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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 16th February, 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}{44}

   \begin{exercise}
     Let $K$ be an algebraically closed field, $X$ a variety over $K$
     and $x\in X$ a regular point.\\
     Show that there is a canonical surjective map
     $$\alpha:\Der_K(\ko_{X,x})\rightarrow \Der_K(\ko_{X,x},K)=T_xX$$ 
     which is as well $\ko_{X,x}$-linear as $K$-linear.\\
     Moreover, show that this induces an isomorphism
     $$\Der_K(\ko_{X,x})/\m_x\Der_K(\ko_{X,x})\cong T_xX.$$
     Is this still true if $x$ is a singular point of $X$?
   \end{exercise} 
   \begin{proof}
     The proof uses the fact that the $\ko_{X,x}$-module of differential
     forms on $\ko_{X,x}$ is free. For the case where $X=\A_K^n$, we
     can give a more down to earth proof. 
     \begin{description}
     \item[\underline{$X=\A_K^n$:}] In this case, $\Der_K(\ko_{\A_K^n,x})
       =\bigoplus_{i=1}^n\ko_{\A_K^n,x}\cdot \frac{\partial}{\partial 
         X_i}$. To see this, note that by the product rule a
       derivation on $\ko_{\A_K^n,x}$ is fixed by its values on the
       coordinates $X_i$, i.\ e.\
       $\delta=\delta(X_1)\frac{\partial}{\partial X_1}+\ldots +
       \delta(X_n)\frac{\partial}{\partial X_n}$. Hence
       $\Der_K(\ko_{\A_K^n,x})$ is generated by
       $\menge{\frac{\partial}{\partial X_i}|i=1,\ldots,n}$ as
       $\ko_{\A_K^n,x}$-module. Moreover,
       $\Der_K(\ko_{\A_K^n,x},K)$ is an n-dimensional vector space
       with basis $\menge{\frac{\partial}{\partial
       X_i}_{|x}|i=1,\ldots, n}$. The map
       $$\alpha:\Der_K(\ko_{\A_K^n,x})\rightarrow\Der_K(\ko_{\A_K^n,x},K): 
       \delta\mapsto\delta_{|x},$$ sending a derivation to its point
       derivation is obviously $\ko_{\A_k^n,x}$-linear (and as well
       $K$-linear). \\
       As we have seen, the images of the $\frac{\partial}{\partial
         X_i}$ give a $K$-basis of $\Der_K(\ko_{\A_K^n,x},K)$, so that 
       $\alpha$ is surjective. Moreover,
       $\m_x\Der_K(\ko_{\A_K^n,x})$ is in the kernel of
       $\alpha$, so that we get an induced $K$-linear map 
       $$\bar{\alpha}:\Der_K(\ko_{\A_K^n,x})/\m_x\Der_K(\ko_{\A_K^n,x}) 
       \rightarrow\Der_K(\ko_{\A_K^n,x},K).$$
       But now the residue classes of the $\frac{\partial}{\partial
         X_i}$ generate the first $K$-vector space, so that its
       dimension is at most $n$, and thus, mapping epimorphic to an
       $n$-dimensional vector space, its dimension must be $n$ and the 
       epimorphism an isomorphism.
     \item[\underline{General case:}] \emph{One way to define the $B$-module 
       $\Omega_{B/A}$
       of differential forms on $B/A$ for any $A$-algebra $B$ is by
       a universal property, namely that we have an $A$-derivation
       $d:B\rightarrow \Omega_{B/A}$ such that for any $B$-module $M$
       $d$ induces an isomorphism of $B$-modules:
       $$ \Hom_B(\Omega_{B/A},M)\stackrel{\cong}{\longrightarrow}\Der_A(B,M):
       f\mapsto f\circ d.$$ (I.~e.~the two functors
       $\Hom_B(\Omega_{B/A},\; .\;)$ and $\Der_A(B,\; .\;)$ are naturally
       equivalent via the natural transformation $\circ d$.)} 
       For the existence of this module see
       e.~g.~Hartshorne, II.8. In our case, $A=K$ and
       $B=\ko_{X,x}$. Then the result of theorem II.8.8 in Hartshorne applies,
       which states that \emph{$\Omega_{\ko_{X,x}/K}$ is a free module of
       rank $\dim(X,x)$, since $\ko_{X,x}$ is regular - the converse,
       by the way, is also true}. \\
       We now look at the following exact sequence of
       $\ko_{X,x}$-modules: 
       $$
       0\rightarrow\m_{X,x}\rightarrow\ko_{X,x}
