text:
\section{Introduction}

The \emph{Auslander-Reiten conjecture} \cite{AR} claims that, over an Artin algebra $R$, if $M$ is a finitely generated $R$-module such that $\Ext^{>0}_{R}(M, M\oplus R)=0$, then $M$ is projective.
This long-standing conjecture can be stated over any Noetherian ring $R$ and is known to be true over several classes of algebras.
For example, Auslander and Reiten \cite{AR} proved the conjecture for algebras of finite representation type.
Hoshino \cite{Hos} determined another noteworthy class; it holds for symmetric Artin algebras with radical cube zero.
The Auslander-Reiten conjecture is closely related to other important conjectures such as the \emph{Tachikawa conjecture} \cite{T} and the \emph{finitistic dimension conjecture} \cite{Happel}.
Indeed it is rooted in a conjecture of Nakayama \cite{Nak} and the one known as the \emph{generalized Nakayama conjecture} \cite{AR}: If $R$ is an Artin algebra, then every indecomposable injective $R$-module occurs as a direct summand of one of the terms of the minimal injective resolution of $R$.
We refer the reader to \cite{ABS,Buchweitz2,CH1,Di,DP,Wei2,Wei1,Yamagata} for more information on the relation of these homological conjectures.

Christensen and Holm \cite{CH1} proved that the Auslander-Reiten conjecture holds over rings satisfying the Auslander condition.
More precisely, they proved that the implication
$$
\xymatrix{
\mathsf{(AC)} \ar@{=>}[r] & \mathsf{(ARC)}
}
$$
holds for a (left) Noetherian ring.
Here, $\mathsf{(AC)}$ and $\mathsf{(ARC)}$ denote the {\em Auslander condition} and the {\em Auslander-Reiten condition}, which are conditions defined for a fixed Noetherian ring $R$, as follows.
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(AC)}$]
For every finitely generated $R$-module $M$ there exists a nonnegative integer $b_{M}$ such that for every finitely generated $R$-module $N$ one has: if $\Ext^{\gg 0}_{R}(M,N)=0$, then $\Ext^{>b_{M}}_{R}(M,N)=0$.
\item[$\mathsf{(ARC)}$]
For every finitely generated $R$-module $M$, if $\Ext^{>0}_{R}(M,M\oplus R)=0$, then $M$ is projective.
\end{enumerate}
\end{quote}
Although there exist rings that do not satisfy $\mathsf{(AC)}$ \cite{JoS}, this result gave new insight on the Auslander-Reiten conjecture and other related homological conjectures.

Commutative Noetherian local rings known to satisfy $\mathsf{(ARC)}$ include Gorenstein local rings of codimension at most four \cite{S}.
Recently Araya \cite{A} proved that if all Gorenstein local rings of dimension at most one satisfy $\mathsf{(ARC)}$ so do all Gorenstein local rings.
The main tool he used for the proof of his result is a duality theorem for Gorenstein local rings, which is known as the {\em Auslander-Reiten duality} theorem (cf. \cite{Au2,Buchweitz,Y}).
On the other hand, as a natural generalization of $\mathsf{(AC)}$, the following {\em generalized Auslander-Reiten condition} has also been investigated by several authors \cite{Di,Wei2,Wei1}.
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(GARC)}$]
For every finitely generated $R$-module $M$, if there exists a nonnegative integer $n$ such that $\Ext^{>n}_{R}(M,M\oplus R)=0$, then $M$ has projective dimension at most $n$.
\end{enumerate}
\end{quote}

In this paper, we consider refining the Auslander-Reiten duality theorem and the Christensen-Holm theorem stated above.

First, we shall prove the following theorem, which extends the Auslander-Reiten duality theorem to arbitrary Cohen-Macaulay local rings and deduces Araya's main observation from a more general context.
\begin{thm}
Let $(R,\mathfrak{m},k)$ be a $d$-dimensional Cohen-Macaulay local ring with a canonical module $\omega$.
Let $X$ be a totally reflexive $R$-module, and let $M$ be a maximal Cohen-Macaulay $R$-module that is locally free on the punctured spectrum of $R$.
Then, for all integers $i$ one has an isomorphism
$$
\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)\cong\mathrm{\widehat{Ext}}\mathrm{}_R^{(d-1)-i}(X^\ast,M^\dag)^\vee,
$$
where $(-)^\ast=\Hom_R(-,R)$, $(-)^\dag=\Hom_R(-,\omega)$, $(-)^\vee=\Hom_R(-,\mathrm{E}_R(k))$ and $\mathrm{\widehat{Ext}}\mathrm{}$ denotes Tate cohomology.
\end{thm}

Next, we define the following conditions over a commutative Noetherian local ring $R$.
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(SAC)}$]
For every finitely generated $R$-module $M$, if $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$, then $\Ext^{>0}_{R}(M,M)=0$.
\item[$\mathsf{(SACC)}$]
For every finitely generated $R$-module $M$ that has constant rank, if $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$, then $\Ext^{>0}_{R}(M,M)=0$.
\end{enumerate}
\end{quote}
We shall prove the following, which supplements the implication shown by Christensen and Holm with the conditions introduced above.
\begin{thm}
For each commutative Noetherian local ring $R$, the implications
$$
\xymatrix{
\mathsf{(AC)} \ar@{=>}[r] & \mathsf{(SAC)} \ar@{=>}[r] & \mathsf{(SACC)} \ar@{=>}[r]_{(\ast)} & \mathsf{(ARC)} \\
& \mathsf{(GARC)} \ar@{<=>}[u]
}
$$
hold, where $R$ is assumed to have positive depth for $(\ast)$.
\end{thm}
\noindent
Here, the assumption in $(\ast)$ that $R$ has positive depth is reasonable, because $R$ satisfies $\mathsf{(SACC)}$ trivially if $\depth R=0$.

We also give an application of Araya's result and pose a question concerning the Auslander-Reiten conjecture for modules having bounded Betti numbers.
We finish the article by discussing in Section 5 the relation of the Auslander-Reiten conjecture with a question of Huneke and Wiegand on the tensor products of modules.

\section{Basic definitions}

This section is devoted to giving basic definitions and their basic properties which will often be used later.

\begin{statement}
Throughout this paper, $R$ denotes a left Noetherian ring, and modules considered over $R$ are finitely generated left modules.
Unless otherwise specified, $R$ is assumed to be a commutative local ring with unique maximal ideal $\mathfrak{m}$ and residue field $k=R/\mathfrak m$.
\end{statement}

\begin{statement}
For a given $R$-module $M$, we set
$$
M^\ast=\Hom_R(M,R),\ M^{\vee}=\Hom_{R}(M, \textup{E}_{R}(k)),\ M^{\dag}=\Hom_{R}(M, \omega_{R}),
$$
where $\textup{E}_{R}(k)$ is the injective hull of $k$, and $\omega_{R}$ is a canonical module (in the case where $R$ is a Cohen-Macaulay ring that is a homomorphic image of a Gorenstein local ring).
Clearly, $M^{\dag}=M^{\ast}$ when $R$ is Gorenstein.
\end{statement}

\begin{statement}
We denote by $U_R$ the {\em punctured spectrum} of $R$, i.e., $U_{R}=\Spec R\setminus\{ \mathfrak{m}\}$.
We say that $M$ is {\em locally free on $U_R$} provided that $M_{p}$ is a free $R_{p}$-module for all $p\in U_{R}$.
The ring $R$ is called an {\em isolated singularity} if $R_{p}$ is a regular local ring for all $p\in U_{R}$.
\end{statement}

\begin{statement}
We use the notation $\overset{\rm st}{\cong}$ to denote a \emph{stable isomorphism}, that is, for two $R$-modules $M,N$ we write $M\overset{\rm st}{\cong}N$ if $M\oplus F\cong N\oplus G$ for some free $R$-modules $F,G$.
\end{statement}

\begin{statement}\label{2.5}
Let $M$ be an $R$-module.
For an integer $i>0$, we denote by $\Omega^{i}M$ the $i$-th {\em syzygy} of $M$, namely, the image of the $i$-th differential map in a minimal free resolution of $M$.
As a convention, we set $\Omega^0M=M$.
For an $R$-module $M$, we denote by $\mathrm{Tr} M$ the {\em (Auslander) transpose} of $M$, which is defined as the cokernel of the $R$-dual map $\delta_1^\ast$ of the first differential map $\delta_1$ in a minimal free resolution of $M$.
Here are some remarks:
\begin{enumerate}[(1)]
\item
The modules $\Omega^iM$ and $\mathrm{Tr} M$ are uniquely determined up to isomorphism, since so is a minimal free resolution of $M$.
\item
There is a stable isomorphism $\Omega^2\mathrm{Tr} M\overset{\rm st}{\cong}M^\ast$.
\end{enumerate}
\end{statement}

\begin{statement}
An $R$-module $X$ is called \emph{totally reflexive} if the natural homomorphism $X\to X^{\ast\ast}$ is an isomorphism and $\Ext^i_{R}(X,R)=\Ext_R^i(X^{\ast},R)=0$ for all $i>0$.
Here are several properties of totally reflexive modules which we will use.
For the details, we refer to \cite{AuBr} and \cite{Ch}.
\begin{enumerate}[(1)]
\item
A totally reflexive module over a Cohen-Macaulay local ring is maximal Cohen-Macaulay.
Over a Gorenstein local ring, the totally reflexive modules are precisely the maximal Cohen-Macaulay modules.
\item
For a totally reflexive $R$-module $X$ and an integer $i>0$, we define the $i$-th {\em cosyzygy} of $X$, denoted by $\Omega^{-i}X$, to be the image of the $R$-dual map $\delta_i^\ast$ of the $i$-th differential map in a minimal free resolution of $X^\ast$.
We have $\Omega^i\Omega^jX\overset{\rm st}{\cong}\Omega^{i+j}X$ for all integers $i,j$.
\item
If $X$ is a totally reflexive $R$-module, then so are $X^\ast$, $\mathrm{Tr} X$ and $\Omega^iX$ for all integers $i$.
\end{enumerate}
\end{statement}

\begin{statement}
For $R$-modules $M,N$, we denote by $\operatorname{\underline{Hom}}_R(M,N)$ the residue $R$-module of $\Hom_R(M,N)$ by the $R$-submodule consisting of the homomorphisms from $M$ to $N$ that factor through free modules.
Here are some remarks:
\begin{enumerate}[(1)]
\item
We have $\operatorname{\underline{Hom}}_R(M,N)\cong\Tor_1^R(\mathrm{Tr} M,N)$; see \cite[(3.9)]{Y}.
\item
If $M$ or $N$ is locally free on $U_R$, then the $R$-module $\operatorname{\underline{Hom}}_R(M,N)$ has finite length.
\end{enumerate}
\end{statement}

\begin{statement}\label{2.8}
Let $M$ be an $R$-module, and let $X$ be a totally reflexive $R$-module.
We denote by $\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)$ the $i$-th {\em Tate cohomology} module, which is defined as
$$
\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)=\operatorname{\underline{Hom}}_R(\Omega^iX,M).
$$
We will use the following properties of Tate cohomology.
We can find details of Tate cohomology, for instance, in \cite[\S 7]{catgp} (the paper \cite{catgp} assumes that the base local ring is Henselian, but this assumption is unnecessary in \cite[\S 7]{catgp}).
\begin{enumerate}[(1)]
\item
One has $\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)\cong\Ext_R^i(X,M)$ for all positive integers $i$.
\item
One has $\mathrm{\widehat{Ext}}\mathrm{}_R^i(\Omega^jX,M)\cong\mathrm{\widehat{Ext}}\mathrm{}_R^{i+j}(X,M)$ for all integers $i,j$.
\item
If $M$ is also totally reflexive, then $\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,\Omega^jM)\cong\mathrm{\widehat{Ext}}\mathrm{}_R^{i-j}(X,M)$ for all integers $i,j$.
\end{enumerate}
\end{statement}

\section{A generalization of Auslander-Reiten duality}

Araya \cite[Theorem 8]{A} made use of the Auslander-Reiten duality theorem over Gorenstein local rings \cite{Y} and proved the following theorem:

\begin{thm}[Araya]\label{Araya}
Let $R$ be a $d$-dimensional Gorenstein local ring and let $M$ be a maximal Cohen-Macaulay $R$-module that is locally free on $U_R$. If $\mathrm{\widehat{Ext}}\mathrm{}_{R}^{d-1}(M,M)=0$, then $M$ is free.
\end{thm}

It is a consequence of a result of Araya \cite{A} that if all one dimensional Gorenstein rings satisfy the Auslander-Reiten conjecture discussed in the introduction, so do all Gorenstein local rings.
Our aim in this section is to deduce Theorem \ref{Araya} from a more general result, Theorem \ref{d}, that holds over arbitrary Cohen-Maculay local rings.
Such a generalization could be useful for researchers who wish to further study the Auslander-Reiten conjecture. It also seems interesting in itself, as it concerns the vanishing of cohomology over Cohen-Macaulay local rings. We start with a few preliminary results:

\begin{lem}\label{a}
\begin{enumerate}[\rm(1)]
\item
Let $R$ be a Cohen-Macaulay local ring with a canonical module $\omega$ and let $M$ and $N$ be maximal Cohen-Macaulay $R$-modules.
Then $\Ext_R^i(M,N^\dag)\cong\Ext_R^i(N,M^\dag)$ for all $i\in\mathbb{Z}$.
\item
Let $R$ be a local ring and let $X$ and $Y$ be totally reflexive $R$-modules.
Then $\Ext_R^i(X^\ast,Y^\ast)\cong\Ext_R^i(Y,X)$ for all $i\in\mathbb{Z}$.
\end{enumerate}
\end{lem}

\begin{proof}
(1) The following isomorphisms hold in the derived category of $R$:
\begin{align*}
\RHom_R(M,N^\dag) & \cong \RHom_R(M,\RHom_R(N,\omega))
\cong \RHom_R(M\otimes_R^{\bf L}N,\omega)\\
& \cong \RHom_R(N,\RHom_R(M,\omega))
\cong \RHom_R(N,M^\dag).
\end{align*}
Here, the second and third isomorphisms follow from \cite[(A.4.21)]{Ch}.
One deduces the required isomorphism by taking the $i$-th cohomology.

