text:
\section{Local Moufang conditions}

Before proceeding, we need the condition ``$(M)_x$''\index{$(M)_x$}. Let $\mS$ be a GQ, and let $x$ be a point of $\mS$.
Then $\mS$ satisfies $(M)_x$ if for any i-root $(x,L,y)$ ($x \I L \I y \ne x$), and any line $M \I y$, $M \ne L$,
the automorphism group of $\mS$ that fixes $x$ and $y$ linewise, and $L$ pointwise, $\Aut(\mS)_{[\{ x,L,y \}]}$,
acts transitively on  the points of $M \setminus \{y\}$. In that case, the action is sharply transitive, and the definition
is independent of $M$ (or $y$).  So $(M)_x$ holds if every i-root on $x$ is Moufang (by definition).

\subsection{MSTGQs}

Now let $(\mS^x,L)$ be a finite EGQ  with the following additionial properties:

\begin{itemize}
\item[(MSTGQ1)]
$\mS^x$ satisfies property $(M)_x$;
\item[(MSTGQ2)]
for each point $y \sim x \ne y$, $L_{[\{x,xy,y\}]} =: L(y)$ has the property that if $\alpha \in L(y)^\times$ fixes some affine line
$U$ (which must be concurrent with $xy$ as $\alpha \ne \id$), it also fixes $U \cap xy$ linewise;
\item[(MSTGQ3)]
no line $U \I x$ is the unique center of a triad $\{ V,W,X \}$, where $V \I x$.
\end{itemize}

Payne introduced such EGQs in \cite{9}, and calls them {\em Moufang skew translation generalized quadrangles} (MSTGQs). He shows in {\em loc. cit.}
that the point $x$ is regular, making it a center of symmetry for a subgroup $\mathbb{S} \leq L$, and $\mathbb{S} \leq Z(L)$ is
elementary abelian. Moreover, for any $u \sim x \ne u$, $L(u)$ is also elementary abelian, and $L$ is a $p$-group.

So $\mS^x$  is an STGQ, but it is of a very particular type: not all STGQs satisfy (MSTGQ1) (as we will see further in the next section),
and no known square STGQ of even order has (MSTGQ3).  A more general version of (MSTGQ1) seems to be true (now supposing that $(\mS^x,L)$ is an STGQ), namely:

\begin{itemize}
\item[(MSTGQ1)$^b$]
Each $\alpha \in L^\times$ which fixes some affine line
$U$, also fixes $U \cap \proj_Ux$ linewise.
\end{itemize}

Already from (MSTGQ1)$^b$, one can prove that the symmetry group $\mathbb{S}$ is in the center of $L$, and conversely, $\mathbb{S} \leq Z(L)$
implies (MSTGQ1)$^b$. So this seems the quintessential Moufang property for STGQs to study and obtain.

\subsection{Property (G)}

In this subsection we have a deeper look into the so-called ``Property (G)''.\\

Let $x$ be a point of a GQ $\mS$ of order $(t^2,t)$, and let $U, V$ be distinct lines incident with $x$. Then $\mS$ satisfies
{\em Property (G)} at the pair $\{U,V\}$ if any triad of lines $\{V,W,Z\}$ in $U^{\perp}$ is $3$-regular (which means that $\vert \{V,W,Z\}^{\perp} \vert =  \vert \{V,W,Z\}^{\perp\perp} \vert = t + 1$. (Note that the definition is symmetric in $U$ and $V$.)
The flag $(x,L)$ has {\em Property (G)} if all pairs of distinct lines $\{L,M\}$ ``on $x$'' have (G).
One says that $x$ has {\em Property (G)} if all pairs $\{U,V\}$ ``on $x$'' have Property (G).\\

The following theorem was first obtained in odd characteristic in \cite{III}, answering a fundamental conjecture of Payne's essay \cite{9}.
In the case of even characteristic, it was obtained only much later by M. R. Brown \cite{broeven}. (In odd characteristic, only
one flag was required in \cite{III}.)

\bt[\cite{broeven,III}]
\label{Geven}
A GQ of order $(t^2,t)$ satisfying Property (G) at two distinct flags $(u,L)$ and $(u,M)$ for a point $u$ is isomorphic to a flock GQ.
\et

\bc
\label{subTGQs}
A GQ $\mS$ of order $(t^2,t)$ with $t$ even is a flock GQ if and only if it contains a point $x$ which is in $t^3 + t^2$ distinct subTGQs of order $t$ for which $x$ is
a translation point.
\ec

{\em Proof}.\quad
It is well known that $t^3 + t^2$ is the maximum number of subGQs of order $t$ that contain the same point in a GQ of order $(t^2,t)$. If this is the case for $x$, we have that
for any triad of lines $\{U,V,W\}$ for which $U \I x$ and which has a center $C$ which is incident with $x$, $U, V, W, C$ and $D$ are contained in a subGQ $\mS'$ of order $t$, where $D$
is any center of $\{U,V,W\}$ different from $C$. Since $\mS'$ is a TGQ with translation point $x$, any line incident with $x$ is regular in $\mS'$, and so it easily follows that
$x$ satisfies Property (G). Whence by Theorem \ref{Geven}, $\mS$ is a flock GQ.\\

Conversely, it is well known that a flock GQ of order $(t^2,t)$ with $t$ even has $t^3 + t^2$ subTGQs of order $t$ containing the special point, which is a common translation point for all these
subGQs.
\eop \\

Let $\mS$ be a GQ satisfying the same properties as in the statement of Corollary \ref{subTGQs}, but now with $t$ odd. Then by the same reasoning, $\mS$ is a flock GQ.
The only flock GQs in odd characteristic with ideal subGQs are isomorphic to the Kantor-Knuth flock  quadrangles \cite{TGQ}, and in that case, all ideal subGQs are isomorphic to
$\hW(t)$, contradicting the fact that we assumed the subGQs to be TGQs. So for $t$ odd, Corollary \ref{subTGQs} makes no sense.

\subsection{The Suzuki-Tits quadrangles}

Let $q = 2^e$ be an odd power of $2$ (so that $\sigma$ is a Tits endomorphism), and let $\sigma \in \mathrm{Aut}(\mathbb{F}_q)$ be such that $\sigma^2 = 2$. Now define a map

\begin{equation}
f: \mathbb{F}_q^2 \mapsto \mathbb{F}_q^2: (a,b) \mapsto a^{\sigma + 2} + ab + b^{\sigma}.
\end{equation}

The {\em Tits ovoid} of $\PG(3,q)$ is then given by

\begin{equation}
\mO = \{(0,1,0,0)\} \cup \{(1,f(a,b),a,b) \vert a, b \in \mathbb{F}_q\}.
\end{equation}

It is well known that this ovoid admits the natural $2$-transitive action of the Suzuki group $\mathbf{Sz}(q)$.
Embed $\PG(3,q)$ as a hyperplane in $\PG(4,q)$ by $(x,y,z,w) \longrightarrow (0,x,y,z,w)$.
Construct the TGQ $\Gamma = \hT_3(\mO)$, and call it the Suzuki-Tits quadrangle. Then $\mathrm{Aut}(\Gamma)_{(\infty)}$ (where $(\infty)$ is the special point of
$\hT_3(\mO)$) admits the  natural $2$-transitive action of $\mathbf{Sz}(q)$ on the $q^2 + 1$ lines incident with $x$.
Also, $\hT_3(\mO)$ is classical if and only if $\mO$ is an ellipic quadric if and only if $q = 2$.

Now define  collineations $\theta(a,b,c,d,e): (u,x,y,z,w) \mapsto (u,x,y,z,w)[a,b,c,d,e]$, with $a,b,c,d,e \in \mathbb{F}_q$,  of $\PG(4,q)$ as follows:

\begin{equation}
[a,b,c,d,e] = \left(
 \begin{array}{ccccc}
 1 & 0 & c & d & e\\
 0 & 1& f(a,b) &a & b\\
 0 & 0 & 1&0&0\\
 0 & 0 & a^{\sigma + 1} + b & 1 & a^{\sigma}\\
 0 & 0 & a & 0 & 1\\
 \end{array}
 \right).                                         \end{equation}

Then by \cite{StanPriv}, $G = \{ \theta(a,b,c,d,e) \vert a,b,c,d,e \in \mathbb{F}_q\}$ is a group of order $q^5$ with binary operation
\begin{equation}
\begin{aligned}
&
[a,b,c,d,e][a',b',c',d',e']\\
&= [a + a', b + b' + a{a'}^{\sigma}, c + c' + d({a'}^{\sigma + 1} + b') + ea', d + d', e + e' + d{a'}^{\sigma}].
\end{aligned}
\end{equation}

The group $G$ leaves $\mO$ invariant; it fixes the line $L$ of $\Gamma$  corresponding to $(0,0,1,0,0)$ $\in \mO$ pointwise, and acts sharply transitively on the lines
not concurrent with $L$. So $\Gamma^D$ is an EGQ with elation group $G$. The center of $G$ is $Z(G) = \{ [0,0,c,0,0] \vert c \in \mathbb{F}_q\}$, and is a group
of symmetries about $L$. So the dual of $\Gamma$ is a central STGQ of order $(t^2,t)$.\\

Now define for $t \in \mathbb{F}_q$:
\begin{equation}
A(t) = \{[a,b,tf(a,b),ta,tb] \vert a, b \in \mathbb{F}_q\}, A^*(t) = \{[a,b,c,ta,tb] \vert a, b, c \in \mathbb{F}_q\},
\end{equation}

and also put
\begin{equation}
A(\infty) = \{[0,0,0,d,e] \vert d, e \in \mathbb{F}_q\}, A^*(\infty) = \{[0,0,c,d,e] \vert c, d, e \in \mathbb{F}_q\}.
\end{equation}

Then $\hF = \{ A(t) \vert t \in \mathbb{F}_q \cup \{(\infty)\}\}$ and $\hF^* = \{ A^*(t) \vert t \in \mathbb{F}_q \cup \{(\infty)\}\}$ define the Kantor family
of $G$ corresponding to the point $\langle (1,0,0,0,0),(0,1,0,0,0)\rangle$ of $\Gamma^D$.

\subsection{Some strange properties}

The following structure lemma is easy, but rather mysterious due to (ii), (iii) and (vi), cf. the remark after the observation.

\bo
\label{Structob}
\begin{itemize}
\item[{\rm (i)}]
$Z(G)$ is the symmetry group w.r.t. the elation point, and is elementary abelian.
\item[{\rm (ii)}]
$A^*(\infty)$ and $A(\infty)$ are elementary abelian and $A^*(\infty) \unlhd G$, so that $A^*(\infty)$ fixes $[A(\infty)]$ pointwise.
\item[{\rm (iii)}]
For $t \in \mathbb{F}_q$, $A^*(t)$ and $A(t)$ are non-abelian of exponent $4$; moreover, for $t \ne t'$, $A^*(t) \cong A^*(t')$ and $A(t) \cong A(t')$.
Also, no $A^*(t)$ is normal in $G$ and hence does not fix $[A(t)]$ pointwise.
\item[{\rm (iv)}]
$G$ is nonabelian of exponent $4$ and $G/Z(G)$ is not abelian.
\item[{\rm (v)}]
$G$ is the complete set of elations about $x$.
\item[{\rm (vi)}]
$\Gamma^D$ is not an MSTGQ.
\item[{\rm (vii)}]
$[G,G]$ is strictly contained in $A^*(\infty)$.
\end{itemize}
\eo

{\em Proof}.\quad
(i)-(ii)-(iii) are obvious. The fact that $G/Z(G)$ is not abelian follows from the fact that for any $t \in \mathbb{F}_q$, $A^*(t)/Z(G) \cong A(t)$.
Claim (v) is proved in \cite{Notes}. Claim (vi) follows from (v) and (iii). Finally, clearly $[G,G]$ fixes $[A(\infty)]$ pointwise so that it is a
subgroup of $A^*(\infty)$. As $G/A^*(\infty)$ is isomorphic to $A(t)$ for any $t \in \mathbb{F}_q$, it follows that $A^*(\infty) \ne [G,G]$ by (iii).
\eop \\

Note that by (ii) and (iii), each i-root $(x,[A(\infty)],z)$ is Moufang, while no other i-root of $\Gamma^D$ is.\\

The Kantor family described in the previous paragraph is, by (ii)-(iii), the only known one for which not all elements are isomorphic.
Besides that, $\Gamma^D$ is the only known STGQ which is not an MSTGQ. The elation group  $G$ can not be constructed by the twisting method described in this paper |
essentially there are two reasons; first of all, the fact that $\Gamma^D$ is not an MSTGQ (which is needed for making the twist), and secondly, if $\Gamma^D \not\cong \hH(3,t^2)$,
it cannot contain subGQs of order $t$ which are fixed elementwise by some involution of $\hT_3(\mO)$ (and such involutions are also needed).

