Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
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2%% Notes for Hartshorne's Algebraic Geometry course
3%% William Stein
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131\theoremstyle{plain}
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141\theoremstyle{definition}
142\newtheorem{defn}[thm]{Definition}
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144\theoremstyle{remark}
145\newtheorem{remark}[thm]{Remark}
146\newtheorem{exercise}[thm]{Exercise}
147\newtheorem{example}[thm]{Example}
148
149
150\author{William A. Stein}
151\title{Notes for Algebraic Geometry II}
152
153\begin{document}
154\maketitle
155\tableofcontents
156
157\section{Preface}
158{\bfseries Read at your own risk!}
159
160These are my {\em very rough, error prone} notes of
161a second course on algebraic geometry
162offered at U.C. Berkeley in the Spring of 1996.
163The instructor
164was Robin Hartshorne and the students were Wayne Whitney,
165William Stein, Matt Baker, Janos Csirik, Nghi Nguyen, and Amod.
166I wish to thank Robin Hartshorne for giving this course
167and to Nghi Nguyen for his insightful suggestions and corrections.
168Of course all of the errors are solely my responsibility.
169
170The remarks in brackets [[like this]] are notes that
171I wrote to myself. They are meant as a warning or as a reminder
172of something I should have checked but did not have time for. You
173may wish to view them as exercises.
174
175If you have suggestions, questions, or comments feel free to write to me.
176My email address is {\tt was@math.berkeley.edu}.
177
178% Day 1, 1/17/96
179\section{Ample Invertible Sheaves}
180
181Let $k$ be an algebraically closed field
182and let $X$ be a scheme over $k$.  Let $\phi:X\into \bP^n_k$ be a morphism.
183Then to give $\phi$ is equivalent to giving an invertible sheaf
184$\sL$ on $X$ and sections $s_0,\ldots,s_n\in\Gamma(X,\sL)$
185which generate $\sL$. If $X$ is projective (that is, if there
186is some immersion of $X$ into {\em some} $\bP^m_k$) then $\phi$
187is a closed immersion iff $s_0,\ldots,s_n$ separate points and tangent
188vectors.
189
190\begin{defn}
191Let $X$ be a scheme and $\sL$ an invertible sheaf on $X$.
192Then we say $\sL$ is {\em very ample} if there is an immersion
193$i:X\hookrightarrow \bP_k^n$ such that $\sL\isom i^{*}\sO(1)$.
194\end{defn}
195
196\begin{thm}
197Let $X$ be a closed subscheme of
198$\bP_k^n$
199and
200$\sF$ a coherent sheaf on $X$, then
201$\sF(n)$ is generated by global sections for
202all $n\gg 0$.
203\end{thm}
204
205\begin{cor}
206Let $X$ be any scheme and $\sL$ a very ample coherent sheaf
207on $X$, then for all $n\gg 0$, $\sF\tensor \sL^{\tensor n}$
208is generated by global sections.
209\end{cor}
210
211\begin{defn}
212Let $X$ be a Noetherian scheme and $\sL$ be an invertible sheaf.
213We say that $\sL$ is {\em ample} if for every coherent sheaf
214$\sF$ on $X$, there is $n_0$ such that for all $n\geq n_0$,
215$\sF \tensor \sL^{\tensor n}$ is generated by its global sections.
216\end{defn}
217
218Thus the previous corollary says that a very ample invertible
219sheaf is ample.
220
221\begin{prop}
222Let $X$ and $\sL$ be as above. Then the following are equivalent.
223
2241) $\sL$ is ample,
225
2262) $\sL^n$ is ample for all $n>0$,
227
2283) $\sL^n$ is ample for some $n>0$.
229\end{prop}
230
231\begin{thm}
232Let $X$ be of finite type over a Noetherian ring $A$ and suppose
233$\sL$ is an invertible sheaf on $A$. Then $\sL$ is ample iff
234there exists $n$ such that $\sL^n$ is very ample over $\spec A$.
235\end{thm}
236
237\begin{example}
238Let $X=\bP^1$, $\sL=\sO(\ell)$, some $\ell \in \bZ$.
239If $\ell<0$ then $\Gamma(\sL)=0$. If $\ell=0$ then
240$\sL=\sO_X$ which is not ample since $\sO_X(-1)^n\tensor 241\sO_X\isom \sO_X(-1)^n$ is not generated by global sections
242for any $n$. Note that $\sO_X$ itself is generated by
243global sections. Finally, if $ell>0$ then
244$\sL=\sO_X(\ell)$ is very ample hence ample.
245\end{example}
246
247\begin{example}
248Let $C\subseteq \bP^2$ be a nonsingular cubic curve and
249$\sL$ an invertible sheaf on $C$ defined by $\sL=\sL(D)$,
250where $D=\sum n_i P_i$ is a divisor on $C$. If $\deg D<0$
251then $\sL$ has no global sections so it can't be ample.
252%% Undone.
253\end{example}
254
255% Day 2, 1/19/96
256\section{Introduction to Cohomology}
257We first ask, what is cohomology and where does it arise in nature?
258Cohomology occurs in commutative
259algebra, for example in the $\ext$ and $\tor$ functors, it occurs in
260group theory, topology, differential geometry, and of course in
261algebraic geometry. There are several flavors of cohomology which are
262studied by algebraic geometers. Serre's coherent sheaf
263cohomology has the advantage of being easy to define, but
264has the property that the cohomology groups are vector spaces.
265Grothendieck introduced \{e}tale cohomology and $\ell$-adic
266cohomology. See, for example, Milne's {\em \{E}tale Cohomology}
267and SGA 4$\frac{1}{2}$, 5 and 6. This cohomology theory arose
268from the study of the Weil Conjectures (1949) which deal with
269a deep relationship between the number of points on a variety
270over a finite field and the geometry of the complex analytic variety
271cut out by the same equations in complex projective space. Deligne
272was finally able to resolve these conjectures in the affirmative
273in 1974.
274
275What is cohomology good for? Cohomology allows one to get numerical
276invariants of an algebraic variety. For example, if $X$ is a projective
277scheme defined over an algebraically closed field $k$ then
278$H^i(X,\sF)$ is a finite dimensional $k$-vector space. Thus the
279$h_i=\dim_k H^i(X,\sF)$ are a set of numbers associated
280to $X$. Numbers are useful in all branches of mathematics.''
281
282\begin{example}{Arithmetic Genus}
283Let $X$ be a nonsingular projective curve. Then $\dim H^1(X,\sO_X)$ is
284the arithmetic genus of $X$. If $X\subseteq \bP^n$ is a projective variety
285of dimension $r$ then, if $p_a=\dim H^1(X,\sO_X)$, then
286$1+(-1)^r p_a =$ the constant term of the Hilbert polynomial of $X$.
287\end{example}
288
289\begin{example}
290Let $X$ be a nonsingular projective surface, then
291\begin{equation*}
2921+p_a=h^0(\sO_X)-h^1(\sO_X)+h^2(\sO_x)
293\end{equation*}
294and $1+(-1)^r p_a = \chi(\sO_X)$, the Euler characteristic of $X$.
295\end{example}
296
297\begin{example}
298Let $X$ be an algebraic variety and $\pic X$ the group of
299Cartier divisors modular linear equivalence (which is isomorphic
300to the group of invertible sheaves under tensor product modulo isomorphism).
301Then $\pic X \isom H^1(X,\sO_{X}^{*})$.
302\end{example}
303
304\begin{example}[Deformation Theory]
305Let $X_0$ be a nonsingular projective variety. Then
306the first order infinitesimal deformations are classified
307by $H^1(X_0,T_{X_0})$ where $T_{X_0}$ is the tangent
308bundle of $X_0$. The obstructions are classified by
309$H^2(X_0,T_{X_0})$.
310\end{example}
311
312One can define Cohen-Macaulay rings in terms of cohomology.
313Let $(A,\gm)$ be a local Noetherian ring of dimension $n$,
314let $X = \spec A$, and let $P=\gm\in X$, then we have the
315following.
316\begin{prop}
317Let $A$ be as above. Then $A$ is Cohen-Macaulay iff
318
3191) $H^0(X-P,\sO_{X-P}) = A$ and
320
3212) $H^i(X-P, \sO_{X-P}) = 0$ for $0<i<n-1$.
322\end{prop}
323
324A good place to get the necessary background for the
325cohomology we will study is in Appendices 3 and 4
326from Eisenbud's {\em Commutative Algebra}.
327
328\section{Cohomology in Algebraic Geometry}
329
330For any scheme $X$ and any sheaf $\sF$ of $\sO_X$-modules
331we want to define the groups $H^i(X,\sF)$. We can either {\em define}
332cohomology by listing its properties, then later prove that we can construct
333the $H^i(X,\sF)$ or we can skip the definition and just construct
334the $H^i(X,\sF)$. The first method is more esthetically pleasing,
335but we will choose the second.
336
337We first forget the scheme structure of $X$ and regard $X$ as a
338topological space and $\sF$ as a sheaf of abelian groups (by ignoring
339the ring multiplication). Let $\Ab(X)$ be the category of sheaves of
340abelian groups on $X$. Let $\Gamma=\Gamma(X,\cdot)$ be the global
341section functor from $\Ab(X)$ into $\Ab$, where $\Ab$ is the category
342of abelian groups. Recall that $\Gamma$ is left exact so if
343\begin{equation*}
3440\rightarrow\sF'\rightarrow\sF\rightarrow\sF''\rightarrow 0
345\end{equation*}
346is an exact sequence in $\Ab(X)$ then the following sequence is exact
347\begin{equation*}
3480\rightarrow \Gamma(\sF') \rightarrow \Gamma(\sF) \rightarrow
349\Gamma(\sF'')
350\end{equation*}
351in $\Ab$.
352
353
354\begin{defn}
355We define the cohomology groups $H^i(X,\sF)$ to be the right derived
356functors of $\Gamma$.
357\end{defn}
358
359%% 1/22/96, Lecture 3
360
361\section{Review of Derived Functors}
362
363The situation will often be as follows.
364Let $\sA$ and $\sB$ be abelian categories and
365$$\sA\xrightarrow{F}\sB$$
366a functor. Derived functors are the measure of the non-exactness of
367a functor. Let $X$ be a topological space, $\Ab(X)$ the category
368of sheaves of abelian groups on $X$ and $\Ab$ the category of
369abelian groups. Then $\Gamma(X,\cdot): 370\Ab(X)\rightarrow\Ab$ is a left exact functor. Our cohomology theory
371will turn out to be the right derived function of $\Gamma(X,\cdot)$.
372
373\subsection{Examples of Abelian Categories}
374Although we will not define an abelian category we will give several
375examples and note that an abelian category is a category which has
376the same basic properties as these examples.
377\begin{example}[$A$-Modules] Let $A$ be a fixed commutative ring
378and consider the category $\Mod(A)$ of $A$-modules. Then if
379$M,N$ are any two modules one has
380
3811) $\Hom(M,N)$ is an abelian group,
382
3832) $\Hom(M,N)\times \Hom(N,L)\into \Hom(M,L)$ is a homomorphism
384of abelian groups.
385
3863) there are kernels, cokernels, etc.
387
388$\Mod(A)$ is an abelian category.
389\end{example}
390
391\begin{example}
392Let $A$ be a Noetherian ring and let our category be the collection
393of all finitely generated $A$-modules. Then this category is abelian.
394Note that if the condition
395that $A$ be Noetherian is relaxed we may no longer have an abelian
396category because the kernel of a morphism of finitely generated
397modules over an arbitrary ring need not be finitely generated (for
398example, take the map from a ring to its quotient by an ideal which
399cannot be finitely generated).
