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\topmatter
\title Coefficient and Stable Ideals in Polynomial Rings
\endtitle
\author
William Heinzer and David Lantz
\endauthor
\date
August 30, 1996
\enddate
\endtopmatter
\document
\baselineskip 18 pt
Let $x_1, \dots, x_d$ be indeterminates over an infinite field $F$, let
$R$ denote the polynomial ring $F[x_1, \dots, x_d]$, and
let $M$ denote the maximal ideal $(x_1, \dots, x_d)R$. If
$I$ is an $M$-primary ideal the Hilbert polynomial
$$
P_I(n) = e_0(I)\binom{n+d-1}{d} - e_1(I)\binom{n +d -2}{d-1}
+ \cdots + (-1)^de_d(I)
$$
gives the length of the $R$-module $R/I^n$ for sufficiently
large positive integers $n$. The integral
closure $I'$ of $I$ is the unique largest ideal of $R$
containing $I$ and having
the same coefficient $e_0$ (i.e., multiplicity) as $I$,
and the {\it Ratliff-Rush ideal } $\rr{I}$
of $I$ is the unique largest ideal containing $I$ and having
the same Hilbert polynomial as $I$.
Kishor Shah has shown in \cite{S1}
that there exists a unique chain of ideals
\footnote{The existence of this unique chain of ideals is
shown in \cite{S1, Theorem~1} for an ideal primary for the maximal ideal
of a quasi-unmixed local ring with infinite residue field. Since
$R_M$ is regular and so, in particular, quasi-unmixed, and since the
length of $R/I^n$ is equal to the length of $R_M/I^nR_M$, Shah's
result also applies in our setting.}
$$
I \subseteq \rr{I} = \k{I}{d} \subseteq \cdots \subseteq
\k{I}{k} \subseteq \cdots \subseteq \k{I}{0} = I'\ ,
$$
where, for $0 \le k \le d$, the
ideal $\k{I}{k}$ is maximal with the property of having the same
coefficients
$e_0, \dots, e_k$ of its Hilbert polynomial as those of $I$.
The ideal $\k{I}{k}$ is called the {\it k-th coefficient ideal}
of $I$. If $I = \k{I}{k}$, we say $I$ is an
$e_k$-{\it ideal}.
We are particularly interested in the case where $R$ is of
dimension two. In this setting, an $M$-primary ideal $I$
has reduction
number at most one (i.e., if $J$ is a minimal reduction of $I$,
then $JI = I^2$) if and only if the Rees algebra $R[It]$ is
Cohen-Macaulay \cite{HM, Prop.~2.6},\cite{JV, Theorem~4.1},
or \cite{S2, Corollary~4(f)}. Moreover, the coefficients
$e_1(I)$ and $e_2(I)$ are nonnegative, and
it follows from \cite{Hu, Theorem~2.1}
that $I$ has reduction number at most one if and only if
$\lambda(R/I) = e_0(I) - e_1(I)$, and if this holds, then
$e_2(I) = 0$.
We say that an ideal with
these properties is {\it stable}. Thus, $I$ is stable if and
only if $I = \rr{I}$ and $e_2(I) = 0$.
Stable ideals are $e_1$-ideals, but
it is shown in \cite{HJL, Example~5.4}
that there exist $e_1$-ideals $I$ for which $e_2(I) > 0$.
We are interested in a better understanding of the features of
$e_1$-ideals and the distinguishing aspects between $e_1$-ideals
and the more restrictive subset of stable ideals.
Our purpose in this paper is:
\roster
\item to present examples of first coefficient and stable ideals
in dimension 2,
\item to compare the description of the coefficient ideals given
by Shah in \cite{S1, Theorems~2 and 3} with that given in
\cite{HJLS, Theorem~3.17} involving the blowup of $I$,
\item to present examples of coefficient ideals in higher dimensions,
\item to present two results on the existence of stable ideals
in dimension 2, and to prove the $e_1$-closure of certain
monomial ideals in dimension 2 are stable ideals.
