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\begin{document} 
 
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   UNIVERSITY OF  KAISERSLAUTERN \hfill  Summerterm 1998 \\
   Mathematical Department \hfill 02nd May, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.3cm}

   \begin{center}
     \LARGE \bf Algebraic Geometry \\[0.1cm] \normalsize Exercise set 4
   \end{center}
   \vspace{0.3cm}
   \normalsize
   \parindent0cm
   \textbf{Due date:} 11th May, 1998
   \\[0.2cm]
   
   \parindent0cm

   \setcounter{exercise}{9}

   \begin{exercise}
     Let $(X,\ko_X)$ be a ringed space, $\kf$ an $\ko_X$-module.
     \begin{enumerate}
        \item Show that if $\kf$ is coherent (in the sense of lemma 1 on
          exercise sheet 3), then the following are equivalent:
          \begin{enumerate}
             \item $\kf$ is locally free (of rank $n$),
             \item $\kf_x$  is free (of rank $n$) for every $x\in X$.
          \end{enumerate}
        \item Show that if $\kf$ is locally free of rank $n$, then
          $T^r\kf$ (resp. $S^r\kf$ resp. $\Lambda^r\kf$) is locally free
          of rank $n^r$ (resp. ${n-1+r} \choose {n-1}$ resp. $n
          \choose r$).
        \item Show if  $\kf$ is locally free of rank $n$, then the
          multiplication map
          \begin{displaymath}
            \Lambda^r\kf\otimes\Lambda^{n-r}\kf\rightarrow\Lambda^n\kf
          \end{displaymath}
          is a perfect pairing, i. e. it induces an isomorphism 
          \begin{displaymath}
            \Lambda^r\kf\cong(\Lambda^{n-r}\kf)^\vee\otimes\Lambda^n\kf.
          \end{displaymath}
%        \item If $\kf$ is a locally free $\ko_X$ of rank $n$,
%          we set $\det(\kf):=\Lambda^n\kf$. Given an exact sequence of 
%          $\ko_X$-modules
%          $0\rightarrow\kf'\rightarrow\kf\rightarrow\kf''\rightarrow
%          0$, show that we get an induced isomorphism:
%          \begin{displaymath}
%            \det(\kf)\cong\det(\kf')\otimes\det(\kf'').
%          \end{displaymath}          
     \end{enumerate}
     \small (Hint, deduce from lemma 1 that if $\kf_x\cong\ko^n_{X,x}$,
     then there is an open neighbourhood $U\subseteq X$ of $x$ such
     that $\kf_{|U}\cong\ko^n_{X|U}$. In part b. and c. you may use the
     corresponding facts from
     multilinear algebra on the tensor, symmetric and exterior
     products of free modules.)
   \end{exercise}

   \begin{exercise}
     Let $A$ be a ring, $(X,\ko_X)$ any scheme. \\
     Show that the map
     \begin{displaymath}
         \begin{array}{ccc}       
           \Hom_{\textswab{\scriptsize
               Scheme}}(X,\Spec(A))&\longrightarrow&\Hom_{\textswab{\scriptsize Ring}}(A,\Gamma(X,\ko_X))\\
           (f,f^\#) & \mapsto & f^\#(\Spec(A)),
         \end{array}
     \end{displaymath}
     sending a morphism of schemes to the morphism of global
     sections, is a (natural) bijection.\\
     \small (Hint, cover $X$ by open affine schemes and  use the
     corresponding theorem for affine schemes of the lecture.) 
   \end{exercise}

   \begin{exercise}
     Let $A$ be a ring and $X=\Spec(A)$, $V\subseteq X$ and
     $I(V)=\bigcap_{P\in V}P=\left\{a\in A| a\in P\;\forall P\in V\right\}$ the vanishing ideal 
     of $V$.\footnote{$a\in P$
       means that $a\equiv 0 (\mod P)$. Viewing $a$ as a function from 
       $\Spec(A)$ to the disjoint union of the $A/P$, sending $P$ to
       $a(\mod P)$, we gain the notion of $a$ ``vanishing'' at a point 
       $P$ of $\Spec(A)$.}
     \begin{enumerate}
        \item Show that: $\overline{V}=V(I(V))$.
        \item Let $V\not=\emptyset$ be closed. Show that the following are
          equivalent:
          \begin{enumerate}
             \item $V$ is irreducible,
             \item $I(V)$ is a prime ideal, and
             \item $V$ contains a generic point (which is then unique).
          \end{enumerate}
          In particular, $\Spec(A)$ is irreducible iff $A/\sqrt{(0)}$ is an
          integral domain.
        \item Generalize b. to an arbitrary scheme.
     \end{enumerate}
   \end{exercise}

