可以取 i = 1 i = 1 i = 1 u ( P ) = 0 u\left( P\right) = 0 u ( P ) = 0 r ∈ K ( E ) ∖ { 0 } r \in K\left( E\right) \smallsetminus \{ 0\} r ∈ K ( E ) ∖ { 0 } r ( P ) = 0 r\left( P\right) = 0 r ( P ) = 0 r = f g r = \frac{f}{g} r = g f f ( P ) = 0 f\left( P\right) = 0 f ( P ) = 0 v ( ω 1 ) = 0 v\left( {\omega }_{1}\right) = 0 v ( ω 1 ) = 0 f ( x , y ) = v ( x ) + y w ( x ) f\left( {x,y}\right) = v\left( x\right) + {yw}\left( x\right) f ( x , y ) = v ( x ) + y w ( x ) v v v v ( x ) = ( x − ω 1 ) v 1 ( x ) v\left( x\right) = \left( {x - {\omega }_{1}}\right) {v}_{1}\left( x\right) v ( x ) = ( x − ω 1 ) v 1 ( x ) v 1 {v}_{1} v 1 s s s
f ( x , y ) = ( x − ω 1 ) v 1 ( x ) + y w ( x ) f\left( {x,y}\right) = \left( {x - {\omega }_{1}}\right) {v}_{1}\left( x\right) + {yw}\left( x\right) f ( x , y ) = ( x − ω 1 ) v 1 ( x ) + y w ( x ) = ( x − ω 1 ) ( x − ω 2 ) ( x − ω 3 ) v 1 ( x ) + y w 1 ( x ) ( x − ω 2 ) ( x − ω 3 ) = \frac{\left( {x - {\omega }_{1}}\right) \left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) {v}_{1}\left( x\right) + y{w}_{1}\left( x\right) }{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } = ( x − ω 2 ) ( x − ω 3 ) ( x − ω 1 ) ( x − ω 2 ) ( x − ω 3 ) v 1 ( x ) + y w 1 ( x ) = y 2 v 1 ( x ) + y w 1 ( x ) ( x − ω 2 ) ( x − ω 3 ) = \;\frac{{y}^{2}{v}_{1}\left( x\right) + y{w}_{1}\left( x\right) }{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } = ( x − ω 2 ) ( x − ω 3 ) y 2 v 1 ( x ) + y w 1 ( x ) = y ⋅ y v 1 ( x ) + w 1 ( x ) ( x − ω 2 ) ( x − ω 3 ) = y \cdot \frac{y{v}_{1}\left( x\right) + {w}_{1}\left( x\right) }{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } = y ⋅ ( x − ω 2 ) ( x − ω 3 ) y v 1 ( x ) + w 1 ( x ) = u ( x , y ) ⋅ W ( x , y ) = u\left( {x,y}\right) \cdot W\left( {x,y}\right) = u ( x , y ) ⋅ W ( x , y ) 其中 w 1 ( x ) := w ( x ) ( x − ω 2 ) ( x − ω 3 ) {w}_{1}\left( x\right) \mathrel{\text{:=}} w\left( x\right) \left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) w 1 ( x ) := w ( x ) ( x − ω 2 ) ( x − ω 3 ) W ( x , y ) := y v 1 ( x ) + w 1 ( x ) ( x − ω 2 ) ( x − ω 3 ) W\left( {x,y}\right) \mathrel{\text{:=}} \frac{y{v}_{1}\left( x\right) + {w}_{1}\left( x\right) }{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } W ( x , y ) := ( x − ω 2 ) ( x − ω 3 ) y v 1 ( x ) + w 1 ( x ) W ( P ) ≠ 0 W\left( P\right) \neq 0 W ( P ) = 0 W W W v v v □ \square □
引理 4.5（在 O \mathcal{O} O E E E u ( x , y ) := x y u\left( {x,y}\right) \mathrel{\text{:=}} \frac{x}{y} u ( x , y ) := y x O ∈ E \mathcal{O} \in E O ∈ E
证明。由于 deg ( y ) = 3 > 2 = deg ( x ) \deg \left( y\right) = 3 > 2 = \deg \left( x\right) deg ( y ) = 3 > 2 = deg ( x ) u ( O ) = 0 u\left( \mathcal{O}\right) = 0 u ( O ) = 0 r = g ∈ K ‾ ( E ) ∖ { 0 } r = \overline{g \in K}\left( E\right) \smallsetminus \{ 0\} r = g ∈ K ( E ) ∖ { 0 } r ( O ) = 0 r\left( \mathcal{O}\right) = 0 r ( O ) = 0 O \mathcal{O} O d := deg ( g ) − deg ( f ) ≠ 0 d \mathrel{\text{:=}} \deg \left( g\right) - \deg \left( f\right) \neq 0 d := deg ( g ) − deg ( f ) = 0 s ( x , y ) = ( y x ) d r ( x , y ) s\left( {x,y}\right) = {\left( \frac{y}{x}\right) }^{d}r\left( {x,y}\right) s ( x , y ) = ( x y ) d r ( x , y ) O \mathcal{O} O
r ( x , y ) = ( x y ) d ( ( y x ) d r ( x , y ) ) = u d ( x , y ) s ( x , y ) r\left( {x,y}\right) = {\left( \frac{x}{y}\right) }^{d}\left( {{\left( \frac{y}{x}\right) }^{d}r\left( {x,y}\right) }\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) r ( x , y ) = ( y x ) d ( ( x y ) d r ( x , y ) ) = u d ( x , y ) s ( x , y ) 但因为
\begin{array}{lll} & \deg \left( {{y}^{d}f\left( {x,y}\right) }\right) - \deg \left( {{x}^{d}g\left( {x,y}\right) }\right) & \\ \text{ Prop_ 3.14 } & \left( {\deg \left( {y}^{d}\right) + \deg \left( f\right) }\right) - \left( {\deg \left( {x}^{d}\right) + \deg \left( g\right) }\right) & \\ {}^{\text{Def }} = & {3.11} & {3d} + \deg \left( f\right) - {2d} - \deg \left( g\right) \\ = & d + \left( {\deg \left( f\right) - \deg \left( g\right) }\right) = 0 & \end{array}
这意味着 s ( x , y ) = y d f ( x , y ) x d g ( x , y ) s\left( {x,y}\right) = \frac{{y}^{d}f\left( {x,y}\right) }{{x}^{d}g\left( {x,y}\right) } s ( x , y ) = x d g ( x , y ) y d f ( x , y )
定理 4.6（整除化定理）。椭圆曲线上的每一点都有一个整除化元素，且定义 4.2 中的数 d 不依赖于其选择。
证明。引理 4.3、4.4 和 4.5 一起得出每个点都存在整除化元素。所以只剩下要证明 d d d u u u u ~ \widetilde{u} u P P P u = u ~ a q u = {\widetilde{u}}^{a}q u = u a q u ~ = u b p \widetilde{u} = {u}^{b}p u = u b p a , b ∈ Z a,b \in \mathbb{Z} a , b ∈ Z q , p ∈ K ( E ) q,p \in K\left( E\right) q , p ∈ K ( E ) P P P
u = u ~ a q = ( u b p ) a q = u a b ( p a q ) u = {\widetilde{u}}^{a}q = {\left( {u}^{b}p\right) }^{a}q = {u}^{ab}\left( {{p}^{a}q}\right) u = u a q = ( u b p ) a q = u ab ( p a q ) 我们假设 a b ≠ 1 {ab} \neq 1 ab = 1 u u u 1 = u a b − 1 ( p a q ) 1 = {u}^{{ab} - 1}\left( {{p}^{a}q}\right) 1 = u ab − 1 ( p a q ) P P P 1 = 0 1 = 0 1 = 0 a b = 1 {ab} = 1 ab = 1 a = b = ± 1 a = b = \pm 1 a = b = ± 1 a = b = − 1 a = b = - 1 a = b = − 1
u = u ~ − 1 q ⇔ u u ~ = q u = {\widetilde{u}}^{-1}q \Leftrightarrow u\widetilde{u} = q u = u − 1 q ⇔ u u = q 这在 P P P 0 = u ( P ) u ~ ( P ) = q ( P ) ≠ 0 0 = u\left( P\right) \widetilde{u}\left( P\right) = q\left( P\right) \neq 0 0 = u ( P ) u ( P ) = q ( P ) = 0 a = b = 1 a = b = 1 a = b = 1 r ∈ K ( E ) ∖ { 0 } r \in K\left( E\right) \smallsetminus \{ 0\} r ∈ K ( E ) ∖ { 0 } u u u u ~ \widetilde{u} u d , d ~ ∈ Z d,\widetilde{d} \in \mathbb{Z} d , d ∈ Z s , t ∈ K ( E ) s,t \in K\left( E\right) s , t ∈ K ( E ) P P P r = u d s r = {u}^{d}s r = u d s r = u ~ d ^ t r = {\widetilde{u}}^{\widehat{d}}t r = u d t
现在我们计算
u d s = u ~ d ~ t = ( u p ) d ~ t = u d ~ ( p d ~ t ) {u}^{d}s = {\widetilde{u}}^{\widetilde{d}}t = {\left( up\right) }^{\widetilde{d}}t = {u}^{\widetilde{d}}\left( {{p}^{\widetilde{d}}t}\right) u d s = u d t = ( u p ) d t = u d ( p d t ) 这产生
u d − d ~ = p d ~ t s {u}^{d - \widetilde{d}} = \frac{{p}^{\widetilde{d}}t}{s} u d − d = s p d t 在右侧仅有有理函数，它们在 P P P d − d ~ ≠ 0 d - \widetilde{d} \neq 0 d − d = 0 P P P d = d ~ d = \widetilde{d} d = d
现在我们知道，一个点上的均匀化子总是产生相同的 d d d
定义 4.7（有理函数的阶）。对于一个椭圆曲线 E E E P ∈ E P \in E P ∈ E u u u P P P r ∈ K ( E ) ∖ { 0 } r \in K\left( E\right) \smallsetminus \{ 0\} r ∈ K ( E ) ∖ { 0 } r = u d ⋅ s r = {u}^{d} \cdot s r = u d ⋅ s d d d r r r P P P
ord P ( r ) = : d {\operatorname{ord}}_{P}\left( r\right) = : d ord P ( r ) =: d 零点的重数是该点的阶数，极点的重数
是阶数的相反数。machen!
