\documentclass[10pt]{article}
\usepackage{amsfonts}
\newtheorem{define}{Definition}
\usepackage{psfig}
\oddsidemargin=0.15in
\evensidemargin=0.15in
\topmargin=-.5in
\textheight=9in
\textwidth=6.25in
\usepackage{amsmath}
\usepackage{comment}
\begin{document}
\input{preamble.tex}
\lecture{10}{October 17, 2005}{Madhu Sudan}{Ben Rossman}
\section{Clarification on Resultants}
Consider polynomials $f(x),g(x) \in R[x]$ where $R$ is a unique
factorization domain (with some notion of order on its elements). In
the last lecture, the {\em resultant} of $f$ and $g$ was defined as
the ``minimal'' element in the ideal $I$ of $R[x]$ generated by $f$
and $g$, i.e., the polynomial with smallest leading coefficient
among all nonzero polynomials of lowest degree in $I$. While this
notion of {\em resultant} serves our purposes, be cautioned that
this is not the standard definition.
Notice that $f$ and $g$ have no common factor if and only if there
exist $a(x),b(x) \in R[x]$ such that $\deg a < \deg g$, $\deg b <
\deg f$ and $a(x)f(x) + b(x)g(x) \in R - \{0\}$. The problem of
finding $a$ and $b$ such that $a(x)f(x) + b(x)g(x) = 1$ can be
represented as a linear system of $\deg f + \deg g$ equations (for
the coefficients of $f$ and $g$) in $\deg f + \deg g$ unknowns (the
coefficients of $a$ and $b$). The determinant of the matrix for this
system is what is usually called the {\em resultant}. In particular,
this is an element of $R$ which is zero if and only if $f$ and $g$
have a common factor.
\section{Factoring in $\F_q[x,y]$}
We are given a polynomial $f(x,y) \in \F_q[x,y]$ that is assumed to
be monic in $x$. (We later show how to get rid of this assumption.)
Our goal is to output a nontrivial factor of $f$ if $f$ is
reducible, or else to report that $f$ is irreducible. Our algorithm
consists of five steps. (In the following, where we write ``large
enough'' or ``sufficiently large'', exactly how large will be
specified when we do the analysis.)
\paragraph{Step 1}
Check if $f(x,y)$ and $\frac{\partial f}{\partial x}(x,y)$ are
relatively prime (i.e.\ if $f(x,y)$ is squarefree). If not, then
output $\gcd(f,\frac{\partial f}{\partial x})$ and stop. Otherwise,
proceed to Step 2.
\paragraph{Step 2}
For a large enough extension field $K \supseteq \F_q$, find $y_0 \in
K$ such that $f(x,y_0)$ and $\frac{\partial f}{\partial x}(x,y_0)$
are relatively prime.
\paragraph{Step 3}
Find an extension field $L \supseteq K$, element $\alpha \in L$ and
$h(x) \in L[x]$ such that $f(x,y) = (x-\alpha){\cdot} h(x) \mod
(y-y_0)$.
\paragraph{\normalfont{\em Remark.}}
Note that $(x-\alpha)$ and $h(x)$ are relatively prime, since
$f(x,y_0)$ is squarefree. Therefore, we can apply Hensel's Lifting
Lemma.
\paragraph{Step 4}
Apply Hensel iterations to obtain $A(y) \in L[y]$ and $H(x,y) \in
L[x,y]$ such that
\[
f(x,y) = (x-A(y)){\cdot} H(x,y) \mod (y-y_0)^t
\]
where $A(y_0) = \alpha$ and $H(x,y_0) = h(x)$ for some sufficiently
large $t$ (a power of two).
\paragraph{Step 5}
If there exist nonzero polynomials $g(x,y),H'(x,y) \in L[x,y]$ such
that
\begin{align*}
\deg_x g &< \deg_x f\\
\deg_y g &\le \deg_y f\\
g(x,y) &= (x-A(y)){\cdot} H'(x,y) \mod (y-y_0)^t,
\end{align*}
then find any such $g(x,y)$ and output $\gcd(f,g)$. Otherwise (if no
such $g$ exists), output ``$f$ is irreducible''.
