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\lecture{8}{October 5, 2005}{Madhu Sudan}{Guy Rothblum}
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\begin{abstract}
Today we will complete the description of Berlekump's
deterministic algorithm for efficiently factorizing polynomials
over $\mathbb{F}_q$ (where $q=p^t$ for a prime $p$) in time
polynomial in $\deg(f)$, $t$, $p$.
We will then begin laying the groundwork for algorithms that
factor polynomials over $\mathbb{Z}[x]$ and for factoring
bivariate polynomials. We will see these two problems are related
and introduce Hensel's Lifting, a useful tool for solving them.
\end{abstract}
\section{A deterministic Algorithm---Continued}
Recall that we saw in the last lecture that for any reducible
polynomial $f(x) \in \mathbb{F}_q$ of degree $2d$, there exists a
polynomial $g(x) \in \mathbb{F}_q$ s.t. $f(x) | g(x)^p - g(x)$ and
the degree of $g(x)$ is at most $2d - 1$. We also saw that if we
could find this $g(x)$ efficiently then we could factor
efficiently. We will proceed to show how to find $g$ efficiently.
Recall also that the field $\mathbb{F}_q$ is isomorphic to the
t-dimentional (additive) vector space $\mathbb{F}_p^t$, where the
isomorphism maps every element $\alpha \in \mathbb{F}_q$ to a
vector $v_{\alpha} \in \mathbb{F}_p^t$.
\begin{claim}
The map $A: v_{\alpha} \rightarrowtail v_{\alpha^p}$ is a linear
map.
\end{claim}
\begin{proof}
We only need to verify that:
\begin{enumerate}
\item
$A(v_{\alpha+\beta}) = v_{(\alpha+\beta)^p} = v_{\alpha^p +\beta^p
(\mathrm{mod }\mbox{ } p)} = v_{\alpha^p} + v_{\beta^p} =
A(v_{\alpha}) + A(v_{\beta})$
\item
$A(v_{a \cdot \alpha}) = a \cdot A(v_{\alpha})$
\end{enumerate} \end{proof}
Since $A$ is a linear map, it can be represented by a $t \times t$
matrix $A \in \mathbb{F}_p^{t \times t}$
\begin{fact}
Given one of the ``nice'' representations of $\mathbb{F}_q$ (e.g.
$\mathbb{F}_q$ represented as a vector space or using an
irreducible polynomial), the matrix $A$ can be computed
efficiently.
\end{fact}
How does this fact help us? We want to find $g(x)$ s.t. $f(x) |
g(x)^p - g(x)$, where $f(x)$ is known but $g(x)$ is unknown. We
will use the linear map $A$ to find $g(x)$! We view $g(x)$ as
$g(x) = \sum_{i=0}^{2d - 1} c_i \cdot x^i$, where the $c_i$'s are
unknowns, and get that: \[g(x)^p = ( \sum_{i=0}^{2d - 1} c_i \cdot
x^i)^p = \sum_{i=0}^{2d - 1} c_i^p \cdot x^{i \cdot p}\] To find
$g(x)$ we will construct a system of linear equations. Towards
this end we define two new polynomials $h(x), a(x)$ where $h(x) =
g(x)^p$ and $a(x) \cdot f(x) = g(x)^p - g(x) = h(x) - g(x)$. Let
$\{e_i\}_{i=0}^{p \cdot (2d-1)}$ and $\{a_i\}_{i=0}^{p \cdot
(2d-1) - 2d}$ be the coefficients of $h(x)$ and $a(x)$
respectively. We get that: \[h(x) = \sum_{j=0}^{p \cdot (2d-1)}
e_j \cdot x^j = \sum_{i=0}^{2d - 1} c_i^p \cdot x^{i \cdot p} =
g(x)^p\]
\begin{enumerate}
\item
The first system of constraints will reflect the equality $h(x) =
g(x)^p$:
\begin{enumerate}
\item For any integer $j$ that is not a multiple of $p$: $e_j =
0$.
\item For any integer $i$: $e_{i \cdot p} = c_i^p = A \times c_i$.
\end{enumerate}
\item
The second constraint specifies that $f(x) \cdot a(x) = h(x) -
g(x)$. Looking at the coefficients on both sides and at $f(x)$ as
$\sum_{i=0}^{2d} f_i \cdot x^i$, for the $j$-th coefficient we get
the constraint: \[e_j - c_j = \sum_{i=0}^j f_i \cdot a_{j - i} =
\sum_{i=0}^j M_{f_i} \cdot a_{j - i}\] Where $M_{f_i}$ is the
matrix representation of $f_i$ (recall this matrix representation
supports multiplication).
\item
Finally, we would like for the solution to be non-trivial, and
thus we add the constraint $(c_1, \ldots c_{2d-1}) \neq (0, \ldots
0)$. Note that while this is not a linear equation per se, it can
be incorporated into the algorithm for solving the other linear
equations, so that the algroithm returns a non-zero solution when
one exists.
