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\begin{document}

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\title{PROJECTIVELY FULL IDEALS IN NOETHERIAN RINGS}
\author{Catalin Ciuperca, William J. Heinzer, Louis J. Ratliff Jr., and
David E. Rush}


\date{}
\maketitle

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%13A18, 13A30, 13B02, 13E05;
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\begin{abstract}
Let $R$ be a Noetherian commutative ring with unit $1 \ne 0$, and
let $I$ be a regular proper ideal of $R$. The set $\mathbf P(I)$ of
integrally closed ideals projectively equivalent to $I$ is linearly
ordered by inclusion and discrete. There is naturally associated to
$\mathbf P(I)$ a numerical semigroup $S(I)$; we have  $S(I) = \IN$
if and only if every element of $\mathbf P(I)$ is the integral
closure of a power of the largest element $J$ of $\mathbf P(I)$. If
this holds, the ideal $J$ and the set $\mathbf P(I)$ are said to be
projectively full. If $I$ is invertible and $R$ is integrally
closed, we prove that $\mathbf P(I)$ is projectively full. We
investigate the behavior of projectively full ideals in various
types of ring extensions. We prove that a normal ideal $I$ of a
local ring $(R,M)$ is projectively full if $I \not\subseteq M^2$ and
both the associated graded ring $G(M)$ and the fiber cone ring $F(I)$ are
reduced.  We present examples of normal local domains $(R,M)$ of
altitude two for which the maximal ideal $M$ is not projectively
full.
\end{abstract}


\section{INTRODUCTION.}
All rings in this paper are commutative with a unit  $1$ $\ne$ $0$. Let
$I$ be a regular proper ideal of the Noetherian ring $R$  (that is, $I$
contains
a regular  element of $R$ and $I \ne R$).  The
concept of projective equivalence of ideals and the study of
ideals projectively equivalent to $I$ was introduced by
Samuel in \cite{S} and further developed by Nagata in \cite{N1}.  Making
use of interesting work of Rees in \cite{Re}, McAdam,
Ratliff, and Sally in \cite[Corollary 2.4]{MRS}
prove that the set $\mathbf P(I)$ of integrally closed ideals
projectively equivalent to $I$ is linearly ordered by inclusion (and
discrete). They also prove that if $I$ and $J$ are projectively
equivalent, then the set $\Rees I$ of Rees valuation rings  of $I$ is
equal to the set $\Rees J$ of Rees valuation rings  of $J$ and the values of
$I$ and $J$ with respect to these Rees valuation rings are proportional
\cite[Proposition 2.10]{MRS}.  We observe in \cite{CHRR} that the converse
also holds  and   further develop the connections
between projectively equivalent ideals and their Rees valuation rings. For
this purpose,
we define in \cite{CHRR} the ideal $I$ to be {\bf projectively full} if the
set $\mathbf P(I)$ of integrally closed ideals projectively equivalent to
$I$
is  precisely the set $\{(I^n)_a\}$ consisting of
the integral closures of the powers of $I$.   If there exists a
projectively full  ideal $J$ that is
projectively equivalent to $I$, we say that $\mathbf P(I)$ is {\bf
projectively full}.
As described in \cite{CHRR},
there is naturally associated to $I$ and to the projective equivalence class
of $I$
a numerical semigroup $S(I)$. One has $S(I) = \IN$, the semigroup of
nonnegative
integers under addition,  if and only if $\mathbf P(I)$ is projectively
full.

Our goal in the present paper is to build on the work in \cite{MRS}
and \cite {CHRR} by further developing the concept of projectively
full ideals and examining the numerical semigroup $S(I)$. In Section
2 we present several results relating $\mathbf P(I)$ and $\mathbf
P(IA)$, for certain $R$-algebras $A$. In Section 3, these results
are applied to explore the relationship between the projective
fullness of $\mathbf P(I)$ and $\mathbf P(IA)$, for certain
$R$-algebras $A$. Several methods are given for obtaining
projectively full ideals. We prove that an integrally closed
complete intersection ideal of a local ring is projectively full. We
also prove that if $I$ is a proper invertible ideal of an integrally
closed Noetherian domain, then $\mathbf P(I)$ is projectively full.
In Section 4 we present classes of examples of projectively full
ideals. For instance, we prove that if $R_P$ is a regular local
domain, then $PI$  is projectively full for all regular ideals  $I$
$\nsubseteq$ $P$. In Section 5 we present a family of examples of
integrally closed local domains $(R,M)$ of altitude two for which
$M$, and therefore $\mathbf P(M)$, is not projectively full.

Our notation is as in \cite{N2} and \cite{M}. A ring is said to be
{\bf integrally closed} if it is integrally closed in its total
quotient ring. In particular, a ring that is equal to its total
quotient ring is integrally closed. If we are given a ring
homomorphism from a ring $R$ to a ring $A$, then we say that $A$
with respect to this homomorphism  is an $R$-{\bf algebra.}

We thank the referee for helpful suggestions for revising this
paper.


\bigskip

\section{PROJECTIVELY EQUIVALENT IDEALS.}
In this section we prove several elementary results about projectively
equivalent ideals.
For this, we need the following definitions.
(Throughout, $\IN$  denotes the
set of nonnegative integers,
and  $\IN_+$  (resp., $\IQ_+$, $\IR_+$)  denotes the set of
positive integers (resp., rational numbers, real numbers).)

\begin{defi}
\label{defi1}
{\em Let  $I$  be an ideal in a Noetherian ring  $R$.

\noindent
{\bf{(\ref{defi1}.1)}}
The
{\bf{Rees ring}}
$\mathbf R(R,I)$
{\bf{of}}
$R$
{\bf{with respect to}}
$I$  is the graded subring  $\mathbf R(R,I)$ $=$  $R[u,tI]$  of  $R[u,t]$,
where  $t$  is an indeterminate and  $u$ $=$ $1/t$.

\noindent
{\bf{(\ref{defi1}.2)}}
$R'$  denotes the
{\bf{integral closure}}
of  $R$ in its total quotient ring.

\noindent
{\bf{(\ref{defi1}.3)}}
If  $R$  is local with maximal ideal  $M$, then  $a(I)$ denotes the
{\bf{analytic spread}}
of  $I$  (so  $a(I)$ $=$ $\altitude( \mathbf R(R,I)/((u,M) \mathbf R(R,I))$)
(see (\ref{defi1}.1))).

\noindent
{\bf{(\ref{defi1}.4)}}
$I_a$  denotes the
{\bf{integral closure of}}
$I$
{\bf{in}}
$R$, so  $I_a$ $=$ $\{b \in R \mid b$  satisfies an equation of the form
$b^n +i_1b^{n-1} + \cdots + i_n$ $=$ $0$,
where  $i_k$ $\in$ $I^k$  for  $k$ $=$ $1,\dots,n\}$.  The ideal $I$  is 
said
to be
{\bf{integrally closed}}
in case  $I$ $=$ $I_a$, and  $I$  is
{\bf{normal}}
in case  $(I^i)_a$ $=$ $I^i$  for all  $i$ $\in$ $\IN_+$.

\noindent
{\bf{(\ref{defi1}.5)}}
An ideal  $J$  in  $R$  is a
{\bf{reduction}}
of  $I$  in case  $J$ $\subseteq$ $I$  and  $JI^n$
$=$ $I^{n+1}$  for some  $n$ $\in$ $\IN$.

\noindent
{\bf{(\ref{defi1}.6)}}
An ideal  $J$  in  $R$  is
{\bf{projectively equivalent}}
to  $I$  in case  $(I^i)_a$ $=$ $(J^j)_a$  for some  $i,j$ $\in$ $\IN_+$.}
\end{defi}

Concerning (\ref{defi1}.6),
Samuel introduced projectively equivalent ideals in 1952 in \cite{S},
and a number of properties
of projective equivalence can be found in \cite{KMOR}, \cite{KR},
\cite{MR1}, \cite{MR2}, \cite{MR3}, \cite{MRS}, \cite{Ra2}, \cite{Ra3}.

\begin{rema}
\label{rema3}
{\em  Let  $R$  be a Noetherian ring.  Then

\noindent
{\bf{(\ref{rema3}.1)}}
The relation  ``$I$  is projectively equivalent to  $J$''
is an equivalence relation on  $\mathbf I$ $=$
$\{I \mid I$  is an ideal of  $R\}$.

\noindent
{\bf{(\ref{rema3}.2)}}
\cite[(2.1)(b)]{MRS}:
If  $I$  and  $J$  are ideals in  $R$  and if
$i,j,k,l$ $\in$ $\IN_+$  with  $\frac{i}{j}$ $=$ $\frac{k}{l}$, then
$(I^i)_a$ $=$ $(J^j)_a$  if and only if  $(I^k)_a$ $=$ $(J^l)_a$.

\noindent
{\bf{(\ref{rema3}.3)}}
Assume that  $I$ and $J$  are projectively equivalent in  $R$  and let  $K$
be an ideal
in  $R$.  Then  $(I+K)/K$ and $(J+K)/K$  are projectively equivalent in
$R/K$.

\noindent
{\bf{(\ref{rema3}.4)}}
Let  $A$  be an $R$-algebra.  If  $I,J$  are projectively equivalent
in  $R$, then  $IA,JA$  are projectively equivalent in  $A$. }
\end{rema}


Concerning (\ref{rema3}.4), it is not true in general that  integral
closedness of ideals is
preserved under a faithfully flat ring extension.
This need not be true even
if $(R,M)$ is a
regular local domain and $A$ = $\widehat R$
is the $M$-adic completion of $R$
\cite{HRW}.  Thus
there exists a regular local domain $R$ and an integrally closed ideal $I$
of $R$ such that
$\mathbf P (I \widehat R)$ $\ne$ $\{J\widehat R \mid J \in \mathbf P (I)\}$
(see (\ref{defid(I)}.3)).

With the preceding paragraph in mind, we note in (\ref{comb-prop})
below that  $\mathbf P (I)$,
$\Rees I$  (see
(\ref{defiRees})), and  $d(I)$  (see (\ref{defid(I)}.4))
behave nicely when passing to:  $R[X]$; $R(X)$; $R_S$  (for certain
multiplicatively
closed subsets  $S$  of  $R$); and, $R/K$
(where  $K$ $\subseteq$ $\Rad(R)$, the nilradical of $R$), and we also
show that  $d(I)$ $\le$ $d(IA)$  for certain types of $R$-algebras $A$. 
   For this, we first
define,
for a regular proper ideal $I$ in a Noetherian ring $R$,
the Rees valuation rings of  $I$,
the set $\mathbf P (I)$,  and
the positive integer  $d(I)$.

\begin{defi}
\label{defiRees} {\em Let  $I$  be a regular proper ideal in a
Noetherian ring  $R$, for each $x \in R$   let $v_I(x)$ $=$ $\max
\{k \in \IN  \mid x \in I^k\}$ (as usual, $I^0$ $=$ $R$  and
$v_I(x)$ $=$ $\infty$  in case  $x$ $\in$ $I^k$  for all  $k$ $\in$
$\IN$), and   let  $\overline v _I (x)$ $=$ $\lim _{k \rightarrow
\infty} (\frac{v_I(x^k)}{k})$. Rees shows in \cite{Re} that:  (a)
$\overline v _I(x)$  is well defined;  (b)  for each  $k$ $\in$
$\IN$ and  $x$ $\in$ $R$, $\overline v _I (x)$ $\ge$ $k$  if and
only if $x$ $\in$ $(I^k)_a$  (as usual, $(I^0)_a$ $=$ $R$); and, (c)
there exist valuations  $v_1,\dots,v_g$  defined on  $R$ (with
values in  $\IN \cup \{\infty\}$)  and positive integers
$e_1,\dots,e_g$  such that, for each  $x$ $\in$ $R$, $\overline v
_I(x) = \min \{\frac{v_i(x)}{e_i} \mid i = 1,\dots,g\}$. (These
$v_i$  and  $e_i$  are described as follows: let  $z_1,\dots,z_d$
be the minimal prime ideals  $z$  in  $R$ such that   $z+I$ $\ne$
$R$, for $i$ $=$ $1,\dots,d$, let  $R_i$ $=$ $R/z_i$, let  $F_i$  be
the quotient field of  $R_i$, let  $\mathbf R_i$ $=$ $\mathbf
R(R_i,(I+z_i)/z_i)$, let  $p_{i,1},\dots,p_{i,h_i}$  be the (height
one) prime divisors of  $u {\mathbf {R}_i} '$  (see
(\ref{defi1}.2)), let  $w_{i,j}$ be the valuation of the discrete
valuation ring $W_{i,j}$ $=$ ${{\mathbf {R}_i} '}_{p_{i,j}}$, let
$e_{i,j}$ $=$ $w_{i,j}(u)$, let  $V_{i,j}$ $=$ $W_{i,j} \cap F_i$,
and define $v_{i,j}$  on  $R$  by $v_{i,j}(x)$ $=$ $w_{i,j}(x+z_i)$.
Then  $v_1,\dots,v_g$ are the valuations  $v_{1,1},\dots,v_{d,h_d}$
resubscripted and  $e_1,\dots,e_g$ are the corresponding  $e_{i,j}$
resubscripted, and $\mathbf {Rees~I}$ $=$
$\{(V_1,N_1),\dots,(V_g,N_g)\}$, where $V_i$  is the valuation ring
of the valuation  $v_i$; note that  $IV_i$ $=$ ${N_i}^{e_i}V_i$.}
\end{defi}



\begin{defi}
\label{defid(I)}
{\em Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.}

\noindent
{\bf{(\ref{defid(I)}.1)}}
{\em For  $\alpha$ $\in$ $\mathbf \IR_+$
let  $I_\alpha$ $=$ $\{x$ $\in$ $R \mid \overline v _I(x)$ $\ge$
$\alpha\}$  (see (\ref{defiRees})).}

\noindent
{\bf{(\ref{defid(I)}.2)}}
{\em $\mathbf W(I)$ $=$ $\{\alpha$ $\in$ $\IR_+ \mid$  $\overline v _I(x)$
$=$ $\alpha$
for some  $x$ $\in$ $R\}$.}


\noindent
{\bf{(\ref{defid(I)}.3)}}
{\em $\mathbf U(I)$ $=$ $\{\alpha$ $\in$ $\mathbf W(I) \mid$ $I_\alpha$
is projectively equivalent to  $I\}$  (see (\ref{defid(I)}.1), 
(\ref{defid(I)}.2),
and (\ref{defi1}.6)),
and  $\mathbf P(I)  = \{I_\alpha \mid \alpha \in \mathbf U(I) \}$.}

\noindent
{\bf{(\ref{defid(I)}.4)}}
{\em $d(I)$  is the
smallest positive integer  $d$
such that, for all  $J$ $\in$ $\mathbf P (I)$,
$(J^d)_a$ $=$ $(I^j)_a$  for some  $j$ $\in$ $\IN_+$.
(See \cite[Theorem 2.8]{MRS}.)}
\end{defi}



\begin{rema}
\label{rema2}
{\em Let  $R$  be a Noetherian ring and let  $I$  be a regular proper ideal
in  $R$.

