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\begin{document} 
 
   \pagestyle{plain} 
   \parindent0cm
   UNIVERSITY OF  KAISERSLAUTERN \hfill  Summerterm 1998 \\
   Mathematical Department \hfill 28th June, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.5cm}

   \begin{center}
     \LARGE \bf Algebraic Geometry 
   \end{center}
   \vspace{0.5cm}
   \normalsize
   \parindent0cm

   For the following two glueing propositions see e.\ g.\ Eisenbud-Harris,
   Schemes, Chapter 1. The category $\kc$ shall be one of the
   following $(Sets)$, $(Ab)$, $(R-Mod)$, or $(R-Alg)$. What we
   actually need is the existence of inverse and direct limits.

   \begin{exercise-proposition}[Sheaves given on a basis of topology]\label{basis}
     Let $X$ be a topological space.
     Suppose that $\mathfrak{B}=\left\{B_i\right\}_{i\in I}$ is
     a basis of the topology of $X$, and that for any $i\in I$ we are given
     an object $\kf(B_i)$ in the category $\kc$. Suppose
     moreover that for any $B_j\subseteq B_i$ we have a morphism
     $\rho_{ij}:\kf(B_i)\rightarrow \kf(B_j)$ such that
     $\rho_{ii}=\id_{\kf(B_i)}$ and
     $\rho_{jk}\circ\rho_{ij}=\rho_{ik}$, whenever $B_k\subseteq
     B_j\subseteq B_i$, and suppose finally that the following
     version of the sheaf axiom is fulfilled:
     \begin{displaymath}
       \mbox{if } B_i=\bigcup_{j\in J\subseteq I}B_j \mbox{ and } 
       f_j\in\kf(B_j) : \rho_{jl}(f_j)=\rho_{kl}(f_k) \mbox{ for
         all } B_l\subseteq B_j\cap B_k, 
     \end{displaymath}
     \begin{displaymath}
       \mbox{ then there exists 
         a unique } f\in\kf(B_i) : \rho_{ij}(f)=f_j \mbox{ for all }
       j\in J.
     \end{displaymath}
     I.\ e.\ suppose that on $\mathfrak{B}$ we have presheaf data 
     which moreover fulfill the sheaf axiom.\\
     Then there exists a unique $\kc$-sheaf $\kf$ on $X$
     extending the given data.
   \end{exercise-proposition}

   
   \begin{exercise-proposition}[Sheaves given on an open covering]\label{covering}
     Let $X$ be a topological space.\\
     Suppose that $\mathfrak{U}=\left\{U_i\right\}_{i\in I}$
     is an open covering of $X$, and for any $i\in I$ we are given a
     $\kc$-sheaf $\kf_i$ on $U_i$. Suppose moreover that for any
     $i,j\in I$ we have an isomorphism
     $\psi_{ij}:(\kf_i)_{|U_i\cap U_j}\rightarrow
     (\kf_j)_{|U_i\cap U_j}$ such that
     $\psi_{ii}=\id_{\kf_i}$ and
     $\psi_{jk|}\circ\psi_{ij|}=\psi_{ik|}$ on $U_i\cap U_j\cap U_k$.\\
     Then there exists a $\kc$-sheaf $\kf$ and isomorphisms
     $\psi_i:\kf_{|U_i}\rightarrow \kf_i$ such that on
     $U_i\cap U_j$ we have $\psi_{i|}=\psi_{ji}\circ\psi_{j|}$. 
   \end{exercise-proposition}

   \setcounter{exercise}{31}
   \begin{exercise-break}\label{vector-bundle}
     \begin{enumerate}
        \item Let $\pi:E\rightarrow X$ be a vector bundle of rank $n$ with 
          bundle atlas $(\mathfrak{U},\Psi)$, and $\ks_{E/X}$ be
          the sheaf (of sets) of sections of $\pi:E\rightarrow X$.\\
          Then $(\mathfrak{U},\Psi)$ admits an $\ko_X$-module structure on
          $\ks_{E/X}$ such that $\ks_{E/X}$ is locally free of rank
          $n$.
