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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 09th 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

   \begin{claim}[Exercise 23 for "new" varieties] 
     Let $K$ be any field, $V$ and $W$ be varieties over $K$ in the
     wider sense of I.5, and let
     $\varphi:V\rightarrow W$ be any map. If $\menge{ V_i|i\in I}$ is
     an open covering of $V$ and $\menge{ W_i|i\in I}$ is
     an open covering of $W$ such that $\varphi(V_i) \subseteq W_i$
     for all $i\in I$.\\
     Then the following are equivalent:
     \begin{enumerate}
        \item $\varphi$ is a morphism.
        \item $\varphi_{|}:V_i\rightarrow W_i$ is a morphism for all
          $i\in I$.
     \end{enumerate}
   \end{claim}
   \begin{proof}
    \begin{description}
     \item[a.$\Rightarrow$b.:] By the definition of the subspace
       topology, the restriction of a continous map is continous, thus 
       the $\varphi_{|U_i}$ are continous. Moreover, since
       $V_i\subseteq \varphi^{-1}(W_i)$, we have
       $\varphi^*(\ko_W(W_i))\subseteq\ko_V(\varphi^{-1}(W_i))\subseteq\ko_V(V_i)$. 
     \item[b.$\Rightarrow$a.:] First we show, that $\varphi$ is
       continous. For this let $U\subset W$ be open, then $U\cap W_i$
       is open in $W_i$. Thus $\varphi_|^{-1}(U\cap W_i)$ is open in
       $V_i$ and hence in $V$, 
       but $\varphi^{-1}(U)=\bigcup (\varphi^{-1}(U)\cap V_i)=\bigcup
       \varphi_|^{-1}(U\cap W_i)$, and is thus open in $V$ as the
       union of open sets.\\
       We then want to show, that  $\varphi^*(\ko_W(U))\subseteq
       \ko_V(\varphi^{-1}(U))$ for any open set $U\subset W$. Now,
       given $f=\varphi^*g\in \varphi^*(\ko_W(U))$ and $P\in\varphi^{-1}(U)$,
       then there is some $i$, such that $P\in \varphi^{-1}(U)\cap
       V_i$. Now, $g\in\ko_W(U)\subseteq\ko_W(U\cap W_i)$ and thus
       $f_{|(\varphi^{-1}(U)\cap V_i)}=\varphi^*g_{|(\varphi^{-1}(U)\cap V_i)}
       =\varphi_|^*g\in\ko_{V_i}(\varphi_|^{-1}(U))=
       \ko_V(\varphi^{-1}(U)\cap V_i)$. Hence, $f$ is locally
       regular, and by
       of the sheaf axiom for spaces with functions, this means that
       $f$ is regular. 
    \end{description}
   \end{proof}

   \begin{claim}
     Let $V_{ij}=\menge{X_i\cdot Y_j\not=0}\subset\P_K^r\times\P_K^s$, 
     where $K[X_0,\ldots,X_r]$ and $K[Y_0,\ldots,Y_s]$ are the
     coordinate rings for $\P_K^r$ resp. $\P_K^s$. \\
     Then $V_{ij}$ is an affine variety with coordinate functions
     $\menge{\frac{X_k}{X_i},\frac{Y_l}{Y_j}|k\not=i,l\not=j}$. 
   \end{claim}
   \begin{proof}
     Notation: set $U_i:=\menge{X_i\not=0}\subset\P_K^r$ and
     $V_j:=\menge{Y_j\not=0}\subset\P_K^s$, which are affine varieties 
     with coordinate functions
     $\menge{\frac{X_k}{X_i}|k\not=i}$ and
     $\menge{\frac{Y_l}{Y_j}|l\not=j}$. As sets we have
     $V_{ij}=U_i\times V_j$.\\ 
     \underline{Idea:} Show that the subvariety structure on $V_{ij}$
     induced by $\P_K^r\times\P_K^s$ coincides with the product
     structure on $U_i\times V_j$. 
     \\
     (Then by theorem I.5.10 (ii) $V_{ij}$ is affine with the
     proclaimed coordinate functions.)\\
     We have to show that:
     \begin{itemize}
        \item $W\subset V_{ij}$ is open $\Leftrightarrow$ $W\subset
          U_i\times V_j$ is open.
        \item For $W$ open we have to show that:
          $\ko_{V_{ij}}(W)=\ko_{U_i\times V_j}(W)$.
     \end{itemize}
     We note, that for any variety $X$ and $W\subseteq U\subseteq X$ open, by
     definition of the subvariety structure of $U$, we have
     $\ko_X(W)=\ko_U(W)$! That is all we need.\\
     Let now $U\subseteq U_i$ and $V\subseteq V_j$ be open. Let
     moreover $g_\nu\in\ko_{\P_K^r}(U)=\ko_{U_i}(U)$ and
     $h_\nu\in\ko_{\P_K^s}(V)=\ko_{V_j}(V)$. Set $f=\sum g_\nu\cdot
     h_\nu$.\\
     By the definition (I.5.2) of the product structure on
     $\P_K^r\times\P_K^s$ (and hence on $V_{ij}$), a basic open subset
     in $V_{ij}$ is of the form $D(f):=\menge{R\in U\times 
     V|f(R)\not=0}$ for such an $f$. But the same is true for the 
     basic open sets w.\ r.\ t.\ the product structure on $U_i\times 
     V_j$. Therefore the open sets w.\ r.\ t.\ both structures
     coincide.\\
     Moreover, a function is regular, if it locally is the quotient of 
     functions of this type. So also the regular functions coincide.
   \end{proof}

