---
abstract: 'Let $X$ be a complete simplicial toric variety over a finite field ${\mathbb F}_q$ with homogeneous coordinate ring $S={\mathbb F}_q[x_1,\dots,x_r]$ and split torus $T_X\cong ({\mathbb F}^*_q)^n$. We prove that vanishing ideal of a subset $Y$ of the torus $T_X$ is a lattice ideal if and only if $Y$ is a subgroup. We show that these subgroups are exactly those subsets that are parameterized by Laurents monomials. We give an algorithm for determining this parametrization if the subgroup is the zero locus of a lattice ideal in the torus. We also show that vanishing ideals of subgroups of $T_X$ are radical homogeneous lattice ideals of dimension $r-n$. We identify the lattice corresponding to a degenerate torus in $X$ and completely characterize when its lattice ideal is a complete intersection. We compute dimension and length of some generalized toric codes defined on these degenerate tori.'
address: 'Department of Mathematics, Hacettepe University, Ankara,TURKEY'
author:
- Mesut ŞAHİN
title: Toric Codes and Lattice Ideals
---

[^1]

Introduction
============

Let $X$ be a complete simplicial toric variety over a field ${{\mathbb K}}$, corresponding to a fan ${\Sigma}\subset {\mathbb R}^n$ and let $T_X \cong ({{\mathbb K}}^*)^n$ be split which is the case for instance when ${{\mathbb K}}$ is algebraically closed or finite. Denote by $\rho_1,\dots,\rho_r$ the rays in ${\Sigma}$ and $\vv_1,\dots,\vv_r\in {\mathbb Z}^n$ the corresponding primitive lattice vectors generating them. Given a vector $\uu=(u_1,\dots,u_r)$ we use $\mathbf{t}^\uu$ to denote the monomial $\mathbf{t}^\uu=t_1^{u_1}\dots t_r^{u_r}$. Recall the following dual exact sequences: $${\displaystyle}\xymatrix{ \mathfrak{P}: 0  \ar[r] & {\mathbb Z}^n \ar[r]^{\phi} & {\mathbb Z}^r \ar[r]^{{\bb}} & {{\mathcal{A}}}\ar[r]& 0},$$ where $\phi$ is the matrix with rows $\vv_1,\dots,\vv_r$, and $${\displaystyle}\xymatrix{ \mathfrak{P}^*: 1  \ar[r] & G \ar[r]^{i} & ({{\mathbb K}}^*)^r \ar[r]^{\pi} & T_X \ar[r]& 1},$$ where $\pi:(t_1,\dots, t_r)\mapsto [\mathbf{t}^{\uu_1}: \cdots : \mathbf{t}^{\uu_n}],$ with $\uu_1,\dots, \uu_n$ being the columns of $\phi$ and $G={\operatorname{Ker}}(\pi)$.

Let ${\displaystyle}S={{\mathbb K}}[x_1,\dots, x_r]=\bigoplus_{\aa \in {{\mathcal{A}}}} S_{\aa}$ be the homogeneous coordinate or Cox ring of $X$, multigraded by ${{\mathcal{A}}}\cong {\operatorname{Cl}}(X)$ via $\bb_j:=\deg_{{{\mathcal{A}}}}(x_j):=\bb(e_j)$, where $e_j$ is the standart basis element of ${\mathbb Z}^r$ for each $j=1,\dots,r$. The irrelevant ideal is a monomial ideal defined by $B={\langle}x^{\hat{{\sigma}}} ~:~ {\sigma}\in {\Sigma}{\rangle}$, where ${\displaystyle}x^{\hat{{\sigma}}}=\Pi_{\rho_i \notin {\sigma}}^{}x_i$. Thus, $T_X \cong ({{\mathbb K}}^*)^r /G$ and $X\cong ({ {{\mathbb K}}^r}\setminus V(B)) /G$ as a geometric quotient if we additionally assume that the order of the torsion part of the group ${{\mathcal{A}}}$ is coprime to the characteristic of the field ${{\mathbb K}}$, since $G$ will be reductive in this case, see [@Coxhom; @CLSch]. This assumption is satisfied in particular when ${{\mathcal{A}}}$ is free or ${{\mathbb K}}$ has characteristic zero. The homogeneous polynomials of $S$ are supported in the semigroup ${\mathbb N}\bb$ generated by $\bb_1,\dots,\bb_r$, i.e. $\dim_{{{\mathbb K}}}{S}_{\aa}=0$ when $\aa \notin {\mathbb N}\bb$.

Next, we recall evaluation codes defined on subsets $Y=\{[P_1],\dots,[P_N]\}$ of the torus $T_X$. Fix a degree $\aa\in{\mathbb N}\beta$ and a monomial $F_0=\x^{\phi({\mathfra{{\mathbb K}}{m}}_0)+\a} \in S_{\aa}$, where ${\mathfra{{\mathbb K}}{m}}_0\in {\mathbb Z}^n$, $\a$ is any element of ${\mathbb Z}^r$ with $\deg(\a)=\aa$, and $\phi$ as in the exact sequence $\mathfrak{P}$. This defines the [*evaluation map*]{} $$\label{e:evalmap}
\ev_{Y}:S_\aa\to {\mathbb F}_q^N,\quad F\mapsto \left(\frac{F(P_1)}{F_0(P_1)},\dots,\frac{F(P_N)}{F_0(P_N)}\right).$$ The image ${{\mathcal{C}}}_{\aa,Y}=\text{ev}_{Y}(S_\aa)$ is a linear code, called the [*generalized toric code*]{}. The block-length $N$, the dimension $k=\dim_{{\mathbb F}_q}({{\mathcal{C}}}_{\aa,Y})$, and the minimum distance $d=d({{\mathcal{C}}})$ are three basic parameters of ${{\mathcal{C}}}_{\aa,Y}$. Minimum distance is the minimum of the number of nonzero components of nonzero vectors in ${{\mathcal{C}}}_{\aa,Y}$. Toric codes was introduced for the first time by Hansen in [@Ha0; @Ha1] and studied later in [@Jo; @Ru] for the special case of $Y=T_X$. Clearly, the block-length of ${{\mathcal{C}}}_{\aa,Y}$ equals $N=|T_X|=(q-1)^n$ in this case. But it is not known in the general case. An algebraic way to compute the dimension and length of a generalized toric code is given in [@sasop]. This method is based on the observation that the kernel of the evaluation map above is determined by the vanishing ideal of $Y$ defined as follows. For $Y\subset X$, we define the vanishing ideal $I(Y)$ of $Y$ to be the ideal generated by homogeneous polynomials vanishing on $Y$. $I(Y)$ is a [*complete intersection*]{} if it is generated by a regular sequence of homogeneous polynomials $F_1,\dots, F_k\in S$ where $k$ is the height of $I_Y$. When the vanishing ideal $I_Y$ is a complete intersection, bounds on the minimum distance of ${{\mathcal{C}}}_{\aa,Y}$ is provided in [@sop]. There are interesting results about evaluation codes on complete intersections in literature, e.g. [@DRT; @GLSch; @Ha3]. These results motivate studying vanishing ideals of special subsets of the torus $T_X$ and characterize when they are complete intersections.

