%%%%%%%%%%%%%%%%%%%%%%%%%% %% Homework assignment 2 %%%% \documentclass[12pt]{article} \include{defs} \author{William A. Stein} \title{Homework 2, MAT256B\\Chapter III, 4.8, 4.9, 5.6} \begin{document} \maketitle \section{Homework} \begin{exercise} {\bfseries (4.8)} \quad {\em Cohomological Dimension}. Let $X$ be a neotherian separated scheme. We define the {\em cohomological dimension} of $X$, denoted $\cd(X)$, to be the least integer $n$ such that $H^{i}(X,\F)=0$ for all quasi-coherent sheaves $\F$ and all $i>n$. Thus for example, Serre's theorem (3.7) says that $\cd(X)=0$ if and only if $X$ is affine. Grothendieck's theorem (2.7) implies that $\cd(X)\leq\dim X$. (a) In the definition of $\cd(X)$, show that it is sufficient to consider only coherent sheaves on $X$. (b) If $X$ is quasi-projective over a field $k$, then it is even sufficient to consider only locally free coherent sheaves on $X$. (c) Suppose $X$ has a covering by $r+1$ open affine subsets. Use \cech{} cohomology to show that $\cd(X)\leq r$. (d) If $X$ is a quasi-projective variety of dimension $r$ over a field $k$, then $X$ can be covered by $r+1$ open affine subsets. Conclude that $\cd(X)\leq \dim X$. (e) Let $Y$ be a set-theoretic complete intersection of codimension $r$ in $X=\P_k^n$. Show that $\cd(X-Y)\leq r-1$. \end{exercise} \begin{proof} (a) It suffices to show that if, for some $i$, $H^i(X,\F)=0$ for all coherent sheaves $\F$, then $H^i(X,\F)=0$ for all quasi-coherent sheaves $\F$. Thus suppose the $i$th cohomology of all coherent sheaves on $X$ vanishes and let $\F$ be quasi-coherent. Let $(\F_{\alpha})$ be the collection of coherent subsheaves of $\F$, ordered by inclusion. Then by (II, Ex. 5.15e) $\varinjlim\F_{\alpha}=\F$, so by (2.9) $$H^i(X,\F)=H^i(X,\varinjlim \F_{\alpha})=\varinjlim H^i(X,\F_{\alpha})=0.$$ (b) Suppose $n$ is an integer and $H^i(X,\F)=0$ for all coherent locally free sheaves $\F$ and integers $i>n$. We must show $H^i(X,\F)=0$ for all coherent $\F$ and all $i>n$, then applying (a) gives the desired result. Since $X$ is quasiprojective there is an open immersion $$i:X\hookrightarrow Y\subset \P_k^n$$ with $Y$ a closed subscheme of $\P_k^n$ and $i(X)$ open in $Y$. By (II, Ex. 5.5c) the sheaf $\F$ on $X$ pushes forward to a coherent sheaf on $\F'=i_{*}\F$ on $Y$. By (II, 5.18) we may write $\F'$ as a quotient of a locally free coherent sheaf $\E'$ on $Y$. Letting $\R'$ be the kernel gives an exact sequence $$0\into\R'\into\E'\into\F'\into 0$$ with $R'$ coherent (it's the quotient of coherent sheaves). Pulling back via $i$ to $X$ gives an exact sequence %%%%%%%%%%%%%%%%%%%%%%%%% %% This is probably possible because $i$ is an %% open immersion, but there are definately some %% details to check!! %%%%%%%%%%%%%%%%%%%%%%%%% $$0\into\R\into\E\into\F\into 0$$ of coherent sheaves on $X$ with $E$ locally free. The long exact sequence of cohomology shows that for $i>n$, there is an exact sequence $$0=H^i(X,\E)\into H^i(X,\F) \into H^{i+1}(X,\R) \into H^{i+1}(X,\E)=0.