Sharedwww / hartsoln5.texOpen in CoCalc
Author: William A. Stein
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\documentclass[12pt]{article}
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\include{defs}
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\author{William A. Stein}
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\title{Homework 2, MAT256B\\II.8.4, III.6.8, III.7.1, III.7.3}
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\begin{document}
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\maketitle
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\section{Exercise II.8.4}
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{\em Complete Intersections in $\P^n$.
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A closed subscheme $Y$ of $\P^n_k$ is called a {\em (strict, global)
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complete intersection} if the homogenous ideal $I$ of $Y$ in
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$S=k[x_0,\ldots,x_n]$ can be generated by $r$ elements where
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$r=\codim(Y,\P^n)$.}
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{\em (a) Let $Y$ be a closed subscheme of codimension $r$ in $\P^n$. Then
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$Y$ is a complete intersection iff there are hypersurfaces (i.e., locally
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principal subschemes of codimension $1$) $H_1,\ldots,H_r$, such that
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$Y=H_1\intersect\cdots\intersect H_r$ {\em as schemes}, i.e.,
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$\sI_Y=\sI_{H_1}+\cdots+\sI_{H_r}$.}
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($\Rightarrow$) By (II, Ex 5.14) $I$ is defined to be $\Gamma_{*}(\sI_Y)$.
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By (II, 5.15), $\tilde{I}\isom\sI_Y$. Write $I=(f_1,\ldots,f_r)$, then
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since localization commutes with taking sums,
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$$\sI_Y=(f_1,\ldots,f_r)\tilde{ }=((f_1)+\cdots+(f_r))\tilde{ }=(f_1)\tilde{ }+\cdots+(f_r)\tilde{ }.$$
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Let $H_i$ be the locally principal closed subscheme of codimension
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$1$ determined by the ideal sheaf $(f_i)\tilde{ }$. Then $Y$ is the
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intersection of the $H_i$.
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($\Leftarrow$) Someone suggested I should apply unmixedness and
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primary decomposition to some ideal somewhere and use the fact
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that a saturated ideal doesn't have primary components corresponding
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to the irrelevant ideal or something like that. NOT DONE.
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{\em (b) If $Y$ is a complete intersection of dimension $\geq 1$ in
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$\P^n$, and if $Y$ is normal, then $Y$ is projectively normal (Ex. 5.14).}
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Let $Z$ be the cone over $Y$, then $A(Z)=S/I(Y)$. By
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(I, Ex. 3.17d), $A(Z)$ is integrally closed iff $Z$ is
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normal. By definition $A(Z)$ is integrally closed iff
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$Y$ is projectively normal. Thus we must show that $Z$
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is normal. Since $Y$ is a complete intersection, $I(Y)=(f_1,\ldots,f_r)$
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so $Z$ is a complete intersection subscheme of $\bA^{n+1}$. By
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(II, 8.23) $Z$ is normal iff $Z$ is regular in codimension $1$.
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Also by (II, 8.23) $Y$ is regular in codimension $1$ because we
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have assumed $Y$ is normal. But $Y$ regular in codimension $1$ implies
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$Z$ regular in codimension $1$. [We used this last semester in
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(II, Ex. 6.3d). Intuitively, the only singularity in $Z$ not in $Y$
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is the cone point which has codimension $>1$. This is because $Z$
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is locally $U_i\cross\bA^1$.]
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{\em (c) With the same hypothesis as in (b), conclude that for all
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$\ell\geq 0$, the natural map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$
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is surjective. In particular, taking $\ell=0$, show that $Y$ is connected.}
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That the map $\Gamma(\P^n,\sO_{\P^n}(\ell))\into\Gamma(Y,\sO_Y(\ell))$
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is surjective is just the statement of (II, Ex. 5.14d). When $\ell=0$
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this says that
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$k=\Gamma(\P^n,\sO_{\P^n}(\ell))$ surjects onto $\Gamma(Y,\sO_Y(\ell))$
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so $\dim\Gamma(Y,\sO_Y)\leq 1$ and hence $Y$ is connected. [If $Y$ were
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not connected then $\Gamma(Y,\sO_Y)=k\oplus\cdots\oplus k$ where the
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number of direct summands equals the number of components of $Y$.]