        \item Let $(E,\pi,\mathfrak{U},\Psi)$ and
          $(E',\pi',\mathfrak{U}',\Psi')$ be two isomorphic vector bundles
          on the scheme $X$, 
          then $\ks_{E/X}$ is isomorphic to $\ks_{E'/X}$.
     \end{enumerate}
   \end{exercise-break}

   \begin{proof-named}{of exercise \ref{vector-bundle}, a.}
     The basic idea is to define a sheaf $\ka$ of $\ko_X$-modules which as a 
     sheaf of sets is isomorphic to $\ks_{E/X}$, and to identify these 
     two. We proceed along the following steps:
     \begin{enumerate}
        \item Find a good (affine) covering of $X$.
        \item Define on each open set $U$ of this covering a sheaf of
          $(\ko_X)_{|U}$-modules by defining it on a suitable basis of 
          the topology.
        \item Show that those sheaves are isomorphic to each other on
          the overlaps by defining the necessary isomorphisms again on 
          a basis of the topology for the overlaps.
        \item Conclude finally that these sheaves glue to sheaf of the
          desired form.
     \end{enumerate}
     Let's start!
     \begin{enumerate}
        \item We may assume that $\mathfrak{U}$ consists only of
          affine sets by first refining $\mathfrak{U}$ and afterwards
          throwing away the non-affine sets. (We may even assume that
          $\mathfrak{U}$ contains all open affine subsets of elements
          of $\mathfrak{U}$.) 
        \item Let now $U=\Spec(A)\in \mathfrak{U}$ be given, and let
          $\psi:\pi^{-1}(U)\rightarrow \A^n_U$, with
          $\A^n_U=\Spec(A[X_1,\dots,X_n])$ 
          be the corresponding 
          linearisation isomorphism in $\Psi$, which is indeed an
          isomorphism over $U=\Spec(A)$,
          \begin{displaymath}
            \xymatrix{\pi^{-1}(U)\ar[rr]^{\psi}_{\cong}\ar[dr]_{\pi_|}& 
              &\A^n_U\ar[dl]^{\pr_U}\\  
              &U=\Spec(A)&}.
          \end{displaymath}
          Since $\A^n_U$ is affine, this isomorphism corresponds to an 
          isomorphism of $A$-algebras 
          \begin{displaymath}
            \xymatrix{\ko_E(\pi^{-1}(U))&
              &A[X_1,\dots,X_n]\ar[ll]_{\cong}\\ 
              &A.\ar[ul]^{\alpha}\ar[ur]_{\iota_A}&}            
          \end{displaymath}
          The following observation motivates the definition of the
          sheaf $\ka$ which we will give afterwards:
          \begin{displaymath}
            \ks_{E/X}(U)=\left\{s\in\Hom(U,\pi^{-1}(U))|\pi\circ
              s=\id_U\right\}
          \end{displaymath}
          \begin{displaymath}
            \cong
            \left\{\sigma\in\Hom_{Ring}(\ko_E(\pi^{-1}(U)),A)|
              \sigma\circ \alpha=\id_A\right\}
          \end{displaymath}
          \begin{displaymath}          
            =\Hom_{A-alg}(\ko_E(\pi^{-1}(U)),A) 
          \end{displaymath}
          \begin{displaymath}
            \cong
            \Hom_{A-alg}(A[X_1,\dots,X_n],A) \cong A^n,
          \end{displaymath}
          where the first $\cong$ means a simple bijection, while the
          latter always mean isomorphisms of $A$-modules. Let us call the
          bijection $\ks_{E/X}\rightarrow
          \Hom_{A-alg}(A[\underline{X}],A)$ obtained in that way $\tau_U$.\\
          We know that $\mathfrak{B}=\left\{V|V\subseteq U \mbox{ open 
            and affine}\right\}$ is a basis of the topology of
        $U$. For $V=\Spec(B)\in\mathfrak{B}$ we define:
        \begin{displaymath}
          \ka(V):=\Hom_{B-alg}(B[X_1,\dots,X_n],B)\cong
          \Hom_{B-alg}(\ko_E(\pi^{-1}(V)),B).  
