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\NoBlackBoxes
\rightheadtext{ Domains inside power series rings}\topmatter
\title
Examples of integral domains inside\\
power series rings
\endtitle
\author
William Heinzer, Christel Rotthaus and Sylvia
Wiegand
\endauthor
% \date October 22, 2001 \enddate
\thanks{The authors would like to thank the National Science
Foundation and the National Security Agency for
support for this research. In addition they are grateful for
the hospitality and cooperation of Michigan State, Nebraska
and Purdue, where several work sessions on this research were
conducted.
}
\endthanks
\address{Department of Mathematics, Purdue
University,
West Lafayette, IN 47907-1395}
\endaddress
\address{Department of Mathematics, Michigan State University, East
Lansing,
MI 48824-1027}
\endaddress
\address{Department of Mathematics and Statistics,
University of Nebraska,
Lincoln, NE 68588-0323}
\endaddress
\abstract{We present examples of Noetherian and non-Noetherian
integral domains which can be built inside power series rings.
Given a power series ring $R^*$ over a Noetherian integral domain $R$
and given a subfield $L$ of the total quotient ring of $R^*$ with
$R \subseteq L$, we construct subrings $A$ and $B$ of $L$
such that $B$ is a localization of a nested union of
polynomial rings over $R$ and $B \subseteq A := L \cap R^*$.
We show in certain cases that flatness of a related
map on polynomial rings is equivalent to the
Noetherian property for $B$. Moreover if $B$ is Noetherian,
then $B = A$. We use this construction to obtain for
each positive integer $n$ an explicit example of a
3-dimensional quasilocal unique factorization domain $B$
such that the maximal ideal of $B$ is 2-generated, $B$ has
precisely $n$ prime ideals of height two, and each
prime ideal of $B$ of height two is not finitely generated.}
\endabstract
\endtopmatter
\document
\baselineskip 18pt
\subheading{1. Introduction }
This paper is a continuation of our study of a technique
for constructing integral domains by (1) intersecting a
power series ring with a field to obtain an
integral domain $A$ as in the abstract, and (2) approximating
the domain $A$ with a nested union of localized
polynomial rings to obtain an integral domain $B$ as in
the abstract. Classical
examples such as those of Akizuki \cite{A} and Nagata
\cite{N, pages~209-211} use the second (nested union) description
of this construction. It
is possible to also realize these classical examples as the
intersection domains of the first description \cite{HRW6}.
In this paper we observe that, in certain
applications of this technique, flatness of a
map of associated polynomial rings implies the constructed
domains are Noetherian and that $A = B$. We also in the present paper
apply this observation to the construction of examples of both
Noetherian and non-Noetherian integral domains.
We begin by describing the technique.
\subheading{1.1 General Setting}
Let $R$ be a commutative Noetherian integral domain.
Let $a$ be a nonzero
nonunit of $R$ and let $R^*$ be the $(a)$-adic completion of $R$.
Then $R^*$ is isomorphic to $R[[y]]/(y-a)$, where $y$ is an indeterminate;
thus we consider $R^*$ as $R[[a]]$, the ``power series ring" in $a$ over $R$.
The intersection domain (type 1 above) and the approximation domain
(type 2) of the construction
are inside $R^*$. Let $\tau_1, \dots, \tau_n \in aR^*$ be algebraically independent
over the fraction field $K$ of $R$ and let $\underline\tau$ abbreviate the
list $\tau_1, \dots, \tau_n$.
By Theorem 2.2, also known
as
\cite{HRW1, Theorem 1.1 }, $A_{\underline\tau} := K(\tau_1, \dots,
\tau_n) \cap R^*$
is simultaneously Noetherian and computable as a nested
union $B_{\underline\tau}$ of certain
associated localized polynomial rings over $R$
using ${\underline\tau}$ if and only if
the extension $T := R[\underline\tau]:=R[\tau_1, \dots, \tau_n]
\overset{\psi}\to \hookrightarrow R^*_a$
is flat.
In the case where $\psi : T \hookrightarrow R^*_a$ is flat, so that
the intersection domain $A_{\underline\tau}$ is Noetherian
and computable,
we construct new ``insider'' examples inside $A_{\underline\tau}$.
We choose elements $f_1, \dots, f_m$ of $T$, considered as
polynomials in the $\tau_i$ with coefficients in $R$ and abbreviated by
$\underline f$.
Assume that $f_1, \dots, f_m$ are algebraically independent
over $K$; thus $m \le n$.
If $S := R[\underline f] := R[f_1, \dots, f_m]
\overset{\varphi}\to\hookrightarrow T=R[\underline\tau]$
is flat, we observe in Section 3 that the
``insider ring'' $A_{\underline f} := K(\underline f) \cap R^*$ is
Noetherian and computable; that is, $A_{\underline f}$ is equal to an
approximating union $B_{\underline f}$ of localized polynomial rings
constructed using
the $f_i$. Moreover, we can often identify conditions
on the map $\varphi$ which imply $B_{\underline f}$ and
$A_{\underline f}$
are not Noetherian.
Thus the ``insider'' examples $A_{\underline f}$ and
$B_{\underline f}$ are inside
intersection domains $A_{\underline \tau}$ known to be Noetherian; the
new insider is Noetherian if
the associated extension $S\to T$ of
polynomial rings is flat. The insider examples are examined in
more detail in Section 3.
In Section 2 we give background and notation for the construction and
for flatness of polynomial extensions
in greater generality:
Suppose that $\underline x:=(x_1,\dots,x_n)$ is a tuple of
indeterminates over $R$ and that $\underline f:=(f_1,\dots,f_m)$
consists of elements of the polynomial
ring $R[\underline x]$ that are algebraically independent over $K$.
We consider flatness of the following map of polynomial rings.
$$\varphi: S := R[\underline f]\hookrightarrow T := R[\underline
x].\tag{1.2}$$
In Section 4 we continue the analysis of the flatness of (1.2) and the nonflat
locus. We discuss results of \cite{P}, \cite{W} and others.
In Section 5 we present for each positive integer $n$
an insider example $B$ such that:
\roster
\item $B$ is a
3-dimensional quasilocal unique factorization domain,
\item $B$ is not catenary,
\item the maximal ideal of $B$ is 2-generated,
\item $B$ has precisely $n$ prime ideals of height two,
\item Each prime ideal of $B$ of height two is not
finitely generated,
\item For every non-maximal prime $P$ of $B$ the ring $B_P$ is Noetherian.
\endroster
\subheading{2. Background and Notation}
We begin this section by recalling some details for
the approximation to the intersection domain $A_\tau$ of (1.1).
\subheading{2.1 Notation for approximations}
Assume that $R$, $K$, $a$, $\tau_1, \dots , \tau_n$, $\underline\tau$ and $A_{\underline\tau}$
are as in General Setting 1.1. Then the $(a)$-adic completion of
$R$ is $R^* = R[[x]]/(x-a) = R[[a]]$.
Write each $\tau_i: =\sum^{\infty}_{j=1} b_{ij}a^j$, with
the $b_{ij}\in R$.