When the vanishing ideal is a binomial or a lattice ideal it is relatively easier to characterize whether it is a complete intersection, see [@MT] and references therein. In the case that the toric variety $X$ is a projective space over a finite field, there are interesting results relating lattice ideals and subtori of the torus $T_X$. Vanishing ideals of subgroups of $T_X$ parameterised by Laurent monomials are studied in [@RSV] and shown to be homogeneous lattice ideals of dimension $1$. Binomial ideals that are vanishing ideals are characterised in [@TV]. It was also proven that homogeneous lattice ideals of dimension $1$ correspond to subgroups of the torus $T_X$. In [@NPV], these subgroups are identified as being the subgroups of $T_X$ parameterized by Laurent monomials. In the case that the toric variety $X$ is weighted projective space over a finite field, vanishing ideal of the torus itself is shown to be a homogeneous lattice ideal of dimension $1$ in [@DN].

We use some of the main ideas in these works and extend the main results to more general toric varieties over any field in the present paper. In section $2$, we show that a homogeneous binomial ideal gives a submonoid of $T_X$, see Proposition \[p:monoid\] and Corollary \[c:latticemonoid\]. Conversely, we show that vanishing ideals of submonoids of $T_X$ are lattice ideals in Proposition \[p:binom\]. We prove that vanishing ideal of a subset $Y$ of the torus $T_X$ is a lattice ideal if and only if $Y$ is a subgroup, for ${{\mathbb K}}={\mathbb F}_q$, in Theorem \[t:subgpsLattices\]. In section $3$, we focus on finite submonoids of the torus when ${{\mathbb K}}$ is algebraically closed or finite. We show that these submonoids coincide with the submonoids parameterized by Laurent monomials on a finite subgroup of ${{\mathbb K}}^*$, see Theorem \[t:parametrizingsubmonoids\]. We also give an algorithm for parameterizing the submonoid obtained from a homogeneous lattice ideal, when ${{\mathbb K}}={\mathbb F}_q$, see Proposition \[p:parameterisingV\_L\] and Algorithm $1$. In section $4$, we determine the homogeneous lattice defining the vanishing ideals of subtori of $T_X$ called the degenerate tori which are parameterized by diagonal matrices, see Theorem \[t:I(Y\_A)\]. We also characterize when these are complete intersection in Proposition \[p:ci\]. In section $5$, we list main properties of vanishing ideals of arbitrary subsets of the torus $T_X$ and prove that they must be radical homogeneous ideals of dimension $r-n$, see Theorem \[t:primdec\]. Finally, we give quick applications of these on evaluation codes on degenerate tori, see Corollaries \[c:dimcode\] and \[c:lengthcode\].

Lattice Ideals and Submonoids of the Torus {#S:latticeideals}
==========================================

In this section we explore the connection between homogeneous lattice ideals and submonoids of the torus $T_X$.

Recall that $\x^{\uu}:=x_1^{u_1}\dots x_m^{u_m}$ denotes a Laurent monomial for a vector $\uu\in{\mathbb Z}^r$. A binomial is a difference $\x^{\a}-\x^{{\mathbf b}}$ of two monomials corresponding to the vectors $\a,{\mathbf b}\in {\mathbb N}^r$. An ideal is called a *binomial ideal* if it is generated by binomials. An exact sequence $\mathfrak{P}$ gives a grading on the polynomial ring $S={{\mathbb K}}[x_1,\dots,x_r]$, where $\deg_{{{\mathcal{A}}}}(x_j):=\bb_j$ is the $j$-th column of the degree matrix $\bb$. A homogeneous ideal in $S$ with respect to this grading is called ${{\mathcal{A}}}$-graded or ${{\mathcal{A}}}$-homogeneous. We usually just say that it is *homogeneous* if the group ${{\mathcal{A}}}$ is clear from the context.

Throughout $[P]=G \cdot P$ denotes a point in $X$. We use $[1]$ to denote $[1:\cdots:1]$. Similarly, $[V(B)]$ denotes the set of elements $[P]$ suct that $P\in V(B)\subset {{\mathbb K}}^r$. If $[P],[P']\in X$ then $[P]\cdot[P']:=[PP']$ is a well-defined element of $X\cup [V(B)]$. The set $[V(B)]$ does not include $[1]$ but it is closed with respect to this coordinatewise multiplication operation as $x^{\hat{{\sigma}}} (PP')=x^{\hat{{\sigma}}} (P)x^{\hat{{\sigma}}} (P')=0$ when $[P]\in [V(B)]$ or $[P']\in [V(B)]$. Although $X$ might not be closed with respect to this multiplication, $X\cup [V(B)]$ is a monoid with identity $[1]$. Finally, if $Y\subseteq X$ is a submonoid of $X\cup [V(B)]$, then so is $Y\cup [V(B)]$.

For a homogeneous ideal $J$ of $S$, let $$V_X(J):=\{[P]\in X : F(P)=0, \mbox{for all homogeneous}\: F\in J\}.$$

\[p:monoid\] If $J\subset S$ is a homogeneous binomial ideal then $V_X(J)\cup [V(B)]$ is a submonoid of $X\cup [V(B)]$. If furthermore, $PP'\notin V(B)$, for all $[P],[P']\in V_X(J)$, then $V_X(J)$ is a submonoid of $X\cup [V(B)]$. In particular, $V_X(J)\cap T_X$ is a submonoid of $T_X$.

If $0\neq f=\x^{\a}- \x^{{\mathbf b}}$ is a generator of $J$, then $f([1])=1-1=0$. So, $[1]\in V_X(J)$. If $[P],[P']\in V_X(J)$, then $f(P)=f(P')=0$. This implies that $\x^\a(P)=\x^{\mathbf b}(P)$ and $\x^\a(P')=\x^{\mathbf b}(P')$. So, $\x^{\a}(PP')=\x^{\mathbf b}(PP')$ yielding $f(PP')=0$, i.e. $PP'\in V(J)$. Hence $[V(J)]=V_X(J)\cup [V(B)]$ is a submonoid. If furthermore, $PP'\notin V(B)$, for all $[P],[P']\in V_X(J)$, then $[PP']\in V_X(J)$ and thus $V_X(J)$ is a submonoid. When $[P],[P']\in V_X(J)\cap T_X$, we have $PP'\notin V(B)$ and so $[PP']\in V_X(J)\cap T_X$. Therefore, $V_X(J)\cap T_X$ is a submonoid.

A lattice $L$ is a subgroup of ${\mathbb Z}^r$ and is called [**homogeneous**]{} if $L \subseteq L_{\bb}={\operatorname{Ker}}\bb={\operatorname{Im}}\phi$. Recall that every vector in ${\mathbb Z}^r$ is written as ${\mathfra{{\mathbb K}}{m}}={\mathfra{{\mathbb K}}{m}}^+ -{\mathfra{{\mathbb K}}{m}}^-$, where ${\mathfra{{\mathbb K}}{m}}^+ ,{\mathfra{{\mathbb K}}{m}}^-\in {\mathbb N}^r$. Letting $F_{\mathfra{{\mathbb K}}{m}}=\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}^-} $, the **lattice ideal** $I_L$ is the binomial ideal $I_L=\langle F_{\mathfra{{\mathbb K}}{m}}\: | \: {\mathfra{{\mathbb K}}{m}}\in L \rangle$ generated by special binomials $F_{\mathfra{{\mathbb K}}{m}}$ arising from $L\subset {\mathbb Z}^r$.

\[p:homogeneouslattice\] $L$ is homogeneous if and only if $I_L$ is homogeneous.

