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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 3}
\author{Ravi Vakil}
\date{September 16, 1999.}
\maketitle
\tableofcontents

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

(Before class started, I showed that (finite) Chomp is a first-player win,
without showing what the winning strategy is.)

If you've seen a lot of this before, try to solve:

{\bf ``Fun problem'' 2.} Suppose $f_1(x_1, \dots, x_n)=0$, \dots,
$f_r(x_1, \dots, x_n)=0$ is a system of $r$ polynomial equations in
$n$ unknowns, with integral coefficients, and suppose this system has
a finite number of complex solutions.  Show that each solution is
algebraic, i.e. if $(x_1,
\dots, x_n)$ is a solution, then  $x_i \in \overline{\Q}$ for all $i$.


\section{Where we are}

We now have seen, in gory detail, the correspondence between {\em
radical ideals} in $\R$, and {\em algebraic subsets} of $\A^n(\kb)$.
Inclusions are reversed; in particular, maximal proper ideals
correspond to minimal non-empty subsets, i.e. points.  A key part of
this correspondence involves the {\em Nullstellensatz}.

Explicitly, if $X$ and $Y$ are two algebraic sets corresponding to
ideals $I_X$ and $I_Y$ (so $I_X = I(X)$ and $I_Y = I(Y)$, and $V(I_X)
= X$, and $V(I_Y)=Y$), then $I(X \cup Y) = I_X \cap I_Y$, and $I(X
\cap Y) = \sqrt{(I_X, I_Y)}$.  That ``root'' is necessary.

Some of these links required the following theorem, which I promised you
I would prove later: 

{\em Theorem.} Each algebraic set in $\A^n(\kbar)$ is cut out by a
finite number of equations.

This leads us to our next topic:

\section{Noetherian rings and the Hilbert basis theorem}

By our equivalence between algebraic sets and radical ideals, this is
equivalent to:

Each radical ideal of $\R$ is finitely-generated.

This will follow from:

{\em Theorem.} {\em Every} ideal of $\R$ is finitely-generated.

In proving this, we will come across some important ideas.

{\em Definition.}  A ring is {\em Noetherian} if every {\em ascending} chain of ideals eventually stabilizes.

In other words, if you  have a chain of ideals
$$
I_1 \subset I_2 \subset I_3 \subset \dots \subset R
$$
then for some $n_0$, for $n \geq n_0$, $I_n = I_{n_0}$.

This is {\em equivalent} to: every ideal $I$ of $R$ is
finitely-generated.  (This definition is often much easier to check.)
That will be an {\em exercise}.  This involves {\em Noetherian
induction}.


{\em Examples.}  
(a)  A field $k$.  There are only two ideals!

(b)  The integers $\zed$.  The ideals are of the form $(0)$, and $(n)$ $n \in \zed^+$. 
It is easy to check that every ascending chain of ideals eventually stabilizes, but easier to check the
second condition, that every ideal is finitely-generated:  they are all generated by one term.
(By this argument, all {\em principal ideal domains} are Noetherian.)

(c) The quotient of a Noetherian ring is also Noetherian, i.e. if $R$
is Noetherian, and $I$ is any ideal, then $R/I$ is Noetherian.

Proof:  ideals in $R/I$ correspond precisely to ideals in $R$ containing $I$.  

(d)  The localization of a Noetherian ring is Noetherian.

(e) {\em The Hilbert Basis Theorem.}  If $R$ is Noetherian, then $R[x]$ is Noetherian.  (Proof:  soon.)
Hence by induction $\kb[x_1, \dots, x_n]$ is Noetherian.  Combining (a)--(d), we see that {\em all
quotients of finitely-generated rings over a field (or the integers) are Noetherian.}

(f) {\em Fact.} Essentially every ring you come across in geometry is
Noetherian.  This isn't so true in number theory / algebraic geometry.

(g) {\em But!:} Not all rings are Noetherian.  For example, consider the ring that I'll call $k[x^{\epsilon}]$,
where elements finite sums of the form $a x^b$ where $a \in k$, and $b$ is a non-zero real number.
It's clear how to add and multiply them, so they are really a ring.  Also for every $\epsilon>0$, 
$(x^\epsilon)$ is an ideal.  Then 
$$
(x^1) \subset (x^{1/2}) \subset (x^{1/3}) \subset \dots \subset R$$
is an ascending chain of ideals that doesn't stabilize.

What we'll do next: (1) The proof is short, so we'll prove the Hilbert
basis theorem.  (2) For fun, we'll see the game of Chomp.  (3)
Finally, I'll describe a geometric picture that will make you believe
it, and that picture will motivate some reasonable definitions.

{\bf (1) Proof of the Hilbert basis theorem.}

Some of your intuition from the geometric picture and from chomp will give you a proof.

{\em The statement:} If $R$ is Noetherian, then the ring of
polynomials in one variable in $R$, $R[x]$, is Noetherian.  In other words,
every ideal of $R[x]$ is finitely-generated.