\noindent
{\bf{(\ref{rema2}.1)}}
Concerning (\ref{defid(I)}.1),
for each  $\alpha$ $\in$ $\IR_+$
the ideal  $I_\alpha$ is an integrally
closed ideal ($=$ $(I_\alpha)_a)$ in  $R$,
$I_n$ $=$ $(I^n)_a$  for all  $n$ $\in$ $\IN_+$,
and for all  $k$ $\in$ $\IN_+$ and for all  $I_{\alpha}$
$\in$ $\mathbf{P}(I)$  it holds that
$({I_{\alpha}}^k)_a$ $=$ $I_{k \alpha}$, by \cite[(2.1)(g), (2.1)(c) and
(2.5)]{MRS}.

\noindent
{\bf{(\ref{rema2}.2)}}
The sets  $\mathbf W(I)$  and  $\mathbf U(I)$  of
(\ref{defid(I)}.2) and (\ref{defid(I)}.3) are
discrete subsets of  $\IQ _+$,
by \cite[(1.1) and (2.8)]{MRS}.

\noindent
{\bf{(\ref{rema2}.3)}}
For the set  $\mathbf P (I)$ of (\ref{defid(I)}.3),
$\mathbf P (I)$  $=$ $\{J \mid J$  is an integrally closed ideal in  $R$
that is projectively equivalent to  $I\}$,
and  $\mathbf P (I)$  is linearly ordered by inclusion, by
\cite[(2.4)]{MRS}.

\noindent
{\bf{(\ref{rema2}.4)}}
Concerning (\ref{defid(I)}.4), it is
shown in \cite[(2.8) and (2.9)]{MRS} that
there exists a unique smallest
positive integer  $d(I)$  that is a common divisor
(but not necessarily the greatest common divisor) of the integers
$e_1,\dots,e_g$  of (\ref{defiRees})
such that, for each ideal  $J$
in  $R$  that is projectively equivalent to  $I$,
$(J^{d(I)})_a$ $=$ $(I^j)_a$  for some  $j$ $\in$ $\IN_+$.
Also, $d(I) \alpha$ $\in$ $\IN_+$   for all  $\alpha$ $\in$ $\mathbf U(I)$.
Further, if  $H,J$ $\in$ $\mathbf P (I)$  and if  $d(H), d(J)$  are the
corresponding
unique positive integers for  $H,J$, then  $(H^{d(J)})_a$ $=$
$(J^{d(H)})_a$,
by \cite[(4.8.3)]{CHRR}.
Finally, there exist $n^*(I)$ $\in$ $\IN_+$  such that
$\{\alpha \in \mathbf U(I) \mid \alpha \ge n^*(I)\}$
$=$ $\{n^*(I)+ \frac{k}{d(I)} \mid k \in \IN\}$
(in fact, each large  $n$ $\in$ $\IN_+$  is a suitable choice
for  $n^*(I)$), by \cite[(2.8)]{MRS}.}
\end{rema}

In Theorem \ref{divisor}  (together with its corollary) we show  that
$d(I)$ $\le$ $d(IA)$  for certain standard types of $R$-algebras $A$.


\begin{theo}
\label{divisor}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$  and let  $A$
be a
Noetherian ring that is an   $R$-algebra having the property that:
(a)  $IA$  is regular and proper;
and, (b)  if  $H$ $\subsetneq$ $J$  in  $\mathbf P (I)$, then  $(HA)_a$
$\subsetneq$ $(JA)_a$  in
$\mathbf P (IA)$.  Then  $d(IA)$ $\ge$ $d(I)$.  If  $d(IA)$ $>$ $d(I)$,
then  $d(IA)$ $=$ $kd(I)$  for some  $k$ $\in$ $\IN_+$.
\end{theo}

\begin{proof}
By the last sentence in (\ref{rema2}.4) let  $n$  be
a large enough positive integer so that, for all integers
$m$ $\ge$ $n$,  there are exactly  $d$ $=$ $d(I)$
ideals in  $\mathbf P(I)$  between  $(I^{m+1})_a$  and  $(I^m)_a$ (counting
one endpoint)  and
there are exactly  $d'$ $=$ $d(IA)$ ideals in
$\mathbf P(IA)$  between  $(I^{m+1}A)_a$  and  $(I^mA)_a$ (counting one
endpoint).
Let  $H_0,H_1,\dots,H_d$  in  $\mathbf P (I)$  such that
$(I^{n+1})_a$ $=$ $H_d$ $\subsetneq \cdots \subsetneq$ $H_0$ $=$ $(I^n)_a$.
Then each  $(H_iA)_a$ $\in$ $\mathbf P(IA)$  (by (\ref{rema3}.4)) and
$(I^{n+1}A)_a$ $=$ $(H_dA)_a$ $\subsetneq$ $ \cdots$ $ \subsetneq$ $
(H_0A)_a$ $=$ $(I^nA)_a$,
by (b).  It follows that $d'$ $\ge$ $d$.

Now assume that  $d'$ $>$ $d$.  Then to show that  $d'$ $=$ $kd$  for
some  $k$ $\in$ $\IN_+$  it suffices to show that there are exactly
$\frac{d'}{d} -1$  ideals
in  $\mathbf P (IA)$  that are strictly between  $(H_{i+1}A)_a$  and
$(H_iA)_a$  for  $i$
$\in$ $\{0,1,\dots,d-1\}$.

For this, it may clearly be assumed that  $d$ $\ge$ $2$, so it suffices to
show that if  $H_{i+1}$
$\subsetneq$ $H_i$ $\subsetneq$ $H_{i-1}$  are three consecutive ideals in
$\mathbf P(I)$
between  $(I^{n+1})_a$  and  $(I^n)_a$, then the number of ideals in
$\mathbf P (IA)$  between
$(H_{i+1}A)_a$  and $(H_iA)_a$  is the same as the number of ideals in
$\mathbf P (IA)$  between
$(H_iA)_a$  and $(H_{i-1}A)_a$.

For this, recall that  $H_j$ $=$ $I_{n+\frac{j}{d}}$
(see (\ref{rema2}.4)) and similarly (since  $(I^{n+1}A)_a$
$=$ $(IA)_{n+1}$ $\subseteq$ $(H_{i+1}A)_a$ $\subsetneq$
$(H_iA)_a$ $\subsetneq$ $(H_{i-1}A)_a$ $\subseteq$ $(IA)_n$ $=$
$(I^nA)_a$) it follows that, for  $j$ $\in$ $\{i-1,i,i+1\}$,
$(H_jA)_a$ $=$ $(IA)_{n+\frac{h_j}{d'}}$  for
some integer  $h_j$  between $j$ and  $d'$.
Therefore there exist exactly  $h_{i+1}-h_i$
ideals in  $\mathbf P (IA)$  between  $(H_{i+1}A)_a$  and  $(H_iA)_a$
(counting one endpoint), and
there exist exactly  $h_i-h_{i-1}$  ideals in  $\mathbf P (IA)$
between  $(H_iA)_a$  and  $(H_{i-1}A)_a$
(counting one endpoint),
so it suffices to show that $h_{i+1} - h_i$ $=$ $h_i-h_{i-1}$.


For this, it follows from (\ref{rema2}.1)
that, for all $m$ $\in$ $d\IN_+$,
$({H_j}^m)_a$ $=$ $({I_{n+\frac{j}{d}}} ^m)_a$ $=$ $(I^{nm + \frac{j}{d}
m})_a$
and, for all $m$ $\in$ $d'\IN_+$,
$(({H_jA)}^m)_a$ $=$ $(((IA)_{n+\frac{h_j}{d'}})^m)_a$ $=$ $((IA)^{nm +
\frac{h_j}{d'} m})_a$
$=$ $((I^{nm + \frac{h_j}{d'} m})_aA)_a$.
Therefore with  $m$ $=$ $dd'$  we get
$((I^{ndd' + jd'})_aA)_a$ $=$ $(({H_j}^{dd'})_aA)_a$ $=$ $((H_jA)^{d'd})_a$
$=$
$((I^{ndd' +h_jd})_aA)_a$.
Therefore it follows from (b) that
$(I^{ndd' + jd'})_a$ $=$
$(I^{nd'd +h_jd})_a$, so we get:
(i)  $(i+1)d'$ $=$ $dh_{i+1}$ $\in$ $\IN_+$ (for  $j$ $=$ $i+1$);
(ii)  $id'$ $=$ $dh_i$ $\in$ $\IN_+$ (for  $j$ $=$ $i$); and,
(iii)  $(i-1)d'$ $=$ $dh_{i-1}$ $\in$ $\IN_+$  (for  $j$ $=$ $i-1$).
It follows by subtracting (ii) from (i) and (iii) from (ii)
that  $h_{i+1} - h_i$ $=$ $\frac{d'}{d}$ $=$ $h_i - h_{i-1}$ $\in$ $\IN_+$,
so the number of ideals in  $\mathbf P(IA)$  between  $(H_{i+1}A)_a$  and
$(H_iA)_a$  is
the number of ideals in  $\mathbf P(IA)$  between  $(H_iA)_a$  and
$(H_{i-1}A)_a$.  Therefore
$d(IA)$ $=$ $d'$ $=$ $kd$ $=$ $kd(I)$, where  $k$ $=$
$h_i - h_{i-1}$ $\ge$ $2$.
\end{proof}

\begin{coro}
\label{corox}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.  Then (a) and
(b) of
Theorem \ref{divisor}  hold
for the following types of $R$-algebras $A$:
\begin{enumerate}
\item
$A$  is a Noetherian integral extension ring of  $R$  such that  $IA$  is
regular.
\item
$A$  is a faithfully flat Noetherian extension ring of  $R$.
\item
$A$ $=$ $R_S$  for some multiplicatively closed subset  $S$  of  $R$
such that  $IR_S$ $\ne$ $R_S$.
\end{enumerate}
\noindent
Therefore  $d(IA)$ $=$ $kd(I)$  for some positive integer  $k$
(possibly  $k$ $=$ $1$) for such rings  $A$.
\end{coro}

\begin{proof}
For (1) and (2) it is well known that  $(JA)_a \cap R$ $=$ $J_a$
for all regular proper ideals  $J$  in  $R$, so (b)
holds for such rings  $A$.  Also, (a) holds
for the rings  $A$  in (1) by hypothesis, and (a) holds for
the rings  $A$  in (2) by faithful flatness (since  $I$  is regular in
$R$).

For the rings  $R_S$  as in (3), by hypothesis  $IR_S$  is proper, so
(a) holds by flatness.
For (b), let $d$ $=$ $d(I)$  and let  $n$  and  $H_0,\dots,H_d$  be as in
the first paragraph of
the proof of (\ref{divisor}).  Suppose
that  $H_{i+1}R_S$ $=$ $H_iR_S$  for some  $i$ $\in$ $\{0,1,\dots,d-1\}$.
(Note that
$JR_S$ $=$ $(JR_S)_a$  if  $J$  is an ideal in  $R$  such that  $J$ $=$
$J_a$.)
Let  $H$ $=$ $H_{i+1}$  and  $J$ $=$ $H_i$.  Since  $H,J$ $\in$ $\mathbf P
(I)$, there
exist  $h,j$ $\in$ $\IN_+$  such
that  $(H^d)_a$ $=$ $(I^h)_a$  and  $(J^d)_a$ $=$ $(I^j)_a$  (by
(\ref{rema2}.4)).
Therefore  $(I^hR_S)_a$
$=$ $H^dR_S$ $=$ $J^dR_S$  (since $H_{i+1}R_S$ $=$ $H_iR_S$) $=$
$(I^jR_S)_a$,
hence  $h$ $=$ $j$  (since  $IR_S$  is a regular proper ideal).
Therefore, since $(H^d)_a$ $=$ $(I^h)_a$  and
$(J^d)_a$ $=$ $(I^j)_a$, it follows that  $(H^d)_a$ $=$ $(J^d)_a$, so
$H_a$ $=$ $J_a$, hence  $H_{i+1}$ $=$ $H_i$, and this is a contradiction.
It follows that the ideals  $H_iR_S$  are distinct (so
(b) holds) and are ideals in
$\mathbf P (IR_S)$
between  $(I^{n+1}R_S)_a$  and  $(I^nR_S)_a$, hence  $d(IR_S)$ $\ge$
$d(I)$.

