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\begin{document}

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


\date{}
\maketitle

%\keywords
%{}
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%\subjclass AMS (MOS) Subject Classification Numbers:  Primary:
%13A18, 13A30, 13B02, 13E05;
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%\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
$I$  and 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 $K$ of $\mathbf
P(I)$. If this holds, the ideal $K$ and the set $\mathbf P(I)$ are
said to be projectively full. A special case of the main result in
this paper shows that if  $R$  contains the rational number field
$\IQ$, then there exists  a finite free integral extension ring  $A$
of  $R$  such that $\mathbf P(IA)$  is projectively full. If $R$ is
an integral domain, then the integral extension $A$ has the property
that  $\mathbf P((IA+z^*)/z^*)$  is projectively full for all minimal
prime ideals  $z^*$  in  $A$. Therefore in the case where $R$ is an
integral domain there exists a finite integral extension domain $B =
A/z^*$ of $R$ such that $\mathbf P(IB)$ is 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 $I \ne R$).  Recall that an ideal  $J$
in  $R$  is
{\bf{projectively equivalent}}
to  $I$  in case  $(J^j)_a$ $=$ $(I^i)_a$  for some positive
integers  $i$  and  $j$  (where  $K_a$  denotes the
integral closure in  $R$  of an ideal  $K$  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 $\mathbf P(I)$ of integrally closed ideals
projectively equivalent to $I$ is linearly ordered by inclusion and
is 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.

In \cite[(3.6)]{MRS} and in \cite[(4.13)]{CHRR} it is noted that
$\mathbf P(I)$ is  projectively full for each nonzero ideal $I$  in
a regular local ring of altitude two.  On the other hand, in
\cite{CHRR2} we give an example of an integrally closed local (Noetherian)  domain
$(L,M)$  of altitude two such that  $M$  (and hence $\mathbf P(M)$)
is not projectively full. We mention  in the paragraph just before
Proposition 4.3 of \cite{CHRR2} that a problem we have not been able
to solve is whether, for a given nonzero ideal $I$ of a Noetherian
domain $R$, there always exists a finite integral extension domain
$A$ of $R$ such that $\mathbf P(IA)$ is projectively full. In
\cite[Proposition 4.3]{CHRR2} we give a ``logical candidate'' for
$A$ and prove for this $A$ that there exists an ideal $H$ of $A$
such that every $J \in \mathbf P(I)$ has the property  that $(JA)_a$
is the integral closure of a power of $H$. A special case of Theorem
\ref{theo1} in the present paper shows that if $I$ is a  regular
proper ideal in a Noetherian ring  $R$ that contains the rational
number field, then there exists a finite integral extension ring  $A$ of
$R$ such that $\mathbf P(IA)$ is projectively full. To obtain in
Theorem \ref{theo1} such an  extension ring  $A$ of $R$, the additional
requirement needed in the construction given in Proposition 4.3 of
\cite{CHRR2} is that certain subsets of the Rees valuation rings of
$I$ are unramified with respect to the extension.

We now give a brief summary of the contents of this paper

In Section 2 we show in Theorem \ref{theo1}  that if  $I$ $=$
$(b_1,\dots,b_g)R$ and $\{(V_1,N_1),\dots,(V_n,N_n)\}$  is a
nonempty subset of $\Rees I$  such that: (a) $b_iV_j$ $=$ $IV_j$
($=$ ${N_j}^{e_j}$, say) for  $i$ $=$ $1,\dots,g$  and $j$ $=$
$1,\dots,n$; and, (b) the greatest common divisor  $c$  of
$e_1,\dots,e_n$ is a unit in  $R$; then $A$ $=$ $R[  {x_1},\dots,
{x_g}]$ ($=$ $R[ {X_1},\dots, {X_g}]/(({X_1}^c -
b_1,\dots,{X_g}^c-b_g))$) is a finite free integral extension ring
of  $R$  such that its ideal  $J$ $=$ $( {x_1},\dots, {x_g})A$ is
projectively full and projectively equivalent to  $I$, so $\mathbf
P(IA)$  is projectively full.  Also, if  $R$  is an integral domain
and if ${z_1}^*,\dots,{z_m}^*$  are the minimal prime ideals in $A$,
then $\mathbf P(IB_h)$  is projectively full for $h$ $=$
$1,\dots,m$, where  $B_h$ $=$ $A/{z_h}^*$. Then in Remark
\ref{rema1}.1  and Remark \ref{rema1}.2  it is shown that  $I$  has
a basis $b_1,\dots,b_g$ such that (a) holds if either  $R$  is local
with an infinite residue field,  or $n$ $=$ $1$. In Remark
\ref{rema1}.4  it is shown that (b) may be replaced with the weaker
assumption that  $c$ $\notin$ $(N_1 \cap R) \cup \cdots \cup (N_n
\cap R)$. Corollary \ref{2.7} states that if $R$ is a Noetherian
ring that contains the field of rational numbers, then for each
regular proper ideal $I$ of $R$ there exists a finite free integral
extension ring $A$ of $R$ such that $\mathbf P(IA)$ is projectively
full. If $R$ is an integral domain, there exists a finite integral
extension domain $B = A/z^*$ of $R$ such that $\mathbf P(IB)$ is
projectively full.




In Proposition \ref{remaprops1} of Section 3 we observe   the following:
(i) $R$ and $A$ satisfy the Theorem of Transition as formulated by
Nagata in \cite[Section 19]{N2}; (ii) $A/J$ $=$ $R/I$, so there is a
one-to-one correspondence between the ideals $H$ in $R$  that
contain  $I$ and the ideals $H'$ in $A$ that contain $J$;  (iii) $A$
is Cohen-Macaulay if and only if  $R$  is Cohen-Macaulay; and  (iv)
$b_1,\dots,b_g$ is an $R$-sequence if and only if $x_1,\dots,x_g$ is
an $A$-sequence. The relation between the ideals  $H$ in $\mathbf
P(I)$  and the ideals $(HA)_a$ in $\mathbf P(IA)$  is considered in
Corollary \ref{coro2} and Remark \ref{rema3}. The special case of
Theorem \ref{theo1} where $R$ is local and $I$  is an open ideal is
considered in Corollary \ref{coro3}.

In Section 4 we concentrate on the case of Theorem \ref{theo1} where
$n$ $=$ $1$, that is,  only one Rees valuation ring  $(V,N)$ of $I$
is considered. In this case, (a) of Theorem \ref{theo1}  holds by
Remark \ref{rema1}.2. If the integer $c$  such that  $IV$ $=$ $N^c$
is a unit in $V$, then it is shown in Lemma \ref{lemm1}.3  and Corollary
\ref{coro4} (together with Proposition \ref{remax2}.3)
that there exists a valuation ring $(U,M)$ extending $V$
and a minimal prime ideal  $z^*$  in  $A$
such that $H$  is projectively full for all ideals $H$ in
all Noetherian rings  $B$ such that $A/z^*$ $\subseteq$ $B$ $\subseteq$ $U$ and
$JB$ $\subseteq$ $H$ $\subseteq$ $M \cap B$.  In particular, if  $B$  is
such a ring, then there
exists a prime ideal $P$  containing  $JB$
such that
$JB$, $P$,  $JB_{P}$, and ${P}B_{P}$
are projectively full.

In Example \ref{exam1}.1  of Section 5 it is shown that a regular
ideal $I$ of $R$ is projectively full if the  associated graded ring
$\mathbf G(R,I)$ has a minimal divisor $p$ that is its own $p$-primary
component of $(0)$, while in  Example \ref{exam2} it is shown that
the projectively full ideal $J$  of Theorem \ref{theo1} may have an
embedded prime divisor $P$  that is the center of a Rees valuation
ring $(U,M)$ such that  $JU$ $=$ $M$. Then some cases where $J_a$ is a
prime (resp., radical) ideal are considered in Example \ref{exam3}
(resp., Example \ref{exam4}).

In Example \ref{5.5}  of Section 6, we consider the behavior of the
projectively full property between $R$ and $R^+$, where  $R$  is a
Noetherian domain and  $R^+$  is a Noetherian integral extension
domain of $R$ contained in the field of fractions of $R$. For a
nonzero proper ideal $I$ of $R$, (i) if $IR^+$ is projectively full,
then $I$ is projectively full, but the converse fails, (ii) there
exist examples where  $\mathbf P(I)$ is projectively full and
$\mathbf P(IR^+)$ fails to be projectively full, and examples where,
conversely,  $\mathbf P(I)$ fails to be projectively full and
$\mathbf P(IR^+)$ is projectively full. In Example \ref{rema3exam}
we present several examples of Noetherian domains $R$ that are not
integrally closed and have the property that $\mathbf P(I)$ is
projectively full for each nonzero proper ideal $I$ of $R$. In
Example \ref{exam6} we present a family of examples of Noetherian
domains $R$ for which there exists an integral extension domain $B$
that differs from the integral extension  domain obtained using
Theorem \ref{theo1},  and has the property that $\mathbf P(IB)$ is
projectively full for each nonzero proper ideal $I$ of $R$. In
Example \ref{nonreduced} we present an example of a normal local
domain $(R,M)$ of altitude two such that $M$ is projectively full
and the associated graded ring $\mathbf G(R,M)$ is not reduced. In
Remark \ref{rational}, we present an argument of J. Lipman to show
that if $(R,M)$ is a normal local domain of altitude two that has a
rational singularity, then $\mathbf P(I)$ is projectively full for
each $M$-primary ideal $I$ of $R$.

Our notation is as in \cite{N2} and \cite{Mat}. Thus, for example,
elements $b_1, \ldots, b_g$ in an ideal $I$ form a {\bf basis}  of
$I$ if they generate $I$.


\bigskip

\section{FINITE FREE EXTENSION RINGS A   OF   R  IN WHICH \\ P(IA) IS PROJECTIVELY FULL.}

Projectively full ideals are  introduced in \cite[Section
4]{CHRR}. It is observed in \cite[(4.13)]{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 \cite{CHRR2} a number of basic properties of
a projectively full ideal are developed, and then it is asked if,
for a given regular proper ideal  $I$  in a Noetherian ring  $R$,
there exists a finite integral extension ring  $A$  of  $R$  such
that  $\mathbf P (IA)$  is projectively full.  It follows from
Theorem \ref{theo1} that this is frequently the case.

The following two remarks and definition will be useful in the proof
of Theorem \ref{theo1}.

\begin{rema}
\label{rema0}
{\em
Let  $R$  be a Noetherian ring, let  $I$ $=$
$(b_1,\dots,b_g)R$  be a regular proper ideal
of  $R$, let  $c$  be a positive integer,
let $R_g$ $=$ $R[X_1,\dots,X_g]$,
and let  $K$ $=$ $( {X_1}^c-b_1,\dots,{X_g}^c-b_g)R_g$.
In  Theorem \ref{theo1} (and throughout this paper)
we let $A$ $=$
$R[{x_1},\dots, {x_g}]$
($=$ $R_g/K$)  and  $J$ $=$
$({x_1},\dots, {x_g})A$,
so  $A$  is  a finite free ``root'' (integral)
extension ring of rank  $c^g$  of $R$.  Also,
for  $i$ $=$ $1,\dots,g$  it holds that
${{x_i}}^c$ $=$ ${b_i}$ $\in$ $IA$,
and $IA$ $\subseteq$ $J^c$, so  $(IA)_a$ $=$
$(J^c)_a$, hence  $\mathbf P(IA)$ $=$
$\mathbf P(J)$.  Note that for
each minimal prime ideal  $z^*$  in  $A$  it
holds that
$A/z^*$ $=$ $R_g/P$  (where  $P$  is a
minimal prime divisor of  $K$)  has the
form  $A/z^*$ $=$
$(R/(z^* \cap R))[\overline{x_1},\dots,\overline{x_g}]$,
where  $\overline{x_i}$ $=$ $x_i+z^*$  for  $i$
$=$ $1,\dots,g$.  Since
${{x_i}}^c$ $=$ ${b_i}$ in $A$,
it follows that  $A/z^*$ $=$
$(R/(z^* \cap R))[{\overline{b_1}}^{1/c},\dots,{\overline{b_g}}^{1/c}]$,
where  $\overline{b_i}$ $=$ $b_i + (z^* \cap R)$  (for
$i$ $=$ $1,\dots,g$),
so  $A/z^*$  is generated by $c$-th roots
${\overline{b_1}}^{1/c}, \ldots, {\overline{b_g}}^{1/c}$ of
$\overline{b_1}, \ldots,\overline{ b_g}$, respectively,
in a fixed algebraic closure
of the quotient field of  $R/(z^* \cap R)$.}
\end{rema}

\begin{defi}
\label{defi1}
{\em
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$.
Then
{\bf{Rees I}}
denotes the set of Rees valuation rings
of  $I$, and if  $(V,N)$ $\in$ $\Rees I$, then the
{\bf{Rees integer}}
of  $I$ with respect to  $V$  is the integer  $e$  such
that  $IV$ $=$ $N^e$.
}
\end{defi}



\begin{rema}
\label{remax} {\em  Let  $I$  be a regular proper ideal in a
Noetherian ring  $R$. If the greatest common divisor of the Rees
integers of  $I$ is equal to one, then  $I$  is projectively full,
by \cite[(4.10)]{CHRR}. (The converse is false, by \cite[Example
3.4, page 401]{MRS}.) Therefore if there exists an ideal  $K$ $\in$
$\mathbf P(I)$  whose Rees integers have greatest common divisor
equal to one, then $K$ and  $\mathbf P (I)$  are projectively full.
(If such an ideal  $K$  exists, then since the ordered sets of
Rees integers of  $I$  and  $K$  are proportional, necessarily
$K$  is the largest ideal in the linearly ordered set  $\mathbf P(I)$.)}
\end{rema}



