%From ratliff@math.ucr.edu  Thu Jun 24 19:18:32 2004
%heinzer@math.purdue.edu
%catalin@math.missouri.edu
%rush@math.ucr.edu
%Hi BCD -
%Below is a revised version of our paper - with all
%the changes the referee mentioned.  Bill will be
%sending it to Craig in the near future, so fairly
%soon please let him know of any changes you feel
%should be made.
%Best regards,
%Jack

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\begin{document}

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\title{PROJECTIVELY EQUIVALENT IDEALS AND \\ REES VALUATIONS}
\author{Catalin Ciuperca, William J. Heinzer, Louis J. Ratliff Jr., and David E. Rush} 


\date{}
\maketitle




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\begin{abstract}
Let $R$  be a Noetherian ring. Two ideals  $I$ and $J$ in  $R$ are 
projectively equivalent in case the integral closure of  $I^i$  is
equal to the integral closure of $J^j$ for some $i,j$ $\in$ $\IN_+$. 
It is known 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. We observe that the
converse also holds. 
In particular, if the ideal $I$ has only one Rees valuation
ring $V$, then the ideals $J$ projectively equivalent to $I$ are precisely the
ideals $J$ such that $\Rees J = \{V\}$. 
In certain cases
such as: (i)  $ \dim R = 1$, or (ii) $R$ is a two-dimensional regular local domain, we
observe that if $I$ has more than one Rees valuation ring, then there
exist ideals  $J$ such that $\Rees I = \Rees J$, but $J$ is 
not projectively equivalent to $I$. 
If $I$ and $J$ are regular ideals of $R$,
we prove that $\Rees I \, \cup \Rees J \subseteq \Rees IJ$
with equality holding if $\dim R \le 2$, but not holding 
in general if $\dim R \ge 3$. We associate to $I$ and to the
set $\mathbf P(I)$ of integrally closed ideals projectively
equivalent to $I$ a numerical semigroup $S(I) \subseteq \IN$
such that $S(I) = \IN$ if and only if there exists $J \in \mathbf P(I)$
for which $\mathbf P(I) = \{(J^n)_a : n \in \IN_+ \}$.
\end{abstract}



\section{INTRODUCTION.}


All rings in this paper are commutative with a unit  $1$ $\ne$ $0$. Let
$I$ be a regular ideal of the Noetherian ring $R$  (that is, $I$ contains 
a regular  element of $R$).  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 of integrally closed ideals
projectively equivalent to $I$ is linearly ordered by inclusion and
eventually periodic. They also prove \cite[Proposition 2.10]{MRS}
that if an ideal $J$ is projectively equivalent to $I$, then $I$ and 
$J$ have the same Rees valuations and the values of $I$ and $J$
with respect to these Rees valuations are proportional. Our
goal in the present paper is to build on the work in \cite{MRS} and further
develop the relationship between projective equivalence 
of ideals and Rees valuations. 

\bigskip


\section{THE REES VALUATION RINGS OF AN IDEAL.}


In this section we review  a description of the Rees
valuations (and their valuation rings) associated to an ideal  $I$  
in a Noetherian ring  $R$.
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)}}
$I_a$  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  $I$ $=$ $I_a$.

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


\noindent
{\bf{(\ref{defi1}.3)}}
For each  $x \in R$,   let
$v_I(x)$ $=$ $\max \{k \in \IN  \mid x \in I^k\}$  
(as usual, $I^0$ $=$ $R$).  (Let  $v_I(x)$ $=$ $\infty$  in case  $x$ $\in$
$I^k$  for all  $k$ $\in$ $\IN$.)


\noindent
{\bf{(\ref{defi1}.4)}}
For each  $x\in R$,   let  $\overline v _I (x)$
$=$ $\lim _{k \rightarrow \infty} (\frac{v_I(x^k)}{k})$  
(see (\ref{defi1}.3) and Remark \ref{rema1}). }


\end{defi}

\begin{rema}
\label{rema1}
{\em Concerning (\ref{defi1}.4), 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  $x$ $\in$ $R$,
$\overline v _I(x) = \min \{\frac{v_i(x)}{e_i} \mid i = 1,\dots,g\}$.
In the case where $R$ is not an integral domain, 
we say that $v$ is a {\bf valuation on}  $R$ if 
$\{x \in R \mid v(x) = \infty\}$ is a prime ideal $P$ of $R$,
$v(x) = v(y)$ if $x + P = y + P$, and the induced function
$\overline{v}$ on the integral domain $R/P$ is a valuation. }

\end{rema}

To describe (\ref{rema1})(c) in more detail and to 
define the Rees valuation rings of  $I$ 
we need  the following definition
and notation.


\begin{defi}
\label{defia}
{\em Let  $I$  be an ideal in a Noetherian ring  $R$, 
let  $t$  be an indeterminate, and let  $u$ $=$ $1/t$.
Then the
{\bf{Rees ring}}
$\mathbf R$
{\bf{of}}
$R$
{\bf{with respect to}}
$I$  is the graded subring  $\mathbf R$ $=$  $R[u,tI]$  of  $R[u,t]$.
($\mathbf R$ $=$ $R[u]$, if  $I$ $=$ $(0)$.)}
\end{defi}


\begin{nota}
\label{nota1}
{\em Let  $I$  be an ideal in a Noetherian ring  $R$,
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  $u {\mathbf {R}_i} '$,
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
$v_{i,j}(x)$ $=$ $w_{i,j}(x+z_i)$.}
\end{nota}




With this notation, Rees shows
in \cite{Re}  that  $v_{i,j}$  is a {\bf valuation on}   $R$  in the
sense defined above and that    
$v_{i,j}(x)$ $=$ $\infty$  if and only
if  $x$ $\in$ $z_i$. 
Thus $v_1,\dots,v_g$ (see (\ref{rema1})) 
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.


Recall that if  $I$  is an ideal in  $R$  and  $v$  is a valuation on  $R$,
then  $v(I)$ $=$ $\min \{v(b) \mid b \in I\}$,
so $\overline {R}[\overline {I}/\overline {b}]$ $\subseteq$ $V$  if and
only if  $v(b)$ $=$ $v(I)$  
(where  $V$  is the valuation ring of  $v$  and the overbar denotes residue class
modulo the prime ideal  $\{x \in R \mid v(x) = \infty\}$).
We use this frequently in this paper.

\begin{rema}
\label{rema1a}
{\em  With notation as in  (\ref{nota1}), we have: 

\noindent
(a)  if  $I$ $\nsubseteq$ $z_i$, 
then  $e_{i,j}$ $=$ $w_{i,j}((I+z_i)/z_i)$  ($=$ $v_{i,j}(I)$);

\noindent
(b)  if  $I$ $\subseteq$ $z_i$, then $h_i = 1$,  $e_{i,1}$ $=$ $1$,
and  $w_{i,1}$  is a
{\bf{trivial}}
valuation on  $R$  (defined by:
$w_{i,1}(x)$ $=$ $0$, if   $x$ $\notin$ $z_i$; 
$w_{i,1}(x)$ $=$ $\infty$, if  $x$ $\in$ $z_i$);

\noindent
(c) if $I$ is
a regular ideal of $R$, then the Rees valuations of $I$ are
all nontrivial.}
\end{rema}

\begin{proof}
For (a), if  $I$ $\nsubseteq$ $z_i$,
then  $t((I+z_i)/z_i)$ $\subseteq$ 
${{\mathbf R}_i} ' \setminus p_{i,j}$, by \cite[(3.6)]{Ra1},
so  $t((I+z_i)/z_i)W_{i,j}$ $\subseteq$ $W_{i,j} \setminus p_{i,j}W_{i,j}$, 
so  $w_{i,j}(t((I+z_i)/z_i))$ $=$ $0$.
Therefore  $w_{i,j}(u)$ $=$ $w_{i,j}((I+z_i)/z_i)$ 
(since  $u$ $=$ $1/t$), and  $w_{i,j}(u)$ $=$ $e_{i,j}$
(by (\ref{nota1})), so  $w_{i,j}((I+z_i)/z_i)$ $=$ $e_{i,j}$.

For (b), if  $I$ $\subseteq$ $z_i$, then  $\mathbf{R}_i$ $=$ $(R/z_i)[u]$, so
$p_{i,1}$ $=$ $u{{\mathbf R}_i} '$  is the only prime divisor
of  $u{\mathbf{R}_i}'$  and
$W_{i,1}$ $=$
${{{\mathbf R}_i}'}_{p_{i,1}}$, so
$e_{i,1}$ $=$ $w_{i,1}(u)$ $=$ $1$  and
$w_{i,1}((I+z_i)/z_i)$ $=$ $w_{i,1}(0)$ $=$ $\infty$.  It follows from this
that  $v_{i,1}$  is a trivial valuation on  $R$. 

