%From wiegand@math.msu.edu Tue Dec  9 22:06 EST 1997
%To: gregorio@math.unipd.it
%Subject: eno8.tex

%To Enrico Gregorio <gregorio@math.unipd.it>
%From: William Heinzer <heinzer@MATH.Purdue.EDU>
%Date: Tue, 5 Aug 1997 14:12:45 -0500 (EST)
%To: rotthaus@math.msu.edu, swiegand@math.unl.edu
%Subject: eno8.tex



\documentstyle{amsppt}
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\rightheadtext{Building Noetherian domains }
\NoBlackBoxes
\topmatter
\title
Building  Noetherian domains inside \\
an ideal-adic completion
\endtitle
\author
William Heinzer, Christel Rotthaus and Sylvia Wiegand 
\endauthor
\date December 9, 1997 \enddate

\abstract
Suppose $a$ is a nonzero nonunit of a Noetherian integral
domain $R$. An interesting construction introduced by
Ray Heitmann addresses the question of how ring-theoretically
to adjoin a transcendental power series in $a$ to the ring $R$. 
We apply  this construction,  and its natural generalization  to 
finitely many elements, to 
exhibit Noetherian extension domains of $R$ inside
the $(a)$-adic completion $R^*$ of $R$. 
Suppose $\tau_1, \dots , \tau_s  \in aR^*$ 
are algebraically independent over $K$, the field of fractions of $R$.  
Starting with
$U_0:=R[\tau_1, \dots , \tau_s]$, there is a natural sequence  
of nested polynomial rings $U_n$ between
$R$ and $A:=K(\tau_1, \dots , \tau_s)\cap R^*$. It is
not hard to show  that if $U := \cup_{n=0}^\infty U_n$ is Noetherian,
then $A$ is a localization of $U$ and $R^*[1/a]$ is flat over $U_0$.  
We prove, conversely,  that if $R^*[1/a]$ is flat over $U_0$,
then  $U$ is Noetherian and 
$A := K(\tau_1, \dots , \tau_s  ) \cap R^*$ is 
a  localization of $U$. Thus the flatness of $R^*[1/a]$ 
over $U_0$ implies the intersection domain $A$ is Noetherian.\endabstract

\thanks{The authors would like to thank the National Science 
Foundation and the University of Nebraska Research Council 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
\endtopmatter

\document
\baselineskip 18pt

\subheading{1. Introduction} Suppose $a$ is a nonzero nonunit of a
Noetherian integral domain $R$. 
The $(a)$-adic completion $R^*$ of $R$ is isomorphic to
the ring $R[[x]]/(x-a)$ \cite{N, (17.5), page~55}.
Thus elements of the $(a)$-adic completion may be regarded
as formal power series in $a$.  Of course if $R$ is already
complete in its $(a)$-adic topology, then $R = R^*$,
but often it is the case that there are elements of $R^*$ 
that are transcendental over $R$.  An interesting construction first
introduced by Ray Heitmann in \cite{H, page~126} addresses the
question of how ring-theoretically to adjoin a transcendental 
(over $R$) power series
in $a$  to the ring $R$. 
We have made use of this construction 
of Heitmann in \cite{HRW3} in a local or semilocal context.
Our purpose here is to consider this construction in the
more general context of an arbitrary Noetherian integral domain.

There are numerous articles in the literature that have
relevance for the building of Noetherian domains inside
an ideal-adic completion, for example \cite{BR1}, \cite{BR2},
\cite{HRS}, \cite{H1}, \cite{H2}, \cite{H3}, \cite{L},
\cite{N2}, \cite{O1}, \cite{O2}, \cite{R1}, \cite{R2},
\cite{R3}, and \cite{W}.

Let $R$ be a Noetherian integral domain 
with field of fractions $K$ and let $a$ be a nonzero nonunit of $R$.
We are interested in the structure of certain
intermediate integral domains  between
$R$ and 
$R^*:=\widehat{(R,(a))} = R[[x]]/(x-a)$,  the $(a)$-adic completion of $R$.
We are 
particularly interested in domains of the form $A:=L\cap R^*$,
where $L$ is an intermediate field between $K$ and 
the total ring of fractions of $R^*$.  
It is often difficult to compute this  intersection
ring $A$. Thus we seek
conditions in order that  $A$ be realizable as a
localization of a 
directed union of polynomial ring extensions of $R$.


This intersection construction
inside the completion of $R$ with respect to a principal
ideal yields interesting Noetherian rings which are directed
unions of
localized polynomial rings, as we see below. By contrast, 
taking the analogous construction inside the completion 
with respect to a maximal ideal, even  of an excellent 
local domain seems less likely to give Noetherian intersection domains.
In \cite{HRW1}, it is shown for a
countable excellent local domain $(R,\m)$ of dimension at
least two that there exist infinitely many algebraically
independent elements $\tau_1, \tau_2, \dots  $ in the
$\m$-adic completion $\widehat R$ of $R$ such that the corresponding intersection domain
is
a  localized polynomial ring in infinitely many variables 
over $R$; that is,  
$\widehat R \cap K(\tau_1, \tau_2, \dots )=R[\tau_1, \tau_2, \dots ]_{(\m, 
\tau_1,\tau_2, \dots )}$.