\paragraph{\normalfont{\em Remark.}}
If $g(x,y)$ and $H(x,y)$ exist, they can be found efficiently by
solving a linear system over $L$.
\subsection{Feasibility}
We want to show that the algorithm outlined above is correct as well
as feasible (i.e.\ has polynomial runtime). We first establish
feasibility. There are three parameters in the outline of the
algorithm we need to bound: the sizes of fields $K$ and $L$, and the
size of $t$.
We need $K$ to be large enough so that there is guaranteed to exist
$y_0$ such that $f(x,y_0)$ and $\frac{\partial f}{\partial
x}(x,y_0)$ have no root in common. We claim that it suffices to take
$\|K\| > 2d(d-1)$ where $d = \deg_x f$. To see this, consider the
{\em discriminant} $\Delta(y)$ of $f$ with respect to $x$, defined
as the resultant of $f$ and $\frac{\partial f}{\partial x}$ in the
ring $\F_q[y][x]$. As we have seen, $\Delta(y)$ is a polynomial in
$\F_q[y]$ of degree $\le 2d(d-1)$. Therefore, since $\|K\| >
2d(d-1)$, there exists $y_0 \in K$ that is not a root of
$\Delta(y)$. It follows that $f(x,y_0)$ and $\frac{\partial
f}{\partial x}(x,y_0)$ have no common root.
What about the size of $L$? $L$ is required to contain a root
$f(x,y_0)$, which could be an irreducible polynomial of degree $d$.
So, the most we can say about $L$ is that it is an extension field
of degree at most $d$ over $K$. This is okay, however, since
elements of $L$ have polynomial-size representations (as $d$-tuples
over $K$), and each field operation over $L$ reduces to
polynomially-many field operations over $\F_q$.
We now turn to $t$ (i.e.\ two to the number of Hensel iterations).
We set $t$ equal to the least power of two greater than $2d^2$. (We
will need this assumption in order to prove correctness of the
algorithm.) Thus, our algorithm involves only $2\log_2(d) + 2$
Hensel liftings. Since each Hensel lifting requires
$\mathrm{poly}(d)$ operations over $L$ (as we saw in the previous
lecture), the overall running time is polynomial in $d$ and $q$.
\bigskip\noindent{\em Remark.}
The algorithm presented here is not optimized for efficiency. More
sophisticated algorithms are known that achieve better running
times.
\subsection{Correctness}
Let $f_1(x,y),...,f_k(x,y)$ be the irreducible factors of $f(x,y)$,
all monic in $x$. When we substitute $y_0$ for $y$, the resulting
polynomials $f_i(x,y_0)$ are no longer necessarily irreducible.
However, because $f(x,y_0)$ is squarefree (by our choice of $y_0$),
we know that $(x - \alpha)$ divides exactly one of the $f_i(x,y_0)$,
say $f_1(x,y_0)$, without loss of generality. The following two
claims now establish the correctness of the algorithm.
\begin{claim}
If $k > 1$ then $f_1(x,y)$ is a candidate for $g(x,y)$ in Step 5.
\end{claim}
\begin{proof}
There exists $h_1(x) \in L[x]$ such that
\[
f_1(x,y) = (x-\alpha)h_1(x) \mod (y-y_0).
\]
$(x-\alpha)$ and $h_1(x)$ are relatively prime, since $f(x,y_0)$ is
squarefree. So we can apply Hensel liftings to get polynomials
$A_1(y)$ and $H_1(x,y)$ such that $A_1(y_0) = \alpha$, $H_1(x,y_0) =
h_1(x)$ and
\[
f_1(x,y) = (x - A_1(y))H_1(x,y) \mod (y-y_0)^t.
\]
Setting $\widetilde{H}(x,y) = H_1(x,y){\cdot}\big(f_2(x,y)\cdots
f_k(x,y)\big)$, we have
\[
f(x,y) = (x - A_1(y))\widetilde{H}(x,y) \mod (y-y_0)^t.
