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Author: William A. Stein
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%% Notes for Hartshorne's Algebraic Geometry course
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%% William Stein
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\author{William A. Stein}
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\title{Notes for Algebraic Geometry II}
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\begin{document}
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\maketitle
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\tableofcontents
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\section{Preface}
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{\bfseries Read at your own risk!}
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These are my {\em very rough, error prone} notes of
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a second course on algebraic geometry
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offered at U.C. Berkeley in the Spring of 1996.
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The instructor
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was Robin Hartshorne and the students were Wayne Whitney,
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William Stein, Matt Baker, Janos Csirik, Nghi Nguyen, and Amod.
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I wish to thank Robin Hartshorne for giving this course
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and to Nghi Nguyen for his insightful suggestions and corrections.
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Of course all of the errors are solely my responsibility.
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The remarks in brackets [[like this]] are notes that
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I wrote to myself. They are meant as a warning or as a reminder
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of something I should have checked but did not have time for. You
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may wish to view them as exercises.
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If you have suggestions, questions, or comments feel free to write to me.
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My email address is {\tt was@math.berkeley.edu}.
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% Day 1, 1/17/96
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\section{Ample Invertible Sheaves}
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Let $k$ be an algebraically closed field
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and let $X$ be a scheme over $k$. Let $\phi:X\into \bP^n_k$ be a morphism.
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Then to give $\phi$ is equivalent to giving an invertible sheaf
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$\sL$ on $X$ and sections $s_0,\ldots,s_n\in\Gamma(X,\sL)$
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which generate $\sL$. If $X$ is projective (that is, if there
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is some immersion of $X$ into {\em some} $\bP^m_k$) then $\phi$
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is a closed immersion iff $s_0,\ldots,s_n$ separate points and tangent
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vectors.
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\begin{defn}
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Let $X$ be a scheme and $\sL$ an invertible sheaf on $X$.
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Then we say $\sL$ is {\em very ample} if there is an immersion
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$i:X\hookrightarrow \bP_k^n$ such that $\sL\isom i^{*}\sO(1)$.
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\end{defn}
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\begin{thm}
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Let $X$ be a closed subscheme of
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$\bP_k^n$
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and
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$\sF$ a coherent sheaf on $X$, then
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$\sF(n)$ is generated by global sections for
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all $n\gg 0$.
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\end{thm}
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\begin{cor}
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Let $X$ be any scheme and $\sL$ a very ample coherent sheaf
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on $X$, then for all $n\gg 0$, $\sF\tensor \sL^{\tensor n}$
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is generated by global sections.
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\end{cor}
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\begin{defn}
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Let $X$ be a Noetherian scheme and $\sL$ be an invertible sheaf.
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We say that $\sL$ is {\em ample} if for every coherent sheaf
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$\sF$ on $X$, there is $n_0$ such that for all $n\geq n_0$,
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$\sF \tensor \sL^{\tensor n}$ is generated by its global sections.
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\end{defn}
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Thus the previous corollary says that a very ample invertible
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sheaf is ample.
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\begin{prop}
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Let $X$ and $\sL$ be as above. Then the following are equivalent.
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1) $\sL$ is ample,
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2) $\sL^n$ is ample for all $n>0$,
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3) $\sL^n$ is ample for some $n>0$.
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\end{prop}
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\begin{thm}
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Let $X$ be of finite type over a Noetherian ring $A$ and suppose
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$\sL$ is an invertible sheaf on $A$. Then $\sL$ is ample iff
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there exists $n$ such that $\sL^n$ is very ample over $\spec A$.
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\end{thm}
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\begin{example}
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Let $X=\bP^1$, $\sL=\sO(\ell)$, some $\ell \in \bZ$.
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If $\ell<0$ then $\Gamma(\sL)=0$. If $\ell=0$ then
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$\sL=\sO_X$ which is not ample since $\sO_X(-1)^n\tensor
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\sO_X\isom \sO_X(-1)^n$ is not generated by global sections
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for any $n$. Note that $\sO_X$ itself is generated by
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global sections. Finally, if $ell>0$ then
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$\sL=\sO_X(\ell)$ is very ample hence ample.
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\end{example}
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\begin{example}
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Let $C\subseteq \bP^2$ be a nonsingular cubic curve and
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$\sL$ an invertible sheaf on $C$ defined by $\sL=\sL(D)$,
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where $D=\sum n_i P_i$ is a divisor on $C$. If $\deg D<0$
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then $\sL$ has no global sections so it can't be ample.
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%% Undone.
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\end{example}
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% Day 2, 1/19/96
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\section{Introduction to Cohomology}
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We first ask, what is cohomology and where does it arise in nature?
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Cohomology occurs in commutative
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algebra, for example in the $\ext$ and $\tor$ functors, it occurs in
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group theory, topology, differential geometry, and of course in
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algebraic geometry. There are several flavors of cohomology which are
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studied by algebraic geometers. Serre's coherent sheaf
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cohomology has the advantage of being easy to define, but
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has the property that the cohomology groups are vector spaces.
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Grothendieck introduced \`{e}tale cohomology and $\ell$-adic
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cohomology. See, for example, Milne's {\em \`{E}tale Cohomology}
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and SGA 4$\frac{1}{2}$, 5 and 6. This cohomology theory arose
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from the study of the Weil Conjectures (1949) which deal with
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a deep relationship between the number of points on a variety
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over a finite field and the geometry of the complex analytic variety
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cut out by the same equations in complex projective space. Deligne
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was finally able to resolve these conjectures in the affirmative
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in 1974.
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What is cohomology good for? Cohomology allows one to get numerical
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invariants of an algebraic variety. For example, if $X$ is a projective
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scheme defined over an algebraically closed field $k$ then
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$H^i(X,\sF)$ is a finite dimensional $k$-vector space. Thus the
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$h_i=\dim_k H^i(X,\sF)$ are a set of numbers associated
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to $X$. ``Numbers are useful in all branches of mathematics.''
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\begin{example}{Arithmetic Genus}
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Let $X$ be a nonsingular projective curve. Then $\dim H^1(X,\sO_X)$ is
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the arithmetic genus of $X$. If $X\subseteq \bP^n$ is a projective variety
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of dimension $r$ then, if $p_a=\dim H^1(X,\sO_X)$, then
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$1+(-1)^r p_a = $ the constant term of the Hilbert polynomial of $X$.
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\end{example}
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\begin{example}
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Let $X$ be a nonsingular projective surface, then
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\begin{equation*}
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1+p_a=h^0(\sO_X)-h^1(\sO_X)+h^2(\sO_x)
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\end{equation*}
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and $1+(-1)^r p_a = \chi(\sO_X)$, the Euler characteristic of $X$.
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\end{example}
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\begin{example}
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Let $X$ be an algebraic variety and $\pic X$ the group of
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Cartier divisors modular linear equivalence (which is isomorphic
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to the group of invertible sheaves under tensor product modulo isomorphism).
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Then $\pic X \isom H^1(X,\sO_{X}^{*})$.
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\end{example}
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\begin{example}[Deformation Theory]
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Let $X_0$ be a nonsingular projective variety. Then
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the first order infinitesimal deformations are classified
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by $H^1(X_0,T_{X_0})$ where $T_{X_0}$ is the tangent
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bundle of $X_0$. The obstructions are classified by
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$H^2(X_0,T_{X_0})$.
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\end{example}
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One can define Cohen-Macaulay rings in terms of cohomology.
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Let $(A,\gm)$ be a local Noetherian ring of dimension $n$,
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let $X = \spec A$, and let $P=\gm\in X$, then we have the
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following.
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\begin{prop}
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Let $A$ be as above. Then $A$ is Cohen-Macaulay iff
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1) $H^0(X-P,\sO_{X-P}) = A$ and
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2) $H^i(X-P, \sO_{X-P}) = 0$ for $0<i<n-1$.
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\end{prop}
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A good place to get the necessary background for the
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cohomology we will study is in Appendices 3 and 4
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from Eisenbud's {\em Commutative Algebra}.
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\section{Cohomology in Algebraic Geometry}
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For any scheme $X$ and any sheaf $\sF$ of $\sO_X$-modules
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we want to define the groups $H^i(X,\sF)$. We can either {\em define}
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cohomology by listing its properties, then later prove that we can construct
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the $H^i(X,\sF)$ or we can skip the definition and just construct
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the $H^i(X,\sF)$. The first method is more esthetically pleasing,
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but we will choose the second.
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We first forget the scheme structure of $X$ and regard $X$ as a
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topological space and $\sF$ as a sheaf of abelian groups (by ignoring
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the ring multiplication). Let $\Ab(X)$ be the category of sheaves of
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abelian groups on $X$. Let $\Gamma=\Gamma(X,\cdot)$ be the global
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section functor from $\Ab(X)$ into $\Ab$, where $\Ab$ is the category
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of abelian groups. Recall that $\Gamma$ is left exact so if
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\begin{equation*}
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0\rightarrow\sF'\rightarrow\sF\rightarrow\sF''\rightarrow 0
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\end{equation*}
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is an exact sequence in $\Ab(X)$ then the following sequence is exact
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\begin{equation*}
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0\rightarrow \Gamma(\sF') \rightarrow \Gamma(\sF) \rightarrow
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\Gamma(\sF'')
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\end{equation*}
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in $\Ab$.
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\begin{defn}
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We define the cohomology groups $H^i(X,\sF)$ to be the right derived
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functors of $\Gamma$.
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\end{defn}
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%% 1/22/96, Lecture 3
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\section{Review of Derived Functors}
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The situation will often be as follows.
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Let $\sA$ and $\sB$ be abelian categories and
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$$\sA\xrightarrow{F}\sB$$
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a functor. Derived functors are the measure of the non-exactness of
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a functor. Let $X$ be a topological space, $\Ab(X)$ the category
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of sheaves of abelian groups on $X$ and $\Ab$ the category of
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abelian groups. Then $\Gamma(X,\cdot):
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\Ab(X)\rightarrow\Ab$ is a left exact functor. Our cohomology theory
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will turn out to be the right derived function of $\Gamma(X,\cdot)$.
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\subsection{Examples of Abelian Categories}
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Although we will not define an abelian category we will give several
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examples and note that an abelian category is a category which has
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the same basic properties as these examples.
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\begin{example}[$A$-Modules] Let $A$ be a fixed commutative ring
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and consider the category $\Mod(A)$ of $A$-modules. Then if
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$M,N$ are any two modules one has
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1) $\Hom(M,N)$ is an abelian group,
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2) $\Hom(M,N)\times \Hom(N,L)\into \Hom(M,L)$ is a homomorphism
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of abelian groups.
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3) there are kernels, cokernels, etc.
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$\Mod(A)$ is an abelian category.
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\end{example}
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\begin{example}
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Let $A$ be a Noetherian ring and let our category be the collection
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of all finitely generated $A$-modules. Then this category is abelian.
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Note that if the condition
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that $A$ be Noetherian is relaxed we may no longer have an abelian
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category because the kernel of a morphism of finitely generated
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modules over an arbitrary ring need not be finitely generated (for
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example, take the map from a ring to its quotient by an ideal which
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cannot be finitely generated).
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\end{example}
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\begin{example} Let $X$ be a topological space, then
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$\Ab(X)$ is an abelian category. If $(X,\sO_X)$ is a ringed
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space then the category $\Mod(\sO_X)$ is abelian. If $X$ is
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a scheme then the category of quasi-coherent $\sO_X$-modules
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is abelian, and if $X$ is also Noetherian then the sub-category
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of coherent $\sO_X$-modules is abelian.
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\end{example}
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\begin{example}
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The category of abelian varieties is {\em not} an abelian category
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since the kernel of a morphism of abelian varieties might be
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reducible (for example an isogeny of degree $n$ of elliptic curves
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has kernel $n$ points which is reducible). It may be the case
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that the category of abelian group schemes is abelian but I don't
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know at the moment.
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\end{example}
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\begin{example}
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The category of compact Hausdorff abelian topological groups
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is an abelian category.
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\end{example}
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\subsection{Exactness}
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\begin{defn} A functor $F:\sA\into\sB$ is {\em additive} if for
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all $X,Y\in \sA$, the map
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$$F:\Hom_{\sA}(X,Y)\into\Hom_{\sB}(FX,FY)$$
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is a homomorphism of abelian groups.
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\end{defn}
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\begin{defn} A sequence
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$$A\xrightarrow{f}B\xrightarrow{g}C$$
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is {\em exact} if $\imag(f)=\ker(g)$.
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\end{defn}
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\begin{defn} Let $F:\sA\into\sB$ be a functor and
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$$0\into M'\into M \into M''\into 0$$
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be an exact sequence and consider the sequence
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$$0\into FM'\into FM\into FM''\into 0.$$
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If the second sequence is exact in the middle, then
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$F$ is a called {\em half exact functor}. If the second sequence is exact
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on the left and the middle then $F$ is called a {\em left exact functor}.
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If the second sequence is exact on the right and in the middle
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then we call $F$ a {\em right exact functor}.
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\end{defn}
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\begin{example}
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Let $A$ be a commutative ring and $N$ an $A$-module. Then
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$N\tensor -$ is a right exact functor on the category of
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$A$-modules. To see that $N\tensor -$ is not exact, suppose
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$A,\gm$ is a local ring and $N=k=A/\gm$.
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Then the sequence
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$$0\into\gm\into A\into k\into 0$$
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is exact, but
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$$0\into k\tensor\gm \into k\tensor A \into k\tensor k \into 0$$
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is right exact but not exact.
