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\begin{document} 
 
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   \large UNIVERSITY OF  KAISERSLAUTERN \hfill  Winterterm 1997/98 \\
   Mathematical Department \hfill 15th December, 1997 \\
   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{lemm}\label{artinian-noetherian}
     Let $R$ be a ring where some product of finitely many (not necessarily
     distinct) maximal ideals is zero. Then the $R$ is artinian iff
     $R$ is noetherian.
   \end{lemm}
   \begin{proof}
          Let $\textswab{n}_1,\ldots,\textswab{n}_s$ be maximal ideals 
          with
          $\textswab{n}_1\cdots\textswab{n}_s=0$. To show the claim, 
          it suffices (by the tower law) to show, that for
          $R/\textswab{n}_1$ and for all
          $\textswab{n}_1\cdots\textswab{n}_l/\textswab{n}_1\cdots\textswab{n}_{l+1}$ 
          ($l=1,\ldots,s-1$) the notions of artinian and noetherian
          coincide. 
          But $R/\textswab{n}_1$ 
          is a field (hence artinian and noetherian), and
          $\textswab{n}_1\cdots\textswab{n}_l/\textswab{n}_1\cdots\textswab{n}_{l+1}$ 
          is a vector space over the field
          $R/\textswab{n}_{l+1}$ (and for
          vectorspaces being artinian or noetherian both coincide with
          being finite dimensional).     
   \end{proof}

%   \begin{theore}\label{artinian}
%     For a ring $R$ the following are equivalent:
%     \begin{enumerate}
%        \item $R$ is artinian.
%        \item $R$ is noetherian and $\dim(R)=0$.
%     \end{enumerate}
%   \end{theore}
%   \begin{proof}
%     \begin{description}
%        \item[a.$\Rightarrow$b.:] By the theorem of Hopkins (c.\ f.\
%          Aufgabe 19, Algebraische Geometrie SS '97) every artinian
%          ring is noetherian. It remains to show, that every prime
%          ideal is maximal. If $P\lhd R$ is a prime ideal, then $R/P$
%          is an integral domain. Let $0\not=\bar{x}\in R/P$. By the
%          descending chain condition on $R/P$ we find some $n\in\N$
%          such that $(\bar{x}^n)=(\bar{x}^{n+1})$. Thus
%          there exists an element $\bar{r}\in R/P$ such that $\bar{x}^n=\bar{r}\bar{x}^{n+1}$,
%          and since $R/P$ is an integral domain we have
%          $1=\bar{r}\bar{x}$. Hence $R/P$ is a field and $P$ is
%          maximal.
%        \item[b.$\Rightarrow$a.:] Since $R$ is noetherian, $R$ has
%          only a finite number of minimal prime ideals (corresponding
%          to certain primary components of a primary decomposition of
%          $(0)$), say
%          $\textswab{m}_1,\ldots,\textswab{m}_r$. Since the dimension
%          of $R$ is 0, these are all maximal ideals. Moreover, the
%          nilradical $\sqrt{(0)}=\bigcap_{i=1}^r\textswab{m_i}$ is nilpotent by
%          Levitzki's theorem, i.\ e.\ $\exists n\in\N :
%          \sqrt{(0)}^n=(0)$. But then
%          $\textswab{m}_1^n\cdots\textswab{m}_r^n=0$, since it is
%          contained in the n-th power of $\sqrt{(0)}$. 
%          By lemma \ref{artinian-noetherian} $R$ is artinian.
%     \end{description}
%   \end{proof}

   \begin{lemm}\label{formal-primes}
     Let $R$ be any ring, $P\in\Spec{R}$. Define $P^{(n)}:= P^nR_P\cap 
     R$.
     \\
     Then $P^{(n)}=\menge{r\in R|\exists c\in R\backslash P:cr\in
       P^n}$ and is a $P$-primary ideal.
   \end{lemm}
   \begin{proof}
     Since $R_P$ is the localization of $R$ at the multiplicatively
     closed subset $R\backslash P$, an element $r\in R$ belongs to
     $P^{(n)}$ if and only if there exists an element $\tilde{r}\in
     P^n$ and an element $c\in R\backslash P$ such that
     $r=\frac{\tilde{r}}{c}$. This gives the desired equality.\\
     It remains to show that $P^{(n)}$ is primary with radical
     $P$. For this let $a,b\in R$ with $ab\in
     P^{(n)}$. Thus we have an element $c\in R\backslash P$ with
     $cab\in P^n$. If $a\in P$, then $a^n\in P^n\subseteq P^{(n)}$. And
     if $a\not\in P$, then $ca\not\in P$ and by the above equality $b$ 
     belongs to $P^{(n)}$. Hence $P^{(n)}$ is primary. Moreover,
     $P^n\subseteq P^{(n)}\subseteq P$ implies
     $P=\sqrt{P^n}\subseteq\sqrt{P^{(n)}}\subseteq\sqrt{P}=P$. 
   \end{proof}

   \begin{definitio}
     Let $R$ be a ring, $P\in\Spec{R}$ and $a\in P$.\\
     We say, the prime ideal $P$ is \emph{minimal over $a$} iff 
     any prime ideal $Q$ with $a\in Q\subseteq P$ is indeed equal to
     $P$. 
   \end{definitio}

