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%% Notes for Hartshorne's Algebraic Geometry course
%% William Stein
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\author{William A. Stein}
\title{Notes for Algebraic Geometry II}

\begin{document}
\maketitle
\tableofcontents

\section{Preface}

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 $0<i<n-1$.
\end{prop}

A good place to get the necessary background for the
cohomology we will study is in Appendices 3 and 4
from Eisenbud's {\em Commutative Algebra}.

\section{Cohomology in Algebraic Geometry}

For any scheme $X$ and any sheaf $\sF$ of $\sO_X$-modules
we want to define the groups $H^i(X,\sF)$. We can either {\em define}
cohomology by listing its properties, then later prove that we can construct
the $H^i(X,\sF)$ or we can skip the definition and just construct
the $H^i(X,\sF)$. The first method is more esthetically pleasing,
but we will choose the second.

We first forget the scheme structure of $X$ and regard $X$ as a
topological space and $\sF$ as a sheaf of abelian groups (by ignoring
the ring multiplication). Let $\Ab(X)$ be the category of sheaves of
abelian groups on $X$. Let $\Gamma=\Gamma(X,\cdot)$ be the global
section functor from $\Ab(X)$ into $\Ab$, where $\Ab$ is the category
of abelian groups. Recall that $\Gamma$ is left exact so if
\begin{equation*}
0\rightarrow\sF'\rightarrow\sF\rightarrow\sF''\rightarrow 0
\end{equation*}
is an exact sequence in $\Ab(X)$ then the following sequence is exact
\begin{equation*}
0\rightarrow \Gamma(\sF') \rightarrow \Gamma(\sF) \rightarrow
\Gamma(\sF'')
\end{equation*}
in $\Ab$.

\begin{defn}
We define the cohomology groups $H^i(X,\sF)$ to be the right derived
functors of $\Gamma$.
\end{defn}

%% 1/22/96, Lecture 3

\section{Review of Derived Functors}

The situation will often be as follows.
Let $\sA$ and $\sB$ be abelian categories and
$$\sA\xrightarrow{F}\sB$$
a functor. Derived functors are the measure of the non-exactness of
a functor. Let $X$ be a topological space, $\Ab(X)$ the category
of sheaves of abelian groups on $X$ and $\Ab$ the category of
abelian groups. Then $\Gamma(X,\cdot): \Ab(X)\rightarrow\Ab$ is a left exact functor. Our cohomology theory
will turn out to be the right derived function of $\Gamma(X,\cdot)$.

\subsection{Examples of Abelian Categories}
Although we will not define an abelian category we will give several
examples and note that an abelian category is a category which has
the same basic properties as these examples.
\begin{example}[$A$-Modules] Let $A$ be a fixed commutative ring
and consider the category $\Mod(A)$ of $A$-modules. Then if
$M,N$ are any two modules one has

1) $\Hom(M,N)$ is an abelian group,

2) $\Hom(M,N)\times \Hom(N,L)\into \Hom(M,L)$ is a homomorphism
of abelian groups.

3) there are kernels, cokernels, etc.

$\Mod(A)$ is an abelian category.
\end{example}

\begin{example}
Let $A$ be a Noetherian ring and let our category be the collection
of all finitely generated $A$-modules. Then this category is abelian.
Note that if the condition
that $A$ be Noetherian is relaxed we may no longer have an abelian
category because the kernel of a morphism of finitely generated
modules over an arbitrary ring need not be finitely generated (for
example, take the map from a ring to its quotient by an ideal which
cannot be finitely generated).
\end{example}

\begin{example} Let $X$ be a topological space, then
$\Ab(X)$ is an abelian category. If $(X,\sO_X)$ is a ringed
space then the category $\Mod(\sO_X)$ is abelian. If $X$ is
a scheme then the category of quasi-coherent $\sO_X$-modules
is abelian, and if $X$ is also Noetherian then the sub-category
of coherent $\sO_X$-modules is abelian.
\end{example}

\begin{example}
The category of abelian varieties is {\em not} an abelian category
since the kernel of a morphism of abelian varieties might be
reducible (for example an isogeny of degree $n$ of elliptic curves
has kernel $n$ points which is reducible). It may be the case
that the category of abelian group schemes is abelian but I don't
know at the moment.
\end{example}

\begin{example}
The category of compact Hausdorff abelian topological groups
is an abelian category.
\end{example}

