Open in CoCalc
1%%%%%%%%%%%%%%%%%%%%%%%%%%
2%% Homework assignment 2
3%%%%
4
5\documentclass[12pt]{article}
6\include{defs}
7\author{William A. Stein}
8\title{Homework 2, MAT256B\\II.8.4, III.6.8, III.7.1, III.7.3}
9
10\begin{document}
11\maketitle
12
13\section{Exercise II.8.4}
14{\em Complete Intersections in $\P^n$.
15A closed subscheme $Y$ of $\P^n_k$ is called a {\em (strict, global)
16complete intersection} if the homogenous ideal $I$ of $Y$ in
17$S=k[x_0,\ldots,x_n]$ can be generated by $r$ elements where
18$r=\codim(Y,\P^n)$.}
19
20{\em (a) Let $Y$ be a closed subscheme of codimension $r$ in $\P^n$. Then
21$Y$ is a complete intersection iff there are hypersurfaces (i.e., locally
22principal subschemes of codimension $1$) $H_1,\ldots,H_r$, such that
23$Y=H_1\intersect\cdots\intersect H_r$ {\em as schemes}, i.e.,
24$\sI_Y=\sI_{H_1}+\cdots+\sI_{H_r}$.}
25
26($\Rightarrow$) By (II, Ex 5.14) $I$ is defined to be $\Gamma_{*}(\sI_Y)$.
27By (II, 5.15), $\tilde{I}\isom\sI_Y$. Write $I=(f_1,\ldots,f_r)$, then
28since localization commutes with taking sums,
29$$\sI_Y=(f_1,\ldots,f_r)\tilde{ }=((f_1)+\cdots+(f_r))\tilde{ }=(f_1)\tilde{ }+\cdots+(f_r)\tilde{ }.$$
30Let $H_i$ be the locally principal closed subscheme of codimension
31$1$ determined by the ideal sheaf $(f_i)\tilde{ }$.  Then $Y$ is the
32intersection of the $H_i$.
33
34($\Leftarrow$) Someone suggested I should apply unmixedness and
35primary decomposition to some ideal somewhere and use the fact
36that a saturated ideal doesn't have primary components corresponding
37to the irrelevant ideal or something like that. NOT DONE.
38
39{\em (b) If $Y$ is a complete intersection of dimension $\geq 1$ in
40$\P^n$, and if $Y$ is normal, then $Y$ is projectively normal (Ex. 5.14).}
41
42Let $Z$ be the cone over $Y$, then $A(Z)=S/I(Y)$. By
43(I, Ex. 3.17d), $A(Z)$ is integrally closed iff $Z$ is
44normal. By definition $A(Z)$ is integrally closed iff
45$Y$ is projectively normal. Thus we must show that $Z$
46is normal. Since $Y$ is a complete intersection, $I(Y)=(f_1,\ldots,f_r)$
47so $Z$ is a complete intersection subscheme of $\bA^{n+1}$. By
48(II, 8.23) $Z$ is normal iff $Z$ is regular in codimension $1$.
49Also by (II, 8.23) $Y$ is regular in codimension $1$ because we
50have assumed $Y$ is normal. But $Y$ regular in codimension $1$ implies
51$Z$ regular in codimension $1$. [We used this last semester in
52(II, Ex. 6.3d). Intuitively, the only singularity in $Z$ not in $Y$
53is the cone point which has codimension $>1$. This is because $Z$
54is locally $U_i\cross\bA^1$.]
55
56{\em (c) With the same hypothesis as in (b), conclude that for all
57$\ell\geq 0$, the natural map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$
58is surjective. In particular, taking $\ell=0$, show that $Y$ is connected.}
59
60That the map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$
61is surjective is just the statement of (II, Ex. 5.14d). When $\ell=0$
62this says that
63$k=\Gamma(\P^n,\sO_{\P^n}(\ell))$ surjects onto $\Gamma(Y,\sO_Y(\ell))$
64so $\dim\Gamma(Y,\sO_Y)\leq 1$ and hence $Y$ is connected. [If $Y$ were
65not connected then $\Gamma(Y,\sO_Y)=k\oplus\cdots\oplus k$ where the
66number of direct summands equals the number of components of $Y$.]
