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\centerline{\bf IDEAL THEORY IN TWO-DIMENSIONAL REGULAR}
\centerline{\bf LOCAL DOMAINS AND BIRATIONAL EXTENSIONS}
\vskip 30 pt 
\centerline{\smc William Heinzer and David Lantz}
\vskip 15 pt

{ 
\parindent 0 pt

Department of Mathematics, Purdue University, West Lafayette, 
IN 47907. E-mail: {\tt heinzer\@math.purdue.edu} 
\vskip 3 pt
Department of Mathematics, Colgate University, 13 Oak Drive, 
Hamilton, NY 13346-1398. E-mail: {\tt dlantz\@center.colgate.edu}
}

\vskip 60 pt

\subheading{0. Introduction}
\smallskip

(0.1) Let $(R,\m)$ be a two-dimensional regular local domain 
with infinite residue field $R/\m$. Associated to an $\m$-primary 
ideal $I$ in $R$ is its Hilbert polynomial
$$
P_I(n) = e_0(I)\binom{n+1}{2} - e_1(I)n + e_2(I)\ ,
$$
the integer-valued polynomial giving the
length of the $R$-module $R/I^n$ for sufficiently 
large positive integers $n$.  The coefficient $e_0$ is 
well known to be a positive integer, the {\it multiplicity} of
$I$, and in our context, the coefficients $e_1$ and $e_2$ are
known to be nonnegative integers.

A well-known result of Rees \cite{Re1, Theorem~3.2} implies
that for each $\m$-primary ideal $I$ of $R$ the integral
closure of $I$ is the unique largest 
ideal containing $I$ and having the same multiplicity.
A result of Shah in \cite{Sh, Theorem~1}  implies the
existence of a unique largest ideal
$\k{I}{1}$ containing $I$ and having the same 
coefficients $e_0$ and $e_1$ of its Hilbert polynomial.  We 
call $\k{I}{1}$ the $e_1${\it-ideal associated with} $I$. 
If $I = \k{I}{1}$, we call $I$ a {\it first
coefficient ideal} or an $e_1${\it-ideal}. 

There is an interplay between the internal structure of 
the ideals in $R$ and the external structure of certain
birational extensions of $R$. In this connection, 
for an $\m$-primary ideal $I$, the {\it blowup} of $I$, 
$$
\B(I) =  \Proj(R[It]) = \{R[I/a]_P : a\in I-0,\ P\in \Spec(R[I/a])\}\ ,
$$ 
is the projective model over $R$ (in the sense of \cite{ZS, page~120})
consisting of the local domains containing $R$ that
are minimal with respect to domination among all the local domains 
containing $R$ in which the extension of $I$ is principal. 
There is a nonempty finite subset of $\B(I)$  consisting of local
domains in which $I$ generates an ideal primary for the maximal
ideal; each of these local domains is one-dimensional and their
intersection $D$  is a one-dimensional semilocal 
domain called the {\it first coefficient domain} of $I$. As noted
in \cite{HJL, (1.3) and (3.2)}, we have $ID\cap R = \k{I}{1}$;
indeed, since all powers $I^n$ of $I$ have the same blowup, we have
$I^nD\cap R = \k{(I^n)}{1}$, for each positive integer $n$. 

Our goal in this paper is a better understanding of 
$e_1$-ideals and their first coefficient domains over a 
two-dimensional regular local domain. 
The situation where the first coefficient domain is a 
semilocal PID is well understood in view of the Zariski 
theory concerning complete ideals and prime divisors on $R$ 
(see, e.g., \cite{Z}, \cite{ZS, Appendix~5} or \cite{Hu}).
In particular, if $V$ is
a DVR birationally dominating $R$ which is a spot over $R$ (i.e., 
in Zariski's terminology a {\it prime divisor of the second kind} 
on $R$; in \cite{A2} a {\it hidden prime divisor} of $R$),
then the ideals of $R$ contracted from $V$ form a descending chain
$\m = \a_0> \a_1 > \a_2 > \dots $ of complete $\m$-primary ideals,
the valuation ideals of $R$ with respect to $V$. The Zariski theory 
associates to the prime divisor of the second kind $V$ 
a unique simple (i.e., not factorable into a product of proper
ideals) complete ideal $\b$. One way of characterizing 
$\b$ is that $\b$ is maximal among $\m$-primary ideals $\c$ of 
$R$ with the property that all powers of $\c$ are contracted from $V$.
We have $\b = \a_n$ for some $n$. For example,
$\b = \m$ if and only if $V$ is the ord-valuation domain
$R[y/x]_{\m R[y/x]}$ where $\m = (x,y)R$. 
If $n > 0$, then certain 
of the ideals $\a_0,\dots \a_{n-1}$ are also simple complete
ideals. If we label the simple complete ideals in this chain
as $\b_0 = \m$, $\b_1$, $\dots$, $\b_s = \b = \a_n$, then 
Zariski proves that each of the valuation ideals $\a_i$, $i \ge 0$, 
is a product of powers of $\b_0,\dots,\b_s$ \cite{ZS, page~392}. 
For example, if $\m = (x,y)R$ and $V$ is the integral closure
of $R[x^2/y^3]_{\m R[x^2/y^3]}$, then $\b = (x^2,xy^2,y^3)R$ is
the simple complete ideal associated to $V$,  and 
$\m = \a_0 > \a_1 = \b_1 = (x,y^2)R > \a_2 = \m^2 > \b$
is the beginning of the chain of ideals 
$\{\a_i\}$ of $R$ contracted from $V$. The
result of Zariski just mentioned implies that each $\a_i$ is
a power product of $\m, \b_1$ and $\b$. More detailed information
as to which products of the $\b_j$ are actually contracted from
$V$ is given by Noh in \cite{No, Theorem~3.1}.

(0.2) To describe the same situation from a different starting point,
let $I$ be a complete $\m$-primary ideal of $R$. The first
coefficient domain $D$ of $I$ is then a semilocal PID which is the 
intersection of the Rees valuation domains of $I$, i.e., 
the DVR's on $\B(I)$ that dominate $R$. In this case, $D$ is uniquely 
determined as the largest one-dimensional semilocal subdomain $E$ of 
the fraction field of $R$ having the property that all
the powers of $I$ are contracted from $E$ (see (3.4) below). 
If $V_1,\dots,V_n$ are
the Rees valuation domains of $I$, then the Zariski theory
implies that $I$ has the form
$$
\b_1^{r_1}\dots \b_n^{r_n} \tag{$*$}
$$ 
where the $r_j$ are positive integers and $\b_j$ is the
simple complete ideal of $R$ associated to $V_j$, $j = 1,\dots,n$. 

In the present paper we
pursue the study of $e_1$-ideals and first coefficient
domains begun in \cite{HJL}. In particular,
we consider implications of the Zariski theory 
for these broader classes of ideals and integral domains. 
Our objective, only partially realized, is to identify the first 
coefficient domains over a two-dimensional regular local domain 
and the ideals of which they are first coefficient domains.

