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\title{Weil Pairing - study}
\author{arnaucube}
\date{August 2022}

\begin{document}

\maketitle

\begin{abstract}
  Notes taken from \href{https://sites.google.com/site/matanprasma/artifact}{Matan Prsma} math seminars and also while reading about Bilinear Pairings. Usually while reading papers and books I take handwritten notes, this document contains some of them re-written to $LaTeX$.

	The notes are not complete, don't include all the steps neither all the proofs. I use these notes to revisit the concepts after some time of reading the topic.
\end{abstract}

\tableofcontents

\section{Divisors and rational functions}

\begin{definition}{Divisor}
  $$D= \sum_{P \in E(\mathbb{K})} n_p \cdot [P]$$
\end{definition}

\begin{definition}{Degree \& Sum}
  $$deg(D)= \sum_{P \in E(\mathbb{K})} n_p$$
  $$sum(D)= \sum_{P \in E(\mathbb{K})} n_p \cdot P$$
\end{definition}

\begin{definition}{Principal divisor}
  iff $deg(D)=0$ and $sum(D)=0$
\end{definition}
$D \sim D'$ iff $D - D'$ is principal.


\begin{definition}{Evaluation of a rational function}
  $$r(D)= \prod r(P)^{n_p}$$
\end{definition}

\section{Weil reciprocity}
\begin{theorem}{(Weil reciprocity)}
  Let $E/ \mathbb{K}$ be an e.c. over an alg. closed field. If $r,~s \in \mathbb{K}\setminus \{0\}$ are rational functions whose divisors have disjoint support, then
$$r(div(s)) = s(div(r))$$
\end{theorem}
Proof. (todo)

\section{Generic Weil Pairing}
Let $E(\mathbb{K})$, with $\mathbb{K}$ of char $p$, $n$ s.t. $p \nmid n$.

$\mathbb{K}$ large enough: $E(\mathbb{K})[n] = E(\mathbb{\overline{K}}) = \mathbb{Z}_n \oplus \mathbb{Z}_n$ (with $n^2$ elements).

For $P, Q \in E[n]$,
\begin{align*}
  D_P &\sim [P] - [0]\\
  D_Q &\sim [Q] - [0]
\end{align*}

We need them to have disjoint support:
\begin{align*}
  D_P &\sim [P] - [0]\\
  D_Q' &\sim [Q+T] - [T]
\end{align*}

$$\Delta D = D_Q - D_Q' = [Q] - [0] - [Q+T] + [T]$$


Note that $n D_P$ and $n D_Q$ are principal. Proof:
\begin{align*}
  n D_P &= n [P] - n [O]\\
  deg(n D_P) &= n - n = 0\\
  sum(n D_P) &= nP - nO = 0
\end{align*}
($nP = 0$ bcs. $P$ is n-torsion)

Since $n D_P,~ n D_Q$ are principal, we know that $f_P,~ f_Q$ exist.

Take
\begin{align*}
  f_P &: div(f_P) = n D_P\\
  f_Q &: div(f_Q) = n D_Q
\end{align*}

We define
$$
e_n(P, Q) = \frac{f_P(D_Q)}{f_Q(D_P)}
$$

Remind: evaluation of a rational function over a divisor $D$:
\begin{align*}
  D &= \sum n_P [P]\\
  r(D) &= \prod r(P)^{n_P}
\end{align*}

If $D_P = [P+S] - [S],~~ D_Q=[Q-T]-[T]$ what is $e_n(P, Q)$?

\begin{align*}
  f_P(D_Q) &= f_P(Q+T)^1 \cdot f_P(T)^{-1}\\
  f_Q(D_P) &= f_Q(P+S)^1 \cdot f_Q(S)^{-1}
\end{align*}

$$
e_n(P, Q) = \frac{f_P(Q+T)}{f_P(T)} / \frac{f_Q(P+S)}{f_Q(S)}
$$

with $S \neq \{O, P, -Q, P-Q \}$.


\section{Properties}


\section{Exercises}
\emph{An Introduction to Mathematical Cryptography, 2nd Edition} - Section 6.8. Bilinear pairings on elliptic curves

\begin{solution}{6.29}
  $div(R(x) \cdot S(x)) = div( R(x)) + div( S(x))$, where $R(x), S(x)$ are rational functions.
  \\proof:\\
  \emph{Norm} of $f$: $N_f = f \cdot \overline{f}$, and we know that $N_{fg} = N_f \cdot N_g~\forall~\mathbb{K}[E]$,\\
  then $$deg(f) = deg_x(N_f)$$\\
  and $$deg(f \cdot g) = deg(f) + deg(g)$$

  Proof:
  $$deg(f \cdot g) = deg_x(N_{fg}) = deg_x(N_f \cdot N_g)$$
  $$= deg_x(N_f) + deg_x(N_g) = deg(f) + deg(g)$$

  So, $\forall P \in E(\mathbb{K}),~ ord_P(rs) = ord_P(r) + ord_P(s)$.\\
  As $div(r) = \sum_{P\in E(\mathbb{K})} ord_P(r)[P]$, $div(s) = \sum ord_P(s)[P]$.

  So,
  $$div(rs) = \sum ord_P(rs)[P]$$
  $$= \sum ord_P(r)[P] + \sum ord_P(s)[P] = div(r) + div(s)$$
\end{solution}

\vspace{0.5cm}

\begin{solution}{6.31}
  $$e_m(P, Q) = e_m(Q, P)^{-1} \forall P, Q \in E[m]$$
  Proof:
  We know that $e_m(P, P) = 1$, so:
  $$1 = e_m(P+Q, P+Q) = e_m(P, P) \cdot e_m(P, Q) \cdot e_m(Q, P) \cdot e_m(Q, Q)$$

  and we know that $e_m(P, P) = 1$, then we have:
  $$1 = e_m(P, Q) \cdot e_m(Q, P)$$
  $$\Longrightarrow e_m(P, Q) = e_m(Q, P)^{-1}$$
\end{solution}




\end{document}
