Add Caulk initial notes

2026-01-09 15:31:36 +01:00 · 2023-02-12 18:54:01 +01:00
parent 8dc59ea1b6
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--- a/README.md
+++ b/README.md
@@ -1,3 +1,24 @@
 # math

 Notes, code and documents done while reading books and papers.
+
+### Sage code
+- [blind signatures over elliptic curve](blind-sign-over-ec.sage)
+- [BLS signatures](bls-sigs.sage)
+- [FFT](fft.sage)
+- [IPA](ipa.sage)
+- [KZG commitments](kzg.sage)
+- [Powers of Tau](powersoftau.sage)
+- [Ring signatures](ring-signatures.sage)
+- [Sigma protocol](sigma.sage)
+
+### Notes
+- [Notes on "Abstract Algebra" book, by Charles C. Pinter](abstract-algebra-charles-pinter-notes.pdf)
+- [Notes on Caulk paper](notes_caulk.pdf)
+- [Notes on the DFT & FFT](fft-notes.pdf)
+- [Notes on the BLS signatures](notes_bls-sig.pdf)
+- [Notes on IPA from Halo paper](notes_halo.pdf)
+- [Notes on Sonic paper](notes_sonic.pdf)
+- [Notes on Weil pairing](weil-pairing.pdf)
+- [Notes on Sigma protocol and OR proofs](sigma-or-notes.pdf)
+- [Notes on Reed-Solomon codes](notes_reed-solomon.pdf)
--- a/notes_caulk.pdf
+++ b/notes_caulk.pdf
--- a/notes_caulk.tex
+++ b/notes_caulk.tex
@@ -0,0 +1,347 @@
+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{enumerate}
+\usepackage{hyperref}
+% \hypersetup{
+%     colorlinks,
+%     citecolor=blue,
+%     filecolor=blue,
+%     linkcolor=black,
+%     urlcolor=blue
+% }
+\usepackage{xcolor}
+
+% prevent warnings of underfull \hbox:
+\usepackage{etoolbox}
+\apptocmd{\sloppy}{\hbadness 4000\relax}{}{}
+
+\theoremstyle{definition}
+\newtheorem{definition}{Def}[section]
+\newtheorem{theorem}[definition]{Thm}
+
+% custom lemma environment to set custom numbers
+\newtheorem{innerlemma}{Lemma}
+\newenvironment{lemma}[1]
+{\renewcommand\theinnerlemma{#1}\innerlemma}
+{\endinnerlemma}
+
+
+\title{Notes on Caulk and Caulk+}
+\author{arnaucube}
+\date{February 2023}
+
+\begin{document}
+
+\maketitle
+
+\begin{abstract}
+	Notes taken while reading about Caulk \cite{cryptoeprint:2022/621} and Caulk+ \cite{cryptoeprint:2022/957}.
+
+	Usually while reading papers 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.
+\end{abstract}
+
+\tableofcontents
+
+\section{Preliminaries}
+
+\subsection{Lagrange Polynomials and Roots of Unity}
+
+Let $\omega$ denote a root of unity, such that $\omega^N=1$. Set $\mathbb{H}=\{1, \omega, \omega^2, \ldots, \omega^{N^{-1}}\}$.
+
+Let the $i^{th}$ Lagrange polynomial be $\lambda_i(X)=\prod_{s\neq i-1} \frac{X-\omega^s}{\omega^{i-1} -\omega}$.
+
+Notice that $\lambda_i(\omega^{i-1})=1$ and $\lambda_i(w^j)=0,~\forall j\neq i-1$.
+
+Let the vanishing polynomial of $\mathbb{H}$ be $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$.
+
+\subsection{KZG Commitments}
+KZG as a Vector Commitment.
+
+We have vector $\overrightarrow{c}=\{c_i\}_1^n$, which can be interpolated into $C(X)$ through Lagrange polynomials $\{ \lambda_i(X) \}$:
+$$C(X) = \sum^n_{i=1} c_i \cdot \lambda_i(X)$$
+so, $C(\omega^{i-1})=c_i$.