(2) The conclusion follows from the following isomorphisms:
\begin{align*}
\RHom_R(X^\ast,Y^\ast) & \cong \RHom_R(X^\ast,\RHom_R(Y,R))
\cong \RHom_R(X^\ast\otimes_R^{\bf L}Y,R) \\
& \cong \RHom_R(Y,\RHom_R(X^\ast,R))
\cong \RHom_R(Y,X).
\end{align*}
\end{proof}

\begin{lem}\label{b}
Let $R$ be a local ring and let $X$ be a totally reflexive $R$-module.
Then $\Omega^{-i}(X^\ast)\overset{\rm st}{\cong}(\Omega^iX)^\ast$ for all $i\in\mathbb{Z}$.
\end{lem}

\begin{proof}
Let $i\ge0$.
Taking a minimal free resolution $(\cdots \to F_1 \to F_0 \to 0)$ of $X$, we have an exact sequence $0 \to \Omega^iX \to F_{i-1} \to \cdots \to F_0 \to X \to 0$.
Since $X$ is totally reflexive, the $R$-dual sequence
$$
0 \to X^\ast \to F_0^\ast \to \cdots \to F_{i-1}^\ast \to (\Omega^iX)^{\ast} \to 0
$$
is also exact.
Hence $X^\ast\overset{\rm st}{\cong}\Omega^i((\Omega^iX)^\ast)$.
Applying $\Omega^{-i}$ to this, we get $\Omega^{-i}(X^\ast)\overset{\rm st}{\cong}(\Omega^iX)^\ast$.

Now let $i\le0$.
Set $j=-i$ and $Y=X^\ast$.
Then, since $j\ge0$ and $Y$ is totally reflexive, we have $\Omega^{-j}(Y^\ast)\overset{\rm st}{\cong}(\Omega^jY)^\ast$.
Hence $\Omega^iX\overset{\rm st}{\cong}(\Omega^{-i}(X^\ast))^\ast$.
Applying $(-)^\ast$, we get $(\Omega^iX)^\ast\overset{\rm st}{\cong}\Omega^{-i}(X^\ast)$.
\end{proof}

We are now ready to prove the main result of this section:

\begin{thm}\label{d}
Let $R$ be a $d$-dimensional Cohen-Macaulay local ring with a canonical module $\omega$.
Let $X$ be a totally reflexive $R$-module.
Let $M$ be a maximal Cohen-Macaulay $R$-module that is locally free on $U_{R}$.
Then, for all $i\in \mathbb{Z}$,
$$
\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)\cong\mathrm{\widehat{Ext}}\mathrm{}_R^{(d-1)-i}(X^\ast,M^\dag)^\vee.
$$
\end{thm}

\begin{proof}
We have the following isomorphisms:
\begin{align*}
\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)
& = \operatorname{\underline{Hom}}_R(\Omega^iX,M)
= \mathrm{H}_{\mathfrak m}^0(\operatorname{\underline{Hom}}_R(\Omega^iX,M))\\
& \cong \Ext_R^d(\operatorname{\underline{Hom}}_R(\Omega^iX,M),\omega)^\vee
\cong \Ext_R^1(M,(\Omega^d\mathrm{Tr}\Omega^iX)^\dag)^\vee \\
& \cong \Ext_R^1(\Omega^d\mathrm{Tr}\Omega^iX,M^\dag)^\vee
\cong \mathrm{\widehat{Ext}}\mathrm{}_R^1(\Omega^d\mathrm{Tr}\Omega^iX,M^\dag)^\vee \\
& \cong \mathrm{\widehat{Ext}}\mathrm{}_R^{d-1}(\Omega^2\mathrm{Tr}\Omega^iX,M^\dag)^\vee
\cong \mathrm{\widehat{Ext}}\mathrm{}_R^{d-1}((\Omega^iX)^\ast,M^\dag)^\vee\\
& \cong \mathrm{\widehat{Ext}}\mathrm{}_R^{(d-1)-i}(\Omega^i((\Omega^iX)^\ast),M^\dag)^\vee
\cong \mathrm{\widehat{Ext}}\mathrm{}_R^{(d-1)-i}(X^\ast,M^\dag)^\vee.
\end{align*}
Here, since $M$ is locally free on $U_R$, the $R$-module $\underline{\Hom}_R(\Omega^iX,M)$ has finite length, which shows the second equality.
The first and second isomorphisms are obtained by \cite[(3.5.9)]{BH} and \cite[(3.10)]{Y}, respectively.
Lemmas \ref{a}(1) and \ref{b} give the third and last isomorphisms, respectively.
The other isomorphisms follow from \ref{2.5}(2) and \ref{2.8}.
\end{proof}

The following result is the original Auslander-Reiten duality theorem, which now follows from Theorem \ref{d}.

\begin{cor}[Auslander-Reiten duality]\label{ARD}
Let $R$ be a $d$-dimensional Gorenstein local ring and let $X$ and $M$ be maximal Cohen-Macaulay $R$-modules such that $M$ is locally free on $U_R$.
Then $\mathrm{\widehat{Ext}}\mathrm{}_R^i(X,M)\cong\mathrm{\widehat{Ext}}\mathrm{}_R^{(d-1)-i}(M,X)^\vee$ for all $i\in\mathbb{Z}$.
\end{cor}

\begin{proof}
In general, let $R$ be a local ring and $X,Y$ totally reflexive $R$-modules.
For an integer $i$ there are isomorphisms
\begin{align*}
\mathrm{\widehat{Ext}}\mathrm{}_R^i(X^\ast,Y^\ast)
& \cong \Ext_R^1(\Omega^{i-1}(X^\ast),Y^\ast)
\cong \Ext_R^1((\Omega^{1-i}X)^\ast,Y^\ast) \\
& \cong \Ext_R^1(Y,\Omega^{1-i}X)
\cong \mathrm{\widehat{Ext}}\mathrm{}_R^i(Y,X),
\end{align*}
where the second and third isomorphisms follow from Lemmas \ref{b} and \ref{a}(2), respectively.
Now the corollary follows from Theorem \ref{d}.
\end{proof}

\section{On a conjecture of Auslander and Reiten}

In this section we are concerned with the Auslander-Reiten conjecture (cf. the introduction,  \cite{ADS} or \cite{AR}) for commutative local rings.
Our purpose is to prove that an affirmative answer to a question of Christensen and Holm \cite{CH1} will show that all local rings that have positive depth satisfy the Auslander-Reiten conjecture, cf. Corollary \ref{corARC}.

Before proving the main result of this section stated as Theorem \ref{thmARC}, we will first discuss an application of Theorem \ref{Araya}; a weaker version of the Auslander-Reiten conjecture holds for certain periodic modules over even dimensional Gorenstein isolated singularities.

\begin{prop}\label{APP}
Let $R$ be a $d$-dimensional Gorenstein local ring, and let $M$ be a maximal Cohen-Macaulay $R$-module that is locally free on $U_R$.
Assume that $d$ is a positive even integer and that $M \overset{\rm st}{\cong} \Omega^{2n}M$ for some positive integer $n$.
If $\Ext^{i}_{R}(M,M)=0$ for all $i=1,3,5,\ldots, 2n-1$, then $M$ is free.
\end{prop}

\begin{proof}
First, note by assumption that $\Ext_R^i(M,M)=0$ for all odd integers $i$.
We set $N=\bigoplus_{0\le i\le n-1}\Omega^{2i}M$.
Then $N\overset{\rm st}{\cong} \Omega^{2}N$, and we have
\begin{align*}
\Ext^{1}_{R}(N,N)
& \cong \textstyle\bigoplus_{0\leq i, j \leq n-1}\Ext^{1}_{R}(\Omega^{2i}M, \Omega^{2j}M)
\cong \textstyle\bigoplus_{0\leq i, j \leq n-1}\Ext^{1}_{R}(\Omega^{2i+2n}M, \Omega^{2j}M)\\
& \cong \textstyle\bigoplus_{0\leq i, j \leq n-1}\mathrm{\widehat{Ext}}\mathrm{}^{1+2i+2n-2j}_{R}(M,M)
\cong \textstyle\bigoplus_{0\leq i, j \leq n-1}\Ext^{2(i+n-j+1)-1}_{R}(M,M)=0.
\end{align*}
Here, the fourth isomorphism follows from the fact that $1+2i+2n-2j=2(i+n-j+1)-1>0$.
Since $d$ is positive and even, we have $\mathrm{\widehat{Ext}}\mathrm{}_R^{d-1}(N,N)\cong\Ext_R^{d-1}(N,N)\cong\Ext_R^1(\Omega^{d-2}N,N)\cong\Ext_R^1(N,N)=0$.
Theorem \ref{Araya} implies that $N$, and hence $M$, is free.
\end{proof}

It follows by definition that a module $M$ with $M\overset{\rm st}{\cong}  \Omega^{n}M$ for some $n>0$ has bounded Betti sequence.
Therefore Theorem \ref{Araya} and Proposition \ref{APP} raise the following question:

\begin{ques}\label{qI}
Let $R$ be a $d$-dimensional Gorenstein local ring (that is not a complete intersection) with $d\geq 3$.
Let $M$ be a maximal Cohen-Macaulay $R$-module that is locally free on $U_R$.
Assume that $M$ has bounded Betti sequence.
If $\Ext^{i}_{R}(M,M)=0$ for all $i=1,2, \ldots, d-2$, then is $M$ free?
\end{ques}

There is a negative answer to Question \ref{qI} in case the ring considered is a complete intersection:

\begin{eg}
Let $R=k[[X,Y,Z,U]]/(XZ-YU)$ and $I=(X,Y)R$.
Then $R$ is a three dimensional hypersurface with an isolated singularity, and $I$ is a maximal Cohen-Macaulay $R$-module.
One can easily see that $\Ext^1_{R}(I,I)=0$ but $I$ is not free.
\end{eg}

The Auslander-Reiten conjecture (for commutative local rings) asserts that all commutative local rings satisfy the \emph{Auslander-Reiten condition} $\mathsf{(ARC)}$, which is defined as follows:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(ARC)}$]
For every $R$-module $M$, if $\Ext^{>0}_{R}(M,M\oplus R)=0$, then $M$ is free.
\end{enumerate}
\end{quote}
Some of the well-known examples of local rings that satisfy $\mathsf{(ARC)}$ are complete intersection rings \cite{ADS}, Golod rings \cite{JoS} and Gorenstein rings of codimension at most four \cite{S}.
As discussed in the previous section, a recent result of Araya \cite{A} shows that if all Gorenstein local rings of dimension at most one satisfy the Auslander-Reiten conjecture, so do all Gorenstein local rings.
As complete intersections satisfy $\mathsf{(ARC)}$, Araya's result in particular reproves a result of Huneke and Leuschke \cite{HL}: Gorenstein rings that are complete intersections in codimension one satisfy $\mathsf{(ARC)}$.
Indeed, over local domains, this result is motivated by a question of Huneke and Wiegand \cite{HW} which we will discuss briefly at the end of this article (cf. also \cite{Ce}).
We refer the interested reader to \cite{CH1} for a survey of the Auslander-Reiten and related conjectures.