In Section~1 we illustrate with several examples properties that
one-dimensional spots birationally dominating a two-dimensional 
regular local domain may have or fail to have. We also observe
in Proposition~1.1 that the condition of being a spot descends from
an integral extension. 

In Section~2 we consider implications of
residual transcendence. As part of Theorem~2.2,
we prove  that if $R$ is a 
two-dimensional RLR of characteristic $p > 0$ with algebraically
closed residue field and $D$ is a one-dimensional local domain 
birationally dominating $R$ such that the integral closure of $D$ 
is a prime divisor on $R$, then $D$ is the first coefficient domain
of an ideal of $R$. 

In Section~3 we examine asymptotic behavior of
ideals and implications for first coefficient domains. Suppose 
$(R,\m)$ is a local domain that is the intersection of its localizations
at height-one primes and $D$ is a one-dimensional semilocal domain
birationally dominating $R$. In Theorem~3.3 we prove that if $J$ is
an $\m$-primary ideal of $R$ such that $JD$ is principal and 
$J^nD \cap R = J^n$ for each positive integer $n$, then the first
coefficient domain of $J$ is a localization of $D$. In particular,
if $D$ is local, then $D$ is the first coefficient domain of $J$.

As usual, we abbreviate ``regular local domain'' by RLR and
``rank-one discrete valuation domain'' by DVR. The words
``local'' and ``semilocal'' include the hypothesis of Noetherian.
The symbol $<$ between sets denotes proper inclusion.
For an ideal $I$ in a Noetherian domain $R$ the blowup of $I$ and 
the first coefficient domain of $I$ are defined as in (0.1) above.
The Rees valuation domains of $I$ are the localizations of the
integral closure of the first coefficient domain of $I$ at its
maximal ideals. It is convenient to extend some familiar terminology 
to the case of rings that are not necessarily Noetherian or that have
more than one maximal ideal: A ring $D$ 
containing a domain $R$ having a unique maximal ideal $\m$ 
is said to {\it birationally
dominate} $R$ if $D$ is contained in the fraction field of
$R$ and for each maximal ideal $N$ of $D$, $N \cap R = \m$.
An extension ring $D$ of a ring $R$ is said to be {\it affine}
over $R$ if $D$ is finitely generated as an algebra
over $R$. We say that a ring $D$ with finitely 
many maximal ideals is a {\it semispot} over a subring $R$ if $D$ 
is a ring of fractions of a ring containing and affine over $R$. If
such a $D$ has only one maximal ideal, then we call it a {\it spot}
over $R$.


\subheading{1. One-dimensional birational spots} 
\smallskip

We are interested in considering one-dimensional semilocal
domains $D$ that birationally dominate a two-dimensional
RLR $R$. A DVR $V$ birationally dominating $R$ is a spot 
over $R$ if and only if the the residue field of $V$ is not
algebraic as an extension of $R/\m$ \cite{A1, Proposition~3, page~336}.
An interesting property of such a DVR $V$ (also proved in \cite{A1})
is that the 
residue field $F$ of $V$ is ruled as an extension field of 
$R/\m$, i.e., $F$ is obtained as a
simple transcendental extension of a field intermediate
between $R/\m$ and $F$. In view of the fact that $R$ is a
two-dimensional RLR, it follows 
that $F$ is a simple transcendental
extension of a finite algebraic extension of $R/\m$.

In general, if $D$ is a one-dimensional semilocal domain 
birationally dominating $R$, then
the integral closure $D'$ of $D$ is a semilocal PID birationally
dominating $R$. If $R$ is complete, then $D'$ is 
necessarily a semispot over $R$; but for certain $R$ (such as
$R = k[x,y]_{(x,y)k[x,y]}$ where $x,y$ are indeterminates over the
field $k$) there
exist DVR's birationally dominating $R$ that are not spots over $R$
(cf., e.g., \cite{HRS}). 

We begin by proving a result (Corollary 1.3) that implies that if 
$D$ is a one-dimensional semilocal domain birationally dominating a 
two-dimensional RLR $R$ and if the integral
closure $D'$ of $D$ is a semispot over $R$, then $D$ is 
a semispot over $R$ and $D'$ is a finitely generated $D$-module.

\proclaim{Proposition 1.1} Let $R$ be a Noetherian ring, 
and let $V$ be a semispot over $R$.
Suppose $R \subseteq D \subseteq V$ with $D$ quasilocal and $V$ 
integral over $D$. Then $D$ is a spot over $R$ and $V$ is a
finitely generated $D$-module.
\endproclaim

\demo{Proof} Since $V$ is a semispot over $R$, there exist
elements $a_1,\dots,a_n \in V$ such that $V$ is a ring of fractions 
of $R[a_1,\dots,a_n]$. Let $b_1,\dots,b_m$ be the coefficients 
of monic polynomials over $D$ satisfied by $a_1,\dots,a_n$; 
set $B = R[b_1,\dots,b_m]$ and $A = B[a_1,\dots,a_n]$. Let 
$Q$ be the center of $D$ on $B$, and let $A_1$ and $B_1$ be the 
rings of fractions of $A$ and $B$ at the multiplicative set $B - Q$.
Then $B_1$ is local, with maximal ideal $Q_1 = QB_1$, and $A_1$ 
is a finite integral extension of $B_1$. Hence $A_1$ has only
finitely many maximal ideals.
Let $P_1, \dots, P_r$ denote the centers on $A_1$ of the
maximal ideals of $V$, and let $S = A_1 - (\bigcup_{i=1}^rP_i)$. 
Since $V$ is a ring of fractions of $R[a_1,\dots,a_n]$, we have
$S^{-1}A_1 = V$. Choose $a \in S$ such
that $a$ is in each maximal ideal of $A_1$ distinct from 
$P_1,\dots,P_r$ (if any --- otherwise let $a = 1$).
Then $1/a$ is in $V$
and hence is integral over $D$. Let $c_1,\dots,c_p$ be the
coefficients of a monic polynomial over $D$ satisfied by $1/a$; let
$(B_2,Q_2)$ be the localization of $B_1[c_1,\dots,c_p]$ at the 
center of $D$ on this ring, and set $A_2 = B_2[1/a,A_1]$. 

We claim that
$A_2 = V$. To see this, it suffices to show each $s$ in $S$ is
a unit in $A_2$: Assume by way of contradiction that $s$ in $S$ is 
in a maximal ideal $M$ of
$A_2$. Since $A_2$ is integral over $B_2$, we have
$M \cap B_2 = Q_2$, and so $Q_1 = M \cap B_1 = (M \cap A_1) \cap B_1$.
Since $A_1$ is integral over $B_1$, $M \cap A_1$ is maximal in
$A_1$. Moreover, $M \cap A_1$ survives in $A_2$ , so our choice of
$a$ assures that $M \cap A_1$ is the center on $A_1$ of one of the
maximal ideals of $V$. But this yields 
$s \in S \subseteq A_1 - (M \cap A_1)$, a contradiction. 

