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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 25}
\author{Ravi Vakil}
\date{Thursday, December 9,  1999.}
\maketitle
\tableofcontents

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

PS10 back today; PS11 due today.  PS12 due Monday December 13.

\section{The genus of a nonsingular projective curve}

The definition I'm going to give you isn't the one people would typically start with.
I prefer to introduce this one here, because it is more easily computable.

{\bf Definition.}  The {\em tentative genus} of a nonsingular projective curve $C$ is given by $\deg \Omega^1_C = 2g-2$.

{\em Fact (from Riemann-Roch, later).} $g$ is always a nonnegative
integer, i.e. $\deg K = -2, 0, 2, \dots$.

Complex picture: Riemann-surface with $g$ ``holes''.  

{\em Examples.} Hence $\proj^1$ has genus 0, smooth plane cubics have genus 1, etc.

{\em Exercise: Hyperelliptic curves.} Suppose $f(x_0,x_1)$ is a
polynomial of homogeneous degree $n$ where $n$ is even.  Let $C_0$ be
the affine plane curve given by $y^2 = f(1,x_1)$, with the generically 2-to-1
cover $C_0 \rightarrow U_0$.  Let $C_1$ be the affine plane curve
given by $z^2 = f(x_0,1)$, with the generically 2-to-1 cover $C_1 \rightarrow
U_1$.  Check that $C_0$ and $C_1$ are nonsingular.  Show that you can
glue together $C_0$ and $C_1$ (and the double covers) so as to give a
double cover $C \rightarrow \proj^1$.  (For computational convenience,
you may assume that neither $[0;1]$ nor $[1;0]$ are zeros of $f$.)
What goes wrong if $n$ is odd? Show that the tentative genus of $C$ is $n/2-1$.
(This is a special case of the {\em Riemann-Hurwitz formula}.)
This provides examples of curves of any genus.

{\em Exercise.}  Suppose $C$ is a nonsingular degree $d$ projective plane curve.  Show that
the tentative genus of $C$ is $\binom {d-1} 2$.

\section{The Riemann-Roch Theorem with Applications but No Proof}

{\bf Definition.} If $X$ is a topological space, and $S$ is a sheaf on
$X$, define $H^0(X,S)$ to be the set of global sections of $S$ over
$X$ (what we've called $S(X)$).  If $X$ is a variety, and $S$ is an $\oh_X$-module, then
$H^0(X,S)$ has the structure of a $\kb$-vector space.  In this
situation, define $h^0(X,S)$ to be the dimension of $H^0(X,S)$ as 
a $\kb$-vector space.  

From here on in, $C$ is a nonsingular projective curve, and we'll deal
with the case when $S$ is an invertible sheaf.  In this case, we
have shown that $h^0$ is finite.

{\bf The Celebrated Riemann-Roch Formula.} $\cL$ a line bundle of
degree $d$ on a nonsingular projective curve $C$ of genus $g$.  Then
$h^0( C, \cL) - h^0(C, \Omega^1 \otimes \cL^\vee) = d-g+1$.

{\em Proof.}  Omitted!

{\bf I think you know enough to understand a proof, or at least a proof
under mild assumptions.  If there are 2 or 3 of you interested, I could try
to come up with a proof that you'd understand in half an hour, and present it
during IAP.}   

{\em Example.}  Let's check this out for $\proj^1$.  (Make a table of
degree, $h^0(C,\cL)$, $h^0(C, \Omega^1 \otimes \cL^\vee)$, $d-g+1$.)

Here come lots of appliations.  

{\bf Corollary.}  Plug in $\cL = \oh$.  Then we see that 
$h^0(C, \Omega^1) = g$.  Hence the tentative genus is always a nonnegative integer.

{\bf Definition.}  Define the  genus of a nonsingular projective curve $C$ to
be $h^0(C, \Omega^1_C)$.

{\bf Corollary.} If $d > 2g-2$ (i.e. the degree of $\cL$ is greater
than the sheaf of differentials), then $h^0(C, \Omega^1 \otimes \cL^\vee) = 0$.
Hence $h^0(C,\cL) = d-g+1.$

Reason:  the invertible sheaf  $\Omega^1 \otimes \cL^\vee$ has negative degree.

