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Add weil reciprocity example, weil pairing properties, small typo in hypernova notes

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arnaucube 10 months ago
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cc9bb99ac5
4 changed files with 42 additions and 11 deletions
  1. BIN
      notes_hypernova.pdf
  2. +4
    -6
      notes_hypernova.tex
  3. BIN
      weil-pairing.pdf
  4. +38
    -5
      weil-pairing.tex

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notes_hypernova.pdf


+ 4
- 6
notes_hypernova.tex

@ -349,17 +349,15 @@ $$\theta_{k,j} = \sum_{y \in \{0, 1\}^{s'}} \widetilde{M}_j(r_x', y) \cdot \wide
And in \emph{step 5}, $V$ checks
% TODO check orange gamma^j...
\begin{align*}
c &= \left(\sum_{j \in [t]} \gamma^j \cdot e_1 \cdot \sigma_{1,j}
~\textcolor{orange}{+ \gamma^{t+j} \cdot e_1 \cdot \sigma_{2,j}}\right)\\
&+ \gamma^{2t+1} \cdot e_2 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)
+ \textcolor{cyan}{\gamma^{2t+2} \cdot e_2 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)}
~\textcolor{orange}{+ \gamma^{t+j} \cdot e_2 \cdot \sigma_{2,j}}\right)\\
&+ \gamma^{2t+1} \cdot e_3 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)
+ \textcolor{cyan}{\gamma^{2t+2} \cdot e_4 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)}
\end{align*}
where
% TODO check e_4
$e_1 \leftarrow \widetilde{eq}(r_{1,x}, r_x'),~ e_2 \leftarrow \widetilde{eq}(r_{2,x}, r_x')$, $e_3 \leftarrow \widetilde{eq}(\beta, r_x'),~ e_4 \leftarrow \widetilde{eq}(\beta', r_x')$ (note: wip, pending check for $\beta, \beta'$ used in step 3).
$e_1 \leftarrow \widetilde{eq}(r_{1,x}, r_x'),~ e_2 \leftarrow \widetilde{eq}(r_{2,x}, r_x')$, $e_3, e_4 \leftarrow \widetilde{eq}(\beta, r_x')$.
\vspace{0.5cm}

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weil-pairing.pdf


+ 38
- 5
weil-pairing.tex

@ -151,14 +151,16 @@ Observe that
\end{enumerate}
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}
\begin{definition}{Principal divisor}
iff $deg(D)=0$ and $sum(D)=0$
iff
$$deg(D)=0$$
$$sum(D)=0$$
\end{definition}
$D \sim D'$ iff $D - D'$ is principal.
@ -169,11 +171,24 @@ $D \sim D'$ iff $D - D'$ is principal.
\section{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
$$r(div(s)) = s(div(r))$$
\end{theorem}
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}
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}
\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}

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