Author: William A. Stein

\documentclass[12pt]{article}1\author{William A. Stein}2\title{Algebraic Geometry Homework\\3II.5, 1,2,3,4,5,7,8}45\font\grmn=eufm10 scaled \magstep 167\renewcommand{\a}{\mbox{\bfseries A}}8\newcommand{\gp}{p}9\newcommand{\ga}{a}10\newcommand{\so}{\mathcal{O}}11\newcommand{\sox}{\mathcal{O}_X}12\newcommand{\soy}{\mathcal{O}_Y}13\newcommand{\se}{\mathcal{E}}14\renewcommand{\sf}{\mathcal{F}}15\newcommand{\sg}{\mathcal{G}}16\newcommand{\shom}{\mathcal{H}\mbox{\rm om}}17\renewcommand{\hom}{\mbox{\rm Hom}}18\newcommand{\tensor}{\otimes}19\newcommand{\tens}{\otimes}20\newcommand{\isom}{\cong}21\renewcommand{\dim}{\mbox{\rm dim}}22\newcommand{\z}{\mbox{\bfseries Z}}23\newcommand{\into}{\rightarrow}24\newcommand{\res}{\mbox{\rm res}}25\newcommand{\ru}{|_U}26\newcommand{\ann}{\mbox{\rm Ann}}27\newcommand{\coker}{\mbox{\rm Coker}}282930\newtheorem{prob}{Problem}31\newtheorem{theorem}{Theorem}32\newtheorem{prop}{Proposition}3334\newcommand{\ov}{\overline{\varphi}}35\renewcommand{\phi}{\varphi}36\newcommand{\fh}{f^{\#}}37\newcommand{\proj}{Proj \hspace{.01in}}38\newcommand{\spec}{Spec \hspace{.01in}}39\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}4041\begin{document}42\maketitle4344\begin{prob}[II.5.1]45Let $(X,\sox)$ be a ringed space, and let $\se$ be a locally46free $\sox$-module of finite rank. We define the dual of47$\se$, denoted by $\check{\se}$, to be the sheaf48$\shom_{\sox}(\se,\sox)$.4950(a) Show that $(\check{\se})\check{}\cong\se$.5152(b) For any $\sox$-module $\sf$, $\shom_{\sox}(\se,\sf)53\cong \check{\se} \tensor_{\sox} \sf$.5455(c) For any $\sox$-modules $\sf, \sg,$56$\hom_{\sox}(\se\tensor\sf,\sg)\cong57\hom_{\sox}(\sf,\shom_{\sox}(\se,\sg))$.5859(d) (Projection Formula). If $f:(X,\sox)\rightarrow(Y,\soy)$60is a morphism of ringed spaces, if $\sf$ is an $\sox$-module,61and if $\se$ is a locally free $\soy$-module of finite62rank, then there is a natural isomorphism63$f_{*}(\sf\tensor_{\sox}f^{*}\se)\cong64f_{*}(\sf)\tensor_{\soy}\se$.656667\end{prob}68\proof69(a)70A free module of finite rank is {\em canonically}71isomorphic to its double-dual via72$\check{m}(\lambda)=\lambda(m)$73where $m\in M$, $\lambda\in \check{M}$, and74$\check{m}\in (\check{M})\check{}$.75Let $U$ be an open set on which $\se|U$ is a76free $\sox$-module of finite rank. Define a77map $\phi:\se|_U\into(\check{\se})\check{}|_U$78by, for all $V\subseteq U$, $\se(V)\into79(\check{\se})\check{}(V)$ is the isomorphism described80above.81Since the isomorphisms are canonical, we can patch on82intersections and define a global isomorphism.8384(b) As above we may assume $\se$ is a free $\sox$-module.85Let $e_1,\ldots,e_n$ be a basis for $\sf$ and let86$e_1^{*},\ldots,e_n^{*}$ be the corresponding dual87basis. Let $U$ be an open subset of $X$.88Define $\phi_U:\shom_{\sox}(\se,\sf)|_U\into89(\hom(\se,\sox) \tensor_{\sox}\sf)|_U$ by90$f\mapsto\sum_{i=1}^{n}e_i^{*}\tens f(e_i).