400\end{example}
401
402\begin{example} Let $X$ be a topological space, then
403$\Ab(X)$ is an abelian category. If $(X,\sO_X)$ is a ringed
404space then the category $\Mod(\sO_X)$ is abelian. If $X$ is
405a scheme then the category of quasi-coherent $\sO_X$-modules
406is abelian, and if $X$ is also Noetherian then the sub-category
407of coherent $\sO_X$-modules is abelian.
408\end{example}
409
410\begin{example}
411The category of abelian varieties is {\em not} an abelian category
412since the kernel of a morphism of abelian varieties might be
413reducible (for example an isogeny of degree $n$ of elliptic curves
414has kernel $n$ points which is reducible). It may be the case
415that the category of abelian group schemes is abelian but I don't
416know at the moment.
417\end{example}
418
419\begin{example}
420The category of compact Hausdorff abelian topological groups
421is an abelian category.
422\end{example}
423
424\subsection{Exactness}
425
426\begin{defn} A functor $F:\sA\into\sB$ is {\em additive} if for
427all $X,Y\in \sA$, the map
428$$F:\Hom_{\sA}(X,Y)\into\Hom_{\sB}(FX,FY)$$
429is a homomorphism of abelian groups.
430\end{defn}
431
432\begin{defn} A sequence
433$$A\xrightarrow{f}B\xrightarrow{g}C$$
434is {\em exact} if $\imag(f)=\ker(g)$.
435\end{defn}
436
437\begin{defn} Let $F:\sA\into\sB$ be a functor and
438$$0\into M'\into M \into M''\into 0$$
439be an exact sequence and consider the sequence
440$$0\into FM'\into FM\into FM''\into 0.$$
441If the second sequence is exact in the middle, then
442$F$ is a called {\em half exact functor}. If the second sequence is exact
443on the left and the middle then $F$ is called a {\em left exact functor}.
444If the second sequence is exact on the right and in the middle
445then we call $F$ a {\em right exact functor}.
446\end{defn}
447
448\begin{example}
449Let $A$ be a commutative ring and $N$ an $A$-module. Then
450$N\tensor -$ is a right exact functor on the category of
451$A$-modules. To see that $N\tensor -$ is not exact, suppose
452$A,\gm$ is a local ring and $N=k=A/\gm$.
453Then the sequence
454$$0\into\gm\into A\into k\into 0$$
455is exact, but
456$$0\into k\tensor\gm \into k\tensor A \into k\tensor k \into 0$$
457is right exact but not exact.
458\end{example}
459
460\begin{example}
461The functor $\tor_1(N,\cdot)$ is neither left nor right exact.
462\end{example}
463
464\begin{example}
465The contravarient hom functor, $\Hom(\cdot,N)$ is left exact.
466\end{example}
467
468
469Often the following is useful in work.
470\begin{thm}
471If $$0\into M'\into M\into M''$$ is exact and $F$ is left exact, then
472$$0\into FM'\into FM\into FM''$$
473is exact.
474\end{thm}
475
476\subsection{Injective and Projective Objects}
477
478Let $\sA$ be an abelian category. Then $\Hom_A(P,-):\sA\into\Ab$ is
479left exact.
480
481\begin{defn}
482An $A$ module $P$ is said to be {\em projective} if
483the functor $\Hom_A(P,-)$ is exact. An $A$ module $I$ is
484said to be {\em injective} if the functor
485$\Hom_A(-,I)$ is exact.
486\end{defn}
487
488\begin{defn}
489We say that an abelian category $\sA$ has {\em enough projectives} if
490every $X$ in $\sA$ is the surjective image of a projective
491$P$ in $\sA$. A category is said to have {\em enough injectives} if
492every $X$ in $\sA$ injects into an injective objective of $\sA$.
493\end{defn}
494
495\begin{example}
496Let $A$ be a commutative ring, then $\Mod(A)$ has enough injectives because
497every module is the quotient of a free module and every free module
498is projective. If $X$ is a topological space then $\Ab(X)$ has enough
499injectives. If $X$ is a Noetherian scheme, then the category of quasi-coherent
500sheaves has enough injectives (hard theorem). The category of $\sO_X$-modules
501has enough injectives but the category of coherent sheaves on
502$X$ doesn't have enough injectives or projectives.
503\end{example}
504
505%% 1/24/96, Lecture 4
506
507\section{Derived Functors and Homological Algebra}
508
509Let $F:\sA\into\sB$ be an additive covariant left-exact functor between
510abelian categories, for example $F=\Gamma:\Ab(X)\into\Ab$.
511Assume $\sA$ has enough injectives, i.e., for all $X$ in $\sA$
512there is an injective object $I$ in $\sA$ such that
513$0\into X\hookrightarrow I$. We construct the right derived functors
514of $F$. If
515$$0\into M'\into M\into M''\into 0$$
516is exact in $\sA$ then
517$$0\into F(M')\into F(M)\into F(M'')\into R^{1}F(M^1)\into R^{1}F(M) 518\into \cdots$$
519is exact in $\sB$ where $R^{i}F$ is the right derived functor of $F$.
520
521\subsection{Construction of $R^{i}F$}
522Take any $M$ in $\sA$, then since $\sA$ has enough injectives we
523can construct an exact sequence
524$$0\into M\into I^{0}\into I^{1}\into I^{2}\into \cdots$$
525where each $I^{i}$ is an injective object. (This isn't totally
526obvious, but is a straightforward argument by putting together
527short exact sequences and composing maps.)
528The right part of the above sequence $I^{0}\into I^{1}\into \cdots$
529is called an {\em injective resolution} of $M$.
530Applying $F$ we get a complex
531$$F(I^{0})\xrightarrow{d_0} F(I^{1})\xrightarrow{d_1} F(I^{2}) 532\xrightarrow{d_2} \cdots$$
533in $\sB$ which may not be exact. The objects
534$H^i=\ker(d_2)/ \imag(d_1)$ measure the deviation of this
535sequence from being exact. $H^i$ is called the $i$th {\em cohomology}
536object of the complex.
537
538\begin{defn} For each object in $\sA$ fix an injective resolution.
539The $i$th {\em right derived functor} of $F$ is the functor
540which assigns to an object $M$
541the $i$th cohomology of the complex $F(I^{\cdot})$ where $I^{\cdot}$
542is the injective resolution of $M$.
543\end{defn}
544
545\subsection{Properties of Derived Functors}
546We should now prove the following:
547\begin{enumerate}
548\item If we fix different injective resolutions for all of our objects
549then 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
551that they are really functors.
552\item If $0\into M'\into M\into M''$ is a short exact sequence then there is
553a long exact sequence of cohomology:
554\begin{align*}
5550\into FM'\into FM\into FM'' \into \\
556R^1FM'\into R^1FM\into R^1FM''\into \\
557R^2FM'\into \cdots.
558\end{align*}
559\item If we have two short exact sequences then the induced maps on
560long 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
5661-6 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
572transformation $F\into R^0 F$. A {\em universal augmented
573$\delta$-functor} is an {\em augmented $\delta$-functor} with
574some universal property which I didn't quite catch.
575\end{defn}
576
577\begin{thm} If $\sA$ has enough injectives then the collection of
578derived functors of $F$ is a universal augmented $\delta$-functor.
579\end{thm}
580
581To 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
583hold.
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
590Today I sat in awe as Hartshorne effortless drew hundreds
591of arrows and objects everywhere, chased some elements and
592proved that there is a long exact sequence of cohomology in
59330 seconds. Then he whipped out his colored chalk and things
594really got crazy. Vojta tried to erase Hartshorne's diagrams
595during the next class but only partially succeeded joking
596that the functor was not effaceable'. (The diagrams are
597still not quite gone 4 days later!) Needless to say, I don't
598feel like texing diagrams and element chases... it's all
599trivial anyways, right?''
600
601%% 1/29/96
602\section{Basic Properties of Cohomology}
603Let $X$ be a topological space, $\Ab(X)$ the category
604of sheaves of abelian groups on $X$ and
605$$\Gamma(X,\cdot):\Ab(X)\into\Ab$$
606the covariant, left exact global sections functor.
607Then we have constructed the derived functors
608$H^{i}(X,\cdot)$.
609
610\subsection{Cohomology of Schemes}
611Let $(X,\sox)$ be a scheme and $\sF$ a sheaf of $\sox$-modules.
612To compute $H^{i}(X,\sF)$ forget all extra structure and use
613the above definitions. We may get some extra structure anyways.
614
615\begin{prop}
616Let $X$ and $\sF$ be as above, then the groups $H^{i}(X,\sF)$
617are naturally modules over the ring $A=\Gamma(X,\sox)$.
618\end{prop}
619
620\begin{proof}
621Let $A=H^{0}(X,\sF)=\Gamma(X,\sox)$ and let $a\in A$. Then because
622of the functoriality of $H^{i}(X,\cdot)$ the
623map $\sF\into\sF$ induced by left multiplication by $a$ induces
624a homomorphism
625$$a:H^{i}(X,\sF)\into H^{i}(X,\sF).$$
626%% Need more!!
627\end{proof}
628
629\subsection{Objective}
630Our objective is to compute $H^{i}(\bP^n_k,\sO(\ell))$
631for all $i,n,\ell$. This is enough for most applications
632because if one knows these groups one can, in principle at
633least, computer the cohomology of any projective scheme.
634If $X$ is any projective variety, we embed $X$ in some
635$\bP^n_k$ and push forward the sheaf $\sF$ on $X$. Then
636we construct a resolution of $\sF$ by sheaves of the
637form $\sO(-\ell)^n$. Using Hilbert's syzigy theorem one
638sees that the resolution so constructed is finite and
639so we can put together our knowledge to get the cohomology
640of $X$.
641
642Our plan of attack is as follows.
643\begin{enumerate}
644\item Define flasque sheaves which are acyclic for cohomology, i.e.,
645the cohomology vanishes for $i>0$.
646\item If $X=\spec A$, $A$ Noetherian, and $\sF$ is quasi-coherent,
647show that $H^i(X,\sF)=0$ for $i>0$.
648\item If $X$ is any Noetherian scheme and $\sU=(U_i)$ is an open
649affine cover, find a relationship between the cohomology of $X$
650and 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}
656A {\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}
660Thus in a flasque sheaf, every section extends''.
661
662\begin{example}
663Let $X$ be a topological space, $p\in X$ a point, not necessarily
664closed, and $M$ an abelian group. Let $j:\{P\}\hookrightarrow X$ be the
665inclusion, then $\sF=j_{*}(M)$ is flasque. This follows since
666$$667j_{*}(M)(U)=\begin{cases} M&\text{if p\in U}\\ 668 0&\text{if p\not\in U} 669 \end{cases}. 670$$
671Note that $j_{*}(M)$ is none other than the skyscraper sheaf
672at $p$ with sections $M$.
673\end{example}
674
675\begin{example}
676If $\sF$ is a flasque sheaf on $Y$ and $f:Y\into X$ is a morphism
677then $f_{*}\sF$ is a flasque sheaf on $X$.
678\end{example}
679
680\begin{example}
681If $\sF_i$ are flasque  then $\bigoplus_{i} \sF_i$ is flasque.
682\end{example}
683
684\begin{lem}
685If
686$$0\into\sF'\into\sF\into\sF''\into 0$$
687is exact and $\sF'$ is flasque
688then
689$$\Gamma(\sF)\into\Gamma(\sF'')\into 0$$
690is exact.