It is clear that assumption (a) in Theorem \ref{theo1} holds if $g$
$=$ $1$ (that is, if  $I$ is a regular principal ideal). Additional
comments concerning assumptions (a) and (b) of Theorem \ref{theo1}
are given in Remarks \ref{rema1}.1  - \ref{rema1}.3.



\begin{theo}
\label{theo1}
Let  $I$  be a regular proper  ideal in a Noetherian ring  $R$,
let  $b_1,\dots,b_g$  be a basis of  $I$,
let  $\{(V_1,N_1),\dots,(V_n,N_n)\}$  be a nonempty subset of
$\Rees I$,
and for  $j$ $=$ $1,\dots,n$  let  $e_j$  be the Rees
integer of  $I$  with respect to  $V_j$.  Assume:
\newline
\noindent
(a) $b_iV_j$ $=$ ${N_j}^{e_j}$  for  $i$ $=$ $1,\dots,g$  and
$j$ $=$ $1,\dots,n$; and,
\newline
\noindent
(b) the greatest common divisor  $c$  of  $e_1,\dots,e_n$
is a unit in  $R$.
\newline
\noindent Let  $A$ $=$ $R[ {x_1},\dots, {x_g}]$ and let  $J$ $=$ $(
{x_1},\dots, {x_g})A$ (see Remark \ref{rema0}). Then   $A$  is a
finite free integral extension ring of $R$, $IA$  and  $J$  are
projectively equivalent, and $J$  is projectively full, so $\mathbf
P(IA)$  is projectively full.
\end{theo}

\begin{proof}
If  $c$ $=$ $1$, then  $A$ $=$ $R$  and
$I$  and  $\mathbf P(I)$  are projectively full (by Remark \ref{remax}),
so the conclusion holds in this case.  Therefore it may be
assumed that  $c$ $>$ $1$.

As noted in Remark \ref{rema0},  $A$  is a finite free integral
extension ring of  $R$ and  $(IA)_a$ $=$ $(J^c)_a$, so $IA$  and $J$
are projectively equivalent in  $A$.  Therefore it suffices to show
that $J$  is projectively full.

For this, let $(U_1,M_1),\dots,(U_k,M_k)$  be all
the Rees valuation rings of  $J$, and for  $j$ $=$ $1,\dots,k$  let  $f_j$
be the Rees integer of  $J$  with respect to  $U_j$.
Then by Remark \ref{remax} it suffices to show that
the greatest common divisor of  $f_1,\dots,f_k$  is  $1$.



For this, for  $j$ $=$
$1,\dots,n$ let  $D_j$ $=$
$V_j[{u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}]$,
where  $u_{1,j},\dots,u_{g,j}$ are
units in  $V_j$  determined by  $b_1,\dots,b_g$,
and let  ${V_j}^*$ $=$ $(D_j)_q$ (where
$q$  is a minimal
prime divisor of $N_j D_j$).
Assume it is known that  ${V_j}^*$  is a discrete valuation
ring such that  $q{V_j}^*$ $=$ $N_j{V_j}^*$ and
${V_j}^*$ $=$ $U_h$  for some  $h$ $\in$ $\{1,\dots,k\}$.
Then it follows (after resubscripting  $U_1,\dots,U_k$,
if necessary) that, for  $j$
$=$ $1,\dots,n$, $J^c U_j$ $=$ $IU_j$ $=$
$(IV_j)U_j$ $=$ ${N_j}^{e_j} U_j$ $=$ ${M_j}^{e_j}$,
so  $JU_j$ $=$ ${M_j}^{c_j}$, where  $c_j$  is the
positive integer such that  $c_jc$ $=$ $e_j$.
However, by hypothesis  $JU_j$ $=$ ${M_j}^{f_j}$, so it
follows first that  $f_j$ $=$ $c_j$, and then that
the greatest common divisor of  $f_1,\dots,f_k$  is  $1$
(since  $k$ $\ge$ $n$  and the greatest common divisor
of  $c_1,\dots,c_n$  is  $1$).
Therefore it remains to show that for  $j$
$=$ $1,\dots,n$:  (i)   there exists a prime ideal  $q$
in  $D_j$
$=$ $V_j[{u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}]$
such that  ${V_j}^*$ $=$ $(D_j)_q$  is
a discrete valuation ring whose
maximal ideal is generated by  $N_j$; and,
(ii)  ${V_j}^*$  is a Rees valuation ring of  $J$.

To see that (i) holds, fix  $j$ $\in$ $\{1,\dots,n\}$.
Then by the
construction of Rees valuation
rings (see \cite[(2.9)]{CHRR})
there exists a minimal
prime divisor  $z_j$  of zero in  $R$  such that
$R/z_j$  is a subring of  $V_j$.
Let $c_j$  be the positive integer
defined by  $c_jc$ $=$ $e_j$  (where  $c$  is the greatest
common divisor of  $e_1,\dots,e_n$),
let  $\pi_j$  be a generator of  $N_j$,
and for  $i$ $=$ $1,\dots,g$  let  $b_{i,j}$ $=$
$b_i+z_j$  (so  $b_{i,j}$ $\in$ $R/z_j$ $\subseteq$ $V_j$,
and  $b_{i,j}$ $=$ $b_i$  if  $R$  is
an integral domain).
Then it follows from assumption (a) that, for
$i$ $=$ $1,\dots,g$, there exists a unit  $u_{i,j}$ $\in$ ${V_j}$
such that  $b_{i,j} $ $=$ $u_{i,j} {\pi_j}^{e_j} $ $=$ $u_{i,j} {\pi_j}^{c_jc}$.
Fix  $c$-th roots
${b_{1,j}}^{1/c},\dots,{b_{g,j}}^{1/c}, {u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}$
of  $b_{1,j},\dots,b_{g,j},u_{1,j},\dots,u_{g,j}$, respectively, in an
algebraic closure of the quotient field of  ${V_j}$.
Then since  $b_{i,j}$ $=$ $u_{i,j} {\pi_j}^{c_jc}$,
it follows that
$$(*) \quad  {V_j}[{u_{i,j}}^{1/c}] ~~ and  ~~{V_j}[{b_{i,j}}^{1/c}]~~  have
~~the~~same~~quotient~~field~~for~~i = 1,\dots,g.$$

Let  $X_1,\dots,X_g$  be indeterminates and for  $i$
$=$ $1,\dots,g$  let  $Y_i$
$=$ $\frac{X_i}{{\pi_j}^{c_j}}$.
Now the derivative
of  $f_{i,j}(Y_i)$ $=$ ${Y_i}^c - u_{i,j}$
(with respect to  $Y_i$) is  ${f_{i,j}}'(Y_i)$ $=$
$c{Y_i}^{c-1}$.  Also, the roots of
$f_{i,j}(Y_i)$ $=$ $0$   are  $\omega ^h {u_{i,j}}^{1/c}$  ($h$
$=$ $1,\dots,c$, where  $\omega$  is a primitive  $c$-th root of the
unit element  $1$ $\in$ $V_j$), so
it follows from \cite[(10.17)]{N2} that the
discriminant $\Disc(f_{i,j}(Y_i))$  of  $f_{i,j}(Y_i)$  is
$\pm \Pi_{h=1}^c~{f_{i,j}}'(\omega^h {u_{i,j}}^{1/c})$ $=$
$\pm c^c (\omega ^{1 + \cdots + c})^{c-1}{u_{i,j}}^{c-1}$ $=$
$\pm c^c {u_{i,j}}^{c-1}$.
Therefore, since  $u_{i,j}$  is
a unit in  ${V_j}$, and since  $c$
is  is a unit in  $V_j$
(since, by assumption (b),
$c$  is a unit in  $R$, so  $c$  is a unit in
$R/z_j$ $\subseteq$ $V_j$),
it follows that  $\Disc(f_{i,j}(Y_i))$
$=$ $\pm c^c {u_{i,j}}^{c-1}$  is
a unit in  ${V_j}$.
Therefore it follows from \cite[(38.9)]{N2}
that  ${V_j}[{y_i}]$ $=$ ${V_j}[Y_i]/(f_{i,j}(Y_i){V_j}[Y_i])$
is integrally closed and that
$N_j{V_j}[{y_i}]$ $=$ $\pi_j V_j[y_i]$  is a radical ideal  (so
for each prime divisor  $P$  of  $N_j {V_j}[{y_i}]$,
${V_j}[{y_i}]_P$  is a
discrete valuation ring whose maximal ideal is  $N_j{V_j}[{y_i}]_P$).
Now  ${{y_i}}^c$ $=$ $u_{i,j}$, so it follows
that  ${V_j}[y_i]_{P}$  $=$
${V_j}[{u_{i,j}}^{1/c}]_{P_1}$
for some
height one prime ideal  $P_1$  in  ${V_j}[{u_{i,j}}^{1/c}]$  that
contains  $N_j$,
so  $V_{1,j}$ $=$
${V_j}[{u_{i,j}}^{1/c}]_{P_1}$
is a discrete valuation ring
and $P_1V_{1,j}$ $=$ $N_jV_{1,j}$.
(Note that, since  $V_j[Y_i]$  is a unique factorization
domain, it follows that  $V_j[{u_{i,j}}^{1/c}]$ $=$
$V_j[Y_i]/(\mu_i(Y_i)V_j[Y_i])$, where  $\mu_i(Y_i)$
is the minimal polynomial of  ${u_{i,j}}^{1/c}$
over  $V_j$.)

By repeating much of the previous paragraph
(first with  $V_{1,j}$  and  $u_{h,j}$
(with  $h$ $\in$ $\{1,\dots,g\}$  and  $h$ $\ne$ $j$)  in place
of  ${V_j}$  and  $u_{i,j}$  to get  $V_{2,j}$, then with  $V_{2,j}$
and $u_{m,j}$  (with
$m$ $\in$ $\{1,\dots,g\}$  and  $m$ $\ne$ $j,h$)  in place
of  ${V_{1,j}}$  and  $u_{h,j}$  to get  $V_{3,j}$, etc.),
it follows that, for  $j$ $=$ $1,\dots,n$,  there exists
a chain of discrete valuation rings  $V_{0,j}$ $=$ $V_j$
$\subseteq$ $V_{1,j}$ $=$ ${V_{0,j}}[{u_{1,j}}^{1/c}]_{P_1}$
$\subseteq$ $\cdots$ $\subseteq$ ${V_{g-1,j}}[{u_{g,j}}^{1/c}]_{P_{g}}$  $=$
$V_{g,j}$  such that  $N_j V_{h,j}$  is the maximal ideal of  $V_{h,j}$
for  $h$ $=$ $1,\dots,g$.  Let  $D_j$ $=$
${V_{j}}[{u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}]$, so
it follows that  $V_{g,j}$ $=$ $(D_j)_q$  for
some height one prime ideal  $q$  in  $D_j$ and that
$q(D_j)_q$ $=$ $N_j(D_j)_q$, so (i) holds.


