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%% Notes for Hartshorne's Algebraic Geometry course
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\author{William A. Stein}
\title{Notes for Algebraic Geometry II}
\begin{document}
\maketitle
\tableofcontents
\section{Preface}
{\bfseries Read at your own risk!}
These are my {\em very rough, error prone} notes of
a second course on algebraic geometry
offered at U.C. Berkeley in the Spring of 1996.
The instructor
was Robin Hartshorne and the students were Wayne Whitney,
William Stein, Matt Baker, Janos Csirik, Nghi Nguyen, and Amod.
I wish to thank Robin Hartshorne for giving this course
and to Nghi Nguyen for his insightful suggestions and corrections.
Of course all of the errors are solely my responsibility.
The remarks in brackets [[like this]] are notes that
I wrote to myself. They are meant as a warning or as a reminder
of something I should have checked but did not have time for. You
may wish to view them as exercises.
If you have suggestions, questions, or comments feel free to write to me.
My email address is {\tt was@math.berkeley.edu}.
% Day 1, 1/17/96
\section{Ample Invertible Sheaves}
Let $k$ be an algebraically closed field
and let $X$ be a scheme over $k$. Let $\phi:X\into \bP^n_k$ be a morphism.
Then to give $\phi$ is equivalent to giving an invertible sheaf
$\sL$ on $X$ and sections $s_0,\ldots,s_n\in\Gamma(X,\sL)$
which generate $\sL$. If $X$ is projective (that is, if there
is some immersion of $X$ into {\em some} $\bP^m_k$) then $\phi$
is a closed immersion iff $s_0,\ldots,s_n$ separate points and tangent
vectors.
\begin{defn}
Let $X$ be a scheme and $\sL$ an invertible sheaf on $X$.
Then we say $\sL$ is {\em very ample} if there is an immersion
$i:X\hookrightarrow \bP_k^n$ such that $\sL\isom i^{*}\sO(1)$.
\end{defn}
\begin{thm}
Let $X$ be a closed subscheme of
$\bP_k^n$
and
$\sF$ a coherent sheaf on $X$, then
$\sF(n)$ is generated by global sections for
all $n\gg 0$.
\end{thm}
\begin{cor}
Let $X$ be any scheme and $\sL$ a very ample coherent sheaf
on $X$, then for all $n\gg 0$, $\sF\tensor \sL^{\tensor n}$
is generated by global sections.
\end{cor}
\begin{defn}
Let $X$ be a Noetherian scheme and $\sL$ be an invertible sheaf.
We say that $\sL$ is {\em ample} if for every coherent sheaf
$\sF$ on $X$, there is $n_0$ such that for all $n\geq n_0$,
$\sF \tensor \sL^{\tensor n}$ is generated by its global sections.
\end{defn}
Thus the previous corollary says that a very ample invertible
sheaf is ample.
\begin{prop}
Let $X$ and $\sL$ be as above. Then the following are equivalent.
1) $\sL$ is ample,
2) $\sL^n$ is ample for all $n>0$,
3) $\sL^n$ is ample for some $n>0$.
\end{prop}
\begin{thm}
Let $X$ be of finite type over a Noetherian ring $A$ and suppose
$\sL$ is an invertible sheaf on $A$. Then $\sL$ is ample iff
there exists $n$ such that $\sL^n$ is very ample over $\spec A$.
\end{thm}
\begin{example}
Let $X=\bP^1$, $\sL=\sO(\ell)$, some $\ell \in \bZ$.
If $\ell<0$ then $\Gamma(\sL)=0$. If $\ell=0$ then
$\sL=\sO_X$ which is not ample since $\sO_X(-1)^n\tensor
\sO_X\isom \sO_X(-1)^n$ is not generated by global sections
for any $n$. Note that $\sO_X$ itself is generated by
global sections. Finally, if $ell>0$ then
$\sL=\sO_X(\ell)$ is very ample hence ample.
\end{example}
\begin{example}
Let $C\subseteq \bP^2$ be a nonsingular cubic curve and
$\sL$ an invertible sheaf on $C$ defined by $\sL=\sL(D)$,
where $D=\sum n_i P_i$ is a divisor on $C$. If $\deg D<0$
then $\sL$ has no global sections so it can't be ample.
%% Undone.
\end{example}
% Day 2, 1/19/96
\section{Introduction to Cohomology}
We first ask, what is cohomology and where does it arise in nature?
Cohomology occurs in commutative
algebra, for example in the $\ext$ and $\tor$ functors, it occurs in
group theory, topology, differential geometry, and of course in
algebraic geometry. There are several flavors of cohomology which are
studied by algebraic geometers. Serre's coherent sheaf
cohomology has the advantage of being easy to define, but
has the property that the cohomology groups are vector spaces.
Grothendieck introduced \`{e}tale cohomology and $\ell$-adic
cohomology. See, for example, Milne's {\em \`{E}tale Cohomology}
and SGA 4$\frac{1}{2}$, 5 and 6. This cohomology theory arose
from the study of the Weil Conjectures (1949) which deal with
a deep relationship between the number of points on a variety
over a finite field and the geometry of the complex analytic variety
cut out by the same equations in complex projective space. Deligne
was finally able to resolve these conjectures in the affirmative
in 1974.
What is cohomology good for? Cohomology allows one to get numerical
invariants of an algebraic variety. For example, if $X$ is a projective
scheme defined over an algebraically closed field $k$ then
$H^i(X,\sF)$ is a finite dimensional $k$-vector space. Thus the
$h_i=\dim_k H^i(X,\sF)$ are a set of numbers associated
to $X$. ``Numbers are useful in all branches of mathematics.''
\begin{example}{Arithmetic Genus}
Let $X$ be a nonsingular projective curve. Then $\dim H^1(X,\sO_X)$ is
the arithmetic genus of $X$. If $X\subseteq \bP^n$ is a projective variety
of dimension $r$ then, if $p_a=\dim H^1(X,\sO_X)$, then
$1+(-1)^r p_a = $ the constant term of the Hilbert polynomial of $X$.
\end{example}
\begin{example}
Let $X$ be a nonsingular projective surface, then
\begin{equation*}
1+p_a=h^0(\sO_X)-h^1(\sO_X)+h^2(\sO_x)
\end{equation*}
and $1+(-1)^r p_a = \chi(\sO_X)$, the Euler characteristic of $X$.
\end{example}
\begin{example}
Let $X$ be an algebraic variety and $\pic X$ the group of
Cartier divisors modular linear equivalence (which is isomorphic
to the group of invertible sheaves under tensor product modulo isomorphism).
Then $\pic X \isom H^1(X,\sO_{X}^{*})$.
\end{example}
\begin{example}[Deformation Theory]
Let $X_0$ be a nonsingular projective variety. Then
the first order infinitesimal deformations are classified
by $H^1(X_0,T_{X_0})$ where $T_{X_0}$ is the tangent
bundle of $X_0$. The obstructions are classified by
$H^2(X_0,T_{X_0})$.
\end{example}
One can define Cohen-Macaulay rings in terms of cohomology.
Let $(A,\gm)$ be a local Noetherian ring of dimension $n$,
let $X = \spec A$, and let $P=\gm\in X$, then we have the
following.
\begin{prop}
Let $A$ be as above. Then $A$ is Cohen-Macaulay iff
1) $H^0(X-P,\sO_{X-P}) = A$ and
2) $H^i(X-P, \sO_{X-P}) = 0$ for $0*0$ one has that $R^i F(I)=0$.
\end{enumerate}
\begin{thm} The $R^{i}F$ and etc. are uniquely determined by properties
1-6 above.
\end{thm}
\begin{defn} A {\em $\delta$-functor} is a collection of functors
$\{R^i F\}$ which satisfy 3 and 4 above. An {\em augmented
$\delta$-functor} is a {\em $\delta$-functor} along with a natural
transformation $F\into R^0 F$. A {\em universal augmented
$\delta$-functor} is an {\em augmented $\delta$-functor} with
some universal property which I didn't quite catch.
\end{defn}
\begin{thm} If $\sA$ has enough injectives then the collection of
derived functors of $F$ is a universal augmented $\delta$-functor.
\end{thm}
To construct the $R^i F$ choose once and for all, for each object
$M$ in $\sA$ an injective resolution, then prove the above properties
hold.
%% 1/26/96 -- today's lecture was too ...@$%^(#$&#
\section{Long Exact Sequence of Cohomology and Other Wonders}
``Today I sat in awe as Hartshorne effortless drew hundreds
of arrows and objects everywhere, chased some elements and
proved that there is a long exact sequence of cohomology in
30 seconds. Then he whipped out his colored chalk and things
really got crazy. Vojta tried to erase Hartshorne's diagrams
during the next class but only partially succeeded joking
that the functor was not `effaceable'. (The diagrams are
still not quite gone 4 days later!) Needless to say, I don't
feel like texing diagrams and element chases... it's all
trivial anyways, right?''
%% 1/29/96
\section{Basic Properties of Cohomology}
Let $X$ be a topological space, $\Ab(X)$ the category
of sheaves of abelian groups on $X$ and
$$\Gamma(X,\cdot):\Ab(X)\into\Ab$$
the covariant, left exact global sections functor.
Then we have constructed the derived functors
$H^{i}(X,\cdot)$.
\subsection{Cohomology of Schemes}
Let $(X,\sox)$ be a scheme and $\sF$ a sheaf of $\sox$-modules.
To compute $H^{i}(X,\sF)$ forget all extra structure and use
the above definitions. We may get some extra structure anyways.
\begin{prop}
Let $X$ and $\sF$ be as above, then the groups $H^{i}(X,\sF)$
are naturally modules over the ring $A=\Gamma(X,\sox)$.
\end{prop}
\begin{proof}
Let $A=H^{0}(X,\sF)=\Gamma(X,\sox)$ and let $a\in A$. Then because
of the functoriality of $H^{i}(X,\cdot)$ the
map $\sF\into\sF$ induced by left multiplication by $a$ induces
a homomorphism
$$a:H^{i}(X,\sF)\into H^{i}(X,\sF).$$
%% Need more!!
\end{proof}
\subsection{Objective}
Our objective is to compute $H^{i}(\bP^n_k,\sO(\ell))$
for all $i,n,\ell$. This is enough for most applications
because if one knows these groups one can, in principle at
least, computer the cohomology of any projective scheme.
If $X$ is any projective variety, we embed $X$ in some
$\bP^n_k$ and push forward the sheaf $\sF$ on $X$. Then
we construct a resolution of $\sF$ by sheaves of the
form $\sO(-\ell)^n$. Using Hilbert's syzigy theorem one
sees that the resolution so constructed is finite and
so we can put together our knowledge to get the cohomology
of $X$.
Our plan of attack is as follows.
\begin{enumerate}
\item Define flasque sheaves which are acyclic for cohomology, i.e.,
the cohomology vanishes for $i>0$.
\item If $X=\spec A$, $A$ Noetherian, and $\sF$ is quasi-coherent,
show that $H^i(X,\sF)=0$ for $i>0$.
\item If $X$ is any Noetherian scheme and $\sU=(U_i)$ is an open
affine cover, find a relationship between the cohomology of $X$
and that of each $U_i$. (The ``\v{C}ech process''.)
\item Apply number 3 to $\bP_k^n$ with $U_i=\{x_i\neq 0\}$.
\end{enumerate}
\section{Flasque Sheaves}
\begin{defn}
A {\em flasque sheaf} (also called {\em flabby sheaf}) is a sheaf
$\sF$ on $X$ such that whenever $V\subset U$ are open sets then
$\rho_{U,V}:\sF(U)\into\sF(V)$ is surjective.
\end{defn}
Thus in a flasque sheaf, ``every section extends''.
\begin{example}
Let $X$ be a topological space, $p\in X$ a point, not necessarily
closed, and $M$ an abelian group. Let $j:\{P\}\hookrightarrow X$ be the
inclusion, then $\sF=j_{*}(M)$ is flasque. This follows since
$$
j_{*}(M)(U)=\begin{cases} M&\text{if $p\in U$}\\
0&\text{if $p\not\in U$}
\end{cases}.
$$
Note that $j_{*}(M)$ is none other than the skyscraper sheaf
at $p$ with sections $M$.
\end{example}
\begin{example}
If $\sF$ is a flasque sheaf on $Y$ and $f:Y\into X$ is a morphism
then $f_{*}\sF$ is a flasque sheaf on $X$.
\end{example}
\begin{example}
If $\sF_i$ are flasque then $\bigoplus_{i} \sF_i$ is flasque.
\end{example}
\begin{lem}
If
$$0\into\sF'\into\sF\into\sF''\into 0$$
is exact and $\sF'$ is flasque
then
$$\Gamma(\sF)\into\Gamma(\sF'')\into 0$$
is exact.
\end{lem}
\begin{lem}
If
$$0\into\sF'\into\sF\into\sF''\into 0$$
is exact and $\sF'$ and $\sF$ are both
flasque then
$\sF''$ is flasque.
\end{lem}
\begin{proof}
Suppose $V\subset U$ are open subsets of $X$.
Since $\sF'$ is flasque and the restriction of a
flasque sheaf is flasque and restriction is exact,
lemma 1 implies that the sequence
$$\sF(V)\into\sF''(V)\into 0$$
is exact.
We thus have a commuting diagram
$$
\begin{CD}
\sF(U) @>>> \sF''(U)\\
@VVV @VVV\\
\sF(V) @>>> \sF''(V) @>>> 0
\end{CD}
$$
which, since $\sF(U)\into \sF(V)$ is surjective,
implies $\sF''(U)\into\sF''(V)$ is surjective.
\end{proof}
\begin{lem}
Injective sheaves (in the category of abelian sheaves) are flasque.
\end{lem}
\begin{proof}
Let $\sI$ be an injective sheaf of abelian groups on $X$ and
let $V\subset U$ be open subsets. Let $s\in\sI(V)$, then we
must find $s'\in\sI(U)$ which maps to $s$ under
the map $\sI(U)\into\sI(V)$. Let $\bZ_V$ be the constant sheaf
$\bZ$ on $V$ extended by $0$ outside $V$ (thus $\bZ_V(W)=0$
if $W\not\subset V$). Define a map $\bZ_V\into \sI$ by
sending the section $1\in\bZ_V(V)$ to $s\in\sI(V)$. Then
since $\bZ_V\hookrightarrow\bZ_U$ and $\sI$ is injective
there is a map $\bZ_U\into \sI$ which sends the section $1\in
\bZ_U$ to a section $s'\in\sI(U)$ whose restriction
to $V$ must be $s$.
\end{proof}
\begin{remark} The same proof also shows that injective sheaves in
the category of $\sox$-modules are flasque.
\end{remark}
\begin{cor}
If $\sF$ is flasque then $H^{i}(X,\sF)=0$ for all $i>0$.
\end{cor}
\begin{proof} Page 208 of [Hartshorne].
\end{proof}
% 1/31/96
\begin{cor}
Let $(X,\sox)$ be a ringed space, then the derived functors of
$\Gamma:\Mod\sox\into\Ab$ are equal to $H^{i}(X,\sF)$.
\end{cor}
\begin{proof}
If
$$0\into\sF\into\sI^0\into\sI^1\into\cdots$$
is an injective resolution of $\sF$
in $\Mod\sox$ then, by the above remark, it
is a flasque resolution in the category
$\Ab(X)$ hence we get the regular cohomology.
\end{proof}
\begin{remark}
{\bf Warning!} If $(X,\sox)$ is a scheme and we choose an injective
resolution in the category of quasi-coherent $\sox$-modules
then we are only guaranteed to get the right answer if
$X$ is Noetherian.
\end{remark}
\section{Examples}
\begin{example}
Suppose $C$ is a nonsingular projective curve over an algebraically
closed field $k$. Let $K=K(C)$ be the function field of $C$ and
let $\sK_C$ denote the constant sheaf $K$. Then we have an exact
sequence
$$ 0\into\sO_C\into\sK_C\into
\bigoplus_{\substack{P\in C\\P\text{ closed}}}
K/\sO_P \into 0,$$
where the map $\sK_C\into\bigoplus K/\sO_P$ has
only finitely many components nonzero since a function $f\in K$
has only finitely many poles.
Since $C$ is irreducible $\sK_C$ is flasque and since $\sK/\sO_P$
is a skyscraper sheaf it is flasque so since direct of flasque
sheaves are flasque, $\bigoplus K/\sO_P$ is flasque.
One checks that the sequence is exact and so this is
a flasque resolution of $\sO_C$. Taking global sections and
applying the exact sequence of cohomology gives an exact sequence
$$K\into\bigoplus_{\text{$P$ closed}} K/\sO_P
\into H^1(X,\sO_C)\into 0,$$ and $H^{i}(X,\sO_C)=0$ for
$i\geq 2$. Thus the only interesting information
is $\dim_k H^1(X,\sO_C)$ which is the {\em geometric genus} of $C$.
\end{example}
%% 2/2/96
\section{First Vanishing Theorem}
\begin{quote}
``Anyone who studies algebraic geometry must read French... looking up
the more general version of this proof in EGA would be a good exercise.''
\end{quote}
\begin{thm}
Let $A$ be a Noetherian ring, $X=\spec A$ and $\sF$ a quasi-coherent
sheaf on $X$, then $H^{i}(X,\sF)=0$ for $i>0$.
\end{thm}
\begin{remark}
The theorem is true without the Noetherian hypothesis on $A$, but
the proof uses spectral sequences.
\end{remark}
\begin{remark}
The assumption that $\sF$ is quasi-coherent is essential. For example,
let $X$ be an affine algebraic curve over an infinite field $k$. Then
$X$ is homeomorphic as a topological space to $\P^1_k$ so
the sheaf $\sO(-2)$ on $\P^1_k$ induces a sheaf $\sF$ of abelian groups
on $X$ such that
$$H^1(X,\sF)\isom{}H^1(\P^1_k,\sO(-2))\neq 0.$$
\end{remark}
\begin{remark}
If $I$ is an injective $A$-module then $\tilde{I}$ need {\em not}
be injective in $\Mod(\sox)$ or $\Ab(X)$. For example, let $A=k=\bF_p$ and
$X=\spec A$, then $I=k$ is an injective $A$-module but $\tilde{I}$
is the constant sheaf $k$. But $k$ is
a finite group hence not divisible so $\tilde{I}$ is not injective.
(See Proposition A3.5 in Eisenbud's {\em Commutative Algebra}.)
\end{remark}
\begin{prop}
Suppose $A$ is Noetherian and $I$ is an injective $A$-module, then
$\tilde{I}$ is flasque on $\spec A$.
\end{prop}
The proposition implies the theorem since if $\sF$ is quasi-coherent
then $\sF=\tilde{M}$ for some $A$-module $M$. There is an injective
resolution
$$0\into M\into I^{\bullet}$$
which, upon applying the exact functor $\tilde{ }$,
gives a flasque resolution
$$0\into \tilde{M}=\sF\into \tilde{I}^{\bullet}.$$
Now applying $\Gamma$ gives us back the original resolution
$$\Gamma:\quad 0\into M\into I^{\bullet}$$
which is exact so the cohomology groups vanish for $i>0$.
\begin{proof} Let $A$ be a Noetherian ring and $I$ an injective $A$,
then $\tilde{I}$ is a quasi-coherent sheaf on $X=\spec A$. We must
show that it is flasque. It is sufficient to show that for any
open set $U$, $\Gamma(X)\into\Gamma(U)$ is surjective.
{\em Case 1, special open affine:}
Suppose $U=X_f$ is a special open affine. Then we have a commutative diagram
$$
\begin{CD}
\Gamma(X,\tilde{I}) @>>> \Gamma(X_f,\tilde{I})\\
@V=VV @V=VV\\
I @>\text{surjective?}>> I_f
\end{CD}
$$
To see that the top map is surjective it is equivalent
to show that $I\into I_f$ is surjective. This is a tricky
algebraic lemma (see Hartshorne for proof).
{\em Case 2, any open set:}
Let $U$ be any open set. See Hartshorne for the rest.
\end{proof}
%% 2/5/96
\section{\cech{} Cohomology}
Let $X$ be a topological space, $\sU=(U_i)_{i\in I}$ an
open cover and $\sF$ a sheaf of abelian groups.
We will define groups $\cH^i(\sU,\sF)$ called
\cech{} cohomology groups.
{\bfseries Warning: } $\cH^i(\sU,\cdot)$ is a functor in $\sF$,
but it is {\em not} a $\delta$-functor.
\begin{thm} Let $X$ be a Noetherian scheme, $\sU$ an open cover
and $\sF$ a quasi-coherent sheaf, then $\cH^{i}(\sU,\sF)=H^i(X,\sF)$
for all $i$.
\end{thm}
\subsection{Construction}
Totally order the index set $I$. Let
$$U_{i_0\cdots i_p}=\intersect_{j=0}^p U_{i_j}.$$
For any $p\geq 0$ define
$$C^p(\sU,\sF)=\prod_{i_00$. On the other hand,
the answer given by this resolution is
$$H^p(\Gamma(X,\sC^{\bullet}(\sU,\sF)))=\cH^{p}(\sU,\sF).$$
So we conclude that $\cH^{p}(\sU,\sF)=0$ for $p>0$.
\end{proof}
\begin{lem}
Let $X$ be a topological space, and $\sU$ an open covering. Then
for each $p\geq 0$ there is a natural map, functorial in $\sF$,
$$\cH^p(\sU,\sF)\into H^p(X,\sF).$$
\end{lem}
\begin{thm}
Let $X$ be a Noetherian separated scheme, let $\sU$ be an open
affine cover of $X$, and let $\sF$ be a quasi-coherent sheaf on
$X$. Then for all $p\geq 0$ the natural maps give isomorphisms
$$\cH^{p}(\sU,\sF)\isom H^p(X,\sF).$$
\end{thm}
% 2/7/96
\section{\v{C}ech Cohomology and Derived Functor Cohomology}
Today we prove
\begin{thm}
Let $X$ be a Noetherian, separated scheme, $\sU$ an open cover
and $\sF$ a quasi-coherent sheaf on $X$. Then
$$\cH^{i}(\sU,\sF)=H^{i}(X,\sF).$$
\end{thm}
To do this we introduce a condition (*):
{\bfseries Condition *:} Let $\sF$ be a sheaf of abelian groups
and $\sU=(U_i)_{i\in I}$ an open cover. Then the pair $\sF$ and
$\sU$ satisfy condition (*) if for all $i_0,\ldots,i_p\in I$,
$$H^(U_{i_0,\ldots,i_p},\sF)=0, \text{all} i>0.$$
\begin{lem}
If $0\into\sF'\into\sF\into\sF''\into 0$ is an exact sequence in
$\Ab(X)$ and $\sF'$ satisfies (*) then there is a long exact sequence
for $\cH^{i}(\sU,\cdot)$.
\end{lem}
\begin{proof}
Since the global sections functor is left exact and cohomology
commutes with products, we have an exact sequence
\begin{align*}
0\into C^{p}(\sU,\sF')=\prod_{i_0<\cdots0$ then $H^{i}(X,\sF)=0$ for $i>n$.
\end{cor}
\begin{example}
Let $X=\bP_k^n$, then the existence of the standard affine
cover $U_0,\ldots,U_n$ implies that $H^{i}(X,\sF)=0$ for
$i>n$.
\end{example}
\begin{example}
Let $X$ be a projective curve embedded in $\bP^k_n$.
Let $U_0\subset X$ be open affine, then $X-U_0$ is finite.
Thus $U_0\subset X\subset \bP^n$ and $X-U_0=\{P_1,\ldots,P_r\}$.
In $\bP^n$ there is a hyperplane $H$ such that
$P_1,\ldots,P_r\not\in H$. Then $P_1,\ldots,P_r\in\bP^n-H=\bA^n=V$.
Then $U_1=V\intersect X$ is closed in the affine set $V$, hence affine.
Then $X=U_0\union U_1$ with $U_0$ and $U_1$ both affine.
Thus $H^{i}(X,\sF)=0$ for all $i\geq 2$.
\end{example}
\begin{exercise}
If $X$ is any projective scheme of dimension $n$ then $X$
can be covered by $n+1$ open affines so
$$H^{i}(X,\sF)=0 \text{ for all } i>n.$$
[Hint: Use induction.]
\end{exercise}
Hartshorne was unaware of the answer to the following question
today.
\begin{ques}
If $X$ is a Noetherian scheme of dimension $n$ do there
exist $n+1$ open affines covering $X$.
\end{ques}
\begin{thm}[Grothendieck]
If $\sF\in\Ab(X)$ then $H^{i}(X,\sF)=0$ for all $i>n=\dim X$.
\end{thm}
\begin{example}
Let $k$ be an algebraically closed field.
Then $X=\bA^2_k-\{(0,0)\}$ is not affine since it has global
sections $k[x,y]$.
We compute $H^1(X,\sox)$ by \cech cohomology.
Write $X=U_1\union U_2$ where $U_1=\{x\neq 0\}$ and
$U_2=\{y\neq 0\}$. Then the \cech complex is
$$C^{\cdot}(\sU,\sox): k[x,y,x^{-1}]\oplus{}k[x,y,y^{-1}]
\xrightarrow{d} k[x,x^{-1},y,y^{-1}].$$
Thus one sees with a little thought that
$H^{0}=\ker{d}=k[x,y]$
and
$H^{1}=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\} = E$
as $k$-vector spaces (all sums are finite).
\end{example}
\subsection{History of this Module $E$}
$$E=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\}$$
\begin{enumerate}
\item Macaulay's ``Inverse System'' (1921?)
\item $E$ is an injective $A$-module, in fact, the indecomposable
injective associated to the prime $(x,y)$
\item $E$ is the dualizing module of $A$, thus
$D=\Hom_A(\cdot,E)$ is a dualizing functor
for finite length modules (so doing $D$ twice
gives you back what you started with).
\item Local duality theorem: this is the module you ``hom into''.
\end{enumerate}
% 2/9/96
\section{Cohomology of $\bP_k^n$}
Today we begin to compute $H^{i}(X,\sox(\ell))$ for all $i$ and all
$\ell$.
a) $H^0(X,\sox(\ell))$ is the vector space of forms of degree $\ell$
in $S=k[x_0,\ldots,x_n]$, thus
$$\oplus_{\ell\in\bZ}H^0(\sox(\ell))=H^0_{*}(\sox)=\Gamma_{*}(\sox)=S.$$
\begin{prop}
There is a natural map
$$H^{0}(\sox(\ell))\times H^{i}(\sox(m))\into H^{i}(\sox(\ell+m)).$$
\end{prop}
\begin{proof}
$\alpha\in H^{0}(\sox(\ell))$ defines a map $\sox\into\sox(\ell)$
given by $1\mapsto\alpha$. This defines a map
$$\sox\tensor\sox(m)\xrightarrow{\alpha(m)}\sox(\ell)\tensor\sox(m)$$
which gives a map $\sox(m)\into\sox(\ell+m)$. This induces the desired
map $H^{i}(\sox(m))\into H^{i}(\sox(\ell+m))$.
\end{proof}
b) $H^{i}(\sox(\ell))=0$ when $0**0$ and
for all $n\geq n_0$, $H^{i}(X,\sF(n))=0$.
\end{enumerate}
\end{thm}
The following was difficult to prove last semester and
we were only able to prove it under somewhat restrictive
hypothesis on $A$ (namely, that $A$ is a finitely generated
$k$-algebra).
\begin{cor} $\Gamma(X,\sF)$ is a finitely generated $A$-module.
\end{cor}
\begin{proof}
Set $i=0$ in 1.
\end{proof}
\begin{proof}(of theorem)
I. {\em Reduce to the case $X=\bP^r_A$.}
Use the fact that the push forward of a closed subscheme has
the same cohomology to replace $\sF$ by $i_{*}(\sF)$.
II. {\em Special case, $\sF=\sO_{\bP^r}(\ell)$ any $\ell\in\bZ$.}
1. and 2. both follow immediately from the previous theorem.
This is where we have done the work in explicit calculations.
III. {\em Cranking the Machine of Cohomology}
\end{proof}
\subsection{Application: The Arithmetic Genus}
Let $k$ be an algebraically closed field and $V\subset X=\bP^n_k$ a
projective variety. The arithmetic genus of $V$ is
$$p_a=(-1)^{\dim V}(p_V(0)-1)$$
where $p_V$ is the Hilbert polynomial of $V$, thus
$p_V(\ell)=\dim_k(S/I_V)_{\ell}$ for all $\ell\gg 0$. The
Hilbert polynomial depends on the projective embedding of $V$.
\begin{prop}
$p_V(\ell)=\sum_{i=0}^{\infty}(-1)^{i}\dim_k H^{i}(\sO_V(\ell))$
for {\em all} $\ell\in\bZ$.
\end{prop}
This redefines the Hilbert polynomial. Furthermore,
$$p_a=(-1)^{\dim V}(p_V(0)-1) = (-1)^{\dim V}
\sum_{i=0}^{\infty}(-1)^i\dim_k(H^{i}(\sO_V))$$
which shows that $p_a$ is intrinsic, i.e., it doesn't depend
on the embedding of $V$ in projective space.
\section{Euler Characteristic}
Fix an algebraically closed field $k$, let $X=\P_k^n$. Suppose
$\sF$ is a coherent sheaf on $X$. Then by Serre's theorem
$H^{i}(X,\sF)$ is a finite dimensional $k$-vector space.
Let $$h^{i}(X,\sF)=\dim_k H^{i}(X,\sF).$$
\begin{defn}
The {\bfseries Euler characteristic} of $\sF$ is
$$\chi(\sF)=\sum_{i=0}^{n}(-1)^i h^i(X,\sF).$$
\end{defn}
Thus $\chi$ is a function $\Coh(X)\into\Z$.
\begin{lem}
If $k$ is a field and
$$0\into{}V_1\into{}V_2\into{}\cdots\into{}V_N\into{}0$$
is an exact sequence of finite dimensional vector spaces,
then $\sum_{i=1}^N(-1)^i\dim{}V_i=0$.
\end{lem}
\begin{proof}
Since every short exact sequence of vector spaces splits, the
statement is true when $N=3$. If the statement is true for an
exact sequence of length $N-1$ then, applying it to the exact sequence
$$0\into{}V_2/V_1\into{}V_3\into\cdots\into{}V_N\into{}0,$$
shows that
$\dim{}V_2/V_1-\dim V_3+\cdots\pm\dim V_n = 0$
from which the result follows.
\end{proof}
\begin{lem}
If $0\into\sF'\into\sF\into\sF''\into{}0$ is an exact sequence
of coherent sheaves on $X$, then
$$\chi(\F)=\chi(\F')+\chi(\F'').$$
\end{lem}
\begin{proof}
Apply the above lemma to the long exact sequence of cohomology
taking into account that $H^n(\F'')=0$ by Serre's vanishing theorem.
\end{proof}
More generally, any map $\chi$ from an abelian category to
$\Z$ is called additive if, whenever
$$0\into\F^0\into\F^1\into\cdots\into\F^n\into{}0$$
is exact, then
$$\sum_{i=0}^{n}(-1)^{i}\chi(\F^i)=0.$$
{\bfseries Question.}
Given an abelian category $\sA$ find an abelian
group $A$ and a map $X:\sA\into{}A$ such that every
additive function $\chi:\sA\into{}G$ factor through $\sA\xrightarrow{X}A$.
In the category of coherent sheaves the Grothendieck group
solves this problem.
Let $X=\P_k^n$ and suppose $\sF$ is a coherent sheaf on $X$.
The Euler characteristic induces a map
$$\Z\into\Z: n\mapsto\chi(\F(n)).$$
\begin{thm}
There is a polynomial $p_{\F}\in\Q[z]$ such that
$p_{\F}(n)=\chi(\F(n))$ for all $n\in\Z$.
\end{thm}
The polynomial $p_{\F}(n)$ is called the Hilbert polynomial of $\F$.
Last semester we defined the Hilbert polynomial of a graded module
$M$ over the ring $S=k[x_0,\ldots,x_n]$. Define $\varphi_M:\Z\into\Z$
by $\varphi_M(n)=\dim_k M_n$. Then we showed that there is a unique
polynomial $p_M$ such that $p_M(n)=\varphi_M(n)$ for all $n\gg 0$.
\begin{proof}
We induct on $\dim(\supp\F)$. If $\dim(\supp\F)=0$ then $\supp\F$ is a
union of closed points so $\sF=\oplus_{i=1}^{k}\F_{p_i}$.
Since each $\F_{p_i}$ is a finite
dimensional $k$-vector space and $\sox(n)$ is locally free,
there is a non-canonical isomorphism $\F(n)=\F\tensor\sox(n)\isom\F$.
Thus $$\chi_{\F}(n)=h^0(\F(n))=h^0(\F)=\sum_{i=1}^{k}\dim_k\F_{p_i}$$
which is a constant function, hence a polynomial.
Next suppose $\dim(\supp\F)=s$.
Let $x\in{}S_1=H^0(\sox(1))$ be such that the hyperplane
$H:=\{x=0\}$ doesn't contain any irreducible component
of $\supp\F$. Multiplication by $x$ defines a map
$\sox(-1)\xrightarrow{x}\sox$ which is an isomorphism
outside of $H$. Tensoring with $\F$ gives a map
$\F(-1)\into\F$. Let $\sR$ be the kernel and
$\sQ$ be the cokernel, then there is an exact sequence
$$0\into\sR\into\F(-1)\xrightarrow{x}\F\into\sQ\into{}0.$$
Now $\supp\sR\union\supp\sQ\subset\supp\F\intersect H$ so
$\dim(\supp\sR)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ and
$\dim(\supp\sQ)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ so
by our induction hypothesis $\chi(\sQ(n))$ and $\chi(\sR(n))$
are polynomials. Twisting the above exact sequence by $n$ and
applying $\chi$ yields
$$\chi(\F(n))-\chi(\F(n-1))=\chi(\Q(n))-\chi(\sR(n))=P_{\sQ}(n)-P_{\sR}(N).$$
Thus the first difference function of $\chi(\F(n))$ is a polynomial
so $\chi(\F(n))$ is a polynomial.
\end{proof}
\begin{example}
Let $X=\P^1$ and $\F=\sox$. Then $S=k[x_0,x_1]$, $M=S$ and
$\dim S_n = n+1$. Thus $p_M(z)=z+1$ and $p_M(n)=\varphi(n)$
for $n\geq -1$. Computing the Hilbert polynomial in terms of
the Euler characteristic gives
$$\chi(\F(n))=h^0(\sox(n))- h^1(\sox(n))
= \begin{cases} (n+1) - 0 & n\geq -1\\
0-(-n-1)=n+1 & n\leq -2
\end{cases} $$
Thus $p_{\F}(n)=n+1$.
\end{example}
The higher cohomology corrects the failure of the
Hilbert polynomial in lower degrees.
%% 2/16/96
\section{Correspondence between Analytic and Algebraic Cohomology}
{\bf Homework. } Chapter III, 4.8, 4.9, 5.6.
Look at Serre's 1956 paper {\em Geometrie Algebraique et Geometrie
Analytique} (GAGA). ``What are the prerequisites?'' asks Janos.
``French,'' answers Nghi. ``Is there an English translation'' asks the
class. ``Translation? ... What for? It's so beautiful in the French,''
retorts Hartshorne.
Let $\F$ be a coherent sheaf on $\P^n_{\C}$ with its Zariski topology.
Then we can associate to $\F$ a sheaf
$\F^{\mbox{\rm an}}$
on $\P^n_{\C}$ with its analytic topology. $\F$ is locally a cokernel
of a morphism of free sheaves so we can define
$\F^{\mbox{\rm an}}$
by defining $\sox^{\mbox{\rm an}}$.
The map
$$\Coh(\P^n_{\C})\xrightarrow{\mbox{\rm an}}\Coh^{\mbox{\rm an}}(\P^n_{\C})$$
is an equivalence of categories and
$$H^i(X,\F)\iso{}H^i(X^{\mbox{\rm an}},\F^{\mbox{\rm an}})$$
for all $i$.
If $X/\C$ is affine the corresponding object $X^{\mbox{\rm an}}_{\C}$
is a Stein manifold.
\section{Arithmetic Genus}
Let $X\hookrightarrow{}\P_k^n$ be a projective variety with
$k$ algebraically closed and suppose $\F$ is a coherent sheaf
on $X$. Then $$\chi(\F)=\sum(-1)^i h^i(\F)$$ is the Euler characteristic
of $\F$, $$P_{\F}(n)=\chi(\F(n))$$ gives the Hilbert polynomial of
$\F$ on $X$, and $$p_a(X)=(-1)^{\dim X}(P_{\sox}(0)-1)$$ is the
arithmetic genus of $X$. The arithmetic genus is independent
of the choice of embedding of $X$ into $\P_k^n$.
If $X$ is a curve then $$1-p_a(X)=h^0(\sox)-h^1(\sox)$$.
Thus if $X$ is an integral projective curve then $h^0(\sox)=1$ so
$p_a(X)=h^1(\sox)$. If $X$ is a nonsingular projective curve
then $p_a(X)=h^1(\sox)$ is called {\bfseries the genus} of $X$.