Therefore, $V$ is an affine extension of $B_2$ and 
hence a finitely generated $D$-module. Thus, by Artin-Tate
\cite{Ku, Lemma~3.3, page~16}, $D$ is an affine
extension of $B_2$ and hence a spot over $R$.~\qed 
\enddemo

To extend this result to the case where $D$ has finitely many 
maximal ideals, we use:

\proclaim{Proposition 1.2} Let $R$ be an integral domain.
Suppose $D$ is an extension domain of $R$ having
only finitely many maximal ideals $N_1,\dots,N_r$ and having 
the property that $D_{N_i}$ is a spot over $R$
for each $i = 1,\dots, r$. Then $D$ is a semispot over $R$.
\endproclaim

\demo{Proof} For each maximal ideal $N_i$ of $D$ 
there is a finite subset $T_i$ of $D_{N_i}$ such that
$D_{N_i}$ is a localization of $R[T_i]$. And there is an
element $s_i$ of $D - N_i$ for which $s_iT_i \subseteq D$.
Let $A = R[(\bigcup_{i = 1}^r s_iT_i) \cup \{s_1,\dots,s_r\}]$.
If $P_i$ denotes the center of $D_{N_i}$ on $A$,
then $A \subseteq D \subseteq D_{N_i} = A_{P_i}$;
so $D$ is the ring of fractions of $A$ at the complement of the
union of the $P_i$'s.~\qed
\enddemo

As an immediate corollary of Propositions~1.1 and 1.2, we have:
 
\proclaim{Corollary 1.3} Let $D$ be a semilocal extension domain 
of a Noetherian domain $R$, and let $V$ be a domain integral over
$D$. If $V$ is a semispot over $R$, then $D$ is also a 
semispot over $R$.~\qed
\endproclaim

(1.4) It follows from Corollary~1.3 that a one-dimensional semilocal 
domain $D$ that birationally dominates a two-dimensional RLR
$R$ is a semispot over $R$ if and only 
the integral closure of $D$ is an intersection
of prime divisors of the second kind on $R$, or equivalently,
if and only if each DVR birationally  containing $D$ is a 
prime divisor of the second kind on $R$. 

We are interested in the question of which one-dimensional 
semilocal domains birationally dominating $R$ are first 
coefficient domains of ideals of $R$. The first coefficient
domains of complete ideals of $R$ are well understood. They
are precisely the one-dimensional semilocal PID's birationally
dominating $R$ that are semispots over $R$. Moreover, if $I$
and $J$ are complete $\m$-primary ideals of $R$ with first
coefficient domains $D_I$ and $D_J$, respectively, then 
$D_I \cap D_J$ is a PID semispot over $R$ and is
the first coefficient domain of $IJ$. More generally, by
the Theorem on Independence of Valuations
(e.g., \cite{N, (11.11)} or \cite{ZS, Theorem~18, p.~45}) 
the intersection of two semilocal PID's birationally dominating 
a local domain is again a semilocal PID birationally dominating
the local domain. But for arbitrary $\m$-primary
ideals $I$ and $J$ of $R$, the relation of $D_I$ and $D_J$ with
the first coefficient domain of $IJ$ is more delicate.
It is not necessarily $D_I \cap D_J$; indeed, in Example~1.5 we
show that $D_I \cap D_J$ need not be a first coefficient domain
of $R$. In this example we make use of the description of the
first coefficient domain of an ideal generated by a regular
sequence given in \cite{HJL, (3.8)}.

\medskip

\noindent
{\bf Example 1.5.} Let $k$ be a field of characteristic 0 and $x,y$
be indeterminates over $k$; set $R = k[x,y]_{(x,y)}$. Then the
first coefficient domains of the ideals $(x^2,y^2)R$ and
$(x^2,xy+y^2)R$ are
$$
D_1 = k((y/x)^2) + M \qquad\text{and}\qquad 
D_2 = k((y/x) + (y/x)^2) + M\ ,
$$
respectively, where $M$ is the maximal ideal of the ord-valuation 
domain $V = R[y/x]_{\m R[y/x]} = k(y/x) + M$ over $R$. (The
maximal ideals $M_1$ and $M_2$ of $D_1$ and $D_2$, respectively,
are contained in $M$, and a module basis for $V$ over either $D_1$
or $D_2$ is $1, y/x$. Since $M_i(y/x) \subseteq D_i$ and 
$M_iV = M$, we have $M_i = M$.)
Since $k$ is of characteristic zero, we have
$k((y/x)^2) \cap k((y/x) + (y/x)^2) = k$. It follows that the residue 
field of $D_1\cap D_2$ at the center of $V$ on $D_1 \cap D_2$ is 
not residually transcendental over the 
residue field $k$ of $R$, so $D_1\cap D_2$ is not a semispot over
$R$ by (1.4) and hence is not the first coefficient domain of an ideal of $R$.