To see that (ii) holds (that is, that $V_{g,j}$  is a Rees valuation
ring of  $J$ $=$ $( {x_1},\dots, {x_g})A$), note that  $R/z_j$
$\subseteq$ $V_j$  and by construction (see \cite[(2.9)]{CHRR})
there exists a height one prime divisor  $p$ of  $b_{1,j}{B_j}'$
such that  $V_j$ $=$ $({B_j}')_p$  and  $N_j$ $=$ $p V_j$, where
${B_j}'$  is the integral closure of  $B_j$ $=$
$(R/z_j)[b_{2,j}/b_{1,j},\dots,b_{g,j}/b_{1,j}]$  in its quotient
field (here we use assumption (a) (that  $IV_j$ $=$ $b_iV_j$  for
$j$ $=$ $1,\dots,n$  and  $i$ $=$ $1,\dots,g$)). By integral
dependence, there exists a minimal prime ideal  ${z_j}^*$  in  $A$
$=$ $R[ {x_1},\dots, {x_g}]$ such that  ${z_j}^* \cap R$ $=$ $z_j$;
then  $A/{z_j}^*$ $=$ $(R/z)[
{\overline{x_1}},\dots,{\overline{x_g}}]$ $=$ $(R/z)[
{b_{1,j}}^{1/c},\dots,{b_{g,j}}^{1/c}]$ (see Remark \ref{rema0}).
(Note  that if  $R$  is an integral domain, then each minimal prime
ideal  $z^*$  in   $A$  is a suitable choice for  ${z_j}^*$.) Then,
since $R/z_j$  and  $V_j$  have the same quotient field, it follows
from (*) that  $A/{z_j}^*$  and $D_j$ $=$
${V_{j}}[{u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}]$  have the same
quotient field. Also, $A$  is a finite free integral extension ring
of  $R$ and $b_{i,j}/b_{1,j}$  $\in$ $B_j$ is such that
$b_{i,j}/b_{1,j}$ $=$ $(\overline {x_i}/\overline {x_1})^c$ (for $i$
$=$ $1,\dots,g$), so it follows that  $C_j$ $=$
$(A/{z_j}^*)[\overline{x_{2}}/\overline{x_{1,j}},
\dots,\overline{x_{g,j}}/\overline{x_{1,j}}]$ is a finite integral
extension domain of $B_j$. Therefore  ${C_j}'$ $=$ ${B_j}''$
$\subseteq$ ${V_j}''$, where ${C_j}'$  (resp., ${B_j}''$, ${V_j}''$)
is the integral closure of  $C_j$  (resp., ${B_j}$, ${V_j}$) in the
quotient field  of  $C_j$  (which is the quotient field of
$A/{z_j}^*$  and of  $D_j$). Also, ${u_{i,j}}^{1/c}$ $\in$
${V_j}''$, since $u_{i,j}$ $\in$ $V_j$, so  $D_j$ $=$
${V_{j}}[{u_{1,j}}^{1/c},\dots,{u_{g,j}}^{1/c}]$ $\subseteq$
${V_j}''$, so  ${V_j}''$  is an integral extension domain of  $D_j$.
Let  $q$  be as at the end of the second preceding paragraph, so
$V_{g,j}$ $=$ $(D_j)_q$  is a discrete valuation ring. Therefore it
follows that  $V_{g,j}$ $=$ $({V_j}'')_{q^*}$, where $q^*$ $=$
$qV_{g,j} \cap {V_j}''$. Since  $q^* \cap {B_j}'$ $=$ $(q^* \cap
V_j) \cap {B_j}'$ $=$ $N_j \cap {B_j}'$  $=$ $p$  (where  $p$  is a
height one prime divisor of  $b_{1,j}{B_j}'$  (by the start of this
paragraph)), and since  ${C_j}'$ $=$ ${B_j}''$ $\subseteq$
${V_j}''$, it follows that  $q_j$ $=$ $q^* \cap {C_j}'$  is a prime
ideal in  ${C_j}'$ such that  $q_j \cap {B_j}'$ $=$ $p$.  Therefore
$q_j$  is a height one prime divisor of  $\overline{x_1}{C_j}'$ $=$
${b_{1,j}}^{1/c}{C_j}'$, so  $({C_j}')_{q_j}$ $=$ $V_{g,j}$  is a
Rees valuation ring of  $J$ (by \cite[(2.9)]{CHRR}), hence (ii)
holds.
\end{proof}

It is clear from the preceding proof that the ring $A$ $=$ $R[
{x_1},\dots, {x_g}]$ and the ideal  $J$ $=$ $( {x_1},\dots, {x_g})A$
are not canonical, in that  they  depend on the basis
$b_1,\dots,b_g$  chosen for  $I$. The next two remarks mention
several positive things about the extension ring  $A$, the ideal
$J$, and the proof of Theorem \ref{theo1}.

\begin{rema}
\label{remay} {\bf{(\ref{remay}.1)}} {\em The proof of Theorem
\ref{theo1}  shows the following:  if  $V_j$  is a Rees valuation
ring  of  $I$, if  $e_j$  is the Rees integer of  $I$ with respect
to  $V_j$, and if  $c$  is the greatest common divisor of
$e_1,\dots,e_n$, then $U_j$ $=$ $V_j[{u_{1,j}}^{1/c},
\dots,{u_{g,j}}^{1/c}]_{q}$  is a Rees valuation ring of $J$ $=$ $(
{x_1},\dots, {x_g})R[ {x_1},\dots, {x_g}]$ (for some height one
prime ideal  $q$), the Rees integer of $J$  with respect to  $U_j$
is  $c_j$ $=$ $e_j/c$, and the greatest common divisor of
$c_1,\dots,c_n$  is equal to one. In particular, if  $e_1$ $=$
$\cdots$ $=$ $e_n$ (for example, if  $n$ $=$ $1$), then  $e_1$ $=$
$c$  and $c_1$ $=$ $\cdots$ $=$ $c_n$ $=$ $1$.

\noindent
{\bf{(\ref{remay}.2)}}
If  $R$  is a Noetherian domain in Theorem \ref{theo1},
then it follows from
the last paragraph of the proof
of Theorem \ref{theo1} that,
for each minimal prime
ideal  $z^*$  in  $A$, the ideal  $(IA+z^*)/z^*$
in  $A/z^*$  is such that  $\mathbf P((IA+z^*)/z^*)$
if projectively full (since the proof shows that
$(IA+z^*)/z^*$ has  $n$  Rees valuation rings whose
Rees integers have greatest common divisor equal to one).
Therefore in the
case where $R$ is an integral domain there exists a finite integral
extension domain $B = A/z^*$ of $R$ such
that $\mathbf P(IB)$ is projectively full.
}
\end{rema}

\begin{rema}
\label{rema1} {\bf{(\ref{rema1}.1)}} {\em Concerning assumption (a)
of Theorem \ref{theo1} that ``$b_1,\dots,b_g$  is a basis of  $I$
such that $b_iV_j$ $=$ $IV_j$  for  $i$ $=$ $1,\dots,g$  and $j$ $=$
$1,\dots,n$'', if $R$  is a local ring with maximal ideal $M$ such
that  $R/M$  is infinite, then  there exists such a basis for $I$
for every  nonempty subset  $\{(V_1,N_1),\dots,(V_n,N_n)\}$ of
$\Rees I$.

\noindent {\bf{(\ref{rema1}.2)}} Let  $I$  be a regular proper ideal
in a Noetherian ring  $R$ and let  $(V,N)$ $\in$ $\Rees I$. Then
assumption (a) of Theorem \ref{theo1} holds for  $I$  and  $V$: that
is, $I$  has a basis (say  $b_1,\dots,b_g$)  such that  $b_iV$ $=$
$IV$ for  $i$ $=$ $1,\dots,g$.


\noindent {\bf{(\ref{rema1}.3)}} If $R$ as in Theorem \ref{theo1}
contains a field  $F$  such that either:  $\Char(F)$  is not a
divisor of $c$; or, $\Char(F)$ $=$ $0$; then assumption (b) holds
(since the greatest common divisor  $c$  of  $e_1,\dots,e_g$ is in
$F$).  Of course, the larger  $n$  is chosen (that is, the more Rees
valuation rings of  $I$ that are considered), the more likely it is
that assumption (b) holds.  On the other hand, if  $H$  is any ideal
that is projectively equivalent to $I$, then by \cite[(2.10)]{MRS}
$H$ and  $I$  have the same Rees valuation rings and their
corresponding Rees integers are proportional, so by choosing  $H$ as
the largest ideal in  $\mathbf P(I)$, the more likely it is that
assumption (b) holds (for the greatest common divisor of the Rees
integers of $H$).

\noindent {\bf{(\ref{rema1}.4)}} If $c$ $\notin$ $(N_1 \cap R) \cup
\cdots \cup (N_n \cap R)$, and if assumption (a) of Theorem
\ref{theo1}  holds for  $I$, then there exists a finite free
integral extension ring $A$ of $R$ and an ideal  $J$  in  $A$  such
that $\mathbf P(IA)$ $=$ $\mathbf P(J)$  is projectively full. }
\end{rema}


\begin{proof}
For (\ref{rema1}.1), fix a nonempty subset
$\{(V_1,N_1),\dots,(V_n,N_n)\}$  of
$\Rees I$, and
for  $j$ $=$ $1,\dots,n$  let  $H_j$ $=$ $\{x \in I \mid xV_j \subsetneq IV_j\}$.
Then it is readily checked that each  $H_j$  is an ideal in  $R$  that is properly
contained in  $I$.
Therefore  $\overline{H_j}$ $=$ $(H_j +MI)/(MI)$  is a
proper subspace (over the field  $R/M$)  of  $\overline I$ $=$ $I/(MI)$.  Since
$R/M$  is infinite, it follows that  $\overline I$
has a basis  $\overline{b_1},\dots,\overline{b_g}$
such that no  $\overline{b_i}$  is in
$\overline {H_1} \cup \cdots \cup \overline {H_n}$.
Therefore if  $b_1,\dots,b_g$  are preimages in  $R$  of
$\overline{b_1},\dots,\overline{b_g}$, then it follows that  $b_1,\dots,b_g$  are
a basis of  $I$  such that  $b_iV_j$ $=$ $IV_j$  for  $i$ $=$ $1,\dots,g$  and
$j$ $=$ $1,\dots,n$.

For (\ref{rema1}.2),
let  $c_1,\dots,c_g$  be a basis of  $I$, so  $IV$ $=$
$c_iV$  for some  $i$ $\in$ $\{1,\dots,g\}$.
Resubscript the  $c_i$  so that  $c_hV$ $=$ $IV$  for  $h$
$=$ $1,\dots,f$  and  $c_hV$ $\subsetneq$ $IV$  for  $h$
$=$ $f+1,\dots,g$.  For  $h$ $=$ $1,\dots,f$  let  $b_h$
$=$ $c_h$, and for  $h$ $=$ $f+1,\dots,g$  let  $b_h$
$=$ $b_1+c_h$.  Then it is readily checked that
$b_1,\dots,b_g$  is
a basis of  $I$ such that  $b_iV$ $=$ $IV$  for
$i$ $=$ $1,\dots,g$.

For (\ref{rema1}.4), let $S=R[1/c]$. If $c$ $\notin$ $(N_1 \cap R)
\cup \cdots \cup (N_n \cap R)$, and if assumption (a) holds for $I$,
then assumptions (a) and (b) hold for  $IS$. Therefore there exists
a finite free integral extension ring $B$ $=$
$S[{x_1},\ldots,{x_g}]$ of $S$ such that  $J'$ $=$
$({x_1},\ldots,{x_g})B$ is projectively full (by Theorem
\ref{theo1}, with  $S$  and  $IS$  in place of  $R$  and  $I$). Let
$A=R[{x_1},\ldots,{x_g}]$ and $J=({x_1},\ldots,{x_g})A$, and let $K$
$\in$ $\mathbf P(J)$. Then there exist positive integers  $n,s$ such
that  $(K^n)_{a}=(J^s)_a$, so $((KB)^n)_a$
$=$ $((K^n)_aB)_a=((J^s)_a B)_a$ $=$ $((JB)^s)_a$ $=$ $(J'^s)_a$,
hence $n$ divides $s$, as $JB$ $=$ $J'$  is projectively full. This
implies that $K=(J^{s/n})_a$. It follows that  $\mathbf P(J)$  is
projectively full, and  $\mathbf P(J)$ $=$ $\mathbf P(IA)$, by
Remark \ref{rema0}.
\end{proof}



\begin{coro} \label{2.7} Let $R$ be a Noetherian ring that contains
the field  $\IQ$  of rational numbers. For each regular proper ideal $I$ of
$R$ there exists a finite free integral extension ring $A$ of $R$
such that $\mathbf P(IA)$ is projectively full. If $R$ is an
integral domain, there exists a finite integral extension domain $B
= A/z^*$ of $R$ such that $\mathbf P(IB)$ is projectively full.
\end{coro}

\begin{proof} Apply Remarks \ref{rema1}.2 - \ref{rema1}.3, and
Remark \ref{remay}.2.
\end{proof}







In Corollary \ref{remanew12},   we show that
$\mathbf P(IA^+)$  is projectively full for certain
integral overrings  $A^+$  of the
ring  $A$  constructed in  Theorem \ref{theo1}.
(A related result is considered in Corollary \ref{coro4}
 and Remark \ref{remanormal} below.)

\begin{coro}
\label{remanew12}
{\em
With the notation and assumptions of Theorem \ref{theo1}, let  $A^+$  be a finite integral
extension ring of  $A$  that is contained in
the total quotient ring of  $A$.  Then  $\mathbf P(IA^+)$
is projectively full.
}
\end{coro}

\begin{proof}
The Rees valuation rings of  $IA$  (and of  $J$)  are the Rees
valuation rings of  $IA^+$ (and of  $JA^+$), and by integral
dependence the Rees integers of  $IA^+$  (resp., $JA^+$) with
respect to these valuation rings are the same as for  $IA$ (resp.,
$J$).  Also, $IA^+$  and  $JA^+$  are projectively equivalent (since
$IA$  and  $J$ are projectively equivalent).  The conclusion follows
from this and Remark \ref{remax}, since the greatest common divisor
of these Rees integers of  $J$  is equal to one.
\end{proof}


Corollary \ref{coronew2} extends Theorem \ref{theo1} to certain
finite collections of regular proper ideals of certain
local rings.

\begin{coro}
\label{coronew2}
Let  $(R,M)$  be a local ring and
let  $I_1,\dots,I_m$  be regular proper
ideals of  $R$.
Assume that  $\IQ$ $\subseteq$ $R$
and that there exist nonempby subsets
$\mathbf C_i$  of  $\Rees I_i$  such
that, for  $i$ $\ne$ $j$ in $\{1,\dots,m\}$,
there are no containment relations
between the centers in  $R$  of the valuation
rings in  $\mathbf C_i$  and the centers
in  $R$  of the valuation
rings in  $\mathbf C_j$.
Then there exists a finite free local
integral extension
ring  $A$  of  $R$  such that  $\mathbf P(I_iA)$
is projectively full for  $i$ $=$ $1,\dots,m$.
\end{coro}

\begin{proof}
For  $i$ $=$ $1,\dots,m$  let
$\mathbf C_i$ $=$ $\{(V_{i,1},N_{i,1}),\dots,(V_{i,n_i},N_{i,n_i})\}$,
and for  $h$ $=$
$1,\dots,n_i$  let  $v_{i,h}$  be the
valuation of  $V_{i,h}$, let  $P_{i,h}$ $=$ $N_{i,h} \cap R$
be the center in  $R$   of  $V_{i,h}$, let  $\pi_{i,h}$
$\in$ $V_{i,h}$  such that  $N_{i,h}$ $=$
$\pi_{i,h}V_{i,h}$, let  $e_{i,h}$  be
the Rees integer of  $I_i$  with respect to  $V_{i,h}$,
let  $c_i$  be the greatest common divisor of
$e_{i,1},\dots,e_{i,n_i}$, and define  $c_{i,h}$
by  $c_{i,h}c_i$ $=$ $e_{i,h}$.


