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Author: William A. Stein
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\documentclass[12pt]{article}
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\author{William A. Stein}
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\title{Algebraic Geometry Homework\\
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II.5, 1,2,3,4,5,7,8}
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\font\grmn=eufm10 scaled \magstep 1
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\renewcommand{\a}{\mbox{\bfseries A}}
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\newcommand{\gp}{p}
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\newcommand{\ga}{a}
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\newcommand{\so}{\mathcal{O}}
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\newcommand{\sox}{\mathcal{O}_X}
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\newcommand{\soy}{\mathcal{O}_Y}
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\newcommand{\se}{\mathcal{E}}
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\renewcommand{\sf}{\mathcal{F}}
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\newcommand{\sg}{\mathcal{G}}
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\newcommand{\shom}{\mathcal{H}\mbox{\rm om}}
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\renewcommand{\hom}{\mbox{\rm Hom}}
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\newcommand{\tensor}{\otimes}
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\newcommand{\tens}{\otimes}
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\newcommand{\isom}{\cong}
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\renewcommand{\dim}{\mbox{\rm dim}}
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\newcommand{\z}{\mbox{\bfseries Z}}
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\newcommand{\into}{\rightarrow}
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\newcommand{\res}{\mbox{\rm res}}
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\newcommand{\ru}{|_U}
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\newcommand{\ann}{\mbox{\rm Ann}}
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\newcommand{\coker}{\mbox{\rm Coker}}
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\newtheorem{prob}{Problem}
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\newtheorem{theorem}{Theorem}
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\newtheorem{prop}{Proposition}
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\newcommand{\ov}{\overline{\varphi}}
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\renewcommand{\phi}{\varphi}
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\newcommand{\fh}{f^{\#}}
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\newcommand{\proj}{Proj \hspace{.01in}}
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\newcommand{\spec}{Spec \hspace{.01in}}
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\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}
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\begin{document}
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\maketitle
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\begin{prob}[II.5.1]
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Let $(X,\sox)$ be a ringed space, and let $\se$ be a locally
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free $\sox$-module of finite rank. We define the dual of
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$\se$, denoted by $\check{\se}$, to be the sheaf
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$\shom_{\sox}(\se,\sox)$.
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(a) Show that $(\check{\se})\check{}\cong\se$.
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(b) For any $\sox$-module $\sf$, $\shom_{\sox}(\se,\sf)
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\cong \check{\se} \tensor_{\sox} \sf$.
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(c) For any $\sox$-modules $\sf, \sg,$
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$\hom_{\sox}(\se\tensor\sf,\sg)\cong
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\hom_{\sox}(\sf,\shom_{\sox}(\se,\sg))$.
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(d) (Projection Formula). If $f:(X,\sox)\rightarrow(Y,\soy)$
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is a morphism of ringed spaces, if $\sf$ is an $\sox$-module,
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and if $\se$ is a locally free $\soy$-module of finite
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rank, then there is a natural isomorphism
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$f_{*}(\sf\tensor_{\sox}f^{*}\se)\cong
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f_{*}(\sf)\tensor_{\soy}\se$.
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\end{prob}
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\proof
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(a)
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A free module of finite rank is {\em canonically}
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isomorphic to its double-dual via
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$\check{m}(\lambda)=\lambda(m)$
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where $m\in M$, $\lambda\in \check{M}$, and
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$\check{m}\in (\check{M})\check{}$.
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Let $U$ be an open set on which $\se|U$ is a
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free $\sox$-module of finite rank. Define a
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map $\phi:\se|_U\into(\check{\se})\check{}|_U$
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by, for all $V\subseteq U$, $\se(V)\into
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(\check{\se})\check{}(V)$ is the isomorphism described
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above.
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Since the isomorphisms are canonical, we can patch on
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intersections and define a global isomorphism.
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(b) As above we may assume $\se$ is a free $\sox$-module.
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Let $e_1,\ldots,e_n$ be a basis for $\sf$ and let
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$e_1^{*},\ldots,e_n^{*}$ be the corresponding dual
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basis. Let $U$ be an open subset of $X$.
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Define $\phi_U:\shom_{\sox}(\se,\sf)|_U\into
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(\hom(\se,\sox) \tensor_{\sox}\sf)|_U$ by
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$f\mapsto\sum_{i=1}^{n}e_i^{*}\tens f(e_i).$
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Define $\psi_U:
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(\hom(\se,\sox) \tensor_{\sox}\sf)|_U \into
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\shom_{\sox}(\se,\sf)|_U$ by
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$f\tens a\mapsto(x\mapsto f(x)a)$.
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For convenience of notation write $\phi=\phi_U$ and
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$\psi=\psi_U$.
