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\title[PURE Math 2012 Residents' Program Week 2]
{PURE Math Residents' Program\\
Gr\"obner Bases and Applications\\
Week 2 Lectures}

\author
{John B. Little}

\institute[College of the Holy Cross]
{
  Department of Mathematics and Computer Science\\
  College of the Holy Cross\\
  $\ $\\
}

\date[June 2012]
{June 2012}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\section{Day 1}

\begin{frame}
\frametitle{Leading terms, etc.}
\begin{itemize}
\item<1-> Last week, we introduced monomial orders so that we can
select a \emph{leading term} from each polynomial.
\item<2-> For instance, if $f(x,y,z) = 2 x^3 y^2 + \frac{1}{3} x y^2 z + 4z^5$
and we use $>_{lex}$ (with $x > y > z$), then
\item<3-> $LT_{>_{lex}}(f) = 2 x^3 y^2$ (including the coefficient)
\item<4-> $LM_{>_{lex}}(f) =  x^3 y^2$ (without the coefficient)
\item<5-> $LC_{>_{lex}}(f) = 2$
\item<6-> In text:  ${\rm multideg}(f) = \a$ if $LT(f) = cx^\a$
\item<7-> If order is clear from context we'll often omit it
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Division in $k[x_1,\ldots,x_n]$}
\begin{itemize}
\item<1-> First major difference with 1-variable case -- we'll allow
more than one divisor $f_1,\ldots,f_s$ (reason: not every ideal is
principal). So there will be as many quotients as divisors.
\item<2-> There can be several $LT(f_i)$ that divide
$LT$ of the dividend.  If so, we'll go down the list
of the $f_i$ from the start and use the first one found.
\item<3-> Second major difference with 1-variable case -- when
a term is not divisible by any of the $LT(f_i)$, it goes into the remainder,
but \emph{division is not necessarily finished}.
\end{itemize}
\end{frame}

\begin{frame}[fragile]
\frametitle{The algorithm}
\begin{verbatim}
Input: f_1,...,f_s,f, monomial order >
Output: a_1,...,a_s,r
a_1 := 0; ... a_s := 0; r := 0; p := f;
while p <> 0 do
   divocc := false; i := 1;
   while i <= s and divocc = false do
      if LT(f_i) divides LT(p) then
         a_i := a_i + LT(p)/LT(f_i)
         p := p - (LT(p)/LT(f_i)) f_i
         divocc := true
      else
         i := i + 1
   if divocc = false
      r := r + LT(p)
      p := p - LT(p)
\end{verbatim}
\end{frame}

\begin{frame}
\frametitle{Division theorem}
\begin{theorem}
Given any input $f_1,\ldots,f_s$, $f$, and a monomial order,
the algorithm above terminates and yields an expression
$$f = a_1 f_1 + \cdots + a_s f_s + r$$
where
\begin{enumerate}
\item[i.]  If $a_i f_i \ne 0$, then $LT(a_i f_i) \le LT(f)$
\item[ii.] If $r \ne 0$, then no monomial in $r$ is divisible by $LT(f_i)$ for any $i$, $1 \le i \le s$.
\end{enumerate}
\end{theorem}
(Note: there is a sense in which this expression is unique too,
but it's more subtle than in the 1-variable case.  See Exercise 11 in Chapter 2, \S 3.)
\end{frame}

\begin{frame}
\frametitle{Example}
Here's a first example.  Suppose $f_1 = x z - y^2$, $f_2 = x^3 - y z$ and use {\it lex}
order with $x > y > z$ so the first term in each is the leading term.  Say $f = x^4 + x^3 z$.
(Work out on board).

