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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 1}
\author{Ravi Vakil}
\date{September 9, 1999.}
\maketitle
\tableofcontents

{\parskip=12pt % closing bracket is just before the bibliography 

I'm going to start by telling you about this course, and about the field of algebraic geometry.

Goals:
\begin{itemize}
\item
 geometric insight
\item concrete examples (geometric and arithmetic)
\item hands on calculations (no fear of commutative algebra)
\item no cohomology, flatness, differentials
\end{itemize}


Modern algebraic geometry lies somewhere between differential geometry, number
theory, and topology.   In a loose sense, it is polynomial equations, and
sets defined by polynomial equations.  This seems to be extremely narrow
and low-tech, but it surprisingly ends up being extremely broad, powerful, and
abstract.

Some of the philosophy --- get at geometry via algebra, algebra via ``pictures''.
High school reference.  Here's a high-powered example of the link between geometry and
arithmetic.   $x^n + y^n = z^n$.  Finite number of solutions for each $n>1$:  
the Mordell Conjecture, Faltings' Theorem.  Vojta's conjecture.  Weil conjectures.

{\em Give out handout of motivating problems.}

This will be a tools course:  examples and pictures, but with generality.
\begin{itemize}
\item fast-moving, but grounded by intuition
\item exercises are important
\item concepts really generalize, but become more abstract.
\end{itemize}

{\em Objects:}

smooth varieties over $\com$           (over $k$)

varieties over $\com$               (over $k$)

schemes

(stacks)


We won't be shy about schemes in this course.

{\bf Administrative stuff.}  Give out handout.  Go through it.  Headcount.

Theme:  curves.  Examples:  $\zed$ and $\kbar[t]$.

\section{Commutative algebra}

Don't worry about it.  See books on handout.

Ideas you should know: ring (commutative, has 1), field, integral
domain, has quotient field, prime ideal, maximal ideal.  

Sample problem (to appear on problem set):

Let $A$ be a (commutative) ring.  An element $a \in A$ is nilpotent
(that is, $a^n = 0$ for some $n>0$) if and only if $a$ belongs to 
every prime ideal of $A$.

\section{Algebraic sets}

Throughout this course:  $k$ is a field.  $\kbar$ is an algebraically closed field.

For now we work over $\kbar$.  Feel free to think of this as $\com$ for now.

$\kbar^n$ will be rewritten $\A^n(\kbar)$, affine $n$-space; we'll
often just write $\A^n$ when there's no confusion about the field.  Coordinates $x_1$ to $x_n$.

Algebraic geometry is about functions on the space, which form a ring.  The only functions we will care about 
will be polynomials, i.e. $\kbar[x_1, \dots, x_n]$.  We'll eventually think of that ring as being the same thing
as $\A^n$.

We'll next define subsets of $\A^n$ that we'll be interested in.  Because we're being very restrictive,
we won't take any subsets, or even analytic subsets; we'll only think of subsets that are in some sense
defined in terms of polynomials.
 
Let $S$ be a set of polynomials, and define $V(S)$ to be the locus where these
polynomials are zero.  (``Vanishing set''.)   Definition:  Any subset of $\A^n(\kbar)$
of the form $V(S)$ is an {\em algebraic set}.

{\em Exercise} (to appear on problem set): prove that the points of the form
$(t,t^2,t^3)$ in $\A^3$ form an algebraic set.  In other words, find a
set of functions that vanish on these points, and no others.

Example/definition:  {\em hypersurface}, defined by 1 polynomial.

{\em Facts.}  

\begin{itemize}
\item  If $I = (S)$, then $V(I)= V(S)$.  So we usually will care only about ideals.
  Hence:  subsets of $\kbar[x_1,...,x_n]$ give us subsets of $\A^n$;
  specifically, {\em ideals} give us {\em algebraic sets}
\item $V(\cup I_a) = \cap V(I_a)$  (Say it in english.)
\item $I \subset J$, then $V(I) \supset V(J)$
\item $V(FG) = V(F) \cup V(G)$
\end{itemize}

Note:  Points are algebraic.  Finite unions of points are algebraic.


{\em Definition.}  A radical of an ideal $I \subset R$, denoted $\sqrt{I}$, is defined by
$$
\sqrt{I} = \{ r \in R | r^n \in I \text{ for some } n \}.
$$

{\em Exercise.}  Show that $\sqrt{I}$ is an ideal.

{\em Definition.}  An ideal $I$  is {\em radical} if $I = \sqrt{I}$.

{\em Claim.} $V(\sqrt{I}) = V(I)$.  (Explain why.)

Conversely, subsets of $\A^n$ give us a subset of $\kbar[x_1,...,x_n]$
For each subset X, let $I(X)$ be those polynomials vanishing on $X$.



{\em Claim.}  $I(X)$ is a radical ideal.  (Explain.)