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\end{example}
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\begin{example}
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The functor $\tor_1(N,\cdot)$ is neither left nor right exact.
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\end{example}
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\begin{example}
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The contravarient hom functor, $\Hom(\cdot,N)$ is left exact.
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\end{example}
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Often the following is useful in work.
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\begin{thm}
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If $$0\into M'\into M\into M''$$ is exact and $F$ is left exact, then
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$$0\into FM'\into FM\into FM''$$
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is exact.
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\end{thm}
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\subsection{Injective and Projective Objects}
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Let $\sA$ be an abelian category. Then $\Hom_A(P,-):\sA\into\Ab$ is
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left exact.
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\begin{defn}
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An $A$ module $P$ is said to be {\em projective} if
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the functor $\Hom_A(P,-)$ is exact. An $A$ module $I$ is
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said to be {\em injective} if the functor
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$\Hom_A(-,I)$ is exact.
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\end{defn}
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\begin{defn}
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We say that an abelian category $\sA$ has {\em enough projectives} if
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every $X$ in $\sA$ is the surjective image of a projective
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$P$ in $\sA$. A category is said to have {\em enough injectives} if
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every $X$ in $\sA$ injects into an injective objective of $\sA$.
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\end{defn}
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\begin{example}
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Let $A$ be a commutative ring, then $\Mod(A)$ has enough injectives because
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every module is the quotient of a free module and every free module
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is projective. If $X$ is a topological space then $\Ab(X)$ has enough
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injectives. If $X$ is a Noetherian scheme, then the category of quasi-coherent
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sheaves has enough injectives (hard theorem). The category of $\sO_X$-modules
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has enough injectives but the category of coherent sheaves on
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$X$ doesn't have enough injectives or projectives.
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\end{example}
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%% 1/24/96, Lecture 4
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\section{Derived Functors and Homological Algebra}
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Let $F:\sA\into\sB$ be an additive covariant left-exact functor between
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abelian categories, for example $F=\Gamma:\Ab(X)\into\Ab$.
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Assume $\sA$ has enough injectives, i.e., for all $X$ in $\sA$
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there is an injective object $I$ in $\sA$ such that
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$0\into X\hookrightarrow I$. We construct the right derived functors
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of $F$. If
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$$0\into M'\into M\into M''\into 0$$
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is exact in $\sA$ then
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$$0\into F(M')\into F(M)\into F(M'')\into R^{1}F(M^1)\into R^{1}F(M)
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\into \cdots$$
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is exact in $\sB$ where $R^{i}F$ is the right derived functor of $F$.
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\subsection{Construction of $R^{i}F$}
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Take any $M$ in $\sA$, then since $\sA$ has enough injectives we
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can construct an exact sequence
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$$0\into M\into I^{0}\into I^{1}\into I^{2}\into \cdots$$
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where each $I^{i}$ is an injective object. (This isn't totally
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obvious, but is a straightforward argument by putting together
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short exact sequences and composing maps.)
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The right part of the above sequence $I^{0}\into I^{1}\into \cdots$
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is called an {\em injective resolution} of $M$.
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Applying $F$ we get a complex
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$$F(I^{0})\xrightarrow{d_0} F(I^{1})\xrightarrow{d_1} F(I^{2})
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\xrightarrow{d_2} \cdots$$
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in $\sB$ which may not be exact. The objects
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$H^i=\ker(d_2)/ \imag(d_1)$ measure the deviation of this
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sequence from being exact. $H^i$ is called the $i$th {\em cohomology}
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object of the complex.
537
538
\begin{defn} For each object in $\sA$ fix an injective resolution.
539
The $i$th {\em right derived functor} of $F$ is the functor
540
which assigns to an object $M$
541
the $i$th cohomology of the complex $F(I^{\cdot})$ where $I^{\cdot}$
542
is the injective resolution of $M$.
543
\end{defn}
544
545
\subsection{Properties of Derived Functors}
546
We should now prove the following:
547
\begin{enumerate}
548
\item If we fix different injective resolutions for all of our objects
549
then the corresponding derived functors are, in a suitable sense, isomorphic.
550
\item The $R^{i}F$ can also be defined on morphisms in such a way
551
that they are really functors.
552
\item If $0\into M'\into M\into M''$ is a short exact sequence then there is
553
a long exact sequence of cohomology:
554
\begin{align*}
555
0\into FM'\into FM\into FM'' \into \\
556
R^1FM'\into R^1FM\into R^1FM''\into \\
557
R^2FM'\into \cdots.
558
\end{align*}
559
\item If we have two short exact sequences then the induced maps on
560
long exact sequences are ``$\delta$-compatible''.
561
\item $R^0 F\isom F$.
562
\item If $I$ is injective, then for any $i>0$ one has that $R^i F(I)=0$.
563
\end{enumerate}
564
565
\begin{thm} The $R^{i}F$ and etc. are uniquely determined by properties
566
1-6 above.
567
\end{thm}
568
569
\begin{defn} A {\em $\delta$-functor} is a collection of functors
570
$\{R^i F\}$ which satisfy 3 and 4 above. An {\em augmented
571
$\delta$-functor} is a {\em $\delta$-functor} along with a natural
572
transformation $F\into R^0 F$. A {\em universal augmented
573
$\delta$-functor} is an {\em augmented $\delta$-functor} with
574
some universal property which I didn't quite catch.
575
\end{defn}
576
577
\begin{thm} If $\sA$ has enough injectives then the collection of
578
derived functors of $F$ is a universal augmented $\delta$-functor.
579
\end{thm}
580
581
To construct the $R^i F$ choose once and for all, for each object
582
$M$ in $\sA$ an injective resolution, then prove the above properties
583
hold.
584
585
586
%% 1/26/96 -- today's lecture was too [email protected]$%^(#$&#
587
588
\section{Long Exact Sequence of Cohomology and Other Wonders}
589
590
``Today I sat in awe as Hartshorne effortless drew hundreds
591
of arrows and objects everywhere, chased some elements and
592
proved that there is a long exact sequence of cohomology in
593
30 seconds. Then he whipped out his colored chalk and things
594
really got crazy. Vojta tried to erase Hartshorne's diagrams
595
during the next class but only partially succeeded joking
596
that the functor was not `effaceable'. (The diagrams are
597
still not quite gone 4 days later!) Needless to say, I don't
598
feel like texing diagrams and element chases... it's all
599
trivial anyways, right?''
600
601
%% 1/29/96
602
\section{Basic Properties of Cohomology}
603
Let $X$ be a topological space, $\Ab(X)$ the category
604
of sheaves of abelian groups on $X$ and
605
$$\Gamma(X,\cdot):\Ab(X)\into\Ab$$
606
the covariant, left exact global sections functor.
607
Then we have constructed the derived functors
608
$H^{i}(X,\cdot)$.
609
610
\subsection{Cohomology of Schemes}
611
Let $(X,\sox)$ be a scheme and $\sF$ a sheaf of $\sox$-modules.
612
To compute $H^{i}(X,\sF)$ forget all extra structure and use
613
the above definitions. We may get some extra structure anyways.
614
615
\begin{prop}
616
Let $X$ and $\sF$ be as above, then the groups $H^{i}(X,\sF)$
617
are naturally modules over the ring $A=\Gamma(X,\sox)$.
618
\end{prop}
619
620
\begin{proof}
621
Let $A=H^{0}(X,\sF)=\Gamma(X,\sox)$ and let $a\in A$. Then because
622
of the functoriality of $H^{i}(X,\cdot)$ the
623
map $\sF\into\sF$ induced by left multiplication by $a$ induces
624
a homomorphism
625
$$a:H^{i}(X,\sF)\into H^{i}(X,\sF).$$
626
%% Need more!!
627
\end{proof}
628
629
\subsection{Objective}
630
Our objective is to compute $H^{i}(\bP^n_k,\sO(\ell))$
631
for all $i,n,\ell$. This is enough for most applications
632
because if one knows these groups one can, in principle at
633
least, computer the cohomology of any projective scheme.
634
If $X$ is any projective variety, we embed $X$ in some
635
$\bP^n_k$ and push forward the sheaf $\sF$ on $X$. Then
636
we construct a resolution of $\sF$ by sheaves of the
637
form $\sO(-\ell)^n$. Using Hilbert's syzigy theorem one
638
sees that the resolution so constructed is finite and
639
so we can put together our knowledge to get the cohomology
640
of $X$.
641
642
Our plan of attack is as follows.
643
\begin{enumerate}
644
\item Define flasque sheaves which are acyclic for cohomology, i.e.,
645
the cohomology vanishes for $i>0$.
646
\item If $X=\spec A$, $A$ Noetherian, and $\sF$ is quasi-coherent,
647
show that $H^i(X,\sF)=0$ for $i>0$.
648
\item If $X$ is any Noetherian scheme and $\sU=(U_i)$ is an open
649
affine cover, find a relationship between the cohomology of $X$
650
and that of each $U_i$. (The ``\v{C}ech process''.)
651
\item Apply number 3 to $\bP_k^n$ with $U_i=\{x_i\neq 0\}$.
652
\end{enumerate}
653
654
\section{Flasque Sheaves}
655
\begin{defn}
656
A {\em flasque sheaf} (also called {\em flabby sheaf}) is a sheaf
657
$\sF$ on $X$ such that whenever $V\subset U$ are open sets then
658
$\rho_{U,V}:\sF(U)\into\sF(V)$ is surjective.
659
\end{defn}
660
Thus in a flasque sheaf, ``every section extends''.
661
662
\begin{example}
663
Let $X$ be a topological space, $p\in X$ a point, not necessarily
664
closed, and $M$ an abelian group. Let $j:\{P\}\hookrightarrow X$ be the
665
inclusion, then $\sF=j_{*}(M)$ is flasque. This follows since
666
$$
667
j_{*}(M)(U)=\begin{cases} M&\text{if $p\in U$}\\
668
0&\text{if $p\not\in U$}
669
\end{cases}.
670
$$
671
Note that $j_{*}(M)$ is none other than the skyscraper sheaf
672
at $p$ with sections $M$.
673
\end{example}
674
675
\begin{example}
676
If $\sF$ is a flasque sheaf on $Y$ and $f:Y\into X$ is a morphism
677
then $f_{*}\sF$ is a flasque sheaf on $X$.
678
\end{example}
679
680
\begin{example}
681
If $\sF_i$ are flasque then $\bigoplus_{i} \sF_i$ is flasque.
682
\end{example}
683
684
\begin{lem}
685
If
686
$$0\into\sF'\into\sF\into\sF''\into 0$$
687
is exact and $\sF'$ is flasque
688
then
689
$$\Gamma(\sF)\into\Gamma(\sF'')\into 0$$
690
is exact.
691
\end{lem}
692
693
\begin{lem}
694
If
695
$$0\into\sF'\into\sF\into\sF''\into 0$$
696
is exact and $\sF'$ and $\sF$ are both
697
flasque then
698
$\sF''$ is flasque.
699
\end{lem}
700
701
\begin{proof}
702
Suppose $V\subset U$ are open subsets of $X$.
703
Since $\sF'$ is flasque and the restriction of a
704
flasque sheaf is flasque and restriction is exact,
705
lemma 1 implies that the sequence
706
$$\sF(V)\into\sF''(V)\into 0$$
707
is exact.
708
We thus have a commuting diagram
709
$$
710
\begin{CD}
711
\sF(U) @>>> \sF''(U)\\
712
@VVV @VVV\\
713
\sF(V) @>>> \sF''(V) @>>> 0
714
\end{CD}
715
$$
716
which, since $\sF(U)\into \sF(V)$ is surjective,
717
implies $\sF''(U)\into\sF''(V)$ is surjective.
718
\end{proof}
719
720
721
\begin{lem}
722
Injective sheaves (in the category of abelian sheaves) are flasque.
723
\end{lem}
724
725
\begin{proof}
726
Let $\sI$ be an injective sheaf of abelian groups on $X$ and
727
let $V\subset U$ be open subsets. Let $s\in\sI(V)$, then we
728
must find $s'\in\sI(U)$ which maps to $s$ under
729
the map $\sI(U)\into\sI(V)$. Let $\bZ_V$ be the constant sheaf
730
$\bZ$ on $V$ extended by $0$ outside $V$ (thus $\bZ_V(W)=0$
731
if $W\not\subset V$). Define a map $\bZ_V\into \sI$ by
732
sending the section $1\in\bZ_V(V)$ to $s\in\sI(V)$. Then
733
since $\bZ_V\hookrightarrow\bZ_U$ and $\sI$ is injective
734
there is a map $\bZ_U\into \sI$ which sends the section $1\in
735
\bZ_U$ to a section $s'\in\sI(U)$ whose restriction
736
to $V$ must be $s$.
737
\end{proof}
738
739
\begin{remark} The same proof also shows that injective sheaves in
740
the category of $\sox$-modules are flasque.
741
\end{remark}
742
743
\begin{cor}
744
If $\sF$ is flasque then $H^{i}(X,\sF)=0$ for all $i>0$.
745
\end{cor}
746
\begin{proof} Page 208 of [Hartshorne].
747
\end{proof}
748
749
% 1/31/96
750
751
\begin{cor}
752
Let $(X,\sox)$ be a ringed space, then the derived functors of
753
$\Gamma:\Mod\sox\into\Ab$ are equal to $H^{i}(X,\sF)$.