Fix  $i$ $\in$ $\{1,\dots,m\}$, let  $H_{i,(i,h)}$
$=$ $\{x \in I_i \mid v_{i,h}(x) > v_{i,h}(I_i)\}$
(for  $h$ $=$ $1,\dots,n_i$), and let  $H_{i,(j,h)}$
$=$ $I_i \cap P_{j,h}$  (for  $j$ $\ne$ $i$  in
$\{1,\dots,m\}$  and for  $h$ $\in$ $\{1,\dots,n_j\}$).
Then by the hypothesis concerning the sets  $\mathbf C_i$
and  $\mathbf C_j$  it follows that
each  $H_{i,(j,h)}$  ($j$ $=$ $1,\dots,m$  and
$h$ $\in$ $\{1,\dots,n_j\}$)  is a proper subset of
$I_i$, so (since  $R/M$  is infinite)
there exists a basis  $b_{i,1},\dots,b_{i,g_i}$
of  $I_i$  such that no  $b_{i,k}$  is in any
$H_{i,(j,h)}$.  Therefore:
(i)  for  $k$ $=$ $1,\dots,g_i$  and for  $h$
$=$ $1,\dots,n_i$ it holds that
$b_{i,k}V_{i,h}$ $=$ $I_iV_{i,h}$
(so there exist units  $u_{k,h}$ $\in$ $V_{i,h}$
such that  $b_{i,k}$ $=$
$u_{k,h}{\pi_{i,h}}^{e_{i,h}}$ $=$
$u_{k,h}{\pi_{i,h}}^{c_{i,h}c_i}$, so
${(b_{i,k}/\pi_{i,h})}^{1/c_i}$ $=$
${u_{k,h}}^{1/c_i} {\pi_{i,h}}^{c_{i,h}}$); and,
(ii) for  $k$ $=$ $1,\dots,g_i$, for  $j$
$\ne$ $i$ $\in$ $\{1,\dots,m\}$, and for
$h$ $\in$ $\{1,\dots,n_j\}$  it holds that
$b_{i,k}V_{j,h}$ $=$ $V_{j,h}$.

Since  $\IQ$ $\subseteq$ $R$, it follows that
assumption (b) of Theorem \ref{theo1} is
satisfied for  $I_1$  in place of  $I$,
and assumption (a) of Theorem \ref{theo1}
is satisfied (for  $I_1$  in place of  $I$)
by the preceding paragraph, so let
$A_1$ $=$ $R[ {x_{1,1}},\dots, {x_{g_1,1}}]$
($=$ $R_{g_1}/K_1$, where  $R_{g_1}$
$=$ $R[X_{1,1},\dots,X_{g_1,1}]$  and
$K_1$ $=$ $({X_{1,1}}^{c_1} - b_{1,1},\dots,
{X_{g_1,1}}^{c_1} - b_{g_1,1})R_{g_1}$),
and let  $J_1$ $=$ $(  {x_{1,1}},\dots, {x_{g_1,1}})A_1$.
Then $A_1$  is a local ring, by Proposition \ref{remaprops1}.5
below, and a finite
free integral extension ring of  $R$, by Theorem \ref{theo1}.
Also, using (i) in the preceding paragraph it follows from
Remark \ref{remay}.1 that the greatest common divisor of the
Rees integers of  $J_1$  is equal to one,
and Theorem \ref{theo1} shows
that  $\mathbf P (I_1A_1)$ $=$ $\mathbf P(J_1)$
is projectively full.
Further, by (ii) of the preceding paragraph,
each  $b_{1,k}$  ($k$ $=$
$1,\dots,g_1$)  is a unit in each
$V_{j,h}$  ($j$ $=$ $2,\dots,m$  and
$h$ $\in$ $\{1,\dots,n_j\}$), so by using
\cite[(38.9)]{N2} (as in the proof
of Theorem \ref{theo1}) it follows
that there exists
a height one prime ideal  $q_{j,h}$
in  $V_{j,h}[{u_{1,1}}^{1/c_1},\dots,{u_{1,g_1}}^{1/c_1}]$
such that  $U_{j,h}$ $=$
$V_{j,h}[{u_{1,1}}^{1/c_1},\dots,{u_{1,g_1}}^{1/c_1}]_{q_{j,h}}$
is a Rees valuation ring of  $I_jA_1$  whose maximal ideal
is  $N_{j,h}U_{j,h}$ $=$ $q_{j,h}U_{j,h}$  (so the
Rees integer of  $I_jA_1$  with respect to  $U_{j,h}$
is  $e_{j,h}$  (so the greatest common divisor of
these Rees integers of  $I_jA_1$  is  $c_j$)).

It therefore follows from  iterating
the preceding paragraph (first with  $A_1$
and  $I_2A_1$  in place of  $R$  and  $I_1$,
etc.) that the conclusion holds.
\end{proof}

Before deriving more corollaries of Theorem \ref{theo1}, we first
observe several  properties of the extension  ring  $A$.


\section{PROPERTIES OF THE FREE EXTENSION RING  A.}
In this section we record  some of the properties of the finite free
integral extension ring  $A$  of Theorem \ref{theo1}. Concerning the
Theorem of Transition in Proposition \ref{remaprops1}.1, see \cite[Section
19]{N2}.  Also, for Proposition \ref{remaprops1}.3, recall that the
{\bf{altitude}} of an ideal  $H$  is defined to be the maximum of
the heights of the minimal prime divisors of  $H$.

\begin{prop}
\label{remaprops1}
Assume notation as in Theorem 2.4.

\noindent
{\bf{(\ref{remaprops1}.1)}}
{\em
$R$ and  $A$  satisfy the Theorem of Transition.

\noindent
{\bf{(\ref{remaprops1}.2)}}
For each prime ideal  $p$  in  $R$  and for each
prime ideal  $P$  of  $A$  such that  $P \cap R$
$=$ $p$  it holds that  $R_p$  is a subspace
of  $A_{P}$.

\noindent {\bf{(\ref{remaprops1}.3)}} For each ideal  $H$  in  $R$
it holds that: $\hgt(H)$ $=$ $\hgt(HA)$; $\altitude(H)$ $=$
$\altitude(HA)$; and  $\dim(R/H)$ $=$ $\dim(A/(HA))$.

\noindent
{\bf{(\ref{remaprops1}.4)}}
$A/J$ $=$ $R/I$.

\noindent
{\bf{(\ref{remaprops1}.5)}}
There exists a one-to-one correspondence between the
ideals  $H'$  in  $A$  that contain  $J$  and the ideals
$H$  in  $R$  that contain  $I$  given by  $H$ $=$ $H' \cap R$ and
$H'$ $=$ $(J,H)A$  (so if  $H$  is prime (resp.,
primary), then  $(J,H)A$  is
prime (resp.,
primary), and if  $\cap_{i=1}^k q_i$  is an irredundant
primary decomposition of  $H$, then
$\cap_{i=1}^k (J,q_i)A$  is an irredundant
primary decomposition of  $(J,H)A$).
In particular:  $H$  and  $(J,H)A$  have the same number of
minimal prime divisors; $\hgt((J,H)A)$ $=$
$\hgt(H)$; $A/((J,H)A)$ $=$
$R/H$; and  $A$  has exactly $k$  maximal ideals  containing
$(J,H)A$ if $H$ is contained in exactly $k$  maximal ideals of  $R$.

\noindent {\bf{(\ref{remaprops1}.6)}} $R$  is a Cohen-Macaulay ring
if and only if $A$  is a Cohen-Macaulay ring.

\noindent {\bf{(\ref{remaprops1}.7)}} $b_1,\dots,b_g$  is an
$R$-sequence if and only if $x_1,\dots,x_g$  is an  $A$-sequence.

\noindent
{\bf{(\ref{remaprops1}.8)}}
If  $(V_1,N_1),\dots,(V_n,N_n)$
are all the Rees valuation rings of  $I$
in Theorem \ref{theo1}, then
$\{c_1,\dots,c_n\}$
are all the Rees integers of  $J$,
where  $c_jc$
$=$ $e_j$  for  $j$ $=$ $1,\dots,n$.
}
\end{prop}

\begin{proof}
Since  $A$  is a
finite free integral extension ring of  $R$,
(\ref{remaprops1}.1) follows from
\cite[(19.1)]{N2}, so
(\ref{remaprops1}.2) follows from
\cite[(19.2)(3)]{N2}, and
(\ref{remaprops1}.3) follows from
\cite[(22.9)]{N2}.

For (\ref{remaprops1}.4), as in Remark \ref{rema0}
let $R_g$ $=$ $R[X_1,\dots,X_g]$
and $K$ $=$ $( {X_1}^c-b_1,\dots,{X_g}^c-b_g)R_g$,
so $A$ $=$ $R[{x_1},\dots, {x_g}]$
$=$ $R_g/K$  and  $J$ $=$
$({x_1},\dots, {x_g})A$ $=$ $(X_1,\dots,X_g,K)/K$.
Therefore  $A/J$ $=$ $R_g/((X_1,\dots,X_g,K)R_g)$
$=$ $R_g/((b_1,\dots,b_g,X_1,\dots,X_g)R_g)$
$=$ $R/I$.

(\ref{remaprops1}.5) follows immediately from
(\ref{remaprops1}.4) and (\ref{remaprops1}.3).

For (\ref{remaprops1}.6), apply \cite[Theorem 23.3 and Theorem
17.6]{Mat}.



For (\ref{remaprops1}.7),
since  $A$  is a free  $R$-module, it follows
that  $(H:_RG)A$ $=$ $HA:_AGA$  for all
ideals  $H,G$  in  $R$, so it follows that
$b_1,\dots,b_g$  are an  $R$-sequence
if and only if they are an  $A$-sequence.  Since
${x_i}^c$ $=$ $b_i$  for  $i$ $=$ $1,\dots,g$,
it follows that  $b_1,\dots,b_g$  are an
$A$-sequence if and only if  $x_1,\dots,x_g$
are an  $A$-sequence.  Therefore $b_1,\dots,b_g$
are an  $R$-sequence if and only if
$x_1,\dots,x_g$ are an  $A$-sequence.

For (\ref{remaprops1}.8),
let  $z^*$  be a minimal
prime ideal in  $A$ and let  $z$ $=$ $z^* \cap R$, so
$z$  is a minimal prime ideal in  $R$  (since
$A$  is a  finite free
integral extension ring of  $R$).
Let an overbar denote residue class modulo  $z^*$
and let  $F$  be the quotient field of  $\overline R$, so
the quotient field of  $\overline A$  is
$E$ $=$ $F[{\overline {b_1}}^{1/c},\dots,{\overline {b_g}}^{1/c}]$.
Let  $\overline{\omega}$  be a primitive  $c$-th root of
the unit element  $1$  in  $F$.
Then it is clear that  $F[\overline{\omega}]$  is a Galois
extension field of  $F$, so it follows that
$F[\overline{\omega},{\overline {b_1}}^{1/c},\dots,{\overline {b_g}}^{1/c}]$
is a Galois extension field of both  $F$  and
$E$.  Therefore the Rees
valuation rings of  $JA[\omega]$ $=$
$({x_1},\dots,{x_g}) A[\omega]$  (and of
$\overline  J$ $=$
$(\overline{x_1},\dots,\overline{x_g}) \overline A$ $=$
$({\overline{b_1}}^{1/c},\dots,{\overline{b_g}}^{1/c}) \overline A$
(see Remark \ref{rema0})) that
lie over a given Rees valuation ring  (say, $V_j$)
of  $I$ $=$ $(b_1,\dots,b_g)R$  (and $\overline I$
$=$ $(\overline{b_1},\dots,\overline{b_g}) \overline R$)
are conjugate, so these Rees integers of  $J$ are all
equal to  $c_j$ $=$ $e_j/c$, by
the fourth paragraph
of the proof of Theorem \ref{theo1}.
Thus if $(V_1,N_1),\dots,(V_n,N_n)$
are all the Rees valuation rings of  $I$, then
the Rees integers of  $J$  are $\{c_1, \dots,c_n\}$.
\end{proof}

Since  $IA$ $\subseteq$ $J^c$ (by Remark \ref{rema0}),
since  $J^c$ $\subseteq$ $J^{c-1}$ $\subseteq$
$\cdots$ $\subseteq$  $J$, and since  $J \cap R$ $=$ $I$
(by Proposition \ref{remaprops1}.4), it follows
that if  $J^i$ $=$ $(J^i)_a$  for some  $i$
$\in$ $\{1,\dots,c\}$, then  $I$ $=$ $I_a$.

