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\begin{document}
\title[]{Properties of the fiber cone \\ of ideals in local rings}
\date{\today }
\author{\ }
\address{\ }
\thanks{We thank Bernd Ulrich for
helpful conversations about the topics considered here.}
\thanks{The second author is supported by Korea Research Foundation Grant
(KRF-2000-015-DP0005).}
\subjclass{Primary: 13A30, 13C05; Secondary: 13E05, 13H15}
\keywords{fiber cone, analytic spread , multiplicity,
Cohen-Macaulay ring}
\maketitle
\begin{center}
\begin{tabular}{c}
{\bf William J. Heinzer} \\
Department of Mathematics, Purdue University \\
West Lafayette, Indiana 47907 USA\\
E-mail: heinzer@math.purdue.edu \\
\ \\
{\bf Mee-Kyoung Kim} \\
Department of Mathematics, Sungkyunkwan University \\
Jangangu Suwon 440-746, Korea \\
E-mail: mkkim@math.skku.ac.kr
\end{tabular}
\end{center}
\bigskip
\begin{abstract}
\noindent
For an ideal $I$ of a Noetherian local ring $(R,\m)$ we
consider properties of $I$ and its powers as reflected
in the fiber cone $F(I)$ of $I$.
In particular, we examine behavior of the fiber cone
under homomorphic image $R \to R/J = R'$ as related to
analytic spread and generators for the kernel of the
induced map on fiber cones $\psi_J : F_R(I) \to F_{R'}(IR')$.
We consider the structure of fiber cones $F(I)$ for which
$\ker \psi_J \ne 0$ for each nonzero ideal $J$ of $R$.
If $\dim F(I) = d > 0$, $\mu(I) = d+1$ and there exists
a minimal reduction $J$ of $I$ generated by a regular
sequence, we prove that if $\grade(G_+(I)) \ge d-1$, then
$F(I)$ is Cohen-Macaulay and thus a hypersurface.
\end{abstract}
\baselineskip 17 pt
\section{Introduction}
For an ideal $I$ in a Noetherian local ring $(R,\m)$,
the {\it fiber cone} of $I$ is the graded ring
$$
F(I) = \bigoplus_{n \ge 0}F_n =
\bigoplus_{n \ge 0}I^n/\m I^n
\cong R[It]/\m R[It],
$$
where $R[It]$ is the Rees ring of $I$ and $F_n = I^n/\m I^n$.
We sometimes write $F_R(I)$ to indicate we are considering the
fiber cone of the ideal $I$ of the ring $R$. In terms of the
height, $\hgt(I)$, of $I$ and the dimension, $\dim R$, of $R$, one always
has the inequalities $\hgt(I) \le \dim F(I) \le \dim R$.
For an arbitrary ideal $I \subseteq \m$ of $(R,\m)$,
the fiber cone $F(I)$ has the attractive property of
being a finitely generated graded ring over the
residue field $k := R/\m$ that is generated in degree one,
i.e., $F_n = F_1^n$ for each positive
integer $n$, so $F(I) = k[F_1]$.
It is well known in this setting that the Hilbert function
$H_F(n)$ giving the dimension of $I^n/\m I^n$ as a vector space over $k$
is defined for $n$ sufficiently large by
a polynomial $h_F(X) \in \Q[X]$, the {\em Hilbert polynomial}
of $F(I)$ \cite[Corollary, page 95]{Mat}, \cite[Corollary 11.2]{AM}.
A simple application of Nakayama's lemma,
\cite[Theorem 2.2]{Mat}, shows that
the cardinality of a minimal set of generators of $I^n$,
$\mu(I^n)$, is equal to $\lambda(I^n/\m I^n)$, the
value of the Hilbert function
$H_F(n)$ of $F(I)$.
An interesting invariant of the ideal $I$ is its analytic spread,
denoted $\ell(I)$, where the {\em analytic spread } of $I$
is by definition the dimension of the fiber cone,
$\ell(I) = \dim F(I)$ \cite{NR}. The analytic spread
measures the asymptotic
growth of the minimal number of generators of $I^n$
as a function of $n$. In relation to the
degree of the Hilbert polynomial, we have the
equality $\ell(I) = 1 + \deg h_F(X)$.
An ideal $J \subseteq I$ is said to be a
{\em reduction } of $I$ if there exists a
positive integer $n$ such that $JI^n = I^{n+1}$. It then
follows that $J^iI^n = I^{n+i}$ for every postive integer $i$.
If $J$ is a reduction of $I$, then $J$ requires at least
$\ell(I)$ generators.
If the residue field $R/\m$ is infinite, then
minimal reductions of $I$ correspond to Noether
normalizations of $F(I)$ in the sense that
$a_1, \dots, a_r \in I - I^2$ generate a minimal
reduction of $I$ if and only if their images
$\overline{a_i} \in I/\m I \subseteq F(I)$ are algebraically
independent over $R/\m$ and $F(I)$ is integral
over the polynomial ring $(R/\m)[\overline{a_1}, \dots ,
\overline{a_r}]$.
In particular, if $R/\m$ is infinite, then
there exist $\ell(I)$-generated reductions of $I$,
For a positive integer $s$, the fiber cone $F(I^s)$ of the
ideal $I^s$ embeds in the fiber cone $F(I) = \oplus_{n=0}^\infty F_n$
of $I$ by means of $F(I^s) \cong \oplus_{n=0}^\infty F_{ns}$.
This isomorphism makes $F(I)$ a finitely generated
integral extension of $F(I^s)$.
Thus $\dim F(I) = \dim F(I^s)$ and $\ell(I) = \ell(I^s)$.
We are particularly interested in conditions that imply
the fiber cone $F(I)$ is a hypersurface. Suppose
$\dim F(I) = d > 0$ and $\mu(I) = d+1$. If $I$ has
a reduction generated by a regular sequence and if
$\grade(G_+(I)) \ge d-1$, we prove in Theorem~5.6
that $F(I)$ is a hypersurface. We have learned from
Bernd Ulrich that this result also follows from
results in the paper \cite{CGPU} of Corso-Ghezzi-Polini-Ulrich.
A useful property of the analytic spread $\ell(I)$ is that it gives an upper
bound on the number of elements needed to generate $I$
up to radical. This property of generation up to radical
behaves well with respect to analytic spread of a
homomorphic image in the following sense:
\begin{lem}\label{1.1}
Suppose $I \subseteq \m$ is an ideal of a Noetherian local ring
$(R,\m)$, where $R/\m$ is infinite. Let $a \in I$
and let $R' := R/aR$ and $I' := IR'$.
If $a_1', \dots, a_s' \in R'$ are such that
$\rad(a_1', \dots, a_s')R' = \rad I'$
and if $a_i \in R$ is a preimage of $a_i'$, then
$\rad I = \rad(a_1, \dots, a_s, a)R$.
In particular, if
$\ell(I') = s$, then $I$ can
be generated up to radical by $s+1$ elements.
\end{lem}
\demo
Assume that $\rad (a_1', \dots, a_s')R' = \rad I'$.
If $x \in \rad I$, then for some positive integer
$n$, we have $x^n = y \in I$.
Hence the image $y'$ of $y$ in $R'$ is in $\rad(a_1', \dots, a_s')R'$.