+
+Commitment:
+$$C = [C(X)]_1 = \sum^n_{i=1} c_i \cdot [\lambda_i(X)]_1$$
+
+Proof of \textbf{opening for single value} $v$ at position $i$:
+
+$$Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$
+
+$$\pi_{KZG} = Q =[Q(X)]_1$$
+
+Verification:
+$$e(C - [v]_1, [1]_2) = e(\pi_{KZG},~[X-\omega^{i-1}]_2)$$
+unfold
+$$e([C(X)]_1 - [v]_1, [1]_2) = e([Q(X)]_1,~[X-\omega^{i-1}]_2)$$
+$$C(X) - v = Q(X) \cdot (X-\omega^{i-1}) \Longrightarrow Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$
+
+
+Proof of \textbf{opening for a subset} of positions $I \subset [N]$:
+
+$[H_I]_1$ such that for
+$$C_I(X) = \sum_{i \in I} c_i \cdot \tau_i(X)$$
+$$z_I(X) = \prod_{i \in I} (X - \omega^{i-1})$$
+
+for $\{ \tau_i(X) \}_{i \in I}$ being the Lagrange interpolation polynomials over $\mathbb{H}_I = \{\omega^{i-1}\}_{i \in I}$.
+\emph{(recall, $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$))}
+
+
+$H_I(X)$ can be computed by
+$$H_I(X) = \frac{C(X) - C_I(X)}{z_I(X)}$$
+
+So, prover commits to $C_I(X)$ with $C_I = [C_I(X)]_1$, and computes $\pi_{KZG}$:
+$$\pi_{KZG} = H_I = [H_I(X)]_1$$
+
+Then, verification checks:
+$$e(C - C_I, [1]_2) = e(\pi_{KZG}, [z_I(X)]_2)$$
+unfold
+$$e([C(X)]_1 - [C_I(X)]_1, [1]_2) = e([H_I(X)]_1, [z_I(X)]_2)$$
+$$C(X) - C_I(X) = H_I(X) \cdot z_I(X)$$
+$$C(X) - C_I(X) = \frac{C(X)-C_I(X)}{z_I(X)} \cdot z_I(X)$$
+
+
+
+\subsection{Pedersen Commitments}\label{sec:pedersen}
+Commitment
+$$cm = v [1]_1 + r [h]_1 = [v + hr]_1$$
+
+Prove knowledge of $v,~r$, Verifier sends challenge $\{s_1, s_2\}$. Prover computes:
+$$R = s_1 [1]_1 + s_2 [h]_1 = [s_1 + h s_2]_1$$
+$$c= H(cm, R)$$
+$$t_1 = s_1 + v c, ~~ t_2=s_2 + r c$$
+
+Verification:
+$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1$$
+
+unfold:
+$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1 = [t_1 + h t_2]$$
+$$[s_1 + h s_2]_1 + c \cdot [v + hr]_1 == [s_1 + vc + h(s_2 + rc)]_1$$
+$$[s_1 + h s_2 + cv + rch]_1 == [s_1 + vc + h s_2 + rch]_1$$
+
+
+\section{Caulk}
+\subsection{Blinded Evaluation}
+Main idea: combine KZG commitments with Pedersen commitments to prove knowledge of a value $v$ which Pedersen commitment is committed in the KZG commitment.
+
+Let $C(X) = \sum^N_{i=1} c_i \lambda_i(X)$, where $\overrightarrow{c} = \{c_i\}_{i \in I}$. In normal KZG, prover would compute $Q(X)=\frac{C(X)-v}{X-\omega^{i-1}}$, and  send $[Q(X)]_1$ as proof. We will obfuscate the commitment:
+
+rand $a \in \mathbb{F}$, blind commit to $z(X)=aX - b = a(X - \omega^{i-1})$, where $\omega^{i-1}=b/a$. Denote by $[z]_2$ the commitment to $[z(X)]_2$.