注释 4.8。这个零点阶数的定义与单变量多项式零点阶数的已知定义一致，前提是该零点不对应于阶数为二的点：设 f ∈ K [ X ] f \in K\left\lbrack X\right\rbrack f ∈ K [ X ]
f ( x ) = g ( x ) ⋅ ( x − a ) k f\left( x\right) = g\left( x\right) \cdot {\left( x - a\right) }^{k} f ( x ) = g ( x ) ⋅ ( x − a ) k 对于 g ∈ K [ X ] g \in K\left\lbrack X\right\rbrack g ∈ K [ X ] g ( a ) ≠ 0 , k ∈ N > 0 g\left( a\right) \neq 0,k \in {\mathbb{N}}_{ > 0} g ( a ) = 0 , k ∈ N > 0 a ∈ K a \in K a ∈ K f f f k k k f f f f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ] P = ( a , s ( a ) ) P = \left( {a,s\left( a\right) }\right) P = ( a , s ( a ) ) u u u E E E f f f u ( x , y ) = x − a u\left( {x,y}\right) = x - a u ( x , y ) = x − a f f f
f ( x , y ) = u d ( x , y ) ⋅ s ( x , y ) = ( x − a ) d ⋅ g ( x ) f\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) \cdot s\left( {x,y}\right) = {\left( x - a\right) }^{d} \cdot g\left( x\right) f ( x , y ) = u d ( x , y ) ⋅ s ( x , y ) = ( x − a ) d ⋅ g ( x ) 这意味着 k = d = ord P ( f ) k = d = {\operatorname{ord}}_{P}\left( f\right) k = d = ord P ( f )
然而，当 a = ω i a = {\omega }_{i} a = ω i i = 1 i = 1 i = 1 P = ( a , s ( a ) ) = ( a , 0 ) P = \left( {a,s\left( a\right) }\right) = \left( {a,0}\right) P = ( a , s ( a ) ) = ( a , 0 ) f f f 2 k {2k} 2 k P P P u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y s ( x , y ) := g ( x ) ( x − ω 2 ) k ( x − ω 3 ) k s\left( {x,y}\right) \mathrel{\text{:=}} \frac{g\left( x\right) }{{\left( x - {\omega }_{2}\right) }^{k}{\left( x - {\omega }_{3}\right) }^{k}} s ( x , y ) := ( x − ω 2 ) k ( x − ω 3 ) k g ( x )
f ( x , y ) = u d ( x , y ) ⋅ s ( x , y ) f\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) \cdot s\left( {x,y}\right) f ( x , y ) = u d ( x , y ) ⋅ s ( x , y ) = y d ⋅ g ( x ) ( x − ω 2 ) k ( x − ω 3 ) k = {y}^{d} \cdot \frac{g\left( x\right) }{{\left( x - {\omega }_{2}\right) }^{k}{\left( x - {\omega }_{3}\right) }^{k}} = y d ⋅ ( x − ω 2 ) k ( x − ω 3 ) k g ( x ) = y d ⋅ ( x − a ) k g ( x ) ( x − ω 1 ) k ( x − ω 2 ) k ( x − ω 3 ) k = {y}^{d} \cdot \frac{{\left( x - a\right) }^{k}g\left( x\right) }{{\left( x - {\omega }_{1}\right) }^{k}{\left( x - {\omega }_{2}\right) }^{k}{\left( x - {\omega }_{3}\right) }^{k}} = y d ⋅ ( x − ω 1 ) k ( x − ω 2 ) k ( x − ω 3 ) k ( x − a ) k g ( x ) = y d ⋅ ( x − a ) k g ( x ) y 2 k = {y}^{d} \cdot \frac{{\left( x - a\right) }^{k}g\left( x\right) }{{y}^{2k}} = y d ⋅ y 2 k ( x − a ) k g ( x ) 因此 ( x − a ) k ⋅ g ( x ) = f ( x , y ) = y d ⋅ ( x − a ) k g ( x ) y 2 k {\left( x - a\right) }^{k} \cdot g\left( x\right) = f\left( {x,y}\right) = {y}^{d} \cdot \frac{{\left( x - a\right) }^{k}g\left( x\right) }{{y}^{2k}} ( x − a ) k ⋅ g ( x ) = f ( x , y ) = y d ⋅ y 2 k ( x − a ) k g ( x ) d = 2 k d = {2k} d = 2 k
命题 4.9（有限非根处的阶）。设 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E ) P ∈ E P \in E P ∈ E r ( P ) ≠ 0 r\left( P\right) \neq 0 r ( P ) = 0 r r r P P P
ord P ( r ) = 0 {\operatorname{ord}}_{P}\left( r\right) = 0 ord P ( r ) = 0 证明。在 P P P s ( x , y ) = r ( x , y ) s\left( {x,y}\right) = r\left( {x,y}\right) s ( x , y ) = r ( x , y ) P P P
r ( x , y ) = u 0 ( x , y ) r ( x , y ) = u d ( x , y ) s ( x , y ) r\left( {x,y}\right) = {u}^{0}\left( {x,y}\right) r\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) r ( x , y ) = u 0 ( x , y ) r ( x , y ) = u d ( x , y ) s ( x , y ) 即 ord P ( f ) = d = 0 {\operatorname{ord}}_{P}\left( f\right) = d = 0 ord P ( f ) = d = 0
命题 4.10（非根处多项式的阶）。设 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ] P ∈ P \in P ∈ E ∖ { O } E \smallsetminus \{ \mathcal{O}\} E ∖ { O } f ( P ) ≠ 0 f\left( P\right) \neq 0 f ( P ) = 0
ord P ( f ) = 0 {\operatorname{ord}}_{P}\left( f\right) = 0 ord P ( f ) = 0 证明。由于多项式没有有限根，这由命题 4.9 推出。
□ \square □
命题 4.11（多项式在 O \mathcal{O} O f ∈ K [ E ] ∖ { 0 } f \in K\left\lbrack E\right\rbrack \smallsetminus \{ 0\} f ∈ K [ E ] ∖ { 0 }
ord O ( f ) = − deg ( f ) {\operatorname{ord}}_{\mathcal{O}}\left( f\right) = - \deg \left( f\right) ord O ( f ) = − deg ( f ) 证明。根据引理 4.5，u ( x , y ) = x y u\left( {x,y}\right) = \frac{x}{y} u ( x , y ) = y x O \mathcal{O} O k := deg ( f ) k \mathrel{\text{:=}} \deg \left( f\right) k := deg ( f ) s ( x , y ) = x k y k f ( x , y ) s\left( {x,y}\right) = \frac{{x}^{k}}{{y}^{k}}f\left( {x,y}\right) s ( x , y ) = y k x k f ( x , y ) deg ( x k ⋅ f ( x , y ) ) = Prop. 3.14 2 k + deg ( f ) = 3 k \deg \left( {{x}^{k} \cdot f\left( {x,y}\right) }\right) \overset{\text{ Prop. 3.14 }}{ = }{2k} + \deg \left( f\right) = {3k} deg ( x k ⋅ f ( x , y ) ) = Prop. 3.14 2 k + deg ( f ) = 3 k deg ( y k ) = 3 k \deg \left( {y}^{k}\right) = {3k} deg ( y k ) = 3 k s s s
f ( x , y ) = u d ( x , y ) s ( x , y ) = ( x y ) d x k y k f ( x , y ) f\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = {\left( \frac{x}{y}\right) }^{d}\frac{{x}^{k}}{{y}^{k}}f\left( {x,y}\right) f ( x , y ) = u d ( x , y ) s ( x , y ) = ( y x ) d y k x k f ( x , y ) 这意味着 d = − k = − deg ( f ) d = - k = - \deg \left( f\right) d = − k = − deg ( f )
命题 4.12（有理函数阶数的性质）。对于 r 1 , r 2 ∈ K ( E ) {r}_{1},{r}_{2} \in K\left( E\right) r 1 , r 2 ∈ K ( E ) P ∈ E P \in E P ∈ E
ord P ( r 1 ⋅ r 2 ) = ord P ( r 1 ) + ord P ( r 2 ) {\operatorname{ord}}_{P}\left( {{r}_{1} \cdot {r}_{2}}\right) = {\operatorname{ord}}_{P}\left( {r}_{1}\right) + {\operatorname{ord}}_{P}\left( {r}_{2}\right) ord P ( r 1 ⋅ r 2 ) = ord P ( r 1 ) + ord P ( r 2 ) 证明。设 P ∈ E P \in E P ∈ E P P P u u u d , d 1 , d 2 ∈ Z d,{d}_{1},{d}_{2} \in \mathbb{Z} d , d 1 , d 2 ∈ Z P P P s , s 1 , s 2 ∈ K ( E ) s,{s}_{1},{s}_{2} \in K\left( E\right) s , s 1 , s 2 ∈ K ( E )
r 1 ⋅ r 2 = u d ⋅ s {r}_{1} \cdot {r}_{2} = {u}^{d} \cdot s r 1 ⋅ r 2 = u d ⋅ s r 1 = u d 1 ⋅ s 1 {r}_{1} = {u}^{{d}_{1}} \cdot {s}_{1} r 1 = u d 1 ⋅ s 1 r 2 = u d 2 ⋅ s 2 {r}_{2} = {u}^{{d}_{2}} \cdot {s}_{2} r 2 = u d 2 ⋅ s 2 并且可以计算
u d ⋅ s = r 1 ⋅ r 2 = ( u d 1 ⋅ s 1 ) ⋅ ( u d 2 ⋅ s 2 ) = u d 1 + d 2 ⋅ s 1 ⋅ s 2 {u}^{d} \cdot s = {r}_{1} \cdot {r}_{2} = \left( {{u}^{{d}_{1}} \cdot {s}_{1}}\right) \cdot \left( {{u}^{{d}_{2}} \cdot {s}_{2}}\right) = {u}^{{d}_{1} + {d}_{2}} \cdot {s}_{1} \cdot {s}_{2} u d ⋅ s = r 1 ⋅ r 2 = ( u d 1 ⋅ s 1 ) ⋅ ( u d 2 ⋅ s 2 ) = u d 1 + d 2 ⋅ s 1 ⋅ s 2 由于定理 4.6，因此得出
ord P ( r 1 ⋅ r 2 ) = d = d 1 + d 2 = ord P ( r 1 ) + ord P ( r 2 ) {\operatorname{ord}}_{P}\left( {{r}_{1} \cdot {r}_{2}}\right) = d = {d}_{1} + {d}_{2} = {\operatorname{ord}}_{P}\left( {r}_{1}\right) + {\operatorname{ord}}_{P}\left( {r}_{2}\right) ord P ( r 1 ⋅ r 2 ) = d = d 1 + d 2 = ord P ( r 1 ) + ord P ( r 2 ) 示例 4.13。设 P = ( a , b ) ∈ E P = \left( {a,b}\right) \in E P = ( a , b ) ∈ E b ≠ 0 b \neq 0 b = 0 P P P r ( x , y ) = x − a r\left( {x,y}\right) = x - a r ( x , y ) = x − a Q ∈ E Q \in E Q ∈ E ord Q ( r ) = 0 {\operatorname{ord}}_{Q}\left( r\right) = 0 ord Q ( r ) = 0