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{\em (d) Now suppose given integers $d_1,\ldots,d_r\geq 1$, with
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$r<n$. Use Bertini's theorem (8.18) to show that there exists nonsingular
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hypersurfaces $H_1,\ldots,H_r$ in $\P^n$, with $\deg H_i=d_i$, such that
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the scheme $Y=H_1\intersect\cdots\intersect H_r$ is irreducible
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and nonsingular in codimension $r$ in $\P^n$.}
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[To apply Bertini's theorem we must assume $k$ is algebraically closed.
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I'm going to make this assumption now. Maybe there is a way around this?]
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Let $\P_k^n\hookrightarrow\P_k^{d_1\text{-uple}}$
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be the $d_1$-uple embedding of $\P_k^n$. Use Bertini's theorem to choose
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a hyperplane in
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$\P_k^{d_1\text{-uple}}$ which has nonsingular intersection with
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the image of $\P_k^n$. It pulls back to a degree $d_1$ nonsingular
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hypersurface $H_1$ in $\P_k^n$. If $r>1$ consider the $d_2$-uple embedding
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$\P_k^n\hookrightarrow\P_k^{d_2\text{-uple}}$.
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The image of $H_1$ is a nonsingular variety in $\P_k^{d_2{\text-uple}}$
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of dimension $\geq 2$. By Bertinni's theorem there is a hyperplane
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in $\P_k^{d_2-uple}$ whose intersection with the image of $H_1$ is nonsingular
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and of dimension one less than $H_1$. Pulling back we obtain a hypersurface
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$H_2$ such that $H_1\intersect H_2$ is nonsingular and $H_2$ has degree
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$d_2$. Continuing inductively in this way and noting that
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$\dim H_1\intersect\cdots\intersect H_{r-1}\geq 2$ (since
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$r<n$) completes the proof.
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{\em (e) If $Y$ is a nonsingular complete intersection as in (d)
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show that $\omega_Y\isom\sO_Y(\sum d_i-n-1)$.}
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By (III, 8.20) $\omega_{H_1}\isom\omega_{\P^n}\tensor\sL(H_1)\tensor\sO_{H_1}$.
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By the explicit computation of $Cl\P^n$ (II, 6.17) we know that
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$\sL(H_1)\isom\sO_{\P^n}(d_1)$. Thus
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$$\omega_{H_1}\isom\sO_{\P^n}(-n-1)\tensor\sO_{\P^n}(d_1)\tensor\sO_{H_1}\isom\sO_{H_1}(d_1-n-1).$$
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By (8.20) we have that
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$$\omega_{H_1\intersect H_2}\isom\omega_{H_1}\tensor\sL(H_2.H_1)\tensor\sO_{H_1\intersect H_2}.$$
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We know that $H_2\sim d_2\P^{n-1}$ (linear equivalence) so
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by (II, 6.2b) this implies $H_2.H_1\sim d_2\P^{n-1}.H_1$.
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But $d_2\P^{n-1}.H_1$ corresponds to the invertible sheaf (see (II, Ex 6.8c)
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$\sO_{H_1}(d_2)$. Thus
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$$\omega_{H_1\intersect H_2}\isom\sO_{H_1}(d_1-n-1)\tensor\sO_{H_1}(d_2)\tensor\sO_{H_1\intersect H_2}\isom\sO_{H_1\intersect H_2}(d_1+d_2-n-1).$$
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Repeating this argument inductively yields the desired isomorphism.
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{\em (f) If $Y$ is a nonsingular hypersurface of degree $d$ in $\P^n$,
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use (c) and (e) above to show that $p_g(Y)=\chose{d-1}{n}$. Thus
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$p_g(Y)=p_a(Y)$ (I, Ex. 7.2).}
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By definition $p_g(Y)=\dim_k\Gamma(Y,\omega_Y)$. By (e),
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$\omega_Y\isom\sO(d-n-1)$ and by (c) the natural map
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$$\Gamma(X,\sox(d-n-1))\into\Gamma(Y,\soy(d-n-1))$$
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is surjective. We show that it is also injective.