+
+Prover computes:
+\begin{enumerate}[i.]
+	\item $\pi_{ped}$, Pedersen proof that $cm$ is from $v, r$ (section \ref{sec:pedersen})
+	\item $\pi_{unity}$ (see \ref{sec:pi-unity})
+	\item For random $s$ computes:
+		$$T(X)=\frac{Q(X)}{a} + hs \longrightarrow [T]_1=[T(X)]_1$$
+		$$S(X) = -r - s \cdot z(X) \longrightarrow [S]_2 = [S(X)]_2$$
+\end{enumerate}
+
+i, ii, iii defines the \emph{zk proof of membership}, which proves that $(v, r)$ is a opening of $cm$, and $v$ opens $C$ at $\omega^{i-1}$.
+
+Verifier checks proofs $\pi_{ped},~\pi_{unity}$ (i, ii), and checks
+$$e(C - cm, [1]_2) == e([T]_1, [z]_2) + e([h]_1, [S]_2)$$
+
+unfold:
+\begin{align*}
+	C(X) - cm &== T(X) \cdot z(X) + h \cdot S(X) \\
+	C(X) - v - hr &== (\frac{Q(X)}{a} + s h) \cdot z(X) + h (-r -s \cdot z(X)) \\
+	C(X) - v &== hr + (\frac{Q(X)}{a}) z(X) + sh \cdot z(X) - hr - sh \cdot z(X) \\
+	C(X) - v &== \frac{Q(X)}{a} \cdot z(X) \\
+	C(X) - v &== \frac{Q(X)}{a} \cdot a(X-\omega^{i-1}) \\
+	C(X) -v &== Q(X) \cdot (X - \omega^{i-1})
+\end{align*}
+Which matches with the definition of $Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$.
+
+\subsubsection{\texorpdfstring{Correct computation of $z(x)$, $\pi_{unity}$}%
+				{Correct computation of proof unity}}\label{sec:pi-unity}
+
+Want to prove that prover knows $a, b$ such that $[z]_2 = [a X - b]_2$, and $a^N = b^N$.
+
+To prove $\frac{a}{b}$ is inside the evaluation domain (ie. $\frac{a}{b}$ is a N$^{th}$ root of unity), proves (in $log(N)$ time) that its N$^{th}$ is one ($\frac{a}{b} = 1$).
+
+Conditions:
+\begin{enumerate}[i.]
+	\item $f_0=\frac{a}{b}$
+	\item $f_i = f_{i-1}^2,~\forall~i=1, \ldots, log(N)$
+	\item $f_{log(N)} = 1$
+\end{enumerate}
+
+
+Redefine i, and from there, redefine ii, iii:
+% \begin{minipage}[t]{0.45\textwidth}
+%         \begin{enumerate}[i.]
+%                 \item \begin{align*}
+%                         f_0 &= z(1) = a - b\\
+%                         f_1 &= z(\sigma) a \sigma -b\\
+%                         f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\
+%                         f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\
+%                         f_4 &= \frac{f_2}{f_3} = \frac{a}{b}
+%                 \end{align*}
+%         \end{enumerate}
+%         \end{minipage}
+%                 \begin{enumerate}[i.]
+%                         \item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$
+%                         \item $f_{4+log(N)} = 1$
+%                 \end{enumerate}
+%   \begin{minipage}[t]{0.45\textwidth}
+% \end{minipage}
+
+\begin{enumerate}[i.]
+	\item \begin{align*}
+		f_0 &= z(1) = a - b\\
+		f_1 &= z(\sigma) a \sigma -b\\
+		f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\
+		f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\
+		f_4 &= \frac{f_2}{f_3} = \frac{a}{b}
+	\end{align*}
+	\item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$
+	\item $f_{4+log(N)} = 1$
+\end{enumerate}
+
+\begin{lemma}{1}
+	Let $z(X)$ $deg=1$, $n=log(N)+6$, $\sigma$ such that $\sigma^n =1$.
