\lecture{27 : Bivariate Factoring }{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

We shall now look at Bivariate Factoring. This is an instance where
there are a lot of corner cases that need to be handled but we shall
make some preliminary assumptions on the polynomial. We shall soon see
how we can ensure those properties as well. 

\section{Bivariate Factoring}

We are given a polynomial $f\in K[X,Y]$ that we would like to
factor. Before we go into factorization as such, we first need to
figure out how to do the $\gcd$ algorithm over two variables. 

\subsection{The Euclidian Algorithm for $K[X,Y]$}

Lets say we are given two polynomials $f(X,Y)$ and $g(X,Y).$ How do we
compute their $\gcd$? Remember that when we had univariate factoring,
the euclidian algorithm would divide. But over $K[X,Y]$, how do we do
that? 

Hence, instead of looking at these polynomials over $K[X,Y]$, we look
at them as univariate polynomials rational functions over one
variables, namely $K(X)[Y].$ Thus coefficients could now be of the
form $p(X)/q(X).$ This then give rise to another problem, can the
denominator keep getting larger? That is a serious issue since during
the process, we could potentially be doubling the degree at each step

But clearly, the final answer cannot have a really large $X$-degree;
it clearly must be bounded by the $X$ degrees of $f$ and $g.$ So what
we do instead is do these computations modulo a number of irreducible
polynomials. We just need to make sure that the product of all these
irreducible polynomials have degree larger than the $X$ degree of both
$f$ and $g.$ \\

This would then let us compute the $\gcd$ of two bivariate
polynomials. 

Now we can think of $f$ as an element in $K[X][Y]$, a polynomial in
$Y$ each of whose coefficients are polynomials in $X.$ The first
things we do is pull out the gcd of all the coefficients. And we also
express $f$ as an element of $K[Y][X]$ and repeat the same thing. Then
we shall assume that $f$ is squarefree. This needs justification but
we shall do this later. But for the moment let us assume that this can
be done and proceed. 

Now we have $f(X,Y) \in K[X][Y].$ We shall put $Y = 0$ to get a
univariate polynomial in $X.$ Since we have already done factoring of
univariate polynomials, we shall factorize $f(X,0)$ into coprime
factors $f(X,0) = g_0h_0.$

Note that this is just saying
$$
f = g_0h_0 \bmod{Y}
$$
since $Y=0$ is precisely taking $\bmod{Y}.$ And we further have the
property, because the factors are coprime\footnote{we shall get back
to this} that $sg_0 + th_0 = 1\bmod{Y}.$ Thus we can Hensel Lift this
factorization for $k$ steps to get
$$
f = g_kh_k\bmod{Y^{2^k}}
$$
We would now argue that after sufficient hensel lifting, we can
retrieve a factor of $f.$ 

\begin{claim}
If $f$ was not irreducible, hen there is an irreducible factor of $f$,
say $g$ such that $g = g_0l_0\bmod{Y}$
\end{claim}
\begin{proof}
This is obvious. Suppose $f$ factorizes into irreducible factors as
$f_0f_1\cdots f_l.$ Then clearly, 
$$
f = f_0f_1\cdots f_l \bmod{Y}
$$
And therefore $g_0$ must divide one of these factors and whatever
factor it may be, set that as $g.$ Therefore, $g = g_0l_0\bmod{Y}$
\end{proof}

Now start with $g = g_0l_0\bmod{Y}$ and lift this to $k$ steps. What
we know is that $\deg_X(g) < \deg_X(f)$ and $\deg_Y(g)\leq \deg_Y(f).$
What we have is just $f = g_kh_k\bmod{Y^{2^k}}.$ We shall use this to
try and get hold of a factor. 

Suppose we look at the following equation to solve:
\begin{eqnarray*}
\tilde{g} & = & g_k\tilde{l}\bmod{Y^{2^k}}\\
\deg_X(g) & < & \deg_X(f)\\
\deg_Y(g) & \leq & \deg_Y(f)
\end{eqnarray*}

Firstly, does this system of equations have a solution? Indeed; we
just say that $g = g_0l_0\bmod{Y}$ can be lifted to $k$ steps to give
$g = g_kl_k\bmod{Y^{2^k}}$ and clearly $g$ and $l_k$ satisfy the
system of equations. Therefore, we are guaranteed that at least one
such solution exists if $f$ is not irreducible. Now, how do we compute
such a solution?

Notice that we already know $g_k$ and once we have $g_k$, if we think
of the coefficients of $\tilde{g}$ and $\tilde{l}$ as variables, the
above is just a system of linear equations. Thus, one could solve that
system using gaussian elimination to get some solution $\tilde{g}.$ It
might very well be possible that $\tilde{g} \neq g.$ But from a
solution $\tilde{g}$, how do we retrieve a factor? \\

Look at the two equations $f = g_kh_k\bmod{Y^{2^k}}$ and $\tilde{g} =
g_k\tilde{l}\bmod{Y^{2^k}}.$ And in essence, both $f$ and $\tilde{g}$
share a common factor modulo $Y^{2^k}.$ Therefore, if one were to
consider them as polynomials of $X$ with coefficients coming from
$K[Y]$, this means that the $X$-resultant of $f$ and $\tilde{g}$ is
zero modulo $Y^{2^k}.$ By the $X$ resultant we mean the resultant,
considered as polynomials in $K[Y][X].$

Therefore
$$
\text{Res}_X(f,\tilde{g}) = 0\bmod{Y^{2^k}}
$$
The resultant would be a polynomial in $Y.$ What is the maximum degree
of this polynomial possible? Since we are looking at the
determininant, the size of the sylvester matrix is $2m$ where
$m$ is the $\deg_X(f).$ And each element of this matrix would be a
polynomial in $Y$ of degree at most $n$ where $n = \deg_X(f).$
Therefore, the degree of the resultant polynomial is at most $2mn +
2.$ And if $2^k$ is larger than $2mn+2$, then 
$$
\text{Res}_X(f,\tilde{g}) = 0\bmod{Y^{2^k}} \implies \text{Res}_X(f,\tilde{g}) = 0
$$
and therefore, $f$ and $\tilde{g}$ must infact share a common
factor. Use the euclidian algorithm to extract out this factor. \\

All that is left to do are the corner cases. We shall deal with them
in the next class. 