\endroster
In particular, in connection with (3) we present in more detail
and with typographical corrections
\cite{HJLS, Example~3.22} that establishes the existence of examples
of ideals $I$ in dimension $d$ such that for all sufficiently
large positive integers $n$ one has
$$
I^n = \rr{I^n } = \k{(I^n)}{d} < \cdots <
\k{(I^n)}{k} < \cdots < \k{(I^n)}{0} = (I^n)'\ ,
$$
and thus gives examples where all the associated coefficient
ideals are distinct. Also in Section~3 we use a suggestion
made to us by Karen Smith to observe that if $I$ is a
monomial $M$-primary ideal, then all the associated
coefficient ideals $\k{I}{k}$ of $I$ are monomial ideals.
\subheading{1. Examples of coefficient ideals in dimension 2}
Let $x$ and $y$ be indeterminates over the field $F$,
let $R = F[x,y]$, and let $M = (x,y)R$.
The following examples illustrate the associated
coefficient ideals of various ideals $I$:
Consider the ideal $I = (x^6, x^2y^4, y^6)R$.
The form ring $\G(I)$ of $R$ with respect to $I$ has
depth one (as we have checked with Macaulay via the
Rees algebra $R[It]$), so $I$ and all its powers are
Ratliff-Rush ideals \cite{HLS, (1.2)}.
We have the following data, including
the differences $Hd$ between the Hilbert function and the
Hilbert polynomial as far as we have computed them:
\vskip 3 pt
\settabs\+\ \ &$I'=M^6$\qquad&$I$, all monomials of %
degree 9\quad&216\qquad&66\qquad&88\qquad&32\qquad& \cr
\+&Ideal&Generators&$e_0$&$e_1$&$e_2$& &$Hd$ \cr
\+&$I = \rr{I}$&$x^6, x^2y^4, y^6$&36&12&4& %
&$[0,0,0,0,0,0,0,0,\ldots]$ \cr
\+&$\k{I}{1}$&$I, x^4y^2$&36&12&0& &$[0,0,0,0,0,0,\ldots]$ \cr
\+&$I'=M^6$& &36&15&0& &$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
Since $e_2(\k{I}{1}) = 0$, this ideal is stable.
To obtain a strict inclusion between $I$ and $\rr{I}$, as well as
between the coefficient ideals, we modify as follows: Consider the
ideals
\vskip 3 pt
\+&Ideal&Generators&$e_0$&$e_1$&$e_2$& &$Hd$ \cr
\+&$I$&$x^{12}, x^8y^4, x^6y^6, x^2y^{10}, y^{12}$%
&144&60&4& &$[4,0,0,0,0,0,\ldots]$ \cr
\+&$\rr{I}$&$I, x^4y^8$&144&60&4& &$[0,0,0,0,0,0,0,\ldots]$ \cr
\+&$\k{I}{1}$&$\rr{I}, x^{10}y^2$&144&60&0& %
&$[0,0,0,0,0,0,\ldots]$ \cr
\+&$I'=M^{12}$& &144&66&0& &$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
Note that we have $I^2 = \left(\rr{I}\right)^2 = \rr{(I^2)}$.
Also, we have again that the $e_1$-closure $\k{I}{1}$ of
$I$ is a stable ideal.
Let us turn to an inspection of $M$-primary ideals of $R$
which have $e_1$-closure not stable. Since the set
of stable ideals integral over a given ideal is closed under
intersection \cite{HJL, Corollary~4.4}, we can speak
of the {\it stable closure} of the
ideal $I$, which we denote $s(I)$. Again consider the ideals:
\vskip 3 pt
\+&$I$&$x^7, x^5y^3+x^3y^5, y^7$&49&10&1& %
&$[1,0,0,0,0,0,\ldots]$ \cr
\+&$\rr{I}=\k{I}{1}$&$I,x^6y^4$&49&10&1& %
&$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
Our verification that $\k{I}{1} = (I, x^6y^4)$ is by using
Macaulay, see \cite{HJL, Example~5.4}.
\footnote{In Example~5.4 of
\cite{HJL}, $I = J$.
The element $x^6y^4$ is in $(I:M)$, the preimage of the socle in
$R/I$ (as is $x^4y^6$, with the same image in $R/I$.}
Another
element of $(I:M)-I$ is $x^6y^2 + x^4y^4 + x^2y^6$, not in $\rr{I}$.