{\bf Corollary.} Given $p \in C$.  Then there is a rational
function with a pole only at $p$.

{\em Proof.} Consider the divisor $(2g+1) p$.  Then $\cL := \oh(
(2g+1) p)$ is a degree $2g$ line bundle, and it has $h^0(C, \cL) =
d-g+1 = 2g+1 -g+1=g+2>1$.  Hence there are two linearly independent
sections of $\cL$.  But the global sections of $\cL$ corresponded to
rational functions whose valuation at every point distinct from $p$
was nonnegative, and whose valuation at $p$ was at most $-(2g+1)$.
One such function is the constant function; thus there is a function
that is nonconstant, which must therefore have a pole somewhere; the
pole must be at $p$.  (We also see that the order of pole can be made
to be at most $2g+1$.)

\subsection{A criterion for closed immersions}

Given a nonsingular projective curve $C$, and invertible sheaf $\cL$.
Let $\{ s_0, \dots, s_n \}$ be a basis for the (finite-dimensional)
vector space $H^0(C,\cL)$ of global sections.  If, for each $p \in C$,
there is a section $s$ of $\cL$ such that $s$ doesn't vanish at $p$,
then this basis gives a morphism $$ (s_0, \dots, s_n): C \mapsto
\proj^n.$$ This morphism is denoted $|\cL|$, and is called the {\em
complete linear system} associated with $\cL$.

{\em Remark.}  The sections of $\cL \otimes \oh_C(-p)$ (often written $\cL(-p)$)
over $U$ can be identified with the sections of $\cL$ vanishing at $p$.
(You've seen this before if $\cL$ is $\oh_C$; the proof is the same.)
Hence there is a natural inclusion
$$
H^0(C,\cL(-p)) \hookrightarrow H^0(C,\cL).$$
The cokernel is dimension 0 or 1; it is 1 if there is a global section of $\cL$
that does not vanish at $p$.  Hence we've proved:

{\bf Proposition.} $h^0(C,\cL) - h^0(C,\cL \otimes \oh_C(-p))  \leq 1$, and
if equality holds then then there is a section of $\cL$ that does not vanish at $p$.

{\bf Proposition.} Suppose $p$ and $q$ are distinct points of $C$.  If
$h^0(C,\cL) - h^0(C,\cL \otimes \oh_C(-p-q)) =2$, then $p$ and $q$ are mapped
to distinct points of $\proj^n$ via $|\cL|$.  In other words, $|\cL|$ is injective.

{\em Proof.} (You can check that $h^0(C,\cL) - h^0(C,\cL(-p-q)) \leq
2$, and that the hypotheses of this proposition imply the hypotheses
of the previous one, so that $\cL$ really {\em does} induce a morphism
into projective space.)

It suffices to show that there is a section of $\cL$ that vanishes at $p$ but doesn't vanish at $q$.
(Explain.)
\epf

When the degree of $\cL$ is big, the hypotheses of these propositions are automatically satisfied.

{\bf Proposition.} If $\deg \cL > 2g-1$, then for each $p \in C$,
there is a section $s$ of $\cL$ such that $s$ doesn't vanish at $p$.
If $\deg \cL > 2g$, then the morphism $|\cL|$ is injective.

{\em Proof.}  In this range, $h^0(C, \cL) = d-g+1$, $h^0(C, \cL(-p)) = (d-1)-g+1$,
and (in the second case) $h^0(C, \cL(-p-q)) = (d-1)-g+1$.  \epf

Under the conditions of the second proposition, $|\cL|$ is an
injection of sets.  But it still might get knotted up at some point
(explain).  It would be great to know criteria for when $|\cL|$ gives
a closed immersion.   Here's one:

{\bf Proposition.} Suppose for any pair of points $p$ and $q$ of $C$
({\em not necessarily distinct}), $h^0(C,\cL) - h^0(C,\cL \otimes
\oh_C(-p-q)) =2$.  Then $|\cL|$ is a closed immersion.

