\documentclass[12pt]{article} 

%\usepackage{epsfig}
%\usepackage{latexsym}
%\usepackage{amssymb} 
%\usepackage{theorem}
%\usepackage{euler}
%\usepackage{epic}
%\usepackage{eepic}
%\usepackage{color}
%\usepackage{curves}

\input msc.sty

\setlength{\topmargin}{30mm}     % Grundlinie der 1. Zeile 30mm von Oberkante
\addtolength{\topmargin}{-1in}
\addtolength{\topmargin}{-\headsep}
\addtolength{\topmargin}{-\headheight}
\addtolength{\topmargin}{-\topskip}

\setlength{\textheight}{267mm} % Unterkante der Fussbox 30pt-voffset ueber Rand
\addtolength{\textheight}{\topskip}
\addtolength{\textheight}{-\footskip}
\addtolength{\textheight}{-30pt}

\setlength{\oddsidemargin}{-1in} % Linker Rand 30mm vom Papierrand
\addtolength{\oddsidemargin}{30mm}
\setlength{\evensidemargin}{\oddsidemargin}

\setlength{\textwidth}{150mm} % Rechter Rand 30mm vom Papierrand


 

\begin{document} 
 
   \pagestyle{empty} 
   \parindent0cm
   UNIVERSITY OF  KAISERSLAUTERN \hfill  Summerterm 1998 \\
   Mathematical Department \hfill 18th May, 1998 \\
   Gert-Martin Greuel \hfill Thomas Keilen\\
   \vspace{0.5cm}

   \begin{center}
     \LARGE \bf Algebraic Geometry \\[0.1cm] \normalsize Exercise set 7
   \end{center}
   \vspace{0.5cm}
   \normalsize
   \parindent0cm
   \textbf{Due date:} 8th June, 1998\hfill\textbf{10 a.~m.}
   \\[0.2cm]
   
   \parindent0cm

   \setcounter{exercise}{19}

   \begin{exercise}
     Show that morphisms of finite type are stable under base
     extension, i.~e.~if $f:X\rightarrow S$ is of finite type and
     $\varphi:S'\rightarrow S$ a base change morphism, then
     $f':X'=X\times_S S'\rightarrow S'$ is of finite type.\\
     {\small (Hint, use that for $B$-algebras $A$ and $C$ we have:
       $A\otimes_B C$ is a finitely generated $C$-algebra, if
       $A$ is a finitely generated $B$-algebra.)}
   \end{exercise}

   \begin{exercise-break} 
      \begin{enumerate}
        \item Show that if $f:X\rightarrow Y$ is a finite morphism of schemes,
          then $f$ is quasi-finite, i.~e.~the fibre $f^{-1}(y)$ is a finite set for
          every $y\in Y$. \\
          {\small (Hint, show that the ring defining the fibre is artinian, 
            and then use the fact 
            that an artinian ring has only finitely many prime ideals -
            which are indeed maximal, c.~f.~Atiyah-MacDonald, prop.~8.3.)}
        \item Find an example of a
          surjective, quasi-finite morphism 
          of finite type  which is not finite.\\
          {\small (Hint, you may think of the spectrum of $\Z[X]$ modulo some
            ``bad'' prime ideal over $\Spec(\Z)$, or you may think of
            some non-seperated scheme over $\A_K^1$, or you may think
            of a suitable plane curve projected to $\A_K^1$. For the
            non-finiteness you may use results from commutative
            algebra, e.~g.~Atiyah-MacDonald, cor.~5.2., in the last case.)}
      \end{enumerate}
   \end{exercise-break}
 
   \begin{exercise}
     Let $K$ be an algebraically closed field, and
     $\varphi:K[s]\rightarrow K[s,t]/(s-t^2):s\mapsto s$. Denote by
     $(f,f^\#):X\rightarrow Y=\A_K^1$ the induced morphism of spectra. Show
     the following for the fibres $X_y=f^{-1}(y)$:
     \begin{enumerate}
        \item If $y\in Y$ is a closed point, not equal to the origin,
          then $X_y$ consists of two points with residue field $K$.
        \item If $y\in Y$ is the origin, then $X_y$ is a non-reduced
          one point scheme with residue field $K$.
        \item If $\eta\in Y$ is the generic point, then $X_\eta$ is a
          one point scheme whose residue field is an extension of
          degree two of the residue field of $\eta$.
     \end{enumerate}
     \small (The residue field of a point $p$ of a scheme $X$ is
     $\ko_{X,p}/\m_{X,p}$.)
   \end{exercise}