   \vspace{0.8cm}
   The statement c. in exercise 9 was incorrect. It should have been
   as follows:
   \setcounter{exercise}{8}
   \begin{exercise}
      Let $(X,\ko_X)$ be a ringed space, $\kf$ a locally free
      $\ko_X$-module, $\ke,\kg$ any $\ko_X$-modules.
      \begin{enumerate}
         \item[c.] There is a natural isomorphism of
           $\Gamma(X,\ko_X)$-modules 
           \begin{displaymath}
              \Hom(\ke\otimes_{\ko_X}\kf,\kg)\cong\Hom(\ke,\HOM{13}_{\ko_X}(\kf,\kg)).
          \end{displaymath}
      \end{enumerate}
   \end{exercise}

   We now give a proof of lemma 1 on exercise sheet 3.
   \begin{exercise-lemma}
     Let $(X,\ko_X)$ be a ringed space, $\kf, \kg$ be
     $\ko_X$-modules. Moreover, let $x\in U\subseteq X$ open such that 
     we have an exact sequence
     \begin{equation}\label{blubber}
       \ko_{X|_U}^r\stackrel{\delta}{\longrightarrow}\ko_{X|_U}^s
       \stackrel{\gamma}{\longrightarrow}\kf_{|U}\rightarrow 0.  
     \end{equation}\\
     The natural $\ko_X$-module morphism
     $$\HOM{13}_{\ko_X}(\kf,\kg)_x\stackrel{\alpha_x}{\longrightarrow}
     \Hom_{\ko_{X,x}}(\kf_x,\kg_x)$$ 
     is then an isomorphism. \\
     Note: if every point $x$ of $X$ has a neighbourhood $U$
     such that $\kf_{|U}$ satisfies an equation of type (\ref{blubber}),
     the sheaf $\kf$ is called a 
     $\ko_X$-module \textbf{of finite presentation}, or
     \textbf{coherent} if $(X,\ko_X)$ is a sheme.
   \end{exercise-lemma}
   \begin{proof}
      The same proof as in exercise 3 gives a natural morphism of
     $\ko_{X,x}$-modules (instead of abelian groups)
     $\alpha_x:\HOM{13}_{\ko_X}(\kf,\kg)_x\rightarrow
     \Hom_{\ko_{X,x}}(\kf_x,\kg_x)$. It remains to show, that $\alpha_x$ 
     under the above condition is bijective. W.~l.~o.~g.~$U=X$.
     \begin{enumerate}
        \item \underline{Injectivity:}\footnote{For this we only need the
          surjection $\xymatrix{\gamma:\ko_X^s\ar@{->>}[r]&\kf}$.} Let
          $\varphi\in\HOM{13}_{\ko_X}(\kf,\kg)_x$ be such that
          $\alpha_x(\varphi)\equiv 0$.\\
          Idea: lift $\varphi$ on some neigbourhood $W$ of $x$ via
          $\gamma$ to a free sheaf and
          use the fact that $\gamma$ is an epimorphism (in the
          categorical sense). \\
          Choose a representative $(\tilde{\varphi}, V)$ of $\varphi$
          with
          $\tilde{\varphi}\in\HOM{13}_{\ko_{X|V}}(\kf_{|V},\kg_{|V})$.
          Remind yourself that $\alpha_x([(\tilde{\varphi},V)])$ is
          just the germ $\tilde{\varphi}_x$ of $\tilde{\varphi}$ at
          $x$. Since $\tilde{\varphi}_x=0$, also
          $(\gamma\circ\tilde{\varphi})_x =
          \gamma_x\circ\tilde{\varphi}_x=0$.
          This is equivalent to the fact that the standard basis
          $e_{1,x},\ldots,e_{s,x}$ of $\ko_{X,x}^s$ gets mapped to zero, that is there
          exist open neighbourhoods $V_i(x)$ such that
          $(\gamma\circ\tilde{\varphi})(V_i)(e_{i|V_i})=0$. Setting
          $W:=V_1\cap\ldots\cap V_s$, we get that
          $(\gamma\circ\tilde{\varphi})_{|W}=0$,
          since $\ko_{X|W}^s$ is free with free generators
