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\vskip 4ex
\baselineskip = 12pt
\centerline{\large INVARIANTS OF IDEALS HAVING}
\vskip 3ex
\centerline{\large PRINCIPAL REDUCTIONS}
\vskip 4ex
\centerline{\sc Marco D'Anna, Anna Guerrieri and William Heinzer}
\vskip 1cm
\noindent
{\bf Abstract.} For a regular ideal having a principal reduction
in a Noetherian ring we consider the structural
numbers that arise from taking the Ratliff-Rush closure of
the ideal and its powers. In particular, we analyze the
interconnections among these numbers and the relation type
and reduction number of the ideal. We prove that certain
inequalites hold in general among these invariants,
while for ideals contained in the
conductor of the integral closure of the ring
we obtain sharper results that do not hold in general.
We provide applications to the one-dimensional
local setting and present a sequence of examples in this context.
\document
\baselineskip=17pt
\section{Introduction.}
%\vskip 1.5cm
Given a regular ideal $I$ in a Noetherian ring $R$ it is possible
to construct several filtrations and, in turn, various graded rings
whose structures provide information about the geometrical properties
of $I$. Two very well-known examples are the $I$-adic filtration and
the filtration obtained by taking the Ratliff--Rush closure of
the powers of $I$, the so called {\it Ratliff--Rush filtration}.
A wealth of work has been produced, in the past years, on these filtrations,
especially because of the role they play in the theory
of the Hilbert
function of $I$ when $R$ is a local ring with
maximal ideal $\m$ and $I$ is $\m$-primary.
These two filtrations coincide asymptotically but can differ greatly
in the first steps. To be able to control these differences
means to be able to control the properties of the associated graded ring
of $I$. For example, the associated graded ring of $I$ contains a
regular element if and only if the $I$-adic filtration and the
Ratliff--Rush filtration coincide from the very first step.
In the present work we study the interplay between these two
filtrations when $I$ is a regular ideal having a principal reduction;
if $R$ is local this corresponds to the case in which the special
fiber of the Rees ring of $I$, the {\it fiber cone} of $I$,
is a one-dimensional graded ring. (For a characterization of analytic
spread one ideals in local rings see \cite{RR2}.)
To carry out this work, we analyze the interconnections among some structural
numbers that arise from taking the Ratliff--Rush closure of $I$ and of its
powers and some classical invariants related to $I$: the reduction
number and the relation type.
For example, we investigate the mutual relations
among the following, where $\N$ denotes the nonnegative integers:
$$
\align
r&= r(I)
= \text{min}\{ n \in \N \ | \ I^{n+1}=xI^n \text{ for some x } \in I\} \\
k&= k(I)
= \text{min}\{ n \in \N \ | \ \I= (I^{n+1}: I^n )\} \\
h&= h(I) =
\text{min}\{ n \in \N \ | \ \widetilde {I^m}
= I^m\ \text{for all}\ m\ge n\}.
\endalign
$$
The integer $r$ denotes the {\it reduction number} of $I$.
In defining $k$, we are using the
{\it Ratliff--Rush closure} $\I$ of $I$, which is by
definition
$$
\I := \bigcup_{n \in \N}(I^{n+1}:_R I^n).
$$
We refer to the paper of Ratliff and Rush \cite{RR1} where
this concept was first introduced for a description
of other properties of this object.
We call $\I$ a {\it Ratliff--Rush ideal},
and refer to the integer $k$ as {\it the Ratliff--Rush number of}
$I$.
In \cite{RR1, Remark~(2.3)} it is proved that
$\widetilde {I^m} = I^m$
for all sufficiently large integers $m$, i.e., all sufficiently high
powers of $I$ are Ratliff--Rush ideals. This motivates our
definition of $h$. We call $h$ {\it the asymptotic Ratliff--Rush
number of} $I$.
The starting point of our analysis is the observation that if $I$ has a
principal reduction, then $h \le r$ and $k \le r-1$, see
Proposition~\RRandRed\ and Proposition~\aRRandRed . These are facts that
can be found in the literature (e.g. \cite{RV}) in the case of the
maximal ideal of a one-dimensional local ring.
When $R$ is a reduced Noetherian ring with
total ring of fractions $Q(R)$ and the integral
closure $\intR$ of $R$ in $Q(R)$ is a finitely generated $R$-module,
we obtain sharper results when $I$ is contained
in the conductor of $\intR$ into $R$.
In this case, for example, $k = r-1$ (Theorem~\equality )
and, if $I$ does not coincide with its Ratliff--Rush closure,
$h=r$ (Proposition~\ratleast ). Again, we prove
that for all $n \ge 0$
$$
\widetilde{I^{n+1}} \cap I^n = I^n \I,
$$
see Thorem~\RRfiltration .
We remark that this gives the analogue for the Ratliff--Rush filtration,
under the current hypotheses, to the equalities shown by Itoh, \cite{I},
and by Huneke, \cite{H}, for the filtration of the integral closures
of the powers of $I$.
Always in this context we prove, see Corollary~\RRfilNo ,
that the reduction number of the
filtration $\{\widetilde{I^n}\}_{n \ge 0}$ is less than or
equal to $1$.
If $J \ssq I$ is a {\it reduction of } $I$, i.e., $JI^s = I^{s+1}$
for some positive integer $s$, clearly we have
$$
(I^2:J) \ssq (I^{s+2}:JI^s) = (I^{s+2}:I^{s+1}) \ssq \I.
$$
Furthermore, for each $n \in \N$, $J^n$ is a reduction of $I^n$ and
$(I^{n+1}:J^n) \ssq \I$. It follows that
the ascending chain
$I \ssq (I^2:J)\ssq (I^3:J^2)\ssq \cdots \ssq (I^{m+1}:J^m) \ssq \cdots $
stabilizes with $(I^{m+1}:J^m) = \widetilde{I}$ for
all sufficiently large $m$.
We refer to
$$
k_J = \text{min}\{ n \in \N \ | \ \I= (I^{k{_J}+1} : J^{k_J})\}
$$
as the {\it Ratliff--Rush number of $I$ with respect to $J$}.
It is easy to see that $k_J \le k$.
We give in Corollary~\reductRRno\ conditions under which
$k_J = k$, and illustrate in Example~\eeight\ that
with $J$ a principal reduction of $I$, it
sometimes occurs that $k_J < k$.
Another interesting structural number associated to
the $I$-adic filtration
is the {\it relation type of $I$}, here denoted as $N(I)$.
Going back to ideas utilized by Huckaba in \cite{Hu2} we describe
the relation between the relation type and the reduction
number of $I$. In particular we show that $N(I) \le r+1$
(Proposition~\Nprinc\,) and that the equality holds
when the ideal is $2$-generated (Proposition~\Nred\,).
In Discussion~2.13 we describe a family of $3$-generated ideals
contained in the conductor of a one-dimensional domain
for which the difference between the reduction number and the relation type
grows arbitrarly large.
The results we find can be translated to the case of a one-dimensional Cohen--Macaulay
local ring. In this already well investigated context (see
the analysis in \cite{RV}) we find that,
when $I$ is contained in the conductor, $k \le \la (\I/I)$, see
Proposition~\RRlambda\, .
In section~2 of the present work we study the connections
among the invariants of $I$ we just described, under the assumption
that $I$ has a principal reduction. In section~3
we study the sharper behavior one obtains adding the condition
$I$ contained in the conductor. In section~4 we provide
applications to the one-dimensional setting as well as a set
of examples in this context.
\vskip 3ex
\section{Ideals having principal reductions.}
\vskip 3ex
Suppose $I$ is a regular ideal of a Noetherian ring $R$ having the
property that there exists $x \in I$ and an integer $n \ge 0$ such
that $xI^n = I^{n+1}$. In this situation $xR$ is said to be a
{\it principal reduction of } $I$.
