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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 22nd 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

   \setcounter{exercise}{36}



   \begin{exercise}[Quadratic Cremona Transformation]
     Define the quadratic Cremona transformation
     $\varphi:\P_K^2\dashrightarrow\P_K^2:(x_0:x_1:x_2)\mapsto
     (x_1x_2:x_0x_2:x_0x_1)$.\\
     Find open and dense subsets $U$ and $V$ in $\P_K^2$ such that
     $\varphi_|:U\rightarrow V$ is an isomorphism, i.\ e.\ show that
     $\varphi$ is birational. What is the definition set of $\varphi$?
   \end{exercise}

   \begin{proof}
     We want to show that
     $\tilde{U}=\P_K^2\backslash\menge{P=(1:0:0),Q=(0:1:0),R=(0:0:1)}$ is
     the definition set of $\varphi$. It is obvious, that $\varphi$ is 
     defined on $\tilde{U}$. In order to see that it is a morphism on $\tilde{U}$, we 
     look at the affine open covering of $\tilde{U}$ by $D(X_1X_2)\cup
     D(X_0X_2) \cup D(X_0X_1)$. Let
     $\psi_i:U_i:=\menge{X_i\not=0} \rightarrow
     \A_K^2$ be the i-th coordinate chart. First we look at
     $\varphi_|:D(X_1X_2)\rightarrow U_0$. Now $\psi_{1|}:
     D(X_1X_2)\cap U_2\rightarrow\A_K^2\backslash V(Y_1)$ is an
     isomorphism (where $Y_0$ and $Y_1$ denote the coordinates on
     $\A_K^2$). Therefore we get that
     $\psi_0\circ\varphi_|\circ\psi_2^{-1} : \A_K^2\backslash
     V(Y_1)\rightarrow \A_K^2:(y_0,y_1)\mapsto
     (y_0:y_1:1)\mapsto(y_1:y_0:y_0y_1)\mapsto (\frac{y_0}{y_1},y_0)$, 
     being a morphism (since its coordinate functions are quotients of
     polynomials) implies that also $\varphi_{|D(X_1X_2)}$ is a
     morphism. In the same way the remaining two restrictions
     $\varphi_{|D(X_0X_2)}$ and $\varphi_{|D(X_0X_1)}$ are morphisms,
     which implies by exercise 23, that $\varphi$ is so on $\tilde{U}$. 
     \\
     Now suppose, the definition set $U'$ was
     larger, and $\varphi':U'\rightarrow\P_K^2$ was a morphism
     extending $\varphi$. 
     We may assume that $U'$
     contains $P$. Let $L_1=\menge{X_1=0}\backslash\menge{R} $ and
     $L_2=\menge{X_2=0}\backslash\menge{Q}$, 
     two ``lines'' in $\P_K^2$. Then
     $\varphi'(L_1\backslash\menge{P})=\menge{Q}$ and
     $\varphi'(L_2\backslash\menge{P})=\menge{R}$. But $L_1\cap
     L_2=\menge{P}$, and since $\varphi'$ is continuous,
     ${\varphi'}^{-1}(Q)\cap (L_1\cup L_2)$ is closed in $L_1\cup L_2$ 
     and thus it
     contains $P$, but the same is true for
     ${\varphi'}^{-1}(R)\cap(L_1\cup L_2)$, which leads to a
     contradiction. \\
     Setting $U=D(X_0X_1X_2)$ it is obvious, that $\varphi(U)\subseteq 
     U$ and $\varphi_{|U}^2=\id_U$. Since $U\subset\P_K^2$ is dense,
     this implies that $\varphi$ is birational.
   \end{proof}

%   \begin{claim}
%     If $X\subseteq \P_K^m$ is an irreducible 
%     quasi projective variety over the algebraically
%     closed field $K$ (important), and $f:X\dashrightarrow \P_K^n$ a
%     rational map, then $f$ lifts to a rational map
%     $\tilde{f}:X\dashrightarrow \A_K^{n+1}$, where the latter is the
%     projective cone over $\P_K^n$. 
%   \end{claim}
%   \begin{proof}
%     There exists an open dense subset $V\subseteq X$ such that
%     $f:V\rightarrow \P^n_K:y\mapsto (f^0(y):\ldots:f^n(y))$ is a
%     morphism. Covering $\P_K^n$ by the
%     open affine subspaces $U_i=\menge{X_i\not=0}$ and restricting $f$ 
%     to the $V_i:=f^{-1}(U_i)$, we get morphisms $f_i:=f_|:V_i\rightarrow 
%     U_i:y\mapsto
%     (\frac{f^0(y)}{f^i(y)}:\ldots:1:\ldots:\frac{f^n(y)}{f^i(y)})$ to
%     affine varieties.  
%     Thus the coordinate functions $\frac{f^j}{f^i}$ of $f_i$ are 
%     regular on $V_i$, which means that locally they are of the 
%     form $\frac{g}{h}$ where $g, h\in\K[X_0,\ldots,X_m]$ are
%     homogeneous polynomials of the same degree. Thus the well defined 
%     map $\tilde{f}_i: V_i\rightarrow
%     \A_K^{n+1}:y\mapsto
%     (\frac{f^0(y)}{f^i(y)},\ldots,1,\ldots,\frac{f^n(y)}{f^i(y)})$
%     is also a morphism by exercise 23. Now, some $V_i$ is not empty
%     and is hence dense in $X$ since $X$ is irreducible. Thus
%     $\tilde{f}_i$ defines a rational map $\tilde{f}:Y\dashrightarrow
%     \A_K^{n+1}$ with $\pi\circ\tilde{f}=f$ as rational maps (i.\ e.\
%     on some open dense set), where $\pi:\A_K^{n+1}\backslash
%     \menge{0} \rightarrow \P_K^n$ denotes the quotient map. The last
%     part means $\tilde{f}$ is a lift of $f$. 
%   \end{proof} 


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