Therefore $y$ and hence also $x $
is in $\rad(a_1, \dots, a_s, a)R$. \QED
\medskip
Examples given by Huckaba in \cite[Examples 3.1 and 3.2]{Hu}
establish the surprising fact of the existence of 3-generated
height-2 prime ideals $I$ of a 3-dimensional
regular local ring $R$ for which $\dim F(I) = 3 = \dim R$ and
for which there exists a principal ideal $J = xR \subseteq I$
such that if $R' := R/xR$ and $I' := IR'$, then
$\dim F_{R'}(I') = 1 < \dim R' = \dim R - 1$.
This result of Huckaba shows that a statement
analogous to Lemma~\ref{1.1} for reductions, rather than
generators up to radical, is false, that is, it is possible
that $I' = I/aR$ has an $s$-generated reduction while
every reduction of $I$ requires at least $s+2$ generators.
These interesting
examples are the original motivation for our interest in
the behavior of analytic spread in a homomorphic image.
\vskip 3ex
\section{Behavior of the fiber cone under homomorphic image.}
\vskip 3ex
\begin{setting}\label{2.1}
Let $J \subseteq \m$ be an ideal of a
Noetherian local ring $(R,\m)$, let $R' := R/J$, and
let $\m' = \m/J$ .
For an ideal $I \subseteq \m$ of $R$ let
$I' = (I + J)/J = IR'$ denote the image of $I$ in $R'$.
There is a
canonical surjective ring homomorphism of the fiber
cone $F_R(I)$ of $I$ onto the fiber cone
$F_{R'}(I')$.
\end{setting}
We have $R[It] = \bigoplus_{n \ge 0} I^nt^n$ and
$R'[I't] = \bigoplus_{n \ge 0} (I')^nt^n$. Since
$$
(I')^n = (I^n + J)/J \cong I^n/(I^n \cap J),
$$
there is a
canonical surjective homomorphism of graded rings
$\phi_J : R[It] \to R'[I't]$, with
$\ker \phi_J = \bigoplus_{n \ge 0}(I^n \cap J)t^n$.
Since $F_R(I) = R[It]/\m R[It]$ and $F_{R'}(I') =
R'[I't]/\m' R'[I't]$, the homomorphism $\phi_J : R[It] \to R'[I't]$
induces a surjective homomorphism
$\psi_J : F_R(I) \to F_{R'}(I')$ which preserves grading.
This is displayed in the
following commutative diagram for which the rows are exact
and the column maps are surjective:
$$
\begin{CD}
0 @>>> \oplus_{n \ge 0}(I^n \cap J)t^n @>>> R[It] @>\phi_J>> R'[I't] @>>> 0 \\
@. @VVV @VVV @VVV \\
0 @>>> \oplus_{n \ge 0} \frac{(I^n \cap J) + \m I^n}{\m I^n} @>>>
\frac{R[It]}{\m R[It]} @>\psi_J>> \frac{R'[I't]}{\m'R'[I']} @>>> 0
\end{CD}
$$
Since we are interested in the behavior of the fiber cone
under homomorphic image, we are especially interested in
$$
\ker \psi_J = \oplus_{n \ge 0} \frac{(I^n \cap J) + \m I^n}{\m I^n}
$$
\begin{remark}\label{2.2}
Let $(R,\m)$ be a Noetherian local ring and let $I \subseteq \m$
be an ideal of $R$.
Suppose $J_1 \subseteq J_2 \subseteq \m$ are ideals of $R$.
Let $R_i := R/J_i$, $i = 1,2$, and let $\psi_i : F_R(I) \to F_{R_i}(IR_i)$
denote the canonical surjective homomorphisms on fiber cones as in (2.1). Then
$R_2 \cong R_1/J'$, where $J' = J_2/J_1$, and there exists a
canonical surjective homomorphism $\psi' : F_{R_1}(IR_1) \to
F_{R_2}(IR_2)$ such that $\psi_2 = \psi'\circ \psi_1$.
\end{remark}
\medskip
With notation as in (2.1), if $J$ is a nilpotent ideal of $R$,
then $\ker \psi_J$ is a nilpotent ideal of $F_R(I)$. For
suppose $x \in J$ is such that $x^s = 0$. If $\overline x
\in \frac{(I^n \cap J) + \m I^n}{\m I^n} = F_n$ is the
image of $x$ in $F(I)$, then by definition $\overline{x}^s$ is
the image of $x^s$ in $F_{sn}$, so $\overline{x}^s = 0$. Thus
for $J$ a nilpotent ideal of $R$,
we have $\dim F_R(I) = \dim F_{R'}(I')$ and $\ell(I) = \ell(I')$.
Applying this
to the situation considered in (2.2), if $s$ is a positive
integer, $J_1 = J^s$
and $J_2 = J$, then with $\psi' : F_{R_1}(IR_1) \to F_{R_2}(IR_2)$
as in (2.2), it follows that $\ker \psi'$ is a nilpotent ideal
and in this situation $\dim F_{R_1}(IR_1) = \dim F_{R_2}(IR_2)$.
In particular for the examples of Huckaba \cite[Examples 3.1 and 3.2]{Hu}
mentioned in the end of Section 1, going modulo a power $x^nR$ of the ideal $xR$ also reduces
the dimension of the fiber cone $F(I)$ from 3 to 1.
\medskip
\begin{prop}\label{2.3}
With notation as in Setting 2.1, we have the
following implications of Remark~2.2.
\begin{enumerate}
\item
If $J' \subseteq J$ are
ideals of $R$ and if $\ker \psi_J = 0$, then
$\ker \psi_{J'} = 0$.
\item
$\ker \psi_J = 0$ if and only if
$\ker \psi_{xR} = 0$ for each $x \in J$.
\item
For $x \in \m$, we have $\ker \psi_{xR} = 0$ if and
only if $(I^n : x) = (\m I^n : x)$ for each $n \ge 0$.
\end{enumerate}
\end{prop}
\demo
Statements (1) and (2) are clear in view of (2.2) and
the description of $\ker \psi_J$ given in (2.1).
For statement (3), we use that $I^n \cap xR = x(I^n :x)$.
Thus $0 = \ker \psi_{xR} =
\oplus_{n \ge 0}\frac{(I^n \cap xR) + \m I^n}{\m I^n}
\iff (I^n \cap xR) \subseteq \m I^n $ for each $n$
$\iff x(I^n:x) \subseteq \m I^n$ for each $n$ $\iff
(I^n:x) \subseteq (\m I^n: x)$ for each $n$.
This last statement is equivalent to $(I^n : x) = (\m I^n : x)$
for each $n$. \QED
\begin{prop}\label{2.4}
Let $(R,\m)$ be a Noetherian local ring and let $I \subseteq \m$
be an ideal of $R$.
Suppose $J_1$ and $J_2$ are ideals of $R$
such that $\rad J_1 = \rad J_2$.
Let $R_i := R/J_i$, $i = 1,2$, and let $\psi_i : F_R(I) \to F_{R_i}(IR_i)$
denote the canonical surjective homomorphisms on fiber cones as in (2.1).
Then $\dim F_{R_1}(IR_1) = \dim F_{R_2}(IR_2)$ and
$\ell(IR_1) = \ell(IR_2)$.
\end{prop}
\demo Since $\rad(J_1 + J_2) = \rad J_1 = \rad J_2$, it
suffices to consider the case where $J_1 \subseteq J_2$. With
notation as in (2.2), $\ker \psi'$ is a nilpotent ideal.
Thus $\dim F_{R_1}(IR_1) = \dim F_{R_2}(IR_2)$ and
$\ell(IR_1) = \ell(IR_2)$. \QED
\medskip
As we remarked in Section 1,
the dimension of the fiber cone $F(I)$ of an
ideal $I$ is the same as the dimension
of the fiber cone $F(I^n)$ of a power $I^n$ of $I$.
