text:
\begin{document}

\title[]{The Gorenstein and complete intersection \\
properties of associated graded rings}

\date{\today }

\author{\ }

\address{\ }

\thanks{ Mee-Kyoung Kim is
partially supported by the Korea Science and Engineering
Foundation (R04-2003-000-10113-0). Bernd Ulrich is partially
supported by the National Science Foundation (DMS-0200858). }
\subjclass{Primary: 13A30; Secondary: 13C40, 13H10, 13H15}

\dedicatory{Dedicated to Wolmer Vasconcelos on the occasion of his
65th birthday.}

\keywords{Associated graded ring, fiber cone, reduction number, multiplicity,
Cohen-Macaulay ring, Gorenstein ring}

\maketitle

\begin{center}
\begin{tabular}{c}
{\bf William  Heinzer} \quad and \quad {\bf Bernd Ulrich} \\
Department of Mathematics, Purdue University \\
West Lafayette, Indiana 47907, USA\\
E-mail: heinzer@math.purdue.edu \\
E-mail: ulrich@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
Let $I$ be an $\m$-primary ideal of a
Noetherian local ring $(R,\m)$.  We
consider the Gorenstein and complete
intersection properties   of the associated graded
ring $G(I)$ and the fiber cone $F(I)$ of $I$ as reflected
in their defining ideals as homomorphic images of
polynomial rings over $R/I$ and $R/\m$ respectively.
In case  all the higher conormal modules of $I$ are free over
$R/I$, we observe that: (i) $G(I)$ is Cohen-Macaulay iff $F(I)$ is
Cohen-Macaulay, (ii) $G(I)$ is Gorenstein iff both $F(I)$ and
$R/I$ are Gorenstein, and (iii) $G(I)$ is a relative complete
intersection iff  $F(I)$ is a complete intersection.
In case  $(R,\m)$ is Gorenstein,
we give a necessary and sufficient condition for $G(I)$ to
be Gorenstein in terms of residuation of powers of $I$
with respect to a reduction $J$ of $I$ with $\mu(J) = \dim R$ and the
reduction number $r$  of $I$ with respect to $J$.  We
prove that $G(I)$ is Gorenstein if and only if
$J:I^{r-i} = J + I^{i+1}$  for $0 \leq i \leq r-1$.
If $(R,\m)$ is a Gorenstein local ring and $I \subseteq \m$
is an ideal having a reduction $J$ with reduction
number $r$ such that
$\mu(J) = \hgt(I) = g > 0$, we prove that
the extended Rees algebra $R[It, t^{-1}]$ is
quasi-Gorenstein with {\bf a}-invariant $a$ if and only if
$J^i:I^r = I^{i+a-r+g-1}$ for every $i \in \Z$. If,
in addition, $\dim R = 1$, we show that $G(I)$ is
Gorenstein if and only if $J^i:I^r = I^i$
for $1 \leq i \leq r$.
\end{abstract}

\baselineskip=15pt

\section{Introduction}

For an ideal $I \subseteq \m$ of a Noetherian local ring $(R,\m)$,
several graded rings naturally associated to $I$ are:
\begin{enumerate}
\item
the symmetric algebra $\Sym_R(I)$ and
the Rees algebra $\R = R[It] = \bigoplus_{i \ge 0} I^it^i$ \, (considered as
a subalgebra of the poynomial ring $R[t]$),
\smallskip
\item
the extended symmetric algebra $\Sym_R(I, t^{-1})$ and
the extended Rees algebra $R[It, t^{-1}] = \bigoplus_{i \in \Z}I^it^i$
\, (using the convention that $I^i=R$ for $i \leq 0$),
\smallskip
\item
the symmetric algebra $\Sym_{R/I}(I/I^2)$ and
the associated graded ring $G(I) = R[It, t^{-1}]/(t^{-1}) = R[It]/IR[It] =
\bigoplus_{i \ge 0}I^i/I^{i+1}$, and
\smallskip
\item
the  symmetric algebra $\Sym_{R/\m}(I/\m I)$ and the fiber cone
$F(I) = \newline R[It]/\m R[It] = \bigoplus_{i \ge 0}I^i/\m I^i$.
\end{enumerate}

These graded rings encode information about $I$ and its powers.
The {\em analytic spread } of $I$, denoted $\ell(I)$, is the
dimension of the fiber cone $F(I)$ of $I$. An ideal $J \subseteq
I$ is said to be a {\em reduction } of $I$ if there exists a
nonnegative integer $k$ such that $JI^k = I^{k+1}$. It then
follows that $J^jI^k = I^{k+j}$ for every nonnegative integer $j$.
These concepts were introduced by Northcott and Rees in \cite{NR}.
If $J$ is a reduction of $I$, then $J$  requires at least
$\ell(I)$ generators. Reductions of $I$ with $\ell(I)$
generators are necessarily minimal reductions in the sense that no
properly smaller ideal is a reduction of $I$. They correspond to
Noether normalizations of $F(I)$ in the sense that $a_1, \dots,
a_{\ell}$ generate a reduction of $I$ with $\ell = \ell(I)$
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 $(R/\m)[\overline{a_1}, \dots , \overline{a_{\ell}}]$. In
particular, if $R/\m$ is infinite, then there exist
reductions of $I$ generated by $\ell(I)$ elements.

Suppose $(R,\m)$ is a Gorenstein local ring of dimension $d$ and
$I$ is an $\m$-primary ideal. We are interested in conditions for
the associated graded ring $G(I)$ or the fiber cone $F(I)$ to be
Gorenstein. Assume $J = (a_1, \ldots, a_d)R$ \, is a reduction of
$I$. If $G(I)$ is Cohen-Macaulay, then the images $a_1^*, \ldots,
a_d^*$ of $a_1, \ldots, a_d$ in $I/I^2$ form a regular sequence on
$G(I)$, and $G(I)$ is Gorenstein if and only if $G(I)/(a_1^*,
\ldots, a_d^*)G(I)$ is Gorenstein. Write $\overline{R} = R/(a_1,
\ldots, a_d)R$ and $\overline{I} =I\overline{R}$. Then
$\overline{R}$ is a zero-dimensional Gorenstein local ring and
$G_{\overline{R}}(\overline{I}) = G(I)/(a_1^*, \ldots,
a_d^*)G(I)$. Thus, under the hypothesis that $G(I)$ is
Cohen-Macaulay,
 the question of whether
$G(I)$ is Gorenstein reduces to a zero-dimensional setting.

For $J \subseteq I$ a reduction of $I$,
the {\em reduction number } $r_J(I)$ of $I$ with
respect to $J$ is the smallest nonnegative integer $r$
such that $JI^r = I^{r+1}$. If $I$ is an $\m$-primary
ideal of  a $d$-dimensional Gorenstein local ring $(R,\m)$
and $J$ is a $d$-generated reduction of $I$,
then the reduction number $r = r_J(I)$ plays an
important role in considering the Gorenstein property
of $G(I)$. If $r = 0$, then $J = I$ is generated by
a regular sequence and $G(I)$ is a polynomial ring in $d$
variables over the zero-dimensional Gorenstein local ring
$R/I$. Thus $G(I)$ is Gorenstein in this case.
If $r = 1$ and $d \ge 2$, then a result of Corso-Polini
\cite[Cor.3.2]{CP} states that $G(I)$ is Gorenstein
if and only if
$J:I = I$, that is if and only if the ideal $I$ is self-linked
with respect to the minimal reduction parameter ideal $J$.

In Theorem \ref{3.8} we extend this result of Corso-Polini on the
Gorenstein property of $G(I)$ in case $I$ is an $\m$-primary
ideal. We prove that $G(I)$ is Gorenstein if and only if
$J:I^{r-i} = J + I^{i+1}$ for $0 \leq i \leq r-1$. This also gives
an analogue for Gorenstein rings  to a well-known result about
Cohen-Macaulay rings (see  \cite[Lemma 2.2]{HM})  that asserts:
Suppose $I$ is an $\m$-primary ideal of a Cohen-Macaulay local
ring $(R,\m)$ of dimension $d$ and the elements $a_1, \ldots, a_s$
are a  superficial sequence for $I$. Let $\overline{R} = R/(a_1,
\ldots, a_s)R$ \, and $\overline{I} = I\overline{R}$. If $s \le
d-1$, then $G(\overline{I})$ is Cohen-Macaulay implies $G(I)$ is
Cohen-Macaulay. Corollary~\ref{3.9} says that even for $s = d$ it
holds that $G(\overline{I})$ is Gorenstein implies $G(I)$ is
Gorenstein if one assumes that $I^r \not\subseteq J = (a_1,
\ldots, a_d)R$.

Let  $B = \oplus_{i \in \Z}B_i$ be a Noetherian $\Z$-graded ring which
is *local in the sense that it has a unique maximal homogeneous
ideal $\M$ \cite[(1.5.13), p.35]{BH}. Notice that $R:=B_0$
is a Noetherian local ring and $B$ is finitely generated as an $R$-algebra
\cite[Thm.1.5.5, p.29]{BH}. We assume for simplicity that $R$ is Gorenstein. Consider
a homogeneous presentation $B \cong S/H$ with $S=R[X_1, \ldots, X_n]$
a $\Z$-graded polynomial ring and
$H$ a homogeneous ideal of height $g$. Let $\sigma \in \Z$ be the
sum of the degrees of the variables $X_1, \ldots, X_n$. We write
$\omega_B= \Ext_S^g(B,S)(-\sigma)$ and call this module the
{\it graded canonical module} of $B$. One easily sees that $\omega_B$
is a finitely generated graded $B$-module that is uniquely
determined up to homogeneous $B$-isomorphisms.
The ring $B$ is said to be {\it quasi-Gorenstein} in case $\omega_B \cong B(a)$
for some $a \in \Z$. If the maximal homogeneous ideal $\M$ of $B$
is a maximal ideal, then the integer $a$ is
uniquely determined and is called the {\it {\bf a}-invariant} of
$B$. We will use the following facts, which are
readily deduced from the above definition of graded canonical modules:

$\bullet$ The localization $(\omega_B)_{\M}$ is the canonical module of
the local ring $B_{\M}$.

$\bullet$ The module $\omega_B$ satisfies $S_2$.

$\bullet$ The ring $B$ is (locally) Gorenstein if and only if it
is quasi-Gorenstein and (locally) Cohen-Macaulay.

$\bullet$ Let $A$ be a $\Z$-graded subring of $B$ with unique maximal
homogeneous ideal $\M \cap A$ and $A_0=R=B_0$, so that $A$ is
Cohen-Macaulay and $B$ is a
finitely generated $A$-module; then $\omega_B \cong \Hom_A(B,\omega_A)$
as graded $B$-modules.

Notice that a quasi-Gorenstein ring
necessarily satisfies $S_2$. Thus if $B$ is quasi-Gorenstein and
$\dim B \le 2$, then $B$ is Gorenstein. In higher dimensions there
do exist examples of quasi-Gorenstein rings that are not
Gorenstein. There exists an  example of a prime ideal $P$ of
height two in a $5$-dimensional regular local ring $R$ such that
the extended Rees algebra $R[Pt, t^{-1}]$ is  quasi-Gorenstein but
not Gorenstein \cite[Ex.4.7]{HH}. We are interested in
classifying quasi-Gorenstein extended Rees algebras  and saying
more about when such rings are Gorenstein. In Theorem~\ref{4.1},
we prove that if $(R,\m)$ is a Gorenstein local ring and $I
\subseteq \m$ is an ideal having a reduction $J$ with reduction
number $r$ such that $\mu(J) = \hgt(I) = g > 0$, then the extended
Rees algebra $R[It, t^{-1}]$ is quasi-Gorenstein with {\bf
a}-invariant $a$ if and only if $J^i:I^r = I^{i+a-r+g-1}$ for
every $i \in \Z$. A natural question here that we consider but
resolve only in special cases is whether $R[It, t^{-1}]$ is
Gorenstein if it is  quasi-Gorenstein.  Equivalently, is the
associated graded ring $G(I)$ then Gorenstein. We observe that
this is true if $\dim R = 1$ or if $\dim R = 2$ and $R$ is
regular. If  $\dim R = 1$, Corollary \ref{4.4} implies that $G(I)$
is Gorenstein if and only if $J^i:I^r = I^i$ for $1 \leq i \leq
r$.