Q = P Q = P Q = P Q = P ′ := ( a , − b ) ≠ P Q = {P}^{\prime } \mathrel{\text{:=}} \left( {a, - b}\right) \neq P Q = P ′ := ( a , − b ) = P Q Q Q u u u r r r
均匀化器，因此得出 r = u d ⋅ s = r 1 ⋅ 1 r = {u}^{d} \cdot s = {r}^{1} \cdot 1 r = u d ⋅ s = r 1 ⋅ 1 ord Q ( r ) = d = 1 {\operatorname{ord}}_{Q}\left( r\right) = d = 1 ord Q ( r ) = d = 1
Q = O ‾ \underline{Q = \mathcal{O}} Q = O Q Q Q u ( x , y ) = x y u\left( {x,y}\right) = \frac{x}{y} u ( x , y ) = y x s ( x , y ) = x 3 − a x 2 y 2 s\left( {x,y}\right) = \frac{{x}^{3} - a{x}^{2}}{{y}^{2}} s ( x , y ) = y 2 x 3 − a x 2 s ( Q ) = 1 s\left( Q\right) = 1 s ( Q ) = 1
u d ( x , y ) ⋅ s ( x , y ) = ( x y ) − 2 s ( x , y ) = y 2 x 2 x 3 − a x 2 y 2 = x − a = r ( x , y ) {u}^{d}\left( {x,y}\right) \cdot s\left( {x,y}\right) = {\left( \frac{x}{y}\right) }^{-2}s\left( {x,y}\right) = \frac{{y}^{2}}{{x}^{2}}\frac{{x}^{3} - a{x}^{2}}{{y}^{2}} = x - a = r\left( {x,y}\right) u d ( x , y ) ⋅ s ( x , y ) = ( y x ) − 2 s ( x , y ) = x 2 y 2 y 2 x 3 − a x 2 = x − a = r ( x , y ) 以及 ord Q ( r ) = d = − 2 {\operatorname{ord}}_{Q}\left( r\right) = d = - 2 ord Q ( r ) = d = − 2 r r r
示例 4.14。现在考虑 r ( x , y ) := y r\left( {x,y}\right) \mathrel{\text{:=}} y r ( x , y ) := y u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y ord Ω i ( r ) = 1 {\operatorname{ord}}_{{\Omega }_{i}}\left( r\right) = 1 ord Ω i ( r ) = 1 r r r O \mathcal{O} O u ( x , y ) = x y u\left( {x,y}\right) = \frac{x}{y} u ( x , y ) = y x s ( x , y ) = x 3 y y 3 s\left( {x,y}\right) = \frac{{x}^{3}y}{{y}^{3}} s ( x , y ) = y 3 x 3 y O \mathcal{O} O
u d ( x , y ) ⋅ s ( x , y ) = ( x y ) − 3 ⋅ s ( x , y ) = y 3 x 3 ⋅ x 3 y y 3 = y = r ( x , y ) {u}^{d}\left( {x,y}\right) \cdot s\left( {x,y}\right) = {\left( \frac{x}{y}\right) }^{-3} \cdot s\left( {x,y}\right) = \frac{{y}^{3}}{{x}^{3}} \cdot \frac{{x}^{3}y}{{y}^{3}} = y = r\left( {x,y}\right) u d ( x , y ) ⋅ s ( x , y ) = ( y x ) − 3 ⋅ s ( x , y ) = x 3 y 3 ⋅ y 3 x 3 y = y = r ( x , y ) 以及 ord O ( r ) = d = − 3 {\operatorname{ord}}_{\mathcal{O}}\left( r\right) = d = - 3 ord O ( r ) = d = − 3 r r r
示例 4.15。r ( x , y ) = x y r\left( {x,y}\right) = \frac{x}{y} r ( x , y ) = y x deg ( x ) = 2 < 3 = deg ( y ) \deg \left( x\right) = 2 < 3 = \deg \left( y\right) deg ( x ) = 2 < 3 = deg ( y ) r ( O ) = 0 r\left( \mathcal{O}\right) = 0 r ( O ) = 0 O \mathcal{O} O u ( x , y ) = x y u\left( {x,y}\right) = \frac{x}{y} u ( x , y ) = y x s ( x , y ) = 1 s\left( {x,y}\right) = 1 s ( x , y ) = 1
u d ( x , y ) ⋅ s ( x , y ) = ( x y ) 1 ⋅ 1 = r ( x , y ) {u}^{d}\left( {x,y}\right) \cdot s\left( {x,y}\right) = {\left( \frac{x}{y}\right) }^{1} \cdot 1 = r\left( {x,y}\right) u d ( x , y ) ⋅ s ( x , y ) = ( y x ) 1 ⋅ 1 = r ( x , y ) 并得到 ord O ( r ) = 1 {\operatorname{ord}}_{\mathcal{O}}\left( r\right) = 1 ord O ( r ) = 1
B ≠ 0 B \neq 0 B = 0 P ± := ( 0 , ± B ) {P}_{ \pm } \mathrel{\text{:=}} \left( {0, \pm \sqrt{B}}\right) P ± := ( 0 , ± B ) r r r P ± , s ( x , y ) := 1 y {P}_{ \pm },s\left( {x,y}\right) \mathrel{\text{:=}} \frac{1}{y} P ± , s ( x , y ) := y 1 u ( x , y ) = x u\left( {x,y}\right) = x u ( x , y ) = x y = ± B y = \pm \sqrt{B} y = ± B
x y = r ( x , y ) = u d ( x , y ) s ( x , y ) = x d 1 y \frac{x}{y} = r\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = {x}^{d}\frac{1}{y} y x = r ( x , y ) = u d ( x , y ) s ( x , y ) = x d y 1 为了获得 ord P ± ( r ) = d = 1 {\operatorname{ord}}_{{P}_{ \pm }}\left( r\right) = d = 1 ord P ± ( r ) = d = 1 r r r Ω i {\Omega }_{i} Ω i Ω i , s ( x , y ) := x {\Omega }_{i},s\left( {x,y}\right) \mathrel{\text{:=}} x Ω i , s ( x , y ) := x u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y B ≠ 0 B \neq 0 B = 0 Ω i {\Omega }_{i} Ω i ω i ≠ 0 {\omega }_{i} \neq 0 ω i = 0
并进行计算
x y = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d x \frac{x}{y} = r\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = {y}^{d}x y x = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d x 以得到 ord Ω i ( r ) = d = − 1 {\operatorname{ord}}_{{\Omega }_{i}}\left( r\right) = d = - 1 ord Ω i ( r ) = d = − 1
B = 0 : B = 0 : \; B = 0 : E A , 0 {E}_{A,0} E A , 0
y 2 = x 3 + A x = x ( x − − A ) ( x + − A ) {y}^{2} = {x}^{3} + {Ax} = x\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) y 2 = x 3 + A x = x ( x − − A ) ( x + − A ) 因此我们得到 ω 1 = 0 , ω 2 = − A {\omega }_{1} = 0,{\omega }_{2} = \sqrt{-A} ω 1 = 0 , ω 2 = − A ω 3 = A {\omega }_{3} = \sqrt{A} ω 3 = A Ω 2 {\Omega }_{2} Ω 2 Ω 3 {\Omega }_{3} Ω 3 r r r ω 2 ≠ 0 {\omega }_{2} \neq 0 ω 2 = 0 ω 3 ≠ 0 {\omega }_{3} \neq 0 ω 3 = 0 u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y s ( x , y ) = x s\left( {x,y}\right) = x s ( x , y ) = x
x y = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d x \frac{x}{y} = r\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = {y}^{d}x y x = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d x 以获得 ord Ω 2 ( r ) = ord Ω 3 ( r ) = d = − 1 {\operatorname{ord}}_{{\Omega }_{2}}\left( r\right) = {\operatorname{ord}}_{{\Omega }_{3}}\left( r\right) = d = - 1 ord Ω 2 ( r ) = ord Ω 3 ( r ) = d = − 1
r ( x , y ) = x y = x y y 2 = x y x ( x − − A ) ( x + − A ) = y ( x − − A ) ( x + − A ) r\left( {x,y}\right) = \frac{x}{y} = \frac{xy}{{y}^{2}} = \frac{xy}{x\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) } = \frac{y}{\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) } r ( x , y ) = y x = y 2 x y = x ( x − − A ) ( x + − A ) x y = ( x − − A ) ( x + − A ) y 因此得到，(0,0) 是 r r r u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y
y ( x − − A ) ( x + − A ) = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d 1 ( x − − A ) ( x + − A ) \frac{y}{\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) } = r\left( {x,y}\right) = {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = {y}^{d}\frac{1}{\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) } ( x − − A ) ( x + − A ) y = r ( x , y ) = u d ( x , y ) s ( x , y ) = y d ( x − − A ) ( x + − A ) 1 使用 s ( x , y ) = 1 ( x − − A ) ( x + − A ) s\left( {x,y}\right) = \frac{1}{\left( {x - \sqrt{-A}}\right) \left( {x + \sqrt{-A}}\right) } s ( x , y ) = ( x − − A ) ( x + − A ) 1 ord ( 0 , 0 ) ( r ) = d = 1 {\operatorname{ord}}_{\left( 0,0\right) }\left( r\right) = d = 1 ord ( 0 , 0 ) ( r ) = d = 1
我们得到两个简单零点和两个简单极点。
例子 4.13、4.14 和 4.15 表明所有点的阶数之和为零，这有点像初等的黎曼-罗赫定理。为了证明这一点，我们需要以下引理：
定理4.16（根的重数之和等于次数）。对于 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ]
