\documentclass[11pt]{article}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amsmath}
\oddsidemargin0.30in
\evensidemargin0in
%\topmargin-0.10in
\topmargin0.00in
%\topmargin0.40in
%\textwidth5.6in
%\textwidth5.8in
\textwidth6.0in
%\textheight8.30in
%\textheight8.20in
\textheight8.25in
%\textheight9.25in
\newcommand{\Z}{\mathbb {Z}}
\newcommand{\Q}{\mathbb {Q}}
\newcommand{\C}{\mathbb {C}}
\newcommand{\R}{\mathbb {R}}
%\newcommand{\N}{\mathbb {N}}
\newcommand{ \Spec}{\mbox{\textup{Spec}}}
\newcommand{ \HSpec}{\mbox{\textup{HSpec}}}
\newcommand{ \Cl}{\mbox{\textup{Cl}}}
\newcommand{ \Ass}{\mbox{\textup{Ass}}}
\newcommand{ \Rad}{\mbox{\textup{Rad}}}
\newcommand{ \rad}{\mbox{\textup{rad }}}
\newcommand{ \Ker}{\mbox{\textup{Ker}}}
\newcommand{ \hgt}{\mbox{\textup{ht }}}
\newcommand{ \trd}{\mbox{\textup{trd}}}
\newcommand{ \height}{\mbox{\textup{ht}}}
\newcommand{ \depth}{\mbox{\textup{depth}}}
\newcommand{ \grade}{\mbox{\textup{grade}}}
\newcommand{ \degree}{\mbox{\textup{degree}}}
\newcommand{ \Grade}{\mbox{\textup{Grade}}}
\newcommand{ \Rees}{\mbox{\textup{Rees }}}
\newcommand{ \sdim}{\mbox{\textup{sdim}}}
\newcommand{ \qed}{\mbox{\textup{QED}}}
%\newcommand{ \enddemo}{\mbox{\textup{QED}}}
\newcommand{\demo}[1]{{\bf #1}}
\newcommand{ \altitude}{\mbox{\textup{altitude}}}
\newcommand{ \irr}{\mbox{\textup{irr}}}
\newenvironment{proof}[1][Proof]{\textbf{#1.}
}{\ \rule{0.5em}{0.5em}}
\def\inbar{\,\vrule height1.5ex width.4pt depth0pt}
\def\inbar{\,\vrule height1.5ex width.4pt depth0pt}
\def\IB{\relax{\rm I\kern-.18em B}}
\def\IC{\relax\hbox{$\inbar\kern-.3em{\rm C}$}}
\def\ID{\relax{\rm I\kern-.18em D}}
\def\IE{\relax{\rm I\kern-.18em E}}
\def\IF{\relax{\rm I\kern-.18em F}}
\def\IG{\relax\hbox{$\inbar\kern-.3em{\rm G}$}}
\def\IH{\relax{\rm I\kern-.18em H}}
\def\II{\relax{\rm I\kern-.18em I}}
\def\IK{\relax{\rm I\kern-.18em K}}
\def\IL{\relax{\rm I\kern-.18em L}}
\def\IM{\relax{\rm I\kern-.18em M}}
\def\IN{\relax{\rm I\kern-.18em N}}
\def\IO{\relax\hbox{$\inbar\kern-.3em{\rm O}$}}
\def\IP{\relax{\rm I\kern-.18em P}}
\def\IQ{\relax\hbox{$\inbar\kern-.3em{\rm Q}$}}
\def\IR{\relax{\rm I\kern-.18em R}}
\font\cmss=cmss10 \font\cmsss=cmss10 at 7pt
\def\IZ{\relax\ifmmode\mathchoice
{\hbox{\cmss Z\kern-.4em Z}}{\hbox{\cmss Z\kern-.4em Z}}
{\lower.9pt\hbox{\cmsss Z\kern-.4em Z}}
{\lower1.2pt\hbox{\cmsss
Z\kern-.4em Z}}\else{\cmss Z\kern-.4em Z}\fi}
\def\IGa{\relax\hbox{${\rm I}\kern-.18em\Gamma$}}
\def\IPi{\relax\hbox{${\rm I}\kern-.18em\Pi$}}
\def\ITh{\relax\hbox{$\inbar\kern-.3em\Theta$}}
\def\IOm{\relax\hbox{$\inbar\kern-3.00pt\Omega$}}
\begin{document}
\baselineskip 20pt
%\baselineskip 19pt
\pagenumbering{arabic}
\pagestyle{plain}
\newtheorem{defi}{Definition}[section]
\newtheorem{theo}[defi]{Theorem}
\newtheorem{lemm}[defi]{Lemma}
\newtheorem{prop}[defi]{Proposition}
\newtheorem{note}[defi]{Note}
\newtheorem{nota}[defi]{Notation}
\newtheorem{exam}[defi]{Example}
\newtheorem{coro}[defi]{Corollary}
\newtheorem{rema}[defi]{Remark}
\newtheorem{cons}[defi]{Construction}
\newtheorem{ques}[defi]{Question}
\newcommand{\abs}{${\bar A}^*}
\newcommand{\qbs}{${\bar Q}^*}
\newcommand{\be}{\begin{enumerate}}
\newcommand{\ee}{\end{enumerate}}
%\newcommand{\demo}[1]{{\bf #1}}
\newcommand{\fany}{\rm for\ \ any\ \ }
%\newcommand{\dfrac}{\displaystyle\frac}
\newcommand{\rb}{\overline{R}}
\newcommand{\mt}{\overline{M}}
\newcommand{\nt}{\overline{N}}
\newcommand{\nb}{\widetilde{N}}
\newcommand{\mb}{\widetilde{M}}
\newcommand{\m}{\bf {m}}
%%% Technical Definitions
%%%%Phrases
\def\cm{Cohen-Macaulay}
\def\wrt{with respect to\ }
\def\pni{\par\noindent}
\def\wma{we may assume without loss of generality that\ }