$$ $H^i(X,\E)=H^{i+1}(X,\E)=0$ because we have assumed that, for $i>n$, cohomology vanishes on locally free coherent sheaves. Thus $H^i(X,\F)\isom H^{i+1}(X,\R)$. But if $k=\dim X$, then Grothendieck vanishing (2.7) implies that $H^{k+1}(X,\R)=0$ whence $H^{k}(X,\F)=0$. But then applying the above argument with $\F$ replaced by $\R$ shows that $H^{k}(X,\R)=0$ which implies $H^{k-1}(X,\F)=0$ (so long as $k-1>n$). Again, apply the entire argument with $\F$ replaced by $\R$ to see that $H^{k-1}(X,\R)=0$. We can continue this descent and hence show that $H^{i}(X,\F)=0$ for all $i>n$. (c) By (4.5) we can compute cohomology by using the \cech{} complex resulting from the cover $\sU$ of $X$ by $r+1$ open affines. By definition $\sC^p=0$ for all $p>r$ since there are no intersections of $p+1\geq r+2$ distinct open sets in our collection of $r+1$ open sets. The \cech{} complex is $$\sC^0\into\sC^1\into\cdots\into\sC^r\into\sC^{r+1}=0\into 0\into 0\into \cdots.$$ Thus if $\sF$ is quasicoherent then $\cH^p(\sU,\sF)=0$ for any $p>r$ which implies that $\cd(X)\leq r$. (d) I will first present my solution in the special case that $X$ is projective. The more general case when $X$ is quasi-projective is similiar, but more complicated, and will be presented next. Suppose $X\subset\P^n$ is a projective variety of dimension $r$. We must cover $X$ with $r+1$ open affines. Let $U$ be nonempty open affine subset of $X$. Since $X$ is irreducible, the irreducible components of $X-U$ all have codimension at least one in $X$. Now pick a hyperplane $H$ which doesn't completely contain any irreducible component of $X-U$. We can do this by choosing one point $P_i$ in each of the finitely many irreducible components of $X-U$ and choosing a hyperplane which avoids all the $P_i$. This can be done because the field is infinite (varieties are only defined over algebraically closed fields) so we can always choose a vector not orthogonal to any of a finite set of vectors. Since $X$ is closed in $\P^n$ and $\P^n-H$ is affine, $(\P^n-H)\intersect X$ is an open affine subset of $X$. Because of our choice of $H$, $U\union((\P^n-H)\intersect X)$ is only missing codimension two closed subsets of $X$. Let $H_1=H$ and choose another hyperplane $H_2$ so it doesn't completely contain any of the (codimension two) irreducible components of $X-U-(\P^n-H_1)$. Then $(\P^n-H_2)\intersect X$ is open affine and $U\union((\P^n-H_1)\intersect X)\union((\P^n-H2)\intersect X)$ is only missing codimension three closed subsets of $X$. Repeating this process a few more times yields hyperplanes $H_1,\cdots,H_r$ so that $$U, (\P^n-H_1)\intersect X, \ldots, (\P^n-H_r)\intersect X$$ form an open affine cover of $X$, as desired. Now for the quasi-projective case. Suppose $X\subset\P^n$ is quasi-projective. From (I, Ex. 3.5) we know that $\P^n$ minus a hypersurface $H$ is affine. Note that the same proof works even if $H$ is a union of hypersurfaces. We now proceed with the same sort of construction as in the projective case, but we must choose $H$ more cleverly to insure that $(\P^n-H)\intersect{}X$ is affine. Let $U$ be a nonempty affine open subset of $X$. As before pick a hyperplane which doesn't completey contain any irreducible component of $X-U$. Since $X$ is only quasi-projective we can't conclude that $(\P^n-H)\intersect{}X$ is affine. But we do know that $(\P^n-H)\intersect{}\overline{X}$ is affine. Our strategy is to add some hypersurfaces to $H$ to get a union of hypersurfaces $S$ so that $$(\P^n-S)\intersect{}\overline{X}=(\P^n-S)\intersect{}X.