deg ( f ) = ∑ P ∈ E f ( P ) = 0 ord P ( f ) \deg \left( f\right) = \mathop{\sum }\limits_{\substack{{P \in E} \\ {f\left( P\right) = 0} }}{\operatorname{ord}}_{P}\left( f\right) deg ( f ) = P ∈ E f ( P ) = 0 ∑ ord P ( f ) 证明。定义 n := deg ( f ) n \mathrel{\text{:=}} \deg \left( f\right) n := deg ( f ) n = deg x ( N f ) n = {\deg }_{x}\left( {N}_{f}\right) n = deg x ( N f )
我们可以写成
( f f ˉ ) ( x ) = N f ( x ) = ∏ i = 1 n ( x − a i ) \left( {f\bar{f}}\right) \left( x\right) = {N}_{f}\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - {a}_{i}}\right) ( f f ˉ ) ( x ) = N f ( x ) = i = 1 ∏ n ( x − a i ) 其中 a i s {a}_{i}\mathrm{\;s} a i s ( a i , 0 ) \left( {{a}_{i},0}\right) ( a i , 0 ) ( x − a i ) \left( {x - {a}_{i}}\right) ( x − a i ) E E E ( a i , ± s A , B ( a i ) ) \left( {{a}_{i}, \pm \sqrt{{s}_{A,B}\left( {a}_{i}\right) }}\right) ( a i , ± s A , B ( a i ) ) f f ˉ f\bar{f} f f ˉ E E E 2 n {2n} 2 n f f f f ˉ \bar{f} f ˉ E , f E,f E , f E , f E,f E , f n n n
定理4.17（阶之和为零）。对于 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E )
∑ P ∈ E ord P ( r ) = 0 \mathop{\sum }\limits_{{P \in E}}{\operatorname{ord}}_{P}\left( r\right) = 0 P ∈ E ∑ ord P ( r ) = 0 证明。由于对于 r = h g ∈ K ( E ) r = \frac{h}{g} \in K\left( E\right) r = g h ∈ K ( E )
∑ P ∈ E ord P ( r ) = ∑ P ∈ E ord P ( h ) − ∑ P ∈ E ord P ( g ) \mathop{\sum }\limits_{{P \in E}}{\operatorname{ord}}_{P}\left( r\right) = \mathop{\sum }\limits_{{P \in E}}{\operatorname{ord}}_{P}\left( h\right) - \mathop{\sum }\limits_{{P \in E}}{\operatorname{ord}}_{P}\left( g\right) P ∈ E ∑ ord P ( r ) = P ∈ E ∑ ord P ( h ) − P ∈ E ∑ ord P ( g ) 对于任何 P ∈ E P \in E P ∈ E f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ]
∑ P ∈ E ∖ { O } ord P ( f ) = Prop. 4.10 ∑ P ∈ E f ( P ) = 0 ord P ( f ) = Lemma 4.16 deg ( f ) \mathop{\sum }\limits_{{P \in E\smallsetminus \{ \mathcal{O}\} }}{\operatorname{ord}}_{P}\left( f\right) \overset{\text{ Prop. }{4.10}}{ = }\mathop{\sum }\limits_{\substack{{P \in E} \\ {f\left( P\right) = 0} }}{\operatorname{ord}}_{P}\left( f\right) \overset{\text{ Lemma }{4.16}}{ = }\deg \left( f\right) P ∈ E ∖ { O } ∑ ord P ( f ) = Prop. 4.10 P ∈ E f ( P ) = 0 ∑ ord P ( f ) = Lemma 4.16 deg ( f ) 另一方面，根据命题4.11，f f f O \mathcal{O} O − deg ( f ) - \deg \left( f\right) − deg ( f )
定理4.18。设 f f f E E E f f f E E E
证明。由于 f f f x x x y y y deg ( x ) = 2 \deg \left( x\right) = 2 deg ( x ) = 2 deg ( y ) = 3 \deg \left( y\right) = 3 deg ( y ) = 3
引理 4.19. 如果两个有理函数在 E E E E E E
证明. 设 f , g ∈ K ( E ) f,g \in K\left( E\right) f , g ∈ K ( E ) P ∈ E P \in E P ∈ E f ( P ) = g ( P ) f\left( P\right) = g\left( P\right) f ( P ) = g ( P ) h := f − g h \mathrel{\text{:=}} f - g h := f − g P ∈ E P \in E P ∈ E ord P ( h ) > 0 {\operatorname{ord}}_{P}\left( h\right) > 0 ord P ( h ) > 0
∑ P ∈ E f ( P ) = 0 ord P ( f ) \mathop{\sum }\limits_{\substack{{P \in E} \\ {f\left( P\right) = 0} }}{\operatorname{ord}}_{P}\left( f\right) P ∈ E f ( P ) = 0 ∑ ord P ( f ) 不是有限的。但是，如果 h h h deg ( h ) \deg \left( h\right) deg ( h )
引理 4.20. 没有有限极点的有理函数是多项式。
证明. 将没有极点的 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E ) r ( x , y ) = r\left( {x,y}\right) = r ( x , y ) = a ( x ) + y b ( x ) a\left( x\right) + {yb}\left( x\right) a ( x ) + y b ( x ) a , b ∈ K ( x ) a,b \in K\left( x\right) a , b ∈ K ( x )
r r r
⇒ r ˉ = a − y b \Rightarrow \;\bar{r} = a - {yb} ⇒ r ˉ = a − y b
⇒ r + r ˉ = 2 a \Rightarrow \;r + \bar{r} = {2a} ⇒ r + r ˉ = 2 a
⇒ y b = r − a \Rightarrow \;{yb} = r - a ⇒ y b = r − a
⇒ ( y b ) 2 = s b 2 \Rightarrow \;{\left( yb\right) }^{2} = s{b}^{2} ⇒ ( y b ) 2 = s b 2
如果 b b b b 2 {b}^{2} b 2 s b 2 s{b}^{2} s b 2 s s s
定义 4.21（有理映射）。一对有理函数 ( u , v ) ∈ K ( E A , B ) \left( {u,v}\right) \in K\left( {E}_{A,B}\right) ( u , v ) ∈ K ( E A , B )
v 2 = u 3 + A u + B {v}^{2} = {u}^{3} + {Au} + B v 2 = u 3 + A u + B 备注 4.22。由于有理映射 F = ( u , v ) F = \left( {u,v}\right) F = ( u , v ) u u u v v v P ∈ E P \in E P ∈ E
u ( P ) is (not) finite ⇔ v ( P ) is (not) finite u\left( P\right) \text{is (not) finite} \Leftrightarrow v\left( P\right) \text{is (not) finite} u ( P ) is (not) finite ⇔ v ( P ) is (not) finite 当我们约定 F ( P ) = O F\left( P\right) = \mathcal{O} F ( P ) = O u u u v v v P P P F F F P ↦ ( u ( P ) , v ( P ) ) P \mapsto \left( {u\left( P\right) ,v\left( P\right) }\right) P ↦ ( u ( P ) , v ( P ) ) E → E E \rightarrow E E → E
备注 4.23。给定一个域 K K K E E E
Y 2 = X 3 + A X + B {Y}^{2} = {X}^{3} + {AX} + B Y 2 = X 3 + A X + B 并考虑 E E E K ( E ) K\left( E\right) K ( E ) E ( K ( E ) ) E\left( {K\left( E\right) }\right) E ( K ( E ) ) K ( E ) K\left( E\right) K ( E ) E ( K ( E ) ) E\left( {K\left( E\right) }\right) E ( K ( E ) ) K ( E ) K\left( E\right) K ( E ) E ( K ( E ) ) E\left( {K\left( E\right) }\right) E ( K ( E ) ) K ( E ) K\left( E\right) K ( E ) O M {\mathcal{O}}_{M} O M O \mathcal{O} O
5 除子和直线
几乎所有上一个讲座的结果仅适用于非零函数
为了存储有理函数的零点和极点（以及它们的度数），我们将使用形式和。为此，我们回顾自由阿贝尔群的定义：
定义 5.1（自由阿贝尔群）。设 S S S S S S F S {\mathcal{F}}_{S} F S
∑ s ∈ S λ ( s ) ⋅ ⟨ s ⟩ \mathop{\sum }\limits_{{s \in S}}\lambda \left( s\right) \cdot \langle s\rangle s ∈ S ∑ λ ( s ) ⋅ ⟨ s ⟩ 其中 λ : S → Z \lambda : S \rightarrow \mathbb{Z} λ : S → Z λ ( s ) = 0 \lambda \left( s\right) = 0 λ ( s ) = 0 s ∈ S s \in S s ∈ S s ∈ S s \in S s ∈ S s ∈ S s \in S s ∈ S
定义 5.2（除数）。对于椭圆曲线 E E E E E E
Div ( E ) := F E \operatorname{Div}\left( E\right) \mathrel{\text{:=}} {\mathcal{F}}_{E} Div ( E ) := F E 对于一个除子 Δ = ∑ P ∈ E λ ( P ) ⋅ ⟨ P ⟩ ∈ Div ( E ) \Delta = \mathop{\sum }\limits_{{P \in E}}\lambda \left( P\right) \cdot \langle P\rangle \in \operatorname{Div}\left( E\right) Δ = P ∈ E ∑ λ ( P ) ⋅ ⟨ P ⟩ ∈ Div ( E ) Δ \Delta Δ
deg ( Δ ) := ∑ P ∈ E λ ( P ) \deg \left( \Delta \right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{P \in E}}\lambda \left( P\right) deg ( Δ ) := P ∈ E ∑ λ ( P ) 以及 Δ \Delta Δ
∣ Δ ∣ := ∑ P ∈ E ∖ { O } ∣ λ ( P ) ∣ \left| \Delta \right| \mathrel{\text{:=}} \mathop{\sum }\limits_{{P \in E\smallsetminus \{ \mathcal{O}\} }}\left| {\lambda \left( P\right) }\right| ∣ Δ ∣ := P ∈ E ∖ { O } ∑ ∣ λ ( P ) ∣ 事实 5.3。范数为 1 的除子具有形式 ± ⟨ P ⟩ + n ⟨ O ⟩ \pm \langle P\rangle + n\langle \mathcal{O}\rangle ± ⟨ P ⟩ + n ⟨ O ⟩ n ∈ Z n \in Z n ∈ Z
不做
命题 5.4（除数度的性质）。对于 Δ 1 , Δ 2 ∈ Div ( E ) {\Delta }_{1},{\Delta }_{2} \in \operatorname{Div}\left( E\right) Δ 1 , Δ 2 ∈ Div ( E )
deg ( Δ 1 + Δ 2 ) = deg ( Δ 1 ) + deg ( Δ 2 ) \deg \left( {{\Delta }_{1} + {\Delta }_{2}}\right) = \deg \left( {\Delta }_{1}\right) + \deg \left( {\Delta }_{2}\right) deg ( Δ 1 + Δ 2 ) = deg ( Δ 1 ) + deg ( Δ 2 ) 证明。注意对于 Δ i = ∑ P ∈ E λ i ( P ) ⟨ P ⟩ {\Delta }_{i} = \mathop{\sum }\limits_{{P \in E}}{\lambda }_{i}\left( P\right) \langle P\rangle Δ i = P ∈ E ∑ λ i ( P ) ⟨ P ⟩ Δ 1 + Δ 2 {\Delta }_{1} + {\Delta }_{2} Δ 1 + Δ 2