The last statement is clear from (\ref{divisor}) and what has already been
shown.
\end{proof}

It would be interesting to know whether Corollary \ref{corox} holds in
general for a Noetherian ring $A$ that is a flat $R$-algebra such
that $IA \ne A$.

We next consider classes   of  $R$-algebras $A$   for which  $d(IA)$ $=$ 
$d(I)$.
For these algebras, $\mathbf P (I)$  and  $\Rees I$  also extend nicely.
The proofs of these results ((\ref{comb-prop}.1) - (\ref{comb-prop}.4))
are straightforward, so they are omitted.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{prop}\label{comb-prop}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.

\noindent
{\bf{(\ref{comb-prop}.1)}}
Let  $K$  be an ideal in  $R$  such that  $K$ $\subseteq$ $\Rad(R)$.
Then $\mathbf P((I+K)/K)$
$=$ $\{(J+K)/K  \mid J \in \mathbf P (I)\}$, $\Rees  (I+K)/K$ $=$ $\Rees I$,
and  $d((I+K)/K)$ $=$ $d(I)$.

\noindent
{\bf{(\ref{comb-prop}.2)}}
Let
$P_1,\dots,P_k$
be the centers in  $R$  of the Rees valuations of  $I$, and  let   $S$  be
a multiplicatively closed subset of  the nonzero elements of $R$. Assume 
that
$S \cap P_i$ $=$ $\emptyset$  if and only if  $i$ $=$ $1,\dots,h$  (for some
integer $h$ with  $1 \le h \le$ $k$).
Then  $\Rees IR_S$
$=$ $\{(V,N) \in \Rees I \mid N \cap R \in  \{P_1,\dots,P_h\} \}$, and
$d(IR_S) \ge d(I)$.
Moreover, if  $h$ $=$ $k$,
then $\mathbf P(IR_S)$
$=$ $\{JR_S \mid J \in \mathbf P (I)\}$  and  $d(IR_S)$ $=$ $d(I)$.

\noindent
{\bf{(\ref{comb-prop}.3)}}
Let
$X_1,\dots,X_g$  be independent
indeterminates, and let  $A$ $=$ $R[X_1,\dots,X_g]$.  Then  $\mathbf P(IA)$
$=$ $\{JA \mid J \in \mathbf P (I)\}$, $\Rees  IA$
$=$ $\{V[X_1,\dots,X_g]_{N[X_1,\dots,X_g]} \mid (V,N) \in \Rees I \}$,
and  $d(IA)$ $=$ $d(I)$.

\noindent
{\bf{(\ref{comb-prop}.4)}}
Let
$X_1,\dots,X_g$  be independent
indeterminates.
As in \cite[pp. 17-18]{N2}, let
$A$ $=$ $R(X_1,\dots,X_g)$
  denote the quotient ring of
the polynomial ring
$R[X_1,\dots,X_g]$  with respect to the multiplicatively closed set of
polynomials whose coefficients generate the unit ideal of  $R$. Then 
$\mathbf P(IA)$
$=$ $\{JA \mid J \in \mathbf P (I)\}$, $\Rees IA$
$=$ $\{V[X_1,\dots,X_g]_{N[X_1,\dots,X_g]} \mid (V,N) \in \Rees I \}$,
and  $d(IA)$ $=$ $d(I)$

\end{prop}





%%%%%%%%%%%%%%%%


\begin{rema}
\label{extendedidl}
{\em Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.  Let
$A$  be either
$R[X_1,\dots,X_g]$  or  $R(X_1,\dots,X_g)$  ($g$ $\ge$ $1$), and let  $J$
be an ideal in  $A$.  If there exist  $i,j$ $\in$ $\IN_+$  such that
either  $J^j$ $=$ $I^iA$  or  $(J^j)_a$ $=$ $(I^iA)_a$, then there exists
an ideal  $G$  in  $R$  such that  $(G^j)_a$ $=$ $(I^i)_a$  and
$((GA)^j)_a$ $=$ $((G^j)_a)A$ $=$ $(J^j)_a$, so  $(J^j)_a$  is
the integral closure of the extension of the  $j$-th power of
an ideal in  $R$.}
\end{rema}

\begin{proof}
If either  $J^j$ $=$ $I^iA$  or  $(J^j)_a$ $=$ $(I^iA)_a$, then  $IA,J$  are
projectively equivalent in  $A$, so  $J_a$ $\in$ $\mathbf P(IA)$  (by
(\ref{rema2}.3)).  But by (\ref{comb-prop}.3) (if  $A$ $=$ 
$R[X_1,\dots,X_g]$)  or
by
(\ref{comb-prop}.4) (if  $A$ $=$ $R(X_1,\dots,X_g)$)  $\mathbf P(IA)$ $=$
$\{HA \mid H \in \mathbf P (I)\}$.  It follows that  $J_a$
$=$ $GA$, where  $G$ $=$ $J_a \cap R$.  It therefore follows
that  $((I^i)_a)A$ $=$ $(J^j)_a$ $=$ $((J_a)^j)_a$ $=$ $((GA)^j)_a$ $=$
$((G^j)_a)A$,
so  $(J^j)_a$ $=$ $((G^j)_a)A$  and  $(G^j)_a$ $=$ $(I^i)_a$.
\end{proof}


In  Proposition \ref{propinvert},  we consider projective
equivalence for invertible ideals in an integrally closed Noetherian
ring. Concerning the hypothesis of (\ref{propinvert}), we remark
that an integrally closed local ring $R$ that contains a regular
proper  principal ideal is an integral domain. Moreover, if $R$ is
an integrally closed Noetherian ring with $\Rad(R) = (0)$, then $R$
is a finite product of Noetherian integrally closed domains.


\begin{prop}
\label{propinvert}
Let  $R$  be an integrally closed Noetherian ring and let
$I$ be a regular proper ideal of $R$.
\begin{enumerate}
\item
If $I$ is invertible, then every ideal projectively equivalent
to $I$ is invertible.
\item
If $R$ is local and
$I$ $=$ $bR$ is  principal, then every ideal  $J$  in  $R$  that
is projectively equivalent to  $I$  is principal
and invertible.
\end{enumerate}
\end{prop}

\begin{proof}
Since a regular ideal $I$ of a Noetherian ring  $R$  is invertible if and
only if
$IR_M$ is principal for each maximal ideal $M$ of $R$ and since projective
equivalence behaves well with respect to localization, to prove item 1  it
suffices to prove item 2. Thus we may assume $(R,M)$ is an integrally closed
local ring
and $I = bR$ is a regular proper principal ideal
(so  $R$  is an integral domain).
Let  $J$  be an ideal in  $R$  that is projectively equivalent
to  $bR$, so  $(J^i)_a$ $=$ $(b^m R)_a$  for some $i,m$  $\in$  $\IN_+$.
Then $a(J)$ $=$ $1$  (see (\ref{defi1}.3)),
since  $a(J)$ $=$ $a((J^i)_a)$ $=$ $a((b^mR)_a)$ $=$ $a(bR)$ $=$ $1$.
Assume temporarily that  $R/M$  is infinite and let  $xR$  be a minimal
reduction of  $J$.
Then  $xR$ $=$ $(xR)_a$, since  $R$  is integrally closed,
and  $xR$ $\subseteq$ $J$ $\subseteq$ $(xR)_a$, so  $J$ $=$ $xR$  is
principal.
Now, if  $R/M$  is finite, then  $T$ $=$ $R[X]_{MR[X]}$  is an integrally
closed local domain with infinite residue field and
$JT$  is projectively equivalent to  $bT$, by (\ref{rema3}.4), so  $JT$  is
principal.
Since a minimal generating set for
$J$  is a generating set for  $JT$, it can be reduced to a minimal
(one element) generating set for  $JT$, so  $J$  is principal (since
$J$ $=$ $JT \cap R$).  Since  $J$  is principal, $J$  is invertible.
\end{proof}

\begin{rema}
\label{nonintclosed}
{\em
Without the assumption that $R$ is integrally closed,
easy examples show that  Proposition \ref{propinvert} fails.
For example,
if $(R,M)$ is  a local domain of altitude one that is not integrally closed
and $I = xR$ is  a reduction of $M$, then $M$ is projectively equivalent to
$I$
and $M$ is not invertible. For a specific example, let $t$ be an
indeterminate
over a field $k$ and consider the subring $R = k[[t^2, t^3]]$ of the formal
power series ring $k[[t]]$. Then $I = t^2R$ is a reduction of $M = (t^2,
t^3)R$,
so $M$ is projectively equivalent to $I$.
}
\end{rema}


\bigskip

\section{PROJECTIVELY FULL IDEALS.}

Projectively full ideals are  introduced in \cite[Section 4]{CHRR}.
It is observed in \cite[(4.12)]{CHRR} that $\mathbf P(I)$ is
projectively full for every nonzero proper ideal $I$  in a regular
local domain  of altitude two;  see also \cite[(3.6)]{MRS}). In this
section we develop basic properties of projectively full ideals. We
then determine various classes  of ideals $I$ for which either $I$
or  $\mathbf P (I)$  is projectively full. Concerning the basic
properties, our main results show that: for certain $R$-algebras
$A$, if  $IA$  is projectively full, then  $I$ is projectively full
(see (\ref{fullcontracts})); the converse holds for  $A$  as in
(\ref{comb-prop}) (by (\ref{fullcors})); and, the
converse need not hold, even if  $A$  is a finite free integral
extension domain of  $R$  (see (\ref{remay}.1)). Concerning  ideals
$I$ where  either  $I$  or  $\mathbf P (I)$  is projectively full, we
show that: the integrally closed complete intersection ideals of a
local ring are projectively full (see (\ref{gen-comp}) and
(\ref{comp-int})); and  $\mathbf P (I)$  is projectively full for an
invertible ideal  $I$ in an integrally closed Noetherian ring (see
(\ref{propfull})).



We begin with the following remark.

\begin{rema}
\label{rema1}
{\em Let  $I$  be a regular proper  ideal in a Noetherian ring  $R$.}

\noindent
{\bf{(\ref{rema1}.1)}}
{\em It is immediate from the relevant definitions that  for each  $J$ $\in$
$\mathbf P (I)$
(see (\ref{defiRees}.3)) we have
$\{(J^i)_a  \mid i \in \IN_+\}$ $\subseteq$ $\mathbf P (I)$, and  $\mathbf P
(I)$
is projectively full if and only if there exists  $J$
$\in$ $\mathbf P (I)$  such that  $\{(J^i)_a  \mid i \in \IN_+\}$     $=$
$\mathbf P (I)$.
It is clear that if such an ideal  $J$
exists, then  $J$  must be the largest element
in the linearly ordered (discrete) set  $\mathbf P (I)$.
The numerical semigroup $S(I)$ $=$ $d(I) \mathbf U (I) \cup \{0\}$
(where  $\mathbf U (I)$  is as in (\ref{defiRees}.3))
is the semigroup of nonnegative integers under
addition,  if and only
if $\mathbf P(I)$ is projectively full.

\noindent
{\bf{(\ref{rema1}.2)}}
It follows
from (\ref{rema2}.4)
and (\ref{rema1}.1) that
$\mathbf P (I)$  is projectively full if and only if  $d(K)$ $=$ $1$, where
$K$  is the largest element in  $\mathbf P (I)$; cf \cite[(4.11)]{CHRR}.
Also, $I$ is projectively full if and only if $d(I) = 1$.
}
\end{rema}

Proposition \ref{fullcontracts}  is an immediate corollary of
(\ref{corox}) and (\ref{rema1}.2).

\begin{prop}
\label{fullcontracts}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$ and let  $A$
be
one of the following types of  $R$-algebras:
\begin{enumerate}
\item
$A$  is a Noetherian integral extension ring of  $R$  such that  $IA$  is
regular.
\item
$A$  is a faithfully flat Noetherian extension ring of  $R$.
\item
$A$ $=$ $R_S$  for some multiplicatively closed subset  $S$
of  $R$  such that  $IR_S$ $\ne$ $R_S$.
\end{enumerate}
\noindent
If  $IA$  is projectively full, then  $I$  is projectively full.
\end{prop}

\begin{proof}
If  $IA$  is projectively full, then  $d(IA)$ $=$ $1$, by 
(\ref{rema1}.2), so
$d(I)$ $=$ $1$, by (\ref{corox}), so  $I$  is projectively full, by
(\ref{rema1}.2).
\end{proof}

\begin{rema}
\label{remafull}
{\bf{(\ref{remafull}.1)}}
{\em The converses  of (\ref{fullcontracts})(1) and (\ref{fullcontracts})(2)
are  not true;
in  (\ref{remay}.1) below we give
specific examples  where $A$ is a finite free integral
extension and $I$ is  a projectively full ideal  such that $IA$  is not
projectively full.