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By (III, 5.5a) an element $f\in\Gamma(X,\sox(d-n-1))$ can be represented as
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a homogeneous polynomial of degree $d-n-1$.
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Now $f$ maps to $0$ in $\Gamma(Y,\soy(d-n-1))$ iff
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$f$ vanishes on $Y$, that is to say, $Y\subset Z(f)$. But
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$\deg Y=d>d-n-1=\deg f$ so $Y$ can not be contained in the hypersurface
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$Z(f)$ unless $f=0$. [Proof: $Y=Z(g)\subset Z(f)$ implies
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$(f)\subset (g)$ so $f$ is a multiple of $g$, but $g$ has degree
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strictly greater than $f$ so must be $0$.]
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Thus
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$$\dim_k \Gamma(Y,\soy(d-n-1))=\dim_k \Gamma(X,\sox(d-n-1))
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=\chose{d-1}{n}$$
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since the number of monomoials in $k[x_0,\ldots, x_n]$ of degree
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$d-n-1$ is $\chose{d-1}{n}$
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as desired.
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{\em (g) If $Y$ is a nonsingular curve in $\P^3$, which is a complete
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intersection of nonsingular surfaces of degrees $d$, $e$, then
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$p_g(Y)=\frac{1}{2}de(d+e-4)+1$. Again the geometric genus is
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the same as the arithmetic genus (I, Ex. 7.2).}
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Let $H$ be the hypersurface of degree $d$.
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There is an exact sequence
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$$0\into\sO_{\P^n}(-d)\into\sO_{\P^n}\into\sO_H\into 0.$$
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Twisting by $a$ and computing dimensions we see that
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$$\dim\sO_H(a)=\dim\sO_{\P^n}(a)-\dim\sO_{\P^n}(a-d)=\chose{3+a}{3}-\chose{3+a-d}{3}.$$
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Using reasoning like that in (e) we obtain an exact sequence
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$$0\into\sO_H(-e)\into\sO_H\into\sO_Y\into 0.$$
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Twisting by $e+d-4$ yields the exact sequence
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$$0\into\sO_H(d-4)\into\sO_H(e+d-4)\into\sO_Y(e+d-4)\into 0.$$
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Applying the above explicit computation of $\dim\sO_H(a)$ we see that
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$$\dim\soy(e+d-4)=\chose{e+d-1}{3}-\chose{e-1}{3}-\chose{d-1}{3}+chose{-1}{3}.$$
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After some algebra the latter expression becomes
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$\frac{1}{2}ed(e+d-4)+1$, as desired.
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[Comment 1: We could have also solved (f) using this method.]
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[Comment 2: Serre duality gives another solution. By (III, 7.12.4)
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$p_g(Y)=\dim H^0(Y,\omega_Y)=\dim H^1(Y,\soy)=p_a(Y).$
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But by (I, Ex. 7.2d) $p_a(Y)=\frac{1}{2}de(d+e-4)+1$.]
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\section{Exercise III.6.8}
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{\em Prove the following theorem of
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Kleiman: if $X$ is a noetherian, integral, seperated,
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locally factorial scheme, then every coherent sheaf on
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$X$ is a quotient of a locally free sheaf (of finite rank).}
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{\em (a) First show that open sets of the form $X_s$, for
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various $s\in\Gamma(X,\sL)$ and various invertible sheaves
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$\sL$ on $X$, form a base for the topology of $X$.}
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Let $x\in U\subset X$ with $U$ open.
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{\em Case 1. $W=X-U$ is irreducible.}
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Since $x\not\in W$, $\so_x\not\subset \sO_W$.
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[This assertion is a matter of some difficulty among the
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others working on this problem. It is not hard to see when
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$X$ is a variety in the classical sense. But in the more general
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situation it isn't at all clear and may use the hypothesis that
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$X$ is seperated in an essential way. For example, the affine line
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with a doubled origin has two different local rings which are equal.
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I'm not sure how to resolve this but there was some talk of using
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the valuative criterion for seperatedness. PUT CORRECT SOLUTION HERE
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AFTERWARDS.]