Question:~Is there a way to tell which elements of $(I:M)$ are in
$\rr{I}$? In this example, $(I:M)$ is not even contained in
$s(I)$, since:
\vskip 3 pt
\+&$s(I)$&$\rr{I}$, all monomials of degree 9&49&11&0&%
&$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
It follows from Result~4.1 that this ideal is stable. To
see that this is indeed the stable closure, we note that
there is no intervening ideal with Hilbert coefficients 4,10,0,
since $\k{I}{1}$ is an $e_1$-ideal; so the Hilbert polynomial of
this ideal is the next possible for a stable ideal. (It is also
true that the set of $e_1$-ideals integral over a given ideal is
closed under intersection \cite{HJL, Prop.~4.5}.)
Another ideal having the property that
its Ratliff-Rush closure is an $e_1$-ideal
that is not stable is
\vskip 3 pt
\+&$I$&$x^8, y^8, x^6y^3 + x^3y^6$&64&14&1& %
&$[1,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
If we adjoin to $I$ the element $x^7y^4$, we get the same Hilbert
polynomial; and we can verify that the resulting ideal and all its
powers are Ratliff-Rush. Thus:
\+&$\rr{I}$&$I, x^7y^4$&64&14&1& &$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
As in Example~5.4 of \cite{HJL}, we have shown using Macaulay
that $\rr{I} = \k{I}{1}$. Now we would like to know $s(I)$;
we first thought it might be
\vskip 3 pt
\+&$K$&$\rr{I}, x^5y^5$&64&15&1& &$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
But the ideal $K$ is not stable. However
\vskip 3 pt
\+&$L$&$I,x^3y^7,x^4y^6$&64&15&0& &$[0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
\underbar{is} stable; and as in the last case, in view of the
Hilbert coefficients, $L = s(I)$. In general, we would like
to better understand the process of passing
from an ideal to its stable closure.
\subheading{2. Passing from an ideal to its coefficient ideals}
Let $R$ be a $d$-dimensional Noetherian quasi-unmixed local
ring, and let $I$ be an ideal primary for the maximal ideal of $R$.
Kishor Shah has shown that one way to attain all the coefficient
ideals of $I$ is as follows: For each integer
$k$ in $\{1,2,\ldots,d\}$:
$$
\k{I}{k} = \bigcup (I^{n+1}:B)\ ,
$$
where $n$ varies over the positive integers and $B$
varies over all the $k$-element subsets of sets of $d$ generators
of minimal reductions of $I^n$.
In particular,
if $I$ is such that $I$ and all its powers are Ratliff-Rush,
i.e., $\G(I)$ has positive depth, then taking a minimal
reduction $\bq$ of $I$ and considering $(I^{n+1}:\bq^n)$ does not
give us more than $I$, for the image of
$\bq^n$ in the $n$-th graded piece $I^n/I^{n+1}$ of
$\G(I)$ contains a regular element of $\G(I)$.
Let us see how Shah's description of the coefficient ideals gives
the same results as the description given in \cite{HJLS} involving
the blowup of $I$.