Let $V_1$ and $V_2$ be varieties, thus they are projective integral
schemes over an algebraically closed field $k$. Then $V_1$ and
$V_2$ are {\bfseries birationally equivalent} if and only if $K(V_1)\isom{}K(V_2)$
over $k$, where $K(V_i)$ is the function field of $V_i$. $V$ is
{\bfseries rational} if $V$ is bironational to $\P_k^n$ for some $n$.
Since a rational map on a nonsingular projective curve always extends,
two nonsingular projective curves are birational if and only if
they are isomorphic. Thus for nonsingular projective curves
the genus $g$ is a birational invariant.
\subsection{The Genus of Plane Curve of Degree $d$}
Let $C\subset\P_k^2$ be a curve of degree $d$. Then $C$
is a closed subscheme defined by a single homogeneous polynomial
$f(x_0,x_1,x_2)$ of degree $d$, thus
$$C=\proj(S/(f)).$$
Some possibilities when $d=3$ are:
\begin{itemize}
\item $f: Y^2-X(X^2-1)$, a nonsingular elliptic curve
\item $f: Y^2-X^2(X-1)$, a nodal cubic
\item $f: Y^3$, a tripled $x$-axis
\item $f: Y(X^2+Y^2-1)$, the union of a circle and the $x$-axis
\end{itemize}
Now we compute $p_a(C)$. Let $I=(f)$ with $\deg f=d$.
Then $$1-p_a=h_0(\so_C)-h_1(\so_C)+h_2(\so_C)=\chi(\so_C).$$
We have an exact sequence
$$0\into\sI_C\into\so_{\P^2}\into\so_C\into 0.$$
Now $\sI_C\isom\so_{\P^2}(-d)$ since $\so_{\P^2}(-d)$
can be thought of as being generated by $1/f$ on $D_+(f)$
and by something else elsewhere, and then multiplication
by $f$ gives an inclusion
$so_{\P^2}(-d)|_{D_+(f)}\into\so_{\P^2}|_{D_+(f)}$, etc.
Therefore
$$\chi(\so_C)=\chi(\so_{\P^2})-\chi(\so_{\P^2}(-d)).$$
Now
$$\chi(\so_{\P^2})=h^0(\so_{\P^2})-h^1(\so_{\P^2})+h^2(\so_{\P^2})=1+0+0$$
and
$$\chi(\so_{\P^2}(-d))=h^0(\so_{\P^2}(-d))-h^1(\so_{\P^2}(-d))+h^2(\so_{\P^2}(-d))
=0+0+\frac{1}{2}(d-1)(d-2).$$
For the last computation we used duality (14.1) to see that
$$h^2(\so_{\P^2}(-d))=h^0(\so_{\P^2}(d-3)=\dim S_{d-3}
=\frac{1}{2}(d-1)(d-2).$$
Thus $\chi(\so_C)=1-\frac{1}{2}(d-1)(d-2)$ so
$$p_a(C)=\frac{1}{2}(d-1)(d-2).$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 2/21/96
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Not Enough Projectives}
\begin{exercise} Prove that the category of quasi-coherent
sheaves on $X=\P_k^1$ doesn't have enough projectives.
\end{exercise}
\begin{proof}
We show that there is no projective object $\sP\in\Qco(X)$ along
with a surjection $\sP\into\sox\into 0$.
\begin{lem}
If $\P\xrightarrow{\varphi}\sox$ is surjective and $\sP$ is
quasi-coherent, then there exists $\ell$ such that
$H^0(\P(\ell))\into{}H^0(\sox(\ell))$ is surjective.
\end{lem}
The {\em false} proof of this lemma is to write down an
exact sequence $0\into\sR\into\sP\into\sox\into 0$ then
use the ``fact'' that $H^1(\sR(\ell))=0$ for sufficiently
large $\ell$. This doesn't work because $\sR$ might not
be coherent since it is only the quotient of
quasi-coherent sheaves. A valid way to proceed is to use
(II, Ex. 5.15) to write $\sP$ as an ascending union of its coherent
subsheaves, $\sP=\union_{i}\sP_{i}$.
Then since $\varphi$ is surjective,
$\sox=\union_{i}\varphi(\sP_{i})$, where $\varphi(\sP_{i})$
is the sheaf image. Using the fact that $\varphi(\sP_{i})$ is
the sheaf image, that $\sox$ is coherent and that the union
is ascending, this implies $\sox=\varphi(\sP_i)$ for some $i$.
We now have an exact sequence
$$0\into\sR_i\into\sP_i\into\sox\into{}0$$
with $\sR_i$ coherent since $\sP_i$ and $\sox$ are both coherent.
Thus $H^i(\sR_i(\ell))=0$ for $l\gg 0$ which, upon computing
the long exact sequence of cohomology, gives the lemma.
Now fix such an $\ell$. We have a commutative diagram
$$\begin{CD}
\sP@>>>\sox@>>>0\\
@V{\exists}VV @VVV\\
\sox(-\ell-1)@>>>k(p)@>>>0
\end{CD}$$
Twisting by $\ell$ gives a commutative diagram
$$\begin{CD}
\sP(\ell)@>>>\sox(\ell)@>>>0\\
@VVV @VVV\\
\sox(-1)@>>>k(p)@>>>0
\end{CD}$$
Let $s\in\Gamma(\sox(\ell))$ be a global section which is nonzero at $p$,
then there is $t\in\Gamma(\sP(\ell))$ which maps to $s$.
But then by commutativity $t$ must map to some element of $\Gamma(\sox(-1))=0$
which maps to a nonzero element of $k(p)$, which is absurd.
\end{proof}
\section{Some Special Cases of Serre Duality}
\subsection{Example: $\sox$ on Projective Space}
Suppose $X=\P_k^n$, then there is a perfect pairing
$$H^0(\sox(\ell))\times{}H^n(\sox(-\ell-n-1))\into{}H^n(\sox(-n-1))\isom{}k.$$
For this section let
$$\omega_X=\sox(-n-1).$$
Because the pairing is perfect we have a non-canonical but functorial
isomorphism
$$H^0(\sox(\ell))\isom H^n(\sox(-\ell-n-1))'.$$
(If $V$ is a vector space then $V'$ denotes its dual.)
\subsection{Example: Coherent sheaf on Projective Space}
Suppose $\sF$ is any coherent sheaf on $X=\P_k^r$.
View $\Hom(\sF,\omega)$ as a $k$-vector space.
% WHY DEFINE SHEAF HOM???
%We recall the definition of the sheaf Hom.
%\begin{defn}
%Let $\sF$ and $\sG$ be sheaves of $\sox$-modules.
%Then $\sHom(\sF,\sG)$ is the sheaf which associates to
%an open set the group
%$$Hom_{\Mod(\sox|_U)}(\sF|_U,\sG|_U).$$
%\end{defn}
By functoriality and since $H^n(\omega)=k$ there is a map
$$\varphi:\Hom(\sF,\omega)\into\Hom(H^n(\sF),H^n(\omega))=H^n(\sF)'.$$
\begin{prop} $\varphi$ is an isomorphism for all coherent sheaves $\sF$.
\end{prop}
\begin{proof}
\par {\em Case 1.} If $\sF=\sox(\ell)$ for some $\ell\in\Z$ then this is
just a restatement of the previous example.
\par {\em Case 2.} If $\sE=\oplus_{i=1}^{k}\so(\ell_i)$ is a finite
direct sum, then the statement follows from the
commutativity of the following diagram.
$$\begin{CD}
\Hom(\oplus_{i=1}^{k}\so(\ell_i),\omega)@>>>H^n(\oplus_{i=1}^k\so(\ell_i))'\\
@VV\isom{}V @VV\isom{}V\\
\oplus_{i=1}^k\Hom(\so(\ell_i),\omega)@>>\sim{}>\oplus_{i=1}^k{}H^n(\so(\ell_i))'
\end{CD}$$
\par {\em Case 3.} Now let $\sF$ be an arbitrary coherent sheaf.
View $\varphi$ as a morphism of functors
$$\Hom(\cdot,\omega)\into H^n(\cdot)'.$$
The functor $\Hom(\cdot,\omega)$ is contravarient left exact.
$H^n(\cdot)$ is covariant right exact since $X=\P_k^n$ so
$H^{n+1}(\sF)=0$ for any coherent sheaf $\sF$. Thus $H^n(\cdot)'$
is contravarient left exact.
\begin{lem}
Let $\sF$ be any coherent sheaf. Then there exists a partial resolution
$$\sE_1\into\sE_0\into\sF\into{}0$$
by sheaves of the form $\oplus_{i}\sox(\ell_i)$.
\end{lem}
By (II, 5.17) for $\ell\gg 0$, $\sF(\ell)$ is
generated by its global sections. Thus there is a surjection
$$\sox^m\into\sF(\ell)\into 0$$
which upon twisting by $-\ell$ becomes
$$\sE_0=\sox(-\ell)^m\into\sF\into 0.$$
Let $\sR$ be the kernel so
$$0\into\sR\into\sE\into\sF$$
is exact. Since $\sR$ is coherent, we can repeat the argument
above to find $\sE_1$ surjecting onto $\sR$. This yields
the desired exact sequence.
Now we apply the functors $\Hom(\cdot,\omega)$ and
$H^n(\cdot)'$. This results in a commutative diagram
$$\begin{CD}
0@>>> \Hom(\sF,\omega) @>>> \Hom(\sE_0,\omega) @>>> \Hom(\sE_1,\omega)\\
@VVV @V\varphi(\sF)VV @V\varphi(\sE_0)VV @VV\varphi(\sE_1)V\\
0@>>> H^n(\sF)'@>>>H^n(\sE_0)' @>>> H^n(\sE_1)'
\end{CD}$$
From cases 1 and 2, the maps $\varphi(\sE_0)$ and
$\varphi(\sE_1)$ are isomorphisms so $\varphi(\sF)$ must also be
an isomorphism.
\end{proof}
\subsection{Example: Serre Duality on $\P_k^n$}
Let $X=\P_k^n$ and $\sF$ be a coherent sheaf.
Then for each $i$ there is an isomorphism
$$\varphi^{i}:\ext^{i}_{\sox}(\sF,\omega)\into H^{n-i}(\sF)'.$$
%%%%%%%%%%%%%%%%%%%%
%% 2/23/96
\section{The Functor $\ext$}
Let $(X,\sox)$ be a scheme and $\sF,\sG\in\Mod(\sox)$. Then
$\Hom(\sF,\sG)\in\Ab$. View $\Hom(\sF,\bullet)$ as a
functor $\Mod(\sox)\into\Ab$. Note that $\Hom(\sF,\bullet)$
is left exact and covariant. Since $\Mod(\sox)$ has enough
injectives we can take derived functors.
\begin{defn}
The $\ext$ functors $\ext^i_{\sox}(\sF,\bullet)$ are the right
derived functors of $\Hom_{\sox}(\sF,\bullet)$ in the
category $\Mod(\sox)$.
\end{defn}
Thus to compute $\ext^i_{\sox}(\sF,\sG)$, take an injective
resolution $$0\into\sG\into I^0\into I^1\into \cdots $$
then
$$\ext^i_{\sox}(\sF,\sG)=H^i(\Hom_{\sox(\sF,I^{\bullet})}).$$
\begin{remark} {\bfseries Warning!}
If $i:X\hookrightarrow\P^n$ is a closed subscheme of $P^n$ then
$\ext^i_{\sox}(\sF,\sG)$ need {\em not} equal
$\ext^i_{\P^n}(i_{*}(\sF),i_{*}(\sG))$. With cohomology
these are the same, but not with $\ext$!
\end{remark}
\begin{example}
Suppose $\sF=\sox$, then
$\Hom_{\sox}(\sox,\sG)=\Gamma(X,\sG)$. Thus
$\ext^i_{\sox}(\sox,\bullet)$ are the derived
functors of $\Gamma(X,\bullet)$ in $\Mod(\sox)$.
Since we can computer cohomology using flasque
sheaves this implies $\ext^i_{\sox}(\sox,\bullet)=H^i(X,\bullet)$.
Thus $\ext$ generalizes $H^i$ but we get a lot more besides.
\end{example}
\subsection{Sheaf $\ext$}
Now we define a new kind of $\ext$. The sheaf hom functor
$$\sHom_{\sox}(\sF,\bullet):\Mod(\sox)\into\Mod(\sox)$$
is covariant and left exact. Since $\Mod(\sox)$ has enough
injectives we can defined the derived functors
$\sext^i_{\sox}(\sF,\bullet)$.
\begin{example}
Consider the functor $\ext^i_{\sox}(\sox,\bullet)$.
Since $\sHom_{\sox}(\sox,\sG)=\sG$ this is the identity
functor which is exact so
$$\ext^i_{\sox}(\sox,\sG)=\begin{cases}\sG\quad i=0\\0\quad i>0\end{cases}$$
\end{example}
What if we have a short exact sequence in the first variables, do we
get a long exact sequence?
\begin{prop}
The functors $\ext^i$ and $\sext^i$ are $\delta$-functors
in the first variable. Thus if $$0\into\sF'\into\sF\into\sF''\into 0$$
is exact then there is a long exact sequence
\begin{align*}0\into&\Hom(\sF'',\sG)\into\Hom(\sF,\sG)\into\Hom(\sF',\sG)
\into&\ext^1(\sF'',\sG)\into\ext^1(\sF,\sG)\into\ext^1(\sF,\sG)\into\cdots
\end{align*}
\end{prop}
The conclusion of this proposition is not obvious because we
the $\ext^i$ as derived functors in the second variable, not the first.
\begin{proof}
Suppose we are given $0\into\sF'\into\sF\into\sF''\into 0$ and $\sG$.
Choose an injective resolution $0\into\sG\into{}I^{\bullet}$ of $\sG$.
Since $\Hom(\bullet,I^n)$ is exact (by definition of injective object),
the sequence
$$0\into\Hom(\sF'',I^{\bullet})\into\Hom(\sF,I^{\bullet})\into
\Hom(\sF',I^{\bullet})\into 0$$
is exact. By general homological algebra these give rise a long
exact sequence of cohomology of these complexes. For $\sext^i$
simply scriptify everything!
\end{proof}
\subsection{Locally Free Sheaves}
\begin{prop}
Suppose $\sE$ is a locally free $\sox$-module of finite rank.
Let $\sE^{\dual}=\Hom(\sE,\sO)$. For any sheaves $\sF$, $\sG$,
$$\ext^i(\sF\tensor\sE,\sG)\isom\ext^i(\sF,\sG\tensor\sE^{\dual})$$
and
$$\sext^i(\sF\tensor\sE,\sG)\isom\sext^i(\sF,\sG\tensor\sE^{\dual})
\isom\sext^i(\sF,\sG)\tensor\sE^{\dual}.$$
\end{prop}
\begin{lem}
If $\sE$ is locally free of finite rank and $\sI\in\Mod(\sox)$ is
injective then $\sI\tensor\sE$ is injective.
\end{lem}
\begin{proof}
Suppose $0\into\sF\into\sG$ is an injection and there is a map
$\varphi:\sF\into\sI\tensor\sE$. Tensor everything with $\sE^{\dual}$.
Then we have an injection $0\into\sF\tensor\sE^{\dual}\into\sG\tensor\sE^{\dual}$
and a map $\varphi':\sF\tensor\sE^{\dual}\into\sI$. Since $\sI$ is injective
there is a map $\sG\tensor\sE^{\dual}\into\sI$ which makes
the appropriate diagram commute. Tensoring everything with $\sE$ gives
a map making the original diagram commute.
\end{proof}
\begin{proof}[of proposition] Let $0\into\sG\into\sI^{\bullet}$
by an injective resolution of $\sG$. Since
$$\Hom(\sF\tensor\sE,\sI^{\bullet})=\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual}),$$
we see that
$$0\into\sG\tensor\sE^{\dual}\into\sI^{\cdot}\tensor\sE{\dual}$$
is an injective resolution of $\sG\tensor\sE^{\dual}$.
Thus $\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual})$ computes
$\ext(\sF\tensor\sE,\bullet)$.
\end{proof}
\begin{prop}
If $\sF$ has a locally free resolution
$\sE_{\cdot}\into\sF\into 0$ then
$$\sext^i_{\sox}(\sF,\sG)=H^i(\sHom(\sE_{\bullet},\sG)).$$
\end{prop}
\begin{remark}
Notice that when $i>0$ and $\sE$ is locally free,
$$\sext^i(\sE,\sG)=\sext^i(\sox,\sG\tensor\sE^{\dual})=0.$$
\end{remark}
\begin{proof}
Regard both sides as functors in $\sG$. The left hand side is a
$\delta$-functor and vanishes for $\sG$ injective. I claim that
that right hand side is also a $\delta$-functor and vanishes for
$\sG$-injective.
\begin{lem}
If $\sE$ is locally free, then $\shom(\sE,\bullet)$ is exact.
\end{lem}
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 2/26/96
\section{More Technical Results on $\ext$}
Let $(X,\sox)$ be a scheme and $\sF,\sG$ be sheaves
in the category $\Mod(\sox)$. Then $\ext(\sF,\sG)$
and $\sext(\sF,\sG)$ are the derived functors of $\Hom$, resp.
$\shom$, in the second variable.
\begin{lem}
If $\sF$ and $\sG$ are coherent over a Noetherian scheme
$X$, then $\sext^{i}(\sF,\sG)$ is coherent.
\end{lem}
This lemma would follow immediately from the following fact
which we haven't proved yet.
\begin{fact}
Let $X=\spec A$ with $A$ Noetherian and let $M$ be an $A$-module.
Then $$\ext_{\sox}^i(\tilde{M},\tilde{N})=\ext^i_{A}(M,N)$$ and
$$\sext_{\sox}^i(\tilde{M},\tilde{N})=(\ext^i_{A}(M,N))^{\tilde{}}.$$
\end{fact}
Instead of using the fact we can prove the lemma using
a proposition from yesterday.
\begin{proof}
Choose a locally free resolution
$$\sL_{\bullet}\into\sF\into 0$$
of $\sF$. Then
$$\sext^{i}(\sF,\sG)=H^i(\shom(\sL_{\bullet},\sG)).$$
But all of the kernels and cokernels in
$\shom(\sL_{\bullet},\sG)$
are coherent, so the cohomology is. (We can't
just choose an injective resolution of $\sF$ and apply
the definitions because there is no guarantee that we
can find an injective resolution by coherent sheaves.)
\end{proof}
\begin{prop}
Let $X$ be a Noetherian projective scheme over $k$ and let $\sF$
and $\sG$ be coherent on $X$. Then for each $i$ there exists
an $n_0$, depending on $i$, such that for all $n\geq{}n_0$,
$$\ext^i_{\sox}(\sF,\sG(n))=\Gamma(\sext^i_{\sox}(\sF,\sG(n))).$$
\end{prop}
\begin{proof}
When $i=0$ the assertion is that
$$\Hom(\sF,\sG(n))=\Gamma(\shom(\sF,\sG(n)))$$
which is obvious.
{\em Claim.} Both sides are $\delta$-functors in $\sF$. We
have already showed this for the left hand side. [I don't understand
why the right hand side is, but it is not trivial and it caused
much consternation with the audience.]
To show the functors are isomorphic we just need to show both
sides are coeffaceable. That is, for every coherent sheaf $\sF$
there is a coherent sheaf $\sE$ and a surjection $\sE\into\sF\into{}0$
such that $\ext^i_{\sox}(\sE)=0$ and similarly for the right hand
side. Thus every coherent sheaf is a quotient of an acyclic sheaf.
Suppose $\sF$ is coherent. Then for $\ell\gg 0$, $\sF(\ell)$ is generated
by its global sections, so there is a surjection
$$\sox^a\into\sF(\ell)\into{}0.$$
Untwisting gives a surjection
$$\sox(-\ell)^a\into\sF\into{}0.$$
Let $\sE=\sox(-\ell)^a$, then I claim that $\sE$ is acyclic for both
sides. First consider the left hand side. Then
\begin{align*}\ext^i(\oplus\so(-\ell),\sG(n))&=\oplus\ext^i(\sox(-\ell),\sG(n))\\
&=\oplus\ext^i(\sox,\sG(\ell+n))\\
&=H^i(X,\sG(\ell+n))\end{align*}
By Serre (theorem 5.2 of the book) this is zero for $n$ sufficiently large.
For the right hand side the statement is just that
$$\sext^i(\sE,\sG(n))=0$$
which we have already done since $\sE$ is a locally free sheaf.
Thus both functors are universal since they are coeffaceable. Since universal
functors are completely determined by their zeroth one they must be equal.
\end{proof}
\begin{example}
One might ask if $\ext^i$ necessarily vanishes for sufficiently large $i$.
The answer is no. Here is an algebraic example which can be converted
to a geometric example. Let $A=k[\varepsilon]/(\varepsilon^2)$, then
a projective resolution $L_{\bullet}$ of $k$ is
$$\cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A
\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}k\into 0.$$
Then $Hom(L_{\bullet},k)$ is the complex
$$k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}\cdots$$
Thus $\ext^i_A(k,k)=k$ for all $i\geq 0$.
\end{example}
\section{Serre Duality}
We are now done with technical results on $\ext$'s so we can
get back to Serre duality on $\P^n$.
Let $X=\P^n_k$ and let $\omega=\sox(-n-1)$. Note that this
is an {\em ad hoc} definition of $\omega$ which just happens
to work since $X=\P^n_k$. In the more general situation it
will be an interesting problem just to show the so called
dualizing sheaf $\omega$ actually exists. When our variety
is nonsingular, $\omega$ will be the canonical sheaf.
We have shown that for any coherent sheaf $\sF$ there is a map
$$\Hom(\sF,\omega)\iso H^n(\sF)^{\dual}.$$
The map is constructed by using the fact that $H^n$ is a
functor:
$$\Hom(\sF,\omega)\into\Hom_k(H^n(\sF),H^n(\omega))
= \Hom_k(H^n(\sF),k)=H^n(\sF)^{\dual}.$$
We shall use satellite functors to prove the following theorem.
\begin{thm} Let $\sF$ be a coherent sheaf on $\P^n_k$. Then
there is an isomorphism $$\ext^i(\sF,\omega)\iso H^{n-i}(\sF)^{\dual}.$$
\end{thm}
\begin{proof} Regard both sides as functors in $\sF$.
\par {\em 1. Both sides are $\delta$-functors in $\sF$.}
We have already checked this for $\ext^i$. Since $H^{n-i}$
is a delta functor in $\sF$, so is $(H^{n-i})^{\dual}$.
Note that both sides are contravarient.
\par {\em 2. They agree for $i=0$.} This was proved last time.
\par {\em 3. Now we just need to show both sides are coeffaceable.}
Suppose $\sE\into\sF\into 0$ with $\sE=\sO(-\ell)^{\oplus a}$.
For some reason we can assume $\ell\gg{}0$.
We just need to show both sides vanish
on this $\sE$. First computing the left hand side gives
$$\oplus\ext^i(\sO(-\ell),\omega)=H^i(\omega(\ell))=0$$
for $\ell\gg 0$. Next computing the right hand side we get
$$H^{n-i}(\sO(-\ell))=0$$
by the explicit computations of cohomology of projective
space (in particular, note that $i>0$).
\end{proof}
Next time we will generalize Serre duality to an arbitrary
projective scheme $X$ of dimension $n$.
We will proceed in two steps. The first is to ask,
what is $\omega_X$? Although the answer to this question
is easy on $\P_k^n$ it is not obvious what the suitable
analogy should be for an arbitrary projective variety.
Second we will define natural maps
$$\ext^i_{\sox}(\sF,\omega)\xrightarrow{\varphi^i}H^{n-i}(\sF)^{\dual}$$
where $n=\dim X$.
Unlike in the case when $X=\P^n_k$, these maps are not necessarily
isomorphisms unless $X$ is locally Cohen-Macaulay (the local rings
at each point are Cohen-Macaulay).
\begin{defn}
Let $A$ be a nonzero Noetherian local ring with residue field $k$.
Then the {\bfseries depth} of $A$ is
$$\operatorname{depth} A = \inf\{i:\ext^i_A(k,A)\neq 0\}.$$
$A$ is said to be {\bfseries Cohen-Macaulay} if
$\operatorname{depth} A = \dim A$.
\end{defn}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 2/28/96
\section{Serre Duality for Arbitrary Projective Schemes}
Today we will talk about Serre duality for an arbitrary projective
scheme. We have already talked about Serre duality in the special
case $X=\P^n_k$. Let $\sF$ be a locally free sheaf. We showed
there is an isomorphism
$$\ext^i(\sF,\omega_X)\iso H^{n-i}(\sF)^{\dual}.$$
This was established by noting that
$$\ext^i(\sF,\omega_X)=\ext^i(\sox,\sF^{\dual}\tensor\omega_X)
=H^i(\sF^{\dual}\tensor\omega_X).$$
Another thing to keep in mind is that locally free sheaves correspond
to what, in other branches of mathematics, are vector bundles. They
aren't the same object, but there is a correspondence.
We would like to generalize this to an arbitrary projective scheme $X$.
There are two things we must do.
\begin{enumerate}
\item Figure out what $\omega_X$ is.
\item Prove a suitable duality theorem.
\end{enumerate}
When $X=\P_k^n$ it is easy to find a suitable $\omega_X=\so_{\P^n_k}(-n-1)$
because of the explicit computations we did before. We now define $\omega_X$
to be a sheaf which will do what we hope it will do. Of course existence
is another matter.
\begin{defn}
Let $X$ be a Noetherian scheme of finite type over a field $k$ and
let $n=\dim X$. Then a {\bfseries dualizing sheaf} for $X$ is
a coherent sheaf $\omega_X$ along with a map
$t:H^n(\omega_X)\into{}k$, such that for all coherent sheaves $\sF$ on
$X$, the map
$\Hom_{\sox}(\sF,\ox)\into{}H^n(\sF)^{\dual}$ is an
isomorphism. The latter map is defined by the diagram
$$\begin{array}{ccc}
\Hom_{\sox}(\sF,\ox)\\
\downarrow\\
\Hom(H^n(\sF),H^n(\ox))&\xrightarrow{t}&\Hom(H^n(\sF),k)=H^n(\sF)^{\dual}
\end{array}$$
\end{defn}
Strictly speaking, a dualizing sheaf is a pair $(\omega_X,t)$. Note
that on $\P^n$ we had $H^n(\omega_{\P^n})\isom{}k$, but on an
arbitrary scheme $X$ we only have a map from $H^n(\omega_X)$ to $k$
which need not be an isomorphism. The definition never mentions existence.
\begin{prop}
If $X$ admits a dualizing sheaf $(\ox,t)$ then the pair
$(\ox,t)$ is unique up to unique isomorphism, i.e., if
$(\eta,s)$ is another dualizing sheaf for $X$ then there
is a unique isomorphism $\varphi:\ox\into\eta$ such that
$$\begin{array}{ccc}
H^n(\ox)&\xrightarrow{H^n(\varphi)}&H^n(\eta)\\
t\downarrow&&\downarrow{}s\\
k&=&k\end{array}$$
commutes.
\end{prop}
Before we prove the proposition we make a short digression to
introduce representable functors which give a proof of the uniqueness
part of the above proposition.
\begin{defn}
Let $\sC$ be a category and $\sD$ a category whose objects happen
to be sets. Suppose $T:\sC\into\sD$ is a contravarient functor.
Then $T$ is {\bfseries representable} if there exists an object
$\omega\in\Ob(\sC)$ and an element $t\in T(\omega)$ such that
for all $F\in\Ob(\sC)$ the map $\Hom_{\sC}(F,\omega)\into{}T(F)$
is a bijection of sets. The latter map is defined by the diagram
$$\begin{array}{ccc}
\Hom_{\sC}(F,\omega)&\xrightarrow{\text{bijection of sets}}&T(F)\\
\searrow&&\nearrow\text{ evaluation at $t$}\\
&\Hom_{\sD}(\Gamma(\omega),T(F))\end{array}$$
\end{defn}
Thus there is an isomorphism of functors $\Hom(\bullet,\omega)=T(\bullet)$.
The pair $(t,\omega)$ is said to represent the functor $T$.
The relevant application of this definition is to the case
when $\sC=\Coh(X)$, $\sD=\{\text{ $k$-vector spaces}\}$,
$T$ is the functor $F\mapsto H^n(\sF)^{\dual}$. Then $\omega=\ox$
and
$$t=t\in\Hom(H^n(\omega),k)=H^n(\omega)^{\dual}=T(\omega).$$
\begin{prop}
If $T$ is a representable functor, then the pair
$(\omega,t)$ representing it is unique.
\end{prop}
\begin{proof}
Suppose $(\omega,t)$ and $(\eta,s)$ both represent the functor $T$.
Consider the diagram
$$\begin{array}{cccl}
\Hom(\eta,\omega)&\xrightarrow{T}&\Hom(T(\omega),T(\eta))\\
\searrow&&\swarrow\text{ eval. at $t$}\\
&T(\eta)\end{array}$$
By definition the map $\Hom(\eta,\omega)\into{}T(\eta)$ is
bijective. Since $s\in T(\eta)$, there is $\varphi\in\Hom(\eta,\omega)$
such that $\varphi\mapsto{}s\in{}T(\eta)$. Thus $\varphi$ has
the property that $T(\varphi)(t)=s$. This argument uses the
fact that $(\omega,t)$ represents $T$. Using the fact that
$(\eta,s)$ represents $T$ implies that there exists
$\psi\in\Hom(\omega,\eta)$ such that $T(\psi)(s)=t$.
We have the following pictures
$$\begin{array}{ccc}
&\xrightarrow{\hspace{.2in}\psi\hspace{.2in}}\\
\omega&&\eta\\
&\xleftarrow{\hspace{.2in}\phi\hspace{.2in}}
\end{array}$$
$$\begin{array}{ccc}
&\xleftarrow{\hspace{.2in}T(\psi)=\psi^{*}\hspace{.2in}}\\
t\in{} T(\omega)&&T(\eta) \ni{} s\\
&\xrightarrow{\hspace{.2in}T(\phi)=\phi^{*}\hspace{.2in}}
\end{array}$$
I claim that
$$\psi\circ\varphi=\Id\in\Hom(\eta,\eta).$$
In diagram form we have
$$\eta\xrightarrow{\varphi}\omega\xrightarrow{\psi}\eta$$
which upon applying $T$ gives
\begin{align*}
T(\eta)&\xrightarrow{\psi^{*}}T(\omega)\xrightarrow{\varphi^{*}}T(\eta)\\
s&\mapsto t\mapsto s\end{align*}
Where does $\psi\circ\varphi$ go to
under the map $\Hom(\eta,\eta)\iso T(\eta)$? By definition
$\psi\circ\varphi$ goes to the evaluation of $T(\psi\circ\varphi)$
at $s\in T(\eta)$. But, as indicated above, the evaluation of
$T(\psi\circ\varphi)$ at $s$ is just $s$ again. But the identity
morphism $1_{\eta}\in\Hom(\eta,\eta)$ also maps to $s$
under the map $\Hom(\eta,\eta)\iso T(\eta)$. Since this
map is a bijection this implies that $\psi\circ\varphi=1_{\eta}$,
as desired. Similarly $\varphi\circ\psi=1_{\omega}$. Thus
$\psi$ and $\varphi$ are both isomorphisms.
\end{proof}
\begin{quote}
``When you define something and it is unique up to unique isomorphism,
you know it must be good.''
\end{quote}
We return to the question of existence.
\begin{prop}
If $X$ is a projective scheme over a field $k$ then $(\ox,t)$ exists.
\end{prop}
\begin{lem}
If $X$ is an $n$ dimensional projective scheme over a field $k$, then
there is a finite morphism $f:X\into\P^n_k$.
\end{lem}
\begin{proof}
Embed $X$ in $\P^N$ then choose a linear projection down to
$\P^n$ which is sufficiently general.
$$\begin{array}{ccc}
X&\hookrightarrow&\P^N\\
f\searrow&&\downarrow\\
&&\P^n\end{array}$$
Let $L$ be a linear space of dimension $N-n-1$ not meeting $X$.
Let the map from $\P^N\into\P^n$ be projection through $L$.
By construction $f$ is quasi-finite, i.e., for all $Q\in\P^n$,
$f^{-1}(Q)$ is finite. It is a standard QUALIFYING EXAM problem to
show that if a morphism is quasi-finite and projective then it is
finite. This can be done by applying (II, Ex. 4.6) by covering
$X$ by subtracting off hyperplanes and noting that the correct
things are affine by construction. See also (III, Ex. 11.2) for
the more general case when $f$ is quasi-finite and proper, but not
necessarily projective.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 3/1/96
%%%%%%%%%%%%%%%%%%%%%%%%
\section{Existence of the Dualizing Sheaf on a Projective Scheme}
Let $X$ be a scheme over $k$. Recall that a
dualizing sheaf is a pair $(\omega,t)$ where $\omega$ is a coherent
sheaf on $X$ and
$$t:H^n(X,\omega)\into k$$
is a homomorphism such that for all coherent sheaves $\sF$
the natural map
$$\Hom_X(\sF,\omega)\into{}H^n(\sF)^{\dual}$$
is an isomorphism.
We know that such a dualizing sheaf exists on $\P_k^n$.
\begin{thm}
If $X$ is a projective scheme of dimension $n$ over $k$, then
$X$ has a dualizing sheaf.
\end{thm}
The book's proof takes an embedding
$j:X\hookrightarrow{}\P_k^N$ and works on $X$
as a subscheme of $\P_k^N$. Then the book's proof shows that
$$\omega_X=\sext_{\sO_{\P^N_k}}^{N-n}(\sox,\omega_{\P^N_k}).$$
Today we will use a different method.
\begin{defn}
A {\bfseries finite morphism} is a morphism $f:X\into Y$ of
Noetherian schemes such that for any open affine $U=\spec A\subset X$,
the preimage $f^{-1}(U)\subset Y$ is affine, say $f^{-1}(U)=\spec B$,
and the natural map $A\into B$ turns $B$ into a finitely
generated $A$-module. We call $f$ an {\bfseries affine morphism}
if we just require that $f^{-1}(U)$ is affine
but not that $B$ is a finitely generated $A$-module.
A morphism $f:X\into{}Y$ is {\bfseries quasi-finite} if for all
$y\in Y$ the set $f^{-1}(y)$ is finite.
\end{defn}
\begin{example}
Consider the morphism
$$j:\P^1-\{\pt\}\injects\P^1.$$
Since $\P^1$ minus any nonempty finite set of points is
affine $j$ is affine. But it is not finite. Indeed, let
$a$ be a point different from $\pt$ and let $U=\P^1-\{a\}$.
Then $U=\spec k[x]$ and
$$j^{-1}(U)=\P^1-\{\pt,a\}=\spec k[x,x^{-1}],$$
but $k[x,x^{-1}]$ is not a finitely generated $k[x]$-module.
\end{example}
\begin{exercise}
A morphism can be affine but not finite or even quasi-finite.
For example, let $f$ be the natural map
$$f:\A^{n+1}-\{0\}\into\P^n$$
then show that $f$ is affine.
This is the fiber bundle associated to the invertible
sheaf $\sO(1)$ [[or is it $\sO(-1)$?]]
\end{exercise}
\subsection{Relative Gamma and Twiddle}
We will now define relative versions of global sections and
$\tilde{}$ analogous to the absolute versions.
It is not a generalization of the absolute notion, but a
relativization.
Suppose $X$ is a scheme over $Y$ with structure
map $f:X\into{}Y$ and assume $f$ is affine.
Then the map sending a sheaf
$\sF$ on $X$ to the sheaf $f_{*}\sF$ on
$Y$ is the analog of taking global sections.
Since $f$ is a morphism there is a map $\soy\into{}f_{*}\sox$
so $f_{*}\sox$ is a sheaf of $\soy$-modules. Note that
$f_{*}\sF$ is a sheaf of $f_{*}\sox$-modules. Thus
we have set up a map
$$\Qco(X)\into\{\text{ quasicoherent $f_{*}\sox$-modules on $Y$ }\}.$$
The next natural thing to do is define a map analogous to
$\tilde{}$ which goes the other direction.
Suppose $\sG$ is a quasi coherent sheaf of $f_{*}\sox$-modules on $Y$.
Let $U\subset Y$ be an affine open subset of $Y$. Let $G=\Gamma(U,\sG)$ and
write $U=\spec A$. Then since $f$ is an affine morphism,
$f^{-1}(U)=\spec B$ where $B=\Gamma(f^{-1}(U),\sox)$.
Since $\sG$ is an $f_{*}\sox$-module, and $f_{*}\sox$ over $U$
is just $B$ thought of as an $A$-module, we see that $G$ is a $B$-module.
Thus we can form the sheaf $\tilde{G}$ on $\spec B=f^{-1}(U)$.