\medskip

\section{Defining ideals and freeness of the higher
conormal modules}

Let $(R, \m)$ be a Noetherian local ring and let
$I = (a_1, \ldots, a_n)R \subseteq \m$
be an ideal of $R$. Consider the  presentation of the
Rees algebra $R[It]$ as a homomorphic image of a polynomial
ring over $R$ obtained  by defining an $R$-algebra homomorphism
$\tau : R[X_1, \ldots , X_n] \to R[It]$ such that
$\tau(X_i) = a_it$ for $1 \leq i \leq  n$. Now
define $\psi = \tau \otimes R/I : (R/I)[X_1, \ldots, X_n] \to G(I)$,
where $\psi(X_i) = a_i + I^2 \in I/I^2 = G_1$, and
$\phi = \tau \otimes R/\m : (R/\m)[X_1, \ldots, X_n] \to F(I)$,
where $\phi(X_i) = a_i + \m I \in  I/\m I = F_1$.

We have  the following commutative diagram for which the rows are
exact and the column maps are surjective.

\begin{equation}\label{eq1}\tag{2.1}
\text{$\CD
0 @>>> E := \ker(\tau)  @>>> R[X_1, \ldots, X_n] @>\tau>> R[It] @>>> 0 \\
@. @V\pi_1 VV        @V\pi_2 VV                @V\pi_3 VV \\
0 @>>> L := \ker(\psi)  @>>> (R/I)[X_1, \ldots, X_n] @>\psi>> G(I)
@>>> 0 \\
@. @V\pi_1' VV        @V\pi_2' VV                @V\pi_3' VV \\
0 @>>> K := \ker(\phi) @>>> (R/\m)[X_1, \ldots, X_n] @>\phi>> F(I) @>>> 0 .
\endCD $}
\end{equation}

\vskip 3ex

The ideal $E = 0$ if and only if $n = 1$
and $a_1$ is a regular element of $R$, while
$L  = 0$ if and only if $a_1, \ldots, a_n$
form a regular sequence, see for example
\cite[Cor.5.13, p.154]{K}. A necessary and
sufficient condition for $K = 0$ is
that $a_1, \ldots, a_n$ be what Northcott and Rees
\cite{NR} term {\it analytically independent. }

Let $v$ be the maximal degree of a homogeneous minimal generator
of the ideal $E$. The integer $N_{\R}(I):=\max\{1,v\}$

is called the {\it relation type } of the Rees algebra $\R =
R[It]$ with respect to the given generating set $a_1, \ldots,
a_n$. The relation type of $R[It]$ may also be defined by
considering  the kernel $N$ of the canonical homomorphism from the
symmetric algebra $\Sym_R(I)$ onto $R[It]$; then  $N_{\R}(I) =
\max\{1, w \}$, where $w$ denotes the maximal degree of a
homogeneous minimal generator of $N$.  Since the symmetric algebra
$\Sym_R(I)$ is independent of the choice of generators for $I$, it
follows that the relation type  $N_{\R}(I)$ of $R[It]$ is
independent of the generating set of $I$. The ideal $I$ is said to
be of {\it linear type} if $N_{\R}(I) = 1$. Thus $I$ is of linear
type if and only if $R[It]$ is canonically isomorphic to the
symmetric algebra $\Sym_R(I)$ of $I$.

The relation type $N_G(I)$ of the associated graded ring $G(I)$
is  defined in a similar
way, using the ideal $L$ or the kernel of the
canonical homogeneous epimorphism
$\beta : \Sym_{R/I}(I/I^2)  \to G(I)$.
Likewise, the relation type $N_F(I)$ of the fiber cone $F(I)$
is defined via the ideal $K$ or the kernel of the

canonical homogeneous
epimorphism $\gamma : \Sym_{R/\m}(I/\m I) \to F(I)$.

The surjectivity of the maps $\pi_1$ and $\pi_1'$ in diagram
(\ref{eq1}) imply that the inequalities $N_F(I) \le N_G(I) \le
N_{\R}(I)$ hold in general.

\setcounter{lem}{1}

\begin{dis}\label{2.1} {\em
It is shown by Valla \cite[Thm.1.3]{Va} and by
Herzog-Simis-Vasconcelos \cite[Thm.3.1]{HSV} that if $N_G(I) = 1$,
then also $N_{\R}(I) = 1$, that is, the relation type of $G(I)$ is
one  if and only if $I$  is of linear type. Planas-Vilanova shows
\cite[Prop.3.3]{PV} that the equality $N_{\R}(I) = N_G(I)$ also
holds in the case where $N_G(I) > 1$. We reprove this fact by
using an `extended symmetric algebra' analogue to the extended
Rees algebra $R[It, t^{-1}]$ of the ideal $I$. As an $R$-module we
define the {\it extended symmetric algebra} \, $\Sym_R(I, t^{-1})$
of $I$ to be $\oplus_{i \in \Z}S_i(I)$, where $S_i(I) = t^iR \cong
R$ for $i \le 0$ and $S_i(I)$ is the $i$-th symmetric power of $I$
for $i \ge 1$. We define a multiplication on $\Sym_R(I, t^{-1})$
that extends the ring structure on the standard symmetric algebra
$\Sym_R(I) = \oplus_{i \ge 0}S_i(I)$. To do this, it suffices to
define multiplication by $t^{-1}$ on $S_m(I)$ for $m \ge 1$.
Consider the product $I \times \cdots \times I$ of $m$ copies of
the ideal $I$ and the map $I \times \cdots \times I \to
S_{m-1}(I)$ that takes $(a_1, \ldots, a_m) \mapsto (a_1\cdot
a_2\cdot \ldots \cdot a_{m-1})a_m$, where the multiplication on
$a_1, \ldots ,a_{m-1}$ is multiplication in the symmetric algebra
and where the multiplication with $a_m$ is the scalar
multiplication given by the $R$-module structure of $S_{m-1}(I)$.
This map is $m$-multilinear over $R$ and hence factors through the
tensor power $I \otimes \ldots \otimes I$. Moreover, the map is
symmetric in $a_1, \ldots, a_m$ and hence  induces a map $S_m(I)
\to S_{m-1}(I)$. To verify this symmetry, it suffices to observe
that by associativity and commutativity of the two
multiplications, one has $(a_1\cdot  \ldots \cdot a_{m-2}\cdot
a_{m-1})a_m  = (a_1\cdot  \ldots \cdot a_{m-2})\cdot (a_{m-1}a_m)
= (a_1\cdot  \ldots \cdot a_{m-2})\cdot (a_ma_{m-1})  = (a_1\cdot
\ldots \cdot a_{m-2}\cdot a_m)a_{m-1}$. The multiplication by
$t^{-1}$ on $S_m(I)$ we just defined coincides with the
downgrading homomorphism $\lambda_{m-1}$ introduced by Herzog,
Simis and Vasconcelos \cite[p.471]{HSV}. Assigning to the element
$t^{-1}$ degree $-1$, $\Sym_R(I, t^{-1})$ becomes a $\Z$-graded
$R$-algebra. This algebra is $^*$local \cite[(1.5.13), p.35]{BH}
in the sense that the ideal of $\Sym_R(I, t^{-1})$ generated by
$t^{-1}, \m$ and $\oplus_{i \ge 1}S_i(I)$ is the unique maximal
homogeneous ideal of $\Sym_R(I, t^{-1})$.

The canonical surjective $R$-algebra homomorphism from the
symmetric algebra $\Sym_R(I)$ onto the Rees algebra $R[It]$
extends to a surjective homogeneous $R$-algebra homomorphism
$\alpha : S = \Sym_R(I, t^{-1}) \to R[It, t^{-1}]$, where
$\alpha(t^{-1}) = t^{-1}$. Notice that the two maps have the same
kernel $\A$. Tensoring the short exact sequence
\[
\CD
0 @>>> \A  @>>> S
@>\alpha>> R[It, t^{-1}] @>>> 0 \\
\endCD
\]
with $\otimes_SS/t^{-1}S$ gives the following isomorphisms:
\[
\CD
0 @>>> \A \otimes S/t^{-1}S  @>>> S \otimes S/t^{-1}S
@>>> R[It, t^{-1}]\otimes S/t^{-1}S @>>> 0 \\
@. @V\cong VV        @V\cong VV                @V\cong VV \\
0 @>>> \B: = \ker(\beta)@>>> \Sym_{R/I}(I/I^2) @>\beta>> G(I)  @>>> 0 .\\
\endCD
\]
Here we are using that $t^{-1}$ is a regular element on $R[It,
t^{-1}]$. A graded version of Nakayama's lemma \cite[(1.5.24),
p.39]{BH} now implies that $\A$ is generated in degrees $\leq
N_G(I)$ as a module over $S=\Sym_R(I)[t^{-1}]$ and hence over
$\Sym_R(I)$. Therefore the equality $N_{\R}(I) = N_G(I)$ holds in
general.}
\end{dis}

\medskip

We observe in Corollary \ref{2.6}
a sufficient condition for $N_G(I) = N_F(I)$.
There exist, however,   examples where
$N_G(I) > N_F(I)$. If $(R,\m)$ is a $d$-dimensional Noetherian
local ring and $I = (a_1, \ldots, a_d)R$ is an $\m$-primary ideal,
then $N_F(I) = 1 $ \cite[Thm.14.5,
p.107]{M}; while if $R$ is not Cohen-Macaulay  it may happen
that $N_G(I) > 1$. For example, let $k$ be a field and let
$A = k[x,y] = k[X,Y]/(X^2Y,Y^3)$. Then $R = A_{(x,y)A}$
is a one-dimensional Noetherian local ring
and $I = xR$ is primary to the maximal ideal $\m = (x,y)R$. Let
$U$ be an indeterminate over $R/I$ and consider
the graded homomorphism $\psi: (R/I)[U] \to G(I)$,
where $\psi(U) = x + I^2 \in I/I^2 = G_1$. Let
$L = \oplus_{i \ge 0}L_i = \ker(\psi)$.
Since $R/I \cong k[X, Y]/(X, Y^3) \cong k[Y]/(Y^3)$,
we see that  $\lambda(R/I) = 3$, where $\lambda$ denotes
length. Similarly,
$R/I^2 \cong k[X, Y]/(X^2, Y^3)$ and
$\lambda(R/I^2) = 6$. Thus  $\lambda(I/I^2) = 3$. It
follows that $L_0 = L_1  = 0$. On the other hand,
$0 \neq (y+I)U^2 \in L_2$. Therefore $N_G(I) \ge 2$, while
$N_F(I) = 1$.

In the case where $N_F(I) \ge 2$, it would be
interesting to have necessary and sufficient conditions
for $N_G(I) = N_F(I)$.
That this is not always true is
shown, for example, by taking $k$ to be a field and
$A = k[x, y, z]=k[X,Y,Z]/(X^2,Y^2,XYZ^2)$.
Let $R = A_{(x,y,z)A}$ and let
$I = (y, z)R$. Then $F(I) \cong k[Y, Z]/(Y^2)$
has relation type 2, while
if $\psi : (R/I)[U, V]  \to G(I)$ with $\psi(U) = y +I^2$
and $\psi(V) = z + I^2$ is a presentation
of $G(I)$, then the relation $xyz^2 = 0$ in the
definition of $R$ implies $(x+I)UV^2 \in \ker(\psi)$.
Let $L = \oplus_{i \geq 0} L_i = \ker(\psi)$.
Since $R/I \cong k[X]/(X^2)$, we see that
$\lambda(R/I) = 2$.  We have
$\lambda(I/I^2) = 4$, hence  $L_1 = 0$. Also
$\lambda(I^2/I^3) = 4$, so $\lambda(L_2) = 2$
and $L_2$ is generated by $U^2$. Since
$(x+I)UV^2 \not\in (U^2+I)(R/I)[U, V]$, we see that
$N_G(I) \ge 3$.

\begin{thm}\label{2.3}
Let $(R,\m)$ be a Noetherian local ring and let $I$ be an
$\m$-primary ideal with $\mu(I) = \lambda(I/\m I) = n$. Let  $A =
(R/I)[X_1, \ldots, X_n]$ and let $\psi : A \to G(I)$ be defined as
in diagram $(\ref{eq1})$. The following are equivalent $:$
\begin{enumerate}
\item
$I^i/I^{i+1}$ is free as an $(R/I)$-module for every $i \geq 0$.
\item
$G(I)$ has finite projective dimension as an $A$-module,
i.e., $\pd_AG(I) < \infty$.
\item If $\F_\bullet$ is a homogeneous minimal free resolution of $G(I)$
as an $A$-module and $k = R/\m$,
then $\F_\bullet \otimes \, k$ is a homogeneous minimal free
resolution of $F(I) = G(I) \otimes  k$
as a module over $A \otimes k = k[X_1, \ldots, X_n]$.
\end{enumerate}
If these equivalent conditions hold, then $\depth G(I) = \depth F(I)$.
\end{thm}

\demo
$(1) \implies (3)$: Condition (1) implies that $\Tor^{R/I}_i(G(I), k) = 0$ for
every $i > 0$, and this implies condition (3).

