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\begin{document} 
 
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   \parindent0cm
   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 06th December, 1997 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.3cm}

   \begin{center}
     \LARGE \bf Introduction to Algebraic Geometry \\
   \end{center}
   \vspace{0.3cm}
   \normalsize
   \parindent0cm
   \textbf{Due date:} 05th January 1998
   \\[0.2cm]
   The exercise 26 below should be handed in by January 5th and will
   be discussed in the exampleclasses. For the remaining exercises
   proofs are included. You do not have to hand them in and they will
   not be discussed in the exampleclasses unless you have any
   questions.\\
   In the proof of theorem 4.28 we used implicitly that for any closed
   variety the rational functions may be viewed as elements of the
   quotient ring of the coordinate ring. This is proved in exercise \ref{quotient}.
   The proof of theorem 4.18 contained some remarks on the
   correspondence of regular functions and germs of regular
   functions. This correspondence and their relation to rational
   functions is the content of exercise \ref{regular}.
   \parindent0cm

   \setcounter{exercise}{25}

   \begin{exercise}
     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}
     \small (Hint, for a. look at the points $P_{x,z}=(0:x:0:z)$, for
     c. at $P_{w,z}=(w:0:0:z)$ and at $P_{w,y}=(w:0:y:0)$. Moreover,
     use in a. that $Q$ is 
     irreducible!)\\
     (Remark: $Q$ can be identified with the Segre embedding of
     $\P_1\times\P_1$ in $\P_3$.)
   \end{exercise}

   \begin{exercise}\label{quotient}
     Let $K$ be algebraically closed, and let $X$ be an affine variety. \\
     Then the quotient ring $\Quot(K[X])$ of the coordinate ring
     $K[X]$ 
     is isomorphic to the ring of rational functions $R(X)$ on $X$. 
   \end{exercise}
   \begin{proof}
     Let $I(X)=P_1\cap\ldots\cap P_r$ be the minimal primary
     decomposition of the vanishing ideal of $X$ (i.\ e.\ $P_i=I(X_i)$ 
     where $X_1,\ldots,X_r$ are the irreducible components of $X$). We 
     denote by $\bar{P_i}$ the image $P_i/I(X)$ of $P_i$ in $K[X]$. \\
     We claim that $S:=K[X]\backslash(\bigcup_{i=1}^r \bar{P_i})$ is the
     set of non zero divisors in $K[X]$, which is a consequence of
     lemma \ref{lemma} below.\\
     Thus by definition $\Quot(K[X])$ is just $S^{-1}K[X]$, the
     localisation of $K[X]$ at $S$. Knowing, that  the prime ideals 
     in $S^{-1}K[X]$ are exactly those of the form $S^{-1}\bar{P}$ with
     $\bar{P}\in\Spec(K[X])$ and $P\cap S=\emptyset$, we get that the maximal ideals of
     $S^{-1}K[X]$ are precisely
     $S^{-1}\bar{P_1},\ldots,S^{-1}\bar{P_r}$. Moreover,
     $\bigcap_{i=1}^r S^{-1}\bar{P_i}=0$.\\
     By the chinese remainder theorem we thus get:
     \begin{equation}\label{eins} \Quot(K[X])\cong
     \bigoplus_{i=1}^r S^{-1}K[X]/S^{-1}\bar{P_i}.\end{equation}
     We now claim the following
     isomorphisms (resp. equalities):
     \begin{equation}\label{zwei}
       S^{-1}K[X]/S^{-1}\bar{P_i}\cong 
       S_i^{-1}(K[X]/\bar{P_i}) = \Quot(K[X]/\bar{P_i})\cong R(X_i),
     \end{equation}
     where $S_i:=\menge{s+\bar{P_i}|s\in S}$.
     The first isomorphism comes from the well known fact, that localisation
     commutes with taking quotients, the last isomorphism is just theorem
     4.26, since $K[X_i]\cong K[X]/\bar{P_i}$. For the equality in the
     middle, we note that the elements of
     $S_i=\menge{s+P_i|s\in K[X]\backslash\bigcup \bar{P_j}}$
     are non zero divisors in  the integral domain
     $K[X]/\bar{P_i}$. This gives the inclusion ``$\subseteq$''. 
     But now by the first isomorphism $S_i^{-1}(K[X]/\bar{P_i})$
     is a field, and hence is equal to its quotient field, which in
     turn contains $\Quot(K[X]/\bar{P_i})$. Thus we also have ``$\supseteq$''.\\
     Putting (\ref{eins}) and (\ref{zwei}) together and keeping
     proposition 4.24 in mind we finally get:$$\Quot(K[X])\cong
     \bigoplus_{i=1}^r R(X_i)\cong R(X).$$
   \end{proof}