67
68{\em (d) Now suppose given integers $d_1,\ldots,d_r\geq 1$, with
69$r<n$. Use Bertini's theorem (8.18) to show that there exists nonsingular
70hypersurfaces $H_1,\ldots,H_r$ in $\P^n$, with $\deg H_i=d_i$, such that
71the scheme $Y=H_1\intersect\cdots\intersect H_r$ is irreducible
72and nonsingular in codimension $r$ in $\P^n$.}
73
74[To apply Bertini's theorem we must assume $k$ is algebraically closed.
75I'm going to make this assumption now. Maybe there is a way around this?]
76
77Let $\P_k^n\hookrightarrow\P_k^{d_1\text{-uple}}$
78be the $d_1$-uple embedding of $\P_k^n$.  Use Bertini's theorem to choose
79a hyperplane in
80$\P_k^{d_1\text{-uple}}$ which has nonsingular intersection with
81the image of $\P_k^n$. It pulls back to a degree $d_1$ nonsingular
82hypersurface $H_1$ in $\P_k^n$. If $r>1$ consider the $d_2$-uple embedding
83$\P_k^n\hookrightarrow\P_k^{d_2\text{-uple}}$.
84The image of $H_1$ is a nonsingular variety in $\P_k^{d_2{\text-uple}}$
85of dimension $\geq 2$. By Bertinni's theorem there is a hyperplane
86in $\P_k^{d_2-uple}$ whose intersection with the image of $H_1$ is nonsingular
87and of dimension one less than $H_1$. Pulling back we obtain a hypersurface
88$H_2$ such that $H_1\intersect H_2$ is nonsingular and $H_2$ has degree
89$d_2$. Continuing inductively in this way and noting that
90$\dim H_1\intersect\cdots\intersect H_{r-1}\geq 2$ (since
91$r<n$) completes the proof.
92
93{\em (e) If $Y$ is a nonsingular complete intersection as in (d)
94show that $\omega_Y\isom\sO_Y(\sum d_i-n-1)$.}
95
96By (III, 8.20) $\omega_{H_1}\isom\omega_{\P^n}\tensor\sL(H_1)\tensor\sO_{H_1}$.
97By the explicit computation of $Cl\P^n$ (II, 6.17) we know that
98$\sL(H_1)\isom\sO_{\P^n}(d_1)$. Thus
99$$\omega_{H_1}\isom\sO_{\P^n}(-n-1)\tensor\sO_{\P^n}(d_1)\tensor\sO_{H_1}\isom\sO_{H_1}(d_1-n-1).$$
100
101By (8.20) we have that
102$$\omega_{H_1\intersect H_2}\isom\omega_{H_1}\tensor\sL(H_2.H_1)\tensor\sO_{H_1\intersect H_2}.$$
103We know that $H_2\sim d_2\P^{n-1}$ (linear equivalence) so
104by (II, 6.2b) this implies $H_2.H_1\sim d_2\P^{n-1}.H_1$.
105But $d_2\P^{n-1}.H_1$ corresponds to the invertible sheaf (see (II, Ex 6.8c)
106$\sO_{H_1}(d_2)$. Thus
107$$\omega_{H_1\intersect H_2}\isom\sO_{H_1}(d_1-n-1)\tensor\sO_{H_1}(d_2)\tensor\sO_{H_1\intersect H_2}\isom\sO_{H_1\intersect H_2}(d_1+d_2-n-1).$$
108Repeating this argument inductively yields the desired isomorphism.
109
110{\em (f) If $Y$ is a nonsingular hypersurface of degree $d$ in $\P^n$,
111use (c) and (e) above to show that $p_g(Y)=\chose{d-1}{n}$. Thus
112$p_g(Y)=p_a(Y)$ (I, Ex. 7.2).}
113
114By definition $p_g(Y)=\dim_k\Gamma(Y,\omega_Y)$. By (e),
115$\omega_Y\isom\sO(d-n-1)$ and by (c) the natural map
116$$\Gamma(X,\sox(d-n-1))\into\Gamma(Y,\soy(d-n-1))$$
117is surjective. We show that it is also injective.