        \end{displaymath}
        Since $V\subseteq U$, we have a diagram of morphisms
        \begin{displaymath}
          \xymatrix{\pi^{-1}(V)\ar[r]^{\psi_|}_{\cong}\ar[dr]_{\pi_|} & 
              \A_V^n\ar@{}[r]|{\subseteq}\ar[d]^{\pr_V}& 
            \A_U^n\ar[d]^{\pr_U} & \pi^{-1}(U)\ar[l]^\cong_{\psi}\ar[dl]^{\pi_|}\\
            &V\ar@{}[r]|{\subseteq}& U,&}
        \end{displaymath}
%        \begin{displaymath}
%          \xymatrix{\pi^{-1}(V)\ar[r]^\cong\ar[dr] & \A_V^n\ar@{^{(}_->}[r]\ar[d]&
%            \A_U^n\ar[d] & \pi^{-1}(U)\ar[l]_\cong\ar[dl]\\
%            &V\ar@{^{(}_->}[r]& U,&}
%        \end{displaymath}
        which in turn corresponds to a diagram of ring homomorphisms
        \begin{displaymath}
          \xymatrix{\ko_E(\pi^{-1}(V)&B[X_1,\dots,X_n]\ar[l]_\cong
            &A[X_1,\dots,X_n] \ar[l]\ar[r]^\cong & \ko_E(\pi^{-1}(U)\\
            &B\ar[ul]^{\beta}\ar[u]^{\iota_B}
            &A.\ar[l]^{q=\rho_{U,V}^{\ko_X}}\ar[u]_{\iota_A}\ar[ur]_{\alpha}&} 
        \end{displaymath}
        Here we note that $B[X_1,\dots,X_n]$ is actually
        $A[X_1,\dots,X_n]\otimes_A B$, where $B$ is viewed as an
        $A$-algebra via $q:A\rightarrow B$.\\
        Thus we may define the restriction map $\rho_{U,V}$ in the
        following way:
        \begin{displaymath}
          \xymatrix{\ka(U)\ar[r]^{\rho_{U,V}}\ar@{=}[d]&
            \ka(V)\ar@{=}[d]\\
            \Hom_{A-alg}(A[\underline{X}],A)&\Hom_{B-alg}(B[\underline{X}],B)\\
            \sigma\ar@{|->}[r]\ar@{}[u]|{\mbox{\begin{sideways}
                  $\in$\end{sideways}}} & (a(\underline{X})\otimes b\mapsto
            q(\sigma(a(\underline{X})))\cdot
            b).\ar@{}[u]|{\mbox{\begin{sideways}
                  $\in$\end{sideways}}}}
        \end{displaymath}
        Note that it is enough to define the image of $\sigma$ on the
        pure tensors in such a way that the definition is
        $A$-bilinear. From the definition it is straight forward that
        $\rho_{U,V}$ is $A=\ko_X(U)$-linear. 
%Version 2 - mittels Rueckgriff auf $\ks_{E/X}$:
%        Moreover, we notice that the 
%        restriction map obtained in that way corresponds to the
%        restriction map on $\ks_{E/X}$, more precisely, the following
%        diagram commutes
%        \begin{equation}\label{tau}
%          \xymatrix{\ar@{}[dr]|\square 
%            S_{E/X}(U)\ar[r]^{\tau_U}\ar[d]^{\rho^{\ks}_{U,V}} & 
%            \ka(U)\ar[d]^{\rho^{\ka}_{U,V}}\\
%            S_{E/X}(V)\ar[r]^{\tau_V}& \ka(V).}
%        \end{equation}
%
%Version 1 - direkte Rechnung in den Ringen:
        \\
        Defining the restrictions for $V\subseteq W$, both open in $U$
        and affine, in an analogous way, we see easily that they satisfy
        the requirements $\rho_{V,V}=\id_{\ka(V)}$ and
        $\rho_{V',V''}\circ\rho_{V,V'}=\rho_{V,V''}$. Thus we have
        presheaf data on the basis of topology $\mathfrak{B}$.\\
        According to the glueing proposition \ref{basis} it remains only
        to show that these data satisfy the sheaf axiom.  