$$ But, we must be careful to add these hypersurfaces in such a way that $((\P^n-S)\intersect{}X)\union U$ is missing only codimension two or greater subsets of $X$. We do this as follows. For each irreducible component $Y$ of $\overline{X}-X$ choose a hypersurface $H'$ which completely contains $Y$ but which does not completely contain any irreducible component of $X-U$. That this can be done is the content of a lemma which will be proved later (just pick a point in each irreducible component and avoid it). Let $S$ by the union of all of the $H'$ along with $H$. Then $\P^n-S$ is affine and so $$(\P^n-S)\intersect{}X=(\P^n-S)\intersect{}\overline{X}$$ is affine. Furthermore, $S$ properly intersects all irreducible components of $X-U$, so $((\P^n-S)\intersect{}X)\union U$ is missing only codimension two or greater subsets of $X$. Repeating this process as above several times yields the desired result because after each repetition the codimension of the resulting pieces is reduced by 1. \begin{lem} If $Y$ is a projective variety and $p_1,\ldots,p_n$ is a finite collection of points not on $Y$, then there exists a (possibly reducible) hypersurface $H$ containing $Y$ but not containing any of the $p_i$. \end{lem} By a possibly reducible hypersurface I mean a union of irreducible hypersurfaces, not a hypersurface union higher codimension varieties. \begin{proof} This is obviously true and I have a proof, but I think there is probably a more algebraic proof. Note that $k$ is infinite since we only talk about varieties over algebraically closed fields. Let $f_1,\cdots,f_m$ be defining equations for $Y$. Thus $Y$ is the common zero locus of the $f_i$ and not all $f_i$ vanish on any $p_i$. I claim that we can find a linear combination $\sum a_i f_i$ of the $f_i$ which doesn't vanish on any $p_i$. Since $k$ is infinite and not all $f_i$ vanish on $p_1$, we can easily find $a_i$ so that $\sum a_i f_i(p_1)\neq 0$ and all the $a_i\neq 0$. If $\sum a_i f_i(p_2)=0$ then, once again since $k$ is infinite, we can easily ``jiggle'' the $a_i$ so that $\sum a_i f_i(p_2)\neq 0$ and $\sum a_i f_i(p_1)$ is still nonzero. Repeating this same argument for each of the finitely many points $p_i$ gives a polynomial $f=\sum a_i f_i$ which doesn't vanish on any $p_i$. Of course I want to use $f$ to define our hypersurface, but I can't because $f$ might not be homogeneous. Fortunately, this is easily dealt with by suitably multiplying the various $f_i$ by the defining equation of a hyperplane not passing through any $p_i$, then repeating the above argument. Now let $H$ be the hypersurface defined by $f=\sum a_i f_i$. Then by construction $H$ contains $Y$ and $H$ doesn't contain any $p_i$. \end{proof} (e) Suppose $Y$ is a set-theoretic complete intersection of codimension $r$ in $X=\P_k^n$. Then $Y$ is the