The description of the coefficient ideals given in
\cite{HJLS, Theorem~3.17}
can be phrased as follows: The ideal $\k{I}{k}$ is the
intersection of $R$ and the
extensions of $I$ to the following family of overrings: $R[I/a]_P$,
as $a$ varies over a fixed set $A$ of $d$ generators of a minimal
reduction of $I$ and $P$ varies over the primes of height $\leq k$
in $R[I/a]$ that contain $I$ (or equivalently $a$). Let $P$ be
such a prime in such a ring $R[I/a]$. Then
since $R[I/a]$ is the degree-0 piece of the localization
$R[It][1/(at)]$ of the Rees algebra $R[It]$ of $I$, and this
localization is a Laurent series ring in the indeterminate $at$
over $R[I/a]$, it follows that $PR[It][1/(at)]\cap R[It]$ is a
prime of height $\leq k$ in the Rees algebra. Since $a\in P$,
we have $I\subseteq P$; so the image $Q$ of $P$ in the form ring
$\G(I) = R[It]/IR[It]$ is a prime of
height $\leq k-1$. Now for any set $C$ of $d$ generators of a
minimal reduction of a power $I^n$ of $I$, $\G(I)$ is
integral over $(R/I)[\overline{c}:c\in C]$, and the elements
$\overline{c} = ct+IR[It]$ form a regular sequence $\G(I)$. Thus,
for any $k$-element subset $B$ of $C$, the prime $Q$ cannot
contain every $\overline{b}$ for $b\in B$ (for otherwise
$\hgt(Q)\geq k$). Taking preimage in the Rees algebra, there is
some $b$ in $B$ for which $bt\notin PR[It]$, so that $b/a$ is a
unit in $R[I/a]_P$. Therefore, if $f\in (I^{n+1}:B)$, then
$f = (b/a^n)^{-1}(bf/a^n) \in IR[I/a]_P$. In other words, the
union of Shah's description is contained in the intersection
of \cite{HJLS}.
For the reverse inclusion, given an element
in the intersection of \cite{HJLS}, we must find a set $B$ of
the kind described by Shah so that $(I^{n+1}:B)$ contains
that element. In the Rees algebra $R[It]$, take an irredundant
primary decomposition of $IR[It]$, say $IR[It] =
Q_1\cap Q_2\cap\dots\cap Q_n$, where each $Q_j$ is homogeneous.
For each $k=1,\ldots,d$,
let $J_k$ denote the intersection of those $Q_j$ of which the
radical has height at least $k+1$. Applying the
``refined generalized prime avoidance lemma''
\cite{S1, Lemma~2(F)} to the images $J_k/IR[It]$ in $\G(I)$,
we see that we can find, for some positive integer~$N$,
elements $c_1,\ldots,c_d$ in $I^N$ for which, for each
$k=1,\ldots,d$, we have $c_1t^N,\ldots,c_kt^N\in J_k$ and
$\dim(G(I)/(c_1t^N+IR[It],\ldots,c_kt^N+IR[It])\G(I)) = d-k$.
Moreover, if $J_{k-1} < J_k$, then we must choose $c_nt^N$ from
$J_k-J_{k-1}$. It follows that all minimal primes of
$(I,c_1t^N,\ldots,c_kt^N)R[It]$ have height at least $k+1$.
Suppose $f$ is in the intersection of \cite{HJLS}, i.e., $f$
in $R$ is also in $IR[I/a]_P$ for each $a$ in $A$ and each
prime $P$ in $R[I/a]$ containing $I$ and of height $\leq k$.
Then $f \in R[It] \cap \bigcap\{IR[It]_P: I\subseteq P,\ \hgt(P)
\leq k\} = \bigcap\{Q_j : \hgt(\rad(Q_j))\leq k\}$. Since
$c_1t^N,\ldots,c_kt^N$ are in the remaining $Q_j$ (in fact,
if $J_{k-1} < J_k$, then $c_kt^N$ has this property), we see that
$f(c_1t^N,\ldots,c_kt^N)$ is in the degree-$N$ piece $I^{N+1}t$
of $IR[It]$, i.e., $f \in I^{N+1}:(c_1,\ldots,c_k)$. (In fact,
if $J_{k-1} < J_k$, then $f \in I^{N+1}:c_k$. Thus, the
distinct coefficient ideals of $I$ are in fact colon ideals of
$I^{N+1}$ by a single element of the sequence $c_1,\ldots,c_d$.
The above is a very slight reworking of Shah's proof, but
he does not note this realization of coefficient ideals as
colons by a single element.)
\subheading{3. Coefficient ideals in higher dimensions}
Let $R = F[x,y,z]$ where $x,y,z$ are indeterminates over
an infinite field $F$, and let $M = (x,y,z)R$.