754
\end{cor}
755
\begin{proof}
756
If
757
$$0\into\sF\into\sI^0\into\sI^1\into\cdots$$
758
is an injective resolution of $\sF$
759
in $\Mod\sox$ then, by the above remark, it
760
is a flasque resolution in the category
761
$\Ab(X)$ hence we get the regular cohomology.
762
\end{proof}
763
764
\begin{remark}
765
{\bf Warning!} If $(X,\sox)$ is a scheme and we choose an injective
766
resolution in the category of quasi-coherent $\sox$-modules
767
then we are only guaranteed to get the right answer if
768
$X$ is Noetherian.
769
\end{remark}
770
771
\section{Examples}
772
773
\begin{example}
774
Suppose $C$ is a nonsingular projective curve over an algebraically
775
closed field $k$. Let $K=K(C)$ be the function field of $C$ and
776
let $\sK_C$ denote the constant sheaf $K$. Then we have an exact
777
sequence
778
$$ 0\into\sO_C\into\sK_C\into
779
\bigoplus_{\substack{P\in C\\P\text{ closed}}}
780
K/\sO_P \into 0,$$
781
where the map $\sK_C\into\bigoplus K/\sO_P$ has
782
only finitely many components nonzero since a function $f\in K$
783
has only finitely many poles.
784
Since $C$ is irreducible $\sK_C$ is flasque and since $\sK/\sO_P$
785
is a skyscraper sheaf it is flasque so since direct of flasque
786
sheaves are flasque, $\bigoplus K/\sO_P$ is flasque.
787
One checks that the sequence is exact and so this is
788
a flasque resolution of $\sO_C$. Taking global sections and
789
applying the exact sequence of cohomology gives an exact sequence
790
$$K\into\bigoplus_{\text{$P$ closed}} K/\sO_P
791
\into H^1(X,\sO_C)\into 0,$$ and $H^{i}(X,\sO_C)=0$ for
792
$i\geq 2$. Thus the only interesting information
793
is $\dim_k H^1(X,\sO_C)$ which is the {\em geometric genus} of $C$.
794
\end{example}
795
796
%% 2/2/96
797
798
\section{First Vanishing Theorem}
799
800
\begin{quote}
801
``Anyone who studies algebraic geometry must read French... looking up
802
the more general version of this proof in EGA would be a good exercise.''
803
\end{quote}
804
805
\begin{thm}
806
Let $A$ be a Noetherian ring, $X=\spec A$ and $\sF$ a quasi-coherent
807
sheaf on $X$, then $H^{i}(X,\sF)=0$ for $i>0$.
808
\end{thm}
809
810
\begin{remark}
811
The theorem is true without the Noetherian hypothesis on $A$, but
812
the proof uses spectral sequences.
813
\end{remark}
814
815
\begin{remark}
816
The assumption that $\sF$ is quasi-coherent is essential. For example,
817
let $X$ be an affine algebraic curve over an infinite field $k$. Then
818
$X$ is homeomorphic as a topological space to $\P^1_k$ so
819
the sheaf $\sO(-2)$ on $\P^1_k$ induces a sheaf $\sF$ of abelian groups
820
on $X$ such that
821
$$H^1(X,\sF)\isom{}H^1(\P^1_k,\sO(-2))\neq 0.$$
822
\end{remark}
823
824
\begin{remark}
825
If $I$ is an injective $A$-module then $\tilde{I}$ need {\em not}
826
be injective in $\Mod(\sox)$ or $\Ab(X)$. For example, let $A=k=\bF_p$ and
827
$X=\spec A$, then $I=k$ is an injective $A$-module but $\tilde{I}$
828
is the constant sheaf $k$. But $k$ is
829
a finite group hence not divisible so $\tilde{I}$ is not injective.
830
(See Proposition A3.5 in Eisenbud's {\em Commutative Algebra}.)
831
\end{remark}
832
833
\begin{prop}
834
Suppose $A$ is Noetherian and $I$ is an injective $A$-module, then
835
$\tilde{I}$ is flasque on $\spec A$.
836
\end{prop}
837
838
The proposition implies the theorem since if $\sF$ is quasi-coherent
839
then $\sF=\tilde{M}$ for some $A$-module $M$. There is an injective
840
resolution
841
$$0\into M\into I^{\bullet}$$
842
which, upon applying the exact functor $\tilde{ }$,
843
gives a flasque resolution
844
$$0\into \tilde{M}=\sF\into \tilde{I}^{\bullet}.$$
845
Now applying $\Gamma$ gives us back the original resolution
846
$$\Gamma:\quad 0\into M\into I^{\bullet}$$
847
which is exact so the cohomology groups vanish for $i>0$.
848
849
\begin{proof} Let $A$ be a Noetherian ring and $I$ an injective $A$,
850
then $\tilde{I}$ is a quasi-coherent sheaf on $X=\spec A$. We must
851
show that it is flasque. It is sufficient to show that for any
852
open set $U$, $\Gamma(X)\into\Gamma(U)$ is surjective.
853
854
{\em Case 1, special open affine:}
855
Suppose $U=X_f$ is a special open affine. Then we have a commutative diagram
856
$$
857
\begin{CD}
858
\Gamma(X,\tilde{I}) @>>> \Gamma(X_f,\tilde{I})\\
859
@V=VV @V=VV\\
860
I @>\text{surjective?}>> I_f
861
\end{CD}
862
$$
863
To see that the top map is surjective it is equivalent
864
to show that $I\into I_f$ is surjective. This is a tricky
865
algebraic lemma (see Hartshorne for proof).
866
867
{\em Case 2, any open set:}
868
Let $U$ be any open set. See Hartshorne for the rest.
869
870
\end{proof}
871
872
873
%% 2/5/96
874
875
\section{\cech{} Cohomology}
876
Let $X$ be a topological space, $\sU=(U_i)_{i\in I}$ an
877
open cover and $\sF$ a sheaf of abelian groups.
878
We will define groups $\cH^i(\sU,\sF)$ called
879
\cech{} cohomology groups.
880
881
{\bfseries Warning: } $\cH^i(\sU,\cdot)$ is a functor in $\sF$,
882
but it is {\em not} a $\delta$-functor.
883
884
\begin{thm} Let $X$ be a Noetherian scheme, $\sU$ an open cover
885
and $\sF$ a quasi-coherent sheaf, then $\cH^{i}(\sU,\sF)=H^i(X,\sF)$
886
for all $i$.
887
\end{thm}
888
889
\subsection{Construction}
890
Totally order the index set $I$. Let
891
$$U_{i_0\cdots i_p}=\intersect_{j=0}^p U_{i_j}.$$
892
For any $p\geq 0$ define
893
$$C^p(\sU,\sF)=\prod_{i_0<i_1<\cdots<i_p}\sF(U_{i_0\cdots i_p}).$$
894
Then we get a complex
895
$$C^0(\sU,\sF)\into C^1(\sU,\sF)\into \cdots \into C^p(\sU,\sF)\into \cdots$$
896
by defining a map
897
$$d:C^p(\sU,\sF)\into C^{p+1}(\sU,\sF)$$
898
by, for $\alpha\in C^p(\sU,\sF)$,
899
$$(d\alpha)_{i_0\cdots i_{p+1}} := \sum_{0}^{p+1} (-1)^j
900
\alpha_{i_0\cdots \hat{i_j}\cdots i_{p+1}}|_{U_{i_0\cdots i_{p+1}}}.$$
901
One checks that $d^2=0$.
902
903
\begin{lem}
904
$\cH^{0}(\sU,\sF)=\Gamma(X,\sF)$
905
\end{lem}
906
\begin{proof}
907
Applying the sheaf axioms to the exact sequence
908
$$0\into\Gamma(X,\sF)\into C^0=\prod_{i\in I}\sF(U_i)
909
\xrightarrow{d}C^1=\prod_{i<j}\sF(U_{ij})$$
910
we see that $\cH^0(\sU,\sF)=\ker d = \Gamma(X,\sF)$.
911
\end{proof}
912
913
\subsection{Sheafify}
914
Let $X$ be a topological space, $\sU$ an open cover and
915
$\sF$ a sheaf of abelian groups. Then we define
916
$$\sC^p(\sU,\sF)=\prod_{i_0<\cdots<i_p}j_{*}(\sF|_{U_{i_0\cdots i_p}})$$
917
and define
918
$$d:\sC^p(\sU,\sF)\into \sC^{p+1}(\sU,\sF)$$
919
in terms of the $d$ defined above by, for $V$ an open set,
920
$$\sC^{p}(\sU,\sF)(V)=C^p(\sU|_{V},\sF|_V)\xrightarrow{d}
921
C^{p+1}(\sU|_{V},\sF|_V)
922
=\sC^{p+1}(\sU,\sF)(V).$$
923
924
\begin{remark}
925
$C^p(\sU,\sF)=\Gamma(X,\sC^{p}(\sU,\sF)$
926
\end{remark}
927
928
\begin{lem}
929
The sequence
930
$$ 0\into\sF\into\sC^{0}(\sU,\sF)\into\sC^{1}(\sU,\sF)\into\cdots$$
931
is a resolution of $\sF$, i.e., it is exact.
932
\end{lem}
933
\begin{proof}
934
We define the map $\sF\into\sC^{0}$ by taking the product of the natural maps
935
$\sF\into f_{*}(\sF|_{U_i})$, exactness then follows from the
936
sheaf axioms.
937
938
To show the rest of the sequence is exact it suffices to show
939
exactness at the stalks. So let $x\in X$, and suppose $x\in U_j$.
940
Given $\alpha_x\in\sC_x^p$ it is represented by a section
941
$\alpha\in\Gamma(V,\sC^p(\sU,\sF))$, over a neighborhood $V$ of
942
$x$, which we may choose so small that $V\subset U_j$. Now
943
for any $p$-tuple $i_0<\ldots<i_{p-1}$, we set
944
$$(k\alpha)_{i_0,\ldots,i_{p-1}}=\alpha_{j,i_0,\ldots,i_{p-1}}.$$
945
This makes sense because
946
$$V\intersect U_{i_0,\ldots,i_{p-1}}=V\intersect U_{j,i_0,\ldots,i_{p-1}}.$$
947
Then take the stalk of $k\alpha$ at $x$ to get the required map $k$.
948
949
Now we check that for any $p\geq 1$ and $\alpha\in\sC_x^p$,
950
$$(dk+kd)(\alpha)=\alpha.$$
951
First note that
952
\begin{align*}(dk\alpha)_{i_0,\ldots,i_p} & =
953
\sum_{\ell=0}^{p} (-1)^\ell (k\alpha)_{i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}\\
954
& = \sum (-1)^\ell \alpha_{j,i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}
955
\end{align*}
956
Whereas, on the other hand,
957
\begin{align*}
958
(kd\alpha)_{i_0,\ldots,i_p}
959
& = (d\alpha)_{j,i_0,\ldots,i_p}\\
960
& = (-1)^0\alpha_{i_0,\ldots,i_p} + \sum_{\ell=1}^{p} (-1)^{\ell+1}\alpha_{j,i_0,\ldots,
961
\hat{i_{\ell}},\ldots,i_{p}}
962
\end{align*}
963
Adding these two expressions yields $\alpha_{i_0,\ldots,i_p}$ as claimed.
964
965
Thus $k$ is a homotopy operator for the complex $\sC_x^{\bullet}$, showing
966
that the identity map is homotopic to the zero map. It follows that the
967
cohomology groups $H^{p}(\sC^{\bullet}_x)$ of this complex
968
are $0$ for $p\geq 1$.
969
\end{proof}
970
971
\begin{lem}
972
If $\sF$ is flasque then $\sC^p(\sU,\sF)$ is also flasque.
973
\end{lem}
974
\begin{proof}
975
If $\sF$ is flasque then $\sF|_{U_{i_0,\ldots,i_p}}$ is flasque
976
so $j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque so
977
$\prod j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque.
978
\end{proof}
979
980
\begin{prop}
981
If $\sF$ is flasque then $\cH^{p}(\sU,\sF)=0.$
982
\end{prop}
983
\begin{proof}
984
Consider the resolution $$0\into\sF\into\sC^{\bullet}(\sU,\sF).$$
985
By the above lemma it is flasque, so we can use it to compute the
986
usual cohomology groups of $\sF$. But $\sF$ is flasque, so
987
$H^p(X,\sF)=0$ for $p>0$. On the other hand,
988
the answer given by this resolution is
989
$$H^p(\Gamma(X,\sC^{\bullet}(\sU,\sF)))=\cH^{p}(\sU,\sF).$$
990
So we conclude that $\cH^{p}(\sU,\sF)=0$ for $p>0$.
991
\end{proof}
992
993
\begin{lem}
994
Let $X$ be a topological space, and $\sU$ an open covering. Then
995
for each $p\geq 0$ there is a natural map, functorial in $\sF$,
996
$$\cH^p(\sU,\sF)\into H^p(X,\sF).$$
997
\end{lem}
998
999
\begin{thm}
1000
Let $X$ be a Noetherian separated scheme, let $\sU$ be an open
1001
affine cover of $X$, and let $\sF$ be a quasi-coherent sheaf on
1002
$X$. Then for all $p\geq 0$ the natural maps give isomorphisms
1003
$$\cH^{p}(\sU,\sF)\isom H^p(X,\sF).$$
1004
\end{thm}
1005
1006
% 2/7/96
1007
1008
\section{\v{C}ech Cohomology and Derived Functor Cohomology}
1009
1010
Today we prove
1011
1012
\begin{thm}
1013
Let $X$ be a Noetherian, separated scheme, $\sU$ an open cover
1014
and $\sF$ a quasi-coherent sheaf on $X$. Then
1015
$$\cH^{i}(\sU,\sF)=H^{i}(X,\sF).$$
1016
\end{thm}
1017
1018
To do this we introduce a condition (*):
1019
1020
{\bfseries Condition *:} Let $\sF$ be a sheaf of abelian groups
1021
and $\sU=(U_i)_{i\in I}$ an open cover. Then the pair $\sF$ and
1022
$\sU$ satisfy condition (*) if for all $i_0,\ldots,i_p\in I$,
1023
$$H^(U_{i_0,\ldots,i_p},\sF)=0, \text{all} i>0.$$
1024
1025
\begin{lem}
1026
If $0\into\sF'\into\sF\into\sF''\into 0$ is an exact sequence in
1027
$\Ab(X)$ and $\sF'$ satisfies (*) then there is a long exact sequence
1028
for $\cH^{i}(\sU,\cdot)$.