\]
From the uniqueness of monic Hensel liftings, it follows that
$A_1(y) = A(y) \mod (y-y_0)^t$. Since $k > 1$, we have $\deg_x(f_1)
< \deg_x(f)$. $f_1(x,y)$ is thus a suitable candidate for $g(x,y)$
in Step 5.
\end{proof}
\begin{claim}
$f_1(x,y)$ divides every candidate $g(x,y)$ in Step 5.
\end{claim}
\begin{proof}
Toward a contradiction, assume $f_1$ does not divide $g$. Since
$f_1$ is irreducible, $f_1$ and $g$ have no common factor.
Therefore, by the results of the last lecture, the resultant $R(y) =
\mathrm{Res}_x(f_1,g)$ is a polynomial of positive degree $\le
2d^2$. Write $R(y)$ as $a(x,y)f_1(x,y) + b(x,y)g(x,y)$ for some
polynomials $a(x,y)$ and $b(x,y)$. We have
\begin{align*}
R(y) &= a(x,y)(x-A(y)) H_1(x,y) + b(x,y)(x-A(y)) H'(x,y) \mod (y-y_0)^t\\
&= (x-A(y)){\cdot}(a(x,y)H_1(x,y) + b(x,y)H'(x,y)) \mod (y-y_0)^t.
\end{align*}
By our choice of $t > \deg_y(R)$, we ensure that $R(y) \ne 0 \mod
(y-y_0)^t$. Therefore,
\[
a(x,y)H_1(x,y) + b(x,y)H'(x,y) \ne 0 \mod (y-y_0)^t.
\]
It follows that $(x-A(y)){\cdot}(a(x,y)H_1(x,y) + b(x,y)H'(x,y))$
has nonzero degree in $x$. However, this contradicts that $\deg_x(R)
= 0$, i.e., $R$ is a polynomial in $y$ alone.
\end{proof}
\subsection{Factoring non-monic $f(x,y)$} %% in $\F_q[x,y]$}
In our description of the algorithm, we assume that input polynomial
$f(x,y)$ is monic in $x$. We now show how to eliminate this
assumption. Suppose $f(x,y)$ is not monic in $x$, and rewrite it as
\[
f(x,y) = g(y)x^d + \mathcal{O}(x^{d-1}).
\]
Define a new polynomial $\widetilde{f}(z,y)$ by
\[
\widetilde{f}(z,y) = g(y)^{d-1} {\cdot} f(\textstyle\frac{z}{g(y)},y).
\]
$\widetilde{f}(z,y)$ is clearly monic in $z$, so we apply our
algorithm to factor $\widetilde{f}(z,y) =
\widetilde{h}(z,y)\widetilde{\ell}(z,y)$. We now have
\[
f(x,y) = \textstyle\frac{1}{g(y)^{d-1}}
\widetilde{h}(g(y)x,y) \widetilde{\ell}(g(y)x,y).
\]
\section{A few words on multivariate factoring}
Our algorithm for factoring in $\F_q[x,y]$ generalizes to rings
$R[x,y]$ where $R$ is any ring over which we have an efficient
univariate factorization algorithm. This immediately suggests an
approach to factoring multivariate polynomials in
$\F_q[x_1,...,x_c]$, since $c$-variate factorization over $\F_q$ is
the same as univariate factorization over $\F_q[x_1,...,x_{c-1}]$.
%
%, summarized as $\F_q[x,y,z] = \F_q[z][y][x]$, or more generally,
%$\F_q[x_1,...,x_c] = \F_q[x_c]...[x_1]$.
%
Unfortunately, the algorithm implies by this approach has the
abysmal running time of %%%(in the number $c$ of variable) of
$\poly(d)^{\raisebox{9pt}{$\left.\raisebox{-9pt}{$2^{
2^{{}^{\vdots^{2}}}}$}\right\}\scriptstyle{\text{height }c}$}}$.