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Let $f\tens a\in (\hom(\se,\sox) \tensor_{\sox}\sf)|_U$.
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Then $\phi\circ\psi(f\tens a)
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=\phi(x\mapsto f(x)a)=\sum_{i=1}^{n} e_i^{*}\tens
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f(e_i)a=\sum_{i=1}^{n}e_i^{*}f(e_i)\tens a
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=f\tens a$. Let $f\in\shom_{\sox}(\se,\sf)$, then
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$\psi\circ\phi(f)=\psi(\sum_{i=1}^{n}e_i^{*}\tens f(e_i))
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=(x\mapsto\sum_{i=1}^{n}e_i^{*}(x)f(e_i))
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=(x\mapsto f(\sum_{i=1}^{n}e_i^{*}(x)e_i))
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=(x\mapsto f(x))$.
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Thus $\phi$ and $\psi$ are inverse bijective
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homomorphisms, hence ring isomorphisms, and since they
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respect the restriction maps we see that the
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corresponding sheaves are isomorphic.
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(c) On each open set define $\phi|_U:
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\hom_{\sox}(\se\tensor\sf,\sg)\into
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\hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$ by
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$f\mapsto(a\mapsto(e\mapsto f(e\tens a)))$.
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(For notational convenience we omit the sheaf restrictions.)
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If $\phi(f)=0$ then the map
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$(a\mapsto(e\mapsto f(e\tens a)))$ is $0$
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so $f$ is the zero map, hence $\phi$ is injective.
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Let $f\in \hom_{\sox}(\sf,\hom_{\sox}(\se,sg))$. Define
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$g\in\hom_{\sox}(\se\tensor\sf,\sg)$ by
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$g(a\tens b)=(f(b))(a)$.
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Then $\phi(g)=(a\mapsto (e\mapsto g(e\tens a)))
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=(a\mapsto(e\mapsto(f(a))(e))=(a\mapsto f(a))=f$
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so $\phi$ is surjective. Thus $\phi$ is the
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desired isomorphism which, since $\phi$ evidently
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commutes with the restriction maps, induces an
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isomorphism of sheaves.
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(d)
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First we consider the case when
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$\se\isom\soy^n$.
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One one hand,
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\begin{eqnarray*}
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f_{*}(\sf)\tensor_{\soy}\se
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&\isom& f_{*}(\sf)\tensor_{\soy}\soy^n \\
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&\isom& \oplus_{i=1}^{n} (f_{*}(\sf)\tensor_{\soy}\soy)\\
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&\isom& \oplus_{i=1}^{n} f_{*}(\sf).\\
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\end{eqnarray*}
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On the other hand,
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\begin{eqnarray*}
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\sf\tensor_{\sox}f^{*}\se &=&\sf\tensor_{\sox}f^{*}(\soy^n)\\
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&\isom&\sf\tensor_{\sox}(f^{-1}(\soy^n)\tensor_{f^{-1}\soy}\sox)\\
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&\isom&\sf\tensor_{\sox}((f^{-1}(\soy)^n\tensor_{f^{-1}\soy}\sox)\\
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&\isom&\sf\tensor_{\sox}(\sum_{i=1}^{n}
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f^{-1}\soy\tensor_{f^{-1}\soy}\sox)\\
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&\isom&\sf\tensor_{\sox}(\sox^n)\\
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&\isom&(\sf\tensor_{\sox}\sox)^n=\sf^n
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\end{eqnarray*}
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where $f^{-1}(\soy^n)\isom f^{-1}(\soy)^n$ since
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$f^{-1}$ is a left adjoint functor hence commutes
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with direct sums (which are a right universal
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construction).
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Putting this together we have that
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$$f_{*}(\sf\tensor_{\soy}f^{*}(\se))
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=f_{*}(\sf^n)=f_{*}(\oplus_{i=1}^{n}\sf)
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=\oplus_{i=1}^{n}f_{*}(\sf).$$
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In general, we construct isomorphisms as above
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on an open cover then, since all of the isomorphisms
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are canonical, the isomorphisms match up on
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the intersections so we can glue to obtain an
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isomorphism.
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\begin{prob}[II.5.2]
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Let $R$ be a discrete valuation ring with quotient field
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$K$, and let $X=\spec R$.
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(a) To give an $\sox$-module $\sf$ is equivalent to giving
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an $R$-module $M$, a $K$-vector space $L$, and a
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homomorphism $\rho:M\tensor_R K\rightarrow L$.
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(b) That $\sox$-module is quasi-coherent if and only
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if $\rho$ is an isomorphism.
