1%%%%%%%%%%%%%%%%%%%%%%%%%%
2%% Homework assignment 2
3%%%%
4
5\documentclass[12pt]{article}
6\include{defs}
7\author{William A. Stein}
8\title{Homework 2, MAT256B\\Chapter III, 4.8, 4.9, 5.6}
9
10\begin{document}
11\maketitle
12
13\section{Homework}
14
15\begin{exercise}
16{\bfseries (4.8)} \quad {\em Cohomological Dimension}. Let $X$ be a neotherian separated scheme.
17We define the {\em cohomological dimension} of $X$, denoted $\cd(X)$, to
18be the least integer $n$ such that $H^{i}(X,\F)=0$ for all quasi-coherent
19sheaves $\F$ and all $i>n$. Thus for example, Serre's theorem (3.7) says
20that $\cd(X)=0$ if and only if $X$ is affine. Grothendieck's theorem (2.7)
21implies that $\cd(X)\leq\dim X$.
22
23(a) In the definition of $\cd(X)$, show that it is sufficient to
24consider only coherent sheaves on $X$.
25
26(b) If $X$ is quasi-projective over a field $k$, then it is
27even sufficient to consider only locally free coherent sheaves
28on $X$.
29
30(c) Suppose $X$ has a covering by $r+1$ open affine subsets. Use
31\cech{} cohomology to show that $\cd(X)\leq r$.
32
33(d) If $X$ is a quasi-projective variety of dimension $r$ over a field
34$k$, then $X$ can be covered by $r+1$ open affine subsets. Conclude
35that $\cd(X)\leq \dim X$.
36
37(e) Let $Y$ be a set-theoretic complete intersection of
38codimension $r$ in $X=\P_k^n$. Show that $\cd(X-Y)\leq r-1$.
39\end{exercise}
40
41\begin{proof}
42(a) It suffices to show that if, for some $i$,
43$H^i(X,\F)=0$ for all coherent sheaves $\F$,
44then $H^i(X,\F)=0$ for all quasi-coherent sheaves $\F$.
45Thus suppose the $i$th cohomology of all coherent
46sheaves on $X$ vanishes and let $\F$ be quasi-coherent.
47Let $(\F_{\alpha})$ be the collection of coherent subsheaves
48of $\F$, ordered by inclusion. Then by (II, Ex. 5.15e)
49$\varinjlim\F_{\alpha}=\F$, so by (2.9)
50$$H^i(X,\F)=H^i(X,\varinjlim \F_{\alpha})=\varinjlim H^i(X,\F_{\alpha})=0.$$
51
52(b) Suppose $n$ is an integer and $H^i(X,\F)=0$ for all coherent
53locally free sheaves $\F$ and integers $i>n$. We must show
54$H^i(X,\F)=0$ for all coherent $\F$ and all $i>n$, then applying
55(a) gives the desired result. Since $X$ is quasiprojective
56there is an open immersion
57     $$i:X\hookrightarrow Y\subset \P_k^n$$
58with $Y$ a closed subscheme of $\P_k^n$ and $i(X)$ open in $Y$.
59By (II, Ex. 5.5c) the sheaf $\F$ on $X$ pushes forward to a coherent
60sheaf on $\F'=i_{*}\F$ on $Y$.
61By (II, 5.18) we may write $\F'$ as a quotient of a locally
62free coherent sheaf $\E'$ on $Y$. Letting $\R'$ be the kernel
63gives an exact sequence
64$$0\into\R'\into\E'\into\F'\into 0$$
65with $R'$ coherent (it's the quotient of coherent sheaves).
66Pulling back via $i$ to $X$ gives an exact sequence
67%%%%%%%%%%%%%%%%%%%%%%%%%
68%% This is probably possible because $i$ is an
69%% open immersion, but there are definately some
70%% details to check!!
71%%%%%%%%%%%%%%%%%%%%%%%%%
72$$0\into\R\into\E\into\F\into 0$$
73of coherent sheaves on $X$ with $E$ locally free.
74The long exact sequence of cohomology shows
75that for $i>n$, there is an exact sequence
76$$0=H^i(X,\E)\into H^i(X,\F) \into H^{i+1}(X,\R) \into 77 H^{i+1}(X,\E)=0.$$
78$H^i(X,\E)=H^{i+1}(X,\E)=0$ because we have assumed that, for $i>n$,
79cohomology vanishes on locally free coherent sheaves.