118By (III, 5.5a) an element $f\in\Gamma(X,\sox(d-n-1))$ can be represented as
119a homogeneous polynomial of degree $d-n-1$.
120Now $f$ maps to $0$ in $\Gamma(Y,\soy(d-n-1))$ iff
121$f$ vanishes on $Y$, that is to say, $Y\subset Z(f)$. But
122$\deg Y=d>d-n-1=\deg f$ so $Y$ can not be contained in the hypersurface
123$Z(f)$ unless $f=0$. [Proof: $Y=Z(g)\subset Z(f)$ implies
124$(f)\subset (g)$ so $f$ is a multiple of $g$, but $g$ has degree
125strictly greater than $f$ so must be $0$.]
126Thus
127$$\dim_k \Gamma(Y,\soy(d-n-1))=\dim_k \Gamma(X,\sox(d-n-1)) 128 =\chose{d-1}{n}$$
129since the number of monomoials in $k[x_0,\ldots, x_n]$ of degree
130$d-n-1$ is $\chose{d-1}{n}$
131as desired.
132
133{\em (g) If $Y$ is a nonsingular curve in $\P^3$, which is a complete
134intersection of nonsingular surfaces of degrees $d$, $e$, then
135$p_g(Y)=\frac{1}{2}de(d+e-4)+1$. Again the geometric genus is
136the same as the arithmetic genus (I, Ex. 7.2).}
137
138Let $H$ be the hypersurface of degree $d$.
139There is an exact sequence
140$$0\into\sO_{\P^n}(-d)\into\sO_{\P^n}\into\sO_H\into 0.$$
141Twisting by $a$ and computing dimensions we see that
142$$\dim\sO_H(a)=\dim\sO_{\P^n}(a)-\dim\sO_{\P^n}(a-d)=\chose{3+a}{3}-\chose{3+a-d}{3}.$$
143
144Using reasoning like that in (e) we obtain an exact sequence
145$$0\into\sO_H(-e)\into\sO_H\into\sO_Y\into 0.$$
146Twisting by $e+d-4$ yields the exact sequence
147$$0\into\sO_H(d-4)\into\sO_H(e+d-4)\into\sO_Y(e+d-4)\into 0.$$
148Applying the above explicit computation of $\dim\sO_H(a)$ we see that
149$$\dim\soy(e+d-4)=\chose{e+d-1}{3}-\chose{e-1}{3}-\chose{d-1}{3}+chose{-1}{3}.$$
150After some algebra the latter expression becomes
151$\frac{1}{2}ed(e+d-4)+1$, as desired.
152
153[Comment 1: We could have also solved (f) using this method.]
154
155[Comment 2: Serre duality gives another solution. By (III, 7.12.4)
156$p_g(Y)=\dim H^0(Y,\omega_Y)=\dim H^1(Y,\soy)=p_a(Y).$
157But by (I, Ex. 7.2d) $p_a(Y)=\frac{1}{2}de(d+e-4)+1$.]
158
159
160\section{Exercise III.6.8}
161{\em Prove the following theorem of
162Kleiman: if $X$ is a noetherian, integral, seperated,
163locally factorial scheme, then every coherent sheaf on
164$X$ is a quotient of a locally free sheaf (of finite rank).}
165
166{\em (a) First show that open sets of the form $X_s$, for
167various $s\in\Gamma(X,\sL)$ and various invertible sheaves
168$\sL$ on $X$, form a base for the topology of $X$.}
169
170Let $x\in U\subset X$ with $U$ open.
171
172{\em Case 1. $W=X-U$ is irreducible.}
173Since $x\not\in W$, $\so_x\not\subset \sO_W$.
174[This assertion is a matter of some difficulty among the
175others working on this problem. It is not hard to see when
176$X$ is a variety in the classical sense. But in the more general
177situation it isn't at all clear and may use the hypothesis that
178$X$ is seperated in an essential way. For example, the affine line
179with a doubled origin has two different local rings which are equal.