Hence, with notation as in (2.1), we have
$\dim F_{R'}(IR') = \dim F_{R'}(I^nR')$
and $\ell(IR') = \ell(I^nR')$ for each
positive integer $n$.
\vskip 3ex
\section{The associated graded ring and the fiber cone.}
\vskip 3ex
The associated graded ring of the ideal $I$ plays a role
in the behavior of the fiber cone of the image of $I$ modulo
a principal ideal as we illustrate in Proposition~3.1
and Example~3.2.
\begin{prop}\label{3.1}
Let $I \subseteq \m$ be an ideal of a Noetherian local ring
$(R,\m)$. For $x \in \m$, let $x^*$ denote the
image of $x$ in the associated graded ring
$G(I) = R[It]/IR[It]$ and let $\overline x$ denote the
image of $x$ in the fiber cone $F(I)$. If
$x^*$ is a regular element of $G(I)$, then
$F(I)/\overline{x}F(I) \cong F_{R'}(I')$, where
$R' = R/xR$ and $I' = IR'$.
\end{prop}
\demo There exists a positive integer $s$ such
that $x \in I^s - I^{s+1}$. Since $x^*$ is a regular element
of $G(I)$ with $\deg x^* = s$, we have $(I^n \cap xR) = xI^{n-s}$
for every $n \ge 0$, where $I^{n-s} := R$ if $n - s \le 0$.
Hence we have
$$
[\ker \psi_{xR}]_n = \frac{(I^n \cap xR) + \m I^n}{\m I^n} =
\frac{xI^{n-s} + \m I^n}{\m I^n} = [\overline{x}F(I)]_n,
$$
for every $n \ge 0$. Therefore $F(I)/\overline{x}F(I) \cong F_{R'}(I')$. \QED
\medskip
With notation as in Proposition 3.1, the following example
shows that for $x \in I - \m I$ such that $\overline{x}$ is
a regular element of $F(I)$, it may happen that
$\overline{x}F(I) \subsetneq \ker \psi_{xR}$ and
$F_{R'}(I') \not\cong F_R(I)/\overline{x}F(I)$,
where $R' = R/xR$ and $I' = IR'$. Proposition~3.1 implies that
for such an example $x^* \in G(I)$ is necessarily a zero divisor.
\begin{exam}\label{3.2}{\em
Let $k$ be a field and consider the subring $R := k[[t^3, t^4, t^5]]$
of the formal power series ring $k[[t]]$. Thus
$R = k + t^3k[[t]]$ is a complete Cohen-Macaulay one-dimensional
local domain. Let $I = (t^3, t^4)R$. An easy computation
implies $I^3 = t^3I^2$. Hence $t^3R$ is a principal reduction of $I$.
Since $I$ is 2-generated, it follows from \cite[Proposition 3.5]{DGH}
that $F(I)$ is Cohen-Macaulay and in fact a complete intersection.
Let $X,Y$ be indeterminates over $k$ and define a $k$-algebra
homomorphism
$\phi : k[X,Y] \to F(I)$ by setting $\phi(X) = \overline{t^3}$ and
$\phi(Y) = \overline{t^4}$. Then $\ker \phi = Y^3k[X,Y]$ and
$F(I) \cong k[X, Y]/Y^3k[X, Y]$. Thus
$\overline{t^3}$ is a regular element of $F(I)$ and
$F(I)/\overline{t^3}F(I) \cong k[Y]/Y^3k[Y]$.
Let $J = t^3R$, $R' = R/J$ and $I' = IR'$.
Since $t^8 \in (I^2 \cap J)$, we have $\phi(Y^2) = \overline{t^8} \in \ker \psi_J$
and $F_{R'}(I') \cong k[Y]/Y^2k[Y]$. Thus
$F(I)/\overline{t^3}F(I) \not\cong F_{R'}(I')$. In fact,
we have $\ker \psi_J = (\overline{t^3}, \overline{t^8})F(I)$
and $\overline{t^8} \not\in \overline{t^3}F(I)$.}
\end{exam}
\medskip
We list several observations and questions concerning
the dimension of fiber cones and their behavior
under homomorphic image.
\begin{dis}\label{3.3}{\em
Let $I \subseteq \m$ be an ideal of a
Noetherian local ring $(R, \m)$. If
$J \subseteq \m$ is an ideal of $R$
and $R' = R/J$, then there exists a
surjective ring homomorphism
$\chi_J : G_R(I) = R[It]/IR[It] \to G_{R'}(IR')$
of the
associated graded ring $G_R(I)$ of $I$ onto the associted graded ring
$G_{R'}(IR')$ of $IR'$ \cite[page 150]{K}.}
\end{dis}
We have the following commutative diagram involving
the associated graded rings and fiber cones for which the
vertical maps $\alpha$ and $\beta$ are surjective:
$$
\CD
G_R(I) = R[It]/IR[It] = \oplus_{n \ge 0}I^n/I^{n=1} @>\chi_J>> R'[I't]/I'R'[I't] =
\oplus(I')^n/(I')^{n+1} \\
@V\alpha VV @V\beta VV \\
F_R(I) = R[It]/\m R[It] \oplus_{n \ge 0} I^n/\m I^n @>\psi_J>> R'[I't]/\m'R'[I']
= \oplus_{n \ge 0}(I')^n/\m (I')^n
\endCD
$$
\medskip
If $J$ is nonzero, then $\ker \chi_J \ne 0$. It can
happen, however, that $J$ is nonzero and yet $\ker \psi_J = 0$. This
is possible even in the case where $I$ is $\m$-primary. In
an example exhibiting this behavior, commutativity
of the diagram above implies one must have
$\ker \chi_J \subseteq \ker \alpha$.
\begin{exam}\label{3.4} {\em
Let $k$ be a field and let $R = k[x, y]_{(x,y)}$,
where $x^2 = xy = 0$. Let $I = yR$ and let $J = xR$.
Then $\ker \psi_J =
\bigoplus_{n \ge 0} \frac{(xR \cap y^nR) + \m y^nR}{\m y^nR} = 0$,
but $J = xR \ne 0$.}
\end{exam}
A reason for the existence of examples such as Example 3.4 is
given in Proposition~3.5.
\begin{prop}\label{3.5}
Suppose $(R,\m)$ is a Noetherian
local ring and $I$ is an $\m$-primary ideal.
If the fiber cone $F(I)$ is an integral domain,
then $\ker \psi_J = 0$ for every ideal $J$ of
$R$ such that $\dim(R/J) = \dim R$.
In particular, if $I$ is
$\m$-primary and $F(I)$ is an integral domain,
then there exists a prime
ideal $J$ of $R$ such that $\ker \psi_ J = 0$.
\end{prop}
\demo Let $R' := R/J$.
Since $I$ is $\m$-primary, $\dim F(IR') = \dim R'$.
Thus $\dim R' = \dim R$ implies $\dim F(IR') = \dim F(I)$.
Since $F(I)$ is an integral domain, it follows that
$\ker \psi_J = 0$. The last statement follows becaues
there exists a prime ideal $J$ of $R$ such that
$\dim R = \dim(R/J)$. \QED
\medskip
Propositon 3.5 and Example 3.4 show that with notation
as in (3.1), it can happen that $x^* \in G(I)$ is not
a regular element and yet $\ker \psi_{xR} = \overline{x}F(I)$.