There are natural sequences $\{\tau_{ir}\}_{r=0}^\infty$ of
elements in
$A$, called the
$r^{\text{th}}$ {\it endpiece}s for the $\tau_i$,
which ``approximate" the $\tau_i$, defined by:
$$\text{For each }i\in\{1,\dots,n\}\text{ and } r\ge 0,\quad
\tau_{ir}:=\sum_{j=r+1}^{\infty} (b_{ij}a^j)/a^r.\tag{2.1.1}$$
Now for each $r$,
$U_r:=R[\tau_{1r},\dots,\tau_{nr}]$ and $B_r$ is $U_r$
localized at the multiplicative system $ 1 + aU_r$. Then define
$U_{\underline\tau}:= \cup_{r=1}^\infty U_r$
and $B_{\underline\tau} := \cup_{r=1}^\infty B_r$. Thus $U_{\underline\tau}$ is
a nested union of polynomial rings over $R$ and $B_{\underline\tau}$ is a
nested union of localized polynomial rings over $R$. The
definition of the $U_r$ (and hence also of $B_r$ and $U_{\underline\tau}$
and $B_{\underline\tau}$) are independent of the representation of the
$\tau_i$ as power series with coefficients in $R$
\cite{HRW1, Proposition~2.3}.
The following theorem is the basis for our construction
of examples.
\proclaim{2.2 Theorem } \cite{HRW1, Theorem 1.1 } Let $R$ be a Noetherian
integral domain with fraction field $K$. Let $a$ be a nonzero
nonunit of $R$. Let $\tau_1,\dots, \tau_n\in aR[[a]]=aR^*$ be
algebraically
independent over $K$, abbreviated by ${\underline\tau}$. Let $
U_{\underline\tau}$ and $B_{\underline\tau}$ be as in (2.1).
Then the following statements
are equivalent:\roster\item
$A_{\underline\tau}:=K(\underline\tau)\cap R^*$ is Noetherian and
$A_{\underline\tau}=B_{\underline\tau}$.
\item $U_{\underline\tau}$ is Noetherian.
\item $B_{\underline\tau}$ is Noetherian.
\item $R[{\underline\tau}]\to R^*_a$ is
flat. \endroster
\endproclaim
Since flatness is a local property, the
following two propositions are immediate corollaries of
\cite{HRW5, Theorem 2.1}; see also
\cite{P, Th\'eor\`eme~3.15}.
\proclaim{2.3 Proposition} Let $T$ be a Noetherian ring and
suppose $R \subseteq S$ are Noetherian subrings of $T$.
Assume that $R \to T$ is flat with
Cohen-Macaulay fibers and that $R \to S$ is flat with regular fibers.
Then $S \to T$ is flat if and only if, for each prime ideal $P$ of $T$,
we have $\text{ht}(P)\ge \text{ht}(P\cap S)$.
\endproclaim
As a special case we have:
\proclaim{2.4 Proposition}
Let $R$ be a Noetherian ring and let $x_1,\dots,x_n$
be indeterminates over $R$. Assume that
$f_1,\dots,f_m\in R[x_1,\dots,x_n]$ are algebraically
independent over $R$. Then
\roster
\item
$\varphi : S:=R[f_1,\dots,f_m]\hookrightarrow
T:=R[x_1,\dots,x_n] $ is flat if and only if, for
each prime ideal $ P$ of $T$,
we have $\text{ht}(P)\ge \text{ht}(P\cap S)$.
\item
For $Q \in \Spec T$,
$\varphi_Q : S \to T_Q $ is flat if and only if for
each prime ideal $ P \subseteq Q$ of $T$,
we have $\text{ht}(P)\ge \text{ht}(P\cap S)$.
\endroster
\endproclaim
\subheading{2.5 Definitions and Remarks} (1)
The {\it Jacobian ideal} $J$ of the extension
(1.2) is the ideal generated by the
$m \times m$ minors of the $m \times n$ matrix $
\Cal J$ given below:
$$
\Cal J :=\left( \frac{\partial f_i}{\partial x_j}\right)_{i,j}.
$$
(2) For the extension (1.2),
the {\it nonflat locus } of $\varphi$ is the set $\Cal F$, where
$$
\Cal F := \{Q \in \Spec(T) : \text{the map} \quad \varphi_Q: S \to T_Q
\text{ is not flat } \}.
$$
For convenience, we also define the set $\Cal F_{\text{min}}$ and
the ideal $F$ of $T$:
$$
\Cal F_{\text{min}} := \{\text{ minimal elements of } \Cal F \}
\text{ and }
F := \cap\{ Q : Q \in \Cal F \}.
$$
By \cite{M2, Theorem 24.3}, the set $\Cal F$ is closed in the Zariski
topology and hence is equal to
$\Cal V( F)$, the set of primes of $T$ that
contain the ideal $F$. Thus the set
$\Cal F_{\text{min}}$ is a finite set and consists precisely
of the minimal primes of the ideal $F$.
Moreover, Proposition 2.4 implies
$\Cal F_{\text{min}} \subseteq \{Q \in \Spec T : \hgt Q < \hgt(Q
\cap S) \}$
and for every prime ideal $P \subsetneq Q
\in \Cal F_{\text{min}}, \, \hgt P \ge \hgt(P \cap S)$.
(3) In general for a commutative ring $T$ and a subring $R$, we say
that elements $f_1,\dots,f_m \in T$ are {\it algebraically
independent} over $R$
if, for indeterminates $t_1,\dots,t_m$ over $R$,
the only polynomial $G(t_1,\dots,t_m)\in R[t_1,\dots,t_m]$
with $G(f_1,\dots,f_m)=0$
is the zero polynomial.
\subheading{2.6 Example and Remarks} (1)
Let $k$ be a field, let $x$ and $y$ be indeterminates
over $k$ and set $f=x, \,g=(x-1)y$.
Then $k[f,g] @>{\varphi}>>
k[x,y]$ is not flat.
\demo{Proof} For the prime ideal $P:=(x-1)\in\Spec(k[x,y]) $, we see
that $\hgt(P)=1$,
but $\hgt(P \cap k[f,g])=2 $; thus the extension is not flat by
Proposition~2.4.
\enddemo
(2) The Jacobian ideal $J$ of $f$ and $g$ in (1) is given by:
$$ J= (\text{det}\left(\matrix\frac{\delta f}{\delta x} &
\frac{\delta f}{\delta y}\\ \frac{\delta g}{\delta x} &
\frac{\delta g}{\delta y}\endmatrix\right))\, =
(\text{det}\left(\matrix 1 & 0\\ y & x-1\endmatrix\right)) = (x-1).$$
(3) In this example the nonflat locus is equal to the set of
prime ideals $Q$ of $k[x,y]$ which contain the Jacobian ideal
$(x-1)k[x,y]$, thus $J = F$.
We record in Proposition 2.7 observations about flatness that
follow from well-known properties of the Jacobian.
\proclaim{2.7 Proposition}
Let $R$ be a Noetherian ring, let
$x_1,\dots,x_n$ be
indeterminates over $R$, and let $f_1, \dots,f_m \in R[x_1,\dots,x_n]$
be algebraically independent over $R$. Consider
the embedding
$\varphi: S := R[f_1, \dots, f_m]
\hookrightarrow T := R[x_1,\dots,x_n]$.
Let $J$ denote the Jacobian ideal of $\varphi$ and let $Q \in \Spec T$.
Then
\roster
\item
$Q $ does not contain $J$ if and only if
$\varphi_Q : S \to T_Q$ is essentially smooth.
\item
If $Q$ does not contain $J$, then $\varphi_Q : S \to T_Q$ is flat.
Thus $J \subseteq F$.
