%From DLANTZ@center.colgate.edu Thu Oct 13 10:53:29 1994
%From: IN%"heinzer@math.purdue.edu" 3-AUG-1994 16:30
%Subj: lbir28.tex
\magnification=\magstep1
%\magnification=1200
%\magnification=\magstephalf
\input amstex
\documentstyle{amsppt}
\pagewidth{6.5 true in} \pageheight{9 true in}
\hoffset .25 true in
\tolerance 1000
\NoBlackBoxes
\def\Z{{\Bbb Z}}
%\def\Z{\text{\bf Z}}
\def\htwo{{\Cal H}_2}
\def\Spec{\operatorname{Spec}}
\def\jS{j\text{-\nolinebreak}\operatorname{Spec}}
\def\G{\operatorname{G}}
\def\Le{\operatorname{L_e}}
\def\m{{\text{\bf m}}}
\def\n{{\text{\bf n}}}
\def\q{{\text{\bf q}}}
\def\p{{\text{\bf p}}}
\topmatter
\title Prime Ideals in Birational Extensions
of Polynomial Rings II\endtitle
\author William J. Heinzer, David Lantz and
Sylvia M. Wiegand \endauthor
\thanks{Heinzer and Wiegand gratefully acknowledge the support
of the National Science Foundation. Lantz and Wiegand thank
Purdue University for its hospitality while we were working on
this research.}
\endthanks
\address{Mathematics Department, Mathematical Sciences Building,
Purdue University, West Lafayette, IN 47907-1395}
\endaddress
\email heinzer\@math.purdue.edu
\endemail
\address{Mathematics Department, Colgate University, 13 Oak Drive,
Hamilton, NY 13346-1398}
\endaddress
\email dlantz\@center.colgate.edu
\endemail
\address{Mathematics Department, University of Nebraska, Lincoln,
NE 68588-0323}
\endaddress
\email swiegand\@unlinfo.unl.edu
\endemail
\subjclass{13E05, 13F20, 13G05, 13H99, 13J15}
\endsubjclass
%\date {August 3, 1994}
%\enddate
\endtopmatter
\document
\baselineskip 15pt
\heading 1. Introduction \endheading
Let $R$ be a semilocal Noetherian domain of dimension one, and let $x$
be an indeterminate over $R$. By a {\it birational extension} of the
polynomial ring $R[x]$ we mean an integral domain $B$ containing $R[x]$
and contained in the fraction field of $R[x]$. We are interested in
the prime ideal structure of birational extensions of $R[x]$.
A general objective, of which the present paper is a part, is to
determine which partially ordered sets, or equivalently topological
spaces, can occur as
the prime spectrum, $\Spec(B)$, of a finitely generated birational
extension $B$ of $R[x]$. A related objective is to determine
necessary and sufficient conditions on two finitely generated
birational extensions $B_1$ and $B_2$ of $R[x]$ in order that
$\Spec(B_1)$ and $\Spec(B_2)$ be order-isomorphic as partially
ordered sets.
Let $\m_1,\m_2, \dots,\m_n$ denote the maximal ideals of $R$. In order
to obtain precise results, we restrict
to extensions of the form $B= R[x, \{ g_i/f_i\}_{i=1}^t]$,
where $t$ is a positive integer, each
$f_i\in R[x] -\bigcup^n_{j=1}\m_j[x]$, and $(f_i,g_i)R[x]$
is not contained in any height-one prime ideal of $R[x]$.
For ease of reference we refer to a finitely generated birational
extension of this special form as a {\it ($t$-generated) sfb-extension}
of $R[x]$. We often restrict to the case where $R$ is the
localization $k[y]_{(y)}$ of the polynomial ring in an indeterminate
$y$ over a countable algebraically closed field $k$. This
article is a continuation of our work in \cite{Heinzer et al.
(1994b)} and also relies heavily on results and notation in
\cite{Heinzer and Wiegand (1989)} and \cite{Heinzer et al. (1994a)}.
Let $D$ be a two-dimensional Noetherian domain with infinitely
many height-two maximal ideals and only finitely many height-one prime
ideals that are contained in infinitely many maximal ideals.
We refer to the height-one primes of $D$ that are
contained in infinitely many maximal ideals as the {\it special
primes } of $D$. Note that a height-one prime ideal $P$ of $D$ is
special if and only if it is nonmaximal and is an intersection of
maximal ideals. It is
easy to see that $\Spec(D)$ as a partially ordered set is
determined by the maximal prime ideals of $D$ and the
containments between the nonmaximal height-one primes and the
height-two maximal ideals. This
amounts to considering two things:
\roster
\item The $j$-spectrum of $D$, $\jS(D)$, (a prime $P \in \Spec(D)$
is a $j$-prime if $P$ is an intersection of maximal ideals, and
$\jS(D)$ is the partially ordered set of $j$-primes in $\Spec(D)$) and
\item The ``inverse $j$-radicals" of finite sets $T$ of height-two
maximal ideals of $\Spec(D)$, that
is, the set of all height-one prime ideals that are contained in every
element of $T$ but are in no other maximal ideals.
\endroster
%In keeping with our previous notation for partially ordered sets,
%we refer to the sets in item (2) as the ``exactly-less-than" sets,
%$$
%\Le(T)=\{ w\in\Spec(D)\, \vert \, wu \} \qquad\text{and} \\
\Le(T)&= \{ w \in U \ \vert\ w1$, then $\jS(B)$ is a union of
$\jS((R-\m)^{-1}B$, over all maximal
ideals $\m$ of $R$---the union is disjoint except that the zeros
and the height-one maximal ideals are identified, and no new
inclusions are added.