Since $I_L$ is generated by the set $\{F_{\mathfra{{\mathbb K}}{m}}\: | \: {\mathfra{{\mathbb K}}{m}}\in L\}$, it is homogeneous if and only if $F_{\mathfra{{\mathbb K}}{m}}$ is homogeneous for all ${\mathfra{{\mathbb K}}{m}}\in L$. But $F_{\mathfra{{\mathbb K}}{m}}=\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}^-} $ is homogeneous $\iff \bb({\mathfra{{\mathbb K}}{m}}^+)=\bb({\mathfra{{\mathbb K}}{m}}^+)$ $\iff \bb({\mathfra{{\mathbb K}}{m}})=0$ $\iff {\mathfra{{\mathbb K}}{m}}\in L_{\bb}$. Thus, $I_L$ is homogeneous if and only if ${\mathfra{{\mathbb K}}{m}}\in L_{\bb}$ for all ${\mathfra{{\mathbb K}}{m}}\in L$, that is, $L$ is homogeneous.

\[c:latticemonoid\] If $L$ is homogeneous, $V_X(I_L)\cap T_X$ is a submonoid of $T_X$.

This follows from Propositions \[p:monoid\] and \[p:homogeneouslattice\].

\[D:VanIdeal\] For $Y\subseteq X$, we define the vanishing ideal $I(Y)$ of $Y$ to be the ideal generated by homogeneous polynomials vanishing on $Y$.

\[p:binom\] Let $Y \subseteq X$. If $Y$ is a submonoid of $X\cup [V(B)]$, then $I(Y)$ is a binomial ideal. If $Y \subseteq T_X$ is a submonoid, $I(Y)$ is a lattice ideal.

If $Y$ is a submonoid of $X\cup [V(B)]$, the set $\mathcal{Y}=\{P\in {{\mathbb K}}^r \:|\: [P]\in Y\}$ is a submonoid of ${{\mathbb K}}^r$. Take a homogeneous polynomial $f=\lambda_1 \x^{\a_1}+\cdots+\lambda_k \x^{\a_k}$ from $ I(Y)$ with exactly $k$ terms. As $[1]\in Y$, $f$ can not be a monomial, i.e. $k>1$. Since monomials are characters from $\mathcal{Y}$ to the multiplicative monoid $({{\mathbb K}},.,1)$, it follows from Dedekind’s Theorem that the monomials of $f$ can not be distinct characters as in the proof of [@TV Theorem 3.2]. If, say, $\x^{\a_1}$ and $\x^{\a_2}$ are the same as characters, then $\x^{\a_1}-\x^{\a_2}\in I(Y)$. Thus $f-\lambda_1(\x^{\a_1}-\x^{\a_2})$ lies in $I(Y)$ and has exactly $k-1$ terms. By induction on $k$, $f$ must be written as an algebraic combination of binomials, and hence $I(Y)$ is a binomial ideal. For the second part, take a polynomial $f$ of $S$ such that $x_jf\in I(Y)$ for any $j=1,\dots,r$. By [@binomial Corollary 2.5], it is enough to show that $f\in I(Y)$. Let $f=\bigoplus\limits_{\alpha\in {{\mathbb N}\beta}}f_\alpha$ be the expansion of $f$ as the sum of its homogeneous components where $f_\alpha$ has total degree $\alpha$. Then $x_j{f}=\bigoplus\limits _{\alpha\in {{\mathbb N}\beta}}x_j{f_\alpha}$ and $x_j{f_\alpha}$ is homogeneous of total degree $\alpha+ \beta_j$. As $I(Y)$ is a homogeneous ideal, $x_j{f_\alpha}\in I(Y)$. So, $(x_jf_\alpha)(P)=p_jf_{\alpha}(P)=0$ for any element $[P] \in Y$. Since $P\in({{\mathbb K}}^*)^r$, $p_j\neq 0$, and so $f_{\alpha}(P)=0$. This yields that $f_\alpha\in I(Y)$ for all $\alpha$, consequently $f\in I(Y)$ completing the proof.

It is not clear if the hypothesis of Proposition \[p:binom\] can be replaced with the weaker hypothesis that $Y\cup [V(B)]$ is a submonoid of $X \cup [V(B)]$. When $X$ is a projective space or weighted projective space $V(B)=\{0\}$, so this can be done using the fact that if $f\in I(Y)$ is a homogeneous polynomial then it also vanishes on $V(B)$. See [@TV Theorem 3.2] for the case of projective space.

\[L:Zariski\]We have the following basic properties.

1.  $\emptyset=V_X(1)$ and $X=V_X(0)$.\

2.  $V_X(J_1)\cup V_X(J_2)=V_X(J_1J_2)$ and $\displaystyle \bigcap_{i} V_X(J_i)=V_X(\bigcup_i J_i)$.\

3.  $Y_1\subseteq Y_2 \implies I(Y_2)\subseteq I(Y_1)$ .\

4.  $J_1\subseteq J_2 \implies V_X(J_2)\subseteq V_X(J_1)$.\

5.  If $\bar{Y}$ is the Zariski closure of $Y$, then $ I(Y)= I(\bar{Y})$ .\

6.  $\bar{Y}= V_X(I(Y))$.\

These can be proven using the fact that homogeneous ideals are generated by homogeneous polynomials. For instance, if $[P]\in V_X(J_1J_2)\setminus V_X(J_1)$, then it suffices to show that $G(P)=0$ for every homogeneous $G\in J_2$. Since, there is some homogeneous $F\in J_1$ with $F(P)\neq 0$ and $F(P)G(P)=0$, this follows at once. So, $V_X(J_1J_2) \subseteq V_X(J_1)\cup V_X(J_2)$. The rest is either similar or obvious.

We define Zariski topology on $X$ by taking closed sets to be the zero loci $V_X(J)$ for homogeneous ideals $J$.

We close this section with establishing the following correspondence.

\[t:subgpsLattices\]Let $Y \subseteq T_X$ and ${{\mathbb K}}={\mathbb F}_q$. Then, $I(Y)$ is a lattice ideal if and only if $Y$ is a subgroup.

If $I(Y)$ is a lattice ideal then $V_X(I(Y))\cap T_X$ is a submonoid of $T_X$ by Proposition \[p:monoid\]. Since $V_X(I(Y))=\bar{Y}=Y$ lies in $T_X$ and ${{\mathbb K}}$ is finite $Y$ is a subgroup. The converse follows directly from Proposition \[p:binom\].

Finite submonoids of the torus
==============================

In this section we completely characterize finite submonoids of the torus $T_X$ when ${{\mathbb K}}$ is a finite field or an algebraically closed field by showing that they are exactly those that are parameterized by Laurent monomials. This was done when $X$ is a projective space over a finite field by [@NPV] and over an algebraically closed field by [@TV]. We give an algorithm producing these monomials when ${{\mathbb K}}$ is a finite field and the submonoid corresponds to a homogeneous lattice ideal.

Let $H$ be a subset of ${{\mathbb K}}^*$ and $Q$ be an $s\times r$ integer matrix with entries $q_{ij}$. Then the parameterized set corresponding to $Q$ and $H$ is defined as follows: $$Y_{Q,H}=\{[t_1^{q_{11}}\cdots t_s^{q_{s1}}:\cdots :t_1^{q_{1r}}\cdots t_s^{q_{sr}}]: t_1,\dots,t_s \in H\}.$$

\[t:parametrizingsubmonoids\] Let ${{\mathbb K}}$ be a finite field or an algebraically closed field. Then the following are equivalent:

1.  $Y$ is a **finite** submonoid of $T_X$.\

2.  $Y =Y_{Q,H}$ for a **finite** submonoid $H$ of ${{\mathbb K}}^*$.