\item
$\Cal F_{\text{min}} \subseteq \{ Q' \in \Spec T : J \subseteq Q'
\text{ and } \hgt(Q' \cap S) > \hgt Q' \}$.
\endroster
\endproclaim
\demo{Proof}
For item 1, we observe that our definition of the Jacobian ideal
$J$ given
in (2.5) agrees with the description of the smooth locus
of an extension given in \cite{E}, \cite{S, Section 4}.
To see this, let $u_1,\dots,u_m$ be indeterminates
over $R[x_1,\dots,x_n]$ and identify
$$
R[x_1,\dots,x_n] \quad \text{ with } \quad
\frac{R[u_1,\dots,u_m][x_1,\dots,x_n]}{(\{ u_i-f_i\}_{i=1,\dots,m})}.
$$
Since $u_1,\dots,u_m$ are algebraically independent,
the ideal $J$ generated by the minors of $\Cal J$
is the Jacobian ideal of
the extension (1.2) by means of this identification.
We make this more explicit as follows.
Let $U_1:= R[u_1, \dots, u_m, x_1, \dots, x_n]$ and
$I = (\{ f_i-u_i\}_{i=1,\dots,m})U_1$.
Consider the following commutative diagram
$$\CD
S:=R[f_1,\dots,f_m] @>>> T:= R[x_1,\dots,x_n] \\
@V{\cong}VV @V{\cong}VV\\
S_1 := R[u_1,\dots,u_m]
@>>> T_1 := R[u_1, \dots, u_m, x_1,\dots,x_n]/I
\endCD $$
Define as in \cite{E}, \cite{S, Section 4}
$$H = H_{T_1/S_1}:=\text{the radical of }
\Sigma \Delta(g_1,\dots,g_s)[(g_1,\dots,g_s):I],$$
where the sum is taken over all $s$ with $0\le s\le m$,
for all choices of $s$ polynomials $g_,\dots,g_s$
from $I = (\{ f_1-u_1,\dots,f_m-u_m\})U_1$, where
$\Delta := \Delta(g_1,\dots,g_s)$ is
the ideal of $T \cong T_1$ generated by the $s\times s$-minors of
$\left(\frac{\partial g_i}{\partial x_j}\right)$, and $\Delta=T$ if $s=0$.
To establish (2.7.1), we show that $H = \rad(J)$.
Since $u_i$ is a constant with respect to $x_j$,
we have
$\left(\frac{\partial ( f_i- u_i)}{\partial x_j}\right) =
\left(\frac{\partial f_i}{\partial x_j}\right)$. Thus
$J \subseteq H$.
For
$g_1,\dots,g_s\in I$, the $s \times s$-minors of
$\left(\frac{\partial g_i}{\partial x_j}\right)$ are
contained in the $s\times s$-minors of
$\left(\frac{\partial f_i}{\partial x_j}\right)$.
Thus it suffices to consider $s$ polynomials $g_1,\dots,g_s$
from the set $\{ f_1-u_1,\dots,f_m-u_m\}$. Now
$f_1-u_1,\dots, f_m-u_m$ is a regular sequence in
$R[ u_1\dots u_r,x_1,\dots,x_n]$. Thus for $s \hgt Q
\text{ and } \forall P \subsetneq Q,
\hgt(P) \ge \hgt(P \cap S) \}$.
\noindent
(4)
By (2.8.3), every prime ideal $Q$ of
$\Cal F_{\text{min}}$ contains two primes $P_1\subsetneq P_2$ of
$S $ such that $Q$ is minimal above both $P_1T$ and $P_2T$.
\subheading{3. Explicit constructions inside simpler extensions}
Using Theorem 2.2 and intersection domains inside the
completion which are known to be Noetherian, we formulate
a shortcut method for the
construction of ``insider'' examples.
\subheading{3.1 General Method}
Let $R$ be a Noetherian integral domain.
Let $a$ be a nonzero
nonunit of $R$ and let $R^* = R[[x]]/(x-a)$
be the $(a)$-adic completion of $R$.
Let $\tau_1, \dots, \tau_n \in aR^*$, abbreviated by
${\underline\tau}$, be algebraically independent
over the fraction field $K$ of $R$. Assume that
the extension $T := R[\tau_1, \dots, \tau_n]
\overset{\psi}\to \hookrightarrow R^*_a$
is flat. Thus by Theorem~2.2,
$D := A_{\underline\tau}=K(\tau_1, \dots, \tau_n) \cap R^*$
is Noetherian and computable as a nested
union of localized polynomial rings over $R$
using the $\tau$'s.
Let $f_1, \dots, f_m$ be elements of $T$, abbreviated by
${\underline f}$ and considered as
polynomials in the $\tau_i$ with coefficients in $R$.
Assume that $f_1, \dots, f_m$ are algebraically independent
over $K$; thus $m \le n$.
Let $S := R[\underline f]
\overset{\varphi}\to\hookrightarrow T = R[\underline\tau]$;
%, where
%$\underline\tau=(\tau_1, \dots, \tau_n)$ and
%$\underline f=(f_1, \dots, f_m)$;
put $\alpha := \psi \circ \varphi : S \to R^*_a$. That is, we have:
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\box4
\vskip 10 pt
Using the $f$'s in place of the $\tau$'s,
we define the ring $A:=A_{\underline f} := K(\underline f) \cap R^*$
and the approximation rings $U_r, B_r, U_{\underline f}$ and $
B=B_{\underline f}$, as in (2.1).
Let $$F := \cap\{P \in \Spec (T) \, | \,
\varphi_P: S \to T_P \text{ is not flat } \}.$$ Thus, as
in (2.5.2), the ideal $F$ defines the nonflat
locus of the map $\varphi : S \to T$.
For $Q^* \in \Spec(R^*_a)$, we consider whether the localized map
$\varphi_{Q^*\cap T}$ is flat:
$$
\varphi_{Q^*\cap T}:S \to T_{Q^*\cap T} \quad \tag{3.1.1}
$$
\proclaim{3.2 Theorem }
With the notation of (3.1) we have
\roster
\item
For $Q^* \in \Spec(R^*_a)$, the map $\alpha_{Q^*} : S \to (R^*_a)_{Q^*}$
is flat if and only if the map $\varphi_{Q^*\cap T}$ in (3.1.1)
is flat.
\item
The following are equivalent:
$\phantom{x}$
(i) $A$ is Noetherian and $A = B$.
$\phantom{x}$
(ii) $B$ is Noetherian.
$\phantom{x}$
(iii) The map $\varphi_{Q^*\cap T}$ in (3.1.1)
is flat for every maximal $Q^* \in \Spec(R^*_a)$.
$\phantom{x}$
(iv) $FR^*_a = R^*_a$.
\item $\varphi_a : S \to T_a$ is flat if and only if $FT_a = T_a$.
Moreover, either of these conditions implies $B$ is Noetherian and
$B = A$.
\endroster
\endproclaim
\demo{Proof}
For item (1), we have
$\alpha_{Q^*} = \psi_{Q^*} \circ \varphi_{Q^* \cap T} :
S \to T_{Q^* \cap T} \to (R^*_a)_{Q^*}$. Since
the map ${\psi}_{Q^*}$ is faithfully flat, the
composition $\alpha_{Q^*}$ is flat if and only if
${\varphi}_{Q^* \cap T}$ is flat
\cite{M1, page~27}.