691\end{lem}
692
693\begin{lem}
694If
695$$0\into\sF'\into\sF\into\sF''\into 0$$
696is exact and $\sF'$ and $\sF$ are both
697flasque then
698$\sF''$ is flasque.
699\end{lem}
700
701\begin{proof}
702Suppose $V\subset U$ are open subsets of $X$.
703Since $\sF'$ is flasque and the restriction of a
704flasque sheaf is flasque and restriction is exact,
705lemma 1 implies that the sequence
706   $$\sF(V)\into\sF''(V)\into 0$$
707is exact.
708We thus have a commuting diagram
709$$710\begin{CD} 711\sF(U) @>>> \sF''(U)\\ 712@VVV @VVV\\ 713\sF(V) @>>> \sF''(V) @>>> 0 714\end{CD} 715$$
716which, since $\sF(U)\into \sF(V)$ is surjective,
717implies $\sF''(U)\into\sF''(V)$ is surjective.
718\end{proof}
719
720
721\begin{lem}
722Injective sheaves (in the category of abelian sheaves) are flasque.
723\end{lem}
724
725\begin{proof}
726Let $\sI$ be an injective sheaf of abelian groups on $X$ and
727let $V\subset U$ be open subsets. Let $s\in\sI(V)$, then we
728must find $s'\in\sI(U)$ which maps to $s$ under
729the map $\sI(U)\into\sI(V)$. Let $\bZ_V$ be the constant sheaf
730$\bZ$ on $V$ extended by $0$ outside $V$ (thus $\bZ_V(W)=0$
731if $W\not\subset V$). Define a map $\bZ_V\into \sI$ by
732sending the section $1\in\bZ_V(V)$ to $s\in\sI(V)$. Then
733since $\bZ_V\hookrightarrow\bZ_U$ and $\sI$ is injective
734there 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
736to $V$ must be $s$.
737\end{proof}
738
739\begin{remark} The same proof also shows that injective sheaves in
740the category of $\sox$-modules are flasque.
741\end{remark}
742
743\begin{cor}
744If $\sF$ 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}
752Let $(X,\sox)$ 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}
756If
757$$0\into\sF\into\sI^0\into\sI^1\into\cdots$$
758is an injective resolution of $\sF$
759in $\Mod\sox$ then, by the above remark, it
760is 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
766resolution in the category of quasi-coherent $\sox$-modules
767then 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}
774Suppose $C$ is a nonsingular projective curve over an algebraically
775closed field $k$. Let $K=K(C)$ be the function field of $C$ and
776let $\sK_C$ denote the constant sheaf $K$. Then we have an exact
777sequence
778$$0\into\sO_C\into\sK_C\into 779 \bigoplus_{\substack{P\in C\\P\text{ closed}}} 780 K/\sO_P \into 0,$$
781where the map $\sK_C\into\bigoplus K/\sO_P$ has
782only finitely many components nonzero since a function $f\in K$
783has only finitely many poles.
784Since $C$ is irreducible $\sK_C$ is flasque and since $\sK/\sO_P$
785is a skyscraper sheaf it is flasque so since direct of flasque
786sheaves are flasque, $\bigoplus K/\sO_P$ is flasque.
787One checks that the sequence is exact and so this is
788a flasque resolution of $\sO_C$. Taking global sections and
789applying 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
793is $\dim_k H^1(X,\sO_C)$ 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}
801Anyone who studies algebraic geometry must read French... looking up
802the more general version of this proof in EGA would be a good exercise.''
803\end{quote}
804
805\begin{thm}
806Let $A$ be a Noetherian ring, $X=\spec A$ and $\sF$ a quasi-coherent
807sheaf on $X$, then $H^{i}(X,\sF)=0$ for $i>0$.
808\end{thm}
809
810\begin{remark}
811The theorem is true without the Noetherian hypothesis on $A$, but
812the proof uses spectral sequences.
813\end{remark}
814
815\begin{remark}
816The assumption that $\sF$ is quasi-coherent is essential. For example,
817let $X$ be an affine algebraic curve over an infinite field $k$. Then
818$X$ is homeomorphic as a topological space to $\P^1_k$ so
819the sheaf $\sO(-2)$ on $\P^1_k$ induces a sheaf $\sF$ of abelian groups
820on $X$ such that
821$$H^1(X,\sF)\isom{}H^1(\P^1_k,\sO(-2))\neq 0.$$
822\end{remark}
823
824\begin{remark}
825If $I$ is an injective $A$-module then $\tilde{I}$ need {\em not}
826be 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 $\tilde{I}$
828is the constant sheaf $k$. But $k$ is
829a 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}
834Suppose $A$ is Noetherian and $I$ is an injective $A$-module, then
835$\tilde{I}$ is flasque on $\spec A$.
836\end{prop}
837
838The proposition implies the theorem since if $\sF$ is quasi-coherent
839then $\sF=\tilde{M}$ for some $A$-module $M$. There is an injective
840resolution
841$$0\into M\into I^{\bullet}$$
842which, upon applying the exact functor $\tilde{ }$,
843gives a flasque resolution
844$$0\into \tilde{M}=\sF\into \tilde{I}^{\bullet}.$$
845Now applying $\Gamma$ gives us back the original resolution
846$$\Gamma:\quad 0\into M\into I^{\bullet}$$
847which is exact so the cohomology groups vanish for $i>0$.
848
849\begin{proof} Let $A$ be a Noetherian ring and $I$ an injective $A$,
850then $\tilde{I}$ is a quasi-coherent sheaf on $X=\spec A$. We must
851show that it is flasque. It is sufficient to show that for any
852open set $U$, $\Gamma(X)\into\Gamma(U)$ is surjective.
853
854{\em Case 1, special open affine:}
855Suppose $U=X_f$ is a special open affine. Then we have a commutative diagram
856$$857\begin{CD} 858\Gamma(X,\tilde{I}) @>>> \Gamma(X_f,\tilde{I})\\ 859 @V=VV @V=VV\\ 860 I @>\text{surjective?}>> I_f 861\end{CD} 862$$
863To see that the top map is surjective it is equivalent
864to show that $I\into I_f$ is surjective. This is a tricky
865algebraic lemma (see Hartshorne for proof).
866
867{\em Case 2, any open set:}
868Let $U$ be any open set. See Hartshorne for the rest.
869
870\end{proof}
871
872
873%% 2/5/96
874
875\section{\cech{} Cohomology}
876Let $X$ be a topological space, $\sU=(U_i)_{i\in I}$ an
877open cover and $\sF$ a sheaf of abelian groups.
878We will define groups $\cH^i(\sU,\sF)$ called
879\cech{} cohomology groups.
880
881{\bfseries Warning: } $\cH^i(\sU,\cdot)$ is a functor in $\sF$,
882but it is {\em not} a $\delta$-functor.
883
884\begin{thm} Let $X$ be a Noetherian scheme, $\sU$ an open cover
885and $\sF$ a quasi-coherent sheaf, then $\cH^{i}(\sU,\sF)=H^i(X,\sF)$
886for all $i$.
887\end{thm}
888
889\subsection{Construction}
890Totally order the index set $I$. Let
891$$U_{i_0\cdots i_p}=\intersect_{j=0}^p U_{i_j}.$$
892For any $p\geq 0$ define
893$$C^p(\sU,\sF)=\prod_{i_0<i_1<\cdots<i_p}\sF(U_{i_0\cdots i_p}).$$
894Then we get a complex
895$$C^0(\sU,\sF)\into C^1(\sU,\sF)\into \cdots \into C^p(\sU,\sF)\into \cdots$$
896by defining a map
897$$d:C^p(\sU,\sF)\into C^{p+1}(\sU,\sF)$$
898by, for $\alpha\in C^p(\sU,\sF)$,
899$$(d\alpha)_{i_0\cdots i_{p+1}} := \sum_{0}^{p+1} (-1)^j 900 \alpha_{i_0\cdots \hat{i_j}\cdots i_{p+1}}|_{U_{i_0\cdots i_{p+1}}}.$$
901One checks that $d^2=0$.
902
903\begin{lem}
904$\cH^{0}(\sU,\sF)=\Gamma(X,\sF)$
905\end{lem}
906\begin{proof}
907Applying the sheaf axioms to the exact sequence
908$$0\into\Gamma(X,\sF)\into C^0=\prod_{i\in I}\sF(U_i) 909 \xrightarrow{d}C^1=\prod_{i<j}\sF(U_{ij})$$
910we see that $\cH^0(\sU,\sF)=\ker d = \Gamma(X,\sF)$.
911\end{proof}
912
913\subsection{Sheafify}
914Let $X$ be a topological space, $\sU$ an open cover and
915$\sF$ a sheaf of abelian groups. Then we define
916$$\sC^p(\sU,\sF)=\prod_{i_0<\cdots<i_p}j_{*}(\sF|_{U_{i_0\cdots i_p}})$$
917and define
918$$d:\sC^p(\sU,\sF)\into \sC^{p+1}(\sU,\sF)$$
919in 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}
929The sequence
930$$0\into\sF\into\sC^{0}(\sU,\sF)\into\sC^{1}(\sU,\sF)\into\cdots$$
931is a resolution of $\sF$, i.e., it is exact.
932\end{lem}
933\begin{proof}
934We 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
936sheaf axioms.
937
938To show the rest of the sequence is exact it suffices to show
939exactness at the stalks. So let $x\in X$, and suppose $x\in U_j$.
940Given $\alpha_x\in\sC_x^p$ 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
943for any $p$-tuple $i_0<\ldots<i_{p-1}$, we set
944$$(k\alpha)_{i_0,\ldots,i_{p-1}}=\alpha_{j,i_0,\ldots,i_{p-1}}.$$
945This makes sense because
946$$V\intersect U_{i_0,\ldots,i_{p-1}}=V\intersect U_{j,i_0,\ldots,i_{p-1}}.$$
947Then take the stalk of $k\alpha$ at $x$ to get the required map $k$.
948
949Now we check that for any $p\geq 1$ and $\alpha\in\sC_x^p$,
950$$(dk+kd)(\alpha)=\alpha.$$
951First 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*}
956Whereas, 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*}
963Adding these two expressions yields $\alpha_{i_0,\ldots,i_p}$ as claimed.
964
965Thus $k$ is a homotopy operator for the complex $\sC_x^{\bullet}$, showing
966that the identity map is homotopic to the zero map. It follows that the
967cohomology groups $H^{p}(\sC^{\bullet}_x)$ of this complex
968are $0$ for $p\geq 1$.
969\end{proof}
970
971\begin{lem}
972If $\sF$ is flasque then $\sC^p(\sU,\sF)$ is also flasque.
973\end{lem}
974\begin{proof}
975If $\sF$ is flasque then $\sF|_{U_{i_0,\ldots,i_p}}$ is flasque
976so $j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque so
977$\prod j_{*}(\sF|_{U_{i_0,\ldots,i_p}})$ is flasque.
978\end{proof}
979
980\begin{prop}
981If $\sF$ is flasque then $\cH^{p}(\sU,\sF)=0.$
982\end{prop}
983\begin{proof}
984Consider the resolution $$0\into\sF\into\sC^{\bullet}(\sU,\sF).$$
985By the above lemma it is flasque, so we can use it to compute the
986usual cohomology groups of $\sF$. But $\sF$ is flasque, so
987$H^p(X,\sF)=0$ for $p>0$. On the other hand,
988the answer given by this resolution is
989$$H^p(\Gamma(X,\sC^{\bullet}(\sU,\sF)))=\cH^{p}(\sU,\sF).$$
990So we conclude that $\cH^{p}(\sU,\sF)=0$ for $p>0$.