180I'm not sure how to resolve this but there was some talk of using
181the valuative criterion for seperatedness. PUT CORRECT SOLUTION HERE
182AFTERWARDS.]
183Thus let $h\in K$ be a rational
184function such that $h\not\in\sO_W$ but $h\in\so_x$. Let $(h)=D_1-D_2$
185with $D_1$=zeros of $h$ and $D_2$=poles of $h$. Since $h\in\so_x$, we
186have $x\not\in\Var(D_2)$=the underlying scheme of the effective
187divisor $D_2$. (This is because $h$ can't have a pole at $x$.) Furthermore
188$y\in W$ implies $\so_y\subset\so_W$ so $h\not\in\sO_y$ thus $y\in\Var(D_2)$
189(this is because $X$
190is factorial so $\sO_y$ is integrally closed so $v_y(h)<0$ iff
191$h\not\in\sO_y$.) Thus $W\subset\Var(D_2)$. Since $X$ is factorial and
192$D_2$ is effective (II, 6.11) implies
193$D_2$ corresponds to an effective Cartier divisor and hence there exists
194an open cover $\sU=(U_i)$ of $X$ and rational functions $h_i\in K$
195such that $h_i|U_i\in\sO_{U_i}$ and $(h_i)=D_2$ on $U_i$. Since
196$\frac{h_i}{h_j}\in\so_x(U_i\intersect U_j)^{*}$ and $X$ is normal,
197$(\frac{h_i}{h_j})=0$. Let $\sL$ be the locally free invertible sheaf
198represented by the Cartier divisor $(U_i,h_i)$ (so $\L$ is locally
199generated by $1/h_i$ on $U_i$), and let $u_i:\sL|U_i\into\so_x|U_i$
200be the isomorphism given by multiplication by $h_i$. Define
201$s(y)=u_i^{-1}(h_i(y))$ for $y\in U_i$. By this we mean $u_i^{-1}$ of
202the map $y\mapsto h_i(y)$, i.e., $s$ is the glueing of the inverse images
203of the $h_i\in\so_x(u_i)$. Thus $s$ is a section of $\L$ such that
204$X_s\intersect U_i=U_i-\Var(D_2)$. Thus
205$X_s=X-\Var(D_2)\subset U$.
206
207{\em Case 2. $W=X-U$ is reducible.}
208Using the fact that $X$ is noetherian write
209$W=Z_1\union \cdots \union Z_n$. From case 1 we know that
210there exists invertible sheaves $\L_1,\ldots,\L_n$ and sections
211$s_i\in\Gamma(X,\L_i)$, $i=1,\ldots,n$ such that $x\in X_{s_i}\subset X-Z_i$.
212Let $s=s_1\tensor\cdots\tensor s_n\in\Gamma(X,\L_1\tensor\cdots\tensor\L_n)$.
213Then $X_s=\intersect_{i=1}^n X_{s_i}$ hence $x\in X_s\subset U$.
214
215[[This proof was copied from Borelli's paper with little modification. One
216danger is that the corresponding theorem in Borelli's paper assumes $X$ to be
217a factorial {\em variety}, not a more general scheme as above. Part (b)
218below was not in Borelli.]]
219
220{\em (b) Now use (II, 5.14) to show that any coherent sheaf is a quotient
221of a direct sum $\oplus\sL_i^{n_i}$ for various invertible sheaves
222$\sL_i$ and various integers $n_i$. }
223
224Let $\sF$ be a coherent sheaf on $X$.
225Let $U$ be an open set on which $\sF_{|U}\isom\tilde{M}$.
226Suppose $X_f\subset U$ where $f$ is a global section of some
227invertible sheaf $\sL$. Our strategy is to construct an appropriate
228map $\oplus\sL_i^{n_i}\into \sF$ which is surjective when
229restricted to $X_f$, then use the fact that $X$ is noetherian and
230that the $X_f$ form a basis for the topology on $X$ to cover $X$
231which such $X_f$ and then take the sum of all the resulting maps.
