%%%%%%%%%%%%%%%%%%%%%%%%%%1%% Homework assignment 22%%%%34\documentclass[12pt]{article}5\include{defs}6\author{William A. Stein}7\title{Homework 2, MAT256B\\II.8.4, III.6.8, III.7.1, III.7.3}89\begin{document}10\maketitle1112\section{Exercise II.8.4}13{\em Complete Intersections in $\P^n$.14A closed subscheme $Y$ of $\P^n_k$ is called a {\em (strict, global)15complete intersection} if the homogenous ideal $I$ of $Y$ in16$S=k[x_0,\ldots,x_n]$ can be generated by $r$ elements where17$r=\codim(Y,\P^n)$.}1819{\em (a) Let $Y$ be a closed subscheme of codimension $r$ in $\P^n$. Then20$Y$ is a complete intersection iff there are hypersurfaces (i.e., locally21principal subschemes of codimension $1$) $H_1,\ldots,H_r$, such that22$Y=H_1\intersect\cdots\intersect H_r$ {\em as schemes}, i.e.,23$\sI_Y=\sI_{H_1}+\cdots+\sI_{H_r}$.}2425($\Rightarrow$) By (II, Ex 5.14) $I$ is defined to be $\Gamma_{*}(\sI_Y)$.26By (II, 5.15), $\tilde{I}\isom\sI_Y$. Write $I=(f_1,\ldots,f_r)$, then27since localization commutes with taking sums,28$$\sI_Y=(f_1,\ldots,f_r)\tilde{ }=((f_1)+\cdots+(f_r))\tilde{ }=(f_1)\tilde{ }+\cdots+(f_r)\tilde{ }.$$29Let $H_i$ be the locally principal closed subscheme of codimension30$1$ determined by the ideal sheaf $(f_i)\tilde{ }$. Then $Y$ is the31intersection of the $H_i$.3233($\Leftarrow$) Someone suggested I should apply unmixedness and34primary decomposition to some ideal somewhere and use the fact35that a saturated ideal doesn't have primary components corresponding36to the irrelevant ideal or something like that. NOT DONE.3738{\em (b) If $Y$ is a complete intersection of dimension $\geq 1$ in39$\P^n$, and if $Y$ is normal, then $Y$ is projectively normal (Ex. 5.14).}4041Let $Z$ be the cone over $Y$, then $A(Z)=S/I(Y)$. By42(I, Ex. 3.17d), $A(Z)$ is integrally closed iff $Z$ is43normal. By definition $A(Z)$ is integrally closed iff44$Y$ is projectively normal. Thus we must show that $Z$45is normal. Since $Y$ is a complete intersection, $I(Y)=(f_1,\ldots,f_r)$46so $Z$ is a complete intersection subscheme of $\bA^{n+1}$. By47(II, 8.23) $Z$ is normal iff $Z$ is regular in codimension $1$.48Also by (II, 8.23) $Y$ is regular in codimension $1$ because we49have assumed $Y$ is normal. But $Y$ regular in codimension $1$ implies50$Z$ regular in codimension $1$. [We used this last semester in51(II, Ex. 6.3d). Intuitively, the only singularity in $Z$ not in $Y$52is the cone point which has codimension $>1$. This is because $Z$53is locally $U_i\cross\bA^1$.]5455{\em (c) With the same hypothesis as in (b), conclude that for all56$\ell\geq 0$, the natural map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$57is surjective. In particular, taking $\ell=0$, show that $Y$ is connected.