text:
Quarry's Ex
by Max Allan Collins
Part of the Hard Case Crime Series
Quarry's Ex Synopsis
Even the enigmatic hit man called Quarry had to start somewhere. And for him that was the day he returned from Vietnam to find his young wife cheating. He'd killed plenty overseas, so killing her lover was no big deal. When he was recruited to use his skills as a contract killer, that transition was easy, too. He survived by expecting trouble. What he didn't expect was running into her again.
Quarry's Ex Press Reviews
Praise for Max Allan Collins: Violent and volatile and packed with sexuality...classic pulp fiction. - USA Today Collins' witty, hard-boiled prose would make Raymond Chandler proud. - Entertainment Weekly No one can twist you through a maze with as much intensity and suspense as Max Allan Collins. - Clive Cussler
Publication date: 30th September 2011
Author: Max Allan Collins
Publisher: Hard Case Crime an imprint of Titan Books Ltd
Categories: Crime & mystery,
About Max Allan Collins
Author of Road to Perdition, the acclaimed graphic novel that inspired the movie, and of the multiple-award-winning Nathan Heller series of historical hardboiled mysteries, Max Allan Collins is one of most prolific and popular authors working in the field today. He is also the literary executor of Mickey Spillane.
More About Max Allan Collins
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redpajama_set_name: RedPajamaCommonCrawl


text:
\section*{Introduction}

A tetrahedral curve $C= C(a_1,...,a_6)$ is a curve in ${\mathbb P}^3$ defined by the  ideal
$$I =(x_1,x_2)^{a_1}\cap (x_1,x_3)^{a_2}\cap (x_1,x_4)^{a_3}\cap
(x_2,x_3)^{a_4}\cap (x_2,x_4)^{a_5}\cap (x_3,x_4)^{a_6} $$
of the polynomial ring $R=K[x_1,x_2,x_3,x_4]$ over a field $K$,
where $a_1,...,a_6$ are non-negative integers and not all of  them
are zero. The case $a_2=a_5=0$ was first considered by Schwartau
\cite{S}. He gave  a characterization of the Cohen-Macaulay
property of $C$ in terms of  $a_1,a_3,a_4,a_6$. The general case
of  tetrahedral curves, when $a_2$ and $a_5$ are not necessarily
zero,  was  introduced in \cite {MN}. Using basic double linkage,
Migliore and Nagel gave there an efficient numerical algorithm for
determining when a particular tetrahedral curve is arithmetically
Cohen-Macaulay and asked for an explicit characterization in terms
of $a_1,...,a_6$.  This problem was solved later by Francisco  in
\cite{F}.  Moreover,  it was  shown in the papers \cite{MN,FMN} that
these curves have  many nice properties.

In this paper we study the structure of the first local cohomology module
$H^1_{\mathfrak m}(R/I)$ with the support in the maximal  homogeneous ideal
${\mathfrak m} = (x_1,x_2,x_3,x_4)$.  This study is important because we can characterize many properties,
such as the Cohen-Macaulayness  or the Buchsbaumness,  of $C$  in terms of $H^1_{\mathfrak m}(R/I)$.

Recall that the diameter of a ${\mathbb Z}$-graded
module $M$ of finite length is  the integer $\operatorname{diam}(M) = \max\{n|\
M_n\neq 0\} - \min\{n|\ M_n\neq 0\} +1$ ($\operatorname{diam}(M) := 0$ if $M=0$).
Let $J$ be the defining ideal of an arbitrary projective curve $X$
in ${\mathbb P}^3$. Then the module  $H^1_{\mathfrak m}(R/J)$ is of finite length
and let $ k(R/J)$  be the  smallest non-negative integer $k$ such
that ${\mathfrak m}^k H^1_{\mathfrak m}(R/J)=0$ (see \cite{MM,BCFH}).  It is obvious that  $k(R/J)\leq
\operatorname{diam}(H^1_{\mathfrak m}(R/J))$.
The main result of this paper states that $k(R/I) =  \operatorname{diam}
(H^1_{\mathfrak m}(R/I))$ for an arbitrary  tetrahedral curve (see Theorem
\ref{C5}). Thus our result implies that for all
tetrahedral curves, $\operatorname{diam}
(H^1_{\mathfrak m}(R/I))$ has no gap and $k(R/I)$ is, in this sense, as large as
possible. (Note that  monomial curves in
${\mathbb P}^3$ also have this property, see \cite{BCFH}.) Moreover, we
can  explicitly compute $\operatorname{diam} (H^1_{\mathfrak m}(R/I))$ in terms of
$a_1,...,a_6$ (see Theorem \ref{C2} and Theorem \ref{C5}). Since $C$ is an arithmetically Cohen-Macaulay  curve if and only if $\operatorname{diam}
(H^1_{\mathfrak m}(R/I)) = 0$, this result  is much more general than the Francisco's one in \cite{F}. In particular, it also enables us to determine all arithmetically Buchsbaum tetrahedral curves (Theorem \ref{C7}), thus extending  Corollary 5.4 in \cite{MN}.

Our approach is to reduce the above question to a problem in
integer programming. First, based on a
description of local cohomology modules of monomial ideals given
recently in \cite{T},  we reduce the problem to describing the set of integer solutions of a certain linear constraints.   Then using the well-known Fourier-Motzkin
elimination we can determine when the set of solutions is empty (Theorem \ref{C2}). This is corresponding the case of arithmetically  Cohen-Macaulay curves. If this set is not empty, we can still use it to determine the module structure of the first  local cohomology   (Proposition \ref{C4}). Thus our result is not only   an interesting application of integer programming to Commutative Algebra, but it also shows the usefulness of Takayama's formula in \cite{T}.  We believe that  Takayama's formula, which is a generalization of Hochster's formula, can be applied in many other situations.

The paper has four sections with the current one being an
introduction. In Section \ref{A} we recall  the main result of
Takayama in \cite{T} and relate the  problem of describing
$H^1_{\mathfrak m}(R/I)$ to a problem in integer programming (Lemma
\ref{A4}). In Section \ref{B} we apply the Fourier-Motzkin
elimination to solve that integer programming problem. The structure of
the first local cohomology module is given in the last Section
\ref{C}, where the main Theorem \ref{C5} is proved and some of its
consequences are derived.

\section{Preliminaries}\label{A}

Let $I \subset R= K[x_1,...,x_n]$ be a monomial ideal. Denote by
$G(I)$ the minimal set of monomial  generators of $I$. Let $\Delta
$ be the simplicial complex corresponding to the radical ideal
$\sqrt{I}$, i.e.
$$\Delta = \{ \{i_1,...,i_k\}\subseteq \{1,...,n\}|\ x_{i_1}\cdots x_{i_k} \not\in
\sqrt{I}\}.$$
A simplicial complex is uniquely defined by the set $\operatorname{Max}(\Delta
)$ of its facets.  Following \cite{T}, for $\underline{\alpha}  =
(\alpha_1,...,\alpha_n)\in {\mathbb Z}^n$, we set
$$G_{\underline{\alpha}} = \{ i|\ \alpha_i <0 \},$$
and
$$\begin{array}{ll} \Delta_{\underline{\alpha}} = \{ F\subset  \{1,...,n\}\setminus
G_{\underline{\alpha}} |\  & \text{ for\ all}\ \underline{x}^{\underline{\beta}} = x_1^{\beta_1}\cdots
x_n^{\beta_n} \in G(I) \  \text{there\ exists}
\\ & \ i \not\in F\cup G_{\underline{\alpha}}\ \text{such\ that}\ \beta_i > \alpha_i \geq 0\}.
\end{array}$$

\begin{Lemma}\label{A1} Denote by $I_{(x_{i_1}...x_{i_k})}$ the monomial
ideal generated by $I$
in the localization $K[\underline{x}]_{(x_{i_1}...x_{i_k})}$ w.r.t. the set  of all monomials in the variables
$x_{i_1},...,x_{i_k}$. Then
 $$\Delta_{\underline{\alpha}} = \{ F\subset \{1,...,n\} \setminus  G_{\underline{\alpha}} |\
\prod_{i\not\in F \cup G_{\underline{\alpha}}}
  x_i^{\alpha_i} \not\in I_{ (\prod_{j\in F\cup G_{\underline{\alpha}}} x_j)} \}.$$
 \end{Lemma}

 \begin{pf} For simplicity we may assume that $F\cup G_{\underline{\alpha}}= \{1,...,r\}$.
For a monomial $m\in K[x_1,...,x_n]$
 let $m' \in K[x_{r+1},...,x_n]$ be the monomial obtained from $m$ by deleting all
powers of $x_i,\ i\leq r$.
 Let $G' = \{ m'|\ m\in G(I)\}$. Then $G'$ is a generating set of $I' :=
I_{(x_1...x_r)}$. Note that the monomial
  $\prod_{i>r}x_i^{\alpha_i} \in I'$ if and only if there exists $m' = \prod_{i>r}
x_i^{\beta_i} \in G'$ such
   that $\beta_i \leq \alpha_i$ for all $i>r$, or equivalently, there exists $m
=\prod_{i=1}^n x_i^{\beta_i} \in G(I)$ such
   that $\beta_i \leq \alpha_i$ for all $i>r$. From that we immediately get the claim.
 \hfill $\square$.
 \end{pf}

 Note that all local cohomology modules $H^i_{\mathfrak m}(R/I),\ i\geq 0$, inherit a natural
${\mathbb Z}^n$-grading.  Theorem 1 in \cite{T} can be
 reformulated as follows.

 \begin{Lemma} \label{A2} Let $\rho_i = \max\{ \beta_i|\ \underline{x}^{\underline{\beta}}\in
G(I)\}$. For all $i\geq 0$ and $\underline{\alpha} \in {\mathbb Z}^n$ we have
 $$\dim H^i_{\mathfrak m}(R/I)_{\underline{\alpha}} = \begin{cases} \dim
\tilde{H}_{i-|G_{\underline{\alpha}}| -1}(\Delta_{\underline{\alpha}},K)
 & \text{if}\ G_{\underline{\alpha}} \in \Delta \ \text{and}\ \alpha_j \leq \rho_j-1,\ j\leq n,\\
 0 &\text{otherwise}.
 \end{cases}$$
 \end{Lemma}

 From now on we consider ideals of tetrahedral curves
 $$I =(x_1,x_2)^{a_1}\cap (x_1,x_3)^{a_2}\cap (x_1,x_4)^{a_3}\cap
(x_2,x_3)^{a_4}\cap (x_2,x_4)^{a_5}\cap (x_3,x_4)^{a_6} $$
of the polynomial ring $R=K[x_1,x_2,x_3,x_4]$.

\begin{Lemma} \label{A3} If $H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} \neq 0$, then $\alpha_i
\geq 0$ for all $i\geq 1$ and
$$\operatorname{Max}(\Delta_{\underline{\alpha}}) =\{ \{ 1,i\}, \{j,k\}|\ \{i,j,k\} = \{2,3,4\} \}.$$
\end{Lemma}
\begin{pf} Assume $H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} \neq 0$. By Lemma \ref{A2},
either $G_{\underline{\alpha}} = \emptyset $ and $\Delta_{\underline{\alpha}}$
is disconnected, or $|G_{\underline{\alpha}}| =1$ and $\Delta_{\underline{\alpha}}= \{\emptyset \}$.