\vskip 2ex
In Section 4 we consider fiber cones $F(I)$ such that
$\ker \psi_J \ne 0$ for each nonzero ideal $J$.
\vskip 2ex
\section{Maximal fiber cones with respect to homomorphic image.}
\vskip 2ex
Suppose $(R,\m)$ is a Noetherian
local ring and $I \subseteq \m$ is an ideal of $R$.
If $J$ is a nonzero ideal of $R$ such that
$\ker \psi_J = \bigoplus_{n \ge 0} \frac{(J \cap I^n) + \m I^n}{\m I^n}$
is the zero ideal of $F(I)$, then we
have $F_R(I) = F_{R'}(IR')$, where $R' := R/J$; so
the fiber cone $F(I)$ is realized as a fiber cone of a
proper homomorphic image $R'$ of $R$. If there
fails to exist such an ideal $J$, i.e., if
$\ker \psi_J \ne 0$ for
each nonzero ideal $J$, then we say that
$F(I)$ is a {\em maximal fiber cone } of $R$.
We record in Remark 4.1 some immediate consequences of the
inequality $\dim F_{R'}(IR') \le \dim R'$.
\vskip 2ex
\begin{remark}\label{4.1} {\em
With notation as in (2.1), we have:
\begin{enumerate}
\item
If $J$ is such that $\dim R' < \dim R$ and
if $\dim F_R(I) = \dim R$, then $\ker \psi_J \ne 0$.
\item
If $I$ is $\m$-primary and $J$ is not
contained in a minimal prime of $R$, then
$\ker \psi_J \ne 0$.
\item
If $R$ is an integral domain and
$\dim F(I) = \dim R$, then $F(I)$ is a
maximal fiber cone.
\item
If $R$ is an integral domain, then
$F(I)$ is a maximal fiber cone for every $\m$-primary
ideal $I$ of $R$.
\end{enumerate}}
\end{remark}
We are interested in describing all the
maximal fiber cones of $R$. Thus we are interested
in conditions on $I$ and $R$ in
order that there exist a nonzero ideal $J$ of $R$
such that $\ker \psi_J = 0$.
In considering
this question, by Proposition~2.3, one may assume that $J = xR$ is a nonzero
principal ideal. Thus the question can also be phrased:
\begin{ques}\label{4.2} {\em
Under what conditions on $I$ and $R$ does it follow
for each nonzero element $x \in \m$ that
$\ker \psi_{xR} \ne 0$? }
\end{ques}
\begin{dis}\label{4.3} {\em
Information about Question 4.2 is provided
by the work of Rees in \cite{R}. In particular,
\cite[Theorem 2.1]{R} implies that if $x \in \m$ is
such that $(I^n : x) = I^n$ for each positive integer $n$,
then $F_R(I) = F_{R'}(I')$, where $R' := R/xR$ and
$I' := IR'$. Thus for $x \in \m$ a sufficient condition for
$\ker \psi_{xR} = 0$ is that $(I^n:x) = I^n$ for each
positive integer $n$. It is readily seen that this
colon condition on $x$ is equivalent to $x \not\in I$ and
the image of $x$ in the associated graded ring
$G(I) = R[It]/IR[It]$ is a regular element.
More generally, if $x \in I^s - I^{s+1}$ and
if the image $x^*$ of $x$ in $G(I)$ is a regular
element, then by Proposition~3.1 $\ker \psi_{xR} =
\overline{x}F(I)$. Thus if we also have
$x \in \m I^s$, then $\ker \psi_{xR} = 0$.
Example~3.4 shows that this sufficient
condition for $\ker \psi_{xR} = 0$ is not a necessary condition.}
\end{dis}
\medskip
Proposition 2.3 gives a
necessary and sufficient condition
on a principal ideal $J = xR$ in order
that $\ker \psi_{xR} = 0$, namely
that $(I^n : x) = (\m I^n : x)$ for
each integer $n \ge 0$.
By Proposition~2.3, if $\ker \psi_{xR} = 0$,
then also $\ker \psi_{yxR} = 0$
for every $y \in R$.
\medskip
If $I = yR$ is a non-nilpotent principal ideal
of $R$, we give in Corollary~4.5 necessary and
sufficient conditions for $F(I)$ to be a maximal
fiber cone.
\begin{prop}\label{4.4}
Suppose $(R,\m)$ is a Noetherian
local ring and $I = yR \subseteq \m$ is a
non-nilpotent principal ideal of $R$.
For $x \in \m$, we have $\ker \psi_{xR} = 0
\iff y^n \not\in xR$ for each
positive integer $n$.
\end{prop}
\demo We have
$\ker \psi_{xR} = 0 \iff (y^nR \cap xR) \subseteq \m y^nR$
for each positive integer $n$, and $y^n \not\in xR \iff
(y^nR \cap xR) \subsetneq y^nR \iff (y^nR \cap xR) \subseteq \m y^nR$. \QED
\medskip
\begin{cor}\label{4.5} Let $(R, \m)$ be a Noetherian local ring
and $I = yR \subseteq \m$ be a non-nilpotent
principal ideal of $R$. Then
$F(I)$ is a maximal fiber cone if and only if $R$ is a
one-dimensional integral domain.
\end{cor}
\demo
By Proposition~4.4, for $x \in \m$ we have
$ y \in \rad xR \iff \ker \psi_{xR} \ne 0.$
Suppose $F(I)$ is a maximal fiber cone. Then
by definition, $\ker \psi_{xR} \ne 0$ for
each nonzero $x \in \m$.
Since $y$ is not nilpotent, there exists a
minimal prime $P$ of $R$ such that $y \not\in P$.
It follows that $P = 0$, for if not, then there
exists a nonzero $x \in P$ and $y \in \rad xR \subseteq P$
implies $y \in P$. Thus $R$ is an integral domain.
Moreover, this same argument implies
$y$ is in every nonzero prime of $R$.
Since $R$ is Noetherian, it follows that $\dim R = 1$.
For $yR$ has only finitely many
minimal primes and every minimal prime of $yR$
has height one by the Altitude Theorem of Krull
\cite[page 26]{N} or \cite[page 100]{Mat}.
If there exists $P \in \Spec R$ with $\hgt P > 1$,
then the Altitude Theorem of Krull
implies $P$ is the union of the height-one primes
contained in $P$. This implies there exist
infinitely many height-one primes contained in $P$.
Since $y$ is contained in only finitely many
height-one primes, this is impossible. Thus
$\dim R = 1$. Since $R$ is local, $\m$ is the
only nonzero prime of $R$.
Conversely, if $R$ is a one-dimensional
Noetherian local integral domain, then (4.1) implies
that $F(I)$ is a maximal fiber cone for every non-nilpotent
principal ideal $I = yR \subseteq \m$. \QED
\begin{ques}\label{4.6} {\em
If $F(I)$ is a maximal fiber cone of $R$,
does it follow that
$\dim F(I) = \dim R$?}
\end{ques}
\begin{prop}\label{4.7}
Suppose $(R,\m)$ is a Noetherian
local ring and $I \subseteq \m$ is an ideal of $R$.
If $\dim F(I) := n = \hgt(I) < \dim R$ and if $F(I)$ is
an integral domain, then $F(I)$ is not a maximal
fiber cone. In particular, if $I$ is of the
principal class, i.e.,
$I = (a_1, \dots, a_n)R$, where $\hgt(I) = n$, and
if $\hgt(I) < \dim R$, then
$F(I)$ is not a maximal fiber cone of $R$.