\noindent
{\bf{(\ref{remafull}.2)}}
The converse of (\ref{fullcontracts})(3) is not true;
in fact, it may happen that:
\begin{enumerate}
\item  $I$ is projectively full, but $IR_S$ is not projectively full.

\item $\mathbf P(I)$ is projectively full, but $\mathbf P(IR_S)$ is not
projectively full.
\end{enumerate}
}
\end{rema}

\begin{proof} {\bf (of \ref{remafull})(2)}
For (1), let  $(R,M = (x,y,z)R)$  be a regular local domain of altitude
three,
let  $p$ $=$ $zR$, let  $P$ $=$ $(x,y)R$, and let  $I$ $=$ $pP^2$.
Then  $I$  is projectively full, by (\ref{remax}.3)
below, $p$  and  $P$
are the centers
in  $R$  of two of the Rees valuation rings of  $I$,
and  $IR_P$ $=$ $P^2R_P$  is not projectively
full.  (However,  in this example
$\mathbf P (IR_P)$ $=$ $\{P^iR_P \mid i \in \IN_+\}$  is projectively full.)

For (2), let $A$ be a Noetherian ring having a regular proper ideal $J$ such
that
$\mathbf P(J)$ is not projectively full. For example, let $A$ be the normal
local
domain of \cite[Example 4.14]{CHRR}. Let $X$ be an indeterminate over $A$
and let
$R = A[X]$. Then the ideal $I = XJR$ is projectively full by 
(\ref{remax}.4),
so $\mathbf P(I)$ is
projectively full. Let $S$ be the multiplicative system generated by $X$.
Then
$IR_S = JR_S$,  and $\mathbf P(JR_S)$ is not projectively full.
\end{proof}


In the case where $A$ is a faithfully flat Noetherian extension of $R$,
or the case where $A = R_S$ and $IA \ne A$, it would be  interesting
to know if  $\mathbf P (IA)$
is projectively full implies that  $\mathbf P (I)$  is projectively full.



\begin{rema}
\label{localizeexample}
{\em There are often many ideals that localize to
the same ideal in a localization.  With this in mind,
if  $JR_S$ $=$ $IR_S$  and  $IR_S$  is projectively full, then so are
both  $I$  and  $J$, by (\ref{fullcontracts})(3).  So assume, for
example,
that  $(R,M)$  is a regular local
domain, let  $P \ne M$  be a nonzero prime ideal in  $R$, and let  $b$
$\in$ $R-P$.  Then for all
positive integers  $n$  it holds that
{\it{every}}
ideal between  $b^n P$  and  $P$  is projectively full
(since  $b^nPR_P$ $=$ $PR_P$  is projectively full).
}
\end{rema}


In Proposition \ref{fullcors}  we consider extension rings  $A$  of  $R$
for
which the converse of (\ref{fullcontracts}) holds.  This
result is an immediate corollary
of (\ref{comb-prop}) and (\ref{rema1}.2).

\begin{prop}
\label{fullcors}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.

\noindent
{\bf{(\ref{fullcors}.1)}}
Let  $K$  be an ideal in  $R$  such that  $K$ $\subseteq$ $\Rad(R)$.
Then  $I$
(respectively $\mathbf P (I)$)
  is projectively full in  $R$   if and only if
$(I+K)/K$
(respectively $\mathbf P ((I+K)/K)$)
is projectively full in  $R/K$.

\noindent
{\bf{(\ref{fullcors}.2)}}
Let  $P_1,\dots,P_k$
be the centers in  $R$  of the Rees
valuations of  $R$  and let  $S$  be
a multiplicatively closed subset of
$R$  such that  $S \cap (P_1 \cup \cdots \cup P_k)$ $=$ $\emptyset$.
Then  $I$
(respectively $\mathbf P (I)$)
is projectively full in  $R$   if and only if  $IR_S$
(respectively $\mathbf P (IR_S)$)
is projectively full in  $R_S$.

\noindent
{\bf{(\ref{fullcors}.3)}}
Let  $X_1,\dots,X_g$  be
indeterminates and let  $A$ $=$ $R[X_1,\dots,X_g]$.
Then  $I$
(respectively $\mathbf P (I)$)
is projectively full in  $R$
if and only if  $IA$
(respectively $\mathbf P (IA)$)
is projectively full in  $A$.

\noindent
{\bf{(\ref{fullcors}.4)}}
Let  $X_1,\dots,X_g$  be
indeterminates, and let  $A$ $=$ $R(X_1,\dots,X_g)$.
Then  $I$
(respectively $\mathbf P (I)$)
is projectively full in  $R$
if and only if  $IA$
(respectively $\mathbf P (IA)$)
is projectively full in  $A$.
\end{prop}

\begin{proof}
For (\ref{fullcors}.1), $I$
is projectively full if and only if  $d(I)$ $=$ $1$
(by (\ref{rema1}.2)) if and only if  $d((I+K)/K)$ $=$ $1$
(by (\ref{comb-prop}.1)) if and only if
$(I+K)/K$  is projectively full (by (\ref{rema1}.2)).
Also, $\mathbf P (I)$  is projectively full
if and only if there exists  $J$ $\in$ $\mathbf P (I)$  such
that  $d(J)$ $=$ $1$  (by (\ref{rema1}.2))
if and only if  $d((J+K)/K)$ $=$ $1$  (by (\ref{comb-prop}.1))
if and only if  $\mathbf P ((I+K)/K)$  is projectively full
(by (\ref{rema1}.2), since
$(J+K)/K$ $\in$ $\mathbf P ((I+K)/K)$).

The proof of (\ref{fullcors}.2)
(resp., (\ref{fullcors}.3),
(\ref{fullcors}.4))
is similar
using  (\ref{comb-prop}.2)
(resp., (\ref{comb-prop}.3), (\ref{comb-prop}.4))  in place of 
(\ref{comb-prop}.1).
\end{proof}

In Proposition \ref{gen-comp} (and its corollary) we
show  that the integrally closed complete intersection ideals of a
local ring as classified by Goto are   projectively full.

\begin{prop}
\label{gen-comp}
Let $(R,M)$ be a local ring and $I$ a normal ideal in $R$ with $I \nsubseteq
M^2$.
If both the associated graded ring $G(M)=\bigoplus_{n \geq 0} M^n/M^{n+1}$
and
the fiber cone ring $F(I)=\bigoplus_{n \geq 0} I^n/M I^n$ are
reduced, then  $I$ is projectively full.
\end{prop}

\begin{proof} Since
$G(M)$ is reduced, the maximal ideal $M$ is also normal.  For
it is shown in \cite[Theorem 2.1]{Re2} that
$G(M)$ $=$ $\mathbf R (R,M)/u \mathbf R (R,M)$, so  $G(M)$  is reduced if
and
only if  $u \mathbf R (R,M)$  is the intersection of its minimal prime
divisors
if and only if  $p \mathbf R (R,M)_p$ $=$ $u \mathbf R (R,M)_p$  for all
prime divisors  $p$  of  $u \mathbf R(R,M)$; it follows that each such
$\mathbf R (R,M)_p$  is a discrete valuation ring, so
$u \mathbf R(R,M)$ $=$ $(u \mathbf R (R,M))_a$, by \cite[Theorem 2.10]{Ra1},
so
$M^n$ $=$ $u^n \mathbf R (R,M) \cap R$ $=$ $(u^n \mathbf R (R,M))_a \cap R$
$=$
${M^n}_a$  for all  $n$ $\in$ $\IN_+$, hence  $M$  is normal.
Let  $J$ be an integrally closed ideal such that $(J^t)_a=I^s$ for some
positive integers $t,s$.
Then we  must have $s \geq t$. If not, then  $I^s \subseteq (J^{s+1})_a
\subseteq M^{s+1}$.
Choose $a \in I \setminus M^2$. If $a'$ denotes the image of $a$ in $M/M^2
\subseteq G(M)$,
then $(a')^s=0$, hence $a'=0$, contradicting the choice of $a$.

In particular,
since $(J^t)_a \subseteq I^t$, we have $J \subseteq I$.   Let $k$ be the
positive integer such that $J \subseteq I^k$ and $J \nsubseteq I^{k+1}$.
We will prove that $J=I^k$.

We first show that $J \nsubseteq M I^k$. Assume that  $J \subseteq M I^k$.
Then $I^s = (J^t)_a \subseteq (M^t I^{kt})_a$. This implies
that $kt \leq s-t$, otherwise $I^s \subseteq (M^t I^{s-t+1})_a \subseteq
M^{s+1}$,
which, as shown above, is not true. But then $(J^t)_a=I^s \subseteq
I^{(k+1)t}$
and hence  $J \subseteq I^{k+1}$, contradicting the choice of $k$.

We now consider the fiber cone ring $F(I)=\bigoplus_{n \geq 0} I^n/M I^n$.
Let $x \in J \setminus M I^k$. The image of $x$ in
$\big(J+M I^k/M I^k \big) \subseteq \big(I^k/M I^k \big)$ is
nonzero and, since $F(I)$ is reduced, we  have $x^t \in J^t \setminus M
I^{kt}$.
This shows that $J^t \nsubseteq M I^{kt}$, and since $J^t \subseteq I^s$, we
get $s \leq kt$. On the other hand, since $J \subseteq I^k$, we
have $I^s=(J^t)_a \subseteq I^{kt}$, and therefore  $s \geq kt$. In
conclusion, $s=kt$
and from   $(J^t)_a=I^{kt}$ we obtain  $I^k \subseteq J_a=J$.
\end{proof}


In \cite{G}, Goto described  the $M$-primary integrally closed complete
intersection
ideals in a local ring $(R,M)$. He proves  that such ideals $I$ exist only
when the ring  $R$  is regular, in which case there exist regular
parameters $x_1,\ldots,x_d$ and a positive integer $n$ such
that $I=(x_1^n,x_2,\ldots,x_d)R$. Moreover, all the powers of such an ideal
are integrally closed,
i.e., the ideal is normal. An immediate corollary of Proposition
\ref{gen-comp} shows
that the ideals of this type are also projectively full.

\begin{coro}
\label{comp-int}
Let $(R,M)$ be a regular local domain of altitude $d\geq 2$
and let $x_1,\ldots,x_d$ be a regular system of parameters. Then every ideal
of the
form $I=(x_1^n,x_2,\ldots,x_d)R$  is projectively full.
\end{coro}

\begin{proof} Under these assumptions,
both $G(M)$ $=$ $\mathbf R (R,M)/u \mathbf R (R,M)$ and $F(I)$
$=$ $\mathbf R (R,M)/(u,M) \mathbf R (R,M)$ are polynomial rings
in $d$ variables with coefficients in $R/M$, and therefore reduced.
Also, $I$  is normal, so the conclusion follows from (\ref{gen-comp}).
\end{proof}


In Proposition \ref{propfull}, we show that the projective
equivalence class of each proper invertible ideal in an integrally
closed Noetherian ring is projectively full.


\begin{prop}
\label{propfull}
Let $R$ be an integrally closed Noetherian ring and let $I$ be a nonzero
proper ideal of $R$.
\begin{enumerate}
\item
If $I$ is invertible, then $\mathbf P(I)$ is projectively full.
\item
If $R$ is local and $I$ is principal (or equivalently invertible),
say $I = bR$, then there
exists  $x$ $\in$ $R$  such that  $\mathbf P (bR)$ $=$
$\{x^iR \mid i \in \IN_+\}$.
\end{enumerate}
\end{prop}



\begin{proof}
Assume that $I$ is invertible. By Proposition \ref{propinvert} every ideal
projectively equivalent to $I$ is invertible. Let $K$ be the largest element
of $\mathbf P(I)$
and let  $H$ $\in$ $\mathbf P (I)$.
Since $H$ is an arbitrary element of $\mathbf P(I)$, to prove that $\mathbf
P(I)$ is
projectively full it suffices to prove that $H$ is a power of $K$.  By the
linear order of
$\mathbf P(I)$, there exists $n \in \IN_+$  such that $K^{n+1} \subseteq H
\subseteq K^n$.
If $H = K^n$ we are done. If $H \subsetneq K^n$, then $H = JK^n$ for some
invertible  ideal $J$ such that $K \subseteq J \subsetneq R$.
Since $H$ and $K$ are projectively equivalent and since
invertible ideals of an integrally closed ring are integrally closed, there
exist positive integers $h$ and $k$ such that $H^h = K^k$. Therefore
$K^k = H^h = J^hK^{nh}$. If $k \le nh$, then multiplying this equation by
$K^{-k}$ gives a contradiction to the fact that $J$ is a proper ideal.
Therefore $k > nh$. Multiplying by $K^{-nh}$, we obtain $K^{k-nh} = J^h$.
Thus $K$ and $J$ are projectively equivalent. Since $K$ is the largest ideal
projectively equivalent to $H$, we have $K = J$ and $H = K^{n+1}$. This
proves item 1. Since an invertible ideal of a local domain is principal,
item 2 is an immediate consequence of item 1.
\end{proof}