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Thus let $h\in K$ be a rational
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function such that $h\not\in\sO_W$ but $h\in\so_x$. Let $(h)=D_1-D_2$
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with $D_1$=zeros of $h$ and $D_2$=poles of $h$. Since $h\in\so_x$, we
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have $x\not\in\Var(D_2)$=the underlying scheme of the effective
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divisor $D_2$. (This is because $h$ can't have a pole at $x$.) Furthermore
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$y\in W$ implies $\so_y\subset\so_W$ so $h\not\in\sO_y$ thus $y\in\Var(D_2)$
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(this is because $X$
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is factorial so $\sO_y$ is integrally closed so $v_y(h)<0$ iff
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$h\not\in\sO_y$.) Thus $W\subset\Var(D_2)$. Since $X$ is factorial and
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$D_2$ is effective (II, 6.11) implies
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$D_2$ corresponds to an effective Cartier divisor and hence there exists
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an open cover $\sU=(U_i)$ of $X$ and rational functions $h_i\in K$
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such that $h_i|U_i\in\sO_{U_i}$ and $(h_i)=D_2$ on $U_i$. Since
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$\frac{h_i}{h_j}\in\so_x(U_i\intersect U_j)^{*}$ and $X$ is normal,
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$(\frac{h_i}{h_j})=0$. Let $\sL$ be the locally free invertible sheaf
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represented by the Cartier divisor $(U_i,h_i)$ (so $\L$ is locally
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generated by $1/h_i$ on $U_i$), and let $u_i:\sL|U_i\into\so_x|U_i$
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be the isomorphism given by multiplication by $h_i$. Define
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$s(y)=u_i^{-1}(h_i(y))$ for $y\in U_i$. By this we mean $u_i^{-1}$ of
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the map $y\mapsto h_i(y)$, i.e., $s$ is the glueing of the inverse images
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of the $h_i\in\so_x(u_i)$. Thus $s$ is a section of $\L$ such that
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$X_s\intersect U_i=U_i-\Var(D_2)$. Thus
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$X_s=X-\Var(D_2)\subset U$.
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{\em Case 2. $W=X-U$ is reducible.}
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Using the fact that $X$ is noetherian write
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$W=Z_1\union \cdots \union Z_n$. From case 1 we know that
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there exists invertible sheaves $\L_1,\ldots,\L_n$ and sections
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$s_i\in\Gamma(X,\L_i)$, $i=1,\ldots,n$ such that $x\in X_{s_i}\subset X-Z_i$.
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Let $s=s_1\tensor\cdots\tensor s_n\in\Gamma(X,\L_1\tensor\cdots\tensor\L_n)$.
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Then $X_s=\intersect_{i=1}^n X_{s_i}$ hence $x\in X_s\subset U$.
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[[This proof was copied from Borelli's paper with little modification. One
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danger is that the corresponding theorem in Borelli's paper assumes $X$ to be
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a factorial {\em variety}, not a more general scheme as above. Part (b)
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below was not in Borelli.]]
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{\em (b) Now use (II, 5.14) to show that any coherent sheaf is a quotient
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of a direct sum $\oplus\sL_i^{n_i}$ for various invertible sheaves
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$\sL_i$ and various integers $n_i$. }
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Let $\sF$ be a coherent sheaf on $X$.
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Let $U$ be an open set on which $\sF_{|U}\isom\tilde{M}$.
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Suppose $X_f\subset U$ where $f$ is a global section of some
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invertible sheaf $\sL$. Our strategy is to construct an appropriate
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map $\oplus\sL_i^{n_i}\into \sF$ which is surjective when
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restricted to $X_f$, then use the fact that $X$ is noetherian and
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that the $X_f$ form a basis for the topology on $X$ to cover $X$
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which such $X_f$ and then take the sum of all the resulting maps.