\subsection{Exactness}

\begin{defn} A functor $F:\sA\into\sB$ is {\em additive} if for
all $X,Y\in \sA$, the map
$$F:\Hom_{\sA}(X,Y)\into\Hom_{\sB}(FX,FY)$$
is a homomorphism of abelian groups.
\end{defn}

\begin{defn} A sequence
$$A\xrightarrow{f}B\xrightarrow{g}C$$
is {\em exact} if $\imag(f)=\ker(g)$.
\end{defn}

\begin{defn} Let $F:\sA\into\sB$ be a functor and
$$0\into M'\into M \into M''\into 0$$
be an exact sequence and consider the sequence
$$0\into FM'\into FM\into FM''\into 0.$$
If the second sequence is exact in the middle, then
$F$ is a called {\em half exact functor}. If the second sequence is exact
on the left and the middle then $F$ is called a {\em left exact functor}.
If the second sequence is exact on the right and in the middle
then we call $F$ a {\em right exact functor}.
\end{defn}

\begin{example}
Let $A$ be a commutative ring and $N$ an $A$-module. Then
$N\tensor -$ is a right exact functor on the category of
$A$-modules. To see that $N\tensor -$ is not exact, suppose
$A,\gm$ is a local ring and $N=k=A/\gm$.
Then the sequence
$$0\into\gm\into A\into k\into 0$$
is exact, but
$$0\into k\tensor\gm \into k\tensor A \into k\tensor k \into 0$$
is right exact but not exact.
\end{example}

\begin{example}
The functor $\tor_1(N,\cdot)$ is neither left nor right exact.
\end{example}

\begin{example}
The contravarient hom functor, $\Hom(\cdot,N)$ is left exact.
\end{example}

Often the following is useful in work.
\begin{thm}
If $$0\into M'\into M\into M''$$ is exact and $F$ is left exact, then
$$0\into FM'\into FM\into FM''$$
is exact.
\end{thm}

\subsection{Injective and Projective Objects}

Let $\sA$ be an abelian category. Then $\Hom_A(P,-):\sA\into\Ab$ is
left exact.

\begin{defn}
An $A$ module $P$ is said to be {\em projective} if
the functor $\Hom_A(P,-)$ is exact. An $A$ module $I$ is
said to be {\em injective} if the functor
$\Hom_A(-,I)$ is exact.
\end{defn}

\begin{defn}
We say that an abelian category $\sA$ has {\em enough projectives} if
every $X$ in $\sA$ is the surjective image of a projective
$P$ in $\sA$. A category is said to have {\em enough injectives} if
every $X$ in $\sA$ injects into an injective objective of $\sA$.
\end{defn}

\begin{example}
Let $A$ be a commutative ring, then $\Mod(A)$ has enough injectives because
every module is the quotient of a free module and every free module
is projective. If $X$ is a topological space then $\Ab(X)$ has enough
injectives. If $X$ is a Noetherian scheme, then the category of quasi-coherent
sheaves has enough injectives (hard theorem). The category of $\sO_X$-modules
has enough injectives but the category of coherent sheaves on
$X$ doesn't have enough injectives or projectives.
\end{example}

%% 1/24/96, Lecture 4

\section{Derived Functors and Homological Algebra}

Let $F:\sA\into\sB$ be an additive covariant left-exact functor between
abelian categories, for example $F=\Gamma:\Ab(X)\into\Ab$.
Assume $\sA$ has enough injectives, i.e., for all $X$ in $\sA$
there is an injective object $I$ in $\sA$ such that
$0\into X\hookrightarrow I$. We construct the right derived functors
of $F$. If
$$0\into M'\into M\into M''\into 0$$
is exact in $\sA$ then
$$0\into F(M')\into F(M)\into F(M'')\into R^{1}F(M^1)\into R^{1}F(M) \into \cdots$$
is exact in $\sB$ where $R^{i}F$ is the right derived functor of $F$.

