%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1%% Notes for Hartshorne's Algebraic Geometry course2%% William Stein3%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%4%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%6%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%7%%%%%%%%%%%%%%%%%%%%8%%%%910\documentclass[12pt]{article}11\textwidth=1.2\textwidth12\textheight=1.3\textheight13\hoffset=-.7in14\voffset=-1.28in15\usepackage{amsmath}16\usepackage{amsthm}17\usepackage{amsopn}18\usepackage{amscd}1920\font\bbb=msbm10 scaled \magstep 121\font\german=eufm10 scaled \magstep 122\font\script=rsfs10 scaled \magstep 12324\newcommand{\nd}{\not\!\!|}25\newcommand{\m}{\mathbf{m}}26\newcommand{\cross}{\times}27\newcommand{\diff}{\Omega_{B/A}}28\newcommand{\injects}{\hookrightarrow}29\newcommand{\pt}{\mbox{\rm pt}}30\newcommand{\dual}{\vee}31\newcommand{\bF}{\mathbf{F}}32\newcommand{\bZ}{\mathbf{Z}}33\newcommand{\bR}{\mathbf{R}}34\newcommand{\bQ}{\mathbf{Q}}35\newcommand{\bC}{\mathbf{C}}36\newcommand{\bA}{\mathbf{A}}37\newcommand{\bP}{\mathbf{P}}38\newcommand{\sP}{\mathcal{P}}39\newcommand{\M}{\mathcal{M}}40\newcommand{\sO}{\mathcal{O}}41\newcommand{\sox}{\sO_X}42\newcommand{\soc}{\sO_C}43\newcommand{\soy}{\sO_Y}44\newcommand{\ox}{\omega_X}45\newcommand{\so}{\sO}46%\newcommand{\sF}{\mbox{\script F}}47\newcommand{\sQ}{\mathcal{Q}}48\newcommand{\sR}{\mathcal{R}}49\newcommand{\sE}{\mathcal{E}}50\newcommand{\E}{\mathcal{E}}51\newcommand{\sF}{\mathcal{F}}52%\newcommand{\sL}{\mbox{\script L}}53\newcommand{\sL}{\mathcal{L}}54\newcommand{\sM}{\mathcal{M}}55\newcommand{\sN}{\mathcal{N}}56\newcommand{\sA}{\mathcal{A}}57\newcommand{\sD}{\mathcal{D}}58\newcommand{\sC}{\mathcal{C}}59\newcommand{\sG}{\mathcal{G}}60\newcommand{\sB}{\mathcal{B}}61\newcommand{\sK}{\mathcal{K}}62\newcommand{\sH}{\mathcal{H}}63\newcommand{\sI}{\mathcal{I}}64\newcommand{\sU}{\mbox{\german U}}65\newcommand{\gm}{\mbox{\german m}}66\newcommand{\isom}{\cong}67\newcommand{\tensor}{\otimes}68\newcommand{\into}{\rightarrow}69\newcommand{\soq}{\sO_{Q}}70\newcommand{\Cone}{Q_{\text{cone}}}71\newcommand{\cech}{\v{C}ech}72\newcommand{\cH}{\text{\v{H}}}73\newcommand{\intersect}{\cap}74\newcommand{\union}{\cup}75\newcommand{\iso}{\xrightarrow{\sim}}76\newcommand{\qone}{Q_{\text{one}}}77\newcommand{\qns}{Q_{\text{ns}}}7879\newcommand{\F}{\sF}80\renewcommand{\P}{\bP}81\newcommand{\A}{\bA}82\newcommand{\C}{\bC}83\newcommand{\Q}{\bQ}84\newcommand{\R}{\bR}85\newcommand{\Z}{\bZ}868788\DeclareMathOperator{\cd}{cd}89\DeclareMathOperator{\Ob}{Ob}90\DeclareMathOperator{\Char}{char}91\DeclareMathOperator{\aut}{Aut}92\DeclareMathOperator{\End}{End}93\DeclareMathOperator{\gl}{GL}94\DeclareMathOperator{\slm}{SL}95\DeclareMathOperator{\supp}{supp}96\DeclareMathOperator{\spec}{Spec}97\DeclareMathOperator{\Spec}{Spec}98\DeclareMathOperator{\ext}{Ext}99\DeclareMathOperator{\Ext}{Ext}100\DeclareMathOperator{\tor}{Tor}101\DeclareMathOperator{\Hom}{Hom}102\DeclareMathOperator{\Aut}{Aut}103\DeclareMathOperator{\PGL}{PGL}104\DeclareMathOperator{\shom}{\mathcal{H}om}105\DeclareMathOperator{\sHom}{\mathcal{H}om}106\DeclareMathOperator{\sext}{\mathcal{E}xt}107\DeclareMathOperator{\proj}{Proj}108\DeclareMathOperator{\Pic}{Pic}109\DeclareMathOperator{\pic}{Pic}110\DeclareMathOperator{\pico}{Pic^0}111\DeclareMathOperator{\gal}{Gal}112\DeclareMathOperator{\imag}{Im}113\DeclareMathOperator{\Id}{Id}114\DeclareMathOperator{\Ab}{\mathbf{Ab}}115\DeclareMathOperator{\Mod}{\mathbf{Mod}}116\DeclareMathOperator{\Coh}{\mathbf{Coh}}117\DeclareMathOperator{\Qco}{\mathbf{Qco}}118\DeclareMathOperator{\hd}{hd}119\DeclareMathOperator{\depth}{depth}120\DeclareMathOperator{\trdeg}{trdeg}121\DeclareMathOperator{\rank}{rank}122\DeclareMathOperator{\Tr}{Tr}123\DeclareMathOperator{\length}{length}124\DeclareMathOperator{\Hilb}{Hilb}125\DeclareMathOperator{\Sch}{\mbox{\bfseries Sch}}126\DeclareMathOperator{\Set}{\mbox{\bfseries Set}}127\DeclareMathOperator{\Grp}{\mbox{\bfseries Grp}}128\DeclareMathOperator{\id}{id}129130\theoremstyle{plain}131\newtheorem{thm}{Theorem}[section]132\newtheorem{prop}[thm]{Proposition}133\newtheorem{claim}[thm]{Claim}134\newtheorem{cor}[thm]{Corollary}135\newtheorem{fact}[thm]{Fact}136\newtheorem{lem}[thm]{Lemma}137\newtheorem{ques}[thm]{Question}138\newtheorem{conj}[thm]{Conjecture}139140\theoremstyle{definition}141\newtheorem{defn}[thm]{Definition}142143\theoremstyle{remark}144\newtheorem{remark}[thm]{Remark}145\newtheorem{exercise}[thm]{Exercise}146\newtheorem{example}[thm]{Example}147148149\author{William A. Stein}150\title{Notes for Algebraic Geometry II}151152\begin{document}153\maketitle154\tableofcontents155156\section{Preface}157{\bfseries Read at your own risk!}158159These are my {\em very rough, error prone} notes of160a second course on algebraic geometry161offered at U.C. Berkeley in the Spring of 1996.162The instructor163was Robin Hartshorne and the students were Wayne Whitney,164William Stein, Matt Baker, Janos Csirik, Nghi Nguyen, and Amod.165I wish to thank Robin Hartshorne for giving this course166and to Nghi Nguyen for his insightful suggestions and corrections.167Of course all of the errors are solely my responsibility.168169The remarks in brackets [[like this]] are notes that170I wrote to myself. They are meant as a warning or as a reminder171of something I should have checked but did not have time for. You172may wish to view them as exercises.173174If you have suggestions, questions, or comments feel free to write to me.175My email address is {\tt was@math.berkeley.edu}.176177% Day 1, 1/17/96178\section{Ample Invertible Sheaves}179180Let $k$ be an algebraically closed field181and let $X$ be a scheme over $k$. Let $\phi:X\into \bP^n_k$ be a morphism.182Then to give $\phi$ is equivalent to giving an invertible sheaf183$\sL$ on $X$ and sections $s_0,\ldots,s_n\in\Gamma(X,\sL)$184which generate $\sL$. If $X$ is projective (that is, if there185is some immersion of $X$ into {\em some} $\bP^m_k$) then $\phi$186is a closed immersion iff $s_0,\ldots,s_n$ separate points and tangent187vectors.188189\begin{defn}190Let $X$ be a scheme and $\sL$ an invertible sheaf on $X$.191Then we say $\sL$ is {\em very ample} if there is an immersion192$i:X\hookrightarrow \bP_k^n$ such that $\sL\isom i^{*}\sO(1)$.193\end{defn}194195\begin{thm}196Let $X$ be a closed subscheme of197$\bP_k^n$198and199$\sF$ a coherent sheaf on $X$, then200$\sF(n)$ is generated by global sections for201all $n\gg 0$.202\end{thm}203204\begin{cor}205Let $X$ be any scheme and $\sL$ a very ample coherent sheaf206on $X$, then for all $n\gg 0$, $\sF\tensor \sL^{\tensor n}$207is generated by global sections.208\end{cor}209210\begin{defn}211Let $X$ be a Noetherian scheme and $\sL$ be an invertible sheaf.212We say that $\sL$ is {\em ample} if for every coherent sheaf213$\sF$ on $X$, there is $n_0$ such that for all $n\geq n_0$,214$\sF \tensor \sL^{\tensor n}$ is generated by its global sections.215\end{defn}216217Thus the previous corollary says that a very ample invertible218sheaf is ample.219220\begin{prop}221Let $X$ and $\sL$ be as above. Then the following are equivalent.2222231) $\sL$ is ample,2242252) $\sL^n$ is ample for all $n>0$,2262273) $\sL^n$ is ample for some $n>0$.228\end{prop}229230\begin{thm}231Let $X$ be of finite type over a Noetherian ring $A$ and suppose232$\sL$ is an invertible sheaf on $A$. Then $\sL$ is ample iff233there exists $n$ such that $\sL^n$ is very ample over $\spec A$.234\end{thm}235236\begin{example}237Let $X=\bP^1$, $\sL=\sO(\ell)$, some $\ell \in \bZ$.238If $\ell<0$ then $\Gamma(\sL)=0$. If $\ell=0$ then239$\sL=\sO_X$ which is not ample since $\sO_X(-1)^n\tensor240\sO_X\isom \sO_X(-1)^n$ is not generated by global sections241for any $n$. Note that $\sO_X$ itself is generated by242global sections. Finally, if $ell>0$ then243$\sL=\sO_X(\ell)$ is very ample hence ample.244\end{example}245246\begin{example}247Let $C\subseteq \bP^2$ be a nonsingular cubic curve and248$\sL$ an invertible sheaf on $C$ defined by $\sL=\sL(D)$,249where $D=\sum n_i P_i$ is a divisor on $C$. If $\deg D<0$250then $\sL$ has no global sections so it can't be ample.251%% Undone.252\end{example}253254% Day 2, 1/19/96255\section{Introduction to Cohomology}256We first ask, what is cohomology and where does it arise in nature?257Cohomology occurs in commutative258algebra, for example in the $\ext$ and $\tor$ functors, it occurs in259group theory, topology, differential geometry, and of course in260algebraic geometry. There are several flavors of cohomology which are261studied by algebraic geometers. Serre's coherent sheaf262cohomology has the advantage of being easy to define, but263has the property that the cohomology groups are vector spaces.264Grothendieck introduced \`{e}tale cohomology and $\ell$-adic265cohomology. See, for example, Milne's {\em \`{E}tale Cohomology}266and SGA 4$\frac{1}{2}$, 5 and 6. This cohomology theory arose267from the study of the Weil Conjectures (1949) which deal with268a deep relationship between the number of points on a variety269over a finite field and the geometry of the complex analytic variety270cut out by the same equations in complex projective space. Deligne271was finally able to resolve these conjectures in the affirmative272in 1974.273274What is cohomology good for? Cohomology allows one to get numerical275invariants of an algebraic variety. For example, if $X$ is a projective276scheme defined over an algebraically closed field $k$ then277$H^i(X,\sF)$ is a finite dimensional $k$-vector space. Thus the278$h_i=\dim_k H^i(X,\sF)$ are a set of numbers associated279to $X$. ``Numbers are useful in all branches of mathematics.''280281\begin{example}{Arithmetic Genus}282Let $X$ be a nonsingular projective curve. Then $\dim H^1(X,\sO_X)$ is283the arithmetic genus of $X$. If $X\subseteq \bP^n$ is a projective variety284of dimension $r$ then, if $p_a=\dim H^1(X,\sO_X)$, then285$1+(-1)^r p_a = $ the constant term of the Hilbert polynomial of $X$.286\end{example}287288\begin{example}289Let $X$ be a nonsingular projective surface, then290\begin{equation*}2911+p_a=h^0(\sO_X)-h^1(\sO_X)+h^2(\sO_x)292\end{equation*}293and $1+(-1)^r p_a = \chi(\sO_X)$, the Euler characteristic of $X$.294\end{example}295296\begin{example}297Let $X$ be an algebraic variety and $\pic X$ the group of298Cartier divisors modular linear equivalence (which is isomorphic299to the group of invertible sheaves under tensor product modulo isomorphism).300Then $\pic X \isom H^1(X,\sO_{X}^{*})$.301\end{example}302303\begin{example}[Deformation Theory]304Let $X_0$ be a nonsingular projective variety. Then305the first order infinitesimal deformations are classified306by $H^1(X_0,T_{X_0})$ where $T_{X_0}$ is the tangent307bundle of $X_0$. The obstructions are classified by308$H^2(X_0,T_{X_0})$.309\end{example}310311One can define Cohen-Macaulay rings in terms of cohomology.312Let $(A,\gm)$ be a local Noetherian ring of dimension $n$,313let $X = \spec A$, and let $P=\gm\in X$, then we have the314following.315\begin{prop}316Let $A$ be as above. Then $A$ is Cohen-Macaulay iff3173181) $H^0(X-P,\sO_{X-P}) = A$ and3193202) $H^i(X-P, \sO_{X-P}) = 0$ for $0<i<n-1$.321\end{prop}322323A good place to get the necessary background for the324cohomology we will study is in Appendices 3 and 4325from Eisenbud's {\em Commutative Algebra}.326327\section{Cohomology in Algebraic Geometry}328329For any scheme $X$ and any sheaf $\sF$ of $\sO_X$-modules330we want to define the groups $H^i(X,\sF)$. We can either {\em define}331cohomology by listing its properties, then later prove that we can construct332the $H^i(X,\sF)$ or we can skip the definition and just construct333the $H^i(X,\sF)$. The first method is more esthetically pleasing,334but we will choose the second.335336We first forget the scheme structure of $X$ and regard $X$ as a337topological space and $\sF$ as a sheaf of abelian groups (by ignoring338the ring multiplication). Let $\Ab(X)$ be the category of sheaves of339abelian groups on $X$. Let $\Gamma=\Gamma(X,\cdot)$ be the global340section functor from $\Ab(X)$ into $\Ab$, where $\Ab$ is the category341of abelian groups. Recall that $\Gamma$ is left exact so if342\begin{equation*}3430\rightarrow\sF'\rightarrow\sF\rightarrow\sF''\rightarrow 0344\end{equation*}345is an exact sequence in $\Ab(X)$ then the following sequence is exact346\begin{equation*}3470\rightarrow \Gamma(\sF') \rightarrow \Gamma(\sF) \rightarrow348\Gamma(\sF'')349\end{equation*}350in $\Ab$.351352353\begin{defn}354We define the cohomology groups $H^i(X,\sF)$ to be the right derived355functors of $\Gamma$.356\end{defn}357358%% 1/22/96, Lecture 3359360\section{Review of Derived Functors}361362The situation will often be as follows.363Let $\sA$ and $\sB$ be abelian categories and364$$\sA\xrightarrow{F}\sB$$365a functor. Derived functors are the measure of the non-exactness of366a functor. Let $X$ be a topological space, $\Ab(X)$ the category367of sheaves of abelian groups on $X$ and $\Ab$ the category of368abelian groups. Then $\Gamma(X,\cdot):369\Ab(X)\rightarrow\Ab$ is a left exact functor. Our cohomology theory370will turn out to be the right derived function of $\Gamma(X,\cdot)$.371372\subsection{Examples of Abelian Categories}373Although we will not define an abelian category we will give several374examples and note that an abelian category is a category which has375the same basic properties as these examples.376\begin{example}[$A$-Modules] Let $A$ be a fixed commutative ring377and consider the category $\Mod(A)$ of $A$-modules. Then if378$M,N$ are any two modules one has3793801) $\Hom(M,N)$ is an abelian group,3813822) $\Hom(M,N)\times \Hom(N,L)\into \Hom(M,L)$ is a homomorphism383of abelian groups.3843853) there are kernels, cokernels, etc.386387$\Mod(A)$ is an abelian category.388\end{example}389390\begin{example}391Let $A$ be a Noetherian ring and let our category be the collection392of all finitely generated $A$-modules. Then this category is abelian.393Note that if the condition394that $A$ be Noetherian is relaxed we may no longer have an abelian395category because the kernel of a morphism of finitely generated396modules over an arbitrary ring need not be finitely generated (for397example, take the map from a ring to its quotient by an ideal which398cannot be finitely generated).399\end{example}400401\begin{example} Let $X$ be a topological space, then402$\Ab(X)$ is an abelian category. If $(X,\sO_X)$ is a ringed403space then the category $\Mod(\sO_X)$ is abelian. If $X$ is404a scheme then the category of quasi-coherent $\sO_X$-modules405is abelian, and if $X$ is also Noetherian then the sub-category406of coherent $\sO_X$-modules is abelian.407\end{example}408409\begin{example}410The category of abelian varieties is {\em not} an abelian category411since the kernel of a morphism of abelian varieties might be412reducible (for example an isogeny of degree $n$ of elliptic curves413has kernel $n$ points which is reducible). It may be the case414that the category of abelian group schemes is abelian but I don't415know at the moment.416\end{example}417418\begin{example}419The category of compact Hausdorff abelian topological groups420is an abelian category.421\end{example}422423\subsection{Exactness}424425\begin{defn} A functor $F:\sA\into\sB$ is {\em additive} if for426all $X,Y\in \sA$, the map427$$F:\Hom_{\sA}(X,Y)\into\Hom_{\sB}(FX,FY)$$428is a homomorphism of abelian groups.429\end{defn}430431\begin{defn} A sequence432$$A\xrightarrow{f}B\xrightarrow{g}C$$433is {\em exact} if $\imag(f)=\ker(g)$.434\end{defn}435436\begin{defn} Let $F:\sA\into\sB$ be a functor and437$$0\into M'\into M \into M''\into 0$$438be an exact sequence and consider the sequence439$$0\into FM'\into FM\into FM''\into 0.$$440If the second sequence is exact in the middle, then441$F$ is a called {\em half exact functor}. If the second sequence is exact442on the left and the middle then $F$ is called a {\em left exact functor}.443If the second sequence is exact on the right and in the middle444then we call $F$ a {\em right exact functor}.445\end{defn}446447\begin{example}448Let $A$ be a commutative ring and $N$ an $A$-module. Then449$N\tensor -$ is a right exact functor on the category of450$A$-modules. To see that $N\tensor -$ is not exact, suppose451$A,\gm$ is a local ring and $N=k=A/\gm$.452Then the sequence453$$0\into\gm\into A\into k\into 0$$454is exact, but455$$0\into k\tensor\gm \into k\tensor A \into k\tensor k \into 0$$456is right exact but not exact.457\end{example}458459\begin{example}460The functor $\tor_1(N,\cdot)$ is neither left nor right exact.461\end{example}462463\begin{example}464The contravarient hom functor, $\Hom(\cdot,N)$ is left exact.465\end{example}466467468Often the following is useful in work.469\begin{thm}470If $$0\into M'\into M\into M''$$ is exact and $F$ is left exact, then471$$0\into FM'\into FM\into FM''$$472is exact.473\end{thm}474475\subsection{Injective and Projective Objects}476477Let $\sA$ be an abelian category. Then $\Hom_A(P,-):\sA\into\Ab$ is478left exact.479480\begin{defn}481An $A$ module $P$ is said to be {\em projective} if482the functor $\Hom_A(P,-)$ is exact. An $A$ module $I$ is483said to be {\em injective} if the functor484$\Hom_A(-,I)$ is exact.485\end{defn}486487\begin{defn}488We say that an abelian category $\sA$ has {\em enough projectives} if489every $X$ in $\sA$ is the surjective image of a projective490$P$ in $\sA$. A category is said to have {\em enough injectives} if491every $X$ in $\sA$ injects into an injective objective of $\sA$.492\end{defn}493494\begin{example}495Let $A$ be a commutative ring, then $\Mod(A)$ has enough injectives because496every module is the quotient of a free module and every free module497is projective. If $X$ is a topological space then $\Ab(X)$ has enough498injectives. If $X$ is a Noetherian scheme, then the category of quasi-coherent499sheaves has enough injectives (hard theorem). The category of $\sO_X$-modules500has enough injectives but the category of coherent sheaves on501$X$ doesn't have enough injectives or projectives.502\end{example}503504%% 1/24/96, Lecture 4505506\section{Derived Functors and Homological Algebra}507508Let $F:\sA\into\sB$ be an additive covariant left-exact functor between509abelian categories, for example $F=\Gamma:\Ab(X)\into\Ab$.510Assume $\sA$ has enough injectives, i.e., for all $X$ in $\sA$511there is an injective object $I$ in $\sA$ such that512$0\into X\hookrightarrow I$. We construct the right derived functors513of $F$. If514$$0\into M'\into M\into M''\into 0$$515is exact in $\sA$ then516$$0\into F(M')\into F(M)\into F(M'')\into R^{1}F(M^1)\into R^{1}F(M)517\into \cdots$$518is exact in $\sB$ where $R^{i}F$ is the right derived functor of $F$.519520\subsection{Construction of $R^{i}F$}521Take any $M$ in $\sA$, then since $\sA$ has enough injectives we522can construct an exact sequence523$$0\into M\into I^{0}\into I^{1}\into I^{2}\into \cdots$$524where each $I^{i}$ is an injective object. (This isn't totally525obvious, but is a straightforward argument by putting together526short exact sequences and composing maps.)527The right part of the above sequence $I^{0}\into I^{1}\into \cdots$528is called an {\em injective resolution} of $M$.529Applying $F$ we get a complex530$$F(I^{0})\xrightarrow{d_0} F(I^{1})\xrightarrow{d_1} F(I^{2})531\xrightarrow{d_2} \cdots$$532in $\sB$ which may not be exact. The objects533$H^i=\ker(d_2)/ \imag(d_1)$ measure the deviation of this534sequence from being exact. $H^i$ is called the $i$th {\em cohomology}535object of the complex.536537\begin{defn} For each object in $\sA$ fix an injective resolution.538The $i$th {\em right derived functor} of $F$ is the functor539which assigns to an object $M$540the $i$th cohomology of the complex $F(I^{\cdot})$ where $I^{\cdot}$541is the injective resolution of $M$.542\end{defn}543544\subsection{Properties of Derived Functors}545We should now prove the following:546\begin{enumerate}547\item If we fix different injective resolutions for all of our objects548then the corresponding derived functors are, in a suitable sense, isomorphic.549\item The $R^{i}F$ can also be defined on morphisms in such a way550that they are really functors.551\item If $0\into M'\into M\into M''$ is a short exact sequence then there is552a long exact sequence of cohomology:553\begin{align*}5540\into FM'\into FM\into FM'' \into \\555R^1FM'\into R^1FM\into R^1FM''\into \\556R^2FM'\into \cdots.557\end{align*}558\item If we have two short exact sequences then the induced maps on559long exact sequences are ``$\delta$-compatible''.560\item $R^0 F\isom F$.561\item If $I$ is injective, then for any $i>0$ one has that $R^i F(I)=0$.562\end{enumerate}563564\begin{thm} The $R^{i}F$ and etc. are uniquely determined by properties5651-6 above.566\end{thm}567568\begin{defn} A {\em $\delta$-functor} is a collection of functors569$\{R^i F\}$ which satisfy 3 and 4 above. An {\em augmented570$\delta$-functor} is a {\em $\delta$-functor} along with a natural571transformation $F\into R^0 F$. A {\em universal augmented572$\delta$-functor} is an {\em augmented $\delta$-functor} with573some universal property which I didn't quite catch.574\end{defn}575576\begin{thm} If $\sA$ has enough injectives then the collection of577derived functors of $F$ is a universal augmented $\delta$-functor.578\end{thm}579580To