80Thus $H^i(X,\F)\isom H^{i+1}(X,\R)$. But if $k=\dim X$, then
81Grothendieck vanishing (2.7) implies that
82$H^{k+1}(X,\R)=0$ whence $H^{k}(X,\F)=0$.
83But then applying the above argument with $\F$ replaced
84by $\R$ shows that $H^{k}(X,\R)=0$ which implies $H^{k-1}(X,\F)=0$
85(so long as $k-1>n$). Again, apply the entire argument
86with $\F$ replaced by $\R$ to see that $H^{k-1}(X,\R)=0$.
87We can continue this descent and hence show that
88$H^{i}(X,\F)=0$ for all $i>n$.
89
90(c) By (4.5) we can compute cohomology by using the \cech{}
91complex resulting from the cover $\sU$ of $X$ by $r+1$ open
92affines. By definition $\sC^p=0$ for all $p>r$ since
93there are no intersections of $p+1\geq r+2$ distinct open sets
94in our collection of $r+1$ open sets. The \cech{} complex is
95$$\sC^0\into\sC^1\into\cdots\into\sC^r\into\sC^{r+1}=0\into 0\into 0\into \cdots.$$
96Thus if $\sF$ is quasicoherent then $\cH^p(\sU,\sF)=0$
97for any $p>r$ which implies that $\cd(X)\leq r$.
98
99(d) I will first present my solution in the special
100case that $X$ is projective. The more general case when $X$ is
101quasi-projective is similiar, but more complicated, and will
102be presented next.
103Suppose $X\subset\P^n$ is a projective variety of dimension $r$.
104We must cover $X$ with $r+1$ open affines. Let $U$ be nonempty
105open affine subset of $X$. Since $X$ is irreducible, the irreducible
106components of $X-U$ all have codimension at least one in $X$.
107Now pick a hyperplane $H$ which doesn't
108completely contain any irreducible component of $X-U$. We can do
109this by choosing one point $P_i$ in each of the finitely
110many irreducible components of $X-U$ and choosing a hyperplane
111which avoids all the $P_i$. This can be done because the field
112is infinite (varieties are only defined over algebraically
113closed fields) so we can always choose a vector not orthogonal
114to any of a finite set of vectors.
115Since $X$ is closed in $\P^n$ and $\P^n-H$ is affine,
116$(\P^n-H)\intersect X$ is an open affine subset of $X$.
117Because of our choice of $H$, $U\union((\P^n-H)\intersect X)$ is only
118missing codimension two closed subsets of $X$. Let $H_1=H$
119and choose another hyperplane $H_2$ so it doesn't completely
120contain any of the (codimension two) irreducible components
121of $X-U-(\P^n-H_1)$. Then $(\P^n-H_2)\intersect X$ is open
122affine and $U\union((\P^n-H_1)\intersect X)\union((\P^n-H2)\intersect X)$
123is only missing codimension three closed subsets of $X$.
124Repeating this process a few more times yields
125hyperplanes $H_1,\cdots,H_r$ so that
126$$U, (\P^n-H_1)\intersect X, \ldots, (\P^n-H_r)\intersect X$$
127form an open affine cover of $X$, as desired.
128
129Now for the quasi-projective case. Suppose $X\subset\P^n$ is
130quasi-projective. From (I, Ex. 3.5) we know that $\P^n$ minus
131a hypersurface $H$ is affine. Note that the same proof
132works even if $H$ is a union of hypersurfaces. We now
133proceed with the same sort of construction as in the projective
134case, but we must choose $H$ more cleverly to insure
135that $(\P^n-H)\intersect{}X$ is affine. Let $U$ be a nonempty
136affine open subset of $X$. As before pick a hyperplane which
137doesn't completey contain any irreducible component of $X-U$.
