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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 16th January, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.5cm}

   \begin{center}
     \LARGE \bf Introduction to Algebraic Geometry \\
   \end{center}
   \vspace{0.5cm}
   \normalsize
   \parindent0cm
   \textbf{Due date:} 26th January, 1998 
   \\[0.2cm]
   
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   \setcounter{exercise}{37}

   \begin{exercise}
     Let $K$ be an algebraically closed field. 
     Prove parts c., d., f. and g.
     \begin{enumerate}
        \item Let $(X,\ko_X)$ be any space with functions over
          $K$, $A\subset X$. $A$ together with the
          structure (sheaf) $\ko_A$ defined 
          in the lecture is a space with functions over $K$.
        \item If $A\subseteq\P_K^n$ is a projective variety or if
          $A\subseteq\A_K^n$ is 
          an affine variety, then the structure (sheaf) induced on $A$
          by the coordinate ring $K[A]$ and the structure (sheaf)
          induced by $\P_K^n$ 
          resp. by $\A_K^n$ coincide.
        \item If $X$ is a variety over $K$ and $A$ is locally closed
          in $X$, then
          $(A,\ko_A)$ is a variety over $K$. 
        \item If $f:X\rightarrow Y$ and $g:X'\rightarrow Y'$ are
          morphisms of varieties over $K$, then $f\times g:X\times
          X'\rightarrow Y\times Y'$ is a morphism of varieties over $K$.
        \item If $f:X\rightarrow Y$ is a morphism of varieties over
          $K$ and
          $Y'\subseteq Y$ is a subvariety with $f(X)\subset Y'$, then
          $f:X\rightarrow Y'$ is a morphism.\\
          If moreover $X'\subseteq X$ is a subvariety of $X$, then
          $f_|:X'\rightarrow Y$ is a morphism.
        \item If $f:X\rightarrow Y$ is a morphism of varieties over
          $K$, then the graph $\Gamma_f$ is locally closed in $X\times 
          Y$. In particular, $(\Gamma_f,\ko_{\Gamma_f})$ is a variety.
        \item If $X$ is a variety over $K$, then the map $\gamma_f:X\rightarrow
          \Gamma_f:x\mapsto (x,f(x))$ is 
          an isomorphism of varieties.
     \end{enumerate}
     {\small (Hint, for c. look at an open affine cover of $A$, for f.
       at an open affine cover of $Y$. In part d. and g. you may use
       the universal property of the product of spaces with functions.)}
   \end{exercise}

   \begin{exercise}
     Let $X$ and $Y$ be spaces with functions, $i:X\rightarrow Y$ a
     morphism. Then the following are equivalent:
     \begin{enumerate}
        \item $i$ is an embedding.
        \item $i$ is injective and
          $i^{-1}:(\im(i),\ko_{\im(i)})\rightarrow (X,\ko_X)$ is a
          morphism.
        \item $i$ is a homeomorphism to $\im(i)$ and $i^*:\ko_{\im(U)}(U)\rightarrow
          \ko_X(i^{-1}(U))$ is surjective for all $U\subset Y$
          open.
        \item $i$ is a homeomorphism to $\im(i)$ and $i^*:\ko_{\im(U),i(x)}\rightarrow
          \ko_{X,x}$ is surjective for all $x\in X$.
     \end{enumerate}
   \end{exercise}

   \begin{exercise}
     We call a variety over $K$ \textbf{rational} if it is birationally
     equivalent to some projective space $\P_K^n$. Let now $K$ be
     algebraically closed. Show the following:
     \begin{enumerate}
       \item A variety $X$ over $K$ is rational iff $R(X)$ is a pure
         trancendental extension field of $K$.
       \item Every irreducible conic in $\P_K^2$ is rational.
       \item The nodal cubic $\menge{Y^2Z-X^2(X+Z)=0}\subseteq \P_K^2$ 
         is rational.
     \end{enumerate}
     {\small (Hint, a. and b. are very easy. For c. use the central
       projection of $\P_K^2$ through the singular point of the cubic
       to an appopriate line.)}
   \end{exercise}


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