        \\
        Let
        $\left\{V_\lambda=\Spec(B_\lambda)\right\}_{\lambda\in\Lambda}$
        be an open covering of  $V=\Spec(B)\subseteq U$, and let
        $\sigma_\lambda\in\ka(V_\lambda)=
        \Hom_{B_{\lambda}-alg}(B_\lambda[\underline{X}],B_\lambda)$ such
        that for any $W=\Spec(C)\subseteq V_\lambda\cap V_\mu$ we have
        $\rho_{V_\lambda,W}(\sigma_\lambda)=\rho_{V_\mu,W}(\sigma_\mu)$.  \\
        Our aim is to glue the $\sigma_\lambda$ to a
        $\sigma\in\Hom_{B-alg}(B[\underline{X}],B)$.
        Given $b(\underline{X})\in B[\underline{X}]$ we have to define 
        $\sigma(b(\underline{X}))$, and we want to do so by glueing the 
        $b_\lambda:=\sigma_\lambda(b(\underline{X}))$. (Here we
        consider $b(\underline{X})$ as an element of
        $B_\lambda[\underline{X}]$ via $B\rightarrow B_\lambda$.)\\
        Since
        $\ko_{\Spec(B)}(V_\lambda)=\ko_{\Spec(B_\lambda)}(V_\lambda)
        \cong B_\lambda$, we may 
        view the $b_\lambda$ as sections of $\ko_{\Spec(B)}$ over
        $V_\lambda$  which, moreover by hypothesis, coincide on the 
        overlaps $V_\lambda\cap V_\mu$ (since they coincide on an open 
        affine covering of those!). Thus they glue to a unique
        global section $b\in\ko_{\Spec(B)}(V)\cong B$, whose image in $B_\lambda$
        coincides with $b_\lambda$, by construction. So we define
        $\sigma(b(\underline{X}))$ to be this $b$. The
        fact that $\sigma_\lambda$ is indeed a $B$-algebra
        homomorphism ensures the same for $\sigma$. 
%Version 2 - mittels Rueckgriff auf $\ks_{E/X}$:
%        For this we use
%        again the bijection of
%        $\ks_{E/X}(V)\cong\Hom_{B-alg}(B[\underline{X}],B)$ for any
%        $V=\Spec(B)$ open in $X$, and the commutativity of the diagram 
%        (\ref{tau}). Since $\ks_{E/X}$ is a sheaf (of sets), it certainly
%        satisfies the sheaf axiom on $\mathfrak{B}$, but then also
%        $\ka$ does so.
%%        More precisely this means the following:\\
%%        Let
%%        $\left\{V_\lambda=\Spec(B_\lambda)\right\}_{\lambda\in\Lambda}$
%%        be an open covering of  $V=\Spec(B)\subseteq U$, and let
%%        $\sigma_\lambda\in\ka(V_\lambda)=
%%        \Hom_{B_{\lambda}-alg}(B_\lambda[\underline{X}],B_\lambda)$ such
%%        that for any $W=\Spec(C)\subseteq V_\lambda\cap V_\mu$ we have
%%        $\rho_{V_\lambda,W}(\sigma_\lambda)=\rho_{V_\mu,W}(\sigma_\mu)$.  
%%        Then
%%        $\left\{\tau_{V_\lambda}^{-1}(\sigma_\lambda)\right\}_{\lambda\in\Lambda}$ 
%%        is a family of functions which according to the assumptions
%%        and the commutativity of  the diagram (\ref{tau}) coincide on
%%        (an open covering of) their intersection. Thus they define a
%%        unique (!) function $s:V\rightarrow \pi^{-1}(V)$ which,
%%        moreover, is regular, 
%%        since regularity is a local notion, and which is a section of
%%        $\pi$, since it is so locally. But then
%%        $\tau_{V}(\sigma)\in\ka(V)$ is the unique element restricting
%%        to the $\sigma_\lambda$. 