\noindent
$(3) \implies (2)$: Let $\F_\bullet$ be a homogeneous minimal free resolution of
$G(I)$ over $A$. Condition (3) implies that $\F_\bullet \otimes \, k$ is
a homogeneous minimal free resolution of $F(I)$ over the regular ring $B = k[X_1, \ldots, X_n]$.
Since $B$ is regular, this homogeneous minimal free resolution of $F(I)$ is finite.
Therefore $\F_\bullet$ is finite and $\pd_AG(I) < \infty$.

\noindent $(2) \implies (1)$: Condition (2) implies that
$\pd_{R/I}(I^i/I^{i+1}) < \infty$ for every $i \geq 0$. Since
$\dim(R/I) = 0$, $I^i/I^{i+1}$ is free over $R/I$ by the
Auslander-Buchsbaum formula, see for instance \cite[Thm.1.3.3,
p.17]{BH}.

If these equivalent conditions hold, then we have
$\pd_AG(I) = \pd_BF(I)$ by (3). Hence
$\depth G(I) = \depth F(I)$ by the Auslander-Buchsbaum formula.
\QED

\begin{remark}\label{2.5}{\em
If $(R, \m)$ is a one-dimensional Cohen-Macaulay local ring, $I$
is an $\m$-primary ideal  and $J=xR$ is a principal reduction of
$I$ with reduction number $r = r_J(I)$, then $x^{i-r}I^r = I^i$
for $i \ge r$, so $I^i/I^{i+1} \cong I^r/I^{r+1}$. Thus in this
case a fourth equivalent condition  in Theorem \ref{2.3} is that
$I^i/I^{i+1}$ is free over $R/I$ for every $i \le r$. On the other
hand, if $I \subset \m$ is an ideal of a zero-dimensional
Noetherian local ring $(R,\m)$, then a condition equivalent  to
the conditions of Theorem \ref{2.3} is that $\lambda(R) =
\lambda(R/I)\cdot \lambda(F(I))$. }
\end{remark}

Under the equivalent conditions of Theorem \ref{2.3}, the
ring $R$ is said to be {\it normally flat} along the ideal $I$.
This is a concept introduced by Hironaka in his work on resolution
of singularities \cite[p.188]{M}. A well known
result of Ferrand \cite{F} and Vasconcelos \cite[Cor.1]{V} asserts that
if the conormal module $I/I^2$ is a free module over $R/I$
and if $R/I$ has finite
projective dimension as an $R$-module, then $I$ is generated
by a regular sequence. It then  follows that $N_G(I) = 1$. Thus,
in the case where $R$ is a regular local ring, the equivalent
conditions of Theorem \ref{2.3} imply $\mu(I) = \dim R$.
However, as we indicate in Example \ref{2.9},  there exist  examples of
Gorenstein local rings $(R,\m)$ having
$\m$-primary ideals $I$  such that
$\mu(I) > \dim R$ and the equivalent
conditions of Theorem \ref{2.3} hold.

In Examples \ref{2.9}, \ref{2.13}, \ref{2.10}, \ref{3.5},
\ref{3.11}, \ref{4.5}, and \ref{4.51},
we present examples
involving an additive monoid $S$ of the nonnegative
integers that contains all sufficiently large integers
and a complete one-dimensional local
domain of the form $R = k[[t^s \, | \,  s \in S]]$.
The formal power series ring $k[[t]] = R[t]$ is the
integral closure of $R$ and is a finitely
generated $R$-module. Properties of $R$ are closely
related to properties of the numerical semigroup $S$.
For example, $R$ is Gorenstein if and only if $S$ is
symmetric \cite[Thm.4.4.8, p.178]{BH}.

\begin{exam}\label{2.9}{\em
Let $k$ be a field, $R = k[[t^4, t^9, t^{10}]]$ and $I =
(t^8, t^9, t^{10})R$. Then $R$ is a one-dimensional
Gorenstein local domain, $\mu(I) = 3$ and
$J = t^8R$ is a reduction of $I$.
We have $JI \subsetneq I^2$ and $JI^2 = I^3$.
Hence $r_J(I) = 2$. If  $w$ denotes the image of
$t^4$ in $R/I$, then  $R/I \cong k[w]$, where
$w^2 = 0$. Thus $R/I$ is Gorenstein with
$\lambda(R/I) = 2$.
Let $A = (R/I)[X, Y, Z]$ and define
$\psi : A \to G(I)$ by $\psi(X) = t^8 + I^2,
\psi(Y) = t^9 + I^2$ and $\psi(Z) = t^{10} + I^2$.
Consider the  short exact sequence
$$
\CD
0 @>>> L = \ker(\psi)  @>>> A
@>\psi>> G(I) @>>> 0.
\endCD
$$
The ring $G(I) \cong A/L$ has multiplicity $e(I) = 8$, $A$
is a Cohen-Macaulay ring of multiplicity 2,
and the two quadrics
$XZ - Y^2, wX^2 - Z^2$ form an $A$-regular
sequence contained in $L$.
Hence $L = (XZ - Y^2, wX^2 - Z^2)A$.
Therefore  the equivalent conditions of
Theorem \ref{2.3} are satisfied.  It also follows that
$I^i/I^{i+1}$ is free of rank 4 over $R/I$ for every
$i \ge 2$.}
\end{exam}

\begin{cor}\label{2.6}
Let $(R,\m)$ be a Noetherian  local ring and  let $I$ be an
$\m$-primary ideal with $\mu(I) = \lambda(I/\m I) = n$. With
notation as in Theorem $\ref{2.3}$ and diagram $(\ref{eq1})$, let
$A=(R/I)[X_1, \ldots, X_n]$ and $B=(R/\m)[X_1, \ldots, X_n]$, and
let $\M_G = (\m/I, G(I)_+)$ and $\M_F = F(I)_+$ denote the maximal
homogeneous ideals of $G(I)$ and $F(I)$, respectively. Suppose the
equivalent conditions of Theorem $\ref{2.3}$ hold. Then$:$
\begin{enumerate}
\item
$G(I)$ is Cohen-Macaulay $\Longleftrightarrow$  $F(I)$ is Cohen-Macaulay.

\item
If $G(I)$, or equivalently $F(I)$, is Cohen-Macaulay,
then the type of $G(I)_{\M_G}$ is the type of $R/I$ times the type
of $F(I)_{\M_F}$. In particular,
$G(I)$ is Gorenstein $\Longleftrightarrow$  both $F(I)$ and
$R/I$ are  Gorenstein.

\item
The relation type $N_G(I)$ of the associated graded ring $G(I)$
is equal to the
relation type $N_F(I)$ of the fiber cone $F(I)$.

\item
The defining ideal $L$ of $G(I)$ is generated by a regular sequence
on $A$ if and only if the defining ideal $K$ of $F(I)$ is generated
by a regular sequence on $B$.

\item
The multiplicity of $G(I)$ is $\lambda(R/I)\cdot e(F(I))$,
where $e(F(I))$ denotes the multiplicity of $F(I)$.
\end{enumerate}
\end{cor}

\demo
As $G(I)$ is flat over $R/I$, statements (1) and (2) follow from
\cite[Prop.1.2.16, p.13]{BH}.

Since the relation types of $F(I)$ and $G(I)$ are determined
by the degrees of minimal generators of  first syzygies,
statement (3) follows from part (3) of
Theorem \ref{2.3}.  Indeed, with the notation of diagram (\ref{eq1}),
the relation type of $G(I)$ is $\max\{1, w\}$, where $w$ is the
maximal degree of a homogeneous minimal generator of $\ker(\psi)$.
Since $\F_{\bullet} \otimes k$ is a minimal free resolution of $F(I)$,
$\pi_1'$ maps a set of homogeneous
minimal generators of $L = \ker(\psi)$ onto a set of homogeneous
minimal generators of $K = \ker(\phi)$. Therefore $N_G(I) = N_F(I)$.
Since $A$ and $B$ are Cohen-Macaulay and $\hgt(L) = \hgt(K)$,
it also follows that
$L$ is generated by a regular sequence on $A$  if and only if
$K$ is generated by a regular sequence on $B$, which is part (4). Statement (5) is
clear in view of the freeness of the $I^i/I^{i+1}$ over $R/I$. \QED

\medskip

We observe
in part (1) of Remark \ref{2.4} that $L$ is generated by
a regular sequence implies $K$ is generated by a regular sequence
holds even without the equivalent conditions of Theorem \ref{2.3}.

In Proposition \ref{2.5.5}, we give
a partial converse to part (5) of
Corollary \ref{2.6}. Proposition \ref{2.5.5} is closely
related to results of Shah \cite[Lemma 8 and Thm.8]{Sh}.

\begin{prop} \label{2.5.5} Let $(R,\m)$ be a Noetherian local
ring and let $I$ be an $\m$-primary ideal. With notation
as in Theorem $\ref{2.3}$ assume that
$e(G(I)) = \lambda(R/I) \cdot e(F(I))$.
\begin{enumerate}
\item
If all associated primes of $F(I)$ have the same dimension,
then $I^i/I^{i+1}$ is a free $(R/I)$-module for every $i \ge 0$.
\item
If all relevant associated primes of $F(I)$ have the same dimension,
then $I^i/I^{i+1}$ is a free $(R/I)$-module for all sufficiently
large $i$.
\end{enumerate}
\end{prop}

\demo
A composition series of $R/I$ induces a filtration on $G(I)$
whose factors are homogeneous $F(I)$-modules of the form
$F(I)/{\bf a}_j$ with $1 \le j \le \lambda(R/I)$. Since
$e(G(I)) = \lambda(R/I) \cdot e(F(I))$, these factors all
have the same dimension and the same multiplicity as $F(I)$.
Thus in the setting of part (1), ${\bf a}_j = 0$ for every $j$,
whereas in part (2), ${\bf a}_j$ is concentrated in finitely
many degrees for every $j$. Computing the Hilbert function of
$G(I)$ one sees that $\lambda(I^i/I^{i+1}) = \lambda(R/I) \cdot
\lambda(I^i/\m I^i) = \lambda(R/I) \cdot \mu(I^i)$ for every
$i \ge 0$  in the
setting of part (1), and for every $i >> 0$ in the setting of
part (2). \QED

\medskip

In Example \ref{2.13}, we exhibit an $\m$-primary ideal $I$ of a
one-dimensional Cohen-Macaulay local domain $(R,\m)$ such
that $I^i/I^{i+1}$ is free over $R/I$ for every $i \ge 2$,
while  $I/I^2$ is not free over $R/I$. This example illustrates
that the equality $e(G(I)) = \lambda(R/I) \cdot e(F(I))$

does not imply the equivalent conditions of Theorem \ref{2.3},
even if the assumption of part (2) of Proposition \ref{2.5.5} is
satisfied.

\begin{exam}\label{2.13} {\em
Let $k$ be a field, $R = k[[t^3, t^7, t^{11}]]$ and $I =
(t^6, t^7, t^{11})R$ as in \cite[Ex.6.4]{DV}.
Then  $\mu(I) = 3$ and
$J = t^6R$ is a principal reduction of $I$.
Also  $JI \subsetneq I^2$ and $JI^2 = I^3$, so
$r_J(I) = 2$.  We have
$\lambda(R/I) = 2$ and  $\lambda(R/I^2) = 7$.
Hence $\lambda(I/I^2) = 5$ and $I/I^2$ is not
free over $R/I$.  On the other hand,
$I^2 = (t^{12}, t^{13}, t^{14})R$. Therefore   $I^i/I^{i+1}$ is
generated by 3 elements and $\lambda(I^i/I^{i+1}) = 6$,
so $I^i/I^{i+1}$ is free over $R/I$ of rank 3 for
every $i \ge 2$.}
\end{exam}

By Proposition \ref{2.5.5},  the fiber cone $F(I)$
in Example \ref{2.13} is not Cohen-Macaulay. To
see this explicitly, let  $w$ denote the image of $t^3$ in $R/I$ and let
$A = (R/I)[X, Y, Z]$. Define
$\psi : A \to G(I)$ by $\psi(X) = t^6 + I^2,
\psi(Y) = t^7 + I^2$ and $\psi(Z) = t^{11} + I^2$,  and
consider the  exact sequence
$$
\CD
0 @>>> L = \ker(\psi)  @>>> A
@>\psi>> G(I) @>>> 0.
\endCD
$$
Then  $L = (wZ, XZ - wY^2, YZ, Z^2, Y^3 - wX^3)A$.
Hence
$$
e(G(I))=e(I)=6=\lambda(G(I)/\psi(X)G(I)),
$$
showing that $G(I)$ is
Cohen-Macaulay, see for instance \cite[Thm.17.11, p.138]{M}.
However, the fiber cone $F(I)$ is isomorphic
to $(R/\m)[X, Y, Z]/(XZ, YZ, Z^2, Y^3)$ and is
not Cohen-Macaulay.