$91Define $\psi_U:92(\hom(\se,\sox) \tensor_{\sox}\sf)|_U \into93\shom_{\sox}(\se,\sf)|_U$ by94$f\tens a\mapsto(x\mapsto f(x)a)$.95For convenience of notation write $\phi=\phi_U$ and96$\psi=\psi_U$.97Let $f\tens a\in (\hom(\se,\sox) \tensor_{\sox}\sf)|_U$.98Then $\phi\circ\psi(f\tens a)99=\phi(x\mapsto f(x)a)=\sum_{i=1}^{n} e_i^{*}\tens100f(e_i)a=\sum_{i=1}^{n}e_i^{*}f(e_i)\tens a101=f\tens a$. Let $f\in\shom_{\sox}(\se,\sf)$, then102$\psi\circ\phi(f)=\psi(\sum_{i=1}^{n}e_i^{*}\tens f(e_i))103=(x\mapsto\sum_{i=1}^{n}e_i^{*}(x)f(e_i))104=(x\mapsto f(\sum_{i=1}^{n}e_i^{*}(x)e_i))105=(x\mapsto f(x))$.106Thus $\phi$ and $\psi$ are inverse bijective107homomorphisms, hence ring isomorphisms, and since they108respect the restriction maps we see that the109corresponding sheaves are isomorphic.110111(c) On each open set define $\phi|_U:112\hom_{\sox}(\se\tensor\sf,\sg)\into113\hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$ by114$f\mapsto(a\mapsto(e\mapsto f(e\tens a)))$.115(For notational convenience we omit the sheaf restrictions.)116If $\phi(f)=0$ then the map117$(a\mapsto(e\mapsto f(e\tens a)))$ is $0$118so $f$ is the zero map, hence $\phi$ is injective.119Let $f\in \hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$. Define120$g\in\hom_{\sox}(\se\tensor\sf,\sg)$ by121$g(a\tens b)=(f(b))(a)$.122Then $\phi(g)=(a\mapsto (e\mapsto g(e\tens a)))123=(a\mapsto(e\mapsto(f(a))(e))=(a\mapsto f(a))=f$124so $\phi$ is surjective. Thus $\phi$ is the125desired isomorphism which, since $\phi$ evidently126commutes with the restriction maps, induces an127isomorphism of sheaves.128129(d)130First we consider the case when131$\se\isom\soy^n$.132One one hand,133\begin{eqnarray*}134f_{*}(\sf)\tensor_{\soy}\se135&\isom& f_{*}(\sf)\tensor_{\soy}\soy^n \\136&\isom& \oplus_{i=1}^{n} (f_{*}(\sf)\tensor_{\soy}\soy)\\137&\isom& \oplus_{i=1}^{n} f_{*}(\sf).\\138\end{eqnarray*}139140On the other hand,141\begin{eqnarray*}142\sf\tensor_{\sox}f^{*}\se &=&\sf\tensor_{\sox}f^{*}(\soy^n)\\143&\isom&\sf\tensor_{\sox}(f^{-1}(\soy^n)\tensor_{f^{-1}\soy}\sox)\\144&\isom&\sf\tensor_{\sox}((f^{-1}(\soy)^n\tensor_{f^{-1}\soy}\sox)\\145&\isom&\sf\tensor_{\sox}(\sum_{i=1}^{n}146f^{-1}\soy\tensor_{f^{-1}\soy}\sox)\\147&\isom&\sf\tensor_{\sox}(\sox^n)\\148&\isom&(\sf\tensor_{\sox}\sox)^n=\sf^n149\end{eqnarray*}150where $f^{-1}(\soy^n)\isom f^{-1}(\soy)^n$ since151$f^{-1}$ is a left adjoint functor hence commutes152with direct sums (which are a right universal153construction).154155Putting this together we have that156$$f_{*}(\sf\tensor_{\soy}f^{*}(\se))157=f_{*}(\sf^n)=f_{*}(\oplus_{i=1}^{n}\sf)158=\oplus_{i=1}^{n}f_{*}(\sf).$$159160In general, we construct isomorphisms as above161on an open cover then, since all of the isomorphisms162are canonical, the isomorphisms match up on163the intersections so we can glue to obtain an164isomorphism.165166\begin{prob}[II.5.2]167Let $R$ be a discrete valuation ring with quotient field168$K$, and let $X=\spec R$.169170(a) To give an $\sox$-module $\sf$ is equivalent to giving171an $R$-module $M$, a $K$-vector space $L$, and a172homomorphism $\rho:M\tensor_R K\rightarrow L$.173174(b) That $\sox$-module is quasi-coherent if and only175if $\rho$ is an isomorphism.176\end{prob}177\proof178(a) First suppose we are given an $\sox$-module $\sf$. Since $R$179is a DVR, $X$ has exactly two nonempty open sets,180$X$ and the set consisting of the generic point,181$\{\xi\}$. Let $M=\Gamma(\sf,X)$ and let182$L=\Gamma(\sf,\{\xi\})$.183Since $\Gamma(\sox,X)=R$ and $\Gamma(\sox,\{\xi\})=K$, $M$ is184an $R$-module and $L$ is a $K$-vector space. Let185$g:M\rightarrow