text:
\section{Introduction}
Let $I$ and $J$ be ideals in the polynomial ring over a field $K$.
We say that $I$ and $J$ intersect \textit{transversally} if $I\cap J= IJ$. Our
aim in this article is to derive some constructive criterion
for transversal intersection of polynomial ideals. It is not difficult to prove
(see Lemma 3.7 in \cite{sstsum})
that given two transversally intersecting ideals $I$ and $J$ in a graded
commutative ring with identity $R$, if $\mathbb{F}_{\cdot}$ resolves $R/I$ and
$\mathbb{G}_{\cdot}$ resolves $R/J$ minimally, then the tensor product
complex $\mathbb{F}_{\cdot} \otimes_{R} \mathbb{G}_{\cdot}$ resolves $R/I+J$ minimally.
Consider the short exact sequence $0\longrightarrow I \longrightarrow R\longrightarrow R/I\longrightarrow 0$ and tensor it with $R/J$ over $R$, which gives rise to
the exact sequence $$0\longrightarrow {\rm Tor}^{1}\left( R/I, R/J\right) \longrightarrow I/I\cdot J\longrightarrow R/J\longrightarrow R/I+J \longrightarrow 0 .$$ The terms on the left are $0$ since $R$ is a flat $R$ module. Moreover, the kernel of the map from $I/I\cdot J\longrightarrow R/J$ is $I\cap J/I\cdot J$. Therefore ${\rm Tor}^{1}\left( R/I, R/J\right)=0$ if and only if $I\cap J=I\cdot J$. By the corollary 1 of theorem 3 proved in \cite{lichtenbaum},
${\rm Tor}^{1}\left( R/I, R/J\right)=0$ implies that ${\rm Tor}^{i}\left( R/I, R/J\right)=0$
for all $i\geq 1$. Hence, $\mathbb{F}_{\cdot} \otimes_{R} \mathbb{G}_{\cdot}$
minimally resolves $R/I+J$.
\medskip

The transversal intersection phenomenon for monomial ideals in the polynomial
ring happens to be extremely interesting and it turns out that it is
equivalent to having disjoint supports for their minimal generating sets; see
Theorem \ref{disjoint}. The Taylor complex which resolves (possibly non-minimally)
a monomial ideals has been understood completely for ideals $I+J$, when $I$ and $J$
are monomial ideals and they intersect transversally; see Theorem \ref{tcomplex}.
As an application of this theorem we prove in Corollary \ref{tenminimal} that for two
ideals $I$ and $J$ in $R$ intersecting transversally, the ideal $I+J$ is resolved minimally by the complex $\mathbb{M}(I)\otimes \mathbb{M}(J)$, if $\mathbb{M}(I)$ and
$\mathbb{M}(J)$ denote minimal free resolutions of $I$ and $J$ respectively. Minimal
free resolutions for ideals of the form $I+J$ have an interesting structure when $I$
and $J$ intersect transversally and are supported simplicially; see \ref{simpsupport}.
\medskip

We provide some criterion for transversal intersection of polynomial ideals
in \ref{transideals}, which we
finally use to exhibit some natural classes of ideals in the polynomial ring intersect transversally with the rational normal curves and determinantal ideals of the form $I_{1}(XY)$; see \ref{transversal_IJ}, \ref{transversal_XY}. Recently,
Professor G. Valla pointed out the resemblance of Lemma \ref{transideals}
in this paper and Lemma 1.1 in \cite{robvalla}.
\bigskip

\section{Transversal Intersection of Monomial Ideals}
Let $R=K[x_{1},x_{2},\cdots ,x_{n}]$, where $x_{i}$'s are indeterminates over
the field $K$. Let ${\rm Mon}(R)$ denote the set of all monomials in $R$.
Every nonzero polynomial $f\in R$ is a unique $K$-linear combination
of monomials given by $f=\sum_{v\in {\rm Mon}(R)} a_{v}v$. Let
$m(f):= \{v\in {\rm Mon}(R) \mid a_{v}\neq 0\}$. An ideal
$I$ in $R$ is said to be a \textit{monomial ideal} if it is
generated by monomials of $R$. We list down some standard facts on
monomial ideas.
\medskip

\begin{proposition}
\begin{enumerate}
\item $I$ is a monomial ideal if and only if $f\in  I \Rightarrow m(f)\subset I$.
\medskip

\item Let $\lbrace u_{1},\cdots ,u_{m}\rbrace$ be a monomial generating set of an
ideal $I$, where $u_{i}$'s are monomials. A monomial $v\in I$ if and only if
$v=u_{i}w$, for some $1\leq i\leq m$.
\medskip

\item Each monomial ideal $I$ has a unique minimal monomial set of generators $G(I)$.
\end{enumerate}
\end{proposition}

\proof See  \cite{miller}.\qed
\medskip

\begin{definition}Let $\emptyset\neq T\subset {\rm Mon}(R)$. We define
$$
{\rm supp}(T) = \{i \mid x_{i}\ {\rm divides}\ m \ \text{for\, some}\ m\in T\}.
$$
If $T = \{m\}$, we simply write ${\rm supp}(m)$ instead of ${\rm supp}(\{m\})$.
If $S$ and $T$ are two nonempty subsets of ${\rm Mon}(R)$, then,
${\rm supp}(S)\cap {\rm supp}(T)=\emptyset$ if and only if
${\rm supp}(f)\cap {\rm supp}(g)=\emptyset$ for every $f\in S$ and $g\in T$.
\end{definition}
\medskip

\begin{definition}
We say that the ideals $I$ and $J$ of $R$ intersect transversally
if $I\cap J= IJ$.
\end{definition}
\medskip

\begin{theorem}\label{disjoint}
Let $I$ and $J$ be two monomial ideals of $R$. Then, $I\cap J =IJ$ if and only if
$\mbox{\rm supp}(G(I)) \cap \mbox{\rm supp}(G(J))=\emptyset$.
\end{theorem}
\proof Let $ I\cap J =IJ$. Consider the set
$$S=\{{\rm lcm}(f,g)\mid f\in G(I)\,, g\in G(J) \quad {\rm and}\quad \mbox{\rm supp}(f)\cap \mbox{\rm supp}(g)\neq
\emptyset\}.$$
If $S= \emptyset$, then for every $f \in G(I)$ and $g \in G(J)$ we have
$\mbox{\rm supp}(f)\cap \mbox{\rm supp}(g)=\emptyset$, which proves that
$\mbox{\rm supp}(G(I)) \cap \mbox{\rm supp}(G(J))=\emptyset$.
\medskip

Suppose that $S\neq \emptyset$. We define a partial order $\leq$ on $S$ in the
following way: Given $s,t\in S$, we define $s\leq t$ if and only
$s\mid t$. Let $\mathbf{m}$ be a minimal element (by Zorn's lemma) of $S$; then there
exists $f,g$ such that $f\in G(I)$ and $g\in G(J)$ such that ${\rm \mbox{lcm}}(f,g)=\mathbf{m}$.
Since ${\rm \mbox{lcm}}(f,g)\in I\cap J$ and $I\cap J=IJ$, we have ${\rm \mbox{lcm}}(f,g)=\mathbf{m}\in IJ$. By proposition, there exist $h_{1}\in G(I)$ and $h_{2}\in G(J)$,
such that $h_{1}h_{2}\mid m$, since the generating set of $IJ $ is the set
$G(I)G(J)=\lbrace uv\mid u\in G(I), v\in G(J)\rbrace$.
\medskip

We now claim that $\mbox{\rm supp}(h_{1})\cap \mbox{\rm supp}(h_{2}) = \emptyset$. Suppose not;
then ${\rm \mbox{lcm}}(h_{1}, h_{2})\neq  h_{1}h_{2}$ and
${\rm \mbox{lcm}}(h_{1}, h_{2})< h_{1}h_{2}$ and ${\rm \mbox{lcm}}(h_{1}, h_{2})\mid \mathbf{m}$,
which contradicts minimality of $\mathbf{m}$ in $S$.
\medskip

We now prove that $\mbox{\rm supp}(f)\cap \mbox{\rm supp}(h_{2})\neq \emptyset$. We know that
$h_{1}h_{2}\mid {\rm \mbox{lcm}}(f,g)=\mathbf{m}$. Therefore, if
$\mbox{\rm supp}(f)\cap \mbox{\rm supp}(h_{2})= \emptyset$, then $h_{2}\mid g$; which contradicts
minimality of the generating set $G(J)$. Similarly we can prove that
$\mbox{\rm supp}(g)\cap \mbox{\rm supp}(h_{1})\neq \emptyset$. Now, $h_{1}h_{2}\mid \mathbf{m}={\rm \mbox{lcm}}(f,g)$ implies that $h_{2}\mid \mathbf{m}={\rm \mbox{lcm}}(f,g)$. Moreover,
$f\mid {\rm \mbox{lcm}}(f,g)$. Therefore, ${\rm \mbox{lcm}}(f,h_{2})\mid {\rm \mbox{lcm}}(f,g)$.
Similarly, ${\rm \mbox{lcm}}(g,h_{1})\mid {\rm \mbox{lcm}}(f,g)$. Now if ${\rm \mbox{lcm}}(f,h_{2})< \mathbf{m}$
or ${\rm \mbox{lcm}}(g,h_{1})<{\rm \mbox{lcm}}(f,g)=\mathbf{m}$, then we have a contradiction,
since ${\rm \mbox{lcm}}(f,g)$ and ${\rm \mbox{lcm}}(g,h_{1})$ both are in the set $S$
and $\mathbf{m}$ is a minimal element in $S$. Therefore, we have must have
${\rm \mbox{lcm}}(f,h_{2})={\rm \mbox{lcm}}(g,h_{1})={\rm \mbox{lcm}}(f,g)=\mathbf{m}=h_{1}h_{2}w$(say) and
$\dfrac{fh_{2}}{\rm{gcd}(f,h_{2})}=\dfrac{gh_{1}}{\gcd(g,h_{1})}=h_{1}h_{2}w$,
therefore $\dfrac{f}{\gcd(f,h_{2})}=h_{1}w$. Hence, $h_{1}\mid f$ and that
contradicts minimality of the generating set $G(I)$. Hence,
$S=\emptyset$ and we are done.
\medskip

Conversely, let us assume that $\mbox{\rm supp}(G(I))\cap\mbox{\rm supp}(G(J)) = \emptyset$.
Without loss of generality, we can assume that $\mbox{\rm supp}(G(I)) = \{1,2,\ldots , k\}$ and
$\mbox{\rm supp}(G(J)) = \{k+1, k+2,..., n\}$. Let $f\in I\cap J$, such that
$f=\sum_{v\in Mon(R)} a_{v}v$. We have $v\in I\cap J$ for all $v\in m(f)$.
It is therefore enough to show that if $m$ is a monomial and
$m\in I\cap J$, then $m\in IJ$. Let $m\in I\cap J$; there exist
$m_{I}\in G(I)$ and $m_{J}\in G(J)$ such that $m_{I}\mid m$ and $m_{J}\mid m$.
Let $m_{I}m_{1}=m$; then $m_{J}\mid m=m_{I}m_{1}$. We know that
$\mbox{\rm supp}(m_{I})\cap \mbox{\rm supp}(m_{J}) = \emptyset$, therefore $m_{J}\mid m_{1}$
and it follows that $m_{1}=m_{J}m_{2}$ and $m=m_{I}m_{J}m_{2}$. Therefore,
$m\in IJ$ since $m_{I}m_{J}\in IJ$. \qed
\medskip

\begin{theorem}\label{dissup}
Let $I$ and $J$ be two ideals of $R$ such that w.r.t. some monomial order $\preceq$,
 $\mbox{\rm supp}(\rm{Lt}(I)) \cap \mbox{\rm supp}(\rm{Lt}(J))=\emptyset$. Then $I$ and $J$ intersect transversally.
\end{theorem}

\proof Let $\mathcal{G}_{I}$ and $\mathcal{G}_{I}$ be Gr\"{o}bner basis w.r.t. monomial order $\preceq$ of the ideals $I$ and $J$ respectively. Again we assume that $f\in I\cap J$, then there exist polynomials $p\in \mathcal{G}_{I}$ and $q\in \mathcal{G}_{J}$ such that $\rm{Lt}(p)\mid \rm{Lt}(f)$ and $\rm{Lt}(q)\mid \rm{Lt}(f)$. Since $\mbox{\rm supp}(\rm{Lt}(I)) \cap \mbox{\rm supp}(\rm{Lt}(J))=\emptyset$, we have  $\rm{Lt}(p)\cdot\rm{Lt}(q)\mid \rm{Lt}(f)$. After division we write $f=pq+r$, then $r\in I\cap J$ and $\rm{Lt}(r)\preceq\rm{Lt}(f)$. we can apply same process on $r$, and after finite stage we get $f\in IJ$. \qed
\medskip