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\end{prob}
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\proof
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(a) First suppose we are given an $\sox$-module $\sf$. Since $R$
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is a DVR, $X$ has exactly two nonempty open sets,
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$X$ and the set consisting of the generic point,
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$\{\xi\}$. Let $M=\Gamma(\sf,X)$ and let
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$L=\Gamma(\sf,\{\xi\})$.
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Since $\Gamma(\sox,X)=R$ and $\Gamma(\sox,\{\xi\})=K$, $M$ is
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an $R$-module and $L$ is a $K$-vector space. Let
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$g:M\rightarrow L$ be the restriction map. Define
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$\rho$ by $\rho(m\tens\alpha)=\alpha\cdot{}g(m)$.
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Since $g$ is a homomorphism, so is $\rho$.
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Now suppose we are given an $R$-module $M$, a $K$-vector
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space $L$, and a homomorphism $\rho:M\tensor_R K\rightarrow L$.
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Define an $\sox$-module $\sf$ as follows. Let
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$\Gamma(\sf,X)=M$ and $\Gamma(\sf,\{\xi\})=L$. Define
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the restriction map $g:M\rightarrow L$ by
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$g(m)=\rho(m\tens 1)$. We just need to check that
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$g$ is a valid restriction map. Let $r\in R, m\in M$,
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then $g$ is a valid restriction map iff
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$r\cdot g(m)=g(rm)$, that is, when
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$r\cdot \rho(m\tens 1)=\rho(rm\tens 1)=\rho(m\tens r)$.
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So we must verify that the given homomorphism $\rho$ is
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$R$-linear. But there is absolutely no reason why this should
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be the case! (For example, let $M=L=K$, then
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$\rho:K\rightarrow K$ and it is easy to construct nontrivial
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homomorphisms of the additive group of a field).
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I think the problem is imprecisely stated. It
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should be assumed throughout that $\rho$ is $K$-linear.
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(b) First suppose $\sf$ is quasi-coherent. Then
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$\sf=\tilde{M}$ so
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Proposition 5.1 implies that
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$L=\Gamma(\tilde{M},\{\xi\})=(R-0)^{-1}M\cong
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M\tensor_R K$. Thus $\rho$ must be an isomorphism.
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Conversely, if $\rho$ is an isomorphism, we see that
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$\sf\cong \tilde{M}$ since they are the same on
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each open set and the restriction map is the
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same.
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\begin{prob}[II.5.3]
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Let $X=\spec A$ be an affine scheme. Show that the functors
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$\tilde{}$ and $\Gamma$ are adjoint, in the
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following sense: for any $A$-module $M$, and for any sheaf
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of $\sox$-modules $\sf$, there is a natural isomorphism
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$$\hom_A(M,\Gamma(X,\sf))\cong\hom_{\sox}(\tilde{M},\sf).$$
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\end{prob}
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\proof
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Define a homomorphism $F:\hom_A(M,\Gamma(X,\sf))\into
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\hom_{\sox}(\tilde{M},\sf)$ as follows. Send
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a ring homomorphism $\phi:M\into\Gamma(X,\sf)$
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to the morphism of sheaves $F(\phi):\tilde{M}\into\sf$.
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It suffices to define $F(\phi)$ on distinguished
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open sets (Eisenbud \& Harris, page 13).
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For $f\in A$ let $F(\phi)_{D(f)}$ be the map
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$${m\over f^n}\mapsto {1\over f^n}
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\cdot\res_{X,D(f)}(\phi(m))$$
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where $\res_{X,D(f)}:\sf(X)\into\sf(D(f))$
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is the restriction map of $\sf$.
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$F(\phi)_{D(f)}$ is a well-defined homomorphism since
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both $\phi$ and $\res_{X,D(f)}$ are homomorphisms and
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$\phi$ is an $A$-module homomorphism. Next note
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that $F(\phi)$ commutes with the
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restriction maps since each $\res_{X,D(f)}$ does.
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To see that $F$ is injective suppose $\phi$ and
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$\psi$ are two homomorphisms $M\into\Gamma(X,\sf)$.
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If $F(\phi)=F(\psi)$ then, in particular,
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$\phi=F(\phi)_{X}=F(\psi)_{X}=\psi$.
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To see that $F$ is surjective, let $\phi\in\hom_{\sox}
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(\tilde{M},\sf)$. Define $\psi:M\into\Gamma(X,\sf)$
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by letting $\psi=\phi_X$, that is, by taking the
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induced map on global sections.