L$ be the restriction map. Define186$\rho$ by $\rho(m\tens\alpha)=\alpha\cdot{}g(m)$.187Since $g$ is a homomorphism, so is $\rho$.188189Now suppose we are given an $R$-module $M$, a $K$-vector190space $L$, and a homomorphism $\rho:M\tensor_R K\rightarrow L$.191Define an $\sox$-module $\sf$ as follows. Let192$\Gamma(\sf,X)=M$ and $\Gamma(\sf,\{\xi\})=L$. Define193the restriction map $g:M\rightarrow L$ by194$g(m)=\rho(m\tens 1)$. We just need to check that195$g$ is a valid restriction map. Let $r\in R, m\in M$,196then $g$ is a valid restriction map iff197$r\cdot g(m)=g(rm)$, that is, when198$r\cdot \rho(m\tens 1)=\rho(rm\tens 1)=\rho(m\tens r)$.199So we must verify that the given homomorphism $\rho$ is200$R$-linear. But there is absolutely no reason why this should201be the case! (For example, let $M=L=K$, then202$\rho:K\rightarrow K$ and it is easy to construct nontrivial203homomorphisms of the additive group of a field).204I think the problem is imprecisely stated. It205should be assumed throughout that $\rho$ is $K$-linear.206207(b) First suppose $\sf$ is quasi-coherent. Then208$\sf=\tilde{M}$ so209Proposition 5.1 implies that210$L=\Gamma(\tilde{M},\{\xi\})=(R-0)^{-1}M\cong211M\tensor_R K$. Thus $\rho$ must be an isomorphism.212Conversely, if $\rho$ is an isomorphism, we see that213$\sf\cong \tilde{M}$ since they are the same on214each open set and the restriction map is the215same.216217\begin{prob}[II.5.3]218Let $X=\spec A$ be an affine scheme. Show that the functors219$\tilde{}$ and $\Gamma$ are adjoint, in the220following sense: for any $A$-module $M$, and for any sheaf221of $\sox$-modules $\sf$, there is a natural isomorphism222$$\hom_A(M,\Gamma(X,\sf))\cong\hom_{\sox}(\tilde{M},\sf).$$223\end{prob}224\proof225226Define a homomorphism $F:\hom_A(M,\Gamma(X,\sf))\into227\hom_{\sox}(\tilde{M},\sf)$ as follows. Send228a ring homomorphism $\phi:M\into\Gamma(X,\sf)$229to the morphism of sheaves $F(\phi):\tilde{M}\into\sf$.230It suffices to define $F(\phi)$ on distinguished231open sets (Eisenbud \& Harris, page 13).232For $f\in A$ let $F(\phi)_{D(f)}$ be the map233$${m\over f^n}\mapsto {1\over f^n}234\cdot\res_{X,D(f)}(\phi(m))$$235where $\res_{X,D(f)}:\sf(X)\into\sf(D(f))$236is the restriction map of $\sf$.237$F(\phi)_{D(f)}$ is a well-defined homomorphism since238both $\phi$ and $\res_{X,D(f)}$ are homomorphisms and239$\phi$ is an $A$-module homomorphism. Next note240that $F(\phi)$ commutes with the241restriction maps since each $\res_{X,D(f)}$ does.242To see that $F$ is injective suppose $\phi$ and243$\psi$ are two homomorphisms $M\into\Gamma(X,\sf)$.244If $F(\phi)=F(\psi)$ then, in particular,245$\phi=F(\phi)_{X}=F(\psi)_{X}=\psi$.246To see that $F$ is surjective, let $\phi\in\hom_{\sox}247(\tilde{M},\sf)$. Define $\psi:M\into\Gamma(X,\sf)$248by letting $\psi=\phi_X$, that is, by taking the249induced map on global sections.250Then, for $f\in A$,251$F(\psi)_{D(f)}:\tilde{M}(D(f))\into\sf(D(f))$ is252the map253$({m\over f^n}\mapsto {1\over f^n}\res_{X,D(f)}\circ\phi_X(m))$254which, since $\phi$ is a morphism of sheaves, equals255$=({m\over f^n}\mapsto {1\over f^n}\phi_{D(f)}(m)256=\phi_{D(f)}({m\over f^n})=\phi_{D(f)}$.257(We are just using the fact that $\phi$ commutes with258the appropriate