text:
\begin{document}

\title[Vanishing of local cohomology, set-theoretically Cohen-Macaulayness ]
{Vanishing of local cohomology and set-theoretically Cohen-Macaulay
ideals $$\text{To\ the\ soul\ of\ professor\ Wolfgang\ Vogel}$$}

\begin{abstract}
In this paper, first, we generalize a result of Peskine-Szpiro on
the relation between the cohomological dimension and projective
dimension. Then, we give conditions for the vanishing of local
cohomology from local to global and vice versa. Our final goal in
the present paper is examining the set-theoretically Cohen-Macaulay
ideals to find some cohomological characterization of these kinds of
ideals.
\end{abstract}

 \maketitle

\section{INTRODUCTION}

Throughout this paper, all rings are commutative and Noetherian with identity.
 After Grothendieck's talk \cite{Gro67} in the early 1960s up to now, local cohomology has
 become an essential tool in algebraic geometry and commutative
 algebra. For an ideal $I$ of a ring $R$, the local
cohomology modules $H^i_I(R)$ may be considered as the isomorphism
$$H^i_I(R) = {\vil}_t \Ext^i_R(R/I^t,R) \text{\ for\ } i > 0.$$

 In their landmark paper, Peskin and Szpiro \cite[Proposition 4.1]{PS} proved that
 whenever $(R, \fm)$ is a regular local ring containing a field of positive characteristic and $I \subset R$ is
a Cohen-Macaulay ideal (i. e. the ring $R/I$ is Cohen-Macaulay),
$H^i_{I}(R)=0 \text{\ for\ all \ } i > \dim R-\depth R/I$. An
immediate implication of their result is that the inequality
$\cd(R,I) \leq \pd R/I$ holds, where cohomological dimension $\cd(R,
I)$ resp. projective dimension $\pd R/I$ of $I$ is defined as
$$\cd(R, I) = \min \{i : H^j_I(R) = 0 \text{\ for\ all\ } j > i\},$$
resp.
$$\pd R/I=\sup \{ n|\ \Ext^i_R(R/I,N)=0,\ \text{for\ all\ } R-\text{mod.\ } N\ \text{and\ all\  }i \geq n+1 \}.$$
In this direction, in Section $3$, (cf. Theorem \ref{PS Generalize}) we generalize the mentioned result using an action
of a ring morphism on local cohomology (without any restriction on
the characteristic of the ring), introduced by Singh and Walther
\cite{S-W}.

 The ring endomorphism $\varphi: R \rightarrow R$
induces a natural $\varphi$ action on the local cohomology modules
$\varphi_{\ast}: H^i_{I}(R) \rightarrow H^i_{\varphi(I)R}(R)$ via
$\varphi (r) \varphi_{\ast} (\eta) =\varphi_{\ast} (r \eta)$, where
$r \in R, \eta \in H^i_{I}(R)$ which is an endomorphism of the
underlying Abelian group (details, including notation, are given in
Section $2$). This action as a generalization of Frobenius action,
is an effective tool in the study of local cohomology modules as it
was used recently by \`{A}lvarez Montaner in \cite{Al} and the author
with his colleagues in \cite{EABR}, and with Boix in \cite{B-E}.

From the other point of view, the result of  Peskin and Szpiro says
 that the invariants $\Ht (I)$ (height of ideal $I$) and $\cd (R,I)$ are equal.
  Recently,  M. Varbaro showed that their idea works for all Noetherian rings
  of positive characteristic (cf. \cite[Corollary 2.2]{Var}).
   To do so further and motivated by a conjecture of Hartshorne (see Discussion \ref{Hartshorne})
   we find out the equivalent conditions between Cohen-Macaulayness of $R/I$ and the equality $\Ht (I)=\cd (R,I)$, See Theorem \ref{CM cond}.
   Let us recall that for an ideal $I$, one has $\Ht (I) \leq \cd (R,I) \leq \ara (I)$, where the arithmetical rank of $I$
denoted $\ara (I)$ is the smallest $s$ such that there exist $s$
elements $g_1, \ldots , g_s$ that generate $I$ up to radical. $I$ is
called a set-theoretically complete intersection ideal whenever $\Ht
(I) = \cd (R,I) = \ara (I)$.
   Note that the ideals satisfying the equality $\Ht (I)=\cd (R,I)$ in a Gorenstein ring are called cohomologically
   complete intersection ideals, introduced and
examined by Hellus and Schenzel in \cite{Hel-Sch}. See also \cite{E}
for more information. Notice that a set-theoretically complete
intersection ideal is a cohomologically complete intersection one
but the converse is no longer true.

As for strengthening of the aforementioned results, one may ask the
following question.

\begin{question} \label{local-global}
Let $(R, \fm)$ be a regular local ring and let $I$ be an ideal. For a given integer $i$, under what kind of conditions, the statement
$$H^i_I(R)=0 \text{\ if\ and\ only\ if\ } H^{\dim R-i}_{\fm}(R/I)=0$$
 holds?
\end{question}

An affirmative answer has been given in \cite[Theorem1 (iii)]{Lyu84}, \cite[Corollary 6.7]{Mi} and \cite[Corollary 4.2]{S-W} whenever $R$ is a polynomial ring over a field $k$
and $I$ is a squarefree monomial ideal of $R$. To generalize the mentioned result and as a positive answer to the Question \ref{local-global}, in Section 4, we give the following.

\begin{thrm}
Let $(R, \fm)$ be a regular local ring satisfying the (Inj)
property and $\varphi:R \longrightarrow R$ be a local flat ring
endomorphism. Then for a given integer $i$,
$$H^i_I(R)=0 \text{\ if\ and\ only\ if\ } H^{\dim R-i}_{\fm}(R/I)=0.$$
\end{thrm}

Let $X$ be a Cohen-Macaulay scheme. It fails to be true that  $X_{red}$ the corresponding
 reduced scheme is Cohen-Macaulay. The further step is asking whether for a given scheme $X$
  there is a Cohen-Macaulay ring $R$ such that $X$ is isomorphic to $\Spec R_{red}$. This kind of a scheme called a  set-theoretically Cohen-Macaulay one.

In general, in a regular local ring $R$ an ideal $I$ is
\textit{set-theoretically Cohen-Macaulay} if there exists an ideal
$J \subset R$ with $\rad (I)=\rad (J)$ such that the ring $R/J$ is
Cohen-Macaulay. It is evident that set-theoretically complete
intersection ideals, determinantal ideals and Cohen-Macaulay radical
ideals are set-theoretically Cohen-Macaulay. If $I$ is a
Cohen-Macaulay monomial ideal of a polynomial ring $R=k[x_1, \ldots , x_n]$ over a field $k$,
  then $I$ is set-theoretically Cohen-Macaulay (cf. \cite[Theorem 2.6]{HTT}). For instance,  principal
  and Veronese monomial ideals have such a property, \cite[Theorem 3.2]{HH}.  See \cite{S-W2} for some counterexamples.

It is a natural question to ask on the relation between
Cohen-Macaulay and set-theoretically Cohen-Macaulay ideals.

Our first result in this direction is Proposition \ref{equivalence},
which is a consequence of Theorem \ref{Main vanishing}. It states
that in a regular local ring $(R,\fm)$ with ideals $I$ and $J$,
where $\rad(I) = \rad(J)$ and triples $(R, I, \varphi)$ and $(R, J,
\varphi)$ have both (Flat) and (Inj) properties, then the concepts
of Cohen-Macaulayness and set-theoretically Cohen-Macaulayness are
the same. In fact, Proposition \ref{equivalence} is a generalization
as a result of C. Huneke appeared in \cite[Section 6]{EMS}.

Not so much is known about the cohomological characterization of
set-theoretically Cohen-Macaulay ideals. In the following, we show
that under certain conditions,  set-theoretically Cohen-Macaulay
ideals are the same as cohomologically complete intersections.

\begin{thrm} (cf. Theorem \ref{setCM-setCI}) Let $(R,\fm)$ be a  regular local ring and the triple $(R, I, \varphi)$ has (Flat) property.
 Suppose that $I$ is a set-theoretically Cohen-Macaulay
 ideal, then $\Ht(I)=\cd (R,I)$.
The reverse statement holds true,
 if $\varphi: R \rightarrow R$ is pure.
\end{thrm}

Another way to study set-theoretically Cohen-Macaulayness of a
scheme is the linkage concept. Roughly speaking, the linkage is the
study of two subschemes where their union has nice properties.
Exploiting this concept one may consider a subscheme linked with the
second one which understood well.

 We cite the following
paragraph from a fruitful paper of A. Martsinkovsky and J. R.
Strooker \cite{Ma-St} on the significance of the concept of linkage
theory.