\bo
Let $[\infty]$ be the special line of $\Gamma^D$ corresponding to $\mO$. Then every point incident with $[\infty]$ is an elation point for a unique elation group.
\eo

{\em Proof}.\quad
Follows from Observation \ref{Structob}(v) and the fact that $\mathrm{Aut}(\Gamma)^D$ is transitive on the lines through $x$.
\eop

\medskip
\subsection{SubTGQs of $\Gamma^D$}

Each nontangent plane section $\Pi \cap \mO = \mO'$ defines a $\hT_2(\mO')$ of Tits, and hence a full subTGQ of $\hT_3(\mO)$.
Moreover, any such $\mO'$ is a translation oval, so that $\hT_2(\mO')^D$ also is a subTGQ of $\Gamma^D$. Whence for any choice of $x \in \mO$,
$\Gamma^D$ admits precisely $q^3 + q^2$ subTGQs of order $q$ that contain $x$.
Still, $\Gamma^D$ is not a flock GQ (supposing that it is not classical). The reason why Corollary \ref{subTGQs} cannot be applied is reflected in the following theorem.

\bt
Let $\mO$ be non-classical.
\begin{itemize}
\item[{\rm (i)}]
Then $\Gamma^D$ contains precisely $q^4 + q^2$ distinct subTGQs of order $q$ which all contain $[\infty]$, and for which the (unique) translation point is a point of $[\infty]$.
\item[{\rm (ii)}]
Each point $z \I [\infty]$ is
contained in precisely $q^3 + q^2$ subGQs of order $q$. Precisely $q^2$ of them have $z$ as (unique) translation point. For each other point $z' \I [\infty]$, $z' \ne z$,
there are precisely $q$ subGQs of order $q$ containing $\{z,z'\}$, and having $z'$ as (unique) translation point.\eop
\end{itemize}
\et

(The proof is left to the reader.)


TITLE: Is a set in between two sets of equal measure measurable?
QUESTION [7 upvotes]: The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well.  But I'd like to know if something stronger is true.
Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$.  Then my question is, does $B$ have to be Lebesgue measurable as well?

REPLY [10 votes]: If $A$ and $C$ are each Lebesgue measurable, then so is $C\setminus A$ and so by finite additivity it has measure zero.
But then $B\setminus A$ also has measure zero and hence is measurable, so $$B=A\cup(B\setminus A)$$ is the union of two measurable sets, hence is measurable.


text:
\begin{document}
\openup .5\jot
\maketitle

\begin{abstract}

A square matrix $M$ with real entries is said to be \textit{algebraically positive} (AP) if there exists a real polynomial $p$ such that all entries of the matrix $p(M)>0$. A square sign pattern matrix $S$ is said to \textit{allow} algebraic positivity if there is an algebraically positive matrix $M$ whose sign pattern class is $S$. On the other hand, $S$ is said to \textit{require} algebraic positivity if any matrix $M$, having sign pattern class $S$, is algebraically positive. Motivated by open problems raised in the work of Kirkland, Qiao and Zhan (2016) on AP matrices, we list down all nonequivalent irreducible $3\times 3$ sign pattern matrices and classify each of them into three groups (i) those that require AP, (ii) those that allow but not require AP, or (iii) those that do not allow AP. We also give a necessary condition for an irreducible $n\times n$ sign pattern to allow algebraic positivity.
\end{abstract}

\noindent \textit{Keywords:} Algebraically Positive Matrices, Signed Digraphs, Sign Pattern Matrices\\
\textit{AMS Classification: 15B35, 15B48, 05C50}

\section{Introduction}
We denote the set of $m\times n$ real matrices by $M_{m\times n}$ and the set of all $n\times n$ matrices by $M_n$ and the $(i,j)$ entry of a matrix $A$ by $(A)_{ij}$.

A matrix $A\in M_{m\times n}$ is said to be \textit{positive}, written $A>0$, if all entries of $A$ are positive. Several applications of positive matrices appear in various fields of study such as Markov chains in probability theory, population models, iterative methods in numerical analysis, economic models, epidomiology, low-dimensional topology, physics, and many more (for example, see\cite{MacCluer,Pillai}). Some natural generalizations of the concept of matrix positivity have also been studied extensively such as matrix nonnegativity, matrix primitivity and eventual positivity. In \cite{Kirk1}, Kirkland, Qiao and Zhan presented another generalization called algebraic positivity. A matrix $A\in M_n$ is said to be \textit{algebraically positive} if $p(A)>0$ for some real polynomial $p$. Using the Cayley-Hamilton Theorem, it can be deduced that a matrix $A\in M_n$, where $n\geq 2$, is algebraically positive if and only if there are real numbers $k_1,\ldots, k_{n-1}$ such that the off-diagonal entries of the following matrix are all positive
\[k_1A+\cdots +k_{n-1}A^{n-1}.\]
The authors of \cite{Kirk1} used the Perron-Frobenius Theorem to prove the following characterization of algebraically positive matrices.

\begin{theorem}[Kirkland, Qiao and Zhan, 2016]\label{thm1}
A real matrix is algebraically positive if and only if it has a simple real eigenvalue and corresponding left and right positive eigenvectors.
\end{theorem}

A \textit{sign pattern} is a matrix whose entries are from the set $\{0,+,-\}$. We define the \textit{pattern class} of a sign pattern $S$, denoted by $Q(S)$, as the set
\[Q(S)=\{A\in M_{m,n}\ | \ \mbox{sgn}((A)_{ij})=(S)_{ij}\}\]
of real matrices whose sign pattern is $S$. In this paper, we are interested in determining when a given square sign pattern matrix \textit{allows algebraic positivity}, i.e. there exists an algebraically positive matrix in its sign pattern class; or \textit{requires algebraic positivity}, i.e. all matrices in its sign pattern class are algebraically positive.

The following results from \cite{Kirk1} will be useful in our study:
\begin{theorem}[Kirkland, Qiao and Zhan, 2016]\label{thm2} \phantom{ \ } \
\begin{enumerate}
\item Every algebraically positive matrix is irreducible.
\item If $A$ is an irreducible real matrix all of whose off-diagonal entries are non-negative (or nonpositive), then $A$ is algebraically positive.
\item If a sign pattern allows algebraic positivity, then every row and column contains a $\ +$, or every row and column contains a $\ - $.
\end{enumerate}
\end{theorem}

Recall that a matrix $A\in M_n$ is reducible if there exists a permutation matrix $P$ such that $PAP^T$ is of the form
\[PAP^T=\begin{bmatrix}
A_1 & A_2\\
0 & A_3
\end{bmatrix}\]
for some $A_1\in M_s$ such that $1\leq s<n$. The reducibility of a matrix is easily determined by looking at its associated digraph. The \textit{digraph} of $A$, denoted by $\Gamma(A)$, is the directed graph whose vertex set is $\{1,\ldots,n\}$ and whose edge set is $E_A=\{(i,j) \ | \ (A)_{ij}\neq 0\}$. It is known (see \cite{handbook}) that a matrix is irreducible if and only if its digraph is strongly connected, i.e. for every pair of distinct vertices $i$ and $j$, there is a directed path in $\Gamma(A)$ from $i$ to $j$. Note that the digraph of $A^T$ is obtained from the digraph of $A$ by reversing the direction of all the edges. Meanwhile, if $P$ is a permutation matrix, then the digraph of $PAP^T$ is obtained from the digraph of $A$ by permuting the labels of the vertices of $\Gamma(A)$. Thus, if $\Gamma(A)$ is irreducible, then so are $\Gamma(A^T)$ and $\Gamma(PAP^T)$. For the purpose of this study, we say that two digraphs are \textit{equivalent digraphs} if one can be obtained from the other by (i) reversing the direction of all the edges, (ii) permuting the labels of the vertices, or (iii) doing both. In Appendix A, we list down all 26 nonequivalent irreducible digraphs in with vertex set $\{1,2,3\}$.

\begin{theorem}\label{thm3}
If $A$ is algebraically positive, then the following matrices are also algebraically positive
\begin{multicols}{2}
\begin{enumerate}
\item $A^T$;
\item $-A$;
\item $PAP^T$ for any permutation matrix $P$;
\item $\beta A + \alpha I$ for any $\alpha\in\mathbb{R}$ and any $\beta\in \mathbb{R}\setminus\{0\}$.
\end{enumerate}
\end{multicols}
\end{theorem}
\textit{Proof:} Suppose $p$ is a real polynomial such that $p(A)>0$ and suppose $P$ is a permutation matrix. Then $p(A^T)=p(A)^T>0$ and $p(PAP^T)=Pp(A)P^T>0$. Define $q(x)=p(\frac{1}{\beta}x-\frac{\alpha}{\beta})$, then $q(\beta A + \alpha I)=p(A)>0$. In particular, if $\beta=-1$ and $\alpha=0$, we have $q(-A)>0$. \qed

For the purpose of our study, we say that two sign pattern matrices are \textit{equivalent sign patterns} if one can be obtained from the other by (i) transposition; (ii) permutation similarity; (iii) negation; (iv) any combination of the first three transformations.

\section{Main Results}
The main result of this study gives a necessary condition for sign patterns that allow algebraic positivity.
\begin{theorem} \label{thm4}
	Suppose A is an $n \times n$ sign pattern matrix. We write $A = A_+ + A_-$ where $A_+$ contains only the nonnegative signs of $A$, while $A_-$ contains only non-positive signs of $A$. Define $B_A = A_+ - A_-^T$. If $A$ is irreducible and $B_A$ is reducible, then $A$ does not allow algebraic positivity.
\end{theorem}
\textit{Proof:} Let $A\in M_n(\{+,-,0\})$ and
\[(A_+)_{ij}=\left\{\begin{array}{ll}
+ & \mbox{ if }(A)_{ij}=+\\
0 & \mbox{ otherwise}\\
\end{array}\right. \quad
(A_-)_{ij}=\left\{\begin{array}{ll}
- & \mbox{ if }(A)_{ij}=-\\
0 & \mbox{ otherwise}\\
\end{array}\right. \]
Define $B_A=A = A_+ - A_-^T$. Clearly $B_A\in M_n(\{0,+\})$. Suppose $B_A$ is reducible. Then, we can assume without loss of generality that
\begin{equation}\label{redB}
	B_A = \begin{bmatrix}
	B_1 & B_2 \\
	0   & B_3
	\end{bmatrix}
\end{equation}
for some $B_1\in M_s(\{0,+\})$, $s \times (n-s)$ sign pattern matrix $B_2$, and $(n-s) \times (n-s)$ sign pattern matrix $B_3$. Otherwise, we can replace $A$ by $PAP^T$ wherer $P$ is a permutation matrix such that $PB_AP^T$ is in the said form. From ($\ref{redB}$), it will follow that $A$ is of the form \[A = \begin{bmatrix}
	C - D & E \\
	-F     & G - H
	\end{bmatrix}\]
for some $C,D\in M_s(\{0,+\})$, $G,H\in M_{n-s}(\{0,+\})$ and $E\in M_{s,n-s}(\{0,+\})$ and $F\in M_{n-s,s}(\{0,+\})$. Note also that the entries of $E$ or $F$ cannot all be zero since $A$ is irreducible.