\begin{itemize}
\item<2->Result is
$$x^4 + x^3 z = (x^2+y)(xz - y^2) + (x)(x^3 - yz) + (x^2y^2+y^3)$$
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{More examples}
\begin{itemize}
\item<1-> Let $f_1 = xy + x + 1$, $f_2 = y^2 - x$,  $f = x^2 y^2$, and use
$>_{grlex}$ with $x > y$.
\item<2-> (Work out on board)
\item<3-> Note how the term $x^2$ went into the remainder $r$, but division continued
for one more step:
$$x^2 y^2 = (xy - x - 1) \cdot (xy + x + 1) + 0\cdot (y^2 - x)  + (x^2 + x + 1)$$
\item<4->  If we reorder the divisors we get different quotients and remainder(!)
\item<5->
$$x^2 y^2 = (x^2)(y^2 - x) + 0 \cdot (xy + x + 1) + (x^3)$$
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{Observations}
\begin{itemize}
\item<1-> The quotients and remainder can change if we
just reorder the divisors(!)
\item<2->  Also, if $r = 0$, it follows that $f \in \langle f_1,\ldots,f_s\rangle$.
\item<3->  But the converse \emph{fails}.  Here is an example:
\item<4->  Say $f_i$ are as above: $f_1 = xy + x + 1$, $f_2 = y^2 - x$.  If
we take $f = yf_1 - xf_2 = xy + y + x^2$, and divide by $(f_1,f_2)$ in
that order, we get
$$xy + y + x^2 = (1)(xy + x + 1) + (0)\cdot (y - x^2) + (x^2 + y - x - 1)$$
\item<5->  Doesn't seem especially useful!
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Key idea}
\begin{itemize}
\item<1-> In $k[x]$, we could tell whether $f(x) \in I$ by finding
the (monic) generator $g(x)$ such that $I = \langle g(x)\rangle$.
\item<2->  Then $f(x) \in I \Leftrightarrow r(x) = 0$ in
$f(x) = q(x) g(x) + r(x)$ from the division algorithm
\item<3->  In $k[x_1,\ldots,x_n]$, to ``fix'' the apparent
undesirable properties we saw in the examples above, we have
to find ``good'' sets of generators with the same properties
as the $g(x)$ of minimal degree.
\item<4->  Analogy will be $\{g(x)\} \leftrightarrow $ a Gr\"obner basis
\item<5->  Euclidean algorithm $\leftrightarrow$ Buchberger's algorithm
\end{itemize}
\end{frame}

\section{Day 2}

\begin{frame}
\frametitle{Motivation for definition of Gr\"obner bases}
\begin{itemize}
\item<1-> In examples like this:  $f_1 = x y  + 1$, $f_2 = y^2 - x$,
$f = yf_1 - xf_2 = y - x^2 \in I = \langle f_1,f_2\rangle$
\item<2-> If we use $>_{grlex}$, then $LT(f_1) = xy$, $LT(f_2) = y^2$,
but $LT(f) = -x^2$
\item<2-> If we divide $f$ by $(f_1,f_2)$, then $r \ne 0$, even
though $f \in \langle f_1,f_2\rangle$
\item<3-> The leading terms of the given
generators $f_1,f_2$ don't account for \emph{all possible leading terms} of elements of $I$
\item<4-> Goal: ``good'' generating sets satifying $f \in I \Leftrightarrow r = 0$ on division
\item<5-> Equivalently, we want generators $\{g_1,\ldots,g_t\}$ for $I$ such that 
for every $f \in I$, $LT(f)$ is divisible by $LT(g_i)$ for some $i$.  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Questions}
\begin{itemize}
\item<1-> Given an arbitrary ideal $I \subset k[x_1,\ldots,x_n]$, 
does there always exist $G = \{g_1,\ldots,g_t\} \subset I$ such that
for every $f \in I$, $LT(f)$ is divisible by $LT(g_i)$ for some $i$?
\item<2-> If so, how do we find them?  
\item<3-> For instance, starting from an arbitrary set of 
generators for $I$, how compute a set $G$ with the property above?
\item<4->  Can also ask: To what extent $G$ depends on the choice of 
monomial order?
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The ideal of leading terms}
\begin{itemize}
\item<1-> Start from a given ideal $I$ and a given monomial order $>$
\item<2-> For each $f\in I$, we have $LT(f)$
\item<3-> Define $\langle LT(I)\rangle = \langle LT(f) \mid f \in I \rangle$
\item<4-> That is $\langle LT(I)\rangle$ is the \emph{ideal generated by 
the leading terms of all elements of $I$} according to the given monomial
order.  
\item<5-> An example of a \emph{monomial ideal} -- an ideal generated
by a collection of monomials.  
\item<6-> These have some nice properties, as we'll see next
\end{itemize} 
\end{frame}