{\em Facts.}  If $X \subset Y$, then $I(X) \supset I(Y)$.
$I(\emptyset) = \kbar[x_1,...,x_n]$.  $I(\A^n) = (0)$.

{\em Question.}  What's $I( (a_1,...,a_n) )$?

(Discuss.)

Notice:  ideal is maximal.  Quotient is field.  Quotient map can be
interpreted as ``value of function at that point''.

{\em Exercise.}  
(a)  Let $V$ be an algebraic set in $\A^n$, $P$ a point not in $V$. Show
that there is a polynomial $F$ in $\kbar[x_1,...,x_n]$ such that $F(Q) = 0$ for all $Q$ in
$V$, but $F(P) = 1$.  Hint:  $I(V) \neq I(V \cup {P})$.

 (b) Let $\{ P_1, ..., P_2 \}$ be a finite set of points in $\A^n(\kbar)$.  Show that
there are polynomials $F_1, ..., F_r \in \kbar [x_1,...,x_n]$ such that $F_i(P_j) =
0$ if $i \neq j$, and $F_i(P_i) = 1$.

{\em Exercise.}  Show that for any ideal $I$ in $\kbar [x_1,...,x_n]$, $V(I) = V(\sqrt{I})$, and
$\sqrt{I}$ is contained in $I(V(I))$.

\section{Nullstellensatz (theorem of zeroes)}

Earlier, we had:  algebraic sets $\rightarrow$ radical ideals and ideals $\rightarrow$
algebraic sets.

This theorem makes an equivalence.  In the literature, the word ``nullstellensatz'' is used to apply to 
a large number of results, not all of them equivalent.  

{\bf Nullstellensatz Version 1.}  Suppose $F_1$, \dots, $F_m \in \kbar[x_1, \dots, x_n]$. If
the ideal $(F_1, \dots, F_m) \neq (1) = \kbar[x_1, \dots, x_n]$ then the system of
equations $F_1  = \dots = F_m = 0$ has a solution in $\kbar$.  

Proof next day.  (There is a better version for fields that are not necessarily algebraically closed,
but we're not worrying about that right now.)

{\bf Nullstellensatz Version 2.}  Supopse $\fm$ is a maximal ideal of $\R$.
Then 
$$
\fm = (x_1 - a_1, \dots, x_n - a_n)
$$
for some $a_1$, \dots, $a_n \in \kbar$.

Show that this is equivalent to version 1, modulo fact that ideals are finitely generated.

{\bf Nullstellensatz Version 3 (sometimes called the ``Weak Nullstellensatz'').}
  If $I$ is a proper ideal in $\kbar[x_1,...,x_n]$, then
$V(I)$ is nonempty.  (From Version 2.)

{\bf Nullstellensatz Version 4.} Let $I$ be an ideal in
$\kbar[x_1,...,x_n]$.  Then $I(V(I)) = \sqrt{I}$.  Equivalently:
Radical ideals are in 1-1 correspondence with algebraic sets: If $I$
is a radical ideal in $\kbar[x_1,...,x_n]$ then $I(V(I)) = I$.  So
there is a 1-1 correspondence between radical ideals and algebraic
sets.

{\bf Nullstellensatz Version 5.} A radical ideal of $\kbar[x_1, \dots, x_n]$ is
the intersection of the maximal ideals containing it.  This is the geometric
rewording of 4.  By version 4, a radical ideal is $I(X)$ for some algebraic set $X$.  Functions
vanishing on $X$ are precisely those functions vanishing on all the points of $X$.

{\bf Nullstellensatz Version 6.} If $F_1$, ..., $F_r$, $G$ are in
$\kbar[x_1,...,x_n]$, and $G$ vanishes wherever $F_1$, ..., $F_r$
vanish, then there is an equation $G^N = A_1 F_1 + ...  + A_r F_r$ for
some $N>0$ and some $A_i$ in $\kbar[x_1,...,x_n]$.

This has a cute proof, with a useful trick in it.

{\em Proof.}  The case $G=0$ is obvious, so assume $G \neq 0$.
Introduce a new variable $U$, and consider the polynomials

$$F_1, \dots, F_m, \text{ and } UG-1 \in \kbar[x_1, \dots, x_n, U].$$

They have no common solutions in $\kbar$, so by Version 1 they generate the
unit ideal, so there are polynomials $P_1, \dots, P_m$, $Q \in  \kbar[x_1, \dots, x_n, U]$ such that
$$P_1 F_1 + \dots + P_m F_m + Q(UG-1) = 1.$$
Now set $U=1/G$ in this formula, and multiply by some large power $G^N$ of $G$ to clear denominators.  
Then the right side is $G^N$, and the left side is in $(F_1, \dots, F_m)$.


} % end of parskip; it started just before the introduction
\end{document}