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Then, for $f\in A$,
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$F(\psi)_{D(f)}:\tilde{M}(D(f))\into\sf(D(f))$ is
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the map
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$({m\over f^n}\mapsto {1\over f^n}\res_{X,D(f)}\circ\phi_X(m))$
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which, since $\phi$ is a morphism of sheaves, equals
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$=({m\over f^n}\mapsto {1\over f^n}\phi_{D(f)}(m)
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=\phi_{D(f)}({m\over f^n})=\phi_{D(f)}$.
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(We are just using the fact that $\phi$ commutes with
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the appropriate restriction maps.)
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Thus $F(\psi)$ agrees with $\phi$ on a basis for $X$
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hence $F(\psi)=\phi$. This shows that $F$ is surjective.
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So $F$ is an isomorphism, as required.
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\begin{prob}[II.5.4]
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Show that a sheaf of $\sox$-modules $\sf$ on a scheme $X$
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is quasi-coherent
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if and only if every point of $X$ has a neighborhood $U$, such
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that $\sf|_{U}$ is isomorphic to a cokernel of a morphism of
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free sheaves on $U$. If $X$ is noetherian, then $\sf$ is coherent
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iff it is locally a cokernel of a morphism of free sheaves
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of finite rank.
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\end{prob}
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\proof
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Suppose first that $\sf$ is quasi-coherent. Let $x\in X$.
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Then there is an affine open neighborhood
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$U=\spec A$ of $x$ such that $\sf|_U\isom \tilde{M}$,
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$M$ an $A$-module. It suffices to show that $M$ is
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isomorphic to a cokernel of a morphism of finitely
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generated $A$-algebras. Indeed, if
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$\phi:A^{(I)}\into A^{(J)}$ then
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$\coker(\tilde{\phi})=\tilde{A^{(J)}}/\phi(A^{(I)})\tilde{}
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=(A^{(J)}/\phi(A^{(I)}))\tilde{}$
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since, for all $f\in A$,
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$(A^{(J)})_f/\phi(A^{(I)})_f
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=(A^{(J)}/\phi(A^{(I)}))_f$
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so that they agree on a basis.
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Let $A^{|M|}$ be the free $A$-module on the elements
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of $M$.
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Let $\phi:A^{|M|}\into M$ be the natural map.
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Similiary, let $\psi:A^{|\ker(\phi)|}\into\ker(\phi)
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\subseteq A^{|M|}$ be the natural map.
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Then $\coker(\psi)\isom A^{|M|}/\ker(\phi)\isom M$,
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as required.
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Now assume the $\sf$ is coherent. We proceed as above
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but now $M$ is a finitely generated $A$-module,
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generated by $e_1,\ldots,e_n$, say, and we must
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show that $M$ is the cokernel of a morphism of
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free modules of finite rank. Let $\phi:A^n\into M$
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be the map which takes the $i$th generator
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$(0,\ldots,1,\ldots,0)$ of $A^n$ to $e_i\in M$.
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Then $\ker(\phi)$ is a submodule of $M$ so, since
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$M$ is neotherian (any finitely generated module
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over a noetherian ring is noetherian), $\ker(\phi)$
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is finitely generated. Let $f_1,\ldots,f_m$ be
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a generating set. Let $\psi:A^m\into\ker(\phi)$ be
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the surjection defined by sending the $i$th
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basis element of $A^m$ to $f_i$. Then
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$\coker(\psi)\isom A^n/\psi(A^m)\isom A^n/\ker(\phi)\isom M$.
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Thus $M$ is isomorphic to a cokernel of a morphism
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of free sheaves of finite rank.
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\begin{prob}[II.5.5]
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Let $f:X\rightarrow Y$ be a morphism of schemes.
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(a) Show by example that if $\sf$ is coherent on $X$, then
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$f_{*}\sf$ need not be coherent on $Y$, even if $X$ and $Y$
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are varieties over a field $k$.
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(b) Show that a closed immersion is a finite morphism.
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(c) If $f$ is a finite morphism of neotherian schemes, and if
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$\sf$ is coherent on $X$, then $f_{*}\sf$ is coherent on $Y$.
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\end{prob}
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\proof
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(a) Lex $k$ be a field, let $X=\spec (k[x]_x)$,
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$Y=\spec (k[x])$ and $\sf=\sox$. Then
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$f_{*}(\sf)(U)=\sf(f^{-1}(U))$, so
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$f_{*}(\sf)=(k[x]_x)^{\tilde{}_Y}$.