138Since $X$ is only quasi-projective we can't conclude that
139$(\P^n-H)\intersect{}X$ is affine. But we do know that
140$(\P^n-H)\intersect{}\overline{X}$ is affine. Our strategy is
141to add some hypersurfaces to $H$ to get a union of hypersurfaces $S$ so that
142$$(\P^n-S)\intersect{}\overline{X}=(\P^n-S)\intersect{}X.$$
143But, we must be careful to add these hypersurfaces in such a
144way that $((\P^n-S)\intersect{}X)\union U$ is missing only codimension
145two or greater subsets of $X$. We do this as follows. For each
146irreducible component $Y$ of $\overline{X}-X$ choose a hypersurface
147$H'$ which completely contains $Y$ but which does not completely
148contain any irreducible component of $X-U$. That this can be done
149is the content of a lemma which will be proved later (just pick a point
150in each irreducible component and avoid it). Let $S$
151by the union of all of the $H'$ along with $H$. Then
152$\P^n-S$ is affine and so
153$$(\P^n-S)\intersect{}X=(\P^n-S)\intersect{}\overline{X}$$
154is affine. Furthermore, $S$ properly intersects all
155irreducible components of $X-U$, so
156$((\P^n-S)\intersect{}X)\union U$ is missing only codimension
157two or greater subsets of $X$. Repeating this process
158as above several times yields the desired result because
159after each repetition the codimension of the resulting
160pieces is reduced by 1.
161\begin{lem}
162If $Y$ is a projective variety and $p_1,\ldots,p_n$ is a finite
163collection of points not on $Y$,
164then there exists a (possibly reducible) hypersurface $H$ containing
165$Y$ but not containing any of the $p_i$.
166\end{lem}
167By a possibly reducible hypersurface I mean a union of irreducible
168hypersurfaces, not a hypersurface union higher codimension varieties.
169\begin{proof}
170This is obviously true and I have a proof, but I think there
171is probably a more algebraic proof. Note that $k$ is infinite
172since we only talk about varieties over algebraically closed fields.
173Let $f_1,\cdots,f_m$ be defining equations for $Y$. Thus
174$Y$ is the common zero locus of the $f_i$ and not all $f_i$ vanish
175on any $p_i$. I claim that we can find a linear combination
176$\sum a_i f_i$ of the $f_i$ which doesn't vanish on any $p_i$.
177Since $k$ is infinite and not all $f_i$ vanish on $p_1$,
178we can easily find $a_i$ so that $\sum a_i f_i(p_1)\neq 0$
179and all the $a_i\neq 0$. If $\sum a_i f_i(p_2)=0$ then,
180once again since $k$ is infinite, we can easily jiggle''
181the $a_i$ so that $\sum a_i f_i(p_2)\neq 0$ and
182$\sum a_i f_i(p_1)$ is still nonzero. Repeating this same
183argument for each of the finitely many points $p_i$ gives
184a polynomial $f=\sum a_i f_i$ which doesn't vanish on any $p_i$.
185Of course I want to use $f$ to define our
186hypersurface, but I can't because $f$ might not be homogeneous.
187Fortunately, this is easily dealt with
188by suitably multiplying the various $f_i$ by the defining equation
189of a hyperplane not passing through any $p_i$, then repeating the
190above argument. Now let $H$ be the hypersurface defined by $f=\sum a_i f_i$.
191Then by construction $H$ contains $Y$ and $H$ doesn't
192contain any $p_i$.
193\end{proof}
194
195
196(e) Suppose $Y$ is a set-theoretic complete intersection of codimension
197$r$ in $X=\P_k^n$. Then $Y$ is the intersection of $r$ hypersurfaces,
198so we can write $Y=H_1\intersect\cdots\intersect{}H_r$ where
199each $H_i$ is a hypersurface. By (I, Ex. 3.5) $X-H_i$ is affine for
200each $i$, thus
201$$X-Y=(X-H_1)\union\cdots\union(X-H_r)$$
202can be covered by $r$ open affine subsets. By (c) this
203implies $\cd(X-Y)\leq r-1$ which completes the proof.
204\end{proof}
205
206
207\begin{exercise}
208{\bfseries (4.9)} Let $X=\spec k[x_1,x_2,x_3,x_4]$ be affine four-space
209over a field $k$. Let $Y_1$ be the plane $x_1=x_2=0$ and let $Y_2$
210be the plane $x_3=x_4=0$. Show that $Y=Y_1\union Y_2$
211is not a set-theoretic complete intersection in $X$. Therefore
212the projective closure $\overline{Y}$ in $\P_k^4$ is not
213a set-theoretic complete intersection.
214\end{exercise}
215\begin{proof}
216By (Ex. 4.8e) it suffices to show that
217$H^2(X-Y,\so_{X-Y})\neq 0$.