intersection of $r$ hypersurfaces, so we can write $Y=H_1\intersect\cdots\intersect{}H_r$ where each $H_i$ is a hypersurface. By (I, Ex. 3.5) $X-H_i$ is affine for each $i$, thus $$X-Y=(X-H_1)\union\cdots\union(X-H_r)$$ can be covered by $r$ open affine subsets. By (c) this implies $\cd(X-Y)\leq r-1$ which completes the proof. \end{proof} \begin{exercise} {\bfseries (4.9)} Let $X=\spec k[x_1,x_2,x_3,x_4]$ be affine four-space over a field $k$. Let $Y_1$ be the plane $x_1=x_2=0$ and let $Y_2$ be the plane $x_3=x_4=0$. Show that $Y=Y_1\union Y_2$ is not a set-theoretic complete intersection in $X$. Therefore the projective closure $\overline{Y}$ in $\P_k^4$ is not a set-theoretic complete intersection. \end{exercise} \begin{proof} By (Ex. 4.8e) it suffices to show that $H^2(X-Y,\so_{X-Y})\neq 0$. Suppose $Z$ is a closed subset of $X$, then by (Ex. 2.3d), for any $i\geq 1$, there is an exact sequence $$H^i(X,\sox)\into H^i(X-Z,\so_{X-Z})\into H_{Z}^{i+1}(X,\sox)\into H^{i+1}(X,\sox).$$ By (3.8), $H^i(X,\sox)=H^{i+1}(X,\sox)=0$ so $H^i(X-Z,\so_{X-Z})=H_{Z}^{i+1}(X,\sox)$. Applying this with $Z=Y$ and $i=2$ shows that $$H^2(X-Y,\so_{X-Y})=H_{Y}^3(X,\sox).$$ Thus we just need to show that $H_{Y}^3(X,\sox)\neq 0$. Mayer-Vietoris (Ex. 2.4) yields an exact sequence \begin{align*} H^3_{Y_1}(X,\sox)\oplus{}H^3_{Y_2}(X,\sox)\into{}H^3_Y(X,\sox)\into\\ H^4_{Y_1\intersect{}Y_2}(X,\sox)\into{} H^4_{Y_1}(X,\sox)\oplus{}H^4_{Y_2}(X,\sox) \end{align*} As above, $H^3_{Y_1}(X,\sox)=H^2(X-Y_1,\so_{X-Y_1})$. But $X-Y_1$ is a set-theoretic complete intersection of codimension $2$ so $\cd(X-Y_1)\leq 1$, whence $H^2(X-Y_1,\so_{X-Y_1})=0$. Similiarly $$H^2(X-Y_2,\so_{X-Y_2})=H^3(X-Y_1,\so_{X-Y_1}) =H^3(X-Y_2,\so_{X-Y_2})=0.$$ Thus from the above exact sequence we see that $H^3_Y(X,\sox)=H^4_{Y_1\intersect{}Y_2}(X,\sox).$ Let $P=Y_1\intersect{}Y_2=\{(0,0,0,0)\}$. We have reduced to showing that $H^4_{P}(X,\sox)$ is nonzero. Since $H^4_{P}(X,\sox)=H^3(X-P,\so_{X-P})$ we can do this by a direct computation of $H^3(X-P,\so_{X-P})$ using \cech{} cohomology. Cover $X-P$ by the affine open sets $U_i=\{x_i\neq{}0\}$. Then the \cech{} complex is $$\begin{array}{l} k[x_1,x_2,x_3,x_4,x_1^{-1}]\oplus\cdots\oplus k[x_1,x_2,x_3,x_4,x_4^{-1}]\xrightarrow{d_0}\\ k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1}]\oplus\cdots\oplus k[x_1,x_2,x_3,x_4,x_3^{-1},x_4^{-1}]\xrightarrow{d_1}\\ k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1}]\oplus\cdots\oplus k[x_1,x_2,x_3,x_4,x_2^{-1},x_3^{-1},x_4^{-1}]\xrightarrow{d_2}\\ k[x_1,x_2,x_3,x_4,x_1^{-1},x_2^{-1},x_3^{-1},x_4^{-1}] \end{array}$$ Thus $$H^3(X-P,\so_{X-P}) = \{x_1^ix_2^jx_3^kx_4^{\ell}:i,j,k,\ell<0\}\neq 0.$$ \end{proof} \begin{exercise} {\bfseries (5.6)} {\em Curves on a Nonsingular Quadric Surface.} Let $Q$ be the nonsingular quadric surface $xy=zw$ in $X=\P_k^3$ over a field $k$. We will consider locally principal closed subschemes $Y$ of $Q$. These correspond to Cartier divisors on $Q$ by (II, 6.17.1). On the other hand, we know that $\pic Q\isom\Z\oplus\Z$, so we can talk about the {\em type} (a,b) of $Y$ (II, 6.16) and (II, 6.6.1). Let us denote the invertible sheaf $\sL(Y)$ by $\so_Q(a,b)$. Thus for any $n\in\Z$, $\so_Q(n)=\so_Q(n,n).