If $|G_{\underline{\alpha}}| =1$, w.l.o.g. we may assume that $G_{\underline{\alpha}}=
\{1\}$, i. e. $\alpha_1 <0$ and $\alpha_2, \alpha_3, \alpha_4 \geq
0$.  By Lemma \ref{A1}, $\Delta_{\underline{\alpha}}= \{\emptyset \}$ is
equivalent to the following two conditions

(i) $x_2^{\alpha_2}x_3^{\alpha_3}x_4^{\alpha_4} \not\in I_{(x_1)} =
(x_2,x_3)^{a_4}\cap (x_2,x_4)^{a_5}\cap (x_3,x_4)^{a_6} $, and

(ii) $x_i^{\alpha_i}x_j^{\alpha_j} \in I_{(x_1,x_k)}$ for all $\{i,j,k\} = \{2,3,4\}$.

This is impossible, because
$$ {\rm (i)} \Leftrightarrow \left[ \begin{array}{l} \alpha_2 + \alpha_3 \leq a_4-1, \
\text{or}\\
\alpha_2 + \alpha_4 \leq a_5-1, \ \text{or}\\
\alpha_3 + \alpha_4 \leq a_6-1,
\end{array}\right.
\ \ \text{and}
\ \ {\rm (ii)} \Leftrightarrow \left\{ \begin{array}{l} \alpha_2 + \alpha_3 \geq a_4, \
\text{and}\\
\alpha_2 + \alpha_4 \geq a_5, \ \text{and}\\
\alpha_3 + \alpha_4 \geq a_6.
\end{array}\right.$$
Hence we must have $G_{\underline{\alpha}} = \emptyset $ and
$\Delta_{\underline{\alpha}}$ is disconnected. The first  condition implies
that $\alpha_i \geq 0$ for all $i\geq 1$. Since
$\Delta_{\underline{\alpha}}$ is a disconnected simplicial complex on  a subset of
$\{1,2,3,4\}$, in order to show the second statement of the lemma
  it suffices to show that $\Delta_{\underline{\alpha}}$ does
not contain a facet consisting of a single point. Assume, by
contrary, that $\{1\}$ is a facet of $\Delta_{\underline{\alpha}}$. Then we
again get (i) and (ii) (the only difference now is that all
$\alpha_i \geq 0$ which, however, have no effect on (i) and (ii)).
This is  a contradiction. \hfill $\square$
\end{pf}

As an example let us consider the well-known Buchsbaum curve defined by $I= (x_1,x_2) \cap (x_3,x_4)$. In this case
$H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} \neq 0$ if and only if $\underline{\alpha} = (0,0,0,0)$. We have $\operatorname{Max}(\Delta_{(0,0,0,0)}) =\{ \{ 1,2\}, \{3,4\}\}$.

\begin{Lemma} \label{A4} Fix an integer $d$. Assume that $\operatorname{deg}(\underline{\alpha}) :=
\alpha_1 + \cdots + \alpha_4 = d$.
Then $\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{\{1,2\}, \{3,4\}\}$ if and only if $\underline{\alpha}$
satisfies the following system of inequalities
\begin{equation} \label{EA4}  \begin{array}{rl}
\alpha_1 + \alpha_3 & \geq a_2 \\
\alpha_1 + \alpha_4 & \geq a_3 \\
\alpha_2 + \alpha_3 & \geq a_4 \\
\alpha_2 + \alpha_4 & \geq a_5 \\
\end{array}
\end{equation}
$$ \begin{array}{rl}
\alpha_1 + \alpha_2& \leq a_1-1\\
\alpha_3 + \alpha_4 & \leq a_6-1\\
 \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 & =d\\
 \alpha_1 , \alpha_2 , \alpha_3 , \alpha_4 &\geq 0.
\end{array}
$$
In this case $\dim H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} =1$.
\end{Lemma}

\begin{pf} The condition $\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{\{1,2\}, \{3,4\}\}$  implies
$G_{\underline{\alpha}} = \emptyset$,
i.e. $\alpha_1 , \alpha_2 , \alpha_3 , \alpha_4 \geq 0$. By Lemma
\ref{A1}, $\{1,2\} \in \Delta_{\underline{\alpha}}$ if and only  if
$x_3^{\alpha_3}x_4^{\alpha_4} \not\in (x_3,x_4)^{a_6}$, or
equivalently, $\alpha_3 + \alpha_4 \leq a_6-1$. Similarly,
$\{3,4\} \in \Delta_{\underline{\alpha}}$ if and only if $\alpha_1 +
\alpha_2 \leq a_1-1$. On the other hand, $\{1,3\}, \{1,4\},
\{2,3\}, \{2,4\} \not\in \Delta_{\underline{\alpha}}$ are equivalent to the
first four inequalities given above. Thus,
$\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{\{1,2\}, \{3,4\}\}$  implies
(\ref{EA4}). The converse is also clear from these arguments.

When $\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{\{1,2\}, \{3,4\}\}$ we have
$\tilde{H}_0(\Delta_{\underline{\alpha}},K) \cong K$ and $|G_{\underline{\alpha}}|=0$.
Hence,  by Lemma \ref{A2}, $\dim H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} =1$, as
required. \hfill $\square$
\end{pf}

\section{Fourier-Motzkin elimination} \label{B}

By Lemma \ref{A4} we are interested in finding an integer solution of the
following system of inequalities
\begin{equation} \label{EBI} \begin{array}{rl}
y_1 + y_3 & \geq a_2  \\
y_1 + y_4 & \geq a_3 \\
y_2 + y_3 & \geq a_4 \\
y_2 + y_4 & \geq a_5 \\
y_1 + y_2& \leq a_1-1\\
 y_3 + y_4 & \leq a_6-1\\
 y_1 + y_2 + y_3 + y_4 & =d\\
 y_1 , y_2 , y_3 , y_4 &\geq 0.
\end{array}
\end{equation}
For this purpose we apply the Fourier-Motzkin elimination which at
first enables to find a real solution  of a system of linear
equalities and inequalities, see, e.g. \cite{DT}, Section 2.3. We
sketch here the algorithm by considering a concrete example.

{\it Example}.  Consider the system
\begin{equation} \label{EB1}  \begin{array}{rl}
y_1 + 2y_2 - y_3 + 4 & \geq 0\\
- 2y_1 + y_2 + 3y_3 - 2&\geq 0\\
2y_2 - y_3 &\geq 0\\
y_1 &= y_2+y_3.
\end{array}
\end{equation}
First, replace the equality $y_1 = y_2+y_3$ by two inequalities
$y_1 \geq  y_2+y_3$ and $y_1 \leq  y_2+y_3$.  The obtained system
is not reduced w.r.t. $y_1$, i.e. $y_1$ appears with a non-zero
coefficient in at least one inequality. After dividing by the
absolute value of the coefficient of $y_1$  when nonzero and
rearranging the terms and the order of the constraints, we can
then partition them  in 3 groups, depending on whether in a
particular constraint $y_1$ is on the right or the left hand, or its $y_1$-coefficient
is zero.
$$ \begin{array}{rlr}
\frac{1}{2} y_2 + \frac{3}{2} y_3 - 1 & \geq y_1 & \hskip1cm (E1)\\
y_2 + y_3 &\geq y_1 & (E2)\\
y_1 &\geq  - 2y_2+ y_3 - 4  & \ (E3)\\
y_1 &\geq  y_2 + y_3  & (E4)\\
2y_2 - y_3 &\geq 0. &(E5)
\end{array}$$
Combining each inequality  in the first group  $\{ (E1),\ (E2)\}$
with another one  in the second group  $\{ (E3),\ (E4) \}$  and keep
all inequalities  in the third group ($\{(E5) \}$ in this example), we
obtain a new system of inequalities
$$ \begin{array}{rlr}
\frac{1}{2} y_2 + \frac{3}{2} y_3 - 1 & \geq  - 2y_2+ y_3 - 4  & \hskip1cm
(E1,E3)\\
\frac{1}{2} y_2 + \frac{3}{2} y_3 - 1 & \geq  y_2 + y_3  & (E1,E4)\\
y_2 + y_3 &\geq - 2y_2+ y_3 - 4  &  (E2,E3)\\
y_2 + y_3 &\geq  y_2 + y_3  & (E2,E4)\\
2y_2 - y_3 &\geq 0. &(E5)
\end{array}$$
The temporary label $(E1,E3)$ means that this inequality appears
by combining $(E1)$ and $(E3)$. Note that in  the last system,
$(E1,E3)$ follows from $(E1,E4)$ and $(E2,E3)$.  For short, we
will write this reduction as $(E1,E4) +(E2,E3)\Rightarrow
(E1,E3)$. The  constraint  $(E2,E4)$ trivially holds. We say that
$(E1,E3)$ and $(E2,E4)$ are redundant.  Deleting the redundant
inequalities, we finally get the system
\begin{equation} \label{EB2}  \begin{array}{rl}
\frac{1}{2} y_2 + \frac{3}{2} y_3 - 1 & \geq  y_2 + y_3  \\
y_2 + y_3 &\geq - 2y_2+ y_3 - 4  \\
2y_2 - y_3 &\geq 0.
\end{array}
\end{equation}
Thus  (\ref{EB1}) implies (\ref{EB2}), where $y_1$ appears with
zero coefficient in all inequalities.  We say that $y_1$ has been
"eliminated". The process is repeated with the new system except
now $y_2$ is eliminated.