	Suppose there exists an $n \times n$ matrix $X \in Q(A)$ such that $X$ is algebraically positive. Then we can say that \[X = \begin{bmatrix}
	X_C - X_D & X_E \\
	-X_F       & X_G - X_H
	\end{bmatrix}\in Q(A)\]
	where $X_J\in Q(J)$ for $J=C,D,E,F,G,H$. By Theorem 1, $X$ has a simple real eigenvalue, say $\lambda \in \mathbb{R}$, with corresponding positive left and right eigenvectors \[\begin{bmatrix}
	u^T & v^T \\
	\end{bmatrix}, \begin{bmatrix}
	y\\
	z
	\end{bmatrix} > 0, \mbox{ where } u, y \in M_{s \times 1} \mbox{ and } v, z \in M_{(n-s) \times 1}.\]	Note that
	\[\begin{bmatrix}
	u^T & v^T \\
	\end{bmatrix} (X - \lambda I) = \begin{bmatrix}
	u^T(X_C - X_D - \lambda I) - v^TX_F & \ast
	\end{bmatrix}
	=
	\begin{bmatrix}
	0 & 0 & \cdots & 0
	\end{bmatrix}\]
	and \[(X - \lambda I) \begin{bmatrix}
	y\\
	z
	\end{bmatrix}
	=
	\begin{bmatrix}
	(X_C - X_D - \lambda I)y + X_Ez \\
	\\
	\ast
	\end{bmatrix}
	=
	\begin{bmatrix}
	0 \\ 0 \\ \vdots \\ 0
	\end{bmatrix}\]
	Therefore, $u^T(X_C - X_D - \lambda I) = v^TX_F$ and $ (X_C - X_D - \lambda I)y = -X_Ez$.	Consider the matrix $u^T(X_C - X_D - \lambda I)y = v^TX_Fy = -u^TX_Ez$. Since the entries of $u^T,v^T, y,z$ are all positive and $X_E,X_F$ are nonzero nonnegative matrices, then $v^TX_Fy$ is a positive real number, while $-u^TX_Ez$ is a negative real number. This gives a contradiction. Hence, any $X\in Q(A)$ is not algebraically positive. \qed\\
Here is an example on how to utilize this result.
\begin{example}\label{ex4.2}
	\normalfont Consider the sign pattern $A$ given below. We write $A$ as follows, separating the positive ($A_+$) and the negative $(A_-)$ signs.
	\[A= \begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & - \\
	+ & 0 & +
	\end{bmatrix} = \begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & 0 \\
	+ & 0 & +
	\end{bmatrix} + \begin{bmatrix}
	0 & 0 & 0 \\
	0 & 0 & - \\
	0 & 0 & 0
	\end{bmatrix}\]
	We then write its $B_A$ as follows, negating and transposing $A_-$.
	$$B= \begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & 0 \\
	+ & + & +
	\end{bmatrix} = \begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & 0 \\
	+ & 0 & +
	\end{bmatrix} + \begin{bmatrix}
	0 & 0 & 0 \\
	0 & 0 & 0 \\
	0 & + & 0
	\end{bmatrix}$$

We can treat the negative signs of $A$ as a reversal of the direction on the corresponding edges in the signed digraph of $A$ to obtain the signed digraph of $B_A$ as seen below.
\bigskip\\
\begin{minipage}{0.49\textwidth}
	\centering
	\begin{tikzpicture}[->,>=stealth,shorten >=1pt,auto,node distance=2cm,
	thick,main node/.style={circle,draw,font=\bfseries}]

	\node[main node] (1) {1};
	\node[main node] (2) [below left of=1] {2};
	\node[main node] (3) [below right of=1] {3};

	\path
	(1) edge [transform canvas={xshift=-2pt,yshift=2pt},shorten <= -1pt] node[above]{$+$} (2)

	(2) edge [transform canvas={xshift=2pt,yshift=-2pt},shorten <= -1pt] node[below]{$+$} (1)
	edge [transform canvas={xshift=0pt,yshift=-3pt},shorten <= -1pt] node[below]{$-$} (3)

	(3) edge [loop below] node{$+$} (3)
	edge [transform canvas={xshift=2pt,yshift=2pt},shorten <= -1pt] node[above]{$+$} (1);
	\end{tikzpicture}
	\\Signed Digraph of $A$
\end{minipage}
\begin{minipage}{0.49\textwidth}
	\centering
	\begin{tikzpicture}[->,>=stealth,shorten >=1pt,auto,node distance=2cm,
	thick,main node/.style={circle,draw,font=\bfseries}]

	\node[main node] (1) {1};
	\node[main node] (2) [below left of=1] {2};
	\node[main node] (3) [below right of=1] {3};

	\path
	(1) edge [transform canvas={xshift=-2pt,yshift=2pt},shorten <= -1pt] node[above]{$+$} (2)

	(2) edge [transform canvas={xshift=2pt,yshift=-2pt},shorten <= -1pt] node[below]{$+$} (1)

	(3) edge [loop below] node{$+$} (3)
	edge [transform canvas={xshift=2pt,yshift=2pt},shorten <= -1pt] node[above]{$+$} (1)
	edge [transform canvas={xshift=0pt,yshift=3pt},shorten <= -1pt] node[above]{$+$} (2);
	\end{tikzpicture}
	\\Signed Digraph of $B_A$
\end{minipage}\bigskip\\
Note that $B_A$ is reducible since its corresponding digraph s not strongly connected \cite{handbook}.
\end{example}
We also note that the converse of Theorem 3 is not true, by providing a counterexample below.
\begin{example}
	\normalfont Consider the sign pattern $C$ below. By the Theorem 2.3, $C$ does not allow algebraic positivity. Now, we write $C$ as follows:
	$$C = \begin{bmatrix}
	0 & - & 0 \\
	- & 0 & + \\
	+ & 0 & +
	\end{bmatrix} =
	\begin{bmatrix}
	0 & 0 & 0 \\
	0 & 0 & + \\
	+ & 0 & +
	\end{bmatrix} +
	\begin{bmatrix}
	0 & - & 0 \\
	- & 0 & 0 \\
	0 & 0 & 0
	\end{bmatrix}$$
	However, one can easily check that the matrix $B_C$ below is irreducible.
	\[B_C = \begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & + \\
	+ & 0 & +
	\end{bmatrix} =
	\begin{bmatrix}
	0 & 0 & 0 \\
	0 & 0 & + \\
	+ & 0 & +
	\end{bmatrix} +
	\begin{bmatrix}
	0 & + & 0 \\
	+ & 0 & 0 \\
	0 & 0 & 0
	\end{bmatrix}.\]
\end{example}
Recall that a matrix $X\in M_n$ is algebraically positive if and only if $X+\alpha I_n$ is algebraically positive for any $\alpha\in \mathbb{R}$. We now introduce the notion of a \textit{scalar-shift subclass of a sign pattern matrix}. We say that a sign pattern $B$ is a (scalar-shift) subclass of another sign pattern $A$, denoted by $B\trianglelefteq A$, if for any $X\in Q(B)$, there exists $\alpha\in \mathbb{R}$ such that the sign pattern of $X+\alpha I_n$ is equivalent to $A$.
\begin{example}
Let $B=\begin{bmatrix}
0 & +\\
+ & 0
\end{bmatrix}$ and $A=\begin{bmatrix}
- & + \\
+ & -
\end{bmatrix}$. Then $B\trianglelefteq A$ but $A\not \trianglelefteq B$.
\end{example}
The following theorem follows directly from Theorem 3.4.
\begin{theorem} Suppose $A \trianglelefteq  B$.
\begin{enumerate}
\item If $B$ requires algebraic positivity, then so does $A$.
\item If $A$ allows algebraic positivity, then so does $B$.
\end{enumerate}
\end{theorem}
Finally, in Appendix A, we list down all nonequivalent irreducible $3\times 3$ sign pattern matrices and determine whether they require, allow or do not allow algebraic positivity.

\section{Acknowledgments}
This work is funded by the Natural Sciences Research Institute of the College of Science, University of the Philippines Diliman under the research grant MAT-18-1-03.

\nocite*

config: arxiv


TITLE: Is there a website like OEIS for real constants?
QUESTION [25 upvotes]: I'm hoping an online service exists where I can type in say 3.14159 and it then shows a bunch of 'interesting' (however one would define that) numbers.
Naturally in that instance it would bring up $\pi$ first as the most likely candidate, then $355/113$ then increasingly more bizarre numbers. Ideally with a weighting towards use of irrational/transcendental numbers and simplicity of expression.
I have a very messy expression that is currently graphed and I want to find out if the maximum of the graph is a known number.
Thanks in advance!

REPLY [9 votes]: Wolfram|Alpha (https://www.wolframalpha.com) can return an assortment of information about a decimal number including a list of possible closed forms, e.g.
https://www.wolframalpha.com/input/?i=1.303577269034
https://isc.carma.newcastle.edu.au/ did not do anything useful with this particular number.


TITLE: Is it possible to find a graphable function solution for the Brachistochrone problem without using trigonometric substitution?
QUESTION [1 upvotes]: The solutions of for the Brachistochrone problem make use of trigonometric substitution to generate parametric equations for x and y. Does a solution exist without trigonometric substitution?

REPLY [0 votes]: The differential equation of the cycloid profile (curvature radii ratio) when
$$ \tan \phi = y^{'}  $$
is
$$ \frac{y y ^{''}}{1+y^{'2}}= -\frac12 $$
Its integral
$$ \cos \phi =  \pm \sqrt{y/c}   $$
involves cos function which is trigonometric... so at  property level  itself embedded and trig is unavoidable.


TITLE: "Rectangularity" measure of a polygon
QUESTION [4 upvotes]: For an arbitrary polygon, I want to be able to calculate how "rectangular" it is.
Does any work exist which has tried to do this?
I found An Efficiently Computable Metric for Comparing Polygonal Shapes (Arkin et al., 1991), which develops a metric for comparing arbitrary polygons. So one way to do what I want might be to use this metric and compare the polygon to some rectangle.

REPLY [3 votes]: One way to do it could be to determine the minimum bounding rectangle of the given polygon, then use the "coverage factor" (ratio between the area of the polygon and area of its bounding rectangle) as a measure of similarity. This will be a percentage varying from $0\%$ for a thin star-shaped polygon to $100\%$ for an actual rectangle. For convex polygons, the lower bound is $50\%$ (e.g. triangles) as discussed in Area ratio of a minimum bounding rectangle of a convex polygon on MO. Some links to relevant algorithms can be found at Finding minimum-area-rectangle for given points.


TITLE: $(n!)!>n^{n!} \forall n \in \Bbb N^{\ge 4}$
QUESTION [4 upvotes]: I'm not sure if this has been asked before, but I came across it in the reasoning of solving a problem for my Real Analysis homework. It was not directly part of the problem so it may be a unique question, but I'm sure somebody has thought about it at some point. The problem is in proving that:
$$(n!)!>n^{n!}\quad \forall n\in \Bbb N^{\geq4}$$
I plotted it on Wolfram Alpha, so I know it to be true. Additionally I had some faint trail of reasoning for why it should be so, but I could not figure out a proof.

REPLY [3 votes]: Let $n\ge5$. In the $n!$ terms that multiply together to yield $(n!)!$, the last $n!-n^2$ of them are bigger than $n^2$. Therefore
$$
(n!)! > (n^2)^{n!-n^2} = n^{2(n!)-2n^2} > n^{n!}.
$$
(The last inequality uses $n! > 2n^2$ for $n\ge5$; indeed, even $n! > n(n-1)(n-2) > 2n^2$ for $n\ge5$.) Of course the desired inequality can be verified directly for $n=4$.
As you can see, there's tons of slack in this inequality and argument; mjqxxxx's answer gives an idea of how much.


