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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 7}
\author{Ravi Vakil}
\date{Thursday, September 30, 1999.}
\maketitle
\tableofcontents

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

{\em Problem set 3 is now due one week from today; future problem sets
will be due on Thursdays, and returned the following Tuesday.  
I'll give out copies of parts of the red book.}

{\bf Where we are.}

Last day, we played around with the structure sheaf of the plane
$\A^2$, and observed that the sections of the structure sheaf over
some random open set $U$ was a subring of $\kb(x,y)$. 

We now formally know what a prevariety is:  it is a topological space
with a structure sheaf, that ``locally'' looks like an irreducible algebraic set.
We'll next make this concrete by working out a fair number of examples.

Some of you are wondering about that prefix ``pre''; we'll begin to answer that today.

I was initially going to tell you about the {\em function field} of a
prevariety beforehand, that generalizes the role of $\kb(x,y)$ in the
plane, but instead, I'll say a few words now, and talk about it at
length after the examples.

{\bf Function fields, quickly.} 

To find the function field of a prevariety, take any affine open set, find
its ring of regular functions $R$, and take its field of fractions $K(R)$.
We'll see later that this is well-defined.

Quick comment about sheaves:  gluability and identity are not just for finite covers.

Also, at some point during the lecture, we discussed why two non-empty
open subsets of a prevariety must intersect.  Here's the summary:
(i) This is true for any irreducible topological space.
(ii) Affine varieties are irreducible (by definition).
(iii) If you glue two irreducibles along non-empty (homeomorphic) open sets,
you get an irreducible topological space.
(iv)  A prevariety is a connected union of a finite number of irreducible spaces,
so you can use (iii) inductively to show that a prevariety is irreducible.

\section{Examples of prevarieties}

Let's start with a prevariety that isn't affine.

Last time we showed that an open subset of a prevariety (with the
induced topology and structure sheaf) is also a prevariety.  This is
called an {\em open subprevariety}.

Consider $\A^2 \setminus (0,0)$. 

{\em Exercise.} Prove that $\A^2 \setminus (0,0)$ is not an affine
variety.  (Hint: If it were, then what would its ring of functions
be?)

Hence not all open subsets of affine varieties are affine varieties.


The function field of this prevariety is $\kb(x,y)$.

{\em Example: $\proj^1$.} Take two copies of $\A^1$; call them $X$ and
$Y$, and say their coordinates are $a$ and $b$.  Glue them together
along the open set $X \setminus 0$ and $Y \setminus \{ 0 \}$, where the two
open sets are identified via $a=1/b$.  

You can see these patches in terms of coordinates on $\proj^1$.

As mentioned earlier, if $\kb$ is the complex numbers, you can picture this quite explicitly.
You're gluing two complex planes  together, and you end up with a sphere.

The function field of $k(X)$ is $\kb(a) \cong \kb(b)$, and the
isomorphism is via $a \mapsto 1/b$. (Discuss a bit.)  To see what
happens if we chose the isomorphism $a \mapsto b$, see the line with
the doubled origin below.

{\em Example:}  Global sections of $\oh_{\proj^1}$.    

Notice that $\proj^1$ isn't affine:  the affine algebraic set associated
to the global sections of $\oh_{\proj^1}$ is a point.

{\em A pathological example: the line with the doubled origin.} Now
let's glue our two copies $X$ and $Y$ of $\A^1$ together, along the
same open sets, except this time identify the open sets via $x=y$.
(Explain.)  This is something nasty that we'll want to get rid of when
defining varieties.  (For the same reason, a {\em real manifold} isn't
just defined as a topological space that is locally isomorphic to a
neighborhood of the origin in $\real^n$; an additional condition (the
{\em Hausdorff} condition) is necessary.)

Again, the function field of $k(X)$ is $\kb(x)$.

{\em Example: $\proj^n$.} Recall that projective space (over the field
$\kb$) is usually defined as the set of $(n+1)$-tuples $(x_0; \dots ;
x_n)$ where the $x_i$ are not all zero, under the equivalence $(x_0;
\dots ; x_n) \sim a (y_0; \dots ; y_n)$ where $a \in \kb^*$.  We'll
define $\proj^n$ as an algebraic variety.  We have $n+1$ open sets,
each isomorphic to $\A^n$, and we glue them together just as we did
$\proj^1$.  For convenience, I'll just describe the first two, and
show how they glue together.  The affine open $U_0$ has coordinates $(a_1, \dots, a_n)$,
and the affine open $U_1$ has coordinates $(b_0, b_2, \dots, b_n)$; the open
subset of $U_0$ corresponding to $a_1 \neq 0$ is glued to the open subset of $U_1$
corresponding to $b_0 \neq 0$, and the identification is via:
$$
(1; a_1; a_2; \dots; a_n) \sim (b_0; 1; b_2; \dots ; b_n),
$$
i.e. 
$a_2/a_1 = b_2/b_0$, $a_3/a_1 = b_3/b_0$, \dots, $a_n/a_1 = b_n/b_0$.

In an analogous manner, this tells you how to glue together any $U_i$
and $U_j$.  We still have something further to check: that these
gluings agree.  You can think through yourself that it suffices to
prove that the gluing data of $U_i$ and $U_j$, $U_j$ and $U_k$, and
$U_k$ and $U_i$ all ``agree''.

This also tells you how to think of points in $\proj^n$ with projective coordinates.
For example, consider $\proj^3$, which is covered by $U_0$, \dots, $U_3$. 
Then (4;3;0;1) is a point which corresponds to $(3/4,0,1/4)$ on $U_0$, $(4/3, 0, 1/3)$ on 
$U_1$, and $(4,3,0)$ on $U_3$; it doesn't lie in the open $U_2$.