restriction maps.)259Thus $F(\psi)$ agrees with $\phi$ on a basis for $X$260hence $F(\psi)=\phi$. This shows that $F$ is surjective.261So $F$ is an isomorphism, as required.262263\begin{prob}[II.5.4]264Show that a sheaf of $\sox$-modules $\sf$ on a scheme $X$265is quasi-coherent266if and only if every point of $X$ has a neighborhood $U$, such267that $\sf|_{U}$ is isomorphic to a cokernel of a morphism of268free sheaves on $U$. If $X$ is noetherian, then $\sf$ is coherent269iff it is locally a cokernel of a morphism of free sheaves270of finite rank.271\end{prob}272\proof273Suppose first that $\sf$ is quasi-coherent. Let $x\in X$.274Then there is an affine open neighborhood275$U=\spec A$ of $x$ such that $\sf|_U\isom \tilde{M}$,276$M$ an $A$-module. It suffices to show that $M$ is277isomorphic to a cokernel of a morphism of finitely278generated $A$-algebras. Indeed, if279$\phi:A^{(I)}\into A^{(J)}$ then280$\coker(\tilde{\phi})=\tilde{A^{(J)}}/\phi(A^{(I)})\tilde{}281=(A^{(J)}/\phi(A^{(I)}))\tilde{}$282since, for all $f\in A$,283$(A^{(J)})_f/\phi(A^{(I)})_f284=(A^{(J)}/\phi(A^{(I)}))_f$285so that they agree on a basis.286287Let $A^{|M|}$ be the free $A$-module on the elements288of $M$.289Let $\phi:A^{|M|}\into M$ be the natural map.290Similiary, let $\psi:A^{|\ker(\phi)|}\into\ker(\phi)291\subseteq A^{|M|}$ be the natural map.292Then $\coker(\psi)\isom A^{|M|}/\ker(\phi)\isom M$,293as required.294295Now assume the $\sf$ is coherent. We proceed as above296but now $M$ is a finitely generated $A$-module,297generated by $e_1,\ldots,e_n$, say, and we must298show that $M$ is the cokernel of a morphism of299free modules of finite rank. Let $\phi:A^n\into M$300be the map which takes the $i$th generator301$(0,\ldots,1,\ldots,0)$ of $A^n$ to $e_i\in M$.302Then $\ker(\phi)$ is a submodule of $M$ so, since303$M$ is neotherian (any finitely generated module304over a noetherian ring is noetherian), $\ker(\phi)$305is finitely generated. Let $f_1,\ldots,f_m$ be306a generating set. Let $\psi:A^m\into\ker(\phi)$ be307the surjection defined by sending the $i$th308basis element of $A^m$ to $f_i$. Then309$\coker(\psi)\isom A^n/\psi(A^m)\isom A^n/\ker(\phi)\isom M$.310Thus $M$ is isomorphic to a cokernel of a morphism311of free sheaves of finite rank.312313314315316317\begin{prob}[II.5.5]318Let $f:X\rightarrow Y$ be a morphism of schemes.319320(a) Show by example that if $\sf$ is coherent on $X$, then321$f_{*}\sf$ need not be coherent on $Y$, even if $X$ and $Y$322are varieties over a field $k$.323324(b) Show that a closed immersion is a finite morphism.325326(c) If $f$ is a finite morphism of neotherian schemes, and if327$\sf$ is coherent on $X$, then $f_{*}\sf$ is coherent on $Y$.328\end{prob}329\proof330(a) Lex $k$ be a field, let $X=\spec (k[x]_x)$,331$Y=\spec (k[x])$ and $\sf=\sox$. Then332$f_{*}(\sf)(U)=\sf(f^{-1}(U))$, so333$f_{*}(\sf)=(k[x]_x)^{\tilde{}_Y}$.334But $(k[x]_x)^{\tilde{}_Y}$ is not a coherent sheaf335of $\soy$-modules. Indeed, if it were, there would336be a distinguished neighborhood $D(f)$ of $0$ so that337$(k[x]_x)^{\tilde{}_Y}|_{D(f)}$338is a finitely generated module over $k[x]$ (here I'm339using the fact that over open set is distinguished340in the topology of $\spec(k[x])$). But,341$k[x]_f$ is never a finitely