Notice that if $xR$ is a principal reduction of a regular ideal $I$,
then $x$ is a regular element of $R$, i.e., $x$ is a
nonzerodivisor of $R$. It
follows from \cite{Hu1, page~504} that if $(x)$ and $(y)$
are principal reductions of $I$, then $xI^n = I^{n+1}$ if
and only if $yI^n = I^{n+1}$. Thus the reduction number
$r(I)$ is independent of the
principal reduction.
If $R$ is local with maximal ideal $\m$, and $I$ is an
ideal of $R$ having a principal reduction, then
the fiber cone $F(I) = \bigoplus_{n \ge 0}I^n/\m I^n = \bigoplus_{n \ge 0}F_n$
is a one-dimensional graded ring over the field $K := R/\m$.
An element $x \in I$ is a principal reduction of $I$ if and
only if $F(I)$ is integral over its $K$-subalgebra
generated by the image of $x$ in $F_1 = I/\m I$. In this
local setting, if $I$ has a principal reduction, then
the principal reductions of $I$ are precisely the
minimal reductions. In general, a reduction $J$ of $I$ is
a minimal reduction if and only if $JR_\m$ is a minimal
reduction of $IR_\m$ for each maximal ideal $\m$ of $R$.
Thus if $I$ is a regular ideal having a principal reduction
in a Noetherian ring $R$ and if $J$ is a minimal reduction of
$I$, then $J$ is locally principal and hence invertible.
Also $JI^n = I^{n+1}$ if and only if equality holds
locally at each maximal ideal. Therefore the reduction
number $r(I)$ is independent of the minimal reduction. However
a regular ideal having a principal reduction in a Noetherian
ring may also have minimal reductions which are not principal.
For example, let $t$ be an indeterminate over a field $K$ and
let $R = K[t^2, t^3]$. Then $(t^2)$ is a principal reduction
of $I = (t^2, t^3)$ and $J = (t^2 + t^3, t^4)$ is an
invertible minimal reduction of $I$ that is not principal.
\vskip 3ex
\prop
\label{\RRandRed }
Suppose $R$ is a Noetherian ring
and $I$ is a regular ideal of $R$ having a principal reduction.
Then $k \leq \max\{0, r-1 \}$, where $k$ is the
Ratliff--Rush number of $I$ and $r$ is
the reduction number of $I$.
\endb
\proof
If $r = 0$, then $I$ is principal and $k = 0$.
Assume that $I^{r+1}=xI^r$ with $r \ge 1$. We need to show that
$\I= (I^r: I^{r-1})$. It is enough to
show that $(I^r: I^{r-1})=(I^{r+s}: I^{r+s-1})$
for each integer $s \ge 1$.
Since $I^{r+s}= x^s I^r$ and $I^{r+s-1}= x^{s-1}I^r$,
it suffices to show $(I^r: I^{r-1})=(I^{r+1}: I^r)$.
The inclusion $(I^r: I^{r-1}) \ssq (I^{r+1}: I^r)$ is always true.
To prove the reverse inclusion, let $y \in (I^{r+1}: I^r)$;
since $xI^{r-1} \ssq I^r$, then $yxI^{r-1} \ssq yI^r
\ssq I^{r+1} =xI^r$. Since $x$ is a regular element,
it follows that $yI^{r-1} \ssq I^r$, or equivalently,
$y \in (I^r: I^{r-1})$.
\qed
\vskip 3ex
\prop
\label{\aRRandRed}
Suppose $R$ is a Noetherian ring and $I$ is a regular
ideal of $R$ having a principal reduction. Then $h \leq r$,
where $h$ is the asymptotic Ratliff--Rush number of
$I$ and $r$ is the reduction number of $I$.
\endb
\proof
By assumption there exists $x \in I$ such that $I^{r+1} = xI^r$.
Let $n \in \N, n \ge r$. It is well known,
see \cite{RR, Proposition~2.6},
that $\widetilde{I^n} = (I^{n+i} : I^i)$ for
all sufficient large $i \in \N$. Since $n\ge r$
and $I^{r+1} = xI^r$, it follows that
$I^{n+i} = x^iI^n$ for
$i\ge 1$.
Since $x$ and all its powers are regular elements, we obtain that
$(I^{n+i} : x^i) = I^n$ for all integers $i\ge 1$.
The conclusion now follows because
$(I^{n+i} : I^i) \ssq (I^{n+i} : x^i)$.
\qed
\vskip 3ex
If one wants to consider a higher dimensional version of the previous
results, one needs to deal with the following example due to
Raghavan (Example 1.2 in \cite{HJLS}): Let $K$ be a field and
$R$ be the complete local $2$-dimensional domain
$K[\![x, y^2, y^7, x^2y^5,x^3y]\!]$. The ideal $I=(x,y^2) R$
is a parameter ideal and therefore has reduction number $r=0$.
However, $I$ is not Ratliff--Rush closed. Note
in fact that $x^2y^5$ belongs to $(I^2 : I)$ but not to $I$.
\vskip 2ex
\question
\label{\q1}
In a Noetherian local ring, do the results in Proposition~\RRandRed\
and Proposition~\aRRandRed\ hold for ideals $I$ that have minimal
reductions generated by regular sequences?
\endb
\vskip 3ex
\rmk
\label{\stable }
We call a regular ideal $I$ in a Noetherian
ring $R$ {\it stable} if there exists $x \in I$ with $xI = I^2$.
For a regular ideal in a one-dimensional local ring, this
definition is equivalent to that of Lipman \cite{L, (1.3) and
(1.11)}. Among regular ideals having principal reductions, the
stable ideals are the ideals having reduction number at most one.
It is well-known that if $I$ is stable, then $I$ is a
Ratliff--Rush ideal. Thus $r(I) = 1$ implies $k(I) = 0$, i.e.,
$\I = (I:R) = I$. This is true quite generally. In fact, see \cite{VV}, if $I$ has
reduction number $r \le 1$ then the depth of the associated graded
ring $G(I) = R[It]/IR[It]$ of $I$ is positive, hence $I$ and all
its powers are Ratliff--Rush ideals, cf. \cite{HLS, (1.2)}. Thus
if $I$ is stable, then $h(I) = 0$. It is
also well known, however, that there exist Ratliff--Rush ideals
$I$ that are not stable. We illustrate this fact in our section of
examples (Example~\efive\,).
\vskip 3ex
\dis
\label{\blowup}
Suppose $I$ is a regular ideal of $R$ having a principal
reduction $xR$. Since $I^n$ is a finitely generated faithful
$R$-module, if $z \in Q(R)$ is such that $zI^n \subseteq I^n$,
then $z$ is in the integral closure $\intR$ of $R$. In
this context, the {\it blowing--up ring} of $I$ may be defined as
$$
R^I:= \bigcup_{n \in \N}(I^n:_{Q(R)}I^n) =
\bigcup_{n \in \N}(I^n:_{\intR}I^n).
$$
Note that this is an increasing union.
If $I^{r+1} = xI^r$, then for $n > r$ we have
$$
I(I^n:_{Q(R)}I^n) = I(xI^{n-1}:_{Q(R)}I^n) = x(I^n:_{Q(R)}I^n).
$$
Therefore $IR^I=xR^I$. It is clear that
$(I^n:_{Q(R)}I^n) \subseteq (I^n / x^n)$ for each positive integer $n$.
We also have $(I^r / x^r) \subseteq \cup_{n=1}^\infty (I^n:_{Q(R)}I^n)$,
for if $y = z/x^r$ with $z \in I^r$,
then $(z/x^r)I^{2r} = (z/x^r)x^rI^r = zI^r \subseteq I^{2r}$.
Hence $y \in (I^{2r}:_{Q(R)}I^{2r})$.
Therefore the following equalities hold:
$$
R^I = \bigcup_{n=1}^\infty \frac{I^n}{x^n} =
\frac{I^r}{x^r}
$$
where the second equality is an equality of fractional ideals;
cf. \cite{L} and \cite{HLS}.