\subsection{Construction of $R^{i}F$}
Take any $M$ in $\sA$, then since $\sA$ has enough injectives we
can construct an exact sequence
$$0\into M\into I^{0}\into I^{1}\into I^{2}\into \cdots$$
where each $I^{i}$ is an injective object. (This isn't totally
obvious, but is a straightforward argument by putting together
short exact sequences and composing maps.)
The right part of the above sequence $I^{0}\into I^{1}\into \cdots$
is called an {\em injective resolution} of $M$.
Applying $F$ we get a complex
$$F(I^{0})\xrightarrow{d_0} F(I^{1})\xrightarrow{d_1} F(I^{2}) \xrightarrow{d_2} \cdots$$
in $\sB$ which may not be exact. The objects
$H^i=\ker(d_2)/ \imag(d_1)$ measure the deviation of this
sequence from being exact. $H^i$ is called the $i$th {\em cohomology}
object of the complex.

\begin{defn} For each object in $\sA$ fix an injective resolution.
The $i$th {\em right derived functor} of $F$ is the functor
which assigns to an object $M$
the $i$th cohomology of the complex $F(I^{\cdot})$ where $I^{\cdot}$
is the injective resolution of $M$.
\end{defn}

\subsection{Properties of Derived Functors}
We should now prove the following:
\begin{enumerate}
\item If we fix different injective resolutions for all of our objects
then the corresponding derived functors are, in a suitable sense, isomorphic.
\item The $R^{i}F$ can also be defined on morphisms in such a way
that they are really functors.
\item If $0\into M'\into M\into M''$ is a short exact sequence then there is
a long exact sequence of cohomology:
\begin{align*}
0\into FM'\into FM\into FM'' \into \\
R^1FM'\into R^1FM\into R^1FM''\into \\
R^2FM'\into \cdots.
\end{align*}
\item If we have two short exact sequences then the induced maps on
long exact sequences are $\delta$-compatible''.
\item $R^0 F\isom F$.
\item If $I$ is injective, then for any $i>0$ 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 [email protected]$%^(#$&#

\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
$$C^p(\sU,\sF)=\prod_{i_0<i_1<\cdots<i_p}\sF(U_{i_0\cdots i_p}).$$
Then we get a complex
$$C^0(\sU,\sF)\into C^1(\sU,\sF)\into \cdots \into C^p(\sU,\sF)\into \cdots$$
by defining a map
$$d:C^p(\sU,\sF)\into C^{p+1}(\sU,\sF)$$
by, for $\alpha\in C^p(\sU,\sF)$,
$$(d\alpha)_{i_0\cdots i_{p+1}} := \sum_{0}^{p+1} (-1)^j \alpha_{i_0\cdots \hat{i_j}\cdots i_{p+1}}|_{U_{i_0\cdots i_{p+1}}}.$$
One checks that $d^2=0$.

\begin{lem}
$\cH^{0}(\sU,\sF)=\Gamma(X,\sF)$
\end{lem}
\begin{proof}
Applying the sheaf axioms to the exact sequence
$$0\into\Gamma(X,\sF)\into C^0=\prod_{i\in I}\sF(U_i) \xrightarrow{d}C^1=\prod_{i<j}\sF(U_{ij})$$
we see that $\cH^0(\sU,\sF)=\ker d = \Gamma(X,\sF)$.
\end{proof}