Statement (c) follows from the fact that a regular ideal is
not contained in any minimal prime ideal of the ring.
\end{proof}

In the literature, the valuation rings 
$W_{i,j}$ $=$ ${{\mathbf {R}_i}'}_{p_{i,j}}$ of (\ref{nota1}) are 
sometimes called the Rees valuation rings of  $I$,
and this causes no problems
when only one ideal  $I$  is under consideration.  However, when an ideal $J$ that 
is projectively equivalent to  $I$  (see (\ref{defi3}.1) below) is
also considered, then this definition of the Rees valuation rings 
applied to  $J$, in place of  $I$,
may yield different valuation rings from
the Rees valuation rings  $W_{i,j}$  of  $I$ (see (\ref{exam1}) 
below for a specific example).
However, the rings  $V_{1,1},\dots,V_{d,h_d}$  of (\ref{nota1})  
are the same for all ideals that are
projectively equivalent to  $I$ (as is shown in (\ref{theo2}) below), 
so we make the following definition.

\begin{defi}
\label{defiReesVal}
{\em The valuation rings  $V_{1,1},\dots,V_{d,h_d}$ in (\ref{nota1})  are the
{\bf{Rees valuation rings}}
of  $I$. The set of Rees valuation rings of $I$ is denoted $\Rees I$.}
\end{defi}

\begin{rema}
\label{remaReesVal}
{\em Concerning (\ref{defiReesVal}), notice  that if  $I$ $\subseteq$ $z_i$  for 
some  $i$ $=$ $1,\dots,d$, then it follows from (\ref{rema1a})(b) that
$h_i$ $=$ $1$, $V_{i,h_i}$ $=$ $V_{i,1}$  $=$ $F_i$, and  $e_{i,1}$ $=$ $1$.}
\end{rema}

\begin{rema}
\label{remaReesVal2}
{\em The centers in  $R$  of the Rees valuation rings of  $I$  are the
ideals  ${\phi_i}^{-1}(p_{i,j} \cap (R/z_i))$, where  $\phi_i$  is the
natural homomorphism from  $R$  to  $R/z_i$.
Therefore these centers  correspond to the prime divisors of  
$(u^nR[u,tI])_a \cap R$  for all large  $n$ $\in$ $\IN$, so they are the
asymptotic prime divisors of  $I$  (see \cite{McA}).
Therefore if these centers are the ideals  $P_1,\dots,P_f$, then
$\Ass(R/(I^i)_a)$ $\subseteq$ $\{P_1,\dots,P_f\}$  for all  $i$ $\in$ $\IN_+$  
and equality holds for
all large  $i$ $\in$ $\IN$.}
\end{rema}



In the next section we prove several results concerning the set  of 
Rees valuations of ideals.  Toward this
end, the following alternate construction of the nontrivial 
Rees  valuation rings 
of an ideal is  helpful.

\begin{cons}
\label{const}
{\em With the notation of (\ref{nota1}), let  $z$  be a minimal 
prime ideal in  $R$  such that  $I$ $\nsubseteq$ $z$  and
$z+I$ $\ne$ $R$, let  $b_1,\dots,b_h$  be generators  of  $I$ 
that are not in $z$,
let an overbar denote residue class modulo  $z$, and let  $F$  be the
quotient field of  $\overline R$.  Let  $V$  be a discrete valuation
ring such that  $\overline R$ $\subseteq$ $V$ $\subsetneq$ $F$, 
and let  $N$  be
the maximal ideal of  $V$.  Then  $V$  is a Rees valuation ring of  $I$
if and only if there exists  
$\overline b$ $\in$ $\{\overline {b_1},\dots,\overline {b_h}\}$
such that $N \cap A'$  is a height one prime ideal, 
where  $A$ $=$ $\overline {R}[\overline {I} /\overline {b}]$.
Moreover, if  $V$  is a Rees valuation ring of  $I$  
and if  $B$ $=$ $\overline {R}[\overline{I}/\overline{c}]$ $\subseteq$
$V$, then  $V$ $=$ ${B'}_{N \cap B'}$.}
\end{cons}

\begin{proof}
By considering each of the rings  $R/z$  (with  $z$  a minimal 
prime ideal in  $R$)  separately 
it may be assumed to begin with that  $R$
is a Noetherian integral domain. Therefore 
\cite[Proposition 3.1]{MRS} applies to
establish this equivalent way to define the set
of Rees valuations of $I$. 

For the final statement, it suffices to observe that
if  $V$ is  a valuation domain with maximal ideal $N$ that has 
$R$ as a subring and $b$ and $c$ are elements of $I$
such that $IV = bV = cV$, then 
$R[I/b]_{N \cap R[I/b]} = R[I/c]_{N \cap R[I/c]}$; this equality is
clear since $c/b$ is a unit of $R[I/b]_{N \cap R[I/b]}$ and 
$b/c$ is a unit of $R[I/c]_{N \cap R[I/c]}$.
\end{proof}


\begin{rema}\label{2.10}
{\em
\noindent
(a)
If  $I$ $=$ $bR$  is a regular principal ideal in  $R$, 
then it follows from (\ref{const})
that  $\Rees I = \{{R'}_{p_1}/z_1,\dots,{R'}_{p_g}/z_g\}$, 
where  $p_1,\dots,p_g$  are the prime divisors of  $bR'$
and  $z_i$ $=$ $\rad({R'}_{p_i})$  for $i$ $=$ $1,\dots,g$
(possibly $z_i \cap R'$ $=$ $z_j \cap R'$ for some  $i,j$).

\noindent
(b) Every minimal prime divisor of an ideal $I$ is the 
center of at least one Rees valuation of $I$.
Therefore for ideals $I$ and $J$ of $R$,
if $\Rees I = \Rees J$, then $I$ and $J$ have the
same radical. 

\noindent
(c) For a fixed ideal $I$,
let $\Gamma(I)$ denote the set of ideals $J$
such that $\Rees I = \Rees J$. It would be interesting
to have conditions on $I$ or on the ring $R$ in order
that the set $\Gamma(I)$ have  a unique maximal element 
with respect to inclusion.
This is true for all ideals $I$ of a  one-dimensional Noetherian  integral 
domain $R$, for in this case,  by statement (1)
of Example \ref{exam2},  $\rad(I)$ is the unique largest
ideal of $R$ having the same Rees valuations as $I$. 
It is also true for all ideals $I$ of  a two-dimensional
regular local domain $R$, for in this case,  as discussed
in Example \ref{exam3}, every integrally closed ideal
of $R$ is uniquely a finite product of simple complete
ideals, and the product of the simple complete factors of 
$I_a$ with no repeated factors is the largest ideal
having the same Rees valuations as $I$. It is also true for an  ideal $I$
of a general Noetherian domain $R$ 
if $I$ has only one Rees valuation ring $V$, for the integrally  
closed ideals $J$ such that $\Rees J = \{V\}$ are all contracted
from $V$ and thus are linearly ordered with respect to inclusion.}
\end{rema}


\begin{rema}\label{2.11}
{\em
Let $(R,M)$ be a Noetherian local domain and let $\widehat R$
denote the $M$-adic completion of $R$. It follows from \cite[(3.19)]{McA} and (\ref{remaReesVal2})
that the following are equivalent:  (a)  there exists a valuation domain $V$ dominating $R$ such that 
$V \in \Rees I$ for every nonzero proper ideal $I$ of $R$; (b)
there exists a minimal prime ideal $z$ of $\widehat R$ such
that $\dim(\widehat R/z) = 1$; and, (c)  there exists a height one maximal
ideal in  $R'$.}

\end{rema}



\section{REES VALUATION RINGS AND PROJECTIVELY \\ EQUIVALENT IDEALS.}

It is shown in \cite[Proposition 2.10]{MRS} that if $I$ is a 
regular ideal of a Noetherian ring,  then  every ideal
$J$ projectively equivalent to $I$ satisfies $\Rees I = \Rees J$ and 
the values of $I$ and $J$ with respect to these Rees valuation rings 
are proportional.  We prove in  Theorem \ref{theo2} that 
the converse also holds. 
In particular, if $I$ has only one Rees valuation
ring $V$, then the ideals $J$ projectively equivalent to $I$ are precisely the
ideals $J$ such that $\Rees J = \{V\}$. In Example \ref{exam2}, we consider
projective equivalence and Rees valuation rings of ideals of a one-dimensional
Noetherian integral domain $R$. For an ideal $I$ of $R$ such that 
$\Rees I$ has cardinality greater than one, we prove  there exist
ideals $J$ of $R$ such that $\Rees I = \Rees J$, but $J$ is not
projectively equivalent to $I$. In Proposition \ref{theo3}, we prove
that if $I$ and $J$ are regular ideals of a Noetherian ring,
then $\Rees I \, \cup \Rees J \subseteq \Rees IJ$, with 
equality holding if $\dim R \le 2$. We observe in Remark (\ref{y}.3) 
that equality does not hold 
in general if $\dim R \ge 3$. 



We recall the following definition.

\begin{defi}
\label{defi3}
{\em Let  $I$  be an ideal in a Noetherian ring  $R$.
An ideal  $J$  in  $R$  is
{\bf{projectively equivalent to}}
$I$  in case  $(J^j)_a$ $=$ $(I^i)_a$  (see (\ref{defi1}.1)) 
for some  $i,j$ $\in$ $\IN_+$.}
\end{defi}


Samuel introduced projectively equivalent ideals in 1952 in \cite{S}. 
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}.   In this section we explore the relation between 
projectively equivalent ideals and Rees valuation rings.


\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)}}
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  $(I^k)_a$ $=$ $(J^l)_a$.}
\end{rema}

\begin{proof}
(\ref{rema3}.1) follows readily from basic properties of integral closures of ideals,
and (\ref{rema3}.2) is proved in \cite[(2.1)(b)]{MRS}.
\end{proof}

The following example shows that projectively equivalent 
ideals may yield different valuation
rings  $W_{i,j}$  as in (\ref{nota1}).

\begin{exam}
\label{exam1}
{\em Let  $R$ $=$ $K[[X]]$, where  $K$  is a field and  $X$  is
an indeterminate, let  $I$ $=$ $XR$, and let  $J$ $=$ $X^2 R$, so  $I$  and  $J$  are
projectively equivalent  (since  $(I^2)_a$ $=$ $I^2$ $=$ $J$ $=$ $(J^1)_a$).
In this case, $d$ $=$ $1$, $z_1$ $=$ $(0)$, 
$\mathbf R(R,I)$ $=$ $R[u,tX]$ $=$ $\mathbf R (R,I)'$, $p_{1,1}$ $=$
$uR[u,tX]$.  Thus $W_{1,1}$ $=$ $R[u,tX]_{p_{1,1}}$  is the only
valuation ring $W$  of  $I$ as in (\ref{nota1}),  and  $w_{1,1}(u)$ $=$ $1$.
On the other hand, $\mathbf R(R,J)$ $=$ $R[u,tX^2]$ $=$ $\mathbf R (R,J)'$, 
$p_{1,1}$ $=$ $(u,X)R[u,tX^2]$. Thus  
${W_{1,1}}^*$ $=$ $R[u,tX^2]_{p_{1,1}}$  is the only 
valuation ring  $W^*$ of  $J$ as in (\ref{nota1}), and  ${w_{1,1}}^*(u)$ $=$ $2$, 
since  $tX^2$ $\in$ $R[u,tX^2] \setminus  p_{1,1}$.  Therefore the
valuation rings  $W_{1,1}$  and  ${W_{1,1}}^*$  differ, while 
$V_{1,1} = W_{1,1} \cap K((X)) = R = W_{1,1}^* \cap K((X)) = V_{1,1}^*$,
so $\Rees I = \Rees J$. }
\end{exam}