In connection with Proposition \ref{propfull}, a question we have
considered, but not resolved,  is whether  ${\bf P}(I)$ is always
projectively full if $I$ is a divisorial ideal of an integrally
closed Noetherian domain. A complicating factor here is that the
integral closure of a power of the divisorial ideal $I$ may fail to
be divisorial.  We discuss this in Remark \ref{divisorial}.


\begin{rema} \label{divisorial}
{\em Let $R$ be an integrally closed Noetherian domain. Invertible
ideals of $R$ are divisorial, and a divisorial ideal $I$ of $R$ is
uniquely representable as the intersection of symbolic powers of the
height-one prime ideals that contain it, say
$$I = P_1^{(e_1)} \cap
P_2^{(e_2)} \cap \cdots \cap P_g^{(e_g)}.
$$
Let $v_i$ denote the normalized valuation associated to the
valuation domain $R_{P_i}$. Then $v_i(I) = e_i$. The valuations
$v_i$ and positive integers $e_i$ are a subset of the valuations and
associated positive integers mentioned in Definition \ref{defiRees}.
If $I$ is invertible, then  this subset is all the valuations of
Definition \ref{defiRees}, and $\Rees I = \{R_{P_1}, \ldots, R_{P_g}
\}$ since $I$ is invertible if and only if $IR_P$ has analytic
spread one for every  prime ideal $P$ of $R$ that contains $I$.
Notice that  $(I^n)_a$ is divisorial for all $n \in \mathbb N_+$ if
and only if $\Rees I = \{R_{P_1}, \ldots, R_{P_g} \}$. A prime ideal
$P$ is the center of a Rees valuation ring of $I$ if the analytic
spread of $IR_P$ is equal to the height of $P$, and the converse
holds if $R_P$ is quasi-unmixed \cite[Prop. 4.1]{McA}. If $I \subset
P$ and $P$ is of height 2, then $R_P$ is Cohen-Macaulay and thus
unmixed, so $P$ is the center of a Rees valuation ring of $I$ if and
only if $IR_P$ has analytic spread 2. Thus if $R$ has altitude two,
then the Rees valuation rings of a divisorial ideal $I$ of $R$ are
all centered on height-one primes of $R$ if and only if $I$ is
invertible. However, there exists a normal local domain $(R,M)$  of
altitude 3 that has a divisorial height-one prime ideal $P$ such
that $P$ has analytic spread 2 (so  $P$  is not invertible) and yet
$R_P$ is the unique Rees valuation ring of $P$  (so
$(P^n)_a$ $=$ $P^{(n)}$  is divisorial for all  $n$ $\in$ $\IN_+$).  For a specific
example, let $k$ be a field and let $R = k[[x, y, z, w]]$, where $xy
= zw$. Then $P = (x, z)R$ is a height-one prime of $R$ and $P$ has
analytic spread 2. Since the localization of $R$ at any nonmaximal
prime ideal is a regular local domain, if $P \subset Q$ with $Q$ a
prime of $R$ of height 2, then $PR_Q$ is principal. Therefore $R_P$
is the unique Rees valuation ring of $P$. }
\end{rema}


In connection with Remark \ref{divisorial}, it seems natural to ask:

\begin{ques} {\em If $I$ is a divisorial ideal of an integrally closed
Noetherian domain and if the integral closure of $I^n$ is divisorial
for every $n \in \mathbb N_+$, does it follow that ${\bf P}(I)$ is
projectively full? }
\end{ques}

\begin{rema} \label{nonintclosed2} {\em Without the assumption that
$R$ is integrally closed, easy examples show that  Proposition
\ref{propfull} fails. Indeed, if $(R,M)$ is  a local domain of
altitude one such that the integral closure of $R$ is a valuation
domain, then all the $M$-primary ideals of $R$ are projectively
equivalent,  and given an arbitrary  numerical semigroup $S$,  it is
possible to construct a local domain $(R,M)$ of altitude one such
that the integral closure of $R$ is a valuation domain and such that
$S(M) = S$. Let $b_1 < b_2 < \cdots < b_r$ be positive integers
having greatest common divisor 1, and let $S = \langle b_1, b_2,
\ldots, b_r \rangle$ denote the numerical semigroup determined by
$b_1, \ldots, b_r$. Let $t$ be an indeterminate over the field $k$
and let $R$ be the subring $k[[t^{b_1}, t^{b_2}, \ldots, t^{b_r}]]$
of the formal power series ring $k[[t]]$. Then $R$ is local with
maximal ideal $M = (t^{b_1}, \ldots, t^{b_r})R$ and the integral
closure of $R$ is the valuation domain $k[[t]]$. The integrally
closed $M$-primary ideals of $R$ are precisely the $M$-primary
ideals that are contracted from $k[[t]]$.  If $I$ is $M$-primary,
then $I_a = Ik[[t]] \cap R$ and $Ik[[t]] = t^bk[[t]]$, where $b \in
S$. Thus  the integrally closed $M$-primary ideals are in one-to-one
correspondence with the elements of $S = \langle b_1, \ldots b_r
\rangle$. Therefore $S(M) = S$. We conclude that every numerical
semigroup $S$ is realizable as $S(M)$ for a local domain $(R,M)$ of
altitude one. In particular, $M$ is projectively full if and only if
$S = \IN$, or, equivalently, if and only if $R = k[[t]]$. }
\end{rema}



In Remark \ref{remaz}  we obtain a partial  extension of
Proposition \ref{propfull} to integrally closed regular principal
ideals in Noetherian rings that are not integrally closed.


\begin{rema}
\label{remaz}
{\em Let
$R$  be a Noetherian ring and let  $I$ $=$ $bR$
be an integrally closed regular proper principal ideal.
Then the following hold:

\noindent
{\bf{(\ref{remaz}.1)}}
For each ideal  $H$  in  $R$  that is projectively equivalent to  $I$,
$H^n$  is principal for infinitely many  $n$ $\in$ $\IN_+$,
and  $H^n$ $=$ $(H^n)_a$  for all large  $n$ $\in$ $\IN_+$.

\noindent
{\bf{(\ref{remaz}.2)}}
If  the largest ideal  in  $\mathbf P (I)$  is
principal, say  $xR$, then  $\mathbf P (I)$  is projectively full and each
ideal
projectively equivalent to  $I$  is a power of  $xR$.

\noindent
{\bf{(\ref{remaz}.3)}}
If  $c$  is a regular element in  $R$
such that  $\Rad(cR)$ $=$ $\Rad(bR)$, then  $bR$ $=$
$(bR)_a$
if and only if  $cR$ $=$ $(cR)_a$.  Therefore if  $b$  is a regular element
in $R$  such that  $bR$ $=$ $(bR)_a$, then the conclusion of
(\ref{remaz}.1) holds for all principal ideals  $cR$  such that  $\Rad(cR)$
$=$ $\Rad(bR)$.}
\end{rema}


\begin{proof}
For (\ref{remaz}.1),
let  $H$  be an ideal in  $R$  that is projectively
equivalent to  $I$.  Then  $(H^h)_a$ $=$ $b^dR$  for
some  $d,h$ $\in$ $\IN_+$.  (It follows from \cite[Theorem 2.10]{Ra1}
that  $bR$ $=$ $(bR)_a$  if and
only if  $R_p$  is integrally closed for all prime divisors  $p$  of  $bR$;
it follows from
this (and the fact that  $bR$  and  $b^iR$  have the same prime
divisors for all  $i$ $\in$ $\IN_+$)  that if  $bR$ $=$ $(bR)_a$, then
$b^iR$ $=$ $(b^iR)_a$  for
all  $i$ $\in$ $\IN_+$.)
Therefore  $H^h$ $\subseteq$ $(H^h)_a$ $=$ $b^dR$, so there exists an ideal
$J$
in  $R$  such that  $H^h$ $=$ $b^dJ$.  Therefore
$(b^dR)_a$ $=$ $b^dR$ $=$ $(H^h)_a$ $=$ $(b^dJ)_a$,
hence  $R$ $=$ $b^dR:b^dR$ $=$ $(b^dJ)_a : b^dR$ $=$ $J_a$, so  $J$ $=$ $R$
and  $H^h$ $=$ $b^dR$.
Therefore  $H^{hn}$ $=$ $b^{dn}R$  is integrally closed and principal for
all  $n$ $\in$ $\IN_+$.
It therefore follows from \cite[(11.15)]{McA} that  $H^i$ $=$
$(H^i)_a$  for all large  $i$ $\in$ $\IN_+$.

For (\ref{remaz}.2), assume that  $xR$  is the largest ideal in 
$\mathbf P (I)$
and let  $H$  be an ideal in  $R$  that is projectively equivalent to  $I$.
To complete the proof it must
be shown that  $H$ $=$ $x^iR$  for some  $i$ $\in$ $\IN_+$.

For this, by the linear order of  $\mathbf P (I)$  there exists  $n$ $\in$
$\IN_+$
such that  $x^{n+1}R$
$\subseteq$ $H$ $\subseteq$ $x^nR$.  If  $H$ $=$ $x^nR$, we are done.  If
$H$
$\subsetneq$ $x^nR$, then  $H$ $=$ $x^nJ$  for some ideal  $J$  in  $R$
such
that  $xR$ $\subseteq$ $J$ $\subsetneq$ $R$.  Since  $H$  are  $xR$  are
projectively
equivalent, $(H^h)_a$ $=$ $x^iR$  for some  $h,i$ $\in$ $\IN_+$.  Therefore
$(x^{hn}J^h)_a$ $=$ $x^iR$.  If  $i$ $\le$ $hn$, then
$(x^{hn-i}J^h)_a$ $=$ $(x^{hn}J^h)_a : x^iR$ $=$ $x^iR:x^iR$ $=$ $R$, and
this contradicts
$J$ $\ne$ $R$.  Therefore  $i$ $>$ $hn$, so  $(J^n)_a$ $=$ $(x^{hn}J^n)_a$
$:x^{hn}R$ $=$
$x^iR:x^{hn}R$ $=$ $x^{i-hn}R$, so  $J$  is projectively equivalent to
$xR$.
Since  $J$ $\supseteq$ $xR$  and  $xR$  is the largest ideal in  $\mathbf
P(I)$, it
follows that  $J$ $=$ $xR$, hence  $H$ $=$ $x^nJ$ $=$ $x^{n+1}R$.

(\ref{remaz}.3) follows as in the parenthetical part of the proof
of (\ref{remaz}.1).
\end{proof}

\section{EXAMPLES OF PROJECTIVELY FULL IDEALS.}

We use Remark \ref{rema1}.2 to obtain several classes of examples of
projectively full ideals.

\begin{exam}
\label{remax}
{\bf{(\ref{remax}.1)}}
{\em If the integer  $e_i$ $=$ $1$  for some Rees valuation ring  $V_i$ 
  of  $I$
(see (\ref{defiRees})),
then  $I$  is projectively full. Moreover, if  $J$  is a
regular proper ideal in  $R$  that is not contained in
the center  $P_i$  in  $R$  of  $V_i$,
then  $IJ$  is projectively full.

\noindent
{\bf{(\ref{remax}.2)}}
Let  $P$  be a prime ideal in a Noetherian ring  $R$  such
that  $R_P$  is
a regular local domain.  Then $P$ is projectively full. Moreover,
$PI$  is projectively full for all regular
ideals  $I$  in  $R$  such that  $I$ $\nsubseteq$ $P$.  In particular,
if  $(R,M)$  is a regular local domain, then  $M^nP$  is projectively
full for all  $n$ $\in$ $\IN_+$  and for all nonzero
prime ideals  $P$ $\ne$ $M$.

\noindent
{\bf{(\ref{remax}.3)}}
Let  $x$  be a regular parameter in a regular local domain  $R$  and let
$I$
be an ideal in  $R$  such that  $\height(I)$ $\ge$ $2$.  Then  $xI$  is
projectively full.

\noindent
{\bf{(\ref{remax}.4)}}
Let  $R$  be a Noetherian domain   and let  $X$  be
an indeterminate.  Then  $XIR[X]$  is projectively full in  $R[X]$  for
every  nonzero ideal   $I$  in  $R$.

\noindent
{\bf{(\ref{remax}.5)}}
Let  $R$  be a Noetherian domain, let  $X$  be
an indeterminate, and let  $I$  be a regular proper ideal in  $R$  that is
not
projectively full.  Therefore $d := d(I) > 1$.
For each  $n$ $>$ $1$  in  $\IN_+$  it holds that
$X^nIR[X]$  is projectively full if and only if  $n$ and $d$ are relatively
prime.
In particular, $X^nIR[X]$  is projectively full  for all  $n$ $\in$ $\IN_+$
if
and only if  $I$  is projectively full in  $R$.}
\end{exam}

\begin{proof}
It is noted in (\ref{defid(I)}.4) that the integer  $d$ $=$ $d(I)$  is
a common divisor of the integers  $e_1,\dots,e_g$  of (\ref{defiRees}).
Therefore, since  $I$  is projectively full if and only if  $d(I)$ $=$ $1$
(by (\ref{rema1}.2)), the
first statement in (\ref{remax}.1) is clear.

