\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 9}
\author{Ravi Vakil}
\date{Thursday, October 7, 1999.}
\maketitle
\tableofcontents

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

{\bf From an e-mail of Oct. 6.}

I thought I'd say in detail how you glue two prevarieties together
along isomorphic open subprevarieties.

Suppose we're just gluing two prevarieties $X$ and $Y$ along open
sets $U$ in $X$ and $V$ in $Y$.  In order to glue, we need a choice of
homeomorphism between $U$ and $V$, along with a choice of isomorphism of
sheaves $\oh_X$ restricted to $U$ and $\oh_Y$ restricted to $V$.  (In other words,
we've chosen an isomorphism of $U$ and $V$ as prevarieties.)

{\em i) Sets.} You can probably see how to glue $X$ and $Y$ together along $U=V$ as
sets.  Let $X \cup Y$ be this set; we can consider $X$ and $Y$ as subsets of $X \cup Y$ in a
natural way.

{\em ii) Topological spaces.} A subset of $X \cup Y$ is declared to be open if its
intersection with $X$ is an open subset of $X$, and its intersection with $Y$ is
an open subset of $Y$.  You can check that i) this is a topology, and ii)
the restriction of the topology to $X$ (resp. $Y$) is the original topology on
$X$ (resp. $Y$), and iii) that $X$ (resp. $Y$) is an open set in this topology.
(This is the definition of quotient topology.)

{\em iii) Structure sheaves.}  Suppose $W$ is an open subset of $X \cup Y$ (defined in
ii)).  I'll now tell you what the structure sheaf on $W$ is.  Let $W_X$ be the
intersection of $W$ and $X$, and $W_Y$ similarly, and $W_U$ the intersection of $W$
and $X$ and $Y$. Sections over $W$ correspond to the data of a section $f$ of $\oh_X$
over $W_X$, and a section $g$ of $\oh_Y$ over $W_Y$, such that the restriction of $f$
to $U$ is the same as the restriction of $g$ to $V$ (using the isomorphism of U
and $V$ as prevarieties).  You can check that i) this is a sheaf, and ii)
the restriction of the sheaf to the open subset $X$ is our original sheaf
$\oh_X$ (and the same thing with $X$ replaced by $Y$).

{\bf Class begins.}

{\em Unrelated exercise, also about $C$; I probably won't mention this
in class.} If $p$ is a point of $\proj^1$, show that $\pi^{-1} p$ is
two points unless $p$ is one of two points of $\proj^1$, when
$\pi^{-1} p$ is one point.

Then if $\kb = \com$, we have the following picture.  (Give it.)

We will later see the {\em Riemann-Hurwitz formula}, which will tell
us the following facts (even if $\kb \neq \com$).  Suppose we knew the
genus of the base was 0 (we have to define genus in general), and that
the cover is generally 2-1 (we have to define ``generally''), except
that it was 1-1 at two points.  Then we could compute that the genus of
the top curve was also 0.  

{\em Copies of Eisenbud-Harris are available in my office.}

Schemes next day.  Problem set:  don't do scheme problems.

\section{Projective Prevarieties}

The original motivation for introducing $\proj^n$ was to add some
points to affine space $\A^n$, that correspond to extra ``points at
infinity''.  If $\kb = \com$, then this is a compactification of $\A^n$.  
Here's the classical picture.  

Suppose you were interested in a subset of $\A^2$, for example the
curve $x^3+y^3=xy$, a curve that comes up in Problem Set 4.
Then if I draw it, it might look something vaguely like this.  It's a
real picture; really there are 3 directions in which the curve goes
off to infinity; more about that soon.  We'll embed it in projective
space as follows.  We'll basically add points at infinity
corresponding to directions, i.e. we'll add a $\proj^1$.

(Draw a picture; put $\A^2$ at the plane $x_0=1$.  $(x,y) \mapsto (1,x,y)$.)

Take the closure.  (This is called taking the {\em projective
closure}, which we'll describe algebraically soon.) We get a curve in
projective space, $x_1^3+x_2^3 = x_1x_2 x_0$.  The open set $U_0$ is just
the affine space we started with.  

Now you can see that if you compactify $\A^2$ in this way, you add
points of infinity that are really a $\proj^1$: the points of
projective space $(x_0; x_1; x_2)$ where $x_0 = 0$.  Note that you can
see the 3 points at infinity we've added to this curve, i.e. where
$x_0=0$ (ask them).

The cone I've described is the {\em affine cone over a projective
variety}; I'll rigorously define that soon.