construct the $R^i F$ choose once and for all, for each object581$M$ in $\sA$ an injective resolution, then prove the above properties582hold.583584585%% 1/26/96 -- today's lecture was too [email protected]$%^(#$򏍛\section{Long Exact Sequence of Cohomology and Other Wonders}588589``Today I sat in awe as Hartshorne effortless drew hundreds590of arrows and objects everywhere, chased some elements and591proved that there is a long exact sequence of cohomology in59230 seconds. Then he whipped out his colored chalk and things593really got crazy. Vojta tried to erase Hartshorne's diagrams594during the next class but only partially succeeded joking595that the functor was not `effaceable'. (The diagrams are596still not quite gone 4 days later!) Needless to say, I don't597feel like texing diagrams and element chases... it's all598trivial anyways, right?''599600%% 1/29/96601\section{Basic Properties of Cohomology}602Let $X$ be a topological space, $\Ab(X)$ the category603of sheaves of abelian groups on $X$ and604$$\Gamma(X,\cdot):\Ab(X)\into\Ab$$605the covariant, left exact global sections functor.606Then we have constructed the derived functors607$H^{i}(X,\cdot)$.608609\subsection{Cohomology of Schemes}610Let $(X,\sox)$ be a scheme and $\sF$ a sheaf of $\sox$-modules.611To compute $H^{i}(X,\sF)$ forget all extra structure and use612the above definitions. We may get some extra structure anyways.613614\begin{prop}615Let $X$ and $\sF$ be as above, then the groups $H^{i}(X,\sF)$616are naturally modules over the ring $A=\Gamma(X,\sox)$.617\end{prop}618619\begin{proof}620Let $A=H^{0}(X,\sF)=\Gamma(X,\sox)$ and let $a\in A$. Then because621of the functoriality of $H^{i}(X,\cdot)$ the622map $\sF\into\sF$ induced by left multiplication by $a$ induces623a homomorphism624$$a:H^{i}(X,\sF)\into H^{i}(X,\sF).$$625%% Need more!!626\end{proof}627628\subsection{Objective}629Our objective is to compute $H^{i}(\bP^n_k,\sO(\ell))$630for all $i,n,\ell$. This is enough for most applications631because if one knows these groups one can, in principle at632least, computer the cohomology of any projective scheme.633If $X$ is any projective variety, we embed $X$ in some634$\bP^n_k$ and push forward the sheaf $\sF$ on $X$. Then635we construct a resolution of $\sF$ by sheaves of the636form $\sO(-\ell)^n$. Using Hilbert's syzigy theorem one637sees that the resolution so constructed is finite and638so we can put together our knowledge to get the cohomology639of $X$.640641Our plan of attack is as follows.642\begin{enumerate}643\item Define flasque sheaves which are acyclic for cohomology, i.e.,644the cohomology vanishes for $i>0$.645\item If $X=\spec A$, $A$ Noetherian, and $\sF$ is quasi-coherent,646show that $H^i(X,\sF)=0$ for $i>0$.647\item If $X$ is any Noetherian scheme and $\sU=(U_i)$ is an open648affine cover, find a relationship between the cohomology of $X$649and that of each $U_i$. (The ``\v{C}ech process''.)650\item Apply number 3 to $\bP_k^n$ with $U_i=\{x_i\neq 0\}$.651\end{enumerate}652653\section{Flasque Sheaves}654\begin{defn}655A {\em flasque sheaf} (also called {\em flabby sheaf}) is a sheaf656$\sF$ on $X$ such that whenever $V\subset U$ are open sets then657$\rho_{U,V}:\sF(U)\into\sF(V)$ is surjective.658\end{defn}659Thus in a flasque sheaf, ``every section extends''.660661\begin{example}662Let $X$ be a topological space, $p\in X$ a point, not necessarily663closed, and $M$ an abelian group. Let $j:\{P\}\hookrightarrow X$ be the664inclusion, then $\sF=j_{*}(M)$ is flasque. This follows since665$$666j_{*}(M)(U)=\begin{cases} M&\text{if $p\in U$}\\6670&\text{if $p\not\in U$}668\end{cases}.669$$670Note that $j_{*}(M)$ is none other than the skyscraper sheaf671at $p$ with sections $M$.672\end{example}673674\begin{example}675If $\sF$ is a flasque sheaf on $Y$ and $f:Y\into X$ is a morphism676then $f_{*}\sF$ is a flasque sheaf on $X$.677\end{example}678679\begin{example}680If $\sF_i$ are flasque then $\bigoplus_{i} \sF_i$ is flasque.681\end{example}682683\begin{lem}684If685$$0\into\sF'\into\sF\into\sF''\into 0$$686is exact and $\sF'$ is flasque687then688$$\Gamma(\sF)\into\Gamma(\sF'')\into 0$$689is exact.690\end{lem}691692\begin{lem}693If694$$0\into\sF'\into\sF\into\sF''\into 0$$695is exact and $\sF'$ and $\sF$ are both696flasque then697$\sF''$ is flasque.698\end{lem}699700\begin{proof}701Suppose $V\subset U$ are open subsets of $X$.702Since $\sF'$ is flasque and the restriction of a703flasque sheaf is flasque and restriction is exact,704lemma 1 implies that the sequence705$$\sF(V)\into\sF''(V)\into 0$$706is exact.707We thus have a commuting diagram708$$709\begin{CD}710\sF(U) @>>> \sF''(U)\\711@VVV @VVV\\712\sF(V) @>>> \sF''(V) @>>> 0713\end{CD}714$$715which, since $\sF(U)\into \sF(V)$ is surjective,716implies $\sF''(U)\into\sF''(V)$ is surjective.717\end{proof}718719720\begin{lem}721Injective sheaves (in the category of abelian sheaves) are flasque.722\end{lem}723724\begin{proof}725Let $\sI$ be an injective sheaf of abelian groups on $X$ and726let $V\subset U$ be open subsets. Let $s\in\sI(V)$, then we727must find $s'\in\sI(U)$ which maps to $s$ under728the map $\sI(U)\into\sI(V)$. Let $\bZ_V$ be the constant sheaf729$\bZ$ on $V$ extended by $0$ outside $V$ (thus $\bZ_V(W)=0$730if $W\not\subset V$). Define a map $\bZ_V\into \sI$ by731sending the section $1\in\bZ_V(V)$ to $s\in\sI(V)$. Then732since $\bZ_V\hookrightarrow\bZ_U$ and $\sI$ is injective733there is a map $\bZ_U\into \sI$ which sends the section $1\in734\bZ_U$ to a section $s'\in\sI(U)$ whose restriction735to $V$ must be $s$.736\end{proof}737738\begin{remark} The same proof also shows that injective sheaves in739the category of $\sox$-modules are flasque.740\end{remark}741742\begin{cor}743If $\sF$ is flasque then $H^{i}(X,\sF)=0$ for all $i>0$.744\end{cor}745\begin{proof} Page 208 of [Hartshorne].746\end{proof}747748% 1/31/96749750\begin{cor}751Let $(X,\sox)$ be a ringed space, then the derived functors of752$\Gamma:\Mod\sox\into\Ab$ are equal to $H^{i}(X,\sF)$.753\end{cor}754\begin{proof}755If756$$0\into\sF\into\sI^0\into\sI^1\into\cdots$$757is an injective resolution of $\sF$758in $\Mod\sox$ then, by the above remark, it759is a flasque resolution in the category760$\Ab(X)$ hence we get the regular cohomology.761\end{proof}762763\begin{remark}764{\bf Warning!} If $(X,\sox)$ is a scheme and we choose an injective765resolution in the category of quasi-coherent $\sox$-modules766then we are only guaranteed to get the right answer if767$X$ is Noetherian.768\end{remark}769770\section{Examples}771772\begin{example}773Suppose $C$ is a nonsingular projective curve over an algebraically774closed field $k$. Let $K=K(C)$ be the function field of $C$ and775let $\sK_C$ denote the constant sheaf $K$. Then we have an exact776sequence777$$ 0\into\sO_C\into\sK_C\into778\bigoplus_{\substack{P\in C\\P\text{ closed}}}779K/\sO_P \into 0,$$780where the map $\sK_C\into\bigoplus K/\sO_P$ has781only finitely many components nonzero since a function $f\in K$782has only finitely many poles.783Since $C$ is irreducible $\sK_C$ is flasque and since $\sK/\sO_P$784is a skyscraper sheaf it is flasque so since direct of flasque785sheaves are flasque, $\bigoplus K/\sO_P$ is flasque.786One checks that the sequence is exact and so this is787a flasque resolution of $\sO_C$. Taking global sections and788applying the exact sequence of cohomology gives an exact sequence789$$K\into\bigoplus_{\text{$P$ closed}} K/\sO_P790\into H^1(X,\sO_C)\into 0,$$ and $H^{i}(X,\sO_C)=0$ for791$i\geq 2$. Thus the only interesting information792is $\dim_k H^1(X,\sO_C)$ which is the {\em geometric genus} of $C$.793\end{example}794795%% 2/2/96796797\section{First Vanishing Theorem}798799\begin{quote}800``Anyone who studies algebraic geometry must read French... looking up801the more general version of this proof in EGA would be a good exercise.''802\end{quote}803804\begin{thm}805Let $A$ be a Noetherian ring, $X=\spec A$ and $\sF$ a quasi-coherent806sheaf on $X$, then $H^{i}(X,\sF)=0$ for $i>0$.807\end{thm}808809\begin{remark}810The theorem is true without the Noetherian hypothesis on $A$, but811the proof uses spectral sequences.812\end{remark}813814\begin{remark}815The assumption that $\sF$ is quasi-coherent is essential. For example,816let $X$ be an affine algebraic curve over an infinite field $k$. Then817$X$ is homeomorphic as a topological space to $\P^1_k$ so818the sheaf $\sO(-2)$ on $\P^1_k$ induces a sheaf $\sF$ of abelian groups819on $X$ such that820$$H^1(X,\sF)\isom{}H^1(\P^1_k,\sO(-2))\neq 0.$$821\end{remark}822823\begin{remark}824If $I$ is an injective $A$-module then $\tilde{I}$ need {\em not}825be injective in $\Mod(\sox)$ or $\Ab(X)$. For example, let $A=k=\bF_p$ and826$X=\spec A$, then $I=k$ is an injective $A$-module but $\tilde{I}$827is the constant sheaf $k$. But $k$ is828a finite group hence not divisible so $\tilde{I}$ is not injective.829(See Proposition A3.5 in Eisenbud's {\em Commutative Algebra}.)830\end{remark}831832\begin{prop}833Suppose $A$ is Noetherian and $I$ is an injective $A$-module, then834$\tilde{I}$ is flasque on $\spec A$.835\end{prop}836837The proposition implies the theorem since if $\sF$ is quasi-coherent838then $\sF=\tilde{M}$ for some $A$-module $M$. There is an injective839resolution840$$0\into M\into I^{\bullet}$$841which, upon applying the exact functor $\tilde{ }$,842gives a flasque resolution843$$0\into \tilde{M}=\sF\into \tilde{I}^{\bullet}.$$844Now applying $\Gamma$ gives us back the original resolution845$$\Gamma:\quad 0\into M\into I^{\bullet}$$846which is exact so the cohomology groups vanish for $i>0$.847848\begin{proof} Let $A$ be a Noetherian ring and $I$ an injective $A$,849then $\tilde{I}$ is a quasi-coherent sheaf on $X=\spec A$. We must850show that it is flasque. It is sufficient to show that for any851open set $U$, $\Gamma(X)\into\Gamma(U)$ is surjective.852853{\em Case 1, special open affine:}854Suppose $U=X_f$ is a special open affine. Then we have a commutative diagram855$$856\begin{CD}857\Gamma(X,\tilde{I}) @>>> \Gamma(X_f,\tilde{I})\\858@V=VV @V=VV\\859I @>\text{surjective?}>> I_f860\end{CD}861$$862To see that the top map is surjective it is equivalent863to show that $I\into I_f$ is surjective. This is a tricky864algebraic lemma (see Hartshorne for proof).865866{\em Case 2, any open set:}867Let $U$ be any open set. See Hartshorne for the rest.868869\end{proof}870871872%% 2/5/96873874\section{\cech{} Cohomology}875Let $X$ be a topological space, $\sU=(U_i)_{i\in I}$ an876open cover and $\sF$ a sheaf of abelian groups.877We will define groups $\cH^i(\sU,\sF)$ called878\cech{} cohomology groups.879880{\bfseries Warning: } $\cH^i(\sU,\cdot)$ is a functor in $\sF$,881but it is {\em not} a $\delta$-functor.882883\begin{thm} Let $X$ be a Noetherian scheme, $\sU$ an open cover884and $\sF$ a quasi-coherent sheaf, then $\cH^{i}(\sU,\sF)=H^i(X,\sF)$885for all $i$.886\end{thm}887888\subsection{Construction}889Totally order the index set $I$. Let890$$U_{i_0\cdots i_p}=\intersect_{j=0}^p U_{i_j}.