\vskip 3ex
\begin{remark}\label{2.4}{\em
With notation as in Theorem \ref{2.3} and diagram (\ref{eq1}), we have:
\begin{enumerate}
\item
If $L$ is generated by a regular sequence in $A$,
then $\pd_AG(I)$ is finite

and the equivalent conditions of Theorem \ref{2.3} hold. Hence
by part (4) of Corollary \ref{2.6}, $K$ is also
generated by a regular sequence. However, as we demonstrate in
Example \ref{2.10}, the converse fails in
general, that is, without the hypothesis that $\pd_A G(I) < \infty$,
it can happen that $K$ is generated by a regular sequence,
while  $L$ is not generated by a regular sequence.

\item Suppose  $R/I$ is Gorenstein and  $\mu(I) = \dim R + 2$. If
$G(I)$ is Gorenstein and $\pd_AG(I) < \infty$, then $L$ is
generated by an $A$-regular sequence. For $\mu(I) = \dim R + 2$
implies $\hgt(L) = 2$; then $G(I) = A/L$ Cohen-Macaulay  and
$\pd_A(A/L) < \infty$ imply $\pd_A(A/L) = 2$ by the
Auslander-Buchsbaum formula. Since $A$ and $A/L$ are Gorenstein
rings, the homogeneous minimal free resolution of $A/L$ has the
form
$$
\CD
0 @>>> A  @>>> A^2 @>>> A @>>> A/L @>>> 0.
\endCD
$$
Hence  $\mu(L)=2=\hgt(L)$, and therefore $L$
is generated by a regular sequence.

\item
Assume that $R$ is Gorenstein and
$J$ is a reduction of $I$ with $\mu(J) = \dim R$.
If $r_J(I)\leq 1$ and $\dim R \ge 2$, then it follows from
\cite[Cor.3.2]{CP} that $G(I)$ is Gorenstein if and only
if $I = J:I$. If $R/I$ is Gorenstein and $r_J(I) \le 1$,
 we prove by induction on
$\dim R$ that $G(I)$ is Gorenstein implies $\pd_A G(I) < \infty$.
Suppose $\dim R = 0$, in which case $G(I) =
R/I \oplus I$. If $G(I)$ is Gorenstein and $I \neq 0$, then $0:I=I$ by Theorem \ref{3.1}.
Thus $I$ is principal since $R$ and $R/I$ are Gorenstein,
see for instance \cite[Prop.3.3.11(b), p.114]{BH}. Therefore  $I/I^2 = I \cong
R/I$ and hence $\pd_A G(I) < \infty$. Now suppose $\dim R > 0$. With
the notation of diagram (\ref{eq1}), we may assume that $\psi(X_1) = a^*
\in G(I)$ is a regular element of $G(I)$, where $a^*$ is the
leading form of some $a \in J \backslash {\m}J$.  Therefore  $G(I)/a^*G(I)
\cong G(I/aR)$  and $\pd_A G(I) = \pd_{A/X_1A} G(I)/a^*G(I)$. We
conclude by induction that  $\pd_A G(I) < \infty$. Thus in the
case where $R/I$ is Gorenstein and $r_J(I) \le 1$, if $G(I)$ is
Gorenstein, then the equivalent conditions  of Theorem \ref{2.3}
hold.

\item
There exist
examples where $R$ is Gorenstein of dimension zero, $R/I$
is Gorenstein, $I^3 = 0$ and $G(I)$ is Gorenstein, but
$\pd_A G(I) = \infty$. To obtain an example illustrating
this consider the submaximal Pfaffians $Y^3, XZ, XY^2 + Z^2, YZ, X^2$
of the matrix
$$
\begin{bmatrix}
0  & X & 0 & 0 & Z \\
-X & 0 & Y & Z & 0 \\
0 & -Y & 0 & X & 0 \\
0 & -Z & -X & 0 & Y^2 \\
-Z & 0 & 0 & -Y^2 & 0
\end{bmatrix}.
$$
Let $k$ be a field and let
$H$ denote the ideal of the polynomial  ring
$k[X, Y, Z]$ generated by these Pfaffians. Notice that
$H$ is homogeneous with respect to the grading
that assigns   $\deg X = 0$ and $\deg Y = \deg Z = 1$.
Let $R = k[X, Y, Z]/H$, write $x$, $y$, $z$ for the images of
$X$, $Y$, $Z$ in $R$
and let $I = (y, z)R$.
Then $R$ is an Artinian Gorenstein local ring by \cite[Thm.2.1]{BE}.
Furthermore $R/I=k[X]/(X^2)$ is Gorenstein and $I^3=0$.
Finally $G(I)\cong R$ by our choice of the grading.

With $A = (R/I)[U, V]$
and $\psi : A \to G(I)\cong R$ defined by
$\psi(U) = y$ and
$\psi(V) = z$, we have
$L = \ker(\psi) = (xV, UV, V^2 + xU^2, U^3)A$.
Thus  $I/I^2$ is not free over $R/I$ and then $\pd_A G(I) = \infty$
by Theorem \ref{2.3}.
This even  provides  an example where the associated
graded ring $G(I)$ is Gorenstein and the
fiber cone $F(I) \cong k[U, V]/(UV, V^2, U^3)$ is not Gorenstein.

\item Suppose $R$ is Gorenstein. If $R/I$ is Gorenstein and of
finite projective dimension over $R$, it is shown in
\cite[Thm.2.6]{NV} that
$G(I)$ is Cohen-Macaulay implies $I$ is generated
by a regular sequence.
\end{enumerate}}
\end{remark}

\begin{exam}\label{2.10}{\em
Let $k$ be a field, $R = k[[t^3, t^4, t^5]]$ and $I =
(t^3, t^4)R$. Then $R$ is a one-dimensional
Cohen-Macaulay local domain, $\mu(I) = 2$ and
$J = t^3R$ \, is a reduction of $I$.
We have $JI \subsetneq I^2$ and $JI^2 = I^3$.
Hence $r_J(I) = 2$. If  $w$ denotes the image of
$t^5$ in $R/I$, then  $R/I \cong k[w]$, where
$w^2 = 0$. Thus $R/I$ is Gorenstein with
$\lambda(R/I) = 2$.
Consider the commutative diagram with exact rows and surjective
column maps
$$
\CD
0 @>>> L := \ker(\psi)  @>>> (R/I)[X, Y] @>\psi>> G(I)  @>>> 0 \\
@. @V\pi_1' VV        @V\pi_2' VV                @V\pi_3' VV \\
0 @>>> K := \ker(\phi) @>>> (R/\m)[X, Y] @>\phi>> F(I) @>>> 0,
\endCD
$$
where $\psi(X) = t^3 + I^2$ and $\psi(Y) = t^4 + I^2$.
It is readily checked that $wX, wY$ and $Y^3$ generate
$L = \ker(\psi)$.
In particular  $K = \ker(\phi)$ is generated by $Y^3$, so
$K$ is generated by a regular sequence, while $L$ is not generated
by a regular sequence.
Also notice that $G(I)$ and $F(I)$ both have multiplicity 3,
and  $I/I^2$ has length 2 and is not a  free module over $R/I$.}
\end{exam}

\smallskip

\section{The Gorenstein property for $G(I)$}

In this section, we establish  a necessary and sufficient condition for $G(I)$ to
be Gorenstein in terms of residuation of powers of $I$
with respect to a reduction $J$ of $I$ for which $\mu(J) = \dim R$.
We first state this in dimension zero. Among
the equivalences in Theorem \ref{3.1}, the equivalence of
(1), (3) and (5) are due to Ooishi \cite[Thm.1.5]{O}.
We include elementary direct arguments in the proof.
We use the floor function $\lfloor x \rfloor $ to denote
the largest integer which is less than or equal to $x$.

\begin{thm}\label{3.1}
Let $(R,\m)$ be a zero-dimensional Gorenstein local ring and let
$I \subseteq \m$ be an ideal of $R$. Assume that $I^r \ne 0$ and
$I^{r+1} = 0$. Let $G := G(I) = \oplus_{i=0}^r G_i$  be the
associated graded ring of $I$, and let $S := \Soc(G) =
\oplus_{i=0}^r S_i$ denote the socle of $G$. Then the following
are equivalent $:$
\begin{enumerate}
\item $G$ is a Gorenstein ring. \item $S_i = 0$ \, for $0 \leq i
\leq r-1$. \item $0:_{R}I^{r-i} = I^{i+1}$ \, for $0 \leq i \leq
r-1$. \item $0:_{R}I^{r-i} = I^{i+1}$ \, for $0 \leq i \leq
\lfloor \frac{ r-1}{2} \rfloor$. \item $\lambda(G_i) =
\lambda(G_{r-i})$ \, for $0 \leq i \leq \lfloor \frac{ r-1}{2}
\rfloor$. \item $I^r :_{R} I^{r-i} = I^i$  for $1 \leq i \leq
r-1$, and $0:_{R} I = I^r$. \item $I^{r-i}/I^r$ is a faithful
module over $R/I^i$  for $1 \leq i \leq r-1$, and $I$ is faithful
over $R/I^r$.
\end{enumerate}
\end{thm}

\demo We may assume $r > 0$. Write $k=R/\m$ and let $\mathfrak{M}$
denote the  unique maximal ideal of $G$. We first compute $S_i = 0
:_{G_i} \mathfrak{M}$  \, for $0 \leq i \leq r$. Since
$\mathfrak{M}$ is generated by $\m/I$ and $I/I^2$ it follows that
$$

S_i = (0 :_{I^i/I^{i+1}} \m/I) \cap (0 :_{I^i/I^{i+1}} I/I^2)

    = \frac{I^i}{I^{i+1}} \cap
\frac{(I^{i+1} :_{R} \m)}{I^{i+1}} \cap \frac{(I^{i+2} :_{R}
I)}{I^{i+1}}.

$$
Therefore

\begin{equation}\label{star} \tag{3.2}
S_i = \frac{I^i \cap (I^{i+1} : \m) \cap (I^{i+2} : I)}{I^{i+1}}
\quad  \text{for} \quad 0 \leq i \leq r.
\end{equation}
In particular $S_r = 0:_{I^r}\m$  because  $I^{r+1} = 0$. Note that $S_r \not = 0$.
Since $(R, \m)$ is a zero-dimensional Gorenstein local ring, we have
\begin{equation}\label{starstar} \tag{3.3}
 S_r \cong k.
\end{equation}

\smallskip

\noindent
$(1) \Longleftrightarrow (2)$ : The ring $G$ is Gorenstein if and only if $\dim_k S = 1$
if and only if $S_i = 0$ for $0 \leq i \leq r-1$, by (\ref{starstar}).

\smallskip

\noindent
$(2) \implies (3)$ : Condition (2) implies that $S = S_r \cong k$  by
(\ref{starstar}).
Hence $S = s^{*}k$ for some $0 \not = s^{*} \in S_r$.

Let $0 \leq i \leq r-1$. It is clear that
$0 : I^{r-i} \supseteq I^{i+1}$
because $I^{r+1} = 0$. To show the reverse inclusion
suppose that $0 : I^{r-i} \not\subseteq I^{i+1}$.
In this case there exists an element $z \in 0:I^{r-i}$ with
$z \in I^j \backslash I^{j+1}$ for some $0 \leq j \leq i$. Since
$0 \not = z^* \in I^j/I^{j+1}$ and $S=S_r=s^{*}k$, we can express
$s^* = z^* w^*$ for some $w^* \in I^{r-j}/I^{r-j+1}$. As $s^*
\not = 0$ it follows that $zw \not = 0$. This is impossible since
$z \in 0 : I^{r-i}$ and $w\in I^{r-j} \subseteq I^{r-i}$.