1029
\end{lem}
1030
\begin{proof}
1031
Since the global sections functor is left exact and cohomology
1032
commutes with products, we have an exact sequence
1033
\begin{align*}
1034
0\into C^{p}(\sU,\sF')=\prod_{i_0<\cdots<i_p}\sF'(U_{i_0,\ldots,i_p})
1035
\into C^{p}(\sU,\sF)=\prod_{i_0<\cdots<i_p}\sF(U_{i_0,\ldots,i_p}) \\
1036
\into C^{p}(\sU,\sF'')=\prod_{i_0<\cdots<i_p}\sF''(U_{i_0,\ldots,i_p})
1037
\into \prod_{i_0<\cdots<i_p} H^{1}(\sU_{i_0,\ldots,i_p},\sF')=0
1038
\end{align*}
1039
where the last term is 0 because $\sF'$ satisfies condition (*).
1040
Replacing $p$ by $\cdot$ gives an exact sequence of complexes.
1041
Applying $\cH^{i}(\sU,\cdot)$ then gives the desired result.
1042
\end{proof}
1043
1044
\begin{thm}
1045
Let $X$ be a topological space, $\sU$ an open cover and $\sF\in \Ab(X)$.
1046
Suppose $\sF$ and $\sU$ satisfy (*). Then the maps
1047
$$\varphi^{i}:\cH^i(\sU,\sF)\into H^{i}(X,\sF)$$
1048
are isomorphisms.
1049
\end{thm}
1050
\begin{proof}
1051
The proof is a clever induction.
1052
\end{proof}
1053
1054
\begin{lem}
1055
If $0\into\sF'\into\sF\into\sF''\into 0$ is exact and
1056
$\sF'$ and $\sF$ satisfy (*) then $\sF''$ satisfies (*).
1057
\end{lem}
1058
1059
To prove the main theorem of the section use the fact that
1060
$X$ separated implies any finite intersection of affines
1061
is affine and then use the vanishing theorem for cohomology
1062
of a quasi-coherent sheaf on an affine scheme. The above theorem
1063
then implies the main result. From now on we will always
1064
assume our schemes are separated unless otherwise stated.
1065
1066
\begin{cor}
1067
If $X$ is a (separated) Noetherian scheme and $X$ can be covered
1068
by $n+1$ open affines for some $n>0$ then $H^{i}(X,\sF)=0$ for $i>n$.
1069
\end{cor}
1070
1071
\begin{example}
1072
Let $X=\bP_k^n$, then the existence of the standard affine
1073
cover $U_0,\ldots,U_n$ implies that $H^{i}(X,\sF)=0$ for
1074
$i>n$.
1075
\end{example}
1076
1077
\begin{example}
1078
Let $X$ be a projective curve embedded in $\bP^k_n$.
1079
Let $U_0\subset X$ be open affine, then $X-U_0$ is finite.
1080
Thus $U_0\subset X\subset \bP^n$ and $X-U_0=\{P_1,\ldots,P_r\}$.
1081
In $\bP^n$ there is a hyperplane $H$ such that
1082
$P_1,\ldots,P_r\not\in H$. Then $P_1,\ldots,P_r\in\bP^n-H=\bA^n=V$.
1083
Then $U_1=V\intersect X$ is closed in the affine set $V$, hence affine.
1084
Then $X=U_0\union U_1$ with $U_0$ and $U_1$ both affine.
1085
Thus $H^{i}(X,\sF)=0$ for all $i\geq 2$.
1086
\end{example}
1087
1088
\begin{exercise}
1089
If $X$ is any projective scheme of dimension $n$ then $X$
1090
can be covered by $n+1$ open affines so
1091
$$H^{i}(X,\sF)=0 \text{ for all } i>n.$$
1092
[Hint: Use induction.]
1093
\end{exercise}
1094
1095
Hartshorne was unaware of the answer to the following question
1096
today.
1097
\begin{ques}
1098
If $X$ is a Noetherian scheme of dimension $n$ do there
1099
exist $n+1$ open affines covering $X$.
1100
\end{ques}
1101
1102
\begin{thm}[Grothendieck]
1103
If $\sF\in\Ab(X)$ then $H^{i}(X,\sF)=0$ for all $i>n=\dim X$.
1104
\end{thm}
1105
1106
\begin{example}
1107
Let $k$ be an algebraically closed field.
1108
Then $X=\bA^2_k-\{(0,0)\}$ is not affine since it has global
1109
sections $k[x,y]$.
1110
We compute $H^1(X,\sox)$ by \cech cohomology.
1111
Write $X=U_1\union U_2$ where $U_1=\{x\neq 0\}$ and
1112
$U_2=\{y\neq 0\}$. Then the \cech complex is
1113
$$C^{\cdot}(\sU,\sox): k[x,y,x^{-1}]\oplus{}k[x,y,y^{-1}]
1114
\xrightarrow{d} k[x,x^{-1},y,y^{-1}].$$
1115
Thus one sees with a little thought that
1116
$H^{0}=\ker{d}=k[x,y]$
1117
and
1118
$H^{1}=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\} = E$
1119
as $k$-vector spaces (all sums are finite).
1120
\end{example}
1121
\subsection{History of this Module $E$}
1122
$$E=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\}$$
1123
\begin{enumerate}
1124
\item Macaulay's ``Inverse System'' (1921?)
1125
\item $E$ is an injective $A$-module, in fact, the indecomposable
1126
injective associated to the prime $(x,y)$
1127
\item $E$ is the dualizing module of $A$, thus
1128
$D=\Hom_A(\cdot,E)$ is a dualizing functor
1129
for finite length modules (so doing $D$ twice
1130
gives you back what you started with).
1131
\item Local duality theorem: this is the module you ``hom into''.
1132
\end{enumerate}
1133
1134
1135
% 2/9/96
1136
\section{Cohomology of $\bP_k^n$}
1137
Today we begin to compute $H^{i}(X,\sox(\ell))$ for all $i$ and all
1138
$\ell$.
1139
1140
a) $H^0(X,\sox(\ell))$ is the vector space of forms of degree $\ell$
1141
in $S=k[x_0,\ldots,x_n]$, thus
1142
$$\oplus_{\ell\in\bZ}H^0(\sox(\ell))=H^0_{*}(\sox)=\Gamma_{*}(\sox)=S.$$
1143
1144
\begin{prop}
1145
There is a natural map
1146
$$H^{0}(\sox(\ell))\times H^{i}(\sox(m))\into H^{i}(\sox(\ell+m)).$$
1147
\end{prop}
1148
\begin{proof}
1149
$\alpha\in H^{0}(\sox(\ell))$ defines a map $\sox\into\sox(\ell)$
1150
given by $1\mapsto\alpha$. This defines a map
1151
$$\sox\tensor\sox(m)\xrightarrow{\alpha(m)}\sox(\ell)\tensor\sox(m)$$
1152
which gives a map $\sox(m)\into\sox(\ell+m)$. This induces the desired
1153
map $H^{i}(\sox(m))\into H^{i}(\sox(\ell+m))$.
1154
\end{proof}
1155
1156
b) $H^{i}(\sox(\ell))=0$ when $0<i<n$ and for all $\ell$.
1157
(This doesn't hold for arbitrary quasi-coherent sheaves!)
1158
1159
c) $H_{*}^n(X,\sox)$ is a graded $S$-module which
1160
is $0$ in degrees $\geq -n$, but is nonzero in degrees $\leq n-1$.
1161
As a $k$-vector space it is equal to
1162
$$\{\sum_{i_j<0} a_{i_0,\ldots,i_n}x_0^{i_0}\cdots x_n^{i_n} :
1163
\text{sum is finite}\}.$$
1164
1165
d) For $\ell\geq 0$ the map
1166
$$H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))\isom{}k$$
1167
is a perfect pairing so we have a duality (which is in fact a special
1168
case of Serre Duality).
1169
1170
1171
% 2/12/96
1172
\section{Serre's Finite Generation Theorem}
1173
We relax the hypothesis from the last lecture and claim that the
1174
same results are still true.
1175
\begin{thm}
1176
Let $A$ be a Noetherian ring and $X=\bP^n_A$. Then
1177
\begin{enumerate}
1178
\item $H^0_{*}(\sox)=\oplus_{\ell}H^{0}_{\ell}(\sox(\ell))=S=A[x_0,\ldots,x_n]$
1179
\item $H^i_{*}(\sox)=0$ for all $0<i<n$
1180
\item $H^n_{*}(\sox)=\{\sum_{I=i_0,\ldots,i_n} a_I x_0^{i_0}\cdots x_n^{i_n} : a_I\in A\}$
1181
\item $H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))$ is
1182
a perfect pairing of free $A$-modules. Notice that $H^n(\sox(-n-1))$ is
1183
a free $A$-module of rank 1 so it is isomorphic to $A$, but {\em not} in
1184
a canonical way!
1185
\end{enumerate}
1186
\end{thm}
1187
1188
Although pairing is in general not functorial as a map into $A$,
1189
there is a special situation in which it is. Let $\Omega_{X/k}^1$
1190
be the sheaf of differentials on $X=\bP^n_k$. Let $\omega=\Omega_{X/k}^n
1191
=\Lambda^n\Omega^1$ be the top level differentials (or ``dualizing
1192
module''). Then some map is functorial (??)
1193
\begin{quote}
1194
``Is $\omega$ more important than $\Omega$?'' -- Janos Csirik
1195
1196
``That's a value judgment... you can make your own decision
1197
on that... I won't.'' -- Hartshorne
1198
\end{quote}
1199
1200
\begin{thm}[Serre]
1201
Let $X$ be a projective scheme over a Noetherian ring $A$. Let $\sF$
1202
be any coherent sheaf on $X$. Then
1203
\begin{enumerate}
1204
\item $H^{i}(X,\sF)$ is a finitely generated $A$-module for all $i$
1205
\item for all $\sF$ there exists $n_0$ such that for all $i>0$ and
1206
for all $n\geq n_0$, $H^{i}(X,\sF(n))=0$.
1207
\end{enumerate}
1208
\end{thm}
1209
1210
The following was difficult to prove last semester and
1211
we were only able to prove it under somewhat restrictive
1212
hypothesis on $A$ (namely, that $A$ is a finitely generated
1213
$k$-algebra).
1214
\begin{cor} $\Gamma(X,\sF)$ is a finitely generated $A$-module.
1215
\end{cor}
1216
\begin{proof}
1217
Set $i=0$ in 1.
1218
\end{proof}
1219
1220
\begin{proof}(of theorem)
1221
1222
I. {\em Reduce to the case $X=\bP^r_A$.}
1223
Use the fact that the push forward of a closed subscheme has
1224
the same cohomology to replace $\sF$ by $i_{*}(\sF)$.
1225
1226
II. {\em Special case, $\sF=\sO_{\bP^r}(\ell)$ any $\ell\in\bZ$.}
1227
1. and 2. both follow immediately from the previous theorem.
1228
This is where we have done the work in explicit calculations.
1229
1230
III. {\em Cranking the Machine of Cohomology}
1231
1232
1233
\end{proof}
1234
1235
\subsection{Application: The Arithmetic Genus}
1236
Let $k$ be an algebraically closed field and $V\subset X=\bP^n_k$ a
1237
projective variety. The arithmetic genus of $V$ is
1238
$$p_a=(-1)^{\dim V}(p_V(0)-1)$$
1239
where $p_V$ is the Hilbert polynomial of $V$, thus
1240
$p_V(\ell)=\dim_k(S/I_V)_{\ell}$ for all $\ell\gg 0$. The
1241
Hilbert polynomial depends on the projective embedding of $V$.
1242
\begin{prop}
1243
$p_V(\ell)=\sum_{i=0}^{\infty}(-1)^{i}\dim_k H^{i}(\sO_V(\ell))$
1244
for {\em all} $\ell\in\bZ$.
1245
\end{prop}
1246
This redefines the Hilbert polynomial. Furthermore,
1247
$$p_a=(-1)^{\dim V}(p_V(0)-1) = (-1)^{\dim V}
1248
\sum_{i=0}^{\infty}(-1)^i\dim_k(H^{i}(\sO_V))$$
1249
which shows that $p_a$ is intrinsic, i.e., it doesn't depend
1250
on the embedding of $V$ in projective space.
1251
1252
\section{Euler Characteristic}
1253
Fix an algebraically closed field $k$, let $X=\P_k^n$. Suppose
1254
$\sF$ is a coherent sheaf on $X$. Then by Serre's theorem
1255
$H^{i}(X,\sF)$ is a finite dimensional $k$-vector space.