\medskip

(1.6) Suppose $I$ and $J$ are $\m$-primary ideals of $R$,
where $(R,\m)$ is a two-dimensional RLR, or more generally, a 
quasi-unmixed analytically unramified local domain. We want to
relate the first coefficient domain $D$ of $IJ$ to 
the first coefficient domains $D_I$ and $D_J$ of $I$ and $J$.
A first remark is that since
the set of Rees valuation domains of $IJ$ is the union of
the sets of Rees valuation domains of $I$ and $J$, 
the integral closure of $D$ is the
intersection of the integral closures of $D_I$ and $D_J$.  
With each DVR $V$ that is a localization of the integral 
closure of $D_I$ (of which there are only finitely many)
we associate a one-dimensional 
semilocal domain $D_V = (D_I)_P[B]$, where $P$ is the center of $V$
on $D_I$ and $B$ is the unique local domain on the blowup of $J$ that
is dominated by $V$. In an analogous way we construct
$D_W$ for each DVR $W$ that is a 
localization of the integral closure of $D_J$. 
The first coefficient domain $D$ of $IJ$ is the
the intersection of the one-dimensional semilocal domains $D_V$ and
$D_W$ as $V$ and $W$ vary over the sets of the Rees valuation
domains of $I$ and $J$ respectively. 
\medskip
(1.7)
The proofs of several results below rely on Theorem~3.12 of \cite{HJL};
and on rereading the proof of that result, we feel one point 
deserves a fuller discussion. The relevant hypotheses in that result
are as follows: $R$ is a normal, analytically unramified, 
quasi-unmixed, local domain with infinite residue field,
$I$ is an ideal primary for the maximal ideal of $R$, $D$ is the
first coefficient domain of $I$, $E$ is a domain birational and
integral over $D$, and $a$ is an element of $I$ for which $ID = aD$. 
In the proof, we set $S = R[1/a] \cap D$ and $T = R[1/a] \cap E$,
and we assert that $D,E$ are rings of fractions of $S,T$ 
respectively. This is true under the hypothesis of Theorem~3.12 of
\cite{HJL}, but in Example~1.8 below we show that for 
$a \in \m$ with $aD \ne ID$ it can happen that $D$ is not a ring of fractions
of $S = R[1/a] \cap D$. So we felt these assertions 
should be given a more explicit justification: The hypothesis that $D$ 
is the first coefficient domain of $I$ means that there exists an 
element $b$ of $I$ 
such that $D$ is an intersection of a finite number of one-dimensional
localizations of $R[I/b]$ and hence is itself a ring of fractions of
$R[I/b]$. Moreover, $bD = ID = aD$. Thus, $b/a$ is an element of 
$R[I/a]$ that is not in any of the prime ideals of $D$, so the
ring of fractions of $R[I/a]$ with respect to the complement in
$R[I/a]$ of the union of the primes in $D$ contains $R[I/b]$ and
hence is all of $D$. Since $S = R[1/a]\cap D \supseteq R[I/a]$,
we see that $D$ is also a ring of fractions of $S$. 
Now we turn to $T = R[1/a] \cap E$, which is almost integral over 
$S$ since there is a nonzero conductor from $E$ into $D$ (because $R$
is analytically unramified \cite{Re2, Theorem~1.2}). Since 
$$ 
S = \cap\{R[I/a]_P \ : \ P \text{ is a height-one prime }\}\,
$$
and since $R[I/a]$ is universally catenary, $S$ is contained 
in the integral closure of $R[I/a]$. Moreover, the
fact that $R$ is analytically unramified implies that the
integral closure of $R[I/a]$ is a finitely generated 
$R[I/a]$-module. Therefore $S$ is Noetherian
and hence $T$ is 
integral over $S$. Since  $D$ is a ring of fractions of $S$, 
the maximal ideals of $D$ are centered on height-one primes of
$S$. It follows that the maximal ideals of $E$ are centered on
height-one primes of $T$. Since the essential valuation
domains of $R[1/a]$ are all localizations of $S$ and of $T$,
it follows that $E$ is a ring of fractions of $T$.

\medskip

\noindent
{\bf Example 1.8.} Let $R = k[x,y]_{(x,y)k[x,y]}$, where $k$ is a field
and $x,y$ are indeterminates over $k$. Let $V = k(y/x)[x]_{(x)}$ 
be the ord-valuation domain of $R$. Then $V = k(y/x) + M$,
where $M$ is the maximal ideal of $V$. Let 
$D = k((y^2 + x^2)/xy) + M$. Then $D$ is the first coefficient
domain of the ideal $(xy, y^2+x^2)R$, a one-dimensional
local domain that birationally dominates $R$, and $V$
is the integral closure of $D$.
Let $T = R[1/x] \cap V$ and $S = R[1/x] \cap D$. Then
$T = R[y/x]$, so $S = R[y/x] \cap D$. Using 
that $k[y/x] \cap k((y^2 + x^2)/xy) = k$ and
considering the unique expression of each element of
a subdomain of $V$ as the sum of an element of $k(y/x)$ and
an element of $M$, we see that $S = k + (M \cap R[y/x])$.
Hence $D$ is centered on a maximal ideal of $S$ and is
not a localization of $S$. We also have in this example
that $S$ is not Noetherian and $T$ is 
almost integral but not integral over $S$. The localization
of $S$ at each of its height-one primes contains $R[1/x]$.


\subheading{2. Residually transcendental elements}
\smallskip

Let $(R,\m)$ be a two-dimensional RLR with residue field
$k = R/\m$. A first coefficient domain
of an $\m$-primary ideal of $R$ is a one-dimensional semispot 
birationally dominating $R$. As a partial converse,
we observe in Proposition~2.1 that 
a domain satisfying these hypotheses is at least a ring of 
fractions of a first coefficient domain of $R$.

\proclaim{Proposition 2.1} Let $(R,\m)$ be a two-dimensional
RLR and $E$ be a one-dimensional semispot birationally dominating 
$R$. Then there exists a first coefficient domain $D$ of $R$ 
such that $E$ is a ring of fractions of $D$.
\endproclaim

\demo{Proof} Let $a_1,\dots,a_n,b$ be elements of $R$ such
that $E$ is a ring of fractions of $R[a_1/b,\dots,a_n/b]$.
We may assume that $a_1,\dots,a_n,b$ have no common factor in
$R$, so that the ideal $I = (a_1,\dots,a_n,b)R$ is $\m$-primary.
Let $D_0$ denote the first coefficient domain of $I$. Since
$E$ is a semispot over $R$, the dimension formula \cite{M, (14.D)}
shows that for each maximal ideal $N$ of $E$
the image of at least one of the quotients $a_i/b$ in $E/N$
is transcendental over $R/\m$. Thus, the center of $N$ on
$R[a_1/b,\dots,a_n/b]$ is one-dimensional, so that $D_0\subseteq
E_N$. Since this holds for each maximal ideal $N$ of $E$, 
$D_0\subseteq E$. But there may be prime divisors 
dominating $R$ that contain $D_0$ but not $E$. The intersection 
$D$ of all these prime divisors and $E$ is an integral extension
of $D_0$ and hence a first coefficient domain (of an ideal
integral over a power of $I$) by \cite{HJL, Theorem~3.12}. We 
have $D \subseteq E$ are one-dimensional semilocal domains 
with $E$ birational over $D$ and $D$ integrally closed in $E$.
Forming the ring of fractions of $D$ with respect to the
elements of $D$ that are units of $E$ and applying 
\cite{N, (33.1)}, we see 
that $E$ is a ring of fractions of $D$.~\qed
\enddemo

A variant of the process used in this proof is as follows:
With $R$, $E$, etc. as in Proposition~2.1 and its proof,
let $(c,d)R$ be a reduction of $I = (a_1,\dots,a_n,b)R$
(or of a power of $I$ if the residue field of $R$ is finite and
$I$ fails to have a 2-generated reduction). For each maximal
ideal $N$ of $E$, the image of $c/d$ in $E/N$ is transcendental
over $R/\m$, so $N \cap R[c/d] = \m R[c/d]$. It follows  
that $E$ is a localization of the integral closure of 
$R[c/d]_{\m R[c/d]}$ in $E$. To realize $E$ itself as a first 
coefficient domain in this manner amounts to answering in the
affirmative the following question: Does
there exist a single element $a/b$ of $E$ such that 
$J = (a,b)R$ is a reduction of a complete ideal of the form
$(*)$ in (0.2) above, where the $r_j$ are positive 
integers and the $\b_j$ are the simple complete
ideals corresponding to the DVR localizations of 
the integral closure of $E$? If so, then $E$ and $R[a/b]_{\m R[a/b]}$
have the same integral closure. Thus, $E$ is integral over 
$R[a/b]_{\m R[a/b]}$ and hence a first coefficient domain
in its own right. The proof of Theorem~2.2 below
is essentially the construction of such an element $a/b$ in
a special case.