In \cite{HRW2}, \cite{HRW3} and the present paper, we study the following  
element-wise form of the problem. Let $\tau_1, 
\dots , \tau_s  \in aR^*$ be elements which are
algebraically independent over $K$.
Starting with $U_0:=R[\tau_1, \dots , \tau_s]$, we define a  
sequence of nested polynomial rings $U_n$ in 
$s$ variables over $R$
inside $A:=K(\tau_1, \dots , \tau_s)\cap R^*$.  
In \cite{HRW3} we consider in the case where $R$ is a
semilocal Noetherian integral domain and $a$ is an 
element of the Jacobson radical of $R$ the
condition that the embedding 
$U_0 \to R^*[1/a]$ is flat. Our goal here is to 
examine flatness of the embedding $U_0 \to R^*[1/a]$ in
a more general context, and to
prove the following theorem.
\footnote{This result generalizes \cite{HRW3, Theorem 2.12}.}


\subheading{Theorem 1.1}  Suppose $R$  is a 
Noetherian domain, $a \in R$ is a nonzero nonunit,
and   $\tau_1, \dots , \tau_s$ are elements of the
$(a)$-adic completion $R^*$ of $R$ that are algebraically
independent over $R$. 
\footnote{We say that elements are algebraically
independent over an integral  domain if they are algebraically independent
over its fraction field.} 
Then the following conditions are
equivalent:
\roster
\item The ring $R^*[1/a]$ is flat over 
$U_0 = R[\tau_1, \dots , \tau_s]$.  
\item The directed union  $U:=\cup_{n=0}^\infty U_n$ is Noetherian. 
\footnote{Heitmann in \cite{H, page~126} considers
the case where there is one transcendental element $\tau$
and defines the corresponding extension $U$ to be a 
{\it simple PS-extension of $R$ for $a$.} Heitmann 
proves in this case that a 
certain monomorphism condition on a sequence of maps is equivalent to $U$
being Noetherian \cite{H, Theorem~1.4}.}
\endroster
Moreover, if these equivalent conditions hold, then the 
integral domain  $A := K(\tau_1, \dots , \tau_s) \cap R^*$ is 
a localization of $U$, and hence $A$ is Noetherian.   

\subheading{Remark 1.2} An example given in \cite{HRW3, (4.4)}
shows that it is possible for $A$ to be a localization of $U$
and yet  for $A$, and therefore also $U$, to fail to be Noetherian.  
Thus the equivalent conditions of (1.1) are not implied by the
property that $A$ is a localization of $U$.

We present  in (2.5) an example that is a modification 
of \cite{HRW2, Example~2.1} to
show that $A$ being Noetherian does not imply that $U$ is Noetherian.  

The following diagram
displays the situation concerning possible implications
between $A$ being a localization of $U$ and $A$ or $U$ being Noetherian:
\vskip 18 pt
                                        
\setbox4=\vbox{\hbox{%
   \rlap{\kern1.6in\lower 0in\hbox to .3in{\hss $\boxed{R^*[1/a]\text{ is flat over }
U_0 = R[\tau_1, \dots , \tau_s]}$ \hss}}%
    \rlap{\kern3.8in\lower 0in\hbox to .3in{\hss $\boxed{U \text{ Noetherian }}$\hss}}%
   \rlap{\kern1.6in\lower.7in\hbox to .3in{\hss $\boxed{A\text{ is a localization of }U}$ \hss}}%
   \rlap{\kern3.8in\lower.7in\hbox to .3in{\hss $\boxed{A \text{ Noetherian }}$ \hss}}%
   \rlap{\special{pa 2850 -70} \special{pa 3350 -70} \special{fp}}%1   
  \rlap{\special{pa 2850 -40} \special{pa 3350 -40} \special{fp}}%1   
   \rlap{\special{pa 2880 -100} \special{pa 2830 -55} \special{fp}}%1   
  \rlap{\special{pa 2880 -10} \special{pa 2830 -55} \special{fp}}%1   
  \rlap{\special{pa 3320 -100} \special{pa 3365 -55} \special{fp}}%1   
  \rlap{\special{pa 3320 -10} \special{pa 3365 -55} \special{fp}}%1   