There is another, more efficient generalization of our algorithm. We
outline the trivariant case. Suppose we are given input polynomial
$f(x,y,z) \in \F_q[x,y,z]$ that is monic in $x$. After checking that
$f$ and $\frac{\partial f}{\partial x})$ have no common factor,find
a pair of values $y_0$ and $z_0$ (in a suitable extension field of
$\F_q$) such that $f(x,y_0,z_0)$ and $\frac{\partial f}{\partial
x})(x,y_0,z_0)$ have no common factor. Next find a nice
factorization of $f$ modulo the ideal generated by $(y-y_0)$ and
$(z-z_0)$ in the ring $L[x,y,z]$ for another extension field $L$ of
$\F_q$. Now apply Hensel liftings (Step 4) and look for a polynomial
$g$ with the right profile (Step 5). This method results in an
efficient running time $\mathrm{poly}(d^c)$ where $d = \deg(f)$ and
$c$ is the number of variables.
One might be tempted ask: is there a multivariate factorization
algorithm with running time polynomial in $c$? The question is
absurd if we continue to represent polynomials explicitly (i.e.\ by
enumerating their coefficients), since a polynomial in $c$ variables
of degree $d$ has $\mathcal{O}(d^c)$ coefficients in general. In
order to make the question meaningful, we have to consider other
ways of representing polynomials.
Kaltofen and Trager devised a clever algorithm for finding the
factors of a polynomial $f(x_1,...,x_c)$ that is given as black box
(together with essential information like the degree of $f$). The
output of the algorithm is, of course, not an explicit polynomial,
but a procedure (involving calls to the original black-box function)
for computing the values of some irreducible factor of $f$. The
Kaltofen-Trager algorithm achieves $\mathrm{poly}(d,c)$ running
time. (The bivariate factorization in the previous section is also
due to Kaltofen. Other bivariate factorization algorithms are due to
Lenstra and Grigoriev.)
\section{Factoring in $\Z[x]$}
As out final topic today, we begin outlining an algorithm for
factoring integer polynomials. Let $f \in \Z[x]$ be a monic
polynomial of degree $d$ whose coefficients are between $-2^n$ and
$2^n$. The representation of $f$ is thus a bit-string of length
$\mathcal{O}(dn)$. The procedure for factoring $f$ uses many of the
same ideas as the bivariate factorization algorithm. The five steps
below are adapted from the corresponding steps in the bivariate
algorithm.
\paragraph{Step 1} Check for common factors
of $f$ and $\frac{\partial f}{\partial x}$. If
$\mathrm{gcd}(f,\frac{\partial f}{\partial x})$ is nontrivial, then
we report it and stop. Otherwise, we continue to Step 2.
\paragraph{Step 2} Find prime $p \in \Z$ such that $f$ and
$\frac{\partial f}{\partial x}$ have no common factors modulo $p$.
(We use resultants and the Chinese Remainder Theorem to prove small
$p$ exists.)
\paragraph{Step 3} Factor $f(x)$ modulo $p$ as $f(x) = g(x)h(x) \mod
p$ where $g,h$ are relatively prime and $g$ is irreducible.
\paragraph{Step 4} Apply Hensel liftings to obtain $G(x)$ and $H(x)$
such that
\begin{align*}
G(x) &= g(x) \mod p\\
H(x) &= h(x) \mod p\\
f(x) &= G(x)H(x) \mod p^t.
\end{align*}
Here we have the added burden of proving that the coefficients of
$G$ and $H$ cannot be too large; $\pm 2^{n^2}$ is a rough bound.
\paragraph{Step 5} Find $\widetilde{g}(x)$ such that
$\widetilde{g}(x) = G(x)H'(x) \mod p^t$, $\deg(\widetilde{g}) < d$
and all coefficients of $\widetilde{g}$ have size $\le 2^{n^2}$.
(Note that $\widetilde{g}(x)$ satisfying the first two conditions
can be found by solving a linear system. The third condition, i.e.\
bounding the size of coefficients, is what makes the problem
difficult.)
\paragraph{Exercise} Prove analogues of Claims 1 and 2 in this
setting.
\end{document}