For item (2), the equivalence of (i) and (ii) is part of
Theorem~2.2. The equivalence of (ii) and (iii) follows from
item (1) and Theorem~2.2. For the equivalence of (iii) and (iv),
we use $FR^* \ne R^* \iff F \subseteq Q^* \cap T$, for some
$Q^*$ maximal in $\Spec(R^*)_a) \iff$ the map in (3.1.1) fails to be flat. Item (3) follows from the definition of $F$ and the fact
that the nonflat locus of $\varphi : S \to T$ is closed. \qed
\enddemo
To examine the map $\alpha : S \to R^*_a$ in more detail, we use
the following terminology.
\subheading{3.3 Definition } For an extension of Noetherian
rings $\varphi:A'\hookrightarrow B'$ and for
$d \in \N$, we say that $\varphi:A'\hookrightarrow B',$
{\it satisfies }LF$_d$ if for each $P \in \Spec (B')$ with $\hgt(P)
\le d$, the composite map $A' \to B' \to B'_P$ is flat.
\proclaim
{3.4 Corollary} With the notation of (3.1),
we have $\hgt(FR^*_a)>1 \iff \varphi : S \to R^*_a$ satisfies
LF$_1 \iff B = A$.
\endproclaim
\demo{Proof} The first equivalence follows from the
definition of LF$_1$ and the second equivalence from \cite{HRW4,Theorem~5.5}.
\enddemo
\subheading{3.5 A more concrete situation}
Let $R := k[x, y_1, \dots,
y_s]$, where $k$ is a field and $x, y_1, \dots, y_s$ are
indeterminates over $k$ with the $y_i$ abbreviated by
${\underline y}$.
Let $R^* = k[\underline y][[x]]$, the $(x)$-adic
completion of $R$.
Let
$\tau_1, \dots,\tau_n$, abbreviated by
${\underline\tau}$, be elements of $xk[[x]]$ which
are algebraically independent over $k(x)$.
Let $D := A_ {\underline\tau}:=k(x, {\underline y}, {\underline\tau}) \cap R^*$.
Let $T = R[{\underline\tau}]$.
Then $T \to R^*_x$ is flat,
$D$ is a nested union of localized polynomial rings obtained using
the $\tau_i$ and
$D$ is a Noetherian regular local ring;
moreover, if $\char k = 0$, then $D$ is excellent
\cite{HRW3, Proposition~4.1}.
We now use the procedure of (3.1) to construct examples inside $D$. Let
$f_1, \dots,f_m$, abbreviated by ${\underline f}$, be elements of $T$ considered as
polynomials in $\tau_1, \dots, \tau_n$ with coefficients in $R$,
that are algebraically independent over
$ k(x, {\underline y})$.
We assume the constant terms
in $R = k[x, {\underline y}]$ of the $f_i$ are zero.
Let $S := R[{\underline f}]$.
The inclusion map $S \hookrightarrow
T$ is an injective $R$-algebra homomorphism, and $m \le n$.
Let $A := \Cal Q(S) \cap R^*$ and let $B$ be the nested union domain
associated to the ${\underline f}$,
as in (2.1). By Theorem 2.2, $B$ is Noetherian
and $B=A$ if and only if
the map $\alpha : S \to R^*_x$ is flat. Furthermore, by
Theorem 3.2, we can recover information about flatness of
$\alpha$ by considering the map $\varphi : S \to T$.
The following remark describes how the $f_i$ are chosen in several
classical examples:
\subheading{3.6 Remark } With the notation of (3.5).
\roster
\item Nagata's famous example \cite{N1}, \cite{N2, Example 7, page~209},
\cite{HRW6, Example~3.1}, may
be described by taking $n = s = m = 1, y_1 = y, \tau_1 = \tau$,
and $f_1 = f$ and localizing. Then $R = k[x, y]_{(x,y)}$, $T = k[x,y,
\tau]_{(x, y, \tau)}$, $f = (y + \tau)^2$, $S = k[x, y,
f]_{(x, y, f)}$ and $A = k(x,y,(y + \tau)^2) \cap R^*$.
The Noetherian property of $B$ is implied by the
flatness property of the map $S \to T_x$. Thus $B = A$. In this case,
$T$ is actually a free $S$-module with $<1, y+\tau>$ as a
free basis.
\item An example of Rotthaus \cite{R1},\cite{HRW6, Example~3.3},
may be described by taking
$n = s = 2$, and $m =1$ and localizing. Then $R = k[x, y_1, y_2]_{(x, y_1,
y_2)}$, $T = R[\tau_1, \tau_2]_{(\m, \tau_1, \tau_2)}$, $f_1 =
(y_1 + \tau_1)(y_2 + \tau_2)$, $S = R[f_1]_{(\m, f_1)}$ and
$A = k(x, y_1, y_2, (y_1 + \tau_1)(y_2 + \tau_2)) \cap R^*$.
Since the map from $R[f_1] \to R_x[\tau_1, \tau_2] = T_x$
is flat, the associated nested union domain $B$ is Noetherian.
\item The following example is given in \cite{HRW5, Section~4}.
Let $n = s = m = 2$, let $f_1 = (y_1 + \tau_1)^2$ and $f_2 = (y_1
+ \tau_1)(y_2 + \tau_2)$. It is shown in \cite{HRW5} for this
example that $B \subsetneq A$ and that both $A$ and $B$ are
non-Noetherian.
\endroster
The following lemma follows from \cite{P, Proposition~2.1}
in the case of one indeterminate $x$, so in the
case where $T = R[x]$.
\proclaim{3.7 Lemma} Let $R$ be a Noetherian ring, let
$x_1, \dots,x_n$ be indeterminates over $R$, and let
$T = R[x_1, \dots x_n]$. Suppose $f \in T - R$ is such that
the constant term of $f$ is zero. Then the following are
equivalent:
\roster
\item $R[f] \to T$ is flat.
\item $R[f] \to T$ is faithfully flat.
\item For each maximal ideal $q$ of $R$, we have
$qT \cap R[f] = qR[f]$.
\item The coefficients of $f$ generate the unit
ideal of $R$.
\endroster
\endproclaim
\demo{Proof} (1) $\implies$ (2): It suffices
to show for $P \in \Spec(R[f])$ that $PT \ne T$.
Let $q = P \cap R$ and let $k(q)$ denote the
fraction field of $R/q$.
Since $R[f] \to T$ is flat,
tensoring with $k(q)$ gives injective maps
$$
k(q)\to k(q)\otimes_R R[f] \cong k(q)[f'] @>{\varphi}>> k(q)
\otimes_R T
\cong k(q)[x_1, \dots, x_n],
$$
where $f'$ is the image of $f$ in $k(q)[x_1, \dots, x_n]$.
The injectivity of $\varphi$ implies $f'$ has positive total degree
as a polynomial in $k(q)[x_1, \dots,x_n]$.
The image $p'$ of $P$ in $k(q)[f']$ is either zero or a
maximal ideal of $k(q)[f']$. It suffices to show
$p'k(q)[x_1, \dots,x_n] \ne k(q)[x_1, \dots,x_n]$. If
$p' = 0$, this is clear. Otherwise $p'$ is generated
by a nonconstant polynomial $h(f')$ and $p'k(q)[x_1, \dots,x_n]$
is generated by $h(f'(x_1,\dots,x_n))$ which has total degree
equal to $\deg(h) \deg(f') > 0$.
Thus (1) implies (2).