We close this section with two more corollaries of Theorem
\ref{theo1}. For the first of these, the integer  $d$  in
Corollary \ref{coro2}.2 is the integer  $d$  shown to exist in \cite[(2.8)
and (2.9)]{MRS} (and denoted  $d(I)$  in \cite[Section 4]{CHRR} and
in \cite{CHRR2}). It is a common divisor of the Rees integers of $I$,
and it is the smallest positive integer  $k$ such that, for all
ideals $G$ $\in$ $\mathbf P(I)$, $(G^k)_a$ $=$ $(I^i)_a$ for some
positive integer $i$.

\begin{coro}
\label{coro2} With the notation and assumptions of Theorem
\ref{theo1},  assume that $H$ is an ideal in $R$ that is
projectively equivalent to $I$. Then:

\noindent
{\bf{(\ref{coro2}.1)}}
If $h,i$  are positive integers such that  $(H^h)_a$
$=$ $(I^i)_a$, then  $(HA)_a$ $=$ $(J^{ci/h})_a$  and
$ci/h$  is a positive integer.

\noindent {\bf{(\ref{coro2}.2)}} If  $e_1,\dots,e_n$  are all the
Rees integers of  $I$ in Theorem \ref{theo1}, then there exists a
positive integer  $k$  such that  $(H^d)_a$ $=$ $(I^k)_a$, so
$(HA)_a$ $=$ $(J^{kd^*})_a$, where  $d^*$ is the positive integer
$c/d$.
\end{coro}

\begin{proof}
For (\ref{coro2}.1), if  $H$  is projectively equivalent to  $I$,
then by definition there exist positive integers  $h,i$  such that
$(H^h)_a$ $=$ $(I^i)_a$, and then it follows that $(H^hA)_a$ $=$
$(I^iA)_a$.  By Theorem \ref{theo1}, $(I^iA)_a$ $=$ $(J^{ci})_a$, so
$(HA)_a$ $=$ $J_{ci/h}$ ($=$ $\{x \in A \mid {\overline v}_J(x)$
$\ge$ $ci/h\}$; see \cite[(2.3)]{MRS}). Also,  $HA$  is projectively
equivalent to  $IA$, and  $IA$ is projectively equivalent to  $J$,
so  $HA$  is projectively equivalent to  $J$.  However, $J$  is
projectively full, by Theorem \ref{theo1}, so $(HA)_a$ $=$ $(J^k)_a$
for some positive integer  $k$. It follows that $J_{ci/h}$ $=$
$(HA)_a$ $=$ $(J^k)_a$ $=$ $J_k$ (by \cite[(2.3)]{MRS}), so $ci/h$
$=$ $k$.

For (\ref{coro2}.2),
as noted preceding this corollary,
there exists
a smallest common divisor  $d$  of the Rees integers
$e_1,\dots,e_n$  of  $I$  such
that for all ideals  $G$ that are projectively equivalent
to  $I$  it holds that  $(G^{d})_a$ $=$ $(I^{k})_a$  for some
positive integer  ${k}$.
Let  $k$  be the integer such
that  $(H^{d})_a$ $=$ $(I^{k})_a$,
and let  $c$  be the greatest common divisor
of  $e_1,\dots,e_n$.
Then $c$ $=$ $dd^*$  for some positive integer $d^*$,
so it follows that
$(H^c)_a$ $=$ $(H^{dd^*})_a$ $=$ $(I^{kd^*})_a$,
so  $(HA)_a$ $=$ $(J^{kd^*})_a$ by (\ref{coro2}.1).
\end{proof}


\begin{rema}
\label{rema3}
{\em
It is shown in \cite[Corollary 2.4]{MRS} that  $\mathbf P(I)$
is linearly ordered and discrete, so there exist positive
integers  $c_1$ $<$ $c_2$ $<$ $\cdots$  such that
$\mathbf P(I)$ $=$ $\{(I^{c_i/d})_a \mid i$
is a positive integer$\}$, where  $d$  is as
in Corollary \ref{coro2}.2.  Let  $d^*$ $=$ $c/d$  as in Corollary \ref{coro2}.2,
so  $\mathbf P(I)$ $=$
$\{(I^{c_id^*/c})_a \mid i$
is a positive integer$\}$.
With this in mind, it follows
from Corollary \ref{coro2}.2 that  $(\mathbf P(I))A$
$=$ $\{(J^{c_id^*})_a \mid i$  is a positive
integer$\}$ $\subseteq$ $\mathbf P(IA)$
(and  $\mathbf P(IA)$ $=$ $\{(J^i)_a \mid i$  is
a positive integer$\}$, by Theorem \ref{theo1}).
}
\end{rema}

\begin{coro}
\label{coro3} With the notation and assumptions of Theorem
\ref{theo1}, assume that  $R$  is a local ring with maximal ideal
$M$.  Then:

\noindent
{\bf{(\ref{coro3}.1)}}
If  $I$  is an
open ideal in  $R$, then there exists a finite
free local integral
extension ring  $A$  of  $R$  such that
$\mathbf P(IA)$  is projectively full.

\noindent
{\bf{(\ref{coro3}.2)}}
If  $I$ $=$ $M$ in (\ref{coro3}.1), then  $A$
$=$ $R[{x_1},\dots,{x_g}]$
is a finite
free local integral extension ring
of  $R$  whose maximal ideal
$N$ $=$ $({x_1},\dots,{x_g})A$
is projectively full.

\noindent
{\bf{(\ref{coro3}.3)}}
Assume that  $b_1,\dots,b_f$  ($f$ $\le$ $g$) in
(\ref{coro3}.2) are such that
$X$ $=$ $(b_1,\dots,b_f)R$  is a reduction of  $M$,
let  $A_0$ $=$
$R[{x_1},\dots,{x_f}]$,
and let $C$ $=$
$({x_1},\dots,{x_f})A_0$.
Then  $C$  is a reduction of the maximal ideal
$({x_1},\dots,{x_f},M)A_0$ $=$
$({x_1},\dots,{x_f},b_{f+1},\dots,b_g)A_0$  of
$A_0$, and  $C$  is projectively full.
\end{coro}

\begin{proof}
For (\ref{coro3}.1),
if  $R$  is local, then  $I$ $=$ $(b_1,\dots,b_g)R$ $\subseteq$
$M$, so  $A$
$=$ $R[{x_1},\dots,{x_g}]$  is a
local ring with maximal ideal  $(M,{x_1},\dots,{x_g})A$,
by Proposition \ref{remaprops1}.5, so
the conclusion follows from Theorem \ref{theo1}.

(\ref{coro3}.2) follows from (\ref{coro3}.1), since if  $I$ $=$ $M$,
then  $MA$ $=$ $({b_1},\dots,{b_g})A$
$\subseteq$ $({x_1},\dots,{x_g})A$,
so it follows that $A/(({x_1},\dots,{x_g})A)$
$=$ $R/M$, hence  $N$ $=$
$({x_1},\dots,{x_g})A$.

For (\ref{coro3}.3),
$X$  and  $M$ ($=$ $I$)  have the same Rees
valuation rings and Rees integers, since  $X$
is a reduction of  $M$, so
$C$  is projectively full by Theorem \ref{theo1}.  Also,
it is clear that  $C$ $\subseteq$ $(C,M)A_0$  and that
$(C,M)A_0$ $=$
$({x_1},\dots,{x_f} ,b_{f+1},\dots,b_g)A_0$
is the maximal ideal in  $A_0$.  Further,
$(MA_0)_a$ $=$ $(XA_0)_a$  (since  $X_a$ $=$ $M_a$ $=$ $M$  in  $R$)
$=$ $(C^c)_a$ $\subseteq$ $C_a$,
so  $MA_0$ $\subseteq$ $C_a$.  Therefore
$(C,M)A_0$ $\subseteq$ $C_a$, so  $C_a$
$=$ $(C,M)A_0$  (since  $(C,M)A_0$  is the maximal
ideal in  $A_0$), hence  $C$  is a reduction of
$(C,M)A_0$.
\end{proof}



\section{IDEALS WITH A REES INTEGER EQUAL TO ONE.}
The last part of Remark \ref{remay}.1  shows that if the number of
Rees valuation rings considered in Theorem \ref{theo1}  is one, then
the ideal $J$  of Theorem \ref{theo1}  has a Rees valuation ring $U$
such that the Rees integer of  $J$  with respect to  $U$  is equal
to one.  In this section we consider some consequences of this.

We begin with the following proposition.


\begin{prop}
\label{remax2}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$,
let $\mathbf R$ $=$ $R[u,tI]$, where  $t$  is an
indeterminate and  $u$ $=$ $1/t$, and let
$\mathbf R'$  be the integral closure of
$\mathbf R$  in its total quotient ring.
Then:

\noindent
{\bf{(\ref{remax2}.1)}}
$I$  has a Rees integer equal to one
if and only if
$u \mathbf R'$  has a primary component
that is prime.

\noindent
{\bf{(\ref{remax2}.2)}}
Every Rees integer of  $I$  is equal to one
if and only if
$u \mathbf R'$  is a radical ideal.

\noindent
{\bf{(\ref{remax2}.3)}}
If there exists an ideal  $K$  in  $\mathbf P(I)$  such
that some Rees integer of  $K$ is equal
to one, then $K$  and  $\mathbf P(I)$  are projectively full.
\end{prop}

\begin{proof}
For (\ref{remax2}.1),
it follows  from \cite[(2.3)]{CHRR2} that
the Rees valuation rings  $V$  of  $I$  correspond
to the rings  ${\mathbf R'}_p$, where  $p$
is a (height one) prime divisor of  $u \mathbf R'$,
and the Rees integer  $e$  of  $I$  with
respect to  $V$  is given by  $u {\mathbf R'}_p$
$=$ $p^e {\mathbf R'}_p$.
Therefore it follows that
$I$  has a Rees integer equal to one if and only if
$u \mathbf R'$  has
a (height one) prime divisor  $p$  such that
$u {\mathbf {R}'}_p$ $=$
$p {\mathbf {R}'}_p$.
The conclusion readily follows from this.

(\ref{remax2}.2) follows immediately from (\ref{remax2}.1).

For (\ref{remax2}.3), if some Rees integer of  $K$  is equal
to one, then the greatest common divisor of the Rees
integers of  $K$ is equal to one, so  $K$
and  $\mathbf P(I)$  are projectively full,
by \cite[(4.10)]{CHRR}.
\end{proof}

Concerning Proposition  \ref{remax2}.1,
some properties of a regular ideal
$I$  in a Noetherian ring  $R$  such that
$u \mathbf R$  (rather than  $u \mathbf R'$)
has a primary component that is prime
are noted in Examples \ref{exam1}.1 and \ref{exam1}.2  below.

\begin{lemm}
\label{lemm1}
Let  $I$  be a regular proper ideal in a Noetherian ring  $R$ and let
$e_1,\dots,e_n$  be all the Rees integers of  $I$.  Then:

\noindent
{\bf{(\ref{lemm1}.1)}}
$e_j$ $=$ $1$  for some  $j$ $\in$ $\{1,\dots,n\}$
if and only if there exists a minimal prime ideal  $z$
in  $R$  such that some Rees integer of
$(I+z)/z$  is equal to one.  If these hold,
then  $I$, $\mathbf P(I)$, $(I+z)/z$,
and  $\mathbf P((I+z)/z)$  are projectively full.

\noindent
{\bf{(\ref{lemm1}.2)}}
$e_j$ $=$ $1$    for some  $j$ $\in$ $\{1,\dots,n\}$
if and only if there exists a multiplicatively closed
subset  $S$  in  $R$  such that some Rees integer of
$IR_S$ is equal to one.  If these hold,
then  $I$, $\mathbf P(I)$, $IR_{S'}$,
and  $\mathbf P(IR_{S'})$  are projectively full for
all multiplicatively closed subsets  $S'$  of  $R$
such that  $P \cap S'$ $=$ $\emptyset$  (where  $P$
$=$ $N \cap R$  with  $(V,N)$  a Rees valuation
ring of  $I$  such that  $IV$ $=$ $N$).