       \rightarrow\ko_{X,x}/\m_{X,x}\cong K\rightarrow 0.
       $$ 
       Applying the functor $\Hom_{\ko_{X,x}}(\Omega_{\ko_{X,x}/K},\; .\;)$ to the
       sequence we get the following exact sequence:
       $$
       0\rightarrow\Hom_{\ko_{X,x}}(\Omega_{\ko_{X,x}/K},\m_{X,x})
       \rightarrow\Hom_{\ko_{X,x}}(\Omega_{\ko_{X,x}/K},\ko_{X,x})\rightarrow
       $$
       $$
       \rightarrow\Hom_{\ko_{X,x}}(\Omega_{\ko_{X,x}/K},K)\rightarrow
       \Ext_{\ko_{X,x}}^1(\Omega_{\ko_{X,x}/K},\m_{X,x})=0,$$
       where the last $\ko_{X,x}$-module is zero, since
       $\Omega_{\ko_{X,x}/K}$ is free. But now we can equally apply the
       functor $\Der_K(\ko_{X,x},\; .\;)$ to the above sequence:
       $$
       0\rightarrow\Der_K(\ko_{X,x},\m_{X,x})
       \rightarrow\Der_K(\ko_{X,x},\ko_{X,x})
       \stackrel{\alpha}{\longrightarrow}\Der_K(\ko_{X,x},K)\rightarrow 0.
       $$       
       Using then 
       the natural transformation $\circ d$ we get that this sequence
       is also exact, which shows that the ``point derivation map'' $\alpha$ is
       surjective. \\
       Moreover, since $\Omega_{\ko_{X,x}/K}$ is a free
       $\ko_{X,x}$-module of rank $\dim(X,x)$, the same is true for
       $\Hom_{\ko_{X,x}}(\Omega_{\ko_{X,x}/K},\ko_{X,x})$ and thus for 
       $\Der_K(\ko_{X,x})$. Then we may apply the same arguments as in 
       the case $X=\A_K^n$ to see that the kernel of $\alpha$ is
       indeed $\m_{X,x}\Der_K(\ko_{X,x})$, since $\Der_K(\ko_{X,x},K)$ 
       is a $K$-vector space of dimension $\dim(X,x)$.
     \item[\underline{Singular case:}] In the singular case this
       statement is not in general true.\\
       \textbf{Claim:} \emph{If $X\subseteq\A_K^n$ is affine, then any derivation
       $\bar{\delta}$ of
       $\ko_{X,x}$ is induced by a derivation $\delta$ of
       $\ko_{\A_K^n,x}$ with $\delta(I(X)\ko_{X,x})\subseteq
       I(X)\ko_{X,x}$. \\ In particular, 
       $\Der_K(\ko_{X,x})$ is a finitely generated $\ko_{X,x}$-module.} \\
        (Let
       $a_i\in\ko_{\A_K^n,x}$ be any representative of
       $\bar{\delta}(X_i)$, then $\delta=\sum_{i=1}^n
       a_i\cdot\frac{\partial}{\partial X_i}$ induces
       $\bar{\delta}$. For this it suffices to see, that
       $\delta(I(X)\ko_{X,x})\subseteq I(X)\ko_{X,x}$. But for $f\in I(X)$ we have by
       the product rule
       $0=\bar{\delta}(\bar{f}) = \sum_{i=1}^n \bar{a}_i\overline{\frac{\partial 
         f}{\partial X_i}}=\overline{\sum_{i=1}^n
       a_i\frac{\partial f}{\partial
         X_i}}=\overline{\delta(f)}$, and for
       $\frac{f}{g}\in I(X)\ko_{X,x}$ we use the quotient rule. The
       ``in particular'' follows since $\Der_K(\ko_{\A_K^n,x})$ is
       a finitely generated $\ko_{\A_K^n,x}$-module.) \\ 
       \textbf{Example:}
       Let $X=\menge{X^2-Y^2=0}\subseteq\A_K^2$ and $x=0$. Then
       $\Der_K(\ko_{X,0},K)=T_0X=T_0\A_K^2$. On the other hand any
       derivation on $\ko_{X,0}$ is induced by a $\delta$ on $\ko_{\A_K^2,0}$ 
       with $2(X\delta(X)-Y\delta(Y))=\delta(X^2-Y^2)\in
       (X^2-Y^2)$. This gives $\delta(X),\delta(Y)\in (X,Y)$, and thus 
       $\delta_{|0}=\delta(X)(0)\cdot\frac{\partial}{\partial X}_{|0}+ 
       \delta(Y)(0)\cdot\frac{\partial}{\partial Y}_{|0}=0$. That
       means, the map sending a derivation to its point
       derivation at zero is the zero map. Since the tangent space of $X$ at
       $0$ is not zero, this map is certainly not surjective. \\
       How about the kernel? Since $\Der_K(\ko_{X,0})$ is a finitely
       generated $\ko_{X,0}$-module and not the zero module, by
       Nakayama's lemma $\m_0\Der_K{\ko_{X,0}}$ cannot be equal to
       $\Der_K(\ko_{X,0})$, where the latter is the kernel of the above 
       map. 
     \end{description}
   \end{proof}