  \rlap{\special{pa 1400 150} \special{pa 1400 500} \special{fp}}%1   
  \rlap{\special{pa 1430 150} \special{pa 1430 500} \special{fp}}%1   
  \rlap{\special{pa 1370 180} \special{pa 1415 120} \special{fp}}%1   
  \rlap{\special{pa 1460 180} \special{pa 1415 120} \special{fp}}%1   
  \rlap{\special{pa 1800 150} \special{pa 1800 500} \special{fp}}%1   
  \rlap{\special{pa 1830 150} \special{pa 1830 500} \special{fp}}%1   
  \rlap{\special{pa 1770 470} \special{pa 1815 530} \special{fp}}%1   
  \rlap{\special{pa 1860 470} \special{pa 1815 530} \special{fp}}%1   

  \rlap{\special{pa 3600 150} \special{pa 3600 500} \special{fp}}%1   
  \rlap{\special{pa 3630 150} \special{pa 3630 500} \special{fp}}%1   
  \rlap{\special{pa 3570 180} \special{pa 3615 120} \special{fp}}%1   
  \rlap{\special{pa 3660 180} \special{pa 3615 120} \special{fp}}%1   
  \rlap{\special{pa 4000 150} \special{pa 4000 500} \special{fp}}%1   
  \rlap{\special{pa 4030 150} \special{pa 4030 500} \special{fp}}%1   
  \rlap{\special{pa 3970 470} \special{pa 4015 530} \special{fp}}%1   
  \rlap{\special{pa 4060 470} \special{pa 4015 530} \special{fp}}%1   

  \rlap{\special{pa 2450 700} \special{pa 3250 700} \special{fp}}%1   
    \rlap{\special{pa 2450 730} \special{pa 3250 730} \special{fp}}%1   
  \rlap{\special{pa 2480 670} \special{pa 2430 715} \special{fp}}%1   
   \rlap{\special{pa 2480 760} \special{pa 2430 715} \special{fp}}%1   
     \rlap{\special{pa 2450 600} \special{pa 3250 600} \special{fp}}%1   
      \rlap{\special{pa 2450 570} \special{pa 3250 570} \special{fp}}%1   
     \rlap{\special{pa 3220 540} \special{pa 3265 585} \special{fp}}%1   
     \rlap{\special{pa 3220 630} \special{pa 3265 585} \special{fp}}%1   

     \rlap{\special{pa 1365 350} \special{pa 1465 250} \special{fp}}%1   
     \rlap{\special{pa 3565 350} \special{pa 3665 250} \special{fp}}%1   

     \rlap{\special{pa 2650 635} \special{pa 2750 535} \special{fp}}%1   
      \rlap{\special{pa 2950 765} \special{pa 3050 665} \special{fp}}%1   
       }}

\box4
\vskip 18 pt                                               


\subheading{2. The general setting}

\noindent
{\bf (2.1)}  Let $R$ be a Noetherian  integral domain 
of dimension $d>0$ with fraction field $K$. 
Let $a$ be a nonzero element nonunit of $R$,  
let $R^* := \widehat {(R,(a))}$ 
be the $(a)$-adic completion of $R$ and let
$R^*_a := R^*[1/a]$.
Suppose $\tau_1, \dots , \tau_s \in aR^*$ are 
regular elements
\footnote{We say an element of a ring is a {\it regular element }
if it is not a zero divisor.}
of $R^*$ that are algebraically
independent over $K$.  We consider the  
polynomial ring 
$$
U_0 := R[\tau_1, \dots , \tau_s].
$$

For every $\gamma\in R^* $ and every $n>0$,
we define the $n^{\text{th}}$-endpiece $\gamma_n$ 
with respect to $a$ of
$\gamma$ to be
$$\gamma_n:=\sum_{j=n+1}^\infty c_{j}a^{j-n}, \text{ where }
\gamma:=\sum_{j=1}^\infty c_{j}a^j
\text{ with each }c_j\in R.\tag{2.1.1}$$
In particular, we represent each of the $\tau_i$ by a power series 
expansion in $a$; we use these representations to obtain 
for each positive integer $n$ the 
$n^{\text{th}}$-endpieces $\tau_{in}$ and the corresponding 
$n^{\text{th}}$-polynomial ring $U_n$:
$ \text{ For }1 \le i \le s, \text{ and }\tau_i := \sum_{j=1}^\infty r_{ij}a^j$,
where the $r_{ij} \in R$,  
$\tau_{in}  := \sum_{j=n+1}^{\infty} r_{ij} a^{j-n}, 
U_n := R[\tau_{1n}, \dots , 
\tau_{sn}],  \text{ for each } n\in\N.$
We have a birational inclusion of polynomial rings $U_n \subset U_{n+1}$.
We define
$$
U := \cup_{n=0}^\infty U_n = \varinjlim U_n \qquad 
\text{ and } \qquad  
A := K(\tau_1 , \dots , \tau_s) \cap R^*. \tag{2.1.2}$$
It is readily seen that $A$ is a birational extension of  $U$.
We say that the $\tau_i$ have {\it good limit-intersecting
behavior } if $A$ is a localization of $U$.