λ 1 + Δ 2 ) = deg ( ∑ P ∈ E λ 1 ( P ) ⟨ P ⟩ + ∑ P ∈ E λ 2 ( P ) ⟨ P ⟩ ) {\lambda }_{1} + {\Delta }_{2})\; = \;\deg \left( {\mathop{\sum }\limits_{{P \in E}}{\lambda }_{1}\left( P\right) \langle P\rangle + \mathop{\sum }\limits_{{P \in E}}{\lambda }_{2}\left( P\right) \langle P\rangle }\right) λ 1 + Δ 2 ) = deg ( P ∈ E ∑ λ 1 ( P ) ⟨ P ⟩ + P ∈ E ∑ λ 2 ( P ) ⟨ P ⟩ ) ∑ P ∈ E ( λ 1 ( P ) + λ 2 ( P ) ) \mathop{\sum }\limits_{{P \in E}}\left( {{\lambda }_{1}\left( P\right) + {\lambda }_{2}\left( P\right) }\right) P ∈ E ∑ ( λ 1 ( P ) + λ 2 ( P ) ) ∑ P ∈ E λ 1 ( P ) + ∑ P ∈ E λ 2 ( P ) \mathop{\sum }\limits_{{P \in E}}{\lambda }_{1}\left( P\right) + \mathop{\sum }\limits_{{P \in E}}{\lambda }_{2}\left( P\right) P ∈ E ∑ λ 1 ( P ) + P ∈ E ∑ λ 2 ( P ) 定义 5.5（相关除子）。对于一个有理函数 r ∈ K ( E ) ∖ { 0 } r \in K\left( E\right) \smallsetminus \{ 0\} r ∈ K ( E ) ∖ { 0 }
div ( r ) = ∑ P ∈ E ord P ( r ) ⋅ ⟨ P ⟩ \operatorname{div}\left( r\right) = \mathop{\sum }\limits_{{P \in E}}{\operatorname{ord}}_{P}\left( r\right) \cdot \langle P\rangle div ( r ) = P ∈ E ∑ ord P ( r ) ⋅ ⟨ P ⟩ 注解 5.6。根据引理 4.16，有理函数具有有限个零点和极点，因此相关除子是良定义的。
事实 5.7。常数非零函数的除子为 0。
有理函数的除数是记录有理函数的所有极点和零点信息的一种方式，即位置和重数。
事实 5.8。对于 f ∈ K [ E ] f \in K\left\lbrack E\right\rbrack f ∈ K [ E ]
∣ div ( f ) ∣ = D e f − ∑ P ∈ E ∖ { O } ord P ( f ) = Prop. 4.10 ∑ P ∈ E f ( P ) = 0 ord P ( f ) = Lemma 4.16 deg ( f ) \left| {\operatorname{div}\left( f\right) }\right| \overset{{De}{f}_{ - }}{ = }\mathop{\sum }\limits_{{P \in E\smallsetminus \{ \mathcal{O}\} }}{\operatorname{ord}}_{P}\left( f\right) \overset{\text{ Prop. }{4.10}}{ = }\mathop{\sum }\limits_{\substack{{P \in E} \\ {f\left( P\right) = 0} }}{\operatorname{ord}}_{P}\left( f\right) \overset{\text{ Lemma }{4.16}}{ = }\deg \left( f\right) ∣ div ( f ) ∣ = De f − P ∈ E ∖ { O } ∑ ord P ( f ) = Prop. 4.10 P ∈ E f ( P ) = 0 ∑ ord P ( f ) = Lemma 4.16 deg ( f ) 定义 5.9。令 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E )
lc ( r ) := [ ( x y ) ord O ( r ) ⋅ r ] ( O ) \operatorname{lc}\left( r\right) \mathrel{\text{:=}} \left\lbrack {{\left( \frac{x}{y}\right) }^{{\operatorname{ord}}_{\mathcal{O}}\left( r\right) } \cdot r}\right\rbrack \left( \mathcal{O}\right) lc ( r ) := [ ( y x ) ord O ( r ) ⋅ r ] ( O ) 示例 5.10。令 r ( x , y ) = 2 x 2 + 7 x 3 y x + 2 r\left( {x,y}\right) = \frac{2{x}^{2} + {7x}}{{3yx} + 2} r ( x , y ) = 3 y x + 2 2 x 2 + 7 x O \mathcal{O} O x y \frac{x}{y} y x
并且
( x y ) − 1 r ( x , y ) = 2 y x 2 + 7 y x 3 y x 2 + 2 x {\left( \frac{x}{y}\right) }^{-1}r\left( {x,y}\right) = \frac{{2y}{x}^{2} + {7yx}}{{3y}{x}^{2} + {2x}} ( y x ) − 1 r ( x , y ) = 3 y x 2 + 2 x 2 y x 2 + 7 y x 我们得到 deg O ( r ) = − 1 {\deg }_{\mathcal{O}}\left( r\right) = - 1 deg O ( r ) = − 1 O \mathcal{O} O lc ( r ) = 2 3 \operatorname{lc}\left( r\right) = \frac{2}{3} lc ( r ) = 3 2
命题 5.11。如果两个有理函数具有相同的除数，那么它们的商是常数。
证明。令 r 1 , r 2 ∈ K ( E ) {r}_{1},{r}_{2} \in K\left( E\right) r 1 , r 2 ∈ K ( E ) div ( r 1 ) = div ( r 2 ) \operatorname{div}\left( {r}_{1}\right) = \operatorname{div}\left( {r}_{2}\right) div ( r 1 ) = div ( r 2 ) ( r 1 ) \left( {r}_{1}\right) ( r 1 ) div ( r 2 ) \operatorname{div}\left( {r}_{2}\right) div ( r 2 ) r 1 {r}_{1} r 1 r 2 {r}_{2} r 2 div ( r 1 ) \operatorname{div}\left( {r}_{1}\right) div ( r 1 ) P ∈ E P \in E P ∈ E m m m P P P u u u
并写出
r 1 r 2 = u m s 1 u m s 2 = s 1 s 2 \frac{{r}_{1}}{{r}_{2}} = \frac{{u}^{m}{s}_{1}}{{u}^{m}{s}_{2}} = \frac{{s}_{1}}{{s}_{2}} r 2 r 1 = u m s 2 u m s 1 = s 2 s 1 对于 s 1 , s 2 ∈ K ( X , Y ) {s}_{1},{s}_{2} \in K\left( {X,Y}\right) s 1 , s 2 ∈ K ( X , Y ) P P P s 1 s 2 \frac{{s}_{1}}{{s}_{2}} s 2 s 1 P P P r 1 r 2 \frac{{r}_{1}}{{r}_{2}} r 2 r 1 P P P r 1 r 2 \frac{{r}_{1}}{{r}_{2}} r 2 r 1 r 1 {r}_{1} r 1 r 2 {r}_{2} r 2
因此，我们可以检查两个有理函数是否相等，如果它们具有相同的分母，并且在 E E E O \mathcal{O} O O \mathcal{O} O
引理 5.12。具有相同分母和最高次项系数的两个有理函数是相等的。
证明。设 r 1 , r 2 ∈ K ( E ) {r}_{1},{r}_{2} \in K\left( E\right) r 1 , r 2 ∈ K ( E )
由于命题 5.11，我们有 r 1 = c ⋅ r 2 {r}_{1} = c \cdot {r}_{2} r 1 = c ⋅ r 2
= Prop. 3.18 ( c − 1 ) [ ( x y ) d ⋅ r 2 ] ( O ) \overset{\text{ Prop. }{3.18}}{ = }\left( {c - 1}\right) \left\lbrack {{\left( \frac{x}{y}\right) }^{d} \cdot {r}_{2}}\right\rbrack \left( \mathcal{O}\right) = Prop. 3.18 ( c − 1 ) [ ( y x ) d ⋅ r 2 ] ( O ) 这意味着 c = 1 c = 1 c = 1 r 1 = r 2 {r}_{1} = {r}_{2} r 1 = r 2
示例 5.13。1. 设 P = ( a , b ) , P ′ = ( a , − b ) ∈ E P = \left( {a,b}\right) ,{P}^{\prime } = \left( {a, - b}\right) \in E P = ( a , b ) , P ′ = ( a , − b ) ∈ E b ≠ 0 b \neq 0 b = 0 r ( x , y ) = r\left( {x,y}\right) = r ( x , y ) = x − a x - a x − a
div ( r ) = ⟨ P ⟩ + ⟨ P ′ ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( r\right) = \langle P\rangle + \left\langle {P}^{\prime }\right\rangle - 2\langle \mathcal{O}\rangle div ( r ) = ⟨ P ⟩ + ⟨ P ′ ⟩ − 2 ⟨ O ⟩
设 P i = ( ω i , 0 ) ∈ E {P}_{i} = \left( {{\omega }_{i},0}\right) \in E P i = ( ω i , 0 ) ∈ E r ( x , y ) = y r\left( {x,y}\right) = y r ( x , y ) = y
div ( r ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( r\right) = \left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle + \left\langle {P}_{3}\right\rangle - 3\langle \mathcal{O}\rangle div ( r ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩
设 Q = ( 0 , B ) , Q ′ = ( 0 , − B ) ∈ E A , B Q = \left( {0,\sqrt{B}}\right) ,{Q}^{\prime } = \left( {0, - \sqrt{B}}\right) \in {E}_{A,B} Q = ( 0 , B ) , Q ′ = ( 0 , − B ) ∈ E A , B r = x y r = \frac{x}{y} r = y x B ≠ 0 B \neq 0 B = 0
div ( r ) = ⟨ Q ⟩ + ⟨ Q ′ ⟩ − ⟨ P 1 ⟩ − ⟨ P 2 ⟩ − ⟨ P 3 ⟩ + ⟨ O ⟩ \operatorname{div}\left( r\right) = \langle Q\rangle + \left\langle {Q}^{\prime }\right\rangle - \left\langle {P}_{1}\right\rangle - \left\langle {P}_{2}\right\rangle - \left\langle {P}_{3}\right\rangle + \langle \mathcal{O}\rangle div ( r ) = ⟨ Q ⟩ + ⟨ Q ′ ⟩ − ⟨ P 1 ⟩ − ⟨ P 2 ⟩ − ⟨ P 3 ⟩ + ⟨ O ⟩ 定义 5.14（主分母）。Δ ∈ Div ( E ) \Delta \in \operatorname{Div}\left( E\right) Δ ∈ Div ( E )
∃ r ∈ K ( E ) : Δ = div ( r ) \exists r \in K\left( E\right) : \Delta = \operatorname{div}\left( r\right) ∃ r ∈ K ( E ) : Δ = div ( r ) 此外，我们说 Δ 1 , Δ 2 ∈ Div ( E ) {\Delta }_{1},{\Delta }_{2} \in \operatorname{Div}\left( E\right) Δ 1 , Δ 2 ∈ Div ( E ) Δ 1 − Δ 2 {\Delta }_{1} - {\Delta }_{2} Δ 1 − Δ 2 Δ 1 ∼ Δ 2 {\Delta }_{1} \sim {\Delta }_{2} Δ 1 ∼ Δ 2
以下命题和推论表明 ∼ \sim ∼
事实 5.15。对于 r 1 , r 2 ∈ K ( E ) {r}_{1},{r}_{2} \in K\left( E\right) r 1 , r 2 ∈ K ( E )
div ( r 1 ⋅ r 2 ) = div ( r 1 ) + div ( r 2 ) \operatorname{div}\left( {{r}_{1} \cdot {r}_{2}}\right) = \operatorname{div}\left( {r}_{1}\right) + \operatorname{div}\left( {r}_{2}\right) div ( r 1 ⋅ r 2 ) = div ( r 1 ) + div ( r 2 ) 推论 5.16。对于 r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E )