Consider the ideal
\vskip 3 pt
\+&Ideal&Generators&$e_0$&$e_1$&$e_2$&$e_3$&$Hd$ \cr
\+&$I$&$x^6, x^2y^4, y^6, x^2z^4, z^6$&216&144&88&32%
&$[8,0,0,0,0,0,0,\ldots]$ \cr
\vskip 3 pt
\noindent
Using Macaulay we see that the form ring $\G(I)$ of $R$
with respect to $I$ has depth one, so $I$ and all its
powers are Ratliff-Rush ideals. We claim that
\vskip 3 pt
\+&$\k{I}{2}$&$I, x^4y^2z^2$&216&144&88&40%
&$[8,0,0,0,0,0,0,\ldots]$\cr
\+&$\k{I}{1}$&$(x^2,y^2,z^2)^3$&216&144&8&0%
&$[0,0,0,0,0,0,\ldots]$\cr
\+&$I'= \k{I}{0} = M^6$& &216&180&20&0%
&$[0,0,0,0,0,0,\ldots]$\cr
\vskip 3 pt
\noindent
so that
$$
I = \rr{I} < \k{I}{2} < \k{I}{1} < I' = M^6.
$$
The Hilbert polynomial
$$
P_I(n) = e_0(I)\binom{n + 2}{3} - e_1(I)\binom{n+1}{2}
+ e_2(I)\binom{n}{1} - e_3(I)
$$
gives the Hilbert function for all positive integers $n$,
but not for $n = 0$, i.e., the postulation number of $I$ is
zero (as this term is used in \cite{M}).
To find these coefficient ideals of $I$ we examine the
blowup of $I$. Dividing by $x^6$, we have the affine piece
$$
R_x = F[x,y,z, (y/x)^4, (y/x)^6, (z/x)^4, (z/x)^6]
\subset (R_x)' = F[x, y/x, z/x].
$$
Then $R_x^{(2,x)} = R_x[x^2(y/x)^2(z/x)^2]$ and
$R_x^{(1,x)} = R_x[(y/x)^2, (z/x)^2]$. Both of these
assertions were checked by using Macaulay: The ring
$R_x^{(2,x)}$ has depth~2 (i.e., the maximal ideal is not
an associated prime of a principal ideal), and $R_x^{(1,x)}$
is a complete intersection.
Also, $R_y = F[x,y,z, (x/y)^2, (x/y)^2(z/y)^4, (z/y)^6]$.
In this case $R_y = R_y^{(2,y)}$ (again, by Macaulay, it
has depth~2), and $R_y^{(1)} = R_y[(z/y)^2]$ is a
complete intersection. Similarly for $R_z$.
Thus, $\k{I}{2}$ is the intersection of the contractions of the
extensions of $I$ to $R_x^{(2)}$, $R_y^{(2)}$ and $R_x^{(2)}$.
It is clear that this intersection contains $x^4y^2z^2$ and
that the blowup of $(I,x^4y^2z^2)R$ has these rings as its affine
pieces; checking that $(I,x^4y^2z^2)R$ is Ratliff-Rush, we
conclude that it is $\k{I}{2}$. Similarly for $\k{I}{1}$.
\result{Interlude 3.1.} We thought at first that $\k{I}{1}$ was the
ideal
$(I,x^4y^2,x^4z^2,x^2y^2z^2)R$ --- we missed some of
the contraction from the affine pieces of the model --- so we
wanted to show that this
ideal is Ratliff-Rush. Eureka! After 8.5 hours of Macaulay run
we obtained this verification by computing that the depth of
the Rees algebra of this ideal is 3. Hence by \cite{HM}
the depth of the form ring with respect to this ideal is 2.
The verification that this ideal is Ratliff-Rush using Macaulay
turned out to be very time- and memory-consuming; so we were led
to seek a simplifying approach. We note the following: The
generators of this ideal are all in the subring $A=F[x^2,y^2,z^2]$
of $B=F[x,y,z]$, and $B$ is free over $A$. Let $J$ denote the
$A$-ideal generated by $x^6,y^6,z^6,x^2y^4,x^4y^2,x^2z^4,x^4z^2,
x^2y^2z^2$. Then the ideal above is $JB$, and $J$ is an ideal
generated in degree~3 in the polynomial ring $A$. Since the
form ring of $B$ with respect to $JB$ is free over the form
ring of $A$ with respect to $J$, if we show that the latter
form ring has positive depth, then so has the former, so that
all powers of $JB$ are Ratliff-Rush. Thus, we set Macaulay to
computing the projective dimension of the Rees algebra of
$(x^3,y^3,z^3,xy^2,x^2y,xz^2,x^2z,xyz)B$. But even this turned
out to challenge Macaulay.