1256
Let $$h^{i}(X,\sF)=\dim_k H^{i}(X,\sF).$$
1257
\begin{defn}
1258
The {\bfseries Euler characteristic} of $\sF$ is
1259
$$\chi(\sF)=\sum_{i=0}^{n}(-1)^i h^i(X,\sF).$$
1260
\end{defn}
1261
Thus $\chi$ is a function $\Coh(X)\into\Z$.
1262
\begin{lem}
1263
If $k$ is a field and
1264
$$0\into{}V_1\into{}V_2\into{}\cdots\into{}V_N\into{}0$$
1265
is an exact sequence of finite dimensional vector spaces,
1266
then $\sum_{i=1}^N(-1)^i\dim{}V_i=0$.
1267
\end{lem}
1268
\begin{proof}
1269
Since every short exact sequence of vector spaces splits, the
1270
statement is true when $N=3$. If the statement is true for an
1271
exact sequence of length $N-1$ then, applying it to the exact sequence
1272
$$0\into{}V_2/V_1\into{}V_3\into\cdots\into{}V_N\into{}0,$$
1273
shows that
1274
$\dim{}V_2/V_1-\dim V_3+\cdots\pm\dim V_n = 0$
1275
from which the result follows.
1276
\end{proof}
1277
1278
\begin{lem}
1279
If $0\into\sF'\into\sF\into\sF''\into{}0$ is an exact sequence
1280
of coherent sheaves on $X$, then
1281
$$\chi(\F)=\chi(\F')+\chi(\F'').$$
1282
\end{lem}
1283
\begin{proof}
1284
Apply the above lemma to the long exact sequence of cohomology
1285
taking into account that $H^n(\F'')=0$ by Serre's vanishing theorem.
1286
\end{proof}
1287
1288
More generally, any map $\chi$ from an abelian category to
1289
$\Z$ is called additive if, whenever
1290
$$0\into\F^0\into\F^1\into\cdots\into\F^n\into{}0$$
1291
is exact, then
1292
$$\sum_{i=0}^{n}(-1)^{i}\chi(\F^i)=0.$$
1293
1294
{\bfseries Question.}
1295
Given an abelian category $\sA$ find an abelian
1296
group $A$ and a map $X:\sA\into{}A$ such that every
1297
additive function $\chi:\sA\into{}G$ factor through $\sA\xrightarrow{X}A$.
1298
In the category of coherent sheaves the Grothendieck group
1299
solves this problem.
1300
1301
Let $X=\P_k^n$ and suppose $\sF$ is a coherent sheaf on $X$.
1302
The Euler characteristic induces a map
1303
$$\Z\into\Z: n\mapsto\chi(\F(n)).$$
1304
\begin{thm}
1305
There is a polynomial $p_{\F}\in\Q[z]$ such that
1306
$p_{\F}(n)=\chi(\F(n))$ for all $n\in\Z$.
1307
\end{thm}
1308
The polynomial $p_{\F}(n)$ is called the Hilbert polynomial of $\F$.
1309
Last semester we defined the Hilbert polynomial of a graded module
1310
$M$ over the ring $S=k[x_0,\ldots,x_n]$. Define $\varphi_M:\Z\into\Z$
1311
by $\varphi_M(n)=\dim_k M_n$. Then we showed that there is a unique
1312
polynomial $p_M$ such that $p_M(n)=\varphi_M(n)$ for all $n\gg 0$.
1313
\begin{proof}
1314
We induct on $\dim(\supp\F)$. If $\dim(\supp\F)=0$ then $\supp\F$ is a
1315
union of closed points so $\sF=\oplus_{i=1}^{k}\F_{p_i}$.
1316
Since each $\F_{p_i}$ is a finite
1317
dimensional $k$-vector space and $\sox(n)$ is locally free,
1318
there is a non-canonical isomorphism $\F(n)=\F\tensor\sox(n)\isom\F$.
1319
Thus $$\chi_{\F}(n)=h^0(\F(n))=h^0(\F)=\sum_{i=1}^{k}\dim_k\F_{p_i}$$
1320
which is a constant function, hence a polynomial.
1321
1322
Next suppose $\dim(\supp\F)=s$.
1323
Let $x\in{}S_1=H^0(\sox(1))$ be such that the hyperplane
1324
$H:=\{x=0\}$ doesn't contain any irreducible component
1325
of $\supp\F$. Multiplication by $x$ defines a map
1326
$\sox(-1)\xrightarrow{x}\sox$ which is an isomorphism
1327
outside of $H$. Tensoring with $\F$ gives a map
1328
$\F(-1)\into\F$. Let $\sR$ be the kernel and
1329
$\sQ$ be the cokernel, then there is an exact sequence
1330
$$0\into\sR\into\F(-1)\xrightarrow{x}\F\into\sQ\into{}0.$$
1331
Now $\supp\sR\union\supp\sQ\subset\supp\F\intersect H$ so
1332
$\dim(\supp\sR)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ and
1333
$\dim(\supp\sQ)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ so
1334
by our induction hypothesis $\chi(\sQ(n))$ and $\chi(\sR(n))$
1335
are polynomials. Twisting the above exact sequence by $n$ and
1336
applying $\chi$ yields
1337
$$\chi(\F(n))-\chi(\F(n-1))=\chi(\Q(n))-\chi(\sR(n))=P_{\sQ}(n)-P_{\sR}(N).$$
1338
Thus the first difference function of $\chi(\F(n))$ is a polynomial
1339
so $\chi(\F(n))$ is a polynomial.
1340
\end{proof}
1341
1342
1343
\begin{example}
1344
Let $X=\P^1$ and $\F=\sox$. Then $S=k[x_0,x_1]$, $M=S$ and
1345
$\dim S_n = n+1$. Thus $p_M(z)=z+1$ and $p_M(n)=\varphi(n)$
1346
for $n\geq -1$. Computing the Hilbert polynomial in terms of
1347
the Euler characteristic gives
1348
$$\chi(\F(n))=h^0(\sox(n))- h^1(\sox(n))
1349
= \begin{cases} (n+1) - 0 & n\geq -1\\
1350
0-(-n-1)=n+1 & n\leq -2
1351
\end{cases} $$
1352
Thus $p_{\F}(n)=n+1$.
1353
\end{example}
1354
1355
The higher cohomology corrects the failure of the
1356
Hilbert polynomial in lower degrees.
1357
1358
%% 2/16/96
1359
\section{Correspondence between Analytic and Algebraic Cohomology}
1360
{\bf Homework. } Chapter III, 4.8, 4.9, 5.6.
1361
1362
Look at Serre's 1956 paper {\em Geometrie Algebraique et Geometrie
1363
Analytique} (GAGA). ``What are the prerequisites?'' asks Janos.
1364
``French,'' answers Nghi. ``Is there an English translation'' asks the
1365
class. ``Translation? ... What for? It's so beautiful in the French,''
1366
retorts Hartshorne.
1367
1368
Let $\F$ be a coherent sheaf on $\P^n_{\C}$ with its Zariski topology.
1369
Then we can associate to $\F$ a sheaf
1370
$\F^{\mbox{\rm an}}$
1371
on $\P^n_{\C}$ with its analytic topology. $\F$ is locally a cokernel
1372
of a morphism of free sheaves so we can define
1373
$\F^{\mbox{\rm an}}$
1374
by defining $\sox^{\mbox{\rm an}}$.
1375
The map
1376
$$\Coh(\P^n_{\C})\xrightarrow{\mbox{\rm an}}\Coh^{\mbox{\rm an}}(\P^n_{\C})$$
1377
is an equivalence of categories and
1378
$$H^i(X,\F)\iso{}H^i(X^{\mbox{\rm an}},\F^{\mbox{\rm an}})$$
1379
for all $i$.
1380
If $X/\C$ is affine the corresponding object $X^{\mbox{\rm an}}_{\C}$
1381
is a Stein manifold.
1382
1383
\section{Arithmetic Genus}
1384
Let $X\hookrightarrow{}\P_k^n$ be a projective variety with
1385
$k$ algebraically closed and suppose $\F$ is a coherent sheaf
1386
on $X$. Then $$\chi(\F)=\sum(-1)^i h^i(\F)$$ is the Euler characteristic
1387
of $\F$, $$P_{\F}(n)=\chi(\F(n))$$ gives the Hilbert polynomial of
1388
$\F$ on $X$, and $$p_a(X)=(-1)^{\dim X}(P_{\sox}(0)-1)$$ is the
1389
arithmetic genus of $X$. The arithmetic genus is independent
1390
of the choice of embedding of $X$ into $\P_k^n$.
1391
1392
If $X$ is a curve then $$1-p_a(X)=h^0(\sox)-h^1(\sox)$$.
1393
Thus if $X$ is an integral projective curve then $h^0(\sox)=1$ so
1394
$p_a(X)=h^1(\sox)$. If $X$ is a nonsingular projective curve
1395
then $p_a(X)=h^1(\sox)$ is called {\bfseries the genus} of $X$.
1396
1397
Let $V_1$ and $V_2$ be varieties, thus they are projective integral
1398
schemes over an algebraically closed field $k$. Then $V_1$ and
1399
$V_2$ are {\bfseries birationally equivalent} if and only if $K(V_1)\isom{}K(V_2)$
1400
over $k$, where $K(V_i)$ is the function field of $V_i$. $V$ is
1401
{\bfseries rational} if $V$ is bironational to $\P_k^n$ for some $n$.
1402
Since a rational map on a nonsingular projective curve always extends,
1403
two nonsingular projective curves are birational if and only if
1404
they are isomorphic. Thus for nonsingular projective curves
1405
the genus $g$ is a birational invariant.
1406
1407
\subsection{The Genus of Plane Curve of Degree $d$}
1408
Let $C\subset\P_k^2$ be a curve of degree $d$. Then $C$
1409
is a closed subscheme defined by a single homogeneous polynomial
1410
$f(x_0,x_1,x_2)$ of degree $d$, thus
1411
$$C=\proj(S/(f)).$$
1412
1413
Some possibilities when $d=3$ are:
1414
\begin{itemize}
1415
\item $f: Y^2-X(X^2-1)$, a nonsingular elliptic curve
1416
\item $f: Y^2-X^2(X-1)$, a nodal cubic
1417
\item $f: Y^3$, a tripled $x$-axis
1418
\item $f: Y(X^2+Y^2-1)$, the union of a circle and the $x$-axis
1419
\end{itemize}
1420
1421
Now we compute $p_a(C)$. Let $I=(f)$ with $\deg f=d$.
1422
Then $$1-p_a=h_0(\so_C)-h_1(\so_C)+h_2(\so_C)=\chi(\so_C).$$
1423
We have an exact sequence
1424
$$0\into\sI_C\into\so_{\P^2}\into\so_C\into 0.$$
1425
Now $\sI_C\isom\so_{\P^2}(-d)$ since $\so_{\P^2}(-d)$
1426
can be thought of as being generated by $1/f$ on $D_+(f)$
1427
and by something else elsewhere, and then multiplication
1428
by $f$ gives an inclusion
1429
$so_{\P^2}(-d)|_{D_+(f)}\into\so_{\P^2}|_{D_+(f)}$, etc.
1430
Therefore
1431
$$\chi(\so_C)=\chi(\so_{\P^2})-\chi(\so_{\P^2}(-d)).$$
1432
Now
1433
$$\chi(\so_{\P^2})=h^0(\so_{\P^2})-h^1(\so_{\P^2})+h^2(\so_{\P^2})=1+0+0$$
1434
and
1435
$$\chi(\so_{\P^2}(-d))=h^0(\so_{\P^2}(-d))-h^1(\so_{\P^2}(-d))+h^2(\so_{\P^2}(-d))
1436
=0+0+\frac{1}{2}(d-1)(d-2).$$
1437
For the last computation we used duality (14.1) to see that
1438
$$h^2(\so_{\P^2}(-d))=h^0(\so_{\P^2}(d-3)=\dim S_{d-3}
1439
=\frac{1}{2}(d-1)(d-2).$$
1440
Thus $\chi(\so_C)=1-\frac{1}{2}(d-1)(d-2)$ so
1441
$$p_a(C)=\frac{1}{2}(d-1)(d-2).$$
1442
1443
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1444
%% 2/21/96
1445
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1446
1447
\section{Not Enough Projectives}
1448
\begin{exercise} Prove that the category of quasi-coherent
1449
sheaves on $X=\P_k^1$ doesn't have enough projectives.
1450
\end{exercise}
1451
\begin{proof}
1452
We show that there is no projective object $\sP\in\Qco(X)$ along
1453
with a surjection $\sP\into\sox\into 0$.
1454
\begin{lem}
1455
If $\P\xrightarrow{\varphi}\sox$ is surjective and $\sP$ is
1456
quasi-coherent, then there exists $\ell$ such that
1457
$H^0(\P(\ell))\into{}H^0(\sox(\ell))$ is surjective.
1458
\end{lem}
1459
The {\em false} proof of this lemma is to write down an
1460
exact sequence $0\into\sR\into\sP\into\sox\into 0$ then
1461
use the ``fact'' that $H^1(\sR(\ell))=0$ for sufficiently
1462
large $\ell$. This doesn't work because $\sR$ might not
1463
be coherent since it is only the quotient of
1464
quasi-coherent sheaves. A valid way to proceed is to use
1465
(II, Ex. 5.15) to write $\sP$ as an ascending union of its coherent
1466
subsheaves, $\sP=\union_{i}\sP_{i}$.