218Suppose $Z$ is a closed subset of $X$, then by (Ex. 2.3d),
219for any $i\geq 1$, there is an exact sequence
220$$H^i(X,\sox)\into H^i(X-Z,\so_{X-Z})\into 221 H_{Z}^{i+1}(X,\sox)\into H^{i+1}(X,\sox).$$
222By (3.8), $H^i(X,\sox)=H^{i+1}(X,\sox)=0$ so
223$H^i(X-Z,\so_{X-Z})=H_{Z}^{i+1}(X,\sox)$.
224Applying this with $Z=Y$ and $i=2$ shows that
225$$H^2(X-Y,\so_{X-Y})=H_{Y}^3(X,\sox).$$ Thus
226we just need to show that $H_{Y}^3(X,\sox)\neq 0$.
227
228Mayer-Vietoris (Ex. 2.4) yields an exact sequence
229\begin{align*}
230H^3_{Y_1}(X,\sox)\oplus{}H^3_{Y_2}(X,\sox)\into{}H^3_Y(X,\sox)\into\\
231H^4_{Y_1\intersect{}Y_2}(X,\sox)\into{}
232H^4_{Y_1}(X,\sox)\oplus{}H^4_{Y_2}(X,\sox)
233\end{align*}
234As above, $H^3_{Y_1}(X,\sox)=H^2(X-Y_1,\so_{X-Y_1})$.
235But $X-Y_1$ is a set-theoretic complete intersection
236of codimension $2$ so $\cd(X-Y_1)\leq 1$,
237whence $H^2(X-Y_1,\so_{X-Y_1})=0$. Similiarly
238$$H^2(X-Y_2,\so_{X-Y_2})=H^3(X-Y_1,\so_{X-Y_1}) 239=H^3(X-Y_2,\so_{X-Y_2})=0.$$ Thus from the above exact
240sequence we see that
241$H^3_Y(X,\sox)=H^4_{Y_1\intersect{}Y_2}(X,\sox).$
242
243Let $P=Y_1\intersect{}Y_2=\{(0,0,0,0)\}$.
244We have reduced to showing that
245$H^4_{P}(X,\sox)$ is nonzero. Since $H^4_{P}(X,\sox)=H^3(X-P,\so_{X-P})$
246we can do this by a direct computation of $H^3(X-P,\so_{X-P})$ using
247\cech{} cohomology. Cover $X-P$ by the affine
248open sets $U_i=\{x_i\neq{}0\}$. Then the \cech{} complex is
249$$\begin{array}{l} 250k[x_1,x_2,x_3,x_4,x_1^{-1}]\oplus\cdots\oplus 251k[x_1,x_2,x_3,x_4,x_4^{-1}]\xrightarrow{d_0}\\ 252k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1}]\oplus\cdots\oplus 253k[x_1,x_2,x_3,x_4,x_3^{-1},x_4^{-1}]\xrightarrow{d_1}\\ 254k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1}]\oplus\cdots\oplus 255k[x_1,x_2,x_3,x_4,x_2^{-1},x_3^{-1},x_4^{-1}]\xrightarrow{d_2}\\ 256k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1},x_4^{-1}] 257\end{array}$$
258Thus
259$$H^3(X-P,\so_{X-P}) 260 = \{x_1^ix_2^jx_3^kx_4^{\ell}:i,j,k,\ell<0\}\neq 0.$$
261\end{proof}
262
263
264\begin{exercise}
265{\bfseries (5.6)} {\em Curves on a Nonsingular Quadric Surface.}
266Let $Q$ be the nonsingular quadric surface $xy=zw$ in $X=\P_k^3$
267over a field $k$. We will consider locally principal closed
268subschemes $Y$ of $Q$. These correspond to Cartier divisors on
269$Q$ by (II, 6.17.1). On the other hand, we know that
270$\pic Q\isom\Z\oplus\Z$, so we can talk about the
271{\em type} (a,b) of $Y$ (II, 6.16) and (II, 6.6.1). Let us denote
272the invertible sheaf $\sL(Y)$ by $\so_Q(a,b)$. Thus
273for any $n\in\Z$, $\so_Q(n)=\so_Q(n,n).$
274
275[{\bf Comment!} In my solution, a subscheme $Y$ of type
276$(a,b)$ corresponds to the invertible sheaf $\soq(-a,-b)$.
277I think this is reasonable since then $\soq(-a,-b)=\sL(-Y)=\sI_Y$.
278The correspondence is not clearly stated in the problem,
279but this choice works.]