\emph{ 'It goes back to the late 19th and early 20th century, when
M. Noether, Halphen, and Severi used it to study algebraic curves in
$\mathbb{P}^3$. Linkage allows to pass from a given curve to another
curve, related in a geometric way to the original one. Iterating the
procedure one obtains a whole series of curves in the same linkage
class. The usefulness of this technique is explained by two
observations: (a) certain properties of the curve are preserved
under linkage, and (b) the resulting curves may be simpler, and thus
easier to handle, than the original one.'}

In the same vein, Peskine and Szpiro in \cite[Proposition
1.3]{P-S74} proved that in a Gorenstein local ring if $I$ and $J$
are linked ideals and $I$ is a Cohen-Macaulay ideal then so is $J$.
The same result proved by Schenzel in \cite[Corollary 3.3]{Sch82}
for Buchsbaum ideals. In this direction, in Corollary \ref{setlink}
we give conditions where the set-theoretically Cohen-Macaulay
property is sharable between two linked ideals.

\section{NOTATIONS AND REMARKS}

We start this section with the idea used by Singh and Walther in \cite{S-W}.

Let $A$ be a commutative Noetherian ring with a flat endomorphism
$\varphi: A \rightarrow A$, and let $\fa$ be an ideal of $A$. We
denote by $\varphi_{\ast}A$ the following $(A,A)$-bimodule: for any
$r, r_1, r_2 \in A$,
$$r_1.(\varphi_{\ast}r).r_2:=\varphi_{\ast}(\varphi (r_1)rr_2).$$

Let $\Phi$ be the functor on the category of $A$-modules with $\Phi(M)=\varphi_{\ast} A \otimes_A M$.
The iteration $\Phi^t$ is the functor
$$\Phi^t(M)=\varphi_{\ast} A \otimes_A \Phi^{t-1}M,\ t \geq 1,$$
where $\Phi^0$ is interpreted as the identity functor; the reader will easily note that the flatness of $\varphi$ is equivalent to the exactness of $\Phi$. At once, one can realize that $\Phi^t(M)=\varphi^t_{\ast} A \otimes_A M$.

Let us notice that $\Phi(A) \cong A$ given by $\varphi_{\ast}r' \otimes r \mapsto \varphi (r)r'$. Furthermore, if $M$ and $N$ are $A$-modules, then (see \cite[2.6.1]{S-W}) there are natural isomorphisms
$$\Phi(\Ext^i_A(M,N)) \cong \Ext^i_A(\Phi(M), \Phi(N)),\ \text{\ for\ all\ } i \geq 0.$$
Assume, in addition, that the ideals $\{\varphi^t(\fa)R\}_{t\geq 0}$ form a descending chain cofinal
with the chain $\{\fa^t\}_{t\geq 0}$, where ${\fa}$ is an ideal of $A$. Under these assumptions, one can easily check (see \cite[page 291]{S-W}) that
$$\Phi(H^i_{\fa}(A)) \cong H^i_{\varphi(\fa)}(A)\cong H^i_{\fa}(A),\ \text{\ for\ all\ } i \geq 0.$$

Now, let us fix our notations. Throughout this section, let $R$ be a  commutative Noetherian local
 ring and $\varphi:R \longrightarrow R$ be a local ring endomorphism. Suppose that $I$ is an ideal of
$R$ and $\{\varphi^t(I)R\}_{t \geq 0}$ is a decreasing chain of ideals cofinal with $\{I^t\}_{t \geq 0}$. Keep these assumptions, we introduce the following notations.

\begin{nt} Let  $\varphi:R \longrightarrow R$ be a flat ring endomorphism (a ring map
  $R \longrightarrow S$ is called flat if $S$ is flat as an $R$-module). In this case, we say that triple $(R, I, \varphi)$ has (Flat) property.
\end{nt}

\begin{lem} \label{Matsumura} Let $A,B$ be two rings. If $B$ is a faithful flat $A$-algebra
 then for an $A$-module $M$ one has $B \otimes_A M$ is $B$-flat if and only if $M$
is $A$-flat.
\end{lem}

\begin{proof} \cite[Exercise 7.1]{Mat}.
\end{proof}

In Lemma \ref{Matsumura}, put $B:=\hat{R}$, the $\fm$-adic
completion of local ring $R$ and $M=A:=R$. Then for every flat ring
endomorphism $\varphi:R \longrightarrow R$, the induced ring
endomorphism $\hat{\varphi}:\hat{R} \longrightarrow \hat{R}$ is
flat.

\begin{definition}
For two rings $A$ and $B$, a ring map $f:A \rightarrow B$ has
  the going-down property if, for any two primes $\fp' \subsetneq \fp$ of
   $A$ and any prime $\mathcal{P}$ in $B$ with $f^{-1}(\mathcal{P})=\fp$,
    there is a prime $\mathcal{P'} \subsetneq \mathcal{P}$ of $B$ such that
    $f^{-1}(\mathcal{P'})=\fp'$.
\end{definition}

\begin{nt} Let  $\varphi:R \longrightarrow R$ be a ring endomorphism
satisfying the going down property. Suppose that $I$ is an ideal of
$R$ with finite projective
 dimension.  We say that the triple $(R, I, \varphi)$ has (GD) property.
\end{nt}

\begin{example} \label{flat} Notice that every flat homomorphism implies the going down property. Take into account that the Frobenius endomorphism of a ring of positive characteristic and for any field $k$ and any integer $t \geq 2$, the $k$-linear endomorphism $\varphi (x_i)=x^t_i$ of $k[x_1, \ldots , x_n]$ are the prototypical examples of the flat endomorphism. In general, an extension $A \subset B$ of domains, with $A$ normal and $B$ integral over $A$ satisfies going down property, for a ring homomorphism  $f:A \longrightarrow B$.
\end{example}

\begin{nt} Suppose that
$$\Ext^i_R(R/\varphi^t(I)R,R) \rightarrow \Ext^i_R(R/\varphi^{t+1}(I)R,R),$$
is injective for all $i \geq 0$ and all $t \geq 0$. We say $(R, I,
\varphi)$ has (Inj) property.
\end{nt}

\begin{definition} \label{pure}
A ring homomorphism  $f:A \rightarrow B$ is pure if the map $f
\otimes 1:A \otimes_A M \rightarrow B \otimes_A M$  is injective for
each $A$-module $M$.  If $A$  contains a field of characteristic $p
> 0$, then $A$ is $F$-pure if the Frobenius homomorphism
  $F:A \rightarrow A, a \mapsto a^p$ is pure, where $a \in A$.
\end{definition}

\begin{remark} \label{pure2} Note that by virtue of
 \cite[Lemma 6.2]{Ho-Ro} purity of $\varphi:R \longrightarrow R$
 implies the same property for $\bar{\varphi}:R/I \longrightarrow
 R/I$. It follows from \cite[Theorem
2.8]{S-W}, that in the case $R$ is a regular ring and triple
$(R,I,\varphi)$ satisfies the (Flat) property, purity of
$\bar{\varphi}:R/I \longrightarrow R/I$
 implies that $(R,I,\varphi)$ has (Inj) property.

Let $R=k[x_1, \ldots , x_n]$ be a polynomial ring over a field $k$. There exists a $k$-linear endomorphism
 $\varphi:R \longrightarrow R$ with $\varphi (x_i)=x^t_i$ for $1 \leq i \leq n$ and positive integer $t$.
  If $I \subset R$ is an ideal generated by square-free monomials, then there exists a pure endomorphism
   $\bar{\varphi}:R/I \longrightarrow R/I$, (cf. \cite[Example 2.2]{S-W}).
\end{remark}

\begin{discussion} \label{discuss}
Let $R$ be a normal ring and $I$ an ideal of $R$ with a ring endomorphism $\varphi:R \longrightarrow R$ with $\varphi (I) \subseteq I$. Then $R$ is integral over itself and  $\varphi:R \longrightarrow R$  satisfies going down property. It implies that $\rad(\varphi (I)R) \subseteq \rad(I)$ (for instance let $R$ be either $R=k[x_1, \ldots , x_n]$ a polynomial ring over a field $k$ or a regular local ring of characteristic $p > 0$ and $I$ be an ideal of $R$, where $\varphi:k[x_1, \ldots , x_n] \longrightarrow k[x_1, \ldots , x_n], x_i \mapsto x^t_i$ and $\varphi:R \longrightarrow R, r \mapsto r^p$ and so one has $\varphi (I) \subseteq I$ ). On the other hand, as $\varphi (\rad(I))R \subseteq \rad (\varphi(I)R)$ then,
   $$\rad(I) \subseteq \varphi (\rad(I))R \cap R \subseteq \rad (\varphi(I)R) \cap R=\rad (\varphi(I)R).$$
   Thus, $\rad(I)=\rad (\varphi(I)R)$. It then follows that $\{I^t\}_{t \geq 0}$ and  $\{\varphi^t(I)R\}_{t \geq 0}$ are cofinal. Hence, the triple $(R,I,\varphi)$ satisfies $(GD)$ property.
\end{discussion}

 \section{ON A RESULT OF PESKINE-SZPIRO}

We start by recalling properties of some homological invariants.
Then discuss and examine their relationships using ring
endomorphism. Let $(R,\fm)$ be a local ring and $I$ be an ideal of
$R$ of finite projective dimension ($\pd R/I < \infty$). The
Auslander-Buchsbaum formula says that $\pd_R R/I=\depth R-\depth
R/I$. On the other hand Huneke and Lyubexnik in \cite[Theorem
1.1]{Hun-Lyu} give conditions for $\cd (R,I) \leq \dim R-2$, where
$R$ is a regular local ring containing a filed. To do so further, we
find conditions for the equality $\cd(R,I)=\dim R- \depth R/I$.