A local ring $R$ is said to satisfy the \emph{Auslander condition} $\mathsf{(AC)}$ on the vanishing of cohomology if the following condition holds:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(AC)}$]
For every $R$-module $M$ there exists a nonnegative integer $b_{M}$ such that for every $R$-module $N$ one has: if $\Ext^{\gg 0}_{R}(M,N)=0$, then $\Ext^{>b_{M}}_{R}(M,N)=0$.
\end{enumerate}
\end{quote}
Christensen and Holm proved in \cite{CH1} that $\mathsf{(AC)}$ implies $\mathsf{(ARC)}$.
More precisely, the following result follows from \cite[(2.3)]{CH1}:
\begin{thm}[Christensen-Holm]\label{thmCH}
Let $R$ be a local ring that satisfies $\mathsf{(AC)}$, and let $M$ be an $R$-module.
If $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$, then $M$ is free.
\end{thm}

As the vanishing conditions imposed on the module $M$ considered in Theorem \ref{thmCH} appear to be weaker than those in the Auslander-Reiten Conjecture, Christensen and Holm asked in \cite[2.4]{CH1} whether the following two conditions are equivalent.
\begin{enumerate}
\item $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{>0}_{R}(M,M)=0$.
\item $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$.
\end{enumerate}
An example of Schulz \cite{Sch} yields a self-injective noncommutative Artin algebra over which the conditions (1) and (2) are \emph{not} equivalent.
As stated in \cite{CH1}, for commutative rings, it is not known whether $(2)$ implies $(1)$, in fact not even for Gorenstein rings.
Notice that Theorem \ref{thmCH} shows that if the ring satisfies $\mathsf{(AC)}$, then $(1)$ and $(2)$ are equivalent (cf. also \cite{Di}).
This motivates us to consider the following two conditions.
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(SAC)}$]
For every $R$-module $M$, if $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$, then $\Ext^{>0}_{R}(M,M)=0$.
\item[$\mathsf{(SACC)}$]
For every $R$-module $M$ that has constant rank, if $\Ext^{>0}_{R}(M,R)=0$ and $\Ext^{\gg 0}_{R}(M,M)=0$, then $\Ext^{>0}_{R}(M,M)=0$.
\end{enumerate}
\end{quote}
Here $\mathsf{(SAC)}$ and $\mathsf{(SACC)}$ stand for the \emph{symmetric Auslander condition} and the \emph{symmetric Auslander condition for modules with constant rank}.
Recall that an $R$-module $M$ is said to be of \emph{constant rank} if there exists an integer $r$ such that $M_{p}\cong R^{(r)}_{p}$ for all associated prime ideals $p$ of $R$.

Now we investigate the conditions $\mathsf{(ARC)},\mathsf{(SAC)},\mathsf{(SACC)}$ under modding out by a nonzerodivisor.

\begin{thm}\label{thmARC}
Let $R$ be a local ring and let $x\in R$ be a nonzerodivisor on $R$.
\begin{enumerate}[\rm(1)]
\item
If $R/xR$ satisfies $\mathsf{(SAC)}$/$\mathsf{(ARC)}$, then $R$ satisfies $\mathsf{(SAC)}$/$\mathsf{(ARC)}$.
\item
If $R$ satisfies $\mathsf{(SACC)}$, then $R/xR$ satisfies $\mathsf{(ARC)}$.
\end{enumerate}
\end{thm}

\begin{proof} We set $S=R/xR$.

(1) Let $M$ be an $R$-module with $\Ext_{R}^{>0}(M,R)=0$.
Then we have $\Ext_{R}^{>0}(\Omega_{R} M,R)=0$, and $\Ext_R^i(\Omega_RM,\Omega_RM)\cong\Ext_R^{i+1}(M,\Omega_RM)\cong\Ext_R^i(M,M)$ for $i>0$.
Therefore, to prove that $R$ satisfies $\mathsf{(SAC)}$/$\mathsf{(ARC)}$, we may replace $M$ with $\Omega_{R}M$ and hence assume that $x$ is a nonzerodivisor on both $R$ and $M$.
(Note that, if  $\Ext^{1}_{R}(M,M)=0$, then $\Omega_{R}M$ is free if and only if $M$ is free \cite[Lemma 1(iii), page 154]{Mat}.)

We consider the short exact sequence $0 \to M  \stackrel{x}{\rightarrow} M \to M/xM \to 0$.
Let $n\ge0$.
Nakayama's lemma implies that $\Ext_{R}^{>n}(M,M)=0$ if and only if $\Ext^{>n}_{R}(M,M/xM)=0$.
Since $\Ext^{i}_{R}(M,M/xM)\cong \Ext^{i}_{S}(M/xM,M/xM)$ for $i\ge0$ by \cite[Lemma 2(ii), page 140]{Mat}, we deduce that $\Ext_{R}^{>n}(M,M)=0$ if and only if $\Ext^{>n}_{S}(M/xM,M/xM)=0$.
Similarly, $\Ext^{>0}_{S}(M/xM, S)=0$ since $\Ext^{>0}_R(M,R)=0$.
Therefore, if $S$ satisfies $\mathsf{(SAC)}$, then so does $R$.
Furthermore, if $\Ext_R^{>0}(M,M)=0$ and $S$ satisfies $\mathsf{(ARC)}$, then $M/xM$ is free over $S$ and hence $M$ is free over $R$.
Thus, $R$ satisfies $\mathsf{(ARC)}$.

(2) Let $M$ be an $S$-module such that $\Ext_{S}^{>0}(M,M\oplus S)=0$.
We would like to prove that $M$ is a free $S$-module.
Using the snake lemma, we see that the multiplication by $x$ on the natural exact sequence $0 \to \Omega_{R}M \to R^{(n)} \to M \to 0$ yields an exact sequence $0 \to M \to \Omega_{R}M/x(\Omega_{R}M) \to S^{(n)} \to M \to 0$.
Hence we have a short exact sequence
$$
0 \to M \to \Omega_{R}M/x(\Omega_{R}M) \to \Omega_{S}M \to 0.
$$
This, being an element of $\Ext^{1}_{S}(\Omega_{S}M,M)\cong\Ext^{2}_{S}(M,M)=0$, splits and hence yields $\Omega_{R}M/x(\Omega_{R}M) \cong M \oplus \Omega_{S}M$.
For all positive integers $i$, we have:
\begin{align*}
\Ext^{i}_{S}(M,\Omega_{S}M)
& \cong \Ext^{i}_{S}(M, M \oplus \Omega_{S}M)
\cong \Ext^{i}_{S}(M, \Omega_{R}M/x(\Omega_{R}M))\\
& \cong  \Ext^{i+1}_{R}(M, \Omega_{R}M)
\cong   \Ext^{i}_{R}(\Omega_{R}M, \Omega_{R}M),
\end{align*}
where the third isomorphism holds by \cite[Lemma 2, page 140]{Mat}.
Since $\Ext^{i}_{S}(M,M) \cong \Ext^{i+1}_{S}(M,\Omega_{S}M)$ for all $i>0$, we have that $\Ext^{>1}_{R}(\Omega_{R}M, \Omega_{R}M)=0$.
Note that $\Omega_{R}M$ is of constant rank and $\Ext^{>0}_{R}(\Omega_{R}M, R)=0$. Therefore, since $R$ satisfies $\mathsf{(SACC)}$, $\Ext^{>0}_{R}(\Omega_{R}M, \Omega_{R}M)=0$.
Now $\Ext^{1}_{S}(M, \Omega_{S}M)\cong\Ext_R^1(\Omega_{R}M, \Omega_{R}M)=0$ and hence $M$ is free over $S$.
\end{proof}

As a corollary of Theorem \ref{thmARC} we have:

\begin{cor} \label{corARC} Let $R$ be a local ring that has positive depth. If $R$ satisfies $\mathsf{(SACC)}$, then $R$ satisfies $\mathsf{(ARC)}$.
\end{cor}

\begin{rmk} Assume $(R, \mathfrak{m})$ is a local ring with $\depth R=0$. Then $R$ satisfies $\mathsf{(SACC)}$ trivially; $\mathfrak{m}$ is an associated prime of $R$ and hence any module that has constant rank is free. Therefore, when $\depth R=0$, whether $\mathsf{(SACC)}$ implies $\mathsf{(ARC)}$ or not, is equivalent to the Auslander-Reiten conjecture recorded above.
\end{rmk}

We do not know whether $\mathsf{(SAC)}$ and $\mathsf{(SACC)}$ are equivalent conditions or not, but $\mathsf{(AC)}$ and $\mathsf{(SACC)}$ are not, that is,  $\mathsf{(SACC)}$ is a weaker condition than $\mathsf{(AC)}$. We explain this as follows.
(Recall that it follows from Theorem \ref{thmCH} that $\mathsf{(AC)}$ implies $\mathsf{(SAC)}$.)

Huneke, \c{S}ega and Vraciu \cite[(4.1)(1)]{HSV} proved that local rings $(R,\mathfrak{m})$ with $\mathfrak{m}^3=0$ satisfy $\mathsf{(SAC)}$.
More precisely, they proved that if $R$ is such a local ring and $M$ is an $R$-module with $\Ext^{i}_{R}(M, M\oplus R)=0$ for $i= n+1, n+2, n+3, n+4$ for some positive integer $n$, then $M$ is free.
On the other hand, Jorgensen and \c{S}ega \cite[(3.3)(2)]{JoS} gave an example of a commutative local ring $(S,\mathfrak{n})$ with $\mathfrak{n}^3=0$ that does not satisfy $\mathsf{(AC)}$.
Setting $R=S[[X]]$, we see from \cite[(2.3)]{CH2} that $R$ does not satisfy $\mathsf{(AC)}$.
Moreover, by Theorem \ref{thmARC}(1), $R$ satisfies $\mathsf{(SAC)}$ since $S=R/(X)$ satisfies $\mathsf{(SAC)}$.
These observations show that there exist local rings (of arbitrary depth) that satisfy $\mathsf{(SAC)}$ (and hence $\mathsf{(SACC)}$) but fail to satisfy $\mathsf{(AC)}$.

Recently Diveris \cite{Di} defined the following condition for left Noetherian rings $R$, where $\text{ext.deg}(M)=\sup\{n\mid\Ext^{n}_{R}(M,M)\neq 0 \}$:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(FED)}$]
The supremum of $\text{ext.deg}(M)$ is finite, where $M$ runs through the $R$-modules with $\text{ext.deg}(M)<\infty$.
\end{enumerate}
\end{quote}
He proved that, when $R$ satisfies $\mathsf{(FED)}$, the conditions (1) and (2) recorded in the discussion just after Theorem \ref{thmCH} are equivalent, that is, $\mathsf{(FED)}$ implies $\mathsf{(SAC)}$.
Moreover, he proved that if $R$ satisfies $\mathsf{(FED)}$, then it satisfies the following {\em generalized Auslander-Reiten condition}, which was also studied in \cite{Wei2,Wei1}:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(GARC)}$]
For each $R$-module $M$, if there exists a nonnegative integer $n$ such that $\Ext^{>n}_{R}(M,M\oplus R)=0$, then $\pd_{R}M\leq n$.
\end{enumerate}
\end{quote}
We should note that there exists a self-injective noncommutative Artin algebra that fails to satisfy this condition $\mathsf{(GARC)}$ (cf. \cite{Sch}).
We do not know whether all commutative rings satisfy $\mathsf{(GARC)}$ or not.

It is interesting that $\mathsf{(SAC)}$ and $\mathsf{(GARC)}$ are indeed equivalent statements.
Before recording our observation, we introduce a modified version of $\mathsf{(GARC)}$:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(GARC')}$]
For each $R$-module $M$, if $\Ext^{\gg 0}_{R}(M,M\oplus R)=0$, then $\pd_{R}M<\infty$.
\end{enumerate}
\end{quote}

\begin{thm}\label{obs}
Let $R$ be a local ring.
The following are equivalent.
\begin{enumerate}[\rm(1)]
\item
$R$ satisfies $\mathsf{(GARC')}$.
\item
$R$ satisfies $\mathsf{(GARC)}$.
\item
$R$ satisfies $\mathsf{(SAC)}$.
\end{enumerate}
\end{thm}

\begin{proof}
It is trivial that $\mathsf{(GARC)}$ implies $\mathsf{(GARC')}$.
Let $M$ be an $R$-module with $\pd_RM=n<\infty$.
Then $\Ext^n_{R}(M,R)\ne0$ by \cite[Lemma 1(iii), page 154]{Mat}.
This observation yields that $\mathsf{(GARC')}$ implies $\mathsf{(GARC)}$.

Assume that $R$ satisfies $\mathsf{(GARC)}$ and that $\Ext_{R}^{>0}(M,R)=\Ext_{R}^{\gg0}(M,M)=0$.
Then there is a nonnegative integer $n$ with $\Ext_{R}^{>n}(M,R)=\Ext_{R}^{>n}(M,M)=0$.
The condition $\mathsf{(GARC)}$ implies $\pd_RM\leq n$.
Since $\Ext_{R}^{>0}(M,R)=0$, we have $n=0$ by \cite[Lemma 1(iii), page 154]{Mat}.
Hence $M$ is free, and we have $\Ext_{R}^{>0}(M,M)=0$.
It follows that $R$ satisfies $\mathsf{(SAC)}$.