   \begin{theore}[Krull's Principal Ideal Theorem]\label{krull}
     Let $R$ be a noetherian ring, $a\in R$.\\
     If a prime ideal $P$ is minimal over $a$, then $\height(P)\leq 1$.
   \end{theore}
   \begin{proof}
     Idea of the proof:\\
     We first show the theorem in the special case where $(R,P)$ is a
     local noetherian integral domain by working along the following
     steps:
     \begin{enumerate}
        \item $R/aR$ is artinian.
        \item With the help of Nakayama's lemma we find for any prime
          ideal $Q\subset P$ some $k$ such
          that $Q^{(k)}=Q^{(k+1)}$.
        \item Passing to the localization $R_Q$ and applying
          Nakayama's lemma once more we deduce that then $Q=0$,
          which shows $\height(P)\leq 1$.
     \end{enumerate}
     Finally we can reduce the general case to this one by deviding
     out some prime ideal and then localizing at $P$.\\
     Let us now assume, that $R$ is a noetherian local integral domain 
     with unique maximal ideal $P$.
     \begin{enumerate}
        \item Since $P$ is minimal over $a$, $R/aR$ has only one prime 
          ideal $P/aR$, which then is equal to the nilradical. Since $R/aR$ is
          noetherian, the nilradical (being finitely generated!) is
          nilpotent, and hence some power of $P$ is zero. Since
          moreover $P/aR$ is 
          maximal, by lemma \ref{artinian-noetherian} $R$ is artinian.
        \item Obviously, $Q^{(k+1)}\subseteq Q^{(k)}$ for any $k$. We
          have to show the opposite inclusion for a suitable $k$.
          Because of a., the descending chain of ideals
          $Q^{(1)}+aR/aR\supseteq Q^{(2)}+aR/aR\supseteq
          Q^{(3)}+aR/aR\supseteq \ldots$ gets eventually stationary,
          i.\ e.\ there exists some $k$ such that
          \begin{equation}\label{eins}Q^{(k)}+aR=Q^{(k+1)}+aR.\end{equation} 
          Now let $x\in Q^{(k)}$ be
          given. By
          (\ref{eins}) we get some $r\in R$ and $y\in Q^{(k+1)}$ such
          that $x=y+ar$, and thus $ar=x-y$ belongs to $Q^{(k)}$.\\
          Since $P$ is minimal over $a$, $a\not\in Q=\sqrt{Q^{(k)}}$,
          and by lemma \ref{formal-primes} we therefore have $r\in
          Q^{(k)}$. Thus (\ref{eins}) comes down to 
          $$
            Q^{(k)}=Q^{(k+1)}+aQ^{(k)}.
          $$
          Working modulo $Q^{(k+1)}$ this leads to
          \begin{equation}\label{zwei}
            Q^{(k)}/Q^{(k+1)}=\left(Q^{(k)}/Q^{(k+1)}\right)\cdot a\subseteq
            \left(Q^{(k)}/Q^{(k+1)}\right)\cdot P.
          \end{equation}
          Applying Nakayama's lemma to the (finitely generated)
          $R$-module $Q^{(k)}/Q^{(k+1)}$, we finally get that
          $Q^{(k)}/Q^{(k+1)}=0$, or equivalently $Q^{(k)}=Q^{(k+1)}$.
        \item Since $Q$-primary ideals
          behave well w.\ r.\ t.\ the localization at $Q$, we have for 
          any $l$ $Q^lR_Q=(Q^lR_Q\cap R)R_Q$. Thus by (\ref{zwei}) we get 
          for a suitable $k$
          \begin{displaymath}
            (QR_Q)^k=Q^{(k)}R_Q=Q^{(k+1)}R_Q=(QR_Q)^{k+1}=(QR_Q)^k\cdot QR_Q.
          \end{displaymath}
          Applying Nakayama's lemma to the (finitely generated) module 
          over the noetherian ring $R_Q$, we find
          $Q^kR_Q=(QR_Q)^k=0$. Since $R_Q$ contains a one, this means
          $Q^k=0$, and thus $Q$ is nilpotent. But since $R$ is an
          integral domain, this implies $Q=0$.
     \end{enumerate}
     We can now prove the general case. For this we assume that there
     are prime ideals $Q_0\subseteq Q_1\subset P$. We have to show,
     that then $Q_0=Q_1$. Passing to the localization $S$ of $R/Q_0$ at
     the prime ideal $P/Q_0$, we get the above situation (note, that
     the good behaviour of prime ideals by taking quotients and under
     localization ensures that the image of $P$ in $S$ is 
     still minimal over the image of $a$), and thus
     $(Q_1/Q_0)S=0$. But then $Q_1/Q_0=0$, and hence $Q_1=Q_0$. 
   \end{proof}

   Using the following version of Krull's principal ideal theorem, we
   may shorten the proof considerably.

   \begin{theore}[Krull's Principal Ideal Theorem]\label{weak-krull}
     Let $R$ be a noetherian ring, $a\in R\backslash R^*$ a non zero
     divisor.\\
     Then $\dim(R/aR)=\dim(R)-1$.
   \end{theore}

   \begin{proof}\textbf{[of theorem \ref{krull}]}\\
     Again we suppose first that $(R,P)$ is a local noetherian
     integral domain. \\
     Since $P$ is minimal over $a$, $R/aR$ has exactly one prime
     ideal, namely $P/aR$. Thus the dimension of $R/aR$ is 0, and by
     theorem \ref{weak-krull} the dimension of $R$ is 1. Since $P$ is
     the only maximal ideal of $R$, we get that
     $\height(P)=\dim(R)=1$.\\
     Now we deduce the general case from this special one as in the
     proof of theorem \ref{krull}.
   \end{proof}

\end{document} 