280
281\noindent(a) Use the special case $(q,0)$ and $(0,q)$, with $q>0$, when $Y$
282is a disjoint union of $q$ lines $\P^1$ in $Q$, to show:
283\begin{enumerate}
284\item if $|a-b|\leq 1$, then $H^1(Q,\so_Q(a,b))=0$;
285\item if $a,b<0$, then $H^1(Q,\so_Q(a,b))=0$;
286\item if $a\leq-2$, then $H^1(Q,\so_Q(a,0))\neq 0)$.
287\end{enumerate}
288{\em Solution.}
289First I will prove a big lemma in which I explicitely
290calculate $H^1(Q,\soq(0,-q))$ and some other things
291which will come in useful later. Next I give an independent
292computation of the other cohomology groups (1), (2).
293\par\begin{lem} Let $q>0$, then
294$$\dim_k H^1(Q,\soq(-q,0))=H^1(Q,\soq(0,-q))=q-1.$$
295Furthermore, we know all terms in the long exact
296sequence of cohomology associated with the short exact sequence
297$$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0.$$
298\end{lem}
299\begin{proof}
300We prove the lemma only for $\soq(-q,0)$, since the argument
301for $\soq(0,-q)$ is exactly the same. Suppose $Y$ is the disjoint
302union of $q$ lines $\P^1$ in $Q$ so $\sI_Y=\soq(-q,0)$.
303The sequence
304$$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0$$
305is exact. The associated long exact sequence of cohomology is
306\begin{align*}
3070\into&\Gamma(Q,\soq(-q,0))\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y)\\
308\into& H^1(Q,\soq(-q,0))\into{}H^1(Q,\soq)\into{}H^1(Q,\sO_Y)\\
309\into& H^2(Q,\soq(-q,0))\into{}H^2(Q,\soq)\into{}H^2(Q,\sO_Y)\into 0
310\end{align*}
311We can compute all of the terms in this long exact sequence. For
312the purposes at hand it suffices to view the summands as $k$-vector
313spaces so we systematically do this throughout.
314Since $\soq(-q,0)=\sI_Y$ is the ideal sheaf of $Y$,
315its global sections must vanish on $Y$. But $\sI_Y$ is a subsheaf
316of $\sO_Q$ whose global sections are the constants. Since the
317only constant which vanish on $Y$ is $0$, $\Gamma(Q,\soq(-q,0))=0$.
318By (I, 3.4), $\Gamma(Q,\soq)=k$. Since $Y$ is the disjoint union
319of $q$ copies of $\P^1$ and each copy has global sections $k$,
320$\Gamma(Q,\sO_Y)=k^{\oplus{}q}$. Since $Q$ is a complete intersection of
321dimension 2, (Ex. 5.5 b) implies $H^1(Q,\soq)=0$.
322Because $Y$ is isomorphic to several copies of $\P^1$,
323the general result (proved in class, but not in the book)
324that $H_{*}^n(\sO_{\P^n})=\{\sum a_{I}X_{I}:\text{entries in$I$negative}\}$
325implies $H^1(Q,\sO_Y)=H^1(Y,\sO_Y)=0$. Since $Q$ is a hypersurface
326of degree $2$ in $\P^3$, (I, Ex. 7.2(c)) implies $p_a(Q)=0$. Thus
327by (Ex. 5.5c) we see that $H^2(Q,\soq)=0$. Putting together the above
328facts and some basic properties of exact sequences show that
329$H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}$, $H^2(Q,\soq(-q,0))=0$
330and $H^2(Q,\so_Y)=0$. Our long exact sequence is now
331\begin{align*}
3320\into&\Gamma(Q,\soq(-q,0))=0\into\Gamma(Q,\soq)=k\into\Gamma(Q,\sO_Y)=k^{\oplus{}q}\\
333\into& H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}\into{}H^1(Q,\soq)=0\into{}H^1(Q,\sO_Y)=0\\
334\into& H^2(Q,\soq(-q,0))=0\into{}H^2(Q,\soq)=0\into{}H^2(Q,\sO_Y)=0\into 0
335\end{align*}
336\end{proof}
337
338Number (3) now follows immediately from the lemma because
339$$H^1(Q,\soq(a,0))=k^{\oplus(-a-1)}\neq 0$$ for $a\leq -2.$
340
341Now we compute (1) and (2). Let $a$ be an arbitrary integer.