$$891For any $p\geq 0$ define892$$C^p(\sU,\sF)=\prod_{i_0<i_1<\cdots<i_p}\sF(U_{i_0\cdots i_p}).$$893Then we get a complex894$$C^0(\sU,\sF)\into C^1(\sU,\sF)\into \cdots \into C^p(\sU,\sF)\into \cdots$$895by defining a map896$$d:C^p(\sU,\sF)\into C^{p+1}(\sU,\sF)$$897by, for $\alpha\in C^p(\sU,\sF)$,898$$(d\alpha)_{i_0\cdots i_{p+1}} := \sum_{0}^{p+1} (-1)^j899\alpha_{i_0\cdots \hat{i_j}\cdots i_{p+1}}|_{U_{i_0\cdots i_{p+1}}}.$$900One checks that $d^2=0$.901902\begin{lem}903$\cH^{0}(\sU,\sF)=\Gamma(X,\sF)$904\end{lem}905\begin{proof}906Applying the sheaf axioms to the exact sequence907$$0\into\Gamma(X,\sF)\into C^0=\prod_{i\in I}\sF(U_i)908\xrightarrow{d}C^1=\prod_{i<j}\sF(U_{ij})$$909we see that $\cH^0(\sU,\sF)=\ker d = \Gamma(X,\sF)$.910\end{proof}911912\subsection{Sheafify}913Let $X$ be a topological space, $\sU$ an open cover and914$\sF$ a sheaf of abelian groups. Then we define915$$\sC^p(\sU,\sF)=\prod_{i_0<\cdots<i_p}j_{*}(\sF|_{U_{i_0\cdots i_p}})$$916and define917$$d:\sC^p(\sU,\sF)\into \sC^{p+1}(\sU,\sF)$$918in terms of the $d$ defined above by, for $V$ an open set,919$$\sC^{p}(\sU,\sF)(V)=C^p(\sU|_{V},\sF|_V)\xrightarrow{d}920C^{p+1}(\sU|_{V},\sF|_V)921=\sC^{p+1}(\sU,\sF)(V).$$922923\begin{remark}924$C^p(\sU,\sF)=\Gamma(X,\sC^{p}(\sU,\sF)$925\end{remark}926927\begin{lem}928The sequence929$$ 0\into\sF\into\sC^{0}(\sU,\sF)\into\sC^{1}(\sU,\sF)\into\cdots$$930is a resolution of $\sF$, i.e., it is exact.931\end{lem}932\begin{proof}933We define the map $\sF\into\sC^{0}$ by taking the product of the natural maps934$\sF\into f_{*}(\sF|_{U_i})$, exactness then follows from the935sheaf axioms.936937To show the rest of the sequence is exact it suffices to show938exactness at the stalks. So let $x\in X$, and suppose $x\in U_j$.939Given $\alpha_x\in\sC_x^p$ it is represented by a section940$\alpha\in\Gamma(V,\sC^p(\sU,\sF))$, over a neighborhood $V$ of941$x$, which we may choose so small that $V\subset U_j$. Now942for any $p$-tuple $i_0<\ldots<i_{p-1}$, we set943$$(k\alpha)_{i_0,\ldots,i_{p-1}}=\alpha_{j,i_0,\ldots,i_{p-1}}.$$944This makes sense because945$$V\intersect U_{i_0,\ldots,i_{p-1}}=V\intersect U_{j,i_0,\ldots,i_{p-1}}.$$946Then take the stalk of $k\alpha$ at $x$ to get the required map $k$.947948Now we check that for any $p\geq 1$ and $\alpha\in\sC_x^p$,949$$(dk+kd)(\alpha)=\alpha.$$950First note that951\begin{align*}(dk\alpha)_{i_0,\ldots,i_p} & =952\sum_{\ell=0}^{p} (-1)^\ell (k\alpha)_{i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}\\953& = \sum (-1)^\ell \alpha_{j,i_0,\ldots,\hat{i_{\ell}},\ldots,i_p}954\end{align*}955Whereas, on the other hand,956\begin{align*}957(kd\alpha)_{i_0,\ldots,i_p}958& = (d\alpha)_{j,i_0,\ldots,i_p}\\959& = (-1)^0\alpha_{i_0,\ldots,i_p} + \sum_{\ell=1}^{p} (-1)^{\ell+1}\alpha_{j,i_0,\ldots,960\hat{i_{\ell}},\ldots,i_{p}}961\end{align*}962Adding these two expressions yields $\alpha_{i_0,\ldots,i_p}$ as claimed.963964Thus $k$ is a homotopy operator for the complex $\sC_x^{\bullet}$, showing965that the identity map is homotopic to the zero map. It follows that the966cohomology groups $H^{p}(\sC^{\bullet}_x)$ of this complex967are $0$ for $p\geq 1$.968\end{proof}969970\begin{lem}971If $\sF$ is flasque then $\sC^p(\sU,\sF)$ is also flasque.972\end{lem}973\begin{proof}974If $\sF$ is flasque then $\sF|_{U_{i_0,\ldots,i_p}}$ is flasque975so $j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque so976$\prod j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque.977\end{proof}978979\begin{prop}980If $\sF$ is flasque then $\cH^{p}(\sU,\sF)=0.$981\end{prop}982\begin{proof}983Consider the resolution $$0\into\sF\into\sC^{\bullet}(\sU,\sF).$$984By the above lemma it is flasque, so we can use it to compute the985usual cohomology groups of $\sF$. But $\sF$ is flasque, so986$H^p(X,\sF)=0$ for $p>0$. On the other hand,987the answer given by this resolution is988$$H^p(\Gamma(X,\sC^{\bullet}(\sU,\sF)))=\cH^{p}(\sU,\sF).$$989So we conclude that $\cH^{p}(\sU,\sF)=0$ for $p>0$.990\end{proof}991992\begin{lem}993Let $X$ be a topological space, and $\sU$ an open covering. Then994for each $p\geq 0$ there is a natural map, functorial in $\sF$,995$$\cH^p(\sU,\sF)\into H^p(X,\sF).$$996\end{lem}997998\begin{thm}999Let $X$ be a Noetherian separated scheme, let $\sU$ be an open1000affine cover of $X$, and let $\sF$ be a quasi-coherent sheaf on1001$X$. Then for all $p\geq 0$ the natural maps give isomorphisms1002$$\cH^{p}(\sU,\sF)\isom H^p(X,\sF).$$1003\end{thm}10041005% 2/7/9610061007\section{\v{C}ech Cohomology and Derived Functor Cohomology}10081009Today we prove10101011\begin{thm}1012Let $X$ be a Noetherian, separated scheme, $\sU$ an open cover1013and $\sF$ a quasi-coherent sheaf on $X$. Then1014$$\cH^{i}(\sU,\sF)=H^{i}(X,\sF).$$1015\end{thm}10161017To do this we introduce a condition (*):10181019{\bfseries Condition *:} Let $\sF$ be a sheaf of abelian groups1020and $\sU=(U_i)_{i\in I}$ an open cover. Then the pair $\sF$ and1021$\sU$ satisfy condition (*) if for all $i_0,\ldots,i_p\in I$,1022$$H^(U_{i_0,\ldots,i_p},\sF)=0, \text{all} i>0.$$10231024\begin{lem}1025If $0\into\sF'\into\sF\into\sF''\into 0$ is an exact sequence in1026$\Ab(X)$ and $\sF'$ satisfies (*) then there is a long exact sequence1027for $\cH^{i}(\sU,\cdot)$.1028\end{lem}1029\begin{proof}1030Since the global sections functor is left exact and cohomology1031commutes with products, we have an exact sequence1032\begin{align*}10330\into C^{p}(\sU,\sF')=\prod_{i_0<\cdots<i_p}\sF'(U_{i_0,\ldots,i_p})1034\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 \prod_{i_0<\cdots<i_p} H^{1}(\sU_{i_0,\ldots,i_p},\sF')=01037\end{align*}1038where the last term is 0 because $\sF'$ satisfies condition (*).1039Replacing $p$ by $\cdot$ gives an exact sequence of complexes.1040Applying $\cH^{i}(\sU,\cdot)$ then gives the desired result.1041\end{proof}10421043\begin{thm}1044Let $X$ be a topological space, $\sU$ an open cover and $\sF\in \Ab(X)$.1045Suppose $\sF$ and $\sU$ satisfy (*). Then the maps1046$$\varphi^{i}:\cH^i(\sU,\sF)\into H^{i}(X,\sF)$$1047are isomorphisms.1048\end{thm}1049\begin{proof}1050The proof is a clever induction.1051\end{proof}10521053\begin{lem}1054If $0\into\sF'\into\sF\into\sF''\into 0$ is exact and1055$\sF'$ and $\sF$ satisfy (*) then $\sF''$ satisfies (*).1056\end{lem}10571058To prove the main theorem of the section use the fact that1059$X$ separated implies any finite intersection of affines1060is affine and then use the vanishing theorem for cohomology1061of a quasi-coherent sheaf on an affine scheme. The above theorem1062then implies the main result. From now on we will always1063assume our schemes are separated unless otherwise stated.10641065\begin{cor}1066If $X$ is a (separated) Noetherian scheme and $X$ can be covered1067by $n+1$ open affines for some $n>0$ then $H^{i}(X,\sF)=0$ for $i>n$.1068\end{cor}10691070\begin{example}1071Let $X=\bP_k^n$, then the existence of the standard affine1072cover $U_0,\ldots,U_n$ implies that $H^{i}(X,\sF)=0$ for1073$i>n$.1074\end{example}10751076\begin{example}1077Let $X$ be a projective curve embedded in $\bP^k_n$.1078Let $U_0\subset X$ be open affine, then $X-U_0$ is finite.1079Thus $U_0\subset X\subset \bP^n$ and $X-U_0=\{P_1,\ldots,P_r\}$.1080In $\bP^n$ there is a hyperplane $H$ such that1081$P_1,\ldots,P_r\not\in H$. Then $P_1,\ldots,P_r\in\bP^n-H=\bA^n=V$.1082Then $U_1=V\intersect X$ is closed in the affine set $V$, hence affine.1083Then $X=U_0\union U_1$ with $U_0$ and $U_1$ both affine.1084Thus $H^{i}(X,\sF)=0$ for all $i\geq 2$.1085\end{example}10861087\begin{exercise}1088If $X$ is any projective scheme of dimension $n$ then $X$1089can be covered by $n+1$ open affines so1090$$H^{i}(X,\sF)=0 \text{ for all } i>n.$$1091[Hint: Use induction.]1092\end{exercise}10931094Hartshorne was unaware of the answer to the following question1095today.1096\begin{ques}1097If $X$ is a Noetherian scheme of dimension $n$ do there1098exist $n+1$ open affines covering $X$.1099\end{ques}11001101\begin{thm}[Grothendieck]1102If $\sF\in\Ab(X)$ then $H^{i}(X,\sF)=0$ for all $i>n=\dim X$.1103\end{thm}11041105\begin{example}1106Let $k$ be an algebraically closed field.1107Then $X=\bA^2_k-\{(0,0)\}$ is not affine since it has global1108sections $k[x,y]$.1109We compute $H^1(X,\sox)$ by \cech cohomology.1110Write $X=U_1\union U_2$ where $U_1=\{x\neq 0\}$ and1111$U_2=\{y\neq 0\}$. Then the \cech complex is1112$$C^{\cdot}(\sU,\sox): k[x,y,x^{-1}]\oplus{}k[x,y,y^{-1}]1113\xrightarrow{d} k[x,x^{-1},y,y^{-1}].$$1114Thus one sees with a little thought that1115$H^{0}=\ker{d}=k[x,y]$1116and1117$H^{1}=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\} = E$1118as $k$-vector spaces (all sums are finite).1119\end{example}1120\subsection{History of this Module $E$}1121$$E=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\}$$1122\begin{enumerate}1123\item Macaulay's ``Inverse System'' (1921?)1124\item $E$ is an injective $A$-module, in fact, the indecomposable1125injective associated to the prime $(x,y)$1126\item $E$ is the dualizing module of $A$, thus1127$D=\Hom_A(\cdot,E)$ is a dualizing functor1128for finite length modules (so doing $D$ twice1129gives you back what you started with).1130\item Local duality theorem: this is the module you ``hom into''.1131\end{enumerate}113211331134% 2/9/961135\section{Cohomology of $\bP_k^n$}1136Today we begin to compute $H^{i}(X,\sox(\ell))$ for all $i$ and all1137$\ell$.11381139a) $H^0(X,\sox(\ell))$ is the vector space of forms of degree $\ell$1140in $S=k[x_0,\ldots,x_n]$, thus1141$$\oplus_{\ell\in\bZ}H^0(\sox(\ell))=H^0_{*}(\sox)=\Gamma_{*}(\sox)=S.$$11421143\begin{prop}1144There is a natural map1145$$H^{0}(\sox(\ell))\times H^{i}(\sox(m))\into H^{i}(\sox(\ell+m)).$$1146\end{prop}1147\begin{proof}1148$\alpha\in H^{0}(\sox(\ell))$ defines a map $\sox\into\sox(\ell)$1149given by $1\mapsto\alpha$. This defines a map1150$$\sox\tensor\sox(m)\xrightarrow{\alpha(m)}\sox(\ell)\tensor\sox(m)$$1151which gives a map $\sox(m)\into\sox(\ell+m)$. This induces the desired1152map $H^{i}(\sox(m))\into H^{i}(\sox(\ell+m))$.1153\end{proof}11541155b) $H^{i}(\sox(\ell))=0$ when $0<i<n$ and for all $\ell$.1156(This doesn't hold for arbitrary quasi-coherent sheaves!)11571158c) $H_{*}^n(X,\sox)$ is a graded $S$-module which1159is $0$ in degrees $\geq -n$, but is nonzero in degrees $\leq n-1$.1160As a $k$-vector space it is equal to1161$$\{\sum_{i_j<0} a_{i_0,\ldots,i_n}x_0^{i_0}\cdots x_n^{i_n} :1162\text{sum is finite}\}.