It follows that, if I is a regular ideal with a principal
reduction $xR$, then, for any $s \geq r$
$$
\I = (I^{s+1}:I^s)= (xI^s:_{Q(R)}I^s) \cap R = xR^I \cap R =
I^sx^{-s+1} \cap R
$$
where the first equality follows by the proof of
Proposition~\RRandRed.
\vskip 3ex
\rmk
\label{\index }
If $xR$ is a principal reduction of $I$, then
$x^mR$ is a principal reduction of $I^m$ for each $m \in \N$;
moreover, the reduction
number of $I^m$ is less than or equal to the reduction number of
$I$. Indeed, if $I^{r+1} = xI^r$, then $(I^m)^{r+1} = x^m(I^m)^r$.
Also, $I^{2r} = x^rI^r$ and for each $m \ge r$ we
have $I^{2m} = x^mI^m$. Hence if $I$ is a
regular ideal having a
principal reduction and if $I$ has reduction number $r$,
then $I^m$ is stable for every integer $m \ge r$,
see \cite{SV, Section 2} and \cite{ES, Corollary 1, page~446}.
This gives another proof for Proposition~\aRRandRed\.
Eakin and Sathaye \cite{ES, page~446} define an ideal $I$ in
a semilocal ring to be {\it prestable } if some power of $I$
is stable. In a general ring $R$, they define an ideal $I$ to
be {\it prestable } if for each prime ideal $P$ of $R$,
$IR_P$ is prestable in $R_P$. It is shown in
\cite{ES, Corollary 1, p. 446} that if $I$ is a prestable
ideal of a local ring and if $I^n$ has $n$-generators,
then $I^{n-1}$ is stable. The smallest positive
integer $s$ such that $I^s$ is stable is related to
the asymptotic Ratliff--Rush number $h(I)$ in that
$h(I) \le s$. The distinction here is that all powers of
$I$ being Ratliff--Rush does not imply in general that
$I$ is stable, so sometimes $h(I) < s$.
For a regular ideal $I$ of a
one-dimensional Cohen--Macaulay local ring $(R,\m)$,
Sally and Vasconcelos \cite{SV, Section 2} define
$n(I)$ to be the least positive integer $n$ such that
$I^n$ is stable. They
define the {\it index of stability } $s(R)$ of $R$ to be
the sup of $n(I)$ as $I$ varies over the regular ideals
of $R$ and prove that $s(R) \le \max\{1, e - 1 \}$,
where $e$ is the multiplicity of $R$. These results of
\cite{SV} and \cite{ES} imply that $r = e-1$ is a
global bound for the reduction number of regular ideals
$I$ of a one-dimensional local Cohen--Macaulay ring
$(R,\m)$ having multiplicity $e$.
(This can be deduced too from
the Hilbert function in the one-dimensional case).
In view of
Proposition~\RRandRed\ and \aRRandRed\,
the number $e-1$ also gives a global bound on the Ratliff--Rush number
and asymptotic Ratliff--Rush number of the regular ideals
of a one-dimensional local Cohen--Macaulay
ring $R$ of multiplicity $e$.
In relation to these results,
Huckaba in \cite{Hu1, Question~2.6}
raises the following question: Suppose $(R,\m)$ is a
quasi-unmixed analytically unramified local ring having infinite
residue field and multiplicity $e$. (Assume if needed that $R$
contains a field of characteristic zero.) Is it true that
$r(I) \le e-1$ for every regular ideal $I$ of $R$ of analytic
spread one? It seems to be unknown whether there even
exists a global bound on $r(I)$ for the ideals $I$ of $R$
having a principal reduction. Work of Vasconcelos in
\cite{V} proves the existence of such a global bound
if there exists a bound on the {\it arithmetic degree}
of the fiber cone of $I$ as
$I$ varies over the regular ideals of $R$ that have a
principal reduction.
\vskip 3ex
In the next result we characterize the condition $h = r$.
\vskip 2ex
\prop
\label{\hequalr }
Suppose $I$ is a regular ideal having a principal reduction
$xR$ with reduction number $r$ in a Noetherian ring $R$.
Then
\roster
\item $(I^r:x) = \widetilde{I^{r-1}}$,
\item $h = r$ if and only if $xI^{r-1} \subsetneq I^r \cap xR$.
\endroster
\endb
\proof
To show $\widetilde{I^{r-1}} = (I^r:x)$, it suffices
to observe that $(I^{r+n}:x^{n+1}) = (I^r:x)$ for
all $n \ge 0$. If $ax^{n+1} \in I^{r+n} = x^nI^r$,
then, since $x$ is a regular element, we have
$ax \in I^r$.
By Proposition~\aRRandRed , $h \le r$. Hence $h = r$ if and only if
$I^{r-1} \subsetneq \widetilde{I^{r-1}} = (I^r:x)$. The
result now follows since $I^{r-1} \subsetneq (I^r:x)$
if and only if $xI^{r-1} \subsetneq I^r \cap xR$.
\qed
\vskip 3ex
We now analyze the relation type of ideals having
principal reductions.
The Rees algebra of $I$ is the graded subalgebra
$R[It] = \DS_{i\ge 0} I^i t^i$ of the polynomial ring $R[t]$.
A presentation of the Rees algebra of $I$ is obtained
as follows: if $I=(x_1, \ldots, x_n)R$, let $R[T_1, \ldots, T_n]$
be a graded polynomial ring over $R$, and
consider the graded
$R$-algebra homomorphism $\phi: R[T_1, \ldots, T_n] \lw R[It]$
defined by $\phi (T_i) = x_i t$.
Let $Q=\Ker \phi$. Then
$$
R[T_1, \ldots, T_n]/Q \cong R[It]
$$
and $Q$ is the homogeneous ideal
generated by all forms in $R[T_1, \ldots, T_n]$ that vanish
when evaluated at the generators $(x_1, \ldots, x_n)$ of $I$.
The relation type $N(I)$ of $I$
is the least bound
on the degrees of the polynomials required to generate $Q$.
It is independent of the choice of a generating set for $I$.
The following proof is similar to that given in \cite{W, page~54}
which in turns relies on ideas given in \cite{Hu2}.
\vskip 3ex
\prop
\label{\Nprinc }
Suppose $I$ is a regular ideal having a principal reduction
in a Noetherian ring $R$.
Then $I$ has relation type $N(I) \le r+1$, where $r$ is the
reduction number of $I$.
\endb
\proof
Let $x \in I$ be such that $I^{r+1} = xI^r$.
Extend $x = x_1$ to a system $(x_1, \ldots, x_n)$ of
generators of $I$.
Consider the presentation $R[It] \cong R[T_1, \ldots, T_n]/Q$.
For $m \in \N$, let $Q_m \ssq Q$ be the ideal
generated by all homogeneous forms
$F \in R[T_1, \ldots , T_n]$
such that $F(x_1,\ldots, x_n) = 0$ and $\deg F \le m$.
It suffices to prove for $i$ a positive integer
and $F \in Q$ a homogeneous polynomial with
$\deg F = r + 1 + i$, then
$F \in Q_{r+i}$.
If $F\in (T_1)R[T_1, \ldots, T_n]$, then $F = T_1 G$
where $G$ is a form of degree $r+i$.
We have
$$
0 = F(x_1, \ldots , x_n) = x_1G(x_1, \ldots , x_n).
$$
Since $x = x_1$ is a regular element of $R, \,
G(x_1, \ldots ,x_n) = 0$. Hence $G \in Q_{r+i}$,
so $F = T_1G \in Q_{r+i}$ in this case.
In the general case write $F = T_1 G_1 + \cdots + T_n G_n$ where each
$G_j$ is a homogeneous polynomial of degree $r+i$.
For $j\ge 2$, let $g_j = G_j (x_1, \ldots, x_n)$. Since $i$ is positive
$g_j \in I^{r+i} = x_1 I^{r+i - 1}$.
Therefore $g_j = x_1 h_j$ with $h_j \in I^{r+i - 1}$.