TITLE: Abstract algebra, prove that $(a^m)^n$ =$ a^{mn}$
QUESTION [3 upvotes]: Let $a$ be an element of group $G$. For any integers $m,n \in \mathbb{Z}$ ($m,n$ can be positive and negative). Prove that $(a^{m})^{n}=a^{mn}$, then show that $(a^{-1})^{-1} = a$ by using what we proved so far.
So far we proved:

the Cancellation law: for $a,b,c$ are 3 elements in group $G$, then: $ab=ac$ implies $b=c$ and $ba=ca$ implies $b=c$.
$ab=ca$ implies that $b=c$ for $a,b,c \in G$.
If $ab=e$, then $b=a^{-1}$ for $a,b$ in group $G$, $e$ is an identity element.
We proved the sock-shoe property: If $G$ is a group and $a,b,c$ are elements, then
$(ab)^{-1}=b^{-1} a^{-1}$.
Next we showed that
$(a^{n})^{-1}=(a^{-1})^{n}$ for $(a^{n})^{-1}$ is the inverse of $a^{n}$ and $(a^{-1})^{n}$
is the multiplication of n copies of $a^{-1}$.

I don't know how to use all of then for the task I'm doing. What I am thinking is separate the task into 4 pieces:($m,n >0$), ($m>0, n<0$), ($m<0, n>0$), ($m,n<0$) and do it one by one.
And this is what I did for the first part ($m,n >0$):
$a^m= a.a.a.a.......a$ ($m$ times)
$(a^m)^n = [(a.a.a....a).(a.a.a...a)....(a.a.a...a)]$ $n$ times.
Open the (), we have $a.a.a.a....a.....a.....a$ for ($mn$ times)
so at the end, we have $(a^m)^n = a^{mn}$
Is my way to prove it good or not, any suggestion for the rest of these parts, please help me, thank you very much.

REPLY [3 votes]: Let's prove the claim first for natural $m$ and $n$ (that is, $m\ge0$ and $n\ge0$) by induction on $n$.
The base case, $n=0$, is easily settled: $(a^m)^0=1=a^{m0}$. So, assume it holds for $n$; then
\begin{align}
(a^m)^{n+1}&=(a^m)^n\cdot a^m &&\text{definition of powers}\\
&=a^{mn}\cdot a^m &&\text{induction hypothesis}\\
&=a^{mn+m} &&\text{property of powers}\\
&=a^{m(n+1)} &&\text{distributive property}
\end{align}
Here we assume that $a^{r+s}=a^r\cdot a^s$ has already been proved.
Now we turn to the case where $m\ge0$ and $n<0$; we'll use the fact that $b^n=(b^{-1})^{-n}$ for any integer $n$ (which can be proved easily); so
\begin{align}
(a^m)^n&=((a^m)^{-1})^{-n}\\
&=((a^{-1})^{m})^{-n}\\
&=(a^{-1})^{-mn}\\
&=a^{-(-mn)}\\
&=a^{mn}
\end{align}
The cases $m<0$, $n\ge0$ and $m<0$, $n<0$ can be proved similarly.


TITLE: Is there an equivalent to the DLP with extension fields?
QUESTION [0 upvotes]: For instance, if I have an extension field of $p^n$, is there a way to recover $p$, other than brute force checking?

REPLY [1 votes]: Of course. There is also a way to recover $p$ in fields of prime order other than brute force. All of these algorithms are faster than brute force. Yet, they are still not practical for real-world attacks.
So, if you meant for real world attacks, then then answer is yes, if the field has small characteristic. See for example this paper and this list of papers.
For large characteristic fields, the answer is no (otherwise elliptic curve cryptography would be dead).


TITLE: Linear operators on distributions with different topologies
QUESTION [3 upvotes]: Denote by $\mathscr{D}^\prime$ and $\mathscr{D}^\prime_b$ the space of distributions on $\mathbb{R}^n$ equipped with the weak and the strong topology, respectively. Because the topology of $\mathscr{D}^\prime_b$ has more open sets than the topology of $\mathscr{D}^\prime$, we have
$$L(\mathscr{D}^\prime, F) \subseteq L(\mathscr{D}^\prime_b, F)$$
for any topological vector space $F$. If $F = \mathbb{C}$, then we have equality (this holds for every locally convex space instead of $\mathscr{D}$). If $F = \mathscr{D}^\prime_b$, the left-hand-side is a proper subset because the identity is not continuous from $\mathscr{D}^\prime$ to $\mathscr{D}^\prime_b$ as the latter topology contains strictly more open sets (but the identity is continuous from each topological space to itself).
Question: What if $F = \mathscr{D}$, the space of test functions or $F = \mathscr{D}^\prime$, the space of distributions with the weak topology? Is this a strict inclusion or do we happen to have inequality?
If we don't have equality, what would be an example of a linear map that is contained in the latter space, but not in the first?
I didn't specify the domain yet. Is the answer different when we take the domain of functions in $\mathscr{D}$ to be a compact subset of $\mathbb{R}^n$ or a compact manifold instead of $\mathbb{R}^n$? What if we change the space $\mathscr{D}$ to the spaces $\mathscr{S}$ of Schwartz functions or $\mathscr{E}$ of all smooth functions, with there usual (Fréchet) topologies?

REPLY [4 votes]: The answer to all of your questions is no.  This follows from the simple fact that if a linear mapping from a locally convex space $E$ into a normed space $F$ is continuous for the weak topology on $E$ and the norm on $F$, then its range is finite dimensional.


TITLE: How to do junior math Olympiad question
QUESTION [1 upvotes]: Let $a$ and $b$ be such that $0<a<b$. Suppose that $a^3 = 3a-1$ and $b^3 =  3b-1$.   Find $b^2 -a$.

REPLY [1 votes]: Substituting $a=2a'$ and $b=2b'$ and dividing the equations by $2$  we have $$4a'^3-3a'=4b'^3-3b'=-1/2.$$ Since $4\cos^3x-3\cos x=\cos 3x$ for all $x$, we see that  $\{\cos 40^o,\cos 160^o,\cos 280^o\}$ is the set of  solutions to $4y^3-3y=\cos 120^o=-1/2 .$
So $a=2\cos 280^o=2\cos 80^o$ and $b=2\cos 40^o.$ The rest is in the  last part of the A by Laars Helenius.


TITLE: Need help showing my example does (or doesn't) work.
QUESTION [1 upvotes]: I am trying to find an example of when $(A \bigcup B)^\circ \supset A^\circ \bigcup B^\circ$. Where $^\circ$ denotes the interior of a set. It has been previously shown in the text that:

The interior of the set $\mathbb{Q}$ of rational numbers is empty.

My intuition is telling me that $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$. Since, $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ = \mathbb{R} = (-\infty, \infty)$, but I don't exactly know what the size of $(\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$ is, all I can think of reducing it down to:
$$(\mathbb{R} - \mathbb{Q})^\circ \bigcup \emptyset$$
$$(\mathbb{R} - \mathbb{Q})^\circ$$
Is there a way I can show $\mathbb{R}^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ$? Is it even true?

REPLY [0 votes]: If you're familiar with the Cantor Set, another example that would work would be:

Let $A$ be the Cantor set, and $B = [0, 1] \setminus A$, so that $B$ is the union of the intervals removed from $[0, 1]$ in the construction of $A$.  Then $B$ is open, so $B^\circ = B$.  But $A$ has empty interior, so that $A^\circ \cup B^\circ = B$.  However, $(A\cup B)^\circ = (0, 1) \supset B$.

Letting $A$ be any closed nowhere dense set and $B$ its complement would work similarly.


text:
\begin{document}

\bibliographystyle{amsplain}

\date{}

\author{P. Sahandi, N. Shirmohammadi and S. Sohrabi}

\address{Department of Mathematics, University of Tabriz,
Tabriz, Iran}
\email{sahandi@tabrizu.ac.ir}

\address{Department of Mathematics, University of Tabriz,
Tabriz, Iran} \email{shirmohammadi@tabrizu.ac.ir}

\address{Department of Mathematics, University of Tabriz,
Tabriz, Iran} \email{sohrabil6@yahoo.com}

\keywords{Amalgamated algebra, Cohen-Macaulay ring, Gorenstein ring,
quasi-Gorenstein ring, Serre condition, universally catenary ring}

\subjclass[2010]{Primary 13A15, 13H10, 13C15}

\title[Amalgamated algebra]{Cohen-Macaulay and Gorenstein properties under the amalgamated construction}

\begin{abstract}
Let $A$ and $B$ be commutative rings with unity, $f:A\to B$ a ring
homomorphism and $J$ an ideal of $B$. Then the subring
$A\bowtie^fJ:=\{(a,f(a)+j)|a\in A$ and $j\in J\}$ of $A\times B$ is
called the amalgamation of $A$ with $B$ along with $J$ with respect
to $f$. In this paper, among other things, we investigate the
Cohen-Macaulay and (quasi-)Gorenstein properties on the ring
$A\bowtie^fJ$.
\end{abstract}

\maketitle

\section{Introduction}

In \cite{DFF} and \cite{DFF2}, D'Anna, Finocchiaro, and Fontana have
introduced the following new ring construction. Let $A$ and $B$ be
commutative rings with unity, let $J$ be an ideal of $B$ and let
$f:A\to B$ be a ring homomorphism. They introduced the following
subring
$$A\bowtie^fJ:=\{(a,f(a)+j)|a\in A\text{ and }j\in J\}$$ of $A\times B$,
called the \emph{amalgamation of $A$ with $B$ along $J$ with respect
to $f$}. This construction generalizes the amalgamated duplication
of a ring along an ideal (introduced and studied in \cite{D},
\cite{DF}). Moreover, several classical constructions such as the
Nagata's idealization (cf. \cite[page 2]{Na}, \cite[Chapter VI,
Section 25]{Hu}), the $A + XB[X]$ and the $A+XB[[X]]$ constructions
can be studied as particular cases of this new construction (see
\cite[Examples 2.5 and 2.6]{DFF}).

Let $M$ be an $A$-module. In 1955, Nagata introduced a ring
extension of $A$ called the \emph{trivial extension} of $A$ by $M$
(or the \emph{idealization} of $M$ in $A$), denoted here by
$A\ltimes M$. Now, assume that $A$ is Noetherian local and that $M$
is finitely generated. It is well known that the trivial extension
$A\ltimes M$ is Cohen-Macaulay if and only if $A$ is Cohen-Macaulay
and the module $M$ is maximal Cohen-Macaulay. Furthermore, it is
proved by Reiten \cite{Re} and Foxby \cite{Fo} that $A\ltimes M$ is
Gorenstein if and only if $A$ is Cohen-Macaulay and $M$ is a
canonical module of $A$. Next, in \cite{A1}, Aoyama obtained a
generalization of this result to quasi-Gorenstein rings. Indeed, he
showed that $A\ltimes M$ is a quasi-Gorenstein ring if and only if
the completion $\widehat{A}$ satisfies Serre's condition $(S_2)$ and
$M$ is a canonical module of $A$.

Let $A$ be a Noetherian local ring and $I$ be an ideal of $A$.
Consider the amalgamated duplication $A\bowtie I:=\{(a,a+i)|a\in
A\text{ and }i\in I\}$ as in \cite{D}, \cite{DF}. D'Anna in \cite{D} (see also \cite{Sh})
proved that if $A$ is a Cohen-Macaulay local ring and $\Ann_A(I)=0$,
then the amalgamated duplication $A\bowtie I$ is Gorenstein if and
only if $A$ has a canonical ideal $I$. Next, in \cite{BSTY}, the
authors generalized this result as follows: if $\Ann_A(I)=0$, then
$A\bowtie I$ is a quasi-Gorenstein ring if and only if the
$\widehat{A}$ satisfies Serre's condition $(S_2)$ and $I$ is a
canonical ideal of $A$. Finally, in \cite{SSh}, the authors determined when the amalgamated duplication $A\bowtie I$ is normal and satisfies Serre's conditions $(S_n)$ and $(R_n)$.