$$11631164d) For $\ell\geq 0$ the map1165$$H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))\isom{}k$$1166is a perfect pairing so we have a duality (which is in fact a special1167case of Serre Duality).116811691170% 2/12/961171\section{Serre's Finite Generation Theorem}1172We relax the hypothesis from the last lecture and claim that the1173same results are still true.1174\begin{thm}1175Let $A$ be a Noetherian ring and $X=\bP^n_A$. Then1176\begin{enumerate}1177\item $H^0_{*}(\sox)=\oplus_{\ell}H^{0}_{\ell}(\sox(\ell))=S=A[x_0,\ldots,x_n]$1178\item $H^i_{*}(\sox)=0$ for all $0<i<n$1179\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\}$1180\item $H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))$ is1181a perfect pairing of free $A$-modules. Notice that $H^n(\sox(-n-1))$ is1182a free $A$-module of rank 1 so it is isomorphic to $A$, but {\em not} in1183a canonical way!1184\end{enumerate}1185\end{thm}11861187Although pairing is in general not functorial as a map into $A$,1188there is a special situation in which it is. Let $\Omega_{X/k}^1$1189be the sheaf of differentials on $X=\bP^n_k$. Let $\omega=\Omega_{X/k}^n1190=\Lambda^n\Omega^1$ be the top level differentials (or ``dualizing1191module''). Then some map is functorial (??)1192\begin{quote}1193``Is $\omega$ more important than $\Omega$?'' -- Janos Csirik11941195``That's a value judgment... you can make your own decision1196on that... I won't.'' -- Hartshorne1197\end{quote}11981199\begin{thm}[Serre]1200Let $X$ be a projective scheme over a Noetherian ring $A$. Let $\sF$1201be any coherent sheaf on $X$. Then1202\begin{enumerate}1203\item $H^{i}(X,\sF)$ is a finitely generated $A$-module for all $i$1204\item for all $\sF$ there exists $n_0$ such that for all $i>0$ and1205for all $n\geq n_0$, $H^{i}(X,\sF(n))=0$.1206\end{enumerate}1207\end{thm}12081209The following was difficult to prove last semester and1210we were only able to prove it under somewhat restrictive1211hypothesis on $A$ (namely, that $A$ is a finitely generated1212$k$-algebra).1213\begin{cor} $\Gamma(X,\sF)$ is a finitely generated $A$-module.1214\end{cor}1215\begin{proof}1216Set $i=0$ in 1.1217\end{proof}12181219\begin{proof}(of theorem)12201221I. {\em Reduce to the case $X=\bP^r_A$.}1222Use the fact that the push forward of a closed subscheme has1223the same cohomology to replace $\sF$ by $i_{*}(\sF)$.12241225II. {\em Special case, $\sF=\sO_{\bP^r}(\ell)$ any $\ell\in\bZ$.}12261. and 2. both follow immediately from the previous theorem.1227This is where we have done the work in explicit calculations.12281229III. {\em Cranking the Machine of Cohomology}123012311232\end{proof}12331234\subsection{Application: The Arithmetic Genus}1235Let $k$ be an algebraically closed field and $V\subset X=\bP^n_k$ a1236projective variety. The arithmetic genus of $V$ is1237$$p_a=(-1)^{\dim V}(p_V(0)-1)$$1238where $p_V$ is the Hilbert polynomial of $V$, thus1239$p_V(\ell)=\dim_k(S/I_V)_{\ell}$ for all $\ell\gg 0$. The1240Hilbert polynomial depends on the projective embedding of $V$.1241\begin{prop}1242$p_V(\ell)=\sum_{i=0}^{\infty}(-1)^{i}\dim_k H^{i}(\sO_V(\ell))$1243for {\em all} $\ell\in\bZ$.1244\end{prop}1245This redefines the Hilbert polynomial. Furthermore,1246$$p_a=(-1)^{\dim V}(p_V(0)-1) = (-1)^{\dim V}1247\sum_{i=0}^{\infty}(-1)^i\dim_k(H^{i}(\sO_V))$$1248which shows that $p_a$ is intrinsic, i.e., it doesn't depend1249on the embedding of $V$ in projective space.12501251\section{Euler Characteristic}1252Fix an algebraically closed field $k$, let $X=\P_k^n$. Suppose1253$\sF$ is a coherent sheaf on $X$. Then by Serre's theorem1254$H^{i}(X,\sF)$ is a finite dimensional $k$-vector space.1255Let $$h^{i}(X,\sF)=\dim_k H^{i}(X,\sF).$$1256\begin{defn}1257The {\bfseries Euler characteristic} of $\sF$ is1258$$\chi(\sF)=\sum_{i=0}^{n}(-1)^i h^i(X,\sF).$$1259\end{defn}1260Thus $\chi$ is a function $\Coh(X)\into\Z$.1261\begin{lem}1262If $k$ is a field and1263$$0\into{}V_1\into{}V_2\into{}\cdots\into{}V_N\into{}0$$1264is an exact sequence of finite dimensional vector spaces,1265then $\sum_{i=1}^N(-1)^i\dim{}V_i=0$.1266\end{lem}1267\begin{proof}1268Since every short exact sequence of vector spaces splits, the1269statement is true when $N=3$. If the statement is true for an1270exact sequence of length $N-1$ then, applying it to the exact sequence1271$$0\into{}V_2/V_1\into{}V_3\into\cdots\into{}V_N\into{}0,$$1272shows that1273$\dim{}V_2/V_1-\dim V_3+\cdots\pm\dim V_n = 0$1274from which the result follows.1275\end{proof}12761277\begin{lem}1278If $0\into\sF'\into\sF\into\sF''\into{}0$ is an exact sequence1279of coherent sheaves on $X$, then1280$$\chi(\F)=\chi(\F')+\chi(\F'').$$1281\end{lem}1282\begin{proof}1283Apply the above lemma to the long exact sequence of cohomology1284taking into account that $H^n(\F'')=0$ by Serre's vanishing theorem.1285\end{proof}12861287More generally, any map $\chi$ from an abelian category to1288$\Z$ is called additive if, whenever1289$$0\into\F^0\into\F^1\into\cdots\into\F^n\into{}0$$1290is exact, then1291$$\sum_{i=0}^{n}(-1)^{i}\chi(\F^i)=0.$$12921293{\bfseries Question.}1294Given an abelian category $\sA$ find an abelian1295group $A$ and a map $X:\sA\into{}A$ such that every1296additive function $\chi:\sA\into{}G$ factor through $\sA\xrightarrow{X}A$.1297In the category of coherent sheaves the Grothendieck group1298solves this problem.12991300Let $X=\P_k^n$ and suppose $\sF$ is a coherent sheaf on $X$.1301The Euler characteristic induces a map1302$$\Z\into\Z: n\mapsto\chi(\F(n)).$$1303\begin{thm}1304There is a polynomial $p_{\F}\in\Q[z]$ such that1305$p_{\F}(n)=\chi(\F(n))$ for all $n\in\Z$.1306\end{thm}1307The polynomial $p_{\F}(n)$ is called the Hilbert polynomial of $\F$.1308Last semester we defined the Hilbert polynomial of a graded module1309$M$ over the ring $S=k[x_0,\ldots,x_n]$. Define $\varphi_M:\Z\into\Z$1310by $\varphi_M(n)=\dim_k M_n$. Then we showed that there is a unique1311polynomial $p_M$ such that $p_M(n)=\varphi_M(n)$ for all $n\gg 0$.1312\begin{proof}1313We induct on $\dim(\supp\F)$. If $\dim(\supp\F)=0$ then $\supp\F$ is a1314union of closed points so $\sF=\oplus_{i=1}^{k}\F_{p_i}$.1315Since each $\F_{p_i}$ is a finite1316dimensional $k$-vector space and $\sox(n)$ is locally free,1317there is a non-canonical isomorphism $\F(n)=\F\tensor\sox(n)\isom\F$.1318Thus $$\chi_{\F}(n)=h^0(\F(n))=h^0(\F)=\sum_{i=1}^{k}\dim_k\F_{p_i}$$1319which is a constant function, hence a polynomial.13201321Next suppose $\dim(\supp\F)=s$.1322Let $x\in{}S_1=H^0(\sox(1))$ be such that the hyperplane1323$H:=\{x=0\}$ doesn't contain any irreducible component1324of $\supp\F$. Multiplication by $x$ defines a map1325$\sox(-1)\xrightarrow{x}\sox$ which is an isomorphism1326outside of $H$. Tensoring with $\F$ gives a map1327$\F(-1)\into\F$. Let $\sR$ be the kernel and1328$\sQ$ be the cokernel, then there is an exact sequence1329$$0\into\sR\into\F(-1)\xrightarrow{x}\F\into\sQ\into{}0.$$1330Now $\supp\sR\union\supp\sQ\subset\supp\F\intersect H$ so1331$\dim(\supp\sR)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ and1332$\dim(\supp\sQ)\leq\dim(\supp\F)\intersect H<\dim(\supp\F)$ so1333by our induction hypothesis $\chi(\sQ(n))$ and $\chi(\sR(n))$1334are polynomials. Twisting the above exact sequence by $n$ and1335applying $\chi$ yields1336$$\chi(\F(n))-\chi(\F(n-1))=\chi(\Q(n))-\chi(\sR(n))=P_{\sQ}(n)-P_{\sR}(N).$$1337Thus the first difference function of $\chi(\F(n))$ is a polynomial1338so $\chi(\F(n))$ is a polynomial.1339\end{proof}134013411342\begin{example}1343Let $X=\P^1$ and $\F=\sox$. Then $S=k[x_0,x_1]$, $M=S$ and1344$\dim S_n = n+1$. Thus $p_M(z)=z+1$ and $p_M(n)=\varphi(n)$1345for $n\geq -1$. Computing the Hilbert polynomial in terms of1346the Euler characteristic gives1347$$\chi(\F(n))=h^0(\sox(n))- h^1(\sox(n))1348= \begin{cases} (n+1) - 0 & n\geq -1\\13490-(-n-1)=n+1 & n\leq -21350\end{cases} $$1351Thus $p_{\F}(n)=n+1$.1352\end{example}13531354The higher cohomology corrects the failure of the1355Hilbert polynomial in lower degrees.13561357%% 2/16/961358\section{Correspondence between Analytic and Algebraic Cohomology}1359{\bf Homework. } Chapter III, 4.8, 4.9, 5.6.13601361Look at Serre's 1956 paper {\em Geometrie Algebraique et Geometrie1362Analytique} (GAGA). ``What are the prerequisites?'' asks Janos.1363``French,'' answers Nghi. ``Is there an English translation'' asks the1364class. ``Translation? ... What for? It's so beautiful in the French,''1365retorts Hartshorne.13661367Let $\F$ be a coherent sheaf on $\P^n_{\C}$ with its Zariski topology.1368Then we can associate to $\F$ a sheaf1369$\F^{\mbox{\rm an}}$1370on $\P^n_{\C}$ with its analytic topology. $\F$ is locally a cokernel1371of a morphism of free sheaves so we can define1372$\F^{\mbox{\rm an}}$1373by defining $\sox^{\mbox{\rm an}}$.1374The map1375$$\Coh(\P^n_{\C})\xrightarrow{\mbox{\rm an}}\Coh^{\mbox{\rm an}}(\P^n_{\C})$$1376is an equivalence of categories and1377$$H^i(X,\F)\iso{}H^i(X^{\mbox{\rm an}},\F^{\mbox{\rm an}})$$1378for all $i$.1379If $X/\C$ is affine the corresponding object $X^{\mbox{\rm an}}_{\C}$1380is a Stein manifold.13811382\section{Arithmetic Genus}1383Let $X\hookrightarrow{}\P_k^n$ be a projective variety with1384$k$ algebraically closed and suppose $\F$ is a coherent sheaf1385on $X$. Then $$\chi(\F)=\sum(-1)^i h^i(\F)$$ is the Euler characteristic1386of $\F$, $$P_{\F}(n)=\chi(\F(n))$$ gives the Hilbert polynomial of1387$\F$ on $X$, and $$p_a(X)=(-1)^{\dim X}(P_{\sox}(0)-1)$$ is the1388arithmetic genus of $X$. The arithmetic genus is independent1389of the choice of embedding of $X$ into $\P_k^n$.13901391If $X$ is a curve then $$1-p_a(X)=h^0(\sox)-h^1(\sox)$$.1392Thus if $X$ is an integral projective curve then $h^0(\sox)=1$ so1393$p_a(X)=h^1(\sox)$. If $X$ is a nonsingular projective curve1394then $p_a(X)=h^1(\sox)$ is called {\bfseries the genus} of $X$.13951396Let $V_1$ and $V_2$ be varieties, thus they are projective integral1397schemes over an algebraically closed field $k$. Then $V_1$ and1398$V_2$ are {\bfseries birationally equivalent} if and only if $K(V_1)\isom{}K(V_2)$1399over $k$, where $K(V_i)$ is the function field of $V_i$. $V$ is1400{\bfseries rational} if $V$ is bironational to $\P_k^n$ for some $n$.1401Since a rational map on a nonsingular projective curve always extends,1402two nonsingular projective curves are birational if and only if1403they are isomorphic. Thus for nonsingular projective curves1404the genus $g$ is a birational invariant.14051406\subsection{The Genus of Plane Curve of Degree $d$}1407Let $C\subset\P_k^2$ be a curve of degree $d$. Then $C$1408is a closed subscheme defined by a single homogeneous polynomial1409$f(x_0,x_1,x_2)$ of degree $d$, thus1410$$C=\proj(S/(f)).