\begin{frame}
\frametitle{A technical result}
\begin{lemma}
Let $M$ be a monomial ideal generated by some collection 
of monomials $\{x^\a \mid \a \in A\}$ (possibly infinite).
Let $x^\beta \in M$.  Then $x^\beta$ is a multiple of $x^\a$ for some 
$\a \in A$.
\end{lemma}

\begin{proof} By definition $x^\beta = \sum_\a h_\a x^\a$
(where $h_\a$ are some polynomials, only finitely many of which
are nonzero).  But then $x^\beta = x^\gamma x^\a$ for
some $x^\gamma$ appearing in one of the $h_\a$.
\end{proof}
\end{frame}

\begin{frame}
\frametitle{Dickson's Lemma}
\begin{theorem}[Dickson's Lemma]
Let $M$ be a monomial ideal in $k[x_1,\ldots,x_n]$.
Then $M$ is generated by a finite collection of monomials.
\end{theorem}

\noindent
{\bf Proof.}
\begin{itemize}
\item<1-> By induction on $n$
\item<2-> If $n = 1$, then by what we did
last week, we know $M$ is principal and hence
$M = \langle x^a\rangle$ where $a$
is the smallest nonnegative integer such that $x^a \in M$.  
\item<3-> Now assume that the result is known for all
monomial ideals in $k[x_1,\ldots,x_{n-1}]$ and consider
$M \subset k[x_1,\ldots,x_{n-1},y]$.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Proof of Dickson, continued}
\begin{itemize}
\item<1-> Write monomials as $x^\a y^b$
\item<2-> The projection $M' = \langle \{x^\a \mid x^\a y^b \in M {\rm\ for\ some\ } b\ge 0\}\rangle$
is a monomial ideal in $k[x_1,\ldots,x_{n-1}]$
\item<3->  So induction hypothesis applies, and $M' = \langle x^{\a(1)}, \ldots, x^{\a(s)}\rangle$
for some $s$.
\item<4->  This means that for each $1 \le i \le s$, there is some $b_i$ such that $x^{\a(i)} y^{b_i} \in M$.
\item<5->  Let $b = \max_i \{b_i\}$ so $x^\a(i) y^b \in M$ for all $1 \le i \le s$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Proof of Dickson, continued}
\begin{itemize}
\item<1->  For each $0 \le c < b$, take the ``horizontal slice'' of $M$
at height $c$ and project that to get 
$M'_c = \langle x^\a \mid x^\a y^c \in M\rangle$
\item<2->  The $M'_c$ are also monomial ideals in $k[x_1,\ldots,x_{n-1}]$ 
so induction $\Rightarrow$ $M'_c = \langle x^{\a(c,1)}, \ldots, x^{\a(c,s_c)} \rangle $
\item<3->  \emph{Claim is}:  The $x^{\a(c,1)},\ldots,x^{\a(c,s_c)}$ for $0 \le c < b$ and 
the $x^{\a(1)} y^b, \ldots, x^{\a(s)} y^b$ generate $M$
\item<4-> That follows fairly easily from the construction. QED
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Consequences of Dickson}
\begin{itemize}
\item<1-> Return to the monomial ideal $\langle LT(I)\rangle$
for a given $I$ and a given monomial order.
\item<2-> By Dickson, we know that $\langle LT(I) \rangle = \langle x^{\a(1)}, \ldots, x^{\a(t)}\rangle $
for some finite collection of monomials.
\item<3-> Every monomial in $\langle LT(I)\rangle$ is $LT(g)$ for some $g \in I$ (why?)
\item<4-> (Reason is Lemma 2 from before implies if $x^\beta \in \langle LT(I)\rangle$, then
$x^\beta = x^\gamma LT(f)$ for some $f\in I$.  But then $x^\gamma LT(f) = LT(x^\gamma f)$ by 
properties of monomial orders and $x^\gamma f \in I$ by definition of an ideal.)
\item<5->  Consequence:  There exist $g_i \in I$ such that $LT(g_i) = x^{\a(i)}$ for all $1 \le i \le t$.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Gr\"obner bases defined}
\begin{itemize}
\item<1-> This leads to 
\begin{definition} Let $I$ be a nonzero ideal and $>$ be a monomial order.  
A \emph{Gr\"obner basis} for $I$ with respect to $>$ is a finite set of 
polynomials $G = \{g_1,\ldots,g_t\} \subset I$ such that 
$\langle LT(I)\rangle = \langle LT(g_1),\ldots, LT(g_t)\rangle$.  
\end{definition}
\item<2->  Dickson's Lemma $\Rightarrow$ 
\begin{theorem}  If $I$ is a nonzero ideal and $>$ is a monomial
order, then Gr\"obner bases of $I$ with respect to $>$ exist.
\end{theorem}
\item<3->  Not unique, though, since as we saw, generating
sets for the monomial ideal $\langle LT(I)\rangle$ are not unique.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Consequences of Dickson, continued}
\begin{itemize}
\item<1->  We have 
\begin{theorem} A Gr\"obner basis $G = \{g_1,\ldots,g_t\}$ for $I$ \emph{generates} $I$.\end{theorem}
\item<2-> \begin{proof} Let $f \in I$ and use the division algorithm.  At every stage,
the polynomial $p$ is in $I$, so its leading term is divisible by $LT(g_i)$ for
some $i$.  The algorithm reduces $p$ to $0$ without putting any terms into $r$, so 
$r = 0$ and $f = a_1 g_1 + \cdots a_t g_t$. \end{proof}
\item<3->  This also proves an unexpected ``big theorem!''  \begin{theorem}[Hilbert Basis Theorem]
Every ideal in $k[x_1,\ldots,x_n]$ is finitely generated. \end{theorem}
\end{itemize}
\end{frame}