generated module over342$k[x]$ for any $f$ of degree at least $1$. For $k[x]_f$343contains elements of arbitrarily small degree whereas344the degrees of elements of $k[x]\{\alpha_1,\ldots,\alpha_n\}$345are bounded below. ($k[x]\{\alpha_1,\ldots,\alpha_n\}$346is the $k[x]$-module generated by $\alpha_1,\ldots,\alpha_n$.)347348(b) Let $f:Y\into X$ be a closed immersion. Let $U=\spec(A)349\subseteq X$ and let $W=f^{-1}(U)\subseteq Y$. Then350$W$ is a a scheme (give it the induced scheme structure as351an open subset of $Y$). Furthermore,352$f(W)=U\cap f(Y)$ is a relatively closed subset of $U$,353that is, a closed subset of $\spec(A)$. The354map $\fh:\so_{U}\into\so_{W}$ is surjective since355$\fh:\sox\into\soy$ is surjective and surjectivety356is a local property. Thus $W$ is a closed subscheme357of $\spec(A)$ so, by Corollary 5.10, $W\cong\spec(A/I)$358for some ideal $I$ of $A$. Since $A/I$ is a finitely359generated $A$ module (generated by $1+I$), $f$360is a finite morphism.361362(c) Let $U=\spec(A)\subseteq Y$ be an affine open subset363of $Y$. Then, since $f$ is finite, $f^{-1}(U)=\spec(B)$364is affine with $B$ is a finitely generated $A$-module.365By Proposition 5.4 it suffices to show that366$f_{*}(\sf)|_U$ is $\tilde{M}$ for some finitely367generated $A$-module $M$. Now by Proposition 5.4,368since $f^{-1}(U)$ is affine and $B$ is noetherian,369$\sf|_{f^{-1}(U)}=\tilde{M}$ for some finitely370generated $B$ module $M$.371But372$f_{*}(\sf)|_U=(f|_{f^{-1}(U)})_{*}\sf|_{f^{-1}(U)}373=(f|_{f^{-1}(U)})_{*}(\tilde{M})=\tilde{(A^M)}$374where the last equality follows from Proposition 5.2(d).375Since $B$ is a finite module over $A$ and $M$376is a finite module over $B$, it follows that $M$377is a finite module over $A$ which completes the proof.378379\begin{prob}[II.5.7]380Let $X$ be a noetherian scheme, and let $sf$ be a coherent sheaf.381382(a) If the stalk $\sf_x$ is a free $\sox$-module for some point383$x\in X$, then there is a neighborhood $U$ of $x$ such that $\sf|_U$384is free.385386(b) $\sf$ is locally free iff its stalks $\sf_x$ are free $\sox$-modules387for all $x\in X$.388389(c) $\sf$ is invertible iff there is a coherent sheaf $\sg$ such390that $\sf\tensor\sg\isom\sox$.391\end{prob}392\proof393(a) Let $U=\spec(A)$ be a neighborhood of $x$ so that394$\sf\ru=\tilde{M}$, $M$ a finitely generated $A$-module.395Then $\sf_x=M_x$ so we have reduced the problem to396the following purely algebraic result.397\begin{prop}398If $M$ is a finitely generated free $A$ module and there399is a prime $\wp$ of $A$ such that $M_{\wp}$ is a free400$A_{\wp}$-module, then there exists $f\in A$ such that401$f\not\in\wp$ and $M_f$ is a free $A_f$-module.402\end{prop}403404Once we have proven this we will know that $\sf|_{D_U(f)}$405is free since Proposition 5.1 asserts that406$\tilde{M}(D(f))\cong M_f$.407408\proof (of Proposition) Let $a_1,\ldots,a_n\in M$ be a free409basis of $M_{\wp}$ over $A_{\wp}$ (we can clear denominators410so that we may assume all $a_i$ lie in $M$.) Let411$b_1,\ldots,b_m\in M$ be a generating set for $M$412over $A$. For each $i$ we can write $b_i$ as413an $A_{\wp}$-linear combination of the $a_i$. Clearing414denominators we see that $d_i b_i\in A\{a_1,\ldots,a_n\}$415for some $d_i\not\in\wp$. Let $f=\prod d_i$, then416$f\not\in\wp$ and $a_1,\ldots,a_n$ have417$A_f$-span including all of the $b_i$, and thus418including $M$, and thus including $M_f$. But419$a_1,\ldots,a_n$ is free over $A_{\wp}$ hence over420$A_f$ since $A_f\subseteq A_{\wp}$. [This proposition421is in Bourbaki, {\em Commutative Algebra},422II.5.1, although the proof is more abstract423than mine.]424425(b) ($\Longrightarrow$) Let $x\in X$ and let $U=\spec(A)$ be an open426neighborhood of $x$ such that $\sf\ru=\tilde{M}$427with $M$ a free $A$ module. Suppose $\wp$ is the428prime of $A$ corresponding to $x$. Then $\sf_x=M_\wp$429which is a free $A_{\wp}$-module. Indeed,430if $e_1,\ldots,e_n$ is a free $A$-basis for $M$,431then it is also a free $A_{\wp}$-basis for432$M_{\wp}$. For if433${a_1\over b_1}e_1 + \cdots + {a_n\over b_n}e_n=0,434b_i\not\in \wp$,435then436$a_1{b\over b_1}e_1+\cdots+a_n{b\over b_n}e_n=0, b=\prod b_i$437so $b{a_i\over b_i}=0$ for each $i$,438so ${a_i\over b_i}=0$ in the localization $A_{\wp}$439since $b\not\in\wp$.440441($\Longleftarrow$) By part (a) every point has a neighborhood442on which $\sf$ is free. Therefore $X$ can be covered by443open affines on which $\sf$ is free so $\sf$ is locally free.444445(c) ($\Longrightarrow$)446Let $\sg=\check{\sf}=\shom(\sf,\sox)$, we will show that447$\sf\tensor\sg\isom\sox$. To define a morphism448$\phi:\sf\tensor\shom(\sf,\sox)\rightarrow\sox$449it is enough to define $\phi$ on the presheaf450$(U\mapsto \sf(U)\tensor_{\sox(U)}\shom(\sf(U),\sox(U))$.451Define $\phi$ by $a\tens f\mapsto f(a)$. Thus $\phi$ commutes452with the restrictions so $\phi$ defines a valid morphism453of sheaves.454455Let $U$ be an open affine subset of $X$ such that $\sf(U)$456is a free $\sox(U)$-module of rank 1 with basis $e_0$.457Then $\shom(\sf(U),\sox(U))$ has basis $e_0^{*}$ where458$e_0^{*}(e_0)=1$. Thus $\shom(\sf(U),\sox(U))$ is a free459$\sox(U)$-module of rank 1. Thus every element of460$\sf(U)\tensor \hom(\sf(U),\sox(U))$ can be written in the461form $a\tens f$ (as opposed to as a sum of such products).462Now $\phi_U(a\tens f)=0$ implies $f(a)=0$ which implies463$a=0$ or $f=0$ so $a\tens f=0$, so $\phi_U$ is injective.464Since $\phi_U(ae_0\tens e_0^{*})=a$, $\phi_U$ is surjective.465Thus $\phi_U$ is an isomorphism.466467Use the definition of locally free of rank 1 to cover468$X$ by affine open sets $U$ such that $\sf\ru$ is469a free $\sox\ru$ module of rank 1. Then, by Proposition4705.1 (c) and an argument like that for (b) above, any471distinguished open subset of $U$ has a $\sf$-sections472free of rank 1. Since $\phi$ is an isomorphism on each473of these distinguished open sets (use the argument in the474above paragraph) and these distinguished open sets form475a basis for the topology on $X$ it follows that $\phi$476must be an isomorphism.477478($\Longleftarrow$) Because of part (b) above it suffices479to show that $\sf_x$ is free of rank one for each $x\in X$.480Since $\sf_x\tensor_{\so_x}\sg_x481=(\sf\tensor\sg)_x\isom\so_{X,x}$482the problem is reduced to the following purely483algebraic statement.484485\begin{prop} Let $M$ and $N$ be finitely generated modules486over a local ring $(A,m)$ and suppose that487$M\tensor_A N\isom A$. Then $M$ is free of rank 1.488\end{prop}489\proof490Let $k=A/m$. Let $\nu:M\tensor_A N\into A$ be491the given isomorphism. Then, taking the product492with the identity, we get an isomorphism493$\nu\tensor 1_k:(M\tensor_A N)\tensor_A k\into494A\tensor_A k\cong k$ (it is obvious that495$\nu\tensor 1_k$ is surjective, but it is496not at all obvious that it is injective,497for this see the Bourbaki reference below.)498Thus $k\cong M\tensor_A N\tensor_A k499=M\tensor_A(k\tensor_A k)\tensor_A N500=(M\tensor_A k)\tensor_A(N\tensor_A k)501=(M\tensor_A k)\tensor_k(N\tensor_A k)502=(M/mM)\tensor_k (N/mN)$503so, since the $k$-rank of504$(M/mM)\tensor_k (N/mN)$505is $1$ and is the product of506the ranks of $M/mM$ and $N/mN$, each has507rank 1. In particular, $M/mM$ is monogeneous (generated508by one element) as an $A/m$-module and hence as509an $A$-module so,510by Nakayama's lemma, $M$ is monogeneous511as an $A$-module. Since512$\ann_A(M)$ anihilates $M\tensor_A N=A$ as well,513it is $0$ (any element of $\ann_A(M)$ would have514to annihilate the identity of $A$ and hence be $0$).515Thus $M$ is a free module of rank one over $A$.516[See Bourbaki, {\em Commutative Algebra}, II.5.4, for a517more general theorem.]518519\begin{prob}[II.5.8]520Again let $X$ be a noetherian scheme, and $\sf$ a coherent sheaf521on $X$. We will consider the function522$$\phi(x)=\dim_{k(x)}\sf_x\tensor_{\sox}k(x)$$523where $k(x)=\sox/m_x$ is the residue field at the point $x$. Use524Nakayama's lemma to prove the following results.525526(a) The function $\phi$ is upper semi-continuous, i.e., for any527$n\in \z$, the set $\{x\in X:\phi(x)\geq n\}$ is closed.528529(b) If $\sf$ is locally free, and $X$ is connected, then530$\phi$ is a constant function.531532(c) Conversely, if $X$ is reduced, and $\phi$ is constant,533then $\sf$ is locally free.534\end{prob}535\proof536537(a) We must show that $\{x:\phi(x)\geq n\}$ is closed.538A good way to do this is by showing that539$\{x:\phi(x)<n\}$ is open. To do this we show that540if $\phi(x)=m$ then there is an open neighborhood $U$ of541$x$ so that, for all $y\in U$, $\phi(y)\leq m$.542Since we need only look locally, we can assume that543$X=\spec A$, $\sf=\tilde{M}$, $M$ a finitely generated544$A$-module. Note that $\sf_x\tensor_{\sox}k(x)=545M_{\wp}\tensor_{A_{\wp}}A_{\wp}/{\wp A_{\wp}}=546M_{\wp}/{\wp M_\wp}$.547Let $s_1,\ldots,s_m\in M$ be elements whose images548form a basis for the vector space $M_{\wp}/\wp M_{\wp}$549over $A_{\wp}/\wp A_{\wp}$ (to do this choose a basis550for $M_{\wp}/\wp M_{\wp}$ then clear fractions). Note551that the images of the $s_i$ in fact generate552$M_{\wp}/{\wp M_{\wp}}$ as an $A_{\wp}$-module.553By Nakayama's lemma the $s_i$ generate554$M_{\wp}$ as an $A_{\wp}$-module.555Let $m_1,\ldots,m_k$ be a generating set for556$M$ over $A$. Write $m_j=\sum {a_i\over b_i}s_i$,557$b_i\not\in\wp$, then, if $c_j=\prod b_i$,558$c_jm_j$ is in the $A$-span of the $s_i$. Let559$f=\prod c_j$. Then $\wp\in D(f)$ and if560$q\in D(f)$, then $m_1,\ldots,m_k$ all lie561in the $A_q$-span of $s_1,\ldots,s_m$ (since562$c_jm_j$ is in the $A$-span of the $s_i$563and $c_j$ is inverted in $A_q$. Thus $M$ is spanned564by the $s_i$ over $A_q$, so $M_q$ is spanned by565the $s_i$ over $A_q$. It follows that566$\phi(q)=\dim M_q/qM_q\leq m$ since the images567of