\begin{lem} \label{EABR} (cf. \cite[Lemma 3.4]{EABR})
Let $(R, \fm)$ be a Noetherian local ring and $I \subset R$ an ideal of
finite projective
 dimension. Assume that $\varphi:R \rightarrow R$ is a ring  endomorphism
 satisfying the going down property and $J=\varphi(I)R$. Then $\depth R/I \leq \depth R/J$.
\end{lem}

\begin{cor} \label{depth equality}
Let $(R, \fm)$ be a Cohen-Macaulay complete local ring. Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has both (GD) and (Inj);
\item [(b)]  projective dimension of $R/I$ is finite;
\item [(c)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
\end{enumerate}
 Then, the following statements are true.
\begin{enumerate}
\item [(1)] $\depth_R R/I = \depth_R R/ \varphi^t(I)R,\ \ \text{\ for \ all \ } t \geq 0$.
\item [(2)]  $R/I$ is Cohen-Macaulay if and only if  $R/ \varphi^t(I)R$, is Cohen-Macaulay for all  $t \geq 0$.
\item [(3)] $\Ass_R R/I = \Ass_R R/ \varphi^t(I)R,\ \ \text{\ for \ all \ } t \geq 0$.
\end{enumerate}
\end{cor}

\begin{proof}
(1) Using  lemma \ref{EABR} one may write $$\depth_R R/I \leq
\depth_R R/ \varphi(I)R \leq \depth_R R/ \varphi^2(I)R \leq
\cdots.$$ On the other hand, suppose that $\depth_R R/I =u$. As $(R,
I, \varphi)$ has (Inj) property,  using the Grothendieck's local
duality one has epimorphisms
$$H^u_{\fm}(R/\varphi^{t+1}(I)R,R) \rightarrow H^u_{\fm}(R/\varphi^{t}(I)R,R),$$
of non zero modules, for all $t$. It implies that $\depth_R R/
\varphi^t(I)R=u$ for all $t$.

(2) Note that $\dim_R R/I = \dim_R R/ \varphi^t(I)R$, for all $t$,
as $\rad (I)=\rad (\varphi(I)R)$. Now, the claim follows from part
(a).

(3) Note that for a prime ideal $\fp$ of $R$, $\fp \in \Ass_R M$ if
and only if $\fp R_{\fp} \in \Ass_{R_{\fp}} M_{\fp}$, where $M$ is
an $R$-module. Thus, one may assume that $\fp=\fm$. Hence, it is
enough to prove that $\fm \in \Ass_R R/I$ if and only if $\fm  \in
\Ass_R R/ \varphi^t(I)R$, where the assertion follows from part (a).
\end{proof}

Now is the time to prove the first main result of this section;

\begin{thrm} \label{PS Generalize} Let $(R,\fm)$ be a  Gorenstein complete local ring.
Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has (GD) property;
\item [(b)]  projective dimension of $R/I$ is finite;
\item [(c)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
\end{enumerate}
 Then one has $\cd(R,I) \leq \pd R/I$.

If in addition, the triple $(R, I, \varphi)$ satisfies (Inj)
property, then the equality holds.
\end{thrm}

\begin{proof} Suppose that $\pd R/I=u$ for some integer $u$. By the
Auslander-Buchsbaum formula $\depth R/I=\dim R-u$. As every composition
of maps satisfying going-down property satisfies that property
itself, exploiting Lemma \ref{EABR} one has
$$H^i_{\fm}(R/\varphi^t(I)R)=0 \text{\ for\ all\ } i < \dim R-u \text{\ and\ all\ } t.$$
Taking inverse limit and then using local duality  we conclude that
 $H^{\dim R-i}_I(R)=0$ for
all $\dim R-i > u$, i. e. $\cd(R,I) \leq \pd R/I$. To this end note
that inverse limit commutes with direct limit in the first place of
$\Hom_R (-,-)$.

 In order to prove the equality $\cd(R,I) = \pd R/I$, it is enough to prove $\cd(R,I) \geq \pd R/I$.
 It follows from
the assumptions and Corollary \ref{depth equality} that the
homomorphisms
$$H^{\dim R-u}_{\fm}(R/\varphi^{t+1}(I)R,R) \rightarrow H^{\dim R-u}_{\fm}(R/\varphi^{t}(I)R,R),$$
induced from natural homomorphism $R/\varphi^{t+1}(I)R \rightarrow
R/\varphi^{t}(I)R$, are surjective of nonzero modules for all $t$.
Then  $ {\vpl}_t H^{\dim R-u}_{\fm}(R/I^t) \neq 0$ and $ {\vpl}_t
H^{i}_{\fm}(R/I^t) = 0$ for all $i < \dim R-u$. To this end note
that as $\{\varphi^t(I)R\}_{t \geq 0}$ and $\{I^tR\}_{t \geq 0}$ are
cofinal, one has $ {\vpl}_t H^{\dim R-u}_{\fm}(R/\varphi^t(I)R)
\cong {\vpl}_t H^{\dim R-u}_{\fm}(R/I^t)$.

By virtue of the Grothendieck's Local Duality, from $ {\vpl}_t
H^{\dim R-u}_{\fm}(R/I^t) \neq 0$ we deduce that $H^{u}_{I}(R) \neq
0$. Thus, $u \leq \cd (R,I)$ that is $\cd(R,I) \geq \pd R/I$, as
desired.
\end{proof}

Next, as a by-product of Theorem \ref{PS Generalize} we can recover
a result of Lyubeznik \cite{Lyu84}.

\begin{cor} Let $R=k[x_1,\ldots, x_n]$ be a polynomial ring of $n$ variables
over a field $k$ and $I$ be a square-free monomial ideal. Then $\pd
R/I=\cd(R,I)$.
\end{cor}

\begin{proof}
Define the $k$-linear endomorphism $\varphi:R \longrightarrow R$ by
$x_i \mapsto x^t_i$
 for $1 \leq i \leq n$ and positive integer $t$.
 It is a flat ring endomorphism and without loss of generality we may assume that $R$ is a graded local ring with the maximal ideal
  $\fm=(x_1,\ldots, x_n)$. By passing to the completion one may assume that $R$ is a complete regular local
  ring satisfying the (GD) property. Hence, we are done
by Theorem \ref{PS Generalize}. To complete the proof note that by
what we have indicated at Remark \ref{pure2}, the endomorphism $R/I
\rightarrow R/I$ is pure.
\end{proof}

\begin{discussion}\label{Hartshorne} A special case of a conjecture of Hartshorne \cite[page 126]{Har70} is the following:

\textbf{Conjecture:} Let $C$ be a curve in $\mathbb{P}^3_k$ over a field $k$ of characteristic $0$.
 If $C$ is a set-theoretic complete intersection, the curve $C$ is arithmatically Cohen-Macaulay.
  I.e. the homogeneous coordinate ring $k[x_1, \ldots, x_n]/I(C)$ is  Cohen-Macaulay.

It is known that this conjecture is not true, see for instance
\cite{St-Vo}.  To do so further and motivated by the preceding
conjecture we consider the equivalent property between the
Cohen-Macaulayness of $R/I$ and the equality $\Ht (I) = \cd (R,I)$.
\end{discussion}

\begin{thrm} \label{CM cond} Let $(R,\fm)$ be a Gorenstein complete local ring.
Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has (GD) property;
\item [(b)]  projective dimension of $R/I$ is finite;
\item [(c)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
 \end{enumerate}
 Then the following statements are equivalent.
\begin{itemize}
  \item[(1)] $R/I$ is a Cohen-Macaulay ring.
  \item[(2)] $\Ht(I)=\cd(R,I)$ and $\Ext^r_R(R/\varphi^{t}(I)R,R) \longrightarrow
  \Ext^r_R(R/\varphi^{t+1}(I)R,R)$ are injective of nonzero $R$-modules for all $t$ and $r:=\depth R/I$.
  \end{itemize}
\end{thrm}

\begin{proof}
 $(1) \Rightarrow (2)$ First of all note that $\dim R/I= \dim R/I^t=\dim R/\varphi^{t}(I)R$
for all $t \in \mathbb{N}$. Since $R/I$ is a Cohen-Macaulay ring, by
virtue of Lemma (\ref{EABR}),
 $\dim R/I= \depth R/I=\depth R/\varphi^{t}(I)R $ for all $t \in \mathbb{N}$. The
 first part follows from Theorem (\ref{PS Generalize}) and the Auslander-Buchsbaum Theorem. For the second part consider
  the short exact sequence
  $$0 \rightarrow \varphi^{t}(I)R/\varphi^{t+1}(I)R \rightarrow
   R/\varphi^{t+1}(I)R \rightarrow R/\varphi^{t}(I)R \rightarrow 0.$$ Applying $H^{i}_{\fm}(-)$ and
    use the Grothendieck's Vanishing Theorem and the fact that
    $\dim \varphi^{t}(I)R/\varphi^{t+1}(I)R \leq r$ we obtain an
     epimorphism $H^r_{\fm}(R/\varphi^{t+1}(I)R) \longrightarrow H^r_{\fm}(R/\varphi^t(I)R)$.
     Now we are done by the local duality.