\subsection{Sheafify}
Let $X$ be a topological space, $\sU$ an open cover and
$\sF$ a sheaf of abelian groups. Then we define
$$\sC^p(\sU,\sF)=\prod_{i_0<\cdots<i_p}j_{*}(\sF|_{U_{i_0\cdots i_p}})$$
and define
$$d:\sC^p(\sU,\sF)\into \sC^{p+1}(\sU,\sF)$$
in terms of the $d$ defined above by, for $V$ an open set,
$$\sC^{p}(\sU,\sF)(V)=C^p(\sU|_{V},\sF|_V)\xrightarrow{d} C^{p+1}(\sU|_{V},\sF|_V) =\sC^{p+1}(\sU,\sF)(V).$$

\begin{remark}
$C^p(\sU,\sF)=\Gamma(X,\sC^{p}(\sU,\sF)$
\end{remark}

\begin{lem}
The sequence
$$0\into\sF\into\sC^{0}(\sU,\sF)\into\sC^{1}(\sU,\sF)\into\cdots$$
is a resolution of $\sF$, i.e., it is exact.
\end{lem}
\begin{proof}
We define the map $\sF\into\sC^{0}$ by taking the product of the natural maps
$\sF\into f_{*}(\sF|_{U_i})$, exactness then follows from the
sheaf axioms.

To show the rest of the sequence is exact it suffices to show
exactness at the stalks. So let $x\in X$, and suppose $x\in U_j$.
Given $\alpha_x\in\sC_x^p$ it is represented by a section
$\alpha\in\Gamma(V,\sC^p(\sU,\sF))$, over a neighborhood $V$ of
$x$, which we may choose so small that $V\subset U_j$. Now
for any $p$-tuple $i_0<\ldots<i_{p-1}$, we set
$$(k\alpha)_{i_0,\ldots,i_{p-1}}=\alpha_{j,i_0,\ldots,i_{p-1}}.$$
This makes sense because
$$V\intersect U_{i_0,\ldots,i_{p-1}}=V\intersect U_{j,i_0,\ldots,i_{p-1}}.$$
Then take the stalk of $k\alpha$ at $x$ to get the required map $k$.

Now we check that for any $p\geq 1$ and $\alpha\in\sC_x^p$,
$$(dk+kd)(\alpha)=\alpha.$$
First note that
\begin{align*}(dk\alpha)_{i_0,\ldots,i_p} & =
\sum_{\ell=0}^{p} (-1)^\ell (k\alpha)_{i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}\\
& = \sum (-1)^\ell \alpha_{j,i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}
\end{align*}
Whereas, on the other hand,
\begin{align*}
(kd\alpha)_{i_0,\ldots,i_p}
& = (d\alpha)_{j,i_0,\ldots,i_p}\\
& = (-1)^0\alpha_{i_0,\ldots,i_p} + \sum_{\ell=1}^{p} (-1)^{\ell+1}\alpha_{j,i_0,\ldots,
\hat{i_{\ell}},\ldots,i_{p}}
\end{align*}
Adding these two expressions yields $\alpha_{i_0,\ldots,i_p}$ as claimed.

Thus $k$ is a homotopy operator for the complex $\sC_x^{\bullet}$, showing
that the identity map is homotopic to the zero map. It follows that the
cohomology groups $H^{p}(\sC^{\bullet}_x)$ of this complex
are $0$ for $p\geq 1$.
\end{proof}

\begin{lem}
If $\sF$ is flasque then $\sC^p(\sU,\sF)$ is also flasque.
\end{lem}
\begin{proof}
If $\sF$ is flasque then $\sF|_{U_{i_0,\ldots,i_p}}$ is flasque
so $j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque so
$\prod j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque.
\end{proof}

\begin{prop}
If $\sF$ is flasque then $\cH^{p}(\sU,\sF)=0.$
\end{prop}
\begin{proof}
Consider the resolution $$0\into\sF\into\sC^{\bullet}(\sU,\sF).$$
By the above lemma it is flasque, so we can use it to compute the
usual cohomology groups of $\sF$. But $\sF$ is flasque, so
$H^p(X,\sF)=0$ for $p>0$. 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<\cdots<i_p}\sF'(U_{i_0,\ldots,i_p})
\into C^{p}(\sU,\sF)=\prod_{i_0<\cdots<i_p}\sF(U_{i_0,\ldots,i_p}) \\
\into C^{p}(\sU,\sF'')=\prod_{i_0<\cdots<i_p}\sF''(U_{i_0,\ldots,i_p})
\into \prod_{i_0<\cdots<i_p} H^{1}(\sU_{i_0,\ldots,i_p},\sF')=0
\end{align*}
where the last term is 0 because $\sF'$ satisfies condition (*).
Replacing $p$ by $\cdot$ gives an exact sequence of complexes.
Applying $\cH^{i}(\sU,\cdot)$ then gives the desired result.
\end{proof}