Patching the various sheaves $\tilde{G}$ together as $U$ runs through
an affine open cover of $Y$ gives a sheaf $\tilde{\sG}$ in
$\Qco(X)$.
Let $\sG$ be a quasi-coherent sheaf of $\soy$-modules. We can't
take $\tilde{}$ of $\sG$ because $\sG$ might not be a sheaf of
$f_{*}\sox$-modules. But
$\shom_{\soy}(f_{*}\sox,\sG)$
is a sheaf of $f_{*}\sox$-modules, so we can
form $(\shom_{\soy}(f_{*}\sox,\sG))^{\tilde{}}$. This
is a quasi-coherent sheaf on $X$ which we denote $f^{!}(\sG)$.
\begin{prop}
Suppose $f:X\into{}Y$ is an affine morphism of Noetherian
schemes, $\sF$ is coherent on $X$, and $\sG$ is quasi-coherent on $Y$.
Then
$$f_{*}\shom_{\sox}(\sF,f^{!}\sG)\isom \shom_{\soy}(f_{*}\sF,\sG)$$
and passing to global sections gives an isomorphism
$$\Hom(\sF,f^{!}\sG)\isom \Hom(f_{*}\sF,\sG).$$
Thus $f^{!}$ is a right adjoint for $f_{*}$.
\end{prop}
\begin{proof}
The {\em natural} map is
\begin{align*}f_{*}\shom_{\sox}(\sF,f^{!}\sG)&\into
\shom_{\soy}(f_{*}\sF,f_{*}f^{!}\sG)\\
&=\shom_{\soy}(f_{*}\sF,\shom_{\soy}(f_*\sox,\sG)) \into
\shom_{\soy}(f_{*}\sF,\sG)\end{align*}
where the map $\shom_{\soy}(f_{*}\sox,\sG)\into\sG$
is obtained obtained by evaluation at $1$.
Since the question is local we may assume
$Y=\spec A$ and $X=\spec B$. Then $\sF$ corresponds
to a finitely generated module $M$ over the Noetherian ring $B$
and $\sG$ corresponds to a module $N$ over $A$.
We must show that
$$\Hom_B(M,\Hom_A(B,N))\isom\Hom_A(M,N).$$
When $M$ is free over $B$ so that $M=B^{\oplus{}n}$ the equality holds.
As functors in $M$, both sides are contravarient and
left exact. Now suppose $M$ is an arbitrary finitely generated
$B$-module. Write $M$ as a
quotient $F_0/F_1$ where $F_0$ and $F_1$ are both free of
finite rank. Applying each of the contravarient left-exact functors to the
exact sequence
$$F_1 \into F_0 \into M\into 0$$
and using the fact that equality holds for finite free modules
yields a diagram
$$\begin{array}{ccccccc}
0&\into&\Hom_B(M,\Hom_A(B,N))&\into&\Hom_B(F_0,\Hom_A(B,N))&
\into&\Hom_B(F_1,\Hom_A(B,N))\\
& & & & || & & ||\\
0&\into&\Hom_A(M,N)&\into&\Hom_A(F_0,N)&\into&\Hom_A(F_1,N)\end{array}$$
The $5$-lemma then yields an isomorphism
$$\Hom_B(M,\Hom_A(B,N))\isom\Hom_A(M,N).$$
\end{proof}
\begin{lem}
Suppose $f:X\into Y$ is affine and $\sF$ is a quasi-coherent
sheaf on $X$. Then
$$H^i(X,\sF)\isom{}H^i(Y,f_{*}\sF).$$
\end{lem}
\begin{proof}
The lemma is proved using \cech{} cohomology. If $\{U_i\}$ is
an open affine cover of $Y$ then $\{f^{-1}(U_i)\}$ is an
open affine cover of $X$. But
$$\Gamma(U_i,f_{*}\sF)=\Gamma(f^{-1}(U_i),\sF)$$
so the \cech{} cohomology of $f_{*}\sF$ on $Y$ is the
same as the \cech{} cohomology of $\sF$ on $X$.
\end{proof}
\begin{thm}[Duality for a finite flat morphism]
Suppose $f:X\into{}Y$ is a finite morphism with $X$ and $Y$ Noetherian
and assume every coherent sheaf on $X$ is the quotient of a locally
free sheaf (this is true for almost every scheme arising naturally
in this course).
Assume that $f_{*}\sox$ is a locally free $\soy$-module. Let
$\sF$ be a coherent sheaf on $X$ and $\sG$ be a quasi-coherent sheaf on $Y$.
Then for all $i\geq 0$ there is an isomorphism
$$\ext^i_{\sox}(\sF,f^{!}\sG)\iso\ext^i_{\soy}(f_{*}\sF,\sG).$$
\end{thm}
\begin{proof}
The proposition shows that the theorem is true when $i=0$.
We next show that the statement is true when $\sF=\sox$.
Applying the above lemma we see that
\begin{align*}
\ext_{\sox}^i(\sox,f^{!}\sG)&=H^i(X,f^{!}\sG)=H^i(Y,f_*f^{!}\sG)\\
&=H^i(Y,\shom_{\soy}(f_{*}\sox,\sG))
\end{align*}
Since $f_{*}\sox$ is a locally free $\soy$-module,
\begin{align*}
\ext_{\soy}^i(f_{*}\sox,\sG)&=\ext_{\soy}^i(\soy,(f_{*}\sox)^{\dual}\tensor\sG)\\
&=H^i(Y,(f_{*}\sox)^{\dual}\tensor\sG))\end{align*}
Putting these two computations together by using the fact
that
$$(f_*\sox)^{\dual}\tensor\sG=\shom_{\soy}(f_{*}\sox,\sG)$$
then gives the desired result.
A clever application of the $5$-lemma can be used to obtain
the general case. This will be done in a subsequent lecture.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 3/4/96
\section{Generalized Grothendieck Duality Theory}
If $X$ is a projective scheme over $k$, then
$$\ext^i(\sF,\omega_X)\iso{}H^{n-i}(\sF)^{\dual}.$$
This is a special type of duality.
If $X$ is an affine scheme over $Y$ with structure morphism
$f:X\into Y$, then
$$\ext_{X}^i(\sF,f^{!}\sG)\iso \ext^i_{Y}(f_{*}\sF,\sG).$$
This is another duality.
At Harvard, Hartshorne was the scribe [Lecture Notes in Math
Vol ???] for Grothendieck's seminar on his duality theory.
Suppose $X$ is proper over $Y$ with structure
morphism $f:X\into Y$. Assume furthermore that $f$ satisfies
hypothesis (*):
\begin{center}
(*)\quad $f_*\sox$ is a locally free $\soy$-module
\end{center}
Note that hypothesis start implies $\sHom(f_*\sox,\sG)$ is an exact
functor in $\sG$.
We have defined functors
$f_*:\Coh(X)\into\Coh(Y)$ and $f^{!}:\Qco(Y)\into\Qco(X)$.
Grothendieck's duality compares
$\Hom_X(\sF,f^{!}\sG)$ and its derived functors to
$\Hom_Y(f_{*}\sF,\sG)$ and its derived functors. The two
families of functors are essentially equal.
We are composing functors here. This often gives rise
to spectral sequences.
It is possible to obtain the duality mentioned above as a
special case of the more general Grothendieck duality.
Suppose $X$ is a projective scheme over $Y=\spec k$ with
morphism $f:X\into Y$. Then $f_*(\sF)=\Gamma(X,\sF)$
which has derived functors $H^i(X,\bullet)$. A coherent sheaf
on $Y$ is a finite dimensional $k$-vector space. Let
$\omega_X$ be the sheaf $f^{!}(k)$. Substituting this into
$$\ext^i_X(\sF,f^{!}\sG)\iso\ext^i_Y(f_*\sF,\sG)$$
with $G=k$ gives
$$\ext^i_X(\sF,\omega_X)\iso\ext^i_Y(f_*\sF,k)=H^{n-i}(\sF)^{\dual}$$
[[Do we obtain $H^{n-i}$ instead of $H^i$ since we are considering
the derived functors of $\Hom(f^*(\bullet),\bullet)$?]]
\section{}
\begin{thm}[Duality for a projective scheme] Suppose $X$ is a
projective scheme. Then $X$ has a dualizing
sheaf $\omega_X$ and their is an isomorphism
$$\ext^i_X(\sF,\omega_X)\isom H^{n-i}(X,\sF)^{\dual}.$$
\end{thm}
\begin{proof}
Using linear projections find a finite morphism
$f:X\into\P^n$. Let $\omega_X=f^!\omega_{\P^n}$ where
$$f^!\omega_{\P^n}=\sHom_{\P^n}(f_*\sox,\omega_{\P^n})^{\tilde{}}$$
and $\omega_{\P^n}=\sO_{\P^n}(-n-1)$. Then for any $\sF$,
\begin{align*}
\Hom_X(\sF,\omega_X)&=\Hom_{\P^n}(f_*\sF,\omega_{\P^n})\\
&\iso{}H^n(\P^n,f_*\sF)^{\dual}=H^n(X,\sF)^{\dual}\\
\end{align*}
The last equality holds since $f$ is finite and hence
affine. The second isomorphism comes from the fact that $\omega_X$
is a dualizing sheaf for $X$.
Next we obtain the duality theorem for $X$.
By duality for a finite (*) morphism
$$\ext_X^i(\sF,\omega_X)\isom\ext^i_{\P^n}(f_*\sF,\omega_{\P^n}).$$
By Serre duality on $\P^n$
$$\ext_{\P^n}^i(f_*\sF,\omega_{\P^n})\iso H^{n-i}(\P^n,f_{*}\sF)^{\dual}.$$
But $f$ is affine so
$$H^{n-i}(\P^n,f_*\sF)^{\dual}=H^{n-i}(X,\sF)^{\dual}.$$
Thus
$$\ext_X^i(\sF,\omega_X)\isom H^{n-i}(X,\sF)^{\dual}.$$
\end{proof}
Under what conditions does a finite map
$f:X\into\P^n$ satisfy (*)?
\begin{defn}
A scheme $X$ is {\bfseries Cohen-Macaulay} if for all $x\in X$ the local
ring $\sO_{X,x}$ is Cohen-Macaulay.
\end{defn}
\begin{defn}
The {\bfseries homological dimension} of a module $M$ over a ring
$A$ is the minimum possible length of a projective resolution of
$M$ in the category of $A$-modules. We denote this number by $\hd M$.
\end{defn}
We will need the following theorem from pure algebra.
\begin{thm}
If $A$ is a regular local ring and $M$ a finitely generated $A$-module,
then
$$\hd M+\depth M=\dim A.$$
\end{thm}
\begin{thm}
Suppose $f:X\into Y$ is a finite morphism of Noetherian schemes and
assume $Y$ is nonsingular.
Then $f$ satisfies (*) iff $X$ is Cohen-Macaulay.
\end{thm}
\begin{proof}
The question is local on $Y$.
Suppose $y\in Y$, then $A=\sO_{Y,y}$ is a regular local ring and
$f^{-1}(y)\subset X$ is a finite set. Let $B$ be the
semi local ring of $f^{-1}(y)$. Thus $B=\varinjlim_{U}\sF(U)$
where the injective limit is taken
over all open sets $U$ containing $f^{-1}(y)$.
$B$ has only finitely many maximal ideals so we call $B$ semi local.
Since $f$ is a finite morphism $B$ is a finite $A$-module.
Now $B$ is free as an $A$-module iff
$\hd_AB=0$, where $\hd_AB$ denotes the homological
dimension of $B$ as an $A$-module. This is clear because $\hd_AB$
is the shortest possible length of a free resolution of $B$.
(Since $A$ is local we need only consider free resolutions and not
the more general projective resolutions.)
Now $(f_*\sox)_y=B$ so condition (*) is that $B$ is a free
$A$-module. Thus if $f$ satisfies (*) then $\hd B=0$.
By the theorem from pure algebra and the fact that $f$ is finite we see that
$$\depth B=\dim A=\dim B.$$
This implies $B$ is a Cohen-Macaulay ring and therefore
$X$ is Cohen-Macaulay.
Conversely, if $X$ is Cohen-Macaulay then $B$ is Cohen-Macaulay
so $\depth B=\dim B$. The purely algebraic theorem then implies
that $\hd B=0$ so $B$ is a free $A$-module and hence $f$ satisfies
condition (*).
[[We are tacitly assuming that $B$ is Cohen-Macaulay iff
it's localizations at maximal ideals are. It would
be nice to know this is true.]]
\end{proof}
Now we finish up the proof of duality for a finite morphism.
\begin{proof}[Proof (of duality, continued)]
Suppose $f:X\into Y$ is a finite morphism of Noetherian schemes which
satisfies (*), and assume that $X$ has enough locally free sheaves
(i.e. every coherent sheaf is a quotient of a locally free sheaf).
We showed that
$$\Hom_X(\sF,f^{!}\sG)\isom\Hom_Y(f_*\sF,\sG).$$
This is the $i=0$ case of the theorem and is true even if we
drop the assumption that $f$ satisfies (*).
It can be shown by taking an injective resolution of $f^!\sG$
and computing $\ext^i_X$ using it that there are natural maps
$$\varphi^i:\ext_X^i(\sF,f^!\sG)\into\ext^i_Y(f_*\sF,\sG).$$
We have already shown that this is an isomorphism when $sF=\sox$.
By the same argument one shows that this is an isomorphism
when $\sF$ is just locally free. The key point to notice is
that if $\sF$ is locally free then $\sF$ is locally
free over $f_*\sox$ which is locally free over $\soy$ by
condition (*).
Finally suppose $\sF$ is an arbitrary coherent sheaf on $X$.
Writing $\sF$ as a quotient of of a locally free sheaf $\sE$
gives an exact sequence
$$0\into\sR\into\sE\into\sF\into 0$$
with $\sR$ coherent.
The long exact sequence of $\ext$'s gives a diagram
$$\begin{array}{ccccccccc}
\Hom_X(\sE,f^!\sG)&\into&\Hom_X(\sR,f^!\sG)&\into&\ext^1_X(\sF,f^!\sG)
&\into&\ext^1(\sE,f^!\sG)&\into&\ext^1_X(\sR,f^!\sG)\\
\downarrow \isom&&\downarrow \isom&&\downarrow\text{??}&&\downarrow\isom
&&\downarrow ??\\
\Hom_X(f_*\sE,\sG)&\into&\Hom_X(f_*\sR,\sG)&\into&\ext^1_X(f_*\sF,\sG)
&\into&\ext^1(f_*\sE,\sG)&\into&\ext^1_X(f_*\sR,\sG)
\end{array}
$$
Now apply the {\em subtle} 5-lemma to show that the
map $$\varphi^1:\ext^1_X(\sF,f^!\sG)\into\ext^1_X(f_*\sF,\sG)$$
is injective. Another diagram chase shows that he map
$$\ext^1_X(\sR,f^!\sG)\into\ext^1_X(f_*\sR,\sG)$$
is injective. Then [[I guess??]] the 5-lemma shows that
$\varphi^i$ is an isomorphism. Climbing the sequence inductively shows
that $\varphi^i$ is an isomorphism for all $i$.
\end{proof}
\begin{exercise}
Give an example of a scheme $X$ which does not have enough
locally free sheaves.
[Hints: By (III, 6.8) $X$ has enough locally free sheaves if
$X$ is quasi-projective or nonsingular.
Hartshorne intimated that this problem is {\em hard}, but suggested
one might search for a counterexample by looking for an appropriate
non-projective $3$-fold with $2$ singular points.]
\end{exercise}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 3/7/96
\section{Review of Differentials}
\begin{defn}
Let $B$ be a ring and $M$ a $B$-module. Then
a map $d:B\into M$ is a {\bfseries derivation} if
\begin{align*}
d(b_1+b_2)&=db_1+db_2\\
d(b_1b_2)&=b_1db_2+b_2db_1\end{align*}
If $B$ is an $A$-algebra, then $d$ is a {\bfseries derivation
over $A$} if in addition $da=0$ for all $a\in A$.
\end{defn}
Note that a derivation over $A$ is linear since
$d(ab)=adb+bda=adb+0=adb.$
\begin{example}
Let $k$ be a field and let $B=k[x]$. Let
$d:B\into B$ be the differentian map $f(x)\mapsto f'(x)$.
Then $d$ is a derivation.
\end{example}
\begin{defn}
Let $B$ be an $A$-algebra. Then the {\bfseries module of differentials}
of $B$ over $A$ is a pair $(\diff,d:B\into\diff)$, where
$\diff$ is a $B$-module and $d$ is a derivation of $B$ into
$\diff$ over $A$, which satisfies the following universal
property: if $d':B\into M$ is a derivation over $A$ then
there exists a unique $B$-linear map $\varphi:\diff\into M$
such that $d'=\varphi\circ d$.
\end{defn}
If $(\diff,d)$ exists it is unique as a pair up to unique isomorphism.
We first construct $\diff$ by brute force.
Let $\diff$ be the free module on symbols $db$ (all $b\in B$)
modulo the submodule generated by the relations
$d(b_1+b_2)-db_1-db_2$, $d(b_1b_2)-b_2db_1-b_1db_2$, and
$da$ for all $b_1,b_2\in B$ in $a\in A$.
This is obviously a module of differentials.
\begin{example}
Suppose $A=B$, then $\diff=0$.
\end{example}
\begin{example}
Suppose $k$ is a field of characteristic $p>0$. Let $B=k[x]$
and let $A=k[x^p]$. Then $\diff$ is the free $B$-module of rank
$1$ generated by $dx$.
\end{example}
\begin{example}
Let $B=\Q[\sqrt{2}]$ and $A=\Q$. Then
$0=d(2)=d(\sqrt{2}\sqrt{2})=2\sqrt{2}d(\sqrt{2})$ so
$d(\sqrt{2})=0$. Thus $\Omega_{\Q[\sqrt{2}]/\Q}=0$.
\end{example}
\begin{cor}
The module of differentials $\diff$ is generated
by $\{db : b\in B, b\not\in A\}$.
\end{cor}
Now we will construct $\diff$ in a more eloquent manner.
Suppose $B$ is an $A$-algebra. Consider the exact sequence of $A$-modules
$$0\into I\into B\tensor_{A} B\xrightarrow{\Delta} B\into 0$$
where $\Delta$ is the diagonal map $b_1\tensor b_2\mapsto b_1b_2$
and $I$ is the kernel of $\Delta$.
Make $I/I^2$ into a $B$-module by letting $B$ act on the first
factor (thus $b(x\tensor y)=(bx)\tensor y$) and define
a map $d:B\into I/I^2$ by $db=1\tensor b-b\tensor 1$.
\begin{prop}
The module $I/I^2$ along with the map $d$ is the module
of differentials for $B$ over $A$.
\end{prop}
\begin{proof}
Suppose $b_1,b_2\in B$, then
$$d(b_1b_2)=1\tensor b_1b_2 - b_1b_2\tensor 1.$$
One the other hand,
\begin{align*}
b_1db_2+b_2db_1&=b_1(1\tensor b_2-b_2\tensor 1)+b_2(1\tensor b_1-b_1\tensor 1)\\
&=b_1\tensor b_2-b_1b_2\tensor 1+b_2\tensor b_1-b_1b_2\tensor 1
\end{align*}
Now taking the difference gives
\begin{align*}
1\tensor b_1b_2- b_1b_2\tensor 1 - b_1\tensor b_2+b_1b_2\tensor 1
-b_2\tensor b_1 + b_1b_2\tensor 1 \\
=(1\tensor b_1-b_1\tensor 1)(1\tensor b_2-b_2\tensor 1)\in I^2
\end{align*}
For the universal property see Matsumura.
\end{proof}
\begin{cor}
If $B$ is a finitely generated $A$-algebra and $A$ is Noetherian
then $\diff$ is a finitely generated $B$-module.
\end{cor}
\begin{proof}
If $B$ is a finitely generated $A$-algebra then $B$ is a Noetherian
ring. Since $I$ is a kernel of a ring homomorphism $I$ is
an ideal so $I$ is finitely generated. Thus $I/I^2$ is
finitely generated as a $B$-module.
\end{proof}
\begin{example}
Let $A$ be any ring and let $B=A[x_1,\ldots,x_n]$.
Then $\diff$ is the free $B$-module generated
by $dx_1,\ldots,dx_n$. The derivation $d:B\into\diff$
is the map $f\mapsto \sum \frac{\partial f}{\partial x_i}dx_i$.
Since $B$ is generated as an $A$-algebra by the $x_i$,
$\diff$ is generated as a $B$-module by the $dx_i$ and
there is an epimorphism $B^r\into\diff$ taking
the $i$th basis vector to $dx_i$.
On the other hand, the partial derivative $\partial/\partial x_i$
is an $A$-linear derivation from $B$ to $B$, and thus induces a
$B$-module map $\partial_i:\diff\into B$ carrying
$dx_i$ to $1$ and all the other $x_j$ to $0$. Putting
these maps together we get the inverse map.
This proof is lifted from Eisenbud's {\em Commutative Algebra}.
\end{example}
There are a few nice exact sequences.
\begin{prop}
Suppose $A$, $B$, and $C$ are three rings and
$$A\into B\xrightarrow{g} C$$
is a sequence of maps between them (it needn't
be exact -- in fact it wouldn't make sense to
stipulate that it is exact because kernels don't
exist in the category of commutative rings with $1$).
Then there is an exact sequence of $C$-modules
$$\diff\tensor_B C\into \Omega_{C/A}\into \Omega_{C/B}\into 0.$$
If $d:B\into\diff$ is the derivation associated
with $\diff$ then the first map
is $db\tensor_B c\mapsto cd(g(b))$.
\end{prop}
We won't prove this here, but note that exactness in the
middle is the most interesting.
The functors $T^i$ comes next on the left. See the work
of Schlesinger and Lichenbaum, or Illusi and Andr\'{e}.
\begin{prop}
Suppose $A$, $B$, and $C$ are rings and
$$A\into B\into C$$ is a sequence of maps.
Assume furthermore $I\subset B$ is an ideal,
that $C=B/I$, and the map from $B\into C$ is
the natural surjection. Then there is an exact sequence
of $C$-modules
$$I/I^2\xrightarrow{d}\diff\tensor_{B}C\into\Omega_{C/A}\into\Omega_{C/B}=0$$
\end{prop}
Next we consider what happens for a local ring.
\begin{prop}
Suppose $B,\m$ is a local ring with residue field $k=B/\m$
and there is an injection $k\hookrightarrow B$.
Then there is an exact sequence
$$\m/\m^2\xrightarrow{d}\Omega_{B/k}\tensor_{B}k\into 0$$
and $d$ is actually an isomorphism.
\end{prop}
The exact sequence is obtained from the previous proposition
by letting $A=C=k$. The fact that $d$ is an isomorphism is
supposed to be tricky.
The proposition implies that a lower bound on the number of
generators of $\Omega_{B/k}$ is $\dim_k\m/\m^2$. If $B$ is regular
and local, then $\dim B=\dim_k \m/\m^2$ which implies
$\Omega_{B/k}$ can be generated by $\dim_k B$ elements.
\begin{prop}
Let $B$ be a localization of an algebra of finite
type over a perfect field, let $\m$ be the maximal ideal
and let $k=B/\m$.
Then $B$ is a regular local ring iff
$\Omega_{B/k}$ is a free $B$-module of rank equal to the
dimension of $B$ over $k$.
\end{prop}
\begin{proof}
($\leftarrow$) If $\Omega_{B/k}$ is free of rank $n=\dim_k B$
then the minimum number of generators of $\Omega_{B/k}$
is $n$, so $\dim\Omega_{B/k}\tensor k=n=\dim_k \m/\m^2$.
Thus $\dim B=\dim \m/\m^2$ whence $B$ is regular.
($\rightarrow$)
%Suppose $B$ is a regular local ring. Then
%$B$ is an integral domain. Let $K$ be the quotient field.
%Then $K/k$ is a finitely generated field extension
%of transcendence degree $n$, thus $\dim B=\trdeg_k K$.
%But $k$ is perfect so $K/k$ is separably generated
%so $\dim_k\Omega_{K/k}=\trdeg_k K=n$, as needed
%(this last equality is a theorem in itself).
\end{proof}
\subsection{The Sheaf of Differentials on a Scheme}
Suppose $f:X\into Y$ is a morphism of schemes.
Let $V=\spec B$ be an open affine subset of $X$ and
$U=\spec A$ an open affine subset of $Y$ such that
$f(V)\subset U$. Then $B$ is an algebra over $A$
so we may consider the module $\diff$. Put
$\tilde{\diff}$ on $V$ and glue to get a sheaf $\Omega_{X/Y}$.
We can glue because localization commutes with forming $\Omega$
and the universal property of $\Omega$ makes gluing
isomorphisms canonical and unique.
If $X/k$ is a nonsingular variety of dimension $n$ then
$\Omega_{X/k}$ is locally free of rank $n$. If $X$ is a
curve, then $\Omega_{X/k}$ is locally free of rank 1 so
it is a line bundle.
The sheaf $\Omega_{X/k}$ is important
because it is intrinsically defined and canonically
associated to $X\xrightarrow{f} Y$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
%%3/8/96
% 1. show local definitions of omega agree
% 2. give pullback definition of omega
% 3. review the two nice exact sequences
% 4. example: A^n
% 5. omega for p^n
% with long "proof"
\section{Differentials on $\P^n$}
\begin{remark}
Suppose $X$ is a scheme over $Y$ then one could also define
$\Omega_{X/Y}$ as follows. Let $\Delta:X\into X\cross_Y X$
be the diagonal morphism and let $\sI_{\Delta}$ be the
ideal sheaf of the image of $\Delta$. Define
$\Omega_{X/Y}=\Delta^*(\sI_{\Delta}/\sI_{\Delta}^2)$.
\end{remark}
\begin{prop}
Let $X$, $Y$, and $Z$ be schemes along with maps $X\xrightarrow{f}Y\xrightarrow{g}Z.$
Then there is an exact sequence of $\sox$-modules
$$f^*(\Omega_{Y/Z})\into\Omega_{X/Z}\into\Omega_{X/Y}\into 0.$$
\end{prop}
\begin{prop}
Let $Y$ be a closed subscheme of a scheme $X$ over $S$.
Let $\sI_Y$ be the ideal sheaf of $Y$ on $X$. Then there is an exact
sequence of sheaves of $\sox$-modules
$$\sI_Y/\sI_Y^2\into\Omega_{X/S}\tensor\sO_Y\into\Omega_{Y/S}\into 0.$$
\end{prop}
\begin{example}
Let $S$ be a scheme and let $X=\bA^n_S$ be affine $n$-space over $S$.
Then $\Omega_{X/S}$ is the free $\sox$-module generated
by $dx_1,\ldots,dx_n$.
\end{example}
Projective space is more interesting.
\begin{thm}
Let $X=\P_k^n$, then there is an exact sequence
$$0\into\Omega_{X/k}\into\sO_X(-1)^{n+1}\into\sox\into 0.$$
\end{thm}
The proof in the book is too computational and nobody understands
it therefore we present an ``explanation'' even though it doesn't
quite have the force of proof.
\begin{proof}[Explanation]
Since $U_i=\{x_i\neq 0\}$ is affine, $\Omega_X|_{U_i}=\Omega_{U_i}$
is free of rank $n$ thus $\Omega_X$ is locally free of rank $n$.
Let $W=\bA^{n+1}-\{0\}$ and let $f:W\into X$ be the natural
quotient map. The sequence $W\xrightarrow{f}X\into k$ gives
rise to an exact sequence
$$f^*\Omega_X\into\Omega_W\into\Omega_{W/X}\into 0.$$
Since the open subset $W\subset\bA^{n+1}$ is a nonsingular
variety, $\Omega_W$ is free of rank $n+1$, generated by
$dx_0,\ldots,dx_n$.
The affine subset $U_0=\{x_0\neq 0\}\subset X$ can be represented as
$$U_0=\Spec k[\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0}]=\spec k[y_1,\ldots,y_n].$$
The inverse image is
\begin{align*}
f^{-1}(U_0)=\bA^{n+1}-\{x_0=0\}&=\spec(k[x_0,\ldots,x_n,\frac{1}{x_0}])\\
&=\spec(k[y_1,\ldots,y_n][x_0,\frac{1}{x_0}])\end{align*}
Thus $f^{-1}(U_0)\isom U_0\cross(\bA^1-\{0\})$ which is affine.
Therefore $\Omega_{W/X}|_{f^{-1}(U_0)}$ is locally free of
rank 1 generated by $dx_0$.
Consider again the exact sequence
$$f^*\Omega_X\into\Omega_W\into\Omega_{W/X}\into 0.$$
Since $\Omega_W$ is free of rank $n+1$, $\Omega_{W/X}$ is
locally free of rank $1$, and $f^{*}\Omega_X$ is locally
free of rank $n$ (pullback preserves rank locally), we conclude
that the map $f^*\Omega_X\into\Omega_W$ must be injective.
We thus obtain an exact sequence of sheaves on $W$
$$0\into f^*\Omega_X\into\Omega_W\into\Omega_{W/X}\into 0.$$
Passing to global sections and using the fact that $W$ is affine
we obtain an exact sequence of modules over $S=k[x_0,\ldots,x_n]$
$$0\into \Gamma(f^*\Omega_X)\into S^{n+1} \xrightarrow{\psi} S.$$
The last term is $S$ because any invertible sheaf on
$W=\bA^n-\{0\}$ is isomorphic to $\sO_W$. This follows from
the exact sequence (II, 6.5)
$$0\into\Pic\bA^n\iso\Pic W\into 0$$
and the fact that $\Pic\bA^n=0$.
Take generators $e_0,\ldots,e_n$ of $S^{n+1}$. Then
$\psi$ is the map $e_i\mapsto x_i$, i.e. the multiplication
by $x$ map.
To finish the proof
we need to know that if $\sF$ is a coherent sheaf on $\P^n$, then
$\Gamma(W,f^*\sF)^{\tilde{}}=\sF$.
This assertion is completely natural but {\em doesn't}
carry the force of proof, i.e., Hartshorne gave no proof.
Taking global sections and applying this to the above
sequence yields the exact sequence of sheaves on $X$
$$0\into\Omega_{X/k}\into\Omega_X(-1)^{n+1}\into\sox\into 0$$
which is just what we want.
\end{proof}
\begin{prop}
If $X$ is a nonsingular variety over a field $k$,
then $\Omega_X$ is locally free of rank $n=\dim X$.
\end{prop}
The proof can be found in the book.
\begin{example}
Let $C$ be a nonsingular curve in $\P^2_k$ defined by an equation $f$
of degree $d$. Then $\Omega_{C/k}$ is a locally free sheaf of rank $1$
so it corresponds to a divisor.
Which divisor class will $\Omega_{C/k}$ correspond to?
Let $\sI$ be the ideal sheaf of $C\subset\P^2$. Then we
have an exact sequence
$$0\into\sI/\sI^2\into\Omega_{\P^2}\tensor\sO_C\into\Omega_{C/k}\into 0.$$
The left map is injective since $\sI/\sI^2\isom \sO_C(-d)$ so $\sI/\sI^2$
is locally free of rank $1$. [[Why is $\sI/\sI^2\isom\sO_C(-d)$?]]
Taking the second exterior power gives an isomorphism
$$\Lambda^2(\Omega_{\P^2}\tensor\sO_C)\isom
(\sI/\sI^2)\tensor\Omega_{C/k}.$$
This is a fact from the general theory of locally free sheaves.
We have an exact sequence
$$0\into\Omega_{\P^2}\into\sO(-1)^3\into \sO\into 0$$
thus
$$\Lambda^3(\sO(-1)^3)\isom\Lambda^2\Omega_{\P^2}\tensor\Lambda^1\sO_{\P^2}$$
so $\sO(-3)\isom \Lambda^2\Omega_{\P^2}$.
Thus
$$\sO(-3)\isom\sI/\sI^2\tensor\Omega_{C/k}\isom\sO_C(-d)\tensor\Omega_{C/k}.$$
Tensoring with $\sO(3)$ gives
$\sO_C\isom\sO_C(3-d)\tensor\Omega_{C/k}$
so $\Omega_{C/k}\isom\sO_C(d-3)$.
If $C$ is a cubic then
$$\Omega_{C/k}=\sO_C(3-3)=\sO_C(0)=\sO_C.$$
Furthermore $$\Omega_{\P^1}=\sO(-2)\not\isom\sO(0)=\Omega_{C/k}$$
so a nonsingular plane cubic is not rational.
\end{example}
\begin{prop}
Suppose $0\into\sE'\into\sE\into\sE''\into 0$ is an
exact sequence of locally free sheaves of ranks
$r'$, $r$, and $r''$. Then
$$\Lambda^r\sE\isom\Lambda^{r'}\sE'\tensor\Lambda^{r''}\sE''.$$
\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 3/11/96
\section{Sheaf of Differentials and Canonical Divisor}
\begin{thm}
Let $X$ be a nonsingular projective variety of dimension $n$.
Then $\Omega^n_{X/k}\isom\omega_X$ where $\omega_X$ is
the dualizing sheaf on $X$. Furthermore, $\Omega_{X/k}$
is locally free of rank $n$ and so
$\Omega^n_{X/k}$ is locally free of rank $1$. Thus $\Omega^n_{X/k}$
is an invertible sheaf on $X$.
\end{thm}
Note that $\Omega^n_{X/k}$ is an (abusive) shorthand notation for
$\Lambda^n\Omega_{X/k}$ and that $\Omega^n_{X/k}$ is {\em not}
the direct sum of $n$ copies of $\Omega_{X/k}$.
Recall the construction of the dualizing sheaf $\omega_X$.
Let $f:X\into\P^n$ be a finite morphism. Let
$\omega_{\P^n}=\sO_{\P^n}(-1)$. Then
$$\omega_X=f^{!}\sO_{\P^n}(-1)=f^{!}\omega_{\P^n}
=\sHom(f_*\sox,\omega_{\P^n})^{\tilde{}}.$$
Since $X$ is Cohen-Macaulay, $f_*\sox$ is locally
free so $f^{!}\omega_{\P^n}$ is locally free of rank $1$.
First we show that the theorem holds when $X=\P^n$.
From last time we have an exact sequence
$$0\into\Omega_{\P^n}\into\sO(-1)^{n+1}\into\sO\into 0$$
so taking highest exterior powers gives an isomorphism
$$\Lambda^{n+1}(\sO(-1)^{n+1})\isom\Lambda^n\Omega_{\P^n}\tensor\Lambda^1\sO\isom\Lambda^n\Omega_{\P^n}.$$
For the last isomorphism we used the fact that $\Lambda^1\sO\iso\sO$.
But since the highest exterior power changes a direct sum into a
tensor product
$$\Lambda^{n+1}(\sO(-1)^{n+1})=\sO(-1)^{\tensor n+1}=\sO(-n-1)=\omega_{\P^n}.$$ so
Combining these shows that $\Lambda^n\Omega_{\P^n}=\omega_{\P^n}.$
Now suppose $X$ is an arbitrary nonsingular projective variety
of dimension $n$. As we have done before let
$f:X\into\P^n$ be a finite morphism (do this using suitable linear
projections).
To prove the theorem it is enough to show that
$f^{!}(\Omega^n_{\P^n/k})\isom\Omega^n_{X/k}$
since $f^{!}(\Omega^n_{\P^n/k})=f^{!}(\omega_{\P^n})=\omega_X$.
This is just a statement about differentials.
More generally suppose $f:X\into Y$ is a finite map and that
both $X$ and $Y$ are nonsingular projective varieties of dimension $n$.
Then we have the duality
$$\Hom_X(\sF,f^{!}\sG)=\Hom_Y(f_*\sF,\sG).$$
Thus to give a map
$\varphi:\Omega^n_X\into f^{!}\Omega^n_Y$ is equivalent
to giving a map
$\overline{\varphi}:f_{*}\Omega^n_X\into\Omega^n_Y$.
It is an {\em open problem} to find a natural yet elementary way to
define a map $f_*\Omega_X^n\into\Omega_Y^n$ which corresponds to
an isomorphism $\Omega_X^n\into f^!\Omega_Y^n$. Because the map
sends differentials ``above'' to differentials ``below'' it
should be called a ``trace'' map.
In Hartshorne's book the existence of such a map is proved by embedding
$X$ in some large $\P^N$, showing that $\Omega_X=\sext^{N-n}(\sox,\omega_{\P^n})$
and then applying the ``fundamental local isomorphism''. The approach
is certainly not elementary because it involves the higher $\sext$ groups.
Continue to assume $f:X\into Y$ is a finite morphism of
nonsingular varieties of
dimension $n$. Assume $f$ is separable so that the field extension
$K(X)/K(Y)$ is finite and separable. A theorem proved last time
gives an exact sequence
$$f^*\Omega_Y\into\Omega_X\into\Omega_{X/Y}\into 0.$$
Since $X$ is nonsingular of dimension $n$,
$\Omega_X$ is locally free of rank $n$. Similarly $\Omega_Y$
is locally free of rank $n$ so $f^*\Omega_Y$ is locally free
of rank $n$. (Locally this is just the fact that $A^n\tensor_A B\isom B^n$.)