\end{prop}
\demo
Choose $x \in \m$ such that $x$ is not in
any minimal prime of $I$. Then $L := (I, x)R$ has
height $n+1$. Let $\overline x$ denote the image of
$x$ in the fiber cone $F_R(L)$. Then
$F_R(L)$ is a homomorphic image of a polynomial
ring in one variable $F_R(I)[z]$ over $F_R(I)$ by
means of a homomorphism mapping $z \to \overline x$.
Since $\dim F(I) = n$ and $F(I)$ is an integral
domain, it follows that $F(I)[z] \cong F(L)$
by means of an isomorphism taking $z \to \overline x$.
Let $J = xR$ and $R' : = R/J$.
Then $\hgt(IR') = \hgt(L/xR) = n$, so
$\dim F_{R'}(IR') \ge n$. Since
$\psi_J : F_R(I) \to F_{R'}(IR')$ is
surjective and $F_R(I)$ is an $n$-dimensional
integral domain, it follows that
$\psi_J : F_R(I) \to F_{R'}(IR')$ is
an isomorphism. In particular, if $I$ is of the
principal class, then $F(I)$ is a polynomial
ring in $n$ variables over the field $R/\m$,
so $F(I)$ is an integral domain with
$\dim F(I) = \hgt(I)$. \QED
\medskip
If $I$ is generated by a regular sequence,
then $I$ is of the principal class. Thus if
$F(I)$ is a maximal fiber cone and $I$ is generated
by a regular sequence, then by
Proposition~4.7, $\dim F(I) = \dim R$.
\medskip
We observe in Proposition \ref{4.8} a situation
where the integral domain hypothesis of
Proposition \ref{4.7} applies.
\vskip 2ex
\begin{prop}\label{4.8}
Let $A = k[X_1,X_2, \cdots, X_d] = \bigoplus_{n=0}^\infty A_n$
be a polynomial ring
in $d$ variables over a field $k$ and let
$\m = (X_1,X_2, \cdots, X_d)A$ denote its homogeneous maximal ideal.
Suppose $I = (f_1,f_2, \cdots,
f_n)A$, where $f_1,f_2, \cdots, f_n$ are
homogeneous polynomials all of the same degree $t$.
Let $R = A_{\m}$. Then $F(IR)$ is an integral domain.
Thus if $F(IR)$ is a maximal fiber cone, then
$\dim F(IR) = d$.
\end{prop}
\demo
We have
$$k[f_1,f_2, \cdots,f_n] = k \oplus I_1 \oplus I_2 \oplus \cdots,$$
where $I_i = I^i \cap A_{it}$ for $i > 0$. Since $I^i/{\m}I^i \cong
I^i \cap A_{it}$ for $i \geq 0$, we have the following isomorphisms:
$$
k[f_1,f_2, \cdots,f_n] \cong \oplus_{i=0}^\infty (I^i/\m I^i) \cong
\oplus_{i=0}^\infty(I^iR/\m I^iR) = F(IR).
$$
Therefore $F(IR)$ is an
integral domain. The result now follows from Proposition \ref{4.7}.
\QED
\begin{cor}\label{4.9}
With notation as in Proposition \ref{4.8},
if $\dim F(I) = \hgt I$ and $F(IR)$ is a maximal fiber cone,
then $I$ is $\m$-primary.
\end{cor}
\demo
We have $\dim F(IR) = d$ by Proposition \ref{4.8}.
Since $I$ is homogeneous ideal and $\hgt I = d$,
$\m$ is the unique homogeneous minimal prime of $I$,
Therefore $I$ is
$\m$-primary. \QED
\begin{ques}\label{4.10} {\em
Let $(R, \m)$ be a Noetherian local ring and let $I \subseteq \m$
be an ideal of $R$.
If $\dim F(I) = \hgt I$ and $F(I)$ is a maximal
fiber cone, does it follow that $I$ is
$\m$-primary? }
\end{ques}
\begin{rem}\label{4.11}
Without the assumption in Question \ref{4.10} that
$\dim F(I) = \hgt I$, it is easy to give examples
where $F(I)$ is a maximal fiber cone and yet $I$ is
not $\m$-primary. For example, with notation as
in Proposition \ref{4.8}, if $d > 1$ and
$I = (X_1^2, X_1X_2, \dots, X_1X_d)A$,
then $\hgt(IR) = 1$, but $\dim F(IR) = d$ and $F(IR)$ is
a maximal fiber cone.
\end{rem}
\vskip 2ex
\section{When is the fiber cone a hypersurface?}
\medskip
\begin{setting}\label{5.1} {\em
Let $I \subseteq \m$ be an ideal of a Noetherian
local ring $(R,\m)$.
In this section we consider the structure of the fiber
cone $F(I) = \oplus_{n \ge 0}F_n$ in the case where
$\dim F(I) = d > 0$ and $\mu(I) = d+1$. If $a_1, \ldots,
a_{d+1}$ is a basis for $F_1 = I/\m I$ as a vector space
over the field $k := R/\m$, then there exists
a presentation $\phi : k[X_1, \ldots, X_{d+1}] \to F(I)$ of $F(I)$ as
a graded $k$-algebra homomorphic image of a polynomial ring in $d+1$
variables over $k$ defined by setting $\phi(X_i) = a_i$, for
$i = 1, \ldots, d+1$. Moreover, $F(I)$ is a hypersurface if and only if
$\ker \phi$ is a principal ideal \cite[Examples 1.2]{K}. }
\end{setting}
\begin{lem} \label{5.2}
Let $(R,\m)$ be a Noetherian local ring having
infinite residue field $R/\m := k$, and let
$I \subseteq \m$ be an ideal of $R$ such
that $\dim F(I) = d > 0$ and $\mu(I) = d+1$.
Let $r = r(I)$ denote the reduction number of $I$
and let $\phi : k[X_1, \ldots , X_{d+1}] \to F(I)$ be
a presentation of the fiber cone $F(I)$ as in Setting 5.1.
Then the minimal degree of a nonzero form $f \in \ker \phi$ is
$r+1$.
\end{lem}
\demo
The map $\phi$ from the graded ring
$A = k[X_1, \ldots, X_{d+1}] = \oplus_{n \ge 0} A_n$ onto the
graded ring $F(I) = \oplus_{n \ge 0}F_n = \oplus_{n \ge 0}(I^n/\m I^n)$
is a surjective graded $k$-algebra homomorphism of degree $0$.
Let $K := \ker \phi = \oplus_{n \ge 0}K_n$. For each positive
integer $n$ we have a short exact sequence
$$
0 \to K_n \to A_n \to F_n \to 0
$$
of finite-dimensional vector spaces over $k$. Since
$I$ has reduction number $r$, it follows from
\cite[Theorem, page 440]{ES} that $\dim_k F_i = \mu(I^i) = \binom {i+d}d$
for $i = 0, 1, \ldots, r$ and $\dim_k F_{r+1} = \mu(I^{r+1}) < \binom {r+d+1}d$.
Since $\dim A_i = \binom {i+d}d $ for all $i$, it follows that $K_i = 0$ for
$i = 0, \ldots, r$ and $K_{r+1} \ne 0$. Hence the minimal degree of a
nonzero form $f \in \ker \phi$ is $r+1$. \QED
\begin{remark}\label{5.3} {\em
Let $A = k[X_1, \ldots, X_n]$ be a polynomial ring in
$n$ variables $X_1, \ldots, X_n$ over a field $k$.