Let $H_j \in R[T_1, \ldots, T_n]$ be a form of degree
$r+i - 1$ such that $H_j(x_1, \ldots ,x_n) = h_j$.
It follows that $G_j - T_1 H_j$
is a form of degree $r+i$ that is in $Q$ and hence in $Q_{r+i}$.
Moreover
$$
F - \sum_{j=2}^n T_j(G_j - T_1 H_j) = T_1(G_1 +T_2 H_2 +\ldots + T_n H_n)
$$
is a homogeneous form of degree $r+1 + i$ that is in both
$Q$ and
$(T_1)R[T_1, \ldots, T_n]$. By the previous case
$F - \sum_{j=2}^n T_j (G_j - T_1 H_j) \in Q_{r+i}$. Therefore
$F \in Q_{r+i}$.
\qed
\vskip 3ex
As in Theorem 2.4 of \cite{Hu2} one has:
\vskip 2ex
\prop
\label{\Nred }
Suppose $R$ is a Noetherian ring and
$I = (x,y)R$ is a $2$-generated non-principal regular ideal
having $xR$ as a principal reduction.
Then $I$ has relation type $N(I) = r+1$,
where $r$ is the reduction number of $I$.
\endb
\proof
By Proposition~\Nprinc , $N(I) \le r+1$.
Consider the presentation $R[It] \cong R[T_1,T_2]/Q$,
where $\phi(T_1) = xt$ and $\phi(T_2) = yt$.
For $m \in \N$, let $Q_m \ssq Q$ be the ideal
generated by all homogeneous forms
$F \in R[T_1, T_2]$
such that $F(x, y) = 0$ and $\deg F \le m$.
For $s \in \N$, $I^{s+1} = xI^s$ if and only if
$y^{s+1} \in xI^s$ if and only if
there exists a homogeneous polynomial $F \in R[T_1, T_2]$
such that $\deg F = s + 1$, $F$ is monic as a polynomial in $T_2$
and $F(x,y) = 0$, i.e., $F \in Q_{s+1}$. Since $r$ is the smallest
integer such that $I^{r+1} = xI^r$, there exists such a
polynomial $F$ for $s = r$, but not for $s < r$. Therefore
$Q_r$ contains no forms that are monic in $T_2$, while $Q_{r+1}$
does contain such a form. It follows that $N(I) = r+1$.
\qed
\vskip 3ex
The previous result is actually a special case of
Observation~4.9 in \cite{W}. We gave the proof for its simplicity
and because no particular assumption on the ring is needed.
If $I$ is stable, we show next
that the equality of (\Nred) \, holds with
no restriction on the number of generators of $I$.
\vskip 3ex
\cor
\label{\Nstable }
Suppose $I$ is a stable
non-principal regular ideal having a principal reduction in a
Noetherian ring $R$. Then $I$ has relation type $N(I) = 2$.
\endb
\proof
By Proposition~\Nprinc , $N(I) \le 2$. By assumption,
there exists $x \in I$ such that $xI = I^2$. Let
$x = x_1, x_2, \ldots , x_n$ be generators of $I$ and
let $R[T_1, \ldots , T_n]/Q = R[x_1t, \ldots , x_nt]$
be a presentation of the Rees algebra of $I$,
where $T_i \to x_it$ for $i = 1, \ldots n$.
Suppose $N(I) = 1$.
The equality $I^2 = x_1 I$ implies
$x_n^2 = x_1 g$, where $g\in I = (x_1, \ldots, x_n)R$.
Thus there exists a degree $2$ form $T_n^2 - T_1 G$ which vanishes
at $x_1, \ldots, x_n$. Our assumption that $N(I) = 1$
implies there are
linear forms $H_1, \ldots, H_s \in Q$ such that
$T_n^2 - T_1 G = F_1 H_1 + \cdots + F_s H_s$, where
$F_1, \ldots, F_s$ are linear forms in $R[T_1, \ldots, T_n]$.
Comparing the coefficients of the monomial $T_n^2$, we
deduce that the coefficients of $T_n$ in the $H_i$ generate the
unit ideal of $R$. Hence there exist $a_i \in R$
such that $a_1H_1 + \cdots + a_sH_s$ is monic in $T_n$.
Since this form vanishes at $x_1, \ldots, x_n$, it follows that
$x_n \in (x_1, \ldots, x_{n-1})R$. Therefore $I = (x_1, \ldots, x_{n-1})R$.
If $n - 1 > 1$, a repetition of the argument yields
$x_{n-1} \in (x_1, \ldots , x_{n-2})R$. Therefore
a simple inductive proof gives $I = x_1R$. This
contradicts our hypothesis that $I$ is not principal.
We conclude that $N(I) = 2$.
\qed
\vskip 3ex
\dis
\label{\rela}
Suppose $I = (x,y,z)R$ is an ideal in a
Noetherian domain $R$ such that $xR$ is a principal
reduction of $I$. If $y^2/x^2$ and $z^2/x^2$
are units of $R$, but $yz \notin xI$, then $I$ has
reduction number $r(I) = 2$ and relation
type $N(I) = 2 < r(I) + 1$. Therefore it is not possible
to extend Proposition~\Nred\ to a situation where
$I$ is 3-generated.
We have
$xI = (x^2, xy, xz) \subsetneq I^2$ since
$yz \notin xI$.
However, the fact that the elements
$y^2$ and $z^2$ are in $x^2R \ssq xI$
implies that $I^3 \ssq xI^2$, so $r(I) = 2$.
To see that $N(I) = 2$, consider the presentation
$R[T_1,T_2, T_3]/Q = R[xt, yt, zt]$ of the Rees algebra
of $I$, where $T_1 \to xt, T_2 \to yt$, and $T_3 \to zt$.
Since $r(I) = 2$, $N(I) \le 3$ by Proposition~\Nprinc. To
show $N(I) \le 2$, we show that each
form $F \in Q$ with $\deg F = 3$ is in $Q_2$.
Since $y^2/x^2 := a$ and $z^2/x^2 := b$ are units of $R$,
$T_2^2 - aT_1^2$ and $T_3^2 - bT_1^2$ are in $Q_2$.
Using these relations we can modify $F$ to another form $G$
of degree 3 that is in $Q$ and is a multiple of $T_1$.
Thus $G = T_1H$, where $H \in R[T_1, T_2, T_3]$ is a
form of degree 2. Since $G \in Q$, we
have $0 = G(x,y,z) = xH(x,y,z)$. Since $x$ is a regular
element in $R$, it follows that $H(x,y,z)= 0$. Therefore
$H \in Q_2$, and so also $G$ and $F$ are in $Q_2$.
We conclude that $Q_2 = Q$ and $N(I) = 2$.
\vskip 3ex
\ex
\label{\rel2 }
An example of a one-dimensional
Noetherian domain $R$ having an ideal $I = (x,y,z)R$ with the
properties of (\rela) may be constructed as follows:
Let $F$ be a field having an algebraic extension $K = F(\al, \be)$
such that $[K:F] = 4$ and such that $\al^2 = a \in F$
and $\be^2 = b \in F$. Let $x$ be an indeterminate over $K$ and
let $R = F + xK[x]$. Thus $R$ is the set of polynomials
in $K[x]$ that have their constant term in $F$.
Let $y = \al x$, $z = \be x$ and $I = (x, y, z)R$.
(For example, $F$ could be the field $\Q$ of rational numbers,
$\al$ could be the square root of 2 and $\be$ the
square root of 3.) Since $1, \al, \be, \al\be$
is a vector space basis for $K$ over $F$, $yz = \al\be x^2$
is not in $xI$.
Notice that $K[x]$ is the integral closure of $R$ and
$I$ is contained in $x K[x]$ which is the conductor of
the integral closure of $R$ into $R$. Localization of
$R$ at the maximal ideal $x K[x]$ gives a
one-dimensional local Noetherian domain with the same properties.