\endproclaim
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern.5in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.9in\lower0in\hbox to .3in{\hss$\ldots$\hss}}%
\rlap{\kern1.3in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern1.9in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.4in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern2.9in\lower0in\hbox to .3in{\hss$\ldots$\hss}}%
\rlap{\kern3.5in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern4.0in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern4.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern0in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.5in\lower1in\hbox to .3in{\hss$H_1$\hss}}%
\rlap{\kern.9in\lower1in\hbox to .3in{\hss$\ldots$\hss}}%
\rlap{\kern1.3in\lower1in\hbox to .3in{\hss$H_m$\hss}}%
\rlap{\kern1.9in\lower1in\hbox to .3in{\hss$Q_1$\hss}}%
\rlap{\kern2.9in\lower1in\hbox to .3in{\hss$\ldots$\hss}}%
\rlap{\kern3.5in\lower1in\hbox to .3in{\hss$Q_n$\hss}}%
\rlap{\kern4.6in\lower1in\hbox to .3in{\hss$P$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 550 80} \special{pa 550 870} \special{fp}}%1
\rlap{\special{pa 1350 80} \special{pa 1350 870} \special{fp}}%2
\rlap{\special{pa 1950 80} \special{pa 1950 870} \special{fp}}%3
\rlap{\special{pa 2450 80} \special{pa 1950 870} \special{fp}}%4
\rlap{\special{pa 2450 80} \special{pa 4650 870} \special{fp}}%4
\rlap{\special{pa 3550 80} \special{pa 3550 870} \special{fp}}%5
\rlap{\special{pa 4050 80} \special{pa 3550 870} \special{fp}}%6
\rlap{\special{pa 4050 80} \special{pa 4650 870} \special{fp}}%6
\rlap{\special{pa 4650 80} \special{pa 4650 870} \special{fp}}%7
\rlap{\special{pa 70 1080} \special{pa 2350 1670} \special{fp}}%9
\rlap{\special{pa 550 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 1350 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 1950 1080} \special{pa 2350 1670} \special{fp}}%12
\rlap{\special{pa 3550 1080} \special{pa 2350 1670} \special{fp}}%13
\rlap{\special{pa 4650 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 12 pt
\centerline{Diagram 1.4.1}
\vskip 18 pt
The paper \cite{Heinzer et al. (1994b)} contains the following four
results on $\Spec(B)$ for cyclic sfb-extensions of $R[x]$:
\proclaim{1.5 Theorem} \cite{Heinzer et al. (1994b), Theorem 3.1}
Suppose $R$ is a semilocal
domain of dimension one, $x$ is an indeterminate, and $B$ is a cyclic
sfb-extension of $R[x]$. Then for each height-two maximal ideal
$N$ of $B$, there exist infinitely many height-one primes $P$ such that
$N$ is the only height-two maximal ideal containing $P$. That is, (P6)
or (P6$\,'$) of Theorem~1.2(1) holds for singleton subsets $T$ of the set
of all height-two elements of Spec($B$).
\endproclaim
\proclaim{1.6 Theorem} \cite{Heinzer et al. (1994b), Theorem 4.1}
Let $R$ be a Henselian local domain
of dimension one, and let $B$ be a cyclic sfb-extension of $R[x]$.
Then $\Spec(B)$ satisfies (P6$\,'$) of Theorem~1.2(1). If $R$ is
countable, then $\Spec(B)$ is uniquely determined by the partially
ordered set $\jS(B)$.
\endproclaim
\proclaim{1.7 Theorem} \cite{Heinzer et al. (1994b), Theorem 4.5}
Let $R$ be a semilocal domain of dimension one, $x$ an indeterminate
over $R$, and $B=R[x,g/f]$ a cyclic sfb-extension of $R[x]$. Suppose
there exists a one-dimensional Noetherian domain $D \subset R$
such that $R$ is a localization of $D$ and such
that $U=\Spec(D[x,g/f])$ satisfies (W):
\itemitem{(W)} Let $S$ be a finite set of height-one
elements of $U$ and $T$ a finite subset of $\htwo(U)$. Then there exist
infinitely many height-one elements $w\in U$ such that $\G(s)\cap
\G(w)\subseteq T\subseteq \G(w)$, for all $s\in S$.
\noindent
Then $\Spec(B)$ satisfies (P6) of 1.2.
\endproclaim
The condition (W) is similar to an axiom described by Roger Wiegand in
\cite{R. Wiegand (1978)} and \cite{R. Wiegand (1986)}.
Condition (W) is satisfied by the prime
spectrum of a polynomial ring over the integers, $\Spec(\Z[x])$ and
by the spectrum of every domain of dimension two that is finitely
generated as a $k$-algebra, if $k$ is a field contained in the algebraic
closure of a finite field \cite{R. Wiegand (1986)},
\cite{Heinzer et al. (1994b)}. Thus the
following corollary is obtained in \cite{Heinzer et al. (1994b)}.
\proclaim{1.8 Corollary} \cite{Heinzer et al. (1994b), Corollary 4.6}
Suppose $k$ is a field
algebraic over a finite field. Let $z$ and $x$ be indeterminates
over $k$, $\{ \p_i \ \vert\ 1 \le i\le n \}$ a finite set of nonzero
prime ideals of $k[z]$, $S = k[z] - \bigcup^n_{i=1} \p_i$,
and $R = S^{-1}k[z]$. Let $B = R[x, g/f]$ be a cyclic
sfb-extension of $R[x]$. Then $\Spec(B)$ satisfies (P6) of
Theorem~1.2 and so is uniquely determined by the isomorphism type
of $\jS(B)$.
\endproclaim
\medskip
\heading{2. Further results on cyclic sfb-extensions of $R[x]$}
\endheading
In Theorem 2.2 we extend Corollary 1.8 to the case where
$R$ is a one-dimensional semilocal domain which is a ring
of fractions of a finitely generated algebra
over an arbitrary ground field $k$.
\medskip
\noindent
{\bf 2.1 Remarks.} (1) Corollary 1.8 stated above holds
more generally if $R$ is a semilocal ring of fractions of
a one-dimensional affine domain over a field $k$ which is
algebraic over a finite field and
$B$ is any sfb-extension of $R[x]$ (not necessarily cyclic).
To prove this more general
statement, modify the original proof in \cite{Heinzer et al. (1994b)} by
replacing ``$ g/f$'' by ``$g_1/f_1,\dots,g_t/f_t$''.
(The crucial tool used in the proof of (1.8) is Theorem 2 of
\cite{R. Wiegand (1986)} which holds for all two-dimensional affine
domains over a field algebraic over a finite field.)
(2) If $R$ is a one-dimensional semilocal domain that is a
localization of an affine domain $D$ over a field $k$, then
$R$ is also a localization of a one-dimensional affine domain
over a possibly larger field. This can be seen using
(14.G) and (14.F) of Matsumura (1980).
\medskip
The $j\text{-\nolinebreak}$spectrum of a cyclic sfb-extension
$B$ of $R[x]$ is described in Theorem~1.4. In view of
Theorem 1.2(2), in order to determine the entire spectrum of $B$,
we consider, as in \cite{Heinzer et al. (1994b)}, the
exactly-less-than sets $\Le({\Cal T})$, where ${\Cal T}$ is a finite subset
of $\htwo(\Spec(B))$.
\proclaim{2.2 Theorem} Let $R$ be a semilocal (Noetherian)
domain of dimension one, and let $x$ be an indeterminate over $R$.
Suppose that $R$ is a ring of fractions of an
affine domain over a field and $B$ is a cyclic
sfb-extension of $R[x]$. Then $\Spec(B)$ satisfies (P6) of
Theorem~1.2. In particular, if $R$ is countable, then $\Spec(B)$
is uniquely determined by the type of $\jS(B)$.
\endproclaim
\demo{Proof} By (2.1.2) we may assume that $R$ is a localization of a
one-dimensional affine domain $D = k[d_1,\dots,d_m]$ over a field $k$.