\section{Day 3}

\begin{frame}
\frametitle{The ACC}
\begin{itemize}
\item<1-> No, not the Atlantic Coast Conference(!) 
\item<2-> ACC = ``Ascending Chain Condition'' 
\begin{theorem}  Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots\ $
be an ascending chain of ideals in $k[x_1,\ldots,x_n]$.  Then there exists 
an index $m$ such that $I_m = I_{m+1} = I_{m+2} = \cdots $.  
\end{theorem}
\item<3->  That is, an ascending chain of ideals \emph{cannot strictly
increase forever} -- it must \emph{stabilize} after finitely many steps.
\item<4->  \begin{proof} The union $I = \cup_{i \ge 1} I_i$ is also an 
ideal (why?)  By the HBT, $I = \langle f_1,\ldots,f_s\rangle$ for some 
$f_i$.  Each $f_i$ ``comes from'' some $I_j$; after some
number $m$ of steps, $I_m$ contains all $f_j$, so equals $I$.  
\end{proof}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Comments on ACC}
\begin{itemize}
\item<1->  In fact ACC $\Leftrightarrow$ every ideal is finitely generated
(HBT for the polynomial ring).  Can you see how the 
other implication might go?
\item<2->  Hint: Argue by contraposition.  If there exists an
ideal that is not finitely generated, then ACC cannot hold!
\item<3->  The class of commutative
rings in which ACC holds, and in which all ideals are finitely generated
is known as the class of \emph{Noetherian} rings, after Emmy Noether.
\item<4->  The ACC might seem like a rather arcane theoretical statement, 
but as we'll see shortly, it has a big practical implication for our story(!)
\end{itemize} 
\end{frame}