\medskip
\result{Interlude 3.2.} Another method of simplifying the computations
on the defining ideal of the Rees algebra, to obtain information
on the depth of the form ring and the Cohen-Macaulay property of
the blowup, was shown to us by Craig Huneke. Let $I$ be a
quasi-homogeneous ideal in the polynomial ring
$R = F[x,y, \dots ]$; let $I = (f_1, \dots, f_n)R$ where the
$f_i$'s are quasi-homogeneous. Then Macaulay can compute the
kernel $J$ of the $R$-algebra epimorphism from
$T = R \otimes_F F[a_1, \dots, a_n]$ ($a_i$'s indeterminates) onto
the Rees algebra $R[It]$ defined by $a_i \mapsto f_it$.
The program can then find a minimal projective resolution of $J$
over $T$; the number of matrices in this resolution is the
projective dimension of $R[It]$ over $T$, so that the
Auslander-Buchsbaum formula yields the depth of $R[It]$.
It follows from \cite{HM, Theorem~2.1, page~262} that, if
$G(I)$ is not Cohen-Macaulay, then the depth of $G(I)$ is one
less than that of $R[It]$. Suppose the maximal minors of the last
matrix in the resolution of $J$ generate an ideal primary for
$(x,y, \dots , a_1, \dots , a_n)T$. Then each ring in the blowup
of $I$, i.e., $\Proj(R[It])$, has projective dimension less than
that of $R[It]$. Thus, in this case, if $\depth(R[It]) = \dim(R)$,
then $\Proj(R[It])$ is Cohen-Macaulay.
But the computation of the projective resolution of $J$ may
be very long and difficult, even for Macaulay. Huneke
suggests passing to the quotient ring $T/xT$ of $T$ obtained
by setting $x = 0$. Since $x$ is regular on both $T$ and $R[It]$,
\cite{N, 27.2} the projective dimension of $J$ over $T$ is
equal to the projective dimension of $J/xJ$ over $T/xT$.
With regard to Example~(E5) on page~387 of \cite{HJLS},
we can now confirm that the linear and constant terms of
Hilbert functions of ideals between $I = (x^3,y^3)$ and
its integral closure $I' = (x,y)^3$ are as given there.
Let $R = F[x,y,z]$.
The ideal $I = (x^4, y^4, z^4, x^2y^2z^2, x^3y^3z, x^3yz^3)R$
was pointed out to us by Les Reid to be an example of a
monomial ideal that is 6-generated but has 7
corner elements. Is this property of the ideal $I$
reflected in some way in the associated coefficient ideals
of $I$?
\result{Example 3.22 of \cite{HJLS} revisited.} Let $F$ be an infinite
field and let $x,y_2,\ldots,y_d$ be
indeterminates over~$F$. Consider the affine domain
$S = F[x,\{xy_i,y_i^4,y_i^6\}_{i=2}^d]$.
Then
$$
S^{(1,xS)} = S^{(1)} = F[x,\{xy_i,y_i^2\}_{i=2}^d] <
F[x,\{y_i\}_{i=2}^d] = S'\ .
$$
To see that, for $k$ in $\{1,\ldots,d-1\}$,
we have $S^{(k+1,xS)} < S^{(k,xS)}$, we note that the
product of $x^{2k-2}$ and $y_i^2$, for $k$~distinct values of~$i$,
is an element of $S^{(k,xS)}$, since any prime~$P$ in~$S$ of height at
most~$k$ and containing~$x$ does not contain $y_i^4,y_i^6$ for at least
one of the $y_i$'s that appear in the product, so one of
the factors $y_i^2$ is a unit in $S_P$, and the product of the remaining
factors is an element of~$S$. But this element is not
in~$S^{(k+1,xS)}$ because it is not in the localization of $S$ at the
prime ideal of height~$k+1$ generated by $x$ and $xy_i,y_i^4,y_i^6$
for the $y_i$'s that appear in the product.