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But $(k[x]_x)^{\tilde{}_Y}$ is not a coherent sheaf
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of $\soy$-modules. Indeed, if it were, there would
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be a distinguished neighborhood $D(f)$ of $0$ so that
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$(k[x]_x)^{\tilde{}_Y}|_{D(f)}$
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is a finitely generated module over $k[x]$ (here I'm
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using the fact that over open set is distinguished
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in the topology of $\spec(k[x])$). But,
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$k[x]_f$ is never a finitely generated module over
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$k[x]$ for any $f$ of degree at least $1$. For $k[x]_f$
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contains elements of arbitrarily small degree whereas
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the degrees of elements of $k[x]\{\alpha_1,\ldots,\alpha_n\}$
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are bounded below. ($k[x]\{\alpha_1,\ldots,\alpha_n\}$
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is the $k[x]$-module generated by $\alpha_1,\ldots,\alpha_n$.)
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(b) Let $f:Y\into X$ be a closed immersion. Let $U=\spec(A)
350
\subseteq X$ and let $W=f^{-1}(U)\subseteq Y$. Then
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$W$ is a a scheme (give it the induced scheme structure as
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an open subset of $Y$). Furthermore,
353
$f(W)=U\cap f(Y)$ is a relatively closed subset of $U$,
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that is, a closed subset of $\spec(A)$. The
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map $\fh:\so_{U}\into\so_{W}$ is surjective since
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$\fh:\sox\into\soy$ is surjective and surjectivety
357
is a local property. Thus $W$ is a closed subscheme
358
of $\spec(A)$ so, by Corollary 5.10, $W\cong\spec(A/I)$
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for some ideal $I$ of $A$. Since $A/I$ is a finitely
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generated $A$ module (generated by $1+I$), $f$
361
is a finite morphism.
362
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(c) Let $U=\spec(A)\subseteq Y$ be an affine open subset
364
of $Y$. Then, since $f$ is finite, $f^{-1}(U)=\spec(B)$
365
is affine with $B$ is a finitely generated $A$-module.
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By Proposition 5.4 it suffices to show that
367
$f_{*}(\sf)|_U$ is $\tilde{M}$ for some finitely
368
generated $A$-module $M$. Now by Proposition 5.4,
369
since $f^{-1}(U)$ is affine and $B$ is noetherian,
370
$\sf|_{f^{-1}(U)}=\tilde{M}$ for some finitely
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generated $B$ module $M$.
372
But
373
$f_{*}(\sf)|_U=(f|_{f^{-1}(U)})_{*}\sf|_{f^{-1}(U)}
374
=(f|_{f^{-1}(U)})_{*}(\tilde{M})=\tilde{(A^M)}$
375
where the last equality follows from Proposition 5.2(d).
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Since $B$ is a finite module over $A$ and $M$
377
is a finite module over $B$, it follows that $M$
378
is a finite module over $A$ which completes the proof.
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380
\begin{prob}[II.5.7]
381
Let $X$ be a noetherian scheme, and let $sf$ be a coherent sheaf.
382
383
(a) If the stalk $\sf_x$ is a free $\sox$-module for some point
384
$x\in X$, then there is a neighborhood $U$ of $x$ such that $\sf|_U$
385
is free.
386
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(b) $\sf$ is locally free iff its stalks $\sf_x$ are free $\sox$-modules
388
for all $x\in X$.
389
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(c) $\sf$ is invertible iff there is a coherent sheaf $\sg$ such
391
that $\sf\tensor\sg\isom\sox$.
392
\end{prob}
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\proof
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(a) Let $U=\spec(A)$ be a neighborhood of $x$ so that
395
$\sf\ru=\tilde{M}$, $M$ a finitely generated $A$-module.
396
Then $\sf_x=M_x$ so we have reduced the problem to
397
the following purely algebraic result.
398
\begin{prop}
399
If $M$ is a finitely generated free $A$ module and there
400
is a prime $\wp$ of $A$ such that $M_{\wp}$ is a free
401
$A_{\wp}$-module, then there exists $f\in A$ such that
402
$f\not\in\wp$ and $M_f$ is a free $A_f$-module.
403
\end{prop}
404
405
Once we have proven this we will know that $\sf|_{D_U(f)}$
406
is free since Proposition 5.1 asserts that
407
$\tilde{M}(D(f))\cong M_f$.
408
409
\proof (of Proposition) Let $a_1,\ldots,a_n\in M$ be a free
410
basis of $M_{\wp}$ over $A_{\wp}$ (we can clear denominators
411
so that we may assume all $a_i$ lie in $M$.) Let
412
$b_1,\ldots,b_m\in M$ be a generating set for $M$
413
over $A$. For each $i$ we can write $b_i$ as
414
an $A_{\wp}$-linear combination of the $a_i$. Clearing
415
denominators we see that $d_i b_i\in A\{a_1,\ldots,a_n\}$
416
for some $d_i\not\in\wp$. Let $f=\prod d_i$, then
417
$f\not\in\wp$ and $a_1,\ldots,a_n$ have
418
$A_f$-span including all of the $b_i$, and thus
419
including $M$, and thus including $M_f$. But
420
$a_1,\ldots,a_n$ is free over $A_{\wp}$ hence over
421
$A_f$ since $A_f\subseteq A_{\wp}$. [This proposition
422
is in Bourbaki, {\em Commutative Algebra},
423
II.5.1, although the proof is more abstract
424
than mine.]