          $e_{1|\tilde{W}},\ldots,e_{s|\tilde{W}}$ for
          $\tilde{W}\subseteq W$ open. \\
          Thus we have the following commutative diagram:
          \begin{displaymath}
            \xymatrix{\ko_{X|W}\ar[r]^{\gamma_{|W}} &
              \kf_{|W}\ar@<0.5ex>[r]^{\tilde{\varphi}_{|W}}
              \ar@<-0.5ex>[r]_0 &\kg_{|W}}.
          \end{displaymath}
          The fact that $\gamma$ is an epimorphism\footnote{Since the
            category (Shv/X) is abelian, the exactness of
            $\ko_X^s\stackrel{\gamma}{\rightarrow} \kf\rightarrow 0$
            implies that $\gamma$ is an epimorphism. If you don't
            want to use that (Shv/X) is abelian, replace the above
            diagram by the cokernel 
            diagram of $\gamma$ and use the fact that the cokernel is
            $0$. (See also the following footnote \ref{follow}.)}  implies thus  
          $\tilde{\varphi}_{|W}=0$. 
        \item \underline{Surjectivity:} Let
          $\psi\in\Hom_{\ko_{X,x}}(\kf_x,\kg_x)$.\\
          This gives the following ``cokernel diagram'' of abelian
          groups:
          \begin{displaymath}
            \xymatrix{
              &\ko_{X,x}^r\ar[r]^{\delta_x}\ar@/_/[drr]^{0} &
              \ko_{X,x}^s\ar[r]^{\gamma_x}\ar[dr]^{\Psi_x} & \kf_x
              \ar[r]\ar[d]^\psi & 0\\ 
              & & & \kg_x & }.       
          \end{displaymath}
          Idea: extend $\Psi_x$ to some neighbourhood $V$ of $x$ and
          use the universal property of the sheaf cokernel to get a
          unique extension of $\psi$ making the cokernal diagram
          commute. \\
          Let $e_{1,x},\ldots,e_{s,x}$ be the standard basis of
          $\ko_{X,x}^s$ and set $f_{i,x}:=\Psi_x(e_{i,x})$. Choosing
          representatives $(f_i,V_i)$ for the $f_{i,x}$ and setting
          $W:=V_1\cap\ldots\cap V_s$, we can define a unique
          $\ko_{X|V}$-module morphism $\Psi$ by linear extension of
          $\Psi(\tilde{W}): e_{i|\tilde{W}}\mapsto f_{i|\tilde{W}}$
            (for $\tilde{W}\subseteq W$), since $\ko_{X|W}^s$ is free.
          Moreover, the germ of $\Psi$ at $x$ is indeed
          $\Psi_x$. By (\ref{blubber}) $\kf$ is the cokernel of      
          $\delta:\ko_X^r\rightarrow\ko_X^s$, and using the 
          universal property of the cokernel\footnote{\label{follow}On the level of
            presheaves the universal property of the presheaf cokernel is
            obvious, and it is induced on the sheaf cokernel via the
            universal property of the sheafification.} we get the
          following commutative diagram:
          \begin{displaymath}
            \xymatrix{
              &\ko_{X|W}^r\ar[r]^{\delta_{|W}}\ar@/_/[drr]^{0} &
              \ko_{X|W}^s\ar[r]^{\gamma_{|W}}\ar[dr]^{\Psi_{|W}} & \kf_{|W}
              \ar[r]\ar@{-->}[d]^{\exists_!\tilde{\psi}} & 0\\ 
              & & & \kg_{|W}& }.       
          \end{displaymath}
          Taking limits we see that
          $\alpha_x[(\tilde{\psi},W)]=\tilde{\psi}_x=\psi$. 
     \end{enumerate}
   \end{proof}

\end{document} 