\begin{frame}
\frametitle{Background for Buchberger's algorithm}
\begin{itemize}
\item<1-> Recall the example we discussed earlier:  $f_1 = x y  + 1$, $f_2 = y^2 - x$,
$f = yf_1 - xf_2 = y + x^2 \in I = \langle f_1,f_2\rangle$
\item<2-> If we use $>_{grlex}$, then $LT(f_1) = xy$, $LT(f_2) = y^2$,
but $LT(f) = x^2 \notin \langle LT(f_1),LT(f_2)\rangle$
\item<3->  In other words, $\{f_1,f_2\}$ is \emph{not a Gr\"obner basis}
for $I$ with respect to $>_{grlex}$. 
\item<4-> Note that we ``found'' a new leading term by forming a polynomial
combination of $f_1,f_2$ that was constructed to \emph{cancel leading terms}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{$S$-polynomials}
\begin{itemize}
\item<1->  A general form of this:  \begin{definition} Let $f,g\in k[x_1,\ldots,x_n]$
and $>$ be a monomial order.  The \emph{$S$-polynomial} of $f,g$ is 
$$S(f,g) = \frac{{\rm lcm}(LM(f),LM(g))}{LT(f)} f - \frac{{\rm lcm}(LM(f),LM(g))}{LT(g)} g$$
\end{definition}
\item<2->  This is defined to make the leading terms cancel.  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{A more elaborate example}
\begin{itemize}
\item<1->  Another example:  $f = 2x^2 y + xy$, $g = x y^2 + 2x + y$, using {\it lex} order $x > y$:
\begin{eqnarray*}
S(f,g) &=& \frac{x^2 y^2}{2x^2y}(2x^2 y + x y) - \frac{x^2y^2}{xy^2}(xy^2 + 2x + y)\\
&=& x^2 y^2 + \frac{1}{2} x y^2 - (x^2 y^2 + 2x^2 + x y)\\
&=& \frac{1}{2} x y^2 - 2x^2 - x y
\end{eqnarray*}
\item<2->  In this case, the leading term of the $S$-polynomial
is a multiple of $LT(g)$.   
\item<3-> But we would get something ``new'' if we subtracted $\frac{1}{2}g$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Idea of Buchberger algorithm}
\begin{itemize}
\item<1->  When we find a ``new'' leading term like this, we will just
append the new polynomial to our list of generators(!)
\item<2->  Even if the $S$-polynomial itself does not have a ``new''
leading term, we can still try to ``strip away'' terms we already
know by computing \emph{the remainder on division of the $S$-polynomial}
by the generators of the ideal we already have.  
\item<3->  Note that if 
$$S(f_i,f_j) = a_1 f_1 + \cdots + a_s f_s + r$$
then by definition $r \in I = \langle f_1,\ldots,f_s\rangle$ so 
if $r \ne 0$, then its leading term will be something we want to know(!)
\end{itemize}
\end{frame}

\begin{frame}[fragile]
\frametitle{Buchberger's algorithm -- basic form}
\begin{verbatim}
Input: F = {f_1,...,f_s}
Output: G containing F
G := F
repeat
   G' := G
   for each pair p <> q in G' do
      S := remainder of S(p,q) on division by G'
      if S <> 0 then
         G = G union {S}
until G = G'
\end{verbatim}
\end{frame}