   \begin{claim}[Exercise 26]
          Let $K$ be algebraically closed. $Q$ be the quadric $V(XW-YZ)$
     in $\P_K^3$, and $f\in R(Q)$ be the rational function on $Q$ induced by
     $\frac{X}{Y}$.
     \begin{enumerate}
        \item Given $\emptyset\not=U\subseteq Q$ open and $g,h\in
          K[W,X,Y,Z]_d$ such that $f\equiv \frac{g}{h}$ on $U$. Show,
          that then $$Yg-Xh\equiv 0 \mbox{ on } Q,$$
          and deduce, that there exist polynomials $h_1,h_2\in
          K[W,X,Y,Z]_{d-1}$ such that $$h=Yh_1+Wh_2.$$
        \item Show, the pol set of $f$ is $\menge{(w:x:y:z)\in Q|
            w=y=0}$.
        \item Show, there exist no polynomials $g,h\in K[W,X,Y,Z]$
          such that $f\equiv \frac{g}{h}$ on the whole definition set
          of $f$.
     \end{enumerate}
   \end{claim}
   \begin{proof}
     Let $\emptyset\not=U\subset Q$ open and $f_{|U}\equiv\frac{g}{h}$ 
     for $g,h\in K[W,X,Y,Z]_d$.
     \begin{enumerate}
        \item We have $\tilde{U}:=Q\backslash V(Y)\subset Q$ open and
          $f_{|\tilde{U}}\equiv \frac{X}{Y}$. Since $Q$ is
          irreducible, $U$ and $\tilde{U}$ are dense in $Q$. Moreover, 
          for dimension reasons, $\hat{U}:=U\cap \tilde{U}$ is not empty and
          hence also dense in $Q$. \\
          Now on $\hat{U}$, $\frac{X}{Y}\equiv\frac{g}{h}$, and hence
          $Xh-Yg=0$ on $\hat{U}$. But since $\hat{U}$ is dense in $Q$ and $Xh-Yg$
          is continous, \begin{equation}\label{eins} Xh-Yg=0 \mbox{ on } Q.\end{equation}
          Collecting the monomials of $h$ which are not divisible by
          $Y$ in $\tilde{h_1}$, those not divisble by $Y$ but divisible by $W$ 
          in $\tilde{h_2}$ and the remaining ones in $h_3$, which then
          is an element of $K[X,Z]_d$, and setting
          $h_1:=\frac{\tilde{h_1}}{Y}\in K[W,X,Y,Z]_{d-1}$ and
          $h_2:=\frac{\tilde{h_2}}{W}\in K[W,X,Y,Z]_{d-1}$, we get:
          $$ h=Yh_1+Wh_2+h_3. $$
          Since for any $(x:z)\in\P^1_K$ we have $(0:x:0:z)\in Q$, we
          get by equation (\ref{eins}):
          $$ X\cdot h_3(X,Z)\equiv 0 \mbox{ on } \P_K^1.$$
          But the only (homogeneous) polynomial (in $K[X,Z]$)
          vanishing on the whole of $\P_K^1$ is the zero polynomial,
          since $K$ is algebraically closed. This proves (i).
        \item Given a point $P=(w:x:y:z)\in Q$, such that not both $w$
          and $y$ are zero. If $w\not= 0$, then $f\equiv \frac{Z}{W}$
          locally in $P$. If $y\not= 0$, then $f\equiv \frac{X}{Y}$
          locally in $P$. Hence, $P$ is not in the pole set of $f$.\\
          Given a point $P=(0:x:0:z)\in Q$, then by (i) for every
          local representation $\frac{g}{h}$ of $f$ we have
          $h(P)=0$. Hence, $P$ belongs to the pole set of $f$.
        \item Suppose there were polynomials $g,h\in K[W,X,Y,Z]$ such
          that $f\equiv\frac{g}{h}$ on the whole definition set
          $Q\backslash\menge{W=Y=0}$ . 
          We know that for each $(w:z)\in\P_K^1$ we get
          $(w:0:0:z)\in Q$. By assumption, $h(w:0:0:z)\not= 0$ 
          for all $(w:z)\in\P_K^1$ with $w\not=0$. But since $K$ is algebraically
          closed, by the Hilbert Nullstellensatz a homogeneous
          polynomials having  at most the zero $(0:1)$ on $\P_K^1$ 
          is of the form $b\cdot W^k$ for some integer $k$ and some
          non zero constant $b$. Hence, $h(W,0,0,Z)=b\cdot W^k$.
          Then $h=b\cdot W^k+X\tilde{h}+Y\hat{h}$. Looking now at the
          points $(w:0:y:0)$ which are in the definition set of $f$
          for all $(w:y)\in\P_K^1$, we get
          $0\not=h(w:0:y:0)=bw^k+y\hat{h}(w:0:y:0)$, i.\ e.\ the
          homogeneous polynomial $h(W,0,Y,0)=bW^k+Y\hat{h}(W,0,Y,0)$
          has no zero at all on $\P^1_K$, and is hence a non zero
          constant, which means that $k=0$ and $\hat{h}=0$ as well.  
          But then $h=b+X\tilde{h}$ and is
          homogeneous, so we finally get that $h=b$. Thus also $g$,
          being of the same degree, is constant, and hence $f$ itself
          is a constant function on $Q\backslash\menge{W=Y=0}$, which is a contradiction.
     \end{enumerate}
   \end{proof}


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