991\end{proof}
992
993\begin{lem}
994Let $X$ be a topological space, and $\sU$ an open covering. Then
995for 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}
1000Let $X$ be a Noetherian separated scheme, let $\sU$ be an open
1001affine 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
1010Today we prove
1011
1012\begin{thm}
1013Let $X$ be a Noetherian, separated scheme, $\sU$ an open cover
1014and $\sF$ a quasi-coherent sheaf on $X$. Then
1015$$\cH^{i}(\sU,\sF)=H^{i}(X,\sF).$$
1016\end{thm}
1017
1018To do this we introduce a condition (*):
1019
1020{\bfseries Condition *:} Let $\sF$ be a sheaf of abelian groups
1021and $\sU=(U_i)_{i\in I}$ 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}
1026If $0\into\sF'\into\sF\into\sF''\into 0$ is an exact sequence in
1027$\Ab(X)$ and $\sF'$ satisfies (*) then there is a long exact sequence
1028for $\cH^{i}(\sU,\cdot)$.
1029\end{lem}
1030\begin{proof}
1031Since the global sections functor is left exact and cohomology
1032commutes with products, we have an exact sequence
1033\begin{align*}
10340\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*}
1039where the last term is 0 because $\sF'$ satisfies condition (*).
1040Replacing $p$ by $\cdot$ gives an exact sequence of complexes.
1041Applying $\cH^{i}(\sU,\cdot)$ then gives the desired result.
1042\end{proof}
1043
1044\begin{thm}
1045Let $X$ be a topological space, $\sU$ an open cover and $\sF\in \Ab(X)$.
1046Suppose $\sF$ and $\sU$ satisfy (*). Then the maps
1047$$\varphi^{i}:\cH^i(\sU,\sF)\into H^{i}(X,\sF)$$
1048are isomorphisms.
1049\end{thm}
1050\begin{proof}
1051The proof is a clever induction.
1052\end{proof}
1053
1054\begin{lem}
1055If $0\into\sF'\into\sF\into\sF''\into 0$ is exact and
1056$\sF'$ and $\sF$ satisfy (*) then $\sF''$ satisfies (*).
1057\end{lem}
1058
1059To prove the main theorem of the section use the fact that
1060$X$ separated implies any finite intersection of affines
1061is affine and then use the vanishing theorem for cohomology
1062of a quasi-coherent sheaf on an affine scheme. The above theorem
1063then implies the main result. From now on we will always
1064assume our schemes are separated unless otherwise stated.
1065
1066\begin{cor}
1067If $X$ is a (separated) Noetherian scheme and $X$ can be covered
1068by $n+1$ open affines for some $n>0$ then $H^{i}(X,\sF)=0$ for $i>n$.
1069\end{cor}
1070
1071\begin{example}
1072Let $X=\bP_k^n$, then the existence of the standard affine
1073cover $U_0,\ldots,U_n$ implies that $H^{i}(X,\sF)=0$ for
1074$i>n$.
1075\end{example}
1076
1077\begin{example}
1078Let $X$ be a projective curve embedded in $\bP^k_n$.
1079Let $U_0\subset X$ be open affine, then $X-U_0$ is finite.
1080Thus $U_0\subset X\subset \bP^n$ and $X-U_0=\{P_1,\ldots,P_r\}$.
1081In $\bP^n$ there is a hyperplane $H$ such that
1082$P_1,\ldots,P_r\not\in H$. Then $P_1,\ldots,P_r\in\bP^n-H=\bA^n=V$.
1083Then $U_1=V\intersect X$ is closed in the affine set $V$, hence affine.
1084Then $X=U_0\union U_1$ with $U_0$ and $U_1$ both affine.
1085Thus $H^{i}(X,\sF)=0$ for all $i\geq 2$.
1086\end{example}
1087
1088\begin{exercise}
1089If $X$ is any projective scheme of dimension $n$ then $X$
1090can 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
1095Hartshorne was unaware of the answer to the following question
1096today.
1097\begin{ques}
1098If $X$ is a Noetherian scheme of dimension $n$ do there
1099exist $n+1$ open affines covering $X$.
1100\end{ques}
1101
1102\begin{thm}[Grothendieck]
1103If $\sF\in\Ab(X)$ then $H^{i}(X,\sF)=0$ for all $i>n=\dim X$.
1104\end{thm}
1105
1106\begin{example}
1107Let $k$ be an algebraically closed field.
1108Then $X=\bA^2_k-\{(0,0)\}$ is not affine since it has global
1109sections $k[x,y]$.
1110We compute $H^1(X,\sox)$ by \cech cohomology.
1111Write $X=U_1\union U_2$ where $U_1=\{x\neq 0\}$ and
1112$U_2=\{y\neq 0\}$. Then the \cech complex is
1113$$C^{\cdot}(\sU,\sox): k[x,y,x^{-1}]\oplus{}k[x,y,y^{-1}] 1114 \xrightarrow{d} k[x,x^{-1},y,y^{-1}].$$
1115Thus one sees with a little thought that
1116$H^{0}=\ker{d}=k[x,y]$
1117and
1118$H^{1}=\{\sum_{i,j<0}a_{ij}x^{i}x^{j} : a_{ij}\in k\} = E$
1119as $k$-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
1126injective 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 $\bP_k^n$}
1137Today we begin to compute $H^{i}(X,\sox(\ell))$ for all $i$ and all
1138$\ell$.
1139
1140a) $H^0(X,\sox(\ell))$ 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}
1145There 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 $\sox\into\sox(\ell)$
1150given by $1\mapsto\alpha$. This defines a map
1151$$\sox\tensor\sox(m)\xrightarrow{\alpha(m)}\sox(\ell)\tensor\sox(m)$$
1152which gives a map $\sox(m)\into\sox(\ell+m)$. This induces the desired
1153map $H^{i}(\sox(m))\into H^{i}(\sox(\ell+m))$.
1154\end{proof}
1155
1156b) $H^{i}(\sox(\ell))=0$ when $0<i<n$ and for all $\ell$.
1157   (This doesn't hold for arbitrary quasi-coherent sheaves!)
1158
1159c) $H_{*}^n(X,\sox)$ is a graded $S$-module which
1160is $0$ in degrees $\geq -n$, but is nonzero in degrees $\leq n-1$.
1161As 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
1165d) For $\ell\geq 0$ the map
1166$$H^0(\sox(\ell))\times H^n(\sox(-\ell-n-1))\into H^n(\sox(-n-1))\isom{}k$$
1167is a perfect pairing so we have a duality (which is in fact a special
1168case of Serre Duality).
1169
1170
1171% 2/12/96
1172\section{Serre's Finite Generation Theorem}
1173We relax the hypothesis from the last lecture and claim that the
1174same results are still true.
1175\begin{thm}
1176Let $A$ 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
1182a perfect pairing of free $A$-modules. Notice that $H^n(\sox(-n-1))$ is
1183a free $A$-module of rank 1 so it is isomorphic to $A$, but {\em not} in
1184a canonical way!
1185\end{enumerate}
1186\end{thm}
1187
1188Although pairing is in general not functorial as a map into $A$,
1189there is a special situation in which it is. Let $\Omega_{X/k}^1$
1190be 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
1192module''). Then some map is functorial (??)
1193\begin{quote}
1194Is $\omega$ more important than $\Omega$?'' -- Janos Csirik
1195
1196That's a value judgment... you can make your own decision
1197on that... I won't.'' -- Hartshorne
1198\end{quote}
1199
1200\begin{thm}[Serre]
1201Let $X$ be a projective scheme over a Noetherian ring $A$. Let $\sF$
1202be 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
1206for all $n\geq n_0$, $H^{i}(X,\sF(n))=0$.
1207\end{enumerate}
1208\end{thm}
1209
1210The following was difficult to prove last semester and
1211we were only able to prove it under somewhat restrictive
1212hypothesis 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}
1217Set $i=0$ in 1.
1218\end{proof}
1219
1220\begin{proof}(of theorem)
1221
1222I. {\em Reduce to the case $X=\bP^r_A$.}
1223Use the fact that the push forward of a closed subscheme has
1224the same cohomology to replace $\sF$ by $i_{*}(\sF)$.
1225
1226II. {\em Special case, $\sF=\sO_{\bP^r}(\ell)$ any $\ell\in\bZ$.}
12271. and 2. both follow immediately from the previous theorem.
1228This is where we have done the work in explicit calculations.
1229
1230III. {\em Cranking the Machine of Cohomology}
1231
1232
1233\end{proof}
1234
1235\subsection{Application: The Arithmetic Genus}
1236Let $k$ be an algebraically closed field and $V\subset X=\bP^n_k$ a
1237projective variety. The arithmetic genus of $V$ is
1238$$p_a=(-1)^{\dim V}(p_V(0)-1)$$
1239where $p_V$ is the Hilbert polynomial of $V$, thus
1240$p_V(\ell)=\dim_k(S/I_V)_{\ell}$ for all $\ell\gg 0$. The
1241Hilbert polynomial depends on the projective embedding of $V$.
1242\begin{prop}
1243$p_V(\ell)=\sum_{i=0}^{\infty}(-1)^{i}\dim_k H^{i}(\sO_V(\ell))$
1244for {\em all} $\ell\in\bZ$.
1245\end{prop}
1246This redefines the Hilbert polynomial. Furthermore,
1247$$p_a=(-1)^{\dim V}(p_V(0)-1) = (-1)^{\dim V} 1248 \sum_{i=0}^{\infty}(-1)^i\dim_k(H^{i}(\sO_V))$$
1249which shows that $p_a$ is intrinsic, i.e., it doesn't depend
1250on the embedding of $V$ in projective space.
1251
1252\section{Euler Characteristic}
1253Fix 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.
1256Let $$h^{i}(X,\sF)=\dim_k H^{i}(X,\sF).$$
1257\begin{defn}
1258The {\bfseries Euler characteristic} of $\sF$ is
1259$$\chi(\sF)=\sum_{i=0}^{n}(-1)^i h^i(X,\sF).$$
1260\end{defn}
1261Thus $\chi$ is a function $\Coh(X)\into\Z$.
1262\begin{lem}
1263If $k$ is a field and
1264$$0\into{}V_1\into{}V_2\into{}\cdots\into{}V_N\into{}0$$
1265is an exact sequence of finite dimensional vector spaces,
1266then $\sum_{i=1}^N(-1)^i\dim{}V_i=0$.
1267\end{lem}
1268\begin{proof}
1269Since every short exact sequence of vector spaces splits, the
1270statement is true when $N=3$. If the statement is true for an
1271exact sequence of length $N-1$ then, applying it to the exact sequence
1272$$0\into{}V_2/V_1\into{}V_3\into\cdots\into{}V_N\into{}0,$$
1273shows that
1274$\dim{}V_2/V_1-\dim V_3+\cdots\pm\dim V_n = 0$
1275from which the result follows.
1276\end{proof}
1277
1278\begin{lem}
1279If $0\into\sF'\into\sF\into\sF''\into{}0$ is an exact sequence
1280of coherent sheaves on $X$, then
1281$$\chi(\F)=\chi(\F')+\chi(\F'').$$
1282\end{lem}
1283\begin{proof}
1284Apply the above lemma to the long exact sequence of cohomology
1285taking into account that $H^n(\F'')=0$ by Serre's vanishing theorem.