In \cite[Remark 5.1]{DFF1}, assuming $A$ is a Cohen-Macaulay local
ring, $J$ is finitely generated as an $A$-module, and $J$ is contained in the  Jacobson radical of $B$, it is observed
that $A\bowtie^f J$ is a Cohen-Macaulay ring if and only if it is a
Cohen-Macaulay $A$-module if and only if $J$ is a maximal
Cohen-Macaulay module. Moreover, it is shown in \cite[Remark
5.4]{DFF1} that if $A$ is a Cohen-Macaulay local ring, having a
canonical module isomorphic (as an $A$-module) to $J$, then
$A\bowtie^f J$ is Gorenstein. Also, it is observed, under the
assumption $\Ann_{f(A)+J} (J) = 0$, that if $A$ is a Cohen-Macaulay
local ring and $A\bowtie^f J$ is Gorenstein, then $A$ has a
canonical module isomorphic to $f^{-1}(J)$ \cite[Proposition
5.5]{DFF1}.

The above results lead us to investigate further when the
amalgamated algebra $A\bowtie^fJ$ is Cohen-Macaulay or
(quasi-)Gorenstein.

More precisely, in Section 2, among other things, we prove that, for
a Noetherian local ring $(A,\fm)$, let $f:A\to B$ be a ring
homomorphism, and $J$ be an ideal of $B$, contained in the Jacobson
radical $B$, such that $f^{-1}(\fq)\neq\fm$, for each
$\fq\in\Spec(B)\backslash V(J)$ and that $\depth J<\infty$. Then
$A\bowtie^f J$ is Cohen-Macaulay if and only if $A$ is
Cohen-Macaulay and $J$ is a big Cohen-Macaulay module. In particular
if $J$ is finitely generated $A$-module (with the structure
naturally induced by $f$), then $A\bowtie^f J$ is Cohen-Macaulay if
and only if $A$ is Cohen-Macaulay and $J$ is maximal Cohen-Macaulay.
This improves \cite[Remark 5.1]{DFF1}, and \cite[Discussion 10]{D}.

In Section 3, among other things, we show that, for a local ring
$(A,\fm)$, $f:A\to B$ be a ring homomorphism, and $J$ be an ideal of
$B$ contained in the Jacobson radical $B$, such that $J$ is a
finitely generated $A$-module, if $\widehat{A}$ satisfies Serre's
condition $(S_2)$ and $J$ is a canonical module of $A$, then
$A\bowtie^fJ$ is quasi-Gorenstein.  Further, assume that $J^2=0$ and
that $\Ann_A(J)=0$. If $A\bowtie^fJ$ is quasi-Gorenstein, then $A$
satisfies $(S_2)$ and $J$ is a canonical module of $A$. This
generalizes \cite[Remark 5.4 and Proposition 5.5]{DFF1},
\cite[Theorem 11]{D} and \cite[Theorem 3.3]{BSTY}.

In Section 4, we describe the amalgamated algebra as a quotient of a
polynomial ring. As a consequence, we derive a characterization of
universally catenary property of the amalgamated algebra.

Now, in the following proposition, we collect some of the main
properties of the amalgamated algebra $A\bowtie^fJ$ needed in the
present paper. We use it in this paper without comments.

\begin{prop}\label{P} Let $A, B$ be commutative rings with
unity, $f:A\to B$ be a ring homomorphism, and $J$ be an ideal of
$B$.
\begin{itemize}
\item [(1)] (\cite[Proposition 5.1(1)]{DFF})
Let $\iota_A:A\to A\bowtie^fJ$ be the natural ring
homomorphism defined by $\iota_A(a)=(a,f(a))$,
for all $a\in A$. Then $\iota_A$ is an embedding,
making $A\bowtie^fJ$ a ring extension of $A$.
\item [(2)] (\cite[Proposition 5.7(a)]{DFF})
If $J$ is a finitely generated $A$-module
(with the structure naturally induced by $f$),
then $A\bowtie^fJ$ is Noetherian if and only
if $A$ is Noetherian.
\item [(3)] (\cite[Lemma 2.3(4)]{DFF}) Let
$\iota_J:J\to A\bowtie^fJ$, defined by $j\mapsto(0,j)$
and $P_A:A\bowtie^fJ\to A$, be the natural projection
of $A\bowtie^fJ\subseteq A\times B$ into $A$.
Then the following is a split exact sequence of $A$-modules:
    $$0\longrightarrow J\stackrel{\iota_J}{\longrightarrow}A\bowtie^fJ\stackrel{P_A}{\longrightarrow}A\longrightarrow0.$$
\item [(4)] (\cite[Corollaries 2.5 and 2.7]{DFF1})
For $\fp\in\Spec(A)$ and $\fq\in\Spec(B)\backslash V(J)$, set
\begin{align*}
\fp^{\prime_f}:= & \fp\bowtie^fJ:=\{(p,f(p)+j)|p\in \fp, j\in J\}, \\[1ex]
\overline{\fq}^f:= & \{(a,f(a)+j)|a\in A, j\in J, f(a)+j\in \fq\}.
\end{align*}
Then, the following statements hold.
\begin{itemize}
\item [(i)] The prime ideals of $A\bowtie^fJ$
are of the type $\overline{\fq}^f$ and $\fp^{\prime_f}$, for $\fq$
varying in $\Spec(B)\backslash V(J)$ and $\fp$ in $\Spec(A)$.
\item [(ii)] $\Max(A\bowtie^fJ)=\{\fp^{\prime_f}|\fp\in\Max(A)\}\cup\{\overline{\fq}^f|\fq\in\Max(B)\backslash V(J)\}.$
\item [(iii)] $A\bowtie^fJ$ is a local ring if
and only if $A$ is local and $J\subseteq\Jac(B)$.
\end{itemize}
\item [(5)] (\cite[Proposition 2.9]{DFF1})
The following statements hold.
\begin{itemize}
\item [(i)] For any prime ideal $\fq\in\Spec(B)\backslash V(J)$,
the localization $(A\bowtie^fJ)_{\overline{\fq}^f}$ is canonically
isomorphic to $B_{\fq}$.
\item [(ii)] For any prime ideal
$\fp\in\Spec(A)\backslash V(f^{-1}(J))$, the localization
$(A\bowtie^fJ)_{\fp^{\prime_f}}$ is canonically isomorphic to
$A_{\fp}$.
\item [(iii)] Let $\fp$ be a prime ideal of $A$
containing $f^{-1}(J)$. Consider the multiplicative subset
$S_{\fp}:=f(A\backslash \fp)+J$ of $B$ and set $B_{S_{\fp}}:=S_{\fp}^{-1}B$ and
$J_{S_{\fp}}:=S_{\fp}^{-1}J$. If $f_{\fp}:A_{\fp}\to B_{S_{\fp}}$ is the ring homomorphism
induced by $f$, then the ring $(A\bowtie^fJ)_{\fp^{\prime_f}}$ is
canonically isomorphic to $A_{\fp}\bowtie^{f_{\fp}}J_{S_{\fp}}$.
\end{itemize}
\end{itemize}
\end{prop}

\section{Cohen-Macaulay property under amalgamated construction}

Let us fix some notation which we shall use frequently throughout
this section: $A, B$ are two commutative rings with unity, $A$ is
Noetherian, $f:A\to B$ is a ring homomorphism, and $J$ denotes an
ideal of $B$, such that $A\bowtie^f J$ is Noetherian.

In \cite[Remark 5.1]{DFF1}, assuming $A$ is a Cohen-Macaulay local
ring, $J$ is finitely generated as an $A$-module, and $J$ is contained in the  Jacobson radical of $B$, it is observed
that $A\bowtie^f J$ is a Cohen-Macaulay ring if and only if it is a
Cohen-Macaulay $A$-module if and only if $J$ is a maximal
Cohen-Macaulay module. Also, in \cite[Remark 5.2]{DFF1}, the authors
mentioned that, if $J$ is not finitely generated as $A$-module, it
is more problematic to find conditions implying $A\bowtie^f J$ is
Cohen-Macaulay. Our main result in this section improves their
observation as well as, in case that $J$ is not finitely generated, it provides conditions implying $A\bowtie^f J$ is Cohen-Macaulay.

To state the main result we need to introduce some terminology. Let
$(A,\fm)$ be a Noetherian local ring, and $N$ an $A$-module (not
necessarily finitely generated). The \emph{depth} of $N$ over $A$ is
defined by the non-vanishing of the local cohomology modules
$H^i_{\fm}(N)$, with respect to $\fm$, in the way that
$$\depth N:=\depth_AN:=\inf\{i|H^i_{\fm}(N)\neq0\},$$
see \cite[Section 9.1]{BH} and \cite{FI}.
When $N$ is finitely generated, $\depth N$ coincides with
the usual depth defined by the common length of the maximal
$N$-regular sequences in $\fm$ by \cite[Proposition 3.5.4(b)]{BH}.
By \cite[Exercise 9.1.12]{BH}, if $\fm N\neq N$, then $\depth N$ is
finite, and when $\depth N$ is finite, then $\depth N\leq\dim A$.

It is well known that, over a local ring $(A,\fm)$, a finitely
generated $A$-module $M$ is Cohen-Macaulay if and only if
$H^i_{\fm}(M)=0$ for all $i<\dim M$ (see \cite[Corollary
6.2.8]{BS}).

Recall that a finitely generated module $M$ over a Noetherian local
ring $(A,\fm)$ is called a \emph{maximal Cohen-Macaulay $A$-module} if
$\depth M=\dim A$. An $A$-module $N$ is said to be \emph{big
Cohen-Macaulay} if $\depth N = \dim A$. Such modules were
constructed by Hochster \cite{Ho}, when $A$ contains a field. The
reader is referred to \cite[Chapter 8]{BH} for details on the
existence and properties of big Cohen-Macaulay modules.

The following auxiliary lemma which is interesting in itself is of
fundamental importance in our study of Cohen-Macaulayness.

In the sequel, $\Jac(B)$ will denote the Jacobson radical of $B$.

\begin{lem}\label{dim}
Let $(A,\fm)$ be a local ring, and $J\subseteq\Jac(B)$ be an ideal
of $B$ such that $f^{-1}(\fq)\neq\fm$, for each
$\fq\in\Spec(B)\backslash V(J)$. Then $\dim A\bowtie^f J=\dim A$,
and $\depth A\bowtie^f J=\min\{\depth A,\depth J\}$.
\end{lem}
\begin{proof}
By \cite[Corollary 3.2]{DFF1}, we have
$\fm^{\prime_f}=\sqrt{\fm(A\bowtie^f J)}$. Using this, the
Independence Theorem of local cohomology \cite[Theorem 4.2.1]{BS}
yields the isomorphism $H^i_{\fm^{\prime_f}}(A\bowtie^f J)\cong
H^i_{\fm}(A\bowtie^f J)$ for each $i$. On the other hand, applying
the functor of local cohomology to the split exact sequence of
$A$-modules appeared in Proposition \ref{P}(3), one obtains the
isomorphism $H^i_{\fm}(A\oplus J) \cong H^i_{\fm}(A)\oplus
H^i_{\fm}(J)$ for each $i$. Thus, for each $i$, we have shown the
following isomorphism
$$H^i_{\fm^{\prime_f}}(A\bowtie^f J)\cong
H^i_{\fm}(A)\oplus H^i_{\fm}(J)$$ of $A$-modules. Using above
isomorphism together with Grothendieck's Vanishing Theorem
\cite[Theorem 6.1.2]{BS}, one concludes the first equality.\\
In order to prove the second equality, again, we use above
isomorphism. First, note that $H^i_{\fm}(A)=H^i_{\fm}(J)=0$ for all
$i<\depth A\bowtie^f J$. Hence $\depth A\bowtie^f J\leq\depth A$ and
$\depth A\bowtie^f J\leq\depth J$. Now set $n:=\depth A\bowtie^f J$
and consider
$$H^{n}_{\fm}(A)\oplus H^{n}_{\fm}(J)\cong H^{n}_{\fm^{\prime_f}}(A\bowtie^f J)\neq0$$
to obtain $H^{n}_{\fm}(A)\neq0$ or $H^{n}_{\fm}(J)\neq0$. In the
first case we deduce that $\depth A=\depth A\bowtie^f J\leq\depth
J$. Similarly, in the second case we have $\depth J=\depth
A\bowtie^f J\leq\depth A$. Therefore the second equality is
obtained.
\end{proof}

\begin{rem}
Let $A$ be a (not necessarily Noetherian) ring, and $J$ be an ideal
of $B$. It is observed, in
\cite[Propositions 4.1 and 4.2]{DFF}, that if $f$ is surjective, or $f$ is integral, then $\dim A\bowtie^f J=\dim A$.
\end{rem}