Let $B= R[x,g/f]$, where $f$ is
outside all the extensions to $R[x]$ of the maximal ideals of $R$
and either $(f,g)$ is an $R[x]$-sequence or $(f,g)R[x] = R[x]$.
We first give the proof in the case where $k$ is an
algebraically closed field.
By localizing $D$ at a unit of $R$, we may assume that
$f,g \in D[x]$ and $(f,g)D[x]$ is either of height two or all
of $D[x]$. It follows that $A = D[x,g/f]$ is of the form
a polynomial ring $D[x,y]$ modulo a principal ideal and
hence is Cohen-Macaulay.
Let ${\Cal T}$ be a finite set of height-two maximal ideals of $B$.
The proof of Theorem~1.5 given in \cite{Heinzer et al. (1994b),
Theorem~3.1} shows that, for each height-two maximal
ideal $N$ of $B$, there is an element of $B$ which is
contained in $N$ and in no other height-two maximal ideal of
$B$; and clearly we can choose that element to be in $A$.
So by multiplication we obtain an element $h$ of $A$ that is
contained in all the elements of ${\Cal T}$ and in no other
height-two maximal ideal of $B$. If we choose from $D$ an element
$r$ in the intersection of the maximal ideals of $R$ and outside all
the minimal primes of $hA$, then $h,r$ is an $A$-sequence:
Since $h$ and $r$ are both contained in the elements of
${\Cal T}$, they do not generate the unit ideal even in $B$,
much less in $A$; and our choice of $r$ assures that $(h,r)A$
has height greater than 1.
With $A = k[d_1,\dots,d_m, x, g/f]$ a two-dimensional
affine domain over an algebraically closed field $k$,
by Lemma~4 of \cite{R. Wiegand (1978)}, there is a nonempty
open subset $U$ of $(k)^{m+3}$ such that, for each point
$\text{\bf p} = (\alpha,\beta_1,\ldots,\beta_m,\gamma,\delta)$
in $U$, the radical of the principal ideal in $A$ generated by
the element
$$
a_{\text{\bf p}} =
h + (\alpha + \beta_1 d_1 + \cdots + \beta_m d_m + \gamma x
+ \delta (g/f)) r
$$
is a (height-one) prime ideal $P$. Every height-two maximal
ideal $N$ of $B$ contains a maximal ideal of $R$ and hence
$r$, so if $N$ contains $P$, then it also contains $h$ and hence
is one of the elements of ${\Cal T}$; and clearly the elements of
${\Cal T}$ contain $P$, that is, $P \in \Le({\Cal T})$.
We want to show that varying the elements
$a_{\text{\bf p}}$ over $U$ will give rise to infinitely many such
height-one primes: Select one point {\bf p} in $U$; then
by leaving fixed the values of the $\beta_i$'s, $\gamma$ and
$\delta$, and varying the value of $\alpha$, we get infinitely many
other points in $U$. Assume by way of contradiction that two of
these {\bf p} in $U$, differing only in the first component
$\alpha$, yield elements $a_{\text{\bf p}}$ with the same prime
radical $P$; then the difference between the two $a_{\text{\bf p}}$
is in $P$, and so $r\in P$. But since $P$ is
contained in only finitely many of the height-two maximals that
survive in $B$, it intersects $R$ in $(0)$, the desired
contradiction. This completes the proof in the case where
$k$ is algebraically closed.
Now let us consider the case where $k$ is not algebraically closed
and observe how the proof follows in this case. Again
by localizing $D$ at a unit of $R$ we may assume that $f,g \in D[x]$.
It follows that $A = D[x, g/f]$ is of the form the polynomial
ring $D[x,y]$ modulo a principal ideal and hence is Cohen-Macaulay.
Let $k^*$
denote an algebraic closure of $k$. Let $D^*$ denote a domain quotient
of $k^*\otimes_kD$ obtained by factoring out a minimal prime, and let
$R^*$ denote the result of localizing $D^*$ at the same
multiplicative system at which $D$ was localized
to get $R$, $A = D[x,g/f]$, $A^* = D^*[x,g/f]$, and $B^* = R^*[x,g/f]$.
(Notes: (1)~Though
$k^*$ is usually an infinite algebraic extension of $k$,
the domain $D^*$ is a one-dimensional Noetherian domain of
finite type over $k^*$ and an integral extension of $D$. Hence
there are only finitely many primes of $D^*$ lying over a prime in
$D$ --- namely the minimal primes of the extension to $D^*$ of the
prime in $D$ --- so $R^*$ is still semilocal, as well as an integral
extension of $R$. (2)~ Our hypothesis on $(f,g)$ implies
$A^*$ is still a domain.)
Let ${\Cal T}$ be a finite set of height-two maximal ideals of $B$,
and let ${\Cal T}^*$ denote the set of all primes of $B^*$ lying
over elements of ${\Cal T}$.
As above, we obtain an element $h$ of $A$ that is contained
in all the elements of ${\Cal T}$ and in no other height-two
maximal ideals of $B$ and an element $r \in D$ in the intersection of
the maximal ideals of $R$ and outside all of the minimal primes
of $hA$.
Let $U$ be a nonempty open subset of $(k^*)^{m+3}$ obtained
via an application of Lemma~4 of \cite{R. Wiegand (1978)} where
$R^*$, $A^*$, $B^*$ and ${\Cal T}^*$now play the roles previously
played by $R$, $A$, $B$ and ${\Cal T}$.
We consider the intersections with $R$ of the extensions to $R^*$ of
all the (prime) radicals $P^*$ of the principal ideals generated by
the elements $a_{\text{\bf p}}$ as {\bf p} varies over $U$. Since only
finitely many primes of $R^*$ lie over a single prime in $R$, there
are infinitely many different such intersections, and it is clear that
each is a prime contained in every element of ${\Cal T}$. By the Going
Up Theorem and the fact that each $P^*$ is contained only in elements
of ${\Cal T}^*$, each such intersection is contained in no other
height-two maximal ideal of $B$. Therefore, $\Spec(B)$ satisfies
(P6).~\qed
\enddemo
\medskip
\noindent
{\bf 2.3 Remark.} It would be interesting to know whether
Theorem~2.2 also holds for noncyclic sfb-extensions. The
proof given here does not extend to noncyclic sfb-extensions
because it relies on Lemma~3.5 of \cite{Heinzer et al. (1994b)}
to deduce that for
each height-two maximal ideal $N$ of $B$ there is an element of $B$
which is contained in $N$ and in no other height-two maximal ideal
of $B$. We show in Example~2.4 that this need not hold for
$B$ a noncyclic sfb-extension.