(2) $\implies$ (3): This follows from Theorem 7.5 (ii) of \cite{M2}.
(3) $\implies (4)$ : If the coefficients of $f$ were
contained in a maximal ideal $q$ of $R$, then
$f \in qT \cap R[f]$, but $f \not\in qR[f]$.
(4) $\implies$ (1):
Let $v$ be another indeterminate and consider the commutative
diagram
$$
\CD
R[v] @>>> T[v] = R[x_1, \dots, x_n, v] \\
@V{\pi}VV @V{\pi'}VV \\
R[f] @>{\varphi}>> \frac{R[x_1, \dots,x_n, v] }{(v - f(x_1, \dots,
x_n))}.
\endCD
$$
where $\pi$ maps $v \to f$ and $\pi'$ is the canonical quotient
homomorphism. By \cite{M1, Corollary~2, p.~152}, $\varphi$ is
flat if the coefficients of $f - v$ generate the unit ideal of $R[v]$.
Moreover, the coefficients of $f-v$ as a polynomial in
$x_1, \dots, x_n$ with coefficients in $R[v]$ generate the unit ideal of
$R[v]$ if and only if the nonconstant coefficients of $f$
generate the unit ideal of $R$. \qed
\enddemo
%In connection with Proposition~3.8,
We observe in Proposition~3.8 that one direction of (3.7) holds for more than one polynomial:
see also
\cite{P, Theorem~3.8} for a related result concerning flatness.
\proclaim{3.8 Proposition } Assume the notation of (3.7) except that $f_1,\dots,f_m\in T$ are polynomials in
$x_1,\dots,x_n$ with
coefficients in $R$ and $m\ge 1$.
If
the inclusion map $\varphi:S = R[f_1, \dots f_m] \to T$ is flat, then the nonconstant coefficients of
each of the $f_i$
generate the unit ideal of $R$.
\endproclaim
\demo{Proof} Since $f_1,\hdots,f_m$ are algebraically independent over $\Cal
Q(R) =K$, for every $1\le i\le m$, the inclusion $R[f_i]\hookrightarrow
R[f_1,\hdots, f_m]$ is flat. If $S \longrightarrow T$ is flat, so is the
composition $R[f_i]\longrightarrow S=R[f_1,\hdots, f_m]\longrightarrow
T$ and the statement follows from Proposition
3.7. \qed
\enddemo
\proclaim{3.9 Theorem} Assume the notation of (3.1).
If $m = 1$, that is, there is only one polynomial $f_1 =
f$, then \roster
\item
The map $S \to T_a$ is flat $\iff$
the nonconstant coefficients of $f$ generate the unit ideal in $R_a$,
\item Either of the conditions in (1) implies the constructed ring $A$ is
Noetherian and $A = B$.
\item $B$ is Noetherian and $A=B$ \,$\iff$ for every prime ideal
$Q^*$ in $R^*$
with $a \not\in Q^*$,
the nonconstant coefficients of $f$ generate the unit ideal in
$R_q$, where $q:= Q^* \cap R$.
\item If the nonconstant coefficients of $f_1 = f$
generate an ideal $L$ of $R_a$ of height $d$, then the map $S
\to R^*_a$ satisfies LF$_{d-1}$, but not LF$_d$.
\endroster
\endproclaim
\demo{Proof}
Item (1) follows from Lemma 3.7 for the ring $R_a$ with
$x_i = \tau_i$.
By Theorems 2.2 and 3.2, the first condition in item (1)
implies item (2).
For item (3), suppose the nonconstant coefficients of $f$
generate the unit ideal of $R_q$. Then by Lemma 3.7,
$R_q[f] \to R_q[\tau_1, \dots, \tau_n]$ is flat.
Since
$R_q[\tau_1, \dots, \tau_n] \to R^*_{Q^*}$ is flat,
$R_q[f] \to R^*_{Q^*}$ is also flat.
For the other direction, suppose
there exists $Q^* \in \Spec R^*$ with $a \not\in Q^*$
such that the nonconstant coefficients of $f$
are in $qR_q$, where $q = Q^* \cap R$.
If $R[f] \to R^*_{Q^*}$ were flat, then,
since $qR^*_{Q^*} \neq R^*_{Q^*}$, we
would have $qR^*_{Q^*} \cap R[f] = qR[f]$.
This would imply $f \in qR^*_{Q^*} \cap R[f]$, but $f \not\in qR[f]$,
a contradiction.
For item (4), if $Q^* \in \Spec(R^*_a)$ the map
$S \to (R^*_a)_{Q^*} $ is not flat if and only if $L \subseteq Q^*$.
By hypothesis there exists such a prime ideal of height $d$,
but no such prime ideal of height less than $d$.
\enddemo
\subheading{3.10 Example} With the notation of (3.5), let
$m = 1$ and assume that $n$ and $s$ are each greater than or
equal to $d$. Then $f_1 = f := y_1\tau_1 + \cdots + y_d\tau_d$
gives an example where $S \to T_x$ satisfies LF$_{d-1}$,
but fails to satisfy LF$_d$. For $d \ge 2$ this gives examples
where $A =B$, i.e., $A$ is ``limit-intersecting'', but is not
Noetherian.
The following is a related even simpler
example: In the notation of (3.5), let $m = 1, n = 1$, and $s = 2$;
that is, $R=k[x,y_1,y_2]_{(x,y_1,y_2)}$ and
$\tau\in xk[[x]]$. If $f_1=f = y_1\tau + y_2{\tau}^2$, then
the constructed intersection
domain $A:=R^*\cap k(x,y_1,y_2,f)$ is not Noetherian. Thus
we have a situation where $B = A$ is not Noetherian. This
gives a simpler example of such behavior than the
example given in Section 4 of \cite{HRW2}.
In dimension two (the two variable case), Valabrega
proved the following.
\proclaim {3.11 Proposition \cite{V, Prop. 3}} For $R=k[x,y]_{(x,y)}$
with completion $\widehat R = k[[x, y]]$, if
$L$ is a field between the fraction field of $R$ and the fraction
field $F$
of $k[y]\,[[x]]$, then $A = L \cap \widehat R$ is a
two-dimensional regular local domain with completion $\widehat
R$.\endproclaim
Example 3.10 shows that the dimension three analog to
Valabrega's result fails. With $R=k[x,y_1,y_2]_{(x,y_1,y_2)}$
the field $L= k(x,y_1,y_2,f)$ is
between $k(x,y_1,y_2)$ and the fraction field of $k[y_1,y_2]\,[[x]]$,
but $ L\cap \widehat R=L\cap R^*$ is not Noetherian.
\subheading{3.12 Remark} With the notation of (3.1), it can
happen that $\varphi_a : S \to T_a$ is not flat, but
$\alpha : S \to R^*_a$ is flat. For example, using the notation
of (3.5), let $R : = k[x, y]$, where $k$ is a field and $x, y$ are
indeterminates over $k$. Let $\sigma, \tau \in xk[[x]]$ be such
that $x, \sigma, \tau$ are algebraically independent over $k$,
let $T := R[\sigma, \tau]$, and let $S := R[\sigma, \sigma\tau]$.
Then $\varphi_x : S \to T_x$ is not flat since $\sigma T_x$
is a height-one prime such that
$\sigma T_x \cap S = (\sigma, \sigma \tau)S$ has
height 2. To see that $R^*_x$ is flat over $S$, observe that
$\dim R^*_x = 1$ and if $Q^* \in \Spec R^*_x$, then
$Q^* \cap k[x, \sigma, \sigma \tau] = (0)$. Therefore
$\hgt(Q^* \cap S) \le 1$.