232
233Let $m_1,\ldots,m_r$ generate $M$. Let $t_1,\ldots,t_r$ be the
234restrictions of the $m_i$ to $X_f$. By (II, 5.14b) there exists $n$
235so that
236$$t_1 f^n,\ldots,t_r f^n\in\Gamma(X_{f^n},\sF\tensor\sL^{\tensor n})$$
237extend to global sections $s_1,\ldots,s_n$ of
238$\Gamma(X,\sF\tensor\sL^{\tensor n})$. Define a map
239$$\oplus_{i=1}^n \sox\into\sF\tensor\sL^{\tensor n}$$
240by sending $(0,\ldots,0,1,0,\ldots,0)$ (1 in the $i$th position only)
241to $s_i$. Then tensoring with
242$(\sL^{\tensor n})^{-1} =(\sL^{-1})^{\tensor n}$
243we obtain a map
244$$\Theta:\oplus_{i=1}^n (\sL^{-1})^{\tensor n}\into\sF.$$
245The map $\Theta$ is surjective when restricted to $X_f$.
246To see this let $p$ be a point of $X_f$. The stalk of
247$\sF$ at $p$ is generated by the stalks of $m_1,\ldots,m_n$
248at $p$. Since the $t_i$ are all in the image of the map $\Theta$ and
249the stalks of the $t_i$ at $p$ are the same as the stalks of the $m_i$
250at $p$ it follows that the stalks of the $m_i$ are all in the image
251under $\Theta$ of the stalk of $\oplus_{i=1}^n(\sL^{-1})^{\tensor n}$
252at $p$.
253
254Take the direct sum of all such maps over a suitable open cover
255$(U)$ of $X$ and suitable open covers $(X_f)$ of each $U$. Since
256$X$ is noetherian we can arrange it so this sum is finite.
257
258\section{Exercise III.7.1}
259{\em Let $X$ be an integral projective
260scheme of dimension $\geq 1$ over a field $k$, and let
261$\sL$ be an ample invertible sheaf $X$. Then $H^0(X,\sL^{-1})=0$.}
262
263\begin{lem} If $\M\neq\sox$ is an invertible sheaf which is generated by
264its global sections then $H^0(X,\M^{-1})=0$. \end{lem}
265\begin{proof}
266By the proof of (II 6.12) $\M^{-1}=\M^{\dual}=\shom(\M,\sox)$.
267Thus we must show that $\Gamma(X,\shom(\M,\sox))=0$, i.e., that
268$\hom_{\sox}(\M,\sox)=0$. Since $\M$ is generated by global
269sections $(m_i)$ to give a morphism $f:\M\into\sox$ is the same
270as to give the images $\alpha_i=f(m_i)\in\Gamma(X,\sox)$ of the
271$m_i$. Since $X$ is integral and projective $\Gamma(X,\sox)=k$
272so the $\alpha_i$ all lie in $k$. Thus if $f$
273is nonzero then some $\alpha_i\neq 0$ so $\frac{1}{\alpha_i}m_i\mapsto 1$.
274Let $t=\frac{1}{\alpha_i}m_i\in\Gamma(X,\M)$. Let $p$ be any
275point of $X$. The map $f_p:\M_p\into\soxp$ sends $t_p$ to
276$1$ so it is surjective being a map of free $\soxp$-modules (and
277since $\soxp$ is generated by $1$ as an $\soxp$-module). On the
278other hand $\M_p$ is free of rank $1$ over the integral domain
279$\soxp$ so $f$ must be injective. Indeed, if $\M_p\isom\soxp\cdot g$
280for some $g$ and $ag\mapsto 0$ then $af(g)=0$ so since $\soxp$ is a domain,
281$a=0$ or $f(g)=0$. But $f(g)\neq 0$ since $f$ is surjective so $a=0$
282and so $ag=0$ whence $f$ is injective. Therefore $f$ is an isomorphism
283since it is an isomorphism on stalks. Thus $\M\isom\sox$ contrary
284to our assumption that $\M\not\isom\sox$ so there
285can be no nonzero $f$ in $\hom_{\sox}(\M,\sox)=\Gamma(X,\M^{-1})$,
286as desired.
