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   \Blatt{3}
   \Abgabe{19/11/2001}

   \Nr
   Let $S\subset T$ be ring extension such that $T$ is
   integral over $S$ and finitely generated as a ring over $S$, then
   $T$ is finite over $S$, i.~e.~finitely generated as a module over
   $S$. 

   \Nr
   \begin{enumerate}
   \item Let $K$ be an infinite field and let $f,g\in
     K[x_1,\ldots,x_n]$. Show that if $f(p)=g(p)$ for all $p\in K^n$,
     then $f=g$.
   \item Show that $\A_\R^1$ is a dens subset of $\A_\C^1$.
   \end{enumerate}
   
   \Nr
   Let $K$ be an algebraically closed field and let $V\subseteq\A_K^n$
   be an affine variety. Show that the following statements are
   equivalent: 
   \begin{enumerate}
   \item $|V|=r< \infty$.
   \item $I(V)=\bigcap_{i=1}^r\m_i$, where the $\m_i\subset
     K[x_1,\ldots,x_n]$ are maximal ideals.
   \item $K[V]\cong K^r$, as a $K$-algebras.
   \item $\dim_K\big(K[V]\big)=r<\infty$.
   \end{enumerate}


   \Nr
   Let $K$ be an algebraically closed field, and $V=V(f)$ with $f\in K[x_1,\ldots,x_n]$ be
   a hypersurface in $\A_K^n$. Suppose, moreover, that
   $f=f_1^{\nu_1}\cdots f_r^{\nu_r}$ is the prime 
   factorisation of $f$, i.~e.~$f_i$ is irreducible for
   $i=1,\ldots,r$ and $f_i\not|\; f_j$ for $i\not=j$. Show:
   \begin{enumerate}
   \item $I(V)=(f_1\cdots f_r)$.
   \item $V$ is irreducible if and only if $r=1$.
   \item $V\big(y^2-x^2-x^3\big)$ is irreducible.
   \item Do the statements remain true if we omit the assumption
     $K=\overline{K}$?
   \end{enumerate}


   \newpage

   We here gather some results from commutative algebra which might be
   very useful for solving the above exercises.

   \begin{theorem}[Eisenstein Criterion]
     Let $R$ be an integral domain, $P\in\Spec(R)$ and $f=\sum_{i=0}^n a_ix^i\in R[x]$
     such that $a_n\not\in P$, $a_0,\ldots,a_{n-1}\in P$ and
     $a_0\not\in P^2$, then $f$ possesses no non-constant factors of
     degree less than $\deg(f)$.
   \end{theorem}
   \begin{proof}
     Cf.~G.~Scheija and U.~Storch, Lehrbuch der Algebra, Teil 2,
     Stuttgart 1988, Lemma 60.11.
   \end{proof}

   \begin{theorem}[Chinese Remainder Theorem]
     Let $R$ be a commutative ring with one and $I_1,\ldots,I_r\lhd R$
     pairwise coprime ideals (i.~e.~$I_i+I_j=R$ for $i\not=j$). Then:
     \begin{enumerate}
     \item[(a)\;\;] $I_1\cap \ldots\cap I_r= I_1\cdots I_r$.
     \item[(b)\;\;] $R/(I_1\cap\ldots\cap I_r)\rightarrow \bigoplus_{i=1}^r
       R/I_i\;:\; x (\mod I_1\cap \ldots\cap I_r)\mapsto \big(x(\mod
       I_1),\ldots, x(\mod I_r)\big)$ is an isomorphism of rings. 
     \end{enumerate}
   \end{theorem}
   \begin{proof}
     Cf.~M.~Atiyah and I.~MacDonald, Introduction to Commutative
     Algebra, London 1969, Proposition 1.10.
   \end{proof}

\end{document}


