\documentclass[10pt]{amsart}
\usepackage{epic,eepic,latexsym}

% Picture stuff
 
 \newlength{\baseunit}               % the basic unit length
 \newlength{\templengthhoriz}        % width of the picture
 \newlength{\templengthvert}         % depth of the picture
 \newlength{\temprule}               % with between left margin and picture
 \newcount{\numlines}                % depth of picture (in number of lines)
 \setlength{\baseunit}{0.05ex}
  
  \newcommand{\getfig}[2] {
	   \setlength{\unitlength}{#2\baseunit}
	    \input #1.tex }

			 % A macro to input the picture.  Use
			 % \getfig{name}{scale}

% Begin template
%\documentclass[12pt]{amsart}
%\usepackage{epic,eepic}
%\parskip=12pt

% The next bits are ``borrowed'' from a paper of Dan Abramovich and Kalle Karu.
% sizes of things
%\setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm}
%\setlength{\textwidth}{17cm} \setlength{\topmargin}{-1.3cm}
%\setlength{\textheight}{24cm} \setlength{\headheight}{.1in}
%\setlength{\headsep}{.3in} %\setlength{\baselineskip}{3pt}
%\setlength{\parskip}{.5mm}

\newcommand{\point}{\vspace{3mm}\par\refstepcounter{subsection}{\bf \thesubsection.} }
\newcommand{\tpoint}[1]{\vspace{3mm}\par\refstepcounter{subsection}{\bf \thesubsection.} 
  {\em #1. ---} }
\newcommand{\epoint}[1]{\vspace{3mm}\par\refstepcounter{subsection}{\bf \thesubsection.} 
  {\em #1.} }
\newcommand{\bpoint}[1]{\vspace{3mm}\par\refstepcounter{subsection}{\bf \thesubsection.} 
  {\bf #1.} }
% End template 

%\newlength{\baseunit}               % the basic unit length
%\setlength{\baseunit}{0.05ex}

\newcommand{\bpf}{\noindent {\em Proof.  }}
\newcommand{\epf}{\qed \vspace{+10pt}}

% I haven't really used this.
\newcommand{\theorem}{\noindent {\bf \addtocounter{subsection}{1} \arabic{section}.\arabic{subsection}  Theorem.} }

% I tend not to use these, and instead use tpoint above.
\newtheorem{tm}{Theorem}[subsection]
\newtheorem{pr}[tm]{Proposition}
\newtheorem{lm}[tm]{Lemma}
\newtheorem{co}[tm]{Corollary}
\newtheorem{df}[tm]{Definition}

\newcommand{\kbar}{\overline{k}}
\newcommand{\kb}{\overline{k}}
\newcommand{\zed}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\A}{\mathbb{A}}
\newcommand{\real}{\mathbb{R}}
\newcommand{\com}{\mathbb{C}}
\newcommand{\eff}{\mathbb{F}}
\newcommand{\ka}{\kappa}
\newcommand{\proj}{\mathbb P}
\newcommand{\oh}{{\mathcal{O}}}
\newcommand{\cL}{{\mathcal{L}}}
\newcommand{\fm}{{\mathfrak{m}}}
\newcommand{\fp}{{\mathfrak{p}}}
\newcommand{\al}{\alpha}
\newcommand{\om}{\omega}
\newcommand{\be}{\beta}
\newcommand{\ga}{\gamma}
\newcommand{\De}{\Delta}
\newcommand{\si}{\sigma}
\newcommand{\la}{\lambda}
\newcommand{\Pic}{\operatorname{Pic}}
\newcommand{\ch}{\operatorname{char}}
\newcommand{\res}{\operatorname{res}}
\newcommand{\Spec}{\operatorname{Spec}}
\newcommand{\Bl}{\operatorname{Bl}}
\newcommand{\Sym}{\operatorname{Sym}}
\newcommand{\Aut}{\operatorname{Aut}}
\newcommand{\grass}{\mathbb{G}}
\newcommand{\x}{\underline{t}}
\newcommand{\R}{\kbar[x_1, \dots, x_n]}
\newcommand{\Vbar}{\overline{V}}

\newcommand{\cited}{}
\newcommand{\photocopy}{}
\newcommand{\remind}[1]{{\bf(#1)}}
% \newcommand{\remind}[1]{{}}

\def\atp#1#2{\stackrel{ \scriptstyle{#1}}{\scriptstyle{#2}}}


\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 12}
\author{Ravi Vakil}
\date{Tuesday, October 19, 1999.}
\maketitle
\tableofcontents

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

{\em Problem sets back at end.}

\section{Products of projective varieties; the Segre map}

{\bf Theorem.}  The product of two projective varieties is a projective variety.

{\em Proof.}  Since a closed subvariety of a projective variety is a projective variety, it is enough to show that $\proj^m
 \times \proj^n$ is a projective variety.  So we're done, modulo the following lemma.