\smallskip

\noindent
(3) $\implies$ (4) : This is obvious.

\smallskip

\noindent
(4) $\implies$ (5) : For
$0 \leq i \leq \lfloor \frac{ r-1}{2} \rfloor$  we have
$$
\aligned
\lambda(G_{r-i}) &= \lambda(I^{r-i}/I^{r-i+1})\\
                 &= \lambda(R/I^{r-i+1}) - \lambda(R/I^{r-i}) \\
                 &= \lambda(0 : I^{r-(i-1)}) - \lambda(0 : I^{r-i}))
\quad \text{by \cite[Prop.3.2.12(b), p.103]{BH}} \\
                 &= \lambda(I^i) - \lambda(I^{i+1}) \quad
\text{by condition (4)}\\
                 &= \lambda(I^i/I^{i+1}) = \lambda(G_i).
\endaligned
$$

\smallskip

\noindent (5) $\implies$ (3) : If condition (5) holds for $0 \leq
i \leq \lfloor \frac{ r-1}{2} \rfloor$, then it obviously holds
for every $i$. For $0 \leq i \leq r-1$ we have
$$
\aligned
\lambda(I^{i+1}) &= \lambda(G_{i+1}) + \cdots + \lambda(G_r)\\
                 &= \lambda(G_{r-(i+1)})+ \cdots + \lambda(G_0)
\quad \text{by condition (5)} \\
                 &= \lambda(R/I^{r-i})  \\
                 &= \lambda(0:I^{r-i}). \\
 \endaligned
$$
As $I^{r+1} = 0$, we  have $I^{i+1} \subseteq 0:I^{r-i}$.
Since these two ideals have the same length, we conclude that
$I^{i+1} = 0: I^{r-i}$.

\smallskip

\noindent
$(3) \implies (6)$ : Let $1 \leq i \leq r-1$. The inclusion
$I^r : I^{r-i} \supseteq I^i$ is clear. To show $"\subseteq"$,
observe that
$$
I^r : I^{r-i} \subseteq I^{r+1} : I^{r-i+1} = 0 : I^{r-(i-1)} = I^i,
$$
where the last equality holds by condition (3).

\smallskip

\noindent $(6) \implies (2)$ : From (\ref{star})  we have  for
$0 \leq i \leq r-2$,
$$
S_i \subseteq \frac{I^{i+2} : I}{I^{i+1}} \subseteq
\frac{I^{i+2+(r-i-2)} : I^{1+(r-i-2)}}{I^{i+1}}
= \frac{I^{r} : I^{r-(i+1)}}{I^{i+1}}= \frac{I^{i+1}}{I^{i+1}}, \quad

$$
where the last equality follows from condition (6). Again by
(\ref{star}) and condition (6),
$$
S_{r-1} \subseteq \frac{0:I}{I^r} = \frac{I^r}{I^r}.
$$
Hence $S_i = 0$ for $0 \leq i \leq r-1$.

\smallskip

\noindent
$(6) \Longleftrightarrow (7)$ : This is clear.
                                         \QED

\medskip

\medskip

\setcounter{lem}{3}

\begin{cor}\label{3.3}
Let $(R,\m)$ be a zero-dimensional Gorenstein local ring
and let $I \subseteq \m$ be an ideal of $R$. Assume
that $I^r \ne 0$ and $I^{r+1} = 0$.
If   $G(I)$ is  Gorenstein, then
$I^i/I^{i+1}$ is a faithful module over $R/I$ for $0 \leq i \leq  r$.
\end{cor}

\demo
It suffices to show that $I^{i+1} : I^i = I$ for $0 \leq i \leq r$.
The inclusion $``\supseteq"$ is clear. To show the
inclusion $``\subseteq"$, observe that
we have
$$
I^{i+1} : I^i

\subseteq I^{r+1} : I^r = 0 : I^r = I,
$$
where the last equality follows from condition (3) of Theorem \ref{3.1}.
\QED

\medskip

\begin{remark} \label{3.4}
{\em
\begin{enumerate}
\item
The existence of a zero-dimensional Gorenstein local ring
$(R,\m)$ for which the associated graded ring
$G(\m)$ is not Gorenstein shows that
the converse of Corollary \ref{3.3} is not true. A specific
example illustrating this is given in Example \ref{3.5}.

\item
In general, suppose $I \subseteq \m$ is an ideal of a Noetherian
local ring $(R,\m)$. If the ideal $G(I)_+ = \oplus_{i \ge
1}(I^i/I^{i+1})$ of $G(I)$ has positive grade, then $I^i/I^{i+1}$
is a faithful module over $R/I$ for every $i \ge 0$. For if
$I^i/I^{i+1}$ is not faithful over $R/I$, then there exists $a \in
R \setminus I$ such that $aI^i\subseteq I^{i+1}$. Hence $aI^{i+j}
\subseteq I^{i+j+1}$ and $I^{i+j}/I^{i+j+1}$ is
not faithful over $R/I$ for every integer $j \geq 0$.
On the other hand, if $G(I)_+$ has
positive grade, then there exists a homogeneous $G(I)$-regular
element $b \in I^k/I^{k+1}$ for some positive integer $k$. Then
$b^\ell \in I^{k\ell}/I^{k\ell +1}$ is $G(I)$-regular, and
therefore $I^{k\ell}/I^{k\ell+1}$ is a faithful module over $R/I$
for every positive integer $\ell$.
\end{enumerate}}
\end{remark}

\medskip

\begin{exam}\label{3.5}{\em
Let $k$ be a field and $R = k[[t^5, t^6, t^9]]$. As observed by Sally
in \cite[Ex.3.6]{S},
$$
R \cong k[[X, Y, Z]]/(YZ - X^3, Z^2 - Y^3)
$$
and
$$
G(\m) \cong k[X, Y, Z]/(YZ, Z^2, Y^4 - ZX^3).
$$
Thus $(R,\m)$ is a  one-dimensional Gorenstein local domain
and $G(\m)$ is Cohen-Macaulay, but not Gorenstein.
It is readily seen that $J = t^5R$ \, is a minimal reduction of
$\m$ with $r_J(\m) = 3$. Sally also observes
that $\overline{R} = R/J$ is a zero-dimensional
Gorenstein local ring with maximal ideal
$\n = \m \overline{R}$ such that $G(\n)$ is
not Gorenstein. Indeed, the leading form $(t^5)^*$ of
$t^5$ in $\m/\m^2$ is  a regular element of $G(\m)$.
Therefore $G(\n) \cong G(\m)/(t^5)^*G(\m)$. It follows that
$G(\n)$ is not Gorenstein, $\n^3 \neq 0$ and $\n^4 = 0$.
Since $\n^i/\n^{i+1}$
is a vector space over $\overline{R}/\n$,  $\n^i/\n^{i+1}$
is a nonzero free $(\overline{R}/\n)$-module for $0 \leq i \leq 3$ and
hence in particular a faithful $(\overline{R}/\n)$-module.
The dimensions of the components of $G(\n) = R/\n \oplus
\n/\n^2 \oplus \n^2/\n^3 \oplus \n^3$ are $1, 2, 1, 1$. This
nonsymmetry reflects the fact that $G(\n)$ is not Gorenstein,
see Theorem \ref{3.1}.}
\end{exam}

We now turn to rings of arbitrary dimension. For this the following
lemma is needed.

\begin{lem}\label{3.6}
Let $(R,\m)$ be a $d$-dimensional Gorenstein local ring and let
$I$ be an $\m$-primary ideal. Assume that $J \subseteq I$ is a
reduction of $I$ with $\mu(J) = d$,  let $r_J(I)$ denote the
reduction number of $I$ with respect to $J$ and let $r$ be an
integer with $r \ge r_J(I)$. If $J:_{R} I^{r-i}  = J + I^{i+1}$
for every $i$ with $0 \le i \le r-1$, then $G(I)$ is
Cohen-Macaulay.
\end{lem}

\demo
The assertion is clear for $d = 0$. Next we assume
that  $d = 1$. By  Valabrega-Valla  \cite[Cor.2.7]{VV}, it
suffices  to verify that
$J \cap I^i = JI^{i-1}$ for $1 \le i \le r$. We first
prove that
\begin{equation}\label{claim} \tag{3.8}
I^r : I^{r-i} = I^i  \quad\quad {\rm for} \quad\quad 1 \le i \le r \,.
\end{equation}

To show (\ref{claim})
we proceed by induction on $i$. One has

$$
I^i \subseteq I^r : I^{r-i} \subseteq I^{r+1} : I^{r-i+1}
= JI^r : I^{r-(i-1)} \subseteq J : I^{r-(i-1)}
=J + I^{(i-1)+1}  = J + I^i.
$$

This proves the case $i=1$. Now assume $i \geq 2$. The containment
$I^r:I^{r-i} \subseteq J+I^i$ gives
$$
\aligned
I^r : I^{r-i} &= (J + I^i) \cap (I^r : I^{r-i}) \\
                    &= [J \cap (I^r : I^{r-i})] + I^i \quad \text{since $I^i \subseteq I^r:I^{r-i}$}\\
                    &= J[(I^r : I^{r-i}) : J] + I^i  \quad
\text{since $J$ is principal} \\
                    &= J(I^r :JI^{r-i})  + I^i \\
                    &\subseteq J(I^{r+1} : JI^{r-i+1}) + I^i \\
                    &= J(JI^r : JI^{r-(i-1)}) + I^i \\
                    &= J(I^r : I^{r-(i-1)}) + I^i  \quad
\text {since  $J$ is principal and regular} \\
                    &= JI^{i-1} + I^i \quad \quad \text{by the induction hypothesis}\\
                    &= I^i.
\endaligned
$$
This completes the proof of (\ref{claim}).

Now we have for $1 \leq i \leq r$,
$$
\aligned
J \cap I^i &= J(I^i : J) \quad \text {since $ J$ is principal} \\
           &\subseteq J(I^{r+1} : JI^{r+1-i}) \\
           &= J(JI^r : JI^{r-(i-1)}) \\
           &= J(I^r : I^{r-(i-1)}) \quad \text {since $J$ is principal and regular} \\
           &= JI^{i-1} \qquad \text {by (\ref{claim})}.
\endaligned
$$
Hence by the criterion of Valabrega-Valla, $G(I)$ is Cohen-Macaulay.
This completes the proof of Lemma \ref{3.6} in the case where $d = 1$.

Finally let $d \geq 2$. We may assume that the residue field of
$R$ is infinite. There exist elements $a_1, \ldots ,a_{d-1}$ that form
part of a minimal generating set of $J$ and a superficial sequence
for $I$. Write $\overline R = R/(a_1, \ldots, a_{d-1})R$ \, and
$\overline{I} = I \overline{R}$. Since $\dim \overline{R} = 1$,
we have that $G(\overline I)$ is a Cohen-Macaulay ring,
necessarily of dimension one. Hence by \cite[Lemma
2.2]{HM}, $G(I)$ is Cohen-Macaulay. \QED

\medskip

We are now ready to prove the main result of this section. Notice that we
do not require $G(I)$ to be Cohen-Macaulay in this theorem; instead, the
Cohen-Macaulayness is a consequence of the colon conditions (2) or (3) of
the theorem.