Next assume that $R$ satisfies $\mathsf{(SAC)}$ and that $\Ext_{R}^{>n}(M,M\oplus R)=0$ for some $n\ge0$.
Setting $N=\Omega^nM$, we have $\Ext_{R}^{>0}(N,R)=0$.
There are isomorphisms
$$
\Ext_R^i(N,N)\cong\Ext_R^{i-1}(N,\Omega^{n-1}M)\cong\cdots\cong\Ext_R^{i-n}(N,M)\cong\Ext_R^i(M,M)=0
$$
for all $i>n$.
Hence $\Ext_{R}^{>n}(N,N\oplus R)=0$.
Now put $X=N\oplus\Omega N$.
Then for $i>n$ we have:
\begin{align*}
\Ext_R^{i+1}(X,X)
& \cong \Ext_R^{i+1}(N,N)\oplus\Ext_R^{i+1}(N,\Omega N)\oplus\Ext_R^{i+2}(N,N)\oplus\Ext_R^{i+2}(N,\Omega N)\\
& \cong \Ext_R^{i+1}(N,N)\oplus\Ext_R^i(N,N)\oplus\Ext_R^{i+2}(N,N)\oplus\Ext_R^{i+1}(N,N)=0.
\end{align*}
Since $R$ satisfies $\mathsf{(SAC)}$, we have that $\Ext_{R}^{>0}(X,X)=0$, which implies $\Ext_{R}^{1}(N,\Omega N)=0$.
This shows that $N$ is free, and hence $\pd_RM\leq n$.
Thus $R$ satisfies $\mathsf{(GARC)}$.
\end{proof}

Corollary \ref{corARC}, Theorem \ref{obs} and the results discussed above particularly yield the following diagram for commutative local Noetherian rings, where for the implication $\mathsf{(SACC)}\Longrightarrow\mathsf{(ARC)}$ we assume that the local rings considered have positive depth.
$$
\xymatrix{
& \mathsf{(GARC')} \ar@{<=>}[d] & \\
& \mathsf{(GARC)} \ar@{<=>}[d] & \\
\mathsf{(FED)}  \ar@{=>}[r] & \mathsf{(SAC)} \ar@{=>}[r]\ar@{=>}[d]<1.5ex>|= & \mathsf{(SACC)} \ar@{=>}[r] & \mathsf{(ARC)} \\
& \mathsf{(AC)} \ar@{=>}[u]
}
$$
We refer the reader to \cite{Di} for details on the relation of the conditions $\mathsf{(FED)}$ and $\mathsf{(AC)}$.
It seems worthwhile to pose a question that follows from our previous discussions:

\begin{ques}\label{SACC}
Let $R$ be a local ring that has positive depth.
Assume that $R$ satisfies $\mathsf{(SACC)}$.
Then does $R$ satisfy $\mathsf{(SAC)}$?
\end{ques}

In view of Theorem \ref{thmARC}, Question \ref{SACC} is equivalent to the following: if $x$ is a nonzerodivisor on a local ring $R$ satisfying $\mathsf{(SACC)}$, then does $R/xR$ satisfy $\mathsf{(SAC)}$?

\section{Further remarks on the Auslander-Reiten conjecture}

Huneke and Leuschke \cite[1.3]{HL} studied the Auslander-Reiten conjecture and proved the following result:

\begin{thm}[Huneke-Leuschke]\label{Hu-Le}
Let $R$ be a $d$-dimensional complete local Cohen-Macaulay ring which is locally a complete intersection in codimension one.
Assume that $M$ is a maximal Cohen-Macaulay $R$-module of constant rank such that $\Ext_{R}^{i}(M,M)=\Ext_{R}^{i}(M^{\ast},R)=\Ext_{R}^{i}(M,R)=0$ for all $1\le i\le d$.
In the case $d=1$ assume further that $\Ext_{R}^{2}(M,M)=0$ holds\,\footnote[5]{\,This is a missing condition in their original statement. The authors communicated with Huneke and Leuschke for this typo.}.
If $R$ is Gorenstein or if $R$ contains $\mathbb{Q}$, then $M$ is free.
\end{thm}

Theorem \ref{Hu-Le} provides an important class of Cohen-Macaulay local rings over which the Auslander--Reiten conjecture holds.
Therefore we would like to examine whether the hypothesis that ``$R$ is either Gorenstein or contains $\mathbb{Q}$" is necessary in Theorem \ref{Hu-Le}.
We do not know whether it is possible or not to remove that assumption, however, following the arguments of the proof of Theorem \ref{Hu-Le}, we deduce that understanding the modules $M$ such that $M^{\ast} \cong M^{\dag}$ may shed light on the Auslander--Reiten conjecture over Cohen-Macaulay local rings that are not necessarily Gorenstein (notice, when $R$ is Gorenstein, $M^{\ast}=\Hom(M,R) \cong \Hom(M,\omega)=M^{\dag}$).
Indeed we have the following concrete statement which shows that the main conclusion of Theorem \ref{Hu-Le} holds under weaker hypotheses for modules $M$ with $M^{\ast} \cong M^{\dag}$.

\begin{prop}\label{newprop}
Let $R$ be a Cohen-Macaulay local ring of dimension $d\ge2$ with a canonical module $\omega$.
Let $M$ be an $R$-module such that:
\begin{enumerate}[\rm(1)]
\item
$M$ is reflexive,
\item
$M$ is locally free on $U_{R}$,
\item
$\Ext_{R}^{i}(M^{\ast},R)=0=\Ext_{R}^{i}(M,R)$ for all $i=1, \dots,d$,
\item
$\Ext_{R}^{d}(M,M)=\Ext_{R}^{d-1}(M,M)=0$.
\end{enumerate}
Then $\depth(M\otimes_{R}M^{\ast}\otimes_{R}\omega)\geq 2$ holds, that is, $M\otimes_{R}M^{\ast}\otimes_{R}\omega$ is reflexive.
In particular, if $M^{\ast} \cong M^{\dag}$, then $M$ is free.
\end{prop}

For the convenience of the reader we record the results which we make use of to prove Proposition \ref{newprop}.

\begin{thm}\cite[1.4]{HL}\label{Thm1}
Let $R$ be a $d$-dimensional Cohen-Macaulay local ring with a canonical module $\omega$, and let $T$ be a maximal Cohen-Macaulay $R$-module.
If $\Ext^{i}_{R}(T,R)=0$ for all $i=1, \dots, d$, then $\omega\otimes_{R}T\cong (T^{\ast})^{\dagger}$ is maximal Cohen-Macaulay.
\end{thm}

\begin{thm}\label{Thm2}\cite[3.2]{CeD}
Let $R$ be a $d$-dimensional Cohen-Macaulay local ring with a canonical module $\omega$, and let $M,N$ be $R$-modules.
Assume that $M$ is locally free on $U_{R}$ and that $N$ is maximal Cohen-Macaulay.
Let $n$ be an integer such that $1\leq n \leq \depth M$.
Then $\depth(M\otimes_{R}N)\geq n$ if and only if $\Ext^i_R(M,N^{\dagger})=0$ for all $i=d-n+1,\dots, d-1,d$.
\end{thm}

\begin{thm}\label{Thm3}{\rm(}Auslander \cite{Au}, see also \cite[3.4]{CeD}{\rm)}
Let $R$ be a Cohen-Macaulay local ring, and let $M$ be a torsionfree $R$-module.
Assume that $M_{\mathfrak{p}}$ is free over $R_{\mathfrak{p}}$ for each $\mathfrak{p}\in \Spec R$ with $\htt\mathfrak{p}\leq 1$.
If $M\otimes_{R}M^{\ast}$ is reflexive, then $M$ is free.
\end{thm}

Now we are ready to prove Proposition \ref{newprop}; we thank Graham Leuschke for several discussions on the proof of the proposition.

\begin{proof}[Proof of Proposition \ref{newprop}]
Set $N=M^{\ast}\otimes_{R}\omega$.
It follows from (3) that $M^{\ast}$ is a $d$th syzgy, hence maximal Cohen-Macaulay.
Therefore, setting $T=M^{\ast}$, we have, by Theorem \ref{Thm1}, that $N\cong (M^{\ast\ast})^{\dagger}$ is maximal Cohen-Macaulay. Thus, by (1), $N\cong M^{\dagger}$ so that $N^{\dagger}\cong M^{\dagger\dagger}$. Furthermore, using (1) and (3), we deduce by a similar argument that $M$ is maximal Cohen-Maculay. Therefore $M^{\dagger\dagger}\cong M$ and this yields that $N^{\dagger}\cong M$. Hence, by (4) we have $\Ext^{d}_{R}(M,N^{\dagger})=\Ext^{d-1}_{R}(M,N^{\dagger})=0$.
Theorem \ref{Thm2} shows $\depth(M\otimes_{R}N)\geq 2$.
This and (2) imply that $M\otimes_{R}N=M\otimes_{R}M^{\ast}\otimes_{R}\omega$ is reflexive; see for example \cite[1.4.1(b)]{BH}.

Now assume $M^{\ast} \cong M^{\dag}$. Then $N\cong M^{\dagger}\cong M^{\ast}$ so that $M\otimes_{R}N\cong M\otimes_{R}M^{\ast}$.
Applying Theorem \ref{Thm3}, we conclude that $M$ is free.
\end{proof}

The foregoing discussion shows that the next question is worth recording:

\begin{ques}\label{wm}
Let $R$ be a Cohen-Macaulay local ring with a canonical module.
Assume that there exists a nonfree $R$-module $M$ such that $M^{\ast} \cong M^{\dag}$.
Then is $R$ Gorenstein?
\end{ques}

We are able to give an affirmative answer to Question \ref{wm} in case the module considered is totally reflexive; we believe such a result is also interesting on its own.
However we do not know more about the existence of such modules in general.

\begin{prop}\label{partialanswer}
Let $R$ be a Cohen-Macaulay local ring with a canonical module $\omega$.
If there exists a totally reflexive $R$-module $M\ne0$ such that $M^{\ast} \cong M^{\dag}$, then $R$ is Gorenstein.
\end{prop}

\begin{proof}
We have isomorphisms
\begin{align*}
M\otimes_R^{\bf L}\omega
& \cong M\otimes_R^{\bf L}\RHom_R(R,\omega)
\cong\RHom_R(\RHom_R(M,R),\omega)\\
& \cong\RHom_R(M^\ast,\omega)
\cong M^{\ast\dag}
\cong M^{\dag\dag}
\cong M.
\end{align*}
Here, the second and third isomorphisms are obtained by \cite[(A.4.24)]{Ch} and the total reflexivity of $M$, respectively.
The fourth isomorphism follows from the fact that $M^\ast$ is maximal Cohen-Macaulay since it is totally reflexive.
These isomorphisms especially says $\Tor^{R}_{>0}(M, \omega)=0$, and hence we have an isomorphism
$$
M\otimes_R\omega\cong M.
$$
Comparing the minimal numbers of generators of both sides, we see that $\omega$ is cyclic, that is, $\omega\cong R$.
Hence $R$ is Gorenstein.
\end{proof}

Next we discuss a question of Huneke and Wiegand \cite{HW} advertised in the introduction:

\begin{ques}[Huneke-Wiegand]\label{queHW}
Is there a nonfree and torsionfree $R$-module $M$ over a one-dimensional Gorenstein domain $R$ such that $M\otimes_{R}M^{\ast}$ is torsionfree?
\end{ques}

We refer the interested reader to \cite[page 473]{HW} for the motivation of Question \ref{queHW}.
Inspired by such a question (see also Theorem \ref{Thm3}) we define the following condition for commutative Noetherian local rings $R$:
\begin{quote}
\begin{enumerate}
\item[$\mathsf{(HWC)}$]
For every torsionfree $R$-module $M$, if $M\otimes_{R}M^{\ast}$ is reflexive, then $M$ is free.
\end{enumerate}
\end{quote}

A result of Huneke and Wiegand \cite[(3.7)]{HW} shows that all one-dimensional hypersurface local domains satisfy $\mathsf{(HWC)}$.
(We do not know whether the same result holds for all complete intersection local domains, cf. \cite{Ce}.)
Examples of local rings (with positive dimension) that do not satisfy $\mathsf{(HWC)}$ are abundant; for example, if $R=k[[X,Y]]/(XY)$, then $M=M\otimes_{R}M^{\ast}$ is reflexive for the nonfree module $M=R/(X)$.
Notice that the ring $R$ here is not an integral domain.
Furthermore the reflexivity of $M\otimes_{R}M^{\ast}$ is not a very strong assumption for the condition $\mathsf{(HWC)}$ to hold over local rings:

\begin{prop}\cite[(3.12)]{CeD}\label{p0}
Given an odd integer $d\geq 3$, there exist a $d$-dimensional local hypersurface $R$ with an isolated singularity (hence $R$ is an integral domain) and a nonfree maximal Cohen-Macaulay $R$-module $M$ such that $M\otimes_{R}M^{\ast}$ is torsionfree.
\end{prop}

The following result displays the relation between the condition $\mathsf{(HWC)}$ and the Auslander-Reiten conjecture for Gorenstein domains (cf. \cite[(3.3), (3.4) and (3.14)]{CeD}):

\begin{prop}\label{HWC-ARC}
Consider the following statements.
\begin{enumerate}[\rm(1)]
\item
All Gorenstein local domains satisfy $\mathsf{(HWC)}$.
\item
All one-dimensional Gorenstein local domains satisfy $\mathsf{(HWC)}$.
\item
All Gorenstein local domains satisfy $\mathsf{(ARC)}$.
\end{enumerate}
Then the implications $(1) \Longleftrightarrow (2) \Longrightarrow (3)$ hold.
\end{prop}