We observe the following properties of $(a)$-adic completions
and an implication of this concerning good limit-intersecting behavior. 

\proclaim{Proposition 2.2 (cf. \cite{HRW2},\cite{HRW3, (2.2)})} 
Assume the 
notation and setting of (2.1),
and  let $U^*$ and $A^*$ denote the $(a)$-adic 
completions of $U$ and $A$. Then
\roster
\item  $a^kU = a^kA \cap U=a^kR^* \cap U$ 
for each positive integer $k$.
\item $U^* = A^* = R^*$, so $R/aR = U/aU = A/aA = R^*/aR^*$.
\item If $U$ is Noetherian, then $R^*$ is 
flat over $U$ and $A$ is the localization of $U$ at the
multiplicative system $1 + aU$ of $U$.
\endroster\endproclaim

\demo{Proof} We have $R \subseteq U \subseteq A \subseteq R^*$.
Since $R$ is Noetherian, $R^*$ is flat over $R$
\cite{M1, Theorem~8.8, page~60}. 
Moreover, $a^kR$ is closed in the $(a)$-adic topology on $R$,
so we have $a^kR^* \cap R = a^kR$ for each positive integer $k$
\cite{ZS, Theorem~8, page~261}. 
Furthermore, $A = R^* \cap K(\tau_1, \dots , \tau_s)$ 
implies $a^kA = a^kR^* \cap A$.  It is clear that 
$a^kU \subseteq a^kR^* \cap U$, thus for (1) and (2) it suffices
to show $a^kR^* \cap U \subseteq a^kU$. Moreover,
if $aR^* \cap U = aU$, it follows that $a^kR^* \cap U
= a^kR^* \cap aU = a(a^{k-1}R^* \cap U)$, and by
induction we see that $a^kR^* \cap U = a^kU$.
Thus we show $aR^* \cap U \subseteq aU$.

Let $g\in aR^*\cap U$. Then there is a positive integer $n$ with
$g\in U_n =  R[\tau_{1n}, \dots , \tau_{sn}].$
Write $g = r_0 + g_0$ where $g_0\in (\tau_{1n}, \dots , \tau_{sn})U_n$ 
and $r_0\in R$. From the definition of $\tau_{in}$,
we have  $\tau_{in}  = a\tau_{in+1} + a_{in}a,$ where $a_{in} \in R,$
for each $i$ with 
$ 1 \le i \le s$. 
Thus  $r_0\in aR^*\cap R = aR$,
$\tau_{in}U_n \subseteq aU_{n+1}$  and  $g\in aU$. 
This completes the proof of (1) and (2). 
If $U$ is Noetherian, then $U^* = R^*$ is 
flat over $U$. 
Let $S$ be the multiplicative system $1 + aU$ and let
$B = S^{-1}U$. Then $B$ is Noetherian, the $(a)$-adic
completion of $B$ is $R^*$ and $R^*$ is faithfully flat over $B$
\cite{M1, Theorem~8.14, page~62}. Therefore
$B = K(\tau_1, \dots , \tau_s) \cap R^* = A$. \qed
\enddemo

With the notation and setting of (2.1), the representation of
the $\tau_i$ as power series in $a$ with coefficients in $R$
is, in general, not unique. However, as we observe in (2.3),
the rings $U$ and $U_n$ are uniquely determined by the $\tau_i$.

\proclaim{Proposition 2.3 (cf. \cite{HRW3, (2.3)})} 
Assume the notation and setting
of (2.1). Then $U$ and the $U_n$ are independent of the 
representation of the $\tau_i$ as power series in $a$
with coefficients in $R$.\endproclaim

\demo{Proof}  For $1 \le i \le s$, assume that $\tau_i$
and  $\omega_i = \tau_i$ have representations
$$ 
\tau_i = \sum_{j=1}^\infty a_{ij}a^j \qquad \text{and} \qquad
\omega_i = \sum_{j=1}^\infty b_{ij}a^j,
$$ 
where each $a_{ij}, b_{ij}  \in R$. We define the 
$n^{\text{th}}$-endpieces $\tau_{in}$ and 
$\omega_{in}$ as in (2.1.1):
$$
\tau_{in} = \sum_{j=n+1}^\infty a_{ij}a^{j-n} \qquad
\text{and} \qquad \omega_{in} = 
 \sum_{j=n+1}^\infty b_{ij}a^{j-n}.
$$
Then we have  
$$\tau_i=\Sigma_{j=1}^\infty a_{ij}a^j=\Sigma_{j=1}^n 
a_{ij}a^j+a^n\tau_{in}
  =\Sigma_{j=1}^\infty b_{ij}a^j=\Sigma_{j=1}^n b_{ij}a^j+a^n\omega_{in}
=\omega_i.
$$ 
Therefore, for $1 \le i \le s$ and each positive integer $n$,
$$
a^n\tau_{in} - a^n\omega_{in}=\Sigma_{j=1}^nb_{ij}a^j-
\Sigma_{j=1}^na_{ij}a^j, \quad \text{ and so } \quad
\tau_{in} -\omega_{in}=\frac{\Sigma_{j=1}^n(b_{ij}-a_{ij})a^j}{a^n}.
$$
Since $\Sigma_{j=1}^n(b_{ij} - a_{ij})a^j\in R$ is divisible by $a^n$ 
in $R^*$ and since
$a^nR=R\cap a^nR^*$ because $a^nR$ is closed in the
$(a)$-adic topology, it follows  that $a^n$ divides the sum
$\Sigma_{j=1}^n(b_{ij} - a_{ij})a^j$ in $R$. 
Therefore  $\tau_{in}-\omega_{in}\in R$. 
It follows that $U_n$ and $U = \cup_{n=1}^\infty U_n$ are
independent of the representation of the $\tau_i$. \qed
\enddemo 