For an ideal $K$ of $A$, it is well known that
$\hgt(K) = 1$ if and only if $\dim(A/K) = n-1$
\cite[Corollary 3.6, page 53]{K}. Moreover,
$K$ is principal if and only if $\hgt(P) = 1$
for each associated prime $P$ of $K$. If
$K = (g_1, \ldots, g_m)A$ and $g$ is a greatest
common divisor of $g_1, \ldots, g_m$, then
$K = gJ$, where $\hgt(J) > 1$. Thus $K$ is
principal if and only if $J = A$. If $K$ is
homogeneous, then $g_1, \ldots, g_m$ may be taken
to be homogeneous; it then follows that $g$ is homogeneous and
$K = gJ$, where $J$ is homogeneous with $\hgt(J) > 1$.
If $K = \rad K$, then each associated prime of $K$ is a
minimal prime and $K$ is principal if and only if $\hgt(P) = 1$
for each minimal prime $P$ of $K$.}
\end{remark}
\begin{prop}\label{5.4}
Let $(R,\m)$ be a Noetherian local ring with infinite
residue field $k = R/\m$ and let $I \subseteq \m$ be
an ideal of $R$ such that $\dim F(I) = d > 0$ and
$\mu(I) = d + 1$. Let $\phi : A = k[X_1, \ldots, X_{d+1}]
\to F(I)$ be a presentation of $F(I)$ as a
graded homomorphic image of a polynomial ring as in Setting 5.1.
Let $f \in K := \ker \phi$ be a nonzero homogeneous form of
minimal degree.
Then the following are equivalent.
\begin{enumerate}
\item
$\ker \phi = fA$, i.e., $F(I)$ is a hypersurface.
\item
$\hgt P = 1$ for each $P \in \Ass K$.
\item $F(I)$ is a Cohen-Macaulay ring.
\item
$\deg f = e(F(I))$, the multiplicity of $F(I)$.
\end{enumerate}
\end{prop}
\demo That (1) is equivalent to (2) is
observed in Remark 5.2. It is clear that (1)
implies (3) and it follows from \cite[(2.2.15) and (2.1.14)]{BH}
that (3) implies (2). To see the
equivalence of (3) and (4), we use
\cite[Theorem 2.1]{DRV}. By Lemma~5.2, $\deg f = r+1$,
where $r$ is the reduction number of $I$.
Since $\dim F(I) = d$,
there exists a minimal reduction $J = (x_1, \ldots , x_d)R$
of $I$ and $y \in I$ such that $I = J + yR$. By
\cite[Theorem 2.1]{DRV}, $F(I)$ is
Cohen-Macaulay if and only if
$$
e(F(I)) = \sum_{n=0}^r \lambda(\frac{I^n}{JI^{n-1} + \m I^n}).
$$
Since for $0 \le n \le r$, \ $\lambda(\frac{I^n}{JI^{n-1} + \m I^n}) = 1$,
the sum on the right hand side of the
displayed equation is $r+1 = \deg f$.
This proves the equivalence of (3) and (4). \QED
\medskip
\begin{remark}\label{5.5} {\em
With notation as in Proposition 5.4,
we have the following inequality
$e(F(I)) \le \deg f$, where $e(F(I))$ is the
multiplicity of $F(I)$. Hence by Proposition~5.4,
$F(I)$ is not Cohen-Macaulay $ \iff e(F(I)) < \deg f$. }
\end{remark}
\demo
Let $J = (x_1, \ldots, x_d)R$ be a minimal reduction
of $I$. Then $JF(I)$ is generated by a homogeneous
system of parameters for $F(I)$ and
$$
\lambda(\frac{F(I)}{JF(I)})\quad = \quad
\sum_{n=0}^r \lambda(\frac{I^n}{JI^{n-1} + \m I^n}).
$$
Let $\M$ denote the maximal homogeneous ideal of $F(I)$.
Then
$$
e(F(I)) = e(F(I)_{\M}) \le \lambda(\frac{F(I)_{\M}}{JF(I)_{\M}}) =
\lambda(\frac{F(I)}{JF(I)}).
$$
Thus $e(F(I)) \le \deg f = r + 1$. Hence by Proposition~5.4,
$F(I)$ is not Cohen-Macaulay if and only if $e(F(I)) < \deg f$. \QED
\begin{thm}\label{5.6}
Let $(R,\m)$ be a Noetherian local ring with infinite
residue field $k = R/\m$ and let $I \subseteq \m$ be
an ideal of $R$ such that $\dim F(I) = d > 0$ and
$\mu(I) = d + 1$. Suppose there exists a minimal
reduction $J$ of $I$ generated by a regular sequence.
Assume that $\grade(G_+(I)) \ge d-1$. Then
$F(I)$ is Cohen-Macaulay and thus a hypersurface.
\end{thm}
\demo
For $x \in R$, let $x^*$ denote the image of $x$ in $G(I)$ and
let $\overline{x}$ denote the image of $x$ in $F(I)$.
There exists a minimal reduction $J = (x_1, \ldots, x_d) \subseteq I$
and $x_{d+1} \in I$ such that
\begin{enumerate}
\item[(I)] $\{x_1, \ldots, x_d \}$ is a regular sequence in $R$.
\item[(II)] $\{x_1, \ldots, x_d, x_{d+1} \}$ is a minimal set of
generators of $I$.
\item[(III)] $\{x_1^*, \ldots, x^*_{d-1} \}$ is a regular sequence
in $G(I)$.
\end{enumerate}
Let $R' = R/(x_1, \ldots, x_{d-1})R$, let
$\m' = \m/(x_1, \ldots,x_{d-1})R$ and let $I' = IR'$.
By Condition II, $I'$ is a 2-generated ideal having a
principal reduction generated by the image $x_d'$ of
$x_d$. Condition I implies that $x_d'$ is a regular
element of $R'$. Hence by \cite[Proposition 3.5]{DGH},
$F_{R'}(I')$ is Cohen-Macaulay.
As observed in (2.1), the kernel of the canonical
map $ \psi : F_R(I) \to F_{R'}(I')$ is
\smallskip
$$
\oplus_{n \ge 0}\frac{(I^n \cap (x_1, \ldots,x_{d-1})) + \m I^n}{ \m I^n}.
$$
Condition III and Proposition 3.1 imply
\smallskip
$$
\ker \psi = \oplus_{n \ge 0}\frac{(x_1, \ldots, x_{d-1})I^{n-1} + \m I^n}{\m I^n} =
(\overline{x_1}, \ldots, \overline{x_{d-1}})F(I).
$$
Hence
$$
\frac{F(I)}{(\overline{x_1}, \ldots, \overline{x_{d-1}})} \cong F_{R'}(I')
$$
and to show $F(I)$ is Cohen-Macaulay, it suffices to show
$\{\overline{x_1}, \ldots, \overline{x_{d-1}} \}$ is a regular
sequence in $F(I)$.
By the generalized Vallabrega-Valla criterion of Cortadellas and
Zarzuela \cite[Theorem 2.8]{CZ}, to show
$\{\overline{x_1}, \ldots, \overline{x_{d-1}} \}$
is a regular sequence in $F(I)$, it suffices to show
$$
(x_1, \ldots, x_{d-1}) \cap \m I^{n+1} = (x_1, \ldots, x_{d-1} )\m I^n,
\text{ for all } n \ge 0.
$$
$`` \supseteq''$ is clear. We prove $``\subseteq''$ by induction on
$n$.
