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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 17th January, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.3cm}

   \begin{center}
     \LARGE \bf Introduction to Algebraic Geometry \\
   \end{center}
   \vspace{0.3cm}
   \normalsize
   \parindent0cm

   \setcounter{exercise}{37}

   First of all we recall the definition of a space with functions:
   \begin{defin}\label{raummitfunktionen}
      Let $K$ be a field.
      A space with functions over $K$ is a topological space $X$
      together with the assignment of a $K$-algebra $\ko_X(U)$ of
      $K$-valued functions (called the regular functions on $U$) 
      to each open subset $U\subseteq X$ such
      that the following three conditions are satisfied:
      \begin{enumerate}
         \item[(i)] For $f\in\ko_X(U)$ and $V\subseteq U\subseteq X$
           open, $f_{|V}\in\ko_X(V)$. (We say, $\ko_X$ is a
           presheaf on $X$.)
         \item[(ii)] Given an open covering $U=\bigcup_{i\in I}U_i$ of 
           the open set $U\subseteq X$ and $f_i\in\ko_X(U_i)$ with
           $f_{i|U_{ij}}=f_{j|U_{ij}}$ for all $i,j\in I$, where
             $U_{ij}=U_i\cap U_j$, then there exists a (unique)
             $f\in\ko_X(U)$ such that $f_{|U_i}=f_i$ for all $i\in
             I$. (We call this the sheaf axiom and say, $\ko_X$ is 
             the structure sheaf on $X$.)
           \item[(iii)] For any $f\in\ko_X(U)$ with $U\subseteq X$
             open the set $D(f):=\menge{x\in U|f(x)\not=0}$ is open in 
             $U$ and $\frac{1}{f}\in\ko_X(D(f))$.
      \end{enumerate}
   \end{defin}

   Let us then define the structure sheaf on a topological subspace
   $A\subseteq X$.

   \begin{defin}\label{unterraum}
     Given a space with functions $(X,\ko_X)$ over the field $K$ and a 
     subset $A\subseteq X$ equipped with the subspace topology. For an 
     open subset $U\subseteq A$ we define:
     \begin{displaymath}
       \begin{array}{rcc}
        \ko_A(U) & = & \left\{f:U\rightarrow
        \A_K^1| \forall P\in U\;\exists\; P\in
       V_P\subseteq X \mbox{ open and }\tilde{f}\in\ko_X(V_P):  \right.\\ 
       & &\left.  f_{|V_P\cap
       U}=\tilde{f}_{|V_P\cap U} \right\}.
      \end{array}
     \end{displaymath}
     Note, if $A\subseteq X$ is open, then $\ko_A(U)=\ko_X(U)$ for
     $U\subseteq A$ open by the sheaf axiom on $\ko_X$.
   \end{defin}

   Next we will do some topology.
   \begin{defin}
     Let $X$ be a topological space, $Y\subseteq X$. \\
     We say $Y$ is \textbf{locally closed} in $X$ iff $Y=A\cap U$ for
     some closed subset $A\subseteq X$ and some open subset
     $U\subseteq X$.
   \end{defin}

   \begin{lemm}
     Let $X$ be a topological space, $Y\subseteq X$. Then the
     following are equivalent:
     \begin{enumerate}
        \item $Y$ is locally closed.
        \item $Y$ is closed in some open subspace $U\subseteq X$ of $X$.
        \item $Y$ is open in some closed subspace $A\subseteq X$ of
          $X$.
        \item $Y$ is open in its closure $\overline{Y}\subseteq X$ in
          $X$.
        \item Any point $P\in Y$ has an open neighbourhood
          $V_P\subseteq X$ in $X$ such that $Y\cap V_P$ is closed in
          $V_P$. 
     \end{enumerate}
   \end{lemm}
   \begin{proof}
     The equivalence of a., b. and c. is immediate from the definition 
     of subspace topology. \\
     Using $\overline{Y}$ as $A$, d. implies c. For the implication
     $(c.\Rightarrow d.)$ let $A\subseteq X$ be closed with
     $Y\subseteq A$ open. Since $A$ is closed and contains $Y$, the
     closure $\overline{Y}$ of $Y$ is contained in $A$ and closed
     therein. Now, $\overline{Y}\backslash Y= \overline{Y}\cap
     (A\backslash Y)$, which is closed in $A$ as the intersection of
     two closed subsets. But then it is also closed in $\overline{Y}$, 
     which is a subspace of $A$.
     Using $U$ as $V_P$ for any $P\in Y$, b. implies e., and using
     $\bigcup_{P\in Y}V_P$ as $U$, e. implies b.
   \end{proof}

