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   UNIVERSITY OF  KAISERSLAUTERN \hfill  Summerterm 1998 \\
   Mathematical Department \hfill 15th May, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.5cm}

   \begin{center}
     \LARGE \bf Algebraic Geometry \\[0.1cm] \normalsize Exercise set 6
   \end{center}
   \vspace{0.5cm}
   \normalsize
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   \textbf{Due date:} 25th May, 1998\hfill\textbf{10 a.~m.}
   \\[0.2cm]
   
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   \setcounter{exercise}{15}

   \begin{exercise}
     Let $\varphi:S\rightarrow T$ be a morphism of degree $0$ of
     graded rings.
     \begin{enumerate}
       \item The set $U:=\left\{ P\in\Proj(T) |
           \varphi(S_+)\not\subseteq P\right\}$ is open in $\Proj(T)$.
       \item Show that if $I\unlhd T$ is a homogeneous ideal, then
         $\varphi^{-1}(I)$ is homogeneous.
       \item  Deduce from b. that $f:P\mapsto \varphi^{-1}(P)$
         defines a map from $U\rightarrow\Proj(S)$.\\
         {\footnotesize (Remark: With litteraly the same arguments as in the affine case
         it follows then that $f$ is continous and that $(f,f^\#)$ with 
         \[
            f^\#(V):\left(s:V\rightarrow\coprod_{P\in V}S_{(P)}\right)\mapsto
            \left((\coprod_{Q\in f^{-1}(V)}\varphi_Q)\circ s\circ
            f:f^{-1}(V)\rightarrow\coprod_{Q\in f^{-1}(V)}T_{(Q)}\right)
         \]
         (for $V\subseteq \Proj(S)$ open) is a morphism of schemes -
         just take homogeneous elements of the same degree whenever
         quotients are involved.)}
       \item Show, if $\varphi$ is surjective, then $U=\Proj(T)$.\\
         {\footnotesize (Remark: With the same proof as in exercise 14 in the
           affine case it follows that
           $(f,f^\#)$ is a closed immersion, i.~e.~$\im(f)$ is closed in 
           $\Proj(S)$, $f$ is a homeomorphism onto
           $\im(f)$ and $f^\#$ is surjective.)}
       \item Show that $(f,f^\#)$ may be an isomorphism, even if
         $\varphi$ is not!\\
         \small (Hint, consider an example, where $\varphi$ is
         surjective and 
         $\varphi_d:S_d\rightarrow T_d$ is an isomorphism only for 
         $d$ greater than some fixed $d_0>0$. Then use part d.) 
     \end{enumerate}
   \end{exercise}

   \begin{exercise}
     Let $f:X\rightarrow Y$ be a morphism of schemes.
     Show that the following are equivalent:
     \begin{enumerate}
        \item $f$ is locally of finite type.
        \item For every open affine subset $V=\Spec(B)$ of $Y$
          $f^{-1}(V)$ can be covered by open affine subsets
          $U_j=\Spec(A_j)$, such that $A_j$ is a finitely generated
          $B$-algebra. 
     \end{enumerate}
     {\small (Hint, show first, if $Y=\Spec(B)$ and $X$ has an
       open covering by $\Spec(A_i)$ ($i\in I$) with $A_i$ a finitely
       generated $B$-algebra, then $f^{-1}(D(g))$ is covered by
       the $\Spec((A_i)_g)$ and these are finitely generated
       $B_g$-algebras (for $g\in B$). Deduce that
       $f_|:f^{-1}(V)\rightarrow V$ is locally of finite type, and show
       finally that w.~l.~o.~g.~we may assume $V=\Spec(B)$ is
       covered by basic open sets of the desired form. For the latter 
       part note that for $D(g)\subseteq\Spec(B_j)\subseteq\Spec(B)$
       with $g\in B$ and $\bar{g}$ the image of $g$ in $B_j$ we
       have $D(g)=D(\bar{g})$ and $B_g\cong (B_j)_{\bar{g}}$.)}
   \end{exercise}
   
   \begin{exercise}
      Show that if $f:X\rightarrow Y$ is a finite morphism of schemes,
      then $f$ is closed. \\
      \small (Hint, reduce to the situation $Y=\Spec(B)$ and
      $X=\Spec(A)$ with $B\subseteq A$ a subring, and show for every
      $P\in\Spec(B)$ there is a $Q\in\Spec(A)$ such that $P=Q\cap B$,
      using the lemma of Nakayama and lemma II.1.6, or using ``going
      up'' instead.)
   \end{exercise}
 
   \begin{exercise}
     Let $K$ be any field, $\A_K^n=\Spec(K[X_1,\dots,X_n])$. \\
     Show that $\A_K^2\cong\A_K^1\times_{\Spec(K)}\A_K^1$.
   \end{exercise}

\end{document} 



          \item If $f$ is finite, then $f^{-1}(y)$ is a finite set for
          every $y\in Y$. We say $f$ is \textbf{quasi finite}.\\
          {\small (Hint, reduce to the case $Y=\Spec(B)$ with $B$ a
            field and $X=\Spec(A)$ with $A$ artinian, and use the fact 
            that an artinian ring has only finitely many maximal
            ideals, c.~f.~Atiyah-MacDonald prop.~8.3.)}
        \item Find an example of a surjective, quasi-finite morphism
          of finite type which is not finite.