With  the definition of Rees valuation rings in (\ref{defiReesVal}),
we have the following.   


\begin{theo} \label{theo2}
Let $I$ and $J$ be regular ideals of the Noetherian ring $R$. The 
following are equivalent:
\begin{enumerate}
\item 
$I$ and $J$ are projectively equivalent.
\item
$\Rees I = \Rees J$ and 
the values of $I$ and $J$ with respect to these Rees valuation rings 
are proportional. 
\end{enumerate}
In particular, if the ideal $I$ has only one Rees valuation
ring $V$, then the ideals $J$ projectively equivalent to $I$ are precisely the
ideals $J$ such that $\Rees J = \{V\}$. 
\end{theo}

\begin{proof}
It is shown in \cite[Proposition 2.10]{MRS} that (1) implies (2).
To prove that (2) implies (1), 
notice first that by considering each of the rings  $R/z$  
(with  $z$  a minimal prime ideal in  $R$), it suffices
to prove (2) implies (1)  in the case 
where   $R$  is a Noetherian integral domain.
Since  $V_1,\dots,V_g$  are the Rees valuation rings of  $I$  it follows
from (\ref{remaReesVal2}) that, for all  $i$ $\in$ $\IN_+$,  
$(I^i)_a$ $=$ $\cap \{I^i V_h \cap R \mid h = 1,\dots,g\}$, 
and, similarly,
for all  $j$ $\in$ $\IN_+$, $(J^j)_a$ $=$ $\cap \{J^j V_h \cap R \mid h = 1,\dots,g\}$.
Therefore, if there exist  $i,j$ $\in$ $\IN_+$  
such that  $v_h(I) =(\frac{j}{i})v_h(J)$  for
$h$ $=$ $1,\dots,g$, then it follows that  $v_h(I^i)$ $=$ $v_h(J^j)$  
for  $h$ $=$ $1,\dots,g$,
so  $I^iV_h$ $=$ $J^jV_h$  for  $h$ $=$ $1,\dots,g$,
hence  $(I^i)_a$ $=$ $\cap \{I^i V_h \cap R \mid h = 1,\dots,g\}$
$=$ $\cap \{J^j V_h \cap R \mid h = 1,\dots,g\}$ $=$ $(J^j)_a$.


To prove the last statement, let  
$v$  be the normalized valuation  associated to the valuation ring 
$V$, let  $j$ $=$ $v(I)$, and let  $i$ $=$ $v(J)$.
Then  $v(I^i)$ $=$ $ij$ $=$ $v(J^j)$, so  $I^{i}V$ $=$ $J^jV$, 
so  $v(I) = (\frac{j}{i})v(J)$,
so the conclusion follows from the equivalence of (1) and (2).
\end{proof}

Let  $I$  be a nonzero ideal in a Noetherian domain  $R$.  
If  $V_1,\dots,V_g$  are the Rees
valuation rings of  $I$, then they may also be the Rees valuation rings of another ideal
$J$  of  $R$  such that  $J$  is not projectively equivalent to  $I$.  
To illustrate the concepts of  projective equivalence of ideals and 
Rees valuations rings, we consider in Example \ref{exam2}
the case where $R$ is a Noetherian domain with $\dim R = 1$. 
In particular, Example \ref{exam2}  provides examples of ideals $I$ and $J$ such that 
$\Rees I = \Rees J$, but $I$ and $J$ are not projectively equivalent. 


\begin{exam}
\label{exam2}
{\em  Let $R$ be a Noetherian integral domain with $\dim R = 1$. It is 
well known that the integral closure $R'$ of $R$ is a Dedekind domain. If $I$ is a
nonzero proper ideal of $R$, then $\Rees I = \{R'_P\}$, where $P$ varies over
the maximal ideals of $R'$ such that $I \subseteq P$. Thus for ideals  $I$ and $J$
of $R$, we have 
\begin{enumerate}
\item 
$\Rees I = \Rees J$ if and only if $\rad I = \rad J$. 
\item
$\Rees IJ = \Rees I \cup \Rees J$.
\item
For a nonzero ideal $I$ of $R$, the set $\Rees I$ has cardinality 
greater than one if and only if $I$ is contained in more than 
one maximal ideal of $R'$.
\item 
If $\Rees I$ has cardinality greater than one, then there exist
ideals $J$ of $R$ such that $\Rees I = \Rees J$, but $J$ is not
projectively equivalent to $I$.
\end{enumerate}
To prove this last statement, notice that if $P, Q_1, \ldots, Q_s$
are distinct maximal ideals of $R'$, then $PQ_1\cdots Q_s \cap R$
is the set of elements of $R$ having positive value in the 
normalized valuation $v$ corresponding to $R'_P$ and also 
positive value in the normalized valuation $w_i$ corresponding
to $R'_{Q_i}$, for $i = 1, \ldots, s$; while for each $n$ $\in$ $\IN_+$,
$P^nQ_1\cdots Q_s  \cap R$ is the set of elements of 
$R$ having $v$-value at least $n$ and positive value with respect
to each $w_i$, $i = 1, \ldots, s$. Here we are using that the ideals
$P^n, Q_1, \ldots, Q_s$ are pairwise comaximal in $R'$. Assume that  
$\Rees I = \{R'_P, R'_{Q_1}, \ldots,R'_{Q_s} \}$, with $s$ $\in$ $\IN_+$.
To show there exists an ideal $J$ of $R$ with $\rad I = \rad J$  
(so  Rees $I$ $=$ Rees $J$, by (\ref{exam2}.1))
such that $J$ is not projectively equivalent to $I$, it suffices
to prove there exists $n$ $\in$ $\IN_+$  such that
$P^nQ_1\cdots Q_s \cap R \subsetneq PQ_1\cdots Q_s \cap R$.
Since 
$PQ_1 \cdots Q_s  \cap R \ne (0)$  and  $\cap_{n=2}^\infty ( P^nQ_1 \cdots Q_s \cap R ) = (0)$,
there must exist $n$ $\in$ $\IN_+$  such that 
$P^nQ_1\cdots Q_s  \cap R \subsetneq PQ_1\cdots Q_s \cap R$. 
}
\end{exam}

As suggested by the referee, Example (\ref{exam2}.4) can be
demonstrated concretely by taking $R$ to be a one-dimensional
semilocal normal Noetherian domain with distinct maximal ideals 
$\m_1$ and $\m_2$. Let $0 \ne a \in (\m_1 \cap \m_2)$ 
and $x \in \m_1 \setminus \m_2$ and $y \in \m_2 \setminus \m_1$.
Then $I = axR$ and $J = ayR$ have the same Rees valuations, 
but are not projectively equivalent.

In Proposition \ref{theo3}, we relate the Rees valuation rings of
the ideals $I$ and $J$ with those of their product $IJ$.
In the case where the ambient ring is a two-dimensional 
pseudo-geometric normal Noetherian local domain, Proposition \ref{theo3}
is due to G\"ohner  \cite[Lemma~2.1]{G}. 
The statement and proof of the first part of Proposition \ref{theo3}
are similar in spirit to a theorem of K.Whittington \cite[Prop. 3.26]{McA}.
See also the forthcoming book by Craig Huneke and Irena 
Swanson on integral closure of ideals for related information. 

\begin{prop}
\label{theo3}
Let  $I$  and  $J$  be regular ideals in a Noetherian ring  $R$.  
Then  $\Rees I \,  \cup \Rees J \subseteq \Rees IJ$,
with equality holding  if  $\dim R \le 2$.    
\end{prop}

\begin{proof}
To prove both statements it may be assumed 
(as in the proof of Theorem \ref{theo2}) that  $R$  is a
Noetherian domain. And for the first statement it suffices (by symmetry)
to show that if  $V \in \Rees I$, then  $V \in \Rees IJ$.

For this,
let  $I$ $=$ $(b_1,\dots,b_h)R$
and by (\ref{const}) let $b$ $\in$ $\{b_1,\dots,b_h\}$  
such that  $V$ $=$ ${A'}_{N \cap A'}$, where  $A$ $=$ $R[I/b]$
(so  $v(I)$ $=$ $v(b)$) and  $N$  is the maximal ideal of  $V$.
Let $J$ $=$ $(c_1,\dots,c_k)R$  and let
$c$ $\in$ $\{c_1,\dots,c_k\}$)
such that  $v(c)$ $=$ $v(J)$.
Then  $v(bc)$ $=$ $v(I)+v(J)$ $=$ $v(IJ)$, 
so  $C$ $=$ $R[(IJ)/(bc)]$ $\subseteq$ $V$, hence  ${C'}_{N \cap C'}$ $\subseteq$ $V$.
However, $R[I/b]$ $=$ $A$ $\subseteq$ $C$ $=$
$R[(IJ)/(bc)]$  (since  $IJ$ $=$ $(c_1I,\dots,c_kI)R$
and  $c$ $\in$ $\{c_1,\dots,c_k\}$  imply that  $I/b$ $\subseteq$ $(IJ)/(bc)$),
so it follows that  $V$ $=$ ${C'}_{N \cap C'}$, 
so  $V \in \Rees IJ$ by (\ref{const}).