\medskip
\noindent
{\bf 2.4 Example.} Let $k$ be a field of characteristic zero,
let $R$ be the DVR $k[y]_{(y)}$, and consider the sfb-extension
$B = R[x, (y/x)^2, (y/x)^3]$ of $R[x]$. Then
$B \subseteq R[x, y/x] = C$ and $xC \cap B = P$ is a transient
height-one prime of $B$ such that $B/P \cong k[t^2, t^3]$,
where $t$ is an indeterminate over $k$. Since $k$ is of
characteristic zero, for any nonzero $a \in k$, the maximal
ideal $(t-a)k[t] \cap k[t^2,t^3]$ is not the radical of a principal
ideal. Hence if $N$ is a height-two maximal ideal of $B$ for which
$N/P \cong (t-a)k[t] \cap k[t^2,t^3]$, then there exists no element
of $B$ which is contained in $N$ and in no other height-two
maximal ideal of $B$.
\heading{3. Examples of $j\text{-\nolinebreak}$spectra of finite
birational extensions}
\endheading
This section displays two examples showing that the
$j$-spectra of noncyclic sfb-extensions can be more
complicated that the cyclic case. In both cases, $k$ is
the field $\Z_{31991}$ of integers modulo the prime 31991.
\medskip
\noindent
{\bf 3.1 Example.} An example of a 2-generated sfb-extension
of $k[a]_{(a)}[x]$ (where $a,x$ are indeterminates over $k$)
in which two transient $j$-primes are
not comaximal. This shows that if an sfb-estension is not cyclic,
a picture different from Diagram~1.4.1 can occur.
The example is the ring
$$
B = k[a]_{(a)}[x,x/(x-a),(x-a)^2/x]\ .
$$
To compute the kernel $P$ of the $k[a]_{(a)}[x]$-homomorphism
from the polynomial ring $k[a]_{(a)}[x,u,v]$ onto $B$ determined by
sending $u$ to $x/(x-a)$ and $v$ to $(x-a)^2/x$, we used the
computer algebra program MACAULAY, written by David Bayer and
Michael Stillman. But since MACAULAY is written for
computations with quasihomogeneous polynomial rings, we added a
homogenizing indeterminate $d$; and to be sure that the program found
the entire kernel, and not just the piece generated by
$(x-a)u-x$ and $xv-(x-a)^2$, we added a indeterminate $w$ to
map to $1/x(x-a)$. Thus, we really began with
the polynomial ring $C = k[a,x,u,v,d][w]$ (the
isolation of $w$ is accomplished by placing it first and specifying
the monomial order $1\ 5$) and its ideal $I$ generated
by $(x-a)u-x$, $xv-(x-a)^2$ and $x(x-a)w-d^3$. The ideal
is described by its generators, with the MACAULAY convention that
an integer after a variable is interpreted as an exponent.
(The example is a bit unusual in that we did not have to multiply
some terms of the numerators and denominators of the generators of
our sfb-extension by powers of $d$ to ``homogenize'' them, and that
after we had given weight 1 to each of
$w,a,x,d$, the weights we had to assign to $u,v$ to make the
generators into quasihomogeneous polynomials were also 1.)
Then we replaced the given generating set for $I$ with a
``standard'' (i.e., Gr\"obner) basis, ``eliminated'' $w$ (i.e,
found the intersection $J$ of $I$ with the coefficient ring
$k[a,x,u,v,d]$), and found a standard basis for $J$, of which we
then displayed a minimal generating set. Here is the slightly
edited ``monitor file'' of our computer run; the characters
we entered are in ``typewriter'' typeface:
\vskip 6 pt
{\obeylines
\% {\tt ring C}
! characteristic (if not 31991) ?
! number of variables ? {\tt 6}
! 6 variables, please ? {\tt waxuvd}
! variable weights (if not all 1) ? {\tt 1 1 1 1 1 1}
! monomial order (if not rev. lex.) ? {\tt 1 5}
; largest degree of a monomial : 512 217
\% {\tt ideal I}
! number of generators ? {\tt 3}
! (1,1) ? {\tt (x-a)u-xd}
! (1,2) ? {\tt xv-(x-a)2}
! (1,3) ? {\tt x(x-a)w-d3}
\% {\tt std I I}
\% {\tt elim I J}
\% {\tt std J J}
\% {\tt type J}
; au-xu+xd a2-2ax+x2-xv uvd3+ad4-xd4
}
\vskip 6 pt
\noindent
Setting $d=1$, we obtain a generating
set for the desired kernel $P$: $(x-a)u-x$, $xv-(x-a)^2$, and
$uv-(x-a)$. (The first two, of course, are not new, and the third
is not surprising but is not in the ideal generated by the
first two.) Now since all the height-two and special primes in $B$
contain the extension of the maximal ideal $\m$ of $k[a]_{(a)}$,
we can take their preimages in $k[a]_{(a)}[x,u,v]$ and factor out
the extension of $\m$ to this ring, i.e., set
$a=0$ to get $k[x,u,v]$. The resulting generators of the image
of $P$ are $x(u-1)$, $x(v-x)$ and $uv-x$. The image in $k[x,u,v]$
of the preimage in $k[a]_{(a)}[x,u,v]$ of the
survivor prime in $B$ is the minimal prime of the image of $P$
that does not contain $x$ and hence is generated by
$u-1$ and $v-x$. So the preimage of the survivor in
$k[a]_{(a)}[x,u,v]$ is $(a,u-1,v-x)$. The image of the preimage
of a transient prime contains $x$ and hence contains either $u$ or
$v$; so the preimages of the transients are $(a,x,u)$ and
$(a,x,v)$. These primes are clearly contained in the maximal ideal
$(a,x,u,v)$, so their images in $B$, the transients, are not
comaximal. Moreover, the preimages of the second transient and the
survivor are both contained in the maximal ideal $(a,x,u-1,v)$.
Here is a diagram of the poset $\jS(B)$:
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern.5in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern1.4in\lower0in\hbox to .3in{\hss$(a,x,u,v)$\hss}}%
\rlap{\kern2.3in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern3.2in\lower0in\hbox to .3in{\hss$(a,x,u-1,v)$\hss}}%
\rlap{\kern4.1in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.7in\lower1in\hbox to .3in{\hss$(a,x,u)$\hss}}%
\rlap{\kern1.5in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.3in\lower1in\hbox to .3in{\hss$(a,x,v)$\hss}}%
\rlap{\kern3.9in\lower1in\hbox to .3in{\hss$(a,u-1,v-x)$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 550 80} \special{pa 750 870} \special{fp}}%1
\rlap{\special{pa 950 80} \special{pa 750 870} \special{fp}}%4
\rlap{\special{pa 950 80} \special{pa 2350 870} \special{fp}}%4
\rlap{\special{pa 2350 80} \special{pa 2350 870} \special{fp}}%4
\rlap{\special{pa 3250 80} \special{pa 2350 870} \special{fp}}%7
\rlap{\special{pa 3250 80} \special{pa 3950 870} \special{fp}}%7
\rlap{\special{pa 4150 80} \special{pa 3950 870} \special{fp}}%8
\rlap{\special{pa 750 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 1550 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 2350 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 3950 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 12 pt
\centerline{Diagram 3.1.1}
\vskip 18 pt
\medskip
\noindent
{\bf 3.2 Example.}
An example of a 2-generated sfb-extension of $k[x-y]_{(x-y)}[x]$,
where $x,y$ are indeterminates over the field $k$, in
which two transients are contained
in two maximal ideals. This cannot happen in a cyclic sfb-extension.