\subsection{The Taylor complex} We define the {\em multidegree} of a
monomial $x^{\mathbf{a}} = x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{n}^{a_{n}}$ to be the
$n$-tuple $(a_{1}, \ldots , a_{n})$, denoted by
$\text{mdeg}(\mathbf{x}^{\mathbf{a}})$. Hence $R=\oplus _{m\in {\rm Mon}(R)} Km$
has a direct sum decomposition as $K$-vector spaces $Km$, where $Km = \{cm\mid c\in K\}$.
\medskip

\begin{definition}[\textbf{The Taylor complex}]
Let $M$ be a monomial ideal of $R$ minimally generated by the
monomials $m_{1},m_{2},\cdots , m_{p}$. The Taylor complex
for $M$ is given by $\mathbb{T}(M)_{\cdot}$ as follows: $\mathbb{T}(M)_{i}$ is the
free $R$-module generated by $e_{j_{1}}\wedge e_{j_{2}}\wedge \cdots\wedge e_{j_{i}}$, for all $1\leq j_{1} < \cdots < j_{i}\leq p$, where
$\{e_{1}, \ldots , e_{p}\}$ is the standard basis for the free $R$-module $R^{p}$.
The differential
$\delta(M)_{\cdot}$ is given by
$$\delta(M)_{i}(e_{j_{1}}\wedge \cdots\wedge e_{j_{i}}) = \sum_{1\leq k\leq i}(-1)^{k-1}
\dfrac{{\rm \mbox{lcm}} (m_{j_{1}}, \cdots ,m_{j_{i}})}{{\rm \mbox{lcm}}(m_{j_{1}}, \cdots,\widehat{m_{j_{k}}}, \cdots, m_{j_{i}})}(e_{j_{1}}\wedge \cdots\wedge \widehat{e_{j_{k}}}\wedge\cdots\wedge e_{j_{i}}).$$
We define ${\rm mdeg}(e_{j_{1}}\wedge \cdots\wedge e_{j_{i}})
= {\rm \mbox{mdeg}}({\rm \mbox{lcm}} (m_{j_{1}}\cdots m_{j_{i}}))$.
\end{definition}
\medskip

\begin{theorem}The Taylor complex $(\mathbb{T}(M)_{\cdot}, \delta(M)_{\cdot})$
defined above is a free resolution (though not minimal) of the monomial ideal M;
called the Taylor resolution of $M$.
\end{theorem}

\proof See Theorem 26.7 in \cite{peeva}.\qed
\medskip

\begin{theorem}\label{tcomplex}
Let $I$ and $J$ be two monomial ideals of $R$ intersecting transversally. Let
$\mathbb{T}(I)$ and $\mathbb{T}(J)$ be the Taylor resolutions of $I$ and $J$
respectively. Then, the Taylor's resolution of $I+J$ is isomorphic to the complex
$\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot}$ and hence $\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot}$ is acyclic.
\end{theorem}

\proof We know that $\mathbb{T}(I)_{i}= R^{\binom{p}{i}}$ and $\mathbb{T}(J) = R^{\binom{q}{j}}$, where $p$ and $q$ denote the minimal number of generators
of $I$ and $J$ respectively. Then,
\begin{eqnarray*}
(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r} & = &\oplus_{i+j=r}(\mathbb{T}(I)_{i}\otimes \mathbb{T}(J)_{j})\\
{} &  = & \oplus_{i+j=r} R^{\binom{p}{i}}\otimes R^{\binom{q}{j}}\\
{} &  = & R^{\sum_{i+j=r}\binom{p}{i}\binom{q}{j}}\\
{} & = & R^{\binom{p+q}{r}} = \mathbb{T}(I+J)_{r}.
\end{eqnarray*}

A basis of $(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r}$ is given by $(e_{k_{1}}\wedge \cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})$,
such that $i+j=r$ and $1\leq k_{1} < k_{2} < \cdots < k_{i}\leq p$ and
$1\leq k'_{1} < k'_{2} < \cdots < k'_{j}\leq q$. Now for each $r$, the free
module $(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r}$ can be graded with the help of mdeg as follows. We define
\begin{eqnarray*}
{\rm \mbox{mdeg}}\left((e_{k_{1}}\wedge \cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge \cdots\wedge e'_{k'_{j}})\right)\\
= {\rm \mbox{mdeg}}\left({\rm \mbox{lcm}}( m_{k_{1}},\ldots m_{k_{i}})\cdot {\rm \mbox{lcm}}(m'_{k'_{1}},\ldots , m'_{k'_{j}})\right).
\end{eqnarray*}
This defines a multi-graded structure for the complex $\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot}$.
\medskip

Let $G(I)=\{m_{1},m_{2}\cdots,m_{p}\}$ and $G(J)=\{m'_{1},m'_{2}\cdots,m'_{q}\}$
denote the minimal set of generators for $I$ and $J$ respectively. Then $I+J$ is
minimally generated by $G(I)\cup G(J)$, by Theorem \ref{disjoint}, since $G(I)$
and $G(J)$ are of disjoint support. We will now show that the tensor product complex
$\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot}$is isomorphic to the Taylor
resolution $\mathbb{T}(I+J)_{\cdot}$. We define $\psi_{r}:(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r}\longrightarrow \mathbb{T}(I+J)_{r}$ as
$$(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\mapsto (e_{k_{1}}\wedge\cdots\wedge e_{k_{i}}\wedge e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}).$$
Moreover,
\begin{eqnarray*}
{} & \hspace*{-1in} {\rm \mbox{mdeg}}(e_{k_{1}}\wedge \cdots\wedge e_{k_{i}}\wedge e_{k'_{1}}\wedge\cdots\wedge e_{k'_{j}})\\
= & \hspace*{-1in} {\rm \mbox{mdeg}}\left({\rm \mbox{lcm}}( m_{k_{1}},\ldots , m_{k_{s}}, m^{'}_{k'_{1}},\ldots, m^{'}_{k'_{s'}})\right)\\
= & \hspace*{-6pt} {\rm \mbox{mdeg}}\left({\rm \mbox{lcm}}(m_{k_{1}},\ldots , m_{k_{s}})\right)\cdot {\rm \mbox{mdeg}}\left({\rm \mbox{lcm}}(m^{'}_{k'_{1}},\ldots , m^{'}_{k'_{s'}})\right),
\end{eqnarray*}
since $G(I)$ and $G(J)$ are with disjoint supports. Hence, the map is a
graded isomorphism between the free modules $(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r}$ and $\mathbb{T}(I+J)_{r}$. We therefore have the following
diagram of complexes:
\medskip

$\xymatrix{
\cdots\ar[r]^{d_{r+1}}  & (\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r}\ar[r]^{d_{r}}\ar[d]^{\psi_{r}}&(\mathbb{T}(I)_{\cdot}\otimes \mathbb{T}(J)_{\cdot})_{r-1}\ar[r]^{d_{r-1}}\ar[d]^{\psi_{r-1}}&\cdots\\
\cdots\ar[r]^{\delta(I+J)_{r+1}} & \mathbb{T}(I+J)_{r}\ar[r]^{\delta(I+J)_{r}} & \mathbb{T}(I+J)_{r-1}\ar[r]^{\delta(I+J)_{r-1}}& \cdots}$
\medskip

To show that $\delta_{r} \circ \psi_{r}  =  \psi_{r-1} \circ d_{r}$, let us take an
arbitrary basis element $(e_{k_{1}}\wedge e_{k_{2}}\wedge\cdots\wedge e_{k_{i}}\otimes e_{k'_{1}}\wedge e_{k'_{2}}\wedge\cdots\wedge e_{k'_{j}})$. Suppose that
$i+j=r$. Then,
\begin{eqnarray*}
{} & \hspace*{-2in} (\psi_{r-1} \circ d_{r})\left((e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\right)\\
= & \hspace*{-20pt} \psi_{r-1}\left((\delta(I)_{i}(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\right)\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
{} & \quad\quad + \quad (-1)^{i}(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes\left((\delta(J)_{j}(e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\right).
\end{eqnarray*}
Now we see that
\begin{eqnarray*}
{} & \hspace*{-3in} \left(\delta(I)_{i}(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\right)\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
= & \sum_{t=1}^{i}(-1)^{t}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}},\ldots , m_{k_{i}})}{{\rm \mbox{lcm}}(m_{k_{1}},\ldots , \widehat{m_{k_{t}}},\ldots , m_{k_{i}})}(e_{k_{1}}\wedge\cdots\wedge \widehat{e_{k_{t}}}\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}).
\end{eqnarray*}
Hence,
\begin{eqnarray*}
{} & \hspace*{-3in} \psi_{r-1}\left(\delta(I)_{i}(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\right)\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
= & \hspace*{-24pt} \sum_{t=1}^{i}(-1)^{t}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}},\cdots ,m_{k_{i}})}{{\rm \mbox{lcm}}(m_{k_{1}},\ldots , \widehat{m_{k_{t}}},\ldots , m_{k_{i}})}(e_{k_{1}}\wedge\cdots\wedge \widehat{e_{k_{t}}}\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}))\\
= & \hspace*{-6pt}\sum_{t=1}^{i}(-1)^{t}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}},\ldots , m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots , \widehat{m_{k_{t}}}, \ldots , m_{k_{i}}, m'_{k'_{1}}, \ldots, m'_{k'_{j}})}(e_{k_{1}}\wedge\cdots\wedge \widehat{e_{k_{t}}}\wedge e_{k_{i}}\wedge e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}).
\end{eqnarray*}
where the last equality follows since $G(I)$ and $G(J)$ are of disjoint support. Similarly,
\begin{eqnarray*}
{} & \hspace*{-3in} \psi_{r-1}((e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes(\delta(J)_{j}(e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})))\\
= & \hspace*{-6pt}\sum_{l=1}^{j}(-1)^{l}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots, m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots,  m_{k_{i}}, m'_{k'_{1}},\ldots , \widehat{m'_{k'_{l}}},\ldots, m'_{k'_{j}})}(e_{k_{1}} \wedge\cdots\wedge e_{k_{i}}\wedge e^{'}_{k'_{1}}\wedge\cdots \widehat{e^{'}_{k'_{l}}}\cdots\wedge
e^{'}_{k'_{j}}).
\end{eqnarray*}
Hence,
\begin{eqnarray*}
(*) & \hspace*{-2.5in}(\psi_{r-1} \circ d_{r})((e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}))\\
= & \hspace*{-11pt} \sum_{t=1}^{i}(-1)^{t}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}},\ldots ,m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots , \widehat{m_{k_{t}}},\ldots, m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}(e_{k_{1}}\wedge\cdots\wedge \widehat{e_{k_{t}}}\wedge e_{k_{i}}\wedge e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
+ & \hspace*{-10pt} \sum_{l=1}^{j}(-1)^{i+l}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots, m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots,  m_{k_{i}}, m'_{k'_{1}},\ldots , \widehat{m'_{k'_{l}}}, \ldots, m'_{k'_{j}})}(e_{k_{1}} \wedge\cdots\wedge e_{k_{i}}\wedge e^{'}_{k'_{1}}\wedge\cdots \widehat{e^{'}_{k'_{l}}}\cdots\wedge
e^{'}_{k'_{j}})\\
= & \hspace*{-11pt} \sum_{t=1}^{i}(-1)^{t}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}},\ldots ,m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots ,\hat{m_{k_{t}}}, \ldots ,  m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}(e_{k_{1}}\wedge\cdots\wedge \widehat{e_{k_{t}}}\wedge e_{k_{i}}\wedge e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
+ & \hspace*{-11pt} \sum_{l=1}^{j}(-1)^{i+l}\dfrac{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots, m_{k_{i}}, m'_{k'_{1}},\ldots, m'_{k'_{j}})}{{\rm \mbox{lcm}}(m_{k_{1}}, \ldots,  m_{k_{i}}, m'_{k'_{1}},\ldots \widehat{m'_{k'_{l}}},\ldots,m'_{k'_{j}})}(e_{k_{1}} \wedge\cdots\wedge e_{k_{i}}\wedge
e^{'}_{k'_{1}}\wedge\cdots \widehat{e^{'}_{k'_{l}}}\cdots\wedge e^{'}_{k'_{j}})
\end{eqnarray*}
We here make a change in notation. We define $i+j$ new symbols $w_{s}$, such that
$w_{s}=m_{k_{s}}$ for $s\leq i$ and $w_{s}=m'_{k'_{s-i}}$ for $s>i$. We
further define another set of $i+j$ symbols $E_{s}$, such $E_{s}=e_{k_{s}}$
for $s\leq i$ and $E_{s}=e'_{k'_{s-i}}$ for $s>i$. Hence the latter sum in (*)
becomes
$$\sum_{k=1}^{i+j}(-1)^{k}\dfrac{{\rm \mbox{lcm}}(w_{1}, \ldots, w_{i+j})}{{\rm \mbox{lcm}}(w_{1},\ldots,\widehat{w_{k}},\ldots, w_{i+j})}(E_{1}\wedge\cdots\wedge \widehat{E_{k}}\cdots\wedge E_{i+j}).$$
Now
\begin{eqnarray*}
{} & \hspace*{-1.25in} (\delta_{r} \circ \psi_{r})((e_{k_{1}}\wedge\cdots\wedge e_{k_{i}})\otimes (e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}}))\\
= & \hspace*{-2in} \delta_{r}(e_{k_{1}}\wedge\cdots\wedge e_{k_{i}}\wedge e'_{k'_{1}}\wedge\cdots\wedge e'_{k'_{j}})\\
= & \sum_{k=1}^{i+j}(-1)^{k}\dfrac{{\rm \mbox{lcm}}(w_{1}, \ldots, w_{i+j})}{{\rm \mbox{lcm}}(w_{1},\ldots, \widehat{w_{k}}, \ldots , w_{i+j})}(E_{1}\wedge\cdots\wedge \widehat{E_{k}}\cdots\wedge E_{i+j}).
\end{eqnarray*}
Hence, the diagram is commutative.\qed
\medskip

\begin{corollary}\label{tenminimal}
Let $I$ and $J$ be ideals in $R$ such that $I\cdot J = I\cap J$. Let
$\mathbb{M}(I)$ and $\mathbb{M}(J)$ denote minimal free resolutions
of $I$ and $J$ respectively. Then, $\mathbb{M}(I)\otimes \mathbb{M}(J)$
is a minimal free resolution of the ideal $I+J$.
\end{corollary}