   \newpage

   \begin{exercise}
     Let $K$ be an algebraically closed field, $X$ and $Y$ varieties over
     $K$, $x\in X$, $y\in Y$.
     Show that $$T_{(x,y)}(X\times Y)\cong T_xX\oplus T_yY.$$
   \end{exercise}
   \begin{proof}
     Passing to open affine neighbourhoods of $x$ and $y$ we may
     assume that $X\subseteq\A_K^n$ and $Y\subseteq\A_K^m$ are
     affine and given by $I(X)=(f_1,\ldots,f_r)\unlhd
     K[\underline{X}]$ resp.\ $I(Y)=(g_1,\ldots,g_s)\unlhd
     K[\underline{Y}]$. \\
     Then by lemma 5.11: $$I(X\times
     Y)=I(X)K[\underline{X},\underline{Y}]+I(Y)K[\underline{X},\underline{Y}]  
     = (f_1,\ldots,f_r,g_1,\ldots,g_s)\unlhd
     K[\underline{X},\underline{Y}]$$
     and thus
     $$X\times
     Y=V(f_1,\ldots,f_r,g_1,\ldots,g_s)\subseteq\A_K^n\times\A_K^m.$$ 
     By lemma 7.4 of the lecture we know that the tangent space to an
     affine variety at a certain point is isomorphic to the kernal of the 
     Jacobian matrix (at this point) of any set of generators of the
     vanishing ideal. Thus we finally get:
     $$T_{(x,y)}(X\times Y)\cong
     \ker\left(\begin{array}[c]{cccccc}\frac{\partial f_1}{\partial x_1}(x)
     & \cdots & \frac{\partial f_1}{\partial x_n}(x) & 0 & \cdots & 0 \\
     \vdots & & \vdots & \vdots & & \vdots \\
     \frac{\partial f_r}{\partial x_1}(x)
     & \cdots & \frac{\partial f_r}{\partial x_n}(x) & 0 & \cdots & 0 \\
     0 & \cdots & 0 &\frac{\partial g_1}{\partial y_1}(y)
     & \cdots & \frac{\partial g_1}{\partial y_m}(y) \\
     \vdots & & \vdots & \vdots & & \vdots \\
     0 & \cdots & 0 &\frac{\partial g_s}{\partial y_1}(y)
     & \cdots & \frac{\partial g_s}{\partial y_m}(y) \\
     \end{array}\right) \cong 
     $$
     $$
     \ker\left(\begin{array}[c]{ccc}\frac{\partial f_1}{\partial x_1}(x)
     & \cdots & \frac{\partial f_1}{\partial x_n}(x) \\
     \vdots & & \vdots \\
     \frac{\partial f_r}{\partial x_1}(x)
     & \cdots & \frac{\partial f_r}{\partial x_n}(x) 
     \end{array}\right)\oplus
     \ker\left(
     \begin{array}[c]{ccc}
     \frac{\partial g_1}{\partial y_1}(y)
     & \cdots & \frac{\partial g_1}{\partial y_m}(y) \\
     \vdots & & \vdots \\
     \frac{\partial g_s}{\partial y_1}(y)
     & \cdots & \frac{\partial g_s}{\partial y_m}(y) \\
     \end{array}\right) \cong
     T_xX\oplus T_yY.
     $$
   \end{proof}

   \begin{lemm}[Vermeidungslemma für Primideale]
     Let $R$ be a ring, $P_1,\ldots,P_k\in\Spec(R)$ and $I,I'\unlhd R$ 
     such that $I\not\subseteq I'$ and $I\not\subseteq P_i$ 
     ($i=1,\ldots,k$). 
     Then $I\not\subseteq I'\cup P_1\cup\cdots\cup P_k$.
   \end{lemm}
   \begin{proof}
     W.~l.~o.~g.~the ideals $P_1,\ldots,P_k$ and $I'$ are pairwise not 
     contained in each other, elsewise we may just drop some of them
     (maybe including $I'$).\\
     We show by induction on $k$ that there exists an $x\in I$ not
     contained in $I'$ or in any of the $P_i$.\\
     The induction base $k=0$ (i.~e.~no $P_i$) is trivial, thus we
     may assume $k\geq 1$ and the statement is true for $k-1$ prime
     ideals. 
     Choose elements $r\in I\backslash P_k$, $s\in I'\backslash P_k$
     and $s_i\in P_i\backslash P_k$. Then $x:=r\cdot s\cdot s_1\cdots
     s_{k-1} \in I\cap I'\cap P_1\cap\cdots\cap P_{k-1}$, but
     $x\not\in P_k$.  By
     induction we have an element $y\in I\backslash (I'\cup P_1\cup
     \cdots \cup P_{k-1})$. If $y\not\in P_k$, then we are
     done. Otherwise $x+y\not\in P_k$, since $x$ is not, and
     $x+y\not\in I'\cup P_1\cup\ldots\cup P_{k-1}$, since $y$ is in
     none of them, whereas $x$ is contained in each one. But $x+y\in
     I$, and thus this is the wanted element.
     
   \end{proof} 

\end{document}  