We now apply the Fourier-Motzkin elimination to our system (\ref{EBI}). First
rewrite it in the form
\begin{equation} \label{EBII} \begin{array}{rl}
a_6-1 -y_3 &\geq y_4 \\
y_4 &= d -y_1 - y_2 - y_3 \\
y_4 & \geq a_3 -y_1 \\
y_4  &\geq a_5 -y_2  \\
y_4 &\geq 0\\
y_1 + y_3 & \geq a_2\\
 y_1 + y_2& \leq a_1-1\\
 y_2 + y_3 & \geq a_4\\
 y_1 , y_2 , y_3 &\geq 0.
\end{array}
\end{equation}
Eliminating $y_4$ we then get
\begin{equation} \label{EBIII}  \begin{array}{rl}
d-a_3-y_2 &\geq y_3 \\
d-a_5-y_1& \geq y_3  \\
d-y_1-y_2 & \geq y_3 \\
y_3&\geq 0\\
y_3 & \geq a_2-y_1\\
y_3 & \geq a_4-y_2\\
 y_1+y_2+a_6-d-1 &\geq 0\\
 y_1+y_2 &\leq a_1-1\\
 y_1,y_2&\geq 0.
\end{array}
\end{equation}
Eliminating $y_3$  we now obtain
\newpage
\begin{equation} \label{EBIV}    \begin{array}{rlr}
d-a_3 &\geq y_2 &\hskip1cm (\ref{EBIV}.1) \\
d-a_2-a_3+y_1 &\geq y_2 & (\ref{EBIV}.2)\\
d-y_1 &\geq y_2 & (\ref{EBIV}.3)\\
d-a_2 &\geq y_2 & (\ref{EBIV}.4)\\
a_1-1-y_1 &\geq y_2 & (\ref{EBIV}.5)\\
y_2 &\geq 0 & (\ref{EBIV}.6)\\
y_2 &\geq a_4+a_5 - d+y_1 & (\ref{EBIV}.7)\\
y_2 &\geq d+1-a_6 -y_1 & (\ref{EBIV}.8)\\
d-a_5 &\geq y_1 &\hskip1cm (\ref{EBIV}.9)\\
d-a_4 & \geq y_1 & (\ref{EBIV}.10)\\
y_1 &\geq 0 & (\ref{EBIV}.11)\\
d & \geq a_3+a_4 & (\ref{EBIV}.12)\\
d & \geq a_2+a_5. & (\ref{EBIV}.13)
\end{array}
\end{equation}
By eliminating $y_2$ we get a system of 20 constraints. However 7
of them  are redundant:  $(\ref{EBIV}.12) \Rightarrow
(\ref{EBIV}.1, \ref{EBIV}.6);\ (\ref{EBIV}.9) + (\ref{EBIV}.12)
\Rightarrow (\ref{EBIV}.1, \ref{EBIV}.7);\
(\ref{EBIV}.12)+(\ref{EBIV}.13) \Rightarrow
(\ref{EBIV}.2,\ref{EBIV}.7); \ (\ref{EBIV}.9)+(\ref{EBIV}.10)
\Rightarrow (\ref{EBIV}.3,\ref{EBIV}.6),
(\ref{EBIV}.3,\ref{EBIV}.7), (\ref{EBIV}.4,\ref{EBIV}.7)$ and
$(\ref{EBIV}.13) \Rightarrow (\ref{EBIV}.4,\ref{EBIV}.6)$. Deleting
these redundant  constraints we get
\begin{equation} \label{EBV}  \begin{array}{rlr}
d-a_4 &\geq y_1 &\hskip1cm (\ref{EBV}.1) \\
d-a_5 &\geq y_1 & (\ref{EBV}.2)\\
a_1-1 &\geq y_1 & (\ref{EBV}.3)\\
   \lfloor  \frac{1}{2}(d+a_1-a_4-a_5-1) \rfloor&\geq y_1 & (\ref{EBV}.4)\\
y_1 &\geq 0 & (\ref{EBV}.5)\\
y_1 &\geq \lceil \frac{1}{2}(a_2+a_3-a_6+1)\rceil & (\ref{EBV}.6)\\
y_1 &\geq a_2+a_3 -d & (\ref{EBV}.7)\\
y_1 & \geq a_3 -a_6 +1 & (\ref{EBV}.8)\\
y_1 &\geq a_2-a_6 + 1 & (\ref{EBV}.9)\\
a_1+a_6-2 &\geq d  & (\ref{EBV}.10)\\
d &\geq a_2+a_5 & (\ref{EBV}.11)\\
\end{array}
\end{equation}
$$\begin{array}{rlr}
d &\geq a_3 + a_4 & \hskip1cm  (\ref{EBV}.12)\\
a_6&\geq 1. & (\ref{EBV}.13)
\end{array}$$
Here, for a real number $a$, we set
$$\lceil a \rceil = \min\{ n\in
{\mathbb Z}|\ n\geq a\}\ \text{and}\ \lfloor a \rfloor = \max\{ n\in {\mathbb Z}|\
n\leq a\}.$$
 Eliminating $y_1$ we get a system of 24 constraints.
Among them 14 are redundant: $(\ref{EBV}.12) \Rightarrow (\ref{EBV}.1,
\ref{EBV}.5);\ (\ref{EBV}.1,\ref{EBV}.9)+ (\ref{EBV}.12)
\Rightarrow (\ref{EBV}.1,\ref{EBV}.6);\ (\ref{EBV}.11) +
(\ref{EBV}.12) \Rightarrow  (\ref{EBV}.1, \ref{EBV}.7);\
(\ref{EBV}.12) + \ref{EBV}.13) \Rightarrow
(\ref{EBV}.1,\ref{EBV}.8);\ (\ref{EBV}.11) \Rightarrow
(\ref{EBV}.2,\ref{EBV}.5);\ (\ref{EBV}.2,\ref{EBV}.8) +
(\ref{EBV}.11) \Rightarrow (\ref{EBV}.2,\ref{EBV}.6);\
(\ref{EBV}.11) + (\ref{EBV}.12) \Rightarrow  (\ref{EBV}.2,
\ref{EBV}.7);\ (\ref{EBV}.11) + (\ref{EBV}.13) \Rightarrow
(\ref{EBV}.2, \ref{EBV}.9);\
(\ref{EBV}.3,\ref{EBV}.7)+(\ref{EBV}.10) \Rightarrow
(\ref{EBV}.3,\ref{EBV}.6);\ (\ref{EBV}.10) +
(\ref{EBV}.12)\Rightarrow (\ref{EBV}.3,\ref{EBV}.8);$
$ (\ref{EBV}.10)+(\ref{EBV}.11) \Rightarrow
(\ref{EBV}.3,\ref{EBV}.9);\
(\ref{EBV}.11)+(\ref{EBV}.12)+(\ref{EBV}.3,\ref{EBV}.7)
\Rightarrow (\ref{EBV}.4,\ref{EBV}.7);\
(\ref{EBV}.10)+(\ref{EBV}.12)+(\ref{EBV}.2,\ref{EBV}.8)
\Rightarrow (\ref{EBV}.4,\ref{EBV}.8)$  and $
(\ref{EBV}.10)+(\ref{EBV}.11)+(\ref{EBV}.1,\ref{EBV}.9)
\Rightarrow (\ref{EBV}.4,\ref{EBV}.9).$ Deleting these redundant  constraints, we
finally
get the system
\begin{equation} \label{EBVI}  \begin{array}{rl}
a_1+a_6-2 &\geq d \\
d &\geq a_2+a_5 \\
d&\geq a_3+a_4\\
d &\geq a_2+a_4-a_6 +1\\
d& \geq a_3+a_5-a_6 +1\\
d& \geq a_2+a_3-a_1+1\\
d&\geq a_4+a_5-a_1+1 \\
 \lfloor  \frac{1}{2}(d+a_1-a_4-a_5-1)\rfloor &\geq \lceil
\frac{1}{2}(a_2+a_3-a_6+1)\rceil \\
a_1,a_6 &\geq 1.
\end{array}
\end{equation}

\begin{Lemma} \label{B1} Assume that $a_1+a_6 - 2\geq d \geq \max\{a_2+a_5,
a_3+a_4\}$. Then $\lfloor
\frac{1}{2}(d+a_1-a_4-a_5-1)\rfloor < \lceil
\frac{1}{2}(a_2+a_3-a_6+1)\rceil$ if and only if $a_2+a_3-a_6$ is
even and $a_1+a_6-2 = a_2+a_5 = a_3+a_4$.
\end{Lemma}

\begin{pf} If $a_2+a_3-a_6$ is odd, then
$$\lceil \frac{1}{2}(a_2+a_3-a_6+1)\rceil = \frac{1}{2}(a_2+a_3-a_6+1).$$
Since $a_2+a_5+a_3+a_4 \leq d + a_1+a_6 -2$, we get $d+a_1-a_4-a_5
-1 \geq a_2+a_3-a_6+1$, which yields
$$\frac{1}{2}(d+a_1-a_4-a_5-1) \geq \frac{1}{2}(a_2+a_3-a_6+1).$$
Hence
$$\lfloor \frac{1}{2}(d+a_1-a_4-a_5-1)\rfloor \geq \frac{1}{2}(a_2+a_3-a_6+1) =
\lceil \frac{1}{2}(a_2+a_3-a_6+1)\rceil.$$
If $a_2+a_3-a_6$ is even, then
$$\lceil \frac{1}{2}(a_2+a_3-a_6+1)\rceil = \frac{1}{2}(a_2+a_3-a_6) + 1.$$
In the case  $a_1+a_6-2 > \min\{ a_2+a_5, a_3+a_4\}$, we have
$a_2+a_5+a_3+a_4
\leq d + a_1+a_6 -3$. Hence $d+a_1-a_4-a_5 -1 \geq a_2+a_3-a_6+2$,
which implies
$$\lfloor \frac{1}{2}(d+a_1-a_4-a_5-1)\rfloor \geq \frac{1}{2}(a_2+a_3-a_6)  + 1=
\lceil \frac{1}{2}(a_2+a_3-a_6+1)\rceil.$$
The left case is $a_1+a_6-2 = \min\{ a_2+a_5, a_3+a_4\}$. Since $a_1+a_6-2 \geq
d\geq  \max\{ a_2+a_5, a_3+a_4\}$, we must have $d= a_2+a_5 = a_3+a_4 =
a_1+a_6-2$. Then
$d+a_1-a_4-a_5-1 = a_2+a_3-a_6+1$ is an odd number. Therefore
$$\lfloor \frac{1}{2}(d+a_1-a_4-a_5-1)\rfloor <  \lceil
\frac{1}{2}(a_2+a_3-a_6+1)\rceil.$$
This completes the proof of  the lemma. \hfill $\square$
\end{pf}

Going back from (\ref{EBVI}) to (\ref{EBII}), the Fourier-Motzkin
algorithm gives us in general only a rational solution  of
(\ref{EBI}) if (\ref{EBVI}) holds. However, in our concrete
situation we can already find an integer solution.

\begin{Lemma} \label{B2} Let
$$\begin{array}{ll}
{\mathcal A} = \max \{ & a_2 + a_5, \ a_3+a_4, \ a_2+a_4-a_6+1, \ a_3+a_5-a_6+1, \\
& a_2+a_3-a_1+1, \ a_4+a_5-a_1+1\}.
\end{array}$$
The system (\ref{EBI}) has an integer solution if and only if $a_1,a_6 \geq 1$ and
one of the following conditions holds:
\begin{itemize}
\item[\rm{(i)}] $a_1+a_6 - 2>{\mathcal A}$ and $a_1+a_6 - 2\geq d \geq {\mathcal A}$.
\item[\rm{(ii)}] $a_1+a_6 - 2={\mathcal A} =d$ and $a_1+a_6 - 2 > \min\{a_2+a_5,\
a_3+a_4\}$.
\item[\rm{(iii)}] $a_1+a_6 - 2=a_2+a_5 = a_3+a_4 = {\mathcal A}= d$ and $a_2+a_3-a_6$ is
odd.
\end{itemize}
\end{Lemma}

\begin{pf} If (\ref{EBI}) has an integer   solution, then by Fourier-Motzkin
algorithm, (\ref{EBVI}) holds. Using Lemma \ref{B1} we get the necessity.