Since localization commutes with taking the module of differentials
the stalk at the generic point of $\Omega_{X/Y}$ is
$\Omega_{K(X)/K(Y)}$. This is $0$ since $K(X)/K(Y)$ is finite separable,
thus $\Omega_{X/Y}$ is a torsion sheaf.
By general facts about free modules this implies
the map $f^*\Omega_Y\into\Omega_X$ is injective, i.e.,
the sequence
$$0\into f^*\Omega_Y\into\Omega_X\into\Omega_{X/Y}\into 0$$
is exact.
We pause to consider the simplest illustrative example, namely the
case when $X$ is a parabola and $Y$ is the line.
\begin{example}
Let $A=k[x]$ and let $B=k[x,y]/(x-y^2)\isom k[y]$ where $k$ is a field
of characteristic not equal to $2$.
Let $X=\spec B$ and let $Y=\spec A$. Let
$f:X\into Y$ be the morphism induced by the inclusion
$A\hookrightarrow B$ (thus $x\mapsto y^2$).
Since $B\isom k[y]$ it follows that $\Omega_X=Bdy$,
the free $B$-module generated by $dy$. Similarly
$\Omega_Y=Adx$.
The exact sequence above becomes
$$\begin{array}{cccccccccc}
0&\into&\Omega_Y\tensor_A B&\into&\Omega_X&\into&\Omega_{X/Y}&\into&0\\
& &|| & & || & & \wr| \\
& &Bdx &\into& Bdy &\into& Bdy/B(2ydy)
\end{array}$$
The point is that $\Omega_{X/Y}=(k[y]/(2y))dy$ is a torsion
sheaf supported on the ramification locus of the map $f:X\into Y$.
(The only ramification point is above $0$.)
Note that $\Omega_{X/Y}$ is the quotient of $\Omega_X$ by the
submodule generated by the image of $dx$ in $\Omega_X=Bdy$. The
image of $dx$ is $2ydy$.
\end{example}
More generally we define the ramification divisor as follows.
\begin{defn}
Let $f:X\into Y$ be a finite separable morphism of nonsingular varieties
of dimension $n$. Then the {\bfseries ramification divisor} of $X/Y$ is
$$R=\sum_{\zeta\in Z\subset X}\length_{\sO_{\zeta}}((\Omega_{X/Y})_{\zeta})\cdot Z$$
where the sum is taken over all closed irreducible subsets $Z\subset X$
of codimension $1$ and $\zeta$ is the generic point of $Z$.
\end{defn}
Since the sequence
$$0\into f^*\Omega_Y\into\Omega_X\into\Omega_{X/Y}\into 0$$
is exact it will follow that
$$\Omega^n_X\isom f^*\Omega_Y^n\tensor\sL(R)$$
where $R$ is the ramification divisor of $X/Y$ and
$\sL(R)$ denotes the corresponding invertible sheaf.
[[This is some linear algebra over modules.]]
The next part of the argument is to study $f^{!}$.
As usual let $f:X\into Y$ be a finite morphism of nonsingular
varieties of dimension $n$ and assume furthermore that
$f_*\sox$ is a locally free $\soy$-module.
Define a trace map $\Tr:f_*\sox\into\soy$ locally as follows.
Let $\spec A$ be an open affine subset of $Y$ and let
$\spec B=f^{-1}(\spec A)$. Then $B$ is a free $A$-module
of rank $d=\deg f$. Choose a basis $e_1,\ldots,e_d$ for
$B/A$. Let $b\in B$, and suppose $be_i=\sum_j a_{ij}e_j$. Define
$\Tr(b)=\sum_i a_{ii}$.
Let $\overline{\Tr}\in\Hom_X(\sox,f^!\soy)$ correspond
to $\Tr\in\Hom_Y(f_*\sox,\soy)$ under the isomorphism between
these Hom groups.
{\em Claim.} $f^!\soy\isom\sL(R)$.
Once we have proved the claim, the theorem will follow. To
see this tensor both sides by $f^*\Omega^n_Y$. Then
$$f^!\Omega^n_Y=f^!\soy\tensor f^*\Omega^n_Y
=\sL(R)\tensor f^*\Omega^n_Y=\Omega^n_{X/Y}.$$
We look what happens locally. Let $\spec A\subset Y$ and
$\spec B=f^{-1}(\spec A)\subset X$. We want to show that
$f^!(\soy)=\sL(R)$. Since $f^!\soy = \shom_{\soy}(f_*\sox,\soy)^{\tilde{}}$
we look at $B^*=\Hom_A(B,A)$. First consider the special
case of the parabola investigated above.
Then $B$ has a basis $1,y$ over $A$ and
$B^*$ is spanned by $e_0$ and $e_1$ where
$e_0(1)=1$, $e_0(y)=0$, and $e_1(1)=0$, $e_1(y)=1$.
Thus $ye_0(1)=e_0(y)=0$, $ye_0(y)=e_0(y^2)=e_0(x)=xe_0(1)=x$, and
$ye_1(1)=e_1(y)=1$, $ye_1(y)=e_1(y^2)=e_1(x)=xe_1(1)=0$.
Therefore $ye_1=e_0$ so $e_1$ generates $B^*$ over $A$.
The trace $Tr:B\into A$ is an element of $B^*$. We determine it.
We see that $Tr(1)=2$ and $Tr(y)=0$ since
$$1\leftrightarrow\Bigl(\begin{matrix}1&0\\0&1\end{matrix}\Bigr)
\text{ and }
y\leftrightarrow\Bigl(\begin{matrix}0&x\\1&0\end{matrix}\Bigr)$$
Thus $\Tr=2e_0=2ye_1$ since $ye_1=e_0$ as shown above.
We have shown that the map
$$\overline{\Tr}:B\into B^*:1\mapsto \Tr$$
has image generated by $2ye_1$.
More generally suppose $B=A[y]/(f(y))$ where $f(y)=y^d+a_1y^{d-1}+\cdots+a_d$
is a monic polynomial of degree $d$.
Then $B$ has $A$-basis $1,y,y^2,\ldots,y^{d-1}$ and
$B^*$ is spanned by $e_0,e_1,\ldots,e_{d-1}$. By a similar argument
to the one above, we can show that
$\Tr=f'(y)e_{d-1}$.
Thus locally $f^{!}\soy$ is a free $\sox$-module of rank 1 generated by
$f'(y)e_{d-1}$. But this is really what we defined $\sL(R)$ to be (where
this derivative is zero is where there is ramification). Thus
$f^{!}\soy=\sL(R)$ and we are done.
%%%%%%%%%%%%%%%%%%
%%CURVES
%%%%%%%%%%%%%%%%%%
\begin{center}{\Huge Curves}\end{center}
\section{Definitions}
\begin{defn}
A {\bfseries curve} is a connected nonsingular projective
$1$-dimensional scheme over an algebraically closed field $k$.
\end{defn}
We have proved that if $C_1$ and $C_2$ are curves then
$C_1\isom C_2$ iff $C_1$ is birational to $C_2$ (i.e., they
have isomorphic open subsets) iff $K(C_1)\isom K(C_2)$.
Here $K(C_1)=\sO_{C,\zeta}$ where $\zeta$ is the generic
point of $C_1$. If any of the hypothesis nonsingular, connected,
projective, or one-dimensional is removed then these equivalences
can fail to hold.
\section{Genus}
Let $X\hookrightarrow \P^n$ be an embedded projective variety.
Let $P_X(n)$ be its Hilbert polynomial. The {\bfseries arithmetic
genus} of $X$ is the quantity $p_a$ defined by the equation
$P_X(0)=1+(-1)^{\dim X}p_a$. Thus if $C$ is a curve then
$p_a=1-P_X(0)$. For a nonsingular projective curve we also define the
quantity $g=h^1(\soc)$ and call it the {\bfseries genus}.
The {\bfseries geometric genus} $p_g=h^0(\omega_X)$ is a third notion
of genus for a nonsingular projective variety $X$. For
curves all types of genus coincide.
\begin{thm} If $C$ is a curve then $$p_a=g=p_g.$$ \end{thm}
\begin{proof}
$P_C(n)=\sum(-1)^i h^i(\sox(n))$ so
$P_C(0)=\sum(-1)^i h^i(\sox)$. But $C$ has dimension $1$ so by Grothendieck
vanishing
$$1-p_a=P_C(0)=h^0(\soc)-h^1(\soc)=1-h^1(\soc)=1-g$$
so $p_a=g$.
Serre duality says that if $X$ is nonsingular of dimension $n$ then
$\Ext_X^i(\sF,\omega_X)$ is linearly dual to $H^{n-i}(X,\sF)$.
Thus when $X=C$, $\sF=\soc$, $i=0$, and $n=1$ we see that
$$H^0(\omega_C)=\ext_C^0(\soc,\omega_C)$$
is dual to $H^1(\soc)$. Thus $p_g=h^1(\soc)=h^0(\omega_C)=g$.
\end{proof}
Thus we may speak of {\em the} genus of a curve. For more
general varieties the concepts diverge.
The classification problem is to describe all curves up to isomorphism.
The set of curves is a disjoint union $\sC=\coprod_{g\geq 0}\sC_g$
where $\sC_g$ consists of all curves of genus $g$.
\begin{thm}
Any curve of genus $0$ is isomorphic to $\P^1$.
\end{thm}
\begin{proof}
Let $C$ be a curve of genus $0$ and let $P\in C$ be a closed point.
There is an exact sequence
$$0\into\soc\into\sL(P)\into\kappa(P)\into 0.$$
AS a divisor $P$ corresponds to the invertible sheaf $\sL(P)$. Let
$s\in\Gamma(\sL(P))$ be a global section which generates $\sL(P)$
as an $\soc$-module. Thus $s$ has a pole of order 1 at $P$ and
no other poles. Then $s$ defines a morphism
$$\soc\into\sL(P):1\mapsto s$$
which has cokernel $\kappa(P)$.
Taking cohomology yields
$$0\into H^0(\soc)\into H^0(\sL(P))\into H^0(\kappa(P))\into H^1(\soc)=0.$$
Since $H^0(\soc)=k$ and $H^0(\kappa(P))=k$ it follows that
$H^0(\sL(P))=k\oplus k$. View $\soc\subset\sL(P)\subset K$
where $K=K(X)$ is the constant function field sheaf. Then
$$H^0(\soc)\hookrightarrow H^0(\sL(P))\hookrightarrow K.$$
Since $\dim H^0(\sL(P))>\dim H^0(\soc)$, there is $f\in H^0(\sL(P))-H^0(\soc)$
and $f$ is non constant. Thus $f\in K=K(X)$ and $f\not\in k$.
So $f$ defines a morphism $C\into\P^1$ as follows. If $Q\in C$ and
$f\in\sO_Q$ send $Q$ to $\overline{f}\in\sO_Q/\m_Q=k\subset\P^1_k$.
If $Q\not\in C$ so that $f$ has a pole at $Q$, send
$Q$ to $\infty\in\P^1_k$. Since $f$ lies in $H^0(\sL(P))$ which
has dimension $1$ over $k=H^0(\soc)$, $v_P(f)=-1$. Thus
$\infty$ does not ramify so the degree of the morphism defined by $f$
is the number of points lying over $\infty$ which is $1$. Thus
$\P^1_k\isom C$.
\end{proof}
\begin{remark} This result is false if $k$ is not algebraically closed.
But it is true if $C$ has a rational point, i.e., a point
$P\in C$ so that $\kappa(P)=\sO_P/\m_P=k$.
\end{remark}
\begin{example}
Let $k=\R$ and let $C$ be the curve $x^2+y^2+z^2=0$ on $\P^2_{\R}$. $C$
is nonsingular of degree $2$ so $g=0$. But $C$ is not isomorphic
over $\R$ to $\P^1_{\R}$ since $C$ has no rational points whereas
$\P^1_\R$ does. Let $P=(1,i,0)\in C_{\C}$ and $\overline{P}=(1,-i,0)\in C_{\C}$.
Let $P^*\in C_{\R}$ be the Galois conjugacy class $\{P,\overline{P}\}$.
Then $\kappa(P^*)=C$ is of degree $2$ over $\R$.
\end{example}
\begin{exercise} Over $\R$ an curve of genus $0$ not isomorphic to
$\P^1_{\R}$ is isomorphic to $$C:x^2+y^2+z^2=0$$.
\end{exercise}
[[If the curve is planar this follows from diagonalizability of
quadratic forms.]]
\section{Riemann-Roch Theorem}
The Riemann-Roch theorem is the cornerstone of all of curve theory.
\begin{thm}[Riemann-Roch] Let $C$ be a curve and $D=\sum_{i=1}^{t} n_i P_i$
be a divisor on $C$. Then
$$h^0(\sL(D))-h^1(\sL(D))=\deg D+1 - g$$
where $\deg D=\sum_{i=1}^{t} n_i$.
\end{thm}
\begin{proof}
When $D=0$ the assertion is that
$$h^0(\soc)-h^1(\soc)=1-g$$
which is easily checked.
Next suppose $D$ is any divisor and $P$ any point.
Compare $D$ and $D+P$ as follows. The statement of the theorem for $D$
is $$h^0(\sL(D))-h^1(\sL(D))=\deg D+1-g$$
and for $D+P$ the theorem is
$$h^0(\sL(D+P))-h^1(\sL(D+P))=\deg D+1+1-g.$$
Use the exact sequence
$$0\into\sL(D)\into\sL(D+P)\into\kappa(P)\into 0$$
to compute the Euler characteristic
$\chi=h^0-h^1$ of $\sL(D+P)$.
This yields
$$\chi(\sL(D+P))=\chi(\sL(D))+\chi(\kappa(P))$$
so
$$h^0(\sL(D+P))-h^1(\sL(D+P))=h^0(\sL(D))-h^1(\sL(D))+1.$$
The theorem is thus true for $D+P$ iff it is true for
$D$. Since the theorem is true for $D=0$ and we can obtain
any divisor by adding or subtracting points starting with
$D=0$ the theorem follows.
\end{proof}
\section{Serre Duality}
By Serre Duality
$H^1(\sL(D))$ is dual to
\begin{align*}\ext^0(\sL(D),\omega_C)&=\ext^0(\soc,\omega_C\tensor\sL(D)^{\dual})\\
&=H^0(\omega_C\tensor\sL(-D))=H^0(\sL(K-D))\end{align*}
Thus $h^1(\sL(D))=h^0(\sL(K-D))$
where $K$ is some divisor so that $\sL(K)=\omega$. $K$ is often
called the canonical divisor. Thus we can restate the Riemann-Roch theorem as
$$h^0(\sL(D))-h^0(\sL(K-D))=\deg D+1-g.$$
Sometimes one abbreviates $h^0(\sL(D))$ as $\ell(D)$ and
then Riemann-Roch becomes $$\sL(D)-\sL(K-D)=\deg D+1-g.$$
\begin{cor}$\deg K=2g-2$.\end{cor}
\begin{proof}
By Riemann-Roch,
$$h^0(\sL(K))-h^0(\sL(0))=\deg K+1-g$$
but $h^0(\sL(K))=g$ and $h^0(L(0))=1$.
Thus $\deg K=2g-2$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\Large Homework} II, Exercise 8.4; III, Exercise 6.8, 7.1, 7.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 3/18??/96
\section{A Bird's Eye View of Curves}
$$3g\quad-\quad 3\quad+\quad g\quad+\quad 4(d+1-g-4)\quad+\quad 15\quad = \quad 4d$$
\begin{enumerate}
\item $3g-3$ is the dimension of the moduli space $\M_g$ of curves
of genus $g$.
\item $g$ is the dimension of $\Pic^d C$.
\item $4(d+1-g-4)$ is the number of ways to choose a linear system
$W$ in $W^0(\sL)$.
\item 15 is the dimension of $\Aut\P^3\isom\PGL(4,k)$.
\item $4d$ is the dimension of the Hilbert scheme $H^0_{d,g}$ of
nonsingular curves of genus $g$ and degree $d$.
\end{enumerate}
\section{Moduli Space}
As a set $\M_g$ is the set of curves of genus $g$ modulo isomorphism.
$\M_g$ can be made into a variety in a natural way.
As a variety $\M_g$ is irreducible and
$$\dim \M_g=\begin{cases}0&\text{if $g=0$}\\
1&\text{if $g=1$}\\3g-3&\text{if $g\geq 2$}\end{cases}$$
$\M_g$ is not a projective variety, but its closure $\overline{\M}_g$
is. The points of $\overline{\M}_g$ not in $\M_g$ are called stable curves.
\section{Embeddings in Projective Space}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% hartnotes 3/20/96
\section{Elementary Curve Theory}
\subsection{Definitions}
Let $C$ be a curve thus $C$ is a nonsingular connected projective variety
of dimension $1$ over an algebraically closed field $k$.
Let $g$ be the genus of $C$. A divisor $D$ is a sum $\sum_{i=1}^{k}n_iP_i$
where $n_i\in\Z$ and $P_i$ is a closed point of $C$.
Let $\deg D=\sum_{i=1}^{k}n_i$. A divisor $D$ corresponds to an invertible
sheaf $\sL(D)=\sI_D^{\dual}$ where $\sI_D$ is the ideal sheaf of $D$ and
$\sI_D^{\dual}=\Hom_{C}(\sI_D,\sO_C)$ is the dual of $\sI_D$.
Note that $\sL(nD)=\sL(D)^{\tensor n}$ since $\tensor$ and $\dual$ commute.
Notation: $\sO(D):=\sL(D)$, $\sM(D)=\sM\tensor\sL(D)$.
We say a divisor $\sum n_i D_i$ is effective if all $n_i\geq 0$.
Let $|D|$ denote the complete linear system associated to $D$.
Thus $$|D|=\{D' : D'\text{ is effective and } D'\sim D\}.$$
Here $\sim$ denotes {\em linear equivalence}. Two divisors $D$
and $D'$ are linearly equivalent if there exists $f\in K=K(C)$
(the function field of $C$) so that $D-D'=(f)$ where
$(f)=\sum_{P\in C \text{ closed}}v_P(f)P$.
Since the condition $D+(f)\geq 0$ is exactly the condition that
$f\in H^0(C,\sL(D))$, it follows that
there is a bijection
$$|D|\iso H^0(C,\sL(D))/k^*$$
$$D'=D+(f)\mapsto f.$$
Two functions $f$, $g$ which differ by an element of $k^*$ give the same divisor.
Note that $|D|$ may be empty.
If $s\in H^0(\sL)$ then there is an injection
$0\into\sO\xrightarrow{s}\sL(D)$ given by multiplication
by $s$. Dualizing we obtain
$$\sL(-D)=\sI_D=\sL^{\dual}\xrightarrow{s^{\dual}}\sO\into\sO_D\into 0.$$
Thus $D\neq 0$ is effective iff $D$ corresponds to a closed subscheme
defined locally by $(t_{P_i}^{n_i}\subset\sO_{P_i}$ where
$t_{P_i}\in\m_{P_i}\subset\sO_{P_i}$ is a uniformizing parameter
at $\m_{P_i}$.
$|D|$ can be regarded as a projective space and as such
$$\dim |D|=h^0(\ell(D))-1=\ell(D)-1,$$ where
$\ell(D)=h^0(\sL(D))$. We
can generalize the notion of a complete linear system by considering
linear subspaces of $|D|$.
A {\em (general) linear system} is a linear subspace
$\sD\subset|D|$. Thus $\sD$ corresponds to a vector subspace
$W\subset H^0(C,\sL(D))$.
\subsection{Maps to Projective Space}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% hartnotes 3/22/96
\section{Low Genus Projective Embeddings}
Let $C$ be a curve of genus $g$,
let $K$ be the canonical divisor (which corresponds to the
invertible sheaf $\Omega=\omega$ of differentials) and let $D$ be
a divisor.
\begin{thm}[Riemann-Roch]$\ell(D)-\ell(K-D)=\deg(D)+1-g$\end{thm}
\begin{prop}A divisor $D$ is ample iff $\deg D>0$.\end{prop}
\begin{prop}If $\deg D\geq 2g+1$ then $D$ is very ample.\end{prop}
\subsection{Genus $0$ curves}
Suppose $C$ is a curve of genus $g=0$. If $\deg D\geq 1$ then $D$
is very ample. Suppose $D=P$ is a point. Then $D$ gives rise (after
a choice of basis for the corresponding invertible sheaf) to an embedding
$C\hookrightarrow\P^n$ where $n=\ell(D)-1=1$. When $D=2P$ we obtain the
$2$-uple embedding $C\hookrightarrow\P^2$ which is a conic since $g=0$.
When $D=3P$, $C\hookrightarrow \P^2$ is the twisted cubic. More
generally $D=dP$ gives the rational normal curve $C\hookrightarrow\P^d$
of degree $d$ (which is in fact projectively normal).
\subsection{Genus $1$ curves}
If $C$ is a curve of genus $g=1$ then $\deg D\geq 2g+1=3$
iff $D$ is very ample. Thus the converse to proposition
2 holds when $g=1$. Suppose $D$ is a very ample divisor of degree $2$.
(For degree $1$ or $0$ a similar argument works.) Then $D$ gives
rise to an embedding $C\hookrightarrow\P^1$
since $\ell(D)-1=2-1=1$ which is absurd since $C$ has genus $1$
but $\P^1$ has genus $0$. To see that $\ell(D)=2$ apply Riemann-Roch
to obtain $\ell(D)-\ell(K-D)=2+1-1=2$. Then since
$\deg K=2g-2=0$ we see that $\deg(K-D)<0$ and hence
$\ell(K-D)=0$ so $\ell(D)=2$ as desired.
Suppose $D$ is a divisor of degree $3$ on a genus $1$ curve $C$.
Then $D$ gives rise to an embedding $C\hookrightarrow\P^2$
since $\ell(D)-1=(\deg D+1-g)-1=3-1=2$. Thus any curve of genus $1$
can be embedded as a nonsingular cubic curve in $\P^2$. This
embedding also allows us to show that $K\sim 0$. We showed
before that if $C\subset\P^2$ is of degree $d$ then
$\omega_C\isom\sO_C(d-3)$. Thus choosing an embedding arising from
a divisor of degree $3$ as above we see that $\omega_C\isom\sO_C(0)$,
thus the canonical sheaf corresponds to the trivial divisor class
so $K\sim 0$. Alternatively, we can prove this by using Riemann-Roch
to see that
$$\ell(K)=\ell(0)+\deg(K)+1-g=1+0+1-1=1$$
and thus $K$
is linearly equivalent to an effective divisor of degree $2g-2=0$.
\subsection{Moduli Space}
The moduli spaces of curves of low genus are
$$\M_0=\{\text{ curves of genus $0$ }\}=\{ \P^1 \}$$
$$\M_1=\{\text{ curves of genus $1$ }\}=\A^1$$
A curve $C$ of genus $g=1$ can be given by a degree $3$ embedding
$C\hookrightarrow\P^2$. We will show [[in nice characteristic only?]]
that two embedded curves
$C_1\hookrightarrow\P^2$ and $C_2\hookrightarrow\P^2$ are isomorphic
as abstract curves
iff there is an automorphism in $\Aut\P^2=\PGL(3)$ sending
$C_1$ to $C_2$. Note that $\Aut\P^2$ is a
$3^2-1=8$ dimensional family. A degree $3$ curve $C\hookrightarrow\P^2$
is given by a degree $3$ polynomial
$$F=a_0x^3+\cdots\in H^0(\sO_{\P^2}(3)).$$
Since $\dim H^0(\sO_{\P^2}(3))=10$ we obtain
a $10-1=9$ dimensional projective space of such
curves (including the singular ones). The nonsingular
ones form an open subset $\Delta\neq 0$.
The moduli space $\M_g$ has
dimension $9-8=1$. [[Not clear.]]
How can degree 3 embedded curves $C_1, C_2 \subset\P^2$
be isomorphic? We generalize to arbitrary degree
and ask the following question.
\begin{ques}
Suppose $C_1, C_2\subset\P^2$ are both nonsingular curves of degree
$d$ and suppose $C_1\isom C_2$ as abstract curves.
Does is necessarily follow that there is an automorphism
$g$ of $\P^2$ sending $C_1$ to $C_2$, i.e., such that $g(C_1)=C_2$?
\end{ques}
Suppose $d=1$. Then $C_1$ and $C_2$ are both lines in $\P^2$ so
the answer is yes.
Suppose $d=2$. Then $C_1$ and $C_2$ are both conics. If $k$ is
algebraically closed and $\Char k\neq 2$ the defining
equations $C_1$ and $C_2$ can be transformed into $x^2+y^2+z^2=0$
by an automorphism of $\P^2$ (by ``completing the square'').
Thus in this case the answer to the question is yes. When $\Char k=2$
the answer is no. [[give an easy counterexample here.]]
Suppose $d=3$. Thus $C_1$ and $C_2$ are both cubic curves in
$\P^2$ and $C_1\isom C_2$ as abstract curves. Equivalently
we are given an abstract curve $C$ and two embeddings
$$\varphi_1:C\hookrightarrow\P^2$$
$$\varphi_2:C\hookrightarrow\P^2$$
The question is then: does there exist an automorphism $g$
of $\P^2$ such that $g(\varphi_1(C))=\varphi_2(C)$?
The embedding data giving $\varphi_1$ is a divisor $D$
along with a basis of global sections $s_0,s_1,s_2\in H^0(\sO_C(D))$.
An automorphism of $\P^2$ induces a map on $C$ which preserves
$D$ but changes the basis $s_0,s_1,s_2$. Suppose $\varphi_1$
is given by $D_1$ and global sections $s_0,s_1,s_2$, and that
$\varphi_2$ is given by $D_2$ and $t_0,t_1,t_2$. The automorphism
$g$ induces a map
$$g':C\isom\varphi_1(C)\xrightarrow{g}\varphi_2(C)\isom C.$$
Then $g'(D_1)=D_2$ and $g':s_0,s_1,s_2\mapsto t_0,t_1,t_2$.
This generalizes so that in degree $d\geq 3$ a necessary condition
for a yes answer to the above question is that for any two divisors
$D_1$ and $D_2$ of degree $d$ on $C$ there is $g\in\Aut\P^2$ such
that $g'(D_1)=D_2$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/1/96???
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/3/96
\section{Curves of Genus 3}
Today we will study curves of genus $3$.
\begin{example} A plane curve $C\subset\P^2$ of degree $d=4$ has
genus $\frac{1}{2}(d-1)(d-2)=3$.\end{example}
\begin{example}A curve $C$ on the quadric surface $Q$ in $\P^3$
of type (2,4) has degree $6$ and genus $3$. \end{example}
These two examples are {\em qualitatively} different. The curves
in the first example are ``canonical'' whereas the curves in the
second class are ``hyperelliptic''.
We consider the first example in more detail.
Let $C\subset\P^2$ be a genus $3$ plane curve (so $C$ has degree $d=4$).
Then $$\omega_C=\soc(d-3)=\soc(1)$$
so $\omega_C$ is very ample.
Thus the canonical embedding arising from the canonical divisor
is exactly the given embedding $C\hookrightarrow\P^2$.
Next we consider example 2 in more detail.
Let $C$ be a genus three curve of type (2,4) on the quadric surface
$Q\isom\P^1\cross\P^1\subset\P^3$. The projections $p_1,p_2:Q\into\P^1$
give rise to a degree $2$ and a degree $4$ morphism of $C$ to $\P^1$.
Thus there exists a $2$-to-$1$ morphism $f:C\into\P^1$.
$f$ corresponds to a base point free linear system on $C$ of
degree $2$ and dimension 1. This linear system
in turn corresponds to an effective divisor $D$ of degree $2$ with $\ell(D)=2$
so $|D|=1$. The existence of such a divisor means
there exists $P$, $Q$ such that $|P+Q|$ has dimension 1.
\begin{defn}
A curve $C$ is {\em hyperelliptic} if $g\geq 2$ and there is a base point
free linear system of degree $2$ and dimension $1$.
\end{defn}
It is classical notation that a base point free linear system
of degree $d$ and dimension $r$ is called a $g_d^r$. To say that
a curve is hyperelliptic is to say that it has a $g^1_2$.
%% In the definition of hyperelliptic why do we exclude genus $0$ and $1$?
\begin{itemize}
\item If $g=0$ then there is always a $g^2_2$.
\item If $g=1$ any divisor of degree
$2$ gives a $g_2^1$ by Riemann-Roch. Indeed, if $D$ has degree $2$ then
$$\dim|D|-\dim|K-D|= 2+1-g=2$$
and $\deg(K-D)=-2$ so
$\dim|D|-(-1)= 2$
and hence $\dim|D|\leq 1$.
\item If $g=2$ every curve is hyperelliptic since $|K|$ is a $g_2^1$.
Indeed, applying Riemann-Roch we see that
$$\dim|K|=\dim|K|-\dim|0|= 2-1=1.$$
\end{itemize}
\begin{lem}
Let $C$ be any curve and $D$ any divisor of degree $d>0$. Then
$\dim|D|\leq d$ with equality iff $C$ is a rational curve.
\end{lem}
\begin{proof} This is (IV, Ex. 1.5) in Hartshorne.
Although one might guess that this lemma follows from Riemann-Roch
this is not the case. Riemann-Roch gives a different sort of relationship
between the dimension and degree of a divisor. We induct on $d$.
First suppose $d=1$. First note that
$$\dim|P|=\ell(P)-1=h^0(\soc(P))-1.$$
There is an exact sequence
$$0\into\soc\into\soc(P)\into k(P)\into 0.$$
Now $h^0(\soc)=1$ and $h^0(k(P))=1$ therefore
$h^0(\soc(P))\leq 2$ so $\dim|P|\leq 1$.
If $\dim|P|=1$ then $|P|$ has no base points so we obtain a
morphism $C\into\P^1$ of degree $\deg P = 1$ which
must be an isomorphism so $C$ is rational.
Next suppose $D=P_1+\cdots+P_d$. Let $D'=P_1+\cdots+P_{d-1}$. There
is an exact sequence
$$0\into\soc(D')\into\soc(D)\into k(P_d)\into 0.$$
Now $h^0(\soc(D'))\leq d$ by induction and $h^0(k(P_d))=1$ so
$h^0(\soc(D))\leq d+1$, therefore $\dim|D|\leq d$ with equality
iff $h^0(\soc(D'))=d$. By induction $h^0(\soc(D'))=d$ iff $C$ is rational.
\end{proof}
\begin{thm}
Suppose $C$ is a curve of genus $g\geq 2$. Then $\omega_C$ is
very ample iff $C$ is not hyperelliptic.
\end{thm}
\begin{proof}
Let $K$ be the canonical divisor. Then by a previous result
$K$ is very ample iff for all points $P$, $Q$,
$\dim|K-P-Q|=\dim|K|-2$. By Riemann-Roch,
$$\dim|P+Q|-\dim|K-P-Q|=2+1-g=3-g.$$
Now $K$ is very ample iff $\dim|K-P-Q|=\dim|K|-2=(g-1)-2=g-3$ so
when $K$ is very ample the above becomes $$\dim|P+Q|-(g-3)=3-g.$$
Thus $K$ is very ample iff for all $P$ and $Q$,
$\dim|P+Q|=0$. Thus $K$ is not very ample iff there exists $P$ and $Q$
so that $\dim|P+Q|=1$. But the latter condition occurs precisely when
$C$ is hyperelliptic.
(We can exclude the case $\dim|P+Q|\geq 2$ by using the previous
lemma and the fact that $C$ is not rational.)
\end{proof}
\begin{cor}
If $C$ is a curve of genus $g\geq 1$ then $|K|$ has no base points.
\end{cor}
\begin{proof}
If $g=1$ then $K=0$ so $|K|=\{0\}$ and we are done. If
$g\geq 2$ then $|K|$ is base point free iff
$$\dim|K-P|=\dim|K|-1$$ for all $P$. (If $|K|$ has a base point $P$
then every effective divisor $D$ linearly equivalent to $K$
is such that $D-P$ is effective and linearly equivalent to $K-P$.
If $|K|$ has no base points then the dimension of the space of
effective divisors equivalent to $K-P$ must go down for ever $P$.)
Since
$$\dim|P|=\dim|K-P|+1+1-g$$
and
$$\dim|K|=2g-2+1-g=g-1$$
we see that
$$\dim|K-P|=\dim|K|-1+\dim|P|$$ so $|K|$ is base point free
iff $\dim|P|=1$ for all $P$. But
$\dim|P|\leq 1$ for every $P$ and we have equality iff $C$
is rational (i.e., g=0). Since $C$ is not rational it follows
that $|K|$ is base point free.
\end{proof}
Suppose given an abstract curve $C$ of genus $3$. Then $C$
belongs to one of two disjoint classes. If the canonical sheaf
$\omega_C$ is very ample then we obtain an embedding of $C$
into $\P^{g-1}=\P^2$ as a nonsingular quartic curve. If $\omega_C$
is not very ample then $C$ is hyperelliptic (since the map induced
by $\omega_C$ is $2$-to-$1$). Does every hyperelliptic curve arise as a curve
of type (2,4) on $Q\subset\P^3$? Hartshorne claims to have
three-fourths of a proof.
\begin{lem}
If $C$ is any curve of genus $3$ then there exists a very ample divisor
of degree $6=2g$.
\end{lem}
\begin{proof}
Let $D$ be a divisor of degree $6$. We have shown that $D$ is very ample
iff $$\dim|D-P-Q|=\dim|D|-2$$ for all $P$ and $Q$.
Since $\deg K=2g-2=4$ Riemann-Roch asserts that
$$\dim|D|=\dim|K-D|+6+1-g=-1+6+1-g=6-g=3$$
$$\dim|D-P| =\dim|K-(D-P)|+5+1-g=5-g=2$$
$$\dim|D-P-Q|=\dim|K-(D-P-Q)|+4+1-g.$$
Thus for $D$ to be very ample it must be the case that
$$\dim|K-(D-P-Q)|=-1.$$ Since $\deg(K-(D-P-Q))=0$ it follows that
$\dim|K-(D-P-Q)|=-1$ iff $K$ is not linearly equivalent to
$D-P-Q$. The assertion is thus reduced to showing that among all
divisors of degree $6$ the set with $D-P-Q\sim K$ is a proper
closed subset.
[[I do not understand Hartshorne's proof of this. He says
$D-P-Q\sim K$ iff $D\sim K+P+Q$. He then claims that the
family of $D$ of degree 6 is a 6 dimensional family and that
the family of divisor $K+P+Q$ is a 5 dimensional family.]]
\end{proof}
What about the converse? If $C$ is hyperelliptic must $C$ then have
to lie on a quadric surface?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/5/96
\section{Curves of Genus 4}
Recall that curves of genus $g\geq 2$ split up into two disjoint classes.
\begin{itemize}
\item[(a)] hyperelliptic
\item[(b)] $\omega_X$ is very ample
\end{itemize}
If $g=3$ and $C$ is of type $(2,4)$ on $Q\subset\P^3$ then $C$ is hyperelliptic.
Also, if $g=3$ then $\omega_X$ is very ample iff $C$ is a degree $4$
curve in $\P^2$. Any curve of genus $g=3$ can be embedded as a curve
of degree $6$ in $\P^3$. We do not know whether any such curve can
actually be put on a quadric surface. ``This would make a great homework
problem --- I do not know the answer.''
Next we consider curves of genus $4$.
\begin{example}
Consider a type $(2,5)$ curve $C$ on $Q\subset\P^3$. Then $C$
has degree $7=2+5$ and $C$ is hyperelliptic (because of the
degree $2$ map coming from projection onto the first fact $p_1:Q\into\P^1$).
A type $(3,3)$ curve on $Q$ is also of genus $4$. It is a degree $6$
complete intersection of $Q$ and a cubic surface. Curves of type
$(3,3)$ have at least two $g^1_3$'s.
\end{example}
\subsection{Aside: existence of $g^1_d$'s in general}
Assume for this discussion that $g^1_d$'s are allowed
to have base points. Call such $g^1_d$'s trivial.
Given a $g^1_d$ adding a point $p$ trivially gives
a (trivial) $g^1_{d+1}$.
\begin{ques} Given any curve $C$ what is the least $d$
for which there exists a $g^1_d$?\end{ques}
\begin{itemize}
\item $g=0$ there is a $g_1^1$ coming from the embedding $\P^1\hookrightarrow\P^1$,
\item $g=1$ there are infinitely many $g^1_2$'s,
\item $g=2$ there is a $g^1_2$ namely $\omega_C$,
\item $g\geq 2$] if there is a $g^1_2$ then $C$ is hyperelliptic,
\item $g>2$ there exists nonhyperelliptic curves.
\end{itemize}
If a curve $C$ has a $g^1_3$ it is called {\em trigonal}.
If $g\leq 2$ then there exists a trivial $g^1_3$ (just add a point to a $g^1_2$).