{\bf Lemma.} $\proj^m \times \proj^n$ is a projective variety.

Describe the image of $(x_0; \dots ; x_m) \times (y_0; \dots; y_n)$ in $\proj^{ (m+1)(n+1)-1}$:  $z_{ij} = x_i y_j$.
This is called the {\em Segre embedding}.

The image is in the locus of rank 1 matrices; in fact it is {\em precisely} the rank 1 matrices.

Make clear that you can recover the points of $\proj^m$ and $\proj^n$.  

You have defining equations:  all the $2 \times 2$ minors.  Hence the image $V$ is a projective prevariety.

We've described the map on points one way.  Describe in another way.

Then you have to check on patches.  Not hard, but I'll omit it; you
can read about it in any of the references.  The details of the proof
are important, but more important is the geometric insight into what's
going on.  So let's go through an example.  \qed

{\em Remark (another parallel between affine and projective prevarieties).}
Recall that if $X$ and $Y$ are affine, with coordinate rings $A(X)$ and $A(Y)$,
then the coordinate ring of $X \times Y$ is $A(X) \otimes_{\kb} A(Y)$.  Something 
similar happens with projective prevarieties too.  Suppose $X$ and $Y$ are
projective, and lie in $\proj^m$ and $\proj^n$ respectively, so $X$
has {\em graded} coordinate ring $R(X) = \kb[x_0, \dots, x_m]/I(X)$, where
$I(X)$ is a {\em homogeneous} ideal, and $R(Y) = \kb[y_0, \dots, y_n]/I(Y)$ similarly.
Then under the Segre embedding (so $X \times Y \subset \proj^{(m+1)(n+1)-1}$),
$R(X \times Y) \cong R(X) \otimes R(Y)$
where the grading behaves well under tensor product (i.e. the tensor product of the
$i$th graded piece of $R(X)$
and the $j$th graded piece of $R(Y)$ lies in the $(i+j)$th graded piece of $R(X \times Y)$.

{\em Exercise.} Prove that this is the case if $X=\proj^m$ and
$Y=\proj^n$.  Caution: I've shown that the product is cut out by the
equations of the $2 \times 2$ minors, but I didn't show that the ideal
$I(X \times Y)$ is generated by the $2 \times 2$ minors, although
that's true.  (Analogy: In $\A^2$, the $y$-axis $Y$ is cut out by the
equation $x^2=0$, but $I(Y)$ isn't generated by $x^2$; it's generated
by $x$.)

\subsection{$\proj^1 \times \proj^1$ and the smooth quadric surface}

When you do the numerology with $m=n=1$, we see that $\proj^1 \times \proj^1$ maps into $\proj^3$, and it 
is given by a single equation $wx-yz=0$.  

Draw a picture.

{\em Remark.} Almost all quadric surfaces look the same.  Hence we
know the (classical) topology of almost all quadric surfaces: they are
products of 2-spheres.

\subsection{Rulings of the smooth quadric surface}

Show them the lines in the real picture.  We'll see these
algebraically.  First of all, a line in $\proj^3$ is the intersection
of two distinct hyperplanes, e.g. $w=x=0$.  They are isomorphic to $\proj^1$, e.g
for $w=x=0$, the isomorphism is given by $(0;0;y;z) \leftrightarrow (y;z)$.  (For
hyperplanes with uglier coefficients, just change coordinates!)

Give the projection to the first $\proj^1$, and to the second $\proj^1$:
$(w_0;x_0;y_0;z_0) \mapsto (w_0;y_0)$ or $(z_0;x_0)$   is the map to the first $\proj^1$, and
$(w_0;x_0;y_0;z_0) \mapsto (w_0;z_0)$ or $(y_0;x_0)$   is the  map to the second $\proj^1$.

The two one-parameter family of lines:  first note that the Segre map is 
$$(a;b) \times (c;d) \mapsto (ac;bd;ad;bc).$$  The first family of lines is:
fix $(a;b)= (a_0;b_0)$, and consider:
$(a_0 c;b_0 d;a_0 d;b_0 c)$; this is the line that is the intersection of $b_0 w = a_0 x$ and
$b_0 y = a_0 z$.  The second is similar; just switch the roles of $(a;b)$ and $(c;d)$.

\section{Defining varieties}

{\bf Definition.} A prevariety $X$ is a {\em variety} if for all
prevarieties $Y$ and for all morphisms $f$ and $g$ from $Y$ to $X$,
the locus where they agree $\{ y \in Y | f(y) = g(y) \}$ is a closed
subset of $Y$.  

This is often called the {\em separatedness} condition.
Mumford calls this the {\em Hausdorff} axiom, because it is the analogue
of the Hausdorff condition in the definition of a manifold.

Note that the line with the doubled origin is not a variety.

{\bf Remark.} An open (resp. closed) subprevariety of a variety is a
variety.