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Let $m_1,\ldots,m_r$ generate $M$. Let $t_1,\ldots,t_r$ be the
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restrictions of the $m_i$ to $X_f$. By (II, 5.14b) there exists $n$
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so that
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$$t_1 f^n,\ldots,t_r f^n\in\Gamma(X_{f^n},\sF\tensor\sL^{\tensor n})$$
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extend to global sections $s_1,\ldots,s_n$ of
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$\Gamma(X,\sF\tensor\sL^{\tensor n})$. Define a map
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$$\oplus_{i=1}^n \sox\into\sF\tensor\sL^{\tensor n}$$
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by sending $(0,\ldots,0,1,0,\ldots,0)$ (1 in the $i$th position only)
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to $s_i$. Then tensoring with
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$(\sL^{\tensor n})^{-1} =(\sL^{-1})^{\tensor n}$
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we obtain a map
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$$\Theta:\oplus_{i=1}^n (\sL^{-1})^{\tensor n}\into\sF.$$
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The map $\Theta$ is surjective when restricted to $X_f$.
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To see this let $p$ be a point of $X_f$. The stalk of
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$\sF$ at $p$ is generated by the stalks of $m_1,\ldots,m_n$
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at $p$. Since the $t_i$ are all in the image of the map $\Theta$ and
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the stalks of the $t_i$ at $p$ are the same as the stalks of the $m_i$
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at $p$ it follows that the stalks of the $m_i$ are all in the image
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under $\Theta$ of the stalk of $\oplus_{i=1}^n(\sL^{-1})^{\tensor n}$
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at $p$.
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Take the direct sum of all such maps over a suitable open cover
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$(U)$ of $X$ and suitable open covers $(X_f)$ of each $U$. Since
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$X$ is noetherian we can arrange it so this sum is finite.
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\section{Exercise III.7.1}
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{\em Let $X$ be an integral projective
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scheme of dimension $\geq 1$ over a field $k$, and let
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$\sL$ be an ample invertible sheaf $X$. Then $H^0(X,\sL^{-1})=0$.}
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\begin{lem} If $\M\neq\sox$ is an invertible sheaf which is generated by
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its global sections then $H^0(X,\M^{-1})=0$. \end{lem}
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\begin{proof}
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By the proof of (II 6.12) $\M^{-1}=\M^{\dual}=\shom(\M,\sox)$.
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Thus we must show that $\Gamma(X,\shom(\M,\sox))=0$, i.e., that
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$\hom_{\sox}(\M,\sox)=0$. Since $\M$ is generated by global
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sections $(m_i)$ to give a morphism $f:\M\into\sox$ is the same
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as to give the images $\alpha_i=f(m_i)\in\Gamma(X,\sox)$ of the
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$m_i$. Since $X$ is integral and projective $\Gamma(X,\sox)=k$
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so the $\alpha_i$ all lie in $k$. Thus if $f$
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is nonzero then some $\alpha_i\neq 0$ so $\frac{1}{\alpha_i}m_i\mapsto 1$.
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Let $t=\frac{1}{\alpha_i}m_i\in\Gamma(X,\M)$. Let $p$ be any
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point of $X$. The map $f_p:\M_p\into\soxp$ sends $t_p$ to
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$1$ so it is surjective being a map of free $\soxp$-modules (and
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since $\soxp$ is generated by $1$ as an $\soxp$-module). On the
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other hand $\M_p$ is free of rank $1$ over the integral domain
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$\soxp$ so $f$ must be injective. Indeed, if $\M_p\isom\soxp\cdot g$
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for some $g$ and $ag\mapsto 0$ then $af(g)=0$ so since $\soxp$ is a domain,
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$a=0$ or $f(g)=0$. But $f(g)\neq 0$ since $f$ is surjective so $a=0$
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and so $ag=0$ whence $f$ is injective. Therefore $f$ is an isomorphism
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since it is an isomorphism on stalks. Thus $\M\isom\sox$ contrary
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to our assumption that $\M\not\isom\sox$ so there
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can be no nonzero $f$ in $\hom_{\sox}(\M,\sox)=\Gamma(X,\M^{-1})$,
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as desired.
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\end{proof}
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Suppose that $\L$ is ample. If $\L=\sox$ then $\L$ can not be ample,
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for if $\sox$ is ample then since $\sox^{\tensor n}=\sox$ for any $n\geq 1$
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it follows by (II.7.5) that $\sox$ is very ample. This means that there
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is an immersion $i:X\hookrightarrow\P_k^n$ where $n=\dim\Gamma(X,\sox)-1=0$
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which is impossible because $X$ has dimension at least $1$.