\vskip 3ex
\dis
\label{\re3}
The examples displayed so far describe ideals for which the
relation type is bounded below by the reduction number.
A modification of the construction of (\rel2 ) \, gives
for each integer $n \ge 3$ an ideal
$I = (x,y,z)R$
in a one-dimensional Noetherian domain $R$ having a
principal reduction $xR$ such that
$I$ has reduction number $r = 2n-2$ and relation
type $N(I) = n$. Thus there exist 3-generated ideals having
principal reductions and having the difference between the reduction
number and the relation type
arbitrarily large; in fact $r - N(I) = n-2$.
To construct such examples,
fix $n \ge 3$ and
let $F$ be a field having an algebraic extension
$K = F(\al, \be)$
such that $[K:F] = n^2$ and such that
$\al^n = a \in F$
and $\be^n = b \in F$.
Let $x$ be an indeterminate over $K$ and
let, as before, $R = F + x K[x]$.
Let $y = \al x, z = \be x$ and $I = (x, y, z)R$.
The set $\{ \al^i \be^j \}$,
where $0 \le i,j \le n - 1$, is a vector space basis
for $K$ over $F$ and a module basis over $R$ for the
integral closure $K[x]$ of $R$.
It follows that
$y^{n-1}z^{n-1} = \al^{n-1} \be^{n-1} x^{2n-2}$
does not belong to $xI^{2n-3} = x(x, \al x, \be x)^{2n-3}$.
Thus $xI^{2n-3}$ is properly contained in $I^{2n-2}$.
Moreover, the elements $y^n, z^n$ belong to $x^nR$ and we get
$I^{2n-1} = (x,y,z)^{2n-1} = xI^{2n-2}$.
Therefore $r(I) = 2n-2$.
To see that $N(I) = n$, consider the presentation
$R[T_1,T_2, T_3]/Q = R[xt, yt, zt]$ of $R[It]$.
Since $r(I) = 2n-2$, $N(I) \le 2n-1$ by
Proposition~\Nprinc .
To show $N(I) \le n$, we show that each
form $G \in Q$ of degree $m$ with
$n \le m \le 2n-1$, is in $Q_n$.
Since the relations
$xT_2 - x\al T_1, xT_3 - x\be T_1$ are in $Q_1$,
we may assume that
$$
G(T_1, T_2, T_3) =
\sum_{i+j+k = m} r_{ijk}T_1^iT_2^jT_3^k
\in R[T_1,T_2,T_3]
$$
has the property that $r_{ijk} \in F$ if $j+k > 0$.
Using the relations
$T_2^n - aT_1^n, T_3^n - bT_1^n \in Q_n$,
we may assume $j < n$ and $k < n$
for each nonzero $r_{ijk}$.
Then
$$
G(x, \al x, \be x) =
\sum_{i+j+k = m}r_{ijk}x^i(\al x)^j (\be x)^k = 0
$$
implies that
$$
r_{m00}(0) + \sum_{j + k > 0, j < n, k < n}r_{ijk} \al^j \be^k = 0.
$$
The linear independence over $F$ of
$\{\al^j\be^k \}, \, 0 \le j,k \le n -1$
implies that $r_{m00}$ and each of the $r_{ijk}$ is zero.
It follows that $G(T_1,T_2,T_3) = 0$. Therefore $N(I) \le n$.
On the other hand, the relations $T_2^n - aT_1^n$ and $T_3^n - bT_1^n$
are monic in $T_2$ and $T_3$ and are
easily seen to be in $Q_n$ and not in $Q_{n-1}$.
In conclusion, we obtain $N(I) = n$.
\vskip 2ex
Observe that the ideals constructed in (\re3) are
contained in the conductor of $R$ in its integral closure. We
show, in the next section, that other invariants of ideals
contained in the conductor have
a more predictable behaviour. The relation type however
seems to remain unaffected with respect to
containment in the conductor.
\newpage
\section{Reduced Noetherian rings with finite integral closure.}
\vskip 3ex
\nota
\label{\reduced}
Suppose now that $R$ is a reduced Noetherian ring
having total ring of fractions $Q(R)$. Assume that the integral
closure $\intR$ of $R$ in $Q(R)$ is a finitely generated $R$-module.
Let $\C = (R:_{Q(R)} \intR)$ denote the conductor
of $\intR$ into $R$. Since $\intR$ is a finitely generated $R$-module,
$\C$ is a regular ideal of both $R$ and $\intR$.
\vskip 2ex
In this context,
if $I \ssq \C$ is a regular ideal that is not principal
but has a
principal reduction, we prove in Theorem~3.2
that the Ratliff--Rush
number $k$ of $I$ is precisely $r-1$, where $r$ is
the reduction number of $I$.
What allows us to prove
a sharp equality result in this case
(as contrasted with
only an inequality in Proposition~\RRandRed )
is that, in this case,
$$
\I = I^s x^{-s+1} \cap R = I^s x^{-s+1}
$$
(for $s$ large enough);
the last equality holds since
$I^s x^{-s+1} = IR^I \ssq R^I \ssq \intR$
and $I \ssq \C$ which is an ideal of $\intR$;
therefore $IR^I \ssq \C \ssq R$.
\vskip 3ex
\thm
\label{\equality}
Suppose $R$ is a reduced Noetherian ring as in (\reduced) and $I$ is
a regular ideal of $R$ having a principal reduction.
If $I \ssq \C$ and if $I$
is not principal, then the
reduction number $r$ of $I$ is $k+1$, where $k$ is the
Ratliff--Rush number of $I$.
\endb
\proof
By hypotheses $I^{r+1}=xI^r$ and $I^r \supsetneq xI^{r-1}$.
Since $I$ is a regular ideal contained in $\C$, it follows that
$$
I \subsetneq I^2x^{-1}\subsetneq
\dots \subsetneq I^{r-1}x^{-r+2} \subsetneq I^rx^{-r+1}
= I^{r+1}x^{-r} =\dots
$$
Hence $\I = I^rx^{-r+1}$.
By Proposition~\RRandRed\ , we have $\I =(I^r: I^{r-1})$,
thus it is enough to prove that
$(I^{r-1}: I^{r-2}) \subsetneq (I^r: I^{r-1})$.
Since $xI^{r-1} \subsetneq I^r$, there exists
an element $a \in I^r$ such that $a/x \notin I^{r-1}$
(and hence $a/x^{r-1} \notin I$).
Since $I \ssq \C$, then $a/x^{r-1} \in \I \ssq R$;
moreover, if $b \in I^{r-1}$, then
$ab/x^{r-1} \in I^{2r-1}x^{-(r-1)}=
I^rx^{r-1}x^{-(r-1)}=I^r$.
Therefore $a/x^{r-1} \in (I^r: I^{r-1})$.
On the other hand $(a/x^{r-1})x^{r-2} = a/x \notin I^{r-1}$,
hence $a/x^{r-1} \notin (I^{r-1}: I^{r-2})$.
\qed
\vskip 3ex
\cor
\label{\RRstable}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction.
If $I \ssq \C$, then $I$ is stable if and only if $I=\I$.
\endb
\proof
The assertion is clear if $I$ is a principal ideal.
If $I$ is not principal, then by
Theorem~\equality , the reduction number
$r$ of $I$ is $1$ if and only if the
Ratliff--Rush number
$k$ of $I$ is zero,
i.e., if and only if $\I = (I:R) =I$.
\qed
\vskip 3ex
\cor
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction.
If $I \ssq \C$ is Ratliff--Rush closed and
non-principal, then $I$ has relation type $2$.
\endb
\proof
Apply Corollary~\RRstable\ and Corollary~\Nstable \ .
\qed
\vskip 3ex
\rmk
We would like to take this opportunity to remark that (1.15)
of \cite{HJLS, page~363} is incorrect.
It is asserted there that,
if $R$ is a one-dimensional semilocal domain and
$I$ is an ideal of $R$ contained in the conductor,
then $I$ is stable.