The $U_i$ are often called the standard affine cover of $\proj^n$.  We
will talk about projective varieties at great length soon.

The function field is $\kb(a_1, \dots, a_n)$.

{\em Exercise.}  Prove that the global sections of $\oh_{\proj^n}$ are constant.

{\bf Remark.} You don't have to know much about the structure sheaf to
glue together prevarieties, thanks to the fact that we understand
affine prevarieties well, and distinguished opens of affines are
affines too.

\section{Function fields of prevarieties}

{\bf Definition.}  Given a prevariety $X$, define the {\em function field} $k(X)$
as follows.  (Caution: that $k$ isn't the field $k$.)  Elements of $k(X)$ are
called {\em rational functions} on $X$.

Intuition: Meromorphic functions on $\com$ (i.e. functions defined
almost everywhere).  There's no point where all such functions are
defined, yet they definitely form a ring: you can add two and get
another such, or multiply, etc.

Elements of $k(X)$ is the data $(U,f)$ where $U$ is an open subset of
$X$ and $f \in \oh_X(U)$ is a section of the structure sheaf over $U$,
modulo an equivalence relation: $(U,f) \sim (V,g)$ if there is an open
$W$ in $U$ and $V$ such that $f$ and $g$ agree on $W$, or precisely:
$$
\res_{U,W} f = \res_{V,W} g.
$$

This might remind you of the definition of the stalk of a sheaf.  This
is no coincidence; in the land of schemes, this actually {\em is} a stalk,
albeit at a weird kind of point (called the {\em generic point} of $X$).

It is a complicated way of saying something that turns out to be quite simple:

{\bf Claim.} 
Let $U$ be an affine open set of $X$.  Then $k(X) = K(A(U))$; the function field
is just the ring of fractions of $A(U)$.

Hence to find $k(X)$, you just have to find {\em any} affine open, take the ring
of regular functions on that affine open, and take its quotient field.

{\em Example.}  $k(\A^n) = \kb(x_1, \dots, x_n)$.  In fact, if
$X$ is any  open subset of $\A^n$, then $k(X) = k(\A^n)$.

{\em Proof.} First, we show the result when $X$ is affine; let its
ring of regular functions be $R$.  Then an element of $k(X)$ is of the
form $(U,f \in \oh_X(U))$.  Using the equivalence relation, make $U$
smaller so that it is a distinguished open (possible as distinguished
opens form a base for the topology); this element is of the form
$(D(g), f \in \oh_X(D(g)) = R_{(g)})$.  Thus $f$ can be written as
$a/b$ where $a,b \in R$ (in fact $b$ is a power of $g$).

Conversely, given an element of $K(R)$, write it as $a/b$.  Then this
gives a section of $\oh_X$ over $D(b)$: $a/b \in R_{(b)} =
\oh_X(D(b))$.  Then you just have to check that these two constructions
commute.
\epf

{\bf Consequences (proofs omitted).}
\begin{enumerate}
\item  For any open $U\subset X$, $\oh_X(U)$ is a subring
of $k(X)$. 
\item  The restriction maps are inclusions, i.e. $U \subset V$
implies $\oh_X(V) \subset \oh_X(U) \subset k(X)$.  
\item $\oh_X(U \cup V) = \oh_X(U) \cap \oh_X(V)$ inside $k(X)$.  (The right side wouldn't even
make sense unless $\oh_X(U)$ and $\oh_X(V)$ were subrings of the same thing.)
\item $\oh_X(U \cap V)$ is the subring of $k(X)$ generated by $\oh_X(U)$ and $\oh_X(V)$.  {\em (False!  See problem on Problem Set 4.  If $X$ is separated,
I suspect this is true.)}
\end{enumerate}

\section{Morphisms of prevarieties:  first thoughts}

First, some philosophy.  We know what morphisms from one affine open
to another looks like: it's the same as the maps of rings of regular
functions (as rings over $\kb$) in the opposite direction.

So here is how we might like to think of morphisms of one prevariety
to another.  (Explain: to describe a morphism $X$ to $Y$, we might
write $X$ as a union of affine opens $U_i$, and $Y$ as a union of
affine opens $V_i$, where $U_i$ maps to $V_i$, and everything glues
nicely.) This is actually correct, but what gets confusing is showing
that this description is independent of the choice of cover; plus it is kind of annoying
to check, so it would be good to have a (slightly) more usable definition.

So we'll define morphisms in another way first, that doesn't involving
covers, then show that this is the same as the ``naive'' description I've just given.
Then in real life, we'll use (a variant of) the ``naive'' description.

We begin by recasting morphisms of affines in a different light.

{\bf Theorem.} Let $X \subset \A^m$, $Y \subset \A^n$ be
irreducible algebraic sets, and let $\pi: X \rightarrow Y$ be a
continuous map.  Then the following are equivalent.  (Draw pictures!)

i)  $\pi$ is a morphism of algebraic sets

ii) for all $g \in \oh_Y(Y) = A(Y)$, $g \circ \pi \in \oh_X(X) = A(X)$.

We've already proved this Theorem.

But note that we've described morphisms in a new way: they are (i) continuous
maps, such that (ii) pullback takes regular functions to regular functions.

Coming in the next few lectures (in some order):  
\begin{enumerate}
\item Morphisms of prevarieties.  
\item Examples of morphisms, including open immersions, closed immersions.
\item The isomorphism of $x^2 + y^2 = z^2$ in $\proj^2$ with $\proj^1$.
\item Projective varieties. Rational maps (and maps of function fields?).  
Degree of rational maps.
Birationality.
\end{enumerate}
} % end of parskip; it started just before the introduction
\end{document}