\subheading{Remark 2.4 } With notation as in (2.1), if
the embedding $U_0 = R[\tau_1, \dots,\tau_s] \to R^*[1/a]$ is
flat, then every nonzero element of $U_0$ is a regular element 
of $R^*$.

\subheading{Example 2.5 } (cf. \cite{HRW2, Example~2.1}) 
In $\Q[[x,y]]$, 
the power series ring in the two variables $x$ and 
$y$ over the rational numbers, let $\gamma:=e^x-1$ and  $\tau:=e^y-1$;
take $\gamma_n$ to be the $n^{\text{th}}$-endpiece of $\gamma$ with 
respect to $x$ and take $\tau_n$ to be the 
$n^{\text{th}}$-endpiece of $\tau$ with 
respect to $y$, as described in (2.1). 
Set $R:=\cup_{n\in\N}\Q[x,y,\gamma_n]_{(x,y,\gamma_n)}$. Then 
$R=\Q[y]_{(y)}[[x]]\cap \Q(x,y,\gamma)$ is an  excellent
two-dimensional regular local domain. 
Now define $U$ in the $(y)$-adic 
completion of $R$ using the endpieces $\tau_n$  as above.
Then $U\supseteq V:=\cup_{n\in\N}\Q[x,y,\gamma_n,\tau_n]$. 
The ring $A:=\Q[[x,y]]\cap \Q(x,y,\gamma,\tau)$ is Noetherian 
but is different from  
$B:=\cup\Q[x,y,\gamma_n,\tau_n]_{(x,y,\gamma_n,\tau_n)}$. 
The ring $B$ is the localization of $U$ at the
multiplicative system $1 + yU$, and 
the rings $B$   and $U$ are  not Noetherian. 
It follows that $A$ is not a localization of $U$.
\medskip

\demo{Proof} Consider the element $\theta=\frac{\gamma-\tau}{x-y}\in A$. 
If $\theta$ is an element of $B$, then 
$$\gamma-\tau\in (x-y)B\cap V=(x-y)V.$$
Now 
$$V=\cup_{n\in\N}\Q[x,y,\gamma_n,\tau_n]\subseteq \Q[x,y,\gamma,\tau][1/x,1/y]
\subseteq\Q[x,y,\gamma,\tau]_{(x-y)},$$
and so$$\gamma-\tau\in (x-y)\Q[x,y,\gamma,\tau]_{(x-y)}\cap\Q[x,y,\gamma,\tau]= (x-y)\Q[x,y,\gamma,\tau],$$
but this contradicts the fact that $x,y, \gamma, \tau$ are algebraically
independent over $\Q$.

If $U$ were  Noetherian, then $B$ 
would be Noetherian. But the maximal ideal of $B$ is $(x,y)B$,  
so if $B$ were Noetherian, then it would be a regular local domain 
with completion $\Q[[x,y]]$. Since the completion of a local Noetherian 
ring is a faithfully flat extension of it, and since the fraction field of 
$B$ is  $\Q(x,y,\gamma,\tau)$, then $B$  would equal $ A$. 

That $A$ is Noetherian follows from \cite{V, Proposition~3}.
If $A$ were a localization of $U$, then $A$ would be a 
localization of $B$. But each of $A$ and $B$ has a unique
maximal ideal and the maximal ideal of $A$ contains the
maximal ideal of $B$. Therefore $B \subsetneq A$ implies
that $A$ is not a localization of $B$. \qed
\enddemo

\subheading{3. The proof of the main theorem} 

\demo{Proof of Theorem 1.1} Assume that $U$ is Noetherian.
By (2.2), the $(a)$-adic completion $U^*$ of $U$ is 
equal to $R^*$. Since $U$ is Noetherian, $U^* = R^*$ is
flat over $U$ \cite{M1, Theorem~8.8}. Therefore the
localization $R^*[1/a]$ is flat over $U$. Since
$U[1/a] = U_0[1/a]$,  the
localization $R^*[1/a]$ is also flat over $U_0$.