Suppose that $(1)$ holds. In this case $Y$ is a finite subgroup as being a finite sumbonoid of a group. Thus $Y$ is a product ${\langle}[P_1] {\rangle}\times\cdots \times \langle [P_s] \rangle$ of cyclic subgroups generated by points $[P_1],\dots, [P_s]\in Y$ of orders, say, $c_1,\dots,c_s$. In this case, for every $i=1,\dots,s$, we have $[P_i]^{c_i}=[1]=G$. If ${{\mathbb K}}$ is finite, we define $H$ to be the subgroup generated by the coordinates $p_{ij}$ of $P_1,\dots, P_s$. Then $H$ is a finite subgroup of ${{\mathbb K}}^*$. If ${{\mathbb K}}$ is algebraically closed, then we define $H$ to be the subgroup generated by all $x\in {{\mathbb K}}$ such that $x^{c_i}=1$, for some $i=1,\dots,s$. We show that $H$ is non-empty. Since $(1,\dots,1)\in G=[P_i^{c_i}]$ and $G$ is parameterized by the columns of $\bb$, it follows that $(1,\dots,1)=(\mathbf{t}^{\bb_1}p_{i1}^{c_i},\dots,\mathbf{t}^{\bb_r}p_{ir}^{c_i})$. So, $\mathbf{t}^{\bb_j}p_{ij}^{c_i}=1$, for all $j=1,\dots,r$. Let ${\mu}_{jk}\in {{\mathbb K}}^*$ be such that ${\mu}_{jk}^{c_i}=t_k$, for $j=1,\dots,r$ and $k=1,\dots,r-n$. This leads to $\mu_j^{c_i\bb_j}=\mathbf{t}^{\bb_j}$, where $\mu_j=({\mu}_{i1},\dots,{\mu}_{ik})$. Thus, we have $(\mu_j^{\bb_j}p_{ij})^{c_i}=1$, for all $j=1,\dots,r$. If $p'_{ij}=\mu_j^{\bb_j}p_{ij}$, then $p'_{ij}\in H$ and $[P'_i]=[P_i]$. In order to unify the rest of the proof for both cases that ${{\mathbb K}}$ is finite and algebraically closed, by abusing the notation, let us assume that $p_{ij}\in H$. This implies that $H$ is a non-empty finite subgroup of ${{\mathbb K}}$. Since ${{\mathbb K}}$ is a field and $H$ is finite, $H$ must be cyclic, i.e., $H={\langle}\eta {\rangle}$, for some $\eta\in {{\mathbb K}}^*$. Then $p_{ij}=\eta^{q_{ij}}$ for some $q_{ij}$. Let $Q$ be the $s\times r$ integer matrix with entries $q_{ij}$ and let $$Y_{Q,H}=\{[t_1^{q_{11}}\cdots t_s^{q_{s1}}:\cdots :t_1^{q_{1r}}\cdots t_s^{q_{sr}}]: t_1,\dots,t_s \in H\}.$$

Then $$\begin{array}{lcl}
[P]\in Y &\iff& [P]=[P_1]^{i_1}\cdots[P_s]^{i_s}=[p_{11}^{i_1}\cdots p_{s1}^{i_s}:\dots:p_{1r}^{i_1}\cdots p_{sr}^{i_s}]\\
&\iff&[P]=[\eta^{i_1q_{11}}\cdots \eta^{i_sq_{s1}}:\dots:\eta^{i_1q_{1r}}\cdots \eta^{i_sq_{sr}}]
\end{array}$$ Since $t_j\in H \iff t_j=\mu^{i_j}$, for all $j=1,\dots,s$, it follows that $[P]\in Y \iff [P]\in Y_{Q,H}$. Therefore, $(2)$ holds.

Suppose now that $(2)$ holds. Then, taking $t_i=1\in H$, for all $i=1,\dots,s$, we see that $[1]\in Y$. The product of two elements $[P]=[t_1^{q_{11}}\cdots t_s^{q_{s1}}:\cdots :t_1^{q_{1r}}\cdots t_s^{q_{sr}}]$ and $[P']= [{t'}_1^{q_{11}}\cdots {t'}_s^{q_{s1}}:\cdots :{t'}_1^{q_{1r}}\cdots {t'}_s^{q_{sr}}]$ of $Y$ lies in $Y$ since $t_it'_i\in H$ and $$[P][P']=[(t_1 {t'}_1)^{q_{11}}\cdots (t_s {t'}_s)^{q_{s1}}:\cdots :(t_1 {t'}_1)^{q_{1r}}\cdots (t_s {t'}_s)^{q_{sr}}].$$ Therefore, $(1)$ holds, as $Y$ is finite whenever $H$ is so.

\[l:varietyoflattice\] If $L$ is a homogeneous lattice with basis $\mathbf b_1,\dots,\mathbf b_{\ell}$ then $$V_X(I_L)\cap T_X=\bigcap_{i=1}^{\ell} V_X(F_{{\mathbf b}_i})\cap T_X.$$

Take a point $[P]$ from the intersection $\bigcap_{i=1}^{\ell} V_X(F_{{\mathbf b}_i})\cap T_X$. Then, $F_{{\mathbf b}_i}(P)=\x^{{{\mathbf b}_i}^+}(P)-\x^{{{\mathbf b}_i}^-}(P)=0$, for all $i=1,\dots,\ell$. Since $[P]\in T_X$, its coordinates are not zero, and thus $\x^{{\mathbf b}_i}(P)=1$, for all $i=1,\dots,\ell$. As we have ${\mathfra{{\mathbb K}}{m}}=c_1{\mathbf b}_1+\cdots+c_{\ell}{\mathbf b}_{\ell}$, for any ${\mathfra{{\mathbb K}}{m}}\in L$, it follows that $\x^{{\mathfra{{\mathbb K}}{m}}}(P)=1$. So, $F_{\mathfra{{\mathbb K}}{m}}(P)=0$, for any ${\mathfra{{\mathbb K}}{m}}\in L$. Hence, $[P]\in V_X(I_L)\cap T_X$. The other inclusion is straightforward.

For finite fields Theorem \[t:parametrizingsubmonoids\] has a more precise form given below.

\[p:parameterisingV\_L\] Let ${{\mathbb K}}={\mathbb F}_q$ be a finite field. Let $L\subset {\mathbb Z}^r$ be a homogeneous lattice of rank $\ell$ with basis vectors $\mathbf b_1,\dots,\mathbf b_{\ell}$ and $B_L$ be a matrix with rows of the form $[\mathbf b_j \; \mathbf (q-1) \mathbf e_j]$ where $\mathbf e_j$ is the standard basis vector of ${\mathbb Z}^{\ell}$. If $A_L$ is an integer matrix of size $(r+\ell)\times r$ such that ${\operatorname{Ker}}(B_L)={\operatorname{Im}}(A_L)$, and $A=[a_{ij}]$ is an $r \times r$ matrix consisting of the first $r$ rows of $A_L$, then $V_X(I_L)\cap T_X=Y_{A,{{\mathbb K}}^*}$.