\subheading{4. Flatness of maps of polynomial rings}
\proclaim{4.1 Proposition} Let $k$ be a field, let $x_1,\dots,x_n$ be
indeterminates over $k$, and let $f_1, \dots,f_m \in k[x_1,\dots,x_n]$
be algebraically independent over $k$. Consider
the embedding $\varphi: S := k[f_1, \dots, f_m] \hookrightarrow T:=
k[x_1,\dots,x_n]$
and let $J$ denote the Jacobian ideal of $\varphi$. Then
\roster
\item
$\Cal F_{\text{min}} \subseteq \{ Q \in \Spec T : J \subseteq Q,
\, \hgt Q \le m - 1 \text{ and } \hgt Q < \hgt(Q \cap S) \}$.
\item
$\varphi$ is flat $\iff$ for every
$Q \in \Spec(T)$ such that $\hgt(Q) \le m-1$ and $J \subseteq Q$
we have $\hgt(Q\cap S) \le \hgt(Q)$.
\item If $\hgt J \ge m$, then $\varphi$ is flat.
\endroster
\endproclaim
\demo{Proof} For item 1,
if $\hgt(Q) \ge m$, then $\hgt(Q \cap S) \le \dim(S) = m \le \hgt(Q)$, so
by (2.3) $S \to T_Q$ is flat. Therefore $Q \not\in \Cal F_{\text{min}}$.
Item 1 now follows from (2.7.3).
The ($\implies$) direction of item 2 is clear \cite{M2, Theorem~9.5}.
For ($\impliedby$) of item 2 and for item 3, it suffices to show
$\Cal F_{\text{min}}$ is empty and this
holds by item 1. \qed
\enddemo
The following is an immediate corollary to (4.1).
\proclaim{4.2 Corollary} Let $k$ be a field,
let $x_1,\dots,x_n$ be indeterminates over $k$ and
let $f,g\in k[x_1,\dots,x_n]$ be algebraically
independent over $k$. Consider the
embedding $\varphi: S := k[f,g] \hookrightarrow
T := k[x_1,\dots,x_n]$
and let $J$ be the associated Jacobian ideal.
Then
\roster
\item
$\Cal F_{\text{min}} \subseteq \{ \text{minimal primes $Q$ of $J$ with
$\hgt(Q \cap S) > \hgt Q = 1$ } \}$.
\item
$\varphi$ is flat $\iff$ for every height-one prime
ideal $Q \in \Spec T$ such that
$J\subseteq Q$ we have $\hgt(Q\cap S)\le 1$.
\item If $\hgt(J) \ge 2$, then $\varphi$ is flat.
\endroster
\endproclaim
In the case where $k$ is algebraically closed,
another argument can be used for (4.2.2): Each height-one
prime ideal $Q \in \Spec T$ has the form
$Q = hT$ for some element $h \in T$. If $\hgt(P\cap S)=2$, then
$Q \cap S$ has the form $(f-a,g-b)S$, where $a,b\in k$.
Thus $f-a =f_1h$ and $ g-b = g_1h$ for some $f_1, g_1 \in T$.
Now the Jacobian ideal of $f, g$
is the same as the Jacobian ideal of
$f-a , g-b$ and an easy computation shows this has $h$ as a factor.
Thus $Q$ contains the Jacobian ideal, and so by assumption,
$\hgt(Q\cap S) \le 1$, a contradiction.
\subheading{4.3 Examples} Let $k$ be a field of characteristic
different from 2 and let $x, y, z$ be indeterminates over $k$.
(1) With $f = x$ and $g = xy^2 - y$,
consider $ S := k[f, g] @>{\varphi}>> T := k[x,y]$. Then
$J = (2xy -1)T$. Since $\hgt((2xy - 1)T \cap S) = 1$,
$\varphi$ is flat. Hence $J \subsetneq F = T$.
(2) With $f =x$ and $g = yz$, consider
$ S := k[f, g] @>{\varphi}>> T := k[x,y, z]$. Then
$J = (y, z)T$. Since $\hgt J \ge 2$,
$\varphi$ is flat. Again $J \subsetneq F = T$.
\medskip
We are interested in extending Prop. 4.1 to the case of polynomial
rings over
a Noetherian domain. In this connection we first
consider behavior with respect to prime ideals of $R$ in a
situation where the extension (1.2) is flat.
\proclaim{4.4 Proposition} Let $R$ be a commutative ring, let
$x_1,\dots,x_n$ be
indeterminates over $R$, and let $f_1, \dots,f_m \in R[x_1,\dots,x_n]$
be algebraically independent over $R$. Consider
the embedding $\varphi: S := R[f_1, \dots, f_m]
\hookrightarrow T := R[x_1,\dots,x_n]$.
\roster
\item
If $\p \in \Spec R$ and $\varphi_{\p T} : S \to T_{\p T}$ is
flat, then $\p S = \p T \cap S$ and the images $\overline{f_i}$
of the $f_i$ in
$T/\p T \cong (R/\p)[x_1, \dots, x_n]$ are algebraically independent
over $R/\p$.
\item
If $\varphi$ is flat, then for each $\p \in \Spec(R)$
we have $\p S = \p T \cap S$ and the images $\overline{f_i}$
of the $f_i$ in
$T/\p T \cong (R/\p)[x_1, \dots, x_n]$ are algebraically independent
over $R/\p$.
\endroster
\endproclaim
\demo{Proof} Item 2 follows from item 1, so it
suffices to prove item 1. Assume that $T_{\p T}$ is
flat over $S$. Then $\p T \ne T$ and it follows from
\cite{M2, Theorem~9.5} that $\p T \cap S = \p S$.
If the $\overline{f_i}$ were algebraically dependent
over $R/\p$, then there exist indeterminates $t_1, \dots, t_m$
and a polynomial $G \in R[t_1, \dots, t_m] - \p R[t_1, \dots, t_m]$
such that $G(\overline{f_1}, \dots, \overline{f_m}) \in \p T$.
This implies $G(f_1, \dots, f_m) \in \p T \cap S$.
But $f_1, \dots, f_m$ are algebraically independent over $R$
and $G(t_1, \dots, t_m) \not\in \p R[t_1, \dots, t_m]$ implies
$G(f_1, \dots, f_m) \not \in \p S = \p T \cap S$, a
contradiction. \qed
\enddemo
\proclaim{4.5 Proposition} Let $R$ be a Noetherian
integral domain, let $x_1,\dots,x_n$ be
indeterminates over $R$ and let $f_1, \dots,f_m \in R[x_1,\dots,x_n]$
be algebraically independent over $R$. Consider
the embedding $\varphi: S := R[f_1, \dots, f_m]
\hookrightarrow T := R[x_1,\dots,x_n]$ and let
$J$ denote the Jacobian ideal of $\varphi$. Then
\roster
\item
$\Cal F_{\text{min}} \subseteq \{ Q \in \Spec T : J \subseteq Q, \,
\dim(T/Q) \ge 1 \text{ and } \hgt(Q \cap S) > \hgt Q \}$.
\item
$\varphi$ is flat $\iff$ $\hgt(Q \cap S) \le \hgt(Q)$
for every nonmaximal $Q \in \Spec(T)$ with $J \subseteq Q$.