If $f = a_n x^n + \dots + a_0$ is a polynomial in $R$ (so the $a_i$'s
are in $R$), we say the {\em initial term} of $f$ is $a_n x^n$, and
the {\em initial co-efficient} of $f$ is $a_n$.

{\em Proof.} Let $I$ be an ideal of $R[x]$; we'll show that it is finitely generated.
Choose a sequence $f_1, f_2, \dots \subset I$ as follows:  let $f_1$  be a non-zero element of 
least degree in $I$.  For $i \geq 1$, if $(f_1, \dots, f_i ) \neq I$, then choose
$f_{i+1}$ to be an element of least degree among those in $I$ but not in $(f_1, \dots, f_i)$.
(If $(f_1, \dots, f_i ) = I$, then we've shown $I$ is finitely generated, so stop choosing elements.)

Let $a_j$ be the initial coefficient of $f_j$.  Since $R$ is Noetherian, the ideal $J=(a_1, a_2, \dots)$ of
the $a_i$ produced is finitely generated.  Let $m$ be an integer such that $a_1$, \dots, $a_m$ generate $J$.
We claim that $I=(f_1, \dots, f_m)$.  

Otherwise, consider $f_{m+1}$.  $a_{m+1} \in J$, so we can write $a_{m+1} = \sum_{j=1}^m u_j a_j$ for some $u_j \in R$.
Define
$$
g = \sum_{j=1}^m u_j f_j x^{\deg f_{m+1} - \deg f_j} \in (f_1, \dots, f_m)
$$
and notice that this is of the same degree as $f_{m+1}$, with the same
initial term.  The difference $f_{m+1}-g$ is in $I$ but not in $(f_1,
\dots, f_m)$, and has degree strictly less than that of $f_{m+1}$.  But
$f_{m+1}$ was something of minimal degree with this property, so we
have a contradiction. \qed

{\em Fact.} We won't need this any time soon, but it is worth
mentioning.  If $R$ is a ring, and $M$ is a module, then $M$ is said
to be Noetherian if every ascending chain of submodules of $M$
eventually stabilizes.  Then it isn't hard to prove that if $R$ is a
Noetherian ring, and $M$ is a finitely-generated $R$ module, then $M$
is a Noetherian module.


{\bf (2) The game of Chomp.}

Show that infinite chomp is a finite game, and how this follows from the fact
that $\R$ is Noetherian.

{\bf (3) A geometric picture.}

We now know that every ascending chain of ideals of $\R$ must
eventually stabilize, so every ascending chain of {\em radical} ideals
of $\R$ must eventually stabilize, so by our identification of radical
ideals with algebraic sets, every {\em descending chain} of algebraic
sets in $\A^n$ must eventually stabilize.  Let me do an example to
convince you that this is reasonable, and then you can tell me why you
find it reasonable; these ideas will turn up in the proof of the
Hilbert basis theorem.

(Do that, and see what ideas come up.)

This leads us to our next topic.

\section{Fundamental definitions:  Zariski topology, irreducible, affine variety, dimension, component, etc.}

We've now defined our objects of study.  We now make some more
fundamental definitions; see e.g. Hartshorne I.1.

{\em Definition / way of thinking.} The {\em regular functions} on an
algebraic set $X=V(I)\subset \R$ (where $I$ is {\em radical}) is the
ring $\R/I$.  Call this the {\em ring of regular functions}, or the
{\em affine coordinate ring} of $X$, denoted $A(X)$.  (We'll later see
how to reconstruct the algebraic set from the ring of regular
functions; if you've seen much of this before, then you can think
about this.)

Elements of this ring are called {\em regular functions}.

Example: consider the algebraic set in $\A^2$ defined by the equation
$y^2 = x^3 + x$ (which I should draw as a real cubic, with two
components).    Then the polynomial $x$ is a function on that curve.
But the polynomial $y^2-x^3$ is too, and it's the same function.

{\em Definition.}  Define the {\em Zariski topology} on an
algebraic set $X$ as follows.  Closed sets are defined to be algebraic
subsets of $X$.  

This is an extremely unusual topology.  But it {\em is} a topology.
I'll check the axioms (which, although usually stated for open sets,
I'll state for closed sets).   The empty sets is closed.  All of $X$ is
closed.  The finite union of closed sets is closed:
$$
V(I_1) \cup \dots \cup V(I_n) = V(I_1 \cap \dots \cap I_n).
$$
{\em Any} intersection of closed sets is closed:
$$
\cap_{a \in A} V(I_a) = V( (\sum I_a)).
$$

As an example, consider the affine line $\A^1$, corresponding to the
polynomials in one variable $\kb[x]$.  The closed subsets are
zero-sets of a single polynomial, so a closed subset is either
$\A^1(\kb)$, or a finite union of points.  The open sets are all huge.
In particular, this is not a Hausdorff topology.  So be careful; your
intuition can lead you astray.  The proper way to think of open subsets
is to imagine throwing away a union of algebraic sets.  For example, 
if $X$ is a plane union a line...

{\em Definition.} A non-empty topological space $Y$ is {\em
irreducible} if it cannot be expressed as the union of $Y_1 \cup Y_2$
of two proper subsets, each closed.