For the second statement in (\ref{remax}.1), it is shown in \cite
[(3.6)]{CHRR}
that for all regular proper ideals $I$ and $J$ in $R$ one has
$\Rees I \cup \Rees J \subseteq \Rees IJ$. Therefore
$V_i$  is a Rees valuation ring of  $IJ$, and if  $J$
$\nsubseteq$ $P_i$,  then the integer  $e_i$ of  $V_i$  for  $IJ$  is also
one (since
$(IJ)V_i$ $=$ $IV_i$ $=$ $N^{e_i}$ $=$ $N$, where  $N$  is the maximal ideal
of  $V_i$),
so  $IJ$  is projectively full by the first statement of this remark.

For (\ref{remax}.2), the integer  $e$  of the
order valuation ring $V$ of  $P$  is one, so the first two statements follow
immediately from the second statement
in (\ref{remax}.1), and the third statement is a special case of the second
statement.

For (\ref{remax}.3), if  $x$  is a regular parameter in a regular local
domain  $R$, then
$R_{xR}$  is a regular local domain, so it follows from
(\ref{remax}.2) that if  $\height(I)$ $>$ $1$,
then  $xI$  is projectively full.

For (\ref{remax}.4), $P$ $=$ $XR[X]$  is a prime ideal such that  $R[X]_P$
is a
regular local domain.  Therefore, since  $IR[X]$
$\nsubseteq$ $XR[X]$  for all nonzero ideals  $I$  of  $R$, it follows from
(\ref{remax}.2) that  $XIR[X]$  is projectively full.

For (\ref{remax}.5), assume first that  $e$ $>$ $1$
is a divisor of  $d$ $=$ $d(I)$, say  $d$ $=$ $eq$.
Let  $n^*(I)$ $\in$ $\IN_+$  such that  $\{n^*(I)+ (i/d) \mid i \in \IN\}$
$\subseteq$
$\mathbf U (I)$  (see (\ref{rema2}.4)).
Then  $n^*(I) + (q/d))$ $\in$ $\mathbf U (I)$,
so  the ideal $H$ $:=$ $I_{n^*(I)+(q/d)}$ $\in$ $\mathbf P (I)$,
by (\ref{defid(I)}.3),
hence  $(H^{d})_a$ $=$ $(I^{(n^*(I))d+q})_a$, by \cite[(2.3)(b)]{MRS}.
However, $((n^*(I))d + q)/d)$ $=$ $(n^*(I)e+1)/e$,
so  $(H^e)_a$ $=$ $(I^{n^*(I)e+1})_a$, by (\ref{rema2}.1).
Therefore, for all  $h$ $\in$ $\IN_+$  it holds that

$((X^{(n^*(I)e+1)h}H)^e)_a$ $=$
$((X^{eh(n^*(I)e+1)}H^e))_a$ $=$
$((X^{eh})^{(n^*(I)e+1)}H^e))_a$ $=$

  $((X^{eh}I)^{n^*(I)e+1})_a$,

\noindent
so
$X^{(n^*(I)e+1)h}H$ is projectively equivalent to   $X^{eh}I$,
and it is clear that $(X^{(n^*(I)e+1)h}H)_a$ is not the integral closure of
any power of  $X^{eh}I$, so  $X^{eh} I$  is not projectively full.

Conversely, assume that  $d$ $\in$ $\IN_+$  is such that  $X^dI$  is not
projectively full,
so it must be shown that  $d$ $=$ $d_ie$  for some divisor  $d_i$ $>$ $ 1$
of  $d$ $=$ $d(I)$
and for some  $e$ $\in$ $\IN_+$.
By hypothesis there exists  an ideal  $J$  in  $R[X]$
that is projectively equivalent to  $X^dI$  such that
$J_a$ $\ne$ $((X^dI)^n)_a$  for all  $n$ $\in$ $\IN_+$.  Now  $(J^j)_a$ $=$
$((X^dI)^n)_a$  $=$ $X^{nd}(I^n)_a$, for some  $j,n$ $\in$ $\IN_+$,
and by \cite[(2.1)(b)]{MRS} it may
be assumed that  $j,n$  are relatively prime (and  $j$ $>$ $1$, by
the preceding sentence).  Now  $J^j$ $\subseteq$ $(J^j)_a$ $=$
$X^{nd}(I^n)_a$ $\subseteq$
$X^{nd}R[X]$, hence  $J^j$ $=$ $X^{nd}K$  for some ideal  $K$  in  $R[X]$
such
that  $K$ $\nsubseteq$ $XR[X]$  (since  $IR[X]$ $\nsubseteq$ $XR[X]$).
Let  $m$ $\in$ $\IN$  such that  $J$ $\subseteq$ $X^mR[X]$  and
$J$ $\nsubseteq$ $X^{m+1}R[X]$, so  $J$ $=$ $X^mH$  for some ideal  $H$  in
$R[X]$.
Then  $X^{mj}H^j$ $=$ $J^j$ $=$ $X^{nd}K$.  Therefore  $mj$ $=$ $nd$
($=$ $nd(I)$),
since  $K$ $\nsubseteq$ $XR[X]$
and  $H^j$ $\nsubseteq$ $XR[X]$  (since  $H$ $\nsubseteq$ $XR[X]$  and
$XR[X]$  is a prime ideal).
Therefore  $K$ $=$ $H^j$, so  $J^j$ $=$ $X^{mj}H^j$,
so  $X^{mj}(H^j)_a$ $=$ $(J^j)_a$ $=$ $X^{nd}(I^n)_a$,
so  $(H^j)_a$ $=$ $(I^n)_a$.  Therefore by (\ref{extendedidl}), if we let
$G$ $=$ $H_a \cap R$, then  $H_a$ $=$ $GR[X]$,
$(G^j)_aR[X]$ $=$  $(H^j)_a$,
and  $(G^j)_a$ $=$ $(I^n)_a$  (so  $G$  is
projectively equivalent to  $I$  (in  $R$)).  Therefore  $(G^{d(I)})_a$ $=$
$(I^i)_a$  for some
$i$ $\in$ $\IN_+$, by (\ref{rema2}.4).

Since  $(G^j)_a$ $=$ $(I^n)_a$  and  $(G^{d(I)})_a$ $=$ $(I^i)_a$, it
follows
from \cite[(2.1)(b)]{MRS} that  $ji$ $=$ $nd(I)$.  Therefore
$nd(I)$ $=$ $mj$  (resp.,  $ji$ $=$ $nd(I)$) and  $n,j$
are relatively prime, so it follows that  $m$ $=$ $en$
(resp., $d(I)$ $=$ $jf$)  for some $e$ (resp., $f$) $\in$ $\IN_+$.
Therefore  $j$  is a divisor of  $d(I)$,
and  since  $nd$ $=$ $mj$ $=$ $enj$  it follows
that  $d$ $=$ $je$  for some divisor  $j$  of  $d(I)$  and
for some  $e$ $\in$ $\IN_+$, and it was noted in the preceding paragraph
that  $j$ $>$ $1$.

The final statement is clear from what has already been shown.
\end{proof}

It follows from  (\ref{remax}.2) that every nonzero prime ideal of a
regular local domain is projectively full. If $(R,M)$ is a
regular local domain and $\dim R = 2$,  then
$\mathbf P(I)$ is projectively full for every nonzero proper ideal
$I$ of $R$  \cite[(4.12)]{CHRR}. It would be interesting to know
whether this is also true when $\dim R \ge 3$.


\begin{rema}
\label{remay} {\bf{(\ref{remay}.1)}} {\em A projectively full ideal
may fail to  extend to a projectively full ideal in a finite free
integral extension domain.

\noindent {\bf{(\ref{remay}.2)}} Concerning (\ref{remax}.1), there
exist regular ideals  $bA$  and  $J$  in a Noetherian domain  $A$
and positive integers  $d$ $>$ $1$  such that: $bA$  has a Rees
valuation ring with integer  $e$ $=$ $1$; $b^{d-1}J$, and $b^{d+1}J$
are projectively full; and,  $b^dJ$  is not projectively full.
Therefore:  (a)  the product of an ideal  $H$  that has a Rees
valuation with integer $e$ $=$ $1$  and an ideal which is
projectively full and is contained in  $H$ need not be projectively
full; and, (b)  the product of an ideal  $H$  that has a Rees
valuation with integer $e$ $=$ $1$  and an ideal which is not
projectively full and is contained in $H$ may be projectively full.


\noindent {\bf{(\ref{remay}.3)}} Concerning (\ref{remax}.2), it is
often the case for a height-one prime ideal $P$ of an integrally
closed Noetherian domain $R$ that the rank-one discrete valuation
domain  $V$ $=$ $R_P$ is  not the only Rees valuation ring of $P$.}
\end{rema}

\begin{proof}
For (\ref{remay}.1),
let  $I$  be a nonzero ideal in a
Noetherian domain  $R$  that
is not projectively full (so  $d$ $=$ $d(I)$ $>$ $1$), so  $(X^dI)R[X^d]$
is projectively
full (by (\ref{remax}.4)).  However, (\ref{remax}.5) shows that 
$(X^dI)R[X]$
is not projectively full in the free
(of degree  $d$)   integral extension domain $R[X]$  of  $R[X^d]$.

For (\ref{remay}.2), let  $R$  and  $I$  be as in (\ref{remax}.5),
let  $d$ $>$ $1$ be a divisor of  $d(I)$  such that $d-1$  and
$d+1$  are  relatively prime to   $d(I)$, let $b$ $=$ $X$, let  $A$
$=$ $R[X]$, and let  $J$ $=$ $IA$.  Then $V$ $=$ $A_{bA}$  is a
regular local domain, so  $bA$  has $V$  as a Rees valuation ring
with  $e$ $=$ $1$.  Also, it follows from (\ref{remax}.5) that
$b^{d-1}J$  is projectively full in  $A$, $b^dJ$  is not
projectively full in  $A$, and $b^{d+1}J$  is projectively full in
$A$.  The last statement (concerning (a) and (b)) clearly follows
from this.

For (\ref{remay}.3), let  $P$ be a height-one prime ideal of an
integrally closed local domain $(R, M)$ of altitude 2. Then $V$ $=$
$R_P$ is the unique Rees valuation ring of $P$ if and only if $P$ is
invertible if and only if the analytic spread $a(P)$ of $P$ is one.
With this in mind, let  $x,y$  be independent indeterminates over a
field $k$ and let  $R$ be the subring $k[[x^2, xy, y^2]]$ of the
formal power series ring $k[[x, y]]$.  Then $R$ is a normal local
domain and $P = (x^2, xy)R$ is a height-one prime ideal of $R$ such
that $a(P) = 2$. Therefore $P$ is projectively full and has more
than one Rees valuation ring.
\end{proof}

One problem we have not been able to solve is:  given a nonzero
ideal in a Noetherian domain  $R$, does there always exist a finite
integral extension domain  $A$  of  $R$  such that  $\mathbf P (IA)$
is projectively full?  In Proposition \ref{goodextension}  we give a
``logical'' candidate for $A$ and show that, at least, for each  $J$
$\in$ $\mathbf P (I)$ it holds that  $(JA)_a$  is the power of some
fixed ideal in $\mathbf P (IA)$.



In Proposition \ref{goodextension}, for elements $b_1, \ldots, b_g$
of a Noetherian ring $R$, we let $R[{b_1}^{1/k},\dots,{b_g}^{1/k}]$
denote an integral extension ring of $R$ generated by elements
$b_1^{1/k}, \ldots, b_g^{1/k}$ that are $k$-th roots of $b_1,
\ldots, b_g$, respectively. This integral extension ring of $R$ can
be obtained in several ways. If $R$ is an integral domain, the
extension $A = R[{b_1}^{1/k},\dots,{b_g}^{1/k}]$ can be constructed
to also be an integral domain. On the other hand, one can also
construct $R[{b_1}^{1/k},\dots,{b_g}^{1/k}]$ so that it is a finite
free $R$-module of rank $gk$.  In any case, we note  that if $R$ is
local with maximal ideal $M$, then $A$  is local with maximal ideal
$(M,{b_1}^{1/k},\dots,{b_g}^{1/k})A$.

\begin{prop}
\label{goodextension} Let  $I$  be a regular proper ideal in a
Noetherian ring  $R$ and assume that  $\mathbf P (I)$  is not
projectively full. Let  $K$  be the largest ideal in  $\mathbf P
(I)$, let $b_1,\dots,b_g$  be regular elements in  $A$ that generate
$K$, and let  $d(K)$ $=$ $k$  (so  $k$ $>$ $1$). Let  $A$ $=$
$R[{b_1}^{1/k},\dots,{b_g}^{1/k}]$  and let  $H$ $=$
$({b_1}^{1/k},\dots,{b_g}^{1/k})A$.  Then for each ideal  $J$ $\in$
$\mathbf P (K)$ $=$ $\mathbf P (I)$  it holds that  $(JA)_a$ $=$
$(H^n)_a$  for some  $n$ $\in$ $\IN_+$.
\end{prop}

\begin{proof}
Note first that
$H^{[k]}$ $=$ $(({b_1}^{1/k})^k,({b_2}^{1/k})^k,\dots,({b_g}^{1/k})^k)A$ $=$
$KA$,
and it is clear that  $H^{[k]}$  is a reduction of  $H^k$, so  $(KA)_a$ $=$
$(H^k)_a$.
Now let  $J$ $\in$ $\mathbf P (K)$,
so  $(J^k)_a$ $=$ $(K^n)_a$  for some  $n$ $\in$ $\IN_+$, by
(\ref{rema2}.4).  Therefore
$((JA)^k)_a$ $=$ $((J^k)_aA)_a$ $=$ $((K^n)_aA)_a$ $=$ $(((KA)_a)^n)_a$ $=$
$(((H^k)_a)^n)_a$ $=$ $(H^{kn})_a$.  Thus $((JA)^k)_a$ $=$ $(H^{kn})_a$,
so  $(JA)_a$ $=$ $(H^n)_a$.
\end{proof}

Concerning (\ref{goodextension}), the only relation between
$d(K)$  and  $d(H)$  we have been able to determine is  $d(K)$ $\le$
$kd(H)$.  This
follows from:  $d(K)$ $\le$ $d(KA)$ (by
(\ref{corox})(1)) $=$ $d(H^k)$
(since  $KA$ $=$ $H^h$) $=$ $kd(H)$  (by \cite[(4.8.3)]{CHRR}).


