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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill 
Lecture 13: Primality Testing (Continued)
\\
Lecturer: Manindra Agrawal
\hfill
Notes by: Arun Iyer
%\\
\begin{flushright}
%date
September 8, 2005.
\end{flushright}
}}\end{center}
\vspace{5mm}

\section{Last Lecture Recap}
A potential primality test was proposed which generalised the basic approach of using the idea $a^{n} \equiv a (mod \, n)$ whenever $n$ is prime, over a ring of polynomials. So we have a ring $R$,
\begin{displaymath}
R = \Z_n[X]/(X^{r}-1)
\end{displaymath}
where $n$ is square-free. Also we have a map,
\begin{displaymath}
\psi{(e(X))} = e^n(X), \, e(X) \in  R
\end{displaymath}
The following question was to be answered, "Is $\psi$ an automorphism of R?".

\section{Proceeding towards Proof}
To check if $\psi$ is a linear map : \begin{center}test $\psi{(e(X))} \, = \, e(\psi{(X)}), \, \forall e(X) \, \in \, R$.\end{center} Let $p$ be a prime divisor of $n$. Let $h(x)$ be an  irreducible factor of $x^r-1$ over $F_p$. Let $F = F_p[X]/(h(X))$. Let $deg(h) = d$ then $|F|=p^d$. Field $F$ occurs as one of the components in the direct sum representation of $R$.
\begin{lemma}
\label{First Size Reduction Lemma}
(First Size Reduction Lemma)\\Let $S \subseteq F$ such that,
\begin{enumerate}
\item $\psi(S) \, \subseteq \, S$ 
\item $\forall e(X) \, \in S, \, \psi{(e(X))}\,  = \, e(\psi(X))$
\item $|S| \, > \, n^{2\sqrt{r}}$ 
\end{enumerate}
Then $n \, = \, p^j$ for some $j$.
\end{lemma}
\begin{proof}
Let $\phi(e(X)) \, = \, e^p(X) , \, e(X) \, \in \, F$. Let $G \, = \, \left\{\, \phi^i\psi^j(X) \, | \, i,j \geq 0, \, X \in F \, \right\}$. Let $t = |G|$.\\Choose a pair $(\alpha,\beta),(\gamma,\delta)$ such that,
\begin{enumerate}
\item $(\alpha,\beta) \, \neq \, (\gamma,\delta)$
\item $0 \, \leq \, \alpha,\beta,\gamma,\delta \, \leq \, \sqrt{t}$
\item $\phi^{\alpha}\psi^{\beta}(X) \, = \, \phi^{\gamma}\psi^{\delta}(X)$
\end{enumerate}
Such a pair always exists due to pigeonhole principle.\\ Consider $\phi^{\alpha}\psi^{\beta}(e(X)), e(X) \in S$. Then,\begin{center}
\begin{tabular}{lclc}
$\phi^{\alpha}\psi^{\beta}(e(X))$ & $=$ & $\phi^{\alpha}\psi^{\beta - 1}(e(\psi{(X)}))$ & \ldots [By definition of S]\\
 & $=$ & $\phi^{\alpha}\psi^{\beta - 2}(e(\psi^{2}{(X)}))$ & $\vdots$\\
 &     & $\vdots$ & $\vdots$\\
 & $=$ & $\phi^{\alpha}(e(\psi^{\beta}{(X)}))$ & $\vdots$\\
 & $=$ & $\phi^{\alpha-1}(e(\phi\psi^{\beta}{(X)}))$ & \ldots [Since $\phi$ is linear over $F$]\\
 &     & $\vdots$ & $\vdots$\\
 & $=$ & $e(\phi^{\alpha}\psi^{\beta}(X))$ & $\vdots$
\end{tabular}\end{center}
Similarly, it can be shown that,
\begin{displaymath}
\phi^{\gamma}\psi^{\delta}(e(X)) \, = \, e(\phi^{\gamma}\psi^{\delta}(X))
\end{displaymath}
Hence,
\begin{displaymath}
\phi^{\alpha}\psi^{\beta}(e(X)) \, = \, \phi^{\gamma}\psi^{\delta}(e(X)), \, \forall e(X) \, \in \, S
\end{displaymath}
This implies that, $\phi^{\alpha}\psi^{\beta}(y) - \phi^{\gamma}\psi^{\delta}(y)$ has atleast $|S|$ many roots in $F$.\\ Let $P(y) \, = \, \phi^{\alpha}\psi^{\beta}(y) - \phi^{\gamma}\psi^{\delta}(y) \, = \, y^{n^\beta p^\alpha} - y^{n^\delta p^\gamma}$. \\$deg(P) \, = \, max\left\{n^\beta p^\alpha,n^\delta p^\gamma\right\} \, \leq \, n^{2\sqrt{t}}$.\\ However, since $|S| \, > \, n^{2\sqrt{r}}$ and $t \leq r$, it implies,
\begin{center}\begin{tabular}{rrclc}
$\Longrightarrow$ & $P$ & $=$ & $0$ &\\ 
$\Longrightarrow$ & $n^\beta p^\alpha$ & $=$ & $n^\delta p^\gamma$ &\\ 
$\Longrightarrow$ & $n^{\alpha^{\prime}}$ & $=$ & $p^{\beta^{\prime}}$ & \ldots [for some $\alpha^{\prime},\beta^{\prime}$]\\
$\Longrightarrow$ & $n$ & $=$ & $p^j$ & \ldots [for some $j$]
\end{tabular}\end{center}
Hence Proved.