{\em Proof.}  Omitted.  Shafarevich p. 217; Hartshorne.

{\em Remark.}  By the same argument as above, if $\deg \cL > 2g$, then the hypotheses
are satisfied.

This is powerful.  Here are applications.

{\bf Proposition.}  A projective nonsingular curve $C$ of genus 0 must be isomorphic to $\proj^1$.

{\em Proof.}  Choose any point $p \in C$.  Then $\cL := \oh(p)$ has degree $1>2g$, so $|\cL|$ is a closed
immersion.  But $\cL$ has 2 sections by Riemann-Roch:
$h^0(C,\cL) = \deg \cL -g+1 = 2$.  Thus $|\cL|$ describes a closed immersion of $C$ into $\proj^1$, so
$C$ must be isomorphic to $\proj^1$.
\epf

Thus there is only one projective nonsingular genus 0 curve.

Let's try a similar trick in genus 1.  Let $C$ be a genus 1
(projective nonsingular) curve, and let $\cL$ be any degree 3
invertible sheaf (e.g. $\oh(3p)$).  Then by Riemann-Roch, 
$h^0(C, \cL) = 3-1+1 = 3$.    

Then by the Proposition, $|\cL|$ gives a closed immersion of $C$ into $\proj^2$, and it must be as
a cubic curve.  (In fact the degree of the image is precisely the degree of the line bundle.  We haven't
analyzed this carefully yet, so I'll resort to another argument:  from an exercise in PS12, you can calculate
that the genus of a degree $d$ nonsingular plane curve is $\binom {d-1} 2$.  Hence as the image of $|\cL|$ has
genus 1, we know $d=3$.)

Hence we have proved:

{\bf Proposition.}  Every genus 1 curve can be written as a cubic plane curve.

Thus we've ``seen'' all genus 1 curves.  By special request, we proved that:

{\bf Proposition.}  There is a natural bijection between the degree 1 line bundles on a genus
1 curve $C$ and the points of the curve.

{\em Proof.} The link is as follows.  If $p$ is a point on the curve,
then $\oh_C(p)$ is a degree 1 invertible sheaf. If $\cL$ is a degree 1
invertible sheaf, then by Riemann-Roch, $h^0(C,\cL) = 1$, so there is
(up to multiple) precisely one section of $\cL$.  It's divisor is
necessarily a point.  These two constructions clearly commute.  \epf

In the problem set, you'll see that all genus 2 curves can be written as
double covers of $\proj^1$ branched at six points.  So now we've
``seen'' all genus 2 curves.

In the problem set, you'll see that all genus 3 curves can be written
either as double covers of $\proj^1$ branched at eight points, or as
nonsingular plane quartics.  So now we've ``seen'' all genus 3 curves.

{\bf Proposition.}  A projective nonsingular curve minus a finite number of points is affine.

{\em Proof.} Consider the invertible sheaf $\oh( k p)$ where $k$ is
huge.  (Exercise: how big does $k$ have to be for this to work?)  Then
this gives a morphism into projective space $$ |\oh(kp)|: C
\rightarrow \proj^n$$ expressing $C$ as a closed subvariety.  Now
there is a section $s_0$ of $\oh_C(kp)$ that has only one zero, of
order $k$, at the point $p$.  Take a basis for the sections of
$\oh_C(kp)$: $s_0$, \dots, $s_n$.  So we have a map $$(s_0, \dots,
s_n): C \rightarrow \proj^n$$ $$ q \mapsto (s_0(q); \dots; s_n(q)).$$
Now where does $x_0=0$ intersect $C$?  Answer: where $s_0=0$, i.e.
only at the point $p$.  (Geometry: the curve meets the hyperplane
$x_0=0$ with multiplicity $k$ there.)  Hence the affine open $U_0$ of
$C$ is $C \setminus \{ p \}$, so $C \setminus \{ p \}$ is affine!
\epf

{\bf Corollary.}  Any nonsingular curve that is {\em not} projective is affine.


Let's calculate more things.