Since ${{\mathbb K}}$ is a finite field, ${{\mathbb K}}^*={\langle}\eta {\rangle}$ is cyclic, and so every point in $T_X$ is of the form $[P]=[\eta^{s_1}:\cdots:\eta^{s_r}]$ for some $\mathbf s=(s_1,\dots,s_r)\in {\mathbb N}^r$. Then, by Lemma \[l:varietyoflattice\], we have $[P]\in V_X(I_L)$ if and only if $F_{{\mathbf b}_j}(P)=0$ for all $j=1,\dots,\ell$. This is equivalent to $\eta^{\mathbf s\cdot {\mathbf b}_j}=1$ for all $j=1,\dots,\ell$. Since the order of $\eta $ is $q-1$ the latter is equivalent to $\mathbf s\cdot \mathbf b_j \cong 0 \mod q-1$ for all $j=1,\dots,\ell$. This can be turned into a system of linear equations as $\mathbf s\cdot \mathbf b_j \cong 0 \mod q-1$ if and only if $\mathbf s\cdot \mathbf b_j =(q-1) c_j$ for some $c_j \in {\mathbb Z}$: $$\begin{array}{lcl}
 s_1b_{11}+\cdots+s_rb_{1r}-(q-1)c_1&=&0\\
 &\vdots&\\
  s_1b_{\ell 1}+\cdots+s_rb_{\ell r}-(q-1)c_{\ell}&=&0.
\end{array}$$ Then $B_L$ is the matrix of this system. Thus, $\mathbf s\in {\mathbb Z}^r$ is a common solution of the equations $\mathbf s\cdot \mathbf b_j \cong 0 \mod q-1$ for all $j=1,\dots,\ell$ if and only if $(\mathbf s,\mathbf c)\in {\operatorname{Ker}}(B_L)\cap {\mathbb Z}^{r+\ell}$ for some $\mathbf c=(c_1,\dots,c_{\ell}) \in {\mathbb Z}^{\ell}$. If $A_L$ is an integer matrix of size $(r+\ell)\times r$ such that ${\operatorname{Ker}}(B_L)={\operatorname{Im}}(A_L)$, then the last condition means that there is some $(\lambda_1,\dots,\lambda_{r})\in {\mathbb Z}^{r}$ with $(\mathbf s,\mathbf c)=A_L(\lambda_1,\dots,\lambda_r)$. If $A=[a_{ij}]$ is an $r \times r$ matrix consisting of the first $r$ rows of $A_L$, we have $[P]\in V_X(I_L)$ if and only if $(s_1,\dots,s_{r})=A(\lambda_1,\dots,\lambda_{r})$. In this case, $s_i=a_{i1}\lambda_1+\cdots+a_{ir}\lambda_{r}$ and hence $\eta^{s_i}=(\eta^{\lambda_1})^{a_{i1}}\cdots (\eta^{\lambda_r})^{a_{ir}}$, for all $i=1,\dots,r$. Letting $t_j=\eta^{\lambda_j}$ for any $j=1,\dots,r$, we conclude that $[P]\in V_X(I_L)$ if and only if $[P]=[t_1^{a_{11}}\cdots t_r^{a_{1r}}:\cdots :t_1^{a_{r1}}\cdots t_r^{a_{rr}}]$. So, if $[P]\in V_X(I_L)\cap T_X$ then $[P]\in Y_{A,{{\mathbb K}}^*}$ by above. Converse is straightforward since any $t_j\in {{\mathbb K}}^*$ is of the form $\eta^{\lambda_j}$ for $j=1,\dots,r$.

\[a:algorithm\]

**Input** A matrix L whose rows constitute a basis for a lattice $L\subset {\mathbb Z}^r$,\
**Output** A matrix $A$ parameterizing $V_X(I_L)\cap T_X$.

Form the matrix $B_L$ adding to L an identity matrix of size rank L. Find a matrix $A_L$ with ${\operatorname{Ker}}(B_L)={\operatorname{Im}}(A_L)$ Choose the first $r$ rows of $A_L$

The following will illustrate how to run this algorithm in Macaulay$2$.

\[ex:H2param\] Let $X={\mathcal{H}}_{2}$ be the Hirzebruch surface whose rays are generated by $\vv_1=(1,0)$, $\vv_2=(0,1)$, $\vv_3=(-1,2)$, and $\vv_4=(0,-1)$. Thus we have the short exact sequence $${\displaystyle}\xymatrix{ \mathfrak{P}: 0  \ar[r] & {\mathbb Z}^2 \ar[r]^{\phi} & {\mathbb Z}^4 \ar[r]^{\bb}& {\mathbb Z}^2 \ar[r]& 0},$$ where $$\phi=\begin{bmatrix}
    1 & 0 & -1& ~~0 \\
    0 & 1 & ~~~2& -1
    \end{bmatrix}^T   \quad  \mbox{ and} \quad \bb=\begin{bmatrix}
    1 & -2 & 1& 0 \\
    0 & ~~~1 & 0&1
    \end{bmatrix}.$$ This shows that the class group is ${{\mathcal{A}}}={\mathbb Z}^2$ and the total coordinate ring is $S=K[x_1,x_2,x_3,x_4]$ where $$\deg_{{{\mathcal{A}}}}(x_1)=\deg_{{{\mathcal{A}}}}(x_3)=(1,0), \deg_{{{\mathcal{A}}}}(x_2)=(-2,1), \deg_{{{\mathcal{A}}}}(x_4)=(0,1).$$

We illustrate Proposition \[p:parameterisingV\_L\] taking $L\subset {\mathbb Z}^4$ to be the lattice generated by the rows of the matrix L below.

    q=11;
    beta = matrix {{1, -2, 1, 0}, {0, 1, 0, 1}};
    L=matrix {{10,0,-10,0},{0,5,10,-5}};
    beta*transpose L--L is homogeneous as this is zero
    r=numColumns L;
    l=numRows L;
    BL=L|(q-1)*id_(ZZ^l)
    O = matrix mutableMatrix(ZZ,r,l)--zero matrix
    A=transpose ((id_(ZZ^r)| O)*gens ker BL)

    o1 = |  0  1  0  1 |
         |  1  0  0  0 |
         |  0  2 -1  0 |
         | -1  0 -1  0 |

In this example the zero set $V_X(I_L)$ of the lattice ideal $I_L={\langle}x_1^{10}-x_3^{10}, x_2^{5}x_3^{10}-x_4^{5} {\rangle}$ does lie in the torus, as $V(I_L)\setminus ({\mathbb F}_q^*)^4 \subset V(B)=V(x_1,x_3)\cup V(x_2,x_4)$. Therefore, $V_X(I_L)=Y_{A,{\mathbb F}_{11}^*}$.