In view of Example \ref{exam2} (and the first paragraph of
this proof), the second statement is 
clear if $\dim R = 1$, so we assume $\dim R = 2$.  
Let  $V \in \Rees IJ$ and let $N$ denote the maximal ideal of $V$.  
Also, let  $b_1,\dots,b_h$  (resp.,
$c_1,\dots,c_k$)  be generators of  $I$  (resp., $J$),
where the $b_i$ and $c_j$ may be assumed to be nonzero.  
Then  $b_1c_1,\dots,b_hc_k$  generate 
$IJ$, so by (\ref{const}) there exists  $b_ic_j$  ($i$ $\in$ $\{1,\dots,h\}$  and
$j$ $\in$ $\{1,\dots,k\}$)  such that  $V$ $=$ ${C'}_{p'}$, where
$C$ $=$ $R[(IJ)/(b_ic_j)]$  and  $p'$ $=$ $N \cap C'$. 
Let $b := b_i$ and $c:= c_j$.  Now  $v(IJ)$ $=$ $v(bc)$  
(since  $(IJ)/(bc)$ $\subseteq$ $V$)
$=$ $v(b) + v(c)$ $\ge$ $v(I) + v(J)$ $=$ $v(IJ)$, so it follows
that:  $v(I)$ $=$ $v(b)$; $v(J)$ $=$ $v(c)$; 
$A$ $=$ $R[I/b]$ $\subseteq$ $C$; and, $B$ $=$
$R[J/c]$ $\subseteq$ $C$.  Let  $q$ $=$ $p' \cap A'$  
and  $w$ $=$ $p' \cap B'$, so  ${A'}_q$ $\subseteq$ ${C'}_{p'}$ $=$ $V$  and
${B'}_{w}$ $\subseteq$ ${C'}_{p'}$ $=$ $V$, so: if  $\hgt q = 1$, then  $V$ $=$ ${A'}_q$
$\in \Rees I$  (by (\ref{const}));
and, if  $\hgt w = 1$, 
then  $V$ $=$ ${B'}_{w}$
$\in \Rees J$  (by (\ref{const})).  Therefore to show that   
$\Rees IJ \subseteq \Rees I \, \cup \Rees J$ it suffices to show
that either:   (a)  $\hgt q  = 1$; or, (b)  $\hgt w = 1$.
We now show that either (a) or (b) holds. 

By the Mori-Nagata theorem \cite[(33.12)]{N2}, $A'$, $B'$ and $C'$ are 
normal Noetherian domains of dimension at most two
and hence are Cohen-Macaulay and therefore universally catenary.
Let $D = A'[(IJ)/(bc)] = A'[J/c] = B'[I/b]$. Notice that 
$D' = C'$. Also  $D$ is universally catenary
since it is a   finitely generated integral domain over 
$A'$. Therefore  $A'$, $B'$ and $D$ satisfy the altitude formula,
(or in other terminology the dimension formula \cite[Theorem 15.6]{M}).
Let $p := p' \cap D$. Then $\hgt p = \hgt p' = 1$. 
Assume that $\hgt q = 2$. Since $D = A'[J/c]$, the altitude
formula implies for some $n \in \{1, \dots, k\}$ that the
image of $c_n/c$ in $D/p$ is transcendental over $A'/q$. It 
follows that the image of $c_n/c$ in $B/(p \cap B)$ is 
transcendental over $R/(p \cap R)$. This implies that
$\hgt(p \cap B) < \hgt(p \cap R)$. Therefore 
$\hgt(p \cap B) = 1$. Notice that $p \cap B = w \cap B$. 
Since $\hgt w \le \hgt(w \cap B)$, it follows that
$\hgt w = 1$,   so
(b) of the preceding paragraph holds, hence  $V \in \Rees J$.
A similar argument shows that if  $\hgt w = 2$, 
then (a) of the preceding paragraph holds,
so  $V \in \Rees I$.  It follows that  
$\Rees IJ  \subseteq \Rees I \,  \cup \Rees J$, so
equality holds by the first statement. 
\end{proof}

\begin{rema}
\label{y}
{\em Let  $I$  and  $J$  be regular ideals in a Noetherian ring  $R$.  Then:}

\noindent
{\bf{(\ref{y}.1)}}
{\em If  $V \in \Rees IJ$, and if  $J$ $\nsubseteq$ $N \cap R$, where
$N$  is the maximal ideal in  $V$, then  $V \in \Rees I$.}

\noindent
{\bf{(\ref{y}.2)}}
{\em If $x$ is a regular element of $R$, then  
$\Rees (xI) = \Rees (xR) \cup \Rees I$.}

\noindent
{\bf{(\ref{y}.3)}}
{\em Let $(R, M)$ be a 3-dimensional regular local domain
where $M = (x, y, z)R$ and let $I = (x, y)R$ and $J = (x, z)R$.
Then the ord-valuation domain defined by the powers of $M$ is a
Rees valuation ring of $IJ = (x^2, xy, xz, yz)R$, but is not
in $\Rees I \, \cup \Rees J$. Therefore the equality statement
of Proposition \ref{theo3} fails if $\dim R = 3$.  
}
\end{rema}



\begin{proof}
For (\ref{y}.1), let  $V$ $\in$ Rees $IJ$.  
Then it follows from (\ref{const}) that there exist  $b$ $\in$ $I$  and
$c$ $\in$ $J$  such that  $V$ $=$ ${C'}_{N \cap C'}$, where  $C$ $=$ $R[(IJ)/(bc)]$
(and then  $v(bc)$ $=$ $v(IJ)$ $=$ $v(I)+v(J)$ $\le$ $v(b)+v(c)$ $=$ $v(bc)$,
so  $v(I)$ $=$ $v(b)$  and  $v(J)$ $=$ $v(c)$, where  $v$  is the valuation of  $V$).
It follows that  $A$ $=$ $R[I/b]$ $\subseteq$ $V$.
Now, if  $J$ $\nsubseteq$ $N \cap R$, then  $v(J)$ $=$ $0$ (so $v(c)$ $=$ $0$), so
$J/c$ $\subseteq$ $R_{N \cap R}$, so it follows that  $C$ $\subseteq$ ${A'}_{N \cap A'}$.
Since  $V$ $=$ ${C'}_{N \cap C'}$, it follows that  $V$ $=$ ${A'}_{N \cap A'}$, so  $V$
$\in$ Rees $I$  by (\ref{const}).

For (\ref{y}.2), it suffices, by (\ref{theo3}), to show 
that  Rees $xI$ $\subseteq$ Rees $xR$ $\cup$ Rees $I$.
For this, by (\ref{const}) there exists  $b$ $\in$ $I$  such 
that  $V$ $=$ ${C'}_p$, where  $C$ $=$ $R[(xI)/(xb)]$
and  $p$  is a height one prime divisor of  $xbC'$  
(so  $v(xb)$ $=$ $v(xI)$, so  $v(b)$ $=$ $v(I)$).
If  $b$ $\notin$ $p$, then  $v(I)$ $=$ $v(b)$ $=$ $0$, 
so  $I$ $\nsubseteq$ $p \cap R$, hence
$V$ $\in$ Rees $xR$  by (\ref{y}.1).  On the other hand if  $b$ $\in$ $p$, then
since it is clear that  $C$ $=$ $A$, where  $A$ $=$ $R[I/b]$, 
it follows that  $p$ is a minimal
prime divisor of  $bA'$ $=$ $bC'$,
hence  $V$ $\in$ Rees $I$  by (\ref{const}).

For (\ref{y}.3), notice that 
the powers of $I$ and the 
powers of $J$ each define a valuation and these are the 
unique Rees valuations of $I$ and $J$. Indeed,
$\Rees I = \{ R[y/x]_{xR[y/x]} \}$ and $\Rees J = \{ R[z/x]_{xR[z/x]} \}$.
On the other hand, the ord-valuation domain defined by the powers
of $M$ is $V = R[y/x, z/x]_{xR[y/x, z/x]}$. Since  
$R[xy/x^2, xz/x^2, yz/x^2] = R[y/x, z/x]$, we see that 
$V \in \Rees IJ$.
\end{proof}

\begin{exam}
\label{exam3}
{\em  (cf. \cite[(3.6)]{MRS}.)  Let $R$ be a two-dimensional regular local domain.
Zariski develops in \cite[Appendix 5]{ZS} the theory of 
complete (or integrally closed) ideals of $R$. He proves that
in $R$ a product of complete ideals is again complete, 
and establishes a unique factorization theorem:
every complete ideal of $R$ is uniquely expressible as
a product of simple complete ideals \cite[pages 385-386]{ZS}.
Here an ideal $I$ is said to be simple if it is not the unit
ideal and has no nontrivial factorizations. Since $R$ is a 
unique factorization domain, every nonzero  ideal $I$
of $R$ is of the form $I = xJ$, where either $J = R$ or 
$J$ is primary for the maximal ideal $M$ of $R$. Since
principal ideals of $R$ are complete, the theory reduces to a 
consideration of complete $M$-primary ideals. The simple
complete $M$-primary ideals of $R$ are in one-to-one 
correspondence with the DVRs  that birationally dominate
$R$ and have the property that their residue field as an
extension of $R/M$ is not algebraic. If $I$ is a simple
complete ideal of $R$, then $I$ has a unique 
Rees valuation domain $V$ \cite[Theorem 4.2]{H}.
It follows that the integrally closed ideals that
are projectively equivalent to $I$ are precisely
the ideals $I^n$ for $n \in \IN_+$. In particular,
every simple complete ideal of $R$ is projectively full in 
the sense of  Definition \ref{defifull}. For an
arbitrary nonzero proper ideal $I$ of $R$, the Rees valuation
rings of $I$ are in one-to-one correspondence with
the distinct simple complete ideals that are factors of the
integral closure of $I$. 
We have  $I_a = J_1^{e_1} \cdots J_n^{e_n}$, where the $J_i$ are
simple complete ideals and the $e_i$ are positive integers.
If $J_i$ is a height-one prime of $R$, the Rees valuation ring
associated to $J_i$ is $V_i = R_{J_i}$, while if $J_i$ is 
$M$-primary, then $V_i$ is the DVR that birationally dominates $R$
described above. Let $K$ be a nonzero proper ideal of $R$ 
and let $K_a = L_1^{f_1} \cdots L_m^{f_m}$, where the $L_i$
are distinct simple complete ideals and the $f_i$ are 
positive integers. Then $\Rees I = \Rees K$ if and only 
if $n = m$ and the set of simple complete ideals $\{J_i\}_{i=1}^n$ 
is the same as the set $\{L_i\}_{i=1}^m$.
Necessary and sufficient conditions for $I$ and $K$ to be 
projectively equivalent are that $\Rees I = \Rees K$, so
$I_a = J_1^{e_1} \cdots J_n^{e_n}$ and $K_a = J_1^{f_1} \cdots J_n^{f_n}$, 
and,  in addition,   the $n$-tuples $e_1, \ldots, e_n$
and $f_1, \ldots , f_n$  are proportional, i.e., there exist positive
integers $a$ and $b$ such that $ae_i = bf_i$ for $i = 1, \ldots, n$. 
In particular, if $I$ has more than one Rees valuation
ring and if $J_1$ and $J_2$ are distinct simple complete ideals
of $R$ that are factors of $I_a$,
then $J_1I_a$ and $J_2I_a$ are complete ideals that 
are not projectively equivalent, but have the
same Rees valuation rings. 
}
\end{exam} 

\begin{rema}\label{3.9}
{\em
Since an ideal and its powers have the same blowup, if
$I$ and $J$ are normal projectively equivalent ideals of
a Noetherian domain $R$, then the blowups of $I$ and $J$
are equal. In the case where $I$ and $J$ are normal ideals of 
a two-dimensional regular local domain, then $I$ and $J$
have the same blowup if $\Rees I = \Rees J$. However,
Cutkosky shows in \cite[Example 2]{C} the existence of an
infinite set of normal ideals in a three-dimensional regular
local domain that have the same Rees valuations but have 
pairwise distinct blowups. 
}
\end{rema}