The domain $S$ is an affine piece of the blowup of the ideal
$I$ generated by $x^6, x^2(xy_i)^4, (xy_i)^6$, $i=2,\ldots,d$
in the polynomial ring $R = F[x,\{xy_i\}_{i=2}^d]$. Forming the rings of
fractions of $R,S$ with respect to the complement in $R$ of the
maximal ideal $M = (x,\{xy_i\}_{i=2}^d)R$ yields a regular local
ring~$R_M$, and $(R-M)^{-1}S$ is an affine piece of the blowup of $IR_M$
which retains the properties verified in the last paragraph.
Moreover, since the extensions of the affine piece of the blowup of
$I$ as described in \cite{HJLS} are distinct, it follows that,
for sufficiently high powers of $I$, the contractions of these
powers from the various extensions of the blowup are distinct; i.e.,
for $n$ sufficiently large,
$$
\rr{(I^n)} = \k{(I^n)}{d} < \k{(I^n)}{d-1} < \dots < \k{(I^n)}{0}
= (I^n)'\ .
$$
In fact, we believe that these strict inclusions hold for $n=1$.
\result{Observation 3.3.} Let $R = F[x_1,\dots,x_d]$ be a
polynomial ring in $d$ variables $x_1,\dots,x_d$ over an
infinite field $F$, let $M = (x_1,\dots,x_d)R$ and let
$I$ be an $M$-primary ideal. We say that $I$ is a {\it monomial
ideal} if $I$ is generated by monomials in $x_1,\dots,x_d$.
Karen Smith suggested to us the following argument to show
that if $I$ is a monomial ideal, then all the associated
coefficient ideals $\k{I}{k}$ of $I$ are also monomial ideals.
For each $d$-tuple $a = (a_1,\dots,a_d)$ in the algebraic $d$-torus
$F^* \times \cdots \times F^*$, where $F^* = F - 0$ is the
multiplicative group of units of $F$,
define an $F$-automorphism $\phi_a:R \to R$ by setting
$\phi_a(x_i) = a_ix_i$ for $1 \le i \le d$.
To show $\k{I}{k}$ is a monomial ideal
it suffices to show $\phi_a(\k{I}{k}) = \k{I}{k}$ for each
$a = (a_1,\dots,a_d)$. By assumption $I = (f_1,\dots, f_n)R$,
where each $f_i$ is a nonzero monomial.
Let $R_i = R[f_1/f_i, \dots , f_n/f_i]$. Then $\phi_a$
naturally extends to an $F$-automorphism of $R_i$ which we
continue to call $\phi_a$,
and $IR_i = f_iR_i$ is mapped to itself under
$\phi_a$. The invariance of $f_iR_i$ under $\phi_a$
implies that the union of the associated primes of
$f_iR_i$ of height at most $k$ in $R_i$ is mapped onto
itself under $\phi_a$. Therefore $\phi_a$ extends to an
automorphism of the localization $T_{ik}$ of $R_i$ at the complement of
the union of the associated primes of $f_iR_i$ of height at
most $k$. It follows that $f_iT_{ik} \cap R$ is mapped onto
itself by $\phi_a$. By \cite{HJLS, Theorem~3.17},
$\k{I}{k} = \cap_{i=1}^nf_iT_{ik} \cap R$. Therefore
the ideal $\k{I}{k}$ is mapped onto itself by $\phi_a$, so
$\k{I}{k}$ is a monomial ideal.
\subheading{4. Stable ideals in dimension 2}
Let us examine with $R = F[x,y]$ what ideals between
$I = (x^n, y^n)$ and its integral closure $I' = (x,y)^nR$
are stable. The following two results show that
many of these ideals are stable:
\result{Result 4.1.} Let $a,b$ be a minimal reduction
of the height-2 ideal $A$ in
the ring $R$, and $n$ be a positive integer; set $I = (a^n,b^n)$.
Then $I$ is a minimal reduction of $A^n$;
suppose the reduction number is 1, i.e., $IA^n = A^{2n}$.