}5859That the map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$60is surjective is just the statement of (II, Ex. 5.14d). When $\ell=0$61this says that62$k=\Gamma(\P^n,\sO_{\P^n}(\ell))$ surjects onto $\Gamma(Y,\sO_Y(\ell))$63so $\dim\Gamma(Y,\sO_Y)\leq 1$ and hence $Y$ is connected. [If $Y$ were64not connected then $\Gamma(Y,\sO_Y)=k\oplus\cdots\oplus k$ where the65number of direct summands equals the number of components of $Y$.]6667{\em (d) Now suppose given integers $d_1,\ldots,d_r\geq 1$, with68$r<n$. Use Bertini's theorem (8.18) to show that there exists nonsingular69hypersurfaces $H_1,\ldots,H_r$ in $\P^n$, with $\deg H_i=d_i$, such that70the scheme $Y=H_1\intersect\cdots\intersect H_r$ is irreducible71and nonsingular in codimension $r$ in $\P^n$.}7273[To apply Bertini's theorem we must assume $k$ is algebraically closed.74I'm going to make this assumption now. Maybe there is a way around this?]7576Let $\P_k^n\hookrightarrow\P_k^{d_1\text{-uple}}$77be the $d_1$-uple embedding of $\P_k^n$. Use Bertini's theorem to choose78a hyperplane in79$\P_k^{d_1\text{-uple}}$ which has nonsingular intersection with80the image of $\P_k^n$. It pulls back to a degree $d_1$ nonsingular81hypersurface $H_1$ in $\P_k^n$. If $r>1$ consider the $d_2$-uple embedding82$\P_k^n\hookrightarrow\P_k^{d_2\text{-uple}}$.83The image of $H_1$ is a nonsingular variety in $\P_k^{d_2{\text-uple}}$84of dimension $\geq 2$. By Bertinni's theorem there is a hyperplane85in $\P_k^{d_2-uple}$ whose intersection with the image of $H_1$ is nonsingular86and of dimension one less than $H_1$. Pulling back we obtain a hypersurface87$H_2$ such that $H_1\intersect H_2$ is nonsingular and $H_2$ has degree88$d_2$. Continuing inductively in this way and noting that89$\dim H_1\intersect\cdots\intersect H_{r-1}\geq 2$ (since90$r<n$) completes the proof.9192{\em (e) If $Y$ is a nonsingular complete intersection as in (d)93show that $\omega_Y\isom\sO_Y(\sum d_i-n-1)$.}9495By (III, 8.20) $\omega_{H_1}\isom\omega_{\P^n}\tensor\sL(H_1)\tensor\sO_{H_1}$.96By the explicit computation of $Cl\P^n$ (II, 6.17) we know that97$\sL(H_1)\isom\sO_{\P^n}(d_1)$. Thus98$$\omega_{H_1}\isom\sO_{\P^n}(-n-1)\tensor\sO_{\P^n}(d_1)\tensor\sO_{H_1}\isom\sO_{H_1}(d_1-n-1).$$99100By (8.20) we have that101$$\omega_{H_1\intersect H_2}\isom\omega_{H_1}\tensor\sL(H_2.H_1)\tensor\sO_{H_1\intersect H_2}.$$102We know that $H_2\sim d_2\P^{n-1}$ (linear equivalence) so103by (II, 6.2b) this implies $H_2.H_1\sim d_2\P^{n-1}.H_1$.104But $d_2\P^{n-1}.H_1$ corresponds to the invertible sheaf (see (II, Ex 6.8c)105$\sO_{H_1}(d_2)$. Thus106$$\omega_{H_1\intersect H_2}\isom\sO_{H_1}(d_1-n-1)\tensor\sO_{H_1}(d_2)\tensor\sO_{H_1\intersect H_2}\isom\sO_{H_1\intersect H_2}(d_1+d_2-n-1).$$107Repeating this argument inductively yields the desired isomorphism.108109{\em (f) If $Y$ is a nonsingular hypersurface of degree $d$ in $\P^n$,110use (c) and (e) above to show that $p_g(Y)=\chose{d-1}{n}$. Thus111$p_g(Y)=p_a(Y)$ (I, Ex. 7.2).