\documentclass[10pt]{article}
\usepackage{amsfonts}
\newtheorem{define}{Definition}
%\newcommand{\Z}{{\bf Z}}
\oddsidemargin=0.15in
\evensidemargin=0.15in
\topmargin=-.5in
\textheight=9in
\textwidth=6.25in
\begin{document}
\input{preamble.tex}
\lecture{19}{November 16, 2005}{Madhu Sudan}{Swastik Kopparty}
\section{Today}
\begin{itemize}
\item Finish Groebner Basis (Recognition)
\item Complexity of Ideal Membership
\end{itemize}
\section{Groebner Bases}
Recall that for an ideal $J$, we call $g_1, \ldots , g_t$ a Groebner Basis for $J$ if
\begin{itemize}
\item $\forall i, g_i \in J$
\item $I(LT(g_1), LT(g_2), \ldots, LT(g_t)) = I(LT(J))$
\end{itemize}
We further define two notions.
We say $r$ is a {\em weak remainder} of $f$ w.r.t. $g_1, \ldots, g_t$
if $f = r + \sum g_i q_i$ and $\forall$ monomials $m$ of $r$ and $\forall i$,
$LT(g_i)$ does not divide $m$.
We say $(q_1, \ldots, q_m)$ is a {\em strong quotient} for $f$ w.r.t. $g_1, \ldots, g_t$
if $\forall i, \deg (g_iq_i) \leq deg f$.
Recall that when we run our algorithm $DIVIDE$, we get a weak remainder.
For two polynomials $f,g$, we define the {\em Syzygy} to be the linear combination of them
which cancels leading terms; i.e.
$$ S(f,g) = LC(g) \frac{M}{LM(f)} f - LC(f)\frac{M}{LM(g)} g$$
where $M = LCM (LM(f), LM(g))$.
We can now give the test for a GB:
\begin{itemize}
\item Given $g_1, \ldots g_t$ as input
\item Check that $\forall i,j$, $DIVIDE(S(g_i, g_j), g_1, \ldots, g_t) $ returns $(0,$strong quotient$)$.
\item Then $\{g_i\}$ form a GB iff it passes the check.
\end{itemize}
We now prove the validity of this test:
\begin{proof}
Take $f \in J = I(g_1, \ldots, g_t)$.
We need to show that $LT(f) \in I(LT(g_1), \ldots, LT(g_t))$.
First write $f = \sum{m_j g_{i_j}}$ where $i_j \in \{1, \ldots, t\}$.
Amongst all such representations, pick the {\em reduced form}; i.e. the sequence with the smallest length satisfying $\deg(m_1g_{i_1}) \geq \deg(m_2g_{i_2}) \geq \ldots $ and also, if $\deg(m_j g_{i_j}) = \deg(m_{j+1}g_{i_{j+1}}$, then $i_j < i_{j+1}$.
Claim: $LT(f) = LT(m_1g_{i_1})$.
Wlog, we can take $f = m_1 g_1 + m_2g_2 + \ldots$.
Suppose $\deg (m_1g_1) = \deg (m_2g_2)$.
In this case we want to say that $m_2g_2 = m_1g_1 + $ lower degree terms. We use the Syzygy property:
$$ m_1g_1 = w\frac{M}{LM(g_1)}g_1$$
$$ m_2g_2 = w\frac{M}{LM(g_2)}g_2$$
$$ S(g_1, g_2) = 0 + \sum g_i q_i $$
where degree($g_iq_i$) < degree($\frac{M}{LM(g_1)}g_1$).
So, $m_2g_2 = m_1g_1 + \sum g_iq_i$.
Thus reducedness is violated, and hence $deg(m_1g_1) > deg(m_2g_2)$, thus $LT(f) = LT(m_1g_1)$, as desired.
\end{proof}
\section{Complexity of Ideal Membership Problem}
\begin{itemize}
\item Given $f_0, \ldots f_m \in K[X_1, \ldots, X_n]$ of degree $d$
\item Decide if $\exists q_1, \ldots q_m$ s.t. $f_0 = \sum f_iq_i$.