For a regular proper ideal  $J$ $=$ $(b_1,\dots,b_g)R$  in a
Noetherian ring   $R$, we present in Example \ref{goodextenexample}
a construction for obtaining a finite integral extension ring  $A$
of $R[X]$ such that $\mathbf P (KA)$ is projectively full, where $K$
$=$ $(b_1X^k,\dots,b_gX^k)R[X]$ and  $k$ $=$ $d(J)$.

\begin{exam}
\label{goodextenexample}  {\em Let $J$ be a regular proper ideal in
a Noetherian ring $R$, assume that $J$ is not projectively full and
is the largest ideal in  $\mathbf P (J)$, let $d(J)$ $=$ $k$ (so $k$
$>$ $1$, by (\ref{rema1}.2)), let $b_1,\dots,b_g$  be regular elements
in $R$ that generate  $J$, and let  $B$ $=$
$R[{b_1}^{1/k},\dots,{b_g}^{1/k}]$. Let  $X$  be an indeterminate,
let  $K$ $=$ $(b_1X^k, b_2X^k,\dots,b_gX^k)R[X]$, let $A$ $=$
$R[X,{b_1}^{1/k}X,\dots,{b_g}^{1/k}X]$, and let  $H$ $=$
$({b_1}^{1/k}X,\dots,{b_g}^{1/k}X)A$.  Then  $K$  is not
projectively full in  $R[X]$, $d(K)$ $=$ $k$, $A$  is obtained from
$R[X]$ by adjoining the  $k$-th root  ${b_i}^{1/k}X$ $=$
$(b_iX^k)^{1/k}$  of each generator  ${b_i}X^k$  of  $K$  to $R[X]$,
and  $H$  is projectively full in  $A$  and is projectively
equivalent to  $KA$.}
\end{exam}

\begin{proof}
By (\ref{remax}.5),  $K$  is not projectively full and $d(K)$ $=$
$k$. Let  $C$ $=$ $B[X]$, so $C$ is a finite integral extension ring
of $A$  and $HC$ $=$ $({b_1}^{1/k}X,\dots,{b_g}^{1/k}X)C$ is
projectively full (by (\ref{remax}.4), since
$({b_1}^{1/k},\dots,{b_g}^{1/k})B$  is an ideal in  $B$). Therefore
$H$  is projectively full in  $A$, by (\ref{fullcontracts})(1), and
$H$  is projectively equivalent to $KA$, by (\ref{goodextension}).
\end{proof}


\section{NON-PROJECTIVELY FULL MAXIMAL IDEALS.}
In Sections  3 and 4 a number of examples of regular ideals $I$ of a
Noetherian ring are constructed  for which $I$ or ${\bf P}(I)$ is
projectively full. The main result in this section gives a family of
integrally closed local domains $(R,M)$ of altitude two for which
the maximal ideal $M$ is not projectively full.


\begin{exam}
\label{expx}
{\em
Let $x, y, Z, W$ be independent indeterminates, let
$F$ be a field whose characteristic
is not $2$, and let
$(R_0,M_0)$ denote
the regular local domain
$F[x, y]_{(x, y)}$
  of altitude two.
Let  $k < i \le j$  be positive integers
that are units in  $F$
and set  $R$ $=$
$R_0[z,w]$ =
$ R_0[Z,W]/(Z^k - x^i - y^j,W^k  - x^i +y^j)R_0[Z,W]$.
Then the following hold:

\item
(i)  $R$  is an integrally closed Cohen-Macaulay
local domain (with maximal ideal  $M$ $=$ $(x,y,z,w)R$)  of altitude two.

\item
(ii)  If  $j$ $=$ $i$  and if  $i,k$  are relatively prime,
then $M$  has a unique Rees valuation ring  $V$.

\item
(iii) If  $j$ $=$ $i$ and if  $i,k$  are relatively prime,
then $M$ $=$ $(x,y,z,w)R$
is not projectively full, $d(M)$ $=$ $k$, and  $((z,w)R)_a$ $=$ $M_{i/k}$.}
\end{exam}

\begin{proof}
We first show  that  $R$  is a Cohen-Macaulay
local ring of altitude two. For this,
let  $K$ $=$ $(Z^k - x^i - y^j, W^k - x^i +y^j)R_0[Z,W]$,
let  $z$ $=$ $Z+K$  and  $w$ $=$ $W+K$  in  $R_0[Z,W]/K$,
and let  $R$ $=$ $R_0[z,w]$,
so  $z^k$ $=$ $x^i + y^j$  and  $w^k$ $=$
$x^i - y^j$  are in $R_0$, but possibly
$R$  is not an integral domain
(since we do not yet know that  $K$  is a prime ideal).
Also, it is clear that $R$  is an integral extension ring
of  $R_0$  and that   $M = (x, y, z,w)R$ is a maximal
ideal in  $R$, and since   $z^k$ $=$ $x^i + y^j$ $\in$ $(x,y)R_0$ $=$ $M_0$
and
$w^k$ $=$ $x^i - y^j$ $\in$ $(x,y)R_0$ $=$ $M_0$,  by integral dependence it
follows that every maximal ideal
in  $R$  contains  $(x,y,z,w)R$, hence  $R$  is local.
Further,  $K$  is a height two
(not necessarily prime) ideal in the locally regular UFD $R_0[Z,W]$,
so $R$ is a free (of
degree  $k^2$) integral extension ring  of the altitude two
regular local domain  $R_0$  (with $\{z^m w^n \mid m = 0,\dots,k-1$
and $n = 0,\dots,k-1\}$  as a
free basis), so $\altitude (R) = 2$ and  $R$  is Cohen-Macaulay, by
\cite[(25.16)]{N2}, so  $R$  is a Cohen-Macaulay local ring of altitude two.
In particular, $R$ satisfies  $(S_i)$  for all  $i$ $\in$ $\IN$  (that is,
the maximum length of a prime sequence in  $R_p$  is  $\height(p)$  for
all  $p$ $\in$ $\Spec(R)$.

Since $R$ is $(S_2)$, to show that  $R$  is integrally closed it suffices
(by \cite[(23.8)]{M})
to show that  $R$  satisfies  $(R_1)$  (that is, $R_p$  is a regular local
domain for all
height one prime ideals  $p$  in  $R$).  We do this in the next four
paragraphs.

Let $p$  be a height one prime ideal in  $R$, let  $P$ be the preimage of
$p$ in $R_0[Z,W]$, and let $T = R_0[Z,W]_{P}$.
Then $T$ is a regular local domain of altitude three and $f = Z^k - x^i -
y^j$,
$g$ $=$ $W^k - x^i + y^j$
are in  $Q$ $=$ $P T$.  Also, $f_Z$ $=$ $kZ^{k-1}$, $f_W$ $=$ $0$, $g_Z$ $=$
$0$,
and  $g_W$ $=$ $kW^{k-1}$,
and the determinant ($=$ $k^2Z^{k-1} W^{k-1}$)  of the two by two matrix
consisting
of these four polynomials is
in  $Q$  if and only if either:  (a)  $Z$ $\in$ $Q$;  or, (b)  $W$ $\in$ $Q$
(since  $k$  is a unit in  $R_0$).
That is, if and only if either:  (a$'$)  $z$ $\in$ $p$; or, (b$'$)  $w$ 
$\in$
$p$.
Therefore by \cite[(30.4)]{M},
$R_p$  is a regular local domain except, possibly, in cases (a$'$) or 
(b$'$).

To handle cases (a$'$) and (b$'$), note first that if $z$ $\in$ $p$,
then $z^k$ $=$ $x^i + y^j$ $\in$ $p$, so  $xy$ $\notin$ $p$  (since,
otherwise, the height two  $M$-primary ideal  $(x,y)R$ is contained
in the height one prime ideal  $p$). Similarly, if  $w$ $\in$ $p$,
then  $w^k$ $=$ $x^i - y^j$ $\in$ $p$, so $xy$ $\notin$ $p$.
Therefore if  $q$  is a height one prime ideal in  $R$  that
contains either  $x$  or  $y$, then  $R_q$  is a regular local
domain (by the preceding paragraph).

Next (as just above) let  $T$  be the altitude three regular local
domain  $R_0[Z,W]_{P}$  and let  $Q$ $=$ $P T$ be the maximal ideal
of  $T$  (so  $(f = Z^k - x^i - y^j, ~g = W^k - x^i + y^j)T$
$\subseteq$  $Q$).  Also, $f_x$ $=$ $-ix^{i-1}$, $f_y$ $=$
$-jy^{j-1}$, $g_x$ $=$ $-ix^{i-1}$, $g_y$ $=$ $jy^{j-1}$, and the
two by two determinant ($=$ $-2ijx^{i-1}y^{j-1}$)  consisting of
these four elements is in  $Q$ if and only if either:  (c)  $x$
$\in$ $Q$; or, (d)  $y$ $\in$ $Q$ (since  $2ij$  is a unit in
$R_0$). Modulo  $K$, it follows that either:  (c$'$) $x$ $\in$ $p$;
or, (d$'$)  $y$ $\in$ $p$. Therefore by \cite[(30.4)]{M}, $R_p$  is
a regular local domain except, possibly, in cases (c$'$) or (d$'$).
However, the preceding paragraph shows that  $R_p$  is a regular
local domain in both cases (c$'$) and (d$'$).

It follows that  $R_p$  is a regular local domain for all height one
prime ideals  $p$  in  $R$, so  $R$  is normal (by  $(R_1)$,
$(S_2)$  (see \cite[Theorem 23.8]{M})). Therefore  $R$  is
an integrally closed reduced local
ring, so $R$  is a local domain (by \cite[(9.11)]{M}), hence  $K$  is a 
prime ideal and
$R$ is an integrally closed Cohen-Macaulay local domain. This
completes the proof of (i).

For (ii), assume that  $j$ $=$ $i$   and that  $i,k$  are relatively prime,
and notice  that
$z^k$ $\in$ $(x^i,y^i)R$ $\subseteq$ $(x,y)^kR$
and  $w^k$ $\in$ $(x^i,y^i)R$ $\subseteq$ $(x,y)^kR$, so
$(x,y)R$  is a reduction of  $M$ $=$ $(x,y,z,w)R$.
It follows that each Rees valuation ring of $M$ is an
extension of the order valuation ring
$V_0$ $=$ $R_0[y/x]_{xR_0[y/x]} = R_0[x/y]_{yR_0[x/y]}$  of  $R_0$.
(Notice that  $xV_0$ $=$ $yV_0$  is the maximal ideal  $N_0$ of  $V_0$
and that if we let  $t = y/x$ and $\overline t$ $=$ $t+N_0$  in
the field $V_0/N_0$, then $V_0/N_0 = (R_0/M_0)(\overline t)$
and $\overline t$  is transcendental over $R_0/M_0$.)

With this in mind, it follows from \cite[Theorem 19, p. 55]{ZS} that
$k^2$ $=$ $[R:R_0]$ $\ge$ ${\sum_{i=1}}^g ~e_if_i$,
where  $V_1,\dots,V_g$  are the extensions (to the quotient field
of  $R$)  of  $V_0$, $e_i$  is the ramification index of  $N_0$  in  $V_i$
(so  $N_0V_i$ $=$
${N_i}^{e_i}$, where  $N_i$  is the maximal ideal of  $V_i$), and  $f_i$  is
the
relative degree  $[(V_i/N_i) : (V_0/N_0)]$.
It will now be shown that  $g$ $=$ $1$ and that  $e_1$ $=$ $k$ $=$ $f_1$

For this, if  $v$  is the valuation of any of the valuation rings  $V$ $\in$
$\{V_1,\dots,V_g\}$, then  $v(x^i)$ $=$ $iv(x)$,
$v(y^i)$ $=$ $iv(y)$, and  $kv(z)$ $=$ $v(z^k)$ $=$ $v(x^i + y^i)$ $=$
$iv(x)$  (since $v$  is
an extension of the order valuation  $v_0$).  Therefore  $kv(z)$ $=$ $iv(x)$
and,
similarly, $kv(w)$ $=$ $iv(x)$.  Since  $k,i$  are relatively prime
(by hypothesis), it follows that  $v(z)$ $=$ $v(w)$ $\ge$ $i$  and
$v(x)$ $\ge$ $k$.  Also, $v(x)$ $=$ $v(N_0)$
(since  $xV_0$ $=$ $N_0$, as noted above), so
the ramification index  $e$  of  $N_0$  in  $V$  is at least  $k$.

