@ -151,14 +151,16 @@ Observe that
\end { enumerate}
\end { enumerate}
Observe that
Observe that
$$ \sum { P \in E ( \Bbbk ) } ord _ P ( r ) \cdot P = 0 $$
$$ \sum _ { P \in E ( \Bbbk ) } ord _ P ( r ) \cdot P = 0 $$
\begin { definition} { Support}
$$ \sum _ P n _ P [ P ] , ~ \forall P \in E ( \Bbbk ) \mid n _ P \neq 0 $$
\begin { definition} { Support of a divisor }
$$ \sum _ P n _ P [ P ] , ~ \forall P \in E ( \Bbbk ) ~ \text { s.t. } ~ n _ P \neq 0 $$
\end { definition}
\end { definition}
\begin { definition} { Principal divisor}
\begin { definition} { Principal divisor}
iff $ deg ( D ) = 0 $ and $ sum ( D ) = 0 $
iff
$$ deg ( D ) = 0 $$
$$ sum ( D ) = 0 $$
\end { definition}
\end { definition}
$ D \sim D' $ iff $ D - D' $ is principal.
$ D \sim D' $ iff $ D - D' $ is principal.
@ -169,11 +171,24 @@ $D \sim D'$ iff $D - D'$ is principal.
\section { Weil reciprocity}
\section { Weil reciprocity}
\begin { theorem} { (Weil reciprocity)}
\begin { theorem} { (Weil reciprocity)}
Let $ E / \Bbbk $ be an e.c. over an alg. closed field. If $ r,~s \in \Bbbk \setminus \{ 0 \} $ are rational functions whose divisors have disjoint support, then
Let $ E / \Bbbk $ be an e.c. over an algebraically closed field. If $ r,~s \in \Bbbk \setminus \{ 0 \} $ are rational functions whose divisors have disjoint support, then
$$ r ( div ( s ) ) = s ( div ( r ) ) $$
$$ r ( div ( s ) ) = s ( div ( r ) ) $$
\end { theorem}
\end { theorem}
Proof. (todo)
Proof. (todo)
\paragraph { Example}
\begin { align*}
p(x)=x^ 2 - 1,& ~ q(x)=\frac { x} { x-2} \\
div(p)& = 1 \cdot [1] + 1 \cdot [-1] - 2 \cdot [\infty ]\\
div(q)& = 1 \cdot [0] - 1 \cdot [2]\\
& \text { (they have disjoint support)} \\
p(div(q)) & = p(0)^ 1 \cdot p(2)^ { -1} = (0^ 2 - 1)^ 1 \cdot (2^ 2 - 1)^ { -1} = \frac { -1} { 3} \\
q(div(p)) & = q(1)^ 1 \cdot q(-1)^ 1 - q(\infty )^ 2\\
& = (\frac { 1} { 1-2} )^ 1 \cdot (\frac { -1} { -1-2} )^ 1 \cdot (\frac { \infty } { \infty - 2} )^ 2 = \frac { -1} { 3}
\end { align*}
so, $ p ( div ( q ) ) = q ( div ( p ) ) $ .
\section { Generic Weil Pairing}
\section { Generic Weil Pairing}
Let $ E ( \Bbbk ) $ , with $ \Bbbk $ of char $ p $ , $ n $ s.t. $ p \nmid n $ .
Let $ E ( \Bbbk ) $ , with $ \Bbbk $ of char $ p $ , $ n $ s.t. $ p \nmid n $ .
@ -236,6 +251,24 @@ with $S \neq \{O, P, -Q, P-Q \}$.
\section { Properties}
\section { Properties}
\begin { enumerate} [i.]
\item $ e _ n ( P, Q ) ^ n = 1 ~ \forall P,Q \in E [ n ] $ \\
($ \Rightarrow ~ e _ n ( P,Q ) $ is a $ n ^ { th } $ root of unity)
\item Bilinearity
$$ e _ n ( P _ 1 + P _ 2 , Q ) = e _ n ( P _ 1 , Q ) \cdot e _ n ( P _ 2 , Q ) $$
$$ e _ n ( P, Q _ 1 + Q _ 2 ) = e _ n ( P, Q _ 1 ) \cdot e _ n ( P, Q _ 2 ) $$
\emph { proof:}
recall that $ e _ n ( P,Q ) = \frac { g ( S + P ) } { g ( S ) } $ , then,
\begin { align*}
e_ n(P_ 1, Q) & \cdot e_ n(P_ 2, Q) = \frac { g(P_ 1 + S)} { g(S)} \cdot \frac { g(P_ 2 + P_ 1 + S)} { g(P_ 1 + S)} \\
& \text { (replace $ S $ by $ S + P _ 1 $ )} \\
& = \frac { g(P_ 2 + P_ 1 + S)} { g(S)} = e_ n(P_ 1+P_ 2, Q)
\end { align*}
\item Alternating
$$ e _ n ( P, P ) = 1 ~ \forall P \in E [ n ] $$
\item Nondegenerate
$$ \text { if } ~ e _ n ( P,Q ) = 1 ~ \forall Q \in E [ n ] ,~ \text { then } ~ P = 0 $$
\end { enumerate}
\section { Exercises}
\section { Exercises}