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Thus we may assume $\L\not\isom\sox$ and apply the above lemma.
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There is an $n$
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so that $\L^{\tensor n}$ is generated by its global sections. By the
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above lemma $H^0(X,(\L^{\tensor n})^{\dual})=0$. Since the collection
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of invertible sheaves forms a group and $\dual$ is the inverse operation
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it follows trivially that
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$(\L^{\tensor n})^{\dual}\isom(\L^{\dual})^{\tensor n}$ and
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hence $\Gamma(X,(\L^{\dual})^{\tensor n})=0$.
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Suppose $\L^{\dual}$
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has a nonzero global section $s$. Let $p\in X$ be a point so that
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$s_p\neq 0$. It follows that $s\tensor\cdots\tensor s\neq 0$ in
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$(\L^{\dual}_p)^{\tensor n}$. Thus $s$ defines a nonzero global section
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$s\tensor\cdots\tensor s$ of $(\L^{\dual})^{\tensor n}$. [This last statement
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is a bit subtle because the tensor product is the sheaf associated to
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a certain presheaf so we don't know, {\em a priori}, that $s\tensor\cdots\tensor s$
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maps to something nonzero under the $\theta$ of (II, Defn 1.2). But
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if $\theta(s\tensor\cdots\tensor s)=0$ then $0=\theta(s\tensor\cdots\tensor s)_p=(s\tensor\cdots\tensor s)_p$
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so $\theta$ is not injective on stalks contradicting the comment after
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(II, Defn 1.2).] Thus if $H^0(X,\L^{\dual})\neq 0$ then
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$H^0(X,(\L^{\tensor n})^{\dual}) \neq 0$, a contradiction. It follows
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that $H^0(X,\L^{\dual})=0$, as desired.
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\section{Exercise III.7.3}
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{\em Let $X=\P^n_k$. Show that $H^q(X,\OX^p)=0$
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for $p\neq q$, $k$ for $p=q$, $0\leq p,q\leq n$.}
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Our strategy is to use the exact sequence
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$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0$$
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of (II, 8.13) along with (II, Ex 5.16 d) to reduce the
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computation of the cohomology of $\OX^p$ to the
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computation of the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$.
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We then show inductively
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that the cohomology of $\Lambda^p\sox(-1)^{\oplus n+1}$ vanishes for
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$p\geq 1$ thus completing the proof.
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We compute the cohomology of $\Omega^r$ inductively on $r$.
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{\em Step 1, $r=0$.} Suppose $r=0$ so $\Omega^r=\sox$. Then
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by (III, 5.5) $H^0(X,\sox)=k$ and $H^i(X,\sox)=0$ for $i\geq 1$.
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[Part (a) of (III, 5.5) gives $H^0(X,\sox)=k$, part (b) gives
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$H^i(X,\sox)=0$ for $0<i<n$ and part (d) gives
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$H^n(X,\sox)\isom H^0(X,\sox(-n-1))^{\dual}=0$.]
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{\em Step 2} Show that
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$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$$
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for $r\geq 1$.
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[Matt Baker pointed out to me that
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$$\Lambda^r\sox(-1)^{\oplus n+1}\isom\sox(-1)^{\oplus \chose{n+1}{r}}.$$
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This is reasonable since it is true on stalks. It immediately implies
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the vanishing of the cohomology groups. My original more complicated proof
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of step 2 is included next anyways.]
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{\em Step 2a, $r=1$.} We treat $r=1$ as a special case.
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We must show that $H^i(X,\sox(-1)^{\oplus n+1}=0$ or
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equivalently that $H^i(X,\sox(-1))=0$. This is immediate
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from the explicit computations of (III, 5.5). The argument proceeds
353
exactly as in step 1.
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{\em Step 2b, $r\geq 2$.} We now assume $r\geq 2$ and proceed inductively
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on $n$. Since $r\geq 2$ there is an exact sequence
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$$0\into\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Lambda^r\sox(-1)^{\oplus n}\into 0.$$
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I obtained the map
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$$\Lambda^r\sox(-1)^{\tensor n+1}\into\Lambda^r\sox(-1)^{\tensor n}$$
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by carefully applying (II, Ex 5.16d) to the map
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$\sox(-1)^{\oplus n+1}\into\sox(-1)^{\oplus n}$.