Let
$$
B = k[x-y]_{(x-y)}[x,(x^2+2y^3)/xy,\,(x^2+2y^3-2xy)/y^2] \ .
$$
By using MACAULAY, we find that the kernel of the map from
$k[x-y]_{(x-y)}[x,u,v]$ onto $B$ determined by sending $u$
to $(x^2+2y^3)/xy$ and $v$ to $(x^2+2y^3-2xy)/y^2$ is generated
by
$$\gather x^2 (v+1-2x)\ ,\qquad x^2(u - 1 - 2x)\ ,\qquad
x(u -2 - v)\ , \\
x(u(v-2x) - v + 4x)\ ,\qquad v(u^2 - 2u - v) - 2x(u-2)^2\ .
\endgather$$
In this case, factoring out the extension of $\m$ means replacing $y$
with $x$ in the generators, and we find that the preimage of
the survivor is $(x-y,u-v-2,2x-u+1)$ and those of the transients are
$(x,y,v)$ and $(x,y,u^2-2u-v)$. Thus, the preimages of the transients
are contained in both $(x,y,u,v)$ and $(x,y,u-2,v)$, while the
preimages of the survivor and the first transient are contained
in $(x,y,u-1,v+1)$. The survivor and the other transient are
comaximal.
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern0in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern1in\lower0in\hbox to .3in{\hss$(x,y,u,v)$\hss}}%
\rlap{\kern1.9in\lower0in\hbox to .3in{\hss$(x,y,u-2,v)$\hss}}%
\rlap{\kern2.8in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern3.8in\lower0in\hbox to .3in{\hss$(x,y,u-1,v+1)$\hss}}%
\rlap{\kern4.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.1in\lower1in\hbox to .3in{\hss$(x,y,v)$\hss}}%
\rlap{\kern1in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.3in\lower1in\hbox to .3in{\hss$(x,y,u^2-2u-v)$\hss}}%
\rlap{\kern4.0in\lower1in\hbox to .3in{\hss$(x-y,2x-u+1,u-v-2)$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 50 80} \special{pa 150 870} \special{fp}}%1
\rlap{\special{pa 1050 80} \special{pa 150 870} \special{fp}}%4
\rlap{\special{pa 1050 80} \special{pa 2350 870} \special{fp}}%4
\rlap{\special{pa 1950 80} \special{pa 150 870} \special{fp}}%4
\rlap{\special{pa 1950 80} \special{pa 2350 870} \special{fp}}%4
\rlap{\special{pa 2850 80} \special{pa 2350 870} \special{fp}}%4
\rlap{\special{pa 3850 80} \special{pa 2350 870} \special{fp}}%7
\rlap{\special{pa 3850 80} \special{pa 4050 870} \special{fp}}%7
\rlap{\special{pa 4650 80} \special{pa 4050 870} \special{fp}}%8
\rlap{\special{pa 150 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 1050 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 2350 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 4050 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 12 pt
\centerline{Diagram 3.2.1}
\heading{4. $j$-Spectra of birational extensions} \endheading
Let $B$ be an sfb-extension of $R[x]$ where $R$ is local, so
that there is a unique survivor $P_0$ in $\jS(B)$.
Recall that for $P$ a transient prime of $B$, $P \cap R[x]$ is
a maximal ideal (from the discussion in (1.3.4)). For $P,Q$
transient primes of $B$, we say $P \sim Q$ if and only if
$P \cap R[x] = Q \cap R[x]$; and for maximal ideals $M,N$ of
$B$, we define $M \sim N$ if and only if $M \cap R[x] = N \cap R[x]$.
There are finitely many transients in $\jS(B)$ and therefore
a finite number $s$ of equivalence classes,
$\Cal T_1, \dots \Cal T_s$, of transient primes
under this relation. Considering just the maximal ideals of
$B$ which contain transients, they also fit into $s$ equivalence
classes,
$\Cal M_1,\dots,\Cal M_s$, where
$$
\Cal M_i = \{\ M \in \htwo(\jS(B)) : M \cap R[x] = P \cap R[x],
\ P \in \Cal T_i \ \} \ .
$$
For each $i$, the only maximal ideals of $B$ containing
elements of $\Cal T_i$ are elements of $\Cal M_i$.
If two transients $P$ and $Q$ of $B$ are not equivalent,
they must be comaximal in $B$. By (1.3.4)
there is at most one connection of the survivor prime
$P_0$ to each $\Cal T_i$ via $\Cal M_i$; that is, if there
is a maximal ideal $M$ containing $P_0$ and some
$P \in \Cal T_i$, then $M \in \Cal M_i$ and for each
$Q \in \Cal T_i$ and $N \in \Cal M_i$ with
$P_0 + Q \subseteq N$, we have $N = M$.
Consider the poset diagram
of $\jS(B)$; there is at most one line upward
joining $P_0$ to each set $\Cal M_i$ and thus at most one
connection from $P_0$ to each set $\Cal T_i$.
In other words, suppose we remove from the poset diagram
of $\jS(B)$ the prime (0) and regard the rest as an undirected
graph. Then $P_0$ is adjacent to at most one element of
each $\Cal M_i$; thus a path through $P_0$ cannot be closed
(without passing back through $P_0$).
Thus $U = \jS(B)$ satisfies
\medskip
\itemitem{(P7)} Let $V$ be the graph resulting from taking
the poset diagram of $U$ and removing the unique minimal
element. Among the special elements there is one, $u_0$,
for which there is no closed path containing $u_0$
(without repeated edges) in $V$.
\medskip
In Theorem 4.1, we describe a systematic procedure for
constructing an sfb-extension with $j$-spectrum a given
countable partially ordered set $U$ that is otherwise
reasonable (dimension two, unique minimal element, infinitely
many height-one maximals, finitely many height-two points
that are greater than two given height-one points, etc.)
and satisfies a strong version of (P7):
the special element $u_0$ has degree one in
the graph $V$.
In (4.4), we construct an example in which $u_0$ has
degree two. We believe that this construction can be
generalized to yield an sfb-extension with $j$-spectrum
isomorphic to a given feasible $U$ satisfying (P7).
The local ring $R$ used in our constructions is the DVR $k[y]_{(y)}$
where $k$ is a countable algebraically closed field and
$y$ is an indeterminate over $k$.