\begin{frame}
\frametitle{Comments and questions}
\begin{itemize}
\item<1->  To understand what this is doing note that $G'$ stores a copy of the
collection of polynomials at the start of each pass through the repeat loop.  The
pairs $p,q$ are selected from this copy, which is not changing.  
\item<2->  Any nonzero S-polynomial remainders are adjoined to the \emph{original}
collection of polynomials, which is in $G$.
\item<3->  The algorithm will terminate the first time $G = G'$ (that is when 
all S-polynomial remainders are zero, so no new polynomials are adjoined to $G$)
\item<4->  Question 1:  How do we know this process will ever stop?
\item<5->  Question 2:  If it does stop, is $G$ a Gr\"obner basis?
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Answers to the questions}
\begin{itemize}
\item<1->  Question 2 is answered by the main technical result of 
Buchberger's theory:  
\begin{theorem}[Buchberger's $S$-polynomial Criterion]  
Let $G = \{g_1,\ldots,g_t\}$ be a collection of polynomials.  
Then $G$ is a Gr\"obner basis for the ideal it generates if and 
only if the remainder on division of $S(g_i,g_j)$ by $G$ is zero 
for all pairs $i\ne j$.  \end{theorem}
\item<2->  One implication is easy; the other one says the answer
to Question 2 is \emph{yes}!  This is Theorem 6 in Chapter 2, 
\S 6 of IVA -- a ``hard slog of a proof if there ever was one''
\item<3-> We won't discuss this in ``class''
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Termination of Buchberger's algorithm}
\begin{itemize}
\item<1->  The remaining question is:  Does this always terminate?
\item<2->  Note that if the algorithm does not terminate, it is
because the new $G$ strictly contains $G'$
\item<3->  That means $\langle LT(G') \rangle \subset \langle LT(G)\rangle$
(strict containment)
\item<4->  The ACC implies this cannot go on forever.   Eventually
this increasing chain of monomial ideals must stabilize.
\item<5->  When it does, the algorithm terminates.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{An example, by hand}
\begin{itemize}
\item<1->  Let  $f_1 = x y  + 1$, $f_2 = y^2 - x$, use {\it grlex} order.
We start with $G = \{f_1,f_2\}$.
\item<2->   $S(f_1,f_2) = x^2 + y$, with leading term $x^2$, and
this is its own remainder on division by $f_1,f_2$.  So we update to $G = \{f_1,f_2,f_3 = x^2 + y\}$
\item<3->  Now, the $S$-polynomial $S(f_1,f_2)$ reduces to zero, so we consider $S(f_1,f_3) = x(xy + 1) - y(x^2 + y) = x - y^2 = -f_2$.
This reduces to a remainder of $0$ because we have $f_2$.  
\item<4->  Next, $S(f_2,f_3) = x^2(y^2 - x) - y^2(x^2 - y^2) = -x^3 + y^4$.  Dividing by $G = \{f_1,f_2,f_3\}$ (in that order),
we find 
$$y^4 - x^3 = (y +1) f_1 + y^2 f_2 + (-x) f_3 + (-y - 1)$$
\item<5->  After cleaning up the signs, we adjoin $f_4 = y + 1$ to $G$ and continue.    
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Example, continued}
\begin{itemize}
\item<1->  We have $S(f_1,f_4) = x - 1$ and that is its own remainder on division by 
$\{f_1,f_2,f_3,f_4\}$, so that polynomial must also be adjoined to $G$.
\item<2->  At this point, it can be checked that $G = \{xy + 1, y^2 - x, x^2 + y, y + 1, x - 1\}$ satisfies
Buchberger's Criterion, so it is a Gr\"obner basis for $I = \langle f_1,f_2\rangle$ with respect to the {\it grlex} order.
\item<3->  Note that $LT(f_1),LT(f_2),LT(f_3)$ are multiples of $LT(f_4)$ or $LT(f_5)$ or both.
\item<4->  This says that $\langle LT(I)\rangle$ is generated by $\langle LT(f_4),LT(f_5)\rangle$.
\item<5->  Hence $\{f_4,f_5\}$ is {\it also} a Gr\"obner basis for $I$.  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Reduced Gr\"obner bases}
A useful theoretical result:

\begin{theorem}
Each nonzero ideal $I$ has a unique \emph{reduced Gr\"obner basis} with respect to each
monomial order -- a Gr\"obner basis $G = \{g_1,\ldots, g_t\}$ such that 
\begin{enumerate}
\item[i.] $LC(g_i) = 1$ for all $i$, and
\item[ii.]  No term in $g_i$ is divisible by $LT(g_j)$ for any $j \ne i$.  
\end{enumerate}
\end{theorem}
\end{frame}