342First we show that $\soq(a,a)=0$. We have an exact sequence
343$$0\into\sO_{\P^3}(-2)\into\sO_{\P^3}\into\sO_{Q}\into{}0$$
344where the first map is multiplication by $xy-zw$. Twisting by $a$
345gives an exact sequence
346$$0\into\sO_{\P^3}(-2+a)\into\sO_{\P^3}(a)\into\sO_{Q}(a)\into{}0.$$
347The long exact sequence of cohomology yields an exact sequence
348$$\cdots\into H^1(\sO_{\P^3}(a))\into H^1(\sO_Q(a))\into 349 H^2(\sO_{\P^3}(-2+a))\into\cdots$$
350But from the explicit computations of projective space (5.1)
351it follows that $H^1(\sO_{\P^3}(a))=0$ and $H^2(\sO_{\P^3}(-2+a))=0$
352from which we conclude that $H^1(\sO_Q(a))=0$.
353
354Next we show that $\soq(a-1,a)=0$. Let $Y$ be a single copy
355of $\P^1$ sitting in $Q$ so that $Y$ has type $(1,0)$. Then
356we have an exact sequence
357$$0\into\sI_Y\into\soq\into\sO_Y\into 0.$$
358But $\sI_Y=\soq(-1,0)$ so this becomes
359$$0\into\soq(-1,0)\into\soq\into\sO_Y\into 0.$$
360Now twisting by $a$ yields the exact sequence
361$$0\into\soq(a-1,a)\into\soq(a)\into\sO_Y(a)\into 0.$$
362The long exact sequence of cohomology gives an exact sequence
363$$\cdots\into\Gamma(\soq(a))\into\Gamma(\sO_Y(a))\into 364 H^1(\soq(a-1,a))\into H^1(\soq(a))\into\cdots$$
365We just showed that $H^1(\soq(a))=0$, so to see that
366$H^1(\soq(a-1,a))=0$ it suffices to note that the
367map $\Gamma(\soq(a))\into\Gamma(\sO_Y(a))$ is surjective.
368This can be seen by writing $Q=\proj(k[x,y,z,w]/(xy-zw))$
369and (w.l.o.g.) $Y=\proj(k[x,y,z,w]/(xy-zw,x,z))$ and
370noting that the degree $a$ part of $k[x,y,z,w]/(xy-zw)$
371surjects onto the degree $a$ part of $k[x,y,z,w]/(xy-zw,x,z)$.
372Thus $H^1(\soq(a-1,a))=0$ and exactly the same argument
373shows $H^1(\soq(a,a-1))=0$. This gives (1).
374
375For (2) it suffices to show that for $a>0$,
376$$H^1(\soq(-a,-a-n))=H^1(\soq(-a-n,-a))=0$$
377for all $n>0$. Thus let $n>0$ and suppose $Y$ is a disjoint union
378of $n$ copies of $\P^1$ in such a way that $\sI_Y=\soq(0,-n)$.
379Then we have an exact sequence
380$$0\into\soq(0,-n)\into\soq\into\sO_Y\into 0.$$
381Twisting by $-a$ yields the exact sequence
382$$0\into\soq(-a,-a-n)\into\soq(-a)\into\sO_Y(-a)\into 0.$$
383The long exact sequence of cohomology then gives an exact sequence
384$$\cdots \into\Gamma(\sO_Y(-a))\into H^1(\soq(-a,-a-n)) 385\into H^1(\soq(-a))\into \cdots$$
386As everyone knows, since $Y$ is just several copies of $\P^1$ and $-a<0$,
387$\Gamma(\sO_Y(-a))=0$. Because of our computations above,
388$H^1(\soq(-a))=0$. Thus $H^1(\soq(-a,-a-n))=0$, as desired.
389Showing that $H^1(\soq(-a-n,-a))=0$ is exactly the same.
390
391\noindent (b) Now use these results to show:
392\begin{enumerate}
393\item If $Y$ is a locally principal closed subscheme of
394type $(a,b)$ with $a,b>0$, then $Y$ is connected.
395\begin{proof}
396Computing the long exact sequence associated to the short
397exact sequence
398$$0\into\sI_Y\into\soq\into\sO_Y\into{}0$$
399gives the exact sequence
400$$0\into\Gamma(Q,\sI_Y)\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y) 401\into H^1(Q,\sI_Y)\into\cdots$$
402But, $\Gamma(\sI_Y)=0$, $\Gamma(Q,\soq)=k$, and
403by (a)2 above $H^1(Q,\sI_Y)=H^1(Q,\soq(-a,-b))=0$. Thus we
404have an exact sequence
405$$0\into{}0\into{}k\into\Gamma(\so_Y)\into{}0\into\cdots$$
406from which we conclude that $\Gamma(\so_Y)=k$ which implies
407$Y$ is connected.