\proof The minimal free resolutions $\mathbb{M}(I)$ and $\mathbb{M}(J)$ are direct summands of the Taylor resolutions $\mathbb{T}(I)$ and $\mathbb{T}(J)$
respectively. It follows that $(\mathbb{M}(I)\otimes \mathbb{M}(J))$ is a
direct summand of $\mathbb{T}(I)\otimes \mathbb{T}(J) = \mathbb{T}(I+J)$. Hence, $(\mathbb{M}(I)\otimes \mathbb{M}(J))$ is a free resolution of
$I+J$ and it is minimal since $\mathbb{M}(I)$ and $\mathbb{M}(J)$ are both minimal.\qed
\bigskip

\section{Simplicial characterization of transversal intersection of monomial ideals}
We first introduce the basic definitions of a simplicial complex; see \cite{miller}.
\medskip

\begin{definition}
A \textit{simplicial complex} $\Delta$ on the vertex set $\{1,\ldots,m\}$ is
a collection of subsets called faces or simplices, satisfying the following condition:
$\sigma\in \Delta$ and $\tau\subset\sigma$ implies that $\tau\in \Delta$.
A simplex $\sigma\in \Delta$ of cardinality $|\sigma|=r+1$ has dimension
$r$ and it is called an \textit{$r$ face} of $\Delta$.
\end{definition}
\medskip

\begin{definition}
A \textit{facet} is a  maximal simplex of a simplicial complex $\Delta$.
Let $\Gamma(\Delta)$ denote the set of all facets of $\Delta$. Then, the
vertex set of $\Delta$ and the set $\Gamma(\Delta)$ determine $\Delta$ completely.
\end{definition}
\medskip

\begin{definition}
A \textit{standard  simplicial complex} of dimension $m-1$ on the vertex set
$\{1,\ldots,m\}$ is the simplicial complex whose $\Gamma(\Delta)=\{\{1,\ldots,m\}\}$.
\end{definition}
\medskip

\begin{definition}
Let $\Delta_{1}$ and $\Delta_{2}$ be simplicial complexes on disjoint vertex sets.
Let the vertext set of $\Delta_{1}$ be $\{1,\ldots,m\}$ and that of
$\Delta_{2}$ be $\{m+1,\ldots,m+p\}$. The \textit{join of simplicial complexes}
$\Delta_{1}$ and $\Delta_{2}$ is the simplicial complex
$\Delta_{1}*\Delta_{2}$, whose vertex set is $\{1,\ldots,m+p\}$ and
$$\Gamma(\Delta_{1}*\Delta_{2})=\{\sigma_{1}\cup\sigma_{2}\mid \sigma_{1}\in \Gamma(\Delta_{1}), \, \sigma_{2}\in \Gamma(\Delta_{2}) \}.$$
\end{definition}
\medskip

We now define a \textit{frame} as a complex of $K$-vector spaces with a fixed basis, which encodes the minimal free resolution of a monomial ideal; see \cite{peeva}.
\medskip

\begin{definition}
An $r-$frame $\mathbb{U}_{\centerdot}$ is  a complex of finite $K$ vector spaces with
differential $\partial$ and a fixed basis satisfying the following:
\begin{enumerate}[(i)]
\item $\mathbb{U}_{i}=0$ for $i<0$;
\item $\mathbb{U}_{0}=K$;
\item $\mathbb{U}_{1}=K^{r}$, with basis $\{w_{1}, \ldots , w_{r}\}$;
\item $\partial(w_{j})=1$, for all $j = 1, \ldots , r$.
\end{enumerate}
\end{definition}
\medskip

\noindent\textbf{Construction of a frame of a simplicial complex.} Let $\Delta$ be a simplicial complex on the vertex set $\{1,\ldots,m\}$. For each integer $i$,
let $\Gamma_{i}(\Delta)$ be the set of all $i$ dimensional faces of $\Delta$
and let $K^{\Gamma_{i}(\Delta)}$ denote the $K$-vector space generated by the
basis elements $e_{\sigma}$, for $\sigma\in \Gamma_{i}(\Delta)$. The
\textit{chain complex} of $\Delta$ over $K$ is the complex
$\mathcal{C}(\Delta)_{\centerdot}$,
$$0\longrightarrow K^{\Gamma_{m-1}(\Delta)}\stackrel{\partial_{m-1}}{\longrightarrow}\cdots K^{\Gamma_{i}(\Delta)}\stackrel{\partial_{i}}{\longrightarrow} K^{\Gamma_{i-1}(\Delta)}\stackrel{\partial_{i-1}}{\longrightarrow} \cdots \stackrel{\partial_{0}}{\longrightarrow} K^{\Gamma_{-1}(\Delta)}\longrightarrow 0.$$
The boundary maps $\partial_{i}$ are defined as
$$\partial_{i}(e_{\sigma})=\sum_{j\in\sigma}sign(j,\sigma)e_{\sigma-j},\quad 0\leq i\leq m-1;$$
such that $sign(j,\sigma)=(-1)^{t-1}$, where $j$ is the $t$-th element of the set
$\sigma\subset \{1,\ldots,m\}$ written in increasing order. If $i<-1$ or
$i>m-1$, then $K^{\Gamma_{i}(\Delta)}=0$ and  $\partial_{i}=0$. We see that
the chain complex of a simplicial complex is an $m$-frame.
\medskip

If $\Delta$ is a standard simplicial complex of dimension $m-1$
on the vertex set $\{1, \ldots, m\}$, then
$\mathcal{C}(\Delta)_{\centerdot}$ is acyclic and
$K^{\Gamma_{i}(\Delta)}=K^{\binom{m}{i}}$. Therefore,
the $m$-frame of a standard simplicial complex of
dimension $m-1$ is nothing but the Koszul complex.
\medskip

\begin{definition}
Let $M=\{m_{1},m_{2},\cdots,m_{r}\}$ be a set of monomials in $R$.
An $M-$complex $\mathbb{G}_{\centerdot}$ is a multigraded complex of
finitely generated free multigraded $R$ modules with
differential $\Delta$ and a fixed multihomogeneous
basis with multidegrees that satisfy
\begin{enumerate}[(i)]
\item $\mathbb{G}_{i}=0$ \, for \, $i<0$;
\item $\mathbb{G}_{0}=R$;
\item $\mathbb{G}_{1}=R(m_{1})\oplus \cdots\oplus R(m_{r})$;
\item $\Delta(w_{j})=m_{j}$ for each basis vector $w_{j}\in\mathbb{G}_{1}$.
\end{enumerate}
\end{definition}

\noindent\textbf{Construction.}  Let $\mathbb{U}_{\centerdot}$ be an
$r-$frame and $M=\{m_{1},m_{2},\cdots,m_{r}\}$ be a set of monomials
in $R$. The $M-$\textit{homogenization} of $\mathbb{U}_{\centerdot}$
is defined to be the complex $\mathbb{G}_{\centerdot}$ with
$\mathbb{G}_{0}= R$ and $\mathbb{G}_{1}= R(m_{1})\oplus \cdots\oplus R(m_{r})$.
Let $\bar{v_{1}},\cdots, \bar{v_{p}}$ and $\bar{u_{1}},\cdots, \bar{u_{q}}$
be the given bases of $\mathbb{U}_{i}$ and $\mathbb{U}_{i-1}$
respectively. Let $u_{1},u_{2},\cdots, u_{q}$ be the basis of
$\mathbb{G}_{i-1}$ chosen in the previous step of induction.
We take $v_{1},\cdots,v_{p}$ to be a basis of $\mathbb{G}_{i}$.
If $$\partial(\bar{v_{j}})=\sum_{1\leq s\leq q}\alpha_{sj}\bar{u_{s}}$$
with coefficients $\alpha_{sj}\in k$, then set
$\rm{mdeg}(v_{j})= {\rm \mbox{lcm}} \left(mdeg(u_{s})|\alpha_{sj}\neq 0\right)$,
where ${\rm \mbox{lcm}}(\emptyset)=1$.
$\mathbb{G}_{i}=\oplus_{1\leq i\leq p}R\left(\rm{mdeg}(v_{j})\right)$,
$d(v_{j})= \Sigma_{1\leq s\leq q} \alpha_{sj} \dfrac{\rm{mdeg} (v_{j})}{\rm{mdeg}(u_{s})}u_{s}.$
\medskip

\begin{definition} Let $I\subset R$ be a monomial ideal with $G(I)=\{m_{1},\ldots,m_{r}\}$. We say that a minimal free resolution of $I$ is supported by a simplicial complex
$\Delta$ on the vertex set $\{1,\ldots,r\}$ if it is isomorphic to the $G(I)$ homogenization of the chain complex $\mathcal{C}(\Delta)_{\centerdot}$ of  $\Delta$.
\end{definition}
\medskip

\begin{lemma}
Let $I$ and $J$ be two monomial ideals such that $I\cap J=IJ$, $|G(I)|=r$ and $|G(J)|=s$, and whose minimal free resolutions are supported by standard simplicial complexes $\Delta_{I}$ and $\Delta_{J}$ on the vertex sets $\{1,\ldots,r\}$ and $\{r+1,\ldots,r+s\}$ respectively. Then minimal free resolution of $I+J$ is supported by the standard simplicial complex $\Delta_{I}*\Delta_{J}$.
\end{lemma}

\proof At first we note that if minimal free resolution $\mathbb{M}(I)_{\centerdot}$ of $I$ supported by standard simplicial complex $\Delta_{I}$, then it is actually isomorphic to the Taylor complex $\mathbb{T}(I)_{\centerdot}$. Since  $I\cap J=IJ$, by the Theorem \ref{disjoint} we have $G(I)\cap G(J)=\emptyset$, then  $|G(I)\cup G(J)|=r+s$. Again $\Delta_{I}*\Delta_{J}$ is also standard simplicial complex of dimension $r+s-1$. Therefore $G(I)\cup G(J)$-homogenization of $\mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$ is isomorphic to the Taylor complex $\mathbb{T}(I+J)_{\centerdot}$. Thus by Theorem \ref{tcomplex}, we have $$\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}\cong \mathbb{T}(I)_{\centerdot}\otimes\mathbb{T}(J)_{\centerdot}\cong \mathbb{T}(I+J)_{\centerdot}$$
Again by the Corollary \ref{tenminimal} $\mathbb{M}(I)_{\centerdot}\otimes\mathbb{M}(J)_{\centerdot}$ is minimal free resolution of $I+J$.
Therefore minimal free resolution of $I+J$ is supported by the standard simplicial complex $\Delta_{I}*\Delta_{J}$\qed
\medskip

\begin{lemma}\label{joinframe}
Let $\Delta_{I}$ be a simplicial complex on vertex $\{1,\ldots,r\}$ and $\Delta_{J}$ be another simplicial complex on vertex $\{r+1,\ldots,r+s\}$,i.e, the vertex sets of
$\Delta_{I}$ and $\Delta_{J}$ are disjoint. Then
$$\mathcal{C}(\Delta_{I})_{\centerdot}\otimes \mathcal{C}(\Delta_{J})_{\centerdot}\cong \mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$$
\end{lemma}