\item
If $\dim R = d$ and $\hgt J \ge d + m$, then $\varphi$ is flat.
\endroster
\endproclaim
\demo{Proof}
For item 1, suppose $Q \in \Cal F_{\text{min}}$ is a maximal ideal
of $T$.
Then $\hgt Q < \hgt(Q \cap S)$ by (2.4.2).
By localizing at
$R - (R \cap Q)$, we may assume that $R$ is local
with maximal ideal $Q \cap R := \m$. Since $Q$ is
maximal, $T/Q$ is a field finitely generated
over $R/\m$. By the Hilbert Nullstellensatz \cite{M2, Theorem 5.3},
$T/Q$ is algebraic over $R/\m$ and $\hgt(Q) = \hgt(\m) + n$.
It follows that $Q \cap S = P$ is maximal in $S$ and
$\hgt(P) = \hgt(\m) + m$. But the algebraic independence
hypothesis for the $f_i$ implies $m \le n$. This is a
contradiction. Therefore item 1 follows from (2.7.3).
The ($\implies$) direction of item 2 is clear. For ($\impliedby$)
of item 2 and for item 3, it
suffices to show the set
$\Cal F_{\text{min}}$ is empty, and
this follows from item 1. \qed
\enddemo
As an immediate corollary to (2.7) and (4.5), we have:
\proclaim{4.6 Corollary} Let $R$ be a Noetherian
integral domain, let $x_1,\dots,x_n$ be
indeterminates over $R$ and let $f_1, \dots,f_m \in R[x_1,\dots,x_n]$
be algebraically independent over $R$. Consider
the embedding $\varphi: S := R[f_1, \dots, f_m]
\hookrightarrow T := R[x_1,\dots,x_n]$
and let $J$ be the associated Jacobian ideal.
Then $\varphi$ is flat if for every nonmaximal
$Q \in \Spec(T)$ such that $J \subseteq Q$
we have $\hgt(Q \cap S) \le \hgt(Q)$.
\endproclaim
Also as a corollary of (2.7) and (4.5) we have:
\proclaim{4.7 Corollary } Let $R$ be a Noetherian ring, let
$x_1,\dots,x_n$ be indeterminates over $R$ and
let $f_1,\dots,f_m\in R[x_1,\dots,x_n]$ be algebraically
independent over $R$. Consider the embedding
$\varphi: S := R[f_1,\dots,f_m] \hookrightarrow
T := R[x_1,\dots,x_n] $, let $J$ be the Jacobian
ideal of $\varphi$ and
let $F$ be the (reduced) ideal which describes
the nonflat locus of $\varphi$ as in (2.4.2).
Then $J \subseteq F$ and either $ F=T$, that is, $\varphi$ is flat, or
$\text{dim}(T/Q)\ge 1$, for all
$Q\in\Spec(T)$ which are minimal over $ F$.
\endproclaim
\proclaim{4.8 Proposition} Let $R$ be a Noetherian
integral domain containing a field
of characteristic zero. Let $x_1,\dots,x_n$ be
indeterminates over $R$ and let $f_1, \dots,f_m \in R[x_1,\dots,x_n]$
be algebraically independent over $R$.
Consider the embedding
$\varphi: S := R[f_1, \dots, f_m] \hookrightarrow T:= R[x_1,\dots,x_n]$
and let $J$ be the associated Jacobian ideal. Then
\roster
\item
If $\p \in \Spec R$ and $J \subseteq \p T$, then $\p T \in \Cal F$,
i.e., $\varphi_{\p T} S \to T_{\p T}$ is not flat.
\item
If the embedding
$\varphi: S \hookrightarrow T$
is flat, then for every $\p \in \Spec(R)$ we have
$J \nsubseteq \p T$.
\endroster
\endproclaim
\demo{Proof} Item 2 follows from item 1, so it suffices to prove item 1.
Let
$\p \in \Spec R$ with $J \subseteq \p T$, and suppose
$\varphi_{\p T}$ is flat. Let $\overline{f_i}$ denote the
image of $f_i$ in $T/\p T$. Consider
$$
\overline{\varphi} :\overline S :=
(R/\p)[\overline{f_1}, \dots, \overline{f_m}] \to \overline T
:= (R/\p)[x_1, \dots , x_n].
$$
By Proposition 4.4,
$\overline{f_1}, \dots, \overline{f_m}$ are algebraically
independent over $\overline{R} := R/\p$. Since the Jacobian ideal
commutes with homomorphic images, the Jacobian ideal of
$\overline{\varphi}$ is zero. Thus for each
$Q \in \Spec \overline T$ the map
$\overline \varphi_Q : \overline S \to \overline{T_Q}$ is
not smooth. But taking $Q = (0)$ gives $\overline{T_Q}$
which is a
field separable over the fraction field of $\overline S$
and hence $\overline{\varphi_Q}$ is a smooth map.
This contradiction completes the proof. \qed
\enddemo
\subheading{5. Examples}
\subheading{5.1 Examples}
For each positive integer $n$, we present an example of a
3-dimensional quasilocal unique factorization domain $B$
such that
\roster
\item $B$ is not catenary,
\item the maximal ideal of $B$ is 2-generated,
\item $B$ has precisely $n$ prime ideals of height two,
\item Each prime ideal of $B$ of height two is not
finitely generated,
\item For every non-maximal prime $P$ of $B$ the ring $B_P$ is Noetherian.
\endroster
The notation for this construction is a localized version of
the notation of Section 3.5, with $s = 1$. Thus $k$ is a field,
$R = k[x, y]_{(x,y)}$ is a 2-dimensional regular local ring and
$R^* = k[y]_{(y)}[[x]]$ is the $(x)$-adic completion of $R$.
Let $\tau = \sum_{j=1}^\infty c_jx^j \in xk[[x]]$ be
algebraically independent over $k(x)$.
Let $p_i \in R - xR$ be such that $p_iR^*$ are $n$ distinct prime ideals.
For example, we could take $p_i = y - x^i$.
Let $q = p_1 \cdots p_n$. We set $f := q\tau$ and consider
the injective $R$-algebra homomorphism $S = R[f] \hookrightarrow
R[\tau] = T$.
Let $B$ be the nested union domain associated to $f$
as in (2.1). If $\tau_r = \sum_{j= r+1}^\infty \frac{c_jx^j}{x^r}$
is the $r^{th}$ endpiece of $\tau$, then $\rho_r := q\tau_r$ is
the $r^{th}$ endpiece of $f$. For each $r \in \N$, let
$B_r = R[\rho_r]_{(x, y, \rho_r)}$. Then
each $B_r$ is a 3-dimensional regular local
ring and $B = \bigcup_{r =1}^\infty B_r$.
The map $\alpha : S \to R^*_x$ is not flat since
$p_iR^*_x$ is a height-one prime and $p_iR^*_x \cap S = (p_i, f)S$ is
of height two. By Theorem~2.2, $B$ is not Noetherian.
By \cite{HRW4, Theorem~4.5}, $B$ is a quasilocal unique factorization
domain.
Moreover, by \cite{HRW4, Theorem~4.4},
for each $t \in \N, x^tB = x^tR^* \cap B$ and $R/x^tR = B/x^tB = R^*/x^tR^*$.
It follows that the maximal ideal of $B$ is $(x,y)B$.