\proclaim{4.1 Theorem}
Let $U$ be a countable two-dimensional partially ordered set with
a unique minimal element and precisely $n + 1$ height-one elements
$u_0,u_1,\dots,u_n$, which are not maximal. Assume that
\roster
\item The set of maximal height-one elements of $U$ is infinite,
\item For each $i,\,0\le i\le n$, $\G(u_i)$ is infinite,
\item $|G(u_0) \cap (\cup_{i > 0}G(u_i))| \le 1$,
\item For every $i\ne j, \,1\le i,j\le n$,
$\G(u_i)\cap\G(u_j)$ is finite,
\endroster
Then, for $k$ a countable algebraically closed
field, $y,x$ indeterminates over $k$, and $R=k[y]_{(y)}$,
there exists an sfb-extension $B$ of $R[x]$ for which
$U\cong\jS(B)$. Furthermore the order-isomorphism
$\varphi: U\to \jS(B)$
can be chosen so that $\varphi(u_i)\cap R[x]=(x,y)$,
for each $i>0$, and so that $\varphi(u_0)$ is the
survivor prime of the sfb-extension $B$ of $R[x]$.
\endproclaim
We first prove a version for which the partially ordered set is
closer to a tree.
\proclaim{4.2 Lemma}
Let $U$ be a countable two-dimensional partially ordered set
with a unique minimal element and precisely $n+1$ height-one elements
$u_0,u_1,\dots,u_n$, which are not maximal. Assume that
\roster
\item The set of maximal height-one elements of $U$ is infinite,
\item For each $i,\,0\le i\le n$, $\G(u_i)$ is infinite,
\item For the pair $u_0,u_1$,
$|G(u_0) \cap G(u_1)| \le 1$, while for
every $i\ne j, \{ i,j\} \ne \{0,1\},
\,0\le i,j\le n$, $\G(u_i)\cap\G(u_j)=\emptyset$.
\endroster
Then, for $k$ a countable algebraically closed
field, $y,x$ indeterminates over $k$, and $R=k[y]_{(y)}$,
there exists an sfb-extension $A$ of $R[x]$ so that
$U\cong \jS(A)$. Furthermore the order-isomorphism
$\varphi: U\to \jS(A)$ can be chosen so that
$\varphi(u_i)\cap R[x]=(x,y)$, for each $i>0$, and so that
$\varphi(u_0)$ is the survivor prime of the sfb-extension $A$ of
$R[x]$. Also if $M$ is a maximal ideal of $A$, then $A/M \cong k$.
\endproclaim
\demo{Proof of Lemma 4.2} Choose distinct elements $a_1,\dots,a_n$ of $k$.
If $|G(u_0) \cap G(u_1)| = 1$, take $a_1=0$, while for
$|G(u_0) \cap G(u_1)| =0$, take all $a_i\ne0$.
For $1\le i\le n,$ let $V_i\supset R[x][y/x]_{(x,(y/x)-a_i)}$ be the
valuation domain of the ord valuation $v_i$ (determined by
the powers of the ideal $(x,(y/x)-a_i)\ )$; let
$D=R[x,1/x]$.
The center of $V_i$ on the regular ring $R[x,y/x]$ is
$(x,(y/x) - a_i)R[x,y/x]$ and the residue field of $V_i$ is a simple
transcendental extension of $k$ with the image of $((y/x) - a_i)/x$
as a field generator for this residue field over $k$.
Set $A=D\cap V_1\dots\cap V_n$.
Let $w = (\prod_{i=1}^n((y/x) - a_i))/x$; we show that
$A = R[x, y/x, w]$. It is easily seen that $R[x,y/x, w] \subseteq A$,
so we prove the reverse inclusion. Since $R[x,y/x]$ is regular,
the extension $R[x,y/x][w]$ is locally a complete intersection and
hence Cohen-Macaulay. Thus to show $A \subseteq R[x,y/x,w]$, it
suffices to show for each height-one prime $Q$ of $R[x,y/x,w]$
that $A \subseteq R[x,y/x,w]_Q$. If $x \notin Q$, then
$R[x,y/x,w]_Q$ contains $D$ and hence $A$. Suppose $x \in Q$.
Then $\prod_{i=1}^n((y/x) - a_i) = xw \in Q$, so one factor
$(y/x) - a_i$ is in $Q$ and the rest are not because the
pairwise differences of the factors are units. Thus,
one element of $R[x,y/x,w]_Q$ is
$$
\frac{w}{\prod_{j \ne i}((y/x)-a_j)} = \frac{(y/x) - a_i}{x}\ .
$$
It follows that $R[x, y/x, w]_Q$ contains
$R[x,y/x]_{(x,(y/x) - a_i)}[((y/x) - a_i)/x]$,
of which $V_i$ is a localization, i.e., $R[x,y/x,w]_Q = V_i$.
This completes the proof that $A = R[x,y/x,w]$.
In particular, $A$ is an sfb-extension of $R[x]$.
Setting $P_0 = yD \cap A$, we have $x\notin P_0$, and
$P_0$ is the survivor prime with respect to the
extension $R[x] \subset A \subset R[x,1/x] = D$, i.e., we
have $P_0 = P_0D \cap A$, a height-one prime of $A$.
For $i \ge 1$, let $P_i$ be the center of $V_i$ on $A$;
then the $P_i$ are height-one primes of $A$, and $A_{P_i} = V_i$.
Moreover the $P_i$ have the property that $P_i\cap R[x]=(x,y)R[x]$.
By (1.3.1), $P_0, P_1,\dots,P_n$ are the special elements of $A$.
Define $\varphi$ on the $u_i$ by setting $\varphi(u_i)=P_i$ for $i\ge 0$.
Then $\G(\varphi(u_i))$ is infinite.
Note that $j$-Spec($A$) has a unique minimal element and
infinitely many height-one maximal elements (since $R[x,1/x]$ does).
Thus to see that $\varphi$ can be extended to an
order-isomorphism on all of $U$, it suffices to show that (3) holds
with the $u_i$ replaced by $P_i$. This is done in the following claim:
\proclaim{Claim} $P_0 + P_1 \ne A$ iff $a_1 = 0$. If
$a_1 = 0$, then $P_0+P_1$ is contained in a unique maximal ideal of $A$.
All pairs $P_i,\,P_j$ with $i\ne j, \{ i,j\}\ne \{ 0,1\}$
are comaximal.
\endproclaim
\demo{Proof of claim}
We have $P_0$ contains $y/x$, $P_i $ contains $y/x-a_i$ for
$i>0$, and the $a_i$ are distinct. Thus
the last sentence of the claim is clear, and if $a_1 \ne 0$, then
$P_0 + P_1 = A$.
If we take $a_1 = 0$
and $n = 1$, so that $A = D \cap V_1$, then
$A = R[x, y/x^2]$ by the argument above with $w = y/x^2$.
It follows that $R[x,y/x^2] = D \cap V_1 = A$, $P_0 = (y/x^2)A$
and $P_1 = xA$. Thus $P_0 + P_1 = (x,y/x^2)A$, a maximal ideal
of $A$ with residue field $k$.