\setcounter{lem}{8}

\begin{thm}\label{3.8}
 Let $(R,\m)$ be a $d$-dimensional Gorenstein local
ring  and let $I$ be an $\m$-primary ideal. Assume that $J
\subseteq I$ is a reduction of $I$ with  $\mu(J) = d$, and let $r
= r_J(I)$ be the reduction number of $I$ with respect to $J$. Then
the following are equivalent $:$
\begin{enumerate}
\item $G(I)$ is Gorenstein. \item $J :_{R} I^{r-i} = J + I^{i+1}$
for $0 \leq i \leq r-1$. \item $J :_{R} I^{r-i} = J + I^{i+1}$ for
$0 \leq i \leq \lfloor \frac{ r-1}{2} \rfloor$.
\end{enumerate}
\end{thm}
\demo
The equivalence of items (2) and (3) follows from the double annihilator property
in the zero-dimensional Gorenstein local ring $R/J$, see for
instance \cite[(3.2.15), p.107]{BH}. To
prove the equivalence of (1) and (2),
by Lemma \ref{3.6}, we may assume that $G(I)$ is a Cohen-Macaulay ring.
Write $J=(a_1, \ldots, a_d)R$ and set $\overline R =R/J$,
$\overline I  =I/J$. Since $G(I)$ is Cohen-Macaulay and $J$ is
a minimal reduction of $I$, it follows that $a_1^*, \ldots, a_d^*$ form a regular
sequence on $G(I)$ and therefore
$$
G(I)/(a_1^*, \ldots, a_d^*)G(I) \cong G(I/J).
$$
In particular ${\overline I}^r \neq 0$. Hence $(\overline R, \overline \m)$
is a zero-dimensional Gorenstein local ring with
${\overline I}^r \neq 0$ and ${\overline I}^{r+1} = 0$.
Now
$$
\aligned
G(I) \quad \text {is Gorenstein}
&\Longleftrightarrow G(\overline I) \quad \text{is Gorenstein} \\
&\Longleftrightarrow 0 : {\overline I}^{r-i} = {\overline I}^{i+1}
 \quad \text {for } 0 \leq i \leq r-1  \text { by Theorem 3.1}\\
&\Longleftrightarrow J : I^{r-i} = J + I^{i+1}
\quad \text {for }  0 \leq i \leq r-1.\\
\endaligned
$$
This completes the proof of Theorem \ref{3.8}.
\QED

\medskip

We  record the following corollary to Theorem \ref{3.8}
for the case of reduction number two.

\begin{cor}\label{3.2}
Let $(R,\m)$ be a $d$-dimensional Gorenstein local ring and let
$I$ be an $\m$-primary ideal. Assume that $J \subseteq I$ is a
reduction of $I$ with $\mu(J) = d$ and that $r_J(I) = 2$, i.e.,
$JI \neq I^2$ and $JI^2 = I^3$. Then the following are equivalent
$:$
\begin{enumerate}
\item $ G(I)$ is  Gorenstein. \item $J:_{R} I^2 = I$.
\end{enumerate}
\end{cor}

\smallskip

The next corollary to Theorem \ref{3.8}  deals with the
problem of lifting the Gorenstein property of associated
graded rings.  Notice we are not assuming that $G(I)$ is
Cohen-Macaulay.

\begin{cor}\label{3.9}
Let $(R,\m)$ be a $d$-dimensional Cohen-Macaulay  local ring
and let $I$ be an $\m$-primary ideal. Assume that
$J \subseteq I$ is a reduction of $I$ with $\mu(J) = d$
and that $I^r \not\subseteq J$ for $r = r_J(I)$ the reduction
number of $I$ with respect to $J$. Set $\overline{I}: = I/J
\subseteq \overline{R}: = R/J$.
If $G(\overline{I})$ is Gorenstein, then $G(I)$ is Gorenstein.
\end{cor}
\demo If $G(\overline{I})$ is Gorenstein, then $\overline{R}$ and
hence $R$ are  Gorenstein \cite[p.121]{M}. Since $I^r
\not\subseteq J$, we have that  $r= r_J(I)$ is also the reduction
number of $\overline{I}$ with respect to the zero ideal. Now the
assertion follows from Theorem \ref{3.8}. \QED

\medskip

In Example \ref{3.11} we exhibit a one-dimensional Gorenstein local
domain $(R,\m)$,  an $\m$-primary ideal $I$ and a principal
reduction $J$ of $I$ such that  for
$\overline{I}: = I/J \subseteq \overline{R}: = R/J$, the associated
graded ring $G(\overline{I})$ is Gorenstein, while $G(I)$ is not Gorenstein.

\smallskip

\begin{exam}\label{3.11}{\em
Let $k$ be a field, $R = k[[t^4, t^5, t^6]]$ and $I =
(t^4, t^5)R$. Then $R$ is a one-dimensional
Gorenstein local domain, $\mu(I) = 2$ and
$J = t^4R$ is a principal reduction of $I$.
An easy computation shows that the reduction number $r_J(I) = 3$.
On the other hand,  $I^2 \subseteq J$. Hence the associated
graded ring $G(I)$ is not Cohen-Macaulay. We have
$\lambda(R/I) = \lambda(\overline{R}/\overline{I}) = 2$ and
$\lambda(\overline{R}) = 4$. Therefore,  by Theorem \ref{3.1},
$G(\overline{I}) = \overline{R}/\overline{I} \oplus \overline{I}$ is
Gorenstein.
On the other hand, the
principal reduction $J' = (t^4 - t^5)R$ of $I$ has the
property that $I^2$ is not contained in $J'$. Therefore the  associated
graded ring $G(I/J')$ has Hilbert function $2, 1, 1$ and
thus is not Gorenstein. }
\end{exam}

\medskip

\section{The Quasi-Gorenstein property of the extended Rees algebra }

In Theorem \ref{4.1} we give a general
characterization for when the extended Rees algebra
 $R[It, t^{-1}]$ is   quasi-Gorenstein.
In case $G(I)$ is Cohen-Macaulay, this characterization would also
follow from \cite[Thm.5.3]{GI}.  In fact, if $G(I)$ is
Cohen-Macaulay, then the quasi-Gorensteinness of $R[It, t^{-1}]$
is equivalent to the Gorensteinness of $G(I)$, since a
Cohen-Macaulay quasi-Gorenstein ring is Gorenstein and $G(I) =
R[It, t^{-1}]/(t^{-1})$  is Cohen-Macaulay (resp. Gorenstein)  if
and only if $R[It, t^{-1}]$ is Cohen-Macaulay (resp. Gorenstein).

For an ideal $I$ of a ring $R$ and an integer $i$,
we define $I^i = R$ if $i \le 0$.

\medskip

\begin{thm}\label{4.1} Let $(R,\m)$ be a Gorenstein local ring
and let $I \subseteq \m$ be an ideal with $\hgt(I) = g > 0$.
Assume that $J \subseteq I$ is a reduction of $I$ with
$\mu(J) = g$. Let $k$ be an integer with
$k \ge r:=r_J(I)$,  the reduction number
of $I$ with respect to $J$,  and let $B = R[It, t^{-1}]$ be
the extended Rees algebra of the ideal $I$. Then the graded
canonical module $\omega_B$ of $B$ has the form
$$
\omega_B = \bigoplus_{i \in \Z}(J^{i+k}:_RI^k)t^{i+g-1}.
$$
In particular, for $a \in \Z$,  the following are equivalent $:$
\begin{enumerate}
\item
$R[It, t^{-1}]$ is quasi-Gorenstein with {\bf a}-invariant $a$.
\item
$J^i:_RI^k = I^{i+a-k+g-1}$ for every $i \in \Z$.
\end{enumerate}
\end{thm}

\demo  Let $K = \Quot(R)$ denote the total ring of quotients of $R$
and let $A = R[Jt, t^{-1}] \subseteq B = R[It, t^{-1}]$.
Notice that $A/(t^{-1}) \cong G(J)$, where $t^{-1}$ is a
homogeneous $A$-regular element of degree $-1$. Moreover,
since $J$ is generated by a regular sequence, $G(J)$ is a standard
graded polynomial ring in $g$ variables over the Gorenstein
local ring $R/J$. Thus $A$ is Cohen-Macaulay and
$\omega_A \cong A(-g+1) \cong At^{g-1}$.

The extension $A \subseteq B$ is finite and
$\Quot(A) = \Quot(B) = K(t)$ since $g > 0$. Therefore
\[
\omega_B \cong \Hom_A(B,\omega_A) \cong
At^{g-1}:_{K(t)}B = (A:_{K(t)}B)t^{g-1}
= (A :_{R[t, t^{-1}]}  B)t^{g-1},
\]
where the last equality holds because
$$
A:_{K(t)}B \subseteq A:_{K(t)} 1 \subseteq A
\subseteq R[t,t^{-1}].
$$
We may now make the identification
$\omega_B = (A:_{R[t,t^{-1}]}B)t^{g-1}$.

Let $i$ and $j$ be any integers. Since $J$ is a complete
intersection, it follows that $J^{i+j+1}:_R J = J^{i+j}$.
Hence
$$
J^{i+j}:_R I^j = (J^{i+j+1}:_RJ):_RI^j = J^{i+j+1}:_RJI^j
\supseteq J^{i+j+1}:_RI^{j+1},
$$
where the last inclusion is an equality whenever $j \ge k \ge r_J(I)$. Therefore
$$
\cap_{j \in \Z}(J^{i+j}:_R I^j) = J^{i+k}:_RI^k.
$$

We conclude that

$$
[A:_{R[t,t^{-1}]}B]_i = (\cap_{j \in \Z}(J^{i+j}:_R I^j))t^i
= (J^{i+k}:_RI^k)t^i,
$$
which gives
$$
\omega_B = \bigoplus_{i \in \Z}(J^{i+k}:_RI^k)t^{i+g-1}.
$$
This description shows in particular that $[\omega_B]_i = Rt^i$ for
$i << 0$.

Now $B$ is quasi-Gorenstein with {\bf a}-invariant $a$ if and only
if $\omega_B \cong B(a)$, or equivalently, $\omega_B = \alpha Bt^{-a}$
for some unit $\alpha$ of $K$. As $[\omega_B]_i = Rt^i$ for $i << 0$,
it follows that $\alpha$ is necessarily a unit in $R$. Thus
$\omega_B \cong B(a)$ if and only if
$$
\oplus_{i \in \Z}(J^{i+k}:_R I^k)t^{i+g-1} = \oplus_{i \in \Z}I^it^{i-a},
$$
which means that $J^i:_RI^k = I^{i+a-k+g-1}$ whenever $i \in \Z$.\QED

\medskip

\begin{cor} \label{4.2}
With notation as in Theorem $\ref{4.1}$, the following are
equivalent $:$
\begin{enumerate}
\item
The extended Rees algebra $R[It, t^{-1}]$ is quasi-Gorenstein.

\item There exists an  integer $u$ such that  $J^i:_{R} I^r =
I^{i-u}$ for every $i \in \Z$.
\end{enumerate}
If these equivalent conditions hold, then $u = r -g +1 -a \geq 0$ with
$a$ denoting the $\bf a$-invariant of $R[It,t^{-1}]$.
\end{cor}

\demo  To prove the last assertion notice that $u$ is uniqely determined since
$I$ is not nilpotent. Thus  part (2) of Theorem \ref{4.1} gives $u=r-g+1-a$. The
inequality $u \geq 0$ can be seen by setting $i=0$ in part (2) of the
present corollary.  \QED

\medskip

\begin{cor} \label{4.21}
With notation as in Theorem $\ref{4.1}$, $I^n(J^i:_{R} I^r)
\subseteq J^{n+i}:_{R} I^r$ for all integers $n$, $i$, and
$J^n(J^i:_{R} I^r) = J^{n+i}:_{R} I^r$ for every $n \geq 0$ and $i
>> 0$.
\end{cor}

\demo
This is clear since
$ \omega_B = \oplus_{i \in \Z}(J^{i+r}:_RI^r)t^{i+g-1} $
is a graded module over $B = R[It, t^{-1}]$ that  is finitely
generated over $A = R[Jt, t^{-1}]$. \QED

\medskip

\begin{remark} \label{4.3} {\em
In the setting of Theorem \ref{4.1}  we define the
{\it index of nilpotency } of $I$ with respect to $J$ to be
$s_J(I)  = \min\{i \, | \, I^{i + 1} \subseteq J \}$.