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But this map just turns out to be locally defined by
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$x_n\mapsto 0$ where $x_0,\ldots,x_n$ are local coordinates
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for $\sox(-1)$. Then $x_{i_0}\wedge\cdots\wedge x_{i_r}$
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maps to $0$ if some $i_k=n$ and itself otherwise.
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The map
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$$\Lambda^{r-1}\sox(-1)^{\oplus n}\into\Lambda^r\sox(-1)^{\oplus n+1}$$
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identifies $\Lambda^{r-1}\sox(-1)^{\oplus n}$ with the kernel of the
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next map. The kernel of the next map is locally generated by all
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``monomials'' which contain an $x_n$. Since $r\geq 2$ we can identify
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$\Lambda^{r-1}\sox(-1)^{\oplus n}$ with this kernel by just removing
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the $x_n$ off of the wedge product.
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[This is not rigorous enough!]
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By induction on $n$ we have that
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$$H^i(\Lambda^{r-1}\sox(-1)^{\oplus n})=H^i(\Lambda^r\sox(-1)^{\oplus n})=0$$
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for all $i$ (we will do the base case $n=1$ in just a moment).
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Thus, by the long exact sequence of cohomology we see that
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$H^i(\Lambda^r\sox(-1)^{\oplus n+1})=0$ for all $i$.
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For $n=1$, since $r\geq 2$
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it follows that $\Lambda^{r-1}\sox(-1)=\sox(-1)$ or $0$
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and $\Lambda^r\sox(-1)=0$ and these both have trivial cohomology
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as computed above.
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{\em Step 3.} The final step is to obtain the long exact sequence
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$$\cdots H^i(\Omega^r)\into H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1})\into \cdots$$
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then apply step 2 and the induction hypothesis (we are inducting
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on $r$, the base case was established in step 1) to
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calculate $H^i(\Omega^r)$ for all $i$.
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Suppose $r\geq 1$, then by (II, 8.13) we have an exact sequence
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$$0\into\OX\into\sox(-1)^{\oplus n+1}\into\sox\into 0.$$
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By (II, Ex. 5.16d), $\Lambda^r\sox(-1)^{\oplus n+1}$ has a filtration
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$$\Lambda^r\sox(-1)^{\oplus n+1}=F^0\supseteq F^1\supseteq\cdots\supseteq F^r\supseteq F^{r+1}=0$$
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with quotients
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$$F^p/F^{p+1}\isom\Omega^p\tensor\Lambda^{r-p}\sox=
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\begin{cases} 0&\text{if $r-p\geq 2$}\\
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\Omega^p&\text{if $r-p$ is $0$ or $1$}\end{cases}$$
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Thus $$\Lambda^r\sox(-1)^{\oplus n+1}=F^0=\cdots=F^{r-1}$$ and the filtration
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becomes $$\Lambda^r\sox(-1)^{\oplus n+1}\supset F^r\supset F^{r+1}=0$$
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with $\Lambda^r\sox(-1)^{\oplus n+1}/F^r\isom \Omega^{r-1}$ and
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$F^r\isom\Omega^r$.
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This gives an exact sequence
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$$0\into\Omega^r\into\Lambda^r\sox(-1)^{\oplus n+1}\into\Omega^{r-1}\into 0.$$
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The associated long exact sequence of cohomology gives for each $i$
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an exact sequence
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$$H^i(\Lambda^r\sox(-1)^{\oplus n+1})\into H^i(\Omega^{r-1})
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\into H^{i+1}(\Omega^r)\into H^{i+1}(\Lambda^r\sox(-1)^{\oplus n+1}).$$
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But by step 2 the groups $H^i(\Lambda^r\sox(-1)^{\oplus n+1})$ all vanish.
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Thus $H^i(\Omega^{r-1})\isom H^{i+1}(\Omega^r)$. By induction on
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$r$ this shows that
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$$H^i(\Omega^r)=\begin{cases}0&\text{if $i\neq r$}\\
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k&\text{if $i=r$}\end{cases}.$$
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This completes the proof.
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\end{document}
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