$$14111412Some possibilities when $d=3$ are:1413\begin{itemize}1414\item $f: Y^2-X(X^2-1)$, a nonsingular elliptic curve1415\item $f: Y^2-X^2(X-1)$, a nodal cubic1416\item $f: Y^3$, a tripled $x$-axis1417\item $f: Y(X^2+Y^2-1)$, the union of a circle and the $x$-axis1418\end{itemize}14191420Now we compute $p_a(C)$. Let $I=(f)$ with $\deg f=d$.1421Then $$1-p_a=h_0(\so_C)-h_1(\so_C)+h_2(\so_C)=\chi(\so_C).$$1422We have an exact sequence1423$$0\into\sI_C\into\so_{\P^2}\into\so_C\into 0.$$1424Now $\sI_C\isom\so_{\P^2}(-d)$ since $\so_{\P^2}(-d)$1425can be thought of as being generated by $1/f$ on $D_+(f)$1426and by something else elsewhere, and then multiplication1427by $f$ gives an inclusion1428$so_{\P^2}(-d)|_{D_+(f)}\into\so_{\P^2}|_{D_+(f)}$, etc.1429Therefore1430$$\chi(\so_C)=\chi(\so_{\P^2})-\chi(\so_{\P^2}(-d)).$$1431Now1432$$\chi(\so_{\P^2})=h^0(\so_{\P^2})-h^1(\so_{\P^2})+h^2(\so_{\P^2})=1+0+0$$1433and1434$$\chi(\so_{\P^2}(-d))=h^0(\so_{\P^2}(-d))-h^1(\so_{\P^2}(-d))+h^2(\so_{\P^2}(-d))1435=0+0+\frac{1}{2}(d-1)(d-2).$$1436For the last computation we used duality (14.1) to see that1437$$h^2(\so_{\P^2}(-d))=h^0(\so_{\P^2}(d-3)=\dim S_{d-3}1438=\frac{1}{2}(d-1)(d-2).$$1439Thus $\chi(\so_C)=1-\frac{1}{2}(d-1)(d-2)$ so1440$$p_a(C)=\frac{1}{2}(d-1)(d-2).$$14411442%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1443%% 2/21/961444%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%14451446\section{Not Enough Projectives}1447\begin{exercise} Prove that the category of quasi-coherent1448sheaves on $X=\P_k^1$ doesn't have enough projectives.1449\end{exercise}1450\begin{proof}1451We show that there is no projective object $\sP\in\Qco(X)$ along1452with a surjection $\sP\into\sox\into 0$.1453\begin{lem}1454If $\P\xrightarrow{\varphi}\sox$ is surjective and $\sP$ is1455quasi-coherent, then there exists $\ell$ such that1456$H^0(\P(\ell))\into{}H^0(\sox(\ell))$ is surjective.1457\end{lem}1458The {\em false} proof of this lemma is to write down an1459exact sequence $0\into\sR\into\sP\into\sox\into 0$ then1460use the ``fact'' that $H^1(\sR(\ell))=0$ for sufficiently1461large $\ell$. This doesn't work because $\sR$ might not1462be coherent since it is only the quotient of1463quasi-coherent sheaves. A valid way to proceed is to use1464(II, Ex. 5.15) to write $\sP$ as an ascending union of its coherent1465subsheaves, $\sP=\union_{i}\sP_{i}$.1466Then since $\varphi$ is surjective,1467$\sox=\union_{i}\varphi(\sP_{i})$, where $\varphi(\sP_{i})$1468is the sheaf image. Using the fact that $\varphi(\sP_{i})$ is1469the sheaf image, that $\sox$ is coherent and that the union1470is ascending, this implies $\sox=\varphi(\sP_i)$ for some $i$.1471We now have an exact sequence1472$$0\into\sR_i\into\sP_i\into\sox\into{}0$$1473with $\sR_i$ coherent since $\sP_i$ and $\sox$ are both coherent.1474Thus $H^i(\sR_i(\ell))=0$ for $l\gg 0$ which, upon computing1475the long exact sequence of cohomology, gives the lemma.14761477Now fix such an $\ell$. We have a commutative diagram1478$$\begin{CD}1479\[email protected]>>>\[email protected]>>>0\\1480@V{\exists}VV @VVV\\1481\sox(-\ell-1)@>>>k(p)@>>>01482\end{CD}$$1483Twisting by $\ell$ gives a commutative diagram1484$$\begin{CD}1485\sP(\ell)@>>>\sox(\ell)@>>>0\\1486@VVV @VVV\\1487\sox(-1)@>>>k(p)@>>>01488\end{CD}$$1489Let $s\in\Gamma(\sox(\ell))$ be a global section which is nonzero at $p$,1490then there is $t\in\Gamma(\sP(\ell))$ which maps to $s$.1491But then by commutativity $t$ must map to some element of $\Gamma(\sox(-1))=0$1492which maps to a nonzero element of $k(p)$, which is absurd.1493\end{proof}14941495\section{Some Special Cases of Serre Duality}1496\subsection{Example: $\sox$ on Projective Space}1497Suppose $X=\P_k^n$, then there is a perfect pairing1498$$H^0(\sox(\ell))\times{}H^n(\sox(-\ell-n-1))\into{}H^n(\sox(-n-1))\isom{}k.$$1499For this section let1500$$\omega_X=\sox(-n-1).$$1501Because the pairing is perfect we have a non-canonical but functorial1502isomorphism1503$$H^0(\sox(\ell))\isom H^n(\sox(-\ell-n-1))'.$$1504(If $V$ is a vector space then $V'$ denotes its dual.)15051506\subsection{Example: Coherent sheaf on Projective Space}1507Suppose $\sF$ is any coherent sheaf on $X=\P_k^r$.1508View $\Hom(\sF,\omega)$ as a $k$-vector space.15091510% WHY DEFINE SHEAF HOM???1511%We recall the definition of the sheaf Hom.1512%\begin{defn}1513%Let $\sF$ and $\sG$ be sheaves of $\sox$-modules.1514%Then $\sHom(\sF,\sG)$ is the sheaf which associates to1515%an open set the group1516%$$Hom_{\Mod(\sox|_U)}(\sF|_U,\sG|_U).$$1517%\end{defn}15181519By functoriality and since $H^n(\omega)=k$ there is a map1520$$\varphi:\Hom(\sF,\omega)\into\Hom(H^n(\sF),H^n(\omega))=H^n(\sF)'.$$1521\begin{prop} $\varphi$ is an isomorphism for all coherent sheaves $\sF$.1522\end{prop}1523\begin{proof}1524\par {\em Case 1.} If $\sF=\sox(\ell)$ for some $\ell\in\Z$ then this is1525just a restatement of the previous example.1526\par {\em Case 2.} If $\sE=\oplus_{i=1}^{k}\so(\ell_i)$ is a finite1527direct sum, then the statement follows from the1528commutativity of the following diagram.1529$$\begin{CD}1530\Hom(\oplus_{i=1}^{k}\so(\ell_i),\omega)@>>>H^n(\oplus_{i=1}^k\so(\ell_i))'\\1531@VV\isom{}V @VV\isom{}V\\1532\oplus_{i=1}^k\Hom(\so(\ell_i),\omega)@>>\sim{}>\oplus_{i=1}^k{}H^n(\so(\ell_i))'1533\end{CD}$$1534\par {\em Case 3.} Now let $\sF$ be an arbitrary coherent sheaf.1535View $\varphi$ as a morphism of functors1536$$\Hom(\cdot,\omega)\into H^n(\cdot)'.$$1537The functor $\Hom(\cdot,\omega)$ is contravarient left exact.1538$H^n(\cdot)$ is covariant right exact since $X=\P_k^n$ so1539$H^{n+1}(\sF)=0$ for any coherent sheaf $\sF$. Thus $H^n(\cdot)'$1540is contravarient left exact.1541\begin{lem}1542Let $\sF$ be any coherent sheaf. Then there exists a partial resolution1543$$\sE_1\into\sE_0\into\sF\into{}0$$1544by sheaves of the form $\oplus_{i}\sox(\ell_i)$.1545\end{lem}1546By (II, 5.17) for $\ell\gg 0$, $\sF(\ell)$ is1547generated by its global sections. Thus there is a surjection1548$$\sox^m\into\sF(\ell)\into 0$$1549which upon twisting by $-\ell$ becomes1550$$\sE_0=\sox(-\ell)^m\into\sF\into 0.$$1551Let $\sR$ be the kernel so1552$$0\into\sR\into\sE\into\sF$$1553is exact. Since $\sR$ is coherent, we can repeat the argument1554above to find $\sE_1$ surjecting onto $\sR$. This yields1555the desired exact sequence.15561557Now we apply the functors $\Hom(\cdot,\omega)$ and1558$H^n(\cdot)'$. This results in a commutative diagram1559$$\begin{CD}15600@>>> \Hom(\sF,\omega) @>>> \Hom(\sE_0,\omega) @>>> \Hom(\sE_1,\omega)\\1561@VVV @V\varphi(\sF)VV @V\varphi(\sE_0)VV @VV\varphi(\sE_1)V\\15620@>>> H^n(\sF)'@>>>H^n(\sE_0)' @>>> H^n(\sE_1)'1563\end{CD}$$1564From cases 1 and 2, the maps $\varphi(\sE_0)$ and1565$\varphi(\sE_1)$ are isomorphisms so $\varphi(\sF)$ must also be1566an isomorphism.1567\end{proof}15681569\subsection{Example: Serre Duality on $\P_k^n$}1570Let $X=\P_k^n$ and $\sF$ be a coherent sheaf.1571Then for each $i$ there is an isomorphism1572$$\varphi^{i}:\ext^{i}_{\sox}(\sF,\omega)\into H^{n-i}(\sF)'.$$15731574%%%%%%%%%%%%%%%%%%%%1575%% 2/23/9615761577\section{The Functor $\ext$}1578Let $(X,\sox)$ be a scheme and $\sF,\sG\in\Mod(\sox)$. Then1579$\Hom(\sF,\sG)\in\Ab$. View $\Hom(\sF,\bullet)$ as a1580functor $\Mod(\sox)\into\Ab$. Note that $\Hom(\sF,\bullet)$1581is left exact and covariant. Since $\Mod(\sox)$ has enough1582injectives we can take derived functors.1583\begin{defn}1584The $\ext$ functors $\ext^i_{\sox}(\sF,\bullet)$ are the right1585derived functors of $\Hom_{\sox}(\sF,\bullet)$ in the1586category $\Mod(\sox)$.1587\end{defn}1588Thus to compute $\ext^i_{\sox}(\sF,\sG)$, take an injective1589resolution $$0\into\sG\into I^0\into I^1\into \cdots $$1590then1591$$\ext^i_{\sox}(\sF,\sG)=H^i(\Hom_{\sox(\sF,I^{\bullet})}).$$15921593\begin{remark} {\bfseries Warning!}1594If $i:X\hookrightarrow\P^n$ is a closed subscheme of $P^n$ then1595$\ext^i_{\sox}(\sF,\sG)$ need {\em not} equal1596$\ext^i_{\P^n}(i_{*}(\sF),i_{*}(\sG))$. With cohomology1597these are the same, but not with $\ext$!1598\end{remark}15991600\begin{example}1601Suppose $\sF=\sox$, then1602$\Hom_{\sox}(\sox,\sG)=\Gamma(X,\sG)$. Thus1603$\ext^i_{\sox}(\sox,\bullet)$ are the derived1604functors of $\Gamma(X,\bullet)$ in $\Mod(\sox)$.1605Since we can computer cohomology using flasque1606sheaves this implies $\ext^i_{\sox}(\sox,\bullet)=H^i(X,\bullet)$.1607Thus $\ext$ generalizes $H^i$ but we get a lot more besides.1608\end{example}16091610\subsection{Sheaf $\ext$}1611Now we define a new kind of $\ext$. The sheaf hom functor1612$$\sHom_{\sox}(\sF,\bullet):\Mod(\sox)\into\Mod(\sox)$$1613is covariant and left exact. Since $\Mod(\sox)$ has enough1614injectives we can defined the derived functors1615$\sext^i_{\sox}(\sF,\bullet)$.16161617\begin{example}1618Consider the functor $\ext^i_{\sox}(\sox,\bullet)$.1619Since $\sHom_{\sox}(\sox,\sG)=\sG$ this is the identity1620functor which is exact so1621$$\ext^i_{\sox}(\sox,\sG)=\begin{cases}\sG\quad i=0\\0\quad i>0\end{cases}$$1622\end{example}16231624What if we have a short exact sequence in the first variables, do we1625get a long exact sequence?1626\begin{prop}1627The functors $\ext^i$ and $\sext^i$ are $\delta$-functors1628in the first variable. Thus if $$0\into\sF'\into\sF\into\sF''\into 0$$1629is exact then there is a long exact sequence1630\begin{align*}0\into&\Hom(\sF'',\sG)\into\Hom(\sF,\sG)\into\Hom(\sF',\sG)1631\into&\ext^1(\sF'',\sG)\into\ext^1(\sF,\sG)\into\ext^1(\sF,\sG)\into\cdots1632\end{align*}1633\end{prop}1634The conclusion of this proposition is not obvious because we1635the $\ext^i$ as derived functors in the second variable, not the first.1636\begin{proof}1637Suppose we are given $0\into\sF'\into\sF\into\sF''\into 0$ and $\sG$.1638Choose an injective resolution $0\into\sG\into{}I^{\bullet}$ of $\sG$.1639Since $\Hom(\bullet,I^n)$ is exact (by definition of injective object),1640the sequence1641$$0\into\Hom(\sF'',I^{\bullet})\into\Hom(\sF,I^{\bullet})\into1642\Hom(\sF',I^{\bullet})\into 0$$1643is exact. By general homological algebra these give rise a long1644exact sequence of cohomology of these complexes. For $\sext^i$1645simply scriptify everything!1646\end{proof}16471648\subsection{Locally Free Sheaves}1649\begin{prop}1650Suppose $\sE$ is a locally free $\sox$-module of finite rank.1651Let $\sE^{\dual}=\Hom(\sE,\sO)$. For any sheaves $\sF$, $\sG$,1652$$\ext^i(\sF\tensor\sE,\sG)\isom\ext^i(\sF,\sG\tensor\sE^{\dual})$$1653and1654$$\sext^i(\sF\tensor\sE,\sG)\isom\sext^i(\sF,\sG\tensor\sE^{\dual})1655\isom\sext^i(\sF,\sG)\tensor\sE^{\dual}.