In the situation where $A = D \cap V_1 \cap \dots \cap V_n$,
$n > 1$, and $a_1 = 0$, let $A_1 = D \cap V_1$.
Then $A_1 = R[x,y/x^2],\ y/x \in P_0 \cap P_1$,
and $(y/x) - a_i \in P_i$ for $i > 1$. Hence $P_0 \cap P_1$ and
$P_i$ are comaximal for each $i > 1$.
If $M$ is a maximal ideal of $A$ containing $P_0 + P_1$
and $S = A-M$, then $S$ meets each $P_i$ for $i > 1$,
so $S^{-1}A = S^{-1}A_1$. It follows that $P_0 + P_1$
is contained in a unique maximal ideal of $A$ which is
$(x,y/x^2)A_1 \cap A$. This completes the proof of the claim.
\enddemo
As noted above, for $i > 0$,
$P_i \cap R[x] = (y,x)R[x]$ and $P_0$ is the survivor prime
of the sfb-extension $A$.
If $M$ is a height-two maximal ideal of $A$, then $A/M$ is
an algebraic extension of $k$. Hence if $k$ is algebraically
closed we have $A/M \cong k$, for each height-two maximal ideal $M$ of $A$.
Thus we have completed the proof of the lemma.~\qed
\enddemo
\demo{Proof of Theorem 4.1} We adapt the proof
of \cite{S. Wiegand (1983), Theorem~1}: Suppose
$$
H=\bigcup\{\G(u_i)\cap\G(u_j) : i,j = 1,\ldots,n,\ i\neq j\}
$$
has $t$ elements; use induction on $t$. If $t=0,$
then the theorem holds by the lemma
(possibly by renumbering). Thus we may assume that $t>0$ and that
the theorem holds for every smaller number.
Let $m\in H$. For convenience rearrange the $u_i$'s for $i\ge 1$ so
that $u_i0$ and
$\varphi(u_0)$ is the survivor in $B^*$.
Now for each $i$, $M_i=\varphi(m_i)$ is a maximal ideal of $B^*$.
Since $k$ is algebraically closed,
$B^*/ \varphi(m_i)\cong k$ for each $i,\, 1\le i\le t$; let
$\varepsilon_i:B^*\to k$ be the natural projection with kernel $M_i$.
Let $B=\{b\in B^* : \varepsilon_1(b)=\dots =\varepsilon_t(b)\}$.
Then by \cite{S. Wiegand (1983), Proposition~2}, the conductor of
$B^*$ in $B$ is the maximal ideal $M=\bigcap_{i=1}^r M_i$ of $B$, and
the spectrum of $B$ differs from the spectrum of $B^*$ only in that $M$
replaces $M_1,\dots,M_t$ which are the primes of $B^*$
lying over $M$. Furthermore, $B^*$ is a finitely
generated $B$-module; therefore by the
Artin-Tate Lemma \cite{Kunz (1985), Lemma 3.3, p.~16} and (1.3.3),
$B$ is an sfb-extension of $R[x]$.
(Note that in order to insure that $R[x]\subseteq B$, the $M_i$'s must
intersect to the {\it same }maximal ideal of $R[x]$.)
It is clear that $\varphi: U^*\to \jS(B^*)$ can be adjusted to an
order-isomorphism $\varphi: U\to \jS(B)$ which has the desired properties
regarding maximal elements.~\qed
\enddemo
\noindent
{\bf 4.3 Remark.} The proofs of Theorem~4.1 and Lemma~4.2 also
apply under the assumption that $R$ is a countable excellent
DVR with algebraically closed residue field.
\medskip
\noindent
{\bf 4.4 Generalization.}
The construction in Theorem~4.1 and Lemma~4.2 can achieve only those
$j$-spectra with the survivor in at most one maximal ideal containing
another special prime. In order to produce the $j$-spectrum
shown below, we outline a modification of this construction.
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern.2in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.8in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern1.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern3.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern4.15in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern4.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern0in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.6in\lower1in\hbox to .3in{\hss$P_1$\hss}}%
\rlap{\kern1.6in\lower1in\hbox to .3in{\hss$P_2$\hss}}%
\rlap{\kern2.6in\lower1.0in\hbox to .3in{\hss$Q_1$\hss}}%
\rlap{\kern3.6in\lower1.0in\hbox to .3in{\hss$Q_2$\hss}}%
\rlap{\kern4.5in\lower1in\hbox to .3in{\hss$P_0$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 250 80} \special{pa 650 870} \special{fp}}%1
\rlap{\special{pa 850 40} \special{pa 650 870} \special{fp}}%4
\rlap{\special{pa 850 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 1650 80} \special{pa 1650 870} \special{fp}}%4
\rlap{\special{pa 2650 80} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 3650 80} \special{pa 3650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 4650 80} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 50 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 1650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 2650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 3650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 4550 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 18 pt
Let $R = k[y]_{(y)}$, where
$k$ is a countable algebraically closed field, and let
$D = R[x, 1/x(x-1)]$. Set $z = y/(x(x-1))$ and let
$C = R[x,z]$. Then $C$ is a regular ring. For $a,b$
nonzero elements of $k$,
consider the following valuations defined by ideals of $C$:
$v_1$ the ord valuation defined by powers of $(x,z)C$
$v_2$ the ord valuation defined by powers of $(x,z-a)C$
$w_1$ the ord valuation defined by powers of $(x-1,z)C$ and
$w_2$ the ord valuation defined by powers of $(x-1,z-b)C$;
\noindent
let $V_1, V_2, W_1, W_2$ be the associated valuation rings.
Let $A=D\cap V_1\cap V_2\cap W_1\cap W_2$ and let
$A_0 = C[(z(z-a))/x, (z(z-b))/(x-1)]$. We show that $A = A_0$.
Let us first note that $A_0$ is Cohen-Macaulay.
To see this, observe that at each maximal ideal $Q$ of $C$,
$(C-Q)^{-1}A_0$ is a domain generated by one element over
the regular local ring $C_Q$ and hence the quotient of a
regular ring modulo a principal ideal. Thus $A_0$ is locally
a complete intersection and hence is Cohen-Macaulay.
Since $A_0 \subseteq A$, to show $A_0 = A$ it suffices
to show that $A \subseteq (A_0)_P$ for each
height-one prime $P$ of $A_0$. If $x(x-1) \notin P$,
then $A \subseteq D \subseteq (A_0)_P$, so assume that
$x(x-1) \in P$. Suppose $x \in P$. Then $z$ or $z - a$ is
in $P$; say, $z \in P$. Set $Q = P \cap C$; then since
$z-a$ and $x-1$ are units in $C_Q$, we have
$(C - Q)^{-1}A_0 = C_Q[z/x]$, so
$$
PC_Q[z/x] = xC_Q[z/x] \qquad
\text{ and }\qquad (A_0)_P = (C_Q[z/x])_{xC_Q[z/x]} = V_1.
$$
Similarly, if $x,z-a \in P$, then $(A_0)_P = V_2$;
if $x-1, z \in P$, then $(A_0)_P = W_1$; and if
$x-1,z-b \in P$, then $(A_0)_P = W_2$. It follows that
$A_0 = A$ is an sfb-extension of $k[y]_{(y)}[x]$ with
$j$-Spec($A$) as in the picture above.