425
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(b) ($\Longrightarrow$) Let $x\in X$ and let $U=\spec(A)$ be an open
427
neighborhood of $x$ such that $\sf\ru=\tilde{M}$
428
with $M$ a free $A$ module. Suppose $\wp$ is the
429
prime of $A$ corresponding to $x$. Then $\sf_x=M_\wp$
430
which is a free $A_{\wp}$-module. Indeed,
431
if $e_1,\ldots,e_n$ is a free $A$-basis for $M$,
432
then it is also a free $A_{\wp}$-basis for
433
$M_{\wp}$. For if
434
${a_1\over b_1}e_1 + \cdots + {a_n\over b_n}e_n=0,
435
b_i\not\in \wp$,
436
then
437
$a_1{b\over b_1}e_1+\cdots+a_n{b\over b_n}e_n=0, b=\prod b_i$
438
so $b{a_i\over b_i}=0$ for each $i$,
439
so ${a_i\over b_i}=0$ in the localization $A_{\wp}$
440
since $b\not\in\wp$.
441
442
($\Longleftarrow$) By part (a) every point has a neighborhood
443
on which $\sf$ is free. Therefore $X$ can be covered by
444
open affines on which $\sf$ is free so $\sf$ is locally free.
445
446
(c) ($\Longrightarrow$)
447
Let $\sg=\check{\sf}=\shom(\sf,\sox)$, we will show that
448
$\sf\tensor\sg\isom\sox$. To define a morphism
449
$\phi:\sf\tensor\shom(\sf,\sox)\rightarrow\sox$
450
it is enough to define $\phi$ on the presheaf
451
$(U\mapsto \sf(U)\tensor_{\sox(U)}\shom(\sf(U),\sox(U))$.
452
Define $\phi$ by $a\tens f\mapsto f(a)$. Thus $\phi$ commutes
453
with the restrictions so $\phi$ defines a valid morphism
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of sheaves.
455
456
Let $U$ be an open affine subset of $X$ such that $\sf(U)$
457
is a free $\sox(U)$-module of rank 1 with basis $e_0$.
458
Then $\shom(\sf(U),\sox(U))$ has basis $e_0^{*}$ where
459
$e_0^{*}(e_0)=1$. Thus $\shom(\sf(U),\sox(U))$ is a free
460
$\sox(U)$-module of rank 1. Thus every element of
461
$\sf(U)\tensor \hom(\sf(U),\sox(U))$ can be written in the
462
form $a\tens f$ (as opposed to as a sum of such products).
463
Now $\phi_U(a\tens f)=0$ implies $f(a)=0$ which implies
464
$a=0$ or $f=0$ so $a\tens f=0$, so $\phi_U$ is injective.
465
Since $\phi_U(ae_0\tens e_0^{*})=a$, $\phi_U$ is surjective.
466
Thus $\phi_U$ is an isomorphism.
467
468
Use the definition of locally free of rank 1 to cover
469
$X$ by affine open sets $U$ such that $\sf\ru$ is
470
a free $\sox\ru$ module of rank 1. Then, by Proposition
471
5.1 (c) and an argument like that for (b) above, any
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distinguished open subset of $U$ has a $\sf$-sections
473
free of rank 1. Since $\phi$ is an isomorphism on each
474
of these distinguished open sets (use the argument in the
475
above paragraph) and these distinguished open sets form
476
a basis for the topology on $X$ it follows that $\phi$
477
must be an isomorphism.
478
479
($\Longleftarrow$) Because of part (b) above it suffices
480
to show that $\sf_x$ is free of rank one for each $x\in X$.
481
Since $\sf_x\tensor_{\so_x}\sg_x
482
=(\sf\tensor\sg)_x\isom\so_{X,x}$
483
the problem is reduced to the following purely
484
algebraic statement.
485
486
\begin{prop} Let $M$ and $N$ be finitely generated modules
487
over a local ring $(A,m)$ and suppose that
488
$M\tensor_A N\isom A$. Then $M$ is free of rank 1.