\proof Let $\Gamma(\Delta_{I})=\{\gamma_{1},\ldots,\gamma_{l_{1}}\}$ and
$\Gamma(\Delta_{J})=\{\sigma_{1},\ldots,\sigma_{l_{2}}\}$. We have
$\Gamma(\Delta_{I}*\Delta_{J})= \{\gamma_{i}\cup \sigma_{j}\mid 1\leq i\leq l_{1}, 1\leq j\leq l_{2}\}$. Now assuming $\binom{s}{t}=0$ for $s<t$, we have
$\mid\Gamma_{p}(\Delta_{I})\mid = \sum_{i=1}^{l_{1}}\binom{\mid \gamma_{i}\mid}{p}$, for all $1\leq p\leq l_{1}$ and  $\mid\Gamma_{q}(\Delta_{J})\mid = \sum_{j=1}^{l_{2}}\binom{\mid \sigma_{j}\mid}{q}$ for all $1\leq q\leq l_{2}$. On the other hand,
$\mid\Gamma_{t}(\Delta_{I}*\Delta_{J})\mid=\sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\cup \sigma_{j}\mid}{t}= \sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\mid+\mid\sigma_{j}\mid}{t}=\sum_{p+q=t}\sum_{j=1}^{l_{2}}\sum_{i=1}^{l_{1}}\binom{\mid\gamma_{i}\mid}{p}\binom{\mid\sigma_{j}\mid}{q}$. Therefore the map $$\theta_{t}:\left(\mathcal{C}(\Delta_{I})_{\centerdot}\otimes \mathcal{C}(\Delta_{J})_{\centerdot}\right)_{t}\longrightarrow \left(\mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}\right)_{t}$$ defined by $$\theta_{t}(e_{\gamma}\otimes e_{\sigma})=e_{\gamma\cup \sigma}$$ is an isomorphism, where $\gamma\in \Delta_{I}, \mid\gamma\mid=p $ and $\sigma\in \Delta_{J}, \mid\gamma\mid=q $ and $p+q=t$. We therefore have to show the following
diagram commutes
\medskip

\begin{center}
\begin{tikzcd}[column sep = 3cm]
(\mathcal{C}(\Delta_{I})_{\cdot}\otimes \mathcal{C}(\Delta_{J})_{\cdot})_{t}
\arrow[r, "(\partial(\Delta_{I})\otimes\partial(\Delta_{J}))_{t}"]
\arrow[d,"\theta_{t}"]
& (\mathcal{C}(\Delta_{I})_{\cdot}\otimes \mathcal{C}(\Delta_{J})_{\cdot})_{t-1}
\arrow[d,"\theta_{t-1}"]\\
\mathcal{C}(\Delta_{I}*\Delta_{J})_{t}\arrow[r, "(\partial(\Delta_{I}*\Delta_{J}))_{t}"] & \mathcal{C}(\Delta_{I}*\Delta_{J})_{t-1}
\end{tikzcd}
\end{center}
\medskip

Let $\gamma\in \Delta_{I}, \mid\gamma\mid=p $ and $\sigma\in \Delta_{J}, \mid\gamma\mid=q$
such that $p+q=t$. Then,
\begin{eqnarray*}
{} & \theta_{t-1}\circ \left(\partial(\Delta_{I})\otimes\partial(\Delta_{J})_{t}\right)(e_{\gamma}\otimes e_{\sigma})\\
= & \theta_{t-1}((\partial(\Delta_{I})_{p}e_{\gamma})\otimes e_{\sigma}+(-1)^{p}e_{\gamma}\otimes(\partial(\Delta_{J})_{p}e_{\sigma}))\\
= & \theta_{t-1}(\sum_{j\in \gamma}sign(j,\gamma)e_{\gamma-j}\otimes e_{\sigma}+(-1)^{p}e_{\gamma}\otimes(\sum_{s\in \sigma}sign(s,\sigma)e_{\sigma-s}))\\
= & \sum_{j\in \gamma}sign(j,\gamma)e_{\gamma\cup\sigma-j}+(-1)^{p}(\sum_{s\in \sigma}sign(s,\sigma)e_{\gamma\cup\sigma-s})\\
= & \sum_{j\in \gamma\cup\sigma}sign(j,\gamma\cup\sigma)e_{\gamma\cup\sigma-j}=\partial(\Delta_{I}*\Delta_{J})_{t}\circ \theta_{t}((e_{\gamma}\otimes e_{\sigma}))\qed
\end{eqnarray*}
\medskip

\begin{theorem}\label{simpsupport}
Let $I$ and $J$ be two monomial ideals such that $I\cap J=IJ$, $|G(I)|=r$ and $|G(J)|=s$, and their minimal free resolutions are supported by the simplicial complexes $\Delta_{I}$ and $\Delta_{J}$ on the vertex sets $\{1,\ldots,r\}$ and $\{r+1,\ldots,r+s\}$ respectively. Then minimal free resolution of $I+J$ is supported by the simplicial complex $\Delta_{I}*\Delta_{J}$.
\end{theorem}

\proof We have $I\cap J=IJ$, therefore by Theorem \ref{disjoint}, $\mbox{\rm supp}(G(I)) \cap \mbox{\rm supp}(G(J))=\emptyset$. Again by the lemma \ref{joinframe} $$\mathcal{C}(\Delta_{I})_{\centerdot}\otimes \mathcal{C}(\Delta_{J})_{\centerdot}\cong \mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$$
Since $\mathbb{M}(I)_{\centerdot}$ is the $G(I)$-homogenization of the complex $\mathcal{C}(\Delta_{I})_{\centerdot}$ and $\mathbb{M}(J)_{\centerdot}$ is the $G(J)$-homogenization of the complex $\mathcal{C}(\Delta_{J})_{\centerdot}$, we can
proceed in the same way as lemma \ref{joinframe} to prove that $\mathbb{M}(I)_{\centerdot}\otimes \mathbb{M}(J)_{\centerdot}$ is the $G(I)\cup G(J)$--homogenization of the complex $\mathcal{C}(\Delta_{I}*\Delta_{J})_{\centerdot}$. Again by the Corollary \ref{tenminimal} $\mathbb{M}(I)_{\centerdot}\otimes \mathbb{M}(J)_{\centerdot}$ is minimal free resolution of $I+J$. Therefore minimal free resolution of $I+J$ is supported by the simplicial complex $\Delta_{I}*\Delta_{J}$. \qed
\bigskip

\section{Crietrion for Transversal Intersection of Ideals}
\begin{theorem}[Rees]\label{rees} Let $N$ be an $R$ module, $h_{1},\ldots, h_{k}$
be an $N$-regular sequence in $R$ and $J=\langle h_{1},\ldots, h_{k}\rangle $. Let $Y=y_{1},\ldots,y_{k}$ be indeterminates over $R$. If $F(y_{1},\ldots ,y_{k})\in N[Y]$
is homogeneous of degree $r$ and $F(h_{1},\ldots ,h_{k})\in J^{r+1}N $ then the coefficients of $F$ are in $JN$.
\end{theorem}

\proof See Theorem 1.1.7 in \cite{bh}.\qed
\medskip

\begin{lemma}\label{nzd}
Let $h_{1},\ldots, h_{k}$ be a regular sequence in $R$. Let $J=\langle h_{1},\ldots,h_{k-1}\rangle$. Then $h_{k}$ is not a zero divisor in $R/J^{r}$, for every
$r\geq 1$.
\end{lemma}

\proof We use induction on $r$. For $r=1$, the result follows from the
fact that $h_{1},\ldots, h_{k}$ is a regular sequence in $R$. Let us
assume that $h_{k}$ is not a zero divisor in $R/J^{r-1}$. Let
$h_{k}p\in J^{r}$, hence $h_{k}p\in J^{r-1}$. By the induction hypothesis
we have $p\in J^{r-1}$. We can write
$$p=\sum_{\lambda_{1}+\cdots+\lambda_{k-1}=r-1}\beta_{(\lambda_{1},\cdots,\lambda_{k-1})}h_{1}^{\lambda_{1}}\cdots h_{k-1}^{\lambda_{k-1}}.$$
Let us consider the homogeneous polynomial $F(y_{1},\ldots ,y_{k-1})$
of degree $r-1$ in $R[y_{1},\ldots,y_{k-1}]$, given by
$$F(y_{1},\ldots ,y_{k-1})=\sum_{\lambda_{1}+\cdots+\lambda_{k-1}=r-1}h_{k}\beta_{(\lambda_{1},\cdots,\lambda_{k-1})}y_{1}^{\lambda_{1}}\cdots y_{k-1}^{\lambda_{k-1}}.$$
Then, $F(h_{1},\ldots ,h_{k-1})=h_{k}p\in J^{r}$. By Theorem \ref{rees}, $\{h_{k}\beta_{(\lambda_{1},\cdots,\lambda_{k-1})}\mid \lambda_{1}+\cdots+\lambda_{k-1}=r-1\}\subset J$. Given that $h_{1},\ldots, h_{k}$ be a regular sequence in $R$, we have $\{\beta_{(\lambda_{1},\cdots,\lambda_{k-1})}\mid \lambda_{1}+\cdots+\lambda_{k-1}=r-1\}\subset J$. Hence, $p\in J^{r}$.\qed
\medskip

\begin{theorem}\label{transideals}
Let $I$ and $J$ be ideals in $R$, such that
$J$ is generated by an $R/I$ regular sequence $g_{1},g_{2},\cdots, g_{k}$.
Then,
\begin{enumerate}[(i)]
\item $I\cap J=IJ$.
\item $I\cap J^{r}=IJ^{r}$, for all positive integers $r$.
\end{enumerate}
\end{theorem}

\proof\textbf{(i)} We use induction on the length $k$ of the $R/I$-regular
sequence generating the ideal $J$. For $k=1$, let $\alpha\in I\cap J$.
Then since $\alpha\in J$, we can write $\alpha = g_{1}r_{1}$ for some
$r_{1}\in R$. Therefore,
$r_{1}g_{1}\in I$ and $\overline{r_{1}g_{1}}= 0\in R/I$. The element
$\overline{g_{1}}$ is not a zero divisor in $R/I$, by hypothesis. Therefore,
$\overline{r_{1}}=\overline{0}\in R/I$ and hence $r_{1}\in I$. This shows that $\alpha = r_{1}g_{1}\in IJ$.
We assume that the statement is true for $k-1$. Let
$\alpha\in I\cap J$. Since $\alpha\in J$, we can write
$\alpha = r_{1}g_{1}+r_{2}g_{2}+\cdots+ r_{k}g_{k}$, for some
$\alpha_{1}, \ldots , \alpha_{k}$ in $R$. Now $\alpha\in I$, therefore $\overline{r_{k}g_{k}}\in R/ (I+ \langle g_{1},g_{2},\cdots, g_{k-1}\rangle)$.
The elements $\overline{g_{1}},\overline{g_{2}},\cdots, \overline{g_{k}}$ being
a regular sequence in $R/I$, we have
$\overline{r}_{k}=0$, that is, $r_{k}\in I+ \langle g_{1},g_{2},\cdots, g_{k-1}\rangle$.
Let $r_{k}= r_{1}^{'}g_{1}+r_{2}^{'}g_{2}+\cdots + r_{k-1}^{'}g_{k-1} + h$, where
$h\in I$ and $r^{'}_{1},r^{'}_{2},\cdots, r^{'}_{k-1}\in R$. Therefore,
$\alpha = (r_{1}+r_{1}^{'}g_{k})g_{1}+(r_{2}+r_{2}^{'}g_{k})g_{2}+\cdots + (r_{k-1}+r_{k-1}^{'}g_{k})g_{k-1}+ hg_{k}$. We have $hg_{k}\in I\cap J$. Hence it is
enough to show that $\alpha^{'}=\alpha- hg_{k}\in IJ$. Let $J^{'}= \langle g_{1},g_{2},\cdots, g_{k-1}\rangle$. Then $\{\overline{g_{1}},\overline{g_{2}},\cdots, \overline{g_{k-1}}\}$
being a part of a regular sequence, is a regular sequence in $R/I$. By the induction hypothesis, we have
$I\cap J'=IJ'$. Now $\alpha\in I$ implies that $\alpha -ig_{k}\in I$ and
therefore $\alpha'\in I$. Also, $\alpha'\in \langle g_{1},g_{2},\cdots, g_{k-1}\rangle= J^{'}$. Therefore $\alpha'\in I\cap J'= IJ'\subseteq IJ$.\qed
\medskip