If $P \in \Spec B$ is such that $P \cap R = (0)$, then because
the field of fractions $K(f)$ of $B$ has transcendence degree one
over the field of fractions $K$ of $R$, $\hgt(P) \le 1$ and hence
because $B$ is a UFD, $P$ is principal.
\proclaim{Claim 1} Let $I$ be an ideal of $B$ and let $t \in \N$.
If $x^t \in IR^*$, then $x^t \in I$.
\endproclaim
\demo{Proof} There exist elements $b_1, \ldots, b_s \in I$ such
that $IR^* = (b_1, \ldots,b_s)R^*$. If $x^t \in IR^*$, there
exist $\alpha_i \in R^*$ such that
$$
x^t = \alpha_1b_1 + \cdots + \alpha_sb_s.
$$
We have $\alpha_i = a_i + x^{t+1}\lambda_i$, where $a_i \in B$ and
$\lambda_i \in R^*$. Thus
$$
x^t[1 - x(b_1\lambda_1 + \cdots + b_s\lambda_s)] = a_1b_1 + \cdots + a_sb_s
\in B.
$$
Since $x^tR^* \cap B = x^tB$, $\gamma := 1 - x(b_1\lambda_1 + \cdots +
b_s\lambda_s)
\in B$. Moreover, $\gamma$ is invertible in $R^*$ and hence also in $B$.
It follows that $x^t \in I$. \qed
\enddemo
To examine more closely the prime ideal structure of $B$, it is
useful to consider the inclusion map $B \hookrightarrow A: = R^* \cap K(f)$
and the map $\Spec A \to \Spec B$.
\proclaim{5.2 Proposition } With the notation of Example~5.1
and $A = R^* \cap K(f)$, we have
\roster
\item $A$ is a two-dimensional regular local domain
with maximal ideal $\m_A = (x, y)A$.
\item
$\m_A$ is the unique prime of $A$ lying over
$\m_B = (x,y)B$, the maximal ideal of $B$.
\item If $P \in \Spec B$ is nonmaximal, then
$\hgt(PR^*) \le 1$ and $\hgt(PA) \le 1$. Thus
every nonmaximal prime of $B$ is contained in a nonmaximal
prime of $A$.
\item
If $P \in \Spec B$ and $xq \not\in P$, then $\hgt P \le 1$.
\item If $P \in \Spec B$, $\hgt P = 1$ and $P \cap R \ne 0$,
then $P = (P \cap R)B$.
\endroster
\endproclaim
\demo{Proof}
By Proposition~3.11 (the result of Valabrega)
$A : = R^* \cap K(f)$ is a two-dimensional regular local domain
having the same completion as $R$ and $R^*$. This
proves item 1.
Since $B/xB = A/xA = R^*/xR^*$,
$\m_A = (x, y)A$ is the unique prime of $A$ lying over
$\m_B = (x, y)B$. Thus item 2 holds and also item 3 if $x \in P$.
To see (3), it remains to consider $P \in \Spec B$ with
$x \not\in P$. By Claim 1, for all $t \in \N$,
$x^t \not\in PR^*$. Thus $\hgt(PR^*) \le 1$. Since $A \hookrightarrow R^*$ is
faithfully flat, $\hgt(PA) \le 1$.
For (4), we see by (3) that $\hgt(PA) \le 1$.
Let $Q \in \Spec A$ be a height-one prime
ideal such that $P \subseteq Q$. Since $xq \not\in P$, we have
$B_P = S_{P \cap S} = T_{Q \cap T} = A_Q$, where $S = R[f]$
and $T = R[\tau]$.
Thus $\hgt(P) \le 1$.
For (5), if $x \in P$, then $P = xB$ and the statement
is clear. Assume $x \not\in P$. Since
$B_x$ is a localization of $(B_r)_x$, we have
$(P \cap R)B_r = P \cap B_r$ for all $r \in \N$. Thus $P = (P \cap R)B$. \qed
\enddemo
We observe that the DVRs $B_{xB}$ and $A_{xA}$ are equal.
Moreover,
$A$ is the nested union $\bigcup_{r=1}^\infty R[\tau_n]_{(x,y, \tau_n)}$
of 3-dimensional regular local domains. Since $A$ is a
two-dimension regular local domain each
nonmaximal prime of $A$ is principal. If $pA$ is a
height-one prime of $A$ with $pA \not\in \{p_1A, \ldots p_nA \}$,
then $A_{pA} = B_{pA \cap B}$ and $\hgt(pA \cap B) = 1$.
We observe in Claim 2 that $p_iA \cap B$ has height two
and is not finitely generated.
\proclaim{Claim 2} Let $p_i$ be one of the prime
factors of $q$. Then $p_iB$ is prime in $B$. Moreover
\roster
\item
$p_iB$ and $Q_i := (p_i, \rho_1, \rho_2, \ldots )B = p_iA \cap B$ are the
only primes of $B$ lying over $p_iR$ in $R$,
\item
$Q_i$ is of height two and is not finitely generated.
\endroster
\endproclaim
\demo{Proof}
We use that $B = \bigcup_{r=1}^\infty B_r$, where
$B_r = R[\rho_r]_{(x, y, \rho_r)}$ is a 3-dimensional
regular local ring.
For each $r \in \N$, $p_iB_r$ is prime in $B_r$. Hence
$p_iB$ is a height-one prime ideal of $B$, for $i = 1, \dots, n$.
Since $\rho_r = q\tau_r$, \,
$p_iA \cap B_r = (p_i, \rho_r)B_r$ is a height-two
prime ideal of the 3-dimensional regular local domain $B_r$.
Therefore $Q_i := (p_i, \rho_1, \rho_2, \ldots )B = p_iA \cap B$
is a nested union of prime ideals of height two, so
$\hgt(Q_i) \le 2$. Since $p_iB$ is a nonzero prime ideal
properly contained in $Q_i$, \, $\hgt(Q_i) = 2$.
Moreover $x \not\in (p_i, \rho_r)B_r$ for each $r$,
so $x \not\in Q_i$. Hence for each $r \in\N$, \,
$\rho_{r+1} \not\in (p_i, \rho_r)B$ and $Q_i$ is
not finitely generated. \qed
\enddemo
Since $x \not\in Q_i$ and $B[1/x]$ is a localization
of the Noetherian
domain $B_n[1/x]$, we see that $B_{Q_i}$ is Noetherian.
Since the $Q_i$ are the only prime
ideals of $B$ of height two and $B$ is a UFD, $B_P$ is
Noetherian for every non-maximal prime $P$.
This completes the presentation of Examples 5.1. With
regard to the birational inclusion $B \hookrightarrow A$
and the map $\Spec A \to \Spec B$, we remark that
the following holds:
Each $Q_i$ contains
infinitely many height-one primes of $B$ that are the
contraction of primes of $A$ and infinitely many that
are not. Among the primes that are not contracted from
$A$ are the $p_iB$.
In the terminology of \cite{ZS, page 325},
$P$ is {\it not lost} in $A$ if $PA \cap B = P$. Since
$p_iA \cap B = Q_i$
properly contains $p_iB$, $p_iB$ is lost in $A$. Since
$(x, y)B$ is the maximal ideal of $B$ and $(x, y)A$ is the
maximal ideal of $A$ and $B$ is integrally closed, a version
of Zariski's Main Theorem \cite{Pe}, \cite{Ev}, implies
that $A$ is not essentially
finitely generated as a $B$-algebra.
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\enddocument