        \\
        Thus finally, for any $U\in\mathfrak{U}$, we have defined a
        sheaf $\ka_U:=\ka$ (abelian 
        groups) on $U$  carrying 
        the structure of an 
        $(\ko_X)_{|U}$-algebra.
      \item We have a familiy $\left\{\ka_i:= \ka_{U_i}\right\}_{i\in I}$ of
        sheaves (of abelian groups) where
        $\mathfrak{U}=\left\{U_i\right\}_{i\in I}$ is an open covering 
        of $X$. According to proposition \ref{covering} we have to
        find a ``good'' set of isomorphisms on the overlaps
        $U_{ij}=U_i\cap U_j$, which we may, of course, construct on an
        open affine covering of the $U_{ij}$.\\
        Let $V=\Spec(B)\subseteq U_{ij}$, then the isomorphism
        $(\psi_j\circ\psi_i^{-1})_{|\A_V^n}\in\Hom(\A_V^n,\A_V^n)$
        corresponds, by hypothesis, via the canonical identification of
        $\Hom(\A_V^n,\A_V^n)$ with
        $\Hom_{B-alg}(B[\underline{X}],B[\underline{X}])$ to a linear 
        isomorphism $\theta_V$ of $B[\underline{X}]$. Thus, by
        composition with $\theta_V$,  we get an
        isomorphism 
        \begin{displaymath}
          \xymatrix{\ka_{i|V}=\ka(V)
            \ar[r]^{\gamma_V}_{\cong} &
            \ka(V)=\ka_{j|V}, 
            }
        \end{displaymath}
        where $\ka(V)=\Hom_{B-alg}(B[\underline{X}],B)$.\\
        The fact that the open affine subsets of $U_{ij}$ form a basis of the
        topology of $U_{ij}$, and the fact that these isomorphisms are
        all derived from 
        $\psi_j\circ\psi_i^{-1}$ ensure that they glue to an
        isomorphism  $\gamma_{ij}:\ka_{i|U_{ij}}\rightarrow
        \ka_{j|U_{ij}}$.\\
        The cocycle condition for the $\gamma_{ij}$ is an immediate
        consequence of their definition with the aid of
        $\psi_j\circ\psi_i^{-1}$. \\
        Thus with proposition \ref{covering} we finally get a sheaf $\ka$
        (of abelian groups) carrying the structure of an
        $\ko_X$-module.  
      \item By construction we have an isomorphism of sheaves of sets
        from $\ks_{E/X}$ to $\ka$, and identifying those two via this
        isomorphism, we equipe $\ks_{E/X}$ with an $\ko_X$-module
        structure.         
     \end{enumerate}
     Now, by construction of $\ka$ we have for any open affine
     $U=\Spec(A)\in\mathfrak{U}$ an isomorphism
     $\ka_{|U}\cong(\ko_{\Spec(A)})^n=((\ko_X)_{|U})^n$, since for any
     open affine $V=\Spec(B)\subseteq U$ we had the canonical
     isomorphism $\ka(V)=\Hom_{B-alg}(B[\underline{X}],B)\cong
     B^n=(\ko_X(V))^n$ given by sending $\sigma$ to
     $(\sigma(X_1),\dots,\sigma(X_n))$. Hence, $\ka$ is locally free
     of rank $n$.
   \end{proof-named}

   \begin{exercise-remark}\label{refinement}
     We notice, that in order to construct $\ka$ in the proof of the
     above theorem, we passed from the given bundle atlas
     $(\mathfrak{U},\Psi)$ to a refinement containing all the open
     affine subsets of elements of $\mathfrak{U}$, and passed then on
     to a bundle atlas containing only affine subsets, by forgetting 
     the non-affine elements of $\mathfrak{U}$. Thus, we ended up with 
     a covering which contained indeed a basis of the topology, and we 
     get the following result:\\
     if $\mathfrak{U}\subseteq\mathfrak{U}'$ is an enlarged open covering
     of $X$ with $\Psi\subseteq\Psi'$ an enlarged family of
     linearisations giving (obviously) rise to the same (i.\ e.\
     canonically isomorphic) vector bundle, then the induced
     $\ko_X$-structures on $\ks_{E/X}$ are (precisely) the same.  