$(2) \Rightarrow (1)$ By the assumptions $\Ht(I)=\cd(R,I)$, i.e.
$H^i_I(R)=0$ for all $i \neq \Ht(I)$. By the local duality one has
\begin{eqnarray*}
 \Hom_R(H^i_I(R),E(R/\fm))  & \simeq &\Hom_R({\vil}_t \Ext^i_R(R/I^t,R),E(R/\fm)) \\
   &\simeq &  {\vpl}_t \Hom_R(\Ext^i_R(R/I^t,R),E(R/\fm))  \\
   &\simeq &  {\vpl}_t H^{\dim R-i}_{\fm}(R/I^t).
\end{eqnarray*}
It means that for all $i \neq \Ht (I)$,  $ {\vpl}_t H^{\dim
R-i}_{\fm}(R/I^t) = 0$. Now, it is enough to prove that $ {\vpl}_t
H^{r}_{\fm}(R/I^t) \neq 0$. This follows from the fact that the
homomorphisms
$$H^r_{\fm}(R/\varphi^{t+1}(I)R) \longrightarrow
H^r_{\fm}(R/\varphi^t(I)R)$$ are surjective  of nonzero $R$-modules
for all $t$.
\end{proof}

\section{VANISHING RESULTS }

As mentioned in the Introduction, Question\ref{local-global} is not
true in general. However under some additional properties, at least
one of directions would be true.

\begin{remark}
Let  $(R, \fm)$ be a regular local ring containing a field of characteristic $p >0$.
 From vanishing of $H^{i}_{\fm}(R/I)$ and applying the Frobenius map $F^e$ ($e$ is
  an arbitrary positive integer) on it, one has $H^{i}_{\fm}(R/I^{[p^e]})=0$ for all $e> 0$.
  It implies that $ {\vpl}_t H^{i}_{\fm}(R/I^{[p^e]})=0$. Using the Grothendieck's duality
  and taking account on the substitution of $ {\vpl}$  with $ {\vil}$ under $\Hom$
  one has vanishing of the Matlis dual of $H^{\dim R-i}_I(R)$ and then so is $H^{\dim R-i}_I(R)$, itself.
\end{remark}

In characteristic zero the story is different and in the best
of my knowledge the Question \ref{local-global} is no longer true in general.

\begin{example} let $R=k[[x_1, \ldots, x_6]]$ denote the formal power series ring in
 $6$ variables over a field $k$ of characteristic zero. Let $I$ denote the ideal generated by the $2 \times 2$ minors of the matrix
 \begin{displaymath}
M=\left( \begin{array}{ccc}
x_1 & x_2 & x_3 \\
x_4 & x_5 & x_6
\end{array} \right).
\end{displaymath}
As $R/I$ is a four-dimensional Cohen-Macaulay ring so $H^{3}_{\fm}(R/I)=0$.
On the other hand, U. Walther in \cite[Example 6.1]{W} has shown that
$H^{3}_{I}(R) \cong E_R(k)$. Thus, Question \ref{local-global} is not valid in characteristic zero, in general.
\end{example}

Now is the time to state our main result in this section. It is a
slight strengthening  of \cite[Theorem 4.1]{S-W}, where the
mentioned result of Singh and Walther is a generalization of
\cite[Theorem 1.1]{Lyu06}.

\begin{thrm} \label{Main vanishing}
Let $(R, \fm)$ be a Gorenstein local ring. Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has (Flat) property;
\item [(b)]  the triple $(R, I, \varphi)$ has (Inj) property;
\item [(c)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
\end{enumerate}
 Then for a given
integer $i$,
$$H^i_I(R)=0 \text{\ if\ and\ only\ if\ } H^{\dim R-i}_{\fm}(R/I)=0$$.
\end{thrm}

As an immediate consequence of Theorem (\ref{Main vanishing}), we state the next result.

\begin{cor}  Let $R$ be either an $F$-pure regular local ring containing a field of positive
characteristic with an ideal $I$ or a polynomial ring over an arbitrary field with a square
free monomial ideal $I$. Then for a given integer $i$,  $$H^{\dim R-i}_{\fm}(R/I)=0  \text{\ if\ and\ only\ if\ } H^i_I(R)=0.$$
\end{cor}

Note that the second part of the above Corollary has been proved in \cite[Corollary 4.2]{S-W}

\subsection{PROOF OF THEOREM \ref{Main vanishing}}

In order to prove Theorem \ref{Main vanishing}, it is enough to
prove the following Propositions.

\begin{prop} \label{1} Let $(R,\fm)$ be a Gorenstein local ring.
Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has (Flat) property;
\item [(b)]  the triple $(R, I, \varphi)$ has (Inj) property;
\item [(c)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
\end{enumerate}
Then for a given integer $i$,
 $$H^i_I(R)=0\ \text{\ implies\ that\ } H^{\dim R-i}_{\fm}(R/I)=0.$$
\end{prop}

 \begin{proof} By passing to the completion, we may assume that $R$ is a  Gorenstein complete local ring.
  Suppose that for a fixed integer $i$ one has $H^{\dim R-i}_{\fm}(R/I) \neq 0$. As $(R, I, \varphi)$
  has (Inj) property, local duality implies that the homomorphism
   $$H^{\dim R-i}_{\fm}(R/\varphi(I)R) \rightarrow H^{\dim R-i}_{\fm}(R/I)$$ induced by the natural
    homomorphism $R/\varphi(I)R \rightarrow R/I$, is surjective. It then follows that $H^{\dim R-i}_{\fm}(R/\varphi(I)R) \neq 0$. By
    iterating, we have $H^{\dim R-i}_{\fm}(R/\varphi^t(I)R) \neq 0$ for all $t$.
    Hence, $ {\vpl}_t H^{\dim R-i}_{\fm}(R/\varphi^t(I)R)$ will be nonzero.
    Once again using the local duality one observe that $H^i_I(R)\neq 0$.
\end{proof}

\begin{prop} \label{2} Let $(R,\fm)$ be a Gorenstein local ring.
Suppose that
\begin{enumerate}
\item [(a)] the triple $(R, I, \varphi)$ has (Flat) property;
\item [(b)] $\rad(\varphi(\fm)R)=\rad(\fm)$.
\end{enumerate}
 Then for a given integer $i$,  $$H^{\dim R-i}_{\fm}(R/I)=0  \text{\ implies\ that\ } H^i_I(R)=0.$$
\end{prop}

\begin{proof} By passing to the completion, we may assume that $R$ is a Gorenstein
 complete local ring. Suppose that $H^{\dim R-i}_{\fm}(R/I)=0$
 for a given integer $i$. As $\varphi:R \longrightarrow R$ is a flat
ring endomorphism, then so does all of its iterations. By applying
the functor $\Phi^t$ to $H^{\dim R-i}_{\fm}(R/I)$, one has
 $H^{\dim R-i}_{\fm}(R/\varphi^t(I)R) =0$ for all $t$.
Thus ${\vpl}_t H^{\dim R-i}_{\fm}(R/\varphi^t(I)R) =0$.
Once again with a similar argument given in the proof of Proposition (\ref{1}), one observes that $H^i_I(R)=0$.
\end{proof}

\section{SET-THEORETICALLY COHEN-MACAULAY IDEALS}

It is known that the radical of a Cohen-Macaulay ideal need not to
be Cohen-Macaulay in general. A well-known evidence is an example
due to Hartshorne \cite{Ha} shows that whenever $k$ is a field of
positive characteristic, the ideal $$I= \ker (\varphi: k[x_1, x_2,
x_3, x_4] \rightarrow k[s^4, s^3t,st^3, t^4])$$ via $x_1 \mapsto
s^4, x_2 \mapsto s^3t, x_3 \mapsto st^3, x_4 \mapsto t^4$ is a
non-Cohen-Macaulay set-theoretic complete intersection. However,
there are some Cohen-Macaulay ideals have the same property for
their radicals. For instance, If $I$ is a Cohen-Macaulay monomial
ideal of a polynomial ring $R = k[x_1, \ldots, x_n]$ over a field
$k$, then $\rad (I)$ is Cohen-Macaulay (cf. \cite[Theorem
2.6]{HTT}). Furthermore, principal and Veronese monomial ideals have
such a property, \cite[Theorem 3.2]{HH}.

Let us recall the following definition in order to strengthen the
above results.

\begin{definition}  Let $(R,\fm)$ be a regular local ring. An ideal $I \subset R$ is
called set-theoretically Cohen-Macaulay if there exists an ideal $J
\subset R$ with $\rad(I) = \rad(J)$ such that the ring $R/J$ is
Cohen-Macaulay.
\end{definition}

It is clear that Cohen-Macaulay radical ideals are set-theoretically
Cohen- Macaulay. It should be noted that there exist also
set-theoretic Cohen-Macaulay ideals which are not Cohen-Macaulay.
Suppose $R=k[[x,y]]$ is a formal power series ring over a field $k$
of $x,y$. Put $I=(y) \cap(xy,y^2)$ and $J=\rad(I)$. It is clear that
$R/J$ is Cohen-Macaulay but this is not the case for $R/I$.