We do not know whether $(3)$ implies $(2)$ or not.
Moreover, in view of Proposition \ref{HWC-ARC}, it seems reasonable to conjecture the following.

\begin{conj}
All one-dimensional local domains satisfy $\mathsf{(HWC)}$.
\end{conj}

Let $R$ be a local ring, and let $\mathcal{X}(R)$ denote the category of torsionfree $R$-modules $M$ such that $M\otimes_{R}M^{\ast}$ is reflexive.
If $R$ is a complete intersection domain and $M$ is an $R$-module in $\mathcal{X}(R)$ having bounded Betti numbers, then $M$ is free \cite[(4.17)(2)]{Ce}.
We finish this section by recording one more class that satisfy $\mathsf{(HWC)}$.
This also yields another application of Theorem \ref{Araya} discussed above.

\begin{prop} \label{propHWC}
Assume $R$ is a reduced ring such that $R=S/(x)$ where $(S, \mathfrak{n})$ is a two-dimensional complete Gorenstein normal local ring and $0\neq x \in \mathfrak{n}$. If $M\in \mathcal{X}(R)$ and $\Tor_{1}^{R}(M,M^{\ast})=0$, then $M$ is free.
\end{prop}

We need two lemmas for the proof of Proposition \ref{propHWC}.
The first one follows from a theorem of Huneke and Jorgensen \cite[5.9]{HJ}:

\begin{lem}\label{727}
Let $R$ be a $d$-dimensional Gorenstein local ring, and let $M,N$ be maximal Cohen-Macaulay $R$-modules.
Assume that $\Ext^{i}_{R}(N,M)$ has finite length for all $i=1, \ldots, d$.
Then $M^{\ast}\otimes_{R}N$ is maximal Cohen-Macaulay if and only if $\Ext^{i}_{R}(N,M)=0$ for all $i=1, \ldots, d$.
\end{lem}

\begin{lem}\label{728}
Let $R$ be a $d$-dimensional Gorenstein local ring, and let $M$ be a maximal Cohen-Macaulay $R$-module.
Let $n$ be a nonnegative integer, and assume that $M$ is locally free on $U_{R}$.
If $\Tor_n^R(M,M^*)=0$, then $\Ext^{n+d}_{R}(M,M)=0$.
\end{lem}

\begin{proof}
Without loss of generality we may assume that $R$ is complete. We will use the following spectral sequence \cite[Theorem 10.62]{Roit}:
$$
E_2^{p,q}=\Ext_{R}^p(\Tor^{R}_q(M,M^*),R)\quad\Longrightarrow\quad H^{p+q}=\Ext_{R}^{p+q}(M,M).
$$
Note that $E_2^{p,q}=0$ if $p>d$ since $R$ has injective dimension $d$.
Furthermore, for $q>0$, $\Tor^{R}_q(M,M^*)$ has finite length.
Therefore $E_2^{p,q}=0$ if $q>0$ and $p\ne d$.
Now the required result follows:
$$
\Ext_{R}^{n+d}(M,M)=H^{n+d}\cong E_2^{d,n}=\Ext_{R}^d(\Tor^{R}_n(M,M^*),R)\cong\Tor^{R}_n(M,M^*)^\vee=0,
$$
where the second isomorphism follows from the local duality theorem.
\end{proof}

\begin{proof}[Proof of Proposition \ref{propHWC}]
Since $R$ is an isolated singularity, $M$ is locally free on $U_{R}$.
By Lemma \ref{728}, the vanishing of $\Tor_1^R(M,M^*)$ forces the vanishing of $\Ext^{2}_{R}(M,M)$. Therefore \cite[(1.7)]{ADS} shows that $M$ lifts to $S$, that is, there exists a finitely generated $S$-module $N$ such that $x$ is $N$-regular and that $N/xN\cong M$. Furthermore, applying \cite[Lemma 2(ii), page 140]{Mat}, we see that $\Ext_{S}^{1}(N,M) \cong \Ext_{R}^{1}(M,M)$.

By assumption, $M\otimes_RM^\ast$ is reflexive.
As $R$ is $1$-dimensional, $M\otimes_RM^\ast$ is maximal Cohen-Macaulay over $R$.
Lemma \ref{727} implies $\Ext_{R}^{1}(M,M)=0$, and hence $\Ext_{S}^{1}(N,M)=0$.
Note also that the $S$-module $N$ is maximal Cohen-Macaulay.
Since $S$ is an isolated singularity, $N$ is locally free on $U_{S}$.
From the short exact sequence $0\rightarrow N \stackrel{x}{\rightarrow} N \rightarrow M \rightarrow 0$ we obtain the surjection $\Ext_{S}^{1}(N,N) \stackrel{x}{\rightarrow}\Ext_{S}^{1}(N,N)$.
Nakayama's lemma shows $\Ext_{S}^{1}(N,N)=0$.
By Theorem \ref{Araya} the $S$-module $N$ is free, and so is the $R$-module $M$.
\end{proof}

An example we discussed before shows that the vanishing of $\Tor_{1}^{R}(M,M^{\ast})$ in Proposition \ref{propHWC} is necessary: for $R=k[[X,Y]]/(XY)$ and $M=R/(X)$, setting $S=k[X,Y]$, we have that $R=S/(XY)$, $M\in \mathcal{X}(R)$, $\Tor_{1}^{R}(M,M^{\ast})\cong k\ne0$ and $M$ is nonfree.

\section*{Acknowledgments}
The authors would like to thank Tokuji Araya, Luchezar Avramov, Lars Christensen, Hailong Dao, Kosmas Diveris, Craig Huneke, Graham Leuschke, Greg Piepmeyer and Roger Wiegand for their valuable comments during the preparation of this paper.
This work was done during the visit of the first author to University of Nebraska-Lincoln in May and July, 2011.
He is grateful for their kind hospitality.
The authors also thank the referee for his/her careful reading and useful suggestions.

redpajama_set_name: RedPajamaArXiv


In our fast-paced, technology-based world, our children have little or no connection with our food supply. How food is produced is abstract and distant. Supporting your local farmers' market means you can educate them, and you, about how we get our food and what it takes to produce healthy food of excellent quality. Taking time to talk to the farmers who come to the market is an excellent way to learn more about how our food is produced. Also, local markets sell food that is in season — meaning it reflects authentic flavors and helps demonstrate the cycles of nature behind food production. Enjoying produce at the height of its seasonal production and anticipating the next seasonal crop helps us connect with the rhythms of the earth. What a great way for our children to learn about the changing seasons and how they influence holidays such as Thanksgiving.
At a local market you will find an amazing array of produce that you might not see in the run of the mill grocery store. Red carrots, watermelon radishes, quail eggs, purple cauliflower, and tomatoes of all colors are just a sampling of the culinary diversity awaiting you. Raw milk and fresh eggs are other options often provided at your local farmers' market, choices many families are turning to in response to allergies and other food concerns.
The prices you pay at a local market reflect the fact that you are bypassing the middle man. You are also avoiding the wasteful distribution chain used to deliver produce to consumers and unnecessary packaging. The less distance food has to travel to get to you the better it is for the environment. Did you know that food in the U.S. travels an average of 1,500 miles to get to your plate? All those miles use up fossil fuels, and extra packaging fills up shrinking landfills. Conventional agriculture more often than not pollutes water, land, and air with toxic byproducts. Buying at your local market helps save the planet.
Perhaps the best reason of all for frequenting your local farmers' market is so that you can enjoy the great outdoors and rub elbows with your neighbors. Instead of pushing a metal cart around a freezing grocery store, wouldn't you rather stroll among outdoor stalls full of fresh aromatic produce? Besides, it's a great place to meet with friends, bring the kids, and just get a taste of small town life in the midst of suburbia.
There are so many excellent reasons to support local farmers' markets.
>There are more than 3,000 farmers' markets operating in the United States currently, according to the U.S. Department of Agriculture, generating in excess of $1 billion annually. Is there any difference between buying from a chain supermarket and your local farmers' market? Yes, says E Magazine, the online environmental magazine, for a number of reasons, not the least of which is your family's health.
> Aboutproduce.com has boiled down the benefits of eating more fresh fruits and vegetables: reduced risk of coronary heart disease, stroke, obesity, and some types of cancer, including cancer of the stomach, esophagus, and lung.
> Yes, but…don't we get the same nutrients from the produce sold at chain supermarkets? Not exactly, says Lola O'Rourke, a registered dietician in Seattle and spokesperson for the American Dietetic Association. "Fruits and vegetables contain their highest levels of nutrients when harvested fully ripe and eaten soon afterwards," she says.
> Unfortunately, much of the produce available at national chains is trucked in from all over the country, often picked green, treated with chemicals to retard ripening and dipped in wax. Much of the produce is weeks old when you buy it in your grocery store. In contrast, locally grown food travels only a short distance from farm to table. It's pulled from the ground or plucked from trees and bushes 24 hours before consumers purchase it, and brought to market in reuseable containers.
> The benefits of farmers' markets are far-reaching: They help preserve farmland and the rural landscape; insure the continued economic viability of the small family farm; counter the growth of agribusiness with its devastating impact on people and places, while supporting clean, environmentally sensitive farming practices; conserve energy; help maintain biodiversity in food plants; and contribute to regional prosperity.
>l Buy only the amount you can use in a short period of time to avoid having to throw away spoiled produce.
>l Look for produce that is free from unusual odors or colors and signs of spoilage such as mold.
>l Handle produce gently to reduce bruising. Bacteria can thrive in the bruised areas. At the grocery store, keep fresh produce on top of other foods in a shopping cart — and separate from fresh meat — and set it down gently on the counter at the check-out line.
>l Remember that buying underripe produce isn't always the best option. Peaches, cantaloupe and nectarines are examples of fruits that may soften during storage, but they won't ripen.
>l When buying cut produce, keep it cold during transport. Put it in a cooler with ice if traveling a distance.
>l Wash your hands for at least 20 seconds with soap and water before handling produce and any other food.
>l Wash all fruits and vegetables with cool running tap water right before eating. Don't use dish soap or detergent because the FDA hasn't approved or labeled these products for use on foods.
>l Scrub melons with a brush and running water, because bacteria can be transferred from the outside of the melon to the inside by a knife.
>l Don't cross-contaminate: Use clean utensils and cutting boards when peeling or cutting up produce. Wash cutting boards with soap and water, rinse and sanitize between uses. A solution of 1 teaspoon bleach per quart of water is considered safe and effective.
>l Cut away bruised parts before eating. Remove the outer leaves from lettuce and cabbage.
>l Avoid serving sprouts to at-risk populations like the very young, old, or those whose immune system isn't able to function at normal levels. For example, people undergoing cancer treatment often cannot eat fresh produce.
>l Keep fresh cut produce cold by placing serving containers on ice. Perishable food should spend no more than two hours in the "danger zone" (40 to 140oF).
>l Store produce in containers that are free from excess liquid.
>l Refrigerate cut produce and use within a few days.


Meet Sydney, A Student Training in Massage Therapy!
Welcome to Spotlight! We will feature Dorsey Schools' students, graduates, staff members, and others here from time to time to highlight their hard work and achievements. Today we would like to introduce you to Sydney. She is a student at the Dorsey Schools Madison Heights, MI campus studying in Massage Therapy training program.
Sydney, congratulations on all that you've accomplished! All of us here at the Dorsey Schools Madison Heights, MI campus and throughout Dorsey are so proud of your achievements. Keep up the good work on your journey towards a new career in Massage Therapy! For more information about our graduation rates, the median debt of students who have completed the program, and other important information, please visit www.dorsey.edu/disclosures.
Previous PostWhat Are Two Types of Medical Billing?


text:
Fun Music Information -> Annie Haslam
This is the most recent information about Annie Haslam that has been submitted to amIright. If we have more information about Annie Haslam, then we provide a link to the section where it appears (the actual page whenever possible).
Use a Song Title to Answer a Different Song:
Song & Band Name
"Sailing," Christopher Cross "Going Home," Annie Haslam
"Going Home," Annie Haslam "I'll Be Home For Christmas," Bing Crosby
Song Parody Lyrics:
Original Song Name
Parody Song Name
Parody Author
"Going Home" "Blowing Gnomes" Porfle Popnecker
Song Title Anagrams:
"No Home Gig" originally "Going Home"
Submitted by: Yvette Bristle
"Hog Imogen" originally "Going Home"
"Moon Gig, Eh?" originally "Going Home"
"Oh, Gem, Gino!" originally "Going Home"
"On Home Gig" originally "Going Home"
There are additional Song Title Anagrams available.
Other Artists:
Check the indexes below for other performers on the site, or use google search to search the entire site.
Most Popular | # | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z
amIright Radio Station
amIright Events
amIright Store
Artist Summaries
Funny Music News
Music Poster Books
WhatFreaks (For Adults)
Selected Albums
Annie in Wonderland (Original Recording Remastered + Bonus Track)
Dawn of Ananda
Women Transending
© 2000-2013 by Charles R. Grosvenor Jr.

redpajama_set_name: RedPajamaCommonCrawl


Bitcoin makes the main stream news this week and well money as in cash is always in the news. However, the biggest money news last week was hardly reported, China publically capping its limit on the number of US treasury bill it will purchase / hold. Again, time will tell if this reporting is truthful. Coincidence or cause and effect crypto currencies have all raised sharply in value this week.
Those new to crypto currencies might not know about them or how they gain 'value'? For cash money, we will take gold. A metal that requires mining, so that gives it an intrinsic cost hence a price. Demand and supply bid from there to get its price day-to-day. A cyro currency like bitcoin, uses computers to data mine prime numbers. This has a cost and sets the intrinsic price and again from there day to day prices are set via demand and supply. Both process involve the concept of mining.
Metal mining is about discovery and location of atoms. Data mining prime numbers involves the discovery of new numbers and algorithms to locate them. Money is a social tool. I prefer the term cultural tool where the forces of society play out. However, both gold and bitcoin mining concepts for money have a primitive intellectual model underpinning their intrinsic value and link to day to day life. An attention currency data mines life. All that aspects that can be captured. A mining the uses the same ingredients that day-to-day reality is played out on.