   \begin{exercise}\label{unterraeume}
     Let $K$ be an algebraically closed field. 
     \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, 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 spaces with functions over $K$, then $f\times g:X\times
          X'\rightarrow Y\times Y'$ is a morphism of spaces with functions over $K$.
        \item If $f:X\rightarrow Y$ is a morphism of spaces with functions over
          $K$ and
          $Y'\subseteq Y$ is a subspace with $f(X)\subset Y'$, then
          $f:X\rightarrow Y'$ is a morphism.\\
          If $X'\subseteq X$ is a subspace 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{proof}
     \begin{enumerate}
        \item We now want to show, that $(A,\ko_A)$ is a space with
          functions. \\
          It is obvious, that for any open subset $U\subseteq A$ the set 
          of functions $\ko_A(U)$ becomes a $K$-algebra by the usual
          additon, multiplication, and scalar multiplication of
          $K$-valued functions. {\small (Given $k\in K$ and $f,g\in\ko_A(U)$ represented
          locally in $P\in U$ by $\tilde{f}\in\ko_X(V_P)$ and
          $\tilde{g}\in\ko_X(V'_P)$, then $f+g$, $f\cdot g$ and
          $k\cdot f$ are locally represented by $\tilde{f}_|
          +\tilde{g}_|\in\ko_X(V_P\cap V'_P)$,
          $\tilde{f}_|\cdot\tilde{g}_|\in\ko_X(V_P\cap V'_P)$ and
          $k\cdot \tilde{f}\in\ko_X(V_P)$ respectively, and thus they
          are of the desired
          form.)}\\
          Moreover it is clear, that restricting a function
          $f\in\ko_A(U)$ for an open subset $U\subseteq A$ to an open
          subset $V\subseteq U$ preserves the locally defined
          structure of $f$, i.\ e.\ $f_{|V}\in\ko_A(V)$. This gives
          condition (i) in definition \ref{raummitfunktionen}.\\
          To show (ii), let $U=\bigcup_{i\in I}U_i$ be an open
          covering of the open subset $U\subseteq X$ and let 
          $f_i\in\ko_A(U_i)$ with
          $f_{i|U_{ij}}=f_{j|U_{ij}}$ for all $i,j\in I$. Defining
          $f:U\rightarrow \A_K^1$ by $f(P)=f_i(P)$ if $P\in U_i$ gives 
          a well defined function. For $P\in U_i\subseteq U$ we find
          some open neighbourhood $V_P\subseteq X$ of $P$ in $X$ (w.\
          l.\ o.\ g.\ $V_P\cap U\subseteq U_i$, i.\ e.\ $V_P\cap
          U_i=V_P\cap U$) and a 
          regular function $\tilde{f}\in\ko_X(V_P)$ such that
          $\tilde{f}_{|V_P\cap U}=\tilde{f}_{|V_P\cap
            U_i}=f_{i|V_P\cap U_i}=f_{|V_P\cap U}$, 
          thus $f$  is of the desired
          form.\\
          Knowing, that a regular function $f\in\ko_A(U)$ for
          $U\subseteq A$ open is continous (since it is locally
          continous), we get $D(f)$ is open in
          $U$. On $D(f)$ the function $\frac{1}{f}$ is
          then defined, and if $f$ locally in $P\in D(f)$ is given by
          $\tilde{f}\in\ko_X(V_P)$, then $D(\tilde{f})$ is an open
          neighbourhood of $P$ in $X$,
          $\frac{1}{\tilde{f}}\in\ko_X(D(\tilde{f}))$ and
          $\frac{1}{f}_{|D(f)\cap
            D(\tilde{f})}=\frac{1}{\tilde{f}}_{|D(f)\cap
            D(\tilde{f})}$. Thus once more $\frac{1}{f}$ is 
          of the desired form, i.\ e.\
          $\frac{1}{f}\in\ko_A(D(f))$.
          This finishes the proof.
        \item The arguments for the projective and the affine case are 
          identical, so we may restrict ourselves to
          the projective case. \\
          Let now $U\subset A$ be open. 
          $$
          \begin{array}{ccl}
             f\in\ko_A(U) & \Leftrightarrow & 
             \forall P\in U\;\exists\; P\in V_P \subseteq \P^n_K\mbox{
               open and } \tilde{f}\in\ko_{\P^n_K}(V_P):\\
             & & \tilde{f}_{|V_P\cap
               U}=f_{|V_P\cap U} \\
             & \Leftrightarrow & 
             \forall P\in U\;\exists\; P\in V'_P \subseteq \P^n_K\mbox{
               open and } g,h\in K[\underline{X}]  \\
             & &
               \mbox{homogeneous of the same
               degree }:\frac{g}{h}=f \mbox{ on } V'_P\cap U\\
             & \Leftrightarrow & 