Our first result in this direction is a consequence of Theorem
\ref{Main vanishing} states that under the assumptions are given in
Proposition \ref{equivalence} the concepts of Cohen-Macaulayness and
set-theoretically Cohen-Macaulayness are the same.

\begin{prop} \label{equivalence} Let $(R,\fm)$ be a regular local ring with a flat ring endomorphism $\varphi: R \rightarrow
R$. Further suppose that triples $(R,I,\varphi)$ and $(R,J,\varphi)$
have (Inj) property where $\rad(I)=\rad(J)$. Then $R/I$ is
Cohen-Macaulay if and only if $R/J$ is Cohen-Macaulay.
\end{prop}

\begin{proof} First note that as $(R,\fm)$ is a regular local ring with a flat local ring endomorphism $\varphi: R \rightarrow
R$, the dimension formula $\dim R + \dim R/\varphi(\fm)R = \dim R$
implies that $\varphi(\fm)R$ is $\fm$-primary. Then, since $R$ is a
Noetherian ring and $\rad(I)=\rad(J)$, thus $\{I^t\}_{t \geq 0}$ and
$\{J^t\}_{t \geq 0}$ are cofinal. Now we are done by Theorem
\ref{Main vanishing}.
\end{proof}

 As an immediate consequence of the Proposition \ref{equivalence}, we
may recover \cite[Lemma 3.1]{S-W2}.

\begin{cor} (Huneke) Let $(R,\fm)$ be a regular local ring of characteristic $p>0$ and $I$ an
 ideal of $R$. If the ring  $R/I$ is $F$-pure and not Cohen-Macaulay, then $I$ is not a set-theoretically Cohen-Macaulay ideal.
\end{cor}

 According to the Discussion \ref{Hartshorne} in Section 3, we observe that
Cohen-Macaulayness of $R/I$ is not
  equivalent to the ideal $I$ being set-theoretically complete intersection. On the other
  hand,
  set-theoretically complete intersection ideals are
  set-theoretically Cohen-Macaulay but the converse is no longer true. See for instance \cite{S-W2}.
 We show that instead of set-theoretically complete intersection ideals one may regard cohomologically
 complete intersection ideals.

Next result (Theorem \ref{setCM-setCI}) shows that in a regular
local ring $R$, whenever $\varphi: R \rightarrow R$ is flat and
pure, a set-theoretically Cohen-Macaulay ideal is the same as
cohomologically complete intersection one. Notice that a
set-theoretically complete intersection ideal is a cohomologically
complete intersection one but the converse is no longer true.

\begin{lem} \label{cofinal}  Let $(R,\fm)$ be a local ring and $I$ , $J$ two ideals of $R$
with a ring endomorphism  $\varphi: R \rightarrow R$  such that
$\rad(I)=\rad(J)$. Then $\rad(\varphi(I)R)=\rad(\varphi(J)R)$.
\end{lem}

\begin{proof} In order to prove $\rad(\varphi(I)R)=\rad(\varphi(J)R)$ suppose that $x$ is an arbitrary element in
$\rad(\varphi(I)R)$. Then there exist an integer $n$ and an element
$a \in I$ such that $x^n=\varphi(a)$. As $\rad(I)=\rad(J)$, there is
an integer $t$ such that $a^t \in J$. Therefore,
$x^{nt}=\varphi(a^t) \in \varphi(J)R$, that is $x \in
\rad(\varphi(J)R)$. \

For the reverse inclusion, from $x \in \rad(\varphi(J)R)$ we observe
that $x^n=\varphi(a)$ for some $a \in J$ and $n \in \mathbb{Z}$.
Once again from the equality $\rad(I)=\rad(J)$  one has $a^t \in I$
for some integer $t$. By applying $\varphi(-)$ to it we have
$x^{nt}=\varphi( a^t) \in \varphi(I)R$. It completes the proof.
\end{proof}

\begin{thrm} \label{setCM-setCI} Let $(R,\fm)$ be a  regular local ring and the triple $(R, I, \varphi)$ has (Flat) property.
 Suppose that $I$ is a set-theoretically Cohen-Macaulay
 ideal, then $\Ht(I)=\cd (R,I)$.

 The reverse statement holds true,
 if $\varphi: R \rightarrow R$ is pure.
\end{thrm}

\begin{proof} Suppose that $I$ is a set-theoretically Cohen-Macaulay ideal. Then there is a Cohen-Macaulay ideal $J$ of $R$ with
$\rad(I)=\rad(J)$ and $H^i_{\fm}(R/I)=0$ for all $i \neq \dim R/I$.
By the assumption $(R, I, \varphi)$ has (Flat) property and Lemma
\ref{cofinal} implies that $\rad(J)=\rad(\varphi(J)R)$. Then
$\{\varphi^t(J)R\}_{t \geq 0}$ is a decreasing chain of ideals
cofinal with $\{J^t\}_{t \geq 0}$. It then follows that the triple
$(R,J,\varphi)$ satisfies (Flat) property. Now in the light of
Proposition \ref{2} one has $0=H^{\dim R-i}_{I}(R) \cong H^{\dim
R-i}_{J}(R)$ for all $i \neq \dim R/I$. It implies that $\Ht(I)=\cd
(R,I)$.

To prove the reverse statement, it is enough to show that
$H^i_{\fm}(R/J)=0$ for all $i \neq \dim R/I$, where $J$ is an ideal
of $R$ with the same radical as $I$. As $H^{i}_{J}(R) \cong
H^{i}_{I}(R)=0$ for all $i \neq \dim R/I$. Once again exploiting
Lemma \ref{cofinal} and Proposition \ref{1} the claim is clear. To
this end notice that purity of $\varphi: R \rightarrow R$ implies
the same property for $ R/J \rightarrow R/J$ and it implies that
triple $(R,J,\varphi)$ satisfies the (Inj) property (cf. Remark
\ref{pure2}).
\end{proof}

The assumptions in Theorem \ref{setCM-setCI} are sharp. A suitable
evidence is the ideal $I$ defined by Hartshorne (at the beginning of
this section) which is set-theoretically complete intersection, then
$\Ht(I)=\cd (R,I)$ but $I$ is not set-theoretically Cohen-Macaulay
as $R/I$ is not $F$-pure.

We conclude this section with an investigation on the cohomological
dimension of the intersection of set-theoretically Cohen-Macaulay
ideals. Before it, we need the following Lemma.

\begin{lem} \label{intersec} Let $(R,\fm)$ be a regular local ring and $I$ , $J$
 two ideals of $R$ with a flat ring endomorphism  $\varphi: R \rightarrow R$
  such that $\rad(I)=\rad(\varphi(I)R)$ and $\rad(J)=\rad(\varphi(J)R)$.
  Then $\rad(I \cap J)=\rad(\varphi(I \cap J)R)$.
\end{lem}

\begin{proof}
To prove, it is enough to note that by the flatness of $\varphi$,
one has $\varphi(I \cap J)R=\varphi(I)R \cap \varphi(J)R$.
\end{proof}

\begin{prop}  Let $(R,\fm)$ be a  regular local ring and the triples $(R,I,\varphi)$
and $(R,J,\varphi)$ have  (Flat) property where $\rad(I) ,\rad(J)$
are Cohen-Macaulay ideals with the same dimension $d$. Further
suppose that $\rad(I) \cap \rad(J) =\rad(I) \rad(J)$ and $d >\depth
R/\rad(I) + \rad(J)$.  Then $$\cd (R, I \cap J) \leq
\Ht(I)+\Ht(J)-1.$$
\end{prop}

\begin{proof} As cohomological dimension is stable under taking radical of ideals, Lemma \ref{intersec} and Theorem \ref{PS Generalize} imply that
$$\cd (R, I \cap J) =\cd (R,\rad ( I) \cap \rad (J)) \leq \dim R -
\depth R/\rad(I) \cap \rad(J).$$
It is not difficult to verify that
$$\depth (R/\rad(I) \cap \rad(J) ) \geq  \depth (R/\rad(I) + \rad(J))+1.$$
 Therefore,
\begin{equation}\label{1a}
   \cd (R, I \cap J) \leq \dim R - \depth (R/\rad(I) +\rad(J)) -1.
\end{equation}
On the other hand, by the assumption, $\Tor^R_1(R/\rad(I), R/\rad(J))=0$.
 By rigidity (see \cite{L}) $\Tor^R_i(R/\rad(I), R/\rad(J))=0$ for all $i \geq 1$.
  Hence, by \cite[Theorem 1.2]{Aus} the depth formula holds for $(R/\rad(I), R/\rad(J))$ . Therefore,
\begin{equation}\label{2a}
  \depth (R/\rad(I) + \rad(J)) = -\dim R+ \depth (R/\rad(I))+ \depth (R/\rad(J)).
\end{equation}

 Combining (\ref{1a}) and (\ref{2a}) we conclude that $\cd (R, I \cap J) \leq
\Ht(I)+\Ht(J)-1$.
\end{proof}

\section{LINKAGE}

 In  the present section, we consider the concept of linkage. Roughly speaking, the linkage is the study of two subschemes where their
 union has nice properties. Exploiting this concept one may consider a subscheme linked with the second one which understood well.
To be more precise and from local algebra point of view we recall the definition of two linked ideals.

\begin{definition} Let $I$ and $J$ be two ideals of pure height $g$ of a local Gorenstein ring $(R,\fm)$.
The ideals $I$ and $J$ are (algebraically) linked by a complete intersection
$\underline{x}:=x_1, \ldots, x_g$ with $(\underline{x}) \subseteq I \cap J$  if
 $I/(\underline{x}) \cong \Hom_R(R/J, R/(\underline{x}))$ and $J/(\underline{x}) \cong \Hom_R(R/I, R/(\underline{x}))$.
 We write it as $I \sim_{(\underline{x})} J$. For brevity we often write $I \sim J$ for $I \sim_{(\underline{x})} J$
  when there is no ambiguity about the ideal $(\underline{x})$.
\end{definition}

Ideals $I$ and $J$ are in the same linkage class if there is a
sequence of links $J=I_0 \sim I_1 \sim \ldots \sim I_q= I$. An ideal
$I$ is in the linkage class of a complete intersection if any of the
ideals in the linkage class of $I$ are generated by a regular
sequence. If $q$ is an even integer, we say that $J$ is in the even
linkage class of $I$.