If you thought that Archie's new look was a wild, new approach to the classic comic book character, get ready to see him burst into song. Archie, Jughead, Veronica, and Betty will soon be like you've never seen them before, live and on Broadway!
Adam McKay, the writer/director behind Anchorman and The Other Guys, is working on the project with Archie Comics and Funny Or Die. Yes, I thought this had to be a Funny Or Die sketch at first. However, it turns out that the Riverdale gang is indeed packing up and heading for The Great White Way.
McKay is writing the book for the new show. The rest of the creative team and a start date will be announced at a later date.


/ On : 9:17 PM / Thank you for visiting this website.
Shantaram Athavale-His movie 'Wavtal', based on a novel by Shankar Patil, on village life won him 10 awards from the Government of Maharashtra, which consisted of a jury of stalwarts like P.L. Deshpande, Vishram Bedekar, and Gajanan Jahagirdar (1965).


The Supplier Ethical Data Exchange Database (SEDEX) is a worldwide platform for the exchange of information on ethically sustainable production in the supply chain.
The Sedex Database enables its members to manage their suppliers and reduce their risks by exchanging information on ethical performance. This way improvements regarding ethically sustainable business practices are made throughout the production chain. Enterprises which perform a Sedex-Audit according to the SMETA-Guidelines and share this information in the Sedex-Database can prove to their customers that they respect the principles of ethically and socially sustainable production.
Customers, on the other hand, can effectively manage their suppliers according to these principles, thus reducing their risk.
The basic 2-Pillars of a SMETA audit are governed by the standards in ETI Base Code.
A 4-Pillars Audit adds Environment (extended version) and Business Ethics, governed by standards arrived at by a process of membership and multi-stakeholder consultation.
The audit type will depend on your request, which shall consider your customers' request.
Normally, the organization should complete self-assessment and provide the report before audit.
Depending on the outcome of risk assessments and/or audit planning, prior notice of the audit may or may not be given.
Announced: audit date is agreed with, or disclosed to, the audited site.
Semi-announced: audit date will fall within an agreed 'window'.
Unannounced: no prior notice is given.
Permits, operating licences, certificates of operations, etc.
Government Inspection reports e.g. sanitation, fire safety, structural safety, environmental compliance, etc.
Any appropriate certifications e.g. ISO 45001, OSHAS 18001, ISO 14001, ISO 9001, Chain of Custody etc.
DQS is a member of Associate Auditor Groups (AAG) of SEDEX.


text:
Abel Morgan (1673 – 16 December 1722) was a Welsh Baptist minister, best known for the posthumously published work Cyd-goriad Egwyddorawl o'r Scrythurau (English: The Joint Principles of the Scriptures) the First Biblical concordance to be written in the Welsh language and the second Welsh book printed in British America.

Early life
Morgan was born at Allt-goch in Cwrtnewydd, in the parish of Llanwenog, Cardiganshire in 1673. The son of Morgan Rhydderch, deacon at Rhydwilym, Morgan moved to Abergavenny at an early age to become a member of Llanwenarth's Baptist church. His career as a preacher began in 1692 and he was ordained in Blaenau Gwent in 1700, having received a call to preach in the region, c. 1696.

Emigration to the Americas
Morgan decided to emigrate to the New World in September 1711, though he did not arrive in North America until February the following year. Abel Morgan landed in the Province of Pennsylvania and held the pastorate of the Pennepack Baptist Church from his arrival, an historic Baptist church in Philadelphia that is one of the oldest Baptist congregations in North America. His brother, Enoch Morgan (1676-1740), was already minister for the congregation of the Welsh Tract Baptist Church in Newcastle County, Delaware, which sprung from the Pennepack Baptist Church over a disagreement concerning the practice of the 'laying on of the hands.'

Later life and death
In 1716, Morgan translated a formal confession of faith that had been signed by the Welsh Tract Church's congregants concerning official Baptist doctrine and practices. Morgan was pastor of the Pennepack Baptist Church until his death on 16 December 1722, and his translations of the Biblical concordance were published in Philadelphia in 1730, eight years after his death. He was originally buried at the back of Pennepack Baptist Church in Lagrange Place. His remains were later moved to Mount Moriah Cemetery, Philadelphia.

Personal life and family
Morgan was married three times during his life. His first wife, Priscilla Powell, and their son died during his initial voyage to North America, though their daughter survived. He later remarried Martha Burrows, and after her death married Judith Gooding née Griffiths, the widowed daughter of Thomas Griffiths (1645–1725), who had served as the first pastor for the Welsh Tract Baptist Church for twenty-five years. Morgan fathered three sons and one daughter from Gooding. One of his sons, also named Abel Morgan, became a prominent preacher during the First Great Awakening and the American War of Independence.

Sources
 The Cambrian (1881), pp. 188–190.
 Conrad, Henry C., Records of the Welsh Tract Baptist Meeting: Pencader Hundred, New Castle County, Delaware, 1701 to 1828 in Two Parts - Part One (Wilmington, Delaware: John M. Rogers Press, 1908), pp. 7–9.
 Geiriadur Bywgraffyddol o Enwogion Cymru, Volume II, pp. 276–277.
 Geiter, Mary K., 'Morgan,  Abel  (1673–1722),' Oxford Dictionary of National Biography (Oxford: Oxford University Press, 2004)
 Hanes y Bedyddwyr ymhlith y Cymry, 1885, pp. 355–356.
 Hanes y Bedyddwyr yn Nghymru, 1893-1907, Volume III, p. 108.
 National Library of Wales Journal, Volume II, pp. 116–117.
 National Library of Wales Journal, Volume III, pp. 19–22.
 National Library of Wales, NLW MS 9258
 National Library of Wales, NLW MS 9267
 Rhydwilym church register (access from the National Library of Wales)
 'Y Cenhadwr Americanaidd sef cylchgrawn gwybodaeth fuddiol a dyddorawl i Gymry America,' The American Messenger, 1880 (New York: Utica, 1840-1901), pp. 325–325.

References

External links
 Morgan, Abel (1673 – 1722)
 Griffiths, Thomas (1645 – 1725)

1673 births
1722 deaths
Baptist ministers from the United States
British emigrants to the Thirteen Colonies
People from Ceredigion
18th-century Welsh Baptist ministers
Welsh Baptist missionaries
Welsh Presbyterians
People of colonial Pennsylvania
Baptists from Pennsylvania
Burials at Mount Moriah Cemetery (Philadelphia)
18th-century American clergy

redpajama_set_name: RedPajamaWikipedia


Health Buzz is a quarterly newsletter for parents of elementary school-aged children. Health Buzz is developed and provided to keep parents informed on the best ways to keep their family healthy.
View the Health Buzz Newsletter.


Earlier this year, clapping was banned at one university. Now, capital letters can't be used to avoid upsetting "snowflake" students.
Journalism professors at Leeds Trinity University in the UK have been instructed not to use certain words — in case they frighten sensitive students.
According to UK media reports, the use of capital letters has been banned as well as the "overuse" of the words "do" and "don't".
In an internal staff memo obtained by the Express, staff are told students' "anxiety" can lead to academic failure.
"Despite our best attempts to explain assessment tasks, any lack of clarity can generate anxiety and even discourage students from attempting the assessment at all," it reads.
"Generally, avoid using capital letters for emphasis and the overuse of 'do', and, especially, 'DON'T'.
It also urged staff to be "explicit about any inexplicitness" in assignment requirements and to be aware that "misconceptions or misunderstandings quickly spread" among students.
"This can lead to further confusion and students may even then decide that the assessment is too difficult and not attempt it," the statement reads.
The story quickly spread across the UK media as well as on social media, with many members of the public slamming it as extreme "political correctness" pandering to the "snowflake generation".
It follows similar news in September, when The University of Manchester's student union made global headlines after voting to ban clapping and cheering at certain events in order to avoid triggering those with anxiety or sensory issues.
Instead, students were urged to use "jazz hands".
"It was argued that the loud noise of traditional clapping and whooping pose an issue to students with anxiety or sensory issues. BSL clapping — or, jazz hands — would be a more inclusive form of expression," student newspaper the Mancunion reported.
In a bizarre twist, Leeds Trinity University released a statement claiming it had not banned capital letters — but confirmed "it is best practice not to write in all capital letters".
But many social media users pointed out there was little difference between "banning" words and capital letters and asking staff not to use them.
We're not banning using them but it's best practice not to use them.
In a statement, vice-chancellor Margaret House said the university was committed to supporting students to be "the very best they can be".
"We're proud to offer a personal and inclusive university experience that gives every student the support to realise their potential," she said.
"We follow national best practice teaching guidelines and the memo cited in the press is guidance from a course leader to academic staff, sharing best practice from the latest teaching research to inform their teaching.
"For every assignment, academic staff have an 'unpacking' session with students so the students are clear on what is expected. The majority of universities do this. It is also about good communication and consistent style. For example, it is best practice not to write in all capital letters regardless of the sector.
The public university is located near Leeds in England's West Yorkshire.


text:
FREE UK SHIPPING ON ORDERS OVER £35 | FREE SHIPPING TO EUROPE ON ORDERS OVER £125
Explore Cerus's entire range of innovative and stylish gym clothes for men & women, designed to flatter your physique and help you smash your goals. Our exceptional gym and fitness clothing is made for performance and comfort that won't let you down. Keep checking our newest releases for the latest in men's & women gym clothing.
Fitness & Wellbeing Blogs
© 2021 Cerus. | Privacy Policy | Terms of use
SIZE CHEST CM (IN) WAIST CM (IN) HIPS CM (IN)
S 100 (39) 79 (31) 90 (35)
M 105 (41) 84 (33) 95 (37)
L 110 (43) 89 (35) 100 (39)
XL 115 (45) 94 (37) 105 (41)
XXL 120 (47) 99 (39) 110 (43)
SIZE WAIST CM (IN) HIPS CM (IN) INSIDE LEG CM (IN)
S - 32 81 (32) 99 (39) 79 (31)
M - 34 86 (34) 104 (41) 80 (31)
L - 36 91 (36) 109 (43) 81 (32)
XL - 38 96 (38) 114 (45) 82 (32)
XXL - 40 101 (40) 119 (47) 83 (34)
XS 84 (33) 60 (24) 86 (34)
S 89 (35) 65 (26) 91 (36)
M 94 (37) 70 (28) 96 (38)
L 99 (39) 75 (30) 101 (40)
S 65 (26) 104 (41) 76 (30)
M 70 (28) 109 (43) 77 (30)
L 75 (30) 114 (45) 78 (31)
XL 80 (32) 119 (47) 79 (31)
Women's dress sizes
SIZE UK DRESS SIZE US DRESS SIZE
S 8-10 4-6
M 10-12 6-8
L 12-14 8-10
We'll notify you about exclusive discounts, new releases and promotions.