This assertion is false without the
additional hypothesis that $I$ is
a Ratliff--Rush ideal.
A simple example that illustrates this is to
let $t$ be an indeterminate
over a field $K$ and let
$R = K[t^3, t^4, t^5]_{(t^3, t^4, t^5)}$.
The maximal ideal
$\m = (t^3, t^4, t^5)R$
coincides with the conductor.
Note however that the ideal
$I = (t^3, t^4)R$ is not Ratliff--Rush,
so is not stable.
We also remark that for $R$ a Noetherian domain, Corollary~\RRstable\
follows from \cite{HJL, (4.7)} which asserts that if $D$ is
a Noetherian integral domain and $I$ is an ideal of $D$
that is integral over a principal ideal generated by an
element of the conductor, then the Ratliff--Rush ideal
$\I$ associated to $I$ is stable.
\vskip 3ex
Notice that, if $xR$ is a principal reduction of $I$, then
$$
\widetilde {I^m} = I^{ms}x^{-m(s-1)} \cap R
$$
for every $s \in \N$ such that $s \ge r$,
where $r$ the reduction number of $I$.
Using this observation we investigate conditions sufficient to imply
the equality $\widetilde {I^{n+1}} \cap I^n = I^n \I$
(cf. \cite {HJLS Questions 1.16}). Naturally the inclusion
$\widetilde {I^{n+1}} \cap I^n \supseteq I^n \I$
always holds, since
$I^n \I \ssq \widetilde {I^n} \I$ and
$\{\widetilde {I^n} \}_{n \ge 0}$ is a filtration.
\vskip 3ex
\thm
\label{\RRfiltration}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction
$xR$.
If $I \ssq \C$, then
$
\widetilde {I^{n+1}} =I^n\I,
$
for every $n \geq 0$.
In particular,
since $I^n \I \ssq I^n$,
we have
$$
\widetilde {I^{n+1}} \cap I^n = I^n \I .
$$
\endb
\proof
Since $I \ssq \C$,
we have, for all sufficiently large $s \in \N$,
$$
\widetilde {I^{n+1}} = I^{(n+1)s}x^{-(n+1)(s-1)}=
I^{ns+s}x^{-ns+n-s+1}
$$
and
$$
I^n \I = I^nI^sx^{-s+1}=I^{n+s}x^{-s+1} \ .
$$
Moreover, if $s+n \geq r$, we have
$$
I^{n+s}x^{-s+1}= I^{n+s}I^{n(s-1)}x^{-s+1}x^{-n(s-1)}=
\widetilde {I^{n+1}} \ . \qed
$$
\vskip 3ex
Concerning the relationship of $\widetilde {I^{n+1}} \cap I^n$
with $I^n\I$ and $I^{n+1}$, we note the following:
\vskip 3ex
\cor
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ contained in $\C$
with a principal reduction
$xR$ and reduction number $r$.
Then, for any integer $n \geq r-1$,
$\widetilde {I^{n+1}} \cap I^n = I^n\I = I^{n+1}$.
\endb
\proof
This is immediate from Proposition \aRRandRed \,\, and the
fact that $I^n\I \ssq \widetilde {I^{n+1}}$.
\qed
\vskip 3ex
Example~\esix\ shows $I^n\I$ may be
properly contained in $\widetilde {I^{n+1}} \cap I^n$
for $n < r-1$.
\vskip 3ex
Recall that a filtration $\{F_n\}_{n\ge 0}$
of a ring $R$ is called a {\it good filtration} if there exists an
ideal $I$ of $R$ such that $IF_i \ssq F_{i+1}$, for all $i
\geq 0$, and if $IF_n = F_{n+1}$, for all $n$ large enough.
(cf.\cite{HZ Definition 2.1}). Hoa and Zarzuela in
\cite{HZ, Example 2.3} show that the
Ratliff--Rush filtration is a good filtration.
For a good filtration it is possible to define the notions of
{\it reduction}, {\it minimal reduction} and {\it reduction number}
(cf. \cite{HZ, Definitions 2.5 and 3.1}).
In the hypotheses of
this section, if $I$ has a principal reduction $xR$,
then $\{x^nR\}_{n\ge 0}$ is
a minimal reduction for
$\{\widetilde {I^n}\}_{n\ge 0}$
(cf.\cite{HZ, Proposition 2.6}).
The reduction number of the Ratliff--Rush filtration
$\{\widetilde{I^n}\}_{n\ge 0}$ with respect to
$\{x^nR \}_{n\ge 0}$ is
defined to be the minimum integer $s$ such that
$\widetilde{I^{n+1}} = x\widetilde{I^n}$
for all $n \geq s$.
This integer
$s$ is independent of the principal reduction $xR$.
It may also
be characterized as the smallest nonnegative integer
$s$ such that
$\widetilde{I^{s+1}} \ssq xR$,
for $\widetilde{I^{n+1}} \ssq xR$
implies $\widetilde{I^{n+1}} = x\widetilde{I^n}$.
It is easy to see that if $r$ is the reduction number of $I$ and
$s$ the reduction number of the filtration
$\{\widetilde {I^n}\}_{n\ge 0}$,
then $s \leq r$.
For if $I^{n+1} = xI^n$, then $I^n$ and $I^{n+1}$
are both stable and hence Ratliff--Rush ideals as noted in
Remark~\stable\ ,
so $\widetilde{I^{n+1}} = x\widetilde{I^n}$.
Example~\eseven\ shows that the reduction number $r$ of $I$ may be
strictly bigger than $s$.
We prove that inclusion in the conductor implies a strict
upper bound on the reduction number of a Ratliff--Rush filtration.
\vskip 3ex
\thm
\label{\RRfiltrno}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction
$xR$ and reduction number $r$.
If there exists an integer $m$
such that $x^m \in \C$, then, for any integer $n \geq m$, $\widetilde
{I^{n+1}} = x\widetilde {I^n}$. In particular, if $s$
is the reduction number of
the filtration
$\{\widetilde {I^n}\}_{n\ge 0}$, then $s \leq m$.
\endb
\proof
If $m \ge r$, the assertion is clear, so we assume that $m < r$.
As we mentioned immediately before Theorem~\RRfiltration\
$$
x \widetilde {I^n} =
x(I^{ns}x^{-n(s-1)} \cap R)
$$
and
$$ \widetilde {I^{n+1}} =
I^{(n+1)s}x^{-(n+1)(s-1)} \cap R \ .
$$
Hence, to prove that
$\widetilde{I^{n+1}} \ssq x \widetilde{I^n}$ (the other
inclusion is always true), we have to prove that any element of
$R$ of the form $y/x^{ns-n+s-1}$, where $y \in I^{ns+s}$, belongs
to $xR$. We may assume
$ns+s>r$; hence we have $I^{ns+s}=x^{ns+s-r}I^r$. It follows that
there exists $z \in I^r$ such that $y/x^{ns-n+s}=
z/x^{r-n}=x^nz/x^r$. Since $n \geq m$, $x^n \in
\C$; hence $y/x^{ns-n+s} \in R$ and $y/x^{ns-n+s-1} \in xR$.
\qed
\vskip 3ex
\cor
\label{\RRfilNo}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction
$xR$ and reduction number $r$.
If $I \ssq \C$, then
$\widetilde {I^{n+1}}=x \widetilde {I^n}$ for all $n \geq 1$,
so the reduction number of the
filtration $\{\widetilde{I^n}\}_{n \ge 0}$ is less than or
equal to 1.
\endb
\vskip 3ex
\prop
\label{\ratleast}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction
$xR$ and reduction number $r$.
If $I \ssq \C$ and $I$ is not Ratliff--Rush closed, then the
asymptotic Ratliff--Rush number, $h$, coincides
with $r$.
\endb
\proof
In this case, one has
$\widetilde{I^{h-1}} = (I^h:x)$
where $h\ge 2$ is the asymptotic Ratliff--Rush number of $I$.