1286\end{proof}
1287
1288More generally, any map $\chi$ from an abelian category to
1289$\Z$ is called additive if, whenever
1290$$0\into\F^0\into\F^1\into\cdots\into\F^n\into{}0$$
1291is exact, then
1292$$\sum_{i=0}^{n}(-1)^{i}\chi(\F^i)=0.$$
1293
1294{\bfseries Question.}
1295Given an abelian category $\sA$ find an abelian
1296group $A$ and a map $X:\sA\into{}A$ such that every
1297additive function $\chi:\sA\into{}G$ factor through $\sA\xrightarrow{X}A$.
1298In the category of coherent sheaves the Grothendieck group
1299solves this problem.
1300
1301Let $X=\P_k^n$ and suppose $\sF$ is a coherent sheaf on $X$.
1302The Euler characteristic induces a map
1303$$\Z\into\Z: n\mapsto\chi(\F(n)).$$
1304\begin{thm}
1305There 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}
1308The polynomial $p_{\F}(n)$ is called the Hilbert polynomial of $\F$.
1309Last semester we defined the Hilbert polynomial of a graded module
1310$M$ over the ring $S=k[x_0,\ldots,x_n]$. Define $\varphi_M:\Z\into\Z$
1311by $\varphi_M(n)=\dim_k M_n$. Then we showed that there is a unique
1312polynomial $p_M$ such that $p_M(n)=\varphi_M(n)$ for all $n\gg 0$.
1313\begin{proof}
1314We induct on $\dim(\supp\F)$. If $\dim(\supp\F)=0$ then $\supp\F$ is a
1315union of closed points so $\sF=\oplus_{i=1}^{k}\F_{p_i}$.
1316Since each $\F_{p_i}$ is a finite
1317dimensional $k$-vector space and $\sox(n)$ is locally free,
1318there is a non-canonical isomorphism $\F(n)=\F\tensor\sox(n)\isom\F$.
1319Thus $$\chi_{\F}(n)=h^0(\F(n))=h^0(\F)=\sum_{i=1}^{k}\dim_k\F_{p_i}$$
1320which is a constant function, hence a polynomial.
1321
1322Next suppose $\dim(\supp\F)=s$.
1323Let $x\in{}S_1=H^0(\sox(1))$ be such that the hyperplane
1324$H:=\{x=0\}$ doesn't contain any irreducible component
1325of $\supp\F$. Multiplication by $x$ defines a map
1326$\sox(-1)\xrightarrow{x}\sox$ which is an isomorphism
1327outside 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$$0\into\sR\into\F(-1)\xrightarrow{x}\F\into\sQ\into{}0.$$
1331Now $\supp\sR\union\supp\sQ\subset\supp\F\intersect H$ 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
1334by our induction hypothesis $\chi(\sQ(n))$ and $\chi(\sR(n))$
1335are polynomials. Twisting the above exact sequence by $n$ and
1336applying $\chi$ yields
1337$$\chi(\F(n))-\chi(\F(n-1))=\chi(\Q(n))-\chi(\sR(n))=P_{\sQ}(n)-P_{\sR}(N).$$
1338Thus the first difference function of $\chi(\F(n))$ is a polynomial
1339so $\chi(\F(n))$ is a polynomial.
1340\end{proof}
1341
1342
1343\begin{example}
1344Let $X=\P^1$ 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 $p_M(n)=\varphi(n)$
1346for $n\geq -1$. Computing the Hilbert polynomial in terms of
1347the Euler characteristic gives
1348$$\chi(\F(n))=h^0(\sox(n))- h^1(\sox(n)) 1349 = \begin{cases} (n+1) - 0 & n\geq -1\\ 1350 0-(-n-1)=n+1 & n\leq -2 1351 \end{cases}$$
1352Thus $p_{\F}(n)=n+1$.
1353\end{example}
1354
1355The higher cohomology corrects the failure of the
1356Hilbert 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
1362Look at Serre's 1956 paper {\em Geometrie Algebraique et Geometrie
1363Analytique} (GAGA). What are the prerequisites?'' asks Janos.
1364French,'' answers Nghi. Is there an English translation'' asks the
1365class. Translation? ... What for? It's so beautiful in the French,''
1366retorts Hartshorne.
1367
1368Let $\F$ be a coherent sheaf on $\P^n_{\C}$ with its Zariski topology.
1369Then we can associate to $\F$ a sheaf
1370$\F^{\mbox{\rm an}}$
1371on $\P^n_{\C}$ with its analytic topology. $\F$ is locally a cokernel
1372of a morphism of free sheaves so we can define
1373$\F^{\mbox{\rm an}}$
1374by defining $\sox^{\mbox{\rm an}}$.
1375The map
1376$$\Coh(\P^n_{\C})\xrightarrow{\mbox{\rm an}}\Coh^{\mbox{\rm an}}(\P^n_{\C})$$
1377is an equivalence of categories and
1378$$H^i(X,\F)\iso{}H^i(X^{\mbox{\rm an}},\F^{\mbox{\rm an}})$$
1379for all $i$.
1380If $X/\C$ is affine the corresponding object $X^{\mbox{\rm an}}_{\C}$
1381is a Stein manifold.
1382
1383\section{Arithmetic Genus}
1384Let $X\hookrightarrow{}\P_k^n$ be a projective variety with
1385$k$ algebraically closed and suppose $\F$ is a coherent sheaf
1386on $X$. Then $$\chi(\F)=\sum(-1)^i h^i(\F)$$ is the Euler characteristic
1387of $\F$,  $$P_{\F}(n)=\chi(\F(n))$$ gives the Hilbert polynomial of
1388$\F$ on $X$, and $$p_a(X)=(-1)^{\dim X}(P_{\sox}(0)-1)$$ is the
1389arithmetic genus of $X$. The arithmetic genus is independent
1390of the choice of embedding of $X$ into $\P_k^n$.
1391
1392If $X$ is a curve then $$1-p_a(X)=h^0(\sox)-h^1(\sox)$$.
1393Thus 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
1395then $p_a(X)=h^1(\sox)$ is called {\bfseries the genus} of $X$.
1396
1397Let $V_1$ and $V_2$ be varieties, thus they are projective integral
1398schemes over an algebraically closed field $k$. Then $V_1$ and
1399$V_2$ are {\bfseries birationally equivalent} if and only if $K(V_1)\isom{}K(V_2)$
1400over $k$, 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 $n$.
1402Since a rational map on a nonsingular projective curve always extends,
1403two nonsingular projective curves are birational if and only if
1404they are isomorphic. Thus for nonsingular projective curves
1405the genus $g$ is a birational invariant.
1406
1407\subsection{The Genus of Plane Curve of Degree $d$}
1408Let $C\subset\P_k^2$ be a curve of degree $d$. Then $C$
1409is 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
1413Some 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
1421Now we compute $p_a(C)$. Let $I=(f)$ with $\deg f=d$.
1422Then $$1-p_a=h_0(\so_C)-h_1(\so_C)+h_2(\so_C)=\chi(\so_C).$$
1423We have an exact sequence
1424$$0\into\sI_C\into\so_{\P^2}\into\so_C\into 0.$$
1425Now $\sI_C\isom\so_{\P^2}(-d)$ since $\so_{\P^2}(-d)$
1426can be thought of as being generated by $1/f$ on $D_+(f)$
1427and by something else elsewhere, and then multiplication
1428by $f$ gives an inclusion
1429$so_{\P^2}(-d)|_{D_+(f)}\into\so_{\P^2}|_{D_+(f)}$, etc.
1430Therefore
1431$$\chi(\so_C)=\chi(\so_{\P^2})-\chi(\so_{\P^2}(-d)).$$
1432Now
1433$$\chi(\so_{\P^2})=h^0(\so_{\P^2})-h^1(\so_{\P^2})+h^2(\so_{\P^2})=1+0+0$$
1434and
1435$$\chi(\so_{\P^2}(-d))=h^0(\so_{\P^2}(-d))-h^1(\so_{\P^2}(-d))+h^2(\so_{\P^2}(-d)) 1436 =0+0+\frac{1}{2}(d-1)(d-2).$$
1437For the last computation we used duality (14.1) to see that
1438$$h^2(\so_{\P^2}(-d))=h^0(\so_{\P^2}(d-3)=\dim S_{d-3} 1439 =\frac{1}{2}(d-1)(d-2).$$
1440Thus $\chi(\so_C)=1-\frac{1}{2}(d-1)(d-2)$ 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
1449sheaves on $X=\P_k^1$ doesn't have enough projectives.
1450\end{exercise}
1451\begin{proof}
1452We show that there is no projective object $\sP\in\Qco(X)$ along
1453with a surjection $\sP\into\sox\into 0$.
1454\begin{lem}
1455If $\P\xrightarrow{\varphi}\sox$ is surjective and $\sP$ is
1456quasi-coherent, then there exists $\ell$ such that
1457$H^0(\P(\ell))\into{}H^0(\sox(\ell))$ is surjective.
1458\end{lem}
1459The {\em false} proof of this lemma is to write down an
1460exact sequence $0\into\sR\into\sP\into\sox\into 0$ then
1461use the fact'' that $H^1(\sR(\ell))=0$ for sufficiently
1462large $\ell$. This doesn't work because $\sR$ might not
1463be coherent since it is only the quotient of
1464quasi-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
1466subsheaves, $\sP=\union_{i}\sP_{i}$.
1467Then since $\varphi$ is surjective,
1468$\sox=\union_{i}\varphi(\sP_{i})$, where $\varphi(\sP_{i})$
1469is the sheaf image. Using the fact that $\varphi(\sP_{i})$ is
1470the sheaf image, that $\sox$ is coherent and that the union
1471is ascending, this implies $\sox=\varphi(\sP_i)$ for some $i$.
1472We now have an exact sequence
1473$$0\into\sR_i\into\sP_i\into\sox\into{}0$$
1474with $\sR_i$ coherent since $\sP_i$ and $\sox$ are both coherent.
1475Thus $H^i(\sR_i(\ell))=0$ for $l\gg 0$ which, upon computing
1476the long exact sequence of cohomology, gives the lemma.
1477
1478Now 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}$$
1484Twisting by $\ell$ gives a commutative diagram
1485$$\begin{CD} 1486\sP(\ell)@>>>\sox(\ell)@>>>0\\ 1487@VVV @VVV\\ 1488\sox(-1)@>>>k(p)@>>>0 1489\end{CD}$$
1490Let $s\in\Gamma(\sox(\ell))$ be a global section which is nonzero at $p$,
1491then there is $t\in\Gamma(\sP(\ell))$ which maps to $s$.
1492But then by commutativity $t$ must map to some element of $\Gamma(\sox(-1))=0$
1493which maps to a nonzero element of $k(p)$, which is absurd.
1494\end{proof}
1495
1496\section{Some Special Cases of Serre Duality}
1497\subsection{Example: $\sox$ on Projective Space}
1498Suppose $X=\P_k^n$, then there is a perfect pairing
1499$$H^0(\sox(\ell))\times{}H^n(\sox(-\ell-n-1))\into{}H^n(\sox(-n-1))\isom{}k.$$
1500For this section let
1501$$\omega_X=\sox(-n-1).$$
1502Because the pairing is perfect we have a non-canonical but functorial
1503isomorphism
1504$$H^0(\sox(\ell))\isom H^n(\sox(-\ell-n-1))'.$$
1505(If $V$ is a vector space then $V'$ denotes its dual.)
1506
1507\subsection{Example: Coherent sheaf on Projective Space}
1508Suppose $\sF$ is any coherent sheaf on $X=\P_k^r$.
1509View $\Hom(\sF,\omega)$ as a $k$-vector space.