If $g=3$ and $C$ is hyperelliptic then it is trivial that there is
a $g^1_3$. If $C$ is not hyperelliptic it is not trivial. But projection
through a point $P\in C$ gives a 3-to-1 map to $\P^1$. This corresponds
to a $g^1_3$. Thus a curve of genus $3$ has infinitely many $g^1_3$'s
(one for each point, coming from projection).
If $g=4$ and $C$ is hyperelliptic then it is trivial that there is a
$g^1_3$. In general given any genus $4$ curve one can always
construct a $g^1_3$. When $g\geq 5$
in general there will not be a $g^1_3$.
This pattern repeats itself.
\subsection{Classifying curves of genus $4$}
Start with an abstract curve $C$ of genus $4$.
We do not deal with the case $C$ hyperelliptic now. A related
question is the following.
\begin{ques}
Does every hyperelliptic curve live on the quadric surface?
\end{ques}
We postpone this question or maybe put it on an upcoming
homework assignment. [Everyone shudders.]
If $C$ is not hyperelliptic then $\omega_C$ is very ample.
Therefore we have the canonical embedding $C\hookrightarrow\P^{g-1}=\P^3.$
The degree of the embedded curve is $\deg\omega_C=2g-2=6$.
Thus view $C$ as a degree $6$ genus 4 curve in $\P^3$.
What does $C$ lie on?
There is an exact sequence
$$0\into H^0(\sI_C(2))\into H^0(\sO_{\P^3}(2))\into H^0(\soc(2))\into\cdots$$
Since Riemann-Roch states that
$\ell(D)=\deg D+1-g+h^1(\sO(D))$
we see that $h^0(\soc(2))=12+1-4+0=9$.
[[This is not quite clear.]]
Since $h^0(\sO_{\P^3}(2))=10$ it follows that
the map $H^0(\sO_{\P^3}(2))\into H^0(\soc(2))$ must
have a nontrivial kernel so $h^0(\sI_C(2))>0$.
Therefore $C$ lies in some surface of degree $2$.
Can the surface be twice a hyperplane? No. Can the surface be
the union of two planes? No. Could the surface by the singular
quadric cone $\qone$? Yes. Could the surface be the nonsingular
quadric surface $\qns$? Yes.
If $C$ lies on $\qns$ then it must have a type $(a,b)$ which
must satisfy $a+b=6$ and $(a-1)(b-1)=4$. The only solution
is $a=b=3$.
The other possibility is that $C$ lies on $\qone$. One way to
understand $C$ is to figure out all divisors on $Q$. Another
way is to compute an exact sequence like the one above. We obtain
$$0\into H^0(\sI_C(3))\into H^0(\sO_{\P^3}(3))\into H^0(\soc(3))\into \cdots$$
As before one sees that $h^0(\soc(3))=15$ and $h^0(\sO_{\P^3}(3))=20$.
Thus $h^0(\sI_C(3))\geq 5$. Let $q\in H^0(\sI_C(2))$ be the defining equation
of $\qone$. Then $xq, yq, zq, wq \in H^0(\sI_C(3))$. But $h^0(\sI_C(3))\geq 5$
so there exists an $f\in H^0(\sI_C(3))$ so that the global sections
$xq, yq, zq, wq, f$ are independent. Thus there is an $f$
not in $(q)$. Since $f\not\in(q)$ we see that
$F_3=Z(f)\not\supset Q$ so $C'=F_3\intersect Q$ is a degree $6$ not
necessarily nonsingular or irreducible curve. Since $C\subset F_3$
and $C\subset Q$ it follows that $C\subset C'$. Since
$\deg C=6=\deg C'$ it follows by an easy exercise that $C=C'$.
\begin{lem} Suppose $C\subset C'$ are both closed subschemes of $\P^n$ with
the same Hilbert polynomial. Then $C=C'$.
\end{lem}
Thus in the case that $C$ lies on $\qone$ we see that
$C$ is also a complete intersection $C=\qone\intersect F_3$.
Next we comment on the $g^1_3$ question in this situation. Projection
from the cone point to the conic (the base of the cone) induces a
$g^1_3$ on $C$. So in genus $g=4$ there is a $g^1_3$. There is one
in the case that $C$ lies on $\qone$ (it is not clear that there is {\em
just} one), there are two in the case that $C$ lies on $\qns$, and there
is a trivial one in the case that $C$ is hyperelliptic.
\begin{prop} Suppose $C$ is a genus 4 nonsingular complete intersection
$Q\intersect F_3 \subset \P^3$ with $Q$ of degree $2$ and $F_3$
of degree $3$. Then $\omega_C\isom\soc(1)$. \end{prop}
Recall that if $C$ is of degree $d$ in $\P^2$ then $\omega_C\isom\soc(d-3)$.
This can be seen from an analysis of the exact sequence
$$0\into\sI/\sI^2\into\Omega_{\P^2}|C\into\Omega_C\into 0$$
together with the exact sequence
$$0\into \Omega_{\P^2}|\into\soc(-1)^3\into\sO_{\P^2}|C\into 0.$$
We could do the same thing for $C\subset\P^3$.
\begin{prop}
Suppose $C=F_e\cdot F_f\subset\P^3$ with $F_e$ and $F_f$ nonsingular of
degree $e, f$, respectively. Then $\omega_C=\soc(e+f-4)$.
\end{prop}
This was (II, Ex. 8.4 e) and it can be found in my homework solutions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/8/96
\section{Curves of Genus 5}
There are hyperelliptic curves of genus 5. For example
a curve of type $(2,6)$ on the quadric surface
$Q\subset\P^3$. Are there any more curves of genus 5?
If $C$ is a curve of genus 5 which is not hyperelliptic
then $\omega_C$ is very ample so there is a canonical embedding
$C\hookrightarrow\P^4$ in which $C$ has degree $8$.
How many quadric hypersurfaces does such a $C$ lie on?
\subsection{The space of quadrics containing an embedded genus 5 curve.}
There is an exact sequence
$$0\into H^0(\sI_C(2))\into H^0(\sO_{\P^4}(2))\into H^0(\soc(2))\into\cdots$$
Riemann-Roch implies $h^0(\soc(2))=\deg(\soc(2))+1-g=16+1-5=12$.
Here $\soc(2)$ has degree 16 since the degree is additive on the
class group and $C$ has degree 8 so the divisor $\soc(1)$ has degree 8.
Also $\soc(2)$ is superspecial since the canonical divisor has
degree $2g-2=8$. Combining this with the fact that $h^0(\sO_{\P^4}(2))=15$
implies $h^0(\sI_C(2))\geq 3$. Thus $C$ lies on a 3 dimensional space
of quadric hypersurfaces. Let $F_2, F_2', F_2''$ be linearly independent
quadric hypersurfaces containing $C$. In the genus 4 case $F_2\intersect F_3$
was a curve $C'$ which we were able to show was equal to $C$. But
in our situation it might be possible that
$$F_2\intersect F_2'=H\isom\P^3.$$
But in that case $C\subset F_2\intersect F_2'$
would be contained in $\P^3$. But this is not true since
$C$ is not contained in any linear subspace since the embedding
comes from the complete linear system $|\omega_C|$. (If $C$
were contained in a linear subspace then there would be a dependence relation
between the linearly independent global sections of $\omega_C$ giving
rise to the embedding.)
If $F_2\intersect F_2'$ is a hypersurface then it is defined by
a single polynomial equation $f=0$. Then $f$ divides the quadratic
defining $F_2$ and the quadratic defining $F_2'$ so $f$ must be linear.
But this case was ruled out above. Thus $S=F_2\intersect F_2'$ is a surface.
We cannot conclude that
$F_2\intersect F_2'\intersect F_2''$ is the curve $C$. It could
happen that $S\subset F_2''$ and so
$S=F_2\intersect F_2'\intersect F_2''$. It could also happen that
$F_2\intersect F_2'\intersect F_2''$ is just a part $S_1$ of the surface $S$.
But this gives us a hint as to how to construct curves of genus $5$.
\subsection{Genus 5 curves with very ample canonical divisor}
In $\P^4$ take $F_2, F_2', F_2''$ three quadric hypersurfaces which are
sufficiently general so that
$C=F_2\intersect F_2'\intersect F_2''$ is a nonsingular curve.
(That this can be done appeals to Bertini's theorem.)
Given such a curve $C$ then $\deg C=2^3=8$ and
$\omega_C=\soc(2+2+2-4-1)=\soc(1)$.
Therefore $C$ is the canonical embedding of the abstract
curve hiding in the shadows behind $C$.
The genus of $C$ is 5 since $2g-2=\deg C = 8$.
This construction gives examples of curves of genus $5$ for which
$\omega_C$ is very ample, i.e., curves of genus 5 which are not
hyperelliptic.
Does this type of curve have a $g^1_3$, i.e., a complete linear system
of degree 3 and dimension 1?
Suppose $C$ is an abstract curve of genus 5 which is not hyperelliptic.
Embed $C$ in $\P^4$ by its canonical embedding $C\hookrightarrow\P^4$.
There are two possibilities:
\begin{enumerate}
\item $C=F_2\intersect F_2'\intersect F_2''$
\item $C\subset S=F_2\intersect F_2'\intersect F_2''$
\end{enumerate}
As seen above case 1 can occur. In case 2, $S$ is a surface
of degree $8$ [[I missed the proof.]] We do not yet know if case
2 can occur.
Is $C$ trigonal, that is, does $C$ have a $g^1_3$?
To study this question we introduce a new technique. Suppose
$C\hookrightarrow \P^4$ is the canonical embedding of $C$
into $\P^4$ as a degree $8$ curve. Then there is a $g^1_3$ iff
there exists three points $P,Q,R\in C$ such that
$\dim|P+Q+R|=1$. Given three points $P,Q,R$ Riemann-Roch implies
$$\dim|P+Q+R|=3+1-5+\dim|K-P-Q-R|=-1+\dim|K-P-Q-R|$$
so $\dim|P+Q+R|=1$ iff $\dim|K-P-Q-R|=2$. The condition that
$\dim|K-P-Q-R|=2$ is that there is a 2 dimensional linear system
of effective canonical divisors $K$ which contain $P,Q,R$.
The technique is to translate the condition $\dim|K-P-Q-R|=2$
into a geometric criterion
involving the embedding $C\hookrightarrow\P^4$. Since
the embedding $C\hookrightarrow\P^4$ is canonical every
effective divisor in the canonical divisor class is the
intersection of $C$ with a hyperplane in $\P^4$.
We obtain {\em every} effective divisor because $|K|$ has
dimension 4 and the dimension of the space of lines
in $\P^4$ is $4$.
Thus an effective canonical divisor contains $P,Q,R$ iff there
is a hyperplane in $\P^4$ containing $P,Q,R$. Hence there
is a 2 dimensional linear system in $\P^4$ containing $P,Q,R$
iff $P,Q,R$ are collinear in $\P^4$. We have thus
interpreted $\dim|P+Q+R|$ in terms of the geometry of where
$P,Q,R$ lie on $C$ in the canonical embedding.
The upshot of this is
\begin{prop}
Suppose $C$ is a not a hyperelliptic curve. Then $C$ has a
$g^1_3$ iff there are $3$ points $P,Q,R\in C$ which
are collinear in the canonical embedding.
\end{prop}
Notice that the proposition is even true without
the assumption that $C$ has genus $5$.
We return to our situation. Suppose $C\hookrightarrow \P^4$
is the degree 8 canonical embedding of some nonhyperelliptic curve
$C$ of genus $5$. Suppose $P,Q,R\in C$ are collinear so they all lie
on some line $L$.
First suppose $C$ is the complete intersection
$C=F_2\intersect F_2'\intersect F_2''$. Then $P,Q,R\in F_2$ and
$P,Q,R\in L$ so $L\intersect F_2$ contains at least $3$ points.
If $L$ is not contained in $F_2$ then
$L\intersect F_2$ has $(\deg L)\cdot(\deg F_2)=2$ points so it must
be the case that
$L\subset F_2$. By similar reasoning we conclude that $L\subset F_2'$
and $L\subset F_2''$ so $L\subset C$, a contradiction. So $C$
does not have a $g^1_3$.
Next suppose $C\subset S=F_2\intersect F_2'\intersect F_2''$. Then
it is possible that $L\subset S$ since this leads to no contradiction.
So {\em maybe} there could be a $g^1_3$ on $C$, we still do not know.
Next we show that there do exist genus $5$ trigonal curves.
We construct directly a genus 5 curve $C$ with nontrivial $g^1_3$.
The things to do is look in $\P^2$ for a curve $D$ of degree $5$
which has one node and no other singularities.
As an aside we consider a more general problem. Consider singular
irreducible curves $C$ of degree $d$ in $\P^2$ containing
$r$ nodes and $k$ cusps. A {\em node} looks locally like $xy=0$
and a cusp looks like $y^2=x^3$. What are the possible triples
$(d,r,k)$ which can occur? The answer for small $d$ is
\begin{center}
\begin{tabular}{|l|c|}\hline
$d$ & $(d,r,k)$\\ \hline\hline
1&(1,0,0)\\ \hline
2&(2,0,0)\\ \hline
3&(3,1,0), (3,0,1) \\ \hline
4&(4,3,0),\ldots \\\hline
\end{tabular}
\end{center}
In general this is an open problem. Solving it is a guaranteed thesis,
seven years of good luck, and an... academic position!
The complete answer is probably known up to degree 10.
One constraint comes from the fact that if $\tilde{C}$ is the
normalization of
$C$ then $$g(\tilde{C})=\frac{1}{2}(d-1)(d-2)-r-k\geq 0$$
so $r+k\leq \frac{1}{2}(d-1)(d-2)$.
Finding a $D$ as above means finding a $(5,1,0)$.
Suppose $\Char k\neq 3,5$. Let
$$f=xyz^3+x^5+y^5.$$
The point $x=y=0$ is a nodal singularity. There are no other singularities.
See this by computing $f_x=yz^3+5x^4$, $f_y=xz^3+5y^4$, and
$f_z=3xyz^2$. For these to all vanish it must be the case that
$x$, $y$, or $z$ is $0$. If $x$ or $y$ is 0 then both $x$ and
$y$ are 0 so we recover the nodal singularity.
If $x$ and $y$ are nonzero then $z=0$ so from $f_x=0$ and $f_y=0$
it follows that $x=y=0$, but $x=y=z=0$ is not a point.
Let $C=\tilde{D}$ be the normalization of $C$. Then $C$
is a genus 5 nonsingular curve. (That the normalization of a curve
is nonsingular is a fact from commutative algebra.)
Pick a node $P$ on $D$. Then lines through $P$ in
$\P^2$ give a map $D-\{P\}\into\P^1$.
A point $Q$ in $\P^1$ corresponds to a line $L$
through $P$. Since $P$ is a double point and $D$ has
degree $5$, we see that $L$ intersects $D$ in $3$ other points.
These three points map to $Q\in\P^1$.
This map extends to a map on $C$ which corresponds to a $g^1_3$.
We can now say something about the $3g-3=12$ dimensional
space of genus $5$ curves.
They can be divided into 3 nonempty classes: trigonal,
hyperelliptic and general ones with no $g^1_3$. The
hyperelliptic curves form a $2g-1=9$ dimensional family. [[The
trigonal curves might form an 11 dimensional family??]]
It might be the case that every hyperelliptic curve is trigonal.
Next time we will do the case of genus $6$ since a new phenomenon appears.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/10/96
\section{Homework Assignment}
Do the following from the book: IV Ex. 3.6, 3.12, 5.4.
Do any one of the following problems (and try to do one that someone
else is not doing!)
1. If $C$ is a hyperelliptic curve of genus $g\geq 2$, find the least
possible degree of a very ample divisor on $C$. Is it independent
of the particular hyperelliptic curve chosen?
2. Can every hyperelliptic curve of genus $g\geq 2$ be embedded in $\P^3$,
so as to be a curve of bidegree $(2,g+1)$ on a non-singular quadric surface
$Q$?
3. If $C$ is hyperelliptic of genus $g\geq 3$, then $C$ does not have a
$g^1_3$ (without base points).
4. If $C$ is a non-hyperelliptic curve of genus $g\geq 4$ show that
$C$ has at most a finite number of $g^1_3$'s.
5. If $C$ is a non-hyperelliptic curve of genus $g\geq 3$, then it admits a very ample
divisor of degree $d\leq g+2$.
\section{Curves of genus 6}
Examples of curves of genus 6.
\begin{itemize}
\item[(a)] degree 5 curve in $\P^2$,
\item[(b)] a type $(2,7)$ curve (of degree 9) on the quadric surface $Q$ in $\P^3$,
\item[(c)] a type $(3,4)$ curve (of degree 7) on the quadric surface $Q$ in $\P^3$.
\end{itemize}
\begin{claim} The abstract curves $C$ which can be realized in types
(a), (b), and (c) are mutually disjoint. \end{claim}
Since (b) is hyperelliptic it has a $g^1_2$. Since (c) is trigonal (i.e.
it has a $g^1_3$) by exercise 3 it does not have a $g^1_2$. Thus
(b) and (c) are disjoint classes.
I will next show that (a) has no overlap with (b) and (c).
First note that (a) has infinitely many $g^1_4$'s, one
for each point. This is because projection through a point gives
a degree $4$ map to $\P^1$.
I will prove that (a) has no $g^1_2$ or $g^1_3$.
\begin{lem} If $C$ is a nonsingular curve of $\deg 5$ in $\P^2$ then $C$ does
not have a $g^1_2$ or a $g^1_3$. \end{lem}
The lemma is a special case of a result of Max Noether:
\begin{quote}
``On a plane curve, the only linear systems $g^r_d$ of maximal dimension
(so $r$ is maximal with respect to $d$) are the {\em obvious} ones.''
\end{quote}
The obvious linear systems are the ones arising in a natural way
by intersecting the curve with a straight line, or with a conic
section, or a conic section but fixing one point, and so on. Suppose
$C$ has degree 5 in $\P^2$. Then cutting with a line gives a
$g^2_5$ since the lines in $\P^2$ form a 2 parameter family and
they intersect $C$ in 5 points. Cutting with a conic gives a $g^5_{10}$
since the conics form a 5 parameter family. By fixing two points
one obtains a $g^4_9$, by fixing three points a $g^3_8$ and similarly
a $g^2_7$ and $g^1_6$. All lines with 1 fixed point gives a $g^2_6$.
More generally, if $C$ has degree $n$ in $\P^2$ then there exists a $g^1_{n-1}$
but Noether's result implies that there does not exist a $g^1_d$ for
$d0$. Thus the canonical embedding
is obtained by following $C\hookrightarrow\P^2$ by the $2$-uple embedding.
From last time we know that there exists a $g^1_3$ on $C$
iff there exists points $P,Q,R\in C$ such that $P,Q,R$ are collinear
in the canonical embedding. This might lead to a proof but I can not
think of it right now, so forget it!
$C$ is hyperelliptic iff there exists $P,Q$ such that $\dim|P+Q|=1$.
By Riemann-Roch
$$\dim|P+Q|=2+1-6+\dim|K-P-Q|$$
so in this situation $\dim|K-P-Q|=4$.
Since $C$ has degree $5$ and the canonical divisor has degree $10$,
the canonical divisor is cut by conics. To see this note that $|K|=g-1=5$
and the dimension of the space of conics in $\P^2$ is also $5$.
If $\dim|K-P-Q|=4$ then there exists $P,Q\in C$ such
that the family
$$\{\text{ conics containing $P,Q$ }\}$$
has dimension $4$.
But the family of all conics in $\P^2$ has dimension $5$, the family
of conics through one fixed point has dimension $4$, and the family
of conics through two fixed points has dimension $3$. Thus $C$ can not
be hyperelliptic.
$C$ is trigonal iff there exists points $P,Q,R$ such that
$\dim|P+Q+R|=1$. By Riemann-Roch this latter condition
implies that $\dim|K-P-Q-R|=3$. This would mean that we could
find three points $P,Q,R$ in $C$ such that the dimension
of the family of conics containing $P,Q,R$ is 3. But the
family of conics through $P$ and $Q$ has dimension $3$ and
there are conics passing through $P$ and $Q$ but not through
$R$ so the family of conics through all of $P,Q,R$ must have
dimension less than 3. Thus $C$ is not trigonal and the lemma
is proved. \end{proof}
Can every curve of genus $6$ be realized as one of type
(a), (b), or (c)? The answer is no.
\begin{center}
\begin{tabular}{|l|c|}\hline
$g$ & $g^r_d$'s\\ \hline\hline
$g=2$ & $\exists g^1_2$\\\hline
$g=3$ & $\exists g^1_3$\\\hline
$g=4$ & $\exists$ finitely many $g^1_3$\\\hline
$g=5$ & $\exists$ infinitely many $g^1_4$ (not shown in class)\\\hline
$g=6$ & we should {\em expect} that the general curve\\
& has only finitely many $g^1_4$'s\\\hline
\end{tabular}
\end{center}
Kleiman-Lacksov prove that what we expect is actually the case in general.
Our examples (a), (b), and (c) all have infinitely many $g^1_4$'s so
we suspect that our examples do not cover all genus 6 curves.
There exists a curve of a fourth type (d)=none of the above, and this
will be the general curve. It is hard to get our hands on a general
curve since any time we explicitly make a curve it has special
properties and is thus not general.
We want to find a plane curve of degree $6$ with $4$ nodes and no other
singularities. The space of all curves of degree 6 has dimension
$\frac{1}{2}d(d+3)=27$. It takes 3 linear conditions to force a node
at a particular point, thus it takes 12 linear conditions to get 4
nodes. Since $12<27$ there exist such curves. The curve could
have weird singularities, but there is a way to get the curve we want.
This is like homework IV.5.4. Thus suppose $C_0\subset\P^2$ is
a curve of degree $6$ with exactly 4 singularities which are all nodes.
Let $C=\tilde{C_0}$ be the normalization of $C_0$. Then
$$g=g(C)=\frac{1}{2}(d-1)(d-2)-4=10-4=6.$$
There are five obvious $g^1_4$'s. Four come from projecting
away from any of the double points. To get the fifth consider conics
passing through all four double points. Thus $C$ has $g^1_4$'s but
none of our previous curves did and so the class (d) is nonempty.
Another question is the following.
\begin{ques}
When $g$ is even there are only finitely many $g^1_{g-1}$'s.
For example when $g=4$ there are two $g^1_3$'s, and when $g=6$
there are five $g^1_4$'s. In general how many $g^1_{g-1}$'s are there?
\end{ques}
We could have also approached showing classes (a), (b), (c), and (d)
are disjoint by asking: what is the least degree of a very ample
divisor? For (a) the least degree is 5, for (b) it is 9, for (c) it
is 7, and for (d) it is 8. We give no proof of this here.
This type of classification method should generalize to arbitrary genus.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/12/96
\section{Oral Report Topics}
During the week of the 29th of April oral reports will be presented
by the students. The suggested topics are
\begin{itemize}
\item Curves $/k$ where $k$ is not necessarily algebraically closed and
rational points on curves over finite fields.
\item The variety of moduli $\mathcal{M}_g$.
\item Duality for a finite smooth morphism $X\into Y$.
\item Jacobian variety of a curve. (``what does it really mean?'')
\item Curves on a nonsingular cubic surface in $\P^3$ (ch 5, sec 4).
\item Flat families of curves in $\P^3$ (ch 3, sec 9).
\end{itemize}
Your oral reports must be 20 minutes in length. They should contain
precise definitions, statements of the main theorems, some examples,
and maybe some proofs if there is time. You should consult with me
before you begin.
For the rest of the semester we will be studying curves of genus
$g$, $g$ general and elliptic curves.
\section{Curves of general genus}
Today's lecture contains hints for some of the homework problems.
\begin{prop} If $C$ is hyperelliptic of genus $g\geq 2$, then the
$g^1_2$ is unique. Furthermore, $K\sim (g-1)D$ for any
$D\in g^1_2$. \end{prop}
\begin{proof} The complete linear system $|K|$ has no base points
iff $\dim|K-P|=\dim|K|-1$ for any point $P$ on $C$.
By Riemann-Roch $\dim|K|=2g-2+1-g=g-1$ and $\dim|P|-\dim|K-P|=2-g$
so $\dim|K-P|=\dim|P|+g-2$. Thus $\dim|K-P|=\dim|K|-1$ for all $P$
iff $\dim|P|=0$ for all $P$ which is true since $g\geq 1$. (If there
were a one dimensional space of points linearly equivalent to a given
point $P$ then $C$ would have genus $0$.) Thus $K$ defines a morphism
(which is not necessarily an embedding)
$$C\into\P^{g-1}.$$
The map is into $\P^{g-1}$ since $\dim|K|=g-1$.
We saw before that if $C$ is not hyperelliptic then this
is an embedding. If $C$ is hyperelliptic then the canonical divisor
can not be very ample so this map will not be an embedding.
First consider the special case $g=1$. Then we have a map
$C\into\P^0=\{\text{ point }\}$ which is not interesting.
Next suppose $g=2$. Then the canonical divisor induces a map
$$C\into\P^1$$ and $K$ is the pullback of $\sO(1)$. Since
$\deg K=2g-2=2$ the map $C\into\P^1$ has degree $2$, and
the complete linear system $|K|$ is a $g^1_2$. In fact,
any $|D|$ is a $g^1_2$ then $D\sim K$. To see this suppose
$P, Q$ are two points and $\dim|P+Q|=1$. By Riemann-Roch,
$$1=\dim|P+Q|=2+1-2+\dim|K-P-Q|$$
so $\dim|K-P-Q|=0$. Thus there is an effective divisor
linearly equivalent to the degree $0$ divisor $K-P-Q$. But the
only effective divisor of degree 0 is the 0 divisor. Thus
$K\sim P+Q$. Thus in genus $2$ {\em any} $g^1_2$ is given by
$K$ since any effective $D$ giving a $g^1_2$ is linearly equivalent to $K$.
Suppose now that $g\geq 3$. The complete linear system $|K|$
gives a morphism
$$\varphi:C\into \P^{g-1}.$$
Since $K$ is not very ample, $\varphi$ is not a closed immersion.
Now assume $C$ is hyperelliptic.
Suppose $P$ and $Q$ are two points on $C$ such that
$P+Q$ lies in some $g^1_2$. Then by Riemann-Roch,
$$1=\dim|P+Q|=2+1-g+\dim|K-P-Q|.$$
Thus $\dim|K-P-Q|=g-2$. But $\dim|K|=g-1$ and $|K|$ has no
base points so $\dim|K-P|=g-2$. Thus $\dim|K-P|=\dim|K-P-Q|$
so $Q$ is a base point of $|K-P|$. This means that any canonical
divisor containing $P$ also contains $Q$. Therefore $\varphi(P)=\varphi(Q)$
if we look at this in terms of the morphism. Thus if $P+Q$
is in some $g^1_2$ then $\varphi(P)=\varphi(Q)$.
Thus $\varphi$ is at least two-to-one. Let $C_0=\varphi(C)\subset\P^{g-1}$
be the image curve. Pulling back $\sO(1)$ on $\P^{g-1}$ gives $\sO_{C_0}(1)$
on $C_0$ which pulls back to the canonical divisor $\soc(1)=K$ which
has degree $2g-2$.
There are two numbers to consider. First the degree of the curve $C_0$,
call it $d$. Let $e$ be the degree of the finite morphism $\varphi$.
Then $$2g-2=\deg|K|=de.$$
See this by noting that $\sO_{C_0}(1)$ is obtained by cutting
$C_0$ with a hyperplane and seeing that it intersects in $d$ points
then pulling these points back to their $de$ preimages to obtain
the $de$ degree divisor $\soc=K$.
Since $e\geq 2$ the equality $de=2(g-1)$ implies $d\leq g-1$.
Consider $C_0$ which is a (possibly singular) integral curve.
Let $D_0=\sO_{C_0}(1)$, then $D_0$ has degree $d$ and since
$|D_0|$ gives the embedding of $C_0$ into $\P^{g-1}$ and
$C_0$ lies in no linear subspace, $\dim|D_0|\geq g-1$.
By a previous proposition (IV, Ex. 1.5), we know that
$\dim|D_0|\leq \deg D_0$ with equality iff $C_0\isom \P^1$.
But $$g-1\leq \dim|D_0|\leq \deg D_0=d\leq g-1$$
so we do have equality and $C_0\isom\P^1$ is nonsingular and is
in fact the $d-1$-uple embedding of $\P^1$ into $\P^{g-1}$.
Furthermore, since $d=g-1$ we also see that $e=2$.
The upshot of all this is that if $C$ is hyperelliptic and $g\geq 3$
then $|K|$ gives an embedding $\varphi:C\into \P^{g-1}$ which factors
through the $g-1$-uple embedding of $\P^1$. Thus $\varphi:C\into\P^{g-1}$
can be written as a composition
$$C\xrightarrow{g^1_2} C_0=\P^1\hookrightarrow \P^{g-1}.$$
[[One shows that the $g^1_2$ is uniquely determined, etc. as
on page 343 of the book. WRITE THIS UP HERE!]]\end{proof}
I thought of another proof of the uniqueness of $g^1_2$'s. It is
not in the book. This technique is useful for the exercises.
The drawback is that it does not explicitly give $K$.
\begin{thm} If $C$ is a curve of genus $g\geq 2$ then $C$ can not
have two distinct $g^1_2$'s. \end{thm}
\begin{proof}
The proof is by contradiction. Suppose $C$ has non-linearly equivalent
divisors $D$ and $D'$ such that $|D|$ and $|D'|$ are $g^1_2$'s. Then
$|D|$ and $|D'|$ give morphisms $p$ and $p'$ to $\P^1$
Taking the product gives a morphism $\varphi=p\times p':C\into\P^1\cross\P^1$.
Since $|D|$ and $|D'|$ do not contain any of the same divisors the
morphisms $p$ and $p'$ collapse different points and so they are different.
Let
$$C_0=\varphi(C)\subset \P^1\cross\P^1=Q\subset\P^3$$
then $C_0$ is a possibly singular but still integral curve on the quadric
surface $Q$. By construction each ruling gives a $g^1_2$ on $C$.
Let $(a,b)$ be the bidegree of $C_0$ on $Q$. It
is clear that $C_0$ is not a point since $p$ and $p'$ are not constant.
Let $e$ be the degree of $\varphi:C\into C_0$.
A divisor of type $(1,0)$ on $Q$ pulls back to a divisor on $C_0$
of degree $a$. Since the morphism
$C\into C_0$ has degree $e$ this divisor of degree $a$ pulls
back to a degree $ea$ divisor on $C$. On the other hand
a divisor of type $(1,0)$ on $Q$ pulls back to a divisor of
degree $2$ since it gives rise to $p$. Thus $2=ae$ and similarly
$2=be$.
{\em Case 1:} First suppose $e=2$ and hence $a=b=1$. Then $C_0$ is
of type $(1,1)$ so the normalization of $C_0$ is of genus $0$. But
the arithmetic genus of $C_0$ is $(1-1)(1-1)=0$ so, since the genus
can only go now upon normalization, we see that $C_0$ must already
be nonsingular. Thus $\varphi$ maps $C$ into $\P^1$. But $p$ is
just the composition of $\varphi$ with the first projection $Q\into\P^1$
and this projection is the identity $C_0=\P^1\subset Q\into \P^1$.
Thus $p$ and $p'$ collapse the same points, a contradiction.
{\em Case 2:} Next suppose $e=1$ and hence $a=b=2$. Then
$\varphi:C\into C_0$ induces a birational morphism to the
normalization of $C_0$. But the normalization of $C_0$
has genus less than or equal to $p_a(C_0)=(2-1)(2-1)=1$ so
the genus of $C$ is less than or equal to $1$. In fact, if the
genus of $C$ is one then there are two distinct $g^1_2$'s. But
under our assumption that $g\geq 2$ we have the desired contradiction.
\end{proof}
The above proof probably can be generalized to show that, for large
enough genus, $C$ can not have any $g^1_d$'s.
As in case 1 we saw that a type (1,1) curve $C_0$ can not be singular since
the normalization would then have negative genus. This might help
with problems 3 and 4.
Next we give a statement of the theorem which will be proved next time.
\begin{thm}[Halphen]
Let $C$ be a curve of genus $g\geq 2$. Then there exists a very
ample nonspecial divisor $D$ of degree $d$ iff $d\geq g+3$.
\end{thm}
The existence is actually very strong in the sense that a Zariski
open subset has the property.
This generalizes our proof that $g=3$ implies that there exists a very
ample divisor of degree $6$. Note that $2g+1= 7$.
There can be very ample {\em special} divisors of degree less than
$g+3$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% nh.tex
%% 4/15/96
\section{Halphen's Theorem}
Today we will give one more general result about curves.
\begin{thm}[Halphen] Suppose $C$ is a curve of genus $g\geq 2$. Then
there exists a very ample nonspecial divisor $D$ of degree $d$
iff $d\geq g+3$.\end{thm}
\begin{proof}
($\Rightarrow$)
Suppose $D$ is a very ample nonspecial divisor of degree $d$. We
will show that $d\geq g+3$. By Riemann-Roch $\ell(D)\geq d+1-g$.
Since $D$ is very ample $D$
gives rise to an embedding $C\hookrightarrow \P^n$. If $n=1$
then $C$ has genus 0, a contradiction. If $n=2$ then
somehow [I can not figure this out from my notes],
this implies $g<2$, a contradiction. Thus $D$ gives an embedding
$C\hookrightarrow\P^n$ with $n\geq 3$. Thus $\ell(D)\geq 4$
and so $d+1-g\geq 4$ so $d\geq g+3$ as claimed.
($\Leftarrow$)
This direction is a little harder. Assume $d\geq g+3$ and look
for a very ample nonspecial divisor $D$. Remember our criterion
for when a divisor is very ample? A divisor $D$ is very ample
iff $\dim|D-P-Q|=\dim|D|-2$ for all points $P,Q$.
By Riemann-Roch this is equivalent to the assertion that
for all points $P,Q$,
$$\dim|K-D|=\dim|K-D+P+Q|.$$
As a ``warm-up exercise'' we fix $d$ and ask the following.
Does there exist a nonspecial effective divisor $D$ of degree $d$?
Recall:
\begin{itemize}
\item For $D$ to be {\em nonspecial} means that $\dim|K-D|=-1$.
\item For $D$ to be {\em special} means that $K-D$ is linearly equivalent
to an effective divisor. Thus $D$ is special iff there exists an effective
divisor $E$ such that $D+E\sim K$, i.e., $D+E$ is a canonical divisor.
\end{itemize}
The last condition shows that $D$ is special iff $D$ is
``inside'' {\em some} canonical divisor.
Since $\dim|K|=g-1$ there is a $g-1$ dimensional family of
effective canonical divisors. From each one there are only finitely many
ways to get a special divisor. Thus the dimension of the family of all
special divisors of degree $d$ is at most $g-1$. This shows that if
$d\geq g$ then there exists nonspecial divisors of degree $d$. In fact,
most are nonspecial.
Next we prove something which is more difficult. We want to show
that there exists a nonspecial very ample divisor of
any given degree $d\geq g+3$. It is enough to prove that the
collection of nonspecial not very ample divisors has dimension
$\leq g+2$.
Suppose $D$ is nonspecial. Then for $D$ to be not very ample means
that there exists points $P$ and $Q$ so that
$$\dim|D-P-Q|>\dim|D|-2.$$
A straightforward check using Riemann-Roch shows that this is equivalent
to
$$\dim|K-D|<\dim|K-D+P+Q|.$$
Since $D$ is assumed nonspecial $\dim|K-D|=-1$ thus
$$\dim|K-D+P+Q|\geq 0.$$ So there exists an effective divisor
$E$ which is special and has degree $d-2$ such that
$E\sim D-P-Q$. [[I thought I saw this yesterday but not even
this makes any sense today.]]
By the above work the dimension of
the set of effective special divisors is $\leq g-1$. Thus
the set
$$\{\text{ $E+P+Q$ : $E$ is effective, special, degree $d-2$, $P,Q$ any points }\}$$
has dimension $\leq g+1$.
But there is another wrinkle. We must count all divisors linearly
equivalent to any such $E+P+Q$.
Somehow [[and I haven't figured out how!!]] Hartshorne counts
this and concludes that it has dimension $\leq g+2$. [[I thought
sort of hard about this and can not see it, but I got confused
at this point in class when I was taking notes so that may be
why my notes do not reveal the truth.]]
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
%% 4/17/96 Hartnotes.tex
\section{Hurwitz's Theorem}
Suppose $X$ and $Y$ are curves (nonsingular, projective, over an algebraically
closed field $k$). Suppose $f:X\into Y$ is a finite morphism. Then
$f$ induces a map of function fields $K(Y)\hookrightarrow K(X)$ which
makes $K(X)$ into a finite extension of $K(Y)$. The {\em degree} of
$f$ is the degree of the corresponding extension of function fields
$K(Y)/K(X)$.
We say $f$ is {\em separable} (not to be confused with separated) if
$K(X)$ is a separable field extension of $K(Y)$.