Also, as noted in the preceding paragraph, $v(z)$ $=$ $v(w)$, so
$\frac{w}{z}$  is a unit in  $V$, and
$({\frac{w}{z}})^k$ $=$ $\frac{w^k}{z^k}$ $=$ $\frac {x^i - y^i}{x^i+y^i}$.
As above, let  $t$ $=$ $\frac {y}{x}$, so
$t$  is a unit in  $V_0$  whose residue class
$\overline t$  in  $V_0/N_0$  is transcendental over  $R_0/M_0$
and $\frac {x^i - y^i}{x^i+y^i}$ $=$
$\frac{1-t^i}{1+t^i}$ $\in$ $V_0$,
It follows that the residue
class of  $\frac{w}{z}$  in  $V/N$  is algebraic of degree  $k$  over
$V_0/N_0$,
so  $[V/N : V_0/N_0]$ $\ge$ $k$.

Therefore, by the preceding two paragraphs,
for each  $(V_i,N_i)$ $\in$ $\{(V_1,N_1),\dots,(V_g,N_g)\}$  it holds that
$N_0V_i$ $=$ ${N_i}^{e_i}$ $\subseteq$ ${N_i}^{k}$  (so  $e_i$ $\ge$ $k$)
and  $f_i$ $=$ $[V_i/N_i : V_0/N_0]$ $\ge$ $k$.  But
$k^2$ $=$ $[R:R_0]$ $\ge$ ${\sum_{i=1}}^g ~e_if_i$,
so it follows that  $g$ $=$ $1$ and that  $e_1$ $=$ $k$ $=$ $f_1$.
In what follows, we denote  $(V_1,N_1)$  by  $(V,N)$, $e_1$  by  $e$, and
$f_1$  by  $f$.
After normalizing $v$, we have
$v(x)$ $=$ $v(y)$ $=$ $k$  (so  $v(M)$ $=$ $k$)  and  $v(z)$ $=$ $v(w)$ $=$
$i$.)

For (iii), assume that  $j$ $=$ $i$ and that  $i,k$  are relatively prime.
To see that  $M$  is not projectively full,
note that  $M^k$ $\supseteq$ $(z^k,w^k)R$ $=$ $(x^i + y^i,x^i-y^i)R$  (since
$j$ $=$ $i$),
and  $(x^i + y^i,x^i-y^i)R$ $=$ $(2x^i,2y^i)R$
$=$ $(x^i,y^i)R$  (since  $2$  is a unit in  $R$),
and $(x^i,y^i)R$  is a reduction of  $M^i$.  It follows that
$((z^k,w^k)R)_a$ $=$ $(M^i)_a$,
so  $((z,w)^kR)_a$ $=$
$(M^i)_a$.  Therefore   $(z,w)R$  and  $M$  are projectively equivalent
and  $((z,w)R)_a$ $=$ $M_{i/k}$, by \cite[(2.3)]{MRS}, so  $((z,w)R)_a$
is not the integral closure of
any power of $M$  (since  $k$  and  $i$  are relatively prime),
hence  $M$  is not projectively full.

By the preceding paragraph  $i/k$ $\in$ $\mathbf U (M)$  (see
(\ref{defid(I)}.3)).  Therefore,
since  $i,k$  are relatively
prime and since  $d(M) \mathbf U (M)$ $\subseteq$ $\IN_+$  (by
(\ref{rema2}.4)),
it follows that  $d(M)$  is a
multiple of  $k$.  On the other hand,  $d(M)$  is a divisor
of the integer  $e$  associated to the
Rees valuation ring  $(V,N)$  of  $M$, by (\ref{rema2}.4).  By
(\ref{defiRees}),
this integer $e$ is
given by  $MV$ $=$ ${N}^{e}$, so  $v(M)$ $=$ $e$.  However, $v(M)$ $=$ $k$,
by the second
preceding paragraph, so it follows that  $d(M)$  is a divisor of  $k$.
Therefore  $d(M)$ $=$ $k$.
\end{proof}



In the next remark we note (with brief indications of proofs) several
properties of two rings
related to the rings  $R[z,w]$  of (\ref{expx}).


\begin{rema}
\label{2trts}
{\em With notation as in (\ref{expx}), let  $A$ $=$ $R[M/x]$ $=$
$R[y/x,z/x,w/x]$
and let  $\mathbf R$ $=$
$R[u,tM]$ $=$ $R[u,tx,ty,tz,tw]$  (where  $t$  is an indeterminate
and  $u$ $=$ $1/t$).  Also, with the
assumptions as in (\ref{expx}.2), let  $(V,N)$  be the unique Rees valuation
ring of  $M$.  Then
the following hold:

\noindent
{\bf{(\ref{2trts}.1)}}
$xA'$  is  $N \cap A'$-primary, $A$  is Cohen-Macaulay,
$xA$  is  $p'$-primary, where  $p'$
$=$ $N \cap A$,  $p'$ $=$ $(M,z/x,w/x)A$, and  $A$  is
not integrally closed.  Moreover, if  $i-k$  is a unit in  $R_0$, then
$A_{p'}$
is not integrally closed,
but $A_p$  is a regular local domain for all height one prime ideals  $p$
$\ne$ $p'$.

\noindent
{\bf{(\ref{2trts}.2)}}
$u\mathbf R'$  is primary, $\mathbf R$  is Cohen-Macaulay,
$u \mathbf R$  is  $p^*$-primary, where  $p^*$
$=$ $(u,M,tz,tw)\mathbf R$, and  $\mathbf R$  is
not integrally closed.  Moreover, if  $i-k$  is a unit in  $R_0$,
then  $\mathbf R_{p^*}$  is not integrally closed,
but $\mathbf R_p$  is a regular local domain for all height one prime ideals
$p$ $\ne$ $p^*$.}
\end{rema}

\begin{proof}
(We only sketch the proofs.)

$xA'$  is  $N \cap A'$-primary (by \cite[(2.9)]{MRS}) and  $u \mathbf R'$
is primary (since $M$  has a unique Rees valuation ring).

$A$  (resp., $\mathbf R$) is a free (of degree  $k^2$) integral extension
domain
of the locally regular UFD   $A_0$ $=$ $R_0[M_0/x]$ $=$ $R_0[y/x]$
(resp., $\mathbf R_0$ $=$ $R_0[u,tM_0]$ $=$ $R_0[u,tx,ty]$)
(since  $(z/x)^k = x^{i-k} + y^{i-k}(y/x)^k$  and  $(w/x)^k = x^{i-k} -
y^{i-k}(y/x)^k$
imply $(f_1,g_1)A_0[Z,W]$ $=$ $(Z^k - x^{i-k} - y^{i-k}(y/x)^k,
W^k - x^{i-k} + y^{i-k}(y/x)^k)A_0[Z,W]$  is
a height two prime ideal)
(resp., $(tz)^k$ $=$ $ u^{i-k}(tx)^{i} + u^{i-k}(ty)^{i}$ and $(tw)^k$ $=$
$ u^{i-k}(tx)^{i} - u^{i-k}(ty)^{i}$)
imply $(f_2,g_2) \mathbf R_0[Z,W]$ $=$
$(Z^k - u^{i-k}(tx)^i - u^{i-k}(ty)^i, W^k - u^{i-k}(tx)^i + y^{i-k}(ty)^i)
\mathbf R_0[Z,W]$  is
a height two prime ideal)),
so  it follows from \cite[(25.16)]{N2} that  $A$  (resp., $\mathbf R$)  is
Cohen-Macaulay.

Since  $A$  (resp., $\mathbf R$)  is Cohen-Macaulay and  $xA'$
(resp., $u \mathbf R'$)  has a unique
(height one) prime divisor, it follows from the structure of  $A$  (resp.,
$\mathbf R$)  that
$xA$  (resp., $u \mathbf R$)  has a unique prime divisor, say  $p'$  (resp.,
$p^*$).
And it then follows that
$p'$ $=$ $N \cap A$ $=$ $(M,z/x,w/x)A$  (resp., $p^*$ $=$ $(u,M,tz,tw)
\mathbf R$).

Using the determinant of  $f_{1Z}$ $=$ $kZ^{k-1}$, $f_{1W}$ $=$ $0$,
$g_{1Z}$ $=$ $0$,
and  $g_{1W}$ $=$ $kW^{k-1}$
and then of
$f_{1x}$ $=$ $-(i-k)x^{i-k-1}$, $f_{1 \tau}$ $=$ $-ky^{i-k} \tau ^{k-1}$,
$g_{1x}$ $=$ $-(i-k)x^{i-k-1}$, and  $g_{1 \tau}$ $=$ $ky^{i-k} \tau ^{k-1}$
(with  $\tau$ $=$ $y/x$),
it follows from \cite[(30.4)]{M} that
$A_p$  is a regular local domain for all  $p$ $\ne$ $p'$.
However, $A_{p'}$  is not a regular local domain, since, otherwise,
it would follow that  $V/N$ $=$ $V_0/N_0$  (where  $(V_0,N_0)$  is the order
valuation ring
of  $R_0$), and this contradicts the
fact (shown in the proof of (\ref{expx})(ii)) that  $w/z$
is algebraic of degree  $k$  over  $V_0/N_0$.
It therefore follows that
$A$  is not integrally closed.

It is clear that analogous statements hold for  $B$ $=$ $R[M/y]$ $=$
$R[x/y,z/y,w/y]$
and  $y$  in place of  $A$  and  $x$.
With this in mind, since  $\mathbf R[1/(tx)]$ $=$ $A[tx,1/(tx)]$
(resp.,  $\mathbf R[1/(ty)]$ $=$ $B[ty,1/(ty)]$),
and since  $u A[tx,1/(tx)]$ $=$ $xA[tx,1/(tx)]$ (resp.,  $u A[ty,1/(ty)]$
$=$ $yA[ty,1/(ty)]$),
it follows that $\mathbf R$  is
not integrally closed, that $\mathbf R_{p^*}$  is not integrally closed,
and that  $\mathbf R_p$  is a regular local domain for all height one prime
ideals  $p$ $\ne$ $p^*$.
\end{proof}

In Example \ref{expx}  $M$  has only one Rees valuation ring.
In the final remark in this paper we consider the ideals
in  $\mathbf P (I)$  in the case where  $I$
has only one Rees valuation ring.

\begin{rema}
\label{single}
{\em
Let  $I$  be a
proper ideal in a Noetherian
domain
$R$  and assume that
$I$  has only one Rees valuation ring, say  $(V,N)$.
Then  $\mathbf P (I)$ $\subseteq$
$\{N^i \cap R \mid i \in \IN\}$, so
if $P = N \cap R$ is the center of $V$ on $R$,
then each ideal in  $\mathbf P (I)$  is a $P$-primary
valuation ideal (that is, it is contracted from
a valuation overring of $R$). Assume that $I$ is
maximal in $\mathbf P(I)$.
If $IV = N^e$, then $I = N^e \cap R$ and $I$ is projectively
full if and only if
each $J \in \mathbf P(I)$ has the property that $JV = N^{ne}$ for
some $n \in \IN$.
In general, the inclusion
$\mathbf P (I)$ $\subseteq \{N^i \cap R \mid i \in \IN\}$
need not be an equality.
For example, if $(R,M)$ is  a regular local domain of altitude two and
$M = (x, y)R$, then $I = (x, y^2)R$
has a unique Rees valuation ring. To see this one can
apply \cite[(2.9)]{CHRR} or \cite[(3.1)]{MRS}.  Notice that $MR[x/y^2] =
yR[x/y^2]$ is
a height-one prime ideal and  $V = R[x/y^2]_{yR[x/y^2]}$ is a
valuation domain. Also, $MR[y^2/x]$ is a height-one prime ideal
and $R[y^2/x]_{MR[y^2/x]} = V$. Thus $V$ is the unique Rees valuation
ring of $I = (x,y^2)R$.
The ideal $I$ is projectively
full (by (\ref{comp-int})) and $I \subsetneq M = N \cap R$.
Also, $IV = N^2$, so $\mathbf P(I) = \{N^{2n} \cap R \}_{n=1}^\infty$.
}
\end{rema}



\bigskip


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\begin{flushleft}
Department of Mathematics, University of Missouri, Columbia,
Missouri 65211-0001
{\em E-mail address: catalin@math.missouri.edu}

\vspace{.15in}

Department of Mathematics, Purdue University, West Lafayette,
Indiana 47909-1395
{\em E-mail address: heinzer@math.purdue.edu}

\vspace{.15in}

Department of Mathematics, University of California, Riverside,
California 92521-0135
{\em E-mail address: ratliff@math.ucr.edu}

\vspace{.15in}

Department of Mathematics, University of California, Riverside,
California 92521-0135
{\em E-mail address: rush@math.ucr.edu}

\end{flushleft}

\end{document}