}112113By definition $p_g(Y)=\dim_k\Gamma(Y,\omega_Y)$. By (e),114$\omega_Y\isom\sO(d-n-1)$ and by (c) the natural map115$$\Gamma(X,\sox(d-n-1))\into\Gamma(Y,\soy(d-n-1))$$116is surjective. We show that it is also injective.117By (III, 5.5a) an element $f\in\Gamma(X,\sox(d-n-1))$ can be represented as118a homogeneous polynomial of degree $d-n-1$.119Now $f$ maps to $0$ in $\Gamma(Y,\soy(d-n-1))$ iff120$f$ vanishes on $Y$, that is to say, $Y\subset Z(f)$. But121$\deg Y=d>d-n-1=\deg f$ so $Y$ can not be contained in the hypersurface122$Z(f)$ unless $f=0$. [Proof: $Y=Z(g)\subset Z(f)$ implies123$(f)\subset (g)$ so $f$ is a multiple of $g$, but $g$ has degree124strictly greater than $f$ so must be $0$.]125Thus126$$\dim_k \Gamma(Y,\soy(d-n-1))=\dim_k \Gamma(X,\sox(d-n-1))127=\chose{d-1}{n}$$128since the number of monomoials in $k[x_0,\ldots, x_n]$ of degree129$d-n-1$ is $\chose{d-1}{n}$130as desired.131132{\em (g) If $Y$ is a nonsingular curve in $\P^3$, which is a complete133intersection of nonsingular surfaces of degrees $d$, $e$, then134$p_g(Y)=\frac{1}{2}de(d+e-4)+1$. Again the geometric genus is135the same as the arithmetic genus (I, Ex. 7.2).}136137Let $H$ be the hypersurface of degree $d$.138There is an exact sequence139$$0\into\sO_{\P^n}(-d)\into\sO_{\P^n}\into\sO_H\into 0.$$140Twisting by $a$ and computing dimensions we see that141$$\dim\sO_H(a)=\dim\sO_{\P^n}(a)-\dim\sO_{\P^n}(a-d)=\chose{3+a}{3}-\chose{3+a-d}{3}.$$142143Using reasoning like that in (e) we obtain an exact sequence144$$0\into\sO_H(-e)\into\sO_H\into\sO_Y\into 0.$$145Twisting by $e+d-4$ yields the exact sequence146$$0\into\sO_H(d-4)\into\sO_H(e+d-4)\into\sO_Y(e+d-4)\into 0.$$147Applying the above explicit computation of $\dim\sO_H(a)$ we see that148$$\dim\soy(e+d-4)=\chose{e+d-1}{3}-\chose{e-1}{3}-\chose{d-1}{3}+chose{-1}{3}.$$149After some algebra the latter expression becomes150$\frac{1}{2}ed(e+d-4)+1$, as desired.151152[Comment 1: We could have also solved (f) using this method.]153154[Comment 2: Serre duality gives another solution. By (III, 7.12.4)155$p_g(Y)=\dim H^0(Y,\omega_Y)=\dim H^1(Y,\soy)=p_a(Y).$156But by (I, Ex. 7.2d) $p_a(Y)=\frac{1}{2}de(d+e-4)+1$.]157158159\section{Exercise III.6.8}160{\em Prove the following theorem of161Kleiman: if $X$ is a noetherian, integral, seperated,162locally factorial scheme, then every coherent sheaf on163$X$ is a quotient of a locally free sheaf (of finite rank).}164165{\em (a) First show that open sets of the form $X_s$, for166various $s\in\Gamma(X,\sL)$ and various invertible sheaves167$\sL$ on $X$, form a base for the topology of $X$.}168169Let $x\in U\subset X$ with $U$ open.170171{\em Case 1. $W=X-U$ is irreducible.}172Since $x\not\in W$, $\so_x\not\subset \sO_W$.173[This assertion is a matter of some difficulty among the174others working on this problem. It is not hard to see when175$X$ is a variety in the classical sense. But in the more general176situation it isn't at all clear and may use the hypothesis that177$X$ is seperated in an essential way. For example, the affine line178with a doubled origin has two different local rings which are equal.179I'm not sure how to resolve this but there was some talk of using180the valuative criterion for seperatedness. PUT CORRECT SOLUTION HERE181AFTERWARDS.]182Thus let $h\in K$ be a rational183function such that $h\not\in\sO_W$ but $h\in\so_x$. Let $(h)=D_1-D_2$184with $D_1$=zeros of $h$ and $D_2$=poles of $h$. Since $h\in\so_x$, we185have $x\not\in\Var(D_2)$=the underlying scheme of the effective186divisor $D_2$. (This is because $h$ can't have a pole at $x$.) Furthermore187$y\in W$ implies $\so_y\subset\so_W$ so $h\not\in\sO_y$ thus $y\in\Var(D_2)$188(this is because $X$189is factorial so $\sO_y$ is integrally closed so $v_y(h)<0$ iff190$h\not\in\sO_y$.) Thus $W\subset\Var(D_2)$. Since $X$ is factorial and191$D_2$ is effective (II, 6.11) implies192$D_2$ corresponds to an effective Cartier divisor and hence there exists193an open cover $\sU=(U_i)$ of $X$ and rational functions $h_i\in K$194such that $h_i|U_i\in\sO_{U_i}$ and $(h_i)=D_2$ on $U_i$. Since195$\frac{h_i}{h_j}\in\so_x(U_i\intersect U_j)^{*}$ and $X$ is normal,196$(\frac{h_i}{h_j})=0$. Let $\sL$ be the locally free invertible sheaf197represented by the Cartier divisor $(U_i,h_i)$ (so $\L$ is locally198generated by $1/h_i$ on $U_i$), and let $u_i:\sL|U_i\into\so_x|U_i$199be the isomorphism given by multiplication by $h_i$. Define200$s(y)=u_i^{-1}(h_i(y))$ for $y\in U_i$. By this we mean $u_i^{-1}$ of201the map $y\mapsto h_i(y)$, i.e., $s$ is the glueing of the inverse images202of the $h_i\in\so_x(u_i)$. Thus $s$ is a section of $\L$ such that203$X_s\intersect U_i=U_i-\Var(D_2)$. Thus204$X_s=X-\Var(D_2)\subset U$.205206{\em Case 2. $W=X-U$ is reducible.}207Using the fact that $X$ is noetherian write208$W=Z_1\union \cdots \union Z_n$. From case 1 we know that209there exists invertible sheaves $\L_1,\ldots,\L_n$ and sections210$s_i\in\Gamma(X,\L_i)$, $i=1,\ldots,n$ such that $x\in X_{s_i}\subset X-Z_i$.211Let $s=s_1\tensor\cdots\tensor s_n\in\Gamma(X,\L_1\tensor\cdots\tensor\L_n)$.212Then $X_s=\intersect_{i=1}^n X_{s_i}$ hence $x\in X_s\subset U$.213214[[This proof was copied from Borelli's paper with little modification. One215danger is that the corresponding theorem in Borelli's paper assumes $X$ to be216a factorial {\em variety}, not a more general scheme as above. Part (b)217below was not in Borelli.]]218219{\em (b) Now use (II, 5.14) to show that any coherent sheaf is a quotient220of a direct sum $\oplus\sL_i^{n_i}$ for various invertible sheaves221$\sL_i$ and various integers $n_i$. }222223Let $\sF$ be a coherent sheaf on $X$.224Let $U$ be an open set on which $\sF_{|U}\isom\tilde{M}$.225Suppose $X_f\subset U$ where $f$ is a global section of some226invertible sheaf $\sL$. Our strategy is to construct an appropriate227map $\oplus\sL_i^{n_i}\into \sF$ which is surjective when228restricted to $X_f$, then use the fact that $X$ is noetherian and229that the $X_f$ form a basis for the topology on $X$ to cover $X$230which such $X_f$ and then take the sum of all the resulting maps.231232Let $m_1,\ldots,m_r$ generate $M$. Let $t_1,\ldots,t_r$ be the233restrictions of the $m_i$ to $X_f$. By (II, 5.14b) there exists $n$234so that235$$t_1 f^n,\ldots,t_r f^n\in\Gamma(X_{f^n},\sF\tensor\sL^{\tensor n})$$236extend to global sections $s_1,\ldots,s_n$ of237$\Gamma(X,\sF\tensor\sL^{\tensor n})$. Define a map238$$\oplus_{i=1}^n \sox\into\sF\tensor\sL^{\tensor n}$$239by sending $(0,\ldots,0,1,0,\ldots,0)$ (1 in the $i$th position only)240to $s_i$. Then tensoring with241$(\sL^{\tensor n})^{-1} =(\sL^{-1})^{\tensor n}$242we obtain a map243$$\Theta:\oplus_{i=1}^n (\sL^{-1})^{\tensor n}\into\sF.