\bigskip


\section{NUMERICAL SEMIGROUPS AND PROJECTIVELY \\ FULL IDEALS.}
Let $I$ be a regular ideal of the Noetherian ring $R$. In \cite{MRS},
McAdam, Ratliff and Sally prove  that the set  $\mathbf P (I)$  of integrally closed ideals
projectively equivalent to $I$ is linearly ordered by inclusion and
eventually periodic. ($\mathbf P (I)$  is
{\bf{eventually periodic}}
means there exist  $I_1,\dots,I_h$  $\in$ $\mathbf P (I)$  such that  $\mathbf P (I)$
$=$ $\{({I_1}^{k_1} \cdots {I_h}^{k_h})_a \mid$; each
$k_i$  is a nonnegative integer$\}$.  (It can be shown that this definition
of ``eventually periodic'' is equivalent to the definition, given in \cite{MRS}, that
the set  $\mathbf U$  in (4.1.3) is eventually periodic; see (\ref{rema2})(d).))
They also prove the existence of a fixed
$d$ $\in$ $\IN_+$  such that for every ideal $J$ projectively equivalent to $I$
there exists $n$ $\in$ $\IN_+$  such that  $(I^n)_a  = (J^d)_a$. 
As we note in Remark \ref{rema22}, using 
results proved in \cite{MRS},   there is naturally associated to $I$ or
to the projective equivalence class of $I$ a unique  numerical semigroup. 
Here we are using the term numerical semigroup  in the sense
of Herzog-Kunz \cite{HK} and Watanabe \cite{W}. 

We recall the following definitions from \cite{MRS}.



\begin{defi}
\label{defi4}
Let  $I$  be a regular ideal in a Noetherian ring  $R$.

\noindent
{\bf{(\ref{defi4}.1)}}
{\em For  $\alpha$ $\in$ $\mathbf \IR_+$  
let  $I_\alpha$ $=$ $\{x$ $\in$ $R \mid \overline v _I(x)$ $\ge$ $\alpha\}$.}

\noindent
{\bf{(\ref{defi4}.2)}}
{\em $\mathbf W$ $=$ $\{\alpha$ $\in$ $\IR_+ \mid$  $\overline v _I(x)$ $=$ $\alpha$  
for some  $x$ $\in$ $R\}$ (see
(\ref{defi1}.4)).}

\noindent
{\bf{(\ref{defi4}.3)}}
{\em $\mathbf U$ $=$ $\{\alpha$ $\in$ $\mathbf W \mid$ $I_\alpha$  
is projectively equivalent to  $I\}$  (see (\ref{defi4}.1)
and (\ref{defi3})),
and  $\mathbf P = \mathbf P(I)  = \{I_\alpha \mid \alpha \in \mathbf U \}$.}
\end{defi}





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

\noindent
(a)  for each  $\alpha$ $\in$ $\IR_+$, 
the ideal  $I_\alpha$ of (\ref{defi4}.1) is an integrally
closed ideal ($=$ $(I_\alpha)_a)$ in  $R$,
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) and (2.6)]{MRS}.

\noindent
(b) for the set  $\mathbf P$ of (\ref{defi4}.3), 
$\mathbf P$  $=$ $\{J \mid J$  is an integrally closed ideal in  $R$
that is projectively equivalent to  $I\}$,
and  $\mathbf P$  is linearly ordered by inclusion, by \cite[(2.4)]{MRS};

\noindent
(c)  $\mathbf W$  and  $\mathbf U$  are 
discrete subsets of  $\IQ _+$,
by \cite[(1.1) and (2.8)]{MRS};

\noindent
(d) there exist $n^*$ $\in$ $\IN_+$  and a
{\it{unique}}
$d$ $\in$ $\IN_+$  such that:
$\{\alpha \in \mathbf U \mid \alpha \ge n^*\}$ $=$ $\{n^*+ \frac{k}{d} \mid k \in \IN\}$;
$d$  is a common divisor (but not necessarily the greatest common divisor)
of the integers  $e_1,\dots,e_g$  of (\ref{rema1})(c); $d \alpha$ $\in$ $\IN_+$  
for all  $\alpha$ $\in$ $\mathbf U$; and,
for each $J$ $\in$ $\mathbf{P}$, there exists  $n$ $\in$ $\IN_+$  
such that  $(J^d)_a$ $=$ $(I^n)_a$,
by \cite[(2.8) and (2.9)]{MRS}.}
\end{rema}

\begin{rema}
\label{rema22}
{\em With  $d$  as in (\ref{rema2})(d), the set
of integers  $d\mathbf U \cup \{0\}$ is a numerical
semigroup in the sense of Herzog-Kunz \cite{HK} and Watanabe \cite{W} that is naturally
associated to $I$. We denote this semigroup by $S(I)$. It is an
invariant of the projective equivalence class of $I$ in the sense
that if $J$ is projectively equivalent to $I$, then $S(J) = S(I)$.
Thus $S(I)$ is an invariant of $\mathbf P(I)$.
We are interested in considering properties of this semigroup.}
\end{rema}



In \cite [Section 2] {It}, Itoh gave the following construction, 
that will be used below to gain some
information concerning  $\mathbf W$  and  $\mathbf U$.

\begin{prop}
\label{itoh}
Let  $I$  be a regular ideal in a Noetherian ring  $R$, 
let  $n$  $\in$ $\IN_+$, let  $\mathbf R$
$=$ $R[u,tI]$, let  $\mathbf S$ $=$ $\mathbf {R}[u^{1/n}]$, let  $\mathbf T$ $=$
$\mathbf {S}' \cap R[u^{1/n},t^{1/n}]$, and let  $I_{[k/n]}$ $=$
$u^{k/n} \mathbf {T} \cap R$.  Then
$I_{[k/n]}$
$=$ $I_{k/n}$  for all  $k$ $\in$ $\IN_+$.
\end{prop}

\begin{proof}
$\mathbf {T}$  is a graded subring of  $R[u^{1/n},t^{1/n}]$.
Also, if  $x$ $\in$ $R$  and  $k$ $\in$ $\IN_+$, then  $x$ $\in$ $I_{[k/n]}$
$=$ $u^{k/n} \mathbf T \cap R$  if and only if  $xt^{k/n}$ $\in$ $\mathbf T$
if and only if  $x^nt^k$ $\in$ $\mathbf R' \cap R[u,t]$ (for  $\mathbf T$  is
an integral extension of  $\mathbf R$, so  $x^nt^k$  is integral
over  $\mathbf R$  and is in  $R[u,t]$).
And  $x^nt^k$ $\in$ $\mathbf R' \cap R[u,t]$  if and only 
if  $x^n$ $\in$ $u^k(\mathbf R' \cap R[u,t]) \cap R$ $=$ $(I^k)_a$  if and
only if  $x$ $\in$ $I_{k/n}$  (since  $x^n$ $\in$ $(I^k)_a$ $=$
$I_k$  (by (\ref{rema2})(a)) if and only if  
$\overline v _I(x^n)$ $\ge$ $k$  (by (\ref{defi4}.1)) if and only
if  $n\overline v_I(x)$ $\ge$ $k$  if and only
if  $\overline v_I(x)$ $\ge$ $k/n$  if and only if  
$x$ $\in$ $I_{k/n}$, by (\ref{defi4}.1))).
\end{proof}


\begin{rema}
\label{rema5}
{\em Let  $I$  be a regular ideal in a Noetherian ring  $R$.  Then:

\noindent
{\bf{(\ref{rema5}.1)}}
In  (\ref{itoh})  let  $n$ $=$ $d$ (with  $d$  as  in (\ref{rema2})(d)) 
and let  $\mathbf V_1$ $=$ $\{k/d \mid k \in \IN_+\}$.
Then  $\mathbf U$ $\subseteq$ $\mathbf V_1$  and  
$\mathbf V_1 \setminus  \mathbf U$  is a finite set.

\noindent
{\bf{(\ref{rema5}.2)}}
In  (\ref{itoh})  let  $n$ $=$ $e_1 \cdots e_g$ 
(with  $e_1,\dots, e_g$  as  in (\ref{rema1})(c)) and
let  $\mathbf V_2$ $=$ $\{k/(e_1 \cdots e_g) \mid k \in \IN_+\}$.
Then $\mathbf W$ $\subseteq$ $\mathbf V_2$.

\noindent
{\bf{(\ref{rema5}.3)}}
$\{I_{[k/n]}\}_{k \ge 1}$  is a filtration of integrally closed ideals on  $R$.
Therefore if  $R$  is an analytically unramified semi-local ring, 
then for all large  $k$ $\in$ $\IN_+$  it holds
that  $I_{mk/n}$ $=$ ${I_{k/n}}^m$  for all  $m$ $\in$ $\IN_+$, so  $I_{k/n}$  is
a normal ideal (that is, all powers of  $I_{k/n}$  are integrally closed).}
\end{rema}



\begin{proof}
(\ref{rema5}.1) is clear by (\ref{rema2})(d) together with (\ref{itoh}),
and (\ref{rema5}.2) is clear by the definitions of $e_1,\dots,e_g$ 
and  $\mathbf W$ together with (\ref{itoh}).