div ( − r ) = div ( r ) \operatorname{div}\left( {-r}\right) = \operatorname{div}\left( r\right) div ( − r ) = div ( r )
− div ( r ) = div ( 1 r ) - \operatorname{div}\left( r\right) = \operatorname{div}\left( \frac{1}{r}\right) − div ( r ) = div ( r 1 )
证明。
div ( − r ) = div ( ( − 1 ) ⋅ r ) = Fact 5.15 div ( − 1 ) + div ( r ) = Fact 5.7 div ( r ) \operatorname{div}\left( {-r}\right) = \operatorname{div}\left( {\left( {-1}\right) \cdot r}\right) \overset{\text{ Fact }}{ = }{5.15}\operatorname{div}\left( {-1}\right) + \operatorname{div}\left( r\right) \overset{\text{ Fact }}{ = }{5.7}\operatorname{div}\left( r\right) div ( − r ) = div ( ( − 1 ) ⋅ r ) = Fact 5.15 div ( − 1 ) + div ( r ) = Fact 5.7 div ( r )
0 Fact 5.7 div ( 1 ) = div ( r r ) = div ( 1 r ⋅ r ) Fact 5.15 div ( 1 r ) + div ( r ) {0}^{\text{Fact }}{}^{5.7}\operatorname{div}\left( 1\right) = \operatorname{div}\left( \frac{r}{r}\right) = \operatorname{div}\left( {\frac{1}{r} \cdot r}\right) {}^{\text{Fact }}{}^{5.15}\operatorname{div}\left( \frac{1}{r}\right) + \operatorname{div}\left( r\right) 0 Fact 5.7 div ( 1 ) = div ( r r ) = div ( r 1 ⋅ r ) Fact 5.15 div ( r 1 ) + div ( r )
定义 5.17。我们定义以下两个 Div ( E ) \operatorname{Div}\left( E\right) Div ( E )
Prin ( E ) := { Δ ∈ Div ( E ) ∣ Δ is principal } \operatorname{Prin}\left( E\right) \mathrel{\text{:=}} \{ \Delta \in \operatorname{Div}\left( E\right) \mid \Delta \text{ is principal }\} Prin ( E ) := { Δ ∈ Div ( E ) ∣ Δ is principal } 以及
Div 0 ( E ) := { Δ ∈ Div ( E ) ∣ deg Δ = 0 } {\operatorname{Div}}_{0}\left( E\right) \mathrel{\text{:=}} \{ \Delta \in \operatorname{Div}\left( E\right) \mid \deg \Delta = 0\} Div 0 ( E ) := { Δ ∈ Div ( E ) ∣ deg Δ = 0 } 以及所谓的 Picard 群 E ‾ \underline{E} E E ‾ \underline{E} E
Pic ( E ) := Div ( E ) / Prin ( E ) \operatorname{Pic}\left( E\right) \mathrel{\text{:=}} \operatorname{Div}\left( E\right) /\operatorname{Prin}\left( E\right) Pic ( E ) := Div ( E ) / Prin ( E ) 由于定理 4.17，我们知道 Prin ( E ) ⊆ Div 0 ( E ) \operatorname{Prin}\left( E\right) \subseteq {\operatorname{Div}}_{0}\left( E\right) Prin ( E ) ⊆ Div 0 ( E )
Pic 0 ( E ) := Div 0 ( E ) / Prin ( E ) {\operatorname{Pic}}_{0}\left( E\right) \mathrel{\text{:=}} {\operatorname{Div}}_{0}\left( E\right) /\operatorname{Prin}\left( E\right) Pic 0 ( E ) := Div 0 ( E ) / Prin ( E ) 本章的目标将证明 Pic 0 ( E ) {\operatorname{Pic}}_{0}\left( E\right) Pic 0 ( E ) E E E
定义 5.18（直线）。在 E ‾ \underline{E} E
l ( x , y ) = α x + β y + γ l\left( {x,y}\right) = {\alpha x} + {\beta y} + \gamma l ( x , y ) = αx + β y + γ 其中 α , β , γ ∈ K \alpha ,\beta ,\gamma \in K α , β , γ ∈ K α ≠ 0 \alpha \neq 0 α = 0 β ≠ 0 \beta \neq 0 β = 0 P ∈ E P \in E P ∈ E l l l l l l P ‾ \underline{P} P P ‾ \underline{P} P l l l
命题 5.19。对于 P 1 , P 2 ∈ E {P}_{1},{P}_{2} \in E P 1 , P 2 ∈ E P 1 ≠ P 2 {P}_{1} \neq {P}_{2} P 1 = P 2 P 1 {P}_{1} P 1 P 2 {P}_{2} P 2
证明。通过 P i = ( a i , b i ) {P}_{i} = \left( {{a}_{i},{b}_{i}}\right) P i = ( a i , b i )
l ( x , y ) = { b 2 − b 1 a 2 − a 1 ( x − a 1 ) − ( y − b 1 ) if a 1 ≠ a 2 x − a 1 else l\left( {x,y}\right) = \left\{ \begin{array}{ll} \frac{{b}_{2} - {b}_{1}}{{a}_{2} - {a}_{1}}\left( {x - {a}_{1}}\right) - \left( {y - {b}_{1}}\right) & \text{ if }{a}_{1} \neq {a}_{2} \\ x - {a}_{1} & \text{ else } \end{array}\right. l ( x , y ) = { a 2 − a 1 b 2 − b 1 ( x − a 1 ) − ( y − b 1 ) x − a 1 if a 1 = a 2 else 定义了以 P 1 {P}_{1} P 1 P 2 {P}_{2} P 2
通过任意点绘制的以下特殊直线将非常有用：
命题 5.20。对于 P = ( a , b ) ∈ E P = \left( {a,b}\right) \in E P = ( a , b ) ∈ E
l ( x , y ) = m ( x − a ) − ( y − b ) l\left( {x,y}\right) = m\left( {x - a}\right) - \left( {y - b}\right) l ( x , y ) = m ( x − a ) − ( y − b ) 以 m = 3 a 2 + A 2 b m = \frac{3{a}^{2} + A}{2b} m = 2 b 3 a 2 + A P P P
∃ Q ∈ E : div ( l ) = 2 ⟨ P ⟩ + ⟨ Q ⟩ − 3 ⟨ O ⟩ \exists Q \in E : \operatorname{div}\left( l\right) = 2\langle P\rangle + \langle Q\rangle - 3\langle \mathcal{O}\rangle ∃ Q ∈ E : div ( l ) = 2 ⟨ P ⟩ + ⟨ Q ⟩ − 3 ⟨ O ⟩ 证明。我们明显看到，l ( P ) = 0 l\left( P\right) = 0 l ( P ) = 0 P P P u ( x , y ) = x − a u\left( {x,y}\right) = x - a u ( x , y ) = x − a P P P
l ( x , y ) = ( x − a ) 2 s ( x , y ) l\left( {x,y}\right) = {\left( x - a\right) }^{2}s\left( {x,y}\right) l ( x , y ) = ( x − a ) 2 s ( x , y ) 并且 s ( x , y ) = l ( x , y ) ( x − a ) 2 s\left( {x,y}\right) = \frac{l\left( {x,y}\right) }{{\left( x - a\right) }^{2}} s ( x , y ) = ( x − a ) 2 l ( x , y ) P P P g ( x , y ) := y − b x − a g\left( {x,y}\right) \mathrel{\text{:=}} \frac{y - b}{x - a} g ( x , y ) := x − a y − b b 2 = a 3 + A a + B {b}^{2} = {a}^{3} + {Aa} + B b 2 = a 3 + A a + B
g ( x , y ) = y − b x − a = y 2 − b 2 ( x − a ) ( y + b ) g\left( {x,y}\right) = \frac{y - b}{x - a} = \frac{{y}^{2} - {b}^{2}}{\left( {x - a}\right) \left( {y + b}\right) } g ( x , y ) = x − a y − b = ( x − a ) ( y + b ) y 2 − b 2 = x 3 + A x + B − b 2 ( x − a ) ( y + b ) = \frac{{x}^{3} + {Ax} + B - {b}^{2}}{\left( {x - a}\right) \left( {y + b}\right) } = ( x − a ) ( y + b ) x 3 + A x + B − b 2 = x 2 + a x + A + a 2 y + b = \frac{{x}^{2} + {ax} + A + {a}^{2}}{y + b} = y + b x 2 + a x + A + a 2 g ( a , b ) = 3 a 2 + A 2 b = m g\left( {a,b}\right) = \frac{3{a}^{2} + A}{2b} = m g ( a , b ) = 2 b 3 a 2 + A = m 现在我们使用 2 m b = 3 a 2 + A {2mb} = 3{a}^{2} + A 2 mb = 3 a 2 + A
s ( x , y ) = m ( x − a ) − ( y − b ) ( x − a ) 2 s\left( {x,y}\right) = \frac{m\left( {x - a}\right) - \left( {y - b}\right) }{{\left( x - a\right) }^{2}} s ( x , y ) = ( x − a ) 2 m ( x − a ) − ( y − b ) = m x − a − g ( x , y ) x − a = \frac{m}{x - a} - \frac{g\left( {x,y}\right) }{x - a} = x − a m − x − a g ( x , y ) = m x − a − x 2 + a x + A + a 2 x − a y + b = \frac{m}{x - a} - \frac{\frac{{x}^{2} + {ax} + A + {a}^{2}}{x - a}}{y + b} = x − a m − y + b x − a x 2 + a x + A + a 2 = m y + b x − a y + b − x + 2 a + A + 3 a 2 x − a y + b = \frac{m\frac{y + b}{x - a}}{y + b} - \frac{x + {2a} + \frac{A + 3{a}^{2}}{x - a}}{y + b} = y + b m x − a y + b − y + b x + 2 a + x − a A + 3 a 2 = m y + b x − a − x − 2 a − 2 m b x − a y + b = \frac{m\frac{y + b}{x - a} - x - {2a} - \frac{2mb}{x - a}}{y + b} = y + b m x − a y + b − x − 2 a − x − a 2 mb = m y − b x − a − x − 2 a y + b = \frac{m\frac{y - b}{x - a} - x - {2a}}{y + b} = y + b m x − a y − b − x − 2 a = m ⋅ g ( x , y ) − x − 2 a y + b = \frac{m \cdot g\left( {x,y}\right) - x - {2a}}{y + b} = y + b m ⋅ g ( x , y ) − x − 2 a 这在 P P P s ( a , b ) = m 2 − 3 a 2 b s\left( {a,b}\right) = \frac{{m}^{2} - {3a}}{2b} s ( a , b ) = 2 b m 2 − 3 a P P P
□ \square □
引理 5.21（直线的除数）。设 l l l