1510
1511% WHY DEFINE SHEAF HOM???
1512%We recall the definition of the sheaf Hom.
1513%\begin{defn}
1514%Let $\sF$ and $\sG$ be sheaves of $\sox$-modules.
1515%Then $\sHom(\sF,\sG)$ is the sheaf which associates to
1516%an open set the group
1517%$$Hom_{\Mod(\sox|_U)}(\sF|_U,\sG|_U).$$
1518%\end{defn}
1519
1520By functoriality and since $H^n(\omega)=k$ there is a map
1521$$\varphi:\Hom(\sF,\omega)\into\Hom(H^n(\sF),H^n(\omega))=H^n(\sF)'.$$
1522\begin{prop} $\varphi$ is an isomorphism for all coherent sheaves $\sF$.
1523\end{prop}
1524\begin{proof}
1525\par {\em Case 1.} If $\sF=\sox(\ell)$ for some $\ell\in\Z$ then this is
1526just a restatement of the previous example.
1527\par {\em Case 2.} If $\sE=\oplus_{i=1}^{k}\so(\ell_i)$ is a finite
1528direct sum, then the statement follows from the
1529commutativity of the following diagram.
1530$$\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}$$
1535\par {\em Case 3.} Now let $\sF$ be an arbitrary coherent sheaf.
1536View $\varphi$ as a morphism of functors
1537$$\Hom(\cdot,\omega)\into H^n(\cdot)'.$$
1538The functor $\Hom(\cdot,\omega)$ is contravarient left exact.
1539$H^n(\cdot)$ is covariant right exact since $X=\P_k^n$ so
1540$H^{n+1}(\sF)=0$ for any coherent sheaf $\sF$. Thus $H^n(\cdot)'$
1541is contravarient left exact.
1542\begin{lem}
1543Let $\sF$ be any coherent sheaf. Then there exists a partial resolution
1544$$\sE_1\into\sE_0\into\sF\into{}0$$
1545by sheaves of the form $\oplus_{i}\sox(\ell_i)$.
1546\end{lem}
1547By (II, 5.17) for $\ell\gg 0$, $\sF(\ell)$ is
1548generated by its global sections. Thus there is a surjection
1549$$\sox^m\into\sF(\ell)\into 0$$
1550which upon twisting by $-\ell$ becomes
1551$$\sE_0=\sox(-\ell)^m\into\sF\into 0.$$
1552Let $\sR$ be the kernel so
1553$$0\into\sR\into\sE\into\sF$$
1554is exact. Since $\sR$ is coherent, we can repeat the argument
1555above to find $\sE_1$ surjecting onto $\sR$. This yields
1556the desired exact sequence.
1557
1558Now we apply the functors $\Hom(\cdot,\omega)$ and
1559$H^n(\cdot)'$. This results in a commutative diagram
1560$$\begin{CD} 15610@>>> \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\\ 15630@>>> H^n(\sF)'@>>>H^n(\sE_0)' @>>> H^n(\sE_1)' 1564\end{CD}$$
1565From cases 1 and 2, the maps $\varphi(\sE_0)$ and
1566$\varphi(\sE_1)$ are isomorphisms so $\varphi(\sF)$ must also be
1567an isomorphism.
1568\end{proof}
1569
1570\subsection{Example: Serre Duality on $\P_k^n$}
1571Let $X=\P_k^n$ and $\sF$ be a coherent sheaf.
1572Then for each $i$ there is an isomorphism
1573$$\varphi^{i}:\ext^{i}_{\sox}(\sF,\omega)\into H^{n-i}(\sF)'.$$
1574
1575%%%%%%%%%%%%%%%%%%%%
1576%% 2/23/96
1577
1578\section{The Functor $\ext$}
1579Let $(X,\sox)$ be a scheme and $\sF,\sG\in\Mod(\sox)$. Then
1580$\Hom(\sF,\sG)\in\Ab$. View $\Hom(\sF,\bullet)$ as a
1581functor $\Mod(\sox)\into\Ab$. Note that $\Hom(\sF,\bullet)$
1582is left exact and covariant. Since $\Mod(\sox)$ has enough
1583injectives we can take derived functors.
1584\begin{defn}
1585The $\ext$ functors $\ext^i_{\sox}(\sF,\bullet)$ are the right
1586derived functors of $\Hom_{\sox}(\sF,\bullet)$ in the
1587category $\Mod(\sox)$.
1588\end{defn}
1589Thus to compute $\ext^i_{\sox}(\sF,\sG)$, take an injective
1590resolution $$0\into\sG\into I^0\into I^1\into \cdots$$
1591then
1592$$\ext^i_{\sox}(\sF,\sG)=H^i(\Hom_{\sox(\sF,I^{\bullet})}).$$
1593
1594\begin{remark} {\bfseries Warning!}
1595If $i:X\hookrightarrow\P^n$ is a closed subscheme of $P^n$ then
1596$\ext^i_{\sox}(\sF,\sG)$ need {\em not} equal
1597$\ext^i_{\P^n}(i_{*}(\sF),i_{*}(\sG))$. With cohomology
1598these are the same, but not with $\ext$!
1599\end{remark}
1600
1601\begin{example}
1602Suppose $\sF=\sox$, then
1603$\Hom_{\sox}(\sox,\sG)=\Gamma(X,\sG)$. Thus
1604$\ext^i_{\sox}(\sox,\bullet)$ are the derived
1605functors of $\Gamma(X,\bullet)$ in $\Mod(\sox)$.
1606Since we can computer cohomology using flasque
1607sheaves this implies $\ext^i_{\sox}(\sox,\bullet)=H^i(X,\bullet)$.
1608Thus $\ext$ generalizes $H^i$ but we get a lot more besides.
1609\end{example}
1610
1611\subsection{Sheaf $\ext$}
1612Now we define a new kind of $\ext$. The sheaf hom functor
1613$$\sHom_{\sox}(\sF,\bullet):\Mod(\sox)\into\Mod(\sox)$$
1614is covariant and left exact. Since $\Mod(\sox)$ has enough
1615injectives we can defined the derived functors
1616$\sext^i_{\sox}(\sF,\bullet)$.
1617
1618\begin{example}
1619Consider the functor $\ext^i_{\sox}(\sox,\bullet)$.
1620Since $\sHom_{\sox}(\sox,\sG)=\sG$ this is the identity
1621functor which is exact so
1622$$\ext^i_{\sox}(\sox,\sG)=\begin{cases}\sG\quad i=0\\0\quad i>0\end{cases}$$
1623\end{example}
1624
1625What if we have a short exact sequence in the first variables, do we
1626get a long exact sequence?
1627\begin{prop}
1628The functors $\ext^i$ and $\sext^i$ are $\delta$-functors
1629in the first variable. Thus if $$0\into\sF'\into\sF\into\sF''\into 0$$
1630is exact then there is a long exact sequence
1631\begin{align*}0\into&\Hom(\sF'',\sG)\into\Hom(\sF,\sG)\into\Hom(\sF',\sG)
1632\into&\ext^1(\sF'',\sG)\into\ext^1(\sF,\sG)\into\ext^1(\sF,\sG)\into\cdots
1633\end{align*}
1634\end{prop}
1635The conclusion of this proposition is not obvious because we
1636the $\ext^i$ as derived functors in the second variable, not the first.
1637\begin{proof}
1638Suppose we are given $0\into\sF'\into\sF\into\sF''\into 0$ and $\sG$.
1639Choose an injective resolution $0\into\sG\into{}I^{\bullet}$ of $\sG$.
1640Since $\Hom(\bullet,I^n)$ is exact (by definition of injective object),
1641the sequence
1642$$0\into\Hom(\sF'',I^{\bullet})\into\Hom(\sF,I^{\bullet})\into 1643 \Hom(\sF',I^{\bullet})\into 0$$
1644is exact. By general homological algebra these give rise a long
1645exact sequence of cohomology of these complexes. For $\sext^i$
1646simply scriptify everything!
1647\end{proof}
1648
1649\subsection{Locally Free Sheaves}
1650\begin{prop}
1651Suppose $\sE$ is a locally free $\sox$-module of finite rank.
1652Let $\sE^{\dual}=\Hom(\sE,\sO)$. For any sheaves $\sF$, $\sG$,
1653$$\ext^i(\sF\tensor\sE,\sG)\isom\ext^i(\sF,\sG\tensor\sE^{\dual})$$
1654and
1655$$\sext^i(\sF\tensor\sE,\sG)\isom\sext^i(\sF,\sG\tensor\sE^{\dual}) 1656\isom\sext^i(\sF,\sG)\tensor\sE^{\dual}.$$
1657\end{prop}
1658
1659\begin{lem}
1660If $\sE$ is locally free of finite rank and $\sI\in\Mod(\sox)$ is
1661injective then $\sI\tensor\sE$ is injective.
1662\end{lem}
1663\begin{proof}
1664Suppose $0\into\sF\into\sG$ is an injection and there is a map
1665$\varphi:\sF\into\sI\tensor\sE$. Tensor everything with $\sE^{\dual}$.
1666Then we have an injection $0\into\sF\tensor\sE^{\dual}\into\sG\tensor\sE^{\dual}$
1667and a map $\varphi':\sF\tensor\sE^{\dual}\into\sI$. Since $\sI$ is injective
1668there is a map $\sG\tensor\sE^{\dual}\into\sI$ which makes
1669the appropriate diagram commute. Tensoring everything with $\sE$ gives
1670a map making the original diagram commute.
1671\end{proof}
1672
1673\begin{proof}[of proposition] Let $0\into\sG\into\sI^{\bullet}$
1674by an injective resolution of $\sG$. Since
1675$$\Hom(\sF\tensor\sE,\sI^{\bullet})=\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual}),$$
1676we see that
1677$$0\into\sG\tensor\sE^{\dual}\into\sI^{\cdot}\tensor\sE{\dual}$$
1678is an injective resolution of $\sG\tensor\sE^{\dual}$.
1679Thus $\Hom(\sF,\sI^{\bullet}\tensor\sE^{\dual})$ computes
1680$\ext(\sF\tensor\sE,\bullet)$.
1681\end{proof}
1682
1683\begin{prop}
1684If $\sF$ has a locally free resolution
1685$\sE_{\cdot}\into\sF\into 0$ then
1686$$\sext^i_{\sox}(\sF,\sG)=H^i(\sHom(\sE_{\bullet},\sG)).$$
1687\end{prop}
1688
1689\begin{remark}
1690Notice that when $i>0$ and $\sE$ is locally free,
1691$$\sext^i(\sE,\sG)=\sext^i(\sox,\sG\tensor\sE^{\dual})=0.$$
1692\end{remark}
1693
1694\begin{proof}
1695Regard both sides as functors in $\sG$. The left hand side is a
1696$\delta$-functor and vanishes for $\sG$ injective. I claim that
1697that right hand side is also a $\delta$-functor and vanishes for
1698$\sG$-injective.
1699\begin{lem}
1700If $\sE$ is locally free, then $\shom(\sE,\bullet)$ is exact.
1701\end{lem}
1702\end{proof}
1703
1704%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1705%% 2/26/96
1706\section{More Technical Results on $\ext$}
1707Let $(X,\sox)$ be a scheme and $\sF,\sG$ be sheaves
1708in the category $\Mod(\sox)$. Then $\ext(\sF,\sG)$
1709and $\sext(\sF,\sG)$ are the derived functors of $\Hom$, resp.
1710$\shom$, in the second variable.
1711
1712\begin{lem}
1713If $\sF$ and $\sG$ are coherent over a Noetherian scheme
1714$X$, then $\sext^{i}(\sF,\sG)$ is coherent.