Suppose $P\in X$ maps to $Q\in Y$. Then there is an induced map of
local rings $f^{\#}:\sO_Q\hookrightarrow\sO_P$. Since $X$ and $Y$ are
nonsingular curves this is an extension of discrete valuation rings.
Let $t\in\sO_Q$ be a uniformizing parameter for $\sO_Q$ and
let $u\in\sO_P$ be a uniformizing parameter for $\sO_P$. Then
$f^{\#}(t)\in\m_P\subset\sO_P$. The {\em ramification index}
of $P$ over $Q$ is $e_P:=v_P(f^{\#}(t))\geq 1$.
We say that a point $P$ lying over a point $Q$ is {\em wildly ramified}
if $\Char k=p>0$ and $p|e$. Otherwise the ramification is
called {\em tame}, i.e., when $\Char k=0$ or $\Char k=p\nd e$.
Note that we do not, as in some definitions, need to worry about
the extension of residue fields being separable because $k$ is
algebraically closed.
\begin{thm}[Hurwitz] Suppose $f:X\into Y$ is a finite separable
morphism of curves which is at most tamely ramified. Then
$$2g(X)-2=(deg f)(2g(Y)-2)+\sum_{P\in X}(e_P-1).$$\end{thm}
Note that as a consequence of this theorem, a finite separable
tamely ramified morphism of curves has at most finitely many
points of ramification. ``Any good theorem has a counterexample.''
\begin{example}[Frobenius map]
If the morphism $f$ is not separable then there can be infinitely
many points of ramification.
For example, let $k$ be a field of characteristic $p$.
The divisor corresponding to the invertible sheaf $\sO_{\P^1}$
gives a map from $\P^1\into\P^1$. It is given explicitly
simply by $$(x:y)\mapsto(x^p:y^p).$$
[[Is this true, or do I just want it to be true?]]
For any point $P\in\P^1$
we have that $e_P=p$ since the map corresponds
to the map $$k(t)\into k(t):t\mapsto t^p.$$ Thus this
map is widely ramified everywhere.
\end{example}
\begin{proof} [Hurwitz's theorem]
[[First part omitted.]]
Thus we have the exact sequence
$$0\into f^*\Omega_{Y/k}\into\Omega_{X/k}\into\Omega_{X/Y}\into 0$$
and $\Omega_{X/Y}$ is a torsion sheaf so it equals
$\oplus_{P\in X}(\Omega_{X/Y})_P$.
Next we study $\Omega_{X/Y}$ locally. Let $P$ be a point in
$X$ lying over $Q$ in $Y$. Then there is an exact sequence
$$0\into f^*\Omega_Q\into\Omega_P\into\Omega_{P/Q}\into 0.$$
Here $\Omega_{P/Q}$ is the module of differentials of the local
ring $\sO_P$ over $\sO_Q$. Let $t$ be a uniformizing parameter
for $\sO_Q$ and let $u$ be a uniformizing parameter for $\sO_P$.
Then $\Omega_Q$ is a free $\sO_Q$-module of rank $1$ locally
generated by $dt$.
Next let $e=e_P=v_P(t)$, thus $t=au^e$ in $\sO_P$
(with $a$ a unit in $\sO_P$). Differentiate to see that
$$dt=aeu^{e-1}du+u^eda.$$
The term $u^eda$ is some mysterious element of $\Omega_P$. We
do not know anything about what $a$ looks like so $da$ can
be very strange. If $e=0$ in $k$ this means that $dt$ would
be {\em wild} in the sense that we have no real control over $da$.
Since we are assuming that the ramification is tame,
$\Char k\nd e$, so $e\neq 0$ in $k$. Thus
$aeu^{e-1}du\neq 0$. Define $b\in\sO_Q$ by $u^eda=bdu$. Since
$v_p(b)\geq e$ and $v_p(ae u^{e-1})=e-1$ we see that
if $dt=Adu$ then $v_p(A)=e-1$. This means that $u^{e-1}du$ is
zero in $\Omega_{P/Q}$ [[but why is no lower power of $u$
times $du$ also 0.]] Thus $\Omega_{P/Q}$ is a principal
module generated by $du$ of length $e-1$. Thus noncanonically
$$\Omega_{P/Q}\isom\sO_P/u^{e-1}.$$
We thus have an exact sequence
$$0\into f^*(\Omega_Y)\into \Omega_X\into R\into 0$$
where $R=\oplus_P \sO_P/u_P^{e_P-1}$.
Let $K_X$ denote the canonical divisor on $X$ and $K_Y$
be the canonical divisor on $Y$. Then $\Omega_X=\sL(K_X)$
and $\Omega_Y=\sL(K_Y)$.
\end{proof}
\begin{example}
Assume $\Char k\neq 2$.
Let $C$ be the cubic curve in $\P^2$ defined by $y^2=x(x+1)(x-1)$.
Let $\pi$ by the degree 2 projection of $C$ from $\infty$
onto $\P^1$ (i.e., the $x$-axis). There are $4$ points of ramification,
namely $(-1,0), (0,0), (1,0),$ and $\infty$.
Hurwitz's theorem is satisfied since
$$2\cdot 1 -2 = 2(2\cdot 0 - 2)+\sum_{4\text{ points}}(2-1).$$
\end{example}
\begin{example}
Let $C$ be a cubic curve in $\P^2$. Let $\sO$ be a point not on $C$
and fix a copy of $\P^1\subset \P^2$. Then projection from $\sO$
onto $\P^1$ defines a degree $3$ map $C\into\P^1$. This map
is ramified at a point $P\in C$ exactly when the line
from $\sO$ to $P$ is tangent to $C$. Assume there are no inflectional
tangents so that the ramification degree of ramified points is $2$.
By Hurwitz's theorem,
$$0=3(-2)+\sum(e_P-1)$$
so there are $6$ points of ramification $2$. This means that from
$\sO$ one can draw $6$ tangent lines to $C$.
\end{example}
\begin{example}
Let $C$ be a cubic curve in $\P^2$. Define a
map $\varphi:C\into\P^1$ by sending a point
$P$ to the intersection of the tangent space to $C$ at
$P$ with $\P^1\subset\P^2$. By the above example this
map has degree $6$.
By Hurwitz's theorem,
$$0=6(-2)+\sum(e_P-1)$$
so there are $12$ points of ramification. [[Of course
one must show (or set things up so) that the ramification
of a point is at most $2$.]] There are $3$ ramification
points where $\P^1$ intersects $C$. The other $9$ ramification
points are inflection points. Thus Hurwitz's theorem implies
that there are 9 inflections on a cubic.
\end{example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/21/96
\section{Elliptic Curves}
\begin{defn} An elliptic curve $C$ is a nonsingular projective
curve of genus $1$.\end{defn}
We know that any divisor of degree $d\geq 3$ is very ample. Thus there is
an embedding $C\hookrightarrow\P^2$ of $C$ into $\P^2$ as a degree $3$ curve.
\begin{thm}
Suppose $\Char k\neq 2$ and $k$ is algebraically closed. If $C$
is an elliptic curve then there is an embedding
$C\hookrightarrow\P^2$ with equation
$$y^2=x(x-1)(x-\lambda)$$
for some $\lambda\in k$, $\lambda\neq 1, 0$.
\end{thm}
\begin{proof}
Fix a point $P_0\in C$ and let $D=3P_0$. Then, after choosing section
$x_0, x_1, x_2\in H^0(\sL(3P_0))$ we obtain an embedding $C\hookrightarrow
\P^2$. Construct basis for the global sections of $\sL(nP_0)$ for various
$n$.
\begin{center}
\begin{tabular}{l|c|c|c|c|c|c|c|}
&$\soc$&$\sL(P_0)$&$\sL(2P_0)$&$\sL(3P_0)$&$\sL(4P_0)$&$\sL(5P_0)$&$\sL(6P_0)$\\\hline
$h^0$&1&1&2&3&4&5&6\\\hline
basis of $H^0$ &1&1&$1,x$&$1,x,y$&$1,x,y,x^2$&$1,x,y,x^2,xy$&$1,x,y,x^2,xy,x^3,y^2$\\\hline
\end{tabular}
\end{center}
Since the seven function
$1,x,y,x^2,xy,x^3,y^2$ lie in a six dimensional space there must
be a dependence relation
$$ay^2+bx^3+cxy+dx^2+ey+fx+g=0$$
for some $a,b,c,d,e,f,g\in k$. Furthermore, both $x^3$ and $y^2$ occur
with coefficient not equal to zero, because they are the only functions
with a $6$-fold pole at $P_0$. So replacing $y$ be a suitable scalar
multiple we may assume $a=1$. Preparing to complete the square we rewrite
the relation as
$$y^2+(cxy+ey)+(\frac{1}{2}cx+\frac{1}{2}e)^2-(\frac{1}{2}cx+\frac{1}{2}e)^2
+bx^3+dx^2+fx+g=0.$$
Replacing $y$ by $\frac{1}{2}cx+\frac{1}{2}e$ transforms the equation into
$$y^2=e(x-a)(x-b)(x-c)$$ where
$a,b,c,e\in k$ are new constants.
Next absorb $e$ to obtain
$$y^2=(x-a)(x-b)(x-c).$$
Now translate $x$ by $a$ to obtain an equation of the form
$$y^2=x(x-a)(x-b).$$
Multiply and divide by $a^3$ to obtain
$$y^2=a^3\frac{x}{a}(\frac{x}{a}-1)(\frac{x}{a}-\frac{b}{a}).$$
Replace $x$ by $\frac{x}{a}$ and absorb $a^3$ to get
$$y^2=x(x-1)(x-\lambda)$$
where $\lambda\neq 0, 1$ since $C$ is nonsingular.
\end{proof}
In this proof we definitely used that $k$ is algebraically closed to
absorb constants and that the characteristic is not $2$ in order to
complete the square.
\begin{remark}
Since $A=k[x,y]/(y^2-x(x-1)(x-\lambda))$ is not a UFD
the class group of $C$ is nontrivial.
\end{remark}
How unique is the representation $y^2=x(x-1)(x-\lambda)$? The answer
is that it is not unique. For example, replace $x$ by $x+1$ to obtain
$$y^2=(x+1)x(x+1-\lambda)$$
then divide by $-(1-\lambda)$ to obtain
$$y^2=-(\frac{x}{1-\lambda}+\frac{1}{-\lambda}
\frac{x}{1-\lambda}
(\frac{x}{1-\lambda}+1).$$
Next replace $x$ by $-\frac{x}{1-\lambda}$ and absorb the minus sign to obtain
$$y^2=x(x-1)(x-\frac{1}{1-\lambda}).$$
By similar methods we get any of six choices:
$$\lambda\mapsto\lambda,\frac{1}{\lambda},1-\lambda,\frac{1}{1-\lambda},
\frac{\lambda}{\lambda-1},1-\frac{1}{\lambda}.$$
\begin{prop}
In the representation $C$ in the form
$$y^2=x(x-1)(x-\lambda)$$,
$\lambda$ up to the group $S_3$ depends only on $C$.
\end{prop}
The first thing to struggle with is the dependency
on the choice of $P_0$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%4/22/96, hartnotes.tex
\section{Automorphisms of Elliptic Curves}
Let $C$ be an elliptic curve so $C$ has genus $1$. Suppose
$\Char k\neq 2$. Fix a point $P_0$ on $C$. Then
the linear system $|3P_0|$ gives an embedding
$$|3P_0|:C\hookrightarrow\P^2.$$
(The divisor $3P_0$ is very ample by (3.3.3) and
Riemann-Roch implies $\dim|3P_0|=2$.)
For suitable choice of coordinates $C$ is given by
$$y^2=x(x-1)(x-\lambda),\quad \lambda\in k,\quad \lambda\neq 0,1.$$
The choice of $\lambda$ is not unique since $S_3$ acts
by
$$\lambda\mapsto\lambda,\frac{1}{\lambda},1-\lambda,1-\frac{1}{\lambda},
\frac{1}{1-\lambda},\frac{\lambda}{\lambda-1}.$$
Except for this ambiguity, $\lambda$ is uniquely determined.
Define
$$j(\lambda)=\frac{2^8(\lambda^2-\lambda+1)^3}{\lambda^2(1-\lambda)^2}.$$
Then
\begin{thm} The $j$-invariant has the following properties.
\begin{itemize}
\item $j\in k$,
\item $j$ depends only on $C$,
\item $C_1\isom C_2$ iff $j(C_1)=j(C_2)$, and
\item for all $j\in k$ there is a curve $C$ such that $j(C)=j$.
\end{itemize}
\end{thm}
The theorem implies that there is a bijection of sets
$$\Bigl\{\text{ isomorphism classes of elliptic curves / $k$ }\Bigr\}
\iso \Bigl\{\text{ elements of $k$ }\Bigr\}.$$
This can be given a moduli theoretic interpretation.
\begin{example}
Let $E$ be the elliptic curve defined by
$$y^2=x(x-1)(x+1)=x^3-x.$$
Thus $\lambda=-1$ so
$$j=\frac{2^8(3)^3}{1\cdot 2^2}=2^6\cdot3^3=12^3=1728.$$
\end{example}
If $C$ is an elliptic curve then $\Aut C$ is transitive.
So to study the automorphisms of $C$ we need to cut down
the number under consideration. To do this fix a point $P_0$ and
consider
$$\Aut(C,P_0)=\{\text{ automorphisms of $C$ fixing $P_0$ }\}.$$
Suppose $C$ is given by the equation $$y^2=x(x-1)(x-\lambda)$$ and
$P_0=(0:1:0)$ is the point at infinity. Then the map $\sigma$
defined by $y\mapsto -y$ and $x\mapsto x$ is an element
of $\Aut(C,P_0)$. Note that $\sigma$ is the covering transformation
of some 2-to-1 map $C\into\P^1$.
Is $\sigma$ unique? Let $P$ be any point on $C$ such that
$Q=\sigma(P)\neq P$. Then $\sigma$ is a covering transformation of
the morphism determined by the linear system $|P+Q|$. Associated
to any degree 2 map to $\P^1$ there is a covering transformation
(the hyperelliptic involution). Thus if $R+S$ is an effective divisor
of degree $2$ which is not linearly equivalent to $P+Q$ then
we obtain a different degree $2$ morphism to $\P^1$ and hence a different
automorphism of degree two. The conclusion in class was that
$\sigma$ is {\em not} unique because you obtain many different automorphisms
of degree $2$ in the above manner. But there is no reason any of these should
{\em fix} $P_0$. Furthermore, if $\Aut(C,P_0)$ really turns out to be
the units in an order in a number field then there $\sigma$ must
be unique because there is only one nontrivial square root of $-1$.
Let $C$ be as in the above example, so $C$ is defined by
$$y^2=x(x-1)(x+1).$$ Then the map $\tau$ defined by
$$\tau:\quad \begin{cases}x\mapsto-x\\y\mapsto iy\end{cases}$$
is an automorphism of $C$ fixing $P_0=(0:1:0)$. Furthermore
$\sigma=\tau^2$ so $\Aut(C,P_0)$ contains at least $\Z/4\Z$.
Is $\Aut(C,P_0)$ exactly $\Z/4\Z$? We will come back to this
question later.
The map sending every point to its double under the group law
is not an automorphism although it does fix $P_0$.
Since $k$ is algebraically closed there are $4$ points on $C$ which
map to $P_0$ under multiplication by $2$.
\begin{example}
Let $C$ be the genus $1$ curve in $\P^3$ defined by
$$x^3+y^3+z^3=0.$$
Suppose $\Char k\neq 2, 3$. We want to find an equation for $C$
of the form $$y^2=x(x-1)(x-\lambda).$$ Since
$\infty$ is an inflectional tangent of
$y^2=x(x-1)(x-\lambda)$ we hunt for an inflectional
tangent of $C$. The line $y+z=0$ meets
$C$ in an inflectional tangent since
$$x^3+y^3+z^3=x^3+(y+z)(y^2-yz+z^2)$$
so setting $y+z=0$ gives $x^3=0$.
Next perform the change of variables $z=z'-y$. The reason for doing this
is so that $z'=0$ will give the point of intersection of $C$ with
$y+z=0$. After substitution the equation becomes
$$x^3+y^3+(z'-y)^3=0$$
which is
$$x^3+y^3+{z'}^3-3{z'}^2y+3{z'}y^2-y^3=0.$$
Setting $z'=1$ we obtain
$$x^3-3y+3y^2+1=0$$
or equivalently, after factoring the -1 into $x^3$,
$$y^2-y=x^3-\frac{1}{3}.$$
Next working modulo linear changes of variables gives
$$y^2-y+\frac{1}{4} = x^3-\frac{1}{3}+\frac{1}{4},$$
$$(y-\frac{1}{2})^2 = x^3-\frac{1}{12},$$
$$y^2 = x^3-\frac{1}{12},$$
$$y^2 = x^3-1,$$
$$y^2 = (x-1)(x-\omega)(x-\omega^2),\quad \text{where $\omega^3=1$}.$$
In general
$$y^2=(x-a)(x-b)(x-c)$$
is equivalent to $y^2=x(x-1)(x-\lambda)$
where $\lambda=\frac{a-c}{b-c}$. In our situation this means
$$\lambda=\frac{1-\omega}{\omega^2-\omega}
=\frac{1-\omega}{\omega(\omega-1)}=-\frac{1}{\omega}
= -\omega^2=\frac{1}{2}+\frac{1}{2}\sqrt{-3}. $$
Thus $j(\lambda)=0$ and $C$ is defined by
$$y^2=x(x-1)(x+\omega^2).$$
At this point Hartshorne tried to explicitly describe a degree
3 automorphism of $C$ which fixes infinity. But try as he may it
did not come out right. Note that $x\mapsto x$ and $y\mapsto -y$
defines a degree 2 automorphism of $C$.
To define a degree 3 automorphism look at the Fermat form of $C$
$$x^3+y^3+z^3=0, \qquad P_0=(0:1:-1).$$
Then
$$\sigma:\quad \begin{cases}x\mapsto\omega x\\y\mapsto y\\z\mapsto z
\end{cases}$$
is an automorphism of degree $3$ fixing $P_0$.
Interchanging $y$ and $z$ yields an automorphism of degree $2$ fixing
$P_0$ and which commutes with $\sigma$.
Thus $\Aut(C,P_0)$ is at least $\Z/6\Z$.
\end{example}
Now we approach the automorphism group abstractly.
\begin{thm}
If $C$ is an elliptic curve and $P_0$ is a fixed point then
$$\Aut(C,P_0)=\begin{cases}
\Z/2\Z & \text{if $j\neq 0, 12^3$}\\
\Z/4\Z & \text{if $j=12^3\neq 0$, $\Char k\neq 3$}\\
\Z/6\Z & \text{if $j=0\neq 12^3$, $\Char k\neq 3$}\\
\Z/12\Z & \text{if $j=0=12^3$, $\Char k=3$}\\
\Z/24\Z & \text{if $j=0=12^3$, $\Char k=2$}
\end{cases}$$
\end{thm}
We do not prove the theorem now but the idea is to stare at the
following square
$$\begin{array}{ccccc}
&C&\xrightarrow{\text{ $\sigma$ }}&C\\
|2P_0|&\downarrow&&\downarrow&|2P_0|\\
&\P^1&\xrightarrow{ \text{ $\exists$ $\tau$ }}&\P^1\end{array}.$$
\begin{example}
Consider
$$y^2=x(x-1)(x+1)$$
in characteristic 3.
Define automorphisms
$$\tau:\quad\begin{cases}x\mapsto -x\\y\mapsto iy\end{cases}
\quad \rho:\quad\begin{cases}x\mapsto x+1\\y\mapsto y\end{cases}$$
of orders $4$ and $3$.
Clearly $\tau\rho=\rho\tau$\ldots but upon checking this,
clearly $\tau\rho\neq\rho\tau$. Thus $\Aut(C,P_0)$ is obviously
{\em not} cyclic of order 12, in fact it is an extension of $S_3$.
\end{example}
Let $C$ be an elliptic curve and $P_0$ a fixed point on $C$.
Then we associate to the pair $C$, $P_0$ the algebraic objects
\begin{itemize}
\item $\Aut(C,P_0)$,
\item a group structure on $C$ with $P_0$ as the identity, and
\item the endomorphism ring $\End(C,P_0)$.
\end{itemize}
The ring $\End(C,P_0)$ is defined to be the set of morphisms $\theta:C\into C$
fixing $P_0$. By general facts about abelian varieties any such
$\theta$ is in fact a group endomorphism. In characteristic $0$
this endomorphism ring is either $\Z$ or an order in an
imaginary quadratic number field. In characteristic $p$ the endomorphism
ring can be an order in a rank $4$ quaternionic extension of $\Z$.
In characteristic $p$ the endomorphism ring can not be just $\Z$
because of the Frobenius endomorphism which verifies a certain quadratic
equation. [[How does this last statement work?]]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/24/96, hartnotes.tex
\section{Moduli Spaces}
For the rest of the semester I am going to talk about Jacobian varieties,
variety of moduli and flat families. In each of these situations we
parameterize ``something or others'' by points on a variety then
apply algebraic geometry to the variety.
Here are some examples.
\begin{center}
\begin{tabular}{|c|l|l|}\hline
Start&Set&Parameter Space\\\hline
A curve $C$& $\Pic^0 C$ = \{ divisors of degree & Closed points of \\
& 0 modulo linear equivalence \} & Jacobian variety $J$.\\
& Not just a set but a functor & $J$ represents the\\
& $\Sch\into\Set$ & functor. \\\hline
$\P_k^n$ and & set of closed subschemes & The Hilbert scheme\\
$P(z)\in\Q[z]$ & $Z\subset\P^n_k$ with Hilbert & $\Hilb^P(\P^n)$ \\
& polynomial $P(z)$ & \\\hline
fix $g>0$ & isomorphism classes of curves & The variety $M_g$ of\\
& of genus $g$ & moduli. Fine if reps. the\\
& & functor, coarse if not.\\\hline
$\P^n_k$, & set of cycles of dim. $r$ and & The Chow scheme.\\
degree $d$, and & degree $d$ in $\P^n$ modulo & It does not represent\\
dimension $r$ & rational equivalence. & a functor.\\\hline
\end{tabular}
\end{center}
A cycle of dimension $r$ is a $\Z$-linear combination of
subschemes of dimension $r$. If
$Z=\sum n_i Y_i$ is a cycle of dimension $r$ then
$\deg Z=\sum n_i \deg(Y_i)$.
\section{The Jacobian Variety}
Fix a curve $C$ of genus $g$. Then
$\Pic^0 C$ is the group of divisors of degree 0 modulo linear
equivalence. The Jacobian is going to be a variety $J$ whose
closed points correspond to divisors $D\in\Pic^0 C$.
A closed point of $J$ is a map $\varphi: \Spec k\into J$.
Taking the product (over $\Spec k$) of this map with the
identity map $C\into C$ gives a commuting square
$$\begin{CD}
C @>\varphi'>> C\cross J\\
@VVV @VVV\\
\Spec k @>\varphi>> J
\end{CD}$$
Given an invertible sheaf $\sD$ on $C\cross J$ and a closed point
$t$ of $J$ (given by a morphism $\varphi:\Spec k\into J$)
define an invertible sheaf $\sD_t$ on $C$ by
$$\sD_t={\varphi'}^*\sD.$$
In this way we obtain a map from the closed points on $J$ to
$\Pic(C)$.
We first strengthen our requirement for the Jacobian of $C$ by
asking for a divisor $\sD$ on $C\cross J$ such that the map
$\sD\mapsto\sD_t$ gives a correspondence
$$\text{closed points}(J)\iso \Pic^0(C).$$
Grothendieck's genius was to generalize all of this.
The point is to replace $\Spec k$-valued points of $J$ with
$T$-valued points of $J$ where $T$ is any scheme over $k$.
Taking the product of $T\xrightarrow{\varphi} J$ with
$C\xrightarrow{\id_C}C$ we obtain a diagram
$$\begin{CD}
C\cross T @>\varphi'>> C\cross J\\
@VVV @VVV\\
T @>\varphi>> J
\end{CD}$$
If $\sL$ is an invertible sheaf on $C\cross J$ then associate
to $\sL$ the invertible sheaf $\sM={\varphi'}^*\sL$.
Next we make a stronger requirement for the Jacobian of $C$.
We require that there exist an invertible sheaf with the following
universal property. For any scheme $T$ and any invertible sheaf $\sM$
on $C\cross T$ which is ``of degree 0 along the fibers'' there
exists a unique morphism $\varphi:T\into J$ such that
$\sM={\varphi'}^*\sL$.
The point is that we are parameterizing families of divisor
classes. We make the following tentative definition. It will
{\em not} turn out to be the right one.
\begin{defn} The {\em Jacobian variety} is a pair $(J,\sL)$ with
$\sL$ an invertible sheaf on $C\cross J$ such that
for all schemes $T$ and for all invertible sheaves $\sM$ on
$C\cross T$ of degree $0$ on the fibers,
there exists a unique morphism $\varphi:T\into J$ such that
$\sM\isom {\varphi'}^*\sL$. \end{defn}
What does ``degree 0 on the fibers'' mean?
Suppose $\sM$ is an invertible sheaf on $C\cross T$. Let $t$ be a
point in $T$. Thus $t$ can be thought of as a map
$$\psi:\Spec \kappa(t)\into T$$ where
$\kappa(t)$ is the residue field at $t$. Tensoring this map with the
identity map $\id: C\into C$ yields a map
$$\psi':C_t=C\cross_k \Spec\kappa(t)\into C\cross T.$$
If $\sM$ is an invertible sheaf, define $\sM_t$ to be the
pullback ${\psi'}^*\sM$. The degree of $\sM_t$ as a divisor on the curve
$C_t$ makes sense. Although $\kappa(t)$ may not be algebraically
closed we can still define the degree of a divisor in a natural
way.
\begin{defn}
$\sM$ on $C\cross T$ is said to be {\em of degree $0$ along the fibers}
if for all $t\in T$, $\deg \sM_t=0$.
\end{defn}
\begin{remark}
One can show that the map $t\mapsto\deg(\sM_t)$
is a continuous function $T\into\Z$.
\end{remark}
There is still one subtlety. Our tentative definition of the Jacobian
is bad since it obviously cannot exist.
Consider the diagram
$$\begin{CD}
C\cross T @>\psi'>> C=C_t @>\varphi'>> C\cross J\\
@VpVV @VVV @VVV\\
T @>\psi>> \Spec k @>\varphi>> J\end{CD}$$
Pullback $\sL$ on $C\cross J$ via $\varphi'$ to obtain
$\sL_t={\varphi'}^*\sL$.
Then pullback $\sL_t$ to obtain
${\psi'}^*\sL_t$ on $C\cross T$. Let $\sN$ be any invertible
sheaf on $T$. Then $\sF=p^*\sN\tensor{\psi'}^*\sL_t$ is an
invertible sheaf on $C\cross T$. Furthermore
$p^*\sN$ is trivial on the fibers so the fibers of $\sF$ do
not depend on $\sN$.
[[I can not seem to figure out why
this all leads to a contradiction!]]
For some reason
the map from $T\into J$ which corresponds to $\sF$ must
be the constant map sending everything to the point
corresponding to $\varphi$. Since in general there are
many possibilities for $p^*\sN$ this gives a contradiction.
[[No matter what, I can not seem to understand this. I do not
see why the map from $T$ to $J$ corresponding to $\sF$
must be anything in particular.]]
So to the correct definition is that
we require $\sM$ to equal ${\varphi'}^*\sL$ in the quotient group
$$\Pic^0(C\cross T/T) := \Pic^0(C\cross T)/p^*\Pic T.$$
Now we have the correct definition of the Jacobian variety.
This says the pair $(J,\sL)$ represents the functor
$$(F:\Sch/k)^{o}\into \Set$$
$$T\mapsto \Pic^0(C\cross T/T).$$
Here $(\Sch/k)^{o}$ is the opposite category of the category
of schemes over $k$ obtained by reversing all the arrows.
If $\psi:T'\into T$ is a morphism then $F$ sends $\psi$
to the group homomorphism
$$\Pic^0(C\cross T/T)\into \Pic^0(C\cross T'/T')$$
defined by $\sM\mapsto {\psi'}^*\sM$. Here
$\psi'$ is the map defined by the diagram
$$\begin{CD}
C\cross T'@>\psi'>> C\cross T\\
@VVV @VVV\\
T' @>\psi>> T
\end{CD}$$
To say that $(J,\sL)$ {\em represents} $F$ is to say that
$\Hom(T,J)=F(T)$
in the sense that the map
$\varphi\mapsto \varphi^*\sL$ is a bijection.
Existence of the Jacobian variety in general is difficult. When the genus
is 1 we are lucky because the Jacobian is just the curve itself. For higher
genus we must do something nontrivial. Let me give some consequences
of existence.
\subsection{Consequence of existence of the Jacobian}
{\em 1. $J$ is automatically an abelian group scheme.}
\begin{defn}
If $C$ is any category with finite products then a group
object is $G\in\Ob(C)$ given with morphisms
\begin{eqnarray*}
\mu:G\cross G\into G&\text{(multiplication)}\\
\rho:G\into G &\text{(inverse)}\\
\varepsilon:\{e\}\into G & \text{(identity)}
\end{eqnarray*}
\end{defn}
For any $G\in\Ob C$ there exists a natural representable functor
$h_G=\Hom(\cdot, G)$. To say that the functor $h_G$ factors
through the category $\Grp$ of groups is equivalent to saying
that $G$ is a group object. In particular since $J$ represents
a functor to the category of abelian groups it is an abelian
group scheme.
{\em 2. The Zariski tangent space to $J$ at $0$.}
Let $P_0$ be the zero point of $J$. Let $\m_0\subset\sO_{P_0}$ be
the maximal ideal of the local ring at $P_0$. Then the
Zariski tangent space at $P_0$ is the dual vector space
$(\m_0/\m_0^2)'$. [[I will figure this out later.]]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/26/96 hartnotes.tex
\section{The Jacobian}
In this lecture we reconsider more directly the Jacobian in
greater depth.
Let $C$ be a curve of genus $g$. How can we parameterize divisors
of degree 0 on $C$? The following definition gives a notion of a
family of divisors of degree 0.
\begin{defn}
If $T$ is any scheme, a {\em family of divisor classes on $C$ of degree 0}
parameterized by $T$ is an element of
$$\pico(C\cross T/T):=\pico(C\cross T)/p^*\pico T$$
where $p:C\cross T\into T$ is projection.
\end{defn}
\begin{defn}
If $\sM$ is an invertible sheaf on $C\cross T$ then the {\em fiber $\sM_t$
of $\sM$ at $t$} is the pullback of $\sM$ by the map $C_t\into C\cross T$.
The diagram is
$$
\begin{CD}
C_t @>>> C\cross T\\
@VVV @VVpV\\
\Spec \kappa(t) @>>> T
\end{CD}$$
\end{defn}
\begin{defn}
The {\em Jacobian variety} is a pair $(J,\sL)$ where $J$ is a scheme over
$k$ and $\sL\in\pico(C\cross J/J)$ such that the pair
$(J,\sL)$ represents the functor
$$T\mapsto \pico(C\cross T/T).$$
In other words,
for every family $\sM$ on $C\cross T$ there
exists a unique $\varphi:T\into J$ such that
$\sM={\varphi'}^*\sL$.
$$
\begin{CD}
C\cross T @>\varphi'>> C\cross J\\
@VVV @VVV\\
T @>\varphi>> J
\end{CD}$$
\end{defn}
\subsection{Consequences of existence}
\subsubsection{Group structure}
$J$ is automatically an abelian group scheme. In particular,
$$\mu:J\cross J\into J$$
is a morphism.
\begin{example}
When $g=1$ we obtain a group structure on the elliptic curve $C$.
Let $\sL$ be any divisor class of degree $1$. By Riemann-Roch
$h^0(\sL)=1+1-1=1$ so there is some $s$ which spans $H^0(\sL)$.
This means that there is exactly one effective divisor of degree
$1$ corresponding to $\sL$, namely a point. Thus for any such
$\sL$ of degree 1 there is a unique point $P\in C$ such that
$\sL\isom\sL(P)$.
Fix a point $P_0\in C$. Consider the map
$C\cross C\into C$ defined as follows. Send
the pair $\langle P,Q\rangle$ to the point $R$ corresponding
to the degree 1 divisor
$$\sL(P+Q-P_0)\isom \sL(R).$$
The map $\langle P,Q\rangle\mapsto R$ defines
a group structure on $C$. It is {\em not} obvious
that this rule gives a morphism $C\cross C\into C$.
[[We have only defined a map on closed points. Hmm.]]
\end{example}
\subsubsection{Natural fibration, dimension}
Fix $d>0$. Pick a basepoint $P_0\in C$. Consider the product
$$C^d=C\cross \cdots \cross C.$$
Its closed points are $(P_1,\ldots,P_d)$ with $P_i\in C$.
\begin{prop}
There exists a natural morphism $C^d\into J$. On closed points
it is
$$(P_1,\ldots, P_d)\mapsto
\sum_{i=1}^{d} P_i - dP_0.$$
\end{prop}
\begin{proof}
To give a morphism $C^d\into J$ is equivalent to
giving an appropriate family on $C\cross T/T$ where
$T=C^d$. Let
$$D=\{ (R,P_1,\ldots,P_d) : R \text{ is one of the } P_i \}.$$
Since $D$ is the sum of pullbacks of various diagonals it defines
a divisor of degree $d$. Let
$p:C\cross T\into C$ be projection onto $C$. Let
$$\sM=\sL(D) - d (p^*\sL(P_0)).$$
[[Probably if someone thinks about it for awhile she sees that
$\sM$ gives the desired family.]]
Check that $\sM=\varphi^*\sL$ where $\varphi$ is the map we want
on closed points.
\end{proof}
The first interesting case to consider is when $d>2g-2$.
Then the fiber of a point $j\in J$ of the map
$$\begin{array}{c}C^d\\\downarrow\\J\end{array}$$
is $$C_j^d=\{(P_1,\ldots,P_d) : \sum P_i-dP_0=j \text{ in $J$ }\}
= | j+ dP_0|.$$
By Riemann-Roch the size of a fiber is thus
$$\dim |j+dP_0| = d +1 -g -1=d-g.$$
Let $C^{(d)}$ by $C^d$ modulo the action of $S_d$. Then
$C^{(d)}$ is of dimension $d$ and the above map factors
as a surjection $C^{(d)}\into J$ with fibers of dimension
$d-g$. Thus $J$ has dimension $g$.
A second interesting special case is when $d=g$.
If $P_1+\cdots +P_g$ is chosen sufficiently generally then
$|P_1+\cdots +P_g|=\{P_1+\cdots+P_g\}$. To see this
choose each $P_i$ so that it is not a basepoint of
$|K-P_1-\cdots-P_{i-1}|$. Then $\ell(K-P_1-\cdots-P_g)=0$ so
by Riemann-Roch
$$\ell(P_1+\cdots+P_g)=g+1-g+\ell(K-P_1-\cdots-P_g)=1.$$
This means that the general fiber of $C^{(g)}\into J$ has
degree $1$. Thus it is a birational map, but it is not necessarily
an isomorphism since there can exist special divisors of degree $g$.
Let $U$ be the open subset of nonspecial divisors on $J$.
This map hints at the construction of the Jacobian since it
gives rise to a ``germ of the group law'' or a ``group chunk''
$$U\cross U\into U$$
$$\langle D,E\rangle \mapsto D+E-gP_0.$$
The next step is to try to fill out the group law to get a group
law on $J$. This is Weil's method.
\subsubsection{The Zariski tangent space}
We compute the Zariski tangent space to $J$ at $0$.
\begin{defn}
The {\em Zariski tangent space} to $0\in J$ is
$$T_{J,0}=(\m_0/\m_0^2)^{\dual}.$$
\end{defn}
Let $\varepsilon$ be such that $\varepsilon^2=0$. The elements
of $T_{J,0}$ are in one-to-one correspondence with morphisms
$$\spec k[\varepsilon]\into J$$
sending the point of $\spec k[\varepsilon]$ to $0\in J$.
The corresponding map $\sO_0\into k[\varepsilon]$ sends
$\m_0$ to $(\varepsilon)$. Since $\varepsilon^2=0$ this map
factors through $\m_0/\m_0^2$ giving an element
of $T_{J,0}$.
Now rewrite $T_{J,0}$ as
$$T_{J,0}=\Hom_0(\spec k[\varepsilon],J)$$
where $\Hom_0$ means all homomorphisms sending the point
of $\spec k[\varepsilon]$ to $0\in J$.
The next step is to use the fact that $J$ represents a certain functor.
The point is that
$$\Hom_0(\spec k[\varepsilon],J)=
\ker (\pico(C\cross T)\into \pico(C)).$$
[[This is a direct check using the fact that $0\in J$, by representability,
must correspond to $\sO_C$ on $C$.]]