\smallskip
\noindent
(Case i ) $n = 0$ : Let $u \in (x_1, \ldots, x_{d-1}) \cap \m I$.
Thus $u = \sum_{i=1}^{d-1} r_ix_i = \sum_{j=1}^{d+1}\alpha_jx_j$,
where $r_i \in R$ and $\alpha_j \in \m$. Therefore
$$
(r_1 - \alpha_1)x_1 + \cdots + (r_{d-1} - \alpha_{d-1})x_{d-1}
- \alpha_dx_d - \alpha_{d+1}x_{d+1} = 0.
$$
Since $\{x_1, \ldots, x_{d+1} \}$ is a minimal generating set
for $I$, each $r_i - \alpha_i \in \m$. Since $\alpha_i \in \m$,
$r_i \in \m$. Hence $u = \sum_{i=1}^{d-1} r_ix_i \in \m (x_1, \ldots, x_{d-1})$.
\smallskip
\noindent
(Case ii ) $1 \le n < r$, where $r = r_J(I)$ is the reduction number of
$I$ with respect to $J$:
We have
$(x_1, \ldots, x_{d-1}) \cap \m I^{n+1} = (x_1, \ldots, x_{d-1}) \cap
(I^{n+1} \cap \m I^{n+1})$
$= ((x_1, \ldots, x_{d-1}) \cap I^{n+1}) \cap \m I^{n+1}
= ((x_1, \ldots, x_{d-1})I^n \cap \m I^{n+1} $, the last
equality by Condition III.
Hence
$u \in (x_1, \ldots, x_{d-1}) \cap \m I^{n+1}$ implies
$u \in ((x_1, \ldots, x_{d-1})I^n \cap \m I^{n+1}$.
Thus $u = \sum_{i=1}^{d-1}x_ig_i$, where $g_i \in I^n$ and
$u = H(x_1, \ldots, x_{d+1})$, where $H(X_1, \ldots , X_{d+1})
\in R[X_1, \ldots , X_{d+1}]$ is a homogeneous polynomial
with coefficients in $\m$
of degree $n+1$. Let $G_i(X_1, \ldots, X_{d+1}) \in R[X_1, \ldots , X_{d+1}]$
be a homogeneous polynomial of degree $n$ such that
$G_i(x_1, \ldots, x_{d+1}) = g_i$.
Let $\tau : R[X_1, \ldots ,X_{d+1}] \to R[It]$,
where $\tau(X_i) = x_it$ be a presentation of the Rees algebra
$R[It]$. Consider the following commutative diagram.
$$
\CD
0 @>>> \ker(\tau) @>>> R[X_1, \ldots, X_{d+1}] @>\tau>> R[It] @>>> 0 \\
@. @V\pi_1 VV @V\pi_2 VV @V\pi_3 VV \\
0 @>>> \ker(\phi) @>>> (R/\m)[X_1, \ldots, X_{d+1}] @>\phi>> F(I) @>>> 0
\endCD %\tag{*}
$$
\medskip
\noindent
Since
$\sum_{i=1}^{d-1}x_ig_i - H(x_1, \ldots, x_{d+1}) = 0$, the
homogeneous polynomial
$$
\sum_{i=1}^{d-1}X_iG_i(X_1, \ldots, X_{d+1}) - H(X_1, \ldots, X_{d+1}) \in \ker \tau.
$$
Since $H(X_1, \ldots, X_{d+1})$ has coefficients in $\m$, we have
$$
0 = \pi_3\tau(\sum_{i=1}^{d-1}X_iG_i - H) = \phi\pi_2(\sum_{i=1}^{d-1}X_iG_i - H)
= \phi\pi_2(\sum_{i=1}^{d-1}X_iG_i).
$$
Hence $\pi_2(\sum_{i=1}^{d-1}X_iG_i) \in \ker \phi$.
Since $\sum_{i=1}^{d-1}X_iG_i $ is of degree $n+1 \le r$,
Lemma 5.2 implies $\pi_2(\sum_{i=1}^{d-1}X_iG_i) = 0$.
Therefore the coefficients of $\sum_{i=1}^{d-1}X_iG_i$ are
in $\m$. Evaluating this polynomial by
mapping $X_i \mapsto x_i$ gives
$u = \sum_{i=1}^{d-1}x_ig_i \in (x_1, \ldots,x_{d_1})\m I^n$.
\smallskip
\noindent
( Case iii ) $n \ge r$ : Since $n \ge r$, we have $I^{n+1} = JI^n =
(x_1, \ldots, x_d)I^n$. \newline
Let $u \in (x_1, \ldots, x_{d-1}) \cap \m I^{n+1}
= (x_1, \ldots, x_{d-1}) \cap \m (x_1, \ldots, x_d)I^n$.
Thus $u = \sum_{i=1}^{d-1}r_ix_i = \sum_{j=1}^d\alpha_jx_j$,
where each $r_i \in R$ and each $\alpha_j \in \m I^n$.
Hence $\alpha_dx_d = \sum_{i=1}^{d-1}(r_i - \alpha_i)x_i$
and this implies $\alpha_d \in ((x_1, \ldots, x_{d-1}): x_d) =
(x_1, \ldots, x_{d-1})$, the last equality because of
Condition I. Hence
$$
\alpha_d \in (x_1, \ldots, x_{d-1}) \cap \m I^n
= (x_1, \ldots, x_{d-1})\m I^{n-1},
$$
the last equality because of our inductive hypothesis.
Thus $u = \sum_{j=1}^d \alpha_jx_j = \sum_{j=1}^{d-1}\alpha_jx_j + \alpha_dx_d
\in (x_1, \ldots,x_{d-1})\m I^n + (x_1, \ldots, x_{d-1})\m I^{n-1}I =
(x_1, \ldots, x_{d-1})\m I^n$. This completes the proof that
$\{\overline{x_1}, \ldots, \overline{x_{d-1}} \}$ is a regular
sequence in $F(I)$, and thus the proof of Theorem~5.6
\QED
\vskip 2ex
\section{The Cohen-Macaulay property of one-dimensional fiber cones}
\medskip
We record in this short section several consequences of a
result of D'Cruz, Raghavan and Verma \cite[Theorem 2.1]{DRV}
for the Cohen-Macaulay
property of the fiber cone of a regular ideal having
a principal reduction.
\vskip 2ex
\begin{prop}\label{6.1}
Let $(R, \m)$ be a Noetherian local ring and let $I \subseteq \m$ be a
regular ideal having a principal reduction $aR$. Let
$r=r_{aR} (I)$ be the reduction number of $I$ with respect to $aR$.
Then the following are equivalent.
\begin{enumerate}
\item[(1)] $F(I)$ is a Cohen-Macaulay ring.
\smallskip
\item[(2)] $\lambda \big( \frac{aI^n +\m I^{n+1}}{\m I^{n+1}}\big) =
\lambda \big( \frac{I^n}{\m I^n} \big)$ \quad for $1 \leq n \leq r-1$.
\end{enumerate}
\end{prop}
\demo
(1) $\Rightarrow$ (2). Suppose $F(I)$ is a Cohen-Macaulay ring.
Then $\overline{a}(= a + \m I)$ is a regular element of $F(I)$
with $\deg \overline{a} = 1$.