1715\end{lem}
1716This lemma would follow immediately from the following fact
1717which we haven't proved yet.
1718\begin{fact}
1719Let $X=\spec A$ with $A$ Noetherian and let $M$ be an $A$-module.
1720Then $$\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{}}.$$
1722\end{fact}
1723Instead of using the fact we can prove the lemma using
1724a proposition from yesterday.
1725\begin{proof}
1726Choose a locally free resolution
1727$$\sL_{\bullet}\into\sF\into 0$$
1728of $\sF$. Then
1729$$\sext^{i}(\sF,\sG)=H^i(\shom(\sL_{\bullet},\sG)).$$
1730But all of the kernels and cokernels in
1731$\shom(\sL_{\bullet},\sG)$
1732are coherent, so the cohomology is. (We can't
1733just choose an injective resolution of $\sF$ and apply
1734the definitions because there is no guarantee that we
1735can find an injective resolution by coherent sheaves.)
1736\end{proof}
1737
1738\begin{prop}
1739Let $X$ be a Noetherian projective scheme over $k$ and let $\sF$
1740and $\sG$ be coherent on $X$. Then for each $i$ there exists
1741an $n_0$, depending on $i$, such that for all $n\geq{}n_0$,
1742$$\ext^i_{\sox}(\sF,\sG(n))=\Gamma(\sext^i_{\sox}(\sF,\sG(n))).$$
1743\end{prop}
1744\begin{proof}
1745When $i=0$ the assertion is that
1746$$\Hom(\sF,\sG(n))=\Gamma(\shom(\sF,\sG(n)))$$
1747which is obvious.
1748
1749{\em Claim.} Both sides are $\delta$-functors in $\sF$. We
1750have already showed this for the left hand side. [I don't understand
1751why the right hand side is, but it is not trivial and it caused
1752much consternation with the audience.]
1753
1754To show the functors are isomorphic we just need to show both
1755sides are coeffaceable. That is, for every coherent sheaf $\sF$
1756there is a coherent sheaf $\sE$ and a surjection $\sE\into\sF\into{}0$
1757such that $\ext^i_{\sox}(\sE)=0$ and similarly for the right hand
1758side. Thus every coherent sheaf is a quotient of an acyclic sheaf.
1759
1760Suppose $\sF$ is coherent. Then for $\ell\gg 0$, $\sF(\ell)$ is generated
1761by its global sections, so there is a surjection
1762$$\sox^a\into\sF(\ell)\into{}0.$$
1763Untwisting gives a surjection
1764$$\sox(-\ell)^a\into\sF\into{}0.$$
1765Let $\sE=\sox(-\ell)^a$, then I claim that $\sE$ is acyclic for both
1766sides. First consider the left hand side. Then
1767\begin{align*}\ext^i(\oplus\so(-\ell),\sG(n))&=\oplus\ext^i(\sox(-\ell),\sG(n))\\
1768&=\oplus\ext^i(\sox,\sG(\ell+n))\\
1769&=H^i(X,\sG(\ell+n))\end{align*}
1770By Serre (theorem 5.2 of the book) this is zero for $n$ sufficiently large.
1771For the right hand side the statement is just that
1772$$\sext^i(\sE,\sG(n))=0$$
1773which we have already done since $\sE$ is a locally free sheaf.
1774
1775Thus both functors are universal since they are coeffaceable. Since universal
1776functors are completely determined by their zeroth one they must be equal.
1777\end{proof}
1778
1779\begin{example}
1780One might ask if $\ext^i$ necessarily vanishes for sufficiently large $i$.
1781The answer is no. Here is an algebraic example which can be converted
1782to a geometric example. Let $A=k[\varepsilon]/(\varepsilon^2)$, then
1783a projective resolution $L_{\bullet}$ of $k$ is
1784$$\cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A 1785 \xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}k\into 0.$$
1786Then $Hom(L_{\bullet},k)$ is the complex
1787$$k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}k\xrightarrow{0}\cdots$$
1788Thus $\ext^i_A(k,k)=k$ for all $i\geq 0$.
1789\end{example}
1790
1791\section{Serre Duality}
1792We are now done with technical results on $\ext$'s so we can
1793get back to Serre duality on $\P^n$.
1794Let $X=\P^n_k$ and let $\omega=\sox(-n-1)$. Note that this
1795is an {\em ad hoc} definition of $\omega$ which just happens
1796to work since $X=\P^n_k$. In the more general situation it
1797will be an interesting problem just to show the so called
1798dualizing sheaf $\omega$ actually exists. When our variety
1799is nonsingular, $\omega$ will be the canonical sheaf.
1800We have shown that for any coherent sheaf $\sF$ there is a map
1801$$\Hom(\sF,\omega)\iso H^n(\sF)^{\dual}.$$
1802The map is constructed by using the fact that $H^n$ is a
1803functor:
1804$$\Hom(\sF,\omega)\into\Hom_k(H^n(\sF),H^n(\omega)) 1805 = \Hom_k(H^n(\sF),k)=H^n(\sF)^{\dual}.$$
1806
1807We shall use satellite functors to prove the following theorem.
1808\begin{thm} Let $\sF$ be a coherent sheaf on $\P^n_k$. Then
1809there is an isomorphism $$\ext^i(\sF,\omega)\iso H^{n-i}(\sF)^{\dual}.$$
1810\end{thm}
1811\begin{proof} Regard both sides as functors in $\sF$.
1812\par {\em 1. Both sides are $\delta$-functors in $\sF$.}
1813We have already checked this for $\ext^i$. Since $H^{n-i}$
1814is a delta functor in $\sF$, so is $(H^{n-i})^{\dual}$.
1815Note that both sides are contravarient.
1816\par {\em 2. They agree for $i=0$.} This was proved last time.
1817\par {\em 3. Now we just need to show both sides are coeffaceable.}
1818Suppose $\sE\into\sF\into 0$ with $\sE=\sO(-\ell)^{\oplus a}$.
1819For some reason we can assume $\ell\gg{}0$.
1820We just need to show both sides vanish
1821on this $\sE$. First computing the left hand side gives
1822$$\oplus\ext^i(\sO(-\ell),\omega)=H^i(\omega(\ell))=0$$
1823for $\ell\gg 0$. Next computing the right hand side we get
1824$$H^{n-i}(\sO(-\ell))=0$$
1825by the explicit computations of cohomology of projective
1826space (in particular, note that $i>0$).
1827\end{proof}
1828
1829Next time we will generalize Serre duality to an arbitrary
1830projective scheme $X$ of dimension $n$.
1831We will proceed in two steps. The first is to ask,
1832what is $\omega_X$? Although the answer to this question
1833is easy on $\P_k^n$ it is not obvious what the suitable
1834analogy should be for an arbitrary projective variety.
1835Second we will define natural maps
1836$$\ext^i_{\sox}(\sF,\omega)\xrightarrow{\varphi^i}H^{n-i}(\sF)^{\dual}$$
1837where $n=\dim X$.
1838Unlike in the case when $X=\P^n_k$, these maps are not necessarily
1839isomorphisms unless $X$ is locally Cohen-Macaulay (the local rings
1840at each point are Cohen-Macaulay).
1841\begin{defn}
1842Let $A$ be a nonzero Noetherian local ring with residue field $k$.
1843Then the {\bfseries depth} of $A$ is
1844$$\operatorname{depth} A = \inf\{i:\ext^i_A(k,A)\neq 0\}.$$
1845$A$ is said to be {\bfseries Cohen-Macaulay} if
1846$\operatorname{depth} A = \dim A$.
1847\end{defn}
1848
1849%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1850%% 2/28/96
1851
1852\section{Serre Duality for Arbitrary Projective Schemes}
1853Today we will talk about Serre duality for an arbitrary projective
1854scheme. We have already talked about Serre duality in the special
1855case $X=\P^n_k$. Let $\sF$ be a locally free sheaf. We showed
1856there is an isomorphism
1857$$\ext^i(\sF,\omega_X)\iso H^{n-i}(\sF)^{\dual}.$$
1858This was established by noting that
1859$$\ext^i(\sF,\omega_X)=\ext^i(\sox,\sF^{\dual}\tensor\omega_X) 1860=H^i(\sF^{\dual}\tensor\omega_X).$$
1861Another thing to keep in mind is that locally free sheaves correspond
1862to what, in other branches of mathematics, are vector bundles. They
1863aren't the same object, but there is a correspondence.
1864
1865We would like to generalize this to an arbitrary projective scheme $X$.
1866There are two things we must do.
1867\begin{enumerate}
1868\item Figure out what $\omega_X$ is.
1869\item Prove a suitable duality theorem.
1870\end{enumerate}
1871When $X=\P_k^n$ it is easy to find a suitable $\omega_X=\so_{\P^n_k}(-n-1)$
1872because of the explicit computations we did before. We now define $\omega_X$
1873to be a sheaf which will do what we hope it will do. Of course existence
1874is another matter.
1875\begin{defn}
1876Let $X$ be a Noetherian scheme of finite type over a field $k$ and
1877let $n=\dim X$. Then a {\bfseries dualizing sheaf} for $X$ is
1878a coherent sheaf $\omega_X$ along with a map
1879$t:H^n(\omega_X)\into{}k$, such that for all coherent sheaves $\sF$ on
1880$X$, the map
1881$\Hom_{\sox}(\sF,\ox)\into{}H^n(\sF)^{\dual}$ is an
1882isomorphism. The latter map is defined by the diagram
1883$$\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}$$
1888\end{defn}
1889Strictly speaking, a dualizing sheaf is a pair $(\omega_X,t)$. Note
1890that on $\P^n$ we had $H^n(\omega_{\P^n})\isom{}k$, but on an
1891arbitrary scheme $X$ we only have a map from $H^n(\omega_X)$ to $k$
1892which need not be an isomorphism. The definition never mentions existence.
1893
1894\begin{prop}
1895If $X$ admits a dualizing sheaf $(\ox,t)$ then the pair
1896$(\ox,t)$ is unique up to unique isomorphism, i.e., if
1897$(\eta,s)$ is another dualizing sheaf for $X$ then there
1898is a unique isomorphism $\varphi:\ox\into\eta$ such that
1899$$\begin{array}{ccc} 1900H^n(\ox)&\xrightarrow{H^n(\varphi)}&H^n(\eta)\\ 1901t\downarrow&&\downarrow{}s\\ 1902k&=&k\end{array}$$
1903commutes.
1904\end{prop}
1905Before we prove the proposition we make a short digression to
1906introduce representable functors which give a proof of the uniqueness
1907part of the above proposition.
1908\begin{defn}
1909Let $\sC$ be a category and $\sD$ a category whose objects happen
1910to be sets. Suppose $T:\sC\into\sD$ is a contravarient functor.
1911Then $T$ is {\bfseries representable} if there exists an object
1912$\omega\in\Ob(\sC)$ and an element $t\in T(\omega)$ such that
1913for all $F\in\Ob(\sC)$ the map $\Hom_{\sC}(F,\omega)\into{}T(F)$
1914is a bijection of sets. The latter map is defined by the diagram
1915$$\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}$$
1919\end{defn}
1920Thus there is an isomorphism of functors $\Hom(\bullet,\omega)=T(\bullet)$.
1921The pair $(t,\omega)$ is said to represent the functor $T$.
