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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill
Lecture 35
\\
Lecturer: Manindra Agrawal
\hfill
Notes by: Ashwini Aroskar
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\begin{flushright}
%date
November 18, 2005.
\end{flushright}
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\vspace{5mm}

\section*{Elliptic Curves on Finite Fields (contd)}
%The text of the notes goes here.
Let $E(F_p)$ be the elliptic curve $y^2 = x^3 + Ax + B (mod$ $p)$\\
We also use $E(F_p)$ to denote the set of points on the curve over $F_p$.\\
\\
We will study $E(\bar{F}_p)$ where $\bar{F}_p$ is the algebraic closure of $F_p$.\\
We want to characterize points of $E(F_p)$ in $E(\bar{F}_p)$. As $\bar{F}_p$ is a field of characteristic $p$, we can raise the coordinates to the $p_{th}$ power and we derive the following test.\\
\\
Define $\phi_p(x,y) = (x^p,y^p)$\\
\textit{Observation:} $\phi_p$ is identity on $(x,y)$ \textit{iff} $(x,y) \in E(F_p)$\\
Equivalently, $\phi_p -1$ is $0$ on $(x,y)$ \textit{iff} $(x,y) \in E(F_p)$.\\
\\
Note: When we write $\phi_p -1$, the subtraction is the one defined on the elliptic curve.\\
\\
\textit{Observation:} $\phi_p -1$ is a homomorphism from $E(\bar{F}_p)$ to $E(\bar{F}_p)$.\\
\\
\textit{Proof:}\\
$\phi_p - 1 [(x_1,y_1) + (x_2,y_2)] = \phi_p[(x_1,y_1) + (x_2,y_2)] - (x_1,y_1) - (x_2,y_2)$\\
Denote the sum of the two points on the curve as $(x_3,y_3)$\\
Let $m = \frac{y_2^p - y_1^p}{x_2^p - x_1^p}$
\begin{eqnarray*}
\phi_p(x_3,y_3)& = & (x_3^p,y_3^p) \\
&=& [(m^2-x_1-x_2)^p,-y_1-m(x_3-x_1)^p]\\
&=& [m^{2p}-x_1^p-x_2^p, -y_1^p-m^p(x_3^p-x_1^p)] \quad (mod  p)\\
&=& \phi_p(x_1,y_1) + \phi_p(x_2,y_2)
\end{eqnarray*}
Also note that each coordinate of the image of a point under this homomorphism is a rational function in $x$ and $y$.
\begin{eqnarray*}
\phi_p-1 (x,y) &=& (x^p,y^p)-(x,y)\\
&=& (x^p,y^p)+(x,-y)\\
&=& (\frac{y^p+y}{x^p-x}^2-x^p-x,...)\\
\end{eqnarray*}
Let $\psi : E(\bar{F}_p) \longrightarrow E(\bar{F}_p)$ be a homomorphism given by $\psi(x,y) = (r_1(x,y),r_2(x,y))$, where $r_1$ and $r_2$ are rational functions in $x$ and $y$.\\
Such a $\psi$ is called an endomorphism.\\
Thus, $\phi_p-1$ is called an endomorphism.\\
\\
So, we can state the previous observation as follows:\\
\textit{Observation:} $Ker(\phi_p-1) = E(F_p)$\\
\\
Let $\psi(x,y) = (\frac{p(x,y)}{q(x,y)},\frac{r(x,y)}{s(x,y)})$\quad $p, q, r, s$ polynomials in x and y\\
Replacing $y^2 = x^3 +Ax +B$, we get\\
$\psi(x,y)= (\frac{p_1(x)+y p_2(x)}{q_1(x)+y q_2(x)},\frac{r_1(x)+y r_2(x)}{s_1(x) + y s_2(x)})$\\
For the first coordinate, multiply both numerator and denominator by $q_1(x)-y q_2(x)$ and replace $y^2$ in terms of $x$ again. Similarly for thr second coordinate. So we get\\
$\psi(x,y) = (\frac{p_3(x)+y p_4(x)}{q_3(x)}, \frac{r_3(x)+y r_4(x)}{s_3(x)})$\\
\\
If $\psi(x,y)=(u,v)$. As $\psi(O)=O$, $\psi(x,-y)=(u,-v)$\\
Therefore $y\longrightarrow -y$ must leave $u$ changed and change the sign of $v$.\\
$\Longrightarrow p_4(x)=0, r_3(x)=0$\\
\\
Thus we can write $\psi$ in the standard or normal form as\\
$\psi(x,y)=(\frac{p(x)}{q(x)},y\frac{r(x)}{s(x)})$\\
\\
Further we may also assume $(p(x),q(x))=1=(r(x),s(x))$\\
\\
Therefore we have $\frac{y^2r^2}{s^2} = \frac{p^3}{q^3} + A\frac{p}{q} +B$\\
\\
So, $s^2(p^3+Apq^2+Bq^3) = q^3r^2(x^3+Ax+B)$\\
\\
A root of $q$ is obviously a root of $s$ since it cannot be a root if $p$.\\
\\
Conversely let $\alpha$ be a root of $s$.\\ Assume the elliptic curve is non-singular, that is, $x^3+Ax+B$ has no repeated root.\\
$(x-\alpha)^2$ divides the R.H.S. $r$ cannot have $\alpha$ as a root. As the curve is non-singular, $(x-\alpha)^2 \vert q^3(x)$, $\alpha$ is a root of $q$.\\
So, $q$ and $s$ have the same roots.\\
\\
The curve is singular is $x^3+Ax+B$ has repeated roots, that is,\\
$\Longleftrightarrow(x^3+Ax+B,3x^2+A) \neq 1$\\
$\Longleftrightarrow(\frac{2}{3}Ax+B,3x^2+A) \neq 1$\\
$\Longleftrightarrow(\frac{2}{3}Ax+B,\frac{27B^2}{4A^2}+A) \neq 1$\\
\\
The curve is singular \textit{iff} $4A^3+27B^2=0$\\
\\
We define the degree of $\psi$ as the maximum of the degree of $p$ and $q$.\\
\\
Endomorphism $\psi$ is said to be \textit{separable} if $p'q-pq'$ is not identically zero.\\
Obviously $\phi_p(x,y)$ is not separable as the computations are modulo $p$.\\
\\
\textbf{Theorem:} Let $\psi(x,y)=(\frac{p(x)}{q(x)},y\frac{r(x)}{s(x)})$ be any separable endomorphism.\\
Then $\vert Ker \psi \vert = deg(\psi)$\\
\\
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