In the proof of Theorem 2.2 is a reference to $R(t)$, where
$t$ is an indeterminate over $R$.
In general, for a ring $A$, the symbol $A(t)$ denotes the ring of 
fractions of the polynomial ring $A[t]$ with respect to the 
multiplicative system of polynomials
whose coefficients generate the unit ideal 
in $A$ (cf. \cite{N, page~18}).  In the present
local case, this means only that not all of the coefficients of
the polynomial are in $\m$. There is a natural
epimorphism from $R(t)$ onto the simple transcendental field
extension $k(t)$ of $k$, with kernel generated by $\m$; images
under this epimorphism (as well as under other extensions of
the epimorphism $R\to k$) are denoted by overbars (vincula).

\proclaim{Theorem 2.2} Let $D$ be a one-dimensional 
local domain birationally dominating a two-dimensional
RLR $R$. Assume that $k = R/\m$
is algebraically closed, that the integral
closure $D'$ of $D$ is a prime divisor on $R$, 
and that either (1)~$R$ has nonzero
characteristic or (2)~$D$ contains the maximal
ideal of $D'$. Then there is an 
$\m$-primary ideal of which $D$ is the first coefficient domain.
\endproclaim

\demo{Proof} By (1.1), $D$ is a spot over $R$ and $D'$ is a 
finitely generated $D$-module.
In view of the last sentence of General Example~3.8
and Theorem~3.12 of \cite{HJL}, it is enough to find a 2-generated
$\m$-primary ideal $(a,b)R$ of $R$ for which $a/b \in D$ and the
integral closure of $R[a/b]_{\m R[a/b]}$ is $D'$. Also,
since $D'$ is a prime divisor of the second kind of
$R$, there is a simple complete $\m$-primary ideal $\b$ with which
$D'$ is associated, in the sense of the Zariski theory. It will
suffice to find elements $a,b$ of $R$ so that $a/b\in D$ and the 
ideal $(a,b)R$ is a reduction of a power of $\b$. 

Let $(c,d)R$ be a minimal reduction of $\b$ (or of a power of $\b$). 
Then the residue field of $D'$ is of transcendence degree
1 over $k$, generated by the image $\overline{c/d}$ of $c/d$
(because $k$ is algebraically closed \cite{HuS, Remark~3.5}), but 
algebraic over the residue field of $D$, and
for any other prime divisor of the second kind of $R$, either
$c/d$ is not in that prime divisor or its image in the residue field
is not transcendental over the image of $k$ (i.e., $c/d$ is not
``residually transcendental'' for any other prime divisor of the
second kind). Thus, for an element $z$ of $D$ of which the image 
$\overline{z}$ in the residue field of $D$ (or $D'$) is 
transcendental over $k$, there is an element $\varphi(t)$
of $R(t)$ such that if 
$\overline{\varphi}(t) \in k(t)$ is the image of
$\varphi(t)$ in $R(t)/\m R(t)$, then
$\overline{z} = \overline{\varphi}(\overline{c/d})$. 
We may assume that the numerator and denominator of 
$\overline{\varphi}(t)$ are relatively prime polynomials over $k$. 
Now $z - \varphi(c/d)$ is in the maximal ideal of $D'$, so under
assumption~(2) of the statement, we immediately have that $\varphi(c/d)
\in D$. To reach a similar (though not identical) conclusion under 
assumption~(1), we note that since $D'$ is local and is a 
finitely generated $D$-module, 
the maximal ideal of $D$ contains a power of the maximal ideal of
$D'$; so we can raise $z - \varphi(c/d)$ to a sufficiently 
high power $q$, a power of the characteristic of $R$, to conclude that
$\varphi(c/d)^q \in D$. Multiplying the numerator and denominator of 
$\varphi$ or $\varphi^q$ by the same power of $d$, we convert them 
into forms $a=a(c,d)$ and $b=b(c,d)$ in $c,d$ of the same degree $n$ 
such that their images in the degree-$n$ piece of the fiber ring 
$F((c,d)) = R[(c,d)t]\otimes_R R/\m$, a polynomial ring in two 
variables over $k$, are relatively prime.

We show that $(a,b)$ is a reduction of $(c,d)^n$, which will complete
the proof. It suffices to show that $(a,b)(c,d)^n = (c,d)^{2n}$, and
by Nakayama's Lemma it suffices to show that the $k$-vector
spaces $[(a,b)(c,d)^n + \m(c,d)^{2n}]/\m(c,d)^{2n}$ and 
$(c,d)^{2n}/\m(c,d)^{2n}$ have the same dimension. The latter is
the degree-$2n$ piece of the fiber ring $F((c,d))$; its dimension 
is $2n+1$. The images of the products $ac^id^{n-i}$, $i=0,\ldots,n$,
span a subspace of the former of dimension $n+1$, and similarly with
$b$ in place of $a$; and since the images of $a,b$ are relatively
prime, the intersection of these two subspaces is spanned by 
the image of $ab$, so it is one-dimensional. Thus, 
$[(a,b)(c,d)^n + \m(c,d)^{2n}]/\m(c,d)^{2n}$ has dimension $2(n+1)-1
= 2n+1$ as required.~\qed
\enddemo
 

\subheading{3. Principal extensions and contracted powers}
\smallskip

(3.1) Suppose $D$ is a one-dimensional semispot birationally 
dominating a quasi-unmixed, analytically unramified,
normal local domain $(R,\m)$. In this
section we seek conditions for $D$ to be the first coefficient 
domain of an ideal $I$ of $R$. If $D$ is the first coefficient
domain of $I$, then $ID$ is principal, and replacing $I$ by 
the associated $e_1$-ideal of a high power of $I$, we obtain
an $\m$-primary ideal $J$ such that $JD$ is principal and 
$J^nD \cap R = J^n$ for each positive integer $n$
\cite{HJLS, Theorem~3.17}. Thus a
necessary condition for $D$ to be a first coefficient domain
is the existence of an $\m$-primary ideal $J$ of $R$ with the two
properties: (1) $JD$ is principal, and (2) $J^nD \cap R = J^n$
for each positive integer $n$. If $D$ is local,
we prove in Theorem~3.3 that this necessary condition 
is also sufficient, and that $D$ is in fact the first 
coefficient domain of each ideal $J$ with these two properties.