Then for every nonnegative integer $j$ we have $IA^{n+j} =
A^{2n+j}$. Set $J = I + A^{n+i+1}$ for a positive
integer $i$; then we have
$$
J^2 = IJ + (A^{n+i+1})^2 = IJ + A^{2n+2i+2} =
IJ + IA^{n+2i+2} \subseteq I(J + A^{n+i+1}) = IJ\ ;
$$
i.e., $J$ is stable. Moreover, for each ideal $K \subseteq A^{n+i}$
the ideal $L = J + K$ is also stable. For, we have
$$\align
L^2 = J^2 + JK + K^2 &\subseteq IJ + (I + A^{n+i+1})K + A^{2n+2i} \\
&\subseteq I(J+K) + A^{2n+2i+1} + A^{2n+2i} =
I(J+K) + A^{2n+2i} \\
&= I(J+K+A^{n+2i}) = I(J+K) = IL
\endalign$$
where we have used $A^{n+2i} \subseteq J$.
\medskip
\result{Result 4.2.} Again, let $R$ be a ring. Let $a,b$ be a
$R$-sequence, and $m,n$ be positive integers; and set
$I = (a^m,b^n)$. Then for any ideal $J$ contained in
$a^{\lceil m/2\rceil}b^{\lceil n/2\rceil}R$, the sum $I+J$
is stable. For, $(a^{\lceil m/2\rceil}b^{\lceil n/2\rceil})^2
\in I^2$, so $J^2 \subseteq I^2$, so $(I+J)^2 = I^2 + IJ + J^2
\subseteq I^2+IJ = I(I+J)$.
\medskip
Applying these paragraphs to $x,y$ in $F[x,y]$, we see that:
(1)~for any integer $n\geq 2$, we can find stable ideals $I$
between $(x^n,y^n)R$ and $(x,y)^nR$ such that $e_0(I) = n^2$
and $e_2(I) = 0$ (both necessarily) and $e_1(I)$ is any integer
from 0 to $(n-2)(n-1)/2$; and (2)~for any positive integers
$m,n$, we can find stable ideals $I$ between $(x^m,y^n)R$ and
its integral closure with $e_0(I) = mn$, $e_2(I) = 0$ (again,
both necessarily) and
$e_1(I)$ is any integer from 0 to $\lceil\frac{m}2\rceil
\lceil\frac{n}2\rceil$.
In (4.3) we prove that
the $e_1$-closure of certain monomial ideals are stable.
This shows that examples such as \cite{HJL, Example~5.4}
are of necessity not generated by monomials.
\result{Observation 4.3.}
Let $R = F[x,y]$ where $x$ and $y$ are indeterminates over the
infinite field $F$, let $m$ and $n$ be positive integers and
suppose $I$ is a monomial ideal of $R$ integral over $(x^m,y^n)R$.
We show that the $e_1$-closure $J$ of $I$ is stable.
By \cite{HJLS, Theorem~3.17} (see also \cite{HJL, (3.2)},
$J = ID \cap R$, where $D$ is the first coefficient domain of $I$.
Then $D$ is also the first coefficient domain of $J$
and each of the rings $R[J/x^m]$ and $R[J/y^n]$ is contained in $D$.
Therefore $JR[J^2/x^my^n] \cap R = J$. It follows that
$(J^3:x^my^n) = J$. To show that $J$ is stable, it suffices
to show that $J^2 \subseteq (x^m, y^n)J$. The Briancon-Skoda
Theorem \cite{LS, Theorem~1} implies $J^2 \subseteq (x^m, y^n)R$.
By (3.3), the ideal $J$ is a monomial ideal. Let $a \in J^2$
with $a$ a monomial. Then $a \in (x^m, y^n)R$ implies either
$a \in x^mR$ or $a \in y^nR$. Suppose $a = x^mb$ with $b \in R$.
Then $b \in (J^2:x^m) \subseteq (J^3:x^my^n) = J$ which means
$a \in (x^m,y^n)J$. A similar arguement applies in case $a \in y^nR$.
Therefore $J$ is stable.
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\enddocument