For (\ref{rema5}.3), $u^{k/n} \mathbf T$  is integrally closed, so  $I_{[k/n]}$  is
integrally closed, so the conclusion follows 
from (\ref{rema2})(a), (\ref{rema5}.1), \cite[(4.4.3)]{OR}, 
and \cite[Theorem (5.2) and (4.5)]{Rax}.
\end{proof}

\begin{rema}
\label{raru}
{\em It is shown in \cite{RR} that the set  $\mathbf{P}(I)$  (together with  $R$) forms
a subfiltration  $f^*$  of the filtration  $e$ $=$ $\{I_{i/d}\}_{i \ge 0}$  
of Proposition \ref{itoh},
as does $f$ $=$ $\{(I^i)_a\}_{i \ge 0}$, and the graded 
subring of  $\mathbf R$ $=$ $R[u,te]$
($=$ $R[u,tI_{1/d},t^2I_{2/d},\dots ]$) generated 
by either of the filtrations  $f^*$
and  $g$  has homogeneous prime spectrum isomorphic to the homogeneous
prime spectrum of  $\mathbf R$; however, if  $I$  is not projectively full 
(see Definition \ref{defifull}),
then the homogeneous prime spectra of the Rees rings of  $f^*$ 
 and  $f$  are not isomorphic.}
\end{rema}


We next note some things concerning  $\mathbf U$  (see (\ref{defi4}.3)) 
and $n^*$ and  $d$  (with  
$n^*$  and  $d$  as in (\ref{rema2})(d)) (recall that  
$\mathbf U$, $n^*$, and  $d$  depend on  $I$).
For this, let  $\mathbf U$ $=$ $\{\alpha_1,\alpha_2,\dots \}$  (with  $\alpha_1$ $<$
$\alpha_2$ $<$ $\cdots$), let  
$\mathbf P$ $=$ $\{I_{\alpha_1},I_{\alpha_2},\dots \}$  (so  $I_{\alpha_1}$
$\supsetneq$ $I_{\alpha_2}$ $\supsetneq$ $\cdots$),
and assume that  $\alpha_i$ $=$ $1$  (that is, assume that  $I_{\alpha_i}$  is
the ideal in  $\mathbf{P}$  that is the integral closure of  $I$).  
Then in what follows  $\mathbf U$, $n^*$, $d$, and  $\mathbf P$
will be denoted  $\mathbf {U}(I)$, $n^*(I)$, $d(I)$, and $\mathbf {P}(I)$
(so  $\{ \alpha \in \mathbf{U}(I) \mid \alpha \ge n^*(I) \}$ 
$=$ $\{n^*(I) + \frac {k}{d(I)} \mid k \in \IN\}$
and for each  $\alpha$ $\in$ $\mathbf{U}(I)$  it holds that  
$({I_\alpha}^{d(I)})_a$ $=$ $(I^n)_a$  for some  $n$ $\in$ $\IN_+$
(by (\ref{rema2})(d)).
For $J$ $\in$ $\mathbf{P}(I)$  let
$\mathbf {U}(J)$ $=$ $\{\beta_1,\beta_2,\dots\}$  
(with  $\beta_1$ $<$ $\beta_2$ $<$ $\cdots$), $n^*(J)$, $d(J)$, and $\mathbf {P}(J)$
$=$ $\{J_{\beta_1},J_{\beta_2},\dots \}$  
(so  $J_{\beta_1}$ $\supsetneq$ $J_{\beta_2}$ $\supsetneq$ $\cdots$) 
be defined analogously
(so  $\{ \beta \in \mathbf{U}(J) \mid \beta \ge n^*(J) \}$ 
$=$ $\{n^*(J) + \frac {k}{d(J)} \mid k \in \IN\}$
and for each  $\beta$ $\in$ $\mathbf{U}(J)$  
it holds that  $({J_\beta}^{d(J)})_a$ $=$ $(J^n)_a$  for some  $n$ $\in$ $\IN_+$).


\begin{rema}
\label{remaU0}
{\bf{(\ref{remaU0}.1)}}
{\em \cite[(2.11)]{MRS}  If  $\delta$ $\in$ $\mathbf{U}(I)$  
and  $J$ $=$ $I_\delta$ $\in$ $\mathbf{P}(I)$,
then  $\mathbf{U}(J)$ $=$ $\{\frac{\alpha}{\delta} \mid \alpha \in \mathbf{U}(I)\}$.}


\noindent
{\bf{(\ref{remaU0}.2)}}
{\em \cite[(2.3)]{MRS}  If  $(I^i)_a$ $=$ $(J^j)_a$, 
then  $\frac{i}{j}$ $\in$ $\mathbf{U}(I)$  and
$J$ $=$ $I_{\frac{i}{j}}$.  Also, if  $m,n$ $\in$ $\IN_+$  and
if  $\frac{m}{n}$ $\in$ $\mathbf{U}(I)$, then  
$({I_{\frac{n}{m}}}^m)_a$ $=$ $(I^n)_a$.}
\end{rema}

\begin{prop}
\label{propU}
Let  $I$  be a regular ideal in a Noetherian ring  $R$  and let  
$J$ $\in$ $\mathbf{P}(I)$.  Then:

\noindent
{\bf{(\ref{propU}.1)}}
$\mathbf {P}(I)$ $=$ $\mathbf{P}(J)$.

\noindent
{\bf{(\ref{propU}.2)}}
$(I^{d(J)})_a$ $=$ $(J^{d(I)})_a$.

\noindent
{\bf{(\ref{propU}.3)}}
If  $H,J$ $\in$ $\mathbf{P}(I)$
and if  $(J^j)_a$ $=$ $(H^h)_a$, then  $jd(J)$ $=$ $hd(H)$
and  $(J^{d(H)})_a$ $=$ $(H^{d(J)})_a$.

\noindent
{\bf{(\ref{propU}.4)}}
If  $\delta$ $\in$ $\mathbf{U}(I)$, then  $\delta$ $=$ $\frac{d(I_\delta)}{d(I)}$.

\noindent
{\bf{(\ref{propU}.5)}}
If  $H,J$ $\in$ $\mathbf{P}(I)$ and if  $J$ $\subsetneq$ $H$, then $d(J)$ $>$ $d(H)$.

\noindent
{\bf{(\ref{propU}.6)}}
If  $H,J$ $\in$ $\mathbf{P}(I)$ and if  $J$ $\subsetneq$ $H$, 
then  $n^*(H)$ $\ge$ $n^*(J)$.  Also,
$n^*(I_{\alpha_j})$ $=$ $1$  for all  $\alpha_j$ $\ge$ $n^*(I)$.

\end{prop}

\begin{proof}
For (\ref{propU}.1), by definition  $\mathbf {P}(I)$  
(resp., $\mathbf{P}(J)$)  is the set of
integrally closed ideals in  $R$  
that are projectively equivalent to  $I$  (resp., $J$).
Since projective equivalence is an equivalence relation, 
and since  $I$  and  $J$  are projectively
equivalent, it follows that  $\mathbf{P}(I)$ $=$ $\mathbf{P}(J)$.

For (\ref{propU}.2), by (\ref{rema2})(d)
it follows that  $(J^{d(I)})_a$ $=$ $(I^n)_a$  for some  $n$ $\in$ $\IN_+$.  
By raising both sides of this
equality to a large power (say the  $k$-th power) it may be assumed 
that  $kd(I)$ $\ge$ $n^*(J)$  and
$kn$ $\ge$ $n^*(I)$.  Therefore if  $m$ $\in$ $\IN_+$  is such that  
$m$ $\ge$ $min \{kd(I),kn\}$,
then there are exactly  $d(J)-1$  ideals in  $\mathbf{P}(J)$  strictly 
between  $(J^m)_a$  and  $(J^{m+1})_a$  (by
(\ref{rema2})(d)) and
there are exactly  $d(I)-1$  ideals in  $\mathbf{P}(I)$  strictly between  
$(I^m)_a$  and  $(I^{m+1})_a$  (by
(\ref{rema2})(d)).
Therefore, it follows that
in the chain  $(J^{kd(I)})_a$
$\supsetneq$ $(J^{kd(I)+1})_a$ $\supsetneq$ $\cdots$ 
$\supsetneq$ $(J^{(k+1)d(I)})_a$
there are exactly $d(I)d(J)-1$ ideals in  $\mathbf{P}(J)$  
strictly between $(J^{kd(I)})_a$  and $(J^{(k+1)d(I)})_a$,
and in the chain
$(I^{kn})_a$ $\supsetneq$ $(I^{kn+1})_a$ 
$\supsetneq$ $\cdots$ $\supsetneq$ $(I^{(k+1)n})_a$
there are exactly $nd(I)-1$ ideals in  $\mathbf{P}(I)$  
strictly between $(I^{kn})_a$  and $(I^{(k+1)n})_a$.
Since  the first and last ideals in these two chains  
are the same (namely, $(I^{kn})_a$ $=$ $(J^{kd(I)})_a$ and  $(I^{(k+1)n})_a$
$=$ $(J^{(k+1)d(I)})_a$),
and since  $\mathbf{P}(J)$ $=$ $\mathbf{P}(I)$  (by (\ref{propU}.1))
(and since  $\mathbf{P}(J)$  and
$\mathbf{P}(I)$  are linearly ordered by inclusion),
it follows that  $nd(I)-1$ $=$ $d(J)d(I)-1$,
hence  $n$ $=$ $d(J)$.