Let $I$ and $J$ be two ideals  linked by a complete intersection $\fc$ of height $g$. Then
$$\Hom_R(R/I, R/\fc) \cong \Hom_{R/\fc}(R/I, R/\fc) \cong \Ext^g_R(R/I,R):=K_{R/I}$$
and
 $$\Hom_R(R/J, R/\fc) \cong \Hom_{R/\fc}(R/J, R/\fc) \cong \Ext^g_R(R/J,R):=K_{R/J}$$
are the canonical modules of $R/I$ resp. $R/J$.

A ring $R$ satisfies Serre's condition $(S_r)$ if for all $\fp \in
\Spec R$,
$$\depth R_{\fp} \geq \min \{r, \dim R_{\fp} \}.$$
For an $n$-dimensional ring, being Cohen-Macaulay is equivalent to
satisfying $(S_n)$.

\begin{lem} \label{Serr charact}  \cite[Theorem 1.14] {Sch98}Let $M$ denote a finitely generated, equidimensional $R$-module with
$d = \dim M$, where $R$ is a factor ring of a Gorenstein ring. Then
for an integer $r \geq 1$ the following statements are equivalent:
\begin{enumerate}
   \item[(a)]  $M$ satisfies condition $(S_r)$.

  \item[(b)]  The natural map $M \rightarrow K_{K_M}$ is bijective (resp. injective for $r = 1$) and
  $$H^i_{\fm}(K_M) = 0 \text{ \ for\  all\ } d - r + 2 \leq i < d.$$
\end{enumerate}
\end{lem}

We will use the following facts on cohomological relations of linked
ideas, later on.

\begin{lem} \label{Schenzel}  Let $I$ and $J$ be two linked ideals of a local Gorenstein ring $(R,\fm)$.
Suppose that $E:=E(R/\fm)$ is the injective hull of $R/\fm$ and
$d=\dim R/I=\dim R/J$.
\begin{enumerate}
   \item[(a)]  There exist a canonical exact sequence
$$0 \rightarrow H^{d-1}_{\fm}(R/J) \rightarrow H^{d}_{\fm}(K_{R/I}) \rightarrow \Hom_R(R/I,E) \rightarrow 0,$$
 and the canonical isomorphisms
$$H^{i-1}_{\fm}(R/J) \cong H^{i}_{\fm}(K_{R/I}),\ i < d,$$
(cf. \cite[Lemma 4.2]{Sch82}).

  \item[(b)]  For an integer $r \geq 2$ the
following statements are equivalent:
\begin{enumerate}
   \item[(1)] $R/I$  satisfies $(S_r)$;
  \item[(2)] $H^{i}_{\fm}(R/J)=0$ for all integers $d-r<i<d$,
  (cf.
\cite[Theorem 4.1]{Sch82}).
\end{enumerate}

\item[(c)]  Assume that the local cohomology
modules $H^{i}_{\fm}(R/I)$ have finite length over $R$ for all
integers $i=0, 1,\ldots, d-1$. Then,
 $$H^{d-i}_{\fm}(R/J)=\Hom_R(H^{i}_{\fm}(R/I),E),\  \text{\ for\ all\ integers\ } i=1,\ldots, d-1,$$
 (cf. \cite[Theorem 1.2, pp. 157]{St-Vo} or \cite[Corollary 3.3]{Sch82}).
\end{enumerate}
\end{lem}

\begin{remark} \label{ass}  Let  $I, J, \fc$ be as before. Then $\fm \nsubseteq \Ass (R/I)$ if and only if $\fm \nsubseteq \Ass
(R/J)$. To see this, without loss of generality assume that $R/I$ is
not Cohen-Macaulay. Suppose that $\fm \nsubseteq \Ass (R/I)$. From
the above descriptions, one can obtain the following exact sequence
\begin{equation}\label{exact}
   0\longrightarrow K_{R/J}\longrightarrow R/\fc\longrightarrow
R/I \longrightarrow0.
\end{equation}
As $\zd(R/I)=\bigcup_{\fp \in \Ass_R(R/I)} \fp$, we observe that
$\fm \nsubseteq \zd(R/I)$ ($\fm$ is not contained in the zero
divisors of $R/I$), that is $H^{0}_{\fm}(R/I)=0$. From the exact
sequence (\ref{exact}), one has $\depth K_{R/J}-1=\depth R/I$ (for
detailed proof see \cite[Proposition 3.3]{egh-sh}). It implies that
$\depth K_{R/J}
>1$. once again, in the light of (\ref{exact}) we have
$H^{0}_{\fm}(R/\fc)=0$. Hence, we conclude that $\fm \nsubseteq \Ass
(R/J)$. To this end note that $ \Ass (R/I) \cup \Ass (R/J)= \Ass
(R/\fc)$.
\end{remark}

Of particular interest is that if $I$ and $J$ are in the same
linkage class then what properties of $I$ is shared by $J$. For
instance, in \cite[Proposition 1.3]{P-S74} Peskine and Szpiro proved
that if $I$ and $J$ are linked ideals and $I$ is a Cohen-Macaulay
ideal then so is $J$. The same result proved by Schenzel in
\cite[Corollary 3.3]{Sch82} for Buchsbaum ideals. In this direction,
using the results of Schenzel (Lemma \ref{Schenzel}), with some mild
assumptions, we show that between evenly linked ideals the Serre's
condition $(S_r)$ is sharable.

\begin{prop} \label{even sr} Let $(R,\fm)$ be a  Gorenstein local ring and $r \geq 2$ be an integer. Suppose
that either
\begin{enumerate}
   \item[(a)] $I, J$ are two evenly linked ideals of $R$, or
  \item[(b)] $I, J$ are two linked ideals of $R$ and $\fm \nsubseteq
\Ass(R/I)$.
\end{enumerate}
Then, $R/I$ satisfies $(S_r)$ if and only if $R/J$ is so.
\end{prop}

\begin{proof} (a) Suppose that $\fa$ is an ideal of $R$ where $I \sim \fa \sim J$. By virtue of Lemma \ref{Serr
charact},
$R/I$ satisfies $(S_r)$ if and only if
\begin{equation}\label{sr1}
   H^d_{\fm}(K_{R/I}) \rightarrow \Hom_R(R/I,E(R/\fm)),\ d= \dim
   R/I,
\end{equation}
 is bijective  and $H^i_{\fm}(K_{R/I})=0,\ d-r+1 < i < d$.
As $I$ is linked to $\fa$, by Lemma \ref{Schenzel}(a)
 $$0=H^{i-1}_{\fm}(R/\fa) \cong H^{i}_{\fm}(K_{R/I}),\ d-r+1 < i < d,$$
 and as $\fa$ is linked to $J$,
 $$H^{i-1}_{\fm}(R/\fa) \cong H^{i}_{\fm}(K_{R/J}),\ d-r+1 < i < d.$$
 From the exact sequence (\ref{sr1}) and Lemma \ref{Schenzel}(a) one has $H^{d-1}_{\fm}(R/\fa)=0$. Thus, by the following exact sequence
 $$0 \rightarrow H^{d-1}_{\fm}(R/\fa) \rightarrow H^{d}_{\fm}(K_{R/J}) \rightarrow \Hom_R(R/J,E(R/\fm)) \rightarrow 0,$$
 one has the homomorphism
 $H^d_{\fm}(K_{R/J}) \rightarrow \Hom_R(R/J,E(R/\fm))$ is bijective. Once again using Lemma \ref{Serr charact}
 it turns out that $R/J$ satisfies $(S_r)$.