\noindent\textbf{(ii)} We use induction on $r$. For $k=1$ the result trivially holds for all $r\geq1$ by (i). We assume by induction that for $I\cap J^{s}=IJ^{s}$
for all $1\leq s<r$. Now we
prove that the result
holds good for $s=r$. Let $x_{k}\in I\cap J^{r}$. Every element of $J^{r}$
can be written in the form
$$\sum_{i_{1}+i_{2}+\cdots+ i_{k}= r}\alpha_{i_{1},i_{2},\cdots, i_{k}}g_{1}^{i_{1}}\cdots g_{k}^{i_{k}}.$$
Therefore, we can write $x_{k} = g_{k}\gamma_{k}+ \beta_{k}$, where
$\beta_{k}\in \langle g_{1},\cdots, g_{k-1}\rangle^{r}$ and
$\gamma_{k}\in J^{r-1}$. We know that $g_{k}$ is not a zero divisor in
$R/(I+\langle g_{1},\cdots, g_{k-1}\rangle )$. It follows by \ref{nzd} that $g_{k}$
is also a non-zero divisor in $R/(I+\langle g_{1},\cdots, g_{k-1}\rangle^{r})$. We know that
$x_{k}\in I\subseteq I+\langle g_{1},\cdots, g_{k-1}\rangle^{r}$. Therefore,
$\overline{x_{k}}=\overline{0}$ in $R/I+\langle g_{1},\cdots, g_{k-1}\rangle^{r}$ and
hence $\overline{g}_{k}\overline{\gamma}_{k}=\overline{0}$. This proves that
$\overline{\gamma}_{k}=\overline{0}$ in $R/I+\langle g_{1},\cdots, g_{k-1}\rangle^{r}$.
Let $\gamma_{1}= i_{k}+\alpha_{k}$, where $i_{k}\in I$ and $\alpha_{k}\in \langle g_{1},\cdots, g_{k-1}\rangle^{r}$. Hence, $i_{k}=\gamma_{k}-\alpha_{k}\in I\cap J^{r-1}$ and therefore $i_{k}\in I\cap J^{r-1}=IJ^{r-1}$, by the induction hypothesis on $r$. It
follows that $g_{k}i\in IJ^{r}$. In order to show that $x_{k} = g_{k}(i_{k}+\alpha_{k})+\beta_{k}\in IJ^{r}$, it is therefore enough to prove that
$x_{k-1}:= g_{k}\alpha_{k} + \beta_{k}\in IJ^{r}$. We have $x_{k-1}\in I \cap \langle g_{1}, \ldots , g_{k-1}\rangle^{r}$. We continue this process to produce $x_{k-2}$,
$x_{k-3}, \ldots, x_{1}$, such that $x_{i}\in I \cap \langle g_{1}, \ldots , g_{k-i+1}\rangle^{r}$, for every $i=1, \ldots, k$. In particular, $x_{1}\in I\cap \langle g_{1}\rangle^{r} = I\cdot \langle g_{1}\rangle^{r}$. Then, it follows that $x_{2}\in
I\cdot \langle g_{1},g_{2}\rangle^{r}$. We can successively go back
and prove that $x_{k}\in IJ^{r}$.\qed
\bigskip

\section{Transversal Intersection with the Rational Normal Curve}
Let $S=K[x_{1},\ldots ,x_{n+1}]$, where $x_{i}$'s are indeterminates over
the field $K$. Let $N =\left(
\begin{array}{ccccc}
x_{1} & x_{2} & x_{3}\cdots &x_{n}\\
x_{2} & x_{3} & x_{4}\cdots &x_{n+1}\\
\end{array}
\right)
$ and let $I_{2}(N)$ denote the ideal generated by the $2\times 2$ minors of the
matrix $N$ in $S$. The ideal $I_{2}(N)$ is the defining ideal of the rational normal
curve in the projective space under the standard parametrization. Our aim
in this section is to show that some natural classes of ideals $J$ in the polynomial
ring intersect transversally with the ideal $I_{2}(N)$. This information helps us
write the minimal free resolution of the sum ideal $I_{2}(N)+J$, since the tensor product complex of the minimal free resolutions of $I_{2}(N)$ and $J$ resolve
$I+J$ minimally; see Lemma 3.7 in \cite{sst}. The main theorems in this section
are Theorems \ref{transversal_IJ} and \ref{transversal_XY}. We first prove the following
Lemmas.
\medskip

\begin{lemma}\label{nzd}
Let $I$ and $J$ be ideals in $S$, such that $J= \langle x \rangle$, where $x$
is a non-zero polynomial in $S$. Then $I\cap J = IJ$ if and only if $\overline{x}$ is not a zero
divisor in $R/I$.
\end{lemma}

\proof Let $\overline{xg}=\overline{0}$ in $R/I$. Then, $xg\in I$ and
also $xg\in J$. Therefore, $xg\in I\cap J =IJ$. We can write $xg=xg^{'}$, for
some $g^{'}\in I$. This shows that $x(g-g')=0$ an hence $g=g'\in I$. The
converse follows from Lemma \ref{transideals}.\qed
\medskip

\begin{lemma}\label{transversal_ab}
The ideals $I_{2}(N)$ and $\langle x_{1}^{a}+x_{n+1}^{b}\rangle $, where $a, b \in \mathbb{N}$, intersects transversally.
\end{lemma}

\proof The ideal $I_{2}(N)$ is kernel of the homomorphism $\zeta:S \rightarrow k[s,t]$ defined as $\zeta(x_{i})=s^{n-i+1}t^{i-1},\quad 1\leq i\leq n+1$. Therefore,
the ideal $I_{2}(N)$ is a prime ideal. Again, $x_{1}^{a}+x_{n+1}^{b}\notin I_{2}(N)$, since $\zeta( x_{1}^{a}+x_{n+1}^{b})=s^{na}+t^{nb}\neq 0$. Therefore, $\overline{x_{1}^{a}+x_{n+1}^{b}}$ is not a zero divisor in $R/I_{2}(N)$. Hence, by lemma \ref{nzd},
the ideals $I_{2}(N)$ and $\langle x_{1}^{a}+x_{n+1}^{b}\rangle$ intersect
transversally.\qed
\medskip

\begin{lemma}\label{nzd3}
Let $a, b$ be natural numbers. Then,
\begin{enumerate}[(i)]
\item The projective dimension of
$R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is $n$.

\item $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is Cohen-Macaulay.
\end{enumerate}
\end{lemma}

\proof \textbf{(i)} The minimal free resolution of $R/I_{2}(N)$
is given by the Eagon-Northcott complex of length $n-1$. The ideals
$I$ and $\langle x_{1}^{a}+x_{n+1}^{b}\rangle $ intersect transversally by
Lemma \ref{transversal_ab}. Therefore, a minimal free resolution of $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is given
by the tensor product complex of the minimal free resolutions of $R/I_{2}(N)$
and $R/\langle x_{1}^{a}+x_{n+1}^{b}\rangle$; see Lemma 3.7 in \cite{sstsum}.
Therefore, it follows that the projective dimension of $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is $n(= n-1+1)$.
\medskip

\noindent\textbf{(ii)} Projective dimension of
$R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is $n$, therefore, by
the Auslander Buchsbaum theorem ${\rm depth}(R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle))=1 $. Again $I_{2}(N)$ is a prime ideal of height
$n-1$ and $\overline{x_{1}^{a}+x_{n+1}^{b}}$ is a non-zero divisor
of $R/I_{2}(N)$, therefore, by the Krull's principal ideal theorem height
of $I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle$ is $n$. Hence,
$\dim(R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle))=1$ and
therefore $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ is Cohen-Macaulay.\qed
\medskip

\begin{lemma}\label{nzd4}
Let $a,b,c$ \, be natural numbers. Let $\mathfrak{p}$ be a prime ideal such that
$I_{2}(N) +\langle x_{1}^{a}+x_{n+1}^{b}, \, x_{n}^{c}\rangle \subset\mathfrak{p}$.
Then, $\mathfrak{p}=\langle x_{1},\ldots,x_{n+1}\rangle$ and hence the
height of the ideal $I_{2}(N) +\langle x_{1}^{a}+x_{n+1}^{b},x_{n}^{c}\rangle$
is $n+1$.
\end{lemma}

\proof We use induction to prove $x_{n-i}\in \mathfrak{p}$, for all $0\leq i\leq n-2$. Since $\mathfrak{p}$ is a prime ideal and $x_{n}^{c}\in I_{2}(N) +\langle x_{1}^{a}+x_{n+1}^{b},x_{n}^{c}\rangle \subset\mathfrak{p}$, therefore $x_{n}\in\mathfrak{p}$. Let us assume that $x_{n-i}\in \mathfrak{p}$ for some $i$. Again
$x_{n-i}x_{n-i-2}-x_{n-i-1}^{2}\in I_{2}(N) +\langle x_{1}^{a}+x_{n+1}^{b},x_{n}^{c}\rangle \subset\mathfrak{p}$. By the induction hypothesis $x_{n-i}\in \mathfrak{p}$,
therefore we have  $x_{n-i-1}\in \mathfrak{p}$. Therefore  $x_{n-i}\in \mathfrak{p}$ for all $0\leq i\leq n-2$.
\medskip

As $x_{1}x_{n+1}-x_{2}x_{n}\in\mathfrak{p}$ and $x_{n}\in\mathfrak{p}$ we have $x_{1}x_{n+1}\in\mathfrak{p}$ hence $ x_{1}$ or $x_{n+1}\in\mathfrak{p}$. If $x_{1}\in \mathfrak{p}$, then $x_{1}^{a}\in \mathfrak{p}$ which implies that $x_{n+1}^{b}\in\mathfrak{p}$. Hence $ x_{n+1}\in \mathfrak{p}$.  If $x_{n+1}\in \mathfrak{p}$, then similarly we can show that $x_{1}\in \mathfrak{p}$ and hence $\mathfrak{p}=\langle x_{1},\ldots,x_{n+1}\rangle$. \qed
\medskip

\begin{lemma}\label{nzd5}
$x_{n}^{l}$ is not a zero divisor in $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$.
\end{lemma}

\proof Suppose that it is a zero divisor, then it is contained in an associated prime ideal of $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$. But,
$R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ being Cohen-Macaulay, the
prime ideal has to be minimal. We know that the
height of the ideal $I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle$ is $n$. Hence,
any minimal prime ideal of $R/(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$
has height $0$. In other words, any minimal prime ideal containing $(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ has height n. But any prime ideal containing
both $(I_{2}(N)+\langle x_{1}^{a}+x_{n+1}^{b}\rangle)$ and $x_{n}$ has height
$n+1$ from the previous lemma. Hence it cannot be minimal.\qed
\medskip

\begin{theorem}\label{transversal_IJ}
Let $J$ denote the ideal $\langle x_{1}^{a}+x_{n+1}^{b}, \, x_{n}^{c}\rangle$ in the polynomial ring $S$, such that $a,b,c$ are in $\mathbb{N}$. Then, $I_{2}(N)\cap J = IJ$.
\end{theorem}

\proof We have seen in Lemmas \ref{transversal_ab} and \ref{nzd5} that
$\overline{x_{1}^{a}+x_{n+1}^{b}}, \, \overline{x_{n}^{c}}$ is a regular sequence
in $R/I_{2}(N)$. Hence, by Theorem \ref{transideals}, the ideals
$I_{2}(N)$ and $J$ intersect transversally.\qed
\medskip

\begin{theorem}\label{transversal_XY}
Let $T=k[x_{ij},y_{i}]$, where $i,j\in \{1,\ldots,n\}$  Let
\begin{center}
$X= \begin{pmatrix}
x_{11}&x_{12}&.&.&.&x_{1n} \\
x_{21}&x_{22}&.&.&.&x_{2n} \\
& &. &.& & \\
& &. &.& & \\
& &. &.& & \\
x_{n1}& x_{n2} &.&.&.&x_{nn} \\
\end{pmatrix}$
 and
$Y= \begin{pmatrix}
 y_{1} \\
 y_{2} \\
 . \\
 . \\
 y_{n}
 \end{pmatrix}$
\end{center}
be generic matrices of indeterminates $x_{rs}$ and $y_{k}$.  Let $f_{r} = \sum_{s=1}^{n}x_{rs}y_{s}$, for $1\leq r\leq n$. Then, $I_{1}(XY)=\mathcal{I} = \langle f_{1}, \ldots , f_{n}\rangle$.
\begin{center}
$$H= \begin{pmatrix}
  x_{i1}&x_{i2}&.&.& x_{i(i-1)}&x_{i(i+1)}&.&.&x_{i(n-1)}&x_{in} \\
 x_{i2}&x_{i3}&.&.& x_{i(i+1)}&x_{i(i+2)}&.&.&x_{in}&x_{pq} \\
 \end{pmatrix},$$
 $1\leq p,q,i\leq n$ and $p\neq i,p\neq q$
 \end{center}
Let $\mathcal{J}=I_{2}(H)$ be rational normal curve. Then $\mathcal{I}\cap \mathcal{J}=\mathcal{I}\mathcal{J}$.
\end{theorem}

\proof Let us consider the monomial order
\begin{itemize}
\item $x_{11}> x_{22}> \cdots >x_{nn}$;
\item $x_{rs}, y_{s} < x_{nn}$ for every $1 \leq r \neq s \leq n$.
\end{itemize}
Then $\{f_{1}, \ldots , f_{n}\}$ forms a Gr\"{o}bner basis of the ideal $\mathcal{I}$ and $\rm{Lt}(f_{r})=x_{rr}y_{r}$. Therefore,  $\mbox{\rm supp}(\rm{Lt}(\mathcal{I})) \cap \mbox{\rm supp}(\rm{Lt}(\mathcal{J}))=\emptyset$. Hence by Theorem \ref{dissup} $\mathcal{I}\cap \mathcal{J}=\mathcal{I}\mathcal{J}$. \qed
\bigskip