\noindent
{\bf{(\ref{lemm1}.3)}}
Assume that  $(V,N)$  is a Rees valuation ring of  $I$  such
that the Rees integer of  $I$  with respect to
$V$  is equal to one.  Let  $B$  be a Noetherian domain
such that  $R/z$ $\subseteq$ $B$ $\subseteq$ $V$  for
some minimal prime ideal  $z$  in  $R$  ($z$
$=$ $(0)$, if  $R$  is an integral domain), and let
$K$  be an ideal in  $B$  such that  $IB$ $\subseteq$
$K$ $\subseteq$ $N \cap B$.  Then  $V$  is a Rees
valuation ring of  $K$  such that the Rees integer
of  $K$  with respect to  $V$  is equal to one,
so  $K$  is projectively full.  In particular,
$IB$  is projectively full,
\end{lemm}

\begin{proof}
For (\ref{lemm1}.1), the construction of Rees
valuation rings in \cite[(2.9)]{CHRR} shows that,
for each minimal prime ideal  $z$  in  $R$,
each Rees valuation ring of  $(I+z)/z$  is a Rees valuation ring
$(V,N)$  of  $I$  such that
the Rees integer of  $(I+z)/z$  with respect to  $V$  is the
Rees integer of  $I$  with respect to  $V$.  The same construction
shows that, for each Rees
valuation ring  $(V,N)$  of  $I$, there exists a
minimal prime ideal
$z$  in  $R$  such that  $V$  is
a Rees valuation ring of  $(I+z)/z$ and
the Rees integer of  $(I+z)/z$  with respect to  $V$  is the
Rees integer of  $I$  with respect to  $V$.
The conclusion clearly follows from this and Proposition  \ref{remax2}.3.

The proof of (\ref{lemm1}.2) is similar, so it will
be omitted.

For (\ref{lemm1}.3),
by hypothesis there exists  $b$ $\in$ $I$  such that
$bV$ $=$ $IV$ $=$ $N$.  Therefore  $b$ $\in$ $K$
$\subseteq$ $N$ $=$ $bV$,
so  $D$ $=$ $B[K/b]$ $\subseteq$ $V$.  Also,
$C$ $=$ $R[I/b]$ $\subseteq$ $D$, and  $N \cap C'$  is
a height one prime divisor of  $bC'$  (by \cite[(2.9)]{CHRR}).
Therefore it follows that  $N \cap D'$  is a height one
prime divisor of  $bD'$, so  $V$  is a Rees valuation
ring of $K$  (by \cite[(2.9)]{CHRR}).  Since
$N$ $=$ $bV$ $\subseteq$ $KV$ $\subseteq$ $N$, it
follows that the Rees integer of  $K$  with
respect to  $V$  is equal to one.
The remaining conclusions follow
from this and Proposition  \ref{remax2}.3.
\end{proof}

Example \ref{exam2} below concerns a special case
of Lemma \ref{lemm1}.3

We remark that the hypothesis ``$e$ is a unit in $V$'' in Corollary
\ref{coro4} holds if either: (i) $e$  is not a multiple of
$\Char(V_j/N_j)$; or, (ii) $\Char(V_j/N_j)$ $=$ $0$.

\begin{coro}
\label{coro4} Let  $I$  be a proper nonzero ideal in a Noetherian
ring  $R$ and assume that  $R$  has a Rees valuation ring  $(V,N)$
such that the Rees integer  $e$  of  $I$  with respect to  $V$  is a
unit of $V$. Then there exists a finite free integral extension ring
$A$ $=$ $R[x_1,\dots,x_g]$  of  $R$ and an ideal $J$ $=$
$(x_1,\dots,x_g)A$ in  $A$ such that  $J$  has a Rees integer equal
to one. Therefore
there exists a minimal prime ideal  $z^*$  in  $A$  such that
if  $B$  is a Noetherian domain between  $A/z^*$  and its integral
closure  $(A/z^*)'$, then there
exists a prime ideal  $P$ containing $JB$
such that each of  $P$, $JB$, $PB_P$, and  $JB_P$
has a Rees integer equal to one.
\end{coro}

\begin{proof}
Remark \ref{rema1}.2  shows that assumption (a) of Theorem
\ref{theo1} holds for $I$ with respect to $V$, and Remark
\ref{rema1}.4 shows that $\mathbf P(IA)$ $=$ $\mathbf P(J)$  is
projectively full. It follows from Remark \ref{remay}.1  that  $J$
has a Rees valuation ring $(U,M)$ such that the  Rees integer of $J$
with respect to $U$ is equal to one. The final statement
follows from this and Lemma
\ref{lemm1}.3.
\end{proof}


In Corollary \ref{coro4}, $P$  need  not be a minimal prime divisor
of $JB$; see Example \ref{exam2}  below.


\begin{rema}
\label{remanormal} {\em  It follows immediately from the last part
of Corollary \ref{coro4} (and Proposition \ref{remaprops1}.8)  that if
$R$ is a Noetherian domain, if $\Rad(I)$  is a prime ideal, and if
there exists only one prime ideal in the integral closure  $R'$  of
$R$ that lies over  $\Rad(I)$, then  $P{B}_{P}$  has a Rees integer
equal to one for each prime ideal $P$  in $B$
that lies over  $(J,\Rad(I))A$.}
\end{rema}

\section{EXAMPLES OF IDEALS WITH SOME REES INTEGER \\ EQUAL TO ONE.}
In Proposition \ref{remax2}.3 it was noted that if
$I$  is a regular proper ideal in a Noetherian ring  $R$
such that some Rees integer of  $I$  is equal to one, then
$I$  is projectively full.
In this section we give some examples of such ideals.

Concerning the conclusion of Example \ref{exam1}.2, recall that an
ideal $I$ is {\bf{normal}} in case each power  $I^n$  of  $I$  is
integrally closed.

\begin{exam}
\label{exam1} {\em Let  $I$  be a regular ideal in a Noetherian ring
$R$  and let   $\mathbf G(R,I)$ $=$ ${\Sigma_{i = 0}}^\infty ~I^i
/I^{i+1}$  denote its associated graded ring.

\noindent {\bf  (\ref{exam1}.1)}  If $\mathbf G(R,I)$ has a minimal
prime ideal  $p$ such that $p$ is its own $p$-primary component of $(0)$,
then $I$ has a Rees integer equal to one.

\noindent {\bf (\ref{exam1}.2)}  If $\mathbf G(R,I)$ is reduced,
then $I$ is a radical ideal and a normal ideal, and
each Rees integer of $I$ is equal to one. }
\end{exam}

\begin{proof}
Let  $\mathbf R$ $=$ $R[u,tI]$, where  $t$  is an indeterminate and
$u$ $=$ $1/t$. It is shown in \cite{Re2} that:  $\mathbf G(R,I)$ $=$
$\mathbf R /(u \mathbf R)$; $u$  is a regular element in $\mathbf
R$; and, $u^n \mathbf R \cap R$ $=$ $I^n$  for all positive integers
$n$.

For the proof of (\ref{exam1}.1), observe that $u \mathbf R_p$ $=$
$p \mathbf R_p$  implies that $\mathbf R_p$ is a discrete valuation
ring. It follows that  $p'$ $=$ $p \mathbf R_p \cap \mathbf R'$  is
the  $p'$-primary component of  $u \mathbf R'$,
so one of the Rees integers of $I$ is
equal to one by Proposition \ref{remax2}.1.

For the proof of (\ref{exam1}.2), if $\mathbf G(R,I)$ is
a radical ideal, then
$u \mathbf R$ is a radical ideal.  Therefore
it follows from
\cite[(33.11)]{N2} that  $u \mathbf R'$  is
a radical ideal, so each Rees integer of  $I$  is
equal to one by Proposition \ref{remax2}.2.
Also, $I$ $=$
$u \mathbf R \cap R$  is a radical ideal. Further, $u \mathbf R_q$ $=$
$q \mathbf R_q$  for each (minimal) prime divisor  $q$  of  $u
\mathbf R$, so each  $\mathbf R_q$ is a discrete valuation ring.
It follows that, for all positive
integers $n$, $u^n \mathbf R$ $=$ $\cap \{ u^n \mathbf R_q \cap
\mathbf R \mid q \in \Ass (\mathbf R/(u \mathbf R))\}$ (by
\cite[(12.6)]{N2}) and that each  $u^n \mathbf R_q \cap \mathbf R$
is integrally closed, so $u^n \mathbf R$ $=$ $(u^n \mathbf R)_a$, by
\cite[Lemma 4]{R2}. Therefore  $I^n$ $=$ $u^n \mathbf R \cap R$ $=$
$(u^n \mathbf R)_a \cap R$ $=$ ${I^n}_a$  (by \cite[Lemma 2.5]{R3})
for all positive integers  $n$, so it follows that  $I$  is a normal
ideal.
\end{proof}

Several specific examples of ideals  $I$  as in
Example \ref{exam1}.2 are given in Example \ref{exam6}.
We delay giving these examples till the next section,
since they are also examples of a Noetherian domain  $R$
with a proper finite integral extension domain  $A$  such
that  $\mathbf P(IA)$  is projectively full for all
nonzero ideals  $I$  of  $R$, and since they are also
closely related to Examples \ref{5.5}.4 and \ref{5.5}.5.

\begin{exam}
\label{exam2} {\em Let  $I$  be a regular ideal in a Noetherian ring
$R$  such that the center  $q$  in  $R$  of some Rees valuation ring
$(V,N)$  of $I$  is not a  minimal prime divisor of  $I$  and the
Rees integer $e$  of  $I$  with respect to  $V$  is a unit of $V$.
Let $b_1,\dots,b_g$  be a basis of  $I$  such that  $b_iV$ $=$ $N^e$
for $i$ $=$ $1,\dots,g$ (see Remark \ref{rema1}.2), let  $A$ $=$
$R[{x_1},\dots,{x_g}]$, let  $J$ $=$ $({x_1},\dots,{x_g})A$ be
as in Corollary \ref{coro4}, and let  $(U,M)$  be the extension
of  $V$  to a Rees valuation ring of  $J$  as in the proof
of Theorem \ref{theo1}. Then  $(J,q)A$ $=$ $M \cap A$,
$(J,q)A$  properly contains a
minimal prime divisor of  $J$, and {\it{every}} ideal  $H$  between
$J$ and $(J,q)A$  has Rees integer equal to one with respect to $U$.}
\end{exam}

\begin{proof}
It follows from the hypothesis concerning  $q$  and
Proposition \ref{remaprops1}.5 that  $(J,q)A$  is a
prime ideal that properly contains a minimal prime
divisor of  $J$.  Therefore the conclusion
follows immediately from Corollary \ref{coro4}  and Lemma
\ref{lemm1}.3.
\end{proof}

\begin{exam}
\label{exam3} {\em Let  $I$  be a nonzero ideal in a Noetherian
domain  $R$  such that   $I$  has a unique Rees valuation ring
$(V,N)$ and the Rees integer $e$  of  $I$  with respect to  $V$  is
a unit of $V$.  Let $b_1,\dots,b_g$  be a basis of  $I$  such that
$b_iV$ $=$ $N^e$  for  $i$ $=$ $1,\dots,g$ (see Remark
\ref{rema1}.2) and let $A$ $=$ $R[{x_1},\dots,{x_g}]$ and $J$ $=$
$({x_1},\dots,{x_g})A$ be as in Corollary \ref{coro4}. Then  $J_a$
is a prime ideal.  Also, each prime ideal in each Noetherian
ring  $A^+$  between  $A$  and its integral closure
$A'$ that lies over  $J_a$  has a
Rees integer that is equal to one. }
\end{exam}

\begin{proof}
The hypothesis implies that  $\Rad(I)$  is a prime ideal and that
there exists a unique prime ideal in  $R'$  that lies over
$\Rad(I)$.  Therefore the last statement follows from Corollary
\ref{coro4}.