$$1656\end{prop}16571658\begin{lem}1659If $\sE$ is locally free of finite rank and $\sI\in\Mod(\sox)$ is1660injective then $\sI\tensor\sE$ is injective.1661\end{lem}1662\begin{proof}1663Suppose $0\into\sF\into\sG$ is an injection and there is a map1664$\varphi:\sF\into\sI\tensor\sE$. Tensor everything with $\sE^{\dual}$.1665Then we have an injection $0\into\sF\tensor\sE^{\dual}\into\sG\tensor\sE^{\dual}$1666and a map $\varphi':\sF\tensor\sE^{\dual}\into\sI$. Since $\sI$ is injective1667there is a map $\sG\tensor\sE^{\dual}\into\sI$ which makes1668the appropriate diagram commute. Tensoring everything with $\sE$ gives1669a map making the original diagram commute.1670\end{proof}16711672\begin{proof}[of proposition] Let $0\into\sG\into\sI^{\bullet}$1673by an injective resolution of $\sG$. Since1674$$\Hom(\sF\tensor\sE,\sI^{\bullet})=\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual}),$$1675we see that1676$$0\into\sG\tensor\sE^{\dual}\into\sI^{\cdot}\tensor\sE{\dual}$$1677is an injective resolution of $\sG\tensor\sE^{\dual}$.1678Thus $\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual})$ computes1679$\ext(\sF\tensor\sE,\bullet)$.1680\end{proof}16811682\begin{prop}1683If $\sF$ has a locally free resolution1684$\sE_{\cdot}\into\sF\into 0$ then1685$$\sext^i_{\sox}(\sF,\sG)=H^i(\sHom(\sE_{\bullet},\sG)).$$1686\end{prop}16871688\begin{remark}1689Notice that when $i>0$ and $\sE$ is locally free,1690$$\sext^i(\sE,\sG)=\sext^i(\sox,\sG\tensor\sE^{\dual})=0.$$1691\end{remark}16921693\begin{proof}1694Regard both sides as functors in $\sG$. The left hand side is a1695$\delta$-functor and vanishes for $\sG$ injective. I claim that1696that right hand side is also a $\delta$-functor and vanishes for1697$\sG$-injective.1698\begin{lem}1699If $\sE$ is locally free, then $\shom(\sE,\bullet)$ is exact.1700\end{lem}1701\end{proof}17021703%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1704%% 2/26/961705\section{More Technical Results on $\ext$}1706Let $(X,\sox)$ be a scheme and $\sF,\sG$ be sheaves1707in the category $\Mod(\sox)$. Then $\ext(\sF,\sG)$1708and $\sext(\sF,\sG)$ are the derived functors of $\Hom$, resp.1709$\shom$, in the second variable.17101711\begin{lem}1712If $\sF$ and $\sG$ are coherent over a Noetherian scheme1713$X$, then $\sext^{i}(\sF,\sG)$ is coherent.1714\end{lem}1715This lemma would follow immediately from the following fact1716which we haven't proved yet.1717\begin{fact}1718Let $X=\spec A$ with $A$ Noetherian and let $M$ be an $A$-module.1719Then $$\ext_{\sox}^i(\tilde{M},\tilde{N})=\ext^i_{A}(M,N)$$ and1720$$\sext_{\sox}^i(\tilde{M},\tilde{N})=(\ext^i_{A}(M,N))^{\tilde{}}.$$1721\end{fact}1722Instead of using the fact we can prove the lemma using1723a proposition from yesterday.1724\begin{proof}1725Choose a locally free resolution1726$$\sL_{\bullet}\into\sF\into 0$$1727of $\sF$. Then1728$$\sext^{i}(\sF,\sG)=H^i(\shom(\sL_{\bullet},\sG)).$$1729But all of the kernels and cokernels in1730$\shom(\sL_{\bullet},\sG)$1731are coherent, so the cohomology is. (We can't1732just choose an injective resolution of $\sF$ and apply1733the definitions because there is no guarantee that we1734can find an injective resolution by coherent sheaves.)1735\end{proof}17361737\begin{prop}1738Let $X$ be a Noetherian projective scheme over $k$ and let $\sF$1739and $\sG$ be coherent on $X$. Then for each $i$ there exists1740an $n_0$, depending on $i$, such that for all $n\geq{}n_0$,1741$$\ext^i_{\sox}(\sF,\sG(n))=\Gamma(\sext^i_{\sox}(\sF,\sG(n))).$$1742\end{prop}1743\begin{proof}1744When $i=0$ the assertion is that1745$$\Hom(\sF,\sG(n))=\Gamma(\shom(\sF,\sG(n)))$$1746which is obvious.17471748{\em Claim.} Both sides are $\delta$-functors in $\sF$. We1749have already showed this for the left hand side. [I don't understand1750why the right hand side is, but it is not trivial and it caused1751much consternation with the audience.]17521753To show the functors are isomorphic we just need to show both1754sides are coeffaceable. That is, for every coherent sheaf $\sF$1755there is a coherent sheaf $\sE$ and a surjection $\sE\into\sF\into{}0$1756such that $\ext^i_{\sox}(\sE)=0$ and similarly for the right hand1757side. Thus every coherent sheaf is a quotient of an acyclic sheaf.17581759Suppose $\sF$ is coherent. Then for $\ell\gg 0$, $\sF(\ell)$ is generated1760by its global sections, so there is a surjection1761$$\sox^a\into\sF(\ell)\into{}0.$$1762Untwisting gives a surjection1763$$\sox(-\ell)^a\into\sF\into{}0.$$1764Let $\sE=\sox(-\ell)^a$, then I claim that $\sE$ is acyclic for both1765sides. First consider the left hand side. Then1766\begin{align*}\ext^i(\oplus\so(-\ell),\sG(n))&=\oplus\ext^i(\sox(-\ell),\sG(n))\\1767&=\oplus\ext^i(\sox,\sG(\ell+n))\\1768&=H^i(X,\sG(\ell+n))\end{align*}1769By Serre (theorem 5.2 of the book) this is zero for $n$ sufficiently large.1770For the right hand side the statement is just that1771$$\sext^i(\sE,\sG(n))=0$$1772which we have already done since $\sE$ is a locally free sheaf.17731774Thus both functors are universal since they are coeffaceable. Since universal1775functors are completely determined by their zeroth one they must be equal.1776\end{proof}17771778\begin{example}1779One might ask if $\ext^i$ necessarily vanishes for sufficiently large $i$.1780The answer is no. Here is an algebraic example which can be converted1781to a geometric example. Let $A=k[\varepsilon]/(\varepsilon^2)$, then1782a projective resolution $L_{\bullet}$ of $k$ is1783$$\cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A1784\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}k\into 0.$$1785Then $Hom(L_{\bullet},k)$ is the complex1786$$k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}\cdots$$1787Thus $\ext^i_A(k,k)=k$ for all $i\geq 0$.1788\end{example}17891790\section{Serre Duality}1791We are now done with technical results on $\ext$'s so we can1792get back to Serre duality on $\P^n$.1793Let $X=\P^n_k$ and let $\omega=\sox(-n-1)$. Note that this1794is an {\em ad hoc} definition of $\omega$ which just happens1795to work since $X=\P^n_k$. In the more general situation it1796will be an interesting problem just to show the so called1797dualizing sheaf $\omega$ actually exists. When our variety1798is nonsingular, $\omega$ will be the canonical sheaf.1799We have shown that for any coherent sheaf $\sF$ there is a map1800$$\Hom(\sF,\omega)\iso H^n(\sF)^{\dual}.$$1801The map is constructed by using the fact that $H^n$ is a1802functor:1803$$\Hom(\sF,\omega)\into\Hom_k(H^n(\sF),H^n(\omega))1804= \Hom_k(H^n(\sF),k)=H^n(\sF)^{\dual}.$$18051806We shall use satellite functors to prove the following theorem.1807\begin{thm} Let $\sF$ be a coherent sheaf on $\P^n_k$. Then1808there is an isomorphism $$\ext^i(\sF,\omega)\iso H^{n-i}(\sF)^{\dual}.$$1809\end{thm}1810\begin{proof} Regard both sides as functors in $\sF$.1811\par {\em 1. Both sides are $\delta$-functors in $\sF$.}1812We have already checked this for $\ext^i$. Since $H^{n-i}$1813is a delta functor in $\sF$, so is $(H^{n-i})^{\dual}$.1814Note that both sides are contravarient.1815\par {\em 2. They agree for $i=0$.} This was proved last time.1816\par {\em 3. Now we just need to show both sides are coeffaceable.}1817Suppose $\sE\into\sF\into 0$ with $\sE=\sO(-\ell)^{\oplus a}$.1818For some reason we can assume $\ell\gg{}0$.1819We just need to show both sides vanish1820on this $\sE$. First computing the left hand side gives1821$$\oplus\ext^i(\sO(-\ell),\omega)=H^i(\omega(\ell))=0$$1822for $\ell\gg 0$. Next computing the right hand side we get1823$$H^{n-i}(\sO(-\ell))=0$$1824by the explicit computations of cohomology of projective1825space (in particular, note that $i>0$).1826\end{proof}18271828Next time we will generalize Serre duality to an arbitrary1829projective scheme $X$ of dimension $n$.1830We will proceed in two steps. The first is to ask,1831what is $\omega_X$? Although the answer to this question1832is easy on $\P_k^n$ it is not obvious what the suitable1833analogy should be for an arbitrary projective variety.1834Second we will define natural maps1835$$\ext^i_{\sox}(\sF,\omega)\xrightarrow{\varphi^i}H^{n-i}(\sF)^{\dual}$$1836where $n=\dim X$.1837Unlike in the case when $X=\P^n_k$, these maps are not necessarily1838isomorphisms unless $X$ is locally Cohen-Macaulay (the local rings1839at each point are Cohen-Macaulay).1840\begin{defn}1841Let $A$ be a nonzero Noetherian local ring with residue field $k$.1842Then the {\bfseries depth} of $A$ is1843$$\operatorname{depth} A = \inf\{i:\ext^i_A(k,A)\neq 0\}.$$1844$A$ is said to be {\bfseries Cohen-Macaulay} if1845$\operatorname{depth} A = \dim A$.1846\end{defn}18471848%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%1849%% 2/28/9618501851\section{Serre Duality for Arbitrary Projective Schemes}1852Today we will talk about Serre duality for an arbitrary projective1853scheme. We have already talked about Serre duality in the special1854case $X=\P^n_k$. Let $\sF$ be a locally free sheaf. We showed1855there is an isomorphism1856$$\ext^i(\sF,\omega_X)\iso H^{n-i}(\sF)^{\dual}.$$1857This was established by noting that1858$$\ext^i(\sF,\omega_X)=\ext^i(\sox,\sF^{\dual}\tensor\omega_X)1859=H^i(\sF^{\dual}\tensor\omega_X).