\end{itemize}
We wish to bound the complexity (in operations over $K$) in terms of $n,d,m$.
\begin{theorem} {\bf [Mayr, Meyer '81]} $ IM \in EXPSPACE = SPACE(2^{poly(n,d,m)}) $ and further, $IM$ is $EXPSPACE$ hard!
\end{theorem}
\subsection{Hardness}
The reduction is from the Commutative word equivalence problem (CWEP).
\begin{itemize}
\item Input:
\item $\Sigma $ a finite alphabet, $|\Sigma| = n$.
\item Rules $\alpha_1 = \beta_1, \alpha_2= \beta_2, \ldots, \alpha_m = \beta_m$, $\alpha_i, \beta_i \in \Sigma^*$
\item $\alpha, \beta \in \Sigma^*$
\item Goal:
\item Determine if $\alpha = \beta$.
\item Using given rules and using commutativity of symbols in $\Sigma$.
\end{itemize}
It is known that $CWEP$ is $EXPSPACE$ hard.
The reduction is obvious. Every word is a monomial. Rules are
binomials $f_i(x) = mono(\alpha_i) - mono(\beta_i)$. Membership in
CWEP is asking if $f_0(x) = mono(\alpha_0) - mono(\beta_0) \in I$?
Thus $IM$ is EXPSPACE hard.
\subsection{Upper bound}
This result rests on 2 facts:
\begin{itemize}
\item Inverting a $m \times n$ linear system can be done in $SPACE(polylog(m+n))$.
\item A 1926 result of Hermann that says that there exist $q_i$ with $deg(q_i) \leq D = (md)^{2^n}$
\end{itemize}
Note that finding $q_i$ (if they exist) can be posed as inverting a linear system.
We will prove Hermann's result. We want to get an understanding of
solutions to the following kind of question, a linear equation over
a ring:
\begin{itemize}
\item Determine if $\exists q_1, q_2, \ldots, q_m \in K[X_1, \ldots, X_n]$ s.t. $\sum f_iq_i = f_0$
\end{itemize}
Note that this question can be posed as a linear system over a
field, a kind of question that we do understand:
\begin{itemize}
\item Determine $\exists q_{i,\alpha} \in K$ s.t. $\forall \beta \sum_{i,\alpha + \alpha' = \beta} q_{i,\alpha} f_{i,\alpha'} =
f_{0,\beta}$, where $\beta$ ranges over all multi-indices over $n$
variables of degree $\leq \deg(f_0)$
\end{itemize}
In order to bound the degree, we introduce a common generalization,
the $j$-variable linear system, that will help us make the
transition between the problems
\begin{itemize}
\item Given polynomials $f_{i,\alpha} \in K[X_1, \ldots, X_j]$, $i \in \{0, 1, \ldots, m\}, \alpha \in A$
\item Determine if $\exists q_{i} \in K[X_1, \ldots, X_j] $ s.t. $\forall \alpha \in A, \sum_{i} q_{i} f_{i,\alpha} =
f_{0,\alpha}$
\end{itemize}
The strategy will be to eliminate 1 variable at a time. The crux of
the Hermann result is that a $j$ variable linear system with $M$
equations and $n$ unknowns reduces to a $j-1$ variable linear system
in $poly(M,n,d)$ equations and $poly(M,n,d)$ unknowns.
\begin{lemma}
Let $f_{i} \in K[X_1, \ldots, X_j]$. Suppose $\exists q_i \in K[X_1,
\ldots, X_j]$ with $X_j$ degree $< D$ satisfying $f_0 = \sum_{i=1}^m
f_iq_i$. Then the following system of equations over has a solution
$q_{i,\alpha}' \in K[X_1, \ldots, X_{j-1}]$
$$\forall \gamma < D, \sum_{i, \beta, \alpha, \beta+ \alpha = \gamma}
f_{i,\beta}q_{i,\alpha}' = f_{0,\gamma}$$ where $f_{i,\beta} \in
K[X_1, \ldots, X_{j-1}$ is the coefficient of $X_j^{\beta}$ in
$f_i$. Furthermore, any solution to this system of equations yields
a solution to the original equation with $X_j$ degree $< D$.