We also consider the following typical singular example.

\[ex:1113param\] Let $X=P(1,1,1,3)$ be the weighted projective surface with homogeneous coordinate ring $S={{\mathbb K}}[x_1,x_2,x_3,x_4]$ where $\deg_{{{\mathcal{A}}}}(x_1)=\deg_{{{\mathcal{A}}}}(x_2)=\deg_{{{\mathcal{A}}}}(x_3)=1$, $\deg_{{{\mathcal{A}}}}(x_4)=3$. The same lattice is homogeneous here as in the previous example.

    q=11;
    beta = matrix {{1,1,1,3}};
    Phi=gens ker beta

    o1 =  |  1  -1   0  |
          | -1   0   3  |
          |  0   1   0  |
          |  0   0  -1  |
    L=matrix {{10,0,-10,0},{0,5,10,-5}};
    beta*transpose L--L is homogeneous as this is zero
    r=numColumns L;
    l=numRows L;
    BL=L|(q-1)*id_(ZZ^l)
    O = matrix mutableMatrix(ZZ,r,l)--zero matrix
    A=transpose ((id_(ZZ^r)| O)*gens ker BL)

    o1 = |  0  1  0  1 |
         |  1  0  0  0 |
         |  0  2 -1  0 |
         | -1  0 -1  0 |

Thus we have the short exact sequence $${\displaystyle}\xymatrix{ \mathfrak{P}: 0  \ar[r] & {\mathbb Z}^3 \ar[r]^{\phi} & {\mathbb Z}^4 \ar[r]^{\bb}& {\mathbb Z}\ar[r]& 0},$$ where $$\phi=\begin{bmatrix}
    ~~1 & -1 & ~~0& ~~0 \\
    -1 & ~~0 & ~~1& ~~0 \\
    ~~0 & ~~3 & ~~0& -1
    \end{bmatrix}^T   \quad  \mbox{ and} \quad \bb=\begin{bmatrix}
    1 & 1 & 1& 3 \\
    \end{bmatrix}.$$ In this example the zero set $V_X(I_L)$ of the lattice ideal does not lie in the torus. But if we add to $L$ the subgroup generated by the vector $(0,10,-10,0)$ and denote by $L'$ the new lattice we obtain, then it follows that $V_X(I_{L'})=V_X(I_L)\cap T_X=Y_{A,{\mathbb F}_{11}^*}$.

We close this section with the following observation.

If ${{\mathbb K}}={\mathbb F}_q$, then $Y_{Q,H}=Y_{A,{{\mathbb K}}^*}$ for some square integer matrix $A$.

Let $I_L=I(Y_{Q,H})$ be the lattice ideal of $Y_{Q,H}\subseteq T_X$. Then, by Proposition \[p:parameterisingV\_L\], there is a square integer matrix $A$ such that $V_X(I_L)\cap T_X=Y_{A,{{\mathbb K}}^*}$. Since $Y_{Q,H}=V_X(I_L)$ by Lemma \[L:Zariski\], the claim follows.

Degenerate Tori
===============

In this section, we focus on finite submonoids of the torus $T_X$ which are parameterized by diagonal matrices.

If $H={\langle}\eta {\rangle}$ is a finite submonoid of ${{\mathbb K}}^*$, then the finite submonoid $Y_{A,H} =\{[t_1^{a_1}:\cdots:t_r^{a_r}] \: : \: t_i \in H \}$ of the torus $T_X$ is called a [**degenerate torus**]{}.

Since $H={\langle}\eta {\rangle}$, every $t_i \in H$ is of the form $t_i=\eta^{s_i}$, for some $0 \leq s_i \leq h-1$, where $h=| \eta |$ is the order of $\eta$. Let $d_i=| \eta^{a_i} |$ and $D=diag(d_1,\dots,d_r)$.

\[p:degY\] If $d_1,\dots,d_r$ are coprime to each other then $Y_{A,H}$ is a cyclic group of order $d_1 \cdots d_r$.

Let $[P_i]=[1:\cdots:\eta^{a_i}:\cdots:1]$, for $i=1,\dots,r$. If $[P_i]^{s_i}=[1]$, then $(1,\dots,1,\eta^{a_is_i},1,\dots,1)\in G$. Since $G={\operatorname{Ker}}\pi$, it follows that $\x^{\uu_1}=\cdots=\x^{\uu_n}=1$ at $[P_i]$. So, $\eta^{a_is_iu_{1i}}=\cdots=\eta^{a_is_iu_{ni}}=1$. Since $\vv_i=(u_{1i},\dots,u_{ni})$ is a primitive ray generator, the greatest common divisor of its coordinates is $1$. Thus, $d_i$ divides $s_i$ as $d_i=| \eta^{a_i} |$. This means the point $[P_i]$ generates a cyclic group of order $d_i$, for all $i=1,\dots,r$.

On the other hand, every element $[t_1^{a_1}:\cdots:t_r^{a_r}]$ is a product $[P_1]^{s_1}\cdots [P_r]^{s_r}$ of powers of the points $[P_1], \dots [P_r]$, as $t_i=\eta^{s_i}$, for all $i=1,\dots,r$. This shows that $Y$ is a product of cyclic subgroups generated by $[P_1],\dots, [P_r]$. Under the hypothesis, this product becomes a direct product, completing the proof.

When $X=\mathbb{P}^{r-1}$, $H={\mathbb F}_q^*$, and $\gcd(d_1,\dots,d_r)=1$, the order of $Y_{A,H}$ is $d_1 \cdots d_r$ by [@LVZ]. This does not hold in general, as the following example illustrates.

\[ex:notcyclic\] Take $X={\mathcal{H}}_{2}$ to be the Hirzebruch surface over ${\mathbb F}_{11}$ and consider $(a_1,a_2,a_3,a_4)=(2,5,4,5)$. Then, we have $(d_1,d_2,d_3,d_4)=(5,2,5,2)$ and $\gcd(d_1,d_2,d_3,d_4)=1$. In this case $G$ has points of the form $(g_1,g_2,g_1,g_1^2g_2)$ for $g_1,g_2 \in {{\mathbb K}}^*$. Denote by $C_i$ the cyclic subgroup generated by $[P_i]$, for $i=1,2,3,4$. One can check that $C_1\cap C_3=[1]$, so $C_1C_3=C_1 \times C_3$. Since the orders $25$ of $C_1C_3$ and $2$ of $C_2$ are relatively prime, it follows that $C_1C_2C_3=C_1 \times C_2\times C_3$ having order $50$. Now, $C_4$ is a subgroup of $C_1C_2C_3$, since $[P_4]=[P_2]$ via $(1,1,1,\eta^{5})=(1,\eta^{5},1,\eta^{5})(1,\eta^{5},1,1)$. Hence, $Y_{A,H}=C_1C_2C_3C_4=C_1C_2C_3$ have order $50$.

On the other hand, if we take $(a_1,a_2,a_3,a_4)=(5,2,5,4)$, then $(d_1,d_2,d_3,d_4)=(2,5,2,5)$ and $\gcd(d_1,d_2,d_3,d_4)=1$ as before. We observe that $[P_3]=[P_1]$ and $C_4=C_2$ in this case. So, $Y_{A,H}=C_1C_2C_3C_4=C_1\times C_2$ have order $10$.

As $Y_{A,H}$ is a submonoid of $T_X$, $I(Y_{A,H})$ is a lattice ideal. We are ready to give our first main result specifying the lattice of this ideal. Algorithms for finding general $I(Y)$ are given in [@BaranSahin].

\[t:I(Y\_A)\] If $Y=Y_{A,H}$ then $I(Y)=I_L$ for $L=D(L_{\beta D})$.