Let $F(I) = \oplus_{n \geq 0} F_n$, where $F_n = I^n/\m I^n$,
and consider the graded $k$-algebra homomorphism
$\phi_{\overline{a}} : F_n \to F_{n+1}$
given by $\phi_{\overline{a}} (\overline{x}) = \overline{x} \cdot \overline{a}$,
for every $\overline{x} \in F_n$. Since $\overline{a}$ is a
regular element of $F(I)$, $\dim_k F_n = \dim_k (\overline{a} F_n)$.
\smallskip
\noindent
For $1 \leq n \leq r-1$, we have
$$
\lambda \big( \frac{aI^n + \m I^{n+1}}{\m I^{n+1}} \big) = \lambda \big( \overline{a}
\big( \frac{I^n}{\m I^n} \big) \big) = \dim_k (\overline{a} F_n) = \dim_k (F_n)
=\lambda \big(\frac{I^n}{\m I^n} \big).
$$
(2) $\Rightarrow$ (1). Suppose that
$\lambda \big( \frac{aI^n + \m I^{n+1}}{\m I^{n+1}} \big)
=\lambda \big(\frac{I^n}{\m I^n} \big)$, for $1 \leq n \leq r-1$. Since $a$
is a non-zero-divisor $R$, $I^{n+r}/\m I^{n+r} \cong I^r/\m I^r$, for every $n \geq 1$.
Hence $e(F(I)) = \lambda (I^r/\m I^r)$. To see the Cohen-Macaulay property
of $F(I)$, we use \cite[Theorem~2.1]{DRV}. We have the following:
$$
\aligned
\sum_{n=0}^{r} \lambda \big( \frac{I^n}{aI^{n-1}+\m I^n} \big)
&= \lambda \big( \frac{R}{\m} \big)+ \sum_{n=1}^{r}
\lambda \big( \frac{I^n}{aI^{n-1} + \m I^n} \big)\\
&=\lambda \big( \frac{R}{\m} \big) + \sum_{n=1}^{r}
\big[ \lambda \big( \frac{I^n}{\m I^n} \big)
- \lambda \big( \frac{aI^{n-1} + \m I^n}{\m I^n} \big) \big]\\
&=\lambda \big( \frac{R}{\m} \big) + \sum_{n=1}^{r}
\big[ \lambda \big( \frac{I^n}{\m I^n} \big)
- \lambda \big( \frac{I^{n-1}}{\m I^{n-1}} \big) \big]\\
&=\lambda \big( \frac{I^r}{\m I^r} \big)\\
&=e(F(I)).
\endaligned
$$
Hence by \cite[Theorem 2.1]{DRV}, $F(I)$ is a Cohen-Macaulay ring.
\QED
\smallskip
As an immediate consequence of Proposition \ref{6.1} we have
\smallskip
\begin{cor}\label{6.2}
Let $(R, \m)$ be a Noetherian local ring and $I$ be a regular ideal
having a principal reduction $aR$ with $r_{aR} (I) =2$. If
$\mu(I) = n$, then
$$
F(I) \text{~ is Cohen-Macaulay}
\iff \lambda \big( \frac{aI + \m I^2}{\m I^2} \big) = n.
$$
\end{cor}
\medskip
Example \ref{6.3} shows that Proposition \ref{6.1}
and Corollary \ref{6.2} may
fail to be true without the assumption on
the length of $\frac{aI^n + \m I^{n +1}}{\m I^{n+1}}$.
\smallskip
\begin{exam}\label{6.3}
Let $k$ be a field and consider the subring $R=k[[t^3, t^7, t^{11}]]$ of
the formal power series ring $k[[t]]$. Let $I = (t^6, t^7, t^{11})R$.
An easy computation implies $t^6 I \not = I^2$ and $t^6 I^2 = I^3$.
Hence $r_{t^6R} (I) =2$.
Note that $\overline{t^6}F(I)$ is a homogeneous system of parameter of $F(I)$.
But $\overline{t^6}
\overline{t^{11}} = (t^6 + \m I) (t^{11} + \m I) = t^{17} + \m I^2 = 0$,
and hence $F(I)$ is not a Cohen-Macaulay ring. And
$\lambda \big( \frac{t^6 I + \m I^2}{\m I^2} \big)
=\lambda \big( \frac{I^2}{\m I^2} \big) -
\lambda \big( \frac{I^2}{t^6 I + \m I^2} \big) =2 < 3$.
\end{exam}
\smallskip
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\end{document}
\end{document}
\smallskip
\begin{thm}\label{5.9}
Let $(R, \m)$ be a Noetherain local ring and let $I$ be an $n+1$-generated
regular ideal
having a principal reduction $aR$ with $r_{aR} (I) =2$. Then
$$
F(I) \text{~ is~ Cohen-Macaulay } \iff \lambda
\big( \frac{aI + \m I^2}{\m I^2} \big) = n+1
$$
\end{thm}
\demo
$\Rightarrow$ Suppose $F(I)$ is a Cohen-Macaulay ring. Then
$\dim_k (\overline{a}F_1) = \dim_k (F_1)$, where $F_1 = I/\m I$, since
$\overline{a}(=a + \m I)$ is a non-zero-divisor of $F(I)$.
Hence $\lambda \big( \frac{I}{\m I} \big) = \lambda
\big( \frac{aI + \m I^2}{\m I^2} \big)
= \mu (I) = n+1$.\\
\smallskip
$\Leftarrow$ Suppose $\lambda \big( \frac{aI + \m I^2}{\m I^2} \big) = n+1$.
Then $e(F(I)) = \lambda (I^2/\m I^2)$, since $a$ is an non-zero-divisor of $R$.
\smallskip
\noindent
Claim : $n+2 \leq \lambda \big( \frac{I^2}{\m I^2}\big) \leq
\frac{n^2 + 3n + 2}{2}$.
\noindent
Proof of Claim:
By assumption,
$\lambda \big( \frac{I^2}{\m I^2}\big) \geq \lambda
\big( \frac{aI + \m I^2}{\m I^2} \big)$.
Assume that they are equal. Then $\lambda \big( \frac{I^2}{aI +\m I^2} \big) =0$.
Hence
$aI=I^2$, by NAK Lemma, which is a contradiction to $r_{aR} (I) =2$. Therefore
$\lambda (I^2/\m I^2) \geq n+2$. And we also have
$$
\lambda \big( \frac{I^2}{\m I^2}\big) = \mu (I^2) \leq \frac {n^2 + 3n +2}{2}.
$$
This completes the proof of the Claim.
\smallskip
\noindent
Note that
$$
\lambda \big( \frac{I^2}{aI + \m I^2 } \big)
= \lambda \big( \frac{I^2}{\m I^2} \big) - \lambda \big( \frac{aI + \m I^2}{\m I^2} \big)
= (n + 2 + i) - (n + 1) = i + 1,
$$
for $1 \leq i \leq \frac{n^2 + n - 2}{2}$. Hence we have
$$
\aligned
\sum_{n =0}^2 \lambda \big( \frac{I^n}{aI^{n-1} + \m I^n }\big)
&= \lambda \big( \frac{R}{\m} \big) + \lambda \big( \frac{I}{aR + \m I} \big)
+ \lambda \big( \frac{I^2}{aI + \m I^2} \big) \\
&= 1 + n + (i + 1) \quad \text{for} \quad 1 \leq i \leq \frac{n^2 + n - 2}{2} \\
&=\lambda (I^2/\m I^2) \\
&= e(F(I)).
\endaligned
$$
Therefore by [DRV, Theorem~2.1], $F(I)$ is a Cohen-Macaulay ring.
\QED