What's Cooking at Your Library? Find Out at the Booklist Webinar!
Whet your patrons' appetites with these delicious reads from celebrity chefs, famous restauranteurs, and web stars. Host a pot luck or feature these cookbooks in a display for the epicureans at your library!
Click for a Full Menu of Titles.
And join us on today at 2pm ET for Booklist's free, hour-long webinar, where we'll share a sneak peek at what new titles we're cooking up for fall!
From Coco Chanel to an American woman who secretly worked for the French Resistance during World War II, these historical fiction novels are inspired by the real lives of some of history's most influential women. Click for More Historical Fiction Starring Real Women.
In honor of Hispanic Heritage Month, we're featuring titles by Hispanic authors that your patrons will want to check out. Click for Titles to Celebrate Hispanic Heritage Month.


text:
Commercial Pallets
Landscape Mulch
Wood Fuel Pellets
Wood Waste Recycling
Mulch Dealers
Pellet Dealers
Wood Pallets: Sustainable and Friendly to the Environment
In an era where we often hear about sustainability, global warming, and managing forests, it's important to know some facts about the shipping and container industry and how it uses wood. Today, we'd like to speak to it from the perspective of a pallet manufacturer and specifically as a producer of wood pallets. Pallets do move the world!
So many products travel around the world on a pallet. They move by truck, railroad, flight, and by water. Sometimes it's a manufactured product that is shipped, and other times it's a product for the food or agriculture industry. Wood happens to be one of the most eco-friendly materials that can be used to create a pallet. This is because it can be used multiple times, can be heat treated for a chemical-free way of shipping food products, and then when it's no longer useful as a pallet, the wood can still be used for a byproduct to create new products, such as mulch, bedding, or bioenergy—wood pellets.
Reusing pallets
One glance through Pinterest, and you'll see that making craft projects and furniture from old pallets is popular. But we see a lot more use in a worn or damaged pallet. First, we can refurbish a pallet or repair it. We can add new boards to used stringers to create a "combo" pallet of new and old material. We can repair broken boards, or recondition a pallet to look like new. It can be rebuilt and heat-treated, or torn down to separate the unrepairable parts from the good parts.
Not a scrap is wasted! Any boards that can't be used for a pallet go into our waste wood stash to be used for our other non-pallet products. Nails in the boards? No worries! The Rotochopper separates out the nails as it shreds.
Why wood is sustainable
Despite what we might be led to think, our forests are not depleted. In fact, according to the American Hardwood Export Council, North American forests are now growing at twice the rate of consumption. The timber industry plants 1.7 billion new trees every year, and in the last 35 years, global tree cover has increased 7.1%, according to satellite data.
By comparing daily readings, the researchers were able to see small changes occurring regularly over a relatively long period of time—which added up to large changes. Over the entire span of 35 years, the researchers found that new tree cover had offset tree cover loss by approximately 2.24 million square kilometers—approximately the size of Texas and Alaska combined.
According to the Food and Agriculture Organization (FAO), "Forest growth nationally has exceeded harvest since the 1940s. By 1997, forest growth exceeded harvest by 42 percent and the volume of forest growth was 380 percent greater than it had been in 1920."
What is a carbon footprint?
As they grow, trees use and store CO2, which means they help to fight against climate change. How? Since carbon dioxide is one of the main global warming gases and trees store CO2, they help to reduce its effect on the environment.
When carbon is in the atmosphere it takes the form of carbon dioxide. If too much carbon dioxide (CO2) builds up in the atmosphere, it acts as a warming agent contributing to climate change. Trees can help restore a balance through the carbon cycle by absorbing carbon dioxide, storing it as carbon and releasing oxygen.
Carbon is stored in a tree by the process of photosynthesis. But that isn't where it ends. Wood sinks and stores carbon.
What is the difference in sinking and storing carbon?
A carbon sink is active—a mechanism that actively absorbs carbon dioxide. A carbon store is passive—an object that holds carbon dioxide as carbon as part of its structure and makeup. The carbon stored in the tree continues to be stored in the wood products made from the harvested tree, and beyond, through reuse and recycling.
This means that wood is sustainable (we can grow more trees), friendly to the environment (trees help to produce oxygen and sink carbon dioxide), and as a building material, the wood continues to store carbon dioxide.
Our dedication to the environment
Because of the science behind the benefits of using wood for pallets, we're dedicated to being responsible with the resources we have, and to using every part of the tree for something. We're happy to see reports of an increase in the growth of new trees because we know the process of harvesting mature wood and growing more trees is also good for the environment.
Poorly managed forests can be unhealthy and unproductive because of overcrowding, disease, insects, and competition for light, water, and nutrients. Foresters use a number of management techniques, including harvesting, prescribed burning, and reforestation to work to sustain the health of a forest.
Therefore, we support the practice of carefully managing forests and responsibly using the wood that harvesting provides.
Hay Creek Pallet is a division of Hay Creek Companies and is a full-service pallet manufacturing operation providing businesses in the food industries, the paper industry, and a variety of other industries, with reconditioned, remanufactured, and new pallets. Additionally, as a by-product utilization, the company also manufactures premium wood heat pellets, wood landscape mulch, and shredded wood animal bedding by recycling wood waste material (both new and used).
Hay Creek Companies is a member of The National Wooden Pallet & Container Association (NWPCA), which is the largest organization of wood packaging professionals in the world.
HAY CREEK COMPANIES
5094 2nd Ave • Pittsville, WI 54466
© 2023 Hay Creek Companies | Crafted by North of Eight Design & Marketing


Carpenter, St. Louis, 32; Arenado, Colorado, 29; Harper, Washington, 28; Aguilar, Milwaukee, 28; Suarez, Cincinnati, 26; Muncy, Los Angeles, 25; JBaez, Chicago, 25; Story, Colorado, 24; Goldschmidt, Arizona, 24; TShaw, Milwaukee, 23.
JBaez, Chicago, 88; Suarez, Cincinnati, 88; Aguilar, Milwaukee, 83; Arenado, Colorado, 82; Story, Colorado, 79; Rizzo, Chicago, 75; Markakis, Atlanta, 73; Hoskins, Philadelphia, 72; Harper, Washington, 71; FFreeman, Atlanta, 71.
Scherzer, Washington, 15-5; Mikolas, St. Louis, 12-3; Nola, Philadelphia, 12-3; Lester, Chicago, 12-4; Godley, Arizona, 12-6; Greinke, Arizona, 12-7; Chacin, Milwaukee, 11-4; Newcomb, Atlanta, 10-5; Freeland, Colorado, 10-7; Quintana, Chicago, 10-8.


Specified in advance formats for printing leaflets and business cards, so that they are comfortable and as uniform as possible. Without standards, even leaflets could be completely unreadable and even repulsive. .


It can sometimes be difficult to decide what to keep and what to get rid of. This is true for a lot of situations in life – including genealogy. Things get harder when you are assigned the task of going through a deceased family member's belongings. There are some things that genealogists should not throw away without taking a close look at it first.
Your relative probably had a good reason to keep the newspaper clippings that you found. That clipping could have photo of a relative in it. There was a time when newspapers held recipe contests and posted the photo of the winning recipe creator – and her food – in the local newspaper. A relative who was on a high school sports team could be featured in that clipping. Did you find multiple newspapers from the same date? Check to see if a relative was mentioned in a story in that newspaper.
The family stories that are written down in the annual Christmas letter are a gold mine for genealogists and family historians! These stories can give you a glimpse into what your relatives were like when they were young. Parents often use the Christmas letter to list their children's academic, musical, or athletic achievements. It is unlikely you will find those precious stories in any other resource.
Old baby books hold more than the length and weight of your ancestor when he or she was born. A baby book could have short stories about amusing things your ancestor did or said as a baby or toddler. Some baby books have a family tree in them that could reveal relatives that you never knew about. You might find some old photos of your relative when her or she was a baby or toddler. Other family members might want a copy of those photos.
Today, it is incredibly easy to obtain a Bible. You can buy them at almost any store that sells books, and at many churches. There are plenty of internet sites that focus on Bible verses and stories. You may have a Bible at home and not feel a need to hang on to a deceased relative's Bible.
Before you donate that old Bible (or recycle if it has been severely damaged) take the time to look through it. Many families recorded their family trees on the inside covers of the family Bible. You might find prayer cards that came from the funerals of your relatives and ancestors. There could be small newspaper clippings of the obituaries of family members who have passed away tucked inside the Bible.


text:
01926 886688 CALL BACK
Governance & regulatory
Incentives and share schemes
Compliance and regulatory
Regulatory risk & compliance
Outsourcing and technology
Defamation, libel & slander claims
Employment law and HR
Share schemes and incentives
Development and securitisation
Wills and tax planning
Court of Protection & Powers of Attorney
Rural families
Health and social care law
Housing and regeneration
Finance and treasury
Charities and not-for-profit
We use cookies to track usage of our site. Details of these can be found on our Cookie Policy. You may choose to decline all tracking cookies, but if you do some key features may not work as expected.
Agree & Close
2021-09-07 Tracy Ashby 15 February 2021
OPG guidance on when deputies need special authority to act
Home / Knowledge base / OPG guidance on when deputies need special authority to act
Posted by Tracy Ashby on 15 February 2021
Tracy Ashby Head of Private Client - Partner
Appointing a Court of Protection Deputy
In handing down judgment on three cases relating to expenditure on litigation services by professional deputies, the senior judge of the Court of Protection took the opportunity to clarify when deputies must seek special authority before committing to certain legal services, and how to avoid conflicts of interest. The Office of the Public Guardian, the administrative body responsible for registering powers of attorney and supervising the activities of deputies appointed by the Court of Protection, has issued guidance based on this judgment, setting out what actions fall within the 'general authority' of a deputy and when specific authority is needed. The guidance relates specifically to a Property and Financial Affairs Deputyship, not Health and Welfare.
Special authority required for specialist legal services
Her Honour Judge Hilder made it clear in her judgment that it was not the intention of the court to stop a deputy from seeking specialist legal advice beyond their general authority, but to provide clear parameters to help them avoid conflicts of interest and ensure that their actions were in the vulnerable person's (P) best interests. The OPG's guidance reiterated that a deputy's general authority extended to dealing with general property matters such as conveyancing, tax returns, employment contracts relating to care and so on. However, general authority does not cover litigation proceedings (for instance to pursue an appeal against a ruling determining some aspect of P's care, such as CHC funding) except when conducted in the Court of Protection. Deputies are able to seek advice on whether litigation proceedings might be appropriate, including Counsel's advice on the merits of the case and its likely success. However, they should not go further than issuing a letter of claim (and reviewing the response) without special authority to incur costs if they are likely to exceed £2000 (+VAT).
Transparency is key to avoiding conflicts of interest
If litigation is considered to be in the best interests of P, deputies need to be transparent about the reasons for employing the services of their own firms. To that end, they must provide evidence that their firm is the best placed to provide that advice by obtaining alternative quotes from other providers. They must also demonstrate that instructing their own firm is in P's best interests and must set out the reasons why such an instruction does not amount to a conflict of interest.
It is accepted that there may be occasions where deputies have to act quickly and do not have the time to go out to tender. In that situation, they can apply to the court for retrospective authority to claim for costs incurred but, as HHJ Hilder was careful to point out, each application would be judged on its individual merit and the granting of such authority could not be taken for granted. Deputies must accept that if they proceed with litigation without seeking special authority, they are ultimately responsible for bearing the costs if the court does not agree that their actions were in the best interests of P.
We have considerable experience both of advising deputies on their obligations. If you need any advice on how to deal with a matter that is likely to require specialist legal advice, please do not hesitate to get in touch and we will be delighted to help.
Tags: Court of Protection & Powers of Attorney Court of Protection problems
Tracy Ashby
Head of Private Client - Partner
Since qualification as a solicitor in 2005, Tracy has gained a vast amount of knowledge and experience in all private client related issues. Advising on care fee funding and protecting vulnerable clients is her particular area of expertise.
18 January 2022 A detailed guide to probate
The process of dealing with the death of a loved one is usually talked about in emotional terms, and understandably so. But it is also a hugely complex logistical and administrative task that most people are unaware of.
17 January 2022 Proprietary estoppel claims: A recent case round up
Due to the rise of media coverage relating to this area of law, awareness of proprietary estoppel claims, is increasing.
17 January 2022 New VAT penalty regime
The new penalties will apply to VAT accounting periods beginning on or after 1 January 2023. This is a deferral by the government who had announced in the latest Autumn Statement a start date of 1 April 2022.
Trainee recruitment
Wright Hassall
Olympus Ave
enquiries@wrighthassall.co.uk 01926 886688
© Wright Hassall 2022


The old resurfaced tweets from Guardians of the Galaxy director James Gunn quickly became a big controversy and as a result, Disney officials decided to fire Gunn from the third film of the franchise.
Since his firing, numerous cast and crew members have shown their support for the released director, including the former WWE Star Dave Bautista.
Now it'd be interesting to see how Marvel and Disney officials reply to Batista and if all this support is enough to bring back James Gunn into the MCU fold.
Too many franchises, is it too hard to write original movies today?
I want him to return to WWE and take away one of the world championships from the midgets for as long as possible.
After completely botching the Star Wars franchise and alienating fans with their social justice bullcrap, Disney needs to stick the to straight and narrow and not piss anyone else off.
If he were alone you might be right, but the whole cast is with him, which gives Disney a bad rep. Disney has more to lose in that situation which means they will probably back down. That is my prediction.
Guardians is done… with this much negative crap, another movie franchise bites the dust.
Give 'em hell, Dave. People in this country have a right to stand up for their beliefs. Hopefully, Disney will do the right thing.
Trying to back out of a contract on a Hollywood movie will not go well for him, especially against a juggernaut like Disney. Regardless of how you feel about Gunns dismissal, you cannot just ask to be let out of such a critical contract.