489
\end{prop}
490
\proof
491
Let $k=A/m$. Let $\nu:M\tensor_A N\into A$ be
492
the given isomorphism. Then, taking the product
493
with the identity, we get an isomorphism
494
$\nu\tensor 1_k:(M\tensor_A N)\tensor_A k\into
495
A\tensor_A k\cong k$ (it is obvious that
496
$\nu\tensor 1_k$ is surjective, but it is
497
not at all obvious that it is injective,
498
for this see the Bourbaki reference below.)
499
Thus $k\cong M\tensor_A N\tensor_A k
500
=M\tensor_A(k\tensor_A k)\tensor_A N
501
=(M\tensor_A k)\tensor_A(N\tensor_A k)
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=(M\tensor_A k)\tensor_k(N\tensor_A k)
503
=(M/mM)\tensor_k (N/mN)$
504
so, since the $k$-rank of
505
$(M/mM)\tensor_k (N/mN)$
506
is $1$ and is the product of
507
the ranks of $M/mM$ and $N/mN$, each has
508
rank 1. In particular, $M/mM$ is monogeneous (generated
509
by one element) as an $A/m$-module and hence as
510
an $A$-module so,
511
by Nakayama's lemma, $M$ is monogeneous
512
as an $A$-module. Since
513
$\ann_A(M)$ anihilates $M\tensor_A N=A$ as well,
514
it is $0$ (any element of $\ann_A(M)$ would have
515
to annihilate the identity of $A$ and hence be $0$).
516
Thus $M$ is a free module of rank one over $A$.
517
[See Bourbaki, {\em Commutative Algebra}, II.5.4, for a
518
more general theorem.]
519
520
\begin{prob}[II.5.8]
521
Again let $X$ be a noetherian scheme, and $\sf$ a coherent sheaf
522
on $X$. We will consider the function
523
$$\phi(x)=\dim_{k(x)}\sf_x\tensor_{\sox}k(x)$$
524
where $k(x)=\sox/m_x$ is the residue field at the point $x$. Use
525
Nakayama's lemma to prove the following results.
526
527
(a) The function $\phi$ is upper semi-continuous, i.e., for any
528
$n\in \z$, the set $\{x\in X:\phi(x)\geq n\}$ is closed.
529
530
(b) If $\sf$ is locally free, and $X$ is connected, then
531
$\phi$ is a constant function.
532
533
(c) Conversely, if $X$ is reduced, and $\phi$ is constant,
534
then $\sf$ is locally free.
535
\end{prob}
536
\proof
537
538
(a) We must show that $\{x:\phi(x)\geq n\}$ is closed.
539
A good way to do this is by showing that
540
$\{x:\phi(x)<n\}$ is open. To do this we show that
541
if $\phi(x)=m$ then there is an open neighborhood $U$ of
542
$x$ so that, for all $y\in U$, $\phi(y)\leq m$.
543
Since we need only look locally, we can assume that
544
$X=\spec A$, $\sf=\tilde{M}$, $M$ a finitely generated
545
$A$-module. Note that $\sf_x\tensor_{\sox}k(x)=
546
M_{\wp}\tensor_{A_{\wp}}A_{\wp}/{\wp A_{\wp}}=
547
M_{\wp}/{\wp M_\wp}$.
548
Let $s_1,\ldots,s_m\in M$ be elements whose images
549
form a basis for the vector space $M_{\wp}/\wp M_{\wp}$
550
over $A_{\wp}/\wp A_{\wp}$ (to do this choose a basis
551
for $M_{\wp}/\wp M_{\wp}$ then clear fractions). Note
552
that the images of the $s_i$ in fact generate
553
$M_{\wp}/{\wp M_{\wp}}$ as an $A_{\wp}$-module.
554
By Nakayama's lemma the $s_i$ generate
555
$M_{\wp}$ as an $A_{\wp}$-module.
556
Let $m_1,\ldots,m_k$ be a generating set for
557
$M$ over $A$. Write $m_j=\sum {a_i\over b_i}s_i$,
558
$b_i\not\in\wp$, then, if $c_j=\prod b_i$,
559
$c_jm_j$ is in the $A$-span of the $s_i$. Let
560
$f=\prod c_j$. Then $\wp\in D(f)$ and if
561
$q\in D(f)$, then $m_1,\ldots,m_k$ all lie
562
in the $A_q$-span of $s_1,\ldots,s_m$ (since
563
$c_jm_j$ is in the $A$-span of the $s_i$
564
and $c_j$ is inverted in $A_q$. Thus $M$ is spanned
565
by the $s_i$ over $A_q$, so $M_q$ is spanned by
566
the $s_i$ over $A_q$. It follows that
567
$\phi(q)=\dim M_q/qM_q\leq m$ since the images
568
of the $s_i$ generate $M_q/qM_q$ as a vector
569
space over $A_q/qM_q=k(q)$. Taking $D(f)$ as
570
our open neighborhood completes the proof.