1467
Then since $\varphi$ is surjective,
1468
$\sox=\union_{i}\varphi(\sP_{i})$, where $\varphi(\sP_{i})$
1469
is the sheaf image. Using the fact that $\varphi(\sP_{i})$ is
1470
the sheaf image, that $\sox$ is coherent and that the union
1471
is ascending, this implies $\sox=\varphi(\sP_i)$ for some $i$.
1472
We now have an exact sequence
1473
$$0\into\sR_i\into\sP_i\into\sox\into{}0$$
1474
with $\sR_i$ coherent since $\sP_i$ and $\sox$ are both coherent.
1475
Thus $H^i(\sR_i(\ell))=0$ for $l\gg 0$ which, upon computing
1476
the long exact sequence of cohomology, gives the lemma.
1477
1478
Now fix such an $\ell$. We have a commutative diagram
1479
$$\begin{CD}
1480
\[email protected]>>>\[email protected]>>>0\\
1481
@V{\exists}VV @VVV\\
1482
\sox(-\ell-1)@>>>k(p)@>>>0
1483
\end{CD}$$
1484
Twisting by $\ell$ gives a commutative diagram
1485
$$\begin{CD}
1486
\sP(\ell)@>>>\sox(\ell)@>>>0\\
1487
@VVV @VVV\\
1488
\sox(-1)@>>>k(p)@>>>0
1489
\end{CD}$$
1490
Let $s\in\Gamma(\sox(\ell))$ be a global section which is nonzero at $p$,
1491
then there is $t\in\Gamma(\sP(\ell))$ which maps to $s$.
1492
But then by commutativity $t$ must map to some element of $\Gamma(\sox(-1))=0$
1493
which maps to a nonzero element of $k(p)$, which is absurd.
1494
\end{proof}
1495
1496
\section{Some Special Cases of Serre Duality}
1497
\subsection{Example: $\sox$ on Projective Space}
1498
Suppose $X=\P_k^n$, then there is a perfect pairing
1499
$$H^0(\sox(\ell))\times{}H^n(\sox(-\ell-n-1))\into{}H^n(\sox(-n-1))\isom{}k.$$
1500
For this section let
1501
$$\omega_X=\sox(-n-1).$$
1502
Because the pairing is perfect we have a non-canonical but functorial
1503
isomorphism
1504
$$H^0(\sox(\ell))\isom H^n(\sox(-\ell-n-1))'.$$
1505
(If $V$ is a vector space then $V'$ denotes its dual.)
1506
1507
\subsection{Example: Coherent sheaf on Projective Space}
1508
Suppose $\sF$ is any coherent sheaf on $X=\P_k^r$.
1509
View $\Hom(\sF,\omega)$ as a $k$-vector space.
1510
1511
% WHY DEFINE SHEAF HOM???
1512
%We recall the definition of the sheaf Hom.
1513
%\begin{defn}
1514
%Let $\sF$ and $\sG$ be sheaves of $\sox$-modules.
1515
%Then $\sHom(\sF,\sG)$ is the sheaf which associates to
1516
%an open set the group
1517
%$$Hom_{\Mod(\sox|_U)}(\sF|_U,\sG|_U).$$
1518
%\end{defn}
1519
1520
By functoriality and since $H^n(\omega)=k$ there is a map
1521
$$\varphi:\Hom(\sF,\omega)\into\Hom(H^n(\sF),H^n(\omega))=H^n(\sF)'.$$
1522
\begin{prop} $\varphi$ is an isomorphism for all coherent sheaves $\sF$.
1523
\end{prop}
1524
\begin{proof}
1525
\par {\em Case 1.} If $\sF=\sox(\ell)$ for some $\ell\in\Z$ then this is
1526
just a restatement of the previous example.
1527
\par {\em Case 2.} If $\sE=\oplus_{i=1}^{k}\so(\ell_i)$ is a finite
1528
direct sum, then the statement follows from the
1529
commutativity of the following diagram.
1530
$$\begin{CD}
1531
\Hom(\oplus_{i=1}^{k}\so(\ell_i),\omega)@>>>H^n(\oplus_{i=1}^k\so(\ell_i))'\\
1532
@VV\isom{}V @VV\isom{}V\\
1533
\oplus_{i=1}^k\Hom(\so(\ell_i),\omega)@>>\sim{}>\oplus_{i=1}^k{}H^n(\so(\ell_i))'
1534
\end{CD}$$
1535
\par {\em Case 3.} Now let $\sF$ be an arbitrary coherent sheaf.
1536
View $\varphi$ as a morphism of functors
1537
$$\Hom(\cdot,\omega)\into H^n(\cdot)'.$$
1538
The functor $\Hom(\cdot,\omega)$ is contravarient left exact.
1539
$H^n(\cdot)$ is covariant right exact since $X=\P_k^n$ so
1540
$H^{n+1}(\sF)=0$ for any coherent sheaf $\sF$. Thus $H^n(\cdot)'$
1541
is contravarient left exact.
1542
\begin{lem}
1543
Let $\sF$ be any coherent sheaf. Then there exists a partial resolution
1544
$$\sE_1\into\sE_0\into\sF\into{}0$$
1545
by sheaves of the form $\oplus_{i}\sox(\ell_i)$.
1546
\end{lem}
1547
By (II, 5.17) for $\ell\gg 0$, $\sF(\ell)$ is
1548
generated by its global sections. Thus there is a surjection
1549
$$\sox^m\into\sF(\ell)\into 0$$
1550
which upon twisting by $-\ell$ becomes
1551
$$\sE_0=\sox(-\ell)^m\into\sF\into 0.$$
1552
Let $\sR$ be the kernel so
1553
$$0\into\sR\into\sE\into\sF$$
1554
is exact. Since $\sR$ is coherent, we can repeat the argument
1555
above to find $\sE_1$ surjecting onto $\sR$. This yields
1556
the desired exact sequence.
1557
1558
Now we apply the functors $\Hom(\cdot,\omega)$ and
1559
$H^n(\cdot)'$. This results in a commutative diagram
1560
$$\begin{CD}
1561
0@>>> \Hom(\sF,\omega) @>>> \Hom(\sE_0,\omega) @>>> \Hom(\sE_1,\omega)\\
1562
@VVV @V\varphi(\sF)VV @V\varphi(\sE_0)VV @VV\varphi(\sE_1)V\\
1563
0@>>> H^n(\sF)'@>>>H^n(\sE_0)' @>>> H^n(\sE_1)'
1564
\end{CD}$$
1565
From cases 1 and 2, the maps $\varphi(\sE_0)$ and
1566
$\varphi(\sE_1)$ are isomorphisms so $\varphi(\sF)$ must also be
1567
an isomorphism.
1568
\end{proof}
1569
1570
\subsection{Example: Serre Duality on $\P_k^n$}
1571
Let $X=\P_k^n$ and $\sF$ be a coherent sheaf.
1572
Then for each $i$ there is an isomorphism
1573
$$\varphi^{i}:\ext^{i}_{\sox}(\sF,\omega)\into H^{n-i}(\sF)'.$$
1574
1575
%%%%%%%%%%%%%%%%%%%%
1576
%% 2/23/96
1577
1578
\section{The Functor $\ext$}
1579
Let $(X,\sox)$ be a scheme and $\sF,\sG\in\Mod(\sox)$. Then
1580
$\Hom(\sF,\sG)\in\Ab$. View $\Hom(\sF,\bullet)$ as a
1581
functor $\Mod(\sox)\into\Ab$. Note that $\Hom(\sF,\bullet)$
1582
is left exact and covariant. Since $\Mod(\sox)$ has enough
1583
injectives we can take derived functors.
1584
\begin{defn}
1585
The $\ext$ functors $\ext^i_{\sox}(\sF,\bullet)$ are the right
1586
derived functors of $\Hom_{\sox}(\sF,\bullet)$ in the
1587
category $\Mod(\sox)$.
1588
\end{defn}
1589
Thus to compute $\ext^i_{\sox}(\sF,\sG)$, take an injective
1590
resolution $$0\into\sG\into I^0\into I^1\into \cdots $$
1591
then
1592
$$\ext^i_{\sox}(\sF,\sG)=H^i(\Hom_{\sox(\sF,I^{\bullet})}).$$
1593
1594
\begin{remark} {\bfseries Warning!}
1595
If $i:X\hookrightarrow\P^n$ is a closed subscheme of $P^n$ then
1596
$\ext^i_{\sox}(\sF,\sG)$ need {\em not} equal
1597
$\ext^i_{\P^n}(i_{*}(\sF),i_{*}(\sG))$. With cohomology
1598
these are the same, but not with $\ext$!
1599
\end{remark}
1600
1601
\begin{example}
1602
Suppose $\sF=\sox$, then
1603
$\Hom_{\sox}(\sox,\sG)=\Gamma(X,\sG)$. Thus
1604
$\ext^i_{\sox}(\sox,\bullet)$ are the derived
1605
functors of $\Gamma(X,\bullet)$ in $\Mod(\sox)$.
1606
Since we can computer cohomology using flasque
1607
sheaves this implies $\ext^i_{\sox}(\sox,\bullet)=H^i(X,\bullet)$.
1608
Thus $\ext$ generalizes $H^i$ but we get a lot more besides.
1609
\end{example}
1610
1611
\subsection{Sheaf $\ext$}
1612
Now we define a new kind of $\ext$. The sheaf hom functor
1613
$$\sHom_{\sox}(\sF,\bullet):\Mod(\sox)\into\Mod(\sox)$$
1614
is covariant and left exact. Since $\Mod(\sox)$ has enough
1615
injectives we can defined the derived functors
1616
$\sext^i_{\sox}(\sF,\bullet)$.
1617
1618
\begin{example}
1619
Consider the functor $\ext^i_{\sox}(\sox,\bullet)$.
1620
Since $\sHom_{\sox}(\sox,\sG)=\sG$ this is the identity
1621
functor which is exact so
1622
$$\ext^i_{\sox}(\sox,\sG)=\begin{cases}\sG\quad i=0\\0\quad i>0\end{cases}$$
1623
\end{example}
1624
1625
What if we have a short exact sequence in the first variables, do we
1626
get a long exact sequence?
1627
\begin{prop}
1628
The functors $\ext^i$ and $\sext^i$ are $\delta$-functors
1629
in the first variable. Thus if $$0\into\sF'\into\sF\into\sF''\into 0$$
1630
is exact then there is a long exact sequence
1631
\begin{align*}0\into&\Hom(\sF'',\sG)\into\Hom(\sF,\sG)\into\Hom(\sF',\sG)
1632
\into&\ext^1(\sF'',\sG)\into\ext^1(\sF,\sG)\into\ext^1(\sF,\sG)\into\cdots
1633
\end{align*}
1634
\end{prop}
1635
The conclusion of this proposition is not obvious because we
1636
the $\ext^i$ as derived functors in the second variable, not the first.
1637
\begin{proof}
1638
Suppose we are given $0\into\sF'\into\sF\into\sF''\into 0$ and $\sG$.
1639
Choose an injective resolution $0\into\sG\into{}I^{\bullet}$ of $\sG$.
1640
Since $\Hom(\bullet,I^n)$ is exact (by definition of injective object),
1641
the sequence
1642
$$0\into\Hom(\sF'',I^{\bullet})\into\Hom(\sF,I^{\bullet})\into
1643
\Hom(\sF',I^{\bullet})\into 0$$
1644
is exact. By general homological algebra these give rise a long
1645
exact sequence of cohomology of these complexes. For $\sext^i$
1646
simply scriptify everything!
1647
\end{proof}
1648
1649
\subsection{Locally Free Sheaves}
1650
\begin{prop}
1651
Suppose $\sE$ is a locally free $\sox$-module of finite rank.
1652
Let $\sE^{\dual}=\Hom(\sE,\sO)$. For any sheaves $\sF$, $\sG$,
1653
$$\ext^i(\sF\tensor\sE,\sG)\isom\ext^i(\sF,\sG\tensor\sE^{\dual})$$
1654
and
1655
$$\sext^i(\sF\tensor\sE,\sG)\isom\sext^i(\sF,\sG\tensor\sE^{\dual})
1656
\isom\sext^i(\sF,\sG)\tensor\sE^{\dual}.$$
1657
\end{prop}
1658
1659
\begin{lem}
1660
If $\sE$ is locally free of finite rank and $\sI\in\Mod(\sox)$ is
1661
injective then $\sI\tensor\sE$ is injective.
1662
\end{lem}
1663
\begin{proof}
1664
Suppose $0\into\sF\into\sG$ is an injection and there is a map
1665
$\varphi:\sF\into\sI\tensor\sE$. Tensor everything with $\sE^{\dual}$.
1666
Then we have an injection $0\into\sF\tensor\sE^{\dual}\into\sG\tensor\sE^{\dual}$
1667
and a map $\varphi':\sF\tensor\sE^{\dual}\into\sI$. Since $\sI$ is injective
1668
there is a map $\sG\tensor\sE^{\dual}\into\sI$ which makes
1669
the appropriate diagram commute. Tensoring everything with $\sE$ gives
1670
a map making the original diagram commute.
1671
\end{proof}
1672
1673
\begin{proof}[of proposition] Let $0\into\sG\into\sI^{\bullet}$
1674
by an injective resolution of $\sG$. Since
1675
$$\Hom(\sF\tensor\sE,\sI^{\bullet})=\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual}),$$
1676
we see that
1677
$$0\into\sG\tensor\sE^{\dual}\into\sI^{\cdot}\tensor\sE{\dual}$$
1678
is an injective resolution of $\sG\tensor\sE^{\dual}$.