Assume that $a_1,a_6 \geq 1$ and one of the above conditions
(i)-(iii) holds. Then for any $d$ such that ${\mathcal A} \leq d \leq a_1+a_6-2$, the
system (\ref{EBVI}) holds by Lemma \ref{B1}. Fix such an integer
$d$. Denote by ${\mathcal L}_{\ref{EBV}}$ the minimum of integers in the left
sides of $(\ref{EBV}.1)-(\ref{EBV}.4)$ and ${\mathcal R}_{\ref{EBV}}$ the
maximum of integers in the right sides of
$(\ref{EBV}.5)-(\ref{EBV}.9)$. Then from (\ref{EBVI}) it follows
that ${\mathcal L}_{\ref{EBV}} \geq {\mathcal R}_{\ref{EBV}}$. Hence $y_1 =
{\mathcal R}_{\ref{EBV}}$ is an integer solution of (\ref{EBV}). Putting $y_1
= {\mathcal R}_{\ref{EBV}}$ into (\ref{EBIV})-(\ref{EBII}) and repeating this
process, we can similarly define ${\mathcal L}_{\ref{EBIV}} \geq
{\mathcal R}_{\ref{EBIV}},\  {\mathcal L}_{\ref{EBIII}} \geq {\mathcal R}_{\ref{EBIII}},\
{\mathcal L}_{\ref{EBII}} \geq {\mathcal R}_{\ref{EBII}}$ such that $y_1 =
{\mathcal R}_{\ref{EBV}},\ y_2 = {\mathcal R}_{\ref{EBIV}},\ y_3 = {\mathcal R}_{\ref{EBIII}},\ y_4
 = {\mathcal R}_{\ref{EBII}}$ is an integer solution of (\ref{EBII}),
which is equivalent to (\ref{EBI}) . \hfill $\square$
\end{pf}

\section{Structure of the first local cohomology module} \label{C}

In this section we describe the first local cohomology module of $R/I$. From now
on, w.l.o.g.,  we always assume that $a_1+a_6$ is the maximum among the sums $a_1+a_6,\ a_2+a_5,\ a_3+a_4$. In other words we may assume that the following holds:
$$ (*) \hskip2cm a_1 + a_6 \geq \max\{ a_2+a_5, \ a_3+a_4\}.$$

\begin{Lemma} \label{C1} Under the assumption (*) there exists no $\underline{\alpha} \in
{\mathbb Z}^4$ such that
$\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{ \{1,3\},\ \{2,4\}\}$ or $\operatorname{Max}(\Delta_{\underline{\alpha}}) =
\{ \{1,4\},\ \{2,3\}\}$.
\end{Lemma}

\begin{pf} Assume, w.l.o.g., the existence of $\underline{\alpha} \in {\mathbb Z}^n$ such that
$\operatorname{Max}(\Delta_{\underline{\alpha}}) = \{ \{1,3\},\ \{2,4\}\}$. Then
applying Lemma \ref{A4} and Lemma \ref{B2} to this situation we would get
$a_2+a_5-2 \geq a_1+a_6$, a contradiction to (*). \hfill $\square$
\end{pf}

We can now  explicitly determine all arithmetically Cohen-Macaulay
tetrahedral  curves in terms of $a_i$. This result recovers the
main theorem in \cite{F}.

\begin{Theorem} \label{C2} Let
$$\begin{array}{ll}
{\mathcal A} = \max \{ & a_2 + a_5, \ a_3+a_4, \ a_2+a_4-a_6+1, \ a_3+a_5-a_6+1, \\
& a_2+a_3-a_1+1, \ a_4+a_5-a_1+1\}.
\end{array}$$
Under the assumption (*), a  tetrahedral
curve $C(a_1,...,a_6)$ is arithmetically Cohen-Macaulay if and only if  one of the
following conditions holds:
\begin{itemize}
\item[\rm{(i)}] $a_1=0$ or $a_6 =0$;
\item[\rm{(ii)}] $a_1+a_6 - 2<{\mathcal A} $;
\item[\rm{(iii)}] $a_1+a_6 - 2=a_2+a_5 = a_3+a_4 = {\mathcal A}$ and $a_2+a_3-a_6$ is
even.
\end{itemize}
\end{Theorem}

\begin{pf} By Lemma \ref{C1}, $C= C(a_1,...,a_6)$ is arithmetically
Cohen-Macaulay if and only
if there is no $d$ such that the system (\ref{EBI}) has an integer solution. Hence
the statement follows from Lemma \ref{B2}. \hfill $\square$
\end{pf}

 {\bf Remark}. In \cite{MN}, Question 7.4(5), Migliore and Nagel asked whether
an arithmetically  Cohen-Macaulay
 tetrahedral curve $C= C(a_1,...,a_6)$ can be explicitly identified by the 6-tuples
$a_1,...,a_6$. This question was solved
 by Francisco in \cite{F}. His main result says that under the assumption (*),
$C(a_1,...,a_6)$ is arithmetically Cohen-Macaulay if and
 only if  one of the following conditions holds:
 \begin{itemize}
\item[\rm{(a)}] $a_1=0$ or $a_6 =0$;
\item[\rm{(b)}] $a_1+a_6 = \epsilon + \max\{a_2+a_5,\ a_3+a_4\}$, where
$\epsilon \in \{0,1\}$.
\item[\rm{(c)}] $2a_1 < a_2 + a_3 -a_6 +3$ or $2a_1 < a_4 + a_5 -a_6 +3$ or
$2a_6< a_2 + a_4 -a_1 +3$ or $2a_6 < a_3 + a_5 -a_1 +3$;
\item[\rm{(d)}] All inequalities of (c) fail, $a_1+a_6 =a_2+a_5 +2= a_3+a_4 +2$
and $a_1+a_3+a_5$ is even.
\end{itemize}
 One can easily check that this statement is equivalent to that of Theorem \ref{C2}.

 Assume now that $C$ is not arithmetically Cohen-Macaulay. Then $a_1,a_6\geq
1$ and one of three conditions in Lemma \ref{B2}
 is satisfied. In particular ${\mathcal A} \leq a_1+a_6-2$. Let
$$\begin{array}{ll} T_1  & =\{ \underline{y} \in {\mathbb N}^4|\   y_1 + y_3  \geq a_2  ,\ y_1 +
y_4  \geq a_3 ,\  y_2 + y_3  \geq a_4 ,\  y_2 + y_4  \geq a_5 \},\\
T_2 & = \{ \underline{y} \in T_1|\ y_1 + y_2 \geq a_1\},\end{array}$$
and
$$T_3 = \{ \underline{y} \in T_1|\ y_3 + y_4 \geq a_6\}.$$
Let $S= T_1\setminus (T_2\cup T_3)$. Then the set $S_d$ of all
elements of degree $d$  of $S$ is the set of all solutions  of the system
(\ref{EBI}).  As usual we identify $K[T_i],\ i\leq 3$,  and $K[S]$
with  subsets of $R=K[x_1,...,x_4]$. Note that $K[T_i], \ i\leq
3,$ are ideals  of  $R$. Hence we may consider $K[S]$ as a factor
module $K[T_1]/K[T_2]+K[T_3]$. Thus, the  module structure on
$K[S]$ over $R$ is defined as follows: for $\underline{\alpha} \in S$ and
$\underline{\beta} \in {\mathbb N}^4$,
 $$\underline{x}^{\underline{\beta}}\cdot \underline{x}^{\underline{\alpha}} = \begin{cases} \underline{x}^{\underline{\beta}+
\underline{\alpha}} \ &\text{if} \ \underline{\beta}+ \underline{\alpha} \in S,\\
 0 \ & \text{otherwise}.\end{cases}$$

The following result describes the module structure of $H^1_{\mathfrak m}(R/I)$.

 \begin{Proposition}\label{C4} Under the assumption (*), $$H^1_{\mathfrak m}(R/I) \cong
K[S]$$ as graded modules over $R$.
 \end{Proposition}

 \begin{pf} Let
$${\mathcal C}^\bullet : 0 \rightarrow R/I \rightarrow \oplus_{i=1}^4 (R/I)_{x_i}
\rightarrow \cdots \rightarrow (R/I)_{x_1x_2x_3 x_4}
 \rightarrow 0,$$
be the $\check{C}$ech complex of $R/I$. Then $H^1_{\mathfrak m}(R/I) \cong
H^1({\mathcal C}^\bullet)$. By \cite{T}, Lemma 2, for all
$\underline{\alpha}\in {\mathbb Z}^4$ there is an isomorphism of complexes
$$({\mathcal C}^\bullet_{\underline{\alpha}}) \cong \operatorname{Hom}_{{\mathbb Z}}({\mathcal
C}(\Delta_{\underline{\alpha}})[-j-1],K),$$
where $j= |G_{\underline{\alpha}}|$ and ${\mathcal
C}(\Delta_{\underline{\alpha}})[-j-1]$ means the  shifting  of the augmented
oriented chain complex ${\mathcal C}(\Delta_{\underline{\alpha}})$ by $-j-1$.
Denote by $\pi $ the simplicial complex on $\{1,2,3,4\}$ with
$\operatorname{Max}(\pi) = \{ \{1,2\},\ \{3,4\}\}$. By Lemmas
\ref{A3}, \ref{A4} and \ref{C1} it follows that $H^1({\mathcal
C}^\bullet_{\underline{\alpha}}) \neq 0$ if and only if $\Delta_{\underline{\alpha}} =
\pi $, $G_{\underline{\alpha}} =\emptyset $ and $\underline{\alpha} \in S$. Moreover,
in this case $H^1({\mathcal C}^\bullet_{\underline{\alpha}}) \cong
K\underline{x}^{\underline{\alpha}}$. From this we get $H^1_{\mathfrak m}(R/I) \cong K[S]$,
as required.