             \forall P\in U\;\exists\; P\in U_P \subseteq A\mbox{
               open and } \bar{g},\bar{h} \in K[A] \\
             & & \mbox{homogeneous of the same
               degree }:\frac{\bar{g}}{\bar{h}}=f \mbox{ on } U_P\\
             & \Leftrightarrow & f \mbox{ is regular on } U.
          \end{array}
          $$
        \item If $(X,\ko_X)$ is a variety, then $X$ has an open cover
          $X=X_1\cup\cdots\cup X_r$ by affine varieties
          $(X_i,\ko_{X_i})$ (where $\ko_{X_i}(U)=\ko_X(U)$ for
          $U\subseteq X_i$ open). For a closed subset $A$ we get
          $A_i:=A\cap X_i$ is a closed subset of $X_i$. Thus
          $(A_i,\ko_{A_i})$ is isomorphic to a closed subset of some
          $\A^n_K$. By b. the space with functions
          $(A_i,\ko_{A_i})$ is thus isomorphic to a classical affine
          variety and is hence an affine variety. Therefore
          $A=A_1\cup\ldots\cup A_r$ is a variety.\\
          Since for an open subspace $Y$ of a space with funtions $X$
          we have $\ko_Y(U)=\ko_X(U)$ for any $U\subseteq Y$ open, we
          get in the case where $A$ is open, that the $(A_i,\ko_{A_i})$ defined as 
          above are quasi affine, and thus $X$ is a variety as the
          finite union of quasi affine varieties. \\
          Thus if $(A,\ko_A)$ is open in some closed subspace
          $(A',\ko_{A'})$, then again $(A,\ko_A)$ is a variety.
        \item We know, that the maps $f\circ\pi_X:X\times X'\rightarrow
          Y$ and $g\circ\pi_{X'}:X\times X'\rightarrow Y'$ are
          morphisms. Hence by the universal property of the product
          (see exercise 32), the map $f\times g =
          (f\circ\pi_X,g\circ\pi_{X'})$ is a morphism.
        \item It is clear, that $f:X\rightarrow Y'$ is
          continous. Hence for this to be a morphism, 
          it remains to show, that $f^*\ko_{Y'}(U)\subseteq
          \ko_X(f^{-1}(U))$ for $U\subset Y'$ open. Let
          $h\in\ko_{Y'}(U)$ such that locally in $P\in U$ $h$ is given 
          by some $\tilde{h}\in\ko_Y(V_P)$, then
          $f^*h_{|f^{-1}(V_P)\cap
            f^{-1}(U)}=f^*\tilde{h}_{|f^{-1}(V_P)\cap f^{-1}(U)}$
          where $f^*\tilde{h}\in \ko_X(f^{-1}(V_P))$, thus locally on $f^{-1}(U)$
          $f^*h$ regular, and hence
          regular on $f^{-1}(U)$ by the sheaf axiom on $\ko_X$. \\
          Again it is clear, that $f_|:X'\rightarrow Y$ is
          continous. Moreover, if $h\in\ko_Y(U)$ for some open subset
          $U\subseteq Y$, then $f_|^*h=(f^*h)_{|X'\cap f^{-1}(U)}$,
          which is obviously regular as the restriction of a regular
          function (see definition \ref{unterraum}). Thus $f_|$ is a
          morphism. 
        \item Let $Y=Y_1\cup\cdots\cup Y_r$ be an open covering of
          $Y$ by affine subvarieties. Set $V_i:=f^{-1}(Y_i)\subseteq
          X$ open, and $f_i:=f_{|V_i}:V_i\rightarrow Y_i$ (a morphism!). \\
          Then $\Gamma_f=\bigcup_{i\in I}\Gamma_{f_i}$. Now, since
          $Y_i$ is affine, it is separated, and thus
          $\Gamma_{f_i}\subseteq V_i\times Y_i$ is closed, where
          $V_i\times Y_i$ is open in $X\times Y$. Set $U=\bigcup_{i\in 
            I}V_i\times Y_i\subseteq X\times Y$ open. It suffices to
          show, that $\Gamma_f\subseteq U$ is closed, and for this it
          is sufficient, that $\Gamma_f\cap (V_i\times Y_i)$ is closed 
          in $V_i\times Y_i$ for each $i\in I$, since the latter form
          an open covering of $U$. But $\Gamma_f\cap (V_i\times
          Y_i)=\Gamma_{f_i}$, thus we are done. 
        \item By the universal property of the product,
          $(\id_X,f):X\rightarrow X\times Y$ is a morphism, and thus
          $\gamma_f:X\rightarrow \Gamma_f :x\mapsto (x,f(x))$ is a
          morphism by part e. Moreover, by part f. $\Gamma_f$ is a
          subvariety of $X\times Y$, and thus the restriction of the
          morphism $\pi_X:X\times Y\rightarrow X$ to $\Gamma_f$ is a
          morphism (also by part e.). It is immediate, that $\pi_{X|}$
          is the inverse of $\gamma_f$. 
     \end{enumerate}
   \end{proof}

