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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill Lecture 32
\\\\
Polynomial Factorization over $\mathbb{Q}$
\\\\
Lecturer: Manindra Agrawal \hfill Scribe: Chandan Saha
%\\
\begin{flushright}
%date
November 10, 2005
\end{flushright}
}}\end{center} \vspace{5mm}

\section{Polynomial Factorization over $\mathbb{Q}$ (contd $\ldots$)}
This is a continuation of the previous lecture on Polynomial
Factorization over the field of rationals $\mathbb{Q}$. Let $f =
\hat{f}_1 \hat{f}_2$ over $\mathbb{Q}$ and $f = g_1 g_2
\hspace{0.05in}(mod \hspace{0.05in}p^l)$. Define a lattice
$\mathcal{L}_1$ spanned by $\{ p^l, xp^l, x^2p^l, \ldots,
x^{d_1-1}p^l, g_1, xg_1, \ldots, x^{d-d_1-1}g_1\}$, where $d =
deg(f)$ and $d_1 = deg(g_1)$. The LLL-algorithm gives a short
vector $u$ of $\mathcal{L}_1$ such that $\mid u \mid
\hspace{0.05in} \leq 2^{\frac{d-1}{2}}\sqrt{d}p^{\frac{d_1l}{d}}$.
Note that $u(x)$ is a multiple of $g_1(x) \hspace{0.05in} (mod
\hspace{0.05in} p^l)$ and
\begin{align*}
\mid Res(u, \hat{f}_1) \mid &\leq (2d)! \hspace{0.05in} [max.
\hspace{0.05in} coeff. \hspace{0.05in} of \hspace{0.05in} (u,
\hat{f}_1)]^{2d} \\
&\leq (2d)! \hspace{0.05in} (\mid u \mid \mid \hat{f}_1 \mid)^{2d}
\end{align*}
We state the following lemma without proof.
\begin{lemma}
If $f$ is square free then it has a unique factorization modulo
$p^l$.
\end{lemma}
Therefore without loss in generality we assume that,
\begin{align*}
& \hspace{0.34in }\hat{f}_1 \in \mathcal{L}_1 \\
&\Rightarrow \mid u \mid \hspace{0.05in} \leq 2^{\frac{d-1}{2}}
\mid \hat{f}_1
\mid \\
&\Rightarrow \mid Res(u, \hat{f}_1) \mid \hspace{0.05in} \leq
(2d)! \hspace{0.05in} 2^{d(d-1)} \mid \hat{f}_1 \mid ^{4d}
\end{align*}
\begin{lemma}
$\mid \hat{f}_1 \mid \hspace{0.05in} \leq 2^{(d-1)} \mid f \mid$.
\end{lemma}
\begin{proof}
Let $f$ have roots $\eta_1, \ldots, \eta_d \in \mathbb{C}$. Let
$k$ be such that $\mid \eta_1 \mid, \ldots, \mid \eta_k \mid
\hspace{0.05in}< 1$ and $\mid \eta_{k+1} \mid, \ldots, \mid \eta_d
\mid \hspace{0.05in} \geq 1$. Let $M(f) = \prod_{j=k+1}^{d} \mid
\eta_j \mid$. Therefore,
\begin{align*}
\mid f \mid &= \hspace{0.05in}\left | \prod_{j=1}^{d}(x-\eta_j)
\right |
\\
&= \hspace{0.05in} \left | \sum_{j=0}^{d} (-1)^j \left( \sum_{S
\subseteq [1,d], \mid S \mid = j} \hspace{0.05in} \prod_{i \in S}
\eta_i
\right ) x^{d-j}\right | \\
&\leq \sum_{j=0}^{d}\left[\left | \sum_{S \subseteq [1,d], \mid S
\mid = j} \hspace{0.05in} (\prod_{i \in S} \eta_i )x^{d-j}\right |
\right ] \\
&\leq \sum_{j=0}^{d} (^dC_j)M(f) = 2^dM(f)
\end{align*}
Therefore, $\mid \hat{f}_1 \mid \leq 2^{d-1}M(\hat{f}_1) \leq
2^{d-1}M(f)$. Let,
\begin{align*}
& \hspace{0.4in} \overline{f} = \prod_{j=1}^{k}(\overline{\eta_j}x
- 1)
\prod_{j=k+1}^{d}(x - \eta_j)\\
&\Rightarrow \hspace{0.05in} \mid \overline{f} \mid
\hspace{0.05in}  \geq M(f)
\end{align*}
We claim that, $\mid \overline{f} \mid = \mid f \mid$. This is
because,
\begin{align*}
\mid f \mid &= \left | \prod_{j=1}^{d} (x-\eta_j) \right | \\
&= \left | \overline{f} \prod_{j=1}^{k}
\frac{(x-\eta_j)}{(\overline{\eta_j}x - 1)}\right | \\
&= | \overline{f} | \cdot \prod_{j=1}^{k} \left |
\frac{(x-\eta_j)}{(\overline{\eta_j}x - 1)}\right |
\end{align*}
Since, $\mid (x-\eta_j) \mid = (1+\eta_j
\overline{\eta_j})^{\frac{1}{2}}$ and $\mid \overline{\eta_j}x - 1
\mid = (1+\eta_j \overline{\eta_j})^{\frac{1}{2}}$, the proof
follows.
\end{proof}
\end{document}