\rightline{ $\square$}
\end{pf}

The above description of $S$ allows us  to describe the module structure of $K[S]$ in an obvious way. Of course, $S$ can be
written as:
$$\begin{array}{ll} S= \{ \underline{y} \in {\mathbb N}^4|\  & y_1 + y_3  \geq a_2  ,\ y_1 +
y_4  \geq a_3 ,\ y_2 + y_3  \geq a_4 ,\  y_2 + y_4  \geq a_5, \\
& y_1+y_2 < a_1,\ y_3+y_4 < a_6 \}.
\end{array}$$
It is easy to write a program to compute this set $S$. Hence the module structure of $H^1_{\mathfrak m}(R/I) $ is known once $a_1,...,a_6$ are given.

We say that a non-zero ${\mathbb Z}$-graded module $M$ has no gap if
$M_i \neq 0$ and $M_j\neq 0$ for  some $i\leq j$, then $M_k\neq 0$
for all $i\leq k\leq j$. Recall that the diameter of a module $M$
of finite length is defined as
$$\operatorname{diam} (M) = \operatorname{end}(M) - \operatorname{beg}(M) +1,$$
where $\operatorname{beg}(M) =\min\{i|\ M_i \neq 0\}$ and $\operatorname{end}(M) =\max\{i|\ M_i \neq 0\}$  (if
$M=0$ we set $\operatorname{diam}(M) =0$).

\begin{Theorem} \label{C5} Let
$$\begin{array}{ll}
{\mathcal A} = \max \{ & a_2 + a_5, \ a_3+a_4, \ a_2+a_4-a_6+1, \ a_3+a_5-a_6+1, \\
& a_2+a_3-a_1+1, \ a_4+a_5-a_1+1\}.
\end{array}$$
Assume that (*) holds and the tetrahedral curve $C$
is not arithmetically Cohen-Macaulay. Then
$a_1+a_6-2 \geq {\mathcal A}$ and
$$k(R/I) =  \operatorname{diam} (H^1_{\mathfrak m}(R/I)) = a_1+a_6- {\mathcal A} -1.$$
In particular, $H^1_{\mathfrak m}(R/I)$ has no
gap.
\end{Theorem}

\begin{pf} Since $R/I$ is not a Cohen-Macaulay ring, by Theorem \ref{C2},
$a_1+a_6-2 \geq {\mathcal A}$ and $a_1,a_6\geq 1$.
By Lemma \ref{B2}, for each $d$ such that ${\mathcal A} \leq d \leq a_1+a_6-2$ we have $S_d \neq
\emptyset $. Hence,
 by Proposition \ref{C4},  $H^1_{\mathfrak m}(R/I)$ has no gap, $\operatorname{beg}(H^1_{\mathfrak m}(R/I)) = {\mathcal A}$ and
$\operatorname{end}(H^1_{\mathfrak m}(R/I)) = a_1+a_6-2$,
 which implies $\operatorname{diam} (H^1_{\mathfrak m}(R/I)) = a_1+a_6- {\mathcal A} -1$.

Further, let $\underline{\alpha} = (\alpha_1,\alpha_2,\alpha_3,\alpha_4)\in
S_{\mathcal A}$ be a fixed element. Then  $\alpha_1+ \alpha_2 \leq a_1-1$ and
$\alpha_3+\alpha_4 \leq a_6-1$. Let $\underline{\alpha}^* = (\alpha_1, a_1-1
-\alpha_1,\alpha_3, a_6-1-\alpha_3)$. Since $a_1-1 -\alpha_1\geq
\alpha_2$ and $ a_6-1-\alpha_3\geq \alpha_4$, the condition
$\underline{\alpha}\in T_1$ implies $\underline{\alpha}^* \in T_1$ too. On the other
hand $\underline{\alpha}^* \not\in T_1\cup T_2$. Hence $\underline{\alpha}^* \in
S_{a_1+a_6-2}$. Note that $\underline{\alpha}^* = \underline{\alpha} + \underline{\beta}$, where
$\underline{\beta}= (0, a_1-1 -\alpha_1-\alpha_2, 0, a_6-1-\alpha_3-
\alpha_4)\in {\mathbb N}^4$ and $\operatorname{deg} (\underline{\beta}) = a_1+a_6-{\mathcal A} -2$.
Therefore, by Proposition \ref{C4}, $$\underline{x}^{\underline{\beta}}
H^1_{\mathfrak m}(R/I)_{\underline{\alpha}} \cong H^1_{\mathfrak m}(R/I)_{\underline{\alpha} + \underline{\beta}}
= H^1_{\mathfrak m}(R/I)_{\underline{\alpha}^*} \neq 0,$$ which yields
$$k(R/I) \geq a_1+a_6- {\mathcal A}-1 = \operatorname{diam} (H^1_{\mathfrak m}(R/I)).$$
Since  $\operatorname{diam} (H^1_{\mathfrak m}(R/I)) \geq k(R/I)$, we finally get $k(R/I)  = \operatorname{diam}
(H^1_{\mathfrak m}(R/I))$, as required. \hfill $\square$
\end{pf}

In the above proof we already showed:

\begin{Corollary} Assume that (*) holds and the tetrahedral curve $C$ is not
arithmetically Cohen-Macaulay. Then
$a_1+a_6-2 \geq {\mathcal A}$ and $\operatorname{end}(H^1_{\mathfrak m}(R/I)) = a_1+a_6-2$.
\end{Corollary}

Recall that $C$ is  arithmetically Buchsbaum if and only if
$k(R/I) \leq1$. As an immediate consequence of  Theorem \ref{C5}
we recover Corollary 4 in \cite{MN}.

\begin{Corollary} A tetrahedral curve $C$ is  arithmetically Buchsbaum if and
only if
$$H^1_{\mathfrak m}(R/I) \cong K^m(t),$$
for some non-negative integers $m,t$.
\end{Corollary}

 Migliore and Nagel found all arithmetically Buchsbaum tetrahedral curves which
are so-called minimal (see Corollary \ref{C8} below).
  Using Theorem \ref{C5} and  \ref{C2} we are able to determine all arithmetically
Buchsbaum tetrahedral curves which are not necessarily minimal.

 \begin{Theorem} \label{C7} Under the assumption (*), a tetrahedral curve $C$ is
 arithmetically Buchsbaum if and only if   one of the following conditions is satisfied:

(i) $a_1=0$ or $a_2=0$;

(ii) $a_1+a_6 - 2 \leq {\mathcal A} $.
\end{Theorem}

\begin{pf} If $C$ is arithmetically Cohen-Macaulay, by Theorem \ref{C2},  one of the above condition holds. Assume that $C$ is  not arithmetically Cohen-Macaulay and arithmetically
Buchsbaum. Then $k(R/I) =1$.
By Theorem \ref{C5}, $a_1,a_6\geq 1$ and $a_1+a_6 - 2={\mathcal A} $. Conversely, by Theorem \ref{C2} we may assume from the beginning that $a_1,a_6\geq 1$.  Under
these conditions,  again by Theorem \ref{C5},we immediately
 have $k(R/I) \leq 1$, i.e. $C$ is  arithmetically Buchsbaum.  \hfill $\square$
\end{pf}

Migliore and Nagel introduced the following notion: Assume that
\newline
$a_6= \max\{a_1,...,a_6\}$. A tetrahedral  curve $C$ is said to be
{\it minimal} if $a_1> \max\{ a_2+a_4,\ a_3+a_5\}$ and $a_6>
\max\{ a_2+a_3,\ a_4+a_5\}$ (see \cite{MN}, Definition 3.4 and
Corollary 3.5). Note that in this case we already have $a_1,a_6
\geq 1$ and $a_1+a_6 - 2\geq {\mathcal A}$.

\begin{Corollary} {\rm (\cite{MN}, Corollary 4.3 and Corollary 5.4)}. \label{C8}
Assume that
\newline $a_6= \max\{a_1,...,a_6\}$ and $C$ is a
minimal tetrahedral curve. Then
\begin{itemize}
\item[{\rm  (i)}] $C$ is not arithmetically Cohen-Macaulay.
\item[{\rm  (ii)}] $C$ is arithmetically Buchsbaum if and only if
either $a_2=a_5=0$ and $a_1=a_6 = a_3+1=a_4+1$ or $a_3=a_4=0$  and
$a_1=a_6 = a_2+1=a_5+1$.
\end{itemize}
\end{Corollary}

\begin{pf} Since $a_1> \max\{ a_2+a_4,\ a_3+a_5\}$ and $a_6> \max\{
a_2+a_3,\ a_4+a_5\}$, we have
\begin{equation} \label{EC8} \begin{array}{rl}
a_1+a_6-2 \geq \max\{ &  a_2 + a_5 + 2a_4,\ a_2+a_5 +2a_3,\\
& a_3+a_4+2a_2, \ a_3+a_4+2a_5\} \geq {\mathcal A}.
\end{array}
\end{equation}
If $C$ is arithmetically Buchsbaum, then since $a_1,a_6\geq 1$, by
Theorem \ref{C2} and Theorem \ref{C7}, we must have $a_1+a_6-2
={\mathcal A}$.  Combining with (\ref{EC8}) it implies that either
$a_2=a_5=0$ or $a_3=a_4=0$. W.l.o.g. assume that $a_2=a_5=0$. Then
${\mathcal A} = a_3+a_4$ and $a_1+a_6-2= a_3+a_4$. Since $a_1, a_6>\max\{
a_3,\ a_4\}$, the later equality gives $a_1=a_6 = a_3+1=a_4+1$. In
this case $a_2+a_3-a_6 = -1$ is odd, so $C$ is not arithmetically
Cohen-Macaulay. Thus we have proved (i) and the necessity of (ii).
The sufficiency of (ii) immediately follows from Theorem \ref{C7}.
\hfill $\square$
\end{pf}