408\end{proof}
409
410\item now assume $k$ is algebraically closed. Then for any
411$a,b>0$, there exists an irreducible nonsingular curve
412$Y$ of type $(a,b)$. Use (II, 7.6.2) and (II, 8.18).
413\begin{proof}
414Given $(a,b)$, (II, 7.6.2) gives a closed immersion
415$$Q=\P^1\times\P^1\into\P^a\times\P^b\into\P^n$$
416which corresponds to the invertible sheaf $\soq(-a,-b)$
417of type $(a,b)$. By Bertini's theorem (II, 8.18) there
418is a hyperplane $H$ in $\P^n$ such that the hyperplane section
419of the $(a,b)$ embedding of $Q$ in $\P^n$ is nonsingular.
420Pull this hyperplane section back to a nonsingular curve $Y$ of
421type $(a,b)$ on $Q$ in $\P^3$. By the previous problem, $Y$ is
422connected. Since $Y$ comes from a hyperplane section this implies
423$Y$ is irreducible (see the remark in the statement of Bertini's
424theorem).
425\end{proof}
426
427\item an irreducible nonsingular curve $Y$ of type $(a,b)$,
428$a,b>0$ on $Q$ is projectively normal (II, Ex. 5.14) if
429and only if $|a-b|\leq 1$. In particular, this gives lots
430of examples of nonsingular, but not projectively normal curves
431in $\P^3$. The simplest is the one of type $(1,3)$ which is
432just the rational quartic curve (I, Ex. 3.18).
433\begin{proof}
434Let $Y$ be an irreducible nonsingular curve of type $(a,b)$.
435The criterion we apply comes from (II, Ex 5.14d) which asserts
436that the maps
437$$\Gamma(\P^3,\sO_{\P^3}(n))\into\Gamma(Y,\sO_Y(n))$$
438are surjective for all $n\geq 0$ if and only if
439$Y$ is projectively normal. To determine when this occurs
440we have to replace $\Gamma(\P^3,\sO_{\P^3}(n))$ with
441$\Gamma(Q,\soq(n))$. It is easy to see that the above criterion
442implies we can make this replacement if $Q$ is projectively normal.
443Since $Q\isom\P^1\times\P^1$ is locally isomorphic to
444$\A^1\times\A^1\isom\A^2$ which is normal, we see that $Q$
445is normal. Then since $Q$ is a complete intersection which is
446normal, (II, 8.4b) implies $Q$ is projectively normal.
447
448Consider the exact sequence
449$$0\into\sI_Y\into\soq\into\sO_Y.$$
450Twisting by $n$ gives an exact sequence
451$$0\into\sI_Y(n)\into\soq(n)\into\sO_Y(n).$$
452Taking cohomology yields the exact sequence
453$$\cdots\into\Gamma(Q,\soq(n))\into\Gamma(Q,\sO_Y(n))\into 454 H^1(Q,\sI_Y(n))\into\cdots$$
455Thus $Y$ is projectively normal precisely if
456$H^1(Q,\sI_Y(n))=0$ for all $n\geq 0$. When can
457this happen? We apply our computations from part (a).
458Since $\soq(n)=\soq(n,n)$,
459$$\sI_Y(n)=\soq(-a,-b)(n)=\soq(-a,-b)\tensor_{\soq}\soq(n,n)=\soq(n-a,n-b)$$
460If $|a-b|\leq 1$ then $|(n-a)-(n-b)|\leq 1$ for all $n$ so
461$$H^1(Q,\soq(-a,-b)(n))=0$$ for all $n$ which implies $Y$ is
462projectively normal. On the other hand, if $|a-b|>1$ let
463$n$ be the minimum of $a$ and $b$, without loss assume $b$ is
464the minimum, so $n=b$. Then from (a) we see that
465$$\soq(-a,-b)(n)=\soq(-a,-b)(b)=\soq(-a+b,0)\neq 0$$
466since $-a+b\leq -2$.
467\end{proof}
468\end{enumerate}
469
470
471(c) If $Y$ is a locally principal subscheme of type $(a,b)$
472in $Q$, show that $p_a(Y)=ab-a-b+1.$ [Hint: Calculate the
473Hilbert polynomials of suitable sheaves, and again use the
474special case (q,0) which is a disjoint union of $q$ copies
475of $\P^1$.]