   \end{exercise-remark}

   \begin{proof-named}{of exercise \ref{vector-bundle}, b.}
     By hypothesis we have an isomorphism $\varphi:E\rightarrow E'$
     such that $\pi=\pi'\circ\varphi$.
     This clearly induces an isomorphism of sheaves of sets
     $\Phi:\ks_{E/X}\rightarrow\ks_{E'/X}$, given by the composition
     with $\varphi$. It remains only to check that this isomorphism is 
     an isomorphism of $\ko_X$-modules.\\
     The isomorphism $\varphi$ has the property that
     $\tilde{\mathfrak{U}}=\mathfrak{U}\cup\mathfrak{U}'$ together
     with
     $\tilde{\Psi}=\Psi\cup\left\{\psi'_{U'}\circ\varphi_{|\pi^{-1}(U')}| 
         U'\in\mathfrak{U}'\right\}$  
     gives a bundle atlas on $(E\stackrel{\pi}{\longrightarrow}X)$,
     while $\tilde{\mathfrak{U}}$  
     together with
     $\tilde{\Psi}'=\Psi'\cup\left\{\psi_{U}\circ\varphi^{-1}_{|\pi^{'-1}(U)}|
         U\in\mathfrak{U}\right\}$ 
     gives a bundle atlas on $(E'\stackrel{\pi'}{\longrightarrow}X)$.
     According to above remark \ref{refinement}, we may think of
     $\ks_{E/X}$ and $\ks_{E'/X}$ as being defined by the new bundle
     atlases, and $\tilde{\mathfrak{U}}$ may be thought of as
     consisting only of affine subsets of $X$. \\
     But then for
     $\tilde{U}=\Spec(\tilde{A})\in\tilde{\mathfrak{U}}$ we have the
     following commutative diagram (using the notations from the proof 
     of part a.)
     \begin{displaymath}
       \xymatrix{
         \ks_{E/X}(\tilde{U})\ar[r]^{\Phi(\tilde{U})}\ar[d]^{\vartheta_{\tilde{U}}}
         \ar@/_5pc/[dd]_{\tau_{\tilde{U}}}&
         \ks_{E'/X}(\tilde{U})\ar[d]^{\vartheta'_{\tilde{U}}}
         \ar@/^5pc/[dd]^{\tau'_{\tilde{U}}} \\
         \Hom_{\tilde{A}}(\ko_{\pi^{-1}(\tilde{U})},\tilde{A})
         \ar[r]^{\circ\varphi(\tilde{U})}\ar[d]^{\cong} &
         \Hom_{\tilde{A}}(\ko_{\pi^{'-1}(\tilde{U})},\tilde{A})\ar[d]^{\cong}
         \\
         \Hom_{\tilde{A}}(\tilde{A}[\underline{X}],\tilde{A})
         \ar[r]^{\delta_{\tilde{U}}}&
         \Hom_{\tilde{A}}(\tilde{A}[\underline{X}],\tilde{A}),
         }
     \end{displaymath}
     where $\vartheta_{\tilde{U}}$ resp.\ $\vartheta'_{\tilde{U}}$
     denote the canonical bijections, and where $\delta_{\tilde{U}}$
     is just the isomorphism which makes the lower square
     commute. But since ``$\circ\varphi(\tilde{U})$'' is an
     $\tilde{A}$-morphism, and since the isomorphisms denoted by
     $\cong$ are so, the same is true for $\delta(\tilde{U})$. (Note
     that the linearisation maps of the bundle atlas define just
     the $\cong$ in the diagram.)\\
     Thus the isomorphism induced by $\Phi$ on $\ka$ is an isomorphism 
     of $\ko_X$-modules (we only had to check this on the defining
     open affine covering), and hence $\Phi$ is so.
   \end{proof-named}


\end{document} 
% ~ ^[\underline{X}]