1679
Thus $\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual})$ computes
1680
$\ext(\sF\tensor\sE,\bullet)$.
1681
\end{proof}
1682
1683
\begin{prop}
1684
If $\sF$ has a locally free resolution
1685
$\sE_{\cdot}\into\sF\into 0$ then
1686
$$\sext^i_{\sox}(\sF,\sG)=H^i(\sHom(\sE_{\bullet},\sG)).$$
1687
\end{prop}
1688
1689
\begin{remark}
1690
Notice that when $i>0$ and $\sE$ is locally free,
1691
$$\sext^i(\sE,\sG)=\sext^i(\sox,\sG\tensor\sE^{\dual})=0.$$
1692
\end{remark}
1693
1694
\begin{proof}
1695
Regard both sides as functors in $\sG$. The left hand side is a
1696
$\delta$-functor and vanishes for $\sG$ injective. I claim that
1697
that right hand side is also a $\delta$-functor and vanishes for
1698
$\sG$-injective.
1699
\begin{lem}
1700
If $\sE$ is locally free, then $\shom(\sE,\bullet)$ is exact.
1701
\end{lem}
1702
\end{proof}
1703
1704
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1705
%% 2/26/96
1706
\section{More Technical Results on $\ext$}
1707
Let $(X,\sox)$ be a scheme and $\sF,\sG$ be sheaves
1708
in the category $\Mod(\sox)$. Then $\ext(\sF,\sG)$
1709
and $\sext(\sF,\sG)$ are the derived functors of $\Hom$, resp.
1710
$\shom$, in the second variable.
1711
1712
\begin{lem}
1713
If $\sF$ and $\sG$ are coherent over a Noetherian scheme
1714
$X$, then $\sext^{i}(\sF,\sG)$ is coherent.
1715
\end{lem}
1716
This lemma would follow immediately from the following fact
1717
which we haven't proved yet.
1718
\begin{fact}
1719
Let $X=\spec A$ with $A$ Noetherian and let $M$ be an $A$-module.
1720
Then $$\ext_{\sox}^i(\tilde{M},\tilde{N})=\ext^i_{A}(M,N)$$ and
1721
$$\sext_{\sox}^i(\tilde{M},\tilde{N})=(\ext^i_{A}(M,N))^{\tilde{}}.$$
1722
\end{fact}
1723
Instead of using the fact we can prove the lemma using
1724
a proposition from yesterday.
1725
\begin{proof}
1726
Choose a locally free resolution
1727
$$\sL_{\bullet}\into\sF\into 0$$
1728
of $\sF$. Then
1729
$$\sext^{i}(\sF,\sG)=H^i(\shom(\sL_{\bullet},\sG)).$$
1730
But all of the kernels and cokernels in
1731
$\shom(\sL_{\bullet},\sG)$
1732
are coherent, so the cohomology is. (We can't
1733
just choose an injective resolution of $\sF$ and apply
1734
the definitions because there is no guarantee that we
1735
can find an injective resolution by coherent sheaves.)
1736
\end{proof}
1737
1738
\begin{prop}
1739
Let $X$ be a Noetherian projective scheme over $k$ and let $\sF$
1740
and $\sG$ be coherent on $X$. Then for each $i$ there exists
1741
an $n_0$, depending on $i$, such that for all $n\geq{}n_0$,
1742
$$\ext^i_{\sox}(\sF,\sG(n))=\Gamma(\sext^i_{\sox}(\sF,\sG(n))).$$
1743
\end{prop}
1744
\begin{proof}
1745
When $i=0$ the assertion is that
1746
$$\Hom(\sF,\sG(n))=\Gamma(\shom(\sF,\sG(n)))$$
1747
which is obvious.
1748
1749
{\em Claim.} Both sides are $\delta$-functors in $\sF$. We
1750
have already showed this for the left hand side. [I don't understand
1751
why the right hand side is, but it is not trivial and it caused
1752
much consternation with the audience.]
1753
1754
To show the functors are isomorphic we just need to show both
1755
sides are coeffaceable. That is, for every coherent sheaf $\sF$
1756
there is a coherent sheaf $\sE$ and a surjection $\sE\into\sF\into{}0$
1757
such that $\ext^i_{\sox}(\sE)=0$ and similarly for the right hand
1758
side. Thus every coherent sheaf is a quotient of an acyclic sheaf.
1759
1760
Suppose $\sF$ is coherent. Then for $\ell\gg 0$, $\sF(\ell)$ is generated
1761
by its global sections, so there is a surjection
1762
$$\sox^a\into\sF(\ell)\into{}0.$$
1763
Untwisting gives a surjection
1764
$$\sox(-\ell)^a\into\sF\into{}0.$$
1765
Let $\sE=\sox(-\ell)^a$, then I claim that $\sE$ is acyclic for both
1766
sides. First consider the left hand side. Then
1767
\begin{align*}\ext^i(\oplus\so(-\ell),\sG(n))&=\oplus\ext^i(\sox(-\ell),\sG(n))\\
1768
&=\oplus\ext^i(\sox,\sG(\ell+n))\\
1769
&=H^i(X,\sG(\ell+n))\end{align*}
1770
By Serre (theorem 5.2 of the book) this is zero for $n$ sufficiently large.
1771
For the right hand side the statement is just that
1772
$$\sext^i(\sE,\sG(n))=0$$
1773
which we have already done since $\sE$ is a locally free sheaf.
1774
1775
Thus both functors are universal since they are coeffaceable. Since universal
1776
functors are completely determined by their zeroth one they must be equal.
1777
\end{proof}
1778
1779
\begin{example}
1780
One might ask if $\ext^i$ necessarily vanishes for sufficiently large $i$.
1781
The answer is no. Here is an algebraic example which can be converted
1782
to a geometric example. Let $A=k[\varepsilon]/(\varepsilon^2)$, then
1783
a projective resolution $L_{\bullet}$ of $k$ is
1784
$$\cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A
1785
\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}k\into 0.$$
1786
Then $Hom(L_{\bullet},k)$ is the complex
1787
$$k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}\cdots$$
1788
Thus $\ext^i_A(k,k)=k$ for all $i\geq 0$.
1789
\end{example}
1790
1791
\section{Serre Duality}
1792
We are now done with technical results on $\ext$'s so we can
1793
get back to Serre duality on $\P^n$.
1794
Let $X=\P^n_k$ and let $\omega=\sox(-n-1)$. Note that this
1795
is an {\em ad hoc} definition of $\omega$ which just happens
1796
to work since $X=\P^n_k$. In the more general situation it
1797
will be an interesting problem just to show the so called
1798
dualizing sheaf $\omega$ actually exists. When our variety
1799
is nonsingular, $\omega$ will be the canonical sheaf.
1800
We have shown that for any coherent sheaf $\sF$ there is a map
1801
$$\Hom(\sF,\omega)\iso H^n(\sF)^{\dual}.$$
1802
The map is constructed by using the fact that $H^n$ is a
1803
functor:
1804
$$\Hom(\sF,\omega)\into\Hom_k(H^n(\sF),H^n(\omega))
1805
= \Hom_k(H^n(\sF),k)=H^n(\sF)^{\dual}.$$
1806
1807
We shall use satellite functors to prove the following theorem.
1808
\begin{thm} Let $\sF$ be a coherent sheaf on $\P^n_k$. Then
1809
there is an isomorphism $$\ext^i(\sF,\omega)\iso H^{n-i}(\sF)^{\dual}.$$
1810
\end{thm}
1811
\begin{proof} Regard both sides as functors in $\sF$.
1812
\par {\em 1. Both sides are $\delta$-functors in $\sF$.}
1813
We have already checked this for $\ext^i$. Since $H^{n-i}$
1814
is a delta functor in $\sF$, so is $(H^{n-i})^{\dual}$.
1815
Note that both sides are contravarient.
1816
\par {\em 2. They agree for $i=0$.} This was proved last time.
1817
\par {\em 3. Now we just need to show both sides are coeffaceable.}
1818
Suppose $\sE\into\sF\into 0$ with $\sE=\sO(-\ell)^{\oplus a}$.
1819
For some reason we can assume $\ell\gg{}0$.
1820
We just need to show both sides vanish
1821
on this $\sE$. First computing the left hand side gives
1822
$$\oplus\ext^i(\sO(-\ell),\omega)=H^i(\omega(\ell))=0$$
1823
for $\ell\gg 0$. Next computing the right hand side we get
1824
$$H^{n-i}(\sO(-\ell))=0$$
1825
by the explicit computations of cohomology of projective
1826
space (in particular, note that $i>0$).
1827
\end{proof}
1828
1829
Next time we will generalize Serre duality to an arbitrary
1830
projective scheme $X$ of dimension $n$.
1831
We will proceed in two steps. The first is to ask,
1832
what is $\omega_X$? Although the answer to this question
1833
is easy on $\P_k^n$ it is not obvious what the suitable
1834
analogy should be for an arbitrary projective variety.
1835
Second we will define natural maps
1836
$$\ext^i_{\sox}(\sF,\omega)\xrightarrow{\varphi^i}H^{n-i}(\sF)^{\dual}$$
1837
where $n=\dim X$.
1838
Unlike in the case when $X=\P^n_k$, these maps are not necessarily
1839
isomorphisms unless $X$ is locally Cohen-Macaulay (the local rings
1840
at each point are Cohen-Macaulay).
1841
\begin{defn}
1842
Let $A$ be a nonzero Noetherian local ring with residue field $k$.
1843
Then the {\bfseries depth} of $A$ is
1844
$$\operatorname{depth} A = \inf\{i:\ext^i_A(k,A)\neq 0\}.$$
1845
$A$ is said to be {\bfseries Cohen-Macaulay} if
1846
$\operatorname{depth} A = \dim A$.
1847
\end{defn}
1848
1849
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1850
%% 2/28/96
1851
1852
\section{Serre Duality for Arbitrary Projective Schemes}
1853
Today we will talk about Serre duality for an arbitrary projective
1854
scheme. We have already talked about Serre duality in the special
1855
case $X=\P^n_k$. Let $\sF$ be a locally free sheaf. We showed
1856
there is an isomorphism
1857
$$\ext^i(\sF,\omega_X)\iso H^{n-i}(\sF)^{\dual}.$$
1858
This was established by noting that
1859
$$\ext^i(\sF,\omega_X)=\ext^i(\sox,\sF^{\dual}\tensor\omega_X)
1860
=H^i(\sF^{\dual}\tensor\omega_X).$$
1861
Another thing to keep in mind is that locally free sheaves correspond
1862
to what, in other branches of mathematics, are vector bundles. They
1863
aren't the same object, but there is a correspondence.
1864
1865
We would like to generalize this to an arbitrary projective scheme $X$.
1866
There are two things we must do.
1867
\begin{enumerate}
1868
\item Figure out what $\omega_X$ is.
1869
\item Prove a suitable duality theorem.
1870
\end{enumerate}
1871
When $X=\P_k^n$ it is easy to find a suitable $\omega_X=\so_{\P^n_k}(-n-1)$
1872
because of the explicit computations we did before. We now define $\omega_X$
1873
to be a sheaf which will do what we hope it will do. Of course existence
1874
is another matter.
1875
\begin{defn}
1876
Let $X$ be a Noetherian scheme of finite type over a field $k$ and
1877
let $n=\dim X$. Then a {\bfseries dualizing sheaf} for $X$ is
1878
a coherent sheaf $\omega_X$ along with a map
1879
$t:H^n(\omega_X)\into{}k$, such that for all coherent sheaves $\sF$ on
1880
$X$, the map
1881
$\Hom_{\sox}(\sF,\ox)\into{}H^n(\sF)^{\dual}$ is an
1882
isomorphism. The latter map is defined by the diagram
1883
$$\begin{array}{ccc}
1884
\Hom_{\sox}(\sF,\ox)\\
1885
\downarrow\\
1886
\Hom(H^n(\sF),H^n(\ox))&\xrightarrow{t}&\Hom(H^n(\sF),k)=H^n(\sF)^{\dual}
1887
\end{array}$$
1888
\end{defn}
1889
Strictly speaking, a dualizing sheaf is a pair $(\omega_X,t)$. Note
1890
that on $\P^n$ we had $H^n(\omega_{\P^n})\isom{}k$, but on an
1891
arbitrary scheme $X$ we only have a map from $H^n(\omega_X)$ to $k$
1892
which need not be an isomorphism. The definition never mentions existence.
1893
1894
\begin{prop}
1895
If $X$ admits a dualizing sheaf $(\ox,t)$ then the pair
1896
$(\ox,t)$ is unique up to unique isomorphism, i.e., if
1897
$(\eta,s)$ is another dualizing sheaf for $X$ then there
1898
is a unique isomorphism $\varphi:\ox\into\eta$ such that
1899
$$\begin{array}{ccc}
1900
H^n(\ox)&\xrightarrow{H^n(\varphi)}&H^n(\eta)\\
1901
t\downarrow&&\downarrow{}s\\
1902
k&=&k\end{array}$$
1903
commutes.
1904
\end{prop}
1905
Before we prove the proposition we make a short digression to
1906
introduce representable functors which give a proof of the uniqueness
1907
part of the above proposition.
1908
\begin{defn}
1909
Let $\sC$ be a category and $\sD$ a category whose objects happen
1910
to be sets. Suppose $T:\sC\into\sD$ is a contravarient functor.