 (b) First of all note that as $\fm \nsubseteq
\Ass(R/I)$, by Remark \ref{ass} one has $\fm \nsubseteq \Ass(R/J)$,
that is $H^0_{\fm}(R/I)=0=H^0_{\fm}(R/J)$. Then, we are done by
virtue of Lemma \ref{Schenzel}(b),(c).
\end{proof}

Of a particular interest is examining cohomological dimension of two
linked ideals.

\begin{prop} \label{evevanish} Let $(R,\fm)$ be a  Gorenstein local ring and $I, J$ two evenly linked ideals of $R$.
 Suppose that
 \begin{enumerate}
   \item[(a)] the triples $(R,I, \varphi)$ and $(R,J, \varphi)$ have (Flat) property;
\item[(b)] the triple $(R,I, \varphi)$ has (Inj)
 property;
  \item[(c)] $\rad (\fm)=\rad (\varphi (\fm R)$.
\end{enumerate}
Then for a given integer $i$, $$H^{i}_{I}(R)=0 \text{\ if\ and\
only\ if \ } H^{i}_{J}(R)=0.$$
\end{prop}

\begin{proof}  Note that two linked ideals have the same height so for $i \leq \Ht (I)$ we have nothing to prove.
 Hence, we may assume that $i > \Ht (I)$.

Suppose that $\fa$ is an ideal of $R$ where $I \sim \fa \sim J$. By
what we have seen in the proof of Proposition \ref{even sr}, in the
light of Lemma \ref{Schenzel}(a) one has $H^{j}_{\fm}(R/I)\cong H^{j}_{\fm}(R/J)$
for all $j < \dim R/I$. Hence, the claim follows
from Propositions \ref{1} and \ref{2}.
\end{proof}

\begin{remark} \label{skew lines} It is noteworthy to mention that the Proposition \ref{evevanish} is
no longer true for non-evenly linked ideals. To see this, let
$R=k[x_0,x_1,x_2,x_3]$ be a polynomial ring over an algebraically
closed field $k$. Let $I=(x_0,x_1)\cap(x_2,x_3)$ be the defining
ideal of the union of the two skew lines in ${\mathbb P}^3$ and
$J=(x_0x_3-x_1x_2,x^3_1-x^2_0x_2,x_0x^2_2-x^2_1x_3,x^3_2-x_1x^2_3)$
be the defining ideal of the  twisted quartic curve in ${\mathbb
P}^3$. Then it is not hard to show that
$$I \cap J=(x_0x_3-x_1x_2,x_0x_2^2-x_1^2x_3)$$
is a complete intersection. Therefore $I$ is linked to $J$ by
$\fc:=I \cap J$, where $H^{3}_{I}(R)\neq 0$
 and $H^{3}_{J}(R)= 0$.
\end{remark}

Next, we consider the property of being set-theoretically
Cohen-Macaulay between linked ideals.

\begin{thrm} \label{link1} Let $(R,\fm)$ be a regular local ring and let $I, J$ be two linked ideals of $R$.
 Suppose that
\begin{enumerate}
   \item[(a)] $(R,I, \varphi)$  satisfies (Flat) property,
  \item[(b)] $(R,I, \varphi)$  satisfies (Inj) property,
  \item[(c)] $\fm \nsubseteq \zd(R/I)$.
\end{enumerate}
If $I$ is set-theoretically Cohen-Macaulay, then $J$ is
Cohen-Macaulay.
\end{thrm}

\begin{proof}
Note that,  $d:=\dim R/\varphi(I)R=\dim R/I=\dim R/J= \dim R/\varphi(J)R$, where
 the third equality follows by the fact that two linked ideals have the same
  dimension. We are going to show that $H^{i}_{\fm}(R/J)=0$ for all $i < d$.

  As $I$ is a set-theoretically Cohen-Macaulay ideal, there exists a Cohen-Macaulay ideal $\fb$
 of the same radical as $I$. Then $H^i_{\fm}(R/\fb)=0$ for all $i < d$.
   As $\rad(\varphi(I)R)=\rad(I)=\rad(\fb)$, by virtue of Lemma \ref{cofinal}
   one has $\rad(\varphi(\fb)R)=\rad(\fb)$. It then follows from Proposition
   (\ref{2}) that $H^{\dim R-i}_{I}(R) \cong H^{\dim R-i}_{\fb}(R)=0$ for all $\dim R-i > \dim R- d$.
Now exploiting Proposition (\ref{1}) shows that $H^{i}_{\fm}(R/I)=0$
 for all $i <d$. Hence, Lemma \ref{Schenzel}(c) ensures that $H^{i}_{\fm}(R/J)=0$ for $i=1, \ldots, d-1$.
 In this way, combining the previous information jointly with the Remark \ref{ass} it follows that
 $H^0_{\fm}(R/J)=0$, this completes the proof.
\end{proof}

\begin{cor} \label{setlink} Let $(R,\fm)$ be a regular local ring and let $I, J$ be two linked ideals of
$R$. Suppose that
\begin{enumerate}
 \item[(a)] $(R,I, \varphi)$ and $(R,J, \varphi)$ satisfy (Flat)
 property,
  \item[(b)] $\fm \nsubseteq \zd(R/I)$,
  \item[(c)] the induced ring
 endomorphisms $ R/I \rightarrow R/I$ and $ R/J \rightarrow R/J$ are pure.
\end{enumerate}
  Then,  If $I$ is set-theoretically Cohen-Macaulay then so is $J$.
\end{cor}

\begin{proof}  Theorem \ref{link1} ensures that $J$ is a Cohen-Macaulay ideal.
Then, the equality $\Ht(J)= \cd(R,J)$ holds by Theorem \ref{CM cond}. Now, exploiting
 Theorem \ref{setCM-setCI} one observe that $J$ is a set-theoretically Cohen-Macaulay ideal.
\end{proof}
Notice that one can not remove the purity assumption from the above
result, as the following example demonstrates it.

\begin{remark}  Let $R, I$ and $J$ be as in Remark \ref{skew lines}.
Since $\depth R/I=1$ and $\dim R/I=2$, then $I$ is a
 non-Cohen-Macaulay radical ideal. As $I$ is a square-free ideal, $R/I$ is
  $F$-pure but this is not the case for $R/J$, because it is not reduced.
  On the other hand by virtue of Hartshorne \cite{Ha} the ideal $J$ is
  set-theoretically Complete intersection and then it is set-theoretically Cohen-Macaulay.
 \end{remark}

config: arxiv


TITLE: Divergence of the sequence $a_n= \frac{n^{2} + 2n + 35}{n^2 + n + 1} \sin(n^{2}) $
QUESTION [2 upvotes]: Are there any methods to prove that the sequence $\left\{a_{n} = \dfrac{n^{2} + 2n + 35}{n^2 + n + 1} \sin(n^{2})\right\}$ is divergent?
To be honest, I'm not good at solving any problems relating to trigonometry. So proving the divergence of this trigonometric sequence above is quite hard. I've tried to use contradiction that means suppose the sequence is convergent and then all of its subsequences are convergent, too. Next I try to find one subsequence of $a_{n}$ is divergent and I get contradiction. But this is not easy to find a subsequence like that.
Thank you for your help.

REPLY [0 votes]: For clarity I write out why the proposed duplicate answers the question. Note that
$$ a_n =\frac{n^2+2n+35}{n^2+n+1}\sin(n^2) = \sin(n^2) + \frac{(n+34)\sin(n^2)}{n^2+n+1}$$
Now
$$\left|\frac{(n+34)\sin(n^2)}{n^2+n+1}\right| \le \frac{35n}{n^2} \to 0$$
and by the equidistribution of $n^2 \pmod {2\pi}$ as explained in the linked posts e.g. Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$ or $\sin (n^2)$ diverges , it follows that there are infinitely many $n$ such that $\sin(n^2)>1/2$, and there are infinitely many $n$ such that $\sin(n^2)<-1/2$. It follows that there are infinitely many $n$ such that $a_n>1/4$, and infinitely many $n$ such that $a_n < -1/4$. Hence, $a_n$ has no limit.


TITLE: how to find quotient ring $k[x,y]/(x, y^2)$
QUESTION [1 upvotes]: Let k be a field. I want to find the quotient ring $k[x,y]/(x,y^2)$.
Here is my attempt:
We have $(x) \subset (x,y^2)\subset k(x,y)$ so  $k[x,y]/(x,y^2)\cong (k(x,y)/(x))/((x,y^2)/(x))$ hence I may guess:
$k[x,y]/(x,y^2)\cong k[y]/(y^2) $
but I still dont  have rigorous proof for this formula.
I looking for some hints.

REPLY [0 votes]: You are correct. We apply the third isomorphism theorem:
$$k[x,y]/ (x,y^2) \cong k[x,y]/(x)/(x,y^2)/(x) =k[y]/(y^2).$$
Thus, the quotient consists of just linear terms $1+ay$ for $a \in k$.
Morally, this makes sense since taking the quotient induces the map $x \mapsto 0$, which is linear so it extends linearly: $p[x] \mapsto 0$ for all polynomials $p$ that vanish at the origin (without constant terms.) On the other hand, the same goes for all the quadratic terms in $y$.
Another way to see it is that we want to simultaneously solve $x=0$ and $y^2=0$.