As a good thought experiment, you might think about what happens when you do the 
same process with a parabola, e.g. $y=x^2$; how does it meet the line at $\infty$?  In 
how many points?

{\em Back to prevarieties.}

These are various constructions that it would be good to be familiar with.

{\bf Definition.} A projective prevariety is a prevariety that is
isomorphic to a closed subprevariety of $\proj^n$.  A quasi-projective
prevariety is a prevariety that is isomorphic to an open subprevariety
of a projective prevariety.

There are many analogies between affine varieties and projective
varieties, and we'll explore them now.

Consider a closed subprevariety $X$ of $\proj^n$, and say the
projective coordinates on $\proj^n$ are $x_0$, \dots, $x_n$.  

Consider the ring $R=\kb[x_0, \dots, x_n]$.  Assign a grading to
monomials in this ring, so the degree of $x_0^{a_0} \dots x_n^{a_n}$
is $\sum a_i$.  An element of $R$ is {\em homogeneous of degree $d$} if it
is a linear combination of monomials of degree $d$.

From the problem set, you've seen that a non-zero homogeneous element
of $R$ of degree $d$ cuts out a closed subset of $\proj^n$, and if
this subset is irreducible, we get a closed subprevariety of
$\proj^n$; such a subset (cut out by one equation) is called a {\em
hypersurface of degree $d$}.

There is a projective analog of the correspondence between affine varieties
and coordinate rings.

{\bf Definition.} A {\em homogeneous ideal} of $R$ is an ideal
generated by homogeneous elements of $R$.

{\bf Lemma.}  Any homogeneous ideal $A$ has a finite number of homogeneous generators.

{\em Proof.} (You won't be surprised that at some point we'll use the fact that $R$ is Noetherian.)  $A$ is
generated by some set of homogeneous elements, $(f_a)$.  We'll find a finite set
of generators in this set.  Pick $f_1$ to be any $f_a$ of lowest degree.  If $(f_1) = A$, we're done.
Otherwise, pick $f_2$ of lowest degree in $A \setminus (f_1)$.  If $(f_1, f_2) = A$, we're done.
Otherwise, pick $f_3$ of lowest degree in $A \setminus (f_1, f_2)$.  Etc. etc.

If we eventually stop, we're happy.  But if we never stop, then we have a strictly ascending   chain of ideals
$$
(f_1) \subset (f_1, f_2) \subset (f_1, f_2, f_3) \subset \dots.
$$
But any ascending chain must stabilize because $R$ is Noetherian, giving a contradiction.  \qed

If $A$ is a homogeneous ideal, define $V(A)$ to be the points of
$\proj^n$ where $(x_0; \dots; x_n)$ where $f(x_0, \dots, x_n) = 0$ for
each element $f$ of $A$; it suffices to check the generators, and by the lemma,
there are a finite number of generators.
 
If $S \subset \proj^n$ is a closed algebraic subset, then define
$I(S)$ to be the ideal generated by all the homogeneous polynomials
that vanish identically on $S$.  

{\bf Theorem.} $V$ and $I$ set up a bijection between the set of
closed algebraic subsets of $\proj^n$ and the set of all homogeneous
ideals $A \subset \kb[x_0, \dots, x_n]$ such that $A = \sqrt{A}$,
except for the one ideal $A=(x_0, \dots, x_n)$, called the {\em
irrelevant ideal}.

(Note why the irrelevant ideal defines the empty set in projective space.)

{\em Proof.}
Omitted; it uses the Nullstellensatz; see Mumford p. 13.  \qed

Just as in the affine case, we have:

{\bf Corollary.}  In the bijection of the Theorem above, the irreducible algebraic sets
correspond exactly to the homogeneous prime ideals (excepting the irrelevant ideal).
Moreover, every closed algebraic set $S$ in $\proj^n$ can be written in a unique
way as the union of distinct irreducibles.

{\bf Definition.}  The {\em projective coordinate ring} is the ring $\kb[x_0, \dots, x_n] / A$.

{\em Caution.} In the affine case, the data of an affine variety is
the same as the data of the coordinate ring.  In the projective case,
the data of an abstract projective prevariety is {\em not} the same as
the data of the projective coordinate ring, although the data of a
projective prevariety with the choice of embedding in some $\proj^n$ {\em
is} the same as the data of the projective coordinate ring.

As an example, consider the two projective prevarieties $\proj^1$ (with
projective coordinates $a$ and $b$, say) and the conic $C$ given by
$x^2+y^2=z^2$ in $\proj^2$.  These two are isomorphic, as you'll see
in Problem Set 4.