$ [{\bf Comment!} In my solution, a subscheme $Y$ of type $(a,b)$ corresponds to the invertible sheaf $\soq(-a,-b)$. I think this is reasonable since then $\soq(-a,-b)=\sL(-Y)=\sI_Y$. The correspondence is not clearly stated in the problem, but this choice works.] \noindent(a) Use the special case $(q,0)$ and $(0,q)$, with $q>0$, when $Y$ is a disjoint union of $q$ lines $\P^1$ in $Q$, to show: \begin{enumerate} \item if $|a-b|\leq 1$, then $H^1(Q,\so_Q(a,b))=0$; \item if $a,b<0$, then $H^1(Q,\so_Q(a,b))=0$; \item if $a\leq-2$, then $H^1(Q,\so_Q(a,0))\neq 0)$. \end{enumerate} {\em Solution.} First I will prove a big lemma in which I explicitely calculate $H^1(Q,\soq(0,-q))$ and some other things which will come in useful later. Next I give an independent computation of the other cohomology groups (1), (2). \par\begin{lem} Let $q>0$, then $$\dim_k H^1(Q,\soq(-q,0))=H^1(Q,\soq(0,-q))=q-1.$$ Furthermore, we know all terms in the long exact sequence of cohomology associated with the short exact sequence $$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0.$$ \end{lem} \begin{proof} We prove the lemma only for $\soq(-q,0)$, since the argument for $\soq(0,-q)$ is exactly the same. Suppose $Y$ is the disjoint union of $q$ lines $\P^1$ in $Q$ so $\sI_Y=\soq(-q,0)$. The sequence $$0\into\soq(-q,0)\into\soq\into\sO_Y\into{}0$$ is exact. The associated long exact sequence of cohomology is \begin{align*} 0\into&\Gamma(Q,\soq(-q,0))\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y)\\ \into& H^1(Q,\soq(-q,0))\into{}H^1(Q,\soq)\into{}H^1(Q,\sO_Y)\\ \into& H^2(Q,\soq(-q,0))\into{}H^2(Q,\soq)\into{}H^2(Q,\sO_Y)\into 0 \end{align*} We can compute all of the terms in this long exact sequence. For the purposes at hand it suffices to view the summands as $k$-vector spaces so we systematically do this throughout. Since $\soq(-q,0)=\sI_Y$ is the ideal sheaf of $Y$, its global sections must vanish on $Y$. But $\sI_Y$ is a subsheaf of $\sO_Q$ whose global sections are the constants. Since the only constant which vanish on $Y$ is $0$, $\Gamma(Q,\soq(-q,0))=0$. By (I, 3.4), $\Gamma(Q,\soq)=k$. Since $Y$ is the disjoint union of $q$ copies of $\P^1$ and each copy has global sections $k$, $\Gamma(Q,\sO_Y)=k^{\oplus{}q}$. Since $Q$ is a complete intersection of dimension 2, (Ex. 5.5 b) implies $H^1(Q,\soq)=0$. Because $Y$ is isomorphic to several copies of $\P^1$, the general result (proved in class, but not in the book) that $H_{*}^n(\sO_{\P^n})=\{\sum a_{I}X_{I}:\text{entries in $I$ negative}\}$ implies $H^1(Q,\sO_Y)=H^1(Y,\sO_Y)=0$. Since $Q$ is a hypersurface of degree $2$ in $\P^3$, (I, Ex. 7.2(c)) implies $p_a(Q)=0$. Thus by (Ex. 5.5c) we see that $H^2(Q,\soq)=0$. Putting together the above facts and some basic properties of exact sequences show that $H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}$, $H^2(Q,\soq(-q,0))=0$ and $H^2(Q,\so_Y)=0$. Our long exact sequence is now \begin{align*} 0\into&\Gamma(Q,\soq(-q,0))=0\into\Gamma(Q,\soq)=k\into\Gamma(Q,\sO_Y)=k^{\oplus{}q}\\ \into& H^1(Q,\soq(-q,0))=k^{\oplus(q-1)}\into{}H^1(Q,\soq)=0\into{}H^1(Q,\sO_Y)=0\\ \into& H^2(Q,\soq(-q,0))=0\into{}H^2(Q,\soq)=0\into{}H^2(Q,\sO_Y)=0\into 0 \end{align*} \end{proof} Number (3) now follows immediately from the lemma because $$H^1(Q,\soq(a,0))=k^{\oplus(-a-1)}\neq 0$$ for $a\leq -2.