Also, $J_a$ $=$ $\cap \{JU_i \cap A \mid U_i$ is a Rees
valuation ring of  $J\}$, by \cite[Theorem 4.12, page 61]{REES} (or
by \cite[(2.5)]{R3} together with \cite[(2.3)]{CHRR2}), and each
such $U_i$ is an extension of  $V$, so the maximal ideal $M_i$ of
$U_i$ lies over the maximal ideal  $N$ of $V$  (so $M_i \cap R$ $=$
$\Rad(I)$), and  $JU_i$ $=$ $M_i$ (since the Rees integer of $J$
with respect to  $U_i$  is equal to one (by Proposition
\ref{remaprops1}.8)), so $JU_i \cap A$ $=$ $M_i \cap A$. Further,
there exists a one-to-one correspondence between the minimal prime
divisors of  $I$ and the minimal prime divisors of $J$, by Proposition
\ref{remaprops1}.5, so it follows that  $J_a$  has a unique minimal
prime divisor and that $J_a$  is a prime ideal.
\end{proof}

Example \ref{exam4} generalizes Example \ref{exam3}.


\begin{exam}
\label{exam4} {\em Let  $R$, $I$, $(V_1,N_1),\dots,(V_n,N_n)$,
$e_1,\dots,e_n$, $A$, and  $J$  be as in Theorem \ref{theo1}, and
let $p_1,\dots,p_h$ be the distinct prime ideals in  $\{N_j \cap R
\mid j = 1,\dots,n\}$  (subscripted so that  $p_j$ $=$ $N_j \cap
R$). Assume that  $e_1$ $=$ $\cdots$ $=$ $c_h$ $=$ (say) $e$  is not
in $N_j$  for  $j$ $=$ $1,\dots,n$ and that $p_1,\dots,p_h$  are
minimal prime divisors of  $I$. Then  $J_a$  has  $h$  primary
components that are prime ideals and each of them has a
Rees integer equal to one. In particular, if  $p_1,\dots,p_h$
are all the minimal prime divisors of  $I$, then  $J_a$  is a
radical ideal that is the intersection of  $h$  (and no fewer)
minimal prime divisors. }
\end{exam}

\begin{proof}
It follows from the fourth paragraph of the proof of Theorem
\ref{theo1}  that the Rees integer of  $J$  with respect to each of
its Rees valuation rings  $(U_1,M_1),\dots,(U_h,M_h)$ (with $U_j$
the extension of  $V_j$  constructed in the proof of Theorem
\ref{theo1}) is equal to one.  Therefore  $JU_j$ $=$ $M_j$, so it
follows as in the proof of Example \ref{exam3} that  $M_j \cap A$
$=$ $(J,p_j)A$, that $J_aA_{(J,p_j)A}$ $=$ $(J,p_j)A_{(J,p_j)A}$
for $j$ $=$ $1,\dots,h$, and that each  $(J,p_j)A$  has a Rees
integer equal to one.  Also, there exists a one-to-one
correspondence between the minimal prime divisors $p$  of  $I$  and
the minimal prime divisors  $P$  of $J$  (given by  $P$ $=$
$(J,p)A$), by Proposition \ref{remaprops1}.5.  The conclusions clearly
follow from this.
\end{proof}


\section{EXAMPLES OF PROJECTIVELY FULL IDEALS.}
In \cite[Section 4]{CHRR2} we give a number of examples of projectively
full ideals.
In this section we give some additional examples.


\begin{exam} \label{5.5} {\em
Let $R$ be a Noetherian domain,
let  $R'$  be the integral closure of  $R$  in  its quotient field,
and let $R^+$ $\subseteq$ $R'$  be a Noetherian integral
extension domain of $R$.
Let $I$ be a nonzero proper ideal of $R$.

\noindent {\bf (\ref{5.5}.1)}
We have $\Rees I = \Rees IR^+$. Also,
for each $V \in \Rees I$,  the Rees integer of $I$ with respect to
$V$ is equal to the Rees integer of $IR^+$ with respect to $V$. Thus
the  $\gcd$ of the Rees integers of $I$ is equal to the $\gcd$ of
the Rees integers of $IR^+$.

\noindent {\bf (\ref{5.5}.2)}   If $IR^+$ is projectively full in
$R^+$, then $I$ is projectively full in $R$.

\noindent {\bf (\ref{5.5}.3)}  It is possible for $I$ to be
projectively full, while $IR^+$ is not projectively full.

\noindent {\bf (\ref{5.5}.4)}  It is possible for $\mathbf P(I)$
to be projectively full in $R$, while $\mathbf P(IR^+)$ is not
projectively full in $R^+$.

\noindent {\bf (\ref{5.5}.5)}   It is possible for  $\mathbf
P(IR^+)$ to be projectively full, while $\mathbf P(I)$ is not
projectively full.


}
\end{exam}


\begin{proof} To establish (\ref{5.5}.1),
since  $R^+$  is contained in the quotient field
of  $R$, we have $\Rees I = \Rees IR^+$
and  $R^+ \subseteq V$ for each $V \in \Rees I$.
Also, $IV = (IR^+)V$, so the Rees integer
of $I$ with respect to each $V_i$ is the same as the Rees integer
of $IR^+$ with respect to $V_i$.  The
last statement in (\ref{5.5}.1) is clear from this.

(\ref{5.5}.2) is proved in \cite[(3.2)(1)]{CHRR2}.

For (\ref{5.5}.3), we use \cite[Example 3.4]{MRS}.  Let $X$ and $Y$
be indeterminates over a field $E$, let $R^+ = E[X, Y]$ and let $R =
E[X^2, XY, Y]$  (so  $R^+$ $=$ $R'$). Then $I = X^2R$ is
projectively full, but $X^2R^+$ is not projectively full.

For (\ref{5.5}.4),  let $R = E[X^2, XY, Y]$ as in the proof of (\ref{5.5}.3),
and let $R^+ = R[X^3] = E[X^2, X^3, XY, Y]$. Since $I = X^2R$
is projectively full, $\mathbf P(I)$ is projectively full. However,
$(IR^+)_a = (X^2, X^3)R^+ := J$ is such that $\mathbf P(J)$ is not
projectively full in $R^+ = E[X^2, X^3, XY, Y]$. For if $H := (X^3,
X^4)R^+$, then $J^3 = H^2 = (X^6, X^7)R^+$  (so  $J$  and
$H$  are projectively equivalent), and $J$ and $H$ are not
the integral closure of powers of any ideal of $R^+$.


 For (\ref{5.5}.5), let $X$ be an
indeterminate over a field $E$,  let $R = E[[X^2, X^3]]$,  and let
$I = (X^2, X^3)R$ be the maximal ideal of $R$. Let  $R^+ = E[[X]]$
(so  $R^+$ $=$ $R'$).
Then $R^+$  is a DVR, so $\mathbf P(IR^+)$ is projectively full. Let
$J = (X^3, X^4)R$. Then $J^2 = I^3 = (X^6, X^7)R$, so it follows that
$\mathbf P(I)$ is not projectively full.
\end{proof}

\begin{ques} {\em Does there exist an example of a Noetherian domain
$R$ for which  Example \ref{5.5}.4  holds with $R^+$ taken to be the
integral closure $R'$ of $R$? }
\end{ques}

In Example \ref{rema3exam} we present several examples where $R$ is a
Noetherian domain that is not integrally closed and  $\mathbf P(I)$
is projectively full for all nonzero proper ideals $I$  of $R$.
The following lemma will be used in explaining why
these examples hold.


\begin{lemm}
\label{exam5} Let  $(R,M)$  be a local domain
and let  $R^+$  be a Noetherian integral extension
domain of  $R$.  Assume that  $M$  is the
Jacobson radical of  $R^+$  and that  $\mathbf P(IR^+)$
is projectively full for all nonzero proper ideals
$I$  of  $R$.  Then  $\mathbf P(I)$  is
projectively full for
all nonzero proper ideals
$I$  of  $R$.
\end{lemm}

\begin{proof}
The hypothesis  that  $R$  and  $R^+$  have the same Jacobson
radical implies  that $H$ $\subseteq$ $M$ $\subset$ $R$  for each
ideal $H$ in  $R^+$  that is projectively equivalent to $IR^+$. The
conclusion readily follows from this.
\end{proof}

In the three examples in Example \ref{rema3exam}, the
Noetherian integral extension
domain  $R^+$  of Lemma \ref{exam5}
is chosen to be the integral closure  $R'$  of  $R$.

\begin{exam}
\label{rema3exam}
{\em For the following rings  $R$,
$\mathbf P(IR)$
is projectively full for all nonzero proper ideals
$I$  of  $R$.}

{\bf{(\ref{rema3exam}.1)}}
{\em
Let $E$
be a finite algebraic extension field of a field $F$,
let  $X$  be an indeterminate,
and let $R'$ $=$
$E[[X]]$ and $R = F + XR'$.

\noindent
{\bf{(\ref{rema3exam}.2)}}
Let $F$ $\subset$ $E$  be as in (\ref{rema3exam}.1),
let  $X,Y$  be indeterminates,
and let $R'$ $=$
$E[[X,Y]]$ and $R = F + (X,Y)R'$.

\noindent
{\bf{(\ref{rema3exam}.3)}}
Let  $R$ $\subset$ $R'$  be as in
\cite[Example 2, pp. 203-205]{N2} in the case where
$m$ $=$ $0$  and  $r$ $=$ $2$.}
\end{exam}

\begin{proof}
For (\ref{rema3exam}.1), since
$E[[X]]$  is a
discrete valuation ring,
it follows from Lemma
\ref{exam5} that
$\mathbf P(IR)$
is projectively full for all nonzero proper ideals
$I$  of  $R$.

For (\ref{rema3exam}.2), since $R'$  is a regular local ring of
altitude two, it follows from Lemma \ref{exam5} and either
\cite[(3.6)]{MRS} or \cite[(4.13)]{CHRR} that $\mathbf P(IR)$ is
projectively full for all nonzero proper ideals $I$  of  $R$.

For (\ref{rema3exam}.3),
it is shown in \cite{N2} that:  $\dim(R)$ $=$ $2$; the integral
closure  $R'$  of  $R$  is a unique factorization regular
domain with exactly two maximal ideals  $M$ $=$ $xR'$  and
$N$; ${R'}_M$  is a discrete valuation ring and ${R'}_N$  is a
regular local domain of altitude two; $M \cap N$  is the maximal
ideal of  $R$; and, $R'$ $=$ $R+eR$  for all elements $e$ $\in$ $R'
- R$. Using these it can be shown that,
for each nonzero ideal
$I$ in  $R$, $IR'$ $=$ $x^i q$  ($=$
$I{R'}_M' \cap I{R'}_N$)  for some positive
integer  $i$  and for some ideal  $q$  in  $R'$  such that
$q$ $\subseteq$ $N$  and  $q$ $\nsubseteq$ $M$.
Since  ${R'}_N$  is a regular local domain of altitude
two, it follows that  $q_a$ $=$ ${Q^m}_a$  for some
positive integer  $m$, where  $Q$  is the largest
element in the projectively full projective equivalence
class  $\mathbf P(q)$
(see either  \cite[(3.6)]{MRS}
or \cite[(4.13)]{CHRR}).
Then, since projectively equivalent ideals  $H,K$
have the same Rees valuation rings and proportional
Rees integers (by \cite[Proposition 2.10]{MRS} and
\cite{CHRR}), it follows
that  $\mathbf P(IR')$  is projectively full
with largest ideal  $x^{i/c}(Q^{m/c})_a$, where
$c$  is the greatest common divisor of  $i$ and  $m$.
The conclusion follows from this and
Lemma \ref{exam5}.
\end{proof}


\begin{rema} \label{5.11} {\em

If the Noetherian domain $R$ has a finite integral extension domain
$R^+$  that is a regular local domain of altitude two, then
\cite[(3.6)]{MRS} or \cite[(4.13)]{CHRR} implies that $\mathbf
P(IR^+)$ is projectively full for every nonzero proper ideal $I$ of
$R$. We present in Example \ref{exam6} specific examples of such
rings $R$.}
\end{rema}




\begin{exam}
\label{exam6} {\em  Let  $F$  be a field, let  $X,Y$ be
indeterminates, let  $n$ be a positive integer, let
$R_n$ $=$ $F[[ \{X^{n-i}Y^i\}_{i=0}^n  ]]$, and let
$M_n$ $=$ $( \{X^{n-i}Y^i\}_{i=0}^n  )R_n$.  Then  $R_1$ $=$
$F[[X,Y]]$ is a finite integral extension domain of  $R_n$  and a
regular local domain of altitude two.  Therefore  $\mathbf P(IR_1)$ is
projectively full for each  nonzero proper ideal   $I$  in  $R_n$.
Also, $M_n$  is a projectively full
normal ideal that has
only one Rees valuation ring  $V_n$  and its Rees integer
with respect to  $V_n$  is equal to  one.}
\end{exam}

\begin{proof}
That  $\mathbf P(IR_1)$  is projectively
full is immediate from Remark \ref{5.11}.

For the last statement, note first that  $R_n[M_n/X^n]$
$=$ $R_n[Y/X]$  (since  $\frac{X^{n-i}Y^i}{X^n}$
$=$ $\frac{Y^i}{X^i}$  for  $i$ $=$ $1,\dots,n$).
For each positive integer  $j$
let  $C_j $ $=$  $R_j[M_j/X^j]$  and let
${C_j}'$  be the integral closure of  $C_j$.
Then, in particular, $C_1$ $=$ $R_1[Y/X]$, and
it is well known that  $C_1$ $=$ ${C_1}'$
and that  $XC_1$  is a prime ideal
such that  $(C_1)_{XC_1}$  is the ord valuation
ring of  $M_1$  (and the only Rees valuation
ring of  $M_1$).  Also, $C_n[X]$  (resp.,
${C_n}'[X]$)  is a free integral
extension domain of  $C_n$ (resp.,
${C_n}'$), and  $Y$ $=$ $X(Y/X)$
$\in$ $C_n[X]$ (so $R_1$ $\subset$
$C_n[X]$), so it follows that  $C_1$
$=$ $C_n[X]$ $=$ ${C_n}'[X]$  is
a free integral
extension domain of  $C_n$  and of  ${C_n}'$.
Therefore, since  $C_n$ $\subseteq$
${C_n}'$, it follows that  $C_n$ $=$
${C_n}'$.  Also, $X^nC_1$  is  $XC_1$-primary,
so it follows that  $X^nC_n$  is primary for
$p_n$ $=$ $XC_1 \cap C_n$.  Since the Rees
valuation rings of  $M_n$  are
the rings  ${(C_n}')_{p_i}$, where the
$p_i$  are the (height one) prime divisors of
$X^n{C_n}'$ ($=$ $X^nC_n$),
it follows that $V_n$ $=$ $(C_n)_{p_n}$
is the only Rees valuation ring
of  $M_n$.