571
572
(b) Choose $n$ so that some section of $\sf$
573
has rank $n$. Let $U$ be the union of all open sets $W$
574
such that $\sf|_W\isom\sox^n|_W$. Then $U$ is nonempty.
575
Let $V$ be the
576
union of all open sets $W$ such that
577
$\sf|_W\isom\sox^m|_W$, $m\not=n$. Since $\sf$ is
578
locally free, $U\cup V=X$. Suppose $x\in U\cap V$,
579
then $\sf_x$ has rank $n$ and rank $m\not=n$ (since
580
rank is preserved under localization), a
581
contradiction. Thus $U\cap V=\emptyset$. Since $U$
582
is nonempty and open, $X-U=V$ is open and $X$ is
583
connected, thus we conclude that $V=X-U=\emptyset$. Thus
584
every point is contained in an open set $W$ such
585
that $\sf|_W\isom\sox^n|_W$.
586
587
Let $x\in X$ and let $U=\spec(A)$ be an affine open set
588
containing $x$ such that $\sf|_U\isom\tilde{M}$. By the above
589
argument, $M$ is a free $A$-module of rank $n$.
590
Thus $\phi(x)=\dim_k M_x\tensor_{A_x}A_x/m_x
591
=\dim_k A_x^n\tensor_k A_x/m_x=\dim_k (A_x/m_x)^n
592
=\dim_k k^n=n$, as desired.
593
594
(c) Let $x\in X$. By exercise 5.7b it suffices to
595
show that the stalk $\sf_x$ is free. Since $\sf$
596
is coherent we can find an affine open set
597
$U=\spec(A)$ such that $\sf\ru=\tilde{M}$ for
598
some finitely generated $A$-module, $x\in U$,
599
and $A_f$ is reduced for each $f\in A$. Let
600
$\wp$ be the prime of $A$ corresponding to
601
$x$. We must show that $M_{\wp}$ is free over
602
$A_{\wp}$. Let $s_1,\ldots,s_n\in M$ be preimages
603
of a basis of $M_{\wp}/\wp M_{\wp}$ over
604
$k(x)=A_{\wp}/\wp A_{\wp}$ (find these
605
as in part (a)). Then, by Nakayama's lemma, the
606
$s_i$ generate $M_{\wp}$ over $A_{\wp}$.
607
608
We must show that the $s_i$ are linearly
609
independent over $A_{\wp}$. It will then follow
610
that $M_{\wp}$ is free of rank $n$ over $A_{\wp}$.
611
So suppose
612
$${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$$
613
in $M_{\wp}$ with ${a_i\over b_i}\in A_{\wp}$.
614
Then for each $i$, $b_i\not\in\wp$ and
615
$a_i\in A$. Since the $s_i$ are linearly independent
616
over $A_{\wp}/\wp A_{\wp}$, for each $i$ there
617
exists $c_i\not\in\wp$ such that ${c_i a_i\over b_i}\in\wp$.
618
Thus $c_i a_i\in\wp$ so $a_i\in\wp$.
619
Let $r$ be as in part (a) so that $q\in D(r)$ implies
620
the $s_i$ generate at least $M$ over $A_q$.
621
By definition there exists $c\in A$ such that
622
$$c(b_2\cdots b_n a_1 s_1+\cdots+b_1\cdots b_{n-1} a_n s_n)=0$$
623
in $A$.
624
Let $f=rc\prod b_i$, then if $q\in D(f)$, then
625
$s_1,\ldots,s_n$ generate $M_q/qM_q$ over $A_q$ and,
626
since $M_q/qM_q$ has dimension $n$
627
(since $\phi$ is constant),
628
the $s_i$ are actually a basis for $M_q/qM_q$ over
629
$A_q/qA_q$.
630
631
Since $c|f$, $c\not\in q$ so, as above,
632
${a_1\over b_1}s_1+\cdots+{a_n\over b_n}s_n=0$ in
633
$M_q$, so, as above, $a_i\in q$ for each $i$.
634
Thus, for all $q\in D(f)$, $a_i\in q$, so
635
$a_i$ lies in the nilradical of $A_f$ which, since
636
$A_f$ is reduced, means that $a_i=0$ in $A_f$.
637
So $a_i$ maps to $0$ under the map $A_f\into A_{\wp}$.
638
Thus $s_1,\ldots,s_n$ are linearly independent
639
over $A_{\wp}$ so $M_{\wp}$ is free of rank $n$
640
over $A_{\wp}$. Applying exercise (5.7b) then completes
641
the proof.
642
643
644
645
646
\end{document}
647