\def\Wma{We may assume without loss of generality that\ }
\def\ets{it suffices to show that\ }
\def\bwoc{by way of contradiction}
\def\iff{if and only if\ }
\def\st{such that\ }
\def\fg{finitely generated}
%%%%Letters
\def\R{\Cal R}
\def\G{\Cal G}
\def\F{\Cal F}
\def\a{\goth a}
\def\Z{\Bbb Z}
\def\M{\Cal M}
\def\N{\goth N}
\def\P{\Cal P}
\def\I{\Cal I}
\def\J{\Cal J}
\def\Q{\Cal Q}
\def\p{\mathbf p}
\def\isom{\thinspace \cong\thinspace}
\def\rtar{\rightarrow}
\def\rta{\rightarrow}
\def\l{\lambda}
\def\d{\Delta}
%%%%%Operators
%\baselineskip 20pt
\title{PROJECTIVELY FULL IDEALS IN NOETHERIAN RINGS}
\author{Catalin Ciuperca, William J. Heinzer, Louis J. Ratliff Jr., and
David E. Rush}
\date{}
\maketitle
%\keywords
%{}
%\endkeywords
%\subjclass AMS (MOS) Subject Classification Numbers: Primary:
%13A18, 13A30, 13B02, 13E05;
%\endsubjclass
%\endtopmatter
\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
\begin{thebibliography}{99}
\bibitem{CHRR}
C. Ciuperca, W. J. Heinzer, L. J. Ratliff, Jr., and D. E. Rush, {\em
Projectively equivalent ideals and Rees valuations}, J. Algebra 282
(2004), 140--156.
\bibitem{G}
S. Goto,
{\em Integral closedness of complete-intersection ideals},
J. Algebra 108 (1987), 151--160.
\bibitem{HRW} W. J. Heinzer, C. Rotthaus and S. Wiegand,
{\em Integral closures of ideals in completions of regular local domains},
to appear in the proceedings of Sevilla-Lisboa.
\bibitem{KMOR}
D. Katz, S. McAdam, J. S. Okon, and L. J. Ratliff, Jr.,
{\em Essential prime divisors and projectively equivalent ideals},
J. Algebra 109 (1987), 468-478.
\bibitem{KR}
D. Katz and L. J. Ratliff, Jr.,
{\em Form rings and projective equivalence},
Math. Proc. Cambridge Philos. Soc. 99 (1986), 447-456.
\bibitem{M}
H. Matsumura,
{\em Commutative Ring Theory},
Cambridge Univ. Press, Cambridge, 1986.
\bibitem{McA}
S. McAdam,
{\em Asymptotic Prime Divisors},
Lecture Notes in Math. No. 1023,
Springer-Verlag, New York, 1983.
\bibitem{MR1}
S. McAdam and L. J. Ratliff, Jr.,
{\em Persistent primes and projective extensions of ideals},
Comm. Algebra 16 (1988), 1141-1185.
\bibitem{MR2}
S. McAdam and L. J. Ratliff, Jr.,
{\em Bounds related to projective equivalence classes of ideals},
J. Algebra 119 (1988), 23-33.
\bibitem{MR3}
S. McAdam and L. J. Ratliff, Jr.,
{\em The linearity of projective relations},
Houston J. Math. 16 (1990), 539-548.
\bibitem{MRS}
S. McAdam, L. J. Ratliff, Jr., and J. D. Sally,
{\em Integrally closed projectively equivalent ideals},
in Commutative Algebra,
MSRI Pub. 15, 1988, 391-405.
\bibitem{N1}
M. Nagata,
{\em Note on a paper of Samuel concerning asymptotic properties of ideals},
Mem. Coll. Sci. Univ. Kyoto, Ser. A Math. 30 (1957) 165-175.
\bibitem{N2}
M. Nagata,
{\em Local Rings},
Interscience, John Wiley, New York, 1962.
\bibitem{Ra1}
L. J. Ratliff, Jr.,
{\em On prime divisors of the integral closure of a principal ideal},
J. Reine Angew. Math. 256 (1972), 210-220.
\bibitem{Ra2}
L. J. Ratliff, Jr.,
{\em Essential sequences and projective equivalence},
J. Algebra 109 (1987), 381-393.
\bibitem{Ra3}
L. J. Ratliff, Jr.,
{\em Notes on projectively related ideals and residual division},
J. Algebra 130 (1990), 435-450.
\bibitem{Re}
D. Rees,
{\em Valuations associated with ideals (II)},
J. London Math. Soc. 36 (1956), 221-228.
\bibitem{Re2}
D. Rees,
{\em A note on form rings and ideals},
Mathematika 4 (1957), 51-60.
\bibitem{S}
P. Samuel,
{\em Some asymptotic properties of powers of ideals}
Annals of Math 56 (1952), 11-21.
\bibitem{ZS}
O. Zariski and P. Samuel,
{\em Commutative Algebra vol II},
Van Nostrand, New York, 1960.
\end{thebibliography}
\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}