Take a generator $F_{\mathfra{{\mathbb K}}{m}}=\x^{{\mathfra{{\mathbb K}}{m}}^{+}}-\x^{{\mathfra{{\mathbb K}}{m}}^{-}} $ of $I(Y_{A,H})$. Since $F_{\mathfra{{\mathbb K}}{m}}$ homogeneous, ${\mathfra{{\mathbb K}}{m}}\in L_{\bb}$ and $F_{{\mathfra{{\mathbb K}}{m}}}(P)=0$, for all $[P]\in Y_{A,H}$. So, $\x^{{\mathfra{{\mathbb K}}{m}}}(P)=1$, for all $[P]\in Y_{A,H}$. This is true in particular for points having homogeneous coordinates $[1:\cdots:\eta^{a_i}:\cdots:1]$ with $\eta^{a_i}$ in the $i-th$ component. So, $(\eta^{a_i})^{m_i}=1$ yielding $d_i | m_i$, i.e., ${\mathfra{{\mathbb K}}{m}}=D({\mathfra{{\mathbb K}}{m}}')$, for some ${\mathfra{{\mathbb K}}{m}}'\in {\mathbb Z}^r$. As ${\mathfra{{\mathbb K}}{m}}\in L_{\bb}$, we have $\bb D({\mathfra{{\mathbb K}}{m}}')=\bb {\mathfra{{\mathbb K}}{m}}=0$, so ${\mathfra{{\mathbb K}}{m}}'\in L_{\bb D}$. Therefore, ${\mathfra{{\mathbb K}}{m}}\in D(L_{\beta D})=L$ and $F_{\mathfra{{\mathbb K}}{m}}$ is a binomial generator of $I_{L}$.

Conversely, take a binomial generator $F_{\mathfra{{\mathbb K}}{m}}$ of $I_{L}$. Then ${\mathfra{{\mathbb K}}{m}}\in L$, so ${\mathfra{{\mathbb K}}{m}}=D({\mathfra{{\mathbb K}}{m}}')$, for some ${\mathfra{{\mathbb K}}{m}}'\in L_{\bb D}$. Since $\bb {\mathfra{{\mathbb K}}{m}}=\bb D({\mathfra{{\mathbb K}}{m}}')=0$, $F_{\mathfra{{\mathbb K}}{m}}$ is homogeneous. On the other hand, $\x^{{\mathfra{{\mathbb K}}{m}}}([P])=\eta^{a_1s_1m_1}\cdots \eta^{a_rs_rm_r}=1$ , for all $[P]=[\eta^{a_1s_1}:\cdots:\eta^{a_rs_r}] \in Y_{A,H}$, since $m_i=d_im'_i$ and $(\eta^{a_i})^{d_i}=1$ for all $i \in [r]$. Thus, $\x^{{\mathfra{{\mathbb K}}{m}}^{+}}([P])=\x^{{\mathfra{{\mathbb K}}{m}}^{-}} ([P])$ and $F_{\mathfra{{\mathbb K}}{m}}([P])=0$, for all $[P] \in Y_{A,H}$. Hence, $F_{\mathfra{{\mathbb K}}{m}}$ is a generator of $I(Y_{A,H})$.

As $d_i=| \eta^{a_i} |=h/ \gcd(h,a_i)$, the lattice $L=D(L_{\beta D})$ depends on $H$. Since the lattice $L_{\beta D}$ is saturated, $I_{L_{\beta D}}$ is a prime lattice ideal also known as a toric ideal. There are various algorithms for finding generators of a toric ideal, and these can be used here.

\[a:algorithm2\]

**Input** A matrix A parameterizing the degenerate torus,\
**Output** A matrix $\Gamma$ whose columns generate the lattice $L$.

Find the matrix $D$ using $d_i=h/ \gcd(h,a_i)$. Compute the matrix LbetaD whose columns generate $L_{\beta D}$. Multiply D by LbetaD

If each column of a matrix has both a positive and a negative entry we say that it is [*mixed*]{}. If it does not have a square mixed submatrix, then it is called [*dominating*]{}.

The matrix $\phi=\begin{bmatrix}
 0 & 1 &1& ~~1& ~~0 & -1& -1& -1 \\
 1 & 1 &0& -1 & -1& -1 & ~~0& ~~1
\end{bmatrix}^T$ is a mixed BUT not dominating matrix.

\[D:CI\] For $Y\subset X$, the ideal $I(Y)$ is called a [*complete intersection*]{} if it is generated by a regular sequence of homogeneous polynomials $F_1,\dots, F_k\in S$ where $k$ is the height of $I(Y)$.

\[Morales-Thoma 2005\]\[t:MT\] Let $L$ be a sublattice of ${\mathbb Z}^r$ with $L\cap {\mathbb N}^r=\{0\}$ and $\Gamma$ be a matrix whose columns constitute a basis of $L$. Then $I_L$ is a complete intersection iff $\Gamma$ is mixed dominating.

\[ex:idealnotcyclic\] Take $X={\mathcal{H}}_{2}$ to be the Hirzebruch surface over ${\mathbb F}_{11}$ and consider $(a_1,a_2,a_3,a_4)=(2,5,4,5)$. Then, we have $(d_1,d_2,d_3,d_4)=(5,2,5,2)$ and $\gcd(d_1,d_2,d_3,d_4)=1$. We compute the vanishing ideal in Macaulay2 as follows:

    i1 :  beta = matrix {{1, -2, 1, 0}, {0, 1, 0, 1}}

    o1 = | 1 -2 1 0 |
         | 0 1  0 1 |

    i2 : D=matrix{{5,0,0,0},{0,2,0,0},{0,0,5,0},{0,0,0,2}};

    i3 : LbetaD =gens ker (beta*D);

    i4 : gamma=D*LbetaD

    o4 =  | -5   20  |
          |  0   10  |
          |  5    0  |
          |  0  -10  |

Thus, $I(Y_{A,{{\mathbb K}}^*})={\langle}x_1^5-x_3^5, x_1^{20}x_2^{10}-x_4^{10} {\rangle}$.

On the other hand, if we take $(a_1,a_2,a_3,a_4)=(5,2,5,4)$, then $(d_1,d_2,d_3,d_4)=(2,5,2,5)$ and $\gcd(d_1,d_2,d_3,d_4)=1$ as before. Similarly, the vanishing ideal is found to be $I(Y_{A,{{\mathbb K}}^*})={\langle}x_1^2-x_3^2, x_1^{10}x_2^{5}-x_4^{5} {\rangle}$. Note that the matrix Gamma above is mixed-dominating and that the ideal $I(Y_{A,{{\mathbb K}}^*})$ is a complete intersection, in both cases.

\[p:ci\] A generating system of binomials for $I(Y_{A,H})$ is obtained from that of $I_{L_{\beta D}}$ by replacing $x_i$ with $x_i^{d_i}$. $I(Y_{A,H})$ is a complete intersection if and only if so is the toric ideal $I_{L_{\beta D}}$. In this case, a minimal generating system is obtained from a minimal generating system of $I_{L_{\beta D}}$ this way.