Similarly, using Theorem \ref{C5}, we can quickly get

\begin{Corollary} {\rm (\cite{MN}, Lemma 6.2)}. \label{C9} Assume that $a_6=
\max\{a_1,...,a_6\}$ and $C$ is
a minimal tetrahedral curve. Then $\operatorname{diam} (H^1_{\mathfrak m}(R/I)) = 2$ if and only if after a
suitable permutation of variables we have
$(a_1,...,a_6) = (k,k-1,0,0,k-1,k+1),\ k\geq 1$ or $(a_1,...,a_6) = (k,k-2,0,0,k-1,k),\
k\geq 2$.
\end{Corollary}

redpajama_set_name: RedPajamaArXiv


text:
Grotz: Virtual draft just first sign of a…
SportsNFL
Grotz: Virtual draft just first sign of a fanless future for NFL
MEDIANEWS GROUP PHOTO Behold the view from the press box at Lincoln Financial Field late in the last Eagles game played there, probably for quite some time. This was the fourth quarter of the Eagles' playoff loss to the Seattle Seahawks last November. It's unknown when the next game will be contested at the Linc, but signs aren't good with a second wave of conovavirus being anticipated for the fall.
By Bob Grotz | bgrotz@delcotimes.com | The Times Herald
PUBLISHED: May 12, 2020 at 11:30 p.m. | UPDATED: September 23, 2021 at 7:27 a.m.
It was tough getting inside the head of NFL Commissioner Roger Goodell when he pre-empted the draft to warn the record TV audience about the tough times ahead.
After all, the NFL was back. It was the NBA, the NHL and MLB who were getting slammed by the coronavirus. They were the businesses losing games and revenue.
The draft went on as scheduled, just as free agency shifted into gear without delay. Deadly and debilitating as the coronavirus had been, it was unable to stop those iconic, if not overrated symbols of the ultimate team sport which prided itself on playing through pain.
With the league's regular season months away, the fears of football fans were dispelled shortly after Goodell announced in prime time on a Thursday night that Cincinnati was on the clock. True to their reputation, the Bengals used almost all of their allotted time before selecting consensus first overall pick Joe Burrow, the quarterback out of LSU who grew up in Ohio.
We should have known better.
Goodell was standing in his basement, not on a stage at the Bellagio Hotel in Las Vegas where the lottery was supposed to be. He had time to run upstairs and change from a blue blazer into a V-neck sweater between picks, for crying out loud.
That's because it was a virtual draft where bandwidth and cyber-security were more important than the trade value chart.
Odd as the draft was, at the end of the weekend it was like all the others. There were shocking picks, anticipated picks, stupid picks, trades, and a lot of discussion about why this guy played as fast as his 40-yard dash time, what a risk this other guy was because he didn't play in a major conference and how Bill Belichick was ahead of the curve again by staying out of camera range and seating his pet husky in front of his laptop.
Three weeks after Burrow came off the board, the NFL has slashed salaries of league personnel, Goodell himself stepping entirely away from a salary worth up to $40 million with incentives.
Contributions to the company pension plan were permanently reduced from 15 percent to 10 percent. There also were furloughs. It will be at least a couple of months (best-case scenario) before the latter group is back according to NFL officials.
Teams are scrambling to reduce costs.
On Tuesday it was no longer a secret that the NFL could wind up playing games in empty or relatively empty stadiums or arenas, just as the NBA and MLB are considering. And what a revenue hit that could be.
To prepare for the worst, NFL teams recently agreed to vote on a measure later this month that would increase debt limits for each team from $150 million to $350 million, per reports. That's basically not a good sign unless you're building a new stadium.
Reuters reported that ESPN has figures indicating the NFL could lose about $130 million per game from reductions in ticket and concession revenues without fans, to say nothing of the home field advantage.
If you thought the Sixers were the Philly sports club that benefited most from their fans, you've forgotten how dominant the Eagles' defense has been when Lincoln Financial Field is rocking.
No fans in the stands?
"For sure it will be different, man, if we have to play the game with no fans," Pro Bowl defensive tackle Fletcher Cox said on a teleconference call Tuesday. "At the same time we have to understand again that health is a priority. We have to look at our health. If we have to play a game with no fans, it will be something to get used to. It will be more like a, you know, I don't know how to explain it.
"But it will be weird, for sure."
Barring a dramatic financial reversal there's a chance the NFL's hard salary cap could go down in 2021, not up, for just the second time since its inception in 1994.
Moreover, with the salary cap increasing from 47 to 48 percent of defined gross revenues this year, and possibly 48.5 percent with a media kicker in subsequent years, teams are economizing by not picking up the option years of first-round draft picks for 2021 at an historic level.
Just 18 of the 32 first-round picks in the 2017 draft had their options picked up, included just five of the top 10 picks. Mitchell Trubisky, Solomon Thomas, Leonard Fournette, Corey Davis or Johnny Ross, anyone? FYI: The Eagles picked up the fifth year option of Derrick Barnett.
Did we mention the NFL is negotiating a new TV contract? The current pact expires after the 2021 season.
The tough decisions have only begun for this business full of billionaires challenged more than ever to maintain standing as the premier sport in this country.
Barring a minor miracle, virtual free agency, the draft and the schedule release are the last vestiges of normalcy this league will have with revenues jeopardized by a deadly virus that won't give up.
Virtual drafts are one thing.
Virtual football is a hard sell.
Contact Bob Grotz at bgrotz@21st-centurymedia.com; follow him on Twitter @BobGrotz.
By BOB GROTZ
Bgrotz@21st-centurymedia.com
@bobgrotz on Twitter
It was tough getting inside the head of NFL Commissioner Roger Goodell when he preempted the draft to warn the record TV audience about the tough times ahead.
After all, the NFL was back. It was the NBA, the NHL and MLB who were getting slammed by the coronavirus virus. They were the businesses losing games and revenue.
With the league's regular season months and months away, the fears of football fans were dispelled shortly after Goodell announced in prime time on a Thursday night that Cincinnati was on the clock. True to their reputation, the Bengals used almost all of their allotted time before selecting consensus first overall pick Joe Burrow, the quarterback out of LSU who grew up in Ohio.
Contributions to the company pension plan were permanently reduced from 15 percent to 10 percent. There also were furloughs. It will be at least a couple of months – best-case scenario – before the latter group is back according to NFL officials.
Reuters reported that ESPN has figures indicating the NFL could lose about $130 million per game from reductions in ticket and concession revenues using minus fans, to say nothing of the home field advantage.
If you thought the Sixers were the Philly sports club that benefitted most from their fans, you've forgotten how dominant the Eagles' defense has been when Lincoln Financial Field is rocking.
The tough decisions have only begun for this business full of billionaires challenged more than ever to maintain standing as the premier sport in this country
Bob Grotz | Reporter


What to Bring Paintballing. What Should You Wear?
When coming to San Diego Paintball Park there are a few things to remember and one of those is what to bring for your paintball experience. Below we have compiled a small list of items and clothing suggestions to help you make the most of your time with us playing paintball.
One item that we suggest all players bring is a camera or an action camera if they are able to. Whether you want to use your cell phone camera or a GoPro style action camera we always suggest you bring something to capture all the fun you will have playing paintball. We also allow groups to take photos on the field after they are done playing. We also like when players tag us in their photos so we can share them on our Facebook, Instagram, and Twitter pages!
Paintball is the same as any other physical activity where you are constantly moving around. You need to stay hydrated! Stay hydrated by drinking plenty of fluids before arriving so that your body does not become dehydrated while on the field playing. We do offer drinks for sale on site but we always want to remind our customers of the importance of staying hydrated!
Imagine going to Disneyland and paying the admission then going into the park with no extra money. If you show up to paintball like this you may feel the same disappointment. We have tons of add ons that can not only help make your experience better but also more memorable. Whether its buying some snacks and drinks at our snack bar or buying extra paintballs and paint grenades you should always be prepared for some additional costs. Nothing is mandatory obviously but we truly do believe that our customers will have a better experience if they do keep in mind that we offer these additional items to enhance your experience.
At San Diego Paintball Park we specialize in low-impact paintball which is great for beginner players who are still renting equipment. With this being said our paintballs hurt up to 75% less than a standard sized round you will find at other paintball parks in San Diego. Don't worry about layering up with 3 shirts and a parka made for the Arctic freeze. Instead come in with a t-shirt and have some fun. We also offer short sleeve and long sleeve shirts for sale and padded jerseys for rent if you absolutely need that extra cushion.
Paintball is a physical game no doubt but part of the fun at San Diego Paintball Park is taking advantage of our artificial turf. Our cushioned turf is an awesome playing surface that you can run, slide, dive, and crawl on. Avoid irritating your knees and legs by wearing proper loose fitting long pants or sweatpants. The baggier the better we say.
Sandals are 100% not allowed to be worn during play at San Diego Paintball Park so we implore you to bring in proper footwear. Your standard running shoe or active life shoe will be great for paintball. We do not allow cleats in our facility.
ALL PAINT USED ON SITE IS WATER BASED AND WILL WASH OUT OF YOUR CLOTHING EASILY!