For (\ref{propU}.3), it follows from (\ref{propU}.2) that  
$(J^{d(I)})_a$ $=$ $(I^{d(J)})_a$  and  $(H^{d(I)})_a$
$=$ $(I^{d(H)})_a$.  Also, $(J^j)_a$ $=$ $(H^h)_a$, by hypothesis, so
it follows that  $(I^{jd(J)})_a$ $=$
$(J^{jd(I)})_a$ $=$
$(H^{hd(I)})_a$ $=$
$(I^{hd(H)})_a$.  Therefore  $jd(J)$ $=$ $hd(H)$, and  
$(J^j)_a$ $=$ $(H^h)_a$  (by hypothesis), so (\ref{rema3}.2) 
shows that  $(J^{d(H)})_a$ $=$ $(H^{d(J)})_a$.

For (\ref{propU}.4),
let  $\delta$ $\in$ $\mathbf{U}(I)$, so  
$\delta$ $=$ $\frac{m}{d(I)}$  for some  $m$ $\in$ $\IN_+$,
by (\ref{rema2})(d).
Then  $I_{\delta}$ $=$ $I_{\frac{m}{d(I)}}$ $\in$ $\mathbf{P}(I)$
and  $({I_{\delta}}^{d(I)})_a$ $=$
$(I^m)_a$  (by (\ref{remaU0}.2))  and  
$({I_\delta}^{d(I)})_a$ $=$ $(I^{d(I_\delta)})_a$
(by (\ref{propU}.3)),
so  $m$ $=$ $d(I_\delta)$,
hence  $\delta$ $=$ $\frac{d(I_{\delta})}{d(I)}$.

For (\ref{propU}.5),
since  $J$ $\subsetneq$ $H$, it follows from (\ref{propU}.3) that  
$(H^{d(J)})_a$ $=$ $(J^{d(H)})_a$ $\subsetneq$ $(H^{d(H)})_a$, so
$d(J)$ $>$ $d(H)$.


For (\ref{propU}.6), let
$\delta$ $<$ $\gamma$  in  $\mathbf{U}(I)$  and let  
$H$ $=$ $I_{\delta}$  and  $J$ $=$ $I_{\gamma}$  in  $\mathbf{P}(I)$.
Then  $H$ $\supsetneq$ $J$, $\delta$ $=$ $\frac{d(H)}{d(I)}$  
and  $\gamma$ $=$ $\frac{d(J)}{d(I)}$  (by (\ref{propU}.4)),
and  $d(H)$ $<$ $d(J)$.
Let  $\alpha$ $\in$ $\mathbf{U}(I)$, so  
$\alpha$ $=$ $\frac{d(I_{\alpha})}{d(I)}$  (by (\ref{propU}.4)),
and (\ref{remaU0}.1) shows that
$\frac{\alpha}{\delta}$ $=$ $\frac{(d(I_\alpha))/(d(I))}{(d(H))/(d(I))}$
$=$ $\frac{d(I_{\alpha})}{d(H)}$ $\in$ $\mathbf{U}(H)$ and
$\frac{\alpha}{\gamma}$ $=$ $\frac{(d(I_\alpha))/(d(I))}{(d(J))/(d(I))}$
$=$ $\frac{d(I_{\alpha})}{d(J)}$ $\in$ $\mathbf{U}(J)$.

Now for $\beta$ $\ge$ $n^*(H)$  in  $\mathbf{U}(H)$  there exists  $m$ $\in$
$\IN$  such that  $\beta$ $=$  $n^*(H)+\frac{m}{d(H)}$.  
Therefore for  $m$ $\in$ $\IN$  define
$\beta_m$ by  $\beta_m$
$=$ $n^*(H) + \frac{m}{d(H)}$  $=$ $\frac{n^*(H)d(H)+m}{d(H)}$, 
so  $\beta_m$ $\in$ $\mathbf{U}(H)$
(by the definition of  $n^*(H)$  and  $d(H)$), $\beta_m \frac{\delta}{\gamma}$
$\in$ $\mathbf{U}(J)$  (by (\ref{remaU0}.1)), and $\beta_m \frac{\delta}{\gamma}$ $=$
$\beta_m \frac{d(H)}{d(J)}$ $=$ $\frac{n^*(H)d(H)+m}{d(J)}$.  
Since  $d(H)$ $<$ $d(J)$, let  $z$ $\in$ $\IN_+$
such that  $d(H)$ $=$ $d(J)- z$, so
$\frac{n^*(H)d(H)+m}{d(J)}$ $=$ $\frac{n^*(H)(d(J)-z)+m}{d(J)}$
$=$ $(n^*(H) + \frac{-zd(H)}{d(J)}) + \frac{m}{d(J)}$, 
and this holds for all  $m$ $\in$ $\IN$.
Therefore define  $n^{~}$ to be $n^*(H)-w$, where  $w$  
is defined by  $wd(J)$ $=$ $zn^*(H) - r$  with  $r$ $\in$ $\IN$  such
that  $0$ $\le$ $r$ $<$ $d(J)$  (note that  $n^{~}$ $=$ $n^*(H)-w$ $\le$ $n^*(H)$).
Then it follows (from the preceding computation) 
that if  $m$ $\ge$ $r$  (say  $m$ $=$ $r+k$  with  $k$ $\in$ $\IN$),
then  
$$(*) \quad for~all~k \in \IN,~\beta_{r+k} = 
\frac{n^*(H)d(H)+(r+k)}{d(H)} \in \mathbf{U}(H)~~and~~ \beta_{r+k} 
\frac{\delta}{\gamma} = n^{~}+ \frac{k}{d(J)} \in \mathbf{U}(J).$$
\noindent
Therefore, to show that  $n^{~}$  is the desired  $n^*(J)$, it remains to show that:
(i)  for each  $k$ $\in$ $\IN$, $n^{~}+\frac{k}{d(J)}$ $\in$ $\mathbf{U}(J)$; and,
(ii) if  $\sigma$ $\in$ $\mathbf{U}(J)$  is such that  $\sigma$ $\ge$ $n^{~}$,
then  $\sigma$ $=$ $n^{~} + \frac{k}{d(J)}$  for some  $k$ $\in$ $\IN$.
However, (i) follows immediately from (*).
And for (ii), $\sigma$  can be written in the form  
$\frac{g}{d(J)}$  for some  $g$ $\in$ $\IN_+$,
so since  $\sigma$ $\ge$ $n^{~}$  it follows that  
$g$ $=$ $d(J)n^{~}+k$  for some  $k$ $\in$ $\IN$,
hence  $\sigma$ $=$ $\beta_{r+k} \frac{\delta}{\gamma}$  (by (*)).
Therefore  $n^*(J)$  may be taken to be  $n^{~}$, and then  $n^*(J)$ $\le$ $n^*(H)$.

For the final statement assume that  
$\delta$ $\ge$ $n^*(I)$ $\in$ $\mathbf{U}(I)$  and let
$\beta$ $\ge$ $1$  in  $\mathbf{U}(I_{\delta})$.
Then  $\delta$ $=$ $\frac{d(I_\delta)}{d(I)}$, by (\ref{propU}.4),
and  $\beta$ $=$ $\frac{\alpha}{\delta}$  for some  
$\alpha$ $\ge$ $\delta$ (with  $\alpha$ $\in$
$\mathbf{U}(I)$).
Therefore  $\alpha$ $=$ $\delta + \frac{m}{d(I)}$  
for some  $m$ $\in$ $\IN$  (since  $\delta$ $\ge$ $n^*(I)$),
so  $\beta$ $=$ $\frac{\alpha}{\delta}$ $=$ $1+\frac{m}{\delta d(I)}$ $=$ 
$1 + \frac{m}{d(I_\delta)}$ $\in$ $\mathbf{U}(I_\delta)$.
It follows that  $n^*(\delta)$  may be chosen to be  $1$.
\end{proof}





\begin{defi}
\label{defifull}
{\em A regular ideal $I$  in a Noetherian ring  $R$  is said to be
{\bf{projectively full}}
in case the only integrally closed ideals that are projectively 
equivalent to  $I$  are the ideals  $(I^k)_a$  with  $k$ $\in$ $\IN_+$.
If there exists $J \in \mathbf P(I)$ such that $J$ is projectively full,
then we say that $\mathbf P(I)$ is
{\bf{projectively full}}.  Such an ideal
$J$, if it exists, must be the largest element of  $\mathbf P (I)$.}
\end{defi}

\begin{rema}
\label{remax}
{\em Concerning (\ref{defifull}), note that it 
follows from (\ref{rema2})(d) that if the greatest
common divisor of
the integers $e_1,\dots,e_g$  of (\ref{rema1}) is $1$, 
then  $I$  is projectively full.  In particular,
if  $P$  is a prime ideal in  $R$  such that  $R_P$  is
a regular local ring, then  $P$  is projectively full (since
the integer  $e$  of (\ref{rema1}) is  $1$  for the
order valuation of  $R_P$). }
\end{rema}

\begin{prop}
\label{propfull}
The following are equivalent for a regular ideal  $I$  in a Noetherian ring  $R$:

\noindent
{\bf{(\ref{propfull}.1)}}
There exists  $K$ $\in$ $\mathbf{P}(I)$  that is projectively full.

\noindent
{\bf{(\ref{propfull}.2)}}
There exists  $K$ $\in$ $\mathbf{P}(I)$  such that $\mathbf{U}(K)$ $=$ $\IN_+$.

\noindent
{\bf{(\ref{propfull}.3)}}
There exists  $K$ $\in$ $\mathbf{P}(I)$  such that $d(K)$ $=$ $1$.

\noindent
{\bf{(\ref{propfull}.4)}}
$I_{1/d(I)}$ $\in$ $\mathbf{P}(I)$.

\end{prop}



\begin{proof}
Assume that (\ref{propfull}.1) holds, let  $\beta$ $\in$ $\mathbf{U}(K)$, 
and let  $J$ $=$ $K_{\beta}$.
Then  $J$  is an integrally closed ideal that is projectively 
equivalent to  $K$, so  there exists  $k$ $\in$ $\IN_+$  such
that  $J$ $=$ $(K^k)_a$  (by hypothesis)  and  $(K^k)_a$
$=$ $K_k$ (by (\ref{remaU0}.2)), hence  $\beta$ $=$ $k$  
(since  $J$ $=$ $K_{\beta}$).
Therefore  $\mathbf{U}(K)$ $\subseteq$ $\IN_+$, and the opposite 
inclusion is clear (since  $K_k$ $=$ $(K^k)_a$ $\in$ $\mathbf{U}(K)$  for
all  $k$ $\in$ $\IN_+$), hence (\ref{propfull}.1) $\Rightarrow$ (\ref{propfull}.2).