Suppose $C$ is a genus 1 curve.  Let $\cL$ be any degree 4 invertible
sheaf on $C$, e.g. $4 p$ for some point $p$.  Then $|\cL|$ expresses
$C$ as a closed subvariety of $\proj^3$ (as $\cL$ has 4 sections, and
the degree satisfies the criterion for closed immersions above).  I
claim that it is the intersection of 2 quadrics.  Here is a pretty
good sketch of why.  

Let $\pi$ be the inclusion.  Let $w$, \dots, $z$ be the coordinates on $\proj^3$.
First, note that the image isn't contained in any hyperplane.

Next, compute the number of sections of $\pi^* \oh(1)$, which has
degree 4:  by Riemann-Roch, it has 4 sections, which are precisely
the 4 sections we used to map it to projective space.  They correspond
to the linear functions $w$, $x$, $y$, $z$.

Compute  the number of sections of $\pi^* \oh(2)$, which has
degree 8:  by Riemann-Roch, it has 8 sections.  But here are 10 sections:
the sections $w^2$, \dots, $yz$, $z^2$.    Hence there is a two-dimensional
vector space of polynomials that pullback to zero.  Let $p(w,x,y,z)$ and    $q(w,x,y,z)$
be a general basis of this two-dimensional vector space.  Then the image of $C$
lies on $p=0$ and $q=0$ (explain).  

We know that $p=0$ and $q=0$ must intersect in a curve.  Here's why:
if their intersection included a surface $S$, the surface must be of
degree 2 or degree 1.  If $S$ were degree 2, then it must be the
surface $p=0$ and $q=0$; but then $p$ is a scalar multiple of $q$,
contradicting the fact that $p$ and $q$ formed a basis.  If $S$ were
degree 1, then $p$ and $q$ both consist of 2 planes (explain, and
explain the contradiction).

Hence $p=0$ and $q=0$ intersect in a curve $C'$ (including $C$).  By the exercises on
the Hilbert polynomial, the curve $C'$ has degree 4, which means that
it meets a general hyperplane at 4 points.  

But a general hyperplane corresponds to the zeros of a general section
of $\cL$.  Any nonzero section of $\cL$ has 4 zeros, so $C$ is all of $C'$.
Hence $C$ is the intersection of 2 quadrics in $\proj^3$.

(Conversely, if you take 2 general quadrics in $\proj^3$, it turns out that you
get an elliptic curve.)

{\em Remark.}  You can use this method to understand curves of genus 4 and 5.

For a genus 4 curve $C$, the sheaf of differentials has degree 6, and
has 4 sections.  If $C$ is not hyperlliptic, then $|\Omega^1|$
expresses $C$ as a closed subvariety of $\proj^3$, (after an
application of the crition above).  The methods we just used show that
$C$ is a complete intersection of a quadric and a cubic.  Hence this
``describes all genus 4 curves'': they are either hyperelliptic, or
the intersection of a quadric and a cubic.  (There is a uniform way of
describing them, that doesn't involve these cases.)

For a genus 5 curve $C$, something similar happens.   $C$ is either hyperelliptic,
or it sits in $\proj^4$ as the intersection of 3 quadrics.
{\em Not quite!  Could be trigonal, in which case the intersection
of the 3 quadrics is, not surprisingly, a rational scroll.  Johan showed me a paper
of Rutger Noot.}

\section{Recap of course}

Generalities:
Algebraic sets.  Zariski topology. 
Nullstellensatz.
Hilbert basis theorem.
Sheaves and stalks; structure sheaf.
Prevarieties and morphisms of prevarieties.
Function fields and birational maps.
Projective prevarieties.
Schemes.
Products.
Dimension.  Nonsingularity.

Curves:  Valuation rings.  Completitions.
Integral closure, Dedekind domains, DVRs.
Extending rational maps of nonsingular curves to projective varieties.
Finitely generated fields over $\kb$ of tr. deg. 1 correspond to
nonsingular projective curves.

Line bundles and invertible sheaves.  $\oh(m)$.  Picard group.  Class group.
Degree.  Riemann-Roch.


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