\end{lemma}
\begin{proof}
Simply take $q_{i,\alpha}'$ to be the coefficient of $X_j^{\alpha}$
in $q_i$.
\end{proof}
\begin{definition} Let $R$ be a ring. We call an $r \times s$ matrix
$A$ with entries in $R[z]$ {\em good} if
\begin{itemize}
\item $r < s$
\item There exists an $r\times r$ minor $\tilde{A}$ with
$\det{\tilde{A}}$ monic and nonzero.
\end{itemize}
\end{definition}
\begin{lemma}
Let $R$ be a ring. Let $A$ be a good matrix in $R[z]$ with each
entry having degree $\leq D$. Let $b$ be a vector with entries in
$R[z]$ with each entry having degree $\leq D$. Suppose $Ax = b$ has
a solution in $R[z]$. Then $Ax = b$ has a solution with each entry
having degree $\leq O(MD)$.
\end{lemma}
\begin{proof}
Consider the minor $\tilde{A}$ guaranteed to exist by the goodness
of $A$. We can rearrange the columns and have $A = [\tilde{A}|B]$.
For a vector $w$ with $w^T = (w_1|w_2)$, we have that $Aw =
\tilde{A}w_1 + Bw_2$. Thus, if we pick $w_2$ arbitrarily, then if
$Aw = b$, it must be that $w_1 = \tilde{A}^{-1}(b-Bw_2)$.
Note that $\tilde{A}^{-1} = \frac{Adj(\tilde{A})}{\det(\tilde{A})}$.
Thus if $(x_1, x_2)$ is a solution, then for any vector $c$, so is
$w =(x_1 + Adj(\tilde{A})Bc, x_2 - \det({\tilde{A}})c)$. Now, by the
goodness hypothesis, $\det({\tilde{A}})$ is monic, and since its
degree $\leq O(MD)$, then by choosing $c$ appropriately, make
$\deg(w_2) = O(MD)$. Then, $\deg(w_1) <
\deg\left(\frac{Adj(\tilde{A})}{\det(\tilde{A})} (b - Bw_2) \right)$
which $= O(MD)$, as desired.
\end{proof}
With this lemma in hand, it is essentially clear what to do. Suppose
we are given a system of $M$ equations $Ax = b$ with coefficients in
$R = K[X_1, \ldots, X_j]$ and degree bounded by $D$. Suppose that we
also know that there is a solution to this system. Then by lemma 4,
there is a solution with $X_j$ degree $< O(MD)$. Thus by lemma 2 we
can reduce to $O(M^2D)$ equations in over $K[X_1, \ldots, X_{j-1}]$
with degree at most $D$. Continuing this way, we get a linear system
over $K$ which has a solution, from which we can reconstruct a
solution to the original problem with degree at most $(MD)^{O(2^n)}$
(note that the degree was squaring at each stage).
Actually, to apply lemma 4 we required some goodness from our linear
system at each stage. This can be achieve by doing the following at
every stage: we throw away all row dependencies to make the matrix
of full row rank. Then applying a random linear transformation to
the $X_1, \ldots X_n$, we get that with high probability for any
single polynomial and any fixed variable, the modified polynomial
will be monic in that variable. This holds in particular for the
determinant of a nonsingular $r\times r$ minor of our $A$, thus
making it good.
To see the high probability result, let us be a bit more precise.
Given a polynomial $f(x)$ homogenous of degree $n$, not identically
0. Pick a random orthogonal matrix $P$ (uniform from
$S^{n-1},S^{n-2}, \ldots, S_0$ ) and consider the polynomial $g(x) =
f(Px)$. Then the resulting polynomial is homogenous of degree $n$
and is not monic if and only if $g(1,0, \ldots, 0) = 0$. However
$P\cdot(1,0, \ldots 0)$ is a point uniformly chosen from the surface
of the sphere and by Schwarz Zippel, $f(P\cdot(1,0,\ldots,0))$ is
nonzero almost everywhere. Thus w.h.p. $g$ is monic.
\end{document}