\newpage

   \begin{exercise-theorem}%[$\Spec(\Z[X])$]
     The prime ideals of $\Z[X]$ are precisely the ideals of the
     following form:
     \begin{itemize}
        \item $(0)$,
        \item $(p)$, where $p\in\Z$ is a prime number,
        \item $(f)$, where $f\in\Z[X]$ is an irreducible polynomial of 
          degree $d\geq 1$,
        \item $(p,f)$, where $p\in\Z$ is a prime number and
          $f\in\Z[X]$ is a monic polynomial whose residue class in
          $\Z_p[X]$ is irreducible.
     \end{itemize}
   \end{exercise-theorem}
   \begin{proof}
     Since $\Z[X]$ is a unique factorization domain (UFD), the first three
     types of ideals are precisely the prime ideals among the
     principal ideals. Let thus $P$ be a prime ideal which is not
     principal.\\
     Choose an element $0\not=g\in P$ of minimal degree. Since $P$ is
     a prime ideal, we may assume that $g$ is irreducible, and thus a
     prime element, since $\Z[X]$ is a UFD. By $p$ we denote 
     the leading coefficient of $g$. Next we choose a polynomial $f\in 
     P\setminus (g)$ of minimal degree.
     \begin{description}
     \item[{Show: $g=p$ is a prime number.}] By the
       generalized division with remainder in polynomial rings
       (c.~f.~Scheija-Storch, Lehrbuch der Algebra, Teil 2, Satz 53.5) 
       there exist $q,r\in\Z[X]$ with $\deg(r)<\deg(g)$ and $l\in\N_0$ 
       such that $$p^l\cdot f=q\cdot g+r.$$
       Hence $r=p^lf-qg\in P$, and by the minimality of the degree of
       $g$ we find $r=0$. Thus the prime element $g$ divides $p^l\cdot
       f$, and since $g$ 
       does not divide 
       $f$, $g$ must divide $p^l$. But then $g$ is equal to $p$, a prime
       number.
     \item[{Show: w.~l.~o.~g.~$f$ is monic of degree $d\geq
         1$.}] 
       Suppose $p$ divides the leading coefficient $a$ of $f$, then
       $f-p\cdot\frac{a}{p}\cdot x^d\in P\setminus (p)$, which
       contradicts the minimality of the degree of $f$. 
       Suppose moreover that $f$ was an integer, then $1\in
       (p,f)\subseteq P$, which also gives a contradiction. Thus
       $f\in\Z[X]\setminus \Z$ with $\gcd(a,p)=1$.\\
       Now we find $r,s\in\Z$ such that $$1=ra+sp,$$
       and hence $$r\cdot f+s\cdot p\cdot x^d\in P\setminus (p),$$
       is monic of degree $d=\deg(f)$. Replace $f$ by this polynomial.
     \item[{Show: $P=(f,p)$.}]
       Let $h\in P$ be any element.
       Once more generalized division with remainder gives rise to
       $q,r\in\Z[X]$ with $\deg(r)<\deg(f)$ such that $$h=q\cdot
       f+r.$$
       Hence $r=h-qf$ belongs to $P$ and thus by the minimality
       condition on $\deg(f)$ to $(p)$, which in turn gives $$h=q\cdot 
       f+r\in (f,p).$$
     \end{description}
     Up to now we have shown, that a prime ideal which is not
     principal must be generated by a prime number $p$ and a
     monic polynomial $f$. We show finally that such an ideal is prime 
     if and only if the residue class $\bar{f}$ of $f$ in $\Z_p[X]$ is 
     irreducible:
     $$\Z[X]/(f,p)\cong(\Z[X]/(p))/((f,p)/(p))\cong
     \Z_p[X]/(\bar{f}),$$
     hence $(f,p)$ is prime iff these rings are integral domains iff
     $\bar{f}$ is irreducible.
   \end{proof}


\end{document} 