E.g. $\A^1$, point, not plane union line.

{\em Exercise (on next Tues. problem set).} Any non-empty open subset of an
irreducible space is irreducible and dense.

{\bf Warning:  At this point in the class, I gave a definition of algebraic
variety.  That definition was only tentative; the real definition will come
up in a class or 2.}

{\bf Important exercise.} Irreducible algebraic subsets of $\A^n(\kb)$
correspond (under the bijection radical ideals $\Leftrightarrow$
algebraic sets) to prime ideals.  

(You can just look this up in any introductory book on algebraic
geometry, but you'll really want to think this through.)

{\em Example:} line union a plane.  Recall that this has ideal
$I=(xz,yz)$.  Then $R/I$ isn't prime, which you can see in two ways.
Here's the geometric way, it ``clearly'' isn't irreducible.  (We'll
find a function $f_1$ that vanishes on one component but not the
other.  And we'll find $f_2$, and we'll see that $f_1 f_2 = 0$ in the
ring of regular function.)

{\em Definition.} A topological space is {\em noetherian} if it
satisfies the {\em descending chain condition} for closed subsets: for
any sequence $Y_1 \supset Y_2 \supset \dots$ of closed subsets, there
is an integer $r$ such that $Y_r = Y_{r+1} = \dots$.  Example:  as
$\R$ is noetherian, $\A^n$ is a Noetherian space.

{\em Important exercise.}  If $Y$ is a Noetherian topological space,
it can be expressed as a finite union $Y = Y_1 \cup 
\dots \cup Y_r$ of irreducible closed subsets $Y_i$.
If we require that $Y_i$ is not contained in any $Y_j$ for any $i \neq j$,
then the $Y_i$ are uniquely determined.   (Proof:  ``Noetherian induction.'')

These are the {\em irreducible components} of $Y$. 

So every algebraic set in $\A^n$ can be uniquely expressed as a finite union
of irreducible algebraic sets (without redundancies).

{\em Definition.} The {\em dimension} of a topological space is the
supremum of all integers $n$ such that there is a chain
$\emptyset=Z_0 \subset Z_1 \subset \dots \subset Z_n$ of
distinct irreducible closed subsets of $X$.  

E.g. $\dim \A^1(\kb)  = 1$, but it already isn't obvious why
$\dim \A^2(\kb)=2$.

{\em Definition.} If $\fp$ is a prime ideal of a ring $R$, then its
{\em height} is the supremum of all integers $n$ such that there
exists a chain $\fp_0 \subset \fp_1 \subset \dots \subset \fp_n = \fp$
of distinct prime ideals.  The {\em Krull dimension} of $R$ is the
supremum of the heights of all prime ideals.

{\em Check:} If $Y$ is an algebraic set in $\A^n$, then $\dim Y$ is
the Krull dimension of its affine coordinate ring $A(Y)$.

{\bf Commutative Algebra Fact.}  If $k$ is a field, and $B$ is
an integral domain which is a finitely-generated $k$-algebra.
Then $\dim B = tr.deg_k K(B)$ (where $K(B)$ is the quotient field of
$B$), and for any prime ideal $\fp$ in $B$,
$$
ht \fp + \dim B/\fp = \dim B.$$  

Hence (i) $\dim \A^n \ n$, as the transcendence degree of $\kb(x_1,
\dots, x_n)$ over $\kbar$ is $n$, and (ii) you should think of the
height as codimension.

{\em Example.} The {\em twisted cubic} in $\A^3$.  On the problem set,
you show that the subset $Y$ of points of the form $(t,t^2,t^3)$ of
$\A^3$ is algebraic.  It turns out to be irreducible, not
surprisingly.  So this corresponds to a prime ideal $\fp$.  What is
its height in $\kb[x_1,x_2,x_3]$?

{\em Exercise.} (a)  Show that $A(Y)$ is isomorphic to $\kb[x]$, i.e. the
ring of polynomials in one variable.  Think about what this might mean.

(b) Let $Z$ be the plane curve $xy=1$.  Show that $A(Z)$ is {\em not}
isomorphic to $\kb[t]$.

{\em Exercise.} Let $Y$ be the algebraic set in $\A^3$ defined by the
two polynomials $x^2 - yz$ and $xz-x$.  Show that $Y$ is the union of
three irreducible components.  Describe them, and find their prime
ideals.  (Not on problem set.)

{\em Exercise.} Show that a $\kb$-algebra $B$ is isomorphic to the
affine coordinate ring of some algebraic set in $\A^n(\kb)$ for some
$n$ if and only if $B$ is a finitely generated $k$-algebra with no
nilpotent elements.  (We'll basically prove this next week; not on
problem set.)

Next to do: we've described the objects we're interested in, algebraic
sets.  We'll define morphism.  This is even more important than it sounds. 
For example, we want to say that the line in $\A^2$ and $\A^1$ are really
the same thing, and we can't say that unless we know what ``same'' means.

Then we'll recover affine algebraic sets from their coordinate rings,
and we'll define general varieties by gluing together irreducible affine algebraic sets.

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