The RBI governor Raghuram Rajan announced on Tuesday that there would not be any change in the Repo rate and Reverse Repo rate. He said that the RBI estimates the inflation to be of five percent in 2017 on the basis of the wholesale price index. He said the present Repo rate at 6.75 percent and the Reverse Repo rate at 5.75 percent and the cash reserve ratio at 4 percent continued.


text:
How to have a career within the performing arts
Live performance and theatre have certainly been through a difficult period but despite this, the arts are still a wonderful community to be a part of. Now, one theatre school franchise is doing a major marketing push to recruit more teachers to be part of their award-winning network.
Razzamataz Theatre Schools is a huge provider of job opportunities to people wanting to work in the performing arts industry. Operating for more than 21 years, the franchise also has a rich history of progressing teachers' careers.
"The teachers are the heart and soul of our theatre schools and are responsible for directly influencing the students' experiences," says Denise Hutton-Gosney, MD and Founder of Razzamataz Theatre Schools. "We really invest in our teachers because of the crucial role they play, which is why many stay with us for years and many build their careers with us."
Razzamataz teachers come from a wide range of backgrounds, some work professionally in the performing arts industry and others start their journey when they are very young.
Matty Aitchison-Rayson, franchisee from Razzamataz West Cumbria, started his career with Razzamataz when he had just turned 17 years old as a work experience placement student.
"I absolutely loved the work experience working with Denise and the team of teachers and all the students," says Matty. "I learnt and developed so many key transferable skills from this experience such as how to communicate effectively when working with both staff, students and parents and guardians. I learnt so much about the business and developed my love for the Performing Arts and working with young people."
There are many other examples of teachers who now run their own Razzamataz schools including Michaela Crumpton, multi-franchise owner of Razzamataz Bristol North and Bristol South, Marc Kelly, multi-franchise owner of Razzamataz Hackney and Radlett, Asha Richardson, multi-franchise owner of Razzamataz South Lakes, Penrith, Kendal and Appleby, Debbie Mitchell, multi-franchise owner of Razzamataz Carlisle and Dumfries and Chloe Lee, franchise owner of Razzamataz Chester.
Although this career progression has been happening for a number of years, Razzamataz is supporting it further with the launch of a Teacher Training Academy to guide those looking to take the next steps in their careers.
"We already offer Manager Training, which allows our multi-unit franchisees the option of promoting a teacher to a manager. They then go on to our course and receive the exacting Razzamataz training in customer service, marketing and PR and how to run the school on a day-to-day basis," adds Denise. "`We have really ambitious growth plans, which includes giving teachers a huge range of opportunities within our network with the launch of our Teacher Training Academy."
The multi-award-winning network is currently looking for singing, dance and drama teachers across the UK to inspire children and gain work in a supportive and nurturing environment.
At newly opened Razzamataz Radlett, singing teacher Tasha says: "There are so many things that I love about teaching at Razzamataz. Seeing the smiles on the kids' faces and the way they come into the room a little bit shy and the way they leave the room full of confidence, it is just so lovely."
For teachers looking to grow their career, being part of a large network allows them to share and learn the very best teaching techniques and implement ways in which to maximise every child's potential.
For teaching positions across the UK, contact individual Principals directly via our website or contact our Head Office. Email: franchise@razzamataz.co.uk
Telephone: +44 7821 122242


text:
Vandoorne 'hurt' at being overlooked by Mercedes to replace Hamilton
The Belgian driver has questioned the decision to use George Russell.
Stoffel Vandoorne was hurt and disappointed by Mercedes' decision to name George Russell as Lewis Hamilton's replacement for this weekend's Sakhir Grand Prix.
Seven-time F1® champion Hamilton has been ruled out of the penultimate race of the 2020 season after testing positive for coronavirus on Monday.
Mercedes reserve drivers Vandoorne and Esteban Gutierrez seemed natural fits to stand in for Hamilton, but the seat instead went to Russell, with Williams agreeing to release the 22-year-old Briton.
Vandoorne, who flew to Bahrain on Tuesday to provide cover for Hamilton, took to social media to air his displeasure but says he respects the Silver Arrows' decision.
"Firstly, I want to wish Lewis Hamilton a speedy recovery and hopefully he'll be back in the car asap! Nobody can really replace him!" he posted in a statement on social media.
pic.twitter.com/nFiuB51euf
— Stoffel Vandoorne (@svandoorne) December 3, 2020
"Obviously, I'm disappointed not to get the chance to drive for Mercedes this weekend. After having spent the year travelling to all F1® races and dedicating so much time, physical training, commitment to this programme… it hurts!
"On the other hand, I respect the choice of putting George Russell in the car. He has been one of the standout drivers and he fully deserves this opportunity.
"It's moments like these that make my drive even bigger and I can assure you I'm going to continue giving 100 per cent."
Some have suggested Mercedes development driver Russell is effectively auditioning for a Silver Arrows seat in 2022, but he is unconcerned by that talk.
"There have been no targets expected of me by Mercedes because you can't judge somebody off one race," he said. "It's not in the back of my mind. I'm going to go out there, enjoy it and learn as much as I can.
"I have a job to do on my return to Williams and try and push that team forward. I need to take this opportunity with both hands and maximise the learning.
"The future will be judged over the course of a season, and what's happened in the past, not just one or maybe two races."
Russell will team up with Valtteri Bottas, who is the favourite to win his third race of the season in Hamilton's absence.
Bottas, who has only finished higher than eighth in two of the past five races, is determined to succeed in Bahrain.
Morning pic.twitter.com/dEtfpesRGp
— George Russell (@GeorgeRussell63) December 3, 2020
"For me, it's a great motivation to have a new team-mate," he told Sky F1. "Obviously I'll try to be ahead of him and win the race, that's the only thing on my mind.
"Being a part of the team for a few years, obviously everyone would expect me to be on pole and win the race. That's my goal.
"I'll try to end this season on a positive note. I'll try to make the most out of it and hopefully that can help me for the future.
"I'm sure this young, hungry guy will give it all he's got. He's a talented guy and he's super excited to be in the team."
But Bottas, under contract with Mercedes until the end of next season, acknowledged there is some added pressure to finish ahead of his temporary team-mate.
"If he beats me, for sure it wouldn't look so good on me if I had a normal race fair and square, so I try to avoid that," he said.
"But I'm not a person who thinks like that. I try to turn things into motivation and positive thinking and that's my mindset for this weekend."
Absent Hamilton opens door for Bottas and Russell
The Ferrari Academy graduates who made it to F1®
Piastri moves up to FIA F2 with Prema Racing


text:
Ryan Marshall
Salisbury Daily Times
Indian River leaned on running back duo of George Martin and Gerald Foreman all evening.
But when they needed a big play the most, the Indians pulled the unexpected.
With the ball on the 37 down 15-12 in the fourth quarter to Milford, Martin took the toss but didn't advance forward. The halfback chucked a long pass down the field to a wide open Deshawn Aiken, who hauled it in at the 6-yard line with a minute left in the game.
"My heart was beating as fast as it can," Martin said. "I just threw the ball. I didn't really think about putting on a spot."
After Foreman was stuffed on first down, Martin punched it in from 5 yards out to give Indian River (3-3) the 22-18 win against Milford (4-2).
Martin finished the game with 138 yards and three touchdowns, and Foreman rushed for 104.
But the Indians could not count out the Buccaneers.
After IR took a 15-12 lead with 5:17 left in the game – on a Martin 3-yard touchdown run and Foreman two-point conversion – the Bucs responded.
Milford scored with a three-play, 90-yard drive in one minute to take the 18-15 lead.
The Buccaneers' workhorse David Bowman had two carries, one for 50 and the other for 38 yards, to give Milford the 18-15 lead.
Bowman finished the game with 159 yards and a touchdown for Milford.
"David Bowman is a workhorse," Milford head coach Shaun Strickland said. "When it comes to heart and determination, he's got it. For his size, he's got a bulk of it. You'll never have to second guess him."
After the Indians took the final lead with 1:04 left, Bowman almost pulled it back for the Bucs.
Taking the squib kick on the 20-yard line, Bowman found a hole.
He was gone.
Except for the one foot that stepped out of bounds on the 47-yard line of the Indians.
"He was forced out of bounds," Strickland said. "Hey, they make a great play. He did a great job getting free."
Milford was able to get the 43 but two passes by backup quarterback Marcus Correa fell incomplete.
On the final one, Foreman deflected it away.
Indian River had to run one final play.
The victory formation with senior quarterback Jake Hudson.
The recently crowned homecoming king took a knee to seal the game.
"It means a lot," IR head coach Phillip Townsend said. "Our guys, first three games, I felt like we didn't play 48 minutes of football. …This week, we came back and won 22-18. Our guys are now playing together as a team and four quarters.
"It's going to make the difference."
In the first half, the Indians could have run away with the game. Milford totaled just 36 yards in the first half while IR totaled 229.
It was the mistakes that cost the Bucs. A kickoff return for a touchdown by Rashawn Price was called back for holding in the first quarter. Still with good field position, Bowman fumbled in the red zone to give IR the ball back.
"They didn't hang their heads," Strickland said. "We kept fighting. We're still not there yet where we can come out and make mistakes."
On the Indians' drive, the Bucs caught a break. Attempting a 29-yard field goal, Andrew White's kick was blocked and fell right into the arms of Son Martinez, who scampered down the sideline for the 91-yard score with one minute left tin the first quarter.
The Indians had taken the lead on Martin's first of three touchdowns on the night, but the score stayed idol 7-6 until halftime.
Quarterback William Kimmel opened the scoring in the second half for Milford with a 15-yard touchdown run 1:50 into the half.
But Martin would rally the troops with two more touchdowns to get the win for IR.
Milford will likely have to win every game to make the playoffs.
"All of them," Strickland said. "As I told the kids today, 'We're in a must-win situation if you want to go after the Henlopen South.' "
Indian River improved to 3-0 in the Henlopen South after losing its first three games of the season.
Seaford, Laurel and Friday's Milford win were big for IR.
Now Lake Forest, who beat Laurel on Friday, and IR are both 3-0 in the division.
But the Indians are OK with that.
After Bowman's rushing touchdown to take the lead late in the fourth quarter, Martin and the team talked.
The Indians knew they needed a win. Now, with three more south games against Woodbridge, Delmar and Lake Forest, it's a must-win mentality for IR.
"If we lose any of the south games, our playoff hopes are going to get shot down," Martin said. "We have to win every game here on out. Setting our sights on the next game is the biggest thing for us."
On Twitter/Instagram: @ByRyanMarshall


text:
Date -
Congress117th Congress (2021 - 2022)116th Congress (2019 - 2020)115th Congress (2017 - 2018)114th Congress (2015 - 2016)113th Congress (2013 - 2014)112th Congress (2011 - 2012)111th Congress (2009 - 2010)110th Congress (2007 - 2008)109th Congress (2005 - 2006)108th Congress (2003 - 2004)107th Congress (2001 - 2002)106th Congress (1999 - 2000)105th Congress (1997 - 1998)104th Congress (1995 - 1996)103rd Congress (1993 - 1994)102nd Congress (1991 - 1992)101st Congress (1989 - 1990)100th Congress (1987 - 1988)99th Congress (1985 - 1986)98th Congress (1983 - 1984)97th Congress (1981 - 1982)96th Congress (1979 - 1980)95th Congress (1977 - 1978)94th Congress (1975 - 1976)93rd Congress (1973 - 1974) Year202220212020201920182017201620152014201320122011201020092008200720062005200420032002200120001999199819971996199519941993199219911990198919881987198619851984198319821981198019791978197719761975197419731972197119701969196819671966196519641963196219611960195919581957195619551954195319521951195019491948194719461945194419431942194119401939193819371936193519341933193219311930192919281927192619251924192319221921192019191918191719161915191419131912
Add terms to search
6James D. Wolfensohn
2Bill Gates
1Bill Clinton
1Al Gore Jr.
1Robert Chase
1Kofi Annan
1Gro Harlem Brundtland M.D.
1Carol Bellamy
1Jack Cushman
1Joaquim Chissano
James D. Wolfensohn
Sorted by Most Popular
Sort by Best Match Most Popular Most Views Most Recent Airing Least Recent Airing Most Recent Event Least Recent Event
Search Program Text
Showing 1 - of 6
April 5, 2000 Last Aired April 6, 2000
Closing the Digital Divide
Panelists talked about increasing access to computer technology both domestically and in developing nations. They concentrated on the areas of technology, health and education, saying that…
January 10, 2000 Last Aired January 11, 2000
Impact of AIDS on Africa
Vice President Al Gore (D) presided over a session of the United Nations Security Council called to debate the AIDS crisis in Africa. Participants said that the AIDS epidemic had reached the level of…
January 31, 2000 Last Aired February 18, 2000
Laying the Foundation for Global Health
World business and political leaders discussed the process of globalization. They tried to find solutions to problems in areas of health, agriculture, and poverty around the world. Specific…
April 17, 2000 Last Aired April 18, 2000
IMF/ World Bank News Conference
Mr. Wolfenson and Mr. Nimmananahaeminda talked to reporters about the World Bank meeting. They said that the meeting focused on reducing poverty and dealt with the AIDS epidemic. Following…
March 14, 2000 Last Aired March 14, 2000
The World Bank in the 21st Century
Mr. Wolfensohn spoke about the role of the World Bank. He argued that the policies of his organization were helpful to all parties. He explained the numerous difficulties that the…
Spring Meeting Opening News Conference
Mr. Wolfensohn spoke about the World Bank meeting agenda and about opposition gathered in Washington. Following his remarks he answered questions from reporters.