\end{enumerate}
Now, to find $g(x)$ all that remains is to solve this system of
(linear) equations! Note that this is {\em not} a proof of
existence of a non-trivial $g(x)$, we proved $g(x)$'s existence in
the previous lecture, this is simply an efficient procedure for
finding $g(x)$.
\section{Framework for the Next Talks}
In the next talks we see how to factor bivariate polynomials and
polynomials over the rational numbers $\mathbb{Q}[x]$. We begin by
laying out the framework that we will follow in these
(surprisingly) related results.
\paragraph{Factoring Bivariate Polynomials}
We will see how to go from factoring polynomials to factoring
bivariate polynomials, we will go from factoring $\mathbb{R}[x]$
to factoring $\mathbb{R}[x,y]$. Given $f(x,y) \in
\mathbb{R}[x,y]$, we will factor it using an algorithm for
factoring in $\mathbb{R}[x]$. We proceed in several steps:
\begin{enumerate}
\item Somehow (the details will follow) ``perturb'' $f(x,y)$ into
$\tilde{f}(x,y)$.
\item Begin by factoring $\tilde{f}(x,y) (\mathrm{mod }\mbox{ }
y)$ using the algorithm for factoring in $\mathbb{R}[x]$.
\item Proceed in Hensel iterations, and progressively go from
factoring $\tilde{f}(x,y) (\mathrm{mod }\mbox{ } y^i)$ to
factoring $\tilde{f}(x,y) (\mathrm{mod }\mbox{ } y^{2i})$.
\item From factoring over $\mathbb{R}[x,y] (\mathrm{mod}\mbox{ }
\mbox{ } y^t)$, go to factoring over $\mathbb{R}[x,y]$.
\end{enumerate}
\paragraph{Factoring over $\mathbb{Q}[x]$:}
To factor polynomials over integers $\mathbb{Z}[x]$, we actually
factor over $\mathbb{Q}[x]$ (we couldn't really expect to factor
over $\mathbb{Z}[x]$, since the polynomials of degree 0 there are
integers...). We proceed similarly to the bivariate case:
\begin{enumerate}
\item Somehow pick a ``nice'' prime $p$.
\item Begin by factoring $f(x) (\mathrm{mod }\mbox{ } p)$.
\item Proceed in Hensel iterations, and progressively go from
factoring $f(x) (\mathrm{mod }\mbox{ } p^i)$ to factoring $f(x)
(\mathrm{mod }\mbox{ } p^{2i})$.
\item From factors over $\mathbb{Z}[x] (\mathrm{mod}\mbox{ }
\mbox{ } p^t)$, go to factoring over $\mathbb{Z}[x]$.
\end{enumerate}
As can be seen, the two seemingly unrelated problems of factoring
integers and bivariate polynomials, are actually closely tied
together by our plan of action and its use of Hensel iterations.
\section{Hensel's Lifting Lemma}
We want to go from a factorization $f(x) = g(x) \cdot h(x) \mbox{
}(\mathrm{mod }\mbox{ } p)$ to $f(x) = \tilde{g}(x) \cdot
\tilde{h}(x) \mbox{ }(\mathrm{mod }\mbox{ } p^2)$. One appealing
idea is to take $\tilde{g}(x) = g(x) \mbox{ }(\mathrm{mod }\mbox{
} p)$ and $\tilde{h}(x) = h(x) \mbox{ }(\mathrm{mod }\mbox{ } p)$.
Unfortunately, this natural idea fails, as can be seen in the
simple case:
\[f(x) = x^2 - 2x + 6 = (x-1) \cdot (x-1) \mbox{
}(\mathrm{mod }\mbox{ } 5)\] We want $\tilde{g}(x) = (x-1) + 5
\cdot a(x)$ and $\tilde{h}(x) = (x-1) + 5 \cdot b(x)$, which
implies that modulo 25 we should get: $f(x) = (x - 1)^2 + 5 \cdot
(x-1) \cdot (a(x) + b(x)) + 25 a(x) \cdot b(x)$. Unfortunately,
$f(x)$ isn't of this form modulo 25!
To overcome this obstacle, we observe that our natural idea may
have failed in the example above simply because the factors $g(x),
h(x)$ were not relatively prime. Before stating the Lemma itself,
note that by $J^2$ we refer to the collection of linear
combinations of products of pairs of items in $J$.
\begin{lemma}{Hensel's Lifing Lemma:}
For a ring $R$ and an ideal $J \subseteq R$:
\textbf{If} there exist $f,g,h,a,b \in R$ such that:
\begin{enumerate}
\item $f - g \cdot h \in J$ ($f = g \cdot h \mbox{
}(\mathrm{mod}\mbox{ } \mbox{ } J))$.