   \begin{theor}\label{isomorphismus}
     Let $f:X\rightarrow Y$ be some morphism of spaces with functions
     over $K$. The following are equivalent:
     \begin{enumerate}
        \item $f$ is an isomorphism.
        \item $f$ is a homeomorphism and
          $f^*:\ko_Y(i(U))\rightarrow\ko_X(U)$ is surjective for
          any $U\subseteq X$ open.
        \item $f$ is homeomorphism and
          $f^*:\ko_{Y,i(x)}\rightarrow\ko_{X,x}$ is surjective for all 
          $x\in X$.
     \end{enumerate}
   \end{theor}

   \begin{proof}
     \begin{description}
        \item[$b.\Rightarrow a.$] Given $f\in\ko_X(U)$ for some open
          set $U\subseteq X$. We have to show that $f\circ
          i^{-1}\in\ko_Y(i(U))$.
          Since $i^*:\ko_Y(i(U))\rightarrow \ko_X(U)$ is surjective,
          there exists $g\in\ko_Y(i(U))$ with $g\circ i=f$,  and thus
          $f\circ i^{-1}=g\circ i\circ i^{-1}=g$ is regular. 
          Thus $i^{-1}$ is a morphism.
        \item[$c.\Rightarrow b.$] Given $U\subseteq X$ open and
          $f\in\ko_X(U)$. We have to show that there exists a
          $g\in\ko_Y(i(U))$ such that $g\circ i=f$.\\
          For all $x\in U$ we denote by $f_x\in\ko_{X,x}$ the germ of $f$
          in $x$. Then there exists a $g_{i(x)}\in\ko_{Y,i(x)}$ such
          that $i^*g_{i(x)}=f_x$. We now define $g:i(U)\rightarrow
          \A_K^1$ by $g(y)=g_y(y)$. Then $g\circ i^{-1}=f$ and thus
          $f\circ i = g$. It remains to show, that $g$ is regular on
          $i(U)$, which is a local question. Given any point $y\in i(U)$,
          there exists an open neighbourhood $V_y\subseteq
          Y$ of $y$ and $\tilde{g}\in\ko_Y(V_y)$ such that
          $\tilde{g}\circ i =f$ on $i^{-1}(V_y)$ (that is the
          definition of $i^*g_y=f_{i(y)}$). But then
          $\tilde{g}_{|V_y\cap i(U)}=g_{|V_y\cap i(U)}$, which means that
          $g$ is locally regular, hence regular.
        \item[$a.\Rightarrow c.$] If $i^{-1}$ exists and is a morphism, then of
          course $i$ is a homeomorphism and $i^*$ is bijective with
          the inverse $(i^{-1})^*$, since composition by $i$ and
          $i^{-1}$ cancel each other.
      \end{description}
   \end{proof}

   \begin{defin}
     Let $(X,\ko_X)$ and $(Y,\ko_Y)$ be spaces with functions. A
     morphism $i:X\rightarrow Y$ is said to be an \textbf{embedding}
     iff $i$ induces an isomorphism $(X,\ko_X)\rightarrow
     (\im(i),\ko_{\im(i)})$, where $\ko_{\im(i)}$ denotes the subspace 
     structure.
   \end{defin}

   \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{proof}
     \begin{description}
        \item[$a.\Leftrightarrow b.$] Definition of a morphism. 
        \item[$a.\Leftrightarrow c.\Leftrightarrow d.$] Follows from
          theorem \ref{isomorphismus}.
     \end{description} 
   \end{proof}


\end{document} 