$$244The map $\Theta$ is surjective when restricted to $X_f$.245To see this let $p$ be a point of $X_f$. The stalk of246$\sF$ at $p$ is generated by the stalks of $m_1,\ldots,m_n$247at $p$. Since the $t_i$ are all in the image of the map $\Theta$ and248the stalks of the $t_i$ at $p$ are the same as the stalks of the $m_i$249at $p$ it follows that the stalks of the $m_i$ are all in the image250under $\Theta$ of the stalk of $\oplus_{i=1}^n(\sL^{-1})^{\tensor n}$251at $p$.252253Take the direct sum of all such maps over a suitable open cover254$(U)$ of $X$ and suitable open covers $(X_f)$ of each $U$. Since255$X$ is noetherian we can arrange it so this sum is finite.256257\section{Exercise III.7.1}258{\em Let $X$ be an integral projective259scheme of dimension $\geq 1$ over a field $k$, and let260$\sL$ be an ample invertible sheaf $X$. Then $H^0(X,\sL^{-1})=0$.}261262\begin{lem} If $\M\neq\sox$ is an invertible sheaf which is generated by263its global sections then $H^0(X,\M^{-1})=0$. \end{lem}264\begin{proof}265By the proof of (II 6.12) $\M^{-1}=\M^{\dual}=\shom(\M,\sox)$.266Thus we must show that $\Gamma(X,\shom(\M,\sox))=0$, i.e., that267$\hom_{\sox}(\M,\sox)=0$. Since $\M$ is generated by global268sections $(m_i)$ to give a morphism $f:\M\into\sox$ is the same269as to give the images $\alpha_i=f(m_i)\in\Gamma(X,\sox)$ of the270$m_i$. Since $X$ is integral and projective $\Gamma(X,\sox)=k$271so the $\alpha_i$ all lie in $k$. Thus if $f$272is nonzero then some $\alpha_i\neq 0$ so $\frac{1}{\alpha_i}m_i\mapsto 1$.273Let $t=\frac{1}{\alpha_i}m_i\in\Gamma(X,\M)$. Let $p$ be any274point of $X$. The map $f_p:\M_p\into\soxp$ sends $t_p$ to275$1$ so it is surjective being a map of free $\soxp$-modules (and276since $\soxp$ is generated by $1$ as an $\soxp$-module). On the277other hand $\M_p$ is free of rank $1$ over the integral domain278$\soxp$ so $f$ must be injective. Indeed, if $\M_p\isom\soxp\cdot g$279for some $g$ and $ag\mapsto 0$ then $af(g)=0$ so since $\soxp$ is a domain,280$a=0$ or $f(g)=0$. But $f(g)\neq 0$ since $f$ is surjective so $a=0$281and so $ag=0$ whence $f$ is injective. Therefore $f$ is an isomorphism282since it is an isomorphism on stalks. Thus $\M\isom\sox$ contrary283to our assumption that $\M\not\isom\sox$ so there284can be no nonzero $f$ in $\hom_{\sox}(\M,\sox)=\Gamma(X,\M^{-1})$,285as desired.286\end{proof}287288Suppose that $\L$ is ample. If $\L=\sox$ then $\L$ can not be ample,289for if $\sox$ is ample then since $\sox^{\tensor n}=\sox$ for any $n\geq 1$290it follows by (II.7.5) that $\sox$ is very ample. This means that there291is an immersion $i:X\hookrightarrow\P_k^n$ where $n=\dim\Gamma(X,\sox)-1=0$292which is impossible because $X$ has dimension at least $1$.293294Thus we may assume $\L\not\isom\sox$ and apply the above lemma.295There is an $n$296so that $\L^{\tensor