∣ div ( l ) ∣ ∈ { 2 , 3 } \left| {\operatorname{div}\left( l\right) }\right| \in \{ 2,3\} ∣ div ( l ) ∣ ∈ { 2 , 3 } 证明。 l ( x , y ) = α x + β y + γ l\left( {x,y}\right) = {\alpha x} + {\beta y} + \gamma l ( x , y ) = αx + β y + γ β = 0 \beta = 0 β = 0 β ≠ 0 \beta \neq 0 β = 0 l l l ∣ div ( l ) ∣ \left| {\operatorname{div}\left( l\right) }\right| ∣ div ( l ) ∣
命题5.22。设 l l l P 1 , P 2 , P 3 ∈ E {P}_{1},{P}_{2},{P}_{3} \in E P 1 , P 2 , P 3 ∈ E l l l
div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = \left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle + \left\langle {P}_{3}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩
div ( l ) = 2 ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = 2\left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = 2 ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 3 ⟨ O ⟩
div ( l ) = 3 ⟨ P 1 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = 3\left\langle {P}_{1}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = 3 ⟨ P 1 ⟩ − 3 ⟨ O ⟩
div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( l\right) = \left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle - 2\langle \mathcal{O}\rangle div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 2 ⟨ O ⟩
div ( l ) = 2 ⟨ P 1 ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( l\right) = 2\left\langle {P}_{1}\right\rangle - 2\langle \mathcal{O}\rangle div ( l ) = 2 ⟨ P 1 ⟩ − 2 ⟨ O ⟩
反之，每个这样的除数都有一条直线。
证明。首先我们证明所有可能的除数都由1-5给出。因为 l l l O \mathcal{O} O O \mathcal{O} O ord O ( l ) = − deg ( l ) ∈ { − 2 , − 3 } {\operatorname{ord}}_{\mathcal{O}}\left( l\right) = - \deg \left( l\right) \in \{ - 2, - 3\} ord O ( l ) = − deg ( l ) ∈ { − 2 , − 3 }
情况 ord O ( l ) = − 3 {\operatorname{ord}}_{\mathcal{O}}\left( l\right) = - 3 ord O ( l ) = − 3
三个单根 ( 1 . div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩ ) \left( {1\text{. div}\left( l\right) = \left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle + \left\langle {P}_{3}\right\rangle - 3\langle \mathcal{O}\rangle }\right) ( 1 . div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ + ⟨ P 3 ⟩ − 3 ⟨ O ⟩ )
一个单根和一个双根（2. div ( l ) = 2 ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = 2\left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = 2 ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 3 ⟨ O ⟩
一个三重根（3. div ( l ) = 3 ⟨ P 1 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = 3\left\langle {P}_{1}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = 3 ⟨ P 1 ⟩ − 3 ⟨ O ⟩
情况 ord O ( l ) = − 2 {\operatorname{ord}}_{\mathcal{O}}\left( l\right) = - 2 ord O ( l ) = − 2
两个单根 ( 4 . div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 2 ⟨ O ⟩ ) \left( {4\text{. div}\left( l\right) = \left\langle {P}_{1}\right\rangle + \left\langle {P}_{2}\right\rangle - 2\langle \mathcal{O}\rangle }\right) ( 4 . div ( l ) = ⟨ P 1 ⟩ + ⟨ P 2 ⟩ − 2 ⟨ O ⟩ )
一个双根（5. div ( l ) = 2 ⟨ P 1 ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( l\right) = 2\left\langle {P}_{1}\right\rangle - 2\langle \mathcal{O}\rangle div ( l ) = 2 ⟨ P 1 ⟩ − 2 ⟨ O ⟩
现在我们证明所有这些除数都是可能的。
情况 ord O ( l ) = − 3 {\operatorname{ord}}_{\mathcal{O}}\left( l\right) = - 3 ord O ( l ) = − 3
三个单根 对于 l ( x , y ) = y l\left( {x,y}\right) = y l ( x , y ) = y
div ( l ) = ⟨ Ω 1 ⟩ + ⟨ Ω 2 ⟩ + ⟨ Ω 3 ⟩ − 3 ⟨ O ⟩ \operatorname{div}\left( l\right) = \left\langle {\Omega }_{1}\right\rangle + \left\langle {\Omega }_{2}\right\rangle + \left\langle {\Omega }_{3}\right\rangle - 3\langle \mathcal{O}\rangle div ( l ) = ⟨ Ω 1 ⟩ + ⟨ Ω 2 ⟩ + ⟨ Ω 3 ⟩ − 3 ⟨ O ⟩ 一个单根和一个双根 Prop. 5.20
一个三重根 我们知道 P = ( 0 , B ) ∈ E P = \left( {0,\sqrt{B}}\right) \in E P = ( 0 , B ) ∈ E l ( x , y ) = l\left( {x,y}\right) = l ( x , y ) =
A x − y + B {Ax} - y + \sqrt{B} A x − y + B B ≠ 0 B \neq 0 B = 0 P P P
可以取均匀化子 u ( x , y ) = x u\left( {x,y}\right) = x u ( x , y ) = x
情况 ord O ( l ) = − 2 : Let P = ( a , b ) ∈ E {\operatorname{ord}}_{\mathcal{O}}\left( l\right) = - 2 : \operatorname{Let}P = \left( {a,b}\right) \in E ord O ( l ) = − 2 : Let P = ( a , b ) ∈ E l ( x , y ) := x − a l\left( {x,y}\right) \mathrel{\text{:=}} x - a l ( x , y ) := x − a
P P P P P P L L L
P P P P = Ω 1 P = {\Omega }_{1} P = Ω 1 P P P u ( x , y ) = y u\left( {x,y}\right) = y u ( x , y ) = y d = 2 d = 2 d = 2 s ( x , y ) = 1 ( x − ω 2 ) ( x − ω 3 ) s\left( {x,y}\right) = \frac{1}{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } s ( x , y ) = ( x − ω 2 ) ( x − ω 3 ) 1 P P P
u d ( x , y ) s ( x , y ) = y 2 ( x − ω 2 ) ( x − ω 3 ) = x − ω 1 = r ( x , y ) {u}^{d}\left( {x,y}\right) s\left( {x,y}\right) = \frac{{y}^{2}}{\left( {x - {\omega }_{2}}\right) \left( {x - {\omega }_{3}}\right) } = x - {\omega }_{1} = r\left( {x,y}\right) u d ( x , y ) s ( x , y ) = ( x − ω 2 ) ( x − ω 3 ) y 2 = x − ω 1 = r ( x , y ) 这说明 l l l P P P
记号 5.23。无论何时我们在除子中写一个 ? 作为系数，意味着该除子出现的陈述应该用“存在一个整的 ?”来量化。换句话说：我们不对系数的特殊值感兴趣。
定理 5.24（线性约化）。设 Δ ∈ Div ( E ) \Delta \in \operatorname{Div}\left( E\right) Δ ∈ Div ( E ) Δ ~ ∈ \widetilde{\Delta } \in Δ ∈
- Δ ~ ∼ Δ \text{-}\widetilde{\Delta } \sim \Delta - Δ ∼ Δ - deg ( Δ ~ ) = deg ( Δ ) \text{-}\deg \left( \widetilde{\Delta }\right) = \deg \left( \Delta \right) - deg ( Δ ) = deg ( Δ ) - ∣ Δ ~ ∣ ≤ 1 \text{-}\left| \widetilde{\Delta }\right| \leq 1 - Δ ≤ 1 证明。想法是，给定任意的除数 Δ \Delta Δ Δ 1 {\Delta }_{1} Δ 1
[ Δ 1 = Δ ± div ( l ) ] ⇔ [ Δ 1 − Δ = div ( l ) or Δ − Δ 1 = div ( l ) ] ⇔ [ Δ ∼ Δ 1 ] \left\lbrack {{\Delta }_{1} = \Delta \pm \operatorname{div}\left( l\right) }\right\rbrack \Leftrightarrow \left\lbrack {{\Delta }_{1} - \Delta = \operatorname{div}\left( l\right) \text{ or }\Delta - {\Delta }_{1} = \operatorname{div}\left( l\right) }\right\rbrack \Leftrightarrow \left\lbrack {\Delta \sim {\Delta }_{1}}\right\rbrack [ Δ 1 = Δ ± div ( l ) ] ⇔ [ Δ 1 − Δ = div ( l ) or Δ − Δ 1 = div ( l ) ] ⇔ [ Δ ∼ Δ 1 ] 并得到 Δ \Delta Δ
deg ( Δ 1 ) = Prop. 5.4 deg ( Δ ) + deg ( l ) = Thm. 4.17 deg ( Δ ) \deg \left( {\Delta }_{1}\right) \overset{\text{ Prop. }{5.4}}{ = }\deg \left( \Delta \right) + \deg \left( l\right) \overset{\text{ Thm. }{4.17}}{ = }\deg \left( \Delta \right) deg ( Δ 1 ) = Prop. 5.4 deg ( Δ ) + deg ( l ) = Thm. 4.17 deg ( Δ ) 我们尝试以规范减小的的方式进行。
所以写出 Δ = ∑ P ∈ E λ ( P ) ⟨ P ⟩ \Delta = \mathop{\sum }\limits_{{P \in E}}\lambda \left( P\right) \langle P\rangle Δ = P ∈ E ∑ λ ( P ) ⟨ P ⟩ Δ \Delta Δ Δ ′ {\Delta }^{\prime } Δ ′
Δ ′ = n 1 ⟨ P ⟩ − n 2 ⟨ Q ⟩ + ? ⟨ O ⟩ (1) {\Delta }^{\prime } = {n}_{1}\langle P\rangle - {n}_{2}\langle Q\rangle + ?\langle \mathcal{O}\rangle \tag{1} Δ ′ = n 1 ⟨ P ⟩ − n 2 ⟨ Q ⟩ + ? ⟨ O ⟩ ( 1 ) 其中 n 1 , n 2 ∈ N > 0 {n}_{1},{n}_{2} \in {\mathbb{N}}_{ > 0} n 1 , n 2 ∈ N > 0