The case in which $V$ is a prime divisor birationally 
dominating a two-dimensional RLR $(R,\m)$ is illustrative.
Suppose $a$ is a nonzero element 
of $\m$ and consider the descending chain $J_n = a^nV \cap R$,
$n = 1,2,\dots$, of ideals of $R$. As noted in the introduction,
each $J_n$ is a complete ideal of $R$, and from the Zariski theory 
it follows that $J_n$ is a product of powers of the simple 
complete ideals associated with the finitely many prime divisors
that ``come out'' on the sequence of quadratic transformations of $R$ 
along $V$. Let $\b$ be the simple complete ideal of $R$ associated
to $V$, and suppose the $V$-values of $a$ and $\b$ are $p$ and $q$
respectively. Then $J_q = \b^p$. Since all powers of $\b$ are
contracted from $V$, for
each positive integer $r$ we have $J_q^r = J_{qr}$, or equivalently
the powers of $J_q$ are contracted from $V$. Moreover,
$J_q$ has $V$ as its first coefficient domain.

\medskip  
(3.2) It was noted in \cite{HJL, (3.7)} that the first coefficient
domain of an ideal $I$ of $R$ can be described using the 
minimal primes of $IR[It]$ of the Rees algebra $R[It]$ or
the minimal primes of $t^{-1}R[t^{-1}, It]$ of the extended
Rees algebra $R[t^{-1}, It]$ of $I$ (where $t$ is an
indeterminate over $R$). These primes are in one-to-one
correspondence with the maximal ideals of the first coefficient
domain $D$ of $I$: If $P$ is one of these minimal primes, then
$P$ does not contain the degree-$1$ piece of the Rees algebra (or
extended Rees algebra), say $bt \notin P$ where $b\in I$. 
Then the localization
of the (extended) Rees algebra at $P$ is also a localization of
$R[I/b][bt, (bt)^{-1}]$ and has the form $D_N(bt)$ (cf. the
paragraph before Theorem~2.2) for the 
maximal ideal $N$ of $D$ corresponding to $P$. [Note: The 
$V(t)$ in the equations on the last line of \cite{HJL, (3.7)} 
should be $V(bt)$, for $b$ as above.]

\proclaim{Theorem 3.3} Let $(R,\m)$ be a local domain that is
the intersection of its localizations at height-one primes, and
let $D$ be a one-dimensional semilocal domain that birationally
dominates $R$. Suppose $J$ is an $\m$-primary ideal of $R$ such
that $JD$ is principal and $J^nD \cap R = J^n$ for each
positive integer $n$. Then the first coefficient domain 
of $J$ is a localization of $D$. In particular, if $D$ is local,
then $D$ is the first coefficient domain of $J$. 
\endproclaim
\demo{Proof} Replacing $J$, if necessary, by a power of $J$,
we may assume that $JD = aD$ where $a \in J$.
Let $A = R[t^{-1}, Jt]$ be the extended Rees algebra
of the ideal $J$ of $R$; let $D(at)$ denote the localization
of the polynomial ring $D[at]$ at the complement of the union of
the extension to $D[at]$ of the maximal ideals of $D$; and let
$K$ be the fraction field of $R$. Since $D[at, (at)^{-1}]$ is
Cohen-Macaulay, it is the intersection of its localizations at
height-one primes. It follows that 
$D[at, (at)^{-1}] = K[at, (at)^{-1}] \cap D(at)$, and hence that
$$\align
R[t, t^{-1}] \cap D[at, (at)^{-1}] 
  &= R[t,t^{-1}] \cap K[at, (at)^{-1}] \cap D(at) \\
  &= R[t,t^{-1}] \cap D(at) = A\ . 
\endalign$$
Let $P$ be a minimal prime of $t^{-1}A$ and let $S = A - P$.
Then $A_P = S^{-1}(R[t,t^{-1}] \cap D(at))=
S^{-1}(R[t,t^{-1}]) \cap S^{-1}D(at)$.
Since $R[t,t^{-1}]$ is the locally finite intersection of its
localizations at height-one primes, to show $S^{-1}(R[t,t^{-1}])
= K(t)$, it suffices to show $S$ meets each height-one prime $Q$ of
$R[t,t^{-1}]$: If $Q \cap S = \emptyset$, then $Q \cap A \subseteq P$.
Since $Q \cap A \ne 0$, we must have $Q \cap A = P$. But 
$P \cap R = \m$ and $Q \cap R < \m$, a contradiction. 
Thus $S$ meets each height-one
prime of $R[t,t^{-1}]$, so $A_P = S^{-1}D(at)$.

Let $E$ be the first coefficient domain of $J$. The maximal ideals
$N$ of $E$ are in one-to-one correspondence with the minimal primes
$P$ of $t^{-1}A$, where $A_P = E_N(at)$. Since each $A_P$ is a
localization of $D(at)$, the intersection $E(at)$ of the $A_P$'s
is a ring of fractions of $D(at)$. Intersecting with $K$ shows
that $E$ is a ring of fractions of $D$.~\qed
\enddemo

The following corollary implies the uniqueness
property of the intersection of the Rees valuation domains of
an ideal mentioned in (0.2).

\proclaim{Corollary 3.4} Let $(R,\m)$ be a quasi-unmixed,
analytically unramified, normal local domain, and
let $I$ be an $\m$-primary ideal of $R$. The first coefficient
domain $E$ of $I$ is the unique largest one-dimensional 
semilocal domain $D$ birationally dominating $R$ and having
the properties that $ID$ is principal and $I^nD \cap R$ is 
contained in the $e_1$-ideal of $I^n$ for each positive
integer $n$.
\endproclaim 
\demo{Proof} By \cite{HJLS, Theorem 3.17} for all sufficiently
large positive integers $r$, the ideal  $J = I^rE \cap R$ has the
property that $E$ is the first coefficient domain of $J$ and
for each positive integer $n$ we have $J^nE \cap R = J^n = 
I^{rn}E \cap R$ is the $e_1$-ideal associated to $I^{rn}$.
Therefore $J^nD$ is
principal and $J^nD \cap R = J^n$ for each $n$. By Theorem~3.3,
$E$ is a localization of $D$.~\qed
\enddemo
 
\proclaim{Corollary 3.5} Let $D$ be a one-dimensional spot 
birationally dominating a two-dimensional RLR $(R,\m)$. 
If $J$ is an $\m$-primary ideal in $R$ such that $JD$ is 
principal and all the powers of $J$ are contracted from $D$,
then $D$ is the first coefficient domain of $J$, and the 
integral closure of $J$ is a product of powers of the 
simple complete ideals associated to the localizations 
of the integral closure of $D$.
\endproclaim

\proclaim{Proposition 3.6} Let $(R,\m)$ be a quasi-unmixed 
analytically unramified local domain of dimension $d \ge 2$, 
and let $J$ be an $\m$-primary ideal of $R$. Let $D$ be a 
one-dimensional semilocal domain birationally dominating $R$, and 
let $V$ be a finitely generated birational integral extension of 
$D$. If all the powers of $J$ are contracted from $D$, then 
for each positive integer $n$, $J^nV \cap R$ is integral over $J^n$.
In particular, if $I = JD \cap R$ is a normal ideal (i.e., the
powers of $I$ are integrally closed), then all the powers of
$I$ are contracted from $V$.
\endproclaim

\noindent
Remark. The hypothesis in Proposition 3.6 (and in Proposition~3.7 
below) that $V$ is a finitely generated $D$-module is
necessary (cf., e.g., \cite{HRS, (1.27)}). But if $D$ (or $V$,
by Corollary~1.3) is a (birational) semispot over $R$, the
hypothesis on $R$ assures that $V$ is a finitely generated
$D$-module \cite{Re2, Theorem~1.2}.