{\bf Remark.} An affine variety is a variety.  (Sorry for nasty
notation!)  Reason: Suppose $X$ is affine, and $Y$ is any prevariety,
and $f$, $g$ are two morphisms $Y \rightarrow X$.  Then the subset of
$Y$ where $f(y)=g(y)$,  $\{ y \in Y | f(y) = g(y) \}$ is as follows.

$x_1 = x_2$ in affine $X$ iff for all regular functions $s \in A(X)$,
$s(x_1) = s(x_2)$.  Hence $\{ y \in Y | f(y) = g(y) \}$ is the locus
where all regular functions $s(f(y)) - s(g(y))$ vanish (where $s$ runs
through all of $A(X)$).  This is a closed set.

{\bf Remark.}  We will soon see that projective prevarieties are varieties.

Special case: if $Y = X \times X$, and $f$ and $g$ are
the projections.  $\Delta(X)$ is the locus where $f$ and $g$ agree.  Hence if $X$ 
is separated, then  $\Delta(X)$ is closed in $X \times X$.

This special case is all you need to check:

{\bf Proposition (Criterion for separatedness).} A prevariety $X$ is a variety iff $\De(X)$ is
closed in $X \times X$.

{\em Proof.} Neat trick, which is a recurring theme.  Suppose you have
two $f$ and $g$, which induce a morphism $(f,g): Y \rightarrow X
\times X$.  Then $$\{ y \in Y | f(y) = g(y) \} = (f,g)^{-1}
(\De(X)).$$ \qed

{\em Exercise.} Show that the line with the doubled origin is not
separated, using this condition.

{\bf Proposition (Another criterion for separatedness).}  Let $X$
be a prevariety.  
Assume that for all $x, y \in X$ there is an open affine $U$ containing both 
$x$ and $y$.  Then $X$ is a variety.

{\em Proof.} We use the definition.  Consider two functions $f, g: Y
\rightarrow X$, and let $Z$ be the locus where they agree $Z = \{y \in
Y | f(y) = g(y) \}$.  Let $z$ be in the closure of $Z$, and let
$x_1=f(z)$, $x_2=g(z)$.  We want to show that $Z$ is closed, so we
want to show that $x_1=x_2$.  By assumption, there is an open affine
$V \subset X$ containing $x_1$ and $x_2$.  Let $U = f^{-1}(V) \cap
g^{-1}(V)$; it is an open neighbourhood of $Z$.  Then 
consider the ``restricted morphisms'' $f, g:  U \rightarrow V$; now
$V$ is affine (hence a variety), so 
$$Z \cap U = \{ y \in U | f(y) = g(y) \}$$ 
is closed in $U$. Thus $z \in Z \cap U$, and $Z$ is indeed closed. \qed

{\bf Corollary.} Every quasiprojective prevariety $X$ (i.e. open subset of a projective prevariety) is a variety.

{\em Proof.} As every projective prevariety $X$ is a closed subprevariety of
$\proj^n$ (for some $n$, by definition), we just need to show that
$\proj^n$ is a variety.  So given any two points $y, z\in \proj^n$, we
just need to find an affine open containing both.  Consider any
hyperplane $H$ not meeting $y$ or $z$; there are lots!  (Instead of
proving this, let me just convince you that it is obvious by example.
If $y=(1;0;0)$ and $z =(0;1;1)$, take the hyperplane $x_1-x_2=0$.) 
The complement of a hyperplane is affine (proved earlier), so we're done.
\qed

Another nice property of varieties: the intersection of any two affine
opens is another affine open.  I don't foresee using this, so I won't
prove it, but you can find a proof in Mumford (p.  55) or Hartshorne
(Exercise II.4.4).  

This isn't a criterion (as it also holds for the line with the doubled
origin), but it can be strengthed a little into a criterion.

Here's a prevariety that doesn't have this property: the {\em plane}
with the doubled origin.  The intersection of the two elements of the
``obvious'' affine cover is the plane minus the origin, which you've
shown earlier is not affine.

\section{Rational maps}

I said a few words about rational maps, the topic we'll address on Thursday.

We can reinterpret the definition of separatedness as follows.
Suppose I'm thinking of a morphism $f: Y \rightarrow X$, where $X$ is
a {\em variety}.  And suppose I tell you what the morphism is on a
non-empty open set $U \subset Y$, i.e. I tell you $f|_U: U \rightarrow
X$.  Then there is only one way for you to recover the ``full''
morphism $f$.  Because if you have two different morphisms $f_1$ and
$f_2$ extending $f$, then you have two morphisms $f_1, f_2: Y
\rightarrow X$ which agree on a dense open set (the set $U$; recall that dense means that the 
closure of $U$ is $Y$), and agree on a closed set (as $X$ is
separated).  Hence they have to agree everywhere.

Coming soon: completeness (roughly, compactness), dimension,
smoothness.  Then we can talk about curves.




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