假设 Δ \Delta Δ Δ \Delta Δ ⟨ P ⟩ \langle P\rangle ⟨ P ⟩ P = ( ω , 0 ) P = \left( {\omega ,0}\right) P = ( ω , 0 ) ⟨ P ⟩ \langle P\rangle ⟨ P ⟩ Δ ) \Delta ) Δ ) l ( x , y ) := x − ω l\left( {x,y}\right) \mathrel{\text{:=}} x - \omega l ( x , y ) := x − ω ⟨ l ⟩ = 2 ⟨ P ⟩ − 2 ⟨ O ⟩ \langle l\rangle = 2\langle P\rangle - 2\langle \mathcal{O}\rangle ⟨ l ⟩ = 2 ⟨ P ⟩ − 2 ⟨ O ⟩ P P P P P P l l l
现在假设我们可以取有限的 Q Q Q R R R Q ≠ R Q \neq R Q = R ⟨ Q ⟩ \langle Q\rangle ⟨ Q ⟩ ⟨ R ⟩ \langle R\rangle ⟨ R ⟩ l l l Q Q Q R R R
如果 λ ( Q ) , λ ( R ) < 0 : \lambda \left( Q\right) ,\lambda \left( R\right) < 0 : \; λ ( Q ) , λ ( R ) < 0 : Δ 1 := Δ + div ( l ) {\Delta }_{1} \mathrel{\text{:=}} \Delta + \operatorname{div}\left( l\right) Δ 1 := Δ + div ( l )
如果 λ ( Q ) , λ ( R ) > 0 : \lambda \left( Q\right) ,\lambda \left( R\right) > 0 : \; λ ( Q ) , λ ( R ) > 0 : Δ 1 := Δ − div ( l ) {\Delta }_{1} \mathrel{\text{:=}} \Delta - \operatorname{div}\left( l\right) Δ 1 := Δ − div ( l )
对于 Δ 1 = ∑ P ∈ E μ ( P ) ⟨ P ⟩ {\Delta }_{1} = \mathop{\sum }\limits_{{P \in E}}\mu \left( P\right) \langle P\rangle Δ 1 = P ∈ E ∑ μ ( P ) ⟨ P ⟩ ∣ μ ( Q ) ∣ = ∣ λ ( Q ) ∣ − 1 \left| {\mu \left( Q\right) }\right| = \left| {\lambda \left( Q\right) }\right| - 1 ∣ μ ( Q ) ∣ = ∣ λ ( Q ) ∣ − 1 ∣ μ ( R ) ∣ = ∣ λ ( R ) ∣ − 1 \left| {\mu \left( R\right) }\right| = \left| {\lambda \left( R\right) }\right| - 1 ∣ μ ( R ) ∣ = ∣ λ ( R ) ∣ − 1 ⟨ Q ⟩ \langle Q\rangle ⟨ Q ⟩ ⟨ R ⟩ \langle R\rangle ⟨ R ⟩
在 l l l ⟨ S ⟩ \langle S\rangle ⟨ S ⟩ S ∈ E S \in E S ∈ E ∣ Δ 1 ∣ < ∣ Δ ∣ \left| {\Delta }_{1}\right| < \left| \Delta \right| ∣ Δ 1 ∣ < ∣ Δ ∣
我们可以重复这个过程有限次，直到我们得到除数 Δ ′ {\Delta }^{\prime } Δ ′ Δ \Delta Δ
Δ ′ = n 1 ⟨ P ⟩ − n 2 ⟨ Q ⟩ + ? ⟨ O ⟩ {\Delta }^{\prime } = {n}_{1}\langle P\rangle - {n}_{2}\langle Q\rangle + ?\langle \mathcal{O}\rangle Δ ′ = n 1 ⟨ P ⟩ − n 2 ⟨ Q ⟩ + ? ⟨ O ⟩ 其中 n 1 , n 2 ∈ N > 0 {n}_{1},{n}_{2} \in {\mathbb{N}}_{ > 0} n 1 , n 2 ∈ N > 0
假设 n 1 > 1 {n}_{1} > 1 n 1 > 1
P P P l l l 5.20 ⇒ div ( l ) = 2 ⟨ P ⟩ + ⟨ S ⟩ − 3 ⟨ O ⟩ {5.20} \Rightarrow \operatorname{div}\left( l\right) = 2\langle P\rangle + \langle S\rangle - 3\langle \mathcal{O}\rangle 5.20 ⇒ div ( l ) = 2 ⟨ P ⟩ + ⟨ S ⟩ − 3 ⟨ O ⟩
P = ( ω , 0 ) P = \left( {\omega ,0}\right) P = ( ω , 0 ) l ( x , y ) = x − ω ⇒ ⟨ l ⟩ = 2 ⟨ P ⟩ − 2 ⟨ O ⟩ l\left( {x,y}\right) = x - \omega \Rightarrow \langle l\rangle = 2\langle P\rangle - 2\langle \mathcal{O}\rangle l ( x , y ) = x − ω ⇒ ⟨ l ⟩ = 2 ⟨ P ⟩ − 2 ⟨ O ⟩
减法减少了 n 1 {n}_{1} n 1 ∣ Δ ′ ∣ \left| {\Delta }^{\prime }\right| ∣ Δ ′ ∣ n 2 {n}_{2} n 2
⟨ P ⟩ − ⟨ Q ⟩ + ? ⟨ O ⟩ \langle P\rangle - \langle Q\rangle + ?\langle \mathcal{O}\rangle ⟨ P ⟩ − ⟨ Q ⟩ + ? ⟨ O ⟩ 其中 P = ( a , b ) P = \left( {a,b}\right) P = ( a , b ) l ( x , y ) = x − a l\left( {x,y}\right) = x - a l ( x , y ) = x − a div ( l ) = ⟨ P ⟩ + ⟨ R ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( l\right) = \langle P\rangle + \langle R\rangle - 2\langle \mathcal{O}\rangle div ( l ) = ⟨ P ⟩ + ⟨ R ⟩ − 2 ⟨ O ⟩ div ( l ) = 2 ⟨ P ⟩ − 2 ⟨ O ⟩ \operatorname{div}\left( l\right) = 2\langle P\rangle - 2\langle \mathcal{O}\rangle div ( l ) = 2 ⟨ P ⟩ − 2 ⟨ O ⟩
推论 5.25。对于每个 Δ ∈ Div 0 ( E ) \Delta \in {\operatorname{Div}}_{0}\left( E\right) Δ ∈ Div 0 ( E ) P ∈ E P \in E P ∈ E
Δ ∼ ⟨ P ⟩ − ⟨ O ⟩ \Delta \sim \langle P\rangle - \langle \mathcal{O}\rangle Δ ∼ ⟨ P ⟩ − ⟨ O ⟩ 证明。定理 5.24 告诉我们 Δ \Delta Δ Δ 1 = ± ⟨ P ⟩ + ? ⟨ O ⟩ {\Delta }_{1} = \pm \langle P\rangle + ?\langle \mathcal{O}\rangle Δ 1 = ± ⟨ P ⟩ + ? ⟨ O ⟩ ⟨ P ⟩ \langle P\rangle ⟨ P ⟩ P = ( a , b ) P = \left( {a,b}\right) P = ( a , b ) div ( l ) \operatorname{div}\left( l\right) div ( l ) l ( x , y ) = x − a l\left( {x,y}\right) = x - a l ( x , y ) = x − a
div ( l ) = { ⟨ P ⟩ + ⟨ Q ⟩ if P not of order tw 2 ⟨ P ⟩ − 2 ⟨ O ⟩ if P of order two \operatorname{div}\left( l\right) = \begin{cases} \langle P\rangle + \langle Q\rangle & \text{ if }P\text{ not of order tw } \\ 2\langle P\rangle - 2\langle \mathcal{O}\rangle & \text{ if }P\text{ of order two } \end{cases} div ( l ) = { ⟨ P ⟩ + ⟨ Q ⟩ 2 ⟨ P ⟩ − 2 ⟨ O ⟩ if P not of order tw if P of order two 为了获得 Δ 1 ∼ ⟨ Q ⟩ + ? ⟨ O ⟩ {\Delta }_{1} \sim \langle Q\rangle + ?\langle \mathcal{O}\rangle Δ 1 ∼ ⟨ Q ⟩ + ? ⟨ O ⟩ ⟨ P ⟩ + ? ⟨ O ⟩ ∼ Δ 1 \langle P\rangle + ?\langle \mathcal{O}\rangle \sim {\Delta }_{1} ⟨ P ⟩ + ? ⟨ O ⟩ ∼ Δ 1 Δ ∈ Div 0 ( E ) \Delta \in {\operatorname{Div}}_{0}\left( E\right) Δ ∈ Div 0 ( E ) 0 = deg Δ = deg Δ 1 0 = \deg \Delta = \deg {\Delta }_{1} 0 = deg Δ = deg Δ 1 ⟨ O ⟩ \langle \mathcal{O}\rangle ⟨ O ⟩ Δ ∼ Δ 1 = ⟨ P ⟩ − ⟨ O ⟩ \Delta \sim {\Delta }_{1} = \langle P\rangle - \langle \mathcal{O}\rangle Δ ∼ Δ 1 = ⟨ P ⟩ − ⟨ O ⟩
所以剩下要证明 P P P
⟨ P ⟩ − ⟨ O ⟩ ∼ Δ ∼ ⟨ Q ⟩ − ⟨ O ⟩ \langle P\rangle - \langle \mathcal{O}\rangle \sim \Delta \sim \langle Q\rangle - \langle \mathcal{O}\rangle ⟨ P ⟩ − ⟨ O ⟩ ∼ Δ ∼ ⟨ Q ⟩ − ⟨ O ⟩ 那么 ⟨ Q ⟩ ∼ Δ + ⟨ O ⟩ ∼ ⟨ P ⟩ \langle Q\rangle \sim \Delta + \langle \mathcal{O}\rangle \sim \langle P\rangle ⟨ Q ⟩ ∼ Δ + ⟨ O ⟩ ∼ ⟨ P ⟩ ⟨ Q ⟩ − ⟨ P ⟩ \langle Q\rangle - \langle P\rangle ⟨ Q ⟩ − ⟨ P ⟩ r ∈ K ( E ) r \in K\left( E\right) r ∈ K ( E ) div ( r ) = ⟨ P ⟩ − ⟨ Q ⟩ \operatorname{div}\left( r\right) = \langle P\rangle - \langle Q\rangle div ( r ) = ⟨ P ⟩ − ⟨ Q ⟩ P ≠ Q P \neq Q P = Q r r r div ( r ) = ⟨ S ⟩ − ⟨ O ⟩ \operatorname{div}\left( r\right) = \langle S\rangle - \langle \mathcal{O}\rangle div ( r ) = ⟨ S ⟩ − ⟨ O ⟩ r r r P = Q P = Q P = Q
定义一个映射 σ ˉ : Div 0 ( E ) → E \bar{\sigma } : {\operatorname{Div}}_{0}\left( E\right) \rightarrow E σ ˉ : Div 0 ( E ) → E σ ˉ ( Δ ) = P \bar{\sigma }\left( \Delta \right) = P σ ˉ ( Δ ) = P P P P Δ ∼ ⟨ P ⟩ − ⟨ O ⟩ \Delta \sim \langle P\rangle - \langle \mathcal{O}\rangle Δ ∼ ⟨ P ⟩ − ⟨ O ⟩ div ( r ) ∼ 0 \operatorname{div}\left( r\right) \sim 0 div ( r ) ∼ 0 σ ˉ ( ⟨ r ⟩ ) = O \bar{\sigma }\left( {\langle r\rangle }\right) = \mathcal{O} σ ˉ ( ⟨ r ⟩ ) = O σ ˉ \bar{\sigma } σ ˉ σ : Pic 0 ( E ) → E \sigma : {\operatorname{Pic}}_{0}\left( E\right) \rightarrow E σ : Pic 0 ( E ) → E
推论 5.26。 σ \sigma σ
证明。
满射。设 P ∈ E P \in E P ∈ E σ ( ⟨ P ⟩ − ⟨ O ⟩ ) = P \sigma \left( {\langle P\rangle -\langle \mathcal{O}\rangle }\right) = P σ ( ⟨ P ⟩ − ⟨ O ⟩ ) = P
单射。设 P , Q ∈ E , Δ ∈ Pic 0 ( E ) P,Q \in E,\Delta \in {\operatorname{Pic}}_{0}\left( E\right) P , Q ∈ E , Δ ∈ Pic 0 ( E ) σ ( Δ ) = P \sigma \left( \Delta \right) = P σ ( Δ ) = P σ ( Δ ) = Q \sigma \left( \Delta \right) = Q σ ( Δ ) = Q ∃ ! S ∈ E \exists !S \in E ∃ ! S ∈ E Δ ∼ ⟨ S ⟩ − ⟨ O ⟩ \Delta \sim \langle S\rangle - \langle \mathcal{O}\rangle Δ ∼ ⟨ S ⟩ − ⟨ O ⟩
P = S = Q . P = S = Q. P = S = Q .
参考文献
[CRD] Leonard S. Charlap, David P. Robbins, CRD Expository Report 31 "An Elementary Introduction to Elliptic Curves", December 1988
[WER] Anette Werner, "Elliptische Kurven in der Kryptographie", Springer Verlag 2002, ISBN 3-540-42518-7
