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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 11}
\author{Ravi Vakil}
\date{Thursday, October 14, 1999.}
\maketitle
\tableofcontents

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

{\em Problem sets can be handed in and picked up at the end of class.}

{\em Discuss late policy.}

{\em Aside.}  There were several points last day that I wish I'd described 
differently.  Here is one.  (Again, please tune out.)  

Given a ring $R$, and a prime ideal $\fp$, I described the residue field as:
(i) localize the ring $R$ at $\fp$, and then mod out by the maximal ideal:
$R_{\fp} / \fp R_{\fp}$.  You can quickly check that this is equivalent to:
(ii) Mod out by $\fp$ to get a domain, and then take the fraction field.

The second point of view is often much more convenient.  For example,
an element $f$ of $R$, which you should think of as being a function,
vanishes at a point $\fp$ if $f \in \fp$.  (Describe this in the case
of $\fp = (y-x^2)$ in $\kb[x,y]$.)  In particular, we get the fact
that $V(f)$ is the set of prime ideals containing $f$, and more
generally $V(I)$ is the set of prime ideals containing $I$.

Back to our regularly-scheduled program.

\section{Products}

We know what we mean when we discuss products of sets.  We'll now
define products more generally.

{\em Categorical products.} Suppose you have two sets $X$ and $Y$.
Then the product set $Z$ has two natural ``projection'' maps $p_X$ and $p_Y$
to $X$ and $Y$ respectively.  Moreover, if you have any other set $W$ 
with maps to both $X$ and $Y$, then there is a unique map $W \rightarrow Z$
such that the  maps to $X$ and $Y$ can be obtained by composition with the projections.

{\bf Definition.} Suppose $X$ and $Y$ are objects in a category.  Then
a {\em product} of $X$ and $Y$ is the data of another object $Z$,
along with projection maps $p_X: Z \rightarrow X$ and $p_Y: Z
\rightarrow Y$, such that for any other object $W$, maps from $W$ to
$Z$ correspond to the data of maps from $W$ to $X$, and maps from $W$
to $Y$.  Equivalently, 
$$
Mor(Z,W) \rightarrow Mor(Z,X) \times Mor(Z,Y)
$$
is a bijection.

You should check that this works with sets, i.e. in the category of
sets.

{\em Remark.}  Then the product is defined up to unique isomorphism. 

{\em Products in the category of topological spaces.} Define product
topology.  Explain why this is called the product topology: this is a
product in the category of topological spaces.

{\em Products in the cateogry of prevarieties?}

If there is a product of $X$ and $Y$ in the category of
prevarieties, then the points of the product are the products of the points.
Reason:  take $Z$ to be a point.

To save you the suspense:  products indeed exist, and we'll construct them.

Topologies will be weird.  It will turn out that $\A^1 \times \A^1 =
\A^2$, but the Zariski topology on $\A^2$ isn't the same as the
product of the Zariski topologies on the factor ({\em Exercise}).

\subsection{Products in the category of affine varieties, and in the category of varieties}

Recall that the category of affine varieties (over $\kb$) is just the
category of ``nice'' rings, with the arrows reversed, where nice means
``finitely generated algebra over $\kb$ that is an integral domain''.

Let $X$ and $Y$ be affine varieties (over $\kb$).   Let's diagram-chase.

If we have some ``nice'' ring $A(W)$, the question is:  given maps (ring morphisms over $\kb$) $A(X) \rightarrow A(W)$ and $A(Y) \rightarrow A(W)$, is there
some ring $A(X \times Y)$ along with morphisms $A(X), A(Y) \rightarrow A(X \times Y)$ through which this has to factor?  

Answer: Yes, $A(X ) \otimes_{\kb} A(Y)$, with the morphism $A(X)
\rightarrow A(X) \otimes_{\kb} A(Y)$ given by $x \mapsto x \otimes 1$,
and similarly for $A(Y)$.  (Remind them what tensor product is; show that tensor product has this property.)

All that's left to show is that the tensor product is also ``nice''.