\item $a \cdot g + b \cdot h = 1 \mbox{ }(\mathrm{mod}\mbox{ }
\mbox{ } J)$ ($f$ and $g$ are relatively prime).
\end{enumerate}
\textbf{Then} there exists a lifting: there exist
$\tilde{g},\tilde{h} \in R$ such that:
\begin{enumerate}
\item $\tilde{g} = g \mbox{ }(\mathrm{mod}\mbox{ } \mbox{ } J)$.
\item $\tilde{h} = h \mbox{ }(\mathrm{mod}\mbox{ } \mbox{ } J)$.
\item $f = \tilde{g} \cdot \tilde{h} \mbox{ }(\mathrm{mod}\mbox{
} \mbox{ } J^2)$.
\end{enumerate}
We refer to the set of conditions satisfied by $\tilde{g}$ and
$\tilde{h}$ as $(*)$.
The lift is \textbf{unique}: for any $g^*, h^*$ satisfying $(*)$,
there exists $u \in J$, such that $g^* = \tilde{g} \cdot (1 + u)$
and $h^* = \tilde{h} \cdot (1 - u)$.
Furthermore, for any $\tilde{g}, \tilde{h}$ that satisfy $(*)$,
there exist $\tilde{a}, \tilde{b} \in R$, such that $\tilde{a}
\cdot \tilde{g} + \tilde{b} \cdot \tilde{h} = 1 \mbox{
}(\mathrm{mod}\mbox{ } \mbox{ } J^2)$. Thus the new factors are
also relatively prime and we can continue to activate Hensel's
Lemma.
\end{lemma}
\begin{proof}
We prove each of the guaranteed properties separately:
\paragraph{The existence of a lifting:}
We proceed as before (but with relatively prime factors!).
$f = g \cdot h + q$ for some $q \in J$, $\tilde{g} = g +g_1$,
$\tilde{h} = h +h_1$, where $g_1, h_1 \in J$.
We get that: $\tilde{g} \cdot \tilde{h} = g \cdot h + g_1 \cdot h
+ h_1 \cdot g + h_1 \cdot g_1$.
Since $h_1 \cdot g_1 \in J^2$, it remains to show that $q = g_1
\cdot h + h_1 \cdot g + h_1$.
We still haven't specified $g_1, h_1$, so to satisfy this
condition we take $g_1 = b \cdot q$, $h_1 = a \cdot q$, and get
that $g_1 \cdot h + h_1 \cdot g + h_1 = q \cdot (b \cdot h + a
\cdot g) = q$, as required!
\paragraph{$\tilde{g}$ and $\tilde{h}$ are relatively prime:}
Observe that: $a \cdot \tilde{g} + b \cdot \tilde{h} = a \cdot g +
b \cdot h + r' = 1 + r$, for some $r', r \in J$. Now we can take
$\tilde{a} = a \cdot (1 - r)$ and $\tilde{b} = b \cdot (1 - r)$,
and get that: \[\tilde{a} \cdot \tilde{g} + \tilde{b} \cdot
\tilde{h} = (1-r) \cdot (a \cdot \tilde{g} + b \cdot \tilde{h}) =
(1 - r) (1 + r) = 1 - r^2 = 1 \mbox{ }(\mathrm{mod}\mbox{ } \mbox{
} J^2)\]
\paragraph{Uniqueness:}
Let $g^* = \tilde{g} + g_2$ and $h^* = \tilde{h} + h_2$ for some
$g_2, h_2 \in J$ (because, modulo $J$, we know that $g^* = g =
\tilde{g}$ and $h^* = h = \tilde{h}$). Furthermore, modulo $J^2$,
we know that $g^* \cdot h^* = f = \tilde{g} \cdot \tilde{h}$.
Now we get that: $g^* \cdot h^* = \tilde{g} \cdot \tilde{h} + g_2
\cdot \tilde{h} + h_2 \cdot \tilde{g} + g_2 \cdot h_2$. This
implies that $g_2 \cdot \tilde{h} + h_2 \cdot \tilde{g} \in J^2$
(because $g_2 \cdot h_2 \in J^2$ and $g^* \cdot h^* = \tilde{g}
\cdot \tilde{h} \mbox{ }(\mathrm{mod}\mbox{ } \mbox{ } J^2)$).
\begin{claim}
The only way to get that $g_2 \cdot \tilde{h} + h_2 \cdot
\tilde{g} \in J^2$ is by setting $g_2 = u \cdot \tilde{g}$ and
$h_2 = -u \cdot \tilde{h}$ for some $u \in J$.
\end{claim}
\end{proof}
Note that in class it was pointed out that these are existence
results. We did not reach a definitive conclusion about whether
there is a problem in actually finding $r, q$ etc.
In the next talk we will complete the procedure for factorizing
bivariate polynomials.
\end{document}