287\end{proof}
288
289Suppose that $\L$ is ample. If $\L=\sox$ then $\L$ can not be ample,
290for if $\sox$ is ample then since $\sox^{\tensor n}=\sox$ for any $n\geq 1$
291it follows by (II.7.5) that $\sox$ is very ample. This means that there
292is an immersion $i:X\hookrightarrow\P_k^n$ where $n=\dim\Gamma(X,\sox)-1=0$
293which is impossible because $X$ has dimension at least $1$.
294
295Thus we may assume $\L\not\isom\sox$ and apply the above lemma.
296There is an $n$
297so that $\L^{\tensor n}$ is generated by its global sections. By the
298above lemma $H^0(X,(\L^{\tensor n})^{\dual})=0$. Since the collection
299of invertible sheaves forms a group and $\dual$ is the inverse operation
300it follows trivially that
301$(\L^{\tensor n})^{\dual}\isom(\L^{\dual})^{\tensor n}$ and
302hence $\Gamma(X,(\L^{\dual})^{\tensor n})=0$.
303Suppose $\L^{\dual}$
304has a nonzero global section $s$. Let $p\in X$ be a point so that
305$s_p\neq 0$. It follows that $s\tensor\cdots\tensor s\neq 0$ in
306$(\L^{\dual}_p)^{\tensor n}$. Thus $s$ defines a nonzero global section
307$s\tensor\cdots\tensor s$ of $(\L^{\dual})^{\tensor n}$. [This last statement
308is a bit subtle because the tensor product is the sheaf associated to
309a certain presheaf so we don't know, {\em a priori}, that $s\tensor\cdots\tensor s$
310maps to something nonzero under the $\theta$ of (II, Defn 1.2). But
311if $\theta(s\tensor\cdots\tensor s)=0$ then $0=\theta(s\tensor\cdots\tensor s)_p=(s\tensor\cdots\tensor s)_p$
312so $\theta$ is not injective on stalks contradicting the comment after
313(II, Defn 1.2).] Thus if $H^0(X,\L^{\dual})\neq 0$ then
314$H^0(X,(\L^{\tensor n})^{\dual}) \neq 0$, a contradiction. It follows
315that $H^0(X,\L^{\dual})=0$, as desired.
316
317
318\section{Exercise III.7.3}
319{\em Let $X=\P^n_k$. Show that $H^q(X,\OX^p)=0$
320for $p\neq q$, $k$ for $p=q$, $0\leq p,q\leq n$.}
321
322Our strategy is to use the exact sequence
323$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0$$
324of (II, 8.13) along with (II, Ex 5.16 d) to reduce the
325computation of the cohomology of $\OX^p$ to the
326computation of the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$.
327We then show inductively
328that the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$ vanishes for
329$p\geq 1$ thus completing the proof.
330
331We compute the cohomology of $\Omega^r$ inductively on $r$.
332
333{\em Step 1, $r=0$.} Suppose $r=0$ so $\Omega^r=\sox$. Then
334by (III, 5.5) $H^0(X,\sox)=k$ and $H^i(X,\sox)=0$ for $i\geq 1$.
335[Part (a) of (III, 5.5) gives $H^0(X,\sox)=k$, part (b) gives
336$H^i(X,\sox)=0$ for $0<i<n$ and part (d) gives
337$H^n(X,\sox)\isom H^0(X,\sox(-n-1))^{\dual}=0$.]
338
339{\em Step 2} Show that
340$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$$
341for $r\geq 1$.
342
343[Matt Baker pointed out to me that
344$$\Lambda^r\sox(-1)^{\oplus n+1}\isom\sox(-1)^{\oplus \chose{n+1}{r}}.$$
345This is reasonable since it is true on stalks. It immediately implies
346the vanishing of the cohomology groups. My original more complicated proof
347of step 2 is included next anyways.]
348
349{\em Step 2a, $r=1$.} We treat $r=1$ as a special case.
350We must show that $H^i(X,\sox(-1)^{\oplus n+1}=0$ or
351equivalently that $H^i(X,\sox(-1))=0$. This is immediate
352from the explicit computations of (III, 5.5). The argument proceeds
353exactly as in step 1.