In the first case, the ring is given by $\kb[a,b]$.

In the second case, the ring you have is $\kb[x,y,z]/ (x^2+y^2-z^2)$.

Then if you look at the part of the first ring with grading 1, you
have a vector space of rank 2; in the second case, you have a vector
space of rank 3.  So they aren't the same.


{\bf Classical constructions.}

We can now define the classical constructions I mentioned earlier.  

{\em Affine cone.}  Suppose you have 
a projective prevariety corresponding to an ideal $A \subset \kb[x_0, \dots, x_n]$.
Then the affine cone corresponds precisely to the same thing.

{\em Projectivizing affine varieties.}  

Suppose you have an ideal $I \subset \R$, defining an affine variety $V \subset U_0 \cong \A^n$, 
and you want to find the projective closure $\Vbar$ of $V$ in $\proj^n$.

First note: as $V$ is an irreducible topological space, $\Vbar$ is
also irreducible.  (This is a general topological fact, and not hard
to prove; I might put it on the next problem set.)

So what's the homogeneous ideal $A \subset \kb[x_0, \dots, x_n]$?  Define a ``homogenization operator''
that takes an element $f(x_1, \dots, x_n) \in \R$, and makes it homogeneous in $\kb[x_0, \dots, x_n]$:
$$f(x_1, \dots, x_n) \mapsto x_0^d f(x_1/x_0, \dots, x_n/x_0)
$$
where $d$ is the largest degree that appears in $f$.  (Example:  $x_2^3 + x_1^3 = x_2 x_1$.)

Then to cook up $A$, just take {\em all} the elements of $I$ (not just
generators), homogenize them all, and look that the ideal $A$ they
generate in $\kb[x_0, \dots, x_n]$.  (Question: why can't you just
take generators?  What goes wrong?)

{\bf An affine base of open sets.}

{\bf Theorem.} Suppose $g$ is a homogeneous element of the projective coordinate
ring of a projective prevariety $V \subset \proj^n$, of {\em positive} degree.
Define $D(g)$ to be the locus where $g$ vanishes.  Then $D(g)$ is affine. 
Moreover, these $D(g)$ form a base of the topology.

For simplicity, I'll give the proof in the case where $V = \proj^n$;
but the general proof is actually the same; if you're feeling really secure, then
you should try to understand what I'm saying in general.

{\bf Theorem.}  $\proj^n$ minus a hypersurface of positive degree is affine.

For example, in $\proj^2$ with projective coordinates $x$, $y$, $z$,
if you discard the locus $x=0$, you get the standard open $U_x$, which
is isomorphic to the affine plane.  Ditto for $y=0$ and $z=0$.  But it
turns out that you can discard {\em any} line (e.g. $x+y=0$), and
still get something isomorphic to the affine plane.  The moral reason
is that, given some random line $L$ there is an automorphism taking
$\proj^2$ to itself, and sending $L$ to the $x$-axis.  (In this case,
$(x;y;z) \mapsto (x+y;y;z)$ works.)

{\em Rough sketch of one proof.} You've already done much of the work in
the problem set.  Suppose the hypersurface is defined by $f=0$, and
the prevariety is $W = \proj^n \setminus \{ f=0 \}$.  The global
section of the structure sheaf form a ring, and elements are of the
form $p/f^m$ where $p$ is a polynomial homogeneous of degree $md$.
This is a ring, finitely generated over $\kb$, with no nilpotents,
etc. etc., so it corresponds to an affine variety $V$.  Then you just
have to check that (i) this defines a morphism to $V$, and (ii) that
this is an isomorphism, using affine opens.  \qed

This is annoying.  Far better is to re-define projective space.

Let $R$ be the projective coordinate ring $\kb[x_0, \dots, x_n]$,
which has a grading.  (If you're following this proof for a general
variety, then $R$ is a quotient of this.)  For each element $f \in R$
of positive homogeneous degree, define $(R_f)_0$ to be the zero-th
graded piece of the ring $R_f$.  Explicitly, it is the ring defined
earlier.  $R$ is nilpotent-free finitely-generated domain over $\kb$; you can
check quickly that hence so is $R_f$, hence so is $(R_f)_0$.  Thus
this corresponds to an affine variety which I'll suggestively call
$D(f)$.  Glue together all these affines (as $f$ varies over all homogeneous elements
of $R$ of positive degree) as follows.  $D(f)$ and $D(g)$ are glued together
along open subset $D(fg)$.  