+
+	If $\exists$ $f(X) \in \mathbb{F}[X]$ such that
+	\begin{enumerate}[1.]
+		\item $f(X) = z(X)$, for $1, \sigma$
+		\item $f(\sigma^2)(1-\sigma)=f(1)-f(\sigma)$
+		\item $f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)$
+		\item $f(\sigma^4)f(\sigma^3)=f(\sigma^2)$
+		\item $f(\sigma^{4+i+1})=f(\sigma^{4+i})^2,~~\forall i= 0, \ldots, log(N)-1$
+		\item $f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1$
+	\end{enumerate}
+	then, $z(X)=aX-b$, where $\frac{b}{a}$ is a N$^{th}$ root of unity.
+\end{lemma}
+
+Let's see the relations between the conditions and the Lemma 1:
+\begin{scriptsize}
+\begin{align*}
+	Conditions &\longrightarrow Lemma~1\\
+	\begin{array}{l}
+		f_0 = z(1) = a - b\\ f_1 = z(\sigma) a \sigma -b
+	\end{array} &\longrightarrow 1.~f(X) = z(X), for 1, \sigma\\
+	f_2 = \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a &\longrightarrow 2.~f(\sigma^2)(1-\sigma)=f(1)-f(\sigma)\\
+	f_3 = \sigma f_2 - f_1 = \sigma a - a \sigma + b = b &\longrightarrow 3.~f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)\\
+	f_4 = \frac{f_2}{f_3} = \frac{a}{b} &\longrightarrow 4.~f(\sigma^4)f(\sigma^3)=f(\sigma^2)\\
+	f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1 &\longrightarrow 5.~f(\sigma^{4+i+1})=f(\sigma^{4+i})^2,~\forall i= 0, \ldots, log(N)-1\\
+	f_{4+log(N)} = 1 &\longrightarrow 6.~f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1
+\end{align*}
+\end{scriptsize}
+
+
+For succintness: aggregate $\{f_i\}$ in a polynomial $f(X)$, whose coefficients in Lagrange basis associated to $\mathbb{V}_n$ are the $f_i$ (ie. s.t. $f(\omega^i)=f_i$).
+
+\begin{small}
+\begin{align*}
+	f(X) &= (a-b) \rho_1(X) + (a \sigma - b) \rho_2(X) + a \rho_3(X) + b \rho_4(X) + \sum_{i=0}^{log(N)} (\frac{a}{b})^{2^i} \rho_{5+i}(X)\\
+	     &= f_0 \rho_1(X) + f_1 \rho_2(X) + f_2 \rho_3(X) + f_3 \rho_4(X) + \sum_{i=0}^{log(N)} (f_4)^{2^i} \rho_{5+i}(X)
+\end{align*}
+\end{small}
+
+Prover shows that $f(X)$ by comparing $f(\sigma^i)$ with the corresponding constraints from Lemma 1:
+
+For rand $\alpha$ (set by Verifier), set $\alpha_1 = \sigma^{-1} \alpha,~\alpha_2= \sigma^{-2} \alpha$, and send $v_1=f(\alpha_1),~v_2=f(\alpha_2)$ with corresponding proofs of opening.
+
+Given $v_1, v_2$, shows that $p_{\alpha}(X)$, which proves the constraints from Lemma 1, evaluates to $0$ at $\alpha$ (ie. $p_{\alpha}(\alpha)=0$).