   \begin{lemm}\label{lemma}
     Let $R$ be a noetherian ring, $I$ a radical
     ideal in $R$, and let $P_1\cap\ldots\cap 
     P_r=I$ be the minimal primary decomposition of $I$ (note, this is
     uniquely determined and the $P_i$ are prime ideals, since $I=\sqrt{I}$), then the set of 
     zero divisors in $R/I$ is precisely the set
     $\bar{P_1}\cup\ldots\cup \bar{P_r}$, with $\bar{P_i}=P_i/I$.
   \end{lemm}
   \begin{proof}
     For this note, that if $\bar{a}=a+I\in
     R/I$ is a zero divisor, then there exists an element $b\not\in 
     I$ such that $ab\in I$. Since $I=\bigcap P_i$, there is
     some $i$ such that $b\not\in P_i$, but $ab\in P_i$ and thus $a\in 
     P_i$, since $P_i$ is a prime ideal. Hence $a\in \bigcup P_i$. \\ If
     conversely $a\in\bigcup P_i$, then $a$ belongs to some
     $P_i$. Since the primary decomposition is minimal, there exists
     some $b\in \bigcap_{k\not=i}P_k\backslash I$, and $ab\in
     P_i\cdot \bigcap_{k\not=i}P_k\subset\bigcap_{k=1}^r P_k
     =I$. Hence $\bar{a}=a+I$ is a zero divisor in $R/I$.
   \end{proof}

   \begin{exercise}
     Let $K$ be algebraically closed, $X$ be a variety over $K$ and
     $p\in X$. 
     \begin{enumerate}
        \item Then $p$ has an \emph{affine} neigbourhood.
        \item For each affine neighbourhood $V$ of $p$ we have
          $\ko_{X,p}\cong K[V]_{\textswab{m}_p}.$
%          where $\textswab{m}_p=\menge{f\in K[V]|
%            f(p)=0}=I(\menge{p})$.
     \end{enumerate}
   \end{exercise}
   \begin{proof}
     \begin{enumerate}
        \item 
          If $X\subseteq \P_K^n$ is quasi
          projective. Then $p\in X\cap D(x_i)$ for some $i=1,\ldots, n$,
          which is open in $X$ and quasi affine by exercise 18.
          Thus we may assume, that $X$ is quasi affine, i.\ e.\
          $X=V(I)\backslash V(g_1,\ldots,g_r)=V(I)\cap(D(g_1)\cup\ldots\cup 
          D(g_r))$ for some ideal $I\unlhd
          K[\underline{X}]$ and polynomials $g_1,\ldots,g_r\in
          K[\underline{X}]$. Now, $p\in V(I)\cap D(g_i)\subseteq X$
          for some $i$, which is
          open in $X$ and isomorphic to the affine variety $V(J)$ with
          $J=(I,tg_i-1)\unlhd K[\underline{X},t]$.
        \item If $V$ is an affine neighbourhood of $p$, then $V$ is
          isomorphic to an affine variety $V'$ and we may identify $V$ 
          with $V'$ and $p$ with its image in $V'$. Via this
          identification it makes sense to talk of $K[V]$ and of
          $\textswab{m}_p$. Since any isomorphism between affine
          varieties induces an isomorphism of the corresponding coordinate rings,
          this identification does not depend on the particular choice 
          of $V'$. \\
          Now we have $\ko_{X,p}=\ko_{V,p}$, since $V$ is an open
          neighbourhood of $p$ in $X$. The result then follows by theorem 4.18.
      \end{enumerate}
   \end{proof}