Finitely-generated is easiest: if $f_1$, \dots, $f_m$ are generators
for $A(X)$ and $g_1$, \dots, $g_n$ are generators for $A(Y)$, then
$f_i \otimes 1$ and $1 \otimes g_j$ generate the tensor product.

Then we invoke a fact from commutative algebra: Let $R$ and $S$ be
integral domains over $\kb$.  Then $R \otimes_{\kb} S$ is also an
integral domain.

{\em Remarks.}  Hence
$\otimes$ is the {\em coproduct} in the category of rings.  And we've also shown
that the product in the category of affine schemes is given by tensor product.


{\bf Theorem.}  Let $X$ and $Y$ be affine varieties.  Then there is product prevariety $Z := X \times Y$; it is affine with
coordinate ring $A(X) \otimes_{\kb} A(Y)$.

{\em Proof.} We've shown that this is the product in the category of
affine varieties; what's different is that $W$ may now be {\em any}
prevariety.  Suppose we have morphism $W \rightarrow X$ and $W
\rightarrow Y$.  How do we get the morphism $W \rightarrow Z$?  

(This is where patching arguments make life really easy.)  Cover $W$ with 
affines $U_i$.  We have maps $U_i \rightarrow X, Y$ so (as $Z$ is a product in 
the category of affine varieties) we get map $U_i \rightarrow Z$.  

We now need to show that these maps ``glue together'' to give us a map
from $W$ to $Z$, i.e. that if you consider the overlap $U_{ij} := U_i
\cap U_j$, then the induced morphism $U_{ij} \rightarrow U_i \rightarrow Z$ is the same as the induced morphism $U_{ij} \rightarrow U_j \rightarrow Z$.

Cover $U_{ij}$ with affines $V_k$; then again because $Z$ is a product
in the category of affine varieties, and we have a morphism $V_k
\rightarrow X, Y$, there is only one morphism $V_k \rightarrow Z$ compatible
with them.  
\epf

{\em Examples.} It isn't hard working out what products of affine
varieties actually are.  For example, $\A^1 \times \A^1 \cong \A^2$.
If $X$ is the affine variety in $\A^2$ cut out by $v^3 + w^3=1$, and
$Y$ is the affine variety in $\A^3$ cut out by $xyz=3$, then $X \times
Y$ is the affine variety in $\A^5$ cut out by $v^3+w^3=1$ and $xyz=3$.
Keep these examples in mind.

{\bf Looking at the product of affines more closely.}

{\em i)  Topology.}
Note that the distinguished opens on $X \times Y$ are of the form
$D(\sum f_i \otimes g_i)$ where $f_i \in A(X)$, $g_i \in A(Y)$; this
gives the beginning of an insight as to why the topology on the product is not
the product of the topologies.

{\em ii) Function field.}
Note also that the function field of the product $k(X \times Y)$ is the
{\em quotient field} of the tensor product of the function fields:
$k(X ) \otimes_{\kb} k(Y)$.

{\em iii) Stalks of the structure sheaf.} Let's interpret the local
ring $\oh_{X\times Y, (x,y)}$ in terms of the local rings $\oh_{X,x}$
and $\oh_{Y,y}$.  Let the maximal ideals of these local rings be
$\fm_x$ and $\fm_y$ respectively.  (This is concrete!  The local rings
are quotients of polynomials where the denominator {\em doesn't}
vanish at the point, and the maximal ideal is where the numerator {\em
does} vanish at the point.)

{\bf Lemma.} Then $\oh_{X \times Y, (x,y)}$ is the localization of $\oh_{X,x}
\otimes_{\kb} \oh_{Y,y}$ at the maximal ideal $\fm_x \oh_{Y,y} + \fm_y
\oh_{X,x}$.  

We'll need this technical fact (which isn't too hard) only once,
in a few minutes, and then you can forget about it.

If you try to parse what this means, you'll realize that it's reasonable.