+
+\begin{align*}
+	p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha) + [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\
+			&+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\
+			&+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\
+			&+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\
+			&+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\
+			&+ [f(\alpha_1)-1] \rho_n(\alpha)
+\end{align*}
+
+\subsubsection{\texorpdfstring{NIZK argument of knowledge for $R_{unity}$ and $deg(z)\leq 1$}%
+				{NIZK argument of knowledge for Runity and deg(z)<=1}}
+
+Prover:
+\begin{small}
+\begin{align*}
+	&r_0, r_1, r_2, r_3 \leftarrow^\$ \mathbb{F},~~~ r(X)=r_1+r_2 X + r_3 X^2\\
+	f(X) &= (a-b) \rho_1(X) + (a \sigma - b) \rho_2(X) + a \rho_3(X) + b \rho_4(X) + \sum_{i=0}^{log(N)} (\frac{a}{b})^{2^i} \rho_{5+i}(X)\\
+	     &+ r_0 \rho_{5+log(N)}(X) + r(X) z_{V_n}(X)\\
+	\\
+p(X) &= [f(X) - (aX-b)](\rho_1(X) + \rho_2(X))\\
+     &+[(1-\sigma)f(X) - f(\sigma^{-1}X) + f(\sigma^{-1}X)] \rho_3(X)\\
+     &+ [f(X) + f(\sigma^{-2}X) - \sigma f(\sigma^{-1} X)] \rho_4(X)\\
+     &+ [f(X)f(\sigma^{-1}X)-f(\sigma^{-2}X)] \rho_5(X)\\
+     &+ [f(X)-f(\sigma^{-1}X)f(\sigma^{-1}X)] \prod_{i \notin [5, 4+log(N)]} (X-\sigma^i)\\
+     &+ [f(\sigma^{-1}X)-1] \rho_n(X)
+\end{align*}
+\end{small}
+
+Set
+$$h'(X) = \frac{p(X)}{z_{V_n}(X)},~~h(X)=h'(X) + X^{d-1} z(X)$$
+output $([F]_1=[f(X)]_1, ~ [H]_1=[h(x)]_1)$.
+
+\begin{footnotesize}
+	Note that
+	\begin{align*}
+		h(x)&=h'(X)+X^{d-1}z(X)\\
+		     &=\frac{p(X)}{z_{V_n}(X)} + X^{d-1} z(X) \longrightarrow p(X)+X^{d-1} z(X) = z_{V_n}(X) h(X)
+	\end{align*}
+\end{footnotesize}
+
+Verifier sets challenge $\alpha \in^{\$} \mathbb{F}$ (hash of transcript by Fiat-Shamir).
+
+\begin{align*}
+	p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha)\\
+			&+ [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\
+			&+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\
+			&+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\
+			&+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\
+			&+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\
+			&+ [f(\alpha_1)-1] \rho_n(\alpha)
+\end{align*}
+
+\begin{footnotesize}
+	Note: for the check that $[z]_1$ has degree 1, we add $-h(X) z_{V_n}(\alpha)$, to include the term $X^{d-1} z(X)$ in $h(X)$. Later the Verifier will compute $[P]_1$ without the terms including $z(X)$ (ie. without $-X^{d-1} z(X)z_{V_n}(\alpha)-z(X)[\rho_1(\alpha)+\rho_2(\alpha)]$), which the Verifier will add via the pairing:
+	\begin{align*}
+		-&X^{d-1} z(X)z_{V_n}(\alpha)-z(X)(\rho_1(\alpha)+\rho_2(\alpha))\\
+		=~ &(-X^{d-1} z_{V_n}(\alpha) - (\rho_1(\alpha)+\rho_2(\alpha))) \cdot z(X)\\
+		\longrightarrow~ &e(- (\rho_1(\alpha)+\rho_2(\alpha)) - z_{V_n}(\alpha) [X^{d-1}]_1, [z]_2)
+	\end{align*}
+\end{footnotesize}
+
+
+Prover then generates KZG proofs
+\begin{align*}
+	((v_1, v_2), \pi_1) &\leftarrow KZG.Open(f(X), (\alpha_1, \alpha_2))\\
+	(0, \pi_2) &\leftarrow KZG.Open(p_{\alpha}(X), \alpha)
+\end{align*}
+
+prover's output: $(v_1, v_2, \pi_1, \pi_2)$.