\begin{thm}
Let $X$ be a topological space, $\sU$ an open cover and $\sF\in \Ab(X)$.
Suppose $\sF$ and $\sU$ satisfy (*). Then the maps
$$\varphi^{i}:\cH^i(\sU,\sF)\into H^{i}(X,\sF)$$
are isomorphisms.
\end{thm}
\begin{proof}
The proof is a clever induction.
\end{proof}

\begin{lem}
If $0\into\sF'\into\sF\into\sF''\into 0$ is exact and
$\sF'$ and $\sF$ satisfy (*) then $\sF''$ satisfies (*).
\end{lem}

To prove the main theorem of the section use the fact that
$X$ separated implies any finite intersection of affines
is affine and then use the vanishing theorem for cohomology
of a quasi-coherent sheaf on an affine scheme. The above theorem
then implies the main result. From now on we will always
assume our schemes are separated unless otherwise stated.

\begin{cor}
If $X$ is a (separated) Noetherian scheme and $X$ can be covered
by $n+1$ open affines for some $n>0$ 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 $0<i<n$ and for all $\ell$.
(This doesn't hold for arbitrary quasi-coherent sheaves!)

c) $H_{*}^n(X,\sox)$ is a graded $S$-module which
is $0$ in degrees $\geq -n$, but is nonzero in degrees $\leq n-1$.
As a $k$-vector space it is equal to
$$\{\sum_{i_j<0} a_{i_0,\ldots,i_n}x_0^{i_0}\cdots x_n^{i_n} : \text{sum is finite}\}.$$

d) For $\ell\geq 0$ the map
$$H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))\isom{}k$$
is a perfect pairing so we have a duality (which is in fact a special
case of Serre Duality).

% 2/12/96
\section{Serre's Finite Generation Theorem}
We relax the hypothesis from the last lecture and claim that the
same results are still true.
\begin{thm}
Let $A$ be a Noetherian ring and $X=\bP^n_A$. Then
\begin{enumerate}
\item $H^0_{*}(\sox)=\oplus_{\ell}H^{0}_{\ell}(\sox(\ell))=S=A[x_0,\ldots,x_n]$
\item $H^i_{*}(\sox)=0$ for all $0<i<n$
\item $H^n_{*}(\sox)=\{\sum_{I=i_0,\ldots,i_n} a_I x_0^{i_0}\cdots x_n^{i_n} : a_I\in A\}$
\item $H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))$ is
a perfect pairing of free $A$-modules. Notice that $H^n(\sox(-n-1))$ is
a free $A$-module of rank 1 so it is isomorphic to $A$, but {\em not} in
a canonical way!
\end{enumerate}
\end{thm}

Although pairing is in general not functorial as a map into $A$,
there is a special situation in which it is. Let $\Omega_{X/k}^1$
be the sheaf of differentials on $X=\bP^n_k$. Let $\omega=\Omega_{X/k}^n =\Lambda^n\Omega^1$ be the top level differentials (or dualizing
module''). Then some map is functorial (??)
\begin{quote}
Is $\omega$ more important than $\Omega$?'' -- Janos Csirik

That's a value judgment... you can make your own decision
on that... I won't.'' -- Hartshorne
\end{quote}

\begin{thm}[Serre]
Let $X$ be a projective scheme over a Noetherian ring $A$. Let $\sF$
be any coherent sheaf on $X$. Then
\begin{enumerate}
\item $H^{i}(X,\sF)$ is a finitely generated $A$-module for all $i$
\item for all $\sF$ there exists $n_0$ such that for all $i>0$ 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} \[email protected]>>>\[email protected]>>>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
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}
\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
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=`