The next theorem is the main result of this section. With it, we not
only offer an application of the above lemma, but we also provide
more information about the Cohen-Macaulayness under the amalgamation than was given in \cite[Remark 5.1]{DFF1}.

\begin{thm}\label{cm}
With the assumptions of Lemma \ref{dim}, the following statements hold.
\begin{enumerate}
       \item If $A\bowtie^f J$ is Cohen-Macaulay, then so does $A$.
       \item Further assume that $\depth J<\infty$.
       Then $A\bowtie^f J$ is Cohen-Macaulay if and
only if $A$ is Cohen-Macaulay and $J$ is a big Cohen-Macaulay module.
\end{enumerate}
\end{thm}
\begin{proof}
(1) Using Lemma \ref{dim}, we have $\dim A=\dim A\bowtie^f J=\depth
A\bowtie^f J\leq\depth A\leq\dim A$. Hence $A$ is Cohen-Macaulay.\\
(2) Assume that
$A\bowtie^fJ$ is Cohen-Macaulay. Then, by (1), $A$ is
Cohen-Macaulay. To prove the big Cohen-Macaulayness of $J$, note
that we have $\dim A=\dim A\bowtie^f J=\depth A\bowtie^f J\leq\depth
J\leq\dim A$, where the last inequality comes from \cite[Exercise
9.1.12]{BH}. Thus $\depth J=\dim A$, that is, $J$ is big
Cohen-Macaulay. Conversely, assume that $A$ is Cohen-Macaulay and
that $J$ is a big Cohen-Macaulay module. It follows that
$H^i_{\fm}(A)=H^i_{\fm}(J)=0$ for all $i<\dim A$. Then, by the above
isomorphism, one has $H^i_{\fm^{\prime_f}}(A\bowtie^fJ)=0$ for all
$i<\dim A=\dim A\bowtie^f J$, and hence $A\bowtie^fJ$ is
Cohen-Macaulay.
\end{proof}

The next example shows that, if, in the above theorem, the
hypothesis $f^{-1}(\fq)\neq\fm$, for each $\fq\in\Spec(B)\backslash
V(J)$, is dropped, then the corresponding statement is no longer
always true.

\begin{exam}
Let $k$ be a field and $X,Y$ are algebraically independent
indeterminates over $k$. Set $A:=k[[X]]$, $B:=k[[X,Y]]$ and let
$J:=(X,Y)$. Let $f: A\to B$ be the inclusion. Note that $A$ is
Cohen-Macaulay, and it is not hard to see that $J$ is a big
Cohen-Macaulay $A$-module. However, $A\bowtie^f J$ which is
isomorphic to $k[[X, Y,Z]]/(Y,Z)\cap(X-Y )$ is not Cohen-Macaulay.
It should be noted that, for $\fq:=(X)\in\Spec(B)\backslash V(J)$,
one has $f^{-1}(\fq)=(X)$ which is the maximal ideal of $A$.
\end{exam}

It is clear that a finitely generated big Cohen-Macaulay module is
maximal Cohen-Macaulay. Then we have the following corollary.

\begin{cor}\label{cm1}
Let $(A,\fm)$ be a local ring, and $J\subseteq\Jac(B)$ be an ideal
of $B$ such that $J$ is a finitely generated $A$-module. Then
$A\bowtie^f J$ is Cohen-Macaulay if and only if $A$ is
Cohen-Macaulay and $J$ is a maximal Cohen-Macaulay $A$-module.
\end{cor}
\begin{proof}
By \cite[Remark 3.3]{DFF1}, we see that if $J$ is finitely generated
$A$-module, then $f^{-1}(\fq)\neq\fm$, for each
$\fq\in\Spec(B)\backslash V(J)$. So the assertion holds by Theorem
\ref{cm}.
\end{proof}

Let $M$ be a finitely generated $A$-module. In 1955, Nagata
introduced a ring extension of $A$ called the \emph{trivial
extension} of $A$ by $M$ (or the \emph{idealization} of $M$ in $A$),
denoted here by $A\ltimes M$ (\cite[page 2]{Na}, \cite[Chapter VI,
Section 25]{Hu}). It should be noted that the module $M$ becomes an
ideal in $A\ltimes M$ and $M^2=0$. As in \cite[Example 2.8]{DFF}, if
$B:=A\ltimes M$, $J:=0\ltimes M$, and $f:A\to B$ be the natural embedding,
then $A\bowtie^f J\cong A\ltimes M$. Therefore the following result
follows from Corollary \ref{cm1}.

\begin{cor}\label{tc}
Let $(A,\fm)$ be a local ring, and $M$ be a finitely generated
$A$-module. Then the trivial extension $A\ltimes M$ is
Cohen-Macaulay if and only if $A$ is Cohen-Macaulay and $M$ is a
maximal Cohen-Macaulay $A$-module.
\end{cor}

\begin{cor}\label{cm2}
Assume that $f^{-1}(\fq)\neq\fm$, for each $\fq\in\Spec(B)\backslash
V(J)$ and each $\fm\in\Max(A)$. If $A\bowtie^f J$ is Cohen-Macaulay,
then so does $A$.
\end{cor}
\begin{proof}
Let $\fp\in\Max(A)$. If $\fp\notin V(f^{-1}(J))$, then
$A_{\fp}\cong(A\bowtie^f J)_{\fp^{\prime_f}}$ is Cohen-Macaulay. Now suppose $\fp\in
V(f^{-1}(J))$. Then
$A_{\fp}\bowtie^{f_{\fp}}J_{S_{\fp}}\cong(A\bowtie^f
J)_{\fp^{\prime_f}}$ is Cohen-Macaulay,
where $S_{\fp}:=f(A\backslash \fp)+J$. Note that, by \cite[Remark
2.4]{F}, one has $f_{\fp}^{-1}(\fq
B_{S_{\fp}})=f^{-1}(\fq)A_{\fp}\neq \fp A_{\fp}$, for each $\fq
B_{S_{\fp}}\in\Spec(B_{S_{\fp}})\backslash V(J_{S_{\fp}})$. Hence,
the Cohen-Macaulayness of $A_{\fp}$ results from Theorem \ref{cm}.
Therefore $A$ is Cohen-Macaulay.
\end{proof}

A finitely generated module $M$ over a Noetherian ring $A$ satisfies
Serre's condition $(S_n)$ if $\depth M_{\fp}\geq\min\{n,\dim
M_{\fp}\}$, for all $\fp\in\Spec(A)$. Note that if $M$ is
Cohen-Macaulay, then it satisfies Serre's condition $(S_n)$ for any
integer $n$. Also, when $\dim M=d$ and $M$ satisfies Serre's
condition $(S_d)$, then $M$ is Cohen-Macaulay. In the following
results we investigate the property $(S_n)$ for the amalgamated
algebra.

\begin{lem}\label{2}
Let $\fp\in\Spec(A)$, and set $S_{\fp}:=f(A\backslash \fp)+J$. If
$J^2=0$, then $J_{\fp}\cong J_{S_{\fp}}$ as $A_{\fp}$-modules.
\end{lem}
\begin{proof}
It is straightforward that the mapping $J_{\fp}\to J_{S_{\fp}}$
defined by $x/t\mapsto x/f(t)$, for all $x\in J$ and $t\in
A\backslash \fp$, is an $A_{\fp}$-isomorphism.
\end{proof}

Recall that an ideal is called a \emph{nil ideal} if each of its elements is nilpotent.

\begin{cor}\label{sur}
The following statements hold.
\begin{enumerate}
\item  Assume that for each $\fp\in V(f^{-1}(J))$
and each $\fq\in\Spec(B)\backslash V(J)$, $f^{-1}(\fq)\neq \fp$
(e.g. if $f$ is surjective or $J$ is a nil ideal of $B$). If $A\bowtie^f
J$ satisfies $(S_n)$, then so does $A$.
\item  Assume that $J^2=0$ and that $J$ is a finitely
generated $A$-module. If $A\bowtie^f J$ satisfies $(S_n)$, then so
does $J$.
\end{enumerate}
\end{cor}
\begin{proof}
(1) Let $\fp\in\Spec(A)$. If $\fp\notin V(f^{-1}(J))$, then, by
assumption, the inequality $\depth A_{\fp}\geq\min\{n,\dim
A_{\fp}\}$ holds for $A_{\fp}\cong(A\bowtie^f J)_{\fp^{\prime_f}}$.
Now suppose $\fp\in V(f^{-1}(J))$. This implies the isomorphism
$A_{\fp}\bowtie^{f_{\fp}}J_{S{\fp}}\cong(A\bowtie^f
J)_{\fp^{\prime_f}}$, where $S_{\fp}=f(A\backslash \fp)+J$.
Meanwhile, by \cite[Remark 2.4]{F}, we have the equality
$f_{\fp}^{-1}(\fq B_{S_{\fp}})=f^{-1}(\fq)A_{\fp}$, for each $\fq
B_{S_{\fp}}\in\Spec(B_{S_{\fp}})\backslash V(J_{S_{\fp}})$. Thus,
the assumptions of Lemma \ref{dim} hold. Therefore we obtain
\begin{align*}
\depth A_{\fp}\geq & \depth A_{\fp}\bowtie^{f_{\fp}}J_{S{\fp}} \\[1ex]
  = & \depth(A\bowtie^f J)_{\fp^{\prime_f}} \\[1ex]
  \geq & \min\{n, \dim (A\bowtie^f J)_{\fp^{\prime_f}}\} \\[1ex]
  = & \min\{n, \dim A_{\fp}\bowtie^{f_{\fp}}J_{S{\fp}}\} \\[1ex]
  = & \min\{n, \dim A_{\fp}\}.
\end{align*} \\
(2) Let $\fp\in\Spec(A)$. Then, by Lemma \ref{2}, we have the
isomorphism $J_{\fp}\cong J_{S_{\fp}}$ of $A_{\fp}$-modules. If
$\fp\notin V(f^{-1}(J))$, there is nothing to prove since
$J_{\fp}\cong J_{S_{\fp}}=0$ by \cite[Remark 2.4]{F}. So we can
assume that $\fp\in V(f^{-1}(J))$. Thus, using Lemma \ref{dim}, one
gets
\begin{align*}
\depth J_{\fp}\geq & \depth A_{\fp}\bowtie^{f_{\fp}}J_{\fp} \\[1ex]
  = & \depth(A\bowtie^f J)_{\fp^{\prime_f}} \\[1ex]
  \geq & \min\{n, \dim (A\bowtie^f J)_{\fp^{\prime_f}}\} \\[1ex]
  = & \min\{n, \dim A_{\fp}\bowtie^{f_{\fp}}J_{\fp}\} \\[1ex]
  = & \min\{n, \dim A_{\fp}\} \\[1ex]
  \geq & \min\{n, \dim J_{\fp}\}.
\end{align*}
Therefore $J$ satisfies $(S_n)$.
\end{proof}

One can employ Corollary \ref{sur} to deduce that the property
$(S_n)$ is retained under the trivial extension construction.