text:
Google's Me on the Web feature
Updated • Nov 12, 2018
Online reputation management becomes more important as focus shifts to a world that is connected online. Chance is that someone searching for your name, or a nickname you are known by, will find information about you online unless you have separated your online identity from your real one from the very first steps you made on the web.
Depending on what you do for a living, you may be present online and can't really do anything about that. If you search for my name for instance, you will find lots of information about me and people who share the name.
It is important that the information returned by the search engines is accurate and not misleading in any form or way as others who search for you online may get a wrong impression or worse otherwise.
Google Me on the Web
Update: Google abandoned Me on the Web. Google Alerts is still a thing, however, and can be used for the purpose. End
Google's new Me on the Web feature aims to provide you with a couple of tools to keep things under tight control. You find the new feature under your Google Account settings. To open it either load the following link directly, or click on the profile photo at the top right of a Google property after you have signed in to a Google account and select Account > Profile and Privacy > Me on the Web.
Here you are presented with three options:
Search for yourself - This may be a basic thing to do considering that you can simply fire up Google Search manually and perform a search for your name, but it may still be useful in the context of the feature. It basically displays the ten top search results for your name on Google right on the Me on the Web page. You can change the name that you want to search for in the search form at the top, which may be useful if you want to narrow down results. This can be done for instance by adding a location to the search or a middle name.
Stay up to date with Web Alerts - This is probably the most useful option of the three available ones. It allows you to create Google Alerts for your name or another name you specify. You will be notified by email once per day when new information appear online. You can change that to "as it happens" to receive near instant information, or "once a week" if you want a weekly digest instead. You can modify the name or phrase you want to receive alerts from to narrow down the search or include searches for family members or colleagues as well. This can be an interesting option for parents to keep track of their children's activity on the Internet.
The last option reminds you to check your Google profile to make sure it contains accurate information.
All available options are available in other locations as well. You can configure Google Alerts manually if you for instance. The Me on the Web feature however makes those options available to a larger audience.
It is a bit hidden in the Account options as it only becomes available when you click on the Profile and Privacy link here, but users who discover it may find all the tools they need to quickly check for names online and create alerts to be notified about new information posted on the Internet.
Are you keeping track of your name or other names on the Internet?
Google introduced a new feature called Me on the Web in 2012 to keep track of how your name appears in the search engine.
Google ends the G Suite legacy free edition leaving users worried
WhatsApp Will Consume Your Google Drive Storage Data: Backup Size Feature in the Works?
South Korea is on its way to being the first country to legislate open app store payments
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Martin Watson said on December 17, 2012 at 10:59 am
Interesting but it seems very badly skewed in favour of LinkedIn.
Nebulus said on December 17, 2012 at 11:46 am
Is there a problem with the comments system? Because I wrote a comment, it doesn't show, but when I tried to resubmit it, the site told me that it is a duplicate…
Nebulus said on December 17, 2012 at 5:08 pm
Yes, there is definitely something wrong, it's the third time it happened to me today… :(
Mountainking said on December 17, 2012 at 1:53 pm
Thanks on that one Martin. You just convinced me to delete my google + account. I hate it when my name appears in search engines. Dont like to be stalked by pervy voyeurs.
I'm not sure that deleting your Google+ account will help, other than having nothing more to do with Google… Others will still be able to search you.
Keith said on December 17, 2012 at 8:31 pm
Astounding. I used the direct link from your article and found my name alright. There's a high profile guy with the same name dominating the result so, I remain under the radar to my ex who stalked me through web and social connections for years. Thanks for the tip.
Nebulus said on December 17, 2012 at 10:39 pm
This is something you could've done without Google's new service. Just google for your full name (with or without quotes around it"), that's exactly what Google Me is doing.
Google will soon enforce the use of two-step verification for Google accounts
Google plans to shift advertising from user tracking to group tracking
Google Play Store dominance is being threatened by new Senate bill
How to deal with Google's and YouTube's aggressive popups (before you continue, sign in)
Published in: September 27, 2020 2:41 pm | Updated in: January 1, 2022 8:56 am
Published in: October 14, 2021 5:08 pm | Updated in: October 14, 2021 5:13 pm
Published in: September 6, 2021 2:41 pm | Updated in: September 6, 2021 2:41 pm


Meet the Current-Argus Pet of the Day, Sandy.
Sandy is a large domestic longhair mix. Her animal ID is 31653325.
Sandy, like many other shelter cats, is waiting for a permanent home at Noah's Ark Animal Shelter.


Youth and Children's work Sunday Services Live Wires (for ages up to 11) meets each Sunday morning except for occasional services where we have an All Age Worship service.
The group starts at approximately 10.50am (after a time of worship with the whole church together) when they enjoy a time of crafts and teaching relevant to their age group.
The co-ordinator is Margaret Sankey.
Youth - Our secondary school age children are encouraged to come along to Sunday morning services and occasionally meet on Wedneday evenings. Contact our Minister for more information.
Mid-Week Activities: "Busy Bees" Toddlers group Wednesdays 09:00-11:30am. Busy Bees caters for all pre-school children from 0 to 5 years old. Come and let your babies or children play while you catch up with friends for a chat and a coffee. We have a large variety of toys and crafts each week. Cost: �50 per family per session. See our Busy Bees Page for more details. Messy Church Held every 6 weeks from 3.30-5.30pm on Saturday afternoons.
A fresh, new way of experiencing church for all the family. A fun time of crafts, sports challenges, Bible stories, songs, puppets and a meal for all ages. Everyone is welcome, but all children must be accompanied by at least one adult.
Trekkers Trekkers is a club for children aged 7-11. It meets at Capel Methodist Church after school on Thursday evenings from 4:00pm. Contact Margaret Sankey for further details. The Junction Youth Club The Junction runs on Friday evenings at the church from 7-9pm. It's for those in school years 7-9 and features a wide range of activities including Table Tennis, Pool, Wii, Playstation etc plus tuck shop and space to relax. Cost is �50p/week. Contact our youth worker to sign up or you can register on the door. Life groups We have 2 "Life" groups which run weekly. One group is for school years 7-9 and runs on Thursdays from 7-8:30pm, and the other is for school years 10+ and runs on Wednesdays from 7:30-9pm. These groups are a time for young people to meet up and play games, have food, chat with each other, and look into the Bible, the Christian faith, and some of life抯 big, tough questions. This group is open to anyone at any stage of faith � whether you抳e been going to church for years, or want to look into the idea a bit more. Both groups run in the upper room at Capel Methodist Church. Feel free to pop along to see what the group is like, or if you want to know more, feel free to contact our youth worker, Caitlan.


There are few things, other than whole foods from sustainable local farms, that we promote - but bone broth (ok, we're a LITTLE bias here) and Pu-Erh tea (pronounced: Pew-Air) are two of those things we feel very confident to tell you about. I (Jenny, owner) found out about bone broth about a year ago when trying to lessen my recovery time after training hard for half marathons, triathlons, and coaching my girl's cross country running team (Shadle high school - Go Highlanders!). It's paid huge dividends in my running - anyone can follow the same training plan, run the same workouts, pay an experienced coach, but the person who has put in the work AND has been able to recover from it, will have a better race. Always. Hands down. No 'ifs', 'ands', or 'buts'...UNLESS, the recovered person gets hit along the course...or trips & breaks something...or some other freak accident happens. What I'm saying, is recovering properly is what sets you apart from your competitors. If you want to be set apart and kick bootay, bone broth has a place in your daily routine.
You might be turned off, initially, by the name or what it is, but bone broth can be made in a variety of ways. Check out some of our favorite ways here! What happens when you boil down bones from high quality animals - such as the organic beef and chicken bones we use from Egger's Meats Northside Spokane - is that you create a product that is highly-nutrient-dense - you create collagen. Collagen is the most absorble protein to our bodies and main component of our connective tissue, making up 25% - 35% of our whole body protein content. It's used in the medical industry for injuries and complications to the bones and skin of our bodies. In other words, helps us heal quicker and makes our skin, hair, and fingernails look beautiful and strong! Who doesn't want locks like Troy Polamalu?!
I love Tim Ferriss (yes, two R's and two S's....thanks for the spellcheck, Chelsea ;) looove you! <3). If you don't know who Tim Ferriss is and you care at all about learning, bettering yourself, or being a business guru and learning from the best & most creative, well, you need to know him! Jump on his podcasts. Most famous for "The Four-Hour Work Week", each podcast he interviews a different expert in their specific field. He compresses how to learn a new language in 8 weeks, tells you how to be the best in the world by listening & learning to the best in the world. My man Tim also introduces us to great products - that I'm sure he gets paid for BUT - he uses and stands behind. Pu-Erh tea was first introduced to me by way of this blog. This tea has an earthy, very subtle taste to it - great with a little creamer, or not. I personally drink it alone and love it! Some have been calling it the "skinny tea", but that's just silly. What I can tell you though, is it has some very beneficial medicinal traits that will fire your metabolism and help heal your gut. These two, together, support weight loss. Pu-Erh tea is a fermented tea and anything fermented creates good bacteria in our gut for us - good bacteria in our gut eat the poopy bacteria and waa-laa, you're clean and your digestive track can work as intended. Tea in general is anti-inflammatory because of high antioxidants. Antioxidants, as a refresher to some, are from plant compounds called phytochemicals - protective layers for our cells from oxidation and therefore free radicals (the things that cause cancer and deadly diseases!).
And to that wreath in the picture above...Well, that represents family & love & Christmas for me - and for many others. My grandma (we call her 'naa-na') taught my sister-in-law, my sister, and I how to make the most beautiful REAL holly (from her backyard) wreaths last week. Then we decorated it. This wreath was far more beautiful than any I've seen at a store & made with so much love. It was a reminder to slow down and even though there is always business - a customer emailing, calling...God forbid, complaining - family and those who are there for you are first. Take some time in the coming weeks to breathe in a moment that you are making into a memory; take note of this moment. These are what you'll remember on your deathbed. But, in the meantime, take care of yourself to get the most out of this life - Pantry Fuel is here to help.