the $s_i$ generate $M_q/qM_q$ as a vector568space over $A_q/qM_q=k(q)$. Taking $D(f)$ as569our open neighborhood completes the proof.570571(b) Choose $n$ so that some section of $\sf$572has rank $n$. Let $U$ be the union of all open sets $W$573such that $\sf|_W\isom\sox^n|_W$. Then $U$ is nonempty.574Let $V$ be the575union of all open sets $W$ such that576$\sf|_W\isom\sox^m|_W$, $m\not=n$. Since $\sf$ is577locally free, $U\cup V=X$. Suppose $x\in U\cap V$,578then $\sf_x$ has rank $n$ and rank $m\not=n$ (since579rank is preserved under localization), a580contradiction. Thus $U\cap V=\emptyset$. Since $U$581is nonempty and open, $X-U=V$ is open and $X$ is582connected, thus we conclude that $V=X-U=\emptyset$. Thus583every point is contained in an open set $W$ such584that $\sf|_W\isom\sox^n|_W$.585586Let $x\in X$ and let $U=\spec(A)$ be an affine open set587containing $x$ such that $\sf|_U\isom\tilde{M}$. By the above588argument, $M$ is a free $A$-module of rank $n$.589Thus $\phi(x)=\dim_k M_x\tensor_{A_x}A_x/m_x590=\dim_k A_x^n\tensor_k A_x/m_x=\dim_k (A_x/m_x)^n591=\dim_k k^n=n$, as desired.592593(c) Let $x\in X$. By exercise 5.7b it suffices to594show that the stalk $\sf_x$ is free. Since $\sf$595is coherent we can find an affine open set596$U=\spec(A)$ such that $\sf\ru=\tilde{M}$ for597some finitely generated $A$-module, $x\in U$,598and $A_f$ is reduced for each $f\in A$. Let599$\wp$ be the prime of $A$ corresponding to600$x$. We must show that $M_{\wp}$ is free over601$A_{\wp}$. Let $s_1,\ldots,s_n\in M$ be preimages602of a basis of $M_{\wp}/\wp M_{\wp}$ over603$k(x)=A_{\wp}/\wp A_{\wp}$ (find these604as in part (a)). Then, by Nakayama's lemma, the605$s_i$ generate $M_{\wp}$ over $A_{\wp}$.606607We must show that the $s_i$ are linearly608independent over $A_{\wp}$. It will then follow609that $M_{\wp}$ is free of rank $n$ over $A_{\wp}$.610So suppose611$${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$$612in $M_{\wp}$ with ${a_i\over b_i}\in A_{\wp}$.613Then for each $i$, $b_i\not\in\wp$ and614$a_i\in A$. Since the $s_i$ are linearly independent615over $A_{\wp}/\wp A_{\wp}$, for each $i$ there616exists $c_i\not\in\wp$ such that ${c_i a_i\over b_i}\in\wp$.617Thus $c_i a_i\in\wp$ so $a_i\in\wp$.618Let $r$ be as in part (a) so that $q\in D(r)$ implies619the $s_i$ generate at least $M$ over $A_q$.620By definition there exists $c\in A$ such that621$$c(b_2\cdots b_n a_1 s_1+\cdots+b_1\cdots b_{n-1} a_n s_n)=0$$622in $A$.623Let $f=rc\prod b_i$, then if $q\in D(f)$, then624$s_1,\ldots,s_n$ generate $M_q/qM_q$ over $A_q$ and,625since $M_q/qM_q$ has dimension $n$626(since $\phi$ is constant),627the $s_i$ are actually a basis for $M_q/qM_q$ over628$A_q/qA_q$.629630Since $c|f$, $c\not\in q$ so, as above,631${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$ in632$M_q$, so, as above, $a_i\in q$ for each $i$.633Thus, for all $q\in D(f)$, $a_i\in q$, so634$a_i$ lies in the nilradical of $A_f$ which, since635$A_f$ is reduced, means that $a_i=0$ in $A_f$.636So $a_i$ maps to $0$ under the map $A_f\into A_{\wp}$.637Thus $s_1,\ldots,s_n$ are linearly independent638over $A_{\wp}$ so $M_{\wp}$ is free of rank $n$639over $A_{\wp}$. Applying exercise (5.7b) then completes640the proof.641642643644645\end{document}646647