+
+Verify:
+set $\alpha_1=\sigma^{-1}\alpha, ~\alpha_2=\sigma^{-2}\alpha$,
+
+\begin{footnotesize}(notice that $f(X)\rightarrow [F]_1$, and $f(\alpha_1)=v_1,~f(\alpha_2)=v_2$)\end{footnotesize}
+
+\begin{align*}
+	[P]_1 = &-z_{V_n}(\alpha)[H]_1 + [F]_1 (\rho_1(\alpha) + \rho_2(\alpha))\\
+		&+ [(1-\sigma) [F]_1 - v_2 + v_1] \rho_3(\alpha)\\
+		&+ [[F]_1 + v_2 - \sigma v_1] \rho_4(\alpha)\\
+		&+ [[F]_1 v_1 - v_2] \rho_5(\alpha)\\
+		&+ [[F]_1 - v_1^2] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\
+		&+ [v_1-1] \rho_n(\alpha)
+\end{align*}
+
+$$KZG.Verify((\alpha_1, \alpha_2), (v_1, v_2), \pi_1)$$
+$$e([P]_1, [1]_2) + e(-(\rho_1(\alpha) + \rho_2(\alpha)) - z_{V_n}(\alpha) [x^{d-1}]_1, [z]_2) = e(\pi_2, [x-\alpha]_2)$$
+
+
+\section{Caulk+}
+\emph{WIP}
+
+\bibliography{paper-notes.bib}
+\bibliographystyle{unsrt}
+
+\end{document}
--- a/paper-notes.bib
+++ b/paper-notes.bib
@@ -39,3 +39,21 @@
      note = {\url{https://eprint.iacr.org/2019/1021}},
      url = {https://eprint.iacr.org/2019/1021}
 }
+
+@misc{cryptoeprint:2022/621,
+      author = {Arantxa Zapico and Vitalik Buterin and Dmitry Khovratovich and Mary Maller and Anca Nitulescu and Mark Simkin},
+      title = {Caulk: Lookup Arguments in Sublinear Time},
+      howpublished = {Cryptology ePrint Archive, Paper 2022/621},
+      year = {2022},
+      note = {\url{https://eprint.iacr.org/2022/621}},
+      url = {https://eprint.iacr.org/2022/621}
+}
+
+@misc{cryptoeprint:2022/957,
+      author = {Jim Posen and Assimakis A. Kattis},
+      title = {Caulk+: Table-independent lookup arguments},
+      howpublished = {Cryptology ePrint Archive, Paper 2022/957},
+      year = {2022},
+      note = {\url{https://eprint.iacr.org/2022/957}},
+      url = {https://eprint.iacr.org/2022/957}
+}
--- a/weil-pairing.pdf
+++ b/weil-pairing.pdf
--- a/weil-pairing.tex
+++ b/weil-pairing.tex
@@ -4,6 +4,7 @@
 \usepackage{amsthm}
 \usepackage{amsmath}
 \usepackage{amssymb}
+\usepackage{mathtools}
 \usepackage{enumerate}
 \usepackage{hyperref}
 \hypersetup{
@@ -28,7 +29,7 @@
 {\endinnersolution}


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

@@ -44,9 +45,7 @@

 \tableofcontents

-\section{Weil reciprocity}
-
-\section{Generic Weil Pairing}
+\section{Divisors and rational functions}

 \begin{definition}{Divisor}
  $$D= \sum_{P \in E(\mathbb{K})} n_p \cdot [P]$$
@@ -67,21 +66,77 @@ $D \sim D'$ iff $D - D'$ is principal.
  $$r(D)= \prod r(P)^{n_p}$$
 \end{definition}

-\subsection{Generic Weil Pairing}
+\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).

-$P, Q \in E[n]$:
-$$D_P \sim [P] - [0]$$
-$$D_Q \sim [Q] - [0]$$
+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:
-$$D_P \sim [P] - [0]$$
-$$D_Q \sim [Q+T] - [T]$$
+\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