n}$ is generated by its global sections. By the297above lemma $H^0(X,(\L^{\tensor n})^{\dual})=0$. Since the collection298of invertible sheaves forms a group and $\dual$ is the inverse operation299it follows trivially that300$(\L^{\tensor n})^{\dual}\isom(\L^{\dual})^{\tensor n}$ and301hence $\Gamma(X,(\L^{\dual})^{\tensor n})=0$.302Suppose $\L^{\dual}$303has a nonzero global section $s$. Let $p\in X$ be a point so that304$s_p\neq 0$. It follows that $s\tensor\cdots\tensor s\neq 0$ in305$(\L^{\dual}_p)^{\tensor n}$. Thus $s$ defines a nonzero global section306$s\tensor\cdots\tensor s$ of $(\L^{\dual})^{\tensor n}$. [This last statement307is a bit subtle because the tensor product is the sheaf associated to308a certain presheaf so we don't know, {\em a priori}, that $s\tensor\cdots\tensor s$309maps to something nonzero under the $\theta$ of (II, Defn 1.2). But310if $\theta(s\tensor\cdots\tensor s)=0$ then $0=\theta(s\tensor\cdots\tensor s)_p=(s\tensor\cdots\tensor s)_p$311so $\theta$ is not injective on stalks contradicting the comment after312(II, Defn 1.2).] Thus if $H^0(X,\L^{\dual})\neq 0$ then313$H^0(X,(\L^{\tensor n})^{\dual}) \neq 0$, a contradiction. It follows314that $H^0(X,\L^{\dual})=0$, as desired.315316317\section{Exercise III.7.3}318{\em Let $X=\P^n_k$. Show that $H^q(X,\OX^p)=0$319for $p\neq q$, $k$ for $p=q$, $0\leq p,q\leq n$.}320321Our strategy is to use the exact sequence322$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0$$323of (II, 8.13) along with (II, Ex 5.16 d) to reduce the324computation of the cohomology of $\OX^p$ to the325computation of the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$.326We then show inductively327that the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$ vanishes for328$p\geq 1$ thus completing the proof.329330We compute the cohomology of $\Omega^r$ inductively on $r$.331332{\em Step 1, $r=0$.} Suppose $r=0$ so $\Omega^r=\sox$. Then333by (III, 5.5) $H^0(X,\sox)=k$ and $H^i(X,\sox)=0$ for $i\geq 1$.334[Part (a) of (III, 5.5) gives $H^0(X,\sox)=k$, part (b) gives335$H^i(X,\sox)=0$ for $0<i<n$ and part (d) gives336$H^n(X,\sox)\isom H^0(X,\sox(-n-1))^{\dual}=0$.]337338{\em Step 2} Show that339$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$$340for $r\geq 1$.341342[Matt Baker pointed out to me that343$$\Lambda^r\sox(-1)^{\oplus n+1}\isom\sox(-1)^{\oplus \chose{n+1}{r}}.$$344This is reasonable since it is true on stalks. It immediately implies345the vanishing of the cohomology groups. My original more complicated proof346of step 2 is included next anyways.]347348{\em Step 2a, $r=1$.} We treat $r=1$ as a special case.349We must show that $H^i(X,\sox(-1)^{\oplus n+1}=0$ or350equivalently that $H^i(X,\sox(-1))=0$. This is immediate351from the explicit computations of (III, 5.5). The argument proceeds352exactly as in step 1.353354{\em Step 2b, $r\geq 2$.} We now assume $r\geq 2$ and proceed inductively355on $n$. Since $r\geq 2$ there is an exact sequence356$$0\into\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Lambda^r\sox(-1)^{\oplus n}\into 0.