1911
Then $T$ is {\bfseries representable} if there exists an object
1912
$\omega\in\Ob(\sC)$ and an element $t\in T(\omega)$ such that
1913
for all $F\in\Ob(\sC)$ the map $\Hom_{\sC}(F,\omega)\into{}T(F)$
1914
is a bijection of sets. The latter map is defined by the diagram
1915
$$\begin{array}{ccc}
1916
\Hom_{\sC}(F,\omega)&\xrightarrow{\text{bijection of sets}}&T(F)\\
1917
\searrow&&\nearrow\text{ evaluation at $t$}\\
1918
&\Hom_{\sD}(\Gamma(\omega),T(F))\end{array}$$
1919
\end{defn}
1920
Thus there is an isomorphism of functors $\Hom(\bullet,\omega)=T(\bullet)$.
1921
The pair $(t,\omega)$ is said to represent the functor $T$.
1922
The relevant application of this definition is to the case
1923
when $\sC=\Coh(X)$, $\sD=\{\text{ $k$-vector spaces}\}$,
1924
$T$ is the functor $F\mapsto H^n(\sF)^{\dual}$. Then $\omega=\ox$
1925
and
1926
$$t=t\in\Hom(H^n(\omega),k)=H^n(\omega)^{\dual}=T(\omega).$$
1927
\begin{prop}
1928
If $T$ is a representable functor, then the pair
1929
$(\omega,t)$ representing it is unique.
1930
\end{prop}
1931
\begin{proof}
1932
Suppose $(\omega,t)$ and $(\eta,s)$ both represent the functor $T$.
1933
Consider the diagram
1934
$$\begin{array}{cccl}
1935
\Hom(\eta,\omega)&\xrightarrow{T}&\Hom(T(\omega),T(\eta))\\
1936
\searrow&&\swarrow\text{ eval. at $t$}\\
1937
&T(\eta)\end{array}$$
1938
By definition the map $\Hom(\eta,\omega)\into{}T(\eta)$ is
1939
bijective. Since $s\in T(\eta)$, there is $\varphi\in\Hom(\eta,\omega)$
1940
such that $\varphi\mapsto{}s\in{}T(\eta)$. Thus $\varphi$ has
1941
the property that $T(\varphi)(t)=s$. This argument uses the
1942
fact that $(\omega,t)$ represents $T$. Using the fact that
1943
$(\eta,s)$ represents $T$ implies that there exists
1944
$\psi\in\Hom(\omega,\eta)$ such that $T(\psi)(s)=t$.
1945
We have the following pictures
1946
$$\begin{array}{ccc}
1947
&\xrightarrow{\hspace{.2in}\psi\hspace{.2in}}\\
1948
\omega&&\eta\\
1949
&\xleftarrow{\hspace{.2in}\phi\hspace{.2in}}
1950
\end{array}$$
1951
$$\begin{array}{ccc}
1952
&\xleftarrow{\hspace{.2in}T(\psi)=\psi^{*}\hspace{.2in}}\\
1953
t\in{} T(\omega)&&T(\eta) \ni{} s\\
1954
&\xrightarrow{\hspace{.2in}T(\phi)=\phi^{*}\hspace{.2in}}
1955
\end{array}$$
1956
I claim that
1957
$$\psi\circ\varphi=\Id\in\Hom(\eta,\eta).$$
1958
In diagram form we have
1959
$$\eta\xrightarrow{\varphi}\omega\xrightarrow{\psi}\eta$$
1960
which upon applying $T$ gives
1961
\begin{align*}
1962
T(\eta)&\xrightarrow{\psi^{*}}T(\omega)\xrightarrow{\varphi^{*}}T(\eta)\\
1963
s&\mapsto t\mapsto s\end{align*}
1964
Where does $\psi\circ\varphi$ go to
1965
under the map $\Hom(\eta,\eta)\iso T(\eta)$? By definition
1966
$\psi\circ\varphi$ goes to the evaluation of $T(\psi\circ\varphi)$
1967
at $s\in T(\eta)$. But, as indicated above, the evaluation of
1968
$T(\psi\circ\varphi)$ at $s$ is just $s$ again. But the identity
1969
morphism $1_{\eta}\in\Hom(\eta,\eta)$ also maps to $s$
1970
under the map $\Hom(\eta,\eta)\iso T(\eta)$. Since this
1971
map is a bijection this implies that $\psi\circ\varphi=1_{\eta}$,
1972
as desired. Similarly $\varphi\circ\psi=1_{\omega}$. Thus
1973
$\psi$ and $\varphi$ are both isomorphisms.
1974
\end{proof}
1975
\begin{quote}
1976
``When you define something and it is unique up to unique isomorphism,
1977
you know it must be good.''
1978
\end{quote}
1979
We return to the question of existence.
1980
\begin{prop}
1981
If $X$ is a projective scheme over a field $k$ then $(\ox,t)$ exists.
1982
\end{prop}
1983
\begin{lem}
1984
If $X$ is an $n$ dimensional projective scheme over a field $k$, then
1985
there is a finite morphism $f:X\into\P^n_k$.
1986
\end{lem}
1987
\begin{proof}
1988
Embed $X$ in $\P^N$ then choose a linear projection down to
1989
$\P^n$ which is sufficiently general.
1990
$$\begin{array}{ccc}
1991
X&\hookrightarrow&\P^N\\
1992
f\searrow&&\downarrow\\
1993
&&\P^n\end{array}$$
1994
Let $L$ be a linear space of dimension $N-n-1$ not meeting $X$.
1995
Let the map from $\P^N\into\P^n$ be projection through $L$.
1996
By construction $f$ is quasi-finite, i.e., for all $Q\in\P^n$,
1997
$f^{-1}(Q)$ is finite. It is a standard QUALIFYING EXAM problem to
1998
show that if a morphism is quasi-finite and projective then it is
1999
finite. This can be done by applying (II, Ex. 4.6) by covering
2000
$X$ by subtracting off hyperplanes and noting that the correct
2001
things are affine by construction. See also (III, Ex. 11.2) for
2002
the more general case when $f$ is quasi-finite and proper, but not
2003
necessarily projective.
2004
\end{proof}
2005
2006
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
2007
%% 3/1/96
2008
%%%%%%%%%%%%%%%%%%%%%%%%
2009
\section{Existence of the Dualizing Sheaf on a Projective Scheme}
2010
Let $X$ be a scheme over $k$. Recall that a
2011
dualizing sheaf is a pair $(\omega,t)$ where $\omega$ is a coherent
2012
sheaf on $X$ and
2013
$$t:H^n(X,\omega)\into k$$
2014
is a homomorphism such that for all coherent sheaves $\sF$
2015
the natural map
2016
$$\Hom_X(\sF,\omega)\into{}H^n(\sF)^{\dual}$$
2017
is an isomorphism.
2018
We know that such a dualizing sheaf exists on $\P_k^n$.
2019
\begin{thm}
2020
If $X$ is a projective scheme of dimension $n$ over $k$, then
2021
$X$ has a dualizing sheaf.
2022
\end{thm}
2023
The book's proof takes an embedding
2024
$j:X\hookrightarrow{}\P_k^N$ and works on $X$
2025
as a subscheme of $\P_k^N$. Then the book's proof shows that
2026
$$\omega_X=\sext_{\sO_{\P^N_k}}^{N-n}(\sox,\omega_{\P^N_k}).$$
2027
Today we will use a different method.
2028
\begin{defn}
2029
A {\bfseries finite morphism} is a morphism $f:X\into Y$ of
2030
Noetherian schemes such that for any open affine $U=\spec A\subset X$,
2031
the preimage $f^{-1}(U)\subset Y$ is affine, say $f^{-1}(U)=\spec B$,
2032
and the natural map $A\into B$ turns $B$ into a finitely
2033
generated $A$-module. We call $f$ an {\bfseries affine morphism}
2034
if we just require that $f^{-1}(U)$ is affine
2035
but not that $B$ is a finitely generated $A$-module.
2036
A morphism $f:X\into{}Y$ is {\bfseries quasi-finite} if for all
2037
$y\in Y$ the set $f^{-1}(y)$ is finite.
2038
\end{defn}
2039
\begin{example}
2040
Consider the morphism
2041
$$j:\P^1-\{\pt\}\injects\P^1.$$
2042
Since $\P^1$ minus any nonempty finite set of points is
2043
affine $j$ is affine. But it is not finite. Indeed, let
2044
$a$ be a point different from $\pt$ and let $U=\P^1-\{a\}$.
2045
Then $U=\spec k[x]$ and
2046
$$j^{-1}(U)=\P^1-\{\pt,a\}=\spec k[x,x^{-1}],$$
2047
but $k[x,x^{-1}]$ is not a finitely generated $k[x]$-module.
2048
\end{example}
2049
\begin{exercise}
2050
A morphism can be affine but not finite or even quasi-finite.
2051
For example, let $f$ be the natural map
2052
$$f:\A^{n+1}-\{0\}\into\P^n$$
2053
then show that $f$ is affine.
2054
This is the fiber bundle associated to the invertible
2055
sheaf $\sO(1)$ [[or is it $\sO(-1)$?]]
2056
\end{exercise}
2057
2058
\subsection{Relative Gamma and Twiddle}
2059
We will now define relative versions of global sections and
2060
$\tilde{}$ analogous to the absolute versions.
2061
It is not a generalization of the absolute notion, but a
2062
relativization.
2063
Suppose $X$ is a scheme over $Y$ with structure
2064
map $f:X\into{}Y$ and assume $f$ is affine.
2065
Then the map sending a sheaf
2066
$\sF$ on $X$ to the sheaf $f_{*}\sF$ on
2067
$Y$ is the analog of taking global sections.
2068
Since $f$ is a morphism there is a map $\soy\into{}f_{*}\sox$
2069
so $f_{*}\sox$ is a sheaf of $\soy$-modules. Note that
2070
$f_{*}\sF$ is a sheaf of $f_{*}\sox$-modules. Thus
2071
we have set up a map
2072
$$\Qco(X)\into\{\text{ quasicoherent $f_{*}\sox$-modules on $Y$ }\}.$$
2073
The next natural thing to do is define a map analogous to
2074
$\tilde{}$ which goes the other direction.
2075
Suppose $\sG$ is a quasi coherent sheaf of $f_{*}\sox$-modules on $Y$.
2076
Let $U\subset Y$ be an affine open subset of $Y$. Let $G=\Gamma(U,\sG)$ and
2077
write $U=\spec A$. Then since $f$ is an affine morphism,
2078
$f^{-1}(U)=\spec B$ where $B=\Gamma(f^{-1}(U),\sox)$.
2079
Since $\sG$ is an $f_{*}\sox$-module, and $f_{*}\sox$ over $U$
2080
is just $B$ thought of as an $A$-module, we see that $G$ is a $B$-module.
2081
Thus we can form the sheaf $\tilde{G}$ on $\spec B=f^{-1}(U)$.
2082
Patching the various sheaves $\tilde{G}$ together as $U$ runs through
2083
an affine open cover of $Y$ gives a sheaf $\tilde{\sG}$ in
2084
$\Qco(X)$.
2085
2086
Let $\sG$ be a quasi-coherent sheaf of $\soy$-modules. We can't
2087
take $\tilde{}$ of $\sG$ because $\sG$ might not be a sheaf of
2088
$f_{*}\sox$-modules. But
2089
$\shom_{\soy}(f_{*}\sox,\sG)$
2090
is a sheaf of $f_{*}\sox$-modules, so we can
2091
form $(\shom_{\soy}(f_{*}\sox,\sG))^{\tilde{}}$. This
2092
is a quasi-coherent sheaf on $X$ which we denote $f^{!}(\sG)$.
2093
\begin{prop}
2094
Suppose $f:X\into{}Y$ is an affine morphism of Noetherian
2095
schemes, $\sF$ is coherent on $X$, and $\sG$ is quasi-coherent on $Y$.
2096
Then
2097
$$f_{*}\shom_{\sox}(\sF,f^{!}\sG)\isom \shom_{\soy}(f_{*}\sF,\sG)$$
2098
and passing to global sections gives an isomorphism
2099
$$\Hom(\sF,f^{!}\sG)\isom \Hom(f_{*}\sF,\sG).$$
2100
Thus $f^{!}$ is a right adjoint for $f_{*}$.
2101
\end{prop}
2102
\begin{proof}
2103
The {\em natural} map is
2104
\begin{align*}f_{*}\shom_{\sox}(\sF,f^{!}\sG)&\into
2105
\shom_{\soy}(f_{*}\sF,f_{*}f^{!}\sG)\\
2106
&=\shom_{\soy}(f_{*}\sF,\shom_{\soy}(f_*\sox,\sG)) \into
2107
\shom_{\soy}(f_{*}\sF,\sG)\end{align*}
2108
where the map $\shom_{\soy}(f_{*}\sox,\sG)\into\sG$
2109
is obtained obtained by evaluation at $1$.
2110
Since the question is local we may assume
2111
$Y=\spec A$ and $X=\spec B$. Then $\sF$ corresponds
2112
to a finitely generated module $M$ over the Noetherian ring $B$
2113
and $\sG$ corresponds to a module $N$ over $A$.
2114
We must show that
2115
$$\Hom_B(M,\Hom_A(B,N))\isom\Hom_A(M,N).$$
2116
When $M$ is free over $B$ so that $M=B^{\oplus{}n}$ the equality holds.
2117
As functors in $M$, both sides are contravarient and
2118
left exact. Now suppose $M$ is an arbitrary finitely generated
2119
$B$-module. Write $M$ as a
2120
quotient $F_0/F_1$ where $F_0$ and $F_1$ are both free of