$$1860Another thing to keep in mind is that locally free sheaves correspond1861to what, in other branches of mathematics, are vector bundles. They1862aren't the same object, but there is a correspondence.18631864We would like to generalize this to an arbitrary projective scheme $X$.1865There are two things we must do.1866\begin{enumerate}1867\item Figure out what $\omega_X$ is.1868\item Prove a suitable duality theorem.1869\end{enumerate}1870When $X=\P_k^n$ it is easy to find a suitable $\omega_X=\so_{\P^n_k}(-n-1)$1871because of the explicit computations we did before. We now define $\omega_X$1872to be a sheaf which will do what we hope it will do. Of course existence1873is another matter.1874\begin{defn}1875Let $X$ be a Noetherian scheme of finite type over a field $k$ and1876let $n=\dim X$. Then a {\bfseries dualizing sheaf} for $X$ is1877a coherent sheaf $\omega_X$ along with a map1878$t:H^n(\omega_X)\into{}k$, such that for all coherent sheaves $\sF$ on1879$X$, the map1880$\Hom_{\sox}(\sF,\ox)\into{}H^n(\sF)^{\dual}$ is an1881isomorphism. The latter map is defined by the diagram1882$$\begin{array}{ccc}1883\Hom_{\sox}(\sF,\ox)\\1884\downarrow\\1885\Hom(H^n(\sF),H^n(\ox))&\xrightarrow{t}&\Hom(H^n(\sF),k)=H^n(\sF)^{\dual}1886\end{array}$$1887\end{defn}1888Strictly speaking, a dualizing sheaf is a pair $(\omega_X,t)$. Note1889that on $\P^n$ we had $H^n(\omega_{\P^n})\isom{}k$, but on an1890arbitrary scheme $X$ we only have a map from $H^n(\omega_X)$ to $k$1891which need not be an isomorphism. The definition never mentions existence.18921893\begin{prop}1894If $X$ admits a dualizing sheaf $(\ox,t)$ then the pair1895$(\ox,t)$ is unique up to unique isomorphism, i.e., if1896$(\eta,s)$ is another dualizing sheaf for $X$ then there1897is a unique isomorphism $\varphi:\ox\into\eta$ such that1898$$\begin{array}{ccc}1899H^n(\ox)&\xrightarrow{H^n(\varphi)}&H^n(\eta)\\1900t\downarrow&&\downarrow{}s\\1901k&=&k\end{array}$$1902commutes.1903\end{prop}1904Before we prove the proposition we make a short digression to1905introduce representable functors which give a proof of the uniqueness1906part of the above proposition.1907\begin{defn}1908Let $\sC$ be a category and $\sD$ a category whose objects happen1909to be sets. Suppose $T:\sC\into\sD$ is a contravarient functor.1910Then $T$ is {\bfseries representable} if there exists an object1911$\omega\in\Ob(\sC)$ and an element $t\in T(\omega)$ such that1912for all $F\in\Ob(\sC)$ the map $\Hom_{\sC}(F,\omega)\into{}T(F)$1913is a bijection of sets. The latter map is defined by the diagram1914$$\begin{array}{ccc}1915\Hom_{\sC}(F,\omega)&\xrightarrow{\text{bijection of sets}}&T(F)\\1916\searrow&&\nearrow\text{ evaluation at $t$}\\1917&\Hom_{\sD}(\Gamma(\omega),T(F))\end{array}$$1918\end{defn}1919Thus there is an isomorphism of functors $\Hom(\bullet,\omega)=T(\bullet)$.1920The pair $(t,\omega)$ is said to represent the functor $T$.1921The relevant application of this definition is to the case1922when $\sC=\Coh(X)$, $\sD=\{\text{ $k$-vector spaces}\}$,1923$T$ is the functor $F\mapsto H^n(\sF)^{\dual}$. Then $\omega=\ox$1924and1925$$t=t\in\Hom(H^n(\omega),k)=H^n(\omega)^{\dual}=T(\omega).$$1926\begin{prop}1927If $T$ is a representable functor, then the pair1928$(\omega,t)$ representing it is unique.1929\end{prop}1930\begin{proof}1931Suppose $(\omega,t)$ and $(\eta,s)$ both represent the functor $T$.1932Consider the diagram1933$$\begin{array}{cccl}1934\Hom(\eta,\omega)&\xrightarrow{T}&\Hom(T(\omega),T(\eta))\\1935\searrow&&\swarrow\text{ eval. at $t$}\\1936&T(\eta)\end{array}$$1937By definition the map $\Hom(\eta,\omega)\into{}T(\eta)$ is1938bijective. Since $s\in T(\eta)$, there is $\varphi\in\Hom(\eta,\omega)$1939such that $\varphi\mapsto{}s\in{}T(\eta)$. Thus $\varphi$ has1940the property that $T(\varphi)(t)=s$. This argument uses the1941fact that $(\omega,t)$ represents $T$. Using the fact that1942$(\eta,s)$ represents $T$ implies that there exists1943$\psi\in\Hom(\omega,\eta)$ such that $T(\psi)(s)=t$.1944We have the following pictures1945$$\begin{array}{ccc}1946&\xrightarrow{\hspace{.2in}\psi\hspace{.2in}}\\1947\omega&&\eta\\1948&\xleftarrow{\hspace{.2in}\phi\hspace{.2in}}1949\end{array}$$1950$$\begin{array}{ccc}1951&\xleftarrow{\hspace{.2in}T(\psi)=\psi^{*}\hspace{.2in}}\\1952t\in{} T(\omega)&&T(\eta) \ni{} s\\1953&\xrightarrow{\hspace{.2in}T(\phi)=\phi^{*}\hspace{.2in}}1954\end{array}$$1955I claim that1956$$\psi\circ\varphi=\Id\in\Hom(\eta,\eta).$$1957In diagram form we have1958$$\eta\xrightarrow{\varphi}\omega\xrightarrow{\psi}\eta$$1959which upon applying $T$ gives1960\begin{align*}1961T(\eta)&\xrightarrow{\psi^{*}}T(\omega)\xrightarrow{\varphi^{*}}T(\eta)\\1962s&\mapsto t\mapsto s\end{align*}1963Where does $\psi\circ\varphi$ go to1964under the map $\Hom(\eta,\eta)\iso T(\eta)$? By definition1965$\psi\circ\varphi$ goes to the evaluation of $T(\psi\circ\varphi)$1966at $s\in T(\eta)$. But, as indicated above, the evaluation of1967$T(\psi\circ\varphi)$ at $s$ is just $s$ again. But the identity1968morphism $1_{\eta}\in\Hom(\eta,\eta)$ also maps to $s$1969under the map $\Hom(\eta,\eta)\iso T(\eta)$. Since this1970map is a bijection this implies that $\psi\circ\varphi=1_{\eta}$,1971as desired. Similarly $\varphi\circ\psi=1_{\omega}$. Thus1972$\psi$ and $\varphi$ are both isomorphisms.1973\end{proof}1974\begin{quote}1975``When you define something and it is unique up to unique isomorphism,1976you know it must be good.''1977\end{quote}1978We return to the question of existence.1979\begin{prop}1980If $X$ is a projective scheme over a field $k$ then $(\ox,t)$ exists.1981\end{prop}1982\begin{lem}1983If $X$ is an $n$ dimensional projective scheme over a field $k$, then1984there is a finite morphism $f:X\into\P^n_k$.1985\end{lem}1986\begin{proof}1987Embed $X$ in $\P^N$ then choose a linear projection down to1988$\P^n$ which is sufficiently general.1989$$\begin{array}{ccc}1990X&\hookrightarrow&\P^N\\1991f\searrow&&\downarrow\\1992&&\P^n\end{array}$$1993Let $L$ be a linear space of dimension $N-n-1$ not meeting $X$.1994Let the map from $\P^N\into\P^n$ be projection through $L$.1995By construction $f$ is quasi-finite, i.e., for all $Q\in\P^n$,1996$f^{-1}(Q)$ is finite. It is a standard QUALIFYING EXAM problem to1997show that if a morphism is quasi-finite and projective then it is1998finite. This can be done by applying (II, Ex. 4.6) by covering1999$X$ by subtracting off hyperplanes and noting that the correct2000things are affine by construction. See also (III, Ex. 11.2) for2001the more general case when $f$ is quasi-finite and proper, but not2002necessarily projective.2003\end{proof}20042005%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2006%% 3/1/962007%%%%%%%%%%%%%%%%%%%%%%%%2008\section{Existence of the Dualizing Sheaf on a Projective Scheme}2009Let $X$ be a scheme over $k$. Recall that a2010dualizing sheaf is a pair $(\omega,t)$ where $\omega$ is a coherent2011sheaf on $X$ and2012$$t:H^n(X,\omega)\into k$$2013is a homomorphism such that for all coherent sheaves $\sF$2014the natural map2015$$\Hom_X(\sF,\omega)\into{}H^n(\sF)^{\dual}$$2016is an isomorphism.2017We know that such a dualizing sheaf exists on $\P_k^n$.2018\begin{thm}2019If $X$ is a projective scheme of dimension $n$ over $k$, then2020$X$ has a dualizing sheaf.2021\end{thm}2022The book's proof takes an embedding2023$j:X\hookrightarrow{}\P_k^N$ and works on $X$2024as a subscheme of $\P_k^N$. Then the book's proof shows that2025$$\omega_X=\sext_{\sO_{\P^N_k}}^{N-n}(\sox,\omega_{\P^N_k}).$$2026Today we will use a different method.2027\begin{defn}2028A {\bfseries finite morphism} is a morphism $f:X\into Y$ of2029Noetherian schemes such that for any open affine $U=\spec A\subset X$,2030the preimage $f^{-1}(U)\subset Y$ is affine, say $f^{-1}(U)=\spec B$,2031and the natural map $A\into B$ turns $B$ into a finitely2032generated $A$-module. We call $f$ an {\bfseries affine morphism}2033if we just require that $f^{-1}(U)$ is affine2034but not that $B$ is a finitely generated $A$-module.2035A morphism $f:X\into{}Y$ is {\bfseries quasi-finite} if for all2036$y\in Y$ the set $f^{-1}(y)$ is finite.2037\end{defn}2038\begin{example}2039Consider the morphism2040$$j:\P^1-\{\pt\}\injects\P^1.$$2041Since $\P^1$ minus any nonempty finite set of points is2042affine $j$ is affine. But it is not finite. Indeed, let2043$a$ be a point different from $\pt$ and let $U=\P^1-\{a\}$.2044Then $U=\spec k[x]$ and2045$$j^{-1}(U)=\P^1-\{\pt,a\}=\spec k[x,x^{-1}],$$2046but $k[x,x^{-1}]$ is not a finitely generated $k[x]$-module.2047\end{example}2048\begin{exercise}2049A morphism can be affine but not finite or even quasi-finite.2050For example, let $f$ be the natural map2051$$f:\A^{n+1}-\{0\}\into\P^n$$2052then show that $f$ is affine.2053This is the fiber bundle associated to the invertible2054sheaf $\sO(1)$ [[or is it $\sO(-1)$?]]2055\end{exercise}20562057\subsection{Relative Gamma and Twiddle}2058We will now define relative versions of global sections and2059$\tilde{}$ analogous to the absolute versions.2060It is not a generalization of the absolute notion, but a2061relativization.2062Suppose $X$ is a scheme over $Y$ with structure2063map $f:X\into{}Y$ and assume $f$ is affine.2064Then the map sending a sheaf2065$\sF$ on $X$ to the sheaf $f_{*}\sF$ on2066$Y$ is the analog of taking global sections.2067Since $f$ is a morphism there is a map $\soy\into{}f_{*}\sox$2068so $f_{*}\sox$ is a sheaf of $\soy$-modules. Note that2069$f_{*}\sF$ is a sheaf of $f_{*}\sox$-modules. Thus2070we have set up a map2071$$\Qco(X)\into\{\text{ quasicoherent $f_{*}\sox$-modules on $Y$ }\}.$$2072The next natural thing to do is define a map analogous to2073$\tilde{}$ which goes the other direction.2074Suppose $\sG$ is a quasi coherent sheaf of $f_{*}\sox$-modules on $Y$.2075Let $U\subset Y$ be an affine open subset of $Y$. Let $G=\Gamma(U,\sG)$ and2076write $U=\spec A$. Then since $f$ is an affine morphism,2077$f^{-1}(U)=\spec B$ where $B=\Gamma(f^{-1}(U),\sox)$.2078Since $\sG$ is an $f_{*}\sox$-module, and $f_{*}\sox$ over $U$2079is just $B$ thought of as an $A$-module, we see that $G$ is a $B$-module.2080Thus we can form the sheaf $\tilde{G}$ on $\spec B=f^{-1}(U)$.2081Patching the various sheaves $\tilde{G}$ together as $U$ runs through2082an affine open cover of $Y$ gives a sheaf $\tilde{\sG}$ in2083$\Qco(X)$.20842085Let $\sG$ be a quasi-coherent sheaf of $\soy$-modules. We can't2086take $\tilde{}$ of $\sG$ because $\sG$ might not be a sheaf of2087$f_{*}\sox$-modules. But2088$\shom_{\soy}(f_{*}\sox,\sG)$2089is a sheaf of $f_{*}\sox$-modules, so we can2090form $(\shom_{\soy}(f_{*}\sox,\sG))^{\tilde{}}$. This2091is a quasi-coherent sheaf on $X$ which we denote $f^{!}(\sG)$.2092\begin{prop}2093Suppose $f:X\into{}Y$ is an affine morphism of Noetherian2094schemes, $\sF$ is coherent on $X$, and $\sG$ is quasi-coherent on $Y$.2095Then2096$$f_{*}\shom_{\sox}(\sF,f^{!}\sG)\isom \shom_{\soy}(f_{*}\sF,\sG)$$2097and passing to global sections gives an isomorphism2098$$\Hom(\sF,f^{!}\sG)\isom \Hom(f_{*}\sF,\sG).$$2099Thus $f^{!}$ is a right adjoint for $f_{*}$.2100\end{prop}2101\begin{proof}2102The {\em natural} map is2103\begin{align*}f_{*}\shom_{\sox}(\sF,f^{!}\sG)&\into2104\shom_{\soy}(f_{*}\sF,f_{*}f^{!}\sG)\\2105&=\shom_{\soy}(f_{*}\sF,\shom_{\soy}(f_*\sox,\sG)) \into2106\shom_{\soy}(f_{*}\sF,\sG)\end{align*}2107where the map $\shom_{\soy}(f_{*}\sox,\sG)\into\sG$2108is obtained obtained by evaluation at $1$.2109Since the question is local we may assume2110$Y=\spec A$ and $X=\spec B$. Then $\sF$ corresponds2111to a finitely generated module $M$ over the Noetherian ring $B$2112and $\sG$ corresponds to a module $N$ over $A$.2113We must show that2114$$\Hom_B(M,\Hom_A(B,N))\isom\Hom_A(M,N).$$2115When $M$ is free over $B$ so that $M=B^{\oplus{}n}$ the equality holds.2116As functors in $M$, both sides are contravarient and2117left exact. Now suppose $M$ is an arbitrary finitely generated2118$B$-module. Write $M$ as a2119quotient $F_0/F_1$ where $F_0$ and $F_1$ are both free of2120finite rank. Applying each of2121