To see that  $M_n$  is a normal projectively
full ideal and that the Rees integer
of  $M_n$  with respect to  $V_n$  is
equal to one, it suffices (by Example \ref{exam1}.2)
to show that  $X^n C_n$  is a prime ideal.

For this, since  $X^n,Y^n$  is a system of
parameters in  $R_n$, it is well known
that  $P$ $=$ $M_nR_n[Y^n/X^n]$  is a prime ideal
and that the  $P$-residue class  $T$  of
$Y^n/X^n$  is transcendental over
$F$ $=$ $R_n/M_n$  (so
$R_n[Y^n/X^n]/(M_nR_n[Y^n/X^n])$ $=$
$F[T]$  is a polynomial ring
over  $F$).  Also,
$X^nC_n$ $=$ $M_nC_n$  (since  $X^{n-i}Y^i$
$=$ $X^n(Y^i/X^i)$ $\in$ $X^nC_n$  for  $i$
$=$ $0,1,\dots,n$), so  $C_n/(X^nC_n)$ $=$
$F[\overline{Y/X}]$.
Further,
$C_n$ $=$ $R_n[Y/X]$  is a finite
integral extension ring of  $R_n[Y^n/X^n]$, so
$P$ $=$ $M_nR_n[Y^n/X^n]$ $=$ $M_nC_n \cap R_n[Y^n/X^n]$
$=$ $X^nC_n \cap R_n[Y^n/X^n]$.
It follows that  $F[ \overline{Y/X}]$ $=$
$C_n/(X^nC_n)$  is a finite
integral extension ring of the
polynomial ring
$R_n[Y^n/X^n]/(M_nR_n[Y^n/X^n])$ $=$
$F[T]$, hence  $X^nC_n$  is a
prime ideal.
\end{proof}


\begin{rema} \label{5.12} {\em
If one applies the construction in Theorem \ref{theo1} to the ring
$R_n$ of Example \ref{exam6} and the  set $\{X^{n-i}Y^i\}_{i=0}^n$
of generators of the  ideal $M_n = (\{X^{n-i}Y^i\}_{i=0}^n )R_n$ of
$R_n$, one obtains a finite free integral extension ring $A_n$ of
$R_n$. By Remark \ref{rema0}  there exists a minimal prime ideal
$z^*$ in $A_n$ such that  $A_n/z^*$ $=$ $R_n[(X^n)^{1/n},
(X^{n-1}Y)^{1/n}, \dots,(Y^n)^{1/n}]$  is a proper finite integral
extension domain of $R_1 = F[[X, Y]]$.  However, if instead of
applying the construction in Theorem \ref{theo1} to the ideal $M_n$,
we instead apply it to the generators $X^n, Y^n$ of the reduction
$(X^n,Y^n)R_n$  of  $M_n$, then the free integral extension ring
$A_n$ $=$ $R_n[T_1,T_2]/({T_1}^n - X^n,{T_2}^n-Y^n)$  of Theorem
\ref{theo1} has a minimal prime ideal $z^*$  such that $A_n/z^* =
R_1 = F[[X,Y]]$.

}
\end{rema}

In Example \ref{nonreduced} we present an example of a normal local
domain $(R,M)$ of altitude two such that $M$ is projectively full
and the associated graded ring $\mathbf G(R,M)$ is not reduced.

\begin{exam}
\label{nonreduced} {\em Let $F$ be an algebraically closed field
with $\Char F = 0$, and let $R_0$ be a regular local domain of
altitude two with maximal ideal $M_0 = (x, y)R_0$ and coefficient
field $F$, e.g., $R_0 = F[x, y]_{(x, y)}$, or $R_0 = F[[x, y]]$,
where $x$ and $y$ are indeterminates over $F$. Let $R = R_0[z]$,
where $z^2 = x^3 + y^j$, where $j \ge 3$. It is readily checked that
$R$ is a normal local domain of altitude two with maximal ideal $M =
(x, y, z)R$, and that $\mathbf G(R,M)$ is not reduced. We prove that
$M$ is projectively full.

}
\end{exam}

\begin{proof} The unique Rees valuation ring of $M_0$ is $V_0 =
R_0[y/x]_{xR_0[y/x]}$. Notice that $I = (x, y)R$ is a reduction of
$M$ since $z$ is integral over $I$. It follows that every Rees
valuation ring of $M$ is an extension of $V_0$. Let $V$ be a Rees
valuation ring of $M$ and let $v$ denote the normalized valuation
with value group $\Z$ corresponding to $V$. Then $v(x) = v(y)$ and
the image of $y/x$ in the residue field of $V$ is transcendental
over $F$. Since $z^2 = x^3 + y^j$ and $j \ge 3$, we have
 $$
2v(z) = v(z^2) = v(x^3 + y^j) = 3v(x). $$ It follows  that  $v(x) =
2$ and $v(z) = 3$. Therefore $V$ is ramified over $V_0$.  This
implies  that $V$ is the unique extension of $V_0$ and thus the
unique Rees valuation ring of $M$.

For each positive integer $n$, let $I_n = \{ r \in R \, | \, v(r)
\ge n \}$. Thus $I_2 = M$. Since $V$ is the unique Rees valuation
ring of $M$, we have $I_{2n} = (M^n)_a$ for each $n \in \mathbb N$.
To show $M$ is projectively full, we prove that  $V$ is not the
unique Rees valuation ring of $I_{2n+1}$ for each $n \in \mathbb N$.
Consider the inclusions
$$
M^2 \subseteq I_4 \subset (z, x^2, xy, y^2)R := J \subseteq I_3 \subset
M.
$$
Since $\lambda(M/M^2) = 3$ and since the images of $x$ and $y$ in
$M/M^2$ are $F$-linearly independent, \,\,  $J = I_3$ and $M^2 = I_4
= (M^2)_a$. Since $x^3 = z^2 - y^j$ and $j \ge 3$, $L = (z, y^2)R$
is  a reduction of $I_3 = (z, x^2, xy, y^2)R$. \,\, Indeed, $(x^2)^3
\in L^3$ and $(xy)^3 \in L^3$ implies $x^2$ and $xy$ are integral
over $L$. It follows that $V$ is not a Rees valuation of $I_3$, \,
for $zV \ne y^2V$. Consider $M^3 \subset I_3M \subseteq I_5 \subset
I_4 = M^2$. Since the images of $x^2, xy, y^2, xz, yz$ in $M^2/M^3$
are an $F$-basis, it follows that $I_3M = I_5$ and $M^3 = (M^3)_a =
I_6$. Proceeding by induction, we assume $M^{n+1} = (M^{n+1})_a =
I_{2n+2}$, and consider $$  M^{n+2} \subset I_3M^n \subseteq
I_{2n+3} \subset M^{n+1} = I_{2n+2}.$$ Since the images in
$M^{n+1}/M^{n+2}$ of $\{x^ay^b \, | \, a + b = n+1 \} \cup \{zx^ay^b
\, | \, a + b = n \}$ is an $F$-basis,
 $\lambda(M^{n+1}/M^{n+2}) = 2n+3$, \quad and the inequalities
 $\lambda(M^{n+1}/I_{2n+3}) \ge n+2$ \quad and \quad
$\lambda(I_3M^n/M^{n+2}) \ge n+1$ imply $I_3M^n = I_{2n+3}$ \quad
and \quad $M^{2n+2} = (M^{2n+2})_a$. Therefore the ideal $I_{2n+3}$
has a Rees valuation ring different from $V$, and thus is not
projectively equivalent to $M$. We conclude that $M$ is projectively
full. We have also shown that $M$ is a normal ideal.
\end{proof}




\begin{rema} \label{rational} {\em
In \cite{Lipman}, Joseph Lipman extends Zariski's theory of complete
ideals of a regular local domain of altitude two to a situation
where $R$ is a normal local domain of altitude two that has a
rational singularity. Lipman proves that $R$ satisfies unique
factorization of complete ideals if and only if the completion of
$R$ is a UFD. For $R$ having this property, it follows that $\mathbf
P(I)$ is projectively full each nonzero proper ideal $I$. An example
to which this applies is $R = F[[x, y, z]]$, where $F$ is a field
and $z^2 + y^3 + x^5 = 0$. In \cite[Corollary 3.11]{Gohner}, Hartmut
G\"ohner proves that if $(R,M)$ is a normal local domain of altitude
two that has a rational singularity, then the set of complete
asymptotically irreducible ideals associated to a prime $R$-divisor
$v$ consists of the powers of an ideal $A_v$ which is uniquely
determined by $v$. In our terminology, this says that if $I$ is a
nonzero proper ideal of $R$ having only one Rees valuation ring,
then $\mathbf P(I)$ is projectively full. G\"ohner's proof involves
choosing a desingularization $f : X \to \Spec R$ such that $v$ is
centered on a component $E_1$ of the closed fiber on $X$. Let $E_2,
\ldots , E_n$ be the other components of the closed fiber on $X$.
Let $E_X$ denote the group of divisors having the form
$\sum_{i=1}^nn_iE_i$, where $n_i \in \Z$. Define
$$E_X^+ = \{D \in
E_X \, | \, D \ne 0 \text{ and } (D \cdot E_i) \le 0  \text{ for all
} 1 \le i \le n \}
$$
and $$E_X^{\#} = \{ D \in E_X \, |\, D \ne 0 \text{ and } O(-D)
\text{ is generated  by its sections over } X \}.
$$
Lipman shows in \cite{Lipman} that  $E_X^{\#} \subseteq E_X^+$ and
that  equality holds  if $R$ has a rational singularity. Also, if $D
= \sum_in_iE_i \in E_X^+$, then negative-definiteness of the
intersection matrix $(E_i \cdot E_j)$ \,\, implies $n_i \ge 0$ for
all $i$. For if $D \in E_X^+$ and $D = A - B$, where $A$ and $B$ are
effective, then $(A-B \cdot B) \le 0$ and $(A \cdot B) \ge 0$ imply
$(B \cdot B) \ge 0$, \,\, so $B = 0$.  Let $v = v_1, v_2, \ldots,
v_n$ denote the discrete valuations corresponding to $E_1, \ldots,
E_n$.  Associated with  $D = \sum_in_iE_i \in E_X^{\#}$ one defines
the complete $M$-primary ideal $I_D = \{ r \in R \, |\, v_i(r) \ge
n_i \text{ for } 1 \le i \le n \}$. This sets up  a one-to-one
correspondence between elements of $E_X^{\#}$ and complete
$M$-primary ideals that generate invertible $O_X$-ideals.   Lipman
suggested to us the following proof that $\mathbf P(I)$ is
projectively  full for each complete $M$-primary ideal $I$ if $R$
has a rational singularity. Fix a desingularization $f : X \to \Spec
R$ such that $I$ generates an invertible $O_X$-ideal and let $D =
\sum_in_iE_i \in E_X^{\#}$ be the divisor associated to $I$. Let $g
= \gcd \{ n_i \} $. Since $E^+ = E^{\#}$, $(1/g) D \in E^{\#}$. The
ideals $J \in \mathbf P(I)$ correspond to divisors in $E^{\#}$ that
are integral multiples of $(1/g)D$. Thus if $K$ is the complete
$M$-primary ideal associated to $(1/g)D$, then each $J \in \mathbf
P(I)$ is the integral closure of a power of $K$, so $\mathbf P(I)$
is projectively full. Since the rings
 $R_n$ $=$ $F[[ \{X^{n-i}Y^i\}_{i=0}^n ]]$ as in Example
\ref{exam6} are normal local domains of altitude two that have
rational singularities, it follows that  $\mathbf P(I)$ is
projectively full for each nonzero proper ideal $I$ of $R_n$.


}
\end{rema}








\noindent {\bf{ACKNOWLEDGMENT OF PRIORITY}}:
\newline
(i) In \cite[Remark 4.2(d)]{CHRR}  we noted that it was shown in
\cite[(2.9)]{MRS} that if  $I$  is a regular ideal in a
Noetherian ring  $R$, then there exists a positive
integer  $d$  such that, for all ideals  $J$  in  $R$
that are projectively equivalent to  $I$,
$(J^d)_a$ $=$ $(I^n)_a$  for some positive integer  $n$.
This result was also proved in \cite[(1.4)]{MR}.

\noindent (ii) In \cite[Proposition 3.3]{CHRR} we showed that $\Rees
I \cup \Rees J$ $=$ $\Rees IJ$  if $\dim(R) \leq 2$, and we noted
just prior to \cite[Proposition 3.3]{CHRR} that for the case that
$R$ is a pseudo-geometric normal Noetherian domain, this result
appears in \cite[Lemma 2.1]{Gohner}. The equality $\Rees I \cup
\Rees J$ $=$ $\Rees IJ$ was first proved for an equicharacteristic
integrally closed analytically irreducible local domain of dimension
two in \cite[Theorem 3.17]{P}.


\bigskip


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\end{thebibliography}


\begin{flushleft}
Department of Mathematics, North Dakota State University, Fargo, North Dakota, 58105-5075
{\em E-mail address: catalin.ciuperca@ndsu.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}