\demo{Proof} For the first assertion, it suffices to show that $I$ 
is integral over $J$. Since $J$ is contained in each nonzero prime 
ideal of $D$, there exists a positive integer $c$ such that $J^c$ is
contained in the conductor of $V$ into $D$. Thus,
for all positive integers $n$, we have
$$
I^{n+c} \subseteq I^{n+c}V \cap R = J^{n+c}V \cap R \subseteq 
J^nD \cap R = J^n \subseteq I^n.
$$
It follows that the length of $R/J^n$ is between those of 
$R/I^n$ and $R/I^{n+c}$. Now, for $n$ sufficiently large,
the length of $I^n/I^{n+c}$ is a polynomial in $n$ of
degree $d-1$, while the lengths of $R/I^n$ and $R/J^n$ are polynomials 
in $n$ of degree $d$. Therefore the Hilbert polynomials of $I$ and $J$
have the same highest degree coefficient, i.e., $I$ and $J$ have
the same multiplicity. By \cite{Re1, Theorem~3.2}, $I$ is integral
over $J$.

For the second assertion, note that $J^n \subseteq I^n \subseteq
I^nV\cap R = J^nV\cap R$; the last ideal is integral over $J^n$, so
if $I^n$ is integrally closed, it is equal to $I^nV\cap R$.~\qed
\enddemo

\proclaim{Proposition 3.7} Let $(R,\m)$ be 
a normal, quasi-unmixed, analytically
unramified local domain of dimension $d \ge 2$, and $J$ be an 
$\m$-primary ideal of $R$. Let $D$ be a one-dimensional semilocal
domain birationally dominating $R$ such that the integral
closure $V$ of $D$ is a finitely generated $D$-module.
Suppose that $J^nD \cap R = J^n$ for each positive integer $n$, 
and let $I_n = J^nV \cap R$ for each $n$.
\itemitem{(1)} For sufficiently large $r$, all the powers of $I_r$ are
contracted from $V$.
\itemitem{(2)} Therefore $V$ is contained in each of the Rees 
valuation domains of $J$, and so $D$ is contained in the 
integral closure of the first coefficient domain of $J$. 
\endproclaim
\demo{Proof} (1) By \cite{Re2, Theorem~1.4}, for sufficiently 
large $r$, $I_r$ is a normal ideal, so by Proposition~3.6 all the powers 
of $I_r$ are contracted from $V$. (2)~Since the intersection of the
Rees valuation domains is the unique largest one-dimensional 
semilocal subdomain $E$ of the fraction field of $R$
with the property that the integral closure of $J^n$ is
$J^nE \cap R$ for each $n$, $V$ is contained in each of
the Rees valuation domains of $J$.~\qed
\enddemo

(3.8) Let $(R,\m)$ be a two-dimensional RLR, let 
$D$ be a one-dimensional semispot birationally dominating  
$R$, and let $V$ be the integral closure of $D$.  
Let $\b_1,\dots , \b_n$ be the simple complete ideals 
of $R$ associated to the DVR's which are localizations of $V$. 
Then the associated $e_1$-ideal of an $\m$-primary
ideal $I$ of $R$ has the form $(*)$ as in (0.2) above,
where the $r_j$ are positive integers,
if and only if $V$ is 
the first coefficient domain of $I$. 
By \cite{HJL, Theorem~3.12}, $D$ is a first coefficient domain if
and only if there exists an ideal $J$ of $R$ such that $JD$ is
principal and such that the integral closure of $J$ is of the
form $(*)$. Thus, for example, if $V$ is the ord-valuation domain
of $R$, then $D$ is a first coefficient domain if and only if 
there exists an ideal $J$ such that $JD$ is principal and such
that the integral closure of $J$ is a power of $\m$.

\medskip 
(3.9) With $R, D$ as in Corollary~3.5, there always 
exist $\m$-primary ideals $J$ with the property that all 
their powers are contracted from $D$ (for, if $J$ is the
product of the simple complete ideals associated to the DVR
localizations of the integral closure of $D$, 
then all the powers of $J$ are contracted 
from the integral closure of $D$ and hence also from $D$). 
Thus, in this case
the issue is whether there exists such a $J$ with $JD$ principal.
However, if one passes to a more general situation where $R$ is a 
two-dimensional excellent normal local domain, then there may exist
birationally dominating DVR spots $V$ over $R$ for which there does
not exist an ideal $J$ of $R$ such that all the powers of 
$J$ are contracted from $V$. By definition, an excellent two-dimensional 
normal local domain $(R,\m)$ with the property that each prime
divisor of the second kind on $R$ is the first coefficient domain
of an $\m$-primary ideal is said to satisfy Muhly's
condition $(N)$ (cf. \cite{HL, page~291}). If $R$ is a two-dimensional 
complete normal local domain, Cutkosky proves in \cite{C, Theorem~4} 
that $R$ satisfies condition $(N)$ if and only if $R$ has torsion 
divisor class group. Thus, for example,
$R = {\Bbb C}[[x,y,z]]$, where $x^3 + y^3 + z^3 = 0$, has prime
divisors of the second kind which are not first coefficient domains of
an ideal of $R$.

\medskip
(3.10) Let $(R,\m)$ be a two-dimensional RLR and let $D$ be the
first coefficient domain of an ideal $I$ of $R$. If $D$ is a prime
divisor of $R$ and $a \in \m$ is a nonzero element, then there exists
a positive integer $n$ such that $D$ is the first coefficient domain
of $a^nD \cap R$ (cf. (3.1)). The case of a general first coefficient
domain, however, is different: In Example~1.8, there is no
positive integer $m$ for which $D$ is
the first coefficient domain of $x^mD \cap R$. 
This phenomenon is the reef on which founders the following
naive approach to realizing a one-dimensional semispot $E$ 
birationally dominating $R$ as a first coefficient domain.
Let $\b_1,\dots, \b_s$ be the distinct simple complete ideals of $R$ 
associated with the prime divisors obtained as 
localizations of the integral closure $E'$ of
$E$, and let  $a \in R$ be such that $aE' \cap R = \b_1\dots \b_s$.
Let $A = R[t^{-1}, t] \cap E(at)$. Then
$A = R[t^{-1}, I_1t, I_2t^2, \dots ]$, where $I_n = a^nE \cap R$. 
The integral closure of $A$ is 
$A' = R[t^{-1}, (I_1)'t, (I_2)'t^2, \dots ]$, while 
the domain $A'' = R[t^{-1}, t] \cap E'(at)$ is almost integral
over $A$ since there is a nonzero conductor from $E'$ to $E$.
The following conditions are equivalent: (1)~$A$ is Noetherian.
(2)~$A$ is affine over $R$. (3)~$A' = A''$. When these 
conditions hold,  
$(I_1)' = \b_1 \dots \b_s$ and $E$ is the first
coefficient domain of an ideal integral over a power of $I_1$.
In Example~1.8, however, for $E = D$ and $a = x$, we have
$A' < A''$. When we have $A' < A''$, 
there is no positive integer $m$ for which the powers of $I_m$ 
are contracted from $E$, nor for which 
$E$ is the first coefficient domain of $I_m$.