TITLE: Blackbody entropy
QUESTION [1 upvotes]: There are several ways to find that the entropy of thermal photon gas in a blackbody cavity takes the from
$$
S=V\int d\nu \frac{8\pi\nu^2}{c^3} \left( (1+n_\nu)\log(1+n_\nu) - n_\nu \log n_\nu \right)
$$
where $n_\nu=(\exp(\frac{h\nu}{kT})-1)^{-1}$ is the average number of photons with energy $h\nu$.
From this expression, it should be possible to calculate$$S=\frac{4}{3}\frac{U}{T}$$ where $$
U=V\int d\nu \frac{8\pi\nu^2}{c^3} h\nu . n_\nu
$$ is the internal energy of the gas
However, I don't know how to tacle this derivation. Do you know how to do it, or is there a good reference for it ?

REPLY [1 votes]: Found the answer to my own question :)
We first re-write the entropy density as $$s_{E}(E,\Omega)=\frac{k_{B}}{4\pi^{3}\hbar^{3}c^{3}}\left[\frac{1}{k_{B}T}\frac{E^{3}}{e^{\frac{E}{k_{B}T}}-1}+\frac{1}{k_{B}T}E^{3}+E^{2}\log\left(\frac{1}{e^{\frac{E}{k_{B}T}}-1}\right)\right]$$
The first term is $u(E,\Omega)/T$ and its integration gives $\sigma T^{3}$. The second and third terms can be integrated by parts
$$\int dE\,E^{2}\log\left(\frac{1}{e^{\frac{E}{k_{B}T}}-1}\right)=\left[\frac{E^{3}}{3}\log\left(\frac{1}{e^{\frac{E}{k_{B}T}}-1}\right)\right]_{0}^{\infty}-\int dE\,\frac{E^{3}}{3}\left(1+\frac{1}{e^{\frac{E}{k_{B}T}}-1}\right)$$ which gives an additionnal $\sigma T^{3}/3$ contribution, hence the result.


TITLE: The isolated point in Hausdorff space
QUESTION [0 upvotes]: We call a point $x$ of a topological space isolated if $\{x\}$ is open.
Let $X$ be a nonempty compact Hausdorff space without isolated points.
Prove that $|X| \geq |\mathbb{R}|$.
In this question I dont want to just show that $X$ is uncountable. But I want to prove that the set $S = \{0, 1\}^\mathbb{N}$ that contains of all binary sequences has the same cardinality
as $\mathbb{R}$ suffices to find an injective function $f : S \to X$.
I will be really really thankful for any help.

REPLY [0 votes]: This is a standard idea from the construction of the Cantor set: splitting closed sets into parts. Here we will use open sets (and their closures), as we know a basic fact for a $T_1$ space without isolated points: all open sets are infinite (a finite open set implies an isolated point).
Let $\Sigma^\ast$ be the set of finite sequences of $0$ and $1$ ,including the empty one $\varepsilon$ .If $\sigma = (s_0,\ldots,s_n)$ we denote by $s \smallfrown 0$ the sequence with one extra $0$, so $((s_0,\ldots,s_n,0)$ and similarly for $s \smallfrown 1$, as well.
Now we define for each $\sigma \in \Sigma^\ast$, a non-empty open set $U_\sigma$ such that the following holds for all $\sigma$

$\overline{U_{\sigma \smallfrown 0}} \cup \overline{U_{\sigma \smallfrown 1}} \subset U_\sigma$
$\overline{U_{\sigma \smallfrown 0}} \cap \overline{U_{\sigma \smallfrown 1}} =\emptyset$

It's clear this can be done as we can start with $U_\varepsilon = X$ and each intermediate $U_\sigma$ is infinite, and has two open subsets with disjoint closures by regularity etc. So a recursive construction (having defined $U_\sigma$, find $U$ for its two extensions) never ends.
Now for each $x \in \{0,1\}^\mathbb{N}$ denote by $x|n$ the sequence of the first $n$ coordinates of $x$, $(x_0,\ldots,x_{n-1}), x|0= \varepsilon$.
Now for a fixed $x$ we have that $\overline{U_{x|n}}$ is a decreasing sequence of non-empty closed sets, so it has non-empty intersection in a compact space. Define $f(x)$ to be some chosen point in that intersection:
$$f(x)\in \bigcap \overline {U_{x|n}}$$
If $x \neq y \in \{0,1\}^\mathbb{N}$, there is a first coordinate $m$ where $x_m \neq y_m$, say (WLOG) $x_m =0, y_m=1$, so that $x|m =y|m$.
Then $f(x) \in \overline{U_{x|m \smallfrown 0}}$, $f(y) \in  \overline{U_{y|m \smallfrown 1}}= \overline{U_{x|m \smallfrown 1}}$ and by construction these are disjoint so $f(x) \neq f(y)$.
So $f$ is injective from $\{0,1\}^\mathbb{N}$ into $X$, proving the claim.


TITLE: Integral Table Result
QUESTION [2 upvotes]: To prove the result
$$\int_{1}^{e}\frac{\ln(x)}{(\alpha +\ln(x))^{(\alpha+1)}}\,dx = \frac{e}{(\alpha+1)^\alpha} - \frac{1}{\alpha^\alpha}$$
I have used the substitution $t = \alpha +\ln(x)$ which is somewhat obvious but end up with a very similar result but with the an power not $\alpha$ but ($\alpha+1)$! This may be my algebra failing me ......but I keep arriving at this wrong result. Can someone please elaborate and guide me towards the correct answer?

REPLY [2 votes]: Let $~u=\ln\left(x\right)+{\alpha}~$ , then $~ x~du= dx\implies dx=e^{u-\alpha}~du~$.
when $~x\to 1,~ u\to \alpha~$ and
when $~x\to e, ~u\to1+\alpha~$
Now
$$I=\int_{1}^{e}\frac{\ln(x)}{(\alpha +\ln(x))^{(\alpha+1)}}\,dx $$
$$=\class{steps-node}{\cssId{steps-node-1}{\mathrm{e}^{-{\alpha}}}}{\displaystyle\int_{\alpha}^{\alpha+1}}u^{-{\alpha}-1}\left(u-{\alpha}\right)\mathrm{e}^u\,\mathrm{d}u$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\left(\frac{e^u}{u^{\alpha}}-\alpha~e^u~u^{-{\alpha}-1}\right)du$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\frac{e^u}{u^{\alpha}}~du-\alpha ~e^{-\alpha}~\int_{\alpha}^{\alpha+1}~e^u~u^{-{\alpha}-1}~du$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\frac{e^u}{u^{\alpha}}~du-\alpha ~e^{-\alpha}~\left\{\left[-~\frac{1}{\alpha}~e^u~u^{-\alpha}\right]_{\alpha}^{\alpha+1}+~\frac{1}{\alpha}~\int_{\alpha}^{\alpha+1}~e^u~u^{-{\alpha}}~du\right\}$$
$$= ~e^{-\alpha}~\left[e^u~u^{-\alpha}\right]_{\alpha}^{\alpha+1}$$
$$= ~e^{-\alpha}~[e^{\alpha+1}~({\alpha+1})^{-\alpha}-e^\alpha~\alpha^{-\alpha}]$$
$$= \frac{e}{(\alpha+1)^\alpha} - \frac{1}{\alpha^\alpha}$$


TITLE: Find the power of $P$ in relation the center circumference $O$ below.
QUESTION [0 upvotes]: For reference: In the acute triangle $ABC$ of orthocenter $H$ and circumcenter $O$, the
height $AP$. If the distance from $O$ to $AC$ is $2$ and $PH = 1$; calculate the power of $P$ in relation the center circumference $O$.(Answer:$ -5$)
My progress:

Points inside the circle have negative power;
$P(P) = d^2 - R^2 = PO^2-AO^2 = r^2 - R^2\\
\triangle AOD: R^2 = 4+AD^2\\
 ADE \sim \triangle APC \implies \frac{DE}{CP} = \frac{AD}{AH+1}=\frac{AE}{2AD}
\\BH=2OD (property)= 2.2=4\\
\triangle BPH: 4^2 = BP^2+1^2 \therefore BP = \sqrt15\\
\triangle APC \sim \triangle BPH \implies: \frac{PC}{1}=\frac{2AD}{4}=\frac{AH+1}{\sqrt{15}} \implies AD = 2PC\\
\therefore AE = 4DE$
???...

REPLY [2 votes]: $ - 5$ is not the correct answer.
Note that $BH = 2 OD = 4$ and hence $BP = \sqrt{15}$
We then use the fact that $\triangle ACP \sim \triangle BHP$
$ \displaystyle \implies AC = \frac{4}{\sqrt{15}} (AH+1), ~ PC = \frac{1}{\sqrt{15}} (AH+1)$
If $M$ is the midpoint of $BC$, $ \displaystyle CM = \frac {\sqrt{15} + PC}{2}$
$ \displaystyle PM = CM - PC = \frac {\sqrt{15} - PC}{2} = \frac{14 - AH}{2 \sqrt{15}}$
Power of point $P$ with respect to the circumcenter $O = OP^2 - OA^2$,
$ \displaystyle = OM^2 + PM^2 - (AD^2 + 4)$
$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{AC^2}{4} - 4$
$ \displaystyle = \frac{AH^2}{4} + \frac{(14-AH)^2}{60} - \frac{4 (AH+1)^2}{15} - 4 = - AH - 1$
So it depends on $AH$ and unless $\triangle ABC$ is isosceles with $\angle A = \angle B$, the answer cannot be $ - 5$.