To prove the converse we use results of Heitmann
in \cite{H1, Theorem~1.4}. 
\enddemo



First we show in (3.1) that
the flatness condition for $
R^*[1/a]$  over $U_0$  behaves well
under certain residue class formations.

\proclaim{Proposition 3.1} Let $R$ be  a Noetherian  
domain,  let $a$ be a nonzero nonunit of $R$, let
$R^*$ be the
$y$-adic completion of $R$ and let $\tau_1, \dots , \tau_s  \in aR^*$
be algebraically independent over $R$.  Suppose that $R^*_a:=
R^*[1/a]$ is  
flat over $U_0$, using the notation of (2.1)
and that  $Q$ is a prime ideal of $R$
with $a\notin Q$. Assume that $Q$ is the contraction
of a prime ideal of $R^*$. Let $\bar{\phantom{x}}$ denote  
image  
in $R^*_a/QR^*_a$ and let  $(R/Q)^*$ denote the 
$(\bar a)$-adic completion of $R/Q$. 
Then  $(R/Q)^*_{\bar a}:=(R/Q)^*[\bar 1/{\bar a}]$ is flat over   
$(R/Q)[\bar \tau_1, \dots , \bar \tau_s]$.
\endproclaim

\demo{Proof} The $(\bar a)$-adic completion $(R/Q)^*$  of $R/Q$ is
canonically isomorphic to $R^*/QR^*$. Therefore $\bar \tau_1, \dots ,
\bar \tau_s$ are regular elements of $(R/Q)^*$. 
We show $\bar \tau_1, \dots , \bar \tau_s$ 
are  algebraically
independent over $R/Q$. 
Since $R[\tau_1, \dots , \tau_s ] \longrightarrow R^*_a$ is flat,
$a \not\in Q$, and $Q$ is the contraction of a prime ideal of $R^*$,
we have 
$QR[\tau_1, \dots , \tau_s] = QR^*_a \cap R[\tau_1, \dots , \tau_s]$.
Thus 
$$
R[\tau_1, \dots , \tau_s]/(QR^*_a \cap R[\tau_1, \dots , \tau_s]) 
\cong (R/Q)[\bar \tau_1, \dots , \bar \tau_s] 
$$
is a polynomial ring in $s$ variables 
$\bar \tau_1, \dots , \bar \tau_s$ over $R/Q$.
Therefore $\bar \tau_1, \dots , \bar \tau_s$ 
are  algebraically independent
over $R/Q$.

We show  flatness of the map:
$$
\bar \phi: (R/Q)[\bar \tau_1, \dots , \bar \tau_s] 
\longrightarrow R^*_a/QR^*_a = (R/Q)^*_{\bar a}.$$ 
Let $\bar P$ be a prime ideal of $ R^*/QR^*$ 
with $\bar a\notin \bar P$. The ideal $\bar P$ lifts to a 
prime ideal $P$ of $R^*$ with $a\notin P$ and $QR^* \subseteq P$. 
By assumption the map
$$
\phi_P: R[\tau_1, \dots , \tau_s]
\longrightarrow R^*_P$$
is flat. The map on the residue class rings:
$$
\bar \phi_{\bar P}: (R/Q)[\bar \tau_1, \dots , \bar \tau_s]
\longrightarrow (R^*/QR^*)_{\bar P}$$
is obtained from $\phi_P$ by tensoring with 
$(R/Q)[\tau_1, \dots , \tau_s]$  
over the ring 
$R[\tau_1, \dots , \tau_s]$. 
Hence $\bar \phi$ is flat. \qed
\enddemo

\proclaim{Theorem 3.2} Assume the notation and setting of (2.1).
Also assume that $s = 1$, $\tau:=\tau_1$ and that the
localization $R^*[1/a]$ is flat over $U_0=R[\tau]$. Then 
$U$ is Noetherian and $A=R^*\cap K(\tau)$ is a localization of $U$. 
\endproclaim
We use the
same proof as in \cite{H1, Theorem~1.4} and prove first 
the following lemma.

\proclaim{Lemma 3.3} With notation as in Theorem 3.2, if
$P$ is a nonzero prime ideal of $U$ such that $P \cap R = (0)$,
then there exists $f \in P$, $r \in R$ and a positive integer $N$ 
such that $P = (fU :_U  ra^N)$. 
\endproclaim
\demo{Proof}  The localization $D := (R-\{0\})^{-1}U$
of $U$ at the nonzero 
elements of $R$ is also a localization of the 
polynomial ring $U_0 := R[\tau]$. Hence $PD$ is a  
a principal maximal ideal of $D$ and there 
exists a polynomial $f \in R[\tau]$ such that $PD = fD$.