1922The relevant application of this definition is to the case
1923when $\sC=\Coh(X)$, $\sD=\{\text{$k$-vector spaces}\}$,
1924$T$ is the functor $F\mapsto H^n(\sF)^{\dual}$. Then $\omega=\ox$
1925and
1926$$t=t\in\Hom(H^n(\omega),k)=H^n(\omega)^{\dual}=T(\omega).$$
1927\begin{prop}
1928If $T$ is a representable functor, then the pair
1929$(\omega,t)$ representing it is unique.
1930\end{prop}
1931\begin{proof}
1932Suppose $(\omega,t)$ and $(\eta,s)$ both represent the functor $T$.
1933Consider the diagram
1934$$\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}$$
1938By definition the map $\Hom(\eta,\omega)\into{}T(\eta)$ is
1939bijective. Since $s\in T(\eta)$, there is $\varphi\in\Hom(\eta,\omega)$
1940such that $\varphi\mapsto{}s\in{}T(\eta)$. Thus $\varphi$ has
1941the property that $T(\varphi)(t)=s$. This argument uses the
1942fact that $(\omega,t)$ represents $T$. Using the fact that
1943$(\eta,s)$ represents $T$ implies that there exists
1944$\psi\in\Hom(\omega,\eta)$ such that $T(\psi)(s)=t$.
1945We have the following pictures
1946$$\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}}\\ 1953t\in{} T(\omega)&&T(\eta) \ni{} s\\ 1954&\xrightarrow{\hspace{.2in}T(\phi)=\phi^{*}\hspace{.2in}} 1955\end{array}$$
1956I claim that
1957$$\psi\circ\varphi=\Id\in\Hom(\eta,\eta).$$
1958In diagram form we have
1959$$\eta\xrightarrow{\varphi}\omega\xrightarrow{\psi}\eta$$
1960which upon applying $T$ gives
1961\begin{align*}
1962T(\eta)&\xrightarrow{\psi^{*}}T(\omega)\xrightarrow{\varphi^{*}}T(\eta)\\
1963s&\mapsto t\mapsto s\end{align*}
1964Where does $\psi\circ\varphi$ go to
1965under the map $\Hom(\eta,\eta)\iso T(\eta)$? By definition
1966$\psi\circ\varphi$ goes to the evaluation of $T(\psi\circ\varphi)$
1967at $s\in T(\eta)$. But, as indicated above, the evaluation of
1968$T(\psi\circ\varphi)$ at $s$ is just $s$ again.  But the identity
1969morphism $1_{\eta}\in\Hom(\eta,\eta)$ also maps to $s$
1970under the map $\Hom(\eta,\eta)\iso T(\eta)$. Since this
1971map is a bijection this implies that $\psi\circ\varphi=1_{\eta}$,
1972as desired. Similarly $\varphi\circ\psi=1_{\omega}$. Thus
1973$\psi$ and $\varphi$ are both isomorphisms.
1974\end{proof}
1975\begin{quote}
1976When you define something and it is unique up to unique isomorphism,
1977you know it must be good.''
1978\end{quote}
1979We return to the question of existence.
1980\begin{prop}
1981If $X$ is a projective scheme over a field $k$ then $(\ox,t)$ exists.
1982\end{prop}
1983\begin{lem}
1984If $X$ is an $n$ dimensional projective scheme over a field $k$, then
1985there is a finite morphism $f:X\into\P^n_k$.
1986\end{lem}
1987\begin{proof}
1988Embed $X$ in $\P^N$ then choose a linear projection down to
1989$\P^n$ which is sufficiently general.
1990$$\begin{array}{ccc} 1991X&\hookrightarrow&\P^N\\ 1992f\searrow&&\downarrow\\ 1993&&\P^n\end{array}$$
1994Let $L$ be a linear space of dimension $N-n-1$ not meeting $X$.
1995Let the map from $\P^N\into\P^n$ be projection through $L$.
1996By construction $f$ is quasi-finite, i.e., for all $Q\in\P^n$,
1997$f^{-1}(Q)$ is finite. It is a standard QUALIFYING EXAM problem to
1998show that if a morphism is quasi-finite and projective then it is
1999finite. This can be done by applying (II, Ex. 4.6) by covering
2000$X$ by subtracting off hyperplanes and noting that the correct
2001things are affine by construction. See also (III, Ex. 11.2) for
2002the more general case when $f$ is quasi-finite and proper, but not
2003necessarily projective.
2004\end{proof}
2005
2006%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
2007%% 3/1/96
2008%%%%%%%%%%%%%%%%%%%%%%%%
2009\section{Existence of the Dualizing Sheaf on a Projective Scheme}
2010Let $X$ be a scheme over $k$. Recall that a
2011dualizing sheaf is a pair $(\omega,t)$ where $\omega$ is a coherent
2012sheaf on $X$ and
2013$$t:H^n(X,\omega)\into k$$
2014is a homomorphism such that for all coherent sheaves $\sF$
2015the natural map
2016$$\Hom_X(\sF,\omega)\into{}H^n(\sF)^{\dual}$$
2017is an isomorphism.
2018We know that such a dualizing sheaf exists on $\P_k^n$.
2019\begin{thm}
2020If $X$ is a projective scheme of dimension $n$ over $k$, then
2021$X$ has a dualizing sheaf.
2022\end{thm}
2023The book's proof takes an embedding
2024$j:X\hookrightarrow{}\P_k^N$ and works on $X$
2025as a subscheme of $\P_k^N$. Then the book's proof shows that
2026$$\omega_X=\sext_{\sO_{\P^N_k}}^{N-n}(\sox,\omega_{\P^N_k}).$$
2027Today we will use a different method.
2028\begin{defn}
2029A {\bfseries finite morphism} is a morphism $f:X\into Y$ of
2030Noetherian schemes such that for any open affine $U=\spec A\subset X$,
2031the preimage $f^{-1}(U)\subset Y$ is affine, say $f^{-1}(U)=\spec B$,
2032and the natural map $A\into B$ turns $B$ into a finitely
2033generated $A$-module. We call $f$ an {\bfseries affine morphism}
2034if we just require that $f^{-1}(U)$ is affine
2035but not that $B$ is a finitely generated $A$-module.
2036A morphism $f:X\into{}Y$ is {\bfseries quasi-finite} if for all
2037$y\in Y$ the set $f^{-1}(y)$ is finite.
2038\end{defn}
2039\begin{example}
2040Consider the morphism
2041$$j:\P^1-\{\pt\}\injects\P^1.$$
2042Since $\P^1$ minus any nonempty finite set of points is
2043affine $j$ is affine. But it is not finite. Indeed, let
2044$a$ be a point different from $\pt$ and let $U=\P^1-\{a\}$.
2045Then $U=\spec k[x]$ and
2046$$j^{-1}(U)=\P^1-\{\pt,a\}=\spec k[x,x^{-1}],$$
2047but $k[x,x^{-1}]$ is not a finitely generated $k[x]$-module.
2048\end{example}
2049\begin{exercise}
2050A morphism can be affine but not finite or even quasi-finite.
2051For example, let $f$ be the natural map
2052$$f:\A^{n+1}-\{0\}\into\P^n$$
2053then show that $f$ is affine.
2054This is the fiber bundle associated to the invertible
2055sheaf $\sO(1)$ [[or is it $\sO(-1)$?]]
2056\end{exercise}
2057
2058\subsection{Relative Gamma and Twiddle}
2059We will now define relative versions of global sections and
2060$\tilde{}$ analogous to the absolute versions.
2061It is not a generalization of the absolute notion, but a
2062relativization.
2063Suppose $X$ is a scheme over $Y$ with structure
2064map $f:X\into{}Y$ and assume $f$ is affine.
2065Then the map sending a sheaf
2066$\sF$ on $X$ to the sheaf $f_{*}\sF$ on
2067$Y$ is the analog of taking global sections.
2068Since $f$ is a morphism there is a map $\soy\into{}f_{*}\sox$
2069so $f_{*}\sox$ is a sheaf of $\soy$-modules. Note that
2070$f_{*}\sF$ is a sheaf of $f_{*}\sox$-modules. Thus
2071we have set up a map
2072$$\Qco(X)\into\{\text{ quasicoherent f_{*}\sox-modules on Y }\}.$$
2073The next natural thing to do is define a map analogous to
2074$\tilde{}$ which goes the other direction.
2075Suppose $\sG$ is a quasi coherent sheaf of $f_{*}\sox$-modules on $Y$.
2076Let $U\subset Y$ be an affine open subset of $Y$. Let $G=\Gamma(U,\sG)$ and
2077write $U=\spec A$. Then since $f$ is an affine morphism,
2078$f^{-1}(U)=\spec B$ where $B=\Gamma(f^{-1}(U),\sox)$.
2079Since $\sG$ is an $f_{*}\sox$-module, and $f_{*}\sox$ over $U$
2080is just $B$ thought of as an $A$-module, we see that $G$ is a $B$-module.
2081Thus we can form the sheaf $\tilde{G}$ on $\spec B=f^{-1}(U)$.
2082Patching the various sheaves $\tilde{G}$ together as $U$ runs through
2083an affine open cover of $Y$ gives a sheaf $\tilde{\sG}$ in
2084$\Qco(X)$.
2085
2086Let $\sG$ be a quasi-coherent sheaf of $\soy$-modules. We can't
2087take $\tilde{}$ of $\sG$ because $\sG$ might not be a sheaf of
2088$f_{*}\sox$-modules. But
2089$\shom_{\soy}(f_{*}\sox,\sG)$
2090is a sheaf of $f_{*}\sox$-modules, so we can
2091form $(\shom_{\soy}(f_{*}\sox,\sG))^{\tilde{}}$. This
2092is a quasi-coherent sheaf on $X$ which we denote $f^{!}(\sG)$.
2093\begin{prop}
2094Suppose $f:X\into{}Y$ is an affine morphism of Noetherian
2095schemes, $\sF$ is coherent on $X$, and $\sG$ is quasi-coherent on $Y$.
2096Then
2097$$f_{*}\shom_{\sox}(\sF,f^{!}\sG)\isom \shom_{\soy}(f_{*}\sF,\sG)$$
2098and passing to global sections gives an isomorphism
2099$$\Hom(\sF,f^{!}\sG)\isom \Hom(f_{*}\sF,\sG).$$
2100Thus $f^{!}$ is a right adjoint for $f_{*}$.
2101\end{prop}
2102\begin{proof}
2103The {\em natural} map is
2104\begin{align*}f_{*}\shom_{\sox}(\sF,f^{!}\sG)&\into
2105\shom_{\soy}(f_{*}\sF,f_{*}f^{!}\sG)\\
2106&=\shom_{\soy}(f_{*}\sF,\shom_{\soy}(f_*\sox,\sG)) \into
2107\shom_{\soy}(f_{*}\sF,\sG)\end{align*}
2108where the map $\shom_{\soy}(f_{*}\sox,\sG)\into\sG$
2109is obtained obtained by evaluation at $1$.
2110Since the question is local we may assume
2111$Y=\spec A$ and $X=\spec B$. Then $\sF$ corresponds
2112to a finitely generated module $M$ over the Noetherian ring $B$
2113and $\sG$ corresponds to a module $N$ over $A$.
2114We must show that
2115$$\Hom_B(M,\Hom_A(B,N))\isom\Hom_A(M,N).$$
2116When $M$ is free over $B$ so that $M=B^{\oplus{}n}$ the equality holds.
2117As functors in $M$, both sides are contravarient and
2118left exact.  Now suppose $M$ is an arbitrary finitely generated
2119$B$-module. Write $M$ as a
2120quotient $F_0/F_1$ where $F_0$ and $F_1$ are both free of
2121finite rank. Applying each of`