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text:
Manhattan Center
Best Rate, Guaranteed
Find out what to do and where to go in the greatest city in the world from the hotel that shares its name — The New Yorker.
Top Things to Do in Ne...
Enjoying the Winter in...
The New Yorker Hotel T...
Fun Facts about Radio ...
Popular Winter Events ...
The New Yorker History (11)
New York City Events (39)
Local Tips (50)
Top Events to Watch Out for This March
February 20th, 2018 New York City Events
There's always something happening in the city that never sleeps and March is slated to be an exciting month for the corner of 34th St. If you're planning a stay at The New Yorker Hotel, check out these upcoming events you can't miss.
Calling all basketball and hockey fans out there! The New York Knicks and New York Rangers are taking on their rivals on various dates in March. Purchase your tickets to any or every game from the MSG website. Special events include:
45th Annual Ice Hockey Heroes: FDNY vs NYPD on March 24, 2018
NCAA Big East Men's Basketball Tournament running from March 7-9
NCAA 81st Annual NIT - Semifinal Doubleheader on March 27.
But wrestling and tennis won't stay behind. WWE Live Road to WrestleMania will fire up the arena on March 16, with some of your favorite WWE stars. Last but definitely not least, Tie Break Tens is coming to the US for the first time on March 5. Watch 8 female players battle each other in a winner takes all tournament.
If you're thinking that MSG is all about sports, think again! The arena lights up and turns into a stage for Justin Timberlake's The Man Of The Woods Tour on March 21-22. The one and only Billy Joel will close up the month on March 28 with what looks to be another memorable concert in The World's Greatest Arena.
A short trip will take you to Brooklyn's Barclays Center will open the doors to fast-paced hockey starring the New York Islanders and basketball from the local team, Brooklyn Nets. What's more, the best college basketball teams such as New York's own Syracuse University, University of Pittsburgh, Duke University, University of North Carolina, Virginia Tech and many other talented colleges/universities, will dispute the ACC title during the 2018 ACC Men's Basketball Tournament from March 6-10. Take the 2 or 3 train located just one block from The New Yorker Hotel, on 34th Street and 7th Avenue going downtown to Atlantic Avenue-Barclays Center.
Exclusive Offers: Available Only On Website
But the excitement doesn't stop there. On March 3, Deontay Wilder and Luis Ortiz will finally face each other in what is expected to be a thrilling bloodbath between these top heavyweight champions. Last but certainly not least, Demi Lovato & DJ Khaled will pay a visit to New York City on March 16 as part of their international tour, with special guest Kehlani (subject to change).
Known for its superb technology and astounding acoustics, the Manhattan Center never disappoints. On March 12, A Day To Remember will rock The Hammerstein as part of their 15 Years in the Making Tour. And the past few years have indeed been a rollercoaster for ADTR, originally from Florida. After all, one doesn't get to the No. 1 spot on Billboard's Rock, Indie and/or Alternative charts for nothing!
The Manhattan Center and The New Yorker Hotel have a long history together - we're, literally, next door. Get your tickets, book your room with us and why not enjoy a nightcap after the show in Butcher & Banker.
Jacob Javits Center
With March just days away, the auto universe begins to stir: the New York International Auto Show is back in town from March 30 to April 8.
The largest collection of cars, SUVs and trucks draws thousands of visitors every year. Its popularity stems from the fact that the show is more than just cars on display. You'll discover everything about the latest car technology, while interactive presentations and virtual reality experiences will keep the young ones entertained.
By now you've probably realized that if you're attending any of these events, The New Yorker Hotel is the only hotel that will be equally entertaining. Read about our rich history and follow this blog to stay up-to-date with the latest happenings in NYC.
#newyorkerhotel
481 8th Avenue & 34th Street New York, NY 10001 (212) 971 0101
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text:
Аеродро́м «Шуге Естейт» () — аеродром Свазіленду, розташований поряд з містечком Біг-Бенд.

Посилання
 Pict Aero
 FDBS - Location

Шуге Естейт

redpajama_set_name: RedPajamaWikipedia


For there is one God and one mediator between God and mankind, the man Christ Jesus. Porque hay un solo Dios, y un solo mediador entre Dios y los hombres, Jesucristo hombre.
For I am convinced that neither death nor life, neither angels nor demons, neither the present nor the future, nor any powers, neither height nor depth, nor anything else in all creation, will be able to separate us from the love of God that is in Christ Jesus our Lord. Por lo cual estoy seguro de que ni la muerte, ni la vida, ni ángeles, ni principados, ni potestades, ni lo presente, ni lo por venir, ni lo alto, ni lo profundo, ni ninguna otra cosa creada nos podrá separar del amor de Dios, que es en Cristo Jesús Señor nuestro.
Jesus answered, "I am the way and the truth and the life. No one comes to the Father except through me." Jesús le dijo: Yo soy el camino, y la verdad, y la vida; nadie viene al Padre, sino por mí.
My prayer is not that you take them out of the world but that you protect them from the evil one. No ruego que los quites del mundo, sino que los guardes del mal.
God made him who had no sin to be sin for us, so that in him we might become the righteousness of God. Al que no conoció pecado, por nosotros lo hizo pecado, para que nosotros fuésemos hechos justicia de Dios en él.
And he who searches our hearts knows the mind of the Spirit, because the Spirit intercedes for God's people in accordance with the will of God. Mas el que escudriña los corazones sabe cuál es la intención del Espíritu, porque conforme a la voluntad de Dios intercede por los santos.


Meal plan week 1 (sample menu).
Our philosophy is 'Love Your Health' so let us help you be the healthiest you can be!
Below is a sample menu, that changes weekly. We can cater for food allergies, dietary requirements and any higher calorie meal preferences. Just let us know: contact us or call 029 2049 3348.
We're excited to announce, you can now order our 5-day Pure Kitchen meal plans online.
The 5 day meal plan package runs from Monday to Friday and costs £99 / £125.
If your current weight is below 180lbs (12st 7lbs) the recommended plan is the 1400kcal a day priced at £99 per week, under £20 a day.
If your weight is above 180lbs the 1800kcal plan is recommended, the price is £125 per week, £25 per day.
All our food is freshly prepared by our chef daily and delivered between 7am – 9am each day. There is a delivery charge of £5 a week to anyone within a 5 mile radius. If you live outside the 5 mile radius please give us a ring to discuss further charges and delivery times. If you prefer you can pick up from our restaurant in the bay at a time that is more convenient to you.
We want you to fully enjoy everything you receive from Pure Kitchens so it's important to us to make sure we prepare all of your meals just how you like them. We also need to make sure we cook and prepare your meal away from any ingredients you may have an allergy to. If you don't suffer from any allergies then please tick the box below to confirm that, otherwise please list your allergies in the text field below.
Yes, I have read the allergy statement and confirm that I do not have any allergies or I have listed them below.
Copyright ©2017 Vistaridge Ltd. Company no. 8741738.


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Q: Angular nodejs Getting an Error in adding file Hello I'm trying to add a new file using Angular and Nodejs into my mongodb database
The backend function works fine on postman
But on the frontend it gives me an error and i didnt figure out what's the problem
It used to work up until I made some changes in the backend function
this is my POST API
router.post('/create',multer({ storage : storage}).any(), (req, res) => {
    var sample = fs.readFileSync('./uploads/'+req.files[0].filename,'utf8');
    function Filetostring(){
        let arr = sample.split(/\r?\n/);
        // Find returns the first element that matches our criteria
        const step = arr.filter((step , idx)=> {
            if (step.includes("step")) {
             console.log(step);             //this gives me the first result I need

             return true;

            }
            return false;
         });
       let stepss = JSON.stringify(step)
       let newString = stepss.replace(/[\[\]\'\s]/g, ' ');
       let newString2 = newString.replace(/"/g, '');
       let newString3 = newString2.replace(/#step/g, '');
       let newString4 = newString3.replace(/,/g, '\n');

       return newString4;
     }

     let steps = Filetostring()
     console.log(steps)

    var tc = new Testcase({
        name: req.body.name,
        upload: req.files[0].filename ,

        run : steps,

        modify: req.body.modify,
        delete: req.body.delete,
        step1: req.body.step1,
        step2: req.body.step2,
        step3: req.body.step3,
        step4: req.body.step4,
        step5: req.body.step5,
        step6: req.body.step6,
        step7: req.body.step7,
    });

    console.log(req.files[0].filename);
    tc.save((err, doc) => {
        if (err) { res.status(401).send("errorrrrr") }
        else {
            res.status(200).send(doc)
        }
    });
});

This is my service.ts
  private create = 'http://localhost:5555/testcase/create';

  postTestCase(testcase,file:File) {
    const formData  = new FormData();
    formData .append('file', file);
    formData .append('name', testcase.name);
    formData .append('step1', testcase.step1);
    formData .append('modify', testcase.modify);
    formData .append('run', testcase.run);
    formData .append('delete', testcase.delete);
    formData .append('step2', testcase.step2);
    formData .append('step3', testcase.step3);
    formData .append('step4', testcase.step4);
    formData .append('step5', testcase.step5);
    formData .append('step6', testcase.step6);
    formData .append('step7', testcase.step7);

    return this.http.post<any>(this.create,   formData );
  }

this is my component.ts
  addtestcaseData = {'name': '',
  'step1': '',

  'modify': '',
  'run': '',
  'delete': '',
  'upload': File = null,
  'step2': '',
  'step3': '',
  'step4': '',
  'step5': '',
  'step6': '',
  'step7':''}

  addtestcase(){
    this.testcaseService.postTestCase(this.addtestcaseData,this.addtestcaseData.upload)
    .subscribe(
      res => {
        console.log(res)
        this._router.navigate(['/admin'])
      },
      err => {
        console.log(err);

      }
    )
  };

  handleFileInput(files: FileList) {
    this.addtestcaseData.upload = files.item(0);
  }

And this is my component.html

Add Test Case

            <form class="form" #testcaseForm="ngForm" enctype="multipart/form-data" >

                Name
                <input [(ngModel)]="addtestcaseData.name" class="form-control rounded-0"  name="Name" type="text" required>

                Upload Test Case
                <input type="file"
                       id="file"  name="file"
                       (change)="handleFileInput($event.target.files)">

              <button (click)="addtestcase()"  type="button"  class="btn btn-success float-right">Add</button>
            </form>

A: Try to get file with req.file
        var tc = new Testcase({
        name: req.body.name,
        upload: req.file,

        run : steps,

        modify: req.body.modify,
        delete: req.body.delete,
        step1: req.body.step1,
        step2: req.body.step2,
        step3: req.body.step3,
        step4: req.body.step4,
        step5: req.body.step5,
        step6: req.body.step6,
        step7: req.body.step7,
    });

    console.log(req.files[0].filename);

And for route
router.post('/create',multer({ storage : storage}).single('upload'), (req, res) => {

For frontend change the name of input to
<input type="file"
                   id="file"  name="upload"
                   (change)="handleFileInput($event.target.files)">

redpajama_set_name: RedPajamaStackExchange


By Vivian M. May Tuesday, October 24th 2017 09:43:36 AM .
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By Vivian M. May Tuesday, October 24th 2017 09:22:22 AM .


MapSource City Navigator is designed specifically for the StreetPilot III. The CD contains premium detailed maps of major metropolitan areas in 10 different pre-defined regions for the USA and one region of Canada. If you plan on using your Streetpilot III in many areas, this CD allows you to download any DCA (Detailed Coverage Area) in the country.
Allows download of any detailed coverage area (DCA) anywhere in the U.S.


text:
Trine Schmidt (nascida em 3 de junho de 1988) é uma ciclista dinamarquesa que compete para a equipe feminina britânica Garmin-Cervélo, desde 2011. Tornou-se profissional em 2007 quando alinhava para a equipe Flexpoint.

Carreira profissional
Especialista em provas do ciclismo de pista, Schmidt é medalhista de prata no campeonato mundial de corrida por pontos em 2008.

Participou nos Jogos Olímpicos de 2008 em Pequim, competindo na prova de corrida por pontos e terminando na décima oitava posição.

Naturais de Copenhague
Ciclistas olímpicos da Dinamarca
Ciclistas nos Jogos Olímpicos de Verão de 2008
Ciclistas de pista