We use the fact that $U$ is the directed union of the
polynomial rings
$U_n := R[\tau_{n}]$,  $U = \cup_{n=0}^\infty U_n$.
Let $P_n = P \cap U_n$. Since 
$D_{PD} = (U_{0})_{P_0}$ and $U_{0}$ is
Noetherian, there exists 
$r \in R$ such that  $P_0 = (fU_0 :_{U_0} r)$. Also
for $g \in U$ there exists a positive integer $b(g)$, depending on
$g$, such
that $a^{b(g)}g \in U_0$. Hence 
for $g \in P$ we have $ra^{b(g)}g \in fU_0$.

The Artin-Rees Lemma \cite{N1, (3.7)} applied to the 
ideals $aR^*$ and $fR^*$ of
the Noetherian ring $R^*$ implies the existence of a positive
integer $N$ such that for $m \ge N$ we have
$$
fR^* \cap (aR^*)^m = (aR^*)^{m - N}((fR^* \cap (aR^*)^N) = 
(a^{m-N})R^*
(fR^* \cap a^NR^*).
$$
We may assume that $b(g) \ge N$.

Suppose $g \in P$. Then $ra^{b(g)}g \in fU_0\subseteq fU$, so   
$$
ra^{b(g)}g \in fR^* \cap a^{b(g)}R^* = a^{b(g) - N}R^*(fR^* 
\cap a^NR^*).
$$ 
Since $a$ is not a zero-divisor in $R^*$,
it follows that  $ra^Ng \in fR^* \cap a^NR^*$. Thus
$ra^Ng = ft$, where $t \in R^*$. Since we
also have $ra^{b(g)}g \in fU$, it follows that $a^{b(g) - N}ft 
\in fU$,
and therefore $a^{b(g) - N}t \in U$, as $f$ is not a zero-divisor 
in $R^*$. Therefore $a^{b(g) - N}t \in a^{b(g) - N} R^* \cap U = 
a^{b(g) - N}U$ by (2.2.1) and so $t\in U$. Hence for 
every $g 
\in P$ we have
$g \in (fU :_U  ra^N)$. It follows that $P = (fU :_U ra^N)$. \qed 
\enddemo

As in \cite{H1, Lemma 1.5}, we have:

\proclaim{Lemma 3.4} With notation as in Theorem 3.2, if
each prime ideal $P$ of $U$ such that $P \cap R \ne (0)$
is finitely generated, then $U$ is Noetherian.
\endproclaim
\demo{Proof} By a Theorem of Cohen \cite{N1, (3.4)},
it suffices to show each $P \in \Spec(U)$
such that $P \cap R = (0)$ is finitely generated. Let
$P$ be a nonzero prime ideal of $U$ such that $P \cap R = (0)$.
Since the localization of $U$ at the nonzero elements of $R$ 
is also a localization of the polynomial ring
$U_0 := R[\tau]$,   
every prime ideal of $U$ properly containing $P$ has a nonzero
intersection with $R$. Therefore the hypothesis implies that
$U/P$ is Noetherian. By
(3.3), there exist $r \in R$ and $f \in P$
such that  $P = (fU :_U ra^N)$. Since $ra^N$ is a nonzero
element of $R$, every prime ideal of $U$ containing $ra^N$ is
finitely generated, so $U/ra^NU$ is Noetherian. Therefore
$U/(P \cap ra^NU)$  is Noetherian \cite{N1, (3.16)}.
Since $ra^N \notin P$ and $P$ is prime, we have
$ra^NU \cap P = ra^NP$. Therefore $U/ra^NP$ is Noetherian. 
We have $ra^NP \subseteq fU \subseteq P$. Hence $U/fU$, as a 
homomorphic image of $U/ra^NP$, is
Noetherian,  and $P/fU$ is finitely generated. It follows that
$P$ is finitely generated. \qed
\enddemo

\demo{Proof of Theorem 3.2} Suppose $U$ is not Noetherian and 
let $Q \in \Spec(R)$ be maximal with respect to being the
contraction to $R$ of a non-finitely generated prime ideal of
$U$. Since $R/aR = U/aU = R^*/aR^*$ by (2.2), we have $a \notin Q$. 
Since $U = \cup_{n=0}^\infty U_n$ and $QU_n$ is prime, we have
$QU$ is prime in $U$. We claim that $Q$ is the contraction of
a prime ideal of $R^*$, for otherwise we have $(Q,a)R = R$.
But this means that the image of $a$ in $U/QU$ is a unit
which implies that $U/QU = U_0/QU_0$ is Noetherian, and this
implies that $P$ is finitely generated. Therefore 
$Q$ is the
contraction of a prime of $R^*$, and  (3.1)  implies that, passing to 
the 
image $\bar \tau$ of $\tau$ in $U/QU$,  the
localization $(R/Q)^*_{\bar a}$ is flat over $(R/Q)[\bar \tau]$.
 But Lemma~3.4 then implies that $U/QU$
is Noetherian. This contradicts the existence of a non-finitely
generated prime ideal of $U$ lying over $Q$ in $R$. We
conclude that $U$ is Noetherian. Therefore $U^* = R^*$ is
flat over $U$ and if $S$ is the multiplicative system
$1 + aU$, then $S^{-1}U = R^* \cap K(\tau)$.  \qed
\enddemo