The last step is to write down some exact sequences.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 4/29/96 hartnotes.tex
\section{Flatness}
[[The orals are in 959 1-3pm on Monday and Wednesday. Monday
Janos, William, then Nghi talk. Wednesday Wayne, Amod, then Matt talk.]]
The outline for this lecture is
\begin{enumerate}
\item technical definition
\item significance
\item properties
\item examples
\end{enumerate}
\subsection{Technical definitions}
Let $A$ be a commutative ring with identity.
\begin{defn}
An $A$-module is {\em flat} if the functor
$M\tensor_{A}\cdot$ is exact.
\end{defn}
For $M\tensor\cdot$ to be exact means that whenever
$$0\into N'\into N\into N''\into 0$$
is an exact sequence of $A$-modules then
$$0\into M\tensor N'\into M\tensor N\into M\tensor N''\into 0$$
is exact. Note that $M\tensor \cdot$ is right exact even if $M$ is
not flat. The salient property of a flat module is that it
preserves injectivity.
\subsubsection{General nonsense}
If $A$ is Noetherian then $M$ is flat if the functor
$M\tensor \cdot$ preserves the exactness of any sequence of
finitely generated $A$-modules. Even better, $M$ is flat
if for any ideal $I\subset A$ the map
$M\tensor I\into M\tensor A$ is injective.
\begin{defn}
Suppose $A\into B$ is a morphism of rings. Then $B$ is {\em flat over
$A$} if $B$ is flat as an $A$-module.
\end{defn}
\subsection{Examples}
\begin{example}
Suppose $A$ is a ring and $S$ a multiplicative set. Then $S^{-1}A$
is a flat $A$-module, i.e., the functor $M\mapsto S^{-1}M$ is exact.
\end{example}
\begin{example}
If $A$ is a Noetherian local ring then the completion $\hat{A}$
of $A$ at its maximal ideal is flat.
\end{example}
\begin{example}
The {\em first example} of flatness was in Serre's GAGA. He called
a flat module an ``exact couple''. The example is
$$A=\C[x_1,\ldots,x_n]\hookrightarrow \C\{x_1,\ldots,x_n\}\subset \hat{A}$$
in which $\C\{x_1,\ldots,x_n\}$, the ring of convergent power series,
is flat over $A$.
A module $B\into C$ is faithfully flat if $C\tensor_B M=0$ implies
$M=0$. Suppose $A\into B\into C$ with $C$ flat over $A$ and
$C$ faithfully flat over $B$. Then $B$ is flat over $A$. This is
how Serre proved that his module was flat.
\end{example}
\begin{thm} Suppose $A$, $\m$ is a local Noetherian ring and $M$ is
a finitely generated flat module. Then $M$ is free.
\end{thm}
\begin{proof}
Take a minimal set of generators for $M$. Nakayama's lemma tells us
how this can be done. The quotient $M/\m{}M$ is a vector space
over $A/\m$ so it has a basis, say $\overline{m}_1,\ldots,\overline{m}_r$.
By Nakayama's lemma these generate $M$. (The module $M/(m_1,\ldots,m_r)$
is sent to itself by $\m$ since every $x\in M$ is equivalent to an element
of $(m_1,\ldots,m_r)$ plus something in $\m M$. Thus $M/(m_1,\ldots,m_r)=0.$)
This gives a surjection $A^r\into M$. Let $Q$ be the kernel so we have an
exact sequence
$$0\into Q\into A^r\into M\into 0.$$
It is a general fact that if $A$ is Noetherian, $M$ is flat, and
$$0\into R\into S\into M\into 0$$
is exact then
$$0\into R\tensor N\into S\tensor N\into M\tensor N\into 0$$
is exact for {\em any} $A$-module $N$. This is proved by using
the the fact that flatness implies the vanishing of
$\tor_1^A(M,\cdot)$.
Using the general fact tensor with $k=A/\m$ to obtain an exact sequence
$$0\into Q\tensor k\into k^r \into M/\m{}M \into 0.$$
Since $k^r$ and $M/\m{}M$ are $k$-vector spaces of the same dimension
and the map $k^r\into M/\m{}M$ is surjective it must also be injective.
Thus $Q\tensor k=0$. But $Q$ is finitely generated since $A$
is Noetherian and $Q$ is a submodule of $A^r$. Thus Nakayama's lemma
implies $Q=0$ so $M\isom A^r$ is free.
\end{proof}
\begin{thm}
Suppose $A$ is a d.v.r. with maximal ideal $\m=(t)$. Let $M$
be any $A$-module. Then the following are equivalent.
\begin{itemize}
\item[(i)] $M$ is flat,
\item[(ii)] $M$ is torsion free,
\item[(iii)] $M\xrightarrow{t} M$ is injective.
\end{itemize}
\end{thm}
\begin{proof}
(i)$\Rightarrow$(ii) Since $A$ is a domain
$$0\into A\xrightarrow{x}A$$
is exact for any $x\in A$. By flatness of $M$
$$0\into M\xrightarrow{x}M$$
is exact so $M$ is torsion free.
(ii)$\Rightarrow$(iii) is trivial
(iii)$\Rightarrow$(i) It is enough to check exactness for
any ideal $\mathbf{a}=(t^n)\subset A$. Why is the map
$M\tensor\mathbf{a}\into M\tensor A$ an isomorphism? Because
under the natural identification of both modules with $M$
the map is just multiplication by $t^n$ which is injective
by assumption. The diagram is
\begin{eqnarray*}
M \isom M\tensor\mathbf{a}&\into&M\tensor A\isom M\\
m\mapsto m\tensor t^n&\mapsto&m\tensor t^n \mapsto t^n m
\end{eqnarray*}
\end{proof}
\begin{example}
If $M$ is a projective $A$ module then $M$ is flat.
\end{example}
\subsection{Algebraic geometry definitions}
\begin{defn}
Let $\sF$ be a coherent sheaf on $X$ and let $f:X\into Y$ be
a morphism. Then $\sF$ is {\em flat over $Y$} if for all $x\in X$,
$\sF_x$ is a flat $\sO_{Y,f(x)}$-module under the map
$\sO_{Y,f(x)}\into \sO_{X,x}$.
The morphism $f:X\into Y$ is {\em flat} if $\sox$ is flat over $Y$.
\end{defn}
[[``Even if $f$ is finite, $\sF_x$ might not be finite over $\soy$.
It is only finite over the semilocal ring.'']]
\begin{example}
\begin{enumerate}
\item An open immersion $U\hookrightarrow X$ is flat since the local
rings are the same.
\item A composition of flat morphisms is flat.
\item Base extension preserves flatness. Thus if $X\into Y$ is flat
and $Y'\into Y$ then $X'=X\cross_Y Y'$ is flat over $Y'$.
$$\begin{CD}
X'=X\cross_Y Y' @>>> X\\
@VVV @VVV\\
Y' @>>> Y
\end{CD}$$
This is an exercise in local rings.
\item The product of flat morphisms is flat.
[[What does this mean?]]
\end{enumerate}
\end{example}
\subsection{Families}
Flatness is used for expressing the notion of a family of subschemes.
For example let $T$ be any scheme and let $X\subset\P^n_T$ be a closed
subscheme. Then $\{X_t : t\in T\}$ is the family. Here $X_t$ is defined
by the diagram
$$\begin{CD}
X_t=X\cross_k \spec\kappa(t) @>>> X\\
@VVV @VVV\\
\spec\kappa(t) @>>> T
\end{CD}$$
where $\kappa(t)$ is the residue field of $t\in T$.
This notion of family is bad since it allows for things which we
do not want.
\begin{example}
Let $T=\A^1$ and $X\subset P^1_T=\A^1\cross\P^1$ be
the union of the coordinate axis. Then for any $t\neq 0$, $X_t$ is a point.
But $X_0=\P^1$.
\end{example}
Hard experience has lead the old pros to agree that the notion of family
should require that $X\into T$ is flat. This rules out the above example
since $X\into T$ is not flat. Indeed, $X$ is defined on an open affine
by $xt=0$ so the coordinate ring is $M=k[x,t]/(xt)$. The localization
of $M$ at $(x,t)$ has torsion as a $k[t]_{(t)}$-module. The element $x$
is killed by $t$. Since $k[t]_{(t)}$ is a d.v.r. it follows that
$M$ is not a flat.
A vague generalization is the following. Suppose $T$ is a
nonsingular curve and $X\subset\P^n_T$. Then $X$ is flat
over $T$ iff $X$ has no associated point (i.e., generic points of embedded
and irreducible components) whose image is a closed point of $T$.
Equivalently one could say ``every component of $X$ dominates $T$.''
In our example the $t$-axis dominates $T$ but the $x$-axis does not.
\begin{example}[Bad] Let $X\subset \P^1_T = \A^1\cross\P^1$
be defined by $(x^2,xt)$. When $t\neq 0$ the fiber is $P_0=\spec k[t]$
since $x$ localizes away.
If $t=0$ the fiber is given by the ring
$$k[x,t]/(x^2,xt)\tensor_{k[t]} k[t]/(t)\isom k[x]/(x^2)$$
which is $P_0$ with multiplicity 2 structure.
\end{example}
\begin{example}[Good]
Let $X\subset\P^1_T$ be defined by $x(x-t)$. Then for
$t=a\neq 0$ the fiber is $P_0+P_a$. When $t=0$ the fiber
is $2P_0$. This is good.
\end{example}
\begin{thm}
Suppose $T$ is a connected scheme. [[Must there be a finiteness assumption
on $T$?]] If $X$ is flat over $T$ then the Hilbert polynomials
$P_{X_t}(z)\in\Q[z]$ are independent of $t$.
If $T$ is integral then the converse is also true.
\end{thm}
\begin{proof}
Recall that for any scheme $Y$,
$$P_Y(m)=h^0(\soy(m))\quad\text{for all $m\gg 0$}.$$
We just need to show $h^0(\sO_{X_t}(m))$ is independent of $t$
for all $m$ sufficiently large.
Since $T$ is connected we reduce to the case $T$ is affine.
Furthermore we may assume $T=\spec A$, with $A,\m$ a local
Noetherian ring. [[``Can get from any local ring to any other
by successive specializations and generalizations. Also use the
fact that flatness is preserved by base extension.'']]
{\par\noindent\em Claims.}
\begin{enumerate}
\item For all $m\gg 0$, $H^i(\sox(m))=0$ for $i>0$
and $H^0(\sox(m))$ is a free finitely generated $A$-module.
\item For all points $t\in T$ and all $m\gg 0$,
$$H^0(\sO_{X_t}(m))=H^0(\sox(m))\tensor_A \kappa(t).$$
\end{enumerate}
Together these two claims imply the theorem.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 5/1/96 hartnotes.tex
\section{Theorem about flat families}
\begin{thm} Let $X\subset\P^n_{T}$ be a closed subscheme where $T$ is a
connected scheme of finite type over $k$. If $X$ is flat over $T$,
then the Hilbert polynomial $P_{X_t}$ is independent of $t\in T$ (all
scheme points -- not just closed points!). Conversely, if $T$
is integral and $P_{X_t}$ is independent of $t$ then $X$ is flat over $T$.
\end{thm}
\begin{proof}
See Chapter III, section 9 of Hartshorne.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 5/3/96
\section{Examples of Flat Families}
\begin{example}[Two lines]
Consider the parameterized family of curves defined as follows.
For each $t\in T=\spec k[t]$ associate the union of the $x$-axis and
a line parallel to the $y$-axis which intersects
the $z$-axis at $z=t$. Thus for $t\neq 0$ the fiber $X_t$ consists of
the disjoint union of two lines. For $t=0$ the fiber consists
of two lines meeting at the origin.
A cohomological computation shows that the Hilbert polynomials are
$$H_{X_t}(z)=\begin{cases}2z^2+1-(-1), & \text{for $t\neq 0$}\\
2z^2+1, &\text{for $t=0$}\end{cases}$$
%% Do the cohomological computation here.
What is wrong?
Let $C'$ be the subscheme of $\P^3_{T-0}$ defined by the ideal
$$I_{C'}=(y,z)\intersect (x,z-t) = (xy, xz, yz-tyw, z^2-tzw).$$
(The intersection of the two ideals is the product of the given
generators since they form a regular sequence.)
Since the Hilbert polynomial of $C'$ is constant over the fibers we
know that $C'$ is flat over $T-0$.
We have the following proposition.
\begin{prop}
Let $T=\spec k[t]$ and let $X\subset\P^n_{T-0}$ be
a closed subscheme which is flat over $T-0$. Then there is a unique
closed subscheme $\overline{X}$, flat over $T$, whose restriction
to $\P^n_{T-0}$ is $X$. Furthermore $\overline{X}$ is the scheme-theoretic
closure of $X$.
\end{prop}
\begin{proof}See Chapter III, Section 9 of Hartshorne.\end{proof}
In our situation a natural guess for $\overline{X}$ is the closed
subscheme $C$ defined by the ideal
$$(xy,xz,yz-tyw,z^2-tzw)\subset k[t][x,y,z,w].$$
As a bonus, if we can show $C$ is flat over $T$
then we will know that it is the scheme-theoretic closure of $C'$.
To test if $C$ is flat is suffices to show that the Hilbert polynomials
are constant over the fibers. We know this when $t\neq 0$ so we
need only check that the Hilbert polynomial at $t=0$ is $2z^2+2$.
At $t=0$ the defining ideal becomes
$$(xy,xz,yz,z^2)=(x,z)\intersect(y,z)\intersect(x,y,z)^2.$$
This looks like the union of the $x$ and $y$ with an embedded
point at the origin. The arithmetic genus of the degree $d=2$ plane curve
defined by $(z,xy)$ is
$$p_a=\frac{1}{2}(d-1)(d-2)=0$$ so it has Hilbert
polynomial $2z+1$. The $i$th graded piece of
$k[x,y,z,w]/(xy,xz,yz,z^2)$ has dimension one more than the dimension of the
$i$th graded piece of $k[x,y,z,w]/(z,xy)$. (In degree $i$ the first ring
has $w^{i-1}z$ whereas the latter does not.) The Hilbert polynomial is then
$$P_{C_0}(z)=(2z+1)+1=2z+2.$$ Thus $P_{C_t}(z)=2z+2$ for all $t$
so by the big theorem from the last lecture $C\subset\P^3_T$ is a flat family.
Note that although the Hilbert polynomial is constant along the fibers
the cohomology of $C_0$ is different than that of the other fibers.
This family can sort of be thought of as 2 planes meeting at a point
in 4-dimensional space.
\begin{quote}
``This is an example of the genius of Grothendieck. The Italians
knew that when two planes came together the genus changed but they
were not able to deal with it like Grothendieck did.''
\end{quote}
\end{example}
\begin{example}[The Twisted Cubic] This is the last example of the
course. Some say this is the only example in algebraic geometry.
Let $T=\spec k[t]$. The fiber over $1$ in our family will be
$C_1\subset\P^3_k$ defined by
$$I_{C_1}=(xy-zw,x^2-yw,y^2-xz).$$
How can $C_1$ give rise to a family?
One way is to assign weights to the variables then homogenize
with respect to a new variable $t$. Assign weights as in the following table.
\begin{center}
\begin{tabular}{|l|c|c|c|c|}\hline
variable&$x$&$y$&$z$&$w$\\\hline
weight&$8$&$4$&$2$&$1$\\\hline
\end{tabular}
\end{center}
Introducing a new variable $t$ of weight $1$ and homogenizing we obtain
the ideal
$$I=(xy-t^9zw,x^2-t^{11}yw,xz-t^2y^2)\subset k[t][x,y,z,w].$$
This ideal defines a closed subscheme $C$ of $\P^3_T$ since $I$ is
homogeneous as an ideal in the polynomial ring $k[t][x,y,z,w]$.
Is $C$ a flat family? When $t\neq 0$, $C_t$ is a curve.
But $C_0$ is defined by the ideal
$$(xy,x^2,xz)=(x)\intersect(x,y,z)^2$$
so it is the entire $y$--$z$ plane plus an embedded point. This
is way too large so $C$ is not a flat family.
There must be some torsion. Indeed,
$$xyz-t^9z^2w, xyz-t^2y^3\in I$$
so $$t^2(y^3-t^7z^2w)=t^2y^3-t^9z^2w\in I.$$
Thus $y^3-t^7z^2w$ is a torsion element of
$k[t][x,y,z,w]/I$ over $k[t]$. So now add $y^3-t^7z^2w$ to $I$.
Let $C'$ be the closed subscheme defined by the ideal
$$I'=(y^3-t^7z^2w,xz-t^2y^2,x^2-t''yw,xy-t^9zw).$$
When $t=0$ the fiber $C_0'$ is defined by
$$(xy,x^2,xz,y^3)=(x,y^3)\intersect(x^2,xy,xz,z^2,y^3).$$
One can show $C_0'$ is of degree $3$ and has arithmetic genus $0$.
Thus $C'$ is a flat family. The basis we have given for $I'$ is
called a {\em Gr\"obner basis}.
\end{example}
[[At this point the class came to an end. Everyone clapped, then
clapped some more. It is clear that we really appreciated Hartshorne.
He did a great job.]]
\section{Homework problems}
I did not type up solutions to the first homework set. I did type
of solutions to the other homework sets.
\subsection{Exercise on basic cohomology and abstract nonsense}
%%%%%%%%%%%%%%%%%%%%
%% Homework assignment 1
%%%%
\begin{exercise} For each of the following categories, decide whether
the category has enough projective objects (i.e. every object
is a quotient of a projective object).
a) $\Ab(X)$, where $X$ is a topological space,
b) $\Mod(X)$, where $(X,\sox)$ is a ringed space,
c) $\Qco(X)$, where $X$ is a Noetherian scheme,
d) $\Qco(X)$, where $X$ is a Noetherian affine scheme.
\end{exercise}
\begin{exercise} Let $X=\P_k^1$, over an algebraically closed field $k$.
Using only material from ChIII, \S 1,2 (esp. Exercise 2.2) show
a) For any coherent sheaf $\sF$ on $X$,
$$H^{i}(X,\sF)=0 \quad \text{for all $i\geq 2$}.$$
[Hint: Treat the case $\sF$ torsion and $\sF$ torsion-free separately;
in the torsion-free case, tensor the exact sequence
of Exercise 2.2 by $\sF$.]
b) Show that for all $\ell > 0$, $H^1(X,\sox(\ell))=0$.
\end{exercise}
\begin{exercise}
Let $X$ be an integral Noetherian scheme.
a) Show that the sheaf $\sK^{*}$ (cf. II \S 6) is flasque. Conclude
that $\pic X\isom H^1(X,\sox^{*})$.
b) Give an example of a Noetherian affine scheme with
$H^1(X,\sox^{*})\neq 0$.
\end{exercise}
\begin{exercise}
Let $X$ be a Noetherian scheme.
a) Show that the sheaf $\sG$ constructed in the proof of (3.6) is an
injective object in the category $\Qco(X)$ of quasi-coherent sheaves
on $X$. Thus $\Qco(X)$ has enough injectives.
b) Show that any injective object of $\Qco(X)$ is flasque. [{\em Hints:}
The method of proof of (2.4) will {\em not} work, because $\sO_U$ is
not quasi-coherent on $X$ in general. Instead, use (II, Ex. 5.15) to show
that if $\sI\in\Qco(X)$ is injective, and if $U\subset X$ is an
open subset, then $\sI|_U$ is an injective object of $\Qco(U)$. Then
cover $X$ with open affine...]
c) Conclude that one can compute cohomology as the derived functors
of $\Gamma(X,\cdot)$, considered as a functor from $\Qco(X)$ to $\Ab$.
\end{exercise}
\subsection{Chapter III, 4.8, 4.9, 5.6}
%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Homework assignment 2
%%%%
\subsubsection{Exercise III.4.8: Cohomological Dimension}
{\em Let $X$ be a Noetherian separated scheme.
We define the {\em cohomological dimension} of $X$, denoted $\cd(X)$, to
be the least integer $n$ such that $H^{i}(X,\F)=0$ for all quasi-coherent
sheaves $\F$ and all $i>n$. Thus for example, Serre's theorem (3.7) says
that $\cd(X)=0$ if and only if $X$ is affine. Grothendieck's theorem (2.7)
implies that $\cd(X)\leq\dim X$.
(a) In the definition of $\cd(X)$, show that it is sufficient to
consider only coherent sheaves on $X$.
(b) If $X$ is quasi-projective over a field $k$, then it is
even sufficient to consider only locally free coherent sheaves
on $X$.
(c) Suppose $X$ has a covering by $r+1$ open affine subsets. Use
\cech{} cohomology to show that $\cd(X)\leq r$.
(d) If $X$ is a quasi-projective variety of dimension $r$ over a field
$k$, then $X$ can be covered by $r+1$ open affine subsets. Conclude
that $\cd(X)\leq \dim X$.
(e) Let $Y$ be a set-theoretic complete intersection of
codimension $r$ in $X=\P_k^n$. Show that $\cd(X-Y)\leq r-1$. }
\begin{proof}
(a) It suffices to show that if, for some $i$,
$H^i(X,\F)=0$ for all coherent sheaves $\F$,
then $H^i(X,\F)=0$ for all quasi-coherent sheaves $\F$.
Thus suppose the $i$th cohomology of all coherent
sheaves on $X$ vanishes and let $\F$ be quasi-coherent.
Let $(\F_{\alpha})$ be the collection of coherent subsheaves
of $\F$, ordered by inclusion. Then by (II, Ex. 5.15e)
$\varinjlim\F_{\alpha}=\F$, so by (2.9)
$$H^i(X,\F)=H^i(X,\varinjlim \F_{\alpha})=\varinjlim H^i(X,\F_{\alpha})=0.$$
(b) Suppose $n$ is an integer and $H^i(X,\F)=0$ for all coherent
locally free sheaves $\F$ and integers $i>n$. We must show
$H^i(X,\F)=0$ for all coherent $\F$ and all $i>n$, then applying
(a) gives the desired result. Since $X$ is quasiprojective
there is an open immersion
$$i:X\hookrightarrow Y\subset \P_k^n$$
with $Y$ a closed subscheme of $\P_k^n$ and $i(X)$ open in $Y$.
By (II, Ex. 5.5c) the sheaf $\F$ on $X$ pushes forward to a coherent
sheaf on $\F'=i_{*}\F$ on $Y$.
By (II, 5.18) we may write $\F'$ as a quotient of a locally
free coherent sheaf $\E'$ on $Y$. Letting $\R'$ be the kernel
gives an exact sequence
$$0\into\R'\into\E'\into\F'\into 0$$
with $R'$ coherent (it's the quotient of coherent sheaves).
Pulling back via $i$ to $X$ gives an exact sequence
%%%%%%%%%%%%%%%%%%%%%%%%%
%% This is probably possible because $i$ is an
%% open immersion, but there are definitely some
%% details to check!!
%%%%%%%%%%%%%%%%%%%%%%%%%
$$0\into\R\into\E\into\F\into 0$$
of coherent sheaves on $X$ with $E$ locally free.
The long exact sequence of cohomology shows
that for $i>n$, there is an exact sequence
$$0=H^i(X,\E)\into H^i(X,\F) \into H^{i+1}(X,\R) \into
H^{i+1}(X,\E)=0.$$
$H^i(X,\E)=H^{i+1}(X,\E)=0$ because we have assumed that, for $i>n$,
cohomology vanishes on locally free coherent sheaves.
Thus $H^i(X,\F)\isom H^{i+1}(X,\R)$. But if $k=\dim X$, then
Grothendieck vanishing (2.7) implies that
$H^{k+1}(X,\R)=0$ whence $H^{k}(X,\F)=0$.
But then applying the above argument with $\F$ replaced
by $\R$ shows that $H^{k}(X,\R)=0$ which implies $H^{k-1}(X,\F)=0$
(so long as $k-1>n$). Again, apply the entire argument
with $\F$ replaced by $\R$ to see that $H^{k-1}(X,\R)=0$.
We can continue this descent and hence show that
$H^{i}(X,\F)=0$ for all $i>n$.
(c) By (4.5) we can compute cohomology by using the \cech{}
complex resulting from the cover $\sU$ of $X$ by $r+1$ open
affines. By definition $\sC^p=0$ for all $p>r$ since
there are no intersections of $p+1\geq r+2$ distinct open sets
in our collection of $r+1$ open sets. The \cech{} complex is
$$\sC^0\into\sC^1\into\cdots\into\sC^r\into\sC^{r+1}=0\into 0\into 0\into \cdots.$$
Thus if $\sF$ is quasicoherent then $\cH^p(\sU,\sF)=0$
for any $p>r$ which implies that $\cd(X)\leq r$.
(d) I will first present my solution in the special
case that $X$ is projective. The more general case when $X$ is
quasi-projective is similar, but more complicated, and will
be presented next.
Suppose $X\subset\P^n$ is a projective variety of dimension $r$.
We must cover $X$ with $r+1$ open affines. Let $U$ be nonempty
open affine subset of $X$. Since $X$ is irreducible, the irreducible
components of $X-U$ all have codimension at least one in $X$.
Now pick a hyperplane $H$ which doesn't
completely contain any irreducible component of $X-U$. We can do
this by choosing one point $P_i$ in each of the finitely
many irreducible components of $X-U$ and choosing a hyperplane
which avoids all the $P_i$. This can be done because the field
is infinite (varieties are only defined over algebraically
closed fields) so we can always choose a vector not orthogonal
to any of a finite set of vectors.
Since $X$ is closed in $\P^n$ and $\P^n-H$ is affine,
$(\P^n-H)\intersect X$ is an open affine subset of $X$.
Because of our choice of $H$, $U\union((\P^n-H)\intersect X)$ is only
missing codimension two closed subsets of $X$. Let $H_1=H$
and choose another hyperplane $H_2$ so it doesn't completely
contain any of the (codimension two) irreducible components
of $X-U-(\P^n-H_1)$. Then $(\P^n-H_2)\intersect X$ is open
affine and $U\union((\P^n-H_1)\intersect X)\union((\P^n-H2)\intersect X)$
is only missing codimension three closed subsets of $X$.
Repeating this process a few more times yields
hyperplanes $H_1,\cdots,H_r$ so that
$$U, (\P^n-H_1)\intersect X, \ldots, (\P^n-H_r)\intersect X$$
form an open affine cover of $X$, as desired.
Now for the quasi-projective case. Suppose $X\subset\P^n$ is
quasi-projective. From (I, Ex. 3.5) we know that $\P^n$ minus
a hypersurface $H$ is affine. Note that the same proof
works even if $H$ is a union of hypersurfaces. We now
proceed with the same sort of construction as in the projective
case, but we must choose $H$ more cleverly to insure
that $(\P^n-H)\intersect{}X$ is affine. Let $U$ be a nonempty
affine open subset of $X$. As before pick a hyperplane which
doesn't completely contain any irreducible component of $X-U$.
Since $X$ is only quasi-projective we can't conclude that
$(\P^n-H)\intersect{}X$ is affine. But we do know that
$(\P^n-H)\intersect{}\overline{X}$ is affine. Our strategy is
to add some hypersurfaces to $H$ to get a union of hypersurfaces $S$ so that
$$(\P^n-S)\intersect{}\overline{X}=(\P^n-S)\intersect{}X.$$
But, we must be careful to add these hypersurfaces in such a
way that $((\P^n-S)\intersect{}X)\union U$ is missing only codimension
two or greater subsets of $X$. We do this as follows. For each
irreducible component $Y$ of $\overline{X}-X$ choose a hypersurface
$H'$ which completely contains $Y$ but which does not completely
contain any irreducible component of $X-U$. That this can be done
is the content of a lemma which will be proved later (just pick a point
in each irreducible component and avoid it). Let $S$
by the union of all of the $H'$ along with $H$. Then
$\P^n-S$ is affine and so
$$(\P^n-S)\intersect{}X=(\P^n-S)\intersect{}\overline{X}$$
is affine. Furthermore, $S$ properly intersects all
irreducible components of $X-U$, so
$((\P^n-S)\intersect{}X)\union U$ is missing only codimension
two or greater subsets of $X$. Repeating this process
as above several times yields the desired result because
after each repetition the codimension of the resulting
pieces is reduced by 1.
\begin{lem}
If $Y$ is a projective variety and $p_1,\ldots,p_n$ is a finite
collection of points not on $Y$,
then there exists a (possibly reducible) hypersurface $H$ containing
$Y$ but not containing any of the $p_i$.
\end{lem}
By a possibly reducible hypersurface I mean a union of irreducible
hypersurfaces, not a hypersurface union higher codimension varieties.
\begin{proof}
This is obviously true and I have a proof, but I think there
is probably a more algebraic proof. Note that $k$ is infinite
since we only talk about varieties over algebraically closed fields.
Let $f_1,\cdots,f_m$ be defining equations for $Y$. Thus
$Y$ is the common zero locus of the $f_i$ and not all $f_i$ vanish
on any $p_i$. I claim that we can find a linear combination
$\sum a_i f_i$ of the $f_i$ which doesn't vanish on any $p_i$.
Since $k$ is infinite and not all $f_i$ vanish on $p_1$,
we can easily find $a_i$ so that $\sum a_i f_i(p_1)\neq 0$
and all the $a_i\neq 0$. If $\sum a_i f_i(p_2)=0$ then,
once again since $k$ is infinite, we can easily ``jiggle''
the $a_i$ so that $\sum a_i f_i(p_2)\neq 0$ and
$\sum a_i f_i(p_1)$ is still nonzero. Repeating this same
argument for each of the finitely many points $p_i$ gives
a polynomial $f=\sum a_i f_i$ which doesn't vanish on any $p_i$.
Of course I want to use $f$ to define our
hypersurface, but I can't because $f$ might not be homogeneous.
Fortunately, this is easily dealt with
by suitably multiplying the various $f_i$ by the defining equation
of a hyperplane not passing through any $p_i$, then repeating the
above argument. Now let $H$ be the hypersurface defined by $f=\sum a_i f_i$.
Then by construction $H$ contains $Y$ and $H$ doesn't
contain any $p_i$.
\end{proof}
(e) Suppose $Y$ is a set-theoretic complete intersection of codimension
$r$ in $X=\P_k^n$. Then $Y$ is the intersection of $r$ hypersurfaces,
so we can write $Y=H_1\intersect\cdots\intersect{}H_r$ where
each $H_i$ is a hypersurface. By (I, Ex. 3.5) $X-H_i$ is affine for
each $i$, thus
$$X-Y=(X-H_1)\union\cdots\union(X-H_r)$$
can be covered by $r$ open affine subsets. By (c) this
implies $\cd(X-Y)\leq r-1$ which completes the proof.
\end{proof}
\subsubsection{Exercise III.4.9}
{\em Let $X=\spec k[x_1,x_2,x_3,x_4]$ be affine four-space
over a field $k$. Let $Y_1$ be the plane $x_1=x_2=0$ and let $Y_2$
be the plane $x_3=x_4=0$. Show that $Y=Y_1\union Y_2$
is not a set-theoretic complete intersection in $X$. Therefore
the projective closure $\overline{Y}$ in $\P_k^4$ is not
a set-theoretic complete intersection.}
\begin{proof}
By (Ex. 4.8e) it suffices to show that
$H^2(X-Y,\so_{X-Y})\neq 0$.
Suppose $Z$ is a closed subset of $X$, then by (Ex. 2.3d),
for any $i\geq 1$, there is an exact sequence
$$H^i(X,\sox)\into H^i(X-Z,\so_{X-Z})\into
H_{Z}^{i+1}(X,\sox)\into H^{i+1}(X,\sox).$$
By (3.8), $H^i(X,\sox)=H^{i+1}(X,\sox)=0$ so
$H^i(X-Z,\so_{X-Z})=H_{Z}^{i+1}(X,\sox)$.
Applying this with $Z=Y$ and $i=2$ shows that
$$H^2(X-Y,\so_{X-Y})=H_{Y}^3(X,\sox).$$ Thus
we just need to show that $H_{Y}^3(X,\sox)\neq 0$.
Mayer-Vietoris (Ex. 2.4) yields an exact sequence
\begin{align*}
H^3_{Y_1}(X,\sox)\oplus{}H^3_{Y_2}(X,\sox)\into{}H^3_Y(X,\sox)\into\\
H^4_{Y_1\intersect{}Y_2}(X,\sox)\into{}
H^4_{Y_1}(X,\sox)\oplus{}H^4_{Y_2}(X,\sox)
\end{align*}
As above, $H^3_{Y_1}(X,\sox)=H^2(X-Y_1,\so_{X-Y_1})$.
But $X-Y_1$ is a set-theoretic complete intersection
of codimension $2$ so $\cd(X-Y_1)\leq 1$,
whence $H^2(X-Y_1,\so_{X-Y_1})=0$. Similarly
$$H^2(X-Y_2,\so_{X-Y_2})=H^3(X-Y_1,\so_{X-Y_1})
=H^3(X-Y_2,\so_{X-Y_2})=0.$$ Thus from the above exact
sequence we see that
$H^3_Y(X,\sox)=H^4_{Y_1\intersect{}Y_2}(X,\sox).$
Let $P=Y_1\intersect{}Y_2=\{(0,0,0,0)\}$.
We have reduced to showing that
$H^4_{P}(X,\sox)$ is nonzero. Since $H^4_{P}(X,\sox)=H^3(X-P,\so_{X-P})$
we can do this by a direct computation of $H^3(X-P,\so_{X-P})$ using
\cech{} cohomology. Cover $X-P$ by the affine
open sets $U_i=\{x_i\neq{}0\}$. Then the \cech{} complex is
$$\begin{array}{l}
k[x_1,x_2,x_3,x_4,x_1^{-1}]\oplus\cdots\oplus
k[x_1,x_2,x_3,x_4,x_4^{-1}]\xrightarrow{d_0}\\
k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1}]\oplus\cdots\oplus
k[x_1,x_2,x_3,x_4,x_3^{-1},x_4^{-1}]\xrightarrow{d_1}\\
k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1}]\oplus\cdots\oplus
k[x_1,x_2,x_3,x_4,x_2^{-1},x_3^{-1},x_4^{-1}]\xrightarrow{d_2}\\
k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1},x_4^{-1}]
\end{array}$$
Thus
$$H^3(X-P,\so_{X-P})
= \{x_1^ix_2^jx_3^kx_4^{\ell}:i,j,k,\ell<0\}\neq 0.$$
\end{proof}
\subsubsection{Exercise III.5.6: Curves on a nonsingular quadric surface}
{\em Let $Q$ be the nonsingular quadric surface $xy=zw$ in $X=\P_k^3$
over a field $k$. We will consider locally principal closed
subschemes $Y$ of $Q$. These correspond to Cartier divisors on
$Q$ by (II, 6.17.1). On the other hand, we know that
$\pic Q\isom\Z\oplus\Z$, so we can talk about the
{\em type} (a,b) of $Y$ (II, 6.16) and (II, 6.6.1). Let us denote
the invertible sheaf $\sL(Y)$ by $\so_Q(a,b)$. Thus
for any $n\in\Z$, $\so_Q(n)=\so_Q(n,n).$
[{\bf Comment!} In my solution, a subscheme $Y$ of type
$(a,b)$ corresponds to the invertible sheaf $\soq(-a,-b)$.
I think this is reasonable since then $\soq(-a,-b)=\sL(-Y)=\sI_Y$.
The correspondence is not clearly stated in the problem,
but this choice works.]
\noindent(a) Use the special case $(q,0)$ and $(0,q)$, with $q>0$, when $Y$
is a disjoint union of $q$ lines $\P^1$ in $Q$, to show:
\begin{enumerate}
\item if $|a-b|\leq 1$, then $H^1(Q,\so_Q(a,b))=0$;
\item if $a,b<0$, then $H^1(Q,\so_Q(a,b))=0$;
\item if $a\leq-2$, then $H^1(Q,\so_Q(a,0))\neq 0)$.
\end{enumerate}}
{\em Solution.}
First I will prove a big lemma in which I explicitly
calculate $H^1(Q,\soq(0,-q))$ and some other things
which will come in useful later. Next I give an independent
computation of the other cohomology groups (1), (2).
\par\begin{lem} Let $q>0$, then
$$\dim_k H^1(Q,\soq(-q,0))=H^1(Q,\soq(0,-q))=q-1.$$
Furthermore, we know all terms in the long exact
sequence of cohomology associated with the short exact sequence
$$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0.$$
\end{lem}
\begin{proof}
We prove the lemma only for $\soq(-q,0)$, since the argument
for $\soq(0,-q)$ is exactly the same. Suppose $Y$ is the disjoint
union of $q$ lines $\P^1$ in $Q$ so $\sI_Y=\soq(-q,0)$.
The sequence
$$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0$$
is exact. The associated long exact sequence of cohomology is
\begin{align*}
0\into&\Gamma(Q,\soq(-q,0))\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y)\\
\into& H^1(Q,\soq(-q,0))\into{}H^1(Q,\soq)\into{}H^1(Q,\sO_Y)\\
\into& H^2(Q,\soq(-q,0))\into{}H^2(Q,\soq)\into{}H^2(Q,\sO_Y)\into 0
\end{align*}
We can compute all of the terms in this long exact sequence. For
the purposes at hand it suffices to view the summands as $k$-vector
spaces so we systematically do this throughout.