In fact, by the definition of $h$,
we have
$\widetilde{I^{h-1}} \neq I^{h-1}$ and
$\widetilde{I^n} = I^n $ for all $n\ge h$.
By Corollary~\RRfilNo\
$I^h = \widetilde{I^h} = x\widetilde{I^{h-1}}$ and we get
$\widetilde{I^{h-1}} \ssq (I^h:x)$ from which
the conclusion.
Since $I \subsetneq \widetilde{I}$, we have
$r \ge 2$ and we deduce that
$I^r \ssq xR$. In fact $I^r = \widetilde{I^r}$ by Proposition~\aRRandRed\
and $\widetilde{I^r} = x\widetilde{I^{r-1}}$ by
Corollary~\RRfilNo .
Also by definition of $r$ we have $xI^{r-1} \subsetneq I^r$. Hence
$xI^{r-1} \subsetneq I^r \cap xR$ and $h = r$ follows by
Proposition~\hequalr .
\qed
\vskip 3ex
For $J = xR$ a principal reduction of $I$, we note the following
case where the
integer $k_J$, the Ratliff--Rush number of $I$ with
respect to $J$, is equal to the
Ratliff--Rush number of $I$, and therefore independent of
the minimal reduction chosen.
\vskip 3ex
\cor
\label{\reductRRno}
Suppose $R$ is a reduced Noetherian ring as in (\reduced)
and $I$ is a regular ideal of $R$ having a principal reduction
$xR = J$ and reduction number $r$. If $I \ssq \C$ and
$I \subsetneq \I$, then $k_J = k
= r-1$.
\endb
\proof
By Theorem~\equality , we have $k = r - 1$. Since
$I \subsetneq \I$,
$r \ge 2$. If $r = 2$, then $k = 1$. Since $k_J \le k$ and
$k_J \ne 0$, we have $k_J = k$ in this case.
Suppose $r \ge 3$.
Since $h = r$, we have
$x^{h-2} \I =\widetilde{I^{h-1}} = (I^h:x)$.
To show $k_J = k$, it suffices to show
that $(I^{r-1}:x^{r-2})$ is strictly contained in $(I^r:x^{r-1})$.
This is true since otherwise we would have
$(I^{r-1}:x^{r-2}) = \I$ and we would obtain
$\widetilde{I^{r-1}} = x^{r-2} \I \ssq I^{r-1}$,
which is impossible since $h=r$.
\qed
\vskip 3ex
\section{Applications and examples in the one-dimensional case. }
\vskip 3ex
\nota
\label{\onedim}
Suppose $(R,\m)$ is a one-dimensional Cohen--Macaulay local ring
such that the residue field $R/\m$ of $R$ is infinite and
the integral closure $\intR$ of $R$ in its total ring of fractions
$Q(R)$ is a finitely generated $R$-module. The conductor
$\C = (R:_{Q(R} \intR)$ of $\intR$ into $R$ is a regular ideal.
Note that if $(R, \m)$ is a one-dimensional Cohen-Macaulay local ring
having the property that its integral closure
$\intR$ is a finitely generated $R$-module,
then $R$ is reduced. For if $a \in R$
is a nilpotent element, let $b \in \frak m$ be a regular element. Then
$a/b^n \in Q(R)$ is integral over $R$ for each
positive integer $n$. Hence $a \in \cap_{n=1}^\infty b^n \intR = (0)$,
the last equality because $b$ is in the Jacobson radical of the
Noetherian ring $\intR$.
\vskip 3ex
\prop
\label{\RRlambda}
With notation as in (\onedim), if
$I \ssq \C $ is an $\m$-primary ideal
and $\tilde I \supsetneq I$,
then $I$ has Ratliff--Rush number
$k \leq \la = \l_R(\I/I)$.
\endb
\proof
Since $I \ssq \C$,
by Theorem~\equality , $I$ has reduction number $r = k + 1$.
Moreover, as noted at the beginning of the
previous section, $\I = I^rx^{-r+1}$
and we have the following chain of ideals:
$$
I \subsetneq I^2x^{-1} \subsetneq \dots \subsetneq I^{r-1}x^{-r+2}
\subsetneq I^rx^{-r+1} =\I \ .
$$
In this chain there are $r-1 = k$ steps; hence
$\la = \l_R(\I/I) \geq k$. \qed
\vskip 3ex
With $(R,\m)$ as in (\onedim), if $I$ is an $\m$-primary
ideal of $R$, then the associated graded ring
$G(I) = R[It]/IR[It]$ is a one-dimensional
graded ring having a unique maximal graded ideal. The $0$-th
local cohomology of $G(I)$ with respect to this maximal ideal
(or equivalently
with respect to the ideal $G(I)_{+}$ consisting of elements of
positive degree) is the ideal of $G(I)$ of elements that
annihilate a power of this maximal ideal (or equivalently a
power of $G(I)_{+}$).
This is a graded ideal and is zero if and only if $G(I)$
is Cohen--Macaulay (since we are in dimension one).
Proposition~\RRfiltration \ provides
information on this $0$-th local cohomology.
\vskip 3ex
\prop
\label{\cohom}
Let $R$ and $I \ssq \C$ be as in (\onedim). Then
$$
H^0_{G(I)_{+}}(G(I)) =
\widetilde{I}/I \ds I\widetilde{I}/I^2 \ds \cdots \ds
I^{r-2}\widetilde{I}/I^{r-1}
$$
where $r$ is the reduction number of $I$.
\endb
\proof
Recalling that
$$
H^0_{G(I)_{+}}(G(I)) =
\DS_{n\ge 0} \frac{\widetilde{I^{n+1}} \cap I^n}{I^{n+1}},
$$
the result follows by Proposition~\RRfiltration \ and
Proposition~\equality .\qed
\vskip 3ex
The following corollary is now immediate from Proposition~\RRlambda \
and Proposition~\cohom .
\vskip 3ex
\cor
Let $R$ and $I \ssq \C$ be as in (\onedim).
If $\la = \l_R(\I/I) = 1$, then $I$ has
Ratliff--Rush number $k = 1$, $\I = (I^2:I)$,
the reduction number of $I$ is $r=2$ and $H^0_{G(I)_{+}}(G(I)) = \I /I$.
\endb
\vskip 3ex
In what follows we provide a set of examples regarding
the results given in the previous sections.
The example are all complete
one-dimensional local Noetherian domains of the form
$R = K[\![t^s : s \in S]\!]$, i.e. formal power
series in the indeterminate $t$ with coefficients in
a field $K$ and exponents from an additive submonoid $S$ of
the nonnegative integers that contains all sufficiently large
integers. The formal power series ring
$K[\![t]\!]$ is a finitely generated $R$-module having the
same fraction field as $R$.
Thus, in each of the examples, the notation
in (\onedim) is valid if we assume the field $K$ is
infinite. Since the ideals we consider are generated
by monomials $t^s$, the examples are valid even if
the field $K$ is finite.
\vskip 3ex
\ex
\label{\eone}
{\it An ideal $I$ such that $I=(I^2:I) $,
but $\I \supsetneq (I^2:I)$.}
Set $R=K[\![t^4, t^5, t^6, t^7]\!]$ and $I=(t^4, t^5)$.
Observe that $I^2=(t^8, t^9, t^{10})$ and $I^3=(t^{12}, t^{13}, t^{14}, t^{15})$.
Thus $\widetilde I = \C = \m$ and $\I = (I^3:I^2)$ while $(I^2:I)=I$.
\vskip 3ex
\ex
\label{\etwo}
{\it An ideal $I \not\ssq \C$ such that $k < r-1$.}
(cf. Proposition~\RRandRed and Theorem~\equality).
Let
$R=K[\![t^5, t^7, t^{23}]\!]$ and $I=(t^5, t^7)$.
We have $I^5 = t^5 I^4$, hence $r=4$, while $k=1$, since
$\I = \m = (I,t^{23}) = (I^2 :I)$.