\section{Day 4}

\begin{frame}  
\frametitle{Elimination}
\begin{itemize}
\item<1-> In elementary algebra, linear algebra, etc., 
a standard method for solving simultaneous equations
in several variables is to form polynomial combinations 
that eliminate variables.
\item<2-> Example:  In the system 
\begin{eqnarray*}
2x - 3y &=& 1\\
4x + 5y &=& 3
\end{eqnarray*}
\item<3-> second equation minus $2 \times $ first equation yields
$11y = 1$, so $y = \frac{1}{11}$, and then $x = \frac{7}{11}$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Elimination ideals}
\begin{itemize}
\item<1->  In our terms, 
$$(-2)(2x - 3y - 1) + (1)(4x + 5y - 3) = 11y - 1$$
is in $I = \langle 2x - 3y - 1, 4x + 5y - 3\rangle$, \emph{and} contains
no $x$.
\item<2->  Generalizing this,
\begin{definition}
Let $I \subset k[x_1,\ldots,x_n]$ be an ideal.  If $1 \le \ell \le n-1$, we
define the $\ell$th elimination ideal of $I$ to be 
$$I_\ell = I \cap k[x_{\ell+1},\ldots,x_n]$$
(in which the variables $x_1,\ldots,x_\ell$ have been eliminated).
\end{definition}
\item<3->  For example, $11y - 1 \in I_1 = I \cap \mathbb{Q}[y]$.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Geometry of elimination}
\begin{itemize}
\item<1-> If $I \subset k[x_1,\ldots,x_n]$, then we have the geometric
object $V(I) \subset k^n$
\item<2-> If we then eliminate the first $\ell$ variables, we
can ask, what is the corresponding variety $V(I_\ell)$?
\item<3-> Partial answer -- it's very closely related to 
the projection of $V(I)$ into the coordinate space $k^{n - \ell}$ of
the variables $x_{\ell + 1}, \ldots, x_n$.
\item<4->  Projection of a variety is not always a variety, but
over $\mathbb{C}$ at least, $V(I_\ell)$ is the \emph{smallest variety} 
containing the projection of $V(I)$.  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Lex Gr\"obner bases and elimination}
\begin{itemize}
\item<1-> A special property of lex order:  Say the variables are
ordered $x_1 > x_2 > \cdots > x_n$.  If a monomial contains any positive
power of $x_1$, then it is larger in lex order than all monomials 
that contain only $x_2,\ldots,x_n$.  Similarly, any monomial that contains
a positive power of $x_2$ is larger than all monomials containing only $x_3,\ldots,x_n$, etc.
\item<2->  Suppose $I$ is an ideal for which $I_\ell \ne \{0\}$, and let $f\ne 0$
be an element of $I_\ell$
\item<3->  If $G$ is a lex Gr\"obner basis for $I$, there must be some $g_i \in G$
such that $LT(g_i)$ divides $LT(f)$, hence $LT(g_i)$ contains only $x_{\ell+1}, \ldots, x_n$. 
\item<4->  But then the observation above shows 
$g_i \in I \cap k[x_{\ell+1},\ldots,x_n] = I_\ell$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Elimination Theorem}
This is the key idea in the proof of:
\begin{theorem}[Elimination Theorem]
Let $I$ be an ideal in $k[x_1,\ldots,x_m]$ and let $G$ be a 
Gr\"obner basis for $I$ with respect to lex order with $x_1 > x_2 > \cdots > x_n$.
For all $\ell$ let $G_\ell = G\cap k[x_{\ell + 1}, \ldots, x_n]$.  Then 
$G_\ell$ is a Gr\"obner basis for the elimination ideal $I_\ell$. 
\end{theorem}

(Note:  If $G_\ell = \emptyset$, this says $I_\ell = \{0\}$.)

In other words, \emph{lex Gr\"obner bases systematically eliminate variables ``as much as 
possible''}
\end{frame}