$I(Y_{A,H})=I_L$ for $L=D(L_{\beta D})$. The first part relies on the fact that ${\mathfra{{\mathbb K}}{m}}\in L_{\beta D}$ iff $D{\mathfra{{\mathbb K}}{m}}\in D(L_{\beta D})$ and thus $\x^{{\mathfra{{\mathbb K}}{m}}^{+}}-\x^{{\mathfra{{\mathbb K}}{m}}^{-}} \in I_{L_{\beta D}}$ iff $\x^{D{\mathfra{{\mathbb K}}{m}}^{+}}-\x^{D{\mathfra{{\mathbb K}}{m}}^{-}} \in I(Y_{A,H})$. So, there is a one-to-one correspondence between the binomials with disjoint supports. If $\x^{{\mathfra{{\mathbb K}}{m}}^{+}}-\x^{{\mathfra{{\mathbb K}}{m}}^{-}}=g_1(\x^{{\mathfra{{\mathbb K}}{m}}_1^{+}}-\x^{{\mathfra{{\mathbb K}}{m}}_1^{-}})+\cdots+g_k(\x^{{\mathfra{{\mathbb K}}{m}}_k^{+}}-\x^{{\mathfra{{\mathbb K}}{m}}_k^{-}})$, then $\x^{D{\mathfra{{\mathbb K}}{m}}^{+}}-\x^{D{\mathfra{{\mathbb K}}{m}}^{-}}=D(g_1)(\x^{D{\mathfra{{\mathbb K}}{m}}_1^{+}}-\x^{D{\mathfra{{\mathbb K}}{m}}_1^{-}})+\cdots+D(g_k)(\x^{D{\mathfra{{\mathbb K}}{m}}_k^{+}}-\x^{D{\mathfra{{\mathbb K}}{m}}_k^{-}})$, where $D(g)$ means that every $x_i$ in the polynomial $g$ is replaced with $x_i^{d_i}$.

Let $\Gamma$ be a matrix whose columns constitute a basis for the lattice $L_{\beta D}$. It is clear that $D\Gamma$ is a matrix whose columns form a basis of $L$. Since $D$ is a diagonal matrix with positive entries in the main diagonal, the sign patterns of $\Gamma$ and $D\Gamma$ are the same. By Theorem \[t:MT\] above, $I(Y_{A,H})$ is a complete intersection iff so is the toric ideal $I_{L_{\beta D}}$. In this case we know that the minimal generating sets for both ideals have the same cardinality : height $I(Y_{A,H})=$ height $I_{L_{\beta D}}$. If $f_1,\dots,f_k$ are minimal generators for the ideal $I_{L_{\beta D}}$, with $k=$ height $I_{L_{\beta D}}$, then $D(f_1),\dots,D(f_k)$ are generators for the ideal $I(Y_{A,H})$. Since height $I(Y_{A,H})$ is a lower bound for the number of generators, this set must be a minimal generating set, completing the proof.

The toric ideal $I_{L_{\beta D}}$ depends on $h$ but not on the field ${{\mathbb K}}$.

When ${{\mathcal{A}}}$ is free, the ideal of $\{[1]\}$ is the toric ideal of ${\mathbb N}\beta$.

\[c:torus\]When ${{\mathbb K}}={\mathbb F}_q$ we have the following.\
(**i**) If $Y=\{[1]\}$ then $I(Y)=I_{L_{\beta}}$,\
(**ii**) If $Y=T_X$ then $I(Y)=I_{L}$, for $L=(q-1)L_{\beta}$,\
(**iii**) $I(T_X)$ is a complete intersection iff so is $I_{L_{\beta}}$, which is independent of $q$.

(**i**) If $Y=\{[1]\}$ then $H={\langle}1 {\rangle}$. So, $D=I_r$ and $I(Y)=I_{L_{\beta}}$. This is true for any field ${{\mathbb K}}$.\
(**ii**) If $Y=T_X$ then $H={{\mathbb K}}^*$ and $A=I_r$. So, $D=(q-1)I_r$ and thus $I(Y)=I_{L}$ for $L=DL_{\beta D}=(q-1)L_{\beta}$, since $\beta D=(q-1)I_r\beta=(q-1)\beta$ and $\beta$ have the same kernel: $L_{\beta D}=L_{\beta}$.\
(**iii**) This is a direct consequence of Proposition \[p:ci\].

These consequences generalize some results of [@DN] from weighted projective spaces to a general toric variety. Using the matrix $\phi$ defined by the fan ${\Sigma}$ and the result presented in this section one can easily check whether the vanishing ideal of $T_X$ is a complete intersection.

The matrix $\phi=\begin{bmatrix}
 1 & 0 & -1& ~~0 \\
0 & 1 & ~~~\ell& -1
\end{bmatrix}^T$ is a mixed dominating matrix.

-   So, $I_{L_\bb}={\langle}{x_1}-{x_3},{x_2}{x_3}^{\ell}-{x_4}^{}{\rangle}$ is a complete intersection.

-   Thus, tori $T_X$ of the Hirzebruch surfaces ${\mathcal{H}}_{\ell}$ are all complete intersection for every $q$ and $\ell$: $$I(T_X)={\langle}{x_1}^{q-1}-{x_3}^{q-1},{x_2}^{q-1}{x_3}^{\ell(q-1)}-{x_4}^{q-1}{\rangle}.$$

Primary Decomposition of Vanishing Ideals
=========================================

In this section, we list main properties of vanishing ideals of subsets of the torus $T_X$ over ${\mathbb F}_q$.

\[t:primdec\] Let $Y$ be a subset of $T_X$ and $[P]\in T_X$.\
(**i**) If $I(Y)=I_L$ then $I([P]Y)={\langle}\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}  ~ | ~  {\mathfra{{\mathbb K}}{m}}\in L {\rangle}$.\
(**ii**) $I([P])={\langle}\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}  ~ | ~ {\mathfra{{\mathbb K}}{m}}\in L_{\beta} {\rangle}$ is a prime ideal of height $n$.\
(**iii**) The minimal primary decomposition of $I(Y)$ is as follows: $$\displaystyle I(Y)=\bigcap_{[P] \in Y} I([P]).$$ (**iv**) $I(Y)$ is radical.\
(**v**) ${\operatorname{height}}I(Y)=n$.\
(**vi**) $\dim I(Y)=r-n$.

(**i**) Let $F_p(\x):=F(p^{-1}\x)$, for a polynomial $F\in S$. Then $F_{p}\in I([P]Y)$ if and only if $F\in I(Y)$, since $F_p(p\t)=F(p^{-1}p\t)=F(\t)$ for all $\t\in Y$. If $I(Y)=I_L$ and $F(\x)=\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}^-}$ is a generator for $I_L$, then $$F_p(\x)=\x^{-{\mathfra{{\mathbb K}}{m}}^+}(P)\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{-{\mathfra{{\mathbb K}}{m}}^-}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}=\x^{-{\mathfra{{\mathbb K}}{m}}^+}(P)(\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}),$$ giving a generator $\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}$ for $I([P]Y)$. When $[P]=[P']$, $p=gp'$, for some $g\in G$. Since $\x^{{\mathfra{{\mathbb K}}{m}}}(P)=\x^{{\mathfra{{\mathbb K}}{m}}}(g)\x^{{\mathfra{{\mathbb K}}{m}}}(P')=\x^{{\mathfra{{\mathbb K}}{m}}}(P')$, for all $g\in G$ and ${\mathfra{{\mathbb K}}{m}}\in L \subset L_\bb$, it follows that $\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-}$ is independent of the representative of $[P]$.

(**ii**) Since $\{[P]\}=[P]Y$ for $Y=\{[1]\}$, the claim follows from the previous item and Corollary \[c:torus\] (**i**). As the map sending $F$ to $F_p$ is ring autormorphism, it follows that $I([P])$ is a prime ideal of height $n$ if and only if so is the toric ideal $I([1])=I_{L_\bb}$. By [@MS