Let me make this somewhat explicit.  The ring for $D(f)$ are elements of the
form $p/f^m$, where $\deg p = m \deg f$.  The ring for $D(g)$ are
elements of the form $q/g^n$, where $\deg q = n \deg g$.  The ring for
$D(fg)$ are elements of the form $r/ (fg)^l$, where $\deg r = l \deg
(fg)$.  Clearly the first two rings lie in the third; this defines the
restriction map.

So this explains how to glue together 2 of these affine opens.  Then
you must check that the gluing data for 3 ``agree'', which isn't too
hard for $\proj^n$ (although you think more carefully if you're
doing this for general $V$, even though the argument is the same).  (Do this if enough time.)

Finally, we should check that we get the same prevariety as when we
defined $\proj^n$ before, which involved gluing together $n+1$ opens
$U_0$, \dots, $U_{n+1}$, not an infinite number of affine opens
(including these).  If you think about it, this new definition
involves gluing together these $n+1$, plus a whole lot more, so we
just need to check that we arne't adding any more points than the ones
we already have.  Equivalently, this says that given any $D(g)$, the
open sets $U_i \cap D(g)$ cover it.  

$D(g) = \{ p / g^q \} \Leftrightarrow V$ 

Exercise (not on PS).  Idea why this isn't surprising:  Think of this as lying in $\proj^n$.
\qed 

{\bf Corollary.}  The $D(g)$'s (where $g$ is a homogeneous element of positive degree)
form a base (of affine opens) of the topology.

{\em Proof.} We show it first for $\proj^n$.  Then the Zariski
topology on any closed subprevariety is induced by the topology on
$\proj^n$, so a base restricts to a base.

We show that it includes the base on the opens $U_0$, \dots, $U_n$; this forms a base of the topology on their union.
(Do this in a special case for convenience of exposition.)  \epf


{\bf Describing morphisms between projective varieties.}
Morphisms from projective prevarieties can be
painful to describe, especially if you deal with them purely in terms
of affine opens.  There is often an easier way to describe them.

Suppose $V \subset \proj^n$ is a projective prevariety, where the projective coordinates on
$\proj^n$ are $(x_0; \dots; x_n)$.  Suppose $p_0$, \dots, $p_m$ are homogeneous polynomials
in the $x_i$ of degree $d$, that don't have any common vanishing set on $V$.
Then $(p_0; \dots; p_m)$ describes a morphism from $V$ to $\proj^m$.

The best way to convince you of this is to describe it in a particular case.

For example, consider the projective variety $\proj^1$ with projective
coordinates $x$, $y$.  Then $(x^2; xy; y^2)$ defines a morphism to
$\proj^2$, with projective coordinates $x_0$, $x_1$, $x_2$.  (You
might notice that the image is given by $x_1^2 = x_0 x_2$.)  First,
(i) I'll describe this map as sets.  It's well-defined, because $(x;y)
\rightarrow (x^2; xy; y^2)$ and $(ax;ay) \rightarrow$ the same thing.

(ii) Next, I'll show it is a continuous map.  A base for the topology on
$\proj^2$ is given by $D(f(x_0,x_1, x_2))$ where $f$ is homogeneous,
so we need to show that the pullback of the zero-locus of $f$ is
closed in $\proj^1$ for all $f$.  But the pullback of
$f(x_0,x_1,x_2)=0$ is $f(x^2,xy,y^2)=0$, which is a closed set (as any
set that is the vanishing set of a polynomial is closed).

(iii) Finally, we need to deal with structure sheaves.  It suffices to consider the
cover $\proj^2 = U_0 \cup U_1 \cup U_2$, and to show that regular functions on $U_i$
pullback to regular functions on the preimage of $U_i$.
Regular functions on $U_0$: $p(x_0, x_1, x_2) / x_0^{\deg p}$, where $x_0 \neq 0$.  Pull back
to get $p(x^2, xy, y^2) / (x^2)^{\deg p}$ where $x^2 \neq 0$.


The general argument is precisely the same.  We have a map on sets, we show it is continuous by 
the same argument, and we deal with structure sheaves using the standard affine cover.

{\em Warning:}  Not every morphism of projective prevarieties is of this form.  But it makes
a lot of them easy to describe.

{\bf If time:}  Classify degree 1 hypersurfaces.  Classify degree 2 curves that are varieties.



{\bf End of lecture.}

Collect problem sets.  Hand out new sets.  

Coming in the next few lectures (in some order):  
\begin{enumerate}
\item Tues:  schemes
\item Rational maps (why equivalent to maps of function fields?).  
Degree of rational maps.
Birationality.
\item products
\item separatedness; defining varieties
\item properness?
\end{enumerate}

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