\bibliographystyle{amsalpha}

redpajama_set_name: RedPajamaArXiv


The global "Battery market" research report offers a comprehensive analysis and forecasts of the Battery market at both the global and regional level. It offers an in-depth and unbiased market review. The report highlights the main inclinations and services that play a key role in the development of the global Battery market for the estimated period. It also emphasizes on the market drivers and opportunities of the global Battery market for the upcoming period. In addition, it also highlights the dominating players in the market joined with their market share. The well-established players in the market are GuoXuan High-Tech, BYD, OptimumNano, CATL, PEVE, Panasonic, AESC, Tianjin Lishen, LG Chem, Pride Power, LEJ, CALB, A123 Systems, Samsung SDI.
The research report encompasses an in-depth value chain analysis by giving an extensive vision of the global Battery market. It uses Porter's Five Forces analysis to understand the competitive scenario of the global Battery market. The report describes the market dynamics and market trends, limiting factors, drivers, and opportunities for the current and future global Battery market. Additionally, the global Battery market report covers the major product categories and segments Lithium Battery, Lead-acid Battery along with their sub-segments Passenger Vehicle, Commercial Vehicle, Golf Car, Marine, Lawn & Garden, Recreational Vehicles, Others in detail.
In addition, the global Battery market research report demonstrates an analysis of varied business strategies implemented by leading players in the global Battery market that help for the development and expansion of their businesses. The study offers an entire outlook on the growth of the Battery market in terms of revenue (USD Million) across numerous regions including the Middle East & Africa, Asia Pacific, Europe, North America, and Latin America.
The global Battery market research includes the decisive analysis of the global Battery market by classifying it on the basis of product type, end user, and application segments. The development of every segment is assessed along with the forecast of their expansion in the near future. The pertinent facts and figures gathered from the regulatory authorities are presented in the global Battery research report to review the expansion of each segment.
Furthermore, the global Battery market is bifurcated on the basis of geographical regions such as Latin America, North America, Middle & East Africa, Europe, and the Asia Pacific. Numerous analytical techniques are used to study the expansion of the global Battery market. It also highlights key parameters of the dominating market players influencing the growth of the global Battery market along with their position in the global market.
Chapter 13, 14 and 15, Battery sales channel, distributors, traders, dealers, Research Findings and Conclusion, appendix and data source.


text:
Supporting employment
Adding value to the economy
Enabling trade
When the system stops
Tourism enabler
Community lifeline
Rapid disaster response
Reducing noise
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Aviation's impact on the environment
Building resilience to climate change
Aircraft programmes boost Asian high-tech
Social development Investment
Today's commercial aircraft are a vital catalyst to high-tech skills development throughout the world. The Asia-Pacific region, which leads global air traffic growth, has also been particularly boosted by investment from major aircraft manufacturers.
Boeing's industrial footprint spans major Asia-Pacific economies and is expanding to meet increases in the company's commercial production.
More than 65 Japanese companies support Boeing programmes, accounting for more than 40% – about 22,000 direct and indirect jobs – of Japan's aerospace employment. Japan builds 35% of the 787 Dreamliner airframe and more than 20% of the 777. Boeing collaborates with Japanese industry and universities to develop advanced manufacturing technologies and other innovations that will benefit Japan's current and future workforce.
Chinese companies supply every Boeing commercial programme, from the vertical fin and horizontal stabiliser for Boeing's Next-Generation 737 and 747-8 to the rudder, wing-to-body fairing panels and other components for the 787. Meanwhile, the aircraft maker has trained more than 50,000 pilots, technicians, factory workers and other professionals in China's fast-growing aviation industry.
The US manufacturer continues to expand its South-east Asia supply chain. In 2013, Boeing and Hexcel Corporation expanded their joint venture Aerospace Composites Malaysia (ACM) by 40%. ACM, which makes flight surfaces for Boeing commercial aircraft, is expected to grow its workforce beyond the current 950 employees.
Boeing has a global logistics hub in Singapore for spares distribution, and supply chain management and technical support teams throughout South-east Asia. The plane maker also partners with Indonesia on commercial aviation safety, efficiency and industry development.
Helping protect the World Biosphere Reserve of Sierra Gorda
Promoting gender equality at Airbus
ATR shows passenger inclusion innovation at AIX
Employees with disabilities: Safran harnesses 3D innovation
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This website is an initiative of the commercial aviation industry, represented by the Geneva-based Air Transport Action Group.
© Copyright ATAG. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


PUTRAJAYA: The Government has removed the 30% bumiputra equity requirement for 27 services sub-sectors under a liberalisation of the services sector.
Prime Minister Datuk Seri Najib Tun Razak, who said this would take effect immediately, added that the sub-sectors — which included the health and social services, tourism, transport, business and computer and related services — would have no equity conditions imposed.
More liberalisation measures are in the offing, he added.
Najib said the move was aimed at creating a conducive business environment to attract more investments, bring in more professionals and technology, encourage competitiveness and create higher value employment opportunities.
"We will be progressively undertaking liberalisation of the other services sub-sectors," he said at a press conference at his office, here, yesterday.
Noting that the services sector would become a new growth sector of the economy, Najib said it contributed 55% to the gross domestic product in 2008, and accounted for 57% of the country's total employment.
The Government, he said, intended to tap the sectors' full potential and raise its contribution to 60% of the GDP.
However, he gave an assurance that the liberalisation would not adversely affect the domestic services industry, which would continue to be supported, adding that a RM100mil services sector capacity development fund had been established under the first economic stimulus package.
To facilitate investments into the sector, a National Committee for Approval of Investments in the Services Sector had been established under the Malaysian Industrial Development Authority.
Najib also said the liberalisation was in line with Malaysia's commitment with Asean, adding, however, that some of the measures were better than those undertaken by others in the region.
The Prime Minister said that in 2008, approved investments in the services sector totalled RM50.1bil, exceeding the target of RM45.8bil per annum under the Third Industrial Master Plan. The share of foreign investments was 11% of the total investments.
"With the liberalisation of the services sector, we expect greater inflow of investments," he said.
Najib also said he would announce the liberalisation of the financial sector next week.