By an extension of this argument we can build examples that include
the examples in \cite{Heinzer et al. (1994b)} of cyclic sfb-extensions.
Using the gluing techniques from \cite{S. Wiegand (1983)} as in
the proof of Theorem 4.1
above, any maximal ideals above $P_1\cap P_2$ can be glued and any maximal
ideals above $Q_1\cap Q_2$ can be glued together. However the technique does
not permit gluing of pairs of maximals one above $P_i$ and one above $Q_j$
since the $P_i\cap R[x]$ and $Q_j\cap R[x]$ are comaximal.
Thus we can get this picture:
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern.2in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.8in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern1.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern3.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern4.15in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern4.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern0in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.6in\lower1in\hbox to .3in{\hss$P_1$\hss}}%
\rlap{\kern1.6in\lower1in\hbox to .3in{\hss$P_2$\hss}}%
\rlap{\kern2.6in\lower1.0in\hbox to .3in{\hss$Q_1$\hss}}%
\rlap{\kern3.6in\lower1.0in\hbox to .3in{\hss$Q_2$\hss}}%
\rlap{\kern4.5in\lower1in\hbox to .3in{\hss$P_0$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 250 80} \special{pa 650 870} \special{fp}}%1
\rlap{\special{pa 850 40} \special{pa 650 870} \special{fp}}%4
\rlap{\special{pa 850 40} \special{pa 1650 870} \special{fp}}%4
\rlap{\special{pa 850 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 1650 80} \special{pa 1650 870} \special{fp}}%4
\rlap{\special{pa 2650 80} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 3650 80} \special{pa 3650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 3650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 4650 80} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 50 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 1650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 2650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 3650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 4550 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 18 pt
But we can not get this picture:
\vskip 18 pt
\setbox4=\vbox{\hbox{%
\rlap{\kern.2in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.8in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern1.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern2.1in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern2.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern3.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern4.15in\lower0in\hbox to .3in{\hss$\bullet$\hss}}%
\rlap{\kern4.6in\lower0in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern0in\lower1in\hbox to .3in{\hss\boxed{\infty}\hss}}%
\rlap{\kern.6in\lower1in\hbox to .3in{\hss$P_1$\hss}}%
\rlap{\kern1.6in\lower1in\hbox to .3in{\hss$P_2$\hss}}%
\rlap{\kern2.6in\lower1.0in\hbox to .3in{\hss$Q_1$\hss}}%
\rlap{\kern3.6in\lower1.0in\hbox to .3in{\hss$Q_2$\hss}}%
\rlap{\kern4.5in\lower1in\hbox to .3in{\hss$P_0$\hss}}%
\rlap{\kern2.3in\lower1.8in\hbox to .3in{\hss$(0)$\hss}}%
\rlap{\special{pa 250 80} \special{pa 650 870} \special{fp}}%1
\rlap{\special{pa 850 40} \special{pa 650 870} \special{fp}}%4
\rlap{\special{pa 850 40} \special{pa 1650 870} \special{fp}}%4
\rlap{\special{pa 850 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 1650 80} \special{pa 1650 870} \special{fp}}%4
\rlap{\special{pa 2150 40} \special{pa 650 870} \special{fp}}%4
\rlap{\special{pa 2150 40} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 2650 80} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 3650 80} \special{pa 3650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 2650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 3650 870} \special{fp}}%4
\rlap{\special{pa 4200 40} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 4650 80} \special{pa 4550 870} \special{fp}}%4
\rlap{\special{pa 50 1080} \special{pa 2350 1670} \special{fp}}%10
\rlap{\special{pa 650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 1650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 2650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 3650 1080} \special{pa 2350 1670} \special{fp}}%11
\rlap{\special{pa 4550 1080} \special{pa 2350 1670} \special{fp}}%14
}}
\box4
\vskip 18 pt
That is, for the examples that arise from this construction, if the
transients $T_1$ and $T_2$ are contained in distinct maximals containing the
survivor, then $T_1$ and $T_2$ are comaximal in the glued ring.
\Refs
%\widestnumber\key{HLW2}
\ref \no 1 \by W. Heinzer, D. Lantz, and S. Wiegand
\paper Projective lines over one-dimensional semilocal domains
and spectra of birational extensions
\inbook Algebraic Geometry and Its Applications:
Collection of papers from Abhyankar's 60th birthday conference
\ed C. Bajaj
\publ Springer Verlag \publaddr New York \pages 309--325 \yr 1994
\endref
\ref\no 2 \by W. Heinzer, D. Lantz, and S. Wiegand
\paper Prime ideals in birational extensions of polynomial rings
\pages 73-93
\inbook Commutative Algebra: Syzygies, Multiplicities,
and Birational Algebra: Proceedings of a summer research conference
on commutative algebra held July 4-10, 1992
\ed W. Heinzer, C. Huneke, J.D. Sally
\bookinfo Contemporary Mathematics \vol 159
\publ Amer. Math. Soc. \publaddr Providence \yr 1994
\endref
\ref \no 3 \by W. Heinzer and S. Wiegand
\paper Prime ideals in two-dimensional polynomial rings
\pages 577-586\jour Proc. Amer. Math. Soc. \vol 107(3)\yr 1989
\endref
\ref \no 4 \by E. Kunz
\book Introduction to Commutative Algebra and Algebraic Geometry
\publ Birkha\" user \publaddr Boston \yr 1985
\endref
\ref \no 5 \by H. Matsumura
\book Commutative Algebra, Second Edition
\publ Benjamin \yr 1980
\endref
\ref \no 6 \by H. Matsumura
\book Commutative Ring Theory
\bookinfo Cambridge Studies in Advanced Mathematics {\bf 8}
\publ Cambridge University Press \yr 1989
\endref
\ref \no 7 \by R. Wiegand
\paper Homeomorphisms of affine surfaces over a finite field
\pages 28-32\jour J. London Math. Soc.(2) \vol 18 \yr 1978
\endref
\ref \no 8 \by R. Wiegand
\paper The prime spectrum of a two-dimensional affine domain
\pages 209--214 \jour J. Pure Appl. Alg. \vol 40 \yr 1986
\endref
\ref \no 9 \by S. Wiegand
\paper Intersections of prime ideals in Noetherian rings
\pages 1853-1876 \jour Communications in Algebra \vol 11(16) \yr 1983
\endref
\endRefs
\enddocument