\vskip 18 pt
\centerline{ACKNOWLEDGEMENT}
\vskip 6 pt
This work is the outgrowth of a project
with Bernard Johnston. We would like to acknowledge his 
penetrating questions which initiated the present paper.
We also thank the referee for a careful reading of 
the paper.



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\endref

\ref \key{ZS} \by O. Zariski and P. Samuel
\book Commutative Algebra, Vol.~II
\publ Springer-Verlag \publaddr New York \yr 1975
\endref

\endRefs

\enddocument

Now let us consider a $D$ with integral closure $E$ a finite
intersection of prime divisors dominating $R$. Let us try to
give a sufficient condition in order that $D$ be the first
coefficient domain of an ideal $J$ of $R$. Getting contracted
ideals from $E$ with the property that all their powers are also
contracted from $E$ seems more complicated now than it was for
the case of $E = V$ a single prime divisor. However, if we take
$(a,b)R$ to be a reduction of the product of the simple complete
ideals associated to the prime divisors containing $E$, then
$aE \cap R = I$ has the property that all its powers are
contracted from $E$. So a sufficient condition in order that
$D$ be a first coefficient domain is as follows: let 
$J_n = a^nD \cap R$. If there exists a positive integer $n$
such that $J_n^r$ is contracted from $D$ for all $r$, then
$D$ is the first coefficient domain of $J_n$.

It seems that we might be able to make progress on the general
problem by looking at a pair of smaller rings. For $D$ a spot
birationally dominating $R$ with integral closure a DVR $V$, let
$A$ be an affine extension of $R$ of which $D$ is
a localization. By inverting one more element, if necessary, we
assume that the center of $D$ on $A$ is the only minimal prime
of $\m A$. Let $b$ be a common denominator for the
generators of $A$, and set $S = 
R[1/b]\cap D$ and $T = R[1/b] \cap V$. Then $S$ is the 
$S_1$-ification of $A$, so it is affine over $R$, and there is a
nonzero conductor from $T$ into $S$, so $T$ is also affine. Also,
if we set $J_n = b^nD\cap R$ and $I_n = b_nV^\cap R$, we have
$S = \bigcup_{n=1}^{\infty} J_n/b^n$ and 
$T = \bigcup_{n=1}^{\infty} I_n/b^n$, and since they are affine
we have $S = R[J_n/b^n]$ and $T = R[I_n/b^n]$ for $n$
sufficiently large.

\noindent
{\bf Question} What additional hypotheses are necessary to
conclude that $(J_n)^k = J_{nk}$ for some $n$ and all $k$? 
[This seems to be true if $D$ is a DVR (i.e., it works for $I_n$
in place of $J_n$, e.g., when $n$ is the value in the valuation 
associated to $D$ of the associated simple complete
ideal) or if $b$ is superficial for $J_n$.] 
A related question: How does the blowup $J_n$ 
relate to that of $J_{nk}$?

\demo{Proof} By Proposition~2.3, $D$ is contained in the integral 
closure $E'$ of the first coefficient domain $E$ of $J$. Since $V$ 
is a DVR, we must have $E' = V$; so $E$ is local and is dominated by $V$. 
Since $JD$ is principal, $E$ is dominated by $D$. 
Let $a \in J$ be such that $aE = JE$, and set $S = R[1/a] \cap E$
and $T = R[1/a] \cap D$. Then 
$$
S = \bigcup_{n=1}^\infty\frac{a^nE \cap R}{a^n}
  = \bigcup_{n=1}^\infty\frac{J^n}{a^n}
  = \bigcup_{n=1}^\infty\frac{a^nD \cap R}{a^n} = T\ .
$$
The centers of $E$ and $D$ on $S = T$ are equal, and by the argument
in (1.7) (where the roles of $E$ and $D$ are reversed), $E$ and $D$
are localizations of $S$ and $T$, respectively. Thus $E = D$.~\qed
\enddemo
\ref
\endref

Alternatively,
we could assume the ideal $I$ is such that its blowup is equal to
its $S_1$-ification. Then $S = R[I/a]$. 

And for this it
is enough to find $a/b$ in $D$ with $a,b$ relatively prime in $R$ so 
that the image of $a/b$ in the residue field of $D'$ is transcendental
over the image of $k = R/\m$ (i.e., ``$a/b$ is residually 
transcendental for $D'\,$'') but $a/b$ is not residually transcendental 
for any other prime divisor of $R$ of the second kind. 

Let us show that, if the question has an affirmative answer, i.e.,
(simplifying notation) that $J^k = b^kD\cap R$ and
$I^k = b^kV\cap R$ for all $k$ (the latter \underbar{can} be
arranged; we are merely adjusting notation), then $D$ is a first 
coefficient domain. There is a positive integer $c$ for which $b^c$ 
is an element of the conductor of $V$ into $D$, so we have $I^{k+c} 
= b^{k+c}V\cap R \subseteq b^kD\cap R = J^k \subseteq I^k$ for
all $k$. Examining the Hilbert polynomials for $I$ and $J$, we
find that they have the same multiplicity; so $I$ is integral
over $J$. But this means that $J$ is a reduction of a power
of the simple
complete ideal associated to $V$. Since $D$ contains the first
coefficient domain of $J$, $D$ is itself a first coefficient domain.

If $I = (a,b)$ and $J = (c,d)$ are $\m$-primary ideals one might ask
for which prime divisors $V$ of the second kind on $R$ is
$a/b + c/d = (ad + bc)/bd$ residually transcendental? If the ideals
$(a,b)R$ and $(c,d)R$ have no common Rees valuation domains, then I
was thinking that $(ad + bc)/bd$ would be residually transcendental
at precisely the union of the Rees valuation domains of $I$ and $J$.
But this is not correct. The elements $a/b$ and $c/d$ may have poles
at a prime divisor $V$ of the second kind on $R$, these poles
might cancel out when we take sum, and the sum might be of value
zero and residually transcendental at $V$. 
It is good that this sort of behavior does not happen when we evaluate
$a/b$ at a rational function in one variable over a coefficient field
for $R$.