$$357I obtained the map358$$\Lambda^r\sox(-1)^{\tensor n+1}\into\Lambda^r\sox(-1)^{\tensor n}$$359by carefully applying (II, Ex 5.16d) to the map360$\sox(-1)^{\oplus n+1}\into\sox(-1)^{\oplus n}$.361But this map just turns out to be locally defined by362$x_n\mapsto 0$ where $x_0,\ldots,x_n$ are local coordinates363for $\sox(-1)$. Then $x_{i_0}\wedge\cdots\wedge x_{i_r}$364maps to $0$ if some $i_k=n$ and itself otherwise.365The map366$$\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}$$367identifies $\Lambda^{r-1}\sox(-1)^{\oplus n}$ with the kernel of the368next map. The kernel of the next map is locally generated by all369``monomials'' which contain an $x_n$. Since $r\geq 2$ we can identify370$\Lambda^{r-1}\sox(-1)^{\oplus n}$ with this kernel by just removing371the $x_n$ off of the wedge product.372[This is not rigorous enough!]373374By induction on $n$ we have that375$$H^i(\Lambda^{r-1}\sox(-1)^{\oplus n})=H^i(\Lambda^r\sox(-1)^{\oplus n})=0$$376for all $i$ (we will do the base case $n=1$ in just a moment).377Thus, by the long exact sequence of cohomology we see that378$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$ for all $i$.379For $n=1$, since $r\geq 2$380it follows that $\Lambda^{r-1}\sox(-1)=\sox(-1)$ or $0$381and $\Lambda^r\sox(-1)=0$ and these both have trivial cohomology382as computed above.383384{\em Step 3.} The final step is to obtain the long exact sequence385$$\cdots H^i(\Omega^r)\into H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1})\into \cdots$$386then apply step 2 and the induction hypothesis (we are inducting387on $r$, the base case was established in step 1) to388calculate $H^i(\Omega^r)$ for all $i$.389390Suppose $r\geq 1$, then by (II, 8.13) we have an exact sequence391$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0.$$392By (II, Ex. 5.16d), $\Lambda^r\sox(-1)^{\oplus n+1}$ has a filtration393$$\Lambda^r\sox(-1)^{\oplus n+1}=F^0\supseteq F^1\supseteq\cdots\supseteq F^r\supseteq F^{r+1}=0$$394with quotients395$$F^p/F^{p+1}\isom\Omega^p\tensor\Lambda^{r-p}\sox=396\begin{cases} 0&\text{if $r-p\geq 2$}\\397\Omega^p&\text{if $r-p$ is $0$ or $1$}\end{cases}$$398Thus $$\Lambda^r\sox(-1)^{\oplus n+1}=F^0=\cdots=F^{r-1}$$ and the filtration399becomes $$\Lambda^r\sox(-1)^{\oplus n+1}\supset F^r\supset F^{r+1}=0$$400with $\Lambda^r\sox(-1)^{\oplus n+1}/F^r\isom \Omega^{r-1}$ and401$F^r\isom\Omega^r$.402This gives an exact sequence403$$0\into\Omega^r\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Omega^{r-1}\into 0.$$404The associated long exact sequence of cohomology gives for each $i$405an exact sequence406$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1})407\into H^{i+1}(\Omega^r)\into H^{i+1}(\Lambda^r\sox(-1)^{\oplus n+1}).$$408But by step 2 the groups $H^i(\Lambda^r\sox(-1)^{\oplus n+1})$ all vanish.409Thus $H^i(\Omega^{r-1})\isom H^{i+1}(\Omega^r)$. By induction on410$r$ this shows that411$$H^i(\Omega^r)=\begin{cases}0&\text{if $i\neq r$}\\412k&\text{if $i=r$}\end{cases}.$$413This completes the proof.414\end{document}415416