Suppose $R[It, t^{-1}]$ is quasi-Gorenstein and let $u = r-g+1-a \geq 0$ be
as in Corollary \ref{4.2}.
The corollary implies $J:I^r = I^{1-u}$ and hence $I^{1-u+r} \subseteq
J$. Thus $1-u+r \ge s_J(I) + 1$, so $r_J(I) - s_J(I) \ge u \geq 0$.
Therefore the ${\bf a}$-invariant $a$ of $R[It, t^{-1}]$ satisfies
$s_J(I) - g + 1 \le a \le r_J(I) -g + 1$. Let $w = \max\{n \, | \,
I^r \subseteq J^n \}$. Then $w = u$; for $I^{w-u}=J^w:I^r = R$
implies $u \ge w$, while $I^r \not\subseteq J^{w+1}$ implies
$I^{w+1-u}= J^{w+1}:I^r \subsetneq  R$, so $w \ge u$. It follows
that  $s_J(I) =  r_J(I)$ if and only if  $u = 0$ if and only
if $a = r_J(I) - g +1$. These
equalities hold in case $G(I)$, or equivalently $R[It,t^{-1}]$,
is  Cohen-Macaulay. }
\end{remark}

\medskip

Since in dimension two, quasi-Gorenstein is equivalent to
Gorenstein, we have the following corollary:

\begin{cor} \label{4.4}
Let $(R,\m)$ be a one-dimensional Gorenstein local ring and let
$I$ be an $\m$-primary ideal. Assume that $J$ is a principal
reduction of $I$ and that $r = r_J(I)$ is the reduction number of
$I$ with respect to $J$. Then the following are equivalent $:$
\begin{enumerate}
\item $G(I)$ is Gorenstein. \item $R[It, t^{-1}]$ is Gorenstein.
\item There exists an integer $u$ such that $J^i:_{R} I^r =
I^{i-u}$ for every $i \in \Z$. \item $J^i:_{R} I^r = I^i$ \, for
every $i \in \Z$. \item $J^i:_{R} I^r = I^i$ \, for $1 \leq i \leq
r$.
\end{enumerate}
\end{cor}

\demo
It remains to prove that (5) implies (4). This
follows since for $i > r$
we have $I^i = J^{i-r}I^r = J^{i-r}(J^r:I^r)
= J^i:I^r$,
where the last equality holds because $J$ is
principal generated by a regular element.
\QED

\begin{remark} \label{4.6} {\em
With notation as in Corollary \ref{4.4}, we have:
\begin{enumerate}
\item  As is well-known in the one-dimensional case, the reduction
number $r=r_J(I)$ is independent of the principal reduction
$J=xR$. Furthermore the ideal $J^r:_RI^r$ is also independent of
$J$. Indeed, the subring $R[I/x] = I^r/x^r =
\bigcup_{i=1}^\infty(I^i:_KI^i)$ of $K : =\Quot(R)$ is the blowup
of $I$, so is independent of $J$, and $J^r:_RI^r=J^r:_KI^r$ is the
conductor of $R[I/x]$ into $R$. \item If $I^r \subseteq J$, then
$I^r \subseteq J'$ for every reduction $J'$ of $I$, that is, $I^r$
is contained in the core of $I$. Indeed, we have $r>0$ and then
$I^r \subseteq J \iff J^{r-1}I^r \subseteq J^r \iff I^{2r-1}
\subseteq J^r \iff I^{r-1} \subseteq  J^r:I^r$. Notice that the
latter ideal is independent of $J$ by item (1). Therefore  if the
index of nilpotency $s_J(I) = r$ for one principal reduction $J$,
then $s_{J'}(I) = r$ for every principal reduction $J'$ of $I$. On
the other hand, if $s_J(I) < r$ there may exist a principal
reduction $J'$ of $I$ such that $s_J(I) \ne s_{J'}(I)$ as is
illustrated in Example \ref{3.11}. \item If $J^r:I^r = I^r$, then
$I^r \not\subseteq J$ and $s_J(I) = r$. Indeed, we may assume
$r>0$,  hence $I^{r-1}  \not\subseteq I^r$ and by item (2),
$I^{r-1} \not\subseteq  J^r:I^r \iff I^r \not\subseteq J$.
\end{enumerate}
}
\end{remark}

\medskip

Let $R$ be a Noetherian ring and $I$ an $R$-ideal containing a
regular element. The ideal $\widetilde I =
\bigcup_{i=1}^\infty(I^{i+1}:_RI^i)$ first studied by Ratliff and
Rush in \cite{RR} is called the {\it Ratliff-Rush ideal associated
to}  $I$, and $I$ is said to be a {\it Ratliff-Rush ideal}  if $I
= \widetilde I$. It turns out that the ideal $G(I)_+$ of $G(I)$
has positive grade if and only if all powers of $I$ are
Ratliff-Rush ideals \cite[(1.2)]{HLS}. Thus if $\dim R = 1$, then
$G(I)$ is Cohen-Macaulay if and only if all powers of $I$ are
Ratliff-Rush ideals.

\begin{remark} \label{4.9} {\em
Let $(R,\m)$ be a Gorenstein local ring and $I \subseteq \m$ an
ideal with $\hgt(I) = g > 0$. As in Theorem \ref{4.1}, assume
there exists a reduction $J$  of $I$ with $\mu(J) = g$. Let $C =
\bigoplus_{i \in \Z}\widetilde{I^i}t^i$ denote the extended Rees
algebra of the Ratliff-Rush filtration associated to $I$, and let
$k \geq 0$ be an integer such that $J^j \widetilde{I^k}=
\widetilde{I^{j+k}}$ for every $j \geq 0$. The following are
equivalent $:$
\begin{enumerate}
\item $C$ is quasi-Gorenstein with ${\bf a}$-invariant $b$. \item
$J^i:_{R} \widetilde{I^k} = \widetilde{I^{i+b-k+g-1}}$ for every
$i \in \Z$.
\end{enumerate}
To show this equivalence one

proceeds as in the proof of Theorem \ref{4.1}
}
\end{remark}

\begin{cor} \label{4.61}
With notation as in Corollary  $\ref{4.4}$, if $J^r:_{R} I^r =
I^{r-u}$ for some integer $u \ge 0$, then $u = 0$. If, in
addition, all powers of $I$ are Ratliff-Rush ideals, then $G(I)$
is Gorenstein.
\end{cor}

\demo
The equality
$J^r:I^r = I^{r-u}$ implies $I^{r+1-u} = I(J^r:I^r)$
and by Corollary \ref{4.21}, $I(J^r:I^r)
\subseteq J^{r+1}:I^r$. On the other hand,
as $J$ is generated by a regular element,
$J^{r+1}:I^r =  J(J^r:I^r) = JI^{r-u}$.
Therefore $I^{r+1-u} = JI^{r-u}$,  and $u = 0$
since $r$ is minimal such that $I^{r+1} = JI^r$.
By Corollary \ref{4.4}, to show $G(I)$ is Gorenstein
it suffices to show
$J^i:I^r = I^i$ for $1 \leq i \leq r$.
Again according to Corollary \ref{4.21}, $I^{r-i}(J^i:I^r) \subseteq J^r:I^r$.
Since $J^r:I^r = I^r$, it follows that
$J^i:I^r  \subseteq I^r:I^{r-i}$. We always have
$I^r:I^{r-i} \subseteq \widetilde{I^i}$. Therefore,  if in
addition,  $\widetilde{I^i} = I^i$, then  $G(I)$ is  Gorenstein.
\QED

\medskip

With notation as in Corollary \ref{4.4}, it can happen that
$J^r:_{R} I^r = I^r$ and yet $G(I)$ is  not Cohen-Macaulay.  We
illustrate this in Example \ref{4.5}.

\begin{exam}\label{4.5}{\em
Let $k$ be a field, $R = k[[t^5, t^6, t^7, t^8]]$ and $I =
(t^5, t^6, t^7)R$. Then $R$ is a one-dimensional
Gorenstein local domain with integral closure
$\overline R = k[[t]]$ and
$J = t^5R$ is a reduction of $I$.
We have $JI \subsetneq I^2 = \m^2 = t^{10}\overline R$ and $JI^2 = I^3$.
Hence $r_J(I) = 2$. Since $t^8 \not\in I$, but
$t^8I \subseteq I^2$, $G(I)$ is not Cohen-Macaulay.
To see that $J^2:I^2 = I^2$, observe that
$t^{19} \not\in J^2$ and that for each integer $i$
with $5 \le i \le 8$ we have
$t^{19-i} \in I^2$ and
$t^{19} = t^{19-i}t^i$. Notice also that the
Ratliff-Rush ideal $\widetilde{I}$ associated to
$I$ is $\m = (t^5, t^6, t^7, t^8)R$, $r_J(\m) = r_J(I) = 2$,
$J^i:\m^2 = \m^i$ for every $i$ (equivalently, for
$1 \leq i \leq 2$), and hence $G(\m)$ is
Gorenstein by Corollary \ref{4.4}.
}
\end{exam}

\begin{exam}\label{4.51}{\em
Let $R = k[[t^4, t^5, t^6]]$,  $I = (t^4, t^5)R$ \,
and $J = t^4R$ \, be as
in Example \ref{3.11}.
The Ratliff-Rush ideal
$\widetilde{I}$ associated to $I$ is $\m=(t^4,t^5,t^6)R$,
and in fact
$\widetilde{I^i} = \m^i$ for every $i \in \Z$.
We have $r_J(I) = 3$ while $r_J(\m) = 2$.  Also
$I^3 = \m^3$. We have
$J^i:I^3 = \m^{i-1}$ while $J^i:\m^2 = \m^i$
for every $i \in \Z$. In particular $G(\m)$
is Gorenstein.
}
\end{exam}

\begin{ques}\label{4.7} {\em
With notation as in  Theorem \ref{4.1}, is
the extended Rees algebra $R[It, t^{-1}]$ Gorenstein
if it is quasi-Gorenstein?  Equivalently, is the
associated graded ring $G(I)$ then Gorenstein? }
\end{ques}

If $\dim R = 2$ and $R$ is pseudo-rational in the sense of
\cite[p.102]{LT}, we observe an affirmative answer to
Question~\ref{4.7}.

\begin{cor} \label{4.8}
Let $(R, \m)$  be a $2$-dimensional pseudo-rational Gorenstein
local ring and let $I$ be an $\m$-primary ideal.
If $B = R[It, t^{-1}]$ is quasi-Gorenstein,
then $B$ is Gorenstein. In particular, if $R$
is a $2$-dimensional regular local ring, then every
extended Rees algebra $R[It, t^{-1}]$ that is
quasi-Gorenstein is Gorenstein.
\end{cor}

\demo We may assume that $R/\m$ is infinite. Let
$J \subseteq I$ be a minimal reduction of $I$
and let $r = r_J(I)$. If $J = I$, then $I$ is
generated by a regular sequence and $R[It, t^{-1}]$ is
Gorenstein. Thus we may assume $J \subsetneq I$.
By \cite[Cor.5.4]{LT}, \,
$\overline{I^{r+1}} \subseteq J^r$, where
$\overline{I^{r+1}}$ denotes the integral closure
of $I^{r+1}$. In particular $\overline{I}I^r \subseteq
J^r$, which gives $\overline{I} \subseteq J^r:I^r$.
By Corollary \ref{4.2}, there exists
an integer $u$ such that $J^i:I^r = I^{i-u}$
for every $i \in \Z$. We have $u < r$ according to Remark \ref{4.3}
since $J \neq I$. Thus $J^r:I^r \subseteq J^{u+1}:I^r$.
Therefore $\overline{I} \subseteq J^r:I^r \subseteq J^{u+1}:I^r = I$,
which shows that $I = \overline{I}$ is integrally closed.
Now \cite[Cor.5.4]{LT} implies $r \le 1$. Therefore
$B$ is Cohen-Macaulay by \cite[Prop.3.1]{VV} and hence
Gorenstein. \QED

\bigskip
\bigskip


TITLE: Showing $\int_{\Gamma_R} \frac{\ln(z)}{z^4+1}dz = 0$
QUESTION [0 upvotes]: I need help showing:
$$\int_{\Gamma_R} \frac{\ln(z)}{z^4+1}dz = 0$$
where $\Gamma_R$ is the semi-circle with radius R centered at the origin.
I wanted to use the estimation lemma to prove that this integral goes to zero, but I'm having trouble. I'm aware the lemma is:
$$\left|\int_{\Gamma_R} \frac{\ln(z)}{z^4+1}dz\right| \le M*l(\Gamma_R)$$
I know $l(\Gamma_R)$ is $\pi R$ but I'm not sure how to go about finding M. I've looked at other posts using the triangle inequality, but I'm not sure how to use it here. Any help that is in pretty laymen's-terms would be much appreciative!

