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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill 
Lecture 18: Modified Hensel Lifting
\\
Lecturer: Manindra Agrawal
\hfill
Notes by: Shashi Mittal
%\\
\begin{flushright}
%date
September 16, 2005.
\end{flushright}
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\vspace{5mm}

\section{Introduction}
In previous lecture, we had discussed hensel lifting, and its application in polynomial division. We had also made a remark that during lifting, $deg~h'$ or $deg~g'$ may exceed $deg~f$, which is undesirable. We present an example to show this, and the discuss the modification needed to remove this drawback.

\section{Drawback of hensel lifting : an example}

Consider $f(x) = x^4 + x^3 + 2x + 2$, $g(x) = x-2$ and $h(x) = x^3 - 1$. It can be easily verified that $f \equiv gh~(mod~3)$. If $s(x) = 2x^2 + x + 2$ and $t=1$, then $sg~+~ht~=~1$. Now, if we lift this to $mod~3^2$, then

\begin{eqnarray}
\lefteqn{e = f - gh (mod~9) } \nonumber \\
 & & = (x^4 + x^3 + 2x + 2) - (x^3 - 1)(x - 2)(mod~9) \nonumber \\
 & & =(3x^3 + 3x)(mod~9) \nonumber
\end{eqnarray}

$g'$ and $h'$ are calculated as follows :
\begin{eqnarray}
\lefteqn{g' = h + se} \nonumber \\
& & = (x - 2) + 1.(3x^3 + 3) \nonumber \\
& & = 3x^3 + 4x - 2 \nonumber
\end{eqnarray}

\begin{eqnarray}
\lefteqn{h' = h + se} \nonumber \\
& & = (x^3 - 1) + (2x^2 + x + 2)(3x^3 + 3x) \nonumber \\
& & = 6x^5 + 3x^4 + 4x^3 + 3x^2 -3x -1 \nonumber
\end{eqnarray}

We see that $deg~h' > deg~f$, which is undesirable.

\section{Modified hensel lifting}
Let $f \equiv gh(mod~m)$, and $s,t,e$ as define before. We assume that $g$ is monic, and $deg~f~=~deg~g~+~deg~h$.  
Let,
\begin{eqnarray}
te = qg + r (mod~m^2) \nonumber \\
g' = g + r(mod~m^2) \nonumber \\
h' = h + se + qh(mod~m^2) \nonumber
\end{eqnarray}
Then,

\begin{eqnarray}
\lefteqn{g'h' = (g + r)(h + se + qh)} \nonumber \\
& & = (g + te - qg)(h + se + qh) \nonumber
\end{eqnarray}

Since $te = qg + r(mod~m^2)$, therefore $q \equiv 0 (mod~m)$ and $r \equiv 0 (mod ~m)$. Hence,

\begin{eqnarray}
\lefteqn{g'h' = gh + gse + hte - qe(sg - th) + ste^2 - q^2gh} \nonumber \\
& & = gh + e (mod~m^2) \nonumber \\
& & = f (mod~m^2) \nonumber
\end{eqnarray}

Therefore, $deg~g = deg~g'$, $deg~h' = deg~h$ and $g$ is monic.
\\ \\
\textbf{\underline{Assignment}} : Define $s'$ and $t'$ in such a way that $deg~s' = deg~s$ and $deg~t' = deg~t$.

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