Assume that (\ref{propfull}.2) holds.  Now for all  $k$ $\in$ $\IN_+$  it holds
that  $n^*(K) + \frac {k}{d(K)}$ $\in$ $\mathbf{U}(K)$ $=$ $\IN_+$.
Therefore  $d(K)$ $=$ $1$, so(\ref{propfull}.2) $\Rightarrow$ (\ref{propfull}.3).

Assume that (\ref{propfull}.3) holds.  Then  $(I ^{d(K)})_a$ $=$ $(K^{d(I)})_a$,
by (\ref{propU}.2), so  $I_{1/d(I)}$ $=$ $I_{d(K)/d(I)}$  
(by hypothesis) $=$ $K$, by (\ref{rema2})(a),
and  $K$ $\in$ $\mathbf{P}(I)$, by hypothesis.  Therefore  $I_{1/d(I)}$ $\in$
$\mathbf{P}(I)$, so (\ref{propfull}.3) $\Rightarrow$ (\ref{propfull}.4).

Finally, assume that (\ref{propfull}.4) holds  and let  $J$ $\in$
$\mathbf P(I)$.  Then  $(J^{d(I)})_a$ $=$ $(I^{d(J)})_a$, by (\ref{propU}.2), and  
$(I^{d(J)})_a$ $=$ $I_{d(J)}$, by (\ref{rema2}.2)(a), so  $J$ $=$ $I_{d(J)/d(I)}$
(by (\ref{remaU0}.2)) $=$ $({I_{1/d(I)}}^{d(J)})_a$  (by (\ref{rema2}.2)(a)).
Therefore, since  $I_{1/d(I)}$ $\in$ $\mathbf P(I)$, it follows from (\ref{defifull})
that  $\mathbf {P}(I)$  is projectively full, hence
(\ref{propfull}.4) $\Rightarrow$ (\ref{propfull}.1).
\end{proof}

\begin{rema}{\em Let $I$ be a regular ideal of a 
Noetherian ring $R$.  Then
it follow from (\ref{rema2}.2)(d) that:

\noindent
  (a)  $I$  is projectively full if and only if  $d(I)$ $=$ $1$.

\noindent
      (b)  $I$  is projectively full if and only if there
 exists a large  $k$ $\in$ $\IN_+$  such that
           $I_k$ $\supsetneq$ $I_{k+1}$  are consecutive in  
$\mathbf P(I)$  if and only if this
           holds for all $k$ $\in$ $\IN_+$.}
\end{rema}



\begin{rema} \label{4.13}
{\em   Let $I$ be a nonzero proper ideal of a 
two-dimensional regular local domain $R$
and let $I_a = J_1^{e_1} \cdots J_n^{e_n}$ 
be the factorization of $I_a$ as a product of distinct 
simple complete ideals. 
Let $d$ be the greatest common divisor of 
$e_1, \ldots , e_n$ and let $f_i = e_i/d$
for $i = 1, \ldots , n$. Then 
$K = J_1^{f_1} \cdots J_n^{f_n}$ is projectively
full and $K^d = I_a$, so 
$\mathbf P(I) = \mathbf P(K)$. Therefore  $\mathbf P(I)$ 
is projectively full for every nonzero proper ideal of a 
two-dimensional regular local domain.}



{\em
In general, if $R$ is a 
two-dimensional normal local domain with maximal ideal $M$ 
and $I$ is an $M$-primary ideal, the set $\mathbf P(I)$ need
not contain a projectively full ideal as we illustrate in
Example \ref{4.14}.  In this example, the ideal $I$
has only one Rees valuation and the numerical semigroup $S(I)$
associated to $I$ is $(2, 3)\IN$.}  
\end{rema}  




\begin{exam} \label{4.14}
{\em Let $k$ be an algebraically closed field of characteristic zero
and let $R = k[[x, y, z]]$, where $z^2 = x^3 + y^7$. It is readily 
seen that $R$ is a two-dimensional normal local domain.
Consider the
ideal $I = (x, y^2)R$. It is shown in \cite[Example 16, page 300]{HL}
that $I$ has a unique  Rees valuation ring $V$. Therefore 
$xV = y^2V$, $I_a  = xV \cap R$, 
and the image of $x/y^2$ in the residue field of $V$ is
transcendental over $k$. The equality $(z/y^3)^2 = (x/y^2)^3 + y$
implies that $z/y^3$ is integral over $R[x/y^2]$. It also implies that
$zV = y^3V$ and that the image of $z/y^3$ in the residue field of $V$
is transcendental over $k$. Let $J = y^3V \cap R$. Then $(z, y^3, xy, x^2)R
\subseteq J$. To show $I$ and $J$ are projectively equivalent it suffices
to show that $I^3$ and $J^2$ have the same integral closure. Since 
$I = (x, y^2)R$ and $V$ is the unique Rees valuation of $I$, 
the integral closure of $I^3$ is $I^3V \cap R$ and is
the integral closure of $(x^3, y^6)R$. We have $J^2V = I^3V$. 
Therefore $(J^2)_a \subseteq (I^3)_a$. To show the reverse inclusion,
it suffices to observe that $x^3$ and $y^6$ are in $J^2$. Since 
$y^3 \in J$, it is clear that $y^6 \in J^2$. Also we have 
$x^3 = z^2 - y^7$ and $z^2$ and $y^7$ are in $J^2$, so $x^3 \in J^2$.
Therefore $I$ and $J$ are projectively equivalent. Notice
that $I_a = (y^2, x, z)R \subsetneq M$ and there are no 
ideals properly between $I_a$ and $M$. To complete the proof
that $\mathbf P(I)$ is not projectively full, it suffices to 
observe that $V$ is not a Rees valuation of $M$. Since $(x, y)R$
is  a reduction of $M$, the Rees valuation rings of $M$ are all
extensions of the order valuation defined by the powers of the
maximal ideal of the two-dimensional regular local subdomain
$k[[x, y]]$ of $R$. In particular, if $W$ is a Rees valuation  ring
of $M$, then $xW = yW$. Since $xV = y^2V$, $V$ is not a Rees 
valuation ring of $M$. Let $v$ denote the normalized valuation 
with value group $\mathbb Z$ associated to the valuation domain $V$.
We have $v(y)=1$, $v(x)=2$, $v(z)= 3$,
$v(I)$ $=$ $2$ $=$ $d$, $v(J)$ $=$ $3$, and  
$\mathbf P(I)$ $=$ $\{I,J,(I^2)_a,(IJ)_a,(I^3)_a = (J^2)_a,(I^2J)_a, \dots\}$.
}
\end{exam}

\medskip

An interesting  question we have not been successful 
in answering is 
whether  for a regular ideal   $I$  in a Noetherian ring  $R$ 
there always exists a finite integral extension ring  $T$  of  $R$  
such that  $\mathbf{P}(IT)$
contains a projectively full ideal. If $R$ is a one-dimensional 
Noetherian domain, then the integral closure $R'$ of $R$ is a 
Dedekind domain and it is easily seen that $\mathbf P(IR')$ 
contains a projectively full ideal.



\medskip

\begin{rema}
{\em
Let $I$ be a nonzero proper ideal of a Noetherian
integrally closed domain $R$. 
 With  $d$  as in (\ref{rema2}.2)(d), it is natural to ask if
there exist  $x$ $\in$ $R$  
such that  $\overline v_I(x)$ $=$ $1/d$. 
To illustrate that this
is not true in general, let $s, t$ be algebraically independent 
elements over the field $k$ and let $R = k[s, t]$. 
Consider the ideal $I = (s^2, t^3)R$. We observe that there
exists a unique $V \in \Rees I$. Indeed, for $V \in \Rees I$
we have $s^2V = t^3V$ and the image of $s^2/t^3$ in the residue
field of $V$ is transcendental over $k$. Therefore $z = s/t$ is
in the maximal ideal of $V$. We have $s = tz$ and $R[z] = k[t, z]$.
Moreover, $I_a = (s^2, st^2, t^3)R$ and $V \in \Rees I$ is
centered on a height-one prime ideal of $R[s^2/t^3, st^2/t^3 = s/t]$
that lies over the maximal ideal $(s, t)R$ of $R$.
Since $s^2/t^3 = z^2t^2/t^3 = z^2/t$, we see that 
$V$ is a localization of $k[t, z][z^2/t]$ at a height-one prime
ideal that contains $M = (t, z)k[t, z]$. Since $P = Mk[t, z][z^2/t]$ 
is a height-one prime ideal, we see that $V$ is the localization
of $k[t, z][z^2/t]$ at $P$. Let $v$ denote the normalized valuation
associated to $V$. Then $v(t) = 2$, $v(s) = 3$, and $v(I) = 6$. 
The integer $d$ of (\ref{rema2}.2)(d) is $6$, while for $x \in R$ the smallest 
possible positive value of $\overline v_I(x)$
is $1/3$. Indeed,  
$\overline v_I(t) = 1/3$ and $\overline v_I(s) = 1/2$,
and therefore $d = 6$.
}
\end{rema}

\begin{rema}
{\em
Let $H, I, K$ and $J$ be ideals of a Noetherian domain $R$. 
In analogy to a result that holds for reductions of ideals,
it is natural to ask whether   $H$  is projectively equivalent to  $I$  and  
$K$  is projectively equivalent to  $J$ implies  that 
$H+K$ is  projectively equivalent to  $I+J$.   
To illustrate that this
is not true in general, let $s, t$ be algebraically independent 
elements over the field $k$ and let $R = k[s, t]$. 
Let $H = (s, t^2)R$ and $I = (s^2, t^4)R$. Also
let $K = J = (s^2, t)R$. Then $H$ and $I$ are 
projectively equivalent as are also $K$ and $J$, 
but $H + K = (s, t)R$ is not projectively equivalent 
to $I + J = (s^2, t)R$. 
}
\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}





