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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill 
Lecture 31: Polynomial Factorization over $\Q$
\\
Lecturer: Manindra Agrawal
\hfill
Notes by: Arun Iyer
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November 8, 2005.
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\vspace{5mm}

\section{Polynomial Factorization over $\Q$}
Given a polynomial $f(x)$ of degree $d$ over $\Q$.\\ Assume that $f$ is monic and square-free.

\begin{enumerate}
\item Choose a small prime $p$ such that $f$ remains square-free in $F_p$. 
\subitem[1.1] For making this choice of $p$, we can simply iterate over all primes starting from the smallest prime that is 2. We will now try to derive an upper bound on $p$. 
\subitem[1.2] Let k be the largest coefficient in $f$. Then, the possible largest coefficient in $f^\prime$ would be $kd$. Now, this would imply that $|Res(f,f^\prime)| \leq (2d)!(kd)^{2d} \leq (2kd^2)^{2d} = 2^{2d\log(2kd^2)}$. This would imply that $p = O(d \log(kd) \log(d))$.
\item Factorize $f \; mod \; p$ as $f = f_1f_2$ where $f_1$ is irreducible. 
\subitem[2.1] Recall that polynomial factorization over field which was discussed in the earlier class was randomized. However, owing to the fact that $p$ here is very small, the process can be made deterministic.
\item Use Modified Hensel Lifting to compute $f = g_1g_2(mod \; p^l)$. 
\subitem[3.1] Note that $g_1$ and $f_1$ have same degree. Also, the way modified hensel lifting is done, if $g_1$ were reducible mod $p^l$ then it would be reducible mod $p$ and this would imply $f_1$ to be reducible mod $p$ which is false. Hence $g_1$ above is irreducible.
\subitem[3.2] The choice of value $l$, will be decided later.
\item Let $deg(g_1) = d_1$. Define lattice $\Lat$ as spanned by [$g_1$, $xg_1$, \ldots, $x^{d-d_1}g_1$, $p^l$, $xp^l$, \ldots, $x^{d_1-1}p^l$].
\subitem[4.1] Volume of this lattice, $Vol(\Lat) = p^{ld_1}$.
\item Use LLL-algorithm (Lenstra, Lenstra and Lov\`{a}sz algorithm) to find a short vector in $\Lat$. Let that vector be $\overrightarrow{u}$.
\subitem[5.1] $|u| \leq 2^{\frac{d-1}{2}}(\textrm{length of the actual shortest vector}) \leq 2^{\frac{d-1}{2}}\sqrt{d}p^{\frac{ld_1}{d}}$
\subitem[5.2] Let $u(x)$ be the polynomial give by the $\overrightarrow{u}$. $\overrightarrow{u}$ can written as a linear combination of its basis vectors. Therefore, $u(x)$ can be written as $g_1(x)h(x) (mod \; p^l)$ for some $h(x)$.
\subitem[5.3] Let $f=\hat{f}_1\hat{f}_2$ over $\Q$ with $g_1 | \hat{f}_1 (mod \; p^l)$. $(\hat{f}_1,u(x)) \neq 0 (mod \; p^l)$, this implies $|Res(\hat{f}_1,u(x))| = 0 (mod \; p^l)$
\end{enumerate}
(To be continued \ldots)
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