\end{proof}
\\Let $T=\left\{ \, X^{j} + a \, | \, 0 \leq j \leq r, \, 0 \leq 2\sqrt{r}logn, \, X \in F \, \right\}$. Now $|T| \, \leq \, 2r^{\frac{3}{2}}logn$, which is small if $r$ is small. Let $S$ be the multiplication closure of $T$ in $F$.
\begin{lemma}
\label{Second Size Reduction Lemma}
(Second Size Reduction Lemma)\\ If $p>t>4log^2n$ and $\psi(e(X)) = e(\psi(X)), \forall e(X) \in T$, then
\begin{enumerate}
\item $\psi(S) \, \subseteq \, S$
\item $\psi{(e(X))} \, = \, e(\psi{(X)})$
\item $|S| \, > \, n^{2\sqrt{t}}$
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item Let $e(X) \in S$. Therefore, $e(X) = \prod_{i=1}^{k} e_i(X), \, e_i(X) \in T$.
\begin{center}\begin{tabular}{lcl}
$\psi{(e(X))}$ & $=$ & $\psi{(\prod_{i=1}^{k} e_i(X))}$\\
 & $=$ & $\prod_{i=1}^{k}e_i(\psi{(X)})$
\end{tabular}\end{center}
Since $e_i(\psi{X}) \in T$, $\psi{(e(X))} \in S$.
\item $\psi{(e(X))} = \prod_{i=1}^{k}e_i(\psi{(X)}) = e(\psi{(X)})$
\item Let $Q = \left\{ \, \prod_{i=1}^{t}(y+a_i) \, | \, 0 \leq a_i \leq 2\sqrt{r}logn \, \right\}$.\\ Let $q(y) \, \in \, Q$. Then, $q(X) \, \in \, S$.
\begin{center}\begin{tabular}{lcl}
Number of polynomials in Q & $=$ & $\nchoosek{2\sqrt{r}logn+t}{t}$\\
 & $>$ & $\nchoosek{2\sqrt{r}logn + 2\sqrt{t}logn}{2\sqrt{t}logn}$\\
 & $\geq$ & $\nchoosek{4\sqrt{t}logn}{2\sqrt{t}logn}$\\
 & $>$ & $2^{2\sqrt{t}logn}$\\
 & $=$ & $n^{2\sqrt{t}}$
\end{tabular}\end{center}
Now, we make a claim,\\Claim : The map $y \mapsto X$ is 1-1 on Q\\Proof : Let $q_1(y), q_2(y) \in Q, \, q_1 \neq q_2$. Suppose that $q_1(X) = q_2(X)$ in $F$. We have $q_1(X),q_2(X) \in S$ and,
\begin{center}\begin{tabular}{rrcl}
& $q_1(X)$ & $=$ & $q_2(X)$\\
$\Longrightarrow$ & $\psi{(q_1(X))}$ & $=$ & $\psi{(q_2(X))}$ \\ 
$\Longrightarrow$ & $q_1(\psi{(X)})$ & $=$ & $q_2(\psi{(X)})$ \\ 
$\Longrightarrow$ & $q_1(\psi^j{(X)})$ & $=$ & $q_2(\psi^j{(X)})$ \\ 
$\Longrightarrow$ & $q_1(\phi^i\psi^j{(X)})$ & $=$ & $q_2(\phi^i\psi^j{(X)})$ \\ 
$\Longrightarrow$ & $q_1(e(X))$ & $=$ & $q_2(e(X))$  
\end{tabular}\end{center}
The above implies that every $e(X) \in G$ is a root of the polynomial $q_1(y) - q_2(y)$ in $F$. But $deg(q_1(y)-q_2(y)) \leq t-1$, which gives us a contradiction. Therefore the map is 1-1.\\ The above proof tells us that $|S|=|Q| \geq n^{2\sqrt{t}}$.
\end{enumerate}
Hence Proved.
\end{proof}
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