354
355{\em Step 2b, $r\geq 2$.} We now assume $r\geq 2$ and proceed inductively
356on $n$. Since $r\geq 2$ there is an exact sequence
357$$0\into\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Lambda^r\sox(-1)^{\oplus n}\into 0.$$
358I obtained the map
359$$\Lambda^r\sox(-1)^{\tensor n+1}\into\Lambda^r\sox(-1)^{\tensor n}$$
360by carefully applying (II, Ex 5.16d) to the map
361$\sox(-1)^{\oplus n+1}\into\sox(-1)^{\oplus n}$.
362But this map just turns out to be locally defined by
363$x_n\mapsto 0$ where $x_0,\ldots,x_n$ are local coordinates
364for $\sox(-1)$. Then $x_{i_0}\wedge\cdots\wedge x_{i_r}$
365maps to $0$ if some $i_k=n$ and itself otherwise.
366The map
367$$\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}$$
368identifies $\Lambda^{r-1}\sox(-1)^{\oplus n}$ with the kernel of the
369next map. The kernel of the next map is locally generated by all
370monomials'' which contain an $x_n$. Since $r\geq 2$ we can identify
371$\Lambda^{r-1}\sox(-1)^{\oplus n}$ with this kernel by just removing
372the $x_n$ off of the wedge product.
373[This is not rigorous enough!]
374
375By induction on $n$ we have that
376$$H^i(\Lambda^{r-1}\sox(-1)^{\oplus n})=H^i(\Lambda^r\sox(-1)^{\oplus n})=0$$
377for all $i$ (we will do the base case $n=1$ in just a moment).
378Thus, by the long exact sequence of cohomology we see that
379$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$ for all $i$.
380For $n=1$, since $r\geq 2$
381it follows that  $\Lambda^{r-1}\sox(-1)=\sox(-1)$ or $0$
382and $\Lambda^r\sox(-1)=0$ and these both have trivial cohomology
383as computed above.
384
385{\em Step 3.} The final step is to obtain the long exact sequence
386$$\cdots H^i(\Omega^r)\into H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1})\into \cdots$$
387then apply step 2 and the induction hypothesis (we are inducting
388on $r$, the base case was established in step 1) to
389calculate $H^i(\Omega^r)$ for all $i$.
390
391Suppose $r\geq 1$, then by (II, 8.13) we have an exact sequence
392$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0.$$
393By (II, Ex. 5.16d), $\Lambda^r\sox(-1)^{\oplus n+1}$ has a filtration
394$$\Lambda^r\sox(-1)^{\oplus n+1}=F^0\supseteq F^1\supseteq\cdots\supseteq F^r\supseteq F^{r+1}=0$$
395with quotients
396$$F^p/F^{p+1}\isom\Omega^p\tensor\Lambda^{r-p}\sox= 397\begin{cases} 0&\text{if r-p\geq 2}\\ 398 \Omega^p&\text{if r-p is 0 or 1}\end{cases}$$
399Thus $$\Lambda^r\sox(-1)^{\oplus n+1}=F^0=\cdots=F^{r-1}$$ and the filtration
400becomes $$\Lambda^r\sox(-1)^{\oplus n+1}\supset F^r\supset F^{r+1}=0$$
401with $\Lambda^r\sox(-1)^{\oplus n+1}/F^r\isom \Omega^{r-1}$ and
402$F^r\isom\Omega^r$.
403This gives an exact sequence
404$$0\into\Omega^r\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Omega^{r-1}\into 0.$$
405The associated long exact sequence of cohomology gives for each $i$
406an exact sequence
407$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1}) 408\into H^{i+1}(\Omega^r)\into H^{i+1}(\Lambda^r\sox(-1)^{\oplus n+1}).$$
409But by step 2 the groups $H^i(\Lambda^r\sox(-1)^{\oplus n+1})$ all vanish.
410Thus $H^i(\Omega^{r-1})\isom H^{i+1}(\Omega^r)$. By induction on
411$r$ this shows that
412$$H^i(\Omega^r)=\begin{cases}0&\text{if i\neq r}\\ 413 k&\text{if i=r}\end{cases}.$$
414This completes the proof.
415\end{document}
416