\begin{cor}
Let $(A,\fm)$ be a local ring, and $M$ be a finitely generated
$A$-module. Then the trivial extension $A\ltimes M$ satisfies
$(S_n)$ if and only if $A$ and $M$ satisfy $(S_n)$.
\end{cor}

\begin{prop}\label{sn}
If $A, B$ satisfy $(S_n)$, and $J_{S_{\fp}}$ is a maximal
Cohen-Macaulay $A_{\fp}$-module for each prime ideal $\fp$ of $A$,
where $S_{\fp}=f(A\backslash \fp)+J$, then $A\bowtie^f J$ satisfies
$(S_n)$.
\end{prop}
\begin{proof}
Each prime ideal of $A\bowtie^f J$ are of the type
$\overline{\fq}^f$ or $\fp^{\prime_f}$, for $\fq$ varying in
$\Spec(B)\backslash V(J)$ and $\fp$ in $\Spec(A)$. The localization
$(A\bowtie^f J)_{\overline{\fq}^f}$ is canonically isomorphic to
$B_{\fq}$. Thus one gets
$$
\depth (A\bowtie^f J)_{\overline{\fq}^f}=\depth
B_{\fq}\geq\min\{n,\dim B_{\fq}\}=\min\{n,\dim (A\bowtie^f
J)_{\overline{\fq}^f}\}.
$$
Next, for $\fp\in\Spec(A)\backslash V(f^{-1}(J))$, we have
$(A\bowtie^f J)_{\fp^{\prime_f}}\cong A_{\fp}$, hence, similarly,
the inequality $\depth (A\bowtie^f
J)_{\fp^{\prime_f}}\geq\min\{n,\dim (A\bowtie^f
J)_{\fp^{\prime_f}}\}$ holds. Now suppose $\fp\in V(f^{-1}(J))$.
This implies the isomorphism $(A\bowtie^f J)_{\fp^{\prime_f}}\cong
A_{\fp}\bowtie^{f_{\fp}}J_{S_{\fp}}$, where $S_{\fp}=f(A\backslash
\fp)+J$. Thus, by assumption, one deduces
\begin{align*}
\depth(A\bowtie^f J)_{\fp^{\prime_f}} = & \min\{\depth A_{\fp},\depth J_{S_{\fp}}\} \\[1ex]
  \geq & \min\{n, \dim A_{\fp}, \depth J_{S_{\fp}}\} \\[1ex]
  = & \min\{n, \dim A_{\fp}\} \\[1ex]
  = & \min\{n, \dim(A\bowtie^f J)_{\fp^{\prime_f}}\}.
\end{align*}
Therefore $A\bowtie^f J$ satisfies $(S_n)$.
\end{proof}

In concluding this section, we want to generalize the above theorem
to generalized Cohen-Macaulay rings. To this end, we need an
auxiliary lemma.

Let us recall that a finitely generated module $M$ over a Noetherian
local ring $(A,\fm)$ is said to be a \emph{generalized
Cohen-Macaulay} $A$-module if $H^i_\fm(M)$ is of finite length for
all $i<\dim M$. A local ring is called generalized Cohen-Macaulay if
it is a generalized Cohen-Macaulay module over itself. It is clear
that every Cohen-Macaulay module is a generalized Cohen-Macaulay
module.

In the course of next lemma and its proof, for a finite length
module $M$ over a ring $R$, we use $\ell_R(M)$ to denote the length
of $M$ over $R$.

\begin{lem}
Let $\varphi:(R,\fm)\to(S,\fn)$ be a local homomorphism of local
rings, such that the natural induced homomorphism
$\overline{\varphi}:R/\fm\to S/\fn$ is an isomorphism. Let $M$ be a
$S$-module. If $\ell_S(M)<\infty$, then $\ell_R(M)<\infty$ and
$\ell_S(M)=\ell_R(M)$. Here $M$ is considered as an $R$-module via
$\varphi$.
\end{lem}
\begin{proof}
Set $n:=\ell_S(M)$ and suppose that $0=M_0\subset
M_1\subset\cdots\subset M_n=M$ is a composition series of the
$S$-module $M$. Then $M_i/M_{i-1}\cong S/\fn$ for all
$i=1,\ldots,n$. On the other hand, it is clear that
$\overline{\varphi}$ is $R$-module homomorphism. This means
$S/\fn\cong R/\fm$ as $R$-modules. Thus, $M_i/M_{i-1}\cong R/\fm$
for all $i=1,\ldots,n$. Therefore $0=M_0\subset
M_1\subset\cdots\subset M_n=M$ is a composition series of $M$ as
$R$-module. This yields $\ell_R(M)<\infty$ and $\ell_R(M)=n$.
\end{proof}

\begin{thm}
Let $(A,\fm)$ be a local ring, and $J\subseteq\Jac(B)$ be an ideal
of $B$ such that $J$ is a finitely generated $A$-module. Then
$A\bowtie^fJ$ is a generalized Cohen-Macaulay ring if and only if
$A$ and $J$ are generalized Cohen-Macaulay and $\dim J\in\{0,\dim
A\}$.
\end{thm}
\begin{proof}
Suppose that $A\bowtie^fJ$ is generalized Cohen-Macaulay. Then
$H^i_{\fm^{\prime_f}}(A\bowtie^fJ)$ is of finite length over
$A\bowtie^fJ$ for all $i<\dim(A\bowtie^fJ)$. In conjunction with the
previous lemma, this shows that $H^i_{\fm^{\prime_f}}(A\bowtie^fJ)$
has finite length over $A$ for $i<\dim A$. Notice that, for all $i$,
one has the isomorphism $H^i_{\fm^{\prime_f}}(A\bowtie^fJ)\cong
H^i_{\fm}(A)\oplus H^i_{\fm}(J)$ of $A$-modules. Hence
$H^i_{\fm}(A)$ and $H^i_{\fm}(J)$ have finite length over $A$ for
all $i<\dim A$. Therefore $A$ and $J$ are generalized Cohen-Macaulay
and $\dim J=\dim A$ or 0 by \cite[Corollary 7.3.3]{BS}. Conversely,
suppose that $A$ and $J$ are generalized Cohen-Macaulay and $\dim
J\in\{0,\dim A\}$; so that there exists a positive integer $t$ such
that $\fm^t H^i_{\fm}(A)=0=\fm^tH^i_{\fm}(J)$ for all $i<\dim A$.
Hence $\fm^tH^i_{\fm^{\prime_f}}(A\bowtie^fJ)=0$ for all
$i<\dim(A\bowtie^fJ)$. On the other hand, by \cite[Corollary 3.2 and
Remark 3.3]{DFF1}, we know $\fm^{\prime_f}=\sqrt{\fm(A\bowtie^fJ)}$;
so that there exists a positive integer $s$ such that
$({\fm^{\prime_f}})^s\subseteq\fm(A\bowtie^fJ)$. Consequently
$({\fm^{\prime_f}})^{st}H^i_{\fm^{\prime_f}}(A\bowtie^fJ)=0$ for all
$i<\dim(A\bowtie^fJ)$. Therefore $A\bowtie^fJ$  is generalized
Cohen-Macaulay.
\end{proof}

\begin{cor}\label{tg}
Let $(A,\fm)$ be a local ring, and $M$ be a finitely generated
$A$-module. Then the trivial extension $A\ltimes M$ is generalized
Cohen-Macaulay if and only if $A$ and $M$ are generalized
Cohen-Macaulay and $\dim M\in\{0,\dim A\}$.
\end{cor}

Let $(A,\fm)$ be a Cohen-Macaulay local ring. Then, using
Corollaries \ref{tg} and \ref{tc}, the trivial extension of $A$ by
$A/\fm$ is a generalized Cohen-Macaulay ring which is not
Cohen-Macaulay.

\section{Gorenstein property under amalgamated construction}

Let $A, B$ be commutative rings with unity, $A$ be Noetherian,
$f:A\to B$ be a ring homomorphism, and $J\subseteq\Jac(B)$ denotes
an ideal of $B$, which is finitely generated $A$-module. It is shown
in \cite[Remark 5.4]{DFF1} that if $A$ is a local Cohen-Macaulay
ring, having a canonical module isomorphic (as an $A$-module) to
$J$, then $A\bowtie^f J$ is Gorenstein. Also it is observed, under
the assumption $\Ann_{f(A)+J} (J) = 0$, that if $A$ is a local
Cohen-Macaulay ring and $A\bowtie^f J$ is Gorenstein, then $A$ has a
canonical module isomorphic to $f^{-1}(J)$ \cite[Proposition
5.5]{DFF1}.

In present section, under some assumptions, specified in Theorem \ref{gor}, we give a sufficient
condition and a necessary condition for the ring $A\bowtie^f J$ to
be quasi-Gorenstein (definition recalled below). Consequently we improve the above mentioned
results.

We now outline to recall some terminology. Let $(A,\fm)$ be local. An $A$-module $K$ is
called a canonical module of $A$ if
$$K\otimes_A\widehat{A}\cong\Hom_A(H^{\dim A}_{\fm}(A),E_A(A/\fm)),$$
where $\widehat{A}$ is the $\fm$-adic completion of $A$, and
$E_A(A/\fm)$ is the injective hull of $A/\fm$ over $A$. We say that
$A$ is a \emph{quasi-Gorenstein ring} if a canonical module of $A$
exists and it is a free $A$-module (of rank one). This is equivalent
to saying that $H^{\dim A}_{\fm}(A)\cong E_A(A/\fm)$. It is known
that $A$ is quasi-Gorenstein if and only if $\widehat{A}$ is
quasi-Gorenstein \cite[Page 88]{A1}.

The following lemma is of crucial importance in this section.

\begin{lem}\label{cong}
With the notation of Proposition \ref{P}, the following statements
hold.
\begin{enumerate}
   \item If $\Ann_{f(A)+J}(J)=0$, then there is
   an $A\bowtie^fJ$-isomorphism $$\Hom_{A\bowtie^fJ}(A,A\bowtie^fJ)\cong f^{-1}(J)\times0.$$
   \item If $J^2=0$, then there is an $A$-isomorphism
   $$\Hom_{A\bowtie^fJ}(A,A\bowtie^fJ)\cong \Ann_A(J)\oplus J.$$
\end{enumerate}
\end{lem}
\begin{proof}
It is clear that $A\bowtie^fJ/0\times J\cong A$ as
$A\bowtie^fJ$-modules \cite[Proposition 5.1(2)]{DFF}. Then we have
\begin{align*}
\Hom_{A\bowtie^fJ}(A,A\bowtie^fJ)\cong & \Hom_{A\bowtie^fJ}(A\bowtie^fJ/0\times J,A\bowtie^fJ) \\[1ex]
  \cong & \Ann_{A\bowtie^fJ}(0\times J) \\[1ex]
  = & \{(a,f(a)+j)\in A\bowtie^fJ|(f(a)+j)J=0\} \\[1ex]
  = & \{(a,f(a)+j)\in A\bowtie^fJ|f(a)+j\in \Ann_{f(A)+J}(J)\}=:X.
\end{align*}
Now assume that $\Ann_{f(A)+J}(J)=0$. Then $X=f^{-1}(J)\times0$.
This completes the proof of (1). To prove (2), assume that $J^2=0$.
It follows that in this case $X=\{(a,f(a)+j)\in
A\bowtie^fJ|f(a)J=0\}$. Hence, one can define the $A$-homomorphism
$\psi:X\to\Ann_A(J)\oplus J$ by $\psi((a,f(a)+j))=(a,j)$. It is now
easy to check that $\psi$ is an isomorphism. Therefore (2) is
obtained.
\end{proof}

We are now ready for the proof of the main result of this section.