Notice that, in this case, $\la =1$.
\vskip 3ex
\ex
\label{\ethree}
{\it An ideal $I$ such that $h(I) < r(I)$.} (cf. Proposition~\aRRandRed and
Proposition~\hequalr ).
Let $R=K[\![ t^3, t^5]\!]$ and
$I = \m$. In this case each power of $I$ is a
Ratliff--Rush ideal, so $h(I) = 0$. However,
$\m^2 \supsetneq t^3 \m$, since
$t^{10}t^{-3} \notin \m$. Hence $r(I) \ge 2$.
\vskip 3ex
\ex
\label{\efour}
{\it An ideal $I \subset \C$ such that $N(I) = 3$ and $r(I) = 4$.}
Let $R=K[\![ t^9, t^{10}, t^{11}, t^{12},
t^{13}, t^{14}, t^{15}, t^{16}, t^{17}]\!]$ and
$I = (t^9, t^{10}, t^{12})$. Using the computer algebra program
MACAULAY one sees that $N(I) = 3$ and $r = 4$. These facts
can also be checked by hand.
\vskip 3ex
\ex
\label{\efive}
{\it A nonstable ideal $I \not\ssq \C$ such that
$\I = I$.} (cf. Remark~\stable\ and Corollary~\RRstable).
Set $R=K[\![t^3, t^4]\!]$ and $I=\m$.
$I$ is not stable, since $t^8 \in I^2 \setminus t^3 I$.
In this case $r=2$ and $k=0$.
\vskip 3ex
\ex
\label{\esix}
{\it An example that shows $I^n\I$ may be
properly contained in $\widetilde{I^{n+1}} \cap I^n$.}
(cf. Theorem~\RRfiltration ).
Let $R=K[\![t^6, t^7, t^{15}, t^{16}, t^{17}]\!]$
and $I= \I = \m$.
Then we have $I^2=(t^{12}, t^{13}, t^{14}, t^{22}, t^{23})$,
$I^3=(t^{18}, t^{19}, t^{20}, t^{21}, t^{29})$,
$I^4=(t^{24}, t^{25}, t^{26}, t^{27}, t^{28})$,
$I^5=t^{30} \overline R$ and $I^{5+m}=t^{6m} I^5$
for any $m \geq 0$. Clearly this implies that $r(I)=5$.
Moreover, for each $m \geq 2$, $\widetilde {I^m} = t^{6m} \overline R$
and $\widetilde{I^m} \cap I^{m-1} =\widetilde {I^m}$.
Hence $II \subsetneq \widetilde {I^2}$,
$I^2I \subsetneq \widetilde {I^3}$,
$I^3I \subsetneq \widetilde {I^4}$ and
$I^4I = \widetilde {I^5}$.
Notice also that $I^2 \ssq \C = t^{12}\intR$.
\vskip 3ex
\ex
\label{\eseven}
{\it An ideal $I$ with reduction number greater than the reduction
number of the Ratliff--Rush filtration.}
(cf. Proposition~\RRfiltrno ).
Let $R=K[\![t^4, t^5, t^{11}]\!]$ and $I= \m$. We have
$\m^4 = t^4\m^3$ and $t^{15} \in \m^3 \setminus t^4\m^2$
Hence $r=3$. Moreover $\m$ is Ratliff--Rush closed and
$\widetilde {\m^2}= (t^8, t^9, t^{10}, t^{11})$; hence
$\widetilde {\m^{n+1}} = t^4 \widetilde {\m^n}$ for all $n \geq 2$.
Notice that $(t^4)^2 \in \C = t^8 K[\![t]\!]$.
\vskip 3ex
\ex
\label{\eeight}
{\it An ideal $I \not\subseteq \C$ with a principal reduction
$J=(x)$ such that $k_J < k$.}
(cf. Corollary~\reductRRno).
Let $R= K[\![ t^4, t^5, t^6]\!]$ and
$I = (t^4, t^5)$.
We have: $I^2=t^8 K+t^9 K+t^{10} K + t^{12} K[\![t]\!] \subsetneq \widetilde{I^2}
= t^8 K[\![t]\!]$ and
$I^3 = t^{12}k[\![t]\!]=\widetilde{I^3}$. This implies that
$r=h=3$; moreover $(I^2:I) =I$,
$(I^3:I^2) = \m = \I$ and $(I^2:t^4 R)= \m$.
Hence if $J=t^4 R$, $k_J=1$, while $k=2$.
Notice that $t^{15} \in (I \cap t^4 R)$ does not belong to $t^4 I^2$
(cf. Proposition~\hequalr).
\vskip 3ex
\ex
\label{\enine}
{\it An ideal $I \not\ssq \C$ such that
$k = 3 > \la = 1$.} (cf. Proposition~\RRlambda).
Set $R=K[\![t^5, t^6, t^8]\!]$ and $I=(t^5, t^6)$.
We have $\I = \m = (I^4 :I^3)$;
hence $\la = 1$, while $k=3$
(notice that, in this case, $I^5=t^5 I^4$, hence $r=4=k+1$).
This is also an example of an ideal
such that $I^2 x^{-1} \supsetneq I$, but $I^2x^{-1} \cap R =I$
(cf. proof of Proposition~\RRlambda).
In fact we have $I^2=(t^{10}, t^{11}, t^{12})$,
hence $t^7 \in I^2 t^{-5}$ does not belong to $I^2t^{-5} \cap R$.
\vskip 3ex
\ex
\label{\eten}
{\it An ideal $I \subset \C$ such that $k < \la$.}
(cf. Proposition~\RRlambda).
Set $R=K[\![t^5, t^6, t^7, t^8, t^9]\!]$ and $I=(t^5, t^6, t^7)$.
Since $I^2 = (t^{10}, t^{11}, t^{12}, t^{13}, t^{14})$ and
$\I = \m$, then $\I = (I^2:I)$, $k= 1$ and
$\la = 2$.
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\endRefs
\vskip 4ex
\centerline{\sl Universit\`a di Catania - Dipartimento di Matematica,}
\centerline{\sl Viale Andrea Doria, 6 - 95125 Catania, Italy}
\centerline{\sl E-mail: {mdanna\@dipmat.unict.it}}
\vskip 2ex
\centerline{\sl Universit\`a di L'Aquila - Dipartimento di Matematica,}
\centerline{\sl Via Vetoio, Loc. Coppito - 67100 L'Aquila, Italy}
\centerline{\sl E-mail: {guerran\@univaq.it}}
\vskip 2ex
\centerline{\sl Purdue University - Department of Mathematics,}
\centerline{\sl West Lafayette, Indiana 47907, USA}
\centerline{\sl E-mail: {heinzer\@math.purdue.edu}}
\footnote{ }{{\footitfont 1991 Mathematics Subject Classification.}
13A30, 13C05, 13E05, 13H15}
\footnote{ }{{\footitfont Key words and phrases.}
\footfont Ratliff--Rush filtration, principal reductions, conductor.}
\enddocument
--=====================_937257413==_--
------------------ooOOO---------------OOOoo---------------------------------
Anna Guerrieri tel: +39 0862 433137
Universita' di L'Aquila fax: +39 0862 433180
Dipartimento di Matematica E-mail: guerran@univaq.it
Via Vetoio, Loc. Coppito guerrann@mat.uniroma1.it
67100 L'Aquila - Italy http://univaq.it/~guerran/Welcome.html
------------------ooOOO---------------OOOoo---------------------------------
------------------ooOOO---------------OOOoo---------------------------------
Anna Guerrieri tel: +39 0862 433137
Universita' di L'Aquila fax: +39 0862 433180
Dipartimento di Matematica E-mail: guerran@univaq.it
Via Vetoio, Loc. Coppito guerrann@mat.uniroma1.it
67100 L'Aquila - Italy http://univaq.it/~guerran/Welcome.html
------------------ooOOO---------------OOOoo---------------------------------