Since $\soq(-q,0)=\sI_Y$ is the ideal sheaf of $Y$,
its global sections must vanish on $Y$. But $\sI_Y$ is a subsheaf
of $\sO_Q$ whose global sections are the constants. Since the
only constant which vanish on $Y$ is $0$, $\Gamma(Q,\soq(-q,0))=0$.
By (I, 3.4), $\Gamma(Q,\soq)=k$. Since $Y$ is the disjoint union
of $q$ copies of $\P^1$ and each copy has global sections $k$,
$\Gamma(Q,\sO_Y)=k^{\oplus{}q}$. Since $Q$ is a complete intersection of
dimension 2, (Ex. 5.5 b) implies $H^1(Q,\soq)=0$.
Because $Y$ is isomorphic to several copies of $\P^1$,
the general result (proved in class, but not in the book)
that $H_{*}^n(\sO_{\P^n})=\{\sum a_{I}X_{I}:\text{entries in $I$ negative}\}$
implies $H^1(Q,\sO_Y)=H^1(Y,\sO_Y)=0$. Since $Q$ is a hypersurface
of degree $2$ in $\P^3$, (I, Ex. 7.2(c)) implies $p_a(Q)=0$. Thus
by (Ex. 5.5c) we see that $H^2(Q,\soq)=0$. Putting together the above
facts and some basic properties of exact sequences show that
$H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}$, $H^2(Q,\soq(-q,0))=0$
and $H^2(Q,\so_Y)=0$. Our long exact sequence is now
\begin{align*}
0\into&\Gamma(Q,\soq(-q,0))=0\into\Gamma(Q,\soq)=k\into\Gamma(Q,\sO_Y)=k^{\oplus{}q}\\
\into& H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}\into{}H^1(Q,\soq)=0\into{}H^1(Q,\sO_Y)=0\\
\into& H^2(Q,\soq(-q,0))=0\into{}H^2(Q,\soq)=0\into{}H^2(Q,\sO_Y)=0\into 0
\end{align*}
\end{proof}
Number (3) now follows immediately from the lemma because
$$H^1(Q,\soq(a,0))=k^{\oplus(-a-1)}\neq 0$$ for $a\leq -2.$
Now we compute (1) and (2). Let $a$ be an arbitrary integer.
First we show that $\soq(a,a)=0$. We have an exact sequence
$$0\into\sO_{\P^3}(-2)\into\sO_{\P^3}\into\sO_{Q}\into{}0$$
where the first map is multiplication by $xy-zw$. Twisting by $a$
gives an exact sequence
$$0\into\sO_{\P^3}(-2+a)\into\sO_{\P^3}(a)\into\sO_{Q}(a)\into{}0.$$
The long exact sequence of cohomology yields an exact sequence
$$\cdots\into H^1(\sO_{\P^3}(a))\into H^1(\sO_Q(a))\into
H^2(\sO_{\P^3}(-2+a))\into\cdots$$
But from the explicit computations of projective space (5.1)
it follows that $H^1(\sO_{\P^3}(a))=0$ and $H^2(\sO_{\P^3}(-2+a))=0$
from which we conclude that $H^1(\sO_Q(a))=0$.
Next we show that $\soq(a-1,a)=0$. Let $Y$ be a single copy
of $\P^1$ sitting in $Q$ so that $Y$ has type $(1,0)$. Then
we have an exact sequence
$$0\into\sI_Y\into\soq\into\sO_Y\into 0.$$
But $\sI_Y=\soq(-1,0)$ so this becomes
$$0\into\soq(-1,0)\into\soq\into\sO_Y\into 0.$$
Now twisting by $a$ yields the exact sequence
$$0\into\soq(a-1,a)\into\soq(a)\into\sO_Y(a)\into 0.$$
The long exact sequence of cohomology gives an exact sequence
$$\cdots\into\Gamma(\soq(a))\into\Gamma(\sO_Y(a))\into
H^1(\soq(a-1,a))\into H^1(\soq(a))\into\cdots$$
We just showed that $H^1(\soq(a))=0$, so to see that
$H^1(\soq(a-1,a))=0$ it suffices to note that the
map $\Gamma(\soq(a))\into\Gamma(\sO_Y(a))$ is surjective.
This can be seen by writing $Q=\proj(k[x,y,z,w]/(xy-zw))$
and (w.l.o.g.) $Y=\proj(k[x,y,z,w]/(xy-zw,x,z))$ and
noting that the degree $a$ part of $k[x,y,z,w]/(xy-zw)$
surjects onto the degree $a$ part of $k[x,y,z,w]/(xy-zw,x,z)$.
Thus $H^1(\soq(a-1,a))=0$ and exactly the same argument
shows $H^1(\soq(a,a-1))=0$. This gives (1).
For (2) it suffices to show that for $a>0$,
$$H^1(\soq(-a,-a-n))=H^1(\soq(-a-n,-a))=0$$
for all $n>0$. Thus let $n>0$ and suppose $Y$ is a disjoint union
of $n$ copies of $\P^1$ in such a way that $\sI_Y=\soq(0,-n)$.
Then we have an exact sequence
$$0\into\soq(0,-n)\into\soq\into\sO_Y\into 0.$$
Twisting by $-a$ yields the exact sequence
$$0\into\soq(-a,-a-n)\into\soq(-a)\into\sO_Y(-a)\into 0.$$
The long exact sequence of cohomology then gives an exact sequence
$$\cdots \into\Gamma(\sO_Y(-a))\into H^1(\soq(-a,-a-n))
\into H^1(\soq(-a))\into \cdots$$
As everyone knows, since $Y$ is just several copies of $\P^1$ and $-a<0$,
$\Gamma(\sO_Y(-a))=0$. Because of our computations above,
$H^1(\soq(-a))=0$. Thus $H^1(\soq(-a,-a-n))=0$, as desired.
Showing that $H^1(\soq(-a-n,-a))=0$ is exactly the same.
\noindent (b) Now use these results to show:
\begin{enumerate}
\item If $Y$ is a locally principal closed subscheme of
type $(a,b)$ with $a,b>0$, then $Y$ is connected.
\begin{proof}
Computing the long exact sequence associated to the short
exact sequence
$$0\into\sI_Y\into\soq\into\sO_Y\into{}0$$
gives the exact sequence
$$0\into\Gamma(Q,\sI_Y)\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y)
\into H^1(Q,\sI_Y)\into\cdots$$
But, $\Gamma(\sI_Y)=0$, $\Gamma(Q,\soq)=k$, and
by (a)2 above $H^1(Q,\sI_Y)=H^1(Q,\soq(-a,-b))=0$. Thus we
have an exact sequence
$$0\into{}0\into{}k\into\Gamma(\so_Y)\into{}0\into\cdots$$
from which we conclude that $\Gamma(\so_Y)=k$ which implies
$Y$ is connected.
\end{proof}
\item now assume $k$ is algebraically closed. Then for any
$a,b>0$, there exists an irreducible nonsingular curve
$Y$ of type $(a,b)$. Use (II, 7.6.2) and (II, 8.18).
\begin{proof}
Given $(a,b)$, (II, 7.6.2) gives a closed immersion
$$Q=\P^1\times\P^1\into\P^a\times\P^b\into\P^n$$
which corresponds to the invertible sheaf $\soq(-a,-b)$
of type $(a,b)$. By Bertini's theorem (II, 8.18) there
is a hyperplane $H$ in $\P^n$ such that the hyperplane section
of the $(a,b)$ embedding of $Q$ in $\P^n$ is nonsingular.
Pull this hyperplane section back to a nonsingular curve $Y$ of
type $(a,b)$ on $Q$ in $\P^3$. By the previous problem, $Y$ is
connected. Since $Y$ comes from a hyperplane section this implies
$Y$ is irreducible (see the remark in the statement of Bertini's
theorem).
\end{proof}
\item an irreducible nonsingular curve $Y$ of type $(a,b)$,
$a,b>0$ on $Q$ is projectively normal (II, Ex. 5.14) if
and only if $|a-b|\leq 1$. In particular, this gives lots
of examples of nonsingular, but not projectively normal curves
in $\P^3$. The simplest is the one of type $(1,3)$ which is
just the rational quartic curve (I, Ex. 3.18).
\begin{proof}
Let $Y$ be an irreducible nonsingular curve of type $(a,b)$.
The criterion we apply comes from (II, Ex 5.14d) which asserts
that the maps
$$\Gamma(\P^3,\sO_{\P^3}(n))\into\Gamma(Y,\sO_Y(n))$$
are surjective for all $n\geq 0$ if and only if
$Y$ is projectively normal. To determine when this occurs
we have to replace $\Gamma(\P^3,\sO_{\P^3}(n))$ with
$\Gamma(Q,\soq(n))$. It is easy to see that the above criterion
implies we can make this replacement if $Q$ is projectively normal.
Since $Q\isom\P^1\times\P^1$ is locally isomorphic to
$\A^1\times\A^1\isom\A^2$ which is normal, we see that $Q$
is normal. Then since $Q$ is a complete intersection which is
normal, (II, 8.4b) implies $Q$ is projectively normal.
Consider the exact sequence
$$0\into\sI_Y\into\soq\into\sO_Y.$$
Twisting by $n$ gives an exact sequence
$$0\into\sI_Y(n)\into\soq(n)\into\sO_Y(n).$$
Taking cohomology yields the exact sequence
$$\cdots\into\Gamma(Q,\soq(n))\into\Gamma(Q,\sO_Y(n))\into
H^1(Q,\sI_Y(n))\into\cdots$$
Thus $Y$ is projectively normal precisely if
$H^1(Q,\sI_Y(n))=0$ for all $n\geq 0$. When can
this happen? We apply our computations from part (a).
Since $\soq(n)=\soq(n,n)$,
$$\sI_Y(n)=\soq(-a,-b)(n)=\soq(-a,-b)\tensor_{\soq}\soq(n,n)=\soq(n-a,n-b)$$
If $|a-b|\leq 1$ then $|(n-a)-(n-b)|\leq 1$ for all $n$ so
$$H^1(Q,\soq(-a,-b)(n))=0$$ for all $n$ which implies $Y$ is
projectively normal. On the other hand, if $|a-b|>1$ let
$n$ be the minimum of $a$ and $b$, without loss assume $b$ is
the minimum, so $n=b$. Then from (a) we see that
$$\soq(-a,-b)(n)=\soq(-a,-b)(b)=\soq(-a+b,0)\neq 0$$
since $-a+b\leq -2$.
\end{proof}
\end{enumerate}
(c) If $Y$ is a locally principal subscheme of type $(a,b)$
in $Q$, show that $p_a(Y)=ab-a-b+1.$ [Hint: Calculate the
Hilbert polynomials of suitable sheaves, and again use the
special case (q,0) which is a disjoint union of $q$ copies
of $\P^1$.]
\begin{proof}
The sequence
$$0\into\soq(-a,-b)\into\soq\into\sO_Y\into{}0$$
is exact so
$$\chi(\so_Y)=\chi(\soq)-\chi(\soq(-a,-b))=1-\chi(\soq(-a,-b)).$$
Thus
$$p_a(Y)=1-\chi(\so_Y)=\chi(\soq(-a,-b)).$$
The problem is thus reduced to computing $\chi(\soq(-a,-b))$.
Assume first that $a,b<0$. To compute $\chi(\soq(-a,-b))$
assume $Y=Y_1\union Y_2$ where
$\sI_{Y_1}=\soq(-a,0)$ and $\sI_{Y_2}=\soq(0,-b)$. Thus
we could take $Y_1$ to be $a$ copies of $\P^1$ in one family of lines
and $Y_2$ to be $b$ copies of $\P^1$ in the other family.
Tensoring the exact sequence
$$0\into\sI_{Y_1}\into\soq\into\so_{Y_1}\into 0$$
by the flat module $\sI_{Y_2}$ yields an exact sequence
$$0\into\sI_{Y_1}\tensor\sI_{Y_2}\into\sI_{Y_2}\into\sO_{Y_1}\tensor\sI_{Y_2}$$
[Note: I use the fact that $\sI_{Y_2}$ is flat. This follows
from a proposition in section 9 which we haven't yet reached,
but I'm going to use it anyways. Since $Y_2$ is locally principal,
$\sI_{Y_2}$ is generated locally by a single element and since $Q$ is
a variety it is integral. Thus $\sI_{Y_2}$ is locally free so
by (9.2) $\sI_{Y_2}$ is flat.]
This exact sequence can also be written as
$$0\into\soq(-a,-b)\into\soq(0,-b)\into\sO_Y\tensor\soq(0,-b)\into{}0.$$
The associated long exact sequence of cohomology is
\begin{align*}
0\into&\Gamma(Q,\soq(-a,-b))\into\Gamma(Q,\soq(0,-b))\into
\Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))\\
\into&H^1(Q,\soq(-a,-b))\into H^1(Q,\soq(0,-b))\into H^1(Q,\sO_{Y_1}\tensor\soq(0,-b))\\
\into&H^2(Q,\soq(-a,-b))\into H^2(Q,\soq(0,-b))\into H^2(Q,\sO_{Y_1}\tensor\soq(0,-b))\into 0
\end{align*}
The first three groups of global sections are $0$. Since $a,b<0$,
(a) implies $H^1(Q,\soq(-a,-b))=0$. From the lemma we know
that $H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}$. Also by the lemma
we know that $H^2(Q,\soq(0,-b))=0$. Since
$\so_{Y_1}\tensor\soq(0,-b)$ is isomorphic
to the ideal sheaf of $b-1$ points in each line
of $Y_1$, a similar proof as that used in the
lemma shows that $$H^1(Q,\sO_Y\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}.$$
Plugging all of this information back in yields the exact sequence
\begin{align*}
0\into&\Gamma(Q,\soq(-a,-b))=0\into\Gamma(Q,\soq(0,-b))=0\into
\Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\\
\into&H^1(Q,\soq(-a,-b))=0\into H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}\\
&\hspace{1in}\into{}H^1(Q,\sO_{Y_1}\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}\\
\into&H^2(Q,\soq(-a,-b))\into H^2(Q,\soq(0,-b))=0\\
&\hspace{1in}\into{}H^2(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\into 0
\end{align*}
From this we conclude that
$$\chi(\soq(-a,-b))=0+0+h^2(Q,\soq(-a,-b))=a(b-1)-(b-1)=ab-a-b+1$$
which is the desired result.
Now we deal with the remaining case, when $Y$ is $a$ disjoint copies
of $\P^1$. We have
$$p_a(Y)=1-\chi(\sO_Y)=1-\chi(\sO_{\P^1}^{\oplus a})
=1-a\chi(\sO_{\P^1})=1-a$$
which completes the proof.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{IV, 3.6, 3.13, 5.4, Extra Problems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% William Stein
%% Homework assignment 3
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Exercise IV.3.6: Curves of Degree $4$}
{\em
(a) If $X$ is a curve of degree 4 in some $\P^n$, show that either
\begin{enumerate}
\item $g=0$, in which case $X$ is either the rational normal quartic
in $\P^4$ (Ex. 3.4) or the rational quartic curve in $\P^3$ (II, 7.8.6), or
\item $X\subset\P^2$, in which case $g=3$, or
\item$X\subset\P^3$ and $g=1$.
\end{enumerate}}
{\em (b) In the case $g=1$, show that $X$ is a complete intersection of
two irreducible quadric surfaces in $\P^3$ (I, Ex. 5.11). }
\begin{proof}
(a) First suppose $n\geq 4$. If $X$ is not contained in any $\P^{n-1}$
then since $4\leq n$ (Ex. 3.4b) implies $n=4$,
$g(X)=0$, and $X$ differs from the rational normal curve of degree
4 only by an automorphism of $\P^4$. (I take this to mean that
$X$ {\em is} the rational normal curve of degree $4$.) Thus when
$n\geq 4$ we have proved that case (1) occurs.
Next suppose $n=3$. Then $X$ is a degree $4$ curve in $\P^3$. If
$X$ is contained in some $\P^2$ then $g=\frac{1}{2}(d-1)(d-2)=3$ and
so case (2) occurs. If $X$ is not contained in any $\P^2$ then
(Ex. 3.5 b) implies
$$g < \frac{1}{2}(d-1)(d-2)=3.$$ Thus $g$ is either 0, 1, or 2.
If $g=0$ then $X$ is the rational quartic curve in $\P^3$ which is
case (1). If $g=1$ then $X$ falls into case (3). If $g=2$ then
$\so_X(1)$ is a very ample divisor of degree $4$ (3.3.2) on
a genus 2 curve contrary to (Ex. 3.1) which asserts that the degree
of a very ample divisor on a curve of genus $2$ is at least 5.
Next suppose $n=2$. Then $X$ is a degree $4$ curve in $\P^2$ so
$X$ has genus $3$ and falls into case (2).
The case $n=1$ cannot occur since $\P^1$ contains no curve of degree
4.
(b) A curve $C$ of genus $1$ has $g^1_2$'s since
Riemann-Roch guarantees that the complete linear system
associated to any divisor of degree $2$ is a $g^1_2$.
There are infinitely many divisors of
degree $2$ which is not linearly equivalent.
This is because $P+Q\sim P+R$ iff $Q \sim R$.
Since $C$ is not rational there are infinitely
many points $Q$ and $R$ with $Q\not\sim R$.
Given two distinct $g^1_2$'s take the product of the corresponding
morphisms to obtain a map $\varphi:C\into\Q$ where $Q$ is the quadric
surface in $\P^3$. The diagram is
$$\begin{array}{ccc}
C&\longrightarrow&\P^1\\
\downarrow&\searrow\varphi&\uparrow\\
\P^1&\longleftarrow&C_0\subset Q\end{array}$$
Let $C_0$ denote the image of $C$ under $\varphi$. Note
that $C_0$ is not a point so the type
$(a,b)$ of $C_0$ is defined.
Let $e$ be the degree of $\varphi$. Then
by analyzing how certain divisors pull back one sees that
$ae=2$ and $be=2$.
The only possibilities are that $C_0$ is of type $(1,1)$ or of type
$(2,2)$. If $C_0$ is of type $(1,1)$ then $C_0$ must be nonsingular
because of the relation between the arithmetic genus of $C_0$ and
that of its normalization. So in this case $C_0\isom\P^1$ and the projections
$Q\into\P^1$ are injective when restricted to $C_0$.
This implies that the two maps coming from the distinct $g^1_2$'s
collapse the same points, a contradiction.
Thus $C_0$ is of type $(2,2)$ and $e=1$.
Because $C$ has genus $1$ and $C_0$ has arithmetic genus
$(2-1)(2-1)=1$ it follows that $C\isom C_0$. We now know that
$C$ embeds as a type $(2,2)$ curve on $Q$.
Since a curve of type $(a,a)$ on $Q$ is a complete
intersection we are done.
Another way to do this problem is to somehow embed $X$ into $\P^3$
as a degree 4 curve then consider the exact sequence
$$0\into\sI_X(2)\into\sO_{\P^3}(2)\into\sox(2)\into 0.$$
Taking cohomology yields the exact sequence of vector spaces
$$0\into H^0(\sI_X(2))\into H^0(\sO_{\P^3}(2))\into H^0(\sox(2))\into\cdots.$$
Since $X$ has degree $4$ it follows that $\dim H^0(\sox(2))=8$.
Furthermore $\dim H^0(\sO_{\P^3}(2))=10$ so we
see that $$\dim H^0(\sI_X(2))\geq 2.$$
This means that there exist linearly
independent homogeneous polynomials $f$ and $g$ of degree 2 such
that $X\subset Z(f)$ and $X\subset Z(g)$. Since $X$ has degree 4 and
$Z(f)\intersect Z(g)$ has degree 4 the fact that
$X\subset Z(f)\intersect Z(g)$ implies that $X=Z(f)\intersect Z(g)$.
This is seen by looking at the appropriate Hilbert polynomials.
\end{proof}
\subsubsection{Exercise IV.3.12}
{\em For each value of $d=2,3,4,5$ and $r$ satisfying
$0\leq r\leq\frac{1}{2}(d-1)(d-2)$, show that there exists
an irreducible plane curve of degree $d$ with $r$ nodes and
no other singularities.}
\begin{proof}
I did the first few by finding explicit equations. It might have
been better to do everything by abstract general methods but it
was a good exercise to search for defining equations.
$d=2, r=0$: Take $f=x^2-yz=0$. This works in any characteristic
since the partials are: $f_x=2x, f_y=-z, f_z=-y$ so if $f_y=f_z=0$
then $z=y=0$ so $x=0$. But $(0,0,0)$ is not a point.
$d=3, r=0$: When $\Char k\neq 3$ take $x^3+y^3+z^3=0$ which is clearly
nonsingular. For $\Char k=3$ take $x^2y+z^2y+z^3+y^3=0$. This is nonsingular
as was proved in my solution to (I, Ex. 5.5).
$d=3, r=1$: Let $f=xyz+x^3+y^3$, then $f_x=yz+3x^2$,
$f_y=xz+3y^2$ and $f_z=xy$. Thus a singular point must have $x=0$
or $y=0$. If $x=0$ then from $f=0$ we see that $y=0$. Thus
$(0:0:1)$ is the only singularity and it is clearly nodal.
Note that this curve works in characteristic 3 as well.
$d=4, r=0$: When $\Char k\neq 2$ take $x^4+y^4+z^4=0$. If $\Char k=2$
take $x^3y+z^3y+z^4+y^4=0$ as in (I, Ex. 5.5).
$d=4, r=1$: If $\Char k\neq 2$ take $f=xyz^2+x^4+y^4=0$. Then
$f_x=yz^2+4x^3$, $f_y=xz^2+4y^3$, and $f_z=2xyz$. The only
singular point is $(0:0:1)$ which is a node. When $\Char k=2$
take $f=xyz^2+x^3z+y^4=0$. The partials are
$f_x=yz^2+x^2z$, $f_y=xz^2$, and $f_z=x^3$. Thus a singular point
must satisfy $x=0$. Then $f=0$ implies $y=0$. Thus
the only singular point is $(0:0:1)$ which is a node.
$d=4, r=2$:
Let $C$ be your favorite genus 1 curve. Let $D$ be a divisor of
degree $4$, then by (3.3.3) $D$ is very ample.
By Riemann-Roch,
$$\dim|D|-\dim|K-D|=4+1-1$$
so $|D|$ gives rise to embedding of $C$ as a degree $4$ curve
in $\P^3$. Using (3.11) project $C$ onto the plane to obtain a plane
curve of degree $4$ with only nodal singularities. Since the genus
of the normalization is $1$ it follows that there are exactly 2 nodes.
$d=4, r=3$: Embed $\P^1$ as the degree 4 rational normal curve in $\P^4$.
Then use (3.5) and (3.10) to project into $\P^2$ to get a curve $X$
of degree $4$ in $\P^2$ having only nodes for singularities. Since
$$0=g(\tilde{X})=\frac{1}{2}(4-1)(3-1)-\text{ number of nodes }$$
it follows that $X$ has 3 nodes.
$d=5, r=0$: When $\Char k\neq 5$ take $x^5+y^5+z^5=0$. If $\Char k=5$
take $x^4y+z^4y+z^5+y^5=0$.
$d=5, r=1$: When $\Char k\neq 5$ take $f=xyz^3+x^5+y^5=0$.
Then $f_x=yz^3+5x^4$, $f_y=xz^3+5y^4$, and $f_z=3xyz^2$. The only
singular point is $(0:0:1)$ which is a node.
In characteristic 5 let $f=xyz^3+x^5+y^5+x^3y^2.$ Then
$f_x=yz^3+3x^2y^2$, $f_y=xz^2+2x^3y$, and $f_z=3xyz^2$
so the only singular point is $(0:0:1)$ which is
clearly a nodal singularity.
$d=5, r=2$: By (Ex 5.4 a) a genus $4$ curve which has two distinct
$g^1_3$'s gives rise to a plane quintic with two nodes.
The curve of type $(3,3)$ on the quadric surface is such a curve.
%%%%%%%%%%% ND %%%%%%%%%
$d=5, r=3$. I have not found one yet. If I could find a degree
5 space curve of genus 3 I would win. But by Theorem 6.4 no
such space curve exists.
$d=5, r=4$: Let $C$ be a curve of genus $2$. By Halphen's theorem
there exists a nonspecial very ample divisor $D$ of degree $5$.
Then $$\dim|D|=5+1-1-1=3$$
so $C$ embeds into $\P^3$ as a curve of degree $5$. Project
to $\P^2$ to obtain a curve $X$ of degree $5$ whose
singularities are all nodal and whose normalization has genus 2.
It follows that $X$ has 4 nodes.
$d=5, r=5$: Pick your favorite curve of genus 1. By (3.3.3) there exists
a very ample nonspecial divisor of degree $5$. As usual, project
to obtain a degree 5 plane curve with singularities only nodes
and normalization of genus 1. It follows that there must be
exactly 5 nodes.
$d=5, r=6$: Embed $\P^1$ in $\P^5$ as a curve of degree $5$ (Ex. 3.4),
and then project it into $\P^2$ to get a curve $X$
of degree $5$ in $\P^2$ having only nodes as singularities
Since $$0=g(\tilde{X})=\frac{1}{2}(5-1)(4-1)-\text{ number of nodes }$$
the number of nodes must be $6$.
\end{proof}
We can also obtain the following general result.
\begin{prop}
If $r$ and $d$ are such that
$$\frac{1}{2}(d-1)(d-2)+3-d\leq r \leq\frac{1}{2}(d-1)(d-2)$$
then there exists a plane curve of degree $d$ which has exactly
$r$ singularities all of which are nodes.\end{prop}
\begin{proof}
By (6.2) if $d\geq g+3$ then there exists a curve in $\P^3$ of genus
$g$ and degree $d$. Using (3.11) project this curve onto the plane
to obtain a curve of degree $d$ with
$$r=\frac{1}{2}(d-1)(d-2)-g$$ singularities all of which are nodes.
We can carry out this process so long as $r\leq\frac{1}{2}(d-1)(d-2)$
and $d\geq g+3$, that is, as long as
$$r=\frac{1}{2}(d-1)(d-2)-g\geq \frac{1}{2}(d-1)(d-2)+3-d.$$
\end{proof}
\subsubsection{Exercise IV.5.4}
{\em Another way of distinguishing curves of genus $g$ is to ask, what is
the least degree of a birational plane model with only nodes as singularities
(3.11)? Let $X$ be nonhyperelliptic of genus $4$. Then:
\par (a) if $X$ has two $g^1_3$'s, it can be represented as a plane quintic
with two nodes, and conversely;
\par (b) if $X$ has one $g^1_3$, then it can be represented as a plane quintic
with a tacnode (I, Ex. 5.14d), but the least degree of a plane representation
with only nodes is $6$.}
\begin{proof}
(a)
Summary: If $X$ is nonhyperelliptic with two $g^1_3$'s then
$X$ is type $(3,3)$ in the quadric so projecting through a point
on $X$ gives a plane quintic model with exactly two
nodal singularities.
Suppose $X$ is nonhyperelliptic and $X$ has two $g^1_3$'s.
Let $p$ and $p'$ be the degree $3$ maps $X\into\P^1$ determined by
the two $g^1_3$'s. Let
$$\varphi:X\into\P^1\cross\P^1=Q\subset\P^3$$
be their product.
Let $X_0=\varphi(X)$ be the image of $X$, and let $(a,b)$ be the
type of $X_0$.
Letting $e$ be the degree of $\varphi$ we see that
$ea=3$ and $eb=3$. This implies that either
$a=b=1$ and $e=3$ or $a=b=3$ and $e=1$.
First suppose
$a=b=1$ and $e=3$. Then $X_0$ is nonsingular of genus $0$ and
the two projection maps $X_0\into\P^1$ are injective. This
implies $p$ and $p'$ collapse the same points, a contradiction.
Thus $a=b=3$ and $e=1$. Because the arithmetic genus of $X_0$
is $(a-1)(b-1)=4$ and $X$ has genus $4$ we see that $X_0$ must
be nonsingular. Thus $X$ embeds as a type $(3,3)$ curve on the
quadric surface. Henceforth view $X$ as embedded in $Q$.
Pick a point $P_0$ on $X\subset Q$ and fix a copy of $\P^2$.
Map $X$ to $\P^2$ by projection through $P_0$. If two points
$P, Q$ project to the same point then $P_0, P, Q$ are collinear.
I claim that this implies the line $L$ through $P_0, P, Q$ is contained
in $Q$. To see this let $H$ by a hyperplane containing $L$.
Then either $H.Q$ is two copies of a line or a degree $2$ curve.
If $H.Q$ is a degree $2$ curve then $L.(H.Q)$ consists of at
most $2$ points, a contradiction since the points $P_0,P,Q$ are all
contained in $L.(H.Q)$. Thus $H.Q$ is two copies of a line.
Since $P_0,P,Q$ are contained in $H.Q$ we see that $H.Q=2L$ and
hence that $L$ is contained in $Q$. Since
there are exactly 2 lines on $Q$ through any given point of $Q$
the image of $X$ under projection can have at most 2 singularities.
To show that the singularities are nodes we show that three or
more points can not collapse under projection. [[This is not quite
enough. I do not know how to show that there is not some other unusual
singularity.]] Suppose some three points are collapsed under projection.
Then there exists points $P,Q,R$ on $X$ such that $P_0, P, Q, R$ are
all collinear. Now $X$ is of type $(3,3)$ so $X$ is a complete
intersection $Q.F_3$ where $F_3$ is some degree $3$ surface.
Since $F_3$ has degree $3$ and contains $P_0, P, Q, R$ it must
contain the line through them. (Take a plane containing the line,
and apply an argument as above.) Similarly $Q$ contains the line
through $P_0,P,Q,R$ so $X$ contains this line, a contradiction.
Since the genus of the normalization of the image $X_0$ of $X$ in $\P^2$
is $4$ and $X_0$ has exactly two nodes as singularities it follows
that $X_0$ has degree 5.
{\em The converse. }
Suppose we are given a plane quintic curve $C$ with two nodes and no other
singularities. Then the normalization has genus
$$g(\tilde{C})=\frac{1}{2}(5-1)(4-1)-2=4.$$
Thus $C$ represents a curve $X$ of genus $4$. Let $P_0$ be one of the two
nodes, then since
$C$ has degree 5 a line through $P_0$ intersect $C$ in $3$ other points.
This gives a degree $3$ map from $C$ to $\P^1$.
Since this map is defined on a nonempty open subset of
$X$ it extends to a degree 3 morphism of $X$ into $\P^3$.
This gives a $g^1_3$ on $X$. The other node gives a different $g^1_3$
so $X$ has two distinct $g^1_3$'s.
We must also show that $X$ is not hyperelliptic.
Suppose $X$ is hyperelliptic
so $X$ has a $g^1_2$. Let $p$ be the map to $\P^1$ corresponding to this
$g^1_2$ and let $p'$ be the map to $\P^1$ corresponding to some $g^1_3$.
Let $\varphi=p\cross p':X\into Q\subset\P^3$ be their product.
Let $X_0=\varphi(X)$ be the image of $X$ and suppose $X_0$ has
type $(a,b)$. Let $e$ be the degree of $\varphi$. Then
$ea=2$ and $eb=3$. Thus $e=1$ so $X$ is birational to the normalization
of $X_0$. But $X_0$ has arithmetic genus $(2-1)(3-1)=2<4$ which is
a contradiction.
(b)
Suppose $C$ is a nonhyperelliptic curve with exactly one $g^1_3$.
By looking at twists of certain exact sequences we showed in class
that $C$ lies on the (singular) quadric cone. We showed furthermore
that $C$ is a complete intersection $\Cone.F_3$ of $\Cone$ with some
cubic hypersurface $F_3$.
Pick a point $P_0$ on $C$ and a copy of $\P^2\subset\P^3$.
Projection through $P_0$ defines a map $C\into\P^2$.
If two points $P,Q$ collapse then the three points $P_0, P, Q$ are
collinear. Let $L$ be the line determined by $P_0, P, Q$. Since $\Cone$ has
degree $2$ and $L$ intersects $\Cone$ in at least three points
it follows that $L$ lies on $\Cone.$ Since there is only one line
through $P_0$ which lies on $\Cone$ it follows that projection
defines a birational map of $C$ to plane curve $C_0$ which has exactly
one singularity.
If 3 points $P,Q,R$ are collapsed by projection then the four
points $P_0, P, Q, R$ are collinear. Let $L$ be the line through
them. Then $L$ must lie in $F_3$ and $L$ must lie in $Q$ which
is a contradiction. This shows that the singular point on $C_0$ is
a double point.
I do not know how to show it must be a tacnode.
{\em The Converse:}
If there is a quintic plane representation of degree less than $6$
with only nodes then there must be at least two nodes because of
the formula for the genus of the normalization. But each node
gives rise to a distinct $g^1_3$. Since there is a unique $g^1_3$
on $X$ this is a contradiction.
\end{proof}
\subsubsection{Extra Problem 3, by William Stein}
{\em Suppose $C$ is hyperelliptic and $g\geq 3$. Then there does not
exist a $g^1_3$ on $C$.}
\begin{proof}
Suppose that $C$ has a $g^1_3$ and let $p:C\into\P^1$ be the corresponding
morphism. Let $p':C\into\P^1$ be the morphism corresponding to
a $g^1_2$ on the hyperelliptic curve $C$.
Let $$\varphi=p\cross p':C\into\P^1\cross\P^1=Q\subset\P^3.$$
Let $(a,b)$ be the type of $C_0=\varphi(C)$.
If $e$ denotes the degree of $\varphi$ then $ea=2$ and $eb=3$.
We see this by looking at how divisors of types $(1,0)$ and $(0,1)$ pull
back. Thus $e=1$, $a=2$ and $b=3$. Now $C$ is isomorphic to the
normalization of $C_0$ which has arithmetic genus $(2-1)(3-1)=2<3$
so $C$ has genus less than $3$, a contradiction.
\end{proof}
\subsubsection{Extra Problem 4, by Nghi Nguyen}
{\em If $C$ is a non-hyperelliptic curve of genus $g\geq 4$, show that
$C$ has at most a finite number of $g^1_3$'s.}
\begin{proof}
This proof is the work of Nghi although the write
up is my own. (Therefore any mistakes are my
responsibility.)
{\em Summary.} For a fixed point $P_0$ there are only finitely many
$g^1_3$'s arising from a divisor $D$ of the form $D=P_0+Q+R$. If
some $g^1_3$ is defined by a divisor $D$, then $D-P_0$ is linearly equivalent
to an effective divisor $Q+R$ so $D\sim P_0+Q+R$. Combining this with
the first assertion implies that there are only finitely many $g^1_3$'s.
{\em Step 1.} Fix a point $P_0$, then we show that there are only
finitely many $g^1_3$'s arising from a divisor $D$ of the form $P_0+Q+R$.
Suppose $D=P_0+Q+R$ is a divisor such that $|D|$ is a $g^1_3$.
Since $C$ is non-hyperelliptic the canonical divisor $K$ is very ample.
Thus $K-P_0$ is base point free. Let
$\varphi:X\into\P^{g-2}$ be the morphism determined by $K-P_0$.
By Riemann-Roch
$$\dim|P_0+Q+R|-\dim|K-P_0-Q-R|=3+1-g$$
so since $\dim|P+Q+R|=1$,
$$\dim|K-P_0-Q-R|=g-3.$$
But $K$ is very ample so $\dim|K-P_0-Q|=\dim|K|-2=g-3$.
Thus $$\dim|K-P_0-Q-R|=\dim|K-P_0-Q|$$
so $R$ is a basepoint of $K-P_0-Q$. This means that
$\varphi(R)=\varphi(Q)$.
Let $X_0=\varphi(X)\subset\P^{g-2}$. Let $\mu=\deg \varphi$ and
let $d=\deg X_0$. Then
$$\mu d=\deg(K-P_0)=2g-3.$$
If $\mu>1$ then $d\leq\frac{2g-3}{3}$ since $\mu$ is not $2$
because $2g-3$ is odd. But $\frac{2g-3}{3}1$. Thus $\mu=1$.
Since $\mu=1$ the map $\varphi$ is birational so it can collapse
only finitely many points. This means that there are only finitely
many choices for $D=P_0+Q+R$ so that $|D|$ is a $g^1_3$.
{\em Step 2.} Suppose $|D|$ is a $g^1_3$. Then $\dim|D|=1$ and since
$|D|$ is basepoint free
$$\dim|D-P_0|=\dim|D|-1=0.$$
Thus there exists an effective divisor $Q+R$ such that
$$D-P_0\sim Q+R,$$
so $D\sim P_0+Q+R$. This shows that every $g^1_3$ is defined by an effective
divisor which contains $P_0$. By step 1 we know there are only
finitely many such $g^1_3$ so we conclude that $C$ has only finitely
many $g^1_3$'s.
\end{proof}
\end{document}
*