{\em Proof.} Here's a real explanation.  $\oh_{X \times Y, (x,y)}$ is the localization
of $A(X) \otimes_{\kb} A(Y)$ at the maximal ideal of all functions vanishing
at $(x,y)$.  Now $A(X) \times A(Y) \subset \oh_{X,x} \otimes_{\kb} \oh_{Y,y}
\subset \oh_{X \times Y, x \times y}$, so we can describe the last term by 
localizing at the maximal ideal of all functions vanishing at $(x,y)$, so we need
to check that this really is $\fm_x \oh_{Y,y} + \fm_y \oh_{X,x}$.  One inclusion is clear.  Conversely, if
$$
h = \sum f_i \otimes_{\kb} g_i \in \oh_{X,x} \otimes_{\kb} \oh_{Y,y}$$
vanishes at $(x,y)$, with $f_i(x) = a_i$ and $g_i(y) = b_i$, then
$$h = h - \sum a_i b_i = \sum (f_i -a_i) \otimes g_i + \sum a_i \otimes (g_i - b_i)$$
which lies in $\fm_x \otimes \oh_{Y,y} + \oh_{X,x} \otimes \fm_y$ as desired.
\qed


{\bf Theorem.} Let $X$ and $Y$ be prevarieties over $\kb$.  Then they
have a product.

{\em Proof.} Let's first  build a reasonable candidate from affines, and then
later check that it really is a product.  We'll start with the set, add
the topology, and finally the structure sheaf.

For the underlying set, we just take the product set.

For the topology, we give a base: For all open affines $U \subset X$,
$V \subset Y$, and all finite sets of elements $f_i \in A(X)$, $g_i
\in A(Y)$, consider $D(\sum f_i \otimes g_i) \subset U \times V$.
(Small check required to make sure that this really gives the
topology you want on $U \times V$.)

Now for the structure sheaf.  Let $K$ be the quotient field of $k(X)
\otimes_{\kb} k(Y)$, which is our candidate for the function field of
the product.  For $x \in X$, $y \in Y$, let $\oh_{X \times Y (x,y)}
\subset K$ be the localization of $\oh_{X,x} \otimes_{\kb} \oh_{Y,y}$
at the ideal $\fm_x \otimes \oh_{Y,y} + \oh_{X,x} \otimes \fm_y$, and
define 

$$\oh_{X \times Y}(U) = \bigcap_{(x ,y) \in U} \oh_{X \times Y, (x ,y)}.
$$

This is a sheaf of functions, which coincides on each $U \times V$ ($U$, $V$ affine) with the structure sheaf of $\oh_{U \times V}$ (by our analysis
of the affine case).  

Then this is a prevariety!  (Check: covered by finitely many affines,
and connected.)

Next to check: that this prevariety is a product.  We have our
projection maps $X \times Y$ to $X$ and $Y$.  Suppose we're given some
morphisms $f_X: W \rightarrow X$, $f_Y: W \rightarrow Y$.  Then we
automatically get a map of sets $W \rightarrow X \times Y$, as the
underlying set o f $X \times Y$ is just the product of the underlying
sets of $X$ and $Y$.  We just need to check that this is a morphism.
To do that, we can cover $X$ and $Y$ by affines $U_i$ and $V_j$
respectively, and cover $W$ with $f_X^{-1} U_i \cap f_Y^{-1} V_j$; we
need only show that $f_X^{-1} U_i \cap f_Y^{-1} V_j \rightarrow U_i
\times V_j \subset U \times V$ is a morphism; but it is because
we've already shown that $U_i \times V_j$ is a  product
in the category of prevarieties when $U_i$ and $V_j$ are affines.
\epf

It isn't hard to check:

{\bf Corollary.} If $U$ is an open subprevariety of $X$, then $U
\times Y$ is an open subprevariety of $X \times Y$. If $Z$ is a closed subprevariety of $X$, then
$Z \times Y$ is a closed subprevariety of $X \times Y$.

{\em Remark.}  If $\kb = \com$, then you might reasonably have the classical
topology in mind.  It is true that if $X$ and $Y$ are complex varieties, then
the classical topology on $X \times Y$ is the same as the product of the classical topologies.

\section{Coming soon}
\begin{enumerate}
\item products of projective prevarieties are projective prevarieties;
the Segre map
\item rational maps; open sets of definition; birational maps
\end{enumerate}

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