\begin{thm}\label{gor}
With the notation of Proposition \ref{P}, assume that $A$ is
Noetherian local, and $J\subseteq\Jac(B)$, which is finitely
generated $A$-module. The following statements hold.
\begin{enumerate}
 \item If $\widehat{A}$ satisfies $(S_2)$
and $J$ is a canonical module of $A$, then $A\bowtie^fJ$
is quasi-Gorenstein.
\item Assume that $J^2=0$ and that $\Ann_A(J)=0$. If $A\bowtie^fJ$
is quasi-Gorenstein, then $A$ satisfies $(S_2)$
and $J$ is a canonical module of $A$.
 \item Assume that $\Ann_{f(A)+J}(J)=0$ and that $A\bowtie^fJ$
is quasi-Gorenstein. Then $f^{-1}(J)$ is a canonical module of $A$.
Furthermore, assume that $f$ is surjective, then $\widehat{A}$ satisfies
$(S_2)$.
\item Assume that $J$ is a
  flat $A$-module. If $A\bowtie^fJ$
is quasi-Gorenstein, then $A$ is quasi-Gorenstein.
\end{enumerate}
\end{thm}
\begin{proof}
(1) Assume that $\widehat{A}$ satisfies $(S_2)$ and $J$ is a
canonical module of $A$. Using \cite[Satz 5.12]{HK}, this shows that
$A\bowtie^fJ$ has a canonical module
$K_{A\bowtie^fJ}\cong\Hom_A(A\bowtie^fJ,J)$. On the other hand,
applying the functor $\Hom_A(-,J)$ to the split exact sequence of
$A$-modules considered in Proposition \ref{P}(3), one obtains the
isomorphism $K_{A\bowtie^fJ}\cong \Hom_A(J,J)\oplus J$. Since
$\widehat{A}$ satisfies $(S_2)$, using \cite[Proposition 2]{A}, we
have $\Hom_A(J,J)\cong A$. Thus $K_{A\bowtie^fJ}\cong A\oplus J\cong
A\bowtie^fJ$ as $A$-modules. Note that
$K_{A\bowtie^fJ}\cong\Hom_A(A\bowtie^fJ,J)$ is an
$A\bowtie^fJ$-module by the usual way. Hence $K_{A\bowtie^fJ}\cong
A\bowtie^fJ$ as $A\bowtie^fJ$-modules. This means that $A\bowtie^fJ$
is quasi-Gorenstein.

(2) Assume that $A\bowtie^fJ$ is quasi-Gorenstein. Then by
\cite[Lemma 2.1]{HSZ}, we have $A\bowtie^fJ$ satisfies $(S_2)$.
Therefore by Corollary \ref{sur}, $A$ satisfies $(S_2)$. Since
$A\bowtie^{f}J$ is quasi-Gorenstein, thus a canonical module of
$A\bowtie^{f}J$ exists and it is isomorphic to $A\bowtie^{f}J$. By
\cite[Satz 5.12]{HK}, $A$ has a canonical module
$K_A\cong\Hom_{A\bowtie^fJ}(A,A\bowtie^fJ)$. On the other hand, by
Lemma \ref{cong}(2), we have $\Hom_{A\bowtie^fJ}(A,A\bowtie^fJ)\cong
J$. Therefore $J$ is a canonical module of $A$.

(3) First we show that $f^{-1}(J)$ is a canonical module of $A$. We
may use the same argument as employed in the proof of (2). One might
take into consideration that in the proof we appeal to Lemma
\ref{cong}(1) instead of Lemma \ref{cong}(2) in the context of the
proof of (2). Next, assuming the surjectivity of $f$, we show that
$\widehat{A}$ satisfies $(S_2)$. Since $A\bowtie^fJ$ is
quasi-Gorenstein, hence $\widehat{A\bowtie^fJ}$ is also
quasi-Gorenstein. Since $f$ is surjective, $B$ is a finitely
generated $A$-module, and therefore there is an
$\widehat{A}$-isomorphism
$\widehat{A\bowtie^fJ}\cong\widehat{A}\bowtie^{\widehat{f}}\widehat{J}$,
where $\widehat{f}:\widehat{A}\to\widehat{B}$ be the induced natural
surjective ring homomorphism. So that
$\widehat{A}\bowtie^{\widehat{f}}\widehat{J}$ is quasi-Gorenstein.
Then by \cite[Lemma 2.1]{HSZ}, we have
$\widehat{A}\bowtie^{\widehat{f}}\widehat{J}$ satisfies $(S_2)$.
Therefore by Corollary \ref{sur}, $\widehat{A}$ satisfies $(S_2)$.

(4) Assume that $A\bowtie^fJ$ is quasi-Gorenstein. Consider the
embedding $\iota_A:A\to A\bowtie^fJ$. Since $J$ is a flat
$A$-module, $\iota_A$ is a flat local homomorphism. Using
\cite[Remark 3.3]{DFF1}, the extension ideal $\fm(A\bowtie^fJ)$ is
$\fm^{\prime_f}=\fm\bowtie^fJ$-primary ideal. This in conjunction
with \cite[Theorem 2.3]{A1} implies that $A$ is quasi-Gorenstein.
\end{proof}

It is well known that $A$ is Gorenstein if and only if it is
Cohen-Macaulay and quasi-Gorenstein \cite[Page 88]{A1}. Also, it is
clear that $A$ is Cohen-Macaulay if and only if $\widehat{A}$ is
Cohen-Macaulay. Therefore, one can use Theorem \ref{cm} together
with Theorem \ref{gor} to derive the following corollaries.

\begin{cor}
Keep the assumptions of Theorem \ref{gor}. Assume that $A$ is
Cohen-Macaulay and $J$ is a canonical module of $A$. Then
$A\bowtie^fJ$ is Gorenstein.
\end{cor}

\begin{cor}
Keep the assumptions of Theorem \ref{gor}. Assume that at least one
of the following conditions holds
\begin{enumerate}
  \item $f$ is an isomorphism and $\Ann_B(J)=0$; or
  \item $J^2=0$ and $\Ann_A(J)=0$.
\end{enumerate}
Then $A\bowtie^fJ$ is Gorenstein if and only if $A$ is
Cohen-Macaulay and $J$ is a canonical module of $A$.
\end{cor}

\begin{cor}
Keep the assumptions of Theorem \ref{gor} and assume that
$\Ann_{f(A)+J}(J)=0$. If $A\bowtie^fJ$ is Gorenstein, then $A$ is
Cohen-Macaulay and $f^{-1}(J)$ is a canonical ideal of $A$.
\end{cor}

We recall that, if $\alpha:A\to C$, $\beta:B\to C$ are ring homomorphisms, the subring $D:=\alpha\times_C\beta:=\{(a,b)\in A\times B|\alpha(a)=\beta(b)\}$ of $A\times B$ is called the \emph{pullback} (or \emph{fiber product}) of $\alpha$ and $\beta$. We refer the readers attention to \cite{Fon}, \cite{Facc} and
in the Noetherian setting, to \cite{O}, \cite{O1} for the properties and importance of pullback constructions.
Let $f:A\to B$ be a ring homomorphism and $J$ be an ideal of $B$. If $\pi:B\to B/J$ is the
canonical projection and $\check{f}:=\pi\circ f$ , then $A\bowtie^fJ\cong\breve{f}\times_{B/J}\pi$, see \cite[Proposition 4.2]{DFF}.

 As an application of \cite[Theorem 4]{O1}, with some assumptions on the ideal $f^{-1}(J)$ and on the ring $f(A) + J$, the authors in \cite{DFF1} obtained the following characterization of when $A\bowtie^fJ$ is Gorenstein.

\begin{prop} (\cite[Proposition 5.7]{DFF1})
With the notation of Proposition \ref{P}, assume that $A$ is
Noetherian local, and $J\subseteq\Jac(B)$, which is finitely
generated $A$-module and, moreover, assume that $A$ is a Cohen-Macaulay ring,
$f(A)+ J$ is $(S_1)$ and equidimensional, $J\neq 0$ and that $f^{-1}(J)$ is a regular ideal
of $A$. Then, the following conditions are equivalent.
\begin{enumerate}
  \item $A\bowtie^fJ$ is Gorenstein.
  \item $f(A) + J$ is a Cohen-Macaulay ring, $J$ is a canonical module of $f(A) + J$ and $f^{-1}(J)$
is a canonical module of $A$.
\end{enumerate}
\end{prop}

In concluding this section, as an application, we provide a method to
construct a new quasi-Gorenstein ring from a given one. Let $A$ be a
local ring and $X$ be an indeterminate over $A$. Set
$B:=A[X]/(X^2)$, $J:=XB$, and $f:A\to B$ be the natural embedding.
It is easy to see that $J$ is isomorphic to $A$ as an $A$-module.
Then, using Theorem \ref{gor} (1) and (4), one sees that $A$ is
quasi-Gorenstein if and only if $A\bowtie^fJ$ is quasi-Gorenstein.
Note that we have $A\bowtie^fJ\cong A[X]/(X^2)$.

\section{Universally catenary property under amalgamated construction}

Let us fix some notation which we shall use frequently throughout
this section: $A, B$ are two commutative rings with unity, $f:A\to
B$ is a ring homomorphism, and $J$ denotes an ideal of $B$.

In \cite[Example 3.11]{DF} and \cite[Example 2.6]{DFF1}, the authors
in some special examples illustrate the amalgamated algebras as a
quotient of some polynomial rings. In the following proposition, we
generally describe the amalgamated algebra $A\bowtie^f J$ as a
quotient of a polynomial ring.

\begin{prop}\label{s}
Let the notation just introduced and assume that $J$ as an
$A$-module is generated by $\{j_{\lambda}|\lambda\in\Lambda\}$. Then
$A\bowtie^f J$ is a homomorphic image of
$A[X_{\lambda}|\lambda\in\Lambda]$ the ring of polynomials over $A$
in indeterminates $X_{\lambda}$, for each $\lambda\in\Lambda$.
\end{prop}
\begin{proof}
Consider the ring homomorphism
$\varphi:A[X_{\lambda}|\lambda\in\Lambda]\to A\bowtie^f J$, defined
by $\varphi(a):=(a,f(a))$, for $a\in A$, and
$\varphi(X_{\lambda}):=(0,j_{\lambda})$, for $\lambda\in\Lambda$. It
is not difficult to see that $\varphi$ is surjective.
\end{proof}

\begin{cor}
Assume that $I$ as an ideal of $A$ is generated by
$\{i_{\lambda}|\lambda\in\Lambda\}$. Then the amalgamated
duplication $A\bowtie I:=\{(a,a+i)|a\in A, i\in I\}$ (\cite{D, DF})
is a homomorphic image of $A[X_{\lambda}|\lambda\in\Lambda]$ the
ring of polynomials over $A$ in indeterminates $X_{\lambda}$, for
each $\lambda\in\Lambda$.
\end{cor}

As a consequence of the previous proposition, we provide a partial
characterization of the Noetherianity of $A\bowtie^f J$. It should
be noted that the authors in \cite[Proposition 5.7]{DFF} have
already given a characterization of the Noetherianity of $A\bowtie^f
J$. But, this is an obvious consequence of the above proposition
provided that $J$ is finitely generated as an $A$-module.

\begin{cor}
Assume that $J$ is finitely generated as an
$A$-module.
Then $A\bowtie^f J$ is Noetherian if and only if $A$ is Noetherian.
\end{cor}

The next corollary investigates the behaviour of universally
catenary property under amalgamated formation provided that $J$ is
finitely generated as an $A$-module. One says a locally finite
dimensional ring $A$ is \emph{catenary} if every saturated chain
joining prime ideals $\fp$ and $\fq$, $\fp\subseteq\fq$, has
(maximal) length height $\fq/\fp$; $A$ is \emph{universally
catenary} if all the polynomial rings $A[X_1,\ldots,X_n]$ are
catenary.

\begin{cor}
Assume that $J$ is finitely generated as an $A$-module. Then
$A\bowtie^f J$ is universally catenary if and only if $A$ is
universally catenary.
\end{cor}

\begin{center} {\bf ACKNOWLEDGMENT}

\end{center}
The authors are deeply grateful to the referee for his/her careful reading of the paper and valuable
suggestions.