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text:
Mixed Reader
Category: News & Links
I Wasn't Raised for Revolution
Daddy you were a 29 year old black man in 1968.
You never told me. Where you were then. This year I hold up as a year of death of change of revolution. Why didn't you tell me? I don't know what kind of man you were.
I don't know if you stood up.
I don't know if you protested,
I don't know if you hid.
I don't know if you said yessir and stepped off the sidewalk,
Daddy I don't know these things, where your body was, what your body did.
But I do.
I know because as our people are being killed, my body wants to hide
to look down
to run
to defend myself
to follow all the rules
to collapse onto the ground and pray for mercy.
You gave me fear.
You think you gave me arrogance,
you think you gave me privilege by telling me to stay out of the sun, by white washing my world, but I teach my students everyday that truth is hollow without context.
I couldn't build that truth for myself because I didn't know why it mattered.
I filled the emptiness you gave me with stories of black heroes.
I gave myself the history my teachers didn't give me, and this gave me purpose.
I couldn't be white Daddy because you didn't tell me why that mattered. What you had seen. What you were afraid of.
Daddy, I'm standing here in a present that looks exactly like the past. Except there's more denial, except there's more pain, more at stake. I don't know what to do, or how to do, or if I can.
I am scared I am not enough.
I am scared to stand up alone,
I am scared to march in a crowd,
I am scared my voice is not as loud or powerful as those before me– scared to shout truths at a room full of strangers.
All this fear, and I wonder if it's you I'm afraid of defying. I wasn't raised for a revolution.
jamie-m	News & Links	Leave a comment July 8, 2016 July 8, 2016 1 Minute
4 Days Left! Support Mixed Remixed
Let's go! Donate here: https://www.indiegogo.com/projects/mixed-remixed-festival-2016#/
Reason #2: Listen, I don't know if you know, but Heidi Durrow is a force of nature. She's a woman I'm modeling myself after, she's that cool.
Her own bio can tell you better than I can, but she grew up the daughter of a Danish mother and an African American father. She became a lawyer and a journalist. If that wasn't badass enough, she went to write the novel, The Girl Who Fell From the Sky, which won the 2008 Bellwether Prize for Socially Engaged Fiction. This book is how I was introduced to Heidi.
I will gush about her novel all day and how important is was to me as a college student and a writer to have that book in my hands. It proved to me that there was a space for mixed stories. Her prose was so accessible and affecting, when I became a teacher, it was one of the first books I wanted to put in my students' hands.
Heidi has worked tirelessly to make space and uplift stories about the mixed experience. She founded a festival that invites writers, (both emerging and accomplished), filmmakers and community members to share these stories with each other. Through its evolution, we have the Mixed Remixed Festival. She also hosts a podcast called the Mixed Experience where she chats with mixed writers and analyzes issues in the mixed community.
What I love most about Heidi is her openness and dedication. She is incredibly welcoming and gracious. Mixed Remixed has a team of volunteers, but she is helm of it all. This is a woman to put your dollars behind because she is putting in work to strengthen our community, build understanding and forge connection.
Heidi Durrow is a reason to support the Mixed Remixed Festival.
See, look how hard I was feeling out the first time we met!
jamie-m	News & Links	Leave a comment March 25, 2016 March 25, 2016 1 Minute
5 Days Left to Support Mixed Remixed
We have 5 days and $1500 to raise. For the next five days, I want to tell you why this festival is important to me.
Reason #1: I've always felt like an outsider in my different communities. Those of you who know me know that I proudly identify as POC, as a black woman. However, I've found myself pushed away in from these communities through some words and actions of some people. Maybe becauseof the assumed privileges I have, (by which I do not mean to discount any of those privileges, in fact I work to acknowledge and confront them in my work), my authenticity was/is continuously questioned. This is especially tough for me in writing communities, where we often expose the most vulnerable parts of ourselves. When I was invited to read at the festival in 2011, my work was taken seriously, my experience was valued, and I was able to be surrounded by professional writers who were encouraging and supportive. It was (and still is) an honor to become the Literary Coordinator; to be able to help create/ maintain this space that embraces diverse, mixed experiences. This space has become sacred to me as a mixed person and a Mixed Writer. I love all my literary communities, however, the Mixed Remixed festival gave me the space to me in the truest way; it allowed me the space and support to do the work of loving my whole self.
Please share, please support the Mixed Remixed Festival. #IAmAStory
Donate here & get awesome perks: https://www.indiegogo.com/projects/mixed-remixed-festival-2016#/
jamie-m	News & Links	Leave a comment March 24, 2016 1 Minute
Interview for Our Small Faces
Recently, I was asked some questions about my novella, Our Small Faces, by writer/journalist George Kevin Jordan as part of a conversation with MFA students at Antioch University Los Angeles. Check it out below!
Also, links!
Find OUR SMALL FACES on Amazon.
Visit George Kevin Jordan's website.
Also, help him raise funds for summer writing workshops on Indiegogo!
George Kevin Jordan and I at Antioch University
Did you ever consider highlighting Leroy as one of the alternating points of view? Why/why not?
I did think of using Leroy's voice and tried writing a few chapters from his perspective. However, his voice always felt less emotionally resonant to me. I felt he was more distant, maybe because he felt more confined to South Park than Zeke or Selma did. I think he has less motivation than Selma, and less sensitivity than Zeke. So, I didn't feel his perspective was as compelling as the other two and actually weighed down the story.
Did you write the story from start to finish chronologically or did a particular scene/plot moment inspire the story and then you wrote it from the inside out? Did you always intend on it being a novella?
I did not plan for this to be a novella. My very first draft of this story, I was writing from the perspective of adult Selma looking back on her life. I had planned for it to be a short story, (one that was due for a workshop assignment). But later I woke up with a specific line in my head – one that placed this story in a younger voice and hinted to a larger story – and I couldn't help but follow that bait.
I did not write chronologically. I wrote this story in pieces and through several drafts. Tee's death was the first of the scenes that made it into the novella that I completed. The last one was the scene at the fair with Selma and Leroy. It was both frustrating and interesting to write in short vignettes and piece them together. I'd wait for a moment or line of inspiration and dive in, worrying about where it fit later on.
Is there any intention to make this a longer piece? If not, will these characters recur in other stories?
The novella as it stands is actually one half of my final manuscript from Antioch. What is presented in the book includes all the chapters I wrote from Zeke and Selma's points of view, but they do recur in the other chapters, which are in adult perspectives. (The other two voices in the manuscript are Johnny B and Tessa, who I believe are referenced in one of Selma's chapters). I'm working on shaping this second half, which has more of the historical context of northern California and the African American community there.
Tell us about the impetus for this story?
I grew up in Santa Rosa, the town where the story is based, and always felt an unspoken racial divide. As the town grew in population and areas became gentrified, it was interesting to see all this development happen around South Park, while the people of color were basically forced into this one neighborhood. And while this neighborhood was considered the "bad" part of town, it had a rich history. It was the first area that African American families settled in was South Park. They founded a church and later, began a chapter of the NAACP.
In a larger sense, it also occurred to me that people outside of California thought of Northern CA as this liberal bubble and didn't realize the kind of old-school racism happening here. While the movement of civil rights began in the early 60s in San Francisco (an hour south), it took years for those ideals to trickle up to the north bay.
Most of this history is used more explicitly in the adult point of view chapters that were part of the larger manuscript. From the younger point of view, I wanted to look at the immediate effects of this – the day to day instances of violence, how it begins to change a community, how it affects the sense of hope, etc.
There were hints that the story would tackle some of the issues of a bi-racial child, but other themes seemed to take center stage. Was that your intent?
Yes and no. I did want to include that in the story, but perhaps for selfish reasons. As someone of mixed race, I am really passionate about talking and writing about the mixed experience. So, I wanted this consciousness to be a part of Selma – especially in regards to colorism. I am fascinated by the way kids perceive race. Though I did want to explore that further, other themes took over. If I return to these characters that is a thread I'd like to pick up and follow through.
Tee's death. There was heated debate about the impact of that scene. Some say it happened so fast that it felt a little rushed. Others feel that is the true mirror of life that things like that happen in a blink of an eye. What is your impression of that scene? What was your intent? And how do you feel about that scene now?
I intended for it to happen quickly. I revised that section several times to cut words and be as concise as possible. I wanted to slow the reader down in that moment. When I read this scene aloud, I pause between paragraphs. When I get to the line "Tee tripped on a rock," I pause between sentences. So I wanted to try ti build that tension in a minimalist way. Also, I wanted to explore in the child's POV how a traumatic moment can feel both fast and in slow motion. That is why the focus turns into Zeke's body – his inability to hear, his eyes flashing between his parents. I believe moments like this happen in flashes – like memory building in real time. Re-reading the scene, I still feel that tension, and I'm not sure how I would expand it without losing that feeling. I don't think it would be true to Zeke's point of view.
Tell us a little about the process of getting this work published.
When I returned to the manuscript as a whole, I realized it was a lot to ask a reader to keep track of four points of view. So I began to think about dividing it. As I went in to revise chapters, I realized a few could stand on their own and was able to publish a few sections as short story pieces. "Island," the first chapter, was placed in Emerge Literary Journal. I was really impressed and encouraged by the editor's enthusiastic response, so when she founded a press, ELJ Publications, and put out the call for chapbook length work, I submitted the sections in Zeke and Selma's points of view (organized in a way I hoped made sense). Luckily, she felt excited about the whole manuscript and worked with me to shape it into the novella. I feel really lucky to be part of a small press that is both growing while also doing its best to make each author a priority.
Give us a little of your philosophy about MFA's and how it has or has not impacted your writing process.
One of the best things about the MFA program is that a writer has access to a community of people that understand the struggles of being a writer and generally want you to succeed. I wouldn't have a manuscript without the guidance of Alma Luz Villanueva, who became my mentor at a moment in time I was so full of doubt that I'm surprised I kept going. Even more important is the peer connections. I've taken away a few incredible friends who – because we all understand this literary language and each other's work – serve as the best readers and encourgers.
I don't believe every writer needs an MFA, but I feel like it was an important step for me. I needed that structure and the deadlines to teach me how to move from writing as a hobby to writing as part of my career. It taught me how to take my work seriously and how to engage with the literary community at a deeper and more enriching level.
Being a writer of color, what have been your challenges working on fiction, and what have been some the opportunities that have arisen because of writing about people of color.
I have continuously bee challenged by those who still carry the belief that stories about people of color are "less than" or can only speak to a specific audience. I had a mentor tell me directly that writing about the black community was holding my work back. I couldn't disagree more. I'd had workshop critiques that stuck on topics like, "I don't understand this slang term you used." I think this is because once you thwart the reader's first impression of who the character is (which most readers automatically default to white, male, POV), people become confused. The reading becomes clouded by either an exoticification of the culture or an unwillingness to find the space to relate. Obviously, and thankfully, this is not every reader, but I think it is really important to realize the biases we approach each piece of literature with.
The mainstream literary world continues to pigeonhole writers of color and as a writer of color it is assumed all of your work is about race and/or you speak for everyone of that race. It is frustrating, and I know for many writers, it has limited their ability for their work to reach a wider audience because editors, publishers, etc., can't seem to imagine that there are readers for this work. (Even as they are continuously proved wrong).
One of the benefits is that there is a strong community of writers of color, and spaces are being created for these writers to share work with each other and give critical feedback that moves beyond characterization and race. There are a few workshop programs like this. One of these is Voices of Our Nation Arts (VONA), which is the only multi-genre workshop for writers of color. I was able to attend this workshop when I started this manuscript and get great notes. Another, which I was just accepted to, is Kimbilio Fiction, which is a fellowship program for African American writers that is offers writers to return to the community several times and build relationships with established writers. Others programs like this include Cave Canem and Kundiman (both for poets). I am drawn to these communities to move beyond shallow workshop critique and get to the heart of our stories. I am shaped by the culture and community, as everyone is, and thus, these spaces are necessary to be able to express that.
What was your research process for this novella?
I did some research of local history. I pulled old news articles about the first sit-in protest at a bar downtown, about the founding of the town's historically black Baptist church, and about the founding of the NAACP. I interviewed some of the elders in the community about their experiences growing up in the segregated neighborhood and about being African American when Santa Rosa was still a very rural area. Having grown there, I culled this information together as I wrote.
Finally can you actually read your own work? Have you read this book now that it is in print? How do you connect with the work now?
I can! When the first copy of the novella arrived, I read it all through, suddenly motivated and nervous by the idea that others were going to be able to access this story. Of course, I found small parts I would've still edited and words I'd like to change around, but I was proud of the overall work.
I connect with it now as a reflection of my hometown, and of the way I construct my own childhood memories. I connect to the sense of lost hope, that geography can equal fate. I connect to the desire to find community and love despite frustration and hope, and finding the motivation to push on.
jamie-m	News & Links	Leave a comment May 26, 2014 May 26, 2014 8 Minutes
Zero Fade, Chris L. Terry
Kevin, the main protagonist of Zero Fade, is a middle school kid experiencing the peer pressures and social awkwardness any early teen can relate to. Facing school bullies and crushes on classmates, Kevin is trying to figure out who he is while he questions notions of race, masculinity and class. Chris L. Terry writes from a perspective grounded in urban sensibility and cross-cultural understanding.
Kevin's voice is sharp, funny, insecure and uncensored. He is fully immersed in mid-nineties hip hop and media, which reflects in his language and interactions with friends. Terry is on point with references I recognized from my own childhood, which added a nice layer of surprise while reading.
Kevin's family keeps him in check, including his Uncle Paul. Paul, the most important figure in his life beside his mother, serves as his example of manhood. So when Paul comes out to Kevin, his traditional ideas of manhood are challenged, forcing him to either reject his uncle or to redefine this ideas for himself.
Terry chooses to write from both the perspectives of Kevin and Paul, which provides a more complex portrait of the dynamics of their family and their emotional investment in each other. With Paul's perspective, the reader is allowed to follow the insecurity that anyone embracing a new identity faces. No matter his age, Paul's emotional journey is parallel to Kevin's. This opens up the story beyond the YA audience and asks the reader to remember the moments of vulnerability in their own life.
Instead of major conflict, Terry focuses on the emotional complexities of growing up and claiming self. He makes the characters deal with the interpersonal consequences of honesty. After a long day of 'normal' events, the characters have to face each other, which is maybe the hardest task of all. While things on the surface remain the same, their experiences change them beneath the surface. They can only reconcile differences through dialogue and care for each other.
Overall, this is an entertaining and worthy read that can appeal to both young and adult readers. Chris L. Terry is a mixed writer and educator, and a participant in this year's Mixed Remixed Festival. Find more info on him on his website, or visit him on Twitter. You can purchase the book on Amazon.
Also, Chris will be doing both a reading and a writing workshop at the Mixed Remixed Festival – both of which are FREE and open to the public. Don't miss it!
Lastly: Please donate to the Mixed Remixed Festival on Indiegogo! Only a few days left! Help us promote stories and films about the mixed experience!
jamie-m	News & Links	Leave a comment April 5, 2014 April 5, 2014 2 Minutes
This blog has not disappeared, but thanks to tons of projects, it has been a little neglected lately. But I come with a few updates!
Firstly: Mixed Remixed is a new, reincarnated festival to celebrate the mixed experience. Please visit the new site, and donate if you are able. Mixed Remixed is seeking submissions – calling all writers and artists – get involved! Share you story!
Secondly: My novella, Our Small Faces, will be published in two short weeks! Visit the Facebook page for more info.
I hope to get some essays up here very soon.
jamie-m	News & Links	Leave a comment November 2, 2013 1 Minute
Check out Alien Citizen & these other links
Elizabeth Liang forwarded me the trailer for her show, Alien Citizen. Here's the description:
Alien Citizen is a funny and poignant one-woman show about growing up as a dual citizen of mixed heritage in Central America, North Africa, the Middle East, and New England.
Be sure to check out more excerpts from the show on the website and find her on Twitter.
Friend and fellow writer Tara Dorabji on racial identity.
On Ebony.com, read this story of of balancing the intersections of race, sexuality & other parts of identity.
At the Native Heritage Project, a discussion of genealogy – what are we all really mixed with?
U.S. News writes about physical identification and the growing mixed population.
Intersecting racial and religion issues – how one writer learned about racism in church.
A look at Greg Carter's book The United States of the United Races.
jamie-m	News & Links	Leave a comment August 12, 2013 August 13, 2013 1 Minute
Aisha Sloan, The Fluency of Light
Sloan's collection of essays are written in a sharp, modernist style. Her book explores the complexities of race, culture, society and media. It's packed with close human moments – subtle interactions that have a profound impact on her development.
Find it here. Also check out this Publisher's Weekly review
jamie-m	News & Links, Reviews	Leave a comment June 27, 2013 June 27, 2013 1 Minute
Weekend Links
A new Cheerios ad features a young mixed girl and an interracial family:
Check out the racist-driven backlash on this article at Jezebel. And on Colorlines. And this one on Business Insider.
I'm currently reading The Fluency of Light by Aisha Sabatini Sloan. She was recently interviewed on Trop, an online literary site.
ABC Family's The Fosters premiers on June 3rd, and features a blended family. Check out the trailer below:
Ready for another hair story? Read this article by Damona Hoffman up at Huffington Post.
Here's a story about a father in Virgina who was accused of kidnapping his kids because "they didn't look like him."
Stay tuned for new reviews & reader submissions!
jamie-m	News & Links	Leave a comment May 31, 2013 May 31, 2013 1 Minute
How do you create opportunities for yourself? This question has nagged at me for weeks. Post MFA/grad school life is here and I've got to make something happen.
I have bigger hopes & aspirations for this blog and for my future as a writer/editor. I'm working to revamp this site really soon. I'm thinking about Mixed Reader on the regular: changing it up, making it a collaborative space, telling more stories, celebrating more stories. This is where my heart is, so I know it can be better.
jamie-m	News & Links	1 Comment January 31, 2013 1 Minute
RT @_Pedagogue_: 📚 ➡ ⏱ Book giveaway time! Winner announced Friday, 01/22. 2⃣ rules: 1) Follow us (@_Pedagogue_) 2) RT 🔁 Prison Peda… 3 days ago
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James McBride, The Color of Water
Nella Larsen, Passing
Publication Links
Works Cited & Links


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The Whitlams Quick Links
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Tim Freedman Charms The Love Boat In The Sawtell RSL's First Post-COVID Concert
MUSIC returned to the Sawtell RSL last Saturday night with an engaging performance by Tim Freedman, mainstay of The Whitlams.
Freedman's masterful piano playing, distinctive voice and sense of humour captivated the crowd.
Sawtell's own Kada Miller opened the show with a slow rendition of Men at Work's "Down Under".
She played and sang some original songs and her musicianship and powerful vocals drew an enthusiastic reception from the crowd.
The RSL had produced a grand piano for Freedman to play and he attacked it enthusiastically.
His songs tell incisive and often humorous stories about life and relationships, and his vocal style allowed every word to be clearly understood by listeners.
His 'patter' between songs elicited plenty of laughs.
As well as classics such as "No Aphrodisiac" and "I Make Hamburgers'" he performed The Whitlams' first new song in 14 years, "Ballad of Bertie Kidd" about a Rabbitoh's supporter who told Freedman he was involved in a bungled heist at a Gosford art gallery.
Freedman introduced the song by explaining that the gang of four were intending to steal paintings by Pro Hart, Norman Lindsay and Ken Done, but they put on their balaclavas too early.
Four men in balaclavas in a car on the quiet streets of Gosford were quickly noticed and apprehended.
After singing the song, Freedman said that he tracked down Bertie Kidd, the ringleader, in Hobart, and the 87-year-old appeared briefly in the song's video.
Freedman appeared to regard the Coffs Coast fondly.
He told the audience that The Whitlams last performed on the same stage at 'The Love Boat' fourteen years ago.
Freedman also said, "I wrote "Melbourne" during a sound check at the Hoey Moey in 1993."
The audience sang along to many of the songs and their obvious enjoyment of the entire perfomance indicated that live music from iconic acts has been sorely missed.
By Andrew Vivian
News of the Area – Modern Media
The Whitlams, Tim Freedman
Copyright © 2021 The Whitlams


This image gallery named as Sealy Frugal Hotspot might be ideal solution for your decoration. As the author has insert this image into Sealy Newfield Queen Mattress Set Costco post section, would ease you to find the perfect match of you decor desire.
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So Periodic Table ic table. Interactive periodic table with dynamic layouts showing names, electrons, oxidation, trend visualization, orbitals, isotopes, and compound search. Full descriptions from write-up sources.
The chemical elements of the periodic table sorted by symbol. Chemical elements listed by symbol The elements of the periodic table sorted by symbol. click on any element's name for further chemical properties, environmental data or health effects.