476\begin{proof}
477The sequence
478$$0\into\soq(-a,-b)\into\soq\into\sO_Y\into{}0$$
479is exact so
480$$\chi(\so_Y)=\chi(\soq)-\chi(\soq(-a,-b))=1-\chi(\soq(-a,-b)).$$
481Thus
482$$p_a(Y)=1-\chi(\so_Y)=\chi(\soq(-a,-b)).$$
483The problem is thus reduced to computing $\chi(\soq(-a,-b))$.
484
485Assume first that $a,b<0$. To compute $\chi(\soq(-a,-b))$
486assume $Y=Y_1\union Y_2$ where
487$\sI_{Y_1}=\soq(-a,0)$ and $\sI_{Y_2}=\soq(0,-b)$. Thus
488we could take $Y_1$ to be $a$ copies of $\P^1$ in one family of lines
489and $Y_2$ to be $b$ copies of $\P^1$ in the other family.
490Tensoring the exact sequence
491$$0\into\sI_{Y_1}\into\soq\into\so_{Y_1}\into 0$$
492by the flat module $\sI_{Y_2}$ yields an exact sequence
493$$0\into\sI_{Y_1}\tensor\sI_{Y_2}\into\sI_{Y_2}\into\sO_{Y_1}\tensor\sI_{Y_2}$$
494[Note: I use the fact that $\sI_{Y_2}$ is flat. This follows
495from a proposition in section 9 which we haven't yet reached,
496but I'm going to use it anyways. Since $Y_2$ is locally principal,
497$\sI_{Y_2}$ is generated locally by a single element and since $Q$ is
498a variety it is integral. Thus $\sI_{Y_2}$ is locally free so
499by (9.2) $\sI_{Y_2}$ is flat.]
500This exact sequence can also be written as
501$$0\into\soq(-a,-b)\into\soq(0,-b)\into\sO_Y\tensor\soq(0,-b)\into{}0.$$
502The associated long exact sequence of cohomology is
503\begin{align*}
5040\into&\Gamma(Q,\soq(-a,-b))\into\Gamma(Q,\soq(0,-b))\into
505                \Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))\\
506\into&H^1(Q,\soq(-a,-b))\into H^1(Q,\soq(0,-b))\into H^1(Q,\sO_{Y_1}\tensor\soq(0,-b))\\
507\into&H^2(Q,\soq(-a,-b))\into H^2(Q,\soq(0,-b))\into H^2(Q,\sO_{Y_1}\tensor\soq(0,-b))\into 0
508\end{align*}
509The first three groups of global sections are $0$. Since $a,b<0$,
510(a) implies $H^1(Q,\soq(-a,-b))=0$. From the lemma we know
511that $H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}$. Also by the lemma
512we know that $H^2(Q,\soq(0,-b))=0$. Since
513$\so_{Y_1}\tensor\soq(0,-b)$ is isomorphic
514to the ideal sheaf of $b-1$ points in each line
515of $Y_1$, a similiar proof as that used in the
516lemma shows that $$H^1(Q,\sO_Y\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}.$$
517Plugging all of this information back in yields the exact sequence
518\begin{align*}
5190\into&\Gamma(Q,\soq(-a,-b))=0\into\Gamma(Q,\soq(0,-b))=0\into
520                \Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\\
521\into&H^1(Q,\soq(-a,-b))=0\into H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}\\
522          &\hspace{1in}\into{}H^1(Q,\sO_{Y_1}\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}\\
523\into&H^2(Q,\soq(-a,-b))\into H^2(Q,\soq(0,-b))=0\\
524          &\hspace{1in}\into{}H^2(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\into 0
525\end{align*}
526From this we conclude that
527$$\chi(\soq(-a,-b))=0+0+h^2(Q,\soq(-a,-b))=a(b-1)-(b-1)=ab-a-b+1$$
528which is the desired result.
529
530Now we deal with the remaining case, when $Y$ is $a$ disjoint copies
531of $\P^1$. We have
532$$p_a(Y)=1-\chi(\sO_Y)=1-\chi(\sO_{\P^1}^{\oplus a}) 533 =1-a\chi(\sO_{\P^1})=1-a$$
534which completes the proof.
535\end{proof}
536\end{exercise}
537\end{document}
538