   \begin{exercise}[Regular Functions - Rational Functions]\label{regular}
     Let $K$ be any field, and $X$ be a variety over $K$ with
     irreducible components $X_1, \ldots, X_r$. Fixing a point $p\in
     X$ we get the following diagram of ring homomorphisms:\\
     \unitlength1cm
     \begin{picture}(15,2.2)
       \put(5,1.5){$\ko_X\stackrel{\alpha_p}{\longrightarrow}\ko_{X,p}$}
       \put(6.5,0){$R(X)$}
       \put(5.5,1.0){\begin{turn}{-50}$\hookrightarrow$\end{turn}}
       \put(5.9,0.9){\scriptsize $\alpha$}
       \put(7.1,0.8){\scriptsize $\beta_p$}
       \put(6.8,1.1){\begin{turn}{-90}$\hookrightarrow$\end{turn}}
     \end{picture}
     \\[0.2cm]
     where the maps $\alpha$ and $\alpha_p$ are given by sending a
     regular function $f$  to the corresponding rational function
     $(f,X)$ or to the germ $(f,X)$ at $p$ respectively, and
     where $\beta_p$ is defined by identifying the germ $(f,U)$ of a function
     which is
     regular at $p$ with the rational function
     $(\tilde{f},U_1\cup U_2)$ with $U_1:=U\cap (X\backslash
     (\bigcup_{p\not\in X_i}X_i)$, $U_2:=X\backslash\bigcup_{p\in
       X_i}X_i$, $\tilde{f}_{|U_1}\equiv f_{|U_1}$ and
     $\tilde{f}_{|U_2}\equiv 0$. Note, that $U_1\cup U_2$ is dense in $X$. \\
     For these maps we have the following:
     \begin{enumerate}
        \item $\alpha$ and $\beta_p$ are injective, and
          $\ker(\alpha_p)=\menge{f\in\ko(X)|f_{|X_i}\equiv 0 \mbox{
              whenever } p\in X_i}$.\\
          In particular, if $X$ is irreducible, then $\alpha_p$ is also
          injective.
        \item Note, that the above diagram is \textbf{not} in general
          \textbf{commutative}. However, if $p\in X_i$ for all $i=1,\ldots,r$, then
          the diagram commutes. 
        \item If $X$ is irreducible, $\im(\alpha)=\cap_{q\in
            X}\im(\beta_q)$, and thus,
          identifying $\ko_{X,q}$ with $\im(\beta_q)$, we get:
          $$\alpha:\ko_X\stackrel{\cong}{\longrightarrow}\cap_{q\in X} 
          \ko_{X,q}.$$
     \end{enumerate}
   \end{exercise}
   \begin{proof}
     One easily sees, that $\beta_p$ is welldefined, since any two
     representatives of a germ $(f,U)$ at $p$ coincide on some open
     neighbourhood of $p$.
     \begin{enumerate}
        \item If two germs at $p$ define the same rational function,
          then any two representatives coincide on an open dense subset of $X$, which is
          an open neighbourhood of $p$, since - being 
          continous - they coincide whereever both are defined, in
          particular in $p$. Thus the germs coincide and $\beta_p$ is
          injective.\\
          Analogously, if two regular functions define the same
          rational function, they coincide on an open dense subset of
          $X$, and hence by the identity theorem for morphisms, they
          are the same. This gives the injectivity of $\alpha$.\\
          If $f\in \ko_X$ vanishes constantly on every irreducible
          component containing $p$, then $f$ vanishes on the open
          neighbourhood $X\backslash (\bigcup_{p\not\in X_i}X_i)$ and
          therefore the germ of $f$ at $p$ is zero, i.\ e.\
          $f\in\ker(\alpha_p)$. Conversely, if $f$ is in the kernel of 
          $\alpha_p$, then $f$ vanishes on an open neighbourhood $U$
          of $p$. Since the intersection $U\cap X_i$ is non empty
          whenever $p\in X_i$ and since the $X_i$ are irreducible, $f$ 
          vanishes on an open dense subset of $X_i$ whenever $p\in
          X_i$ and thus on the whole of $X_i$.
        \item Suppose, $p\in X_i$ for all $i=1,\ldots,r$. 
          Given a regular function $f$ on $X$,
          $\alpha_p(f)$ is given by the germ $(f,X)$ and thus
          $\beta_p(\alpha_p(f))$ is given by the rational function
          $(f,X)$, since in its definition we have $U_1=X$ and
          $U_2=\emptyset$. But this $(f,X)$ is just $\alpha(f)$. Thus
          the diagram commutes.\\
          Consider now the example $X=V(xz,yz)\subset \A_K^3$,
          $p=(0,0,1)$ and $f\equiv x\in \ko_X$. Then
          $\alpha_p(f)\equiv 0$, and thus $\beta_p(\alpha_p(f))\equiv
          0$. However, $\alpha(x)$ is non zero on $D(x)\cap X\not=\emptyset$, which 
          is open in $X$, thus $\alpha(x)\not\equiv 0$.
        \item By part b. it follows, that $\im(\alpha)\subseteq
          \bigcap_{q\in X}\ko_{X,q}$. If conversely we have a rational
          function, that belongs to each $\ko_{X,q}$, then it is
          regular in each point of $X$ and thus regular on $X$.
     \end{enumerate}
   \end{proof}

\end{document} 