\begin{frame}
\frametitle{A first example}
\begin{itemize}
\item<1-> Let 
$$I = \langle x^2 y + y^2 + 2, x y - 3 y + 1 \rangle \subset \mathbb{Q}[x,y]$$
\item<2-> If we compute a (reduced) {\it lex} Gr\"obner basis for $I$ with
$x > y$, we get $G_y = $
$$\{y^3 + 9 y^2 - 4 y + 1, x - y^2 - 9 y + 1\} $$
\item<3-> Note that the first polynomial depends only on $y$.  It is the monic
generator for $I_1 = I \cap \mathbb{Q}[y]$.  
\item<4->  The second polynomial contains $x$ too.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Example, continued}
\begin{itemize}
\item<1->Note the form of 
$$G_y = \{y^3 + 9 y^2 - 4 y + 1, x - y^2 - 9 y + 1\} $$
\item<2-> To find the points in $V(I) = V(x^2y + y^2 + 2, xy - 3y + 1)$,
we could solve the one-variable equation $y^3 + 9 y^2 - 4 y + 1 = 0$ (numerically),
\item<3-> Then, substitute the values into the other equation and determine $x$.  
\item<4->  There are
three points in $V(I)$ over $\mathbb{C}$, one with coordinates in $\mathbb{R}$, approx.
$$( -3.10598633669341, -9.43517845033930)$$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Example, continued}
\begin{itemize}
\item<1-> If we reverse the order of the variables (i.e. look at {\it lex} order
with $y > x$), then the reduced Gr\"obner basis changes
\item<2->  Get $G_x = $
$$\{ x^3 - 5 x^2 + 12 x - 19, y + x^2 - 2 x + 6\}$$
\item<3->  Now, the first basis element generates $I \cap \mathbb{Q}[x]$, and the 
second contains $x,y$.
\item<4->  This other basis could be used in the same way to determine $V(I)$
(and would yield the same results!)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{``Implicitization'' $=$ elimination}
\begin{itemize}
\item<1->  In the first week, we briefly discussed how some varieties
can be given in parametric form as well as by implicit equations
\item<2->  The process of deriving implicit equations from a parametrization
is called ``implicitization''
\item<3->  This can also be performed by means of elimination and
{\it lex} Gr\"obner bases, when the coordinate functions are \emph{polynomial}
(or rational) functions
\item<4->  Example: A parametric surface in $\mathbb{R}^3$: 
\begin{eqnarray*}
x &=& u^2 \\
y &=& u + v \\
z &=& u - v^2 
\end{eqnarray*}
\end{itemize}
\end{frame} 

\begin{frame}
\frametitle{Implicitization example, continued}
\begin{itemize}
\item<1->  The ideal $I = \langle x - u^2, y - u - v, z - u + v^2\rangle$
defines the \emph{graph} of the parametrization map (a subset of $\mathbb{R}^5$).  
\item<2->  Geometrically, we want to project that into the $x,y,z$-coordinate
space to find the image of the parametrization map
\item<3->  In algebraic terms, we want to order the variables with $u,v$ bigger than
$x,y,z$ (for instance as $u > v > x > y > z$) and find the elimination ideal 
$I_2 = I \cap \mathbb{R}[x,y,z]$.
\item<4->  Computing a lex Gr\"obner basis we find 5 polynomials in all; only the 
last contain no $u,v$ terms:
$$I_2 = \langle -x + z^2 + 2 x z - 4 y x + x^2 + 2 z y^2 - 2 x y^2 + y^4\rangle$$
\item<5->  This defines a surface in $\mathbb{R}^3$ that contains the image 
of the parametrization.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Implicitization example, continued}
\begin{itemize}
\item<1->  The rest of the Gr\"obner basis is an ``illustrated book''
of exactly the way this parametrization works.
\item<2->  For instance, the next three polynomials in the basis have $x,y,z,v$, but
no $u$, so $I_1 = I \cap \mathbb{R}[v,x,y,z]$ has {\it lex} Gr\"obner basis consisting
of the generator for $I_2$ above, plus
\item<3->
\begin{eqnarray*} 
&&(1 + 2y)v + x - y + z - y^2\\
&&(1 + 4z + 4x)v + 5x - y + z + 2 y x + y^2 - 6 z y - 2 y^3\\
&&v - y + z + v^2
\end{eqnarray*}
\item<4->  Final polynomial is $u - y + v$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Interpreting the basis elements}
\begin{itemize}
\item<1->  The polynomials $v - y + z + v^2$ and $u - y + v$ show that given 
$(x,y,z) \in V(I_2)$, there are never more than $2$ pairs $(u,v)$ that
yield that $(x,y,z)$.  
\item<2->  The polynomials $(1 + 2y) v + x - y + z - y^2$ and $(1 + 4z + 4x)v + \cdots $
show that for ``most'' $(x,y,z)$, there is only one pair $(u,v)$.  
\item<3->  The only possible ``different'' points would come from places
on $V(I_2)$ where $1 + 2y = 0$ and $1 + 4z + 4x = 0$. Those equations define
a straight line that lies on the surface $V(I_2)$.  
\item<4->  Precise statement of all this comes from the Extension Theorem in text.
\end{itemize}
\end{frame}

\end{document} 