REPLY [1 votes]: Here I choose $R>1$ . Notice that for $R_1>1$ the paths $\Gamma_{R}$ is homotopic to $\Gamma_{R_1}$. So that
$$\int_{\Gamma_{R}} \frac{\ln(z)}{z^4+1}dz =\int_{\Gamma_{R_1}} \frac{\ln(z)}{z^4+1}dz $$
Now $$|\int_{\Gamma_{R_1}} \frac{\ln(z)}{z^4+1}dz|≤length(\Gamma_{R_1})\  sup_{z\in \Gamma_{R_1}} |f(z)| $$
Next $$sup_{z\in \Gamma_{R_1}} |f(z)|=sup_{z\in \Gamma_{R_1}}\frac{|ln|z|+\iota arg(z)|}{|z^4+1|} ≤    sup_{z\in \Gamma_{R_1}}\frac{|ln|z||+c}{|z^4|-1}=\frac{ln(R_1)+c}{R_1^4-1}  $$
Therefore  $$    |\int_{\Gamma_{R}} \frac{\ln(z)}{z^4+1}dz| =    |\int_{\Gamma_{R_1}} \frac{\ln(z)}{z^4+1}dz|≤length(\Gamma_{R_1})\  sup_{z\in \Gamma_{R_1}} |f(z)| ≤\pi R_1  \frac{ln(R_1)+c}{R_1^4-1}\rightarrow 0\ as \ R_1\rightarrow \infty$$
Here $$ln:\Bbb C-\{r(cos(\alpha)+\iota sin(\alpha)): r≥0, \alpha=-\pi/2\}\rightarrow \Bbb C,\ ln(z)=ln(|z|)+\iota arg(z)$$ and $$arg:\Bbb C-\{r(cos(\alpha)+\iota sin(\alpha)): r≥0, \alpha=-\pi/2\}\rightarrow (-\pi/2,2\pi-\pi/2)$$  are continuous functions. And $c$ is some fixed constant such that $c≥|arg(z)|$ on $\Gamma_{R_1}$ and for each $R_1>1$. Hence
$$\int_{\Gamma_R} \frac{\ln(z)}{z^4+1}dz = 0$$


TITLE: How to show that..., $A(C_1)=\{z:|z-1|\leq 1, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$?
QUESTION [0 upvotes]: Let $C_1=\{z:|z|\leq 1, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$ and $A(z)=z-1$. Define $A(C_1)$. How to show that $A(C_1)\neq\{z:|z-1|\leq 1, \arg(z)=\theta \in [\frac{3\pi}{4},\frac{5\pi}{4}] \}$, but $A(C_1)=\{z:|z-1|\leq 1, \arg(z)=\theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$?
My solution for $A(C_1)$:
$A(z)$ is translation, so called elementary function, which translates set $C_1$ one unit to the negative direction of real axis. Thus  $A(C_1)=\{z:|z-1|\leq 1, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}] \}$. Comments on my solution? I'm not sure whether my solution is appropriate proof.

REPLY [0 votes]: The function $A$ is a translation, hence it takes any subset of $\mathbb C$ and it moves it in certain direction; in our case it moves our subset (that is the left half of the closed unit circle) horizontally, shifting it backward of exactly one unit. No more: in particular $A$ is not a rotation, hence the left half of the closed unit circle will remain the left half of a unit circle (but not centered in the origin!), so you won't change the argument of the complex number of the image (under $A$) of your subset: this is why $A(C_1)\neq$ the first set you wrote (because of the argument conditions).
But pay attention! $|z-1|\le1$ denotes the point of the complex plane whose distance from $z_0=1$ is less or equal than $r=1$; so you would have translated your circle forward, not backward! But the translation takes the points backward, so you must consider the point whose distance from $z_0=-1$ is less or equal than $r=1$, i.e. $|z-(-1)|\le1$. Finally you have
$$
A(C_1)=\{z\in\mathbb C\;:|z+1|\le1,\;\theta=\arg z\in[\pi/2,3\pi/2]\}\;.
$$


TITLE: Contour integral - Circle instead of a square
QUESTION [0 upvotes]: I would like to solve the integral $\int_{\gamma} \frac{1}{z \bar{z}} dz$. Here $\gamma \subset \mathbb{C}$ is a square centered at the origin and where his vertices are parallel to the axes. Could I use the Cauchy theorem in using the fact we will change the contour to a unit circle center to the origin?

REPLY [1 votes]: The integrand is not holomorphic, you can't use Cauchy integral theorem.
However, if you parametric $\gamma$ by $[0,2\pi] \ni \theta \mapsto z(\theta) = r(\theta) e^{i\theta}$ by a suitable chosen real valued function $r(\theta)$.
The integral becomes
$$\int_{\gamma} \frac{dz}{z\bar{z}} = \int_0^{2\pi} \left( \frac{r'(\theta)+ir(\theta)}{r(\theta)^2} \right) e^{i\theta} d\theta$$
Notice $r(\theta+\pi) = r(\theta)$ while $e^{i(\theta+\pi)} = -e^{i\theta}$. the contribution from $[\pi,2\pi]$ in last integral cancel the one from $[0,\pi]$. So the integral is $0$.


TITLE: Reference request: Partition of unity...
QUESTION [0 upvotes]: I was looking for some material that could help me understand a real analysis course (1st year undergraduate). My teacher treated the following topics:
Partition of unity
Existence of regular functions on compact support
Dyadic covering and Paley Littlewood's partition of unit.
Morse's lemma.
Do you know any book or note where I can find a simple introduction to these topics?
Thank you!

REPLY [1 votes]: Shifrin's differential geometry notes contain most of those and are free on the web. math.uga.edu/~shifrin/ShifrinDiffGeo.pdf


TITLE: Combinatorics problem: Splitting 90 students into 3 classes
QUESTION [1 upvotes]: A group of 90 students is to be split at random into 3 classes of equal size. All partitions are equally likely. Joe and Jane are members of the 90-student group. Find the probability that Joe and Jane end up in the same class.
Answer:

Another way of thinking of this:
The different ways of partitioning into 3 partitions, each of size 30 is:
all = 90! / (30!)^3
The number of ways in which Joe and Jane can be at the same class is:
x = 88! / (30! * 30! * 28!)
and since there are 3 classes, the answer is 3 times the above expression:
y = 3 * x
Final answer:
y / all
This way of thinking works.

Yet, another way:
We can consider both Joe and Jane to be 1 student, in other words, think of them as one unit, and consider the number of students to be 89, and the partitions to be 30, 30, and 29.
This way, the number of ways in which they would be together:
z = 89! / (30! * 30! * 29!)
and the final answer would be z / all.
Unfortunately, this is not true.
you can see that z = x * 89/29 = x * 3.06896551724 which is not equal to x * 3.
What's wrong with the last logic?

REPLY [1 votes]: Whatever class Joe is in, there are 29 other students.  Jane can be selected to be one among these with a probability of $29/89$.
This may also be viewed as: there are $\binom 31\binom{30}2$ ways to select two seats in the same class, and $\binom{90}{2}$ ways to select two seats from anywhere among the three classes. So the probability for obtaining seats in the same class when selecting two from the ninety seats, without bias, for Joe and Jane is: $$\dfrac{3\cdot 30\cdot 29/2}{90\cdot 89/2}$$

What is wrong with the last logic?

We can consider both Joe and Jane to be 1 student,

No.  We cannot.  If they were super-glued together, they would certainly be in the same class.


TITLE: A child sits at a computer keyboard and types each of the 26 letters of the alphabet exactly once, in a random order.
QUESTION [2 upvotes]: A child sits at a computer keyboard and types each of the 26 letters of the alphabet
exactly once, in a random order.
How many independent children typists would you need such that the probability
that the word ‘exam’ appears is at least 0.9?
Probability of getting EXAM is 23!/26!. So I decided to do (23!/26!)^n ≥ 0.9. Is that correct ?

REPLY [0 votes]: You're right. The probability of getting "EXAM" is $p=\frac{23!}{26!}$. Now, suppose, at least $n$ children are needed. If the total number of occurings we get is $X$, then $P(X=r)={n \choose r} \cdot p^r \cdot (1-p)^{(n-r)}$, according to the theory of Barnoulli Trials. Now, $P(X \geq 1) \geq 0.9 \implies P(X=0) \leq 0.1 \implies {n \choose 0} \cdot p^0 \cdot (1-p)^{(n-0)} \leq 0.1 \implies n \log { \left(1-\frac{23!}{26!}\right)} \leq \log 0.1 \implies n \geq 35920$
So, you'll need at least $35920$ children.


TITLE: Integration involving natural logarithm
QUESTION [3 upvotes]: I tried possible way of getting the integral of the following integrand:
$$\int_0^{\infty} \frac{\ln\left(x+\frac{1}{x}\right)}{1+x^2}~dx$$
Firstly, I tried to write $\ln\left(x+\frac{1}{x}\right)$ as $\ln(x^2+1)-\ln(x)$ and tried to apply biparts but couldn't take it to its destination solution.
Secondly, I tried to substitute $t=x+\frac{1}{x}$ and proceeded, but couldn't solve further.
Nothing seems to work. This was a question from my exam. I don't understand how I'm supposed to solve this question in a short time when I can't solve it at home. I'm guessing there has to be a short but smart solution that is pretty much not striking me.

REPLY [1 votes]: First step is put $x=\tan (t) $ that substitution  gets rid of the denominator . Then after that we have the integral as $\int _0 ^{\frac {\pi}{2}} \log ( 1+\tan^2 (t))-\log(tan (t)) $ now use $1+tan^2(t)=sec^2 (t)$ . Now as in $0-\pi/2$ sec is positive we can write $\ln (\sec^2 (t))=2\ln (\sec (t))$ . Thus the integral is $\int _0 ^{\frac{\pi}{2}} log (\frac{\sec(t)}{\tan (t)})=\int _0 ^{\frac {\pi}{2}}-\log(\sin (t)) $ . Now I am sure you can handle this integral. The result should be $\frac {\pi}{}\log (2) $


TITLE: Using the definition of sequence convergence, prove that if $\lim{y_n}=2$, then $\lim{3(y_n)^2−2}=10$
QUESTION [3 upvotes]: Problem: Using the definition of sequence convergence, prove that if $\lim{y_n}=2$, then $\lim{3(y_n)^2−2}=10$.
Note: not allowed to use the Algebraic Limit Theorem, ONLY allowed to use the definition of sequence convergence.
So I understand how to work with the epsilon proof structure when I'm just trying to show one sequence converges, but I'm confused here.
$\lim{3(y_n)^2−2}=10$ in my proof becomes $|{(3(y_n)^2−2)-10|=|3(y_n+2)(y_n-2)|}$.  What I am struggling with is how to go from that to the ending, $< \epsilon$.
I obviously use $\lim{y_n}=2$ so $|y_n - 2|$ < something, although I'm not quite sure what to make it less than.
Can anybody help me with my proof structure here?

REPLY [1 votes]: 0) Since $y_n$ converges to $2$ there is a $n_0$ s.t. for $n \ge n_0$
$|y_n-2|<1$.
1) $|3((y_n-2)+2)^2-2-10|=$
$|3(y_n-2)^2+12(y_n-2)|\le$
$|3(y_n-2)^2|+ 12|(y_n-2)| \le$
$3|y_n-2| +12|y_n-2|=15|y_n-2|$ for $n\ge n_0$.
Let $\epsilon/15$ be given.
Since $y_n$ converges to $2$, there is a $n_1$ s.t. for $n\ge n_1$
$|y_n-2| \lt \epsilon/15.$
For $n \ge \max(n_0,n_1) $
$|3y_n^2-12| \le 15|y_n-2|\lt 15(\epsilon/15) =\epsilon.$
Used: For $|z| <1:$ $|z|^2 \le |z|$.


TITLE: Is an electron occupying the entire space of its cloud/orbital?
QUESTION [2 upvotes]: It is known that the electrons aren't really at any one place at any time at all. Instead they exist as a sort of cloud and "circulate" around the nucleus in their orbitals. The high-school textbooks say that "the electrons don't strictly follow a set path around the nucleus. Electrons move in every direction, but they are limited to their own area".
Assume I am an electron and my house is a nucleus, and I could fly around my house in an similar way the electrons do... I would then try to avoid the chimney-part of my house for example.
What I wonder is:

Will the electron certainly occupy all the physical points of its own cloud/orbital, given long enough time,

or

is the electron moving absolutely chotically in its orbital without any specific trajectory - might even stay and vibrate at the same "place" for some time, and might there be "places"/points in the orbital that the electron would never happen to "visit" in its motion around the nucleus?

I hope the questions are understandable!

REPLY [1 votes]: The cloud (or orbital ) , that you are talking of , is actually the space vector for which the ψ ( of Schrodinger's equation ) has some solutions .The electron wave is shown as somewhat like the dispered wave over that space . However , when we actually do find the position for some instants , the space vector and wavefunction collapses .
Now , as opposite to above , where ψ is orthogonal , that space , plane or point are called nodes .
How does electron traverse between orbitals through nodes , how does it traverse inside the cloud ? Are all ongoing topics going on in the quantum mechanics .