Works in the city's public defense system since 1946. He is considered an outstanding public defense worker countrywide. Specializing in criminal defense cases, Beniamin Vladimirovich proves himself to be a sophisticated law expert, a brilliant master of court eloquence, a wise and courageous defender of human dignity and civil rights.
Beniamin Bril is an officer of one of the units of the Leningrad Fleet, having dozens of military orders and medals.
In 1997 B.V. Bril has been awarded the "Lawyer of the Year" prize by the Presidium of the St. Petersburg City Bar. In 1998 Beniamin Vladimirovich has been awarded the F. Plevako Gold medal "for outstanding accomplishments in the defense of citizens' rights and freedoms" by Russia's community of public defense attorneys.
This year the City Bar has nominated Beniamin Vladimirovich a candidate for a premium "for achievements in law studies and practice" in the "law practice" category. B.V. Bril has also been nominated for an award "for many years of scrupulous work as an attorney specializing in criminal defense, for an important contribution to the development of public defense organizations in St. Petersburg and all Russia".


When you click on the Properties tab at the top of your screen, you will see a yellow bar beneath the top level tabs that reads View Properties for Asset Type (as shown below). It is important to understand how this bar works, as it operates a bit differently than you might expect.
You can filter the properties displayed by selecting one Asset Type from this list. For instance, if you select Short Sale, the list will display only Short Sale properties. You will notice that when you do this, the sub tabs shown at left (Search, New, etc.) will also change, which impacts which properties you can view. This is due to how your property statuses have been set up.
If you elect to view properties for an Asset Type of Short Sale, the sub tabs that display at left will only be those that use an Asset Type compatible with Short Sale (see our sample below). Our system is set up such that the Asset Type of Listed can be used for ALL Asset Types, therefore, this status option would show at left for Short Sale properties, as well as for any other type of property. Because we do not have any other statuses set up to be used with Short Sales, the only three status options that display at left are Search and New (which by default will display for all properties), and Listed, which we set up to show for all properties as well. If we had created an additional status to be used for Short Sale properties, we would also see that status at left.
Once you change the Asset Type in the yellow bar, it will stay that way until you run another filter.
1. Select the desired Asset Type from the View Properties for Asset Type field in the yellow bar (in our example below, we selected Short Sale).
2. Click the Search option from the Side Navigation Bar on the left side of the screen. The Property Search screen shown below will appear.
3. In the Property Status section at right, select the statuses you wish to view.
4. Click the green Search button at the bottom of the screen.


The cloud is becoming a popular tool for businesses across the country, with 88 percent of enterprises saying cloud improves responsiveness to business needs, 69 percent saying it has enabled them to re-engineer one or more of their business processes, and 65 percent saying it improves overall operations, according to Verizon Enterprise Solutions' 2016 State of the Market: Enterprise Cloud report.
In just a couple of years, the report predicts, half of all workloads–across organizations of all kinds–will be running in the cloud, it said.
And the costs are going down. One of the biggest changes in the cloud market is a dramatic fall in the barriers to entry of private cloud. This is largely being driven by advances in technology.
"Lower starting costs mean that private cloud is no longer only suitable for those with huge budgets. Even a relatively small number of servers can be economically viable as a private cloud," the document said in its third annual report on the cloud.
The survey took place over the summer and involved hundreds of Verizon's large enterprise and government customers.
• Cloud supports mission-critical workloads. Eighty-seven percent of enterprises use cloud for mission-critical workloads, up from 60 percent cited in Verizon's first report.
• Private cloud on the rise. Nearly half of enterprises (44 percent) are currently using, or plan to implement, a private cloud solution. Only slightly more than one in three respondents (37 percent) feel the same about public cloud.
• No one cloud fits all. More than half (53 percent) of enterprises use two to four cloud providers. And 26 percent use more than 10. Enterprises are embracing a multicloud strategy.
• Cloud use in the enterprise continues to grow. Eighty-four percent say their cloud use has increased in the past year. Half of enterprises say they will use cloud for at least 75 percent of their workloads by 2018.
• Security isn't an implementation impediment. In fact, 80 percent of enterprises say their cloud environment is as secure (40 percent) or more secure (20 percent say "a bit more secure," 20 percent say "much more secure") compared to their on-premise infrastructure.
• Keep the projects short. Six months, tops.
• Don't try to do it alone. Tap the skills of specialists.


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Elevation of West Front by Michael Craig, 2001
In the early 1790s, King's Inns was obliged to leave the buildings at Inns Quay (now the location of the Four Courts) that it had occupied from 1541. Some years elapsed before James Gandon was commissioned (in 1800) to produce drawings for the dining hall and library at Constitution Hill. This commission became "a favourite study" for Gandon and was to be his last public building in Dublin.
This print (No 5 in a series of five prints) is limited to an edition of 300 with each print individually numbered and signed by the renowned graphic artist Michael Craig.
Michael Craig has exhibited frequently in Dublin and also at the Royal Academy in London. Some of his best known works are the architectural views of buildings in the definitive series of Irish postage stamps as well as the series of stamps depicting the treasures of Ireland. He is the author of two illustrated books, Fish out of Water and, with Maurice Craig, Mausolea Hf'bernica.
Print dimensions: 16inches by 16inches (square).
Size unframed


Shehu Sani, the senator representing Kaduna Central has said that herdsmen responsible for massacre in Benue and other parts of the country should be tagged terrorists.
The Kaduna lawmaker said this in reaction to the recent killings in the country.
He further promised to support the motion to fund the security agencies on the floor of the senate in order to halt killings.
Sani on his Twitter page wrote, "Herdsmen who commit massacre in Benue and Taraba and 'Bandits' who commit massacre in Zamfara and Birnin Gwari in Kaduna state ,are nothing but terrorists;except if we want to call a spade a pestle or a spoon.
Recall that the Bishop of the Diocese of Badagry, Church of Nigeria (Anglican Communion ), Rt. Rev. Babatunde Adeyemi, had criticised President Muhammadu Buhari for not declaring the killer herdsmen as terrorists because they are Fulani.
The cleric called on Buhari to declare the killer herdsmen as terrorists, owing to the number of killings they had perpetrated.


if the Europe financial situzation will not deteriorate.
if QEx can fix the problems, then we don't even need QEII.
so that the economy can have reasonable growth.
the market and do not have any QEIII in any time soon.
But this does not mean the interest rate will go up soon.
My prediction is that the rate will go up from next Summer.


I really appreciate how you are able to get so much in a quick time. I hate long scrolling soliloquies. Great Work!
Well of course EVERYONE WILL one day. Same for me! Just keep SWIMMING!
I have read several good stuff here. Certainly price bookmarking for revisiting. I wonder how so much attempt you put to create the sort of wonderful informative website.
I do believe Abhi's aunt is going to give his tail a crackin in chapter 2!
You're a very clever 13-year-old. My son is 13 and likes to play this Minecraft video game. It would be nice if he would write stories like you. I enjoyed Chapter 1. Well done!
Incredible that you're only 13 years old! Your blog is great; it's a fun read and I'm honestly in awe at how young you are yet how dedicated you are to writing. Keep it up!
I admire your vision and guts, Super-Kam 007 !
I, Kamakhya Deori…..The Writer Of Super-Kam007 gives you permission to Re-blog It.
Hi Super-Kam007, good stuff keep it up.
I just had one question, in your post you talk about the fearsome trio, I guess people are afraid of them then, or did you mess fearless trio that are not afraid of anything? Just wondering!
It is just a phrase!! To spice up the story a bit!!
Hi Kam ! Your blogs are entertaining me. I love it.
Hey Kam, my fingers don't tremble anymore after I open the chapters one by one….. It was because your are still 13 yo. Incredible !


Our top priority has always been to print beautiful letterpress stationery for our clients. It is equally important to us to provide excellent service to everyone we work with. We want our customers to really enjoy the process of ordering custom stationery so it is an honor to receive this Couples' Choice award from Wedding Wire. We appreciate all of the people who have trusted us to print invitations for their special day and we look forward to collaborating with new couples this year!
We are so excited to see one of our favorite couple's wedding featured on Carats and Cake site this week! Claire and Tom were a dream to collaborate with! Their save the date announcements featured a cow letterpress printed in a pale tint of soft blue, a nod to the wedding venue, Blue Hill at Stone Barns. They chose a soft sage colored floral design for their invitations to match the gorgeous flowers that would surround them on their wedding day. We worked closely with them for a year and it is great to see the pictures of this special couple on their wedding day!
Sesame Letterpress featured in Japanese Book!
The Romance of Old Machines: Brooklyn's Sesame Letterpress from ANIMALNewYork.com on Vimeo.
We love the pattern printed on the front and end papers of the book...isn't it pretty?
One of our bespoke, monogrammed coasters has been featured in Martha Stewart Real Weddings! A great complement to any formal dining experience or a reception bar.
A couple of our wedding announcements were recently featured on the Martha Stewart Weddings "inspiration" section. Have a look at some great ideas from us and others!
Not too long ago, we printed some darling notecards that were featured in a photo shoot in Real Simple. Have a look!


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Diego Luna Net Worth, Height, Family, Age, Weight
Diego Luna Family & Relatives
How tall is Diego Luna measurements
Diego Luna Net Worth & Money Factor
Diego Luna Net Worth $1.5 Million To $10 Million (Approximate)
Diego Luna Education
Diego Luna Contact Address
Diego Luna Social Media Profile
Quick Facts: Diego Luna
Diego Luna Biography, Height, Weight, Age, Measurements, Net Worth, Family, Wiki & much more! Diego Luna was born on Toluca, Mexico 29 Dec 1979 in and her current age 40 years 11 months 28 days . Diego Luna Weight 78 KG and height 1.78 m. he is an Actor in Mexico. Her born home city of Toluca, Mexico. Her primary Profession is an Actor. Right now Diego Luna is a famous Actor in the world. And her Nationality is Mexican.
Diego Luna Actor
Age 40 years 11 months 28 days
BirthDate 29 Dec 1979
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Birth Place: Toluca, Mexico
Residence: Mexico, Mexican
Age: 40 years 11 months 28 days
Home Town: Toluca
Birth Name: Diego Luna
Nickname: Diego Luna
Nationality: Mexican
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Height in Feet 1.78 m
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Following to IMDb, Wikipedia, Forbes & Various many Online document resources and report, familiar Actor Diego Luna's net worth is $1.5 Million To $10 Million (Approximate) at the age of 40 years 11 months 28 days . he acquired the most of money being an professional Actor. he is from Mexico.
At present We don't have adequate information about Diego Luna Monthly & Yearly Salary, etcetera. We will update you soon on Our Site.
Net Worth: $1.5 Million To $10 Million (Approximate)
Source of Earnings: Actor
Last Year Earnings: 12X $1.5 Million To $10 Million (Approximate)
Address: Toluca, Mexico
Q: Where is the birthplace of Diego Luna?
A: Toluca, Mexico
Q: What is the nationality of Diego Luna?
A: Mexican.
Q: Which ethnicity does Diego Luna belong to?
A: Mexican
Q: What is the color of Diego Luna's eyes?
Q: How tall is Diego Luna?
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Dear User, We Shared of Actress Diego Luna Wiki, Biography, Age, Height, Husband, Hot Photos details. Do you have any questions about her Biography and Wiki? Let me know in a comment. Please keep visit on this website for more Actress biographical post updates.