\subheading{Remark 3.5} The proof of Theorem 3.2 is essentially 
due to Ray Heitmann. In his paper \cite{H1} Heitmann defines 
{\it simple PS-extensions}. For a regular element $x$ in a ring
$R$ and a formal power series in $x$ transcendental over $R$,
a simple PS-extension of $R$ for $x$ is an infinite direct 
union of simple transcendental extensions of $R$. If $R$ is
Noetherian and $T$ is a simple PS-extension of $R$, 
Heitmann proves in \cite{H1, Theorem~1.4} that a certain
monomorphism condition is equivalent to $T$ being Noetherian.
Heitmann's monomorphism condition insures that 
the element $f$ in the proof of Lemma 3.3 is a regular element  in 
$R^*$. In our situation our flatness condition on the embedding $U_0\longrightarrow 
R^*_a$, and hence on $U\longrightarrow R^*_a$, implies 
the
regularity of $f$ in $R^*$. Thus
Proposition 3.1 yields that if $s=1$ and the embedding $U_0\longrightarrow 
R^*_a$ is flat, then the ring $U=\varinjlim R[\tau_n]$
is a simple PS-extension satisfying the monomorphism condition
of Heitmann. In view of Theorem~1.1, Heitmann's monomorphism 
condition on the PS-extension determined by $\tau$ is equivalent
to $\tau$ yielding a flat extension. 
The flat extension concept 
however extends to more than one element $\tau$. 
 
\demo{Completion of Proof of Theorem 1.1}
If $U$ is Noetherian, we have already shown that   
$R^*[1/a]$ is flat over $U_0$.
Assume, conversely, that $R^*[1/a]$ is flat over $U_0=
R[\tau_1, \dots ,\tau_s]$. It 
follows that $R^*[1/a]$ is flat over $
R[\tau_1]$. By Theorem~3.2, $U(1)$, the directed union ring
constructed with respect to $\tau_1$ in (2.1) is Noetherian
and $R^* \cap K(\tau_1)$ is a localization 
of $U(1)$. It also follows 
that $U(1)^*[1/a]=R^*[1/a]$ is flat over $U(1)[\tau_2, \dots ,\tau_s]$ 
(cf. \cite{HRW2, Proposition~5.10}). 
Hence a simple induction argument implies that $U$ is Noetherian. 
Hence $U^* = R^*$ is flat over $U$ and $A$ is a localization
of $U$. \qed
\enddemo
 

\Refs
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\jour J. Algebra \vol 75 \yr 1982 \pages 386-94
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\paper A peculiar unmixed domain
\jour Proc. AMS \vol 87 \yr 1983 \pages 596-600
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\paper Intermediate rings between a local domain and its
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\paper Examples of noncatenary rings 
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\endref  



\endRefs


\enddocument 




From: William Heinzer <heinzer@MATH.Purdue.EDU>
Date: Thu, 7 Aug 1997 19:51:09 -0500 (EST)
To: swiegand@math.unl.edu
Subject: Re:  Building Noetherian
Cc: heinzer@MATH.Purdue.EDU, rotthaus@math.msu.edu

Dear Sylvia and Christel,  Thanks Sylvia for your e-mail that just
came.  
        On the situation about infinitely many tau_i's, it looks to
me that if for each positive integer n, the elements tau_1, ..., tau_n
are (in our old terminology) primarily limit-intersecting in a over R,
then with U and A defined as usual with respect to the countably 
infinite sequence tau_1, tau_2, ... that it should follow that U and
A are Noetherian. It looks to me that to show this comes down to 
showing that a directed union of Noetherian domains of a certain type
have the property that their union is Noetherian. ... in this
connection I'm reminded of a question Ray Heitmann raised 20 years ago
that Gilmer and I answered. Ray's question was whether for R a 
Noetherian ring and X an infinite set of variables over R is 
R(X) necessarily Noetherian. The answer is "yes" as is also the case
for an appropriately defined R<X>.
        I hope we can investigate the situation on infinitely many
primarily limit-intersecting elements and prove the Noetherian property
for U and A holds in this situation. But since you mention Sylvia that
you will be going to Colorado tomorrow and since I'm to be in W.Va. 
next Wednesday to Saturday and then leave for Chapel Hill August 19,
it seems to me it might be best not to try to include this in the
eno.tex writeup we are now working on.  ... what do you think?

Best regards,  Bill