$ Now we compute (1) and (2). Let $a$ be an arbitrary integer. First we show that $\soq(a,a)=0$. We have an exact sequence $$0\into\sO_{\P^3}(-2)\into\sO_{\P^3}\into\sO_{Q}\into{}0$$ where the first map is multiplication by $xy-zw$. Twisting by $a$ gives an exact sequence $$0\into\sO_{\P^3}(-2+a)\into\sO_{\P^3}(a)\into\sO_{Q}(a)\into{}0.$$ The long exact sequence of cohomology yields an exact sequence $$\cdots\into H^1(\sO_{\P^3}(a))\into H^1(\sO_Q(a))\into H^2(\sO_{\P^3}(-2+a))\into\cdots$$ But from the explicit computations of projective space (5.1) it follows that $H^1(\sO_{\P^3}(a))=0$ and $H^2(\sO_{\P^3}(-2+a))=0$ from which we conclude that $H^1(\sO_Q(a))=0$. Next we show that $\soq(a-1,a)=0$. Let $Y$ be a single copy of $\P^1$ sitting in $Q$ so that $Y$ has type $(1,0)$. Then we have an exact sequence $$0\into\sI_Y\into\soq\into\sO_Y\into 0.$$ But $\sI_Y=\soq(-1,0)$ so this becomes $$0\into\soq(-1,0)\into\soq\into\sO_Y\into 0.$$ Now twisting by $a$ yields the exact sequence $$0\into\soq(a-1,a)\into\soq(a)\into\sO_Y(a)\into 0.$$ The long exact sequence of cohomology gives an exact sequence $$\cdots\into\Gamma(\soq(a))\into\Gamma(\sO_Y(a))\into H^1(\soq(a-1,a))\into H^1(\soq(a))\into\cdots$$ We just showed that $H^1(\soq(a))=0$, so to see that $H^1(\soq(a-1,a))=0$ it suffices to note that the map $\Gamma(\soq(a))\into\Gamma(\sO_Y(a))$ is surjective. This can be seen by writing $Q=\proj(k[x,y,z,w]/(xy-zw))$ and (w.l.o.g.) $Y=\proj(k[x,y,z,w]/(xy-zw,x,z))$ and noting that the degree $a$ part of $k[x,y,z,w]/(xy-zw)$ surjects onto the degree $a$ part of $k[x,y,z,w]/(xy-zw,x,z)$. Thus $H^1(\soq(a-1,a))=0$ and exactly the same argument shows $H^1(\soq(a,a-1))=0$. This gives (1). For (2) it suffices to show that for $a>0$, $$H^1(\soq(-a,-a-n))=H^1(\soq(-a-n,-a))=0$$ for all $n>0$. Thus let $n>0$ and suppose $Y$ is a disjoint union of $n$ copies of $\P^1$ in such a way that $\sI_Y=\soq(0,-n)$. Then we have an exact sequence $$0\into\soq(0,-n)\into\soq\into\sO_Y\into 0.$$ Twisting by $-a$ yields the exact sequence $$0\into\soq(-a,-a-n)\into\soq(-a)\into\sO_Y(-a)\into 0.$$ The long exact sequence of cohomology then gives an exact sequence $$\cdots \into\Gamma(\sO_Y(-a))\into H^1(\soq(-a,-a-n)) \into H^1(\soq(-a))\into \cdots$$ As everyone knows, since $Y$ is just several copies of $\P^1$ and $-a<0$, $\Gamma(\sO_Y(-a))=0$. Because of our computations above, $H^1(\soq(-a))=0$. Thus $H^1(\soq(-a,-a-n))=0$, as desired. Showing that $H^1(\soq(-a-n,-a))=0$ is exactly the same. \noindent (b) Now use these results to show: \begin{enumerate} \item If $Y$ is a locally principal closed subscheme of type $(a,b)$ with $a,b>0$, then $Y$ is connected. \begin{proof} Computing the long exact sequence associated to the short exact sequence $$0\into\sI_Y\into\soq\into\sO_Y\into{}0$$ gives the exact sequence $$0\into\Gamma(Q,\sI_Y)\into\Gamma(Q,\soq)\into\Gamma(Q,\sO_Y) \into H^1(Q,\sI_Y)\into\cdots$$ But, $\Gamma(\sI_Y)=0$, $\Gamma(Q,\soq)=k$, and by (a)2 above $H^1(Q,\sI_Y)=H^1(Q,\soq(-a,-b))=0$. Thus we have an exact sequence $$0\into{}0\into{}k\into\Gamma(\so_Y)\into{}0\into\cdots$$ from which we conclude that $\Gamma(\so_Y)=k$ which implies $Y$ is connected. \end{proof} \item now assume $k$ is algebraically closed. Then for any $a,b>0$, there exists an irreducible nonsingular curve $Y$ of type $(a,b)$. Use (II, 7.6.2) and (II, 8.18). \begin{proof} Given $(a,b)$, (II, 7.6.2) gives a closed immersion $$Q=\P^1\times\P^1\into\P^a\times\P^b\into\P^n$$ which corresponds to the invertible sheaf $\soq(-a,-b)$ of type $(a,b)$. By Bertini's theorem (II, 8.18) there is a hyperplane $H$ in $\P^n$ such that the hyperplane section of the $(a,b)$ embedding of $Q$ in $\P^n$ is nonsingular. Pull this hyperplane section back to a nonsingular curve $Y$ of type $(a,b)$ on $Q$ in $\P^3$. By the previous problem, $Y$ is connected. Since $Y$ comes from a hyperplane section this implies $Y$ is irreducible (see the remark in the statement of Bertini's theorem). \end{proof} \item an irreducible nonsingular curve $Y$ of type $(a,b)$, $a,b>0$ on $Q$ is projectively normal (II, Ex. 5.14) if and only if $|a-b|\leq 1$. In particular, this gives lots of examples of nonsingular, but not projectively normal curves in $\P^3$. The simplest is the one of type $(1,3)$ which is just the rational quartic curve (I, Ex. 3.18). \begin{proof} Let $Y$ be an irreducible nonsingular curve of type $(a,b)$. The criterion we apply comes from (II, Ex 5.14d) which asserts that the maps $$\Gamma(\P^3,\sO_{\P^3}(n))\into\Gamma(Y,\sO_Y(n))$$ are surjective for all $n\geq 0$ if and only if $Y$ is projectively normal. To determine when this occurs we have to replace $\Gamma(\P^3,\sO_{\P^3}(n))$ with $\Gamma(Q,\soq(n))$. It is easy to see that the above criterion implies we can make this replacement if $Q$ is projectively normal. Since $Q\isom\P^1\times\P^1$ is locally isomorphic to $\A^1\times\A^1\isom\A^2$ which is normal, we see that $Q$ is normal. Then since $Q$ is a complete intersection which is normal, (II, 8.4b) implies $Q$ is projectively normal. Consider the exact sequence $$0\into\sI_Y\into\soq\into\sO_Y.$$ Twisting by $n$ gives an exact sequence $$0\into\sI_Y(n)\into\soq(n)\into\sO_Y(n).$$ Taking cohomology yields the exact sequence $$\cdots\into\Gamma(Q,\soq(n))\into\Gamma(Q,\sO_Y(n))\into H^1(Q,\sI_Y(n))\into\cdots$$ Thus $Y$ is projectively normal precisely if $H^1(Q,\sI_Y(n))=0$ for all $n\geq 0$. When can this happen? We apply our computations from part (a). Since $\soq(n)=\soq(n,n)$, $$\sI_Y(n)=\soq(-a,-b)(n)=\soq(-a,-b)\tensor_{\soq}\soq(n,n)=\soq(n-a,n-b)$$ If $|a-b|\leq 1$ then $|(n-a)-(n-b)|\leq 1$ for all $n$ so $$H^1(Q,\soq(-a,-b)(n))=0$$ for all $n$ which implies $Y$ is projectively normal. On the other hand, if $|a-b|>1$ let $n$ be the minimum of $a$ and $b$, without loss assume $b$ is the minimum, so $n=b$. Then from (a) we see that $$\soq(-a,-b)(n)=\soq(-a,-b)(b)=\soq(-a+b,0)\neq 0$$ since $-a+b\leq -2$. \end{proof} \end{enumerate} (c) If $Y$ is a locally principal subscheme of type $(a,b)$ in $Q$, show that $p_a(Y)=ab-a-b+1.$ [Hint: Calculate the Hilbert polynomials of suitable sheaves, and again use the special case (q,0) which is a disjoint union of $q$ copies of $\P^1$.] \begin{proof} The sequence $$0\into\soq(-a,-b)\into\soq\into\sO_Y\into{}0$$ is exact so $$\chi(\so_Y)=\chi(\soq)-\chi(\soq(-a,-b))=1-\chi(\soq(-a,-b)).$$ Thus $$p_a(Y)=1-\chi(\so_Y)=\chi(\soq(-a,-b)).$$ The problem is thus reduced to computing $\chi(\soq(-a,-b))$. Assume first that $a,b<0$. To compute $\chi(\soq(-a,-b))$ assume $Y=Y_1\union Y_2$ where $\sI_{Y_1}=\soq(-a,0)$ and $\sI_{Y_2}=\soq(0,-b)$. Thus we could take $Y_1$ to be $a$ copies of $\P^1$ in one family of lines and $Y_2$ to be $b$ copies of $\P^1$ in the other family. Tensoring the exact sequence $$0\into\sI_{Y_1}\into\soq\into\so_{Y_1}\into 0$$ by the flat module $\sI_{Y_2}$ yields an exact sequence $$0\into\sI_{Y_1}\tensor\sI_{Y_2}\into\sI_{Y_2}\into\sO_{Y_1}\tensor\sI_{Y_2}$$ [Note: I use the fact that $\sI_{Y_2}$ is flat. This follows from a proposition in section 9 which we haven't yet reached, but I'm going to use it anyways. Since $Y_2$ is locally principal, $\sI_{Y_2}$ is generated locally by a single element and since $Q$ is a variety it is integral. Thus $\sI_{Y_2}$ is locally free so by (9.2) $\sI_{Y_2}$ is flat.] This exact sequence can also be written as $$0\into\soq(-a,-b)\into\soq(0,-b)\into\sO_Y\tensor\soq(0,-b)\into{}0.$$ The associated long exact sequence of cohomology is \begin{align*} 0\into&\Gamma(Q,\soq(-a,-b))\into\Gamma(Q,\soq(0,-b))\into \Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))\\ \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))\\ \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 \end{align*} The first three groups of global sections are $0$. Since $a,b<0$, (a) implies $H^1(Q,\soq(-a,-b))=0$. From the lemma we know that $H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}$. Also by the lemma we know that $H^2(Q,\soq(0,-b))=0$. Since $\so_{Y_1}\tensor\soq(0,-b)$ is isomorphic to the ideal sheaf of $b-1$ points in each line of $Y_1$, a similiar proof as that used in the lemma shows that $$H^1(Q,\sO_Y\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}.$$ Plugging all of this information back in yields the exact sequence \begin{align*} 0\into&\Gamma(Q,\soq(-a,-b))=0\into\Gamma(Q,\soq(0,-b))=0\into \Gamma(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\\ \into&H^1(Q,\soq(-a,-b))=0\into H^1(Q,\soq(0,-b))=k^{\oplus(b-1)}\\ &\hspace{1in}\into{}H^1(Q,\sO_{Y_1}\tensor\soq(0,-b))=k^{\oplus{}a(b-1)}\\ \into&H^2(Q,\soq(-a,-b))\into H^2(Q,\soq(0,-b))=0\\ &\hspace{1in}\into{}H^2(Q,\sO_{Y_1}\tensor\soq(0,-b))=0\into 0 \end{align*} From this we conclude that $$\chi(\soq(-a,-b))=0+0+h^2(Q,\soq(-a,-b))=a(b-1)-(b-1)=ab-a-b+1$$ which is the desired result. Now we deal with the remaining case, when $Y$ is $a$ disjoint copies of $\P^1$. We have $$p_a(Y)=1-\chi(\sO_Y)=1-\chi(\sO_{\P^1}^{\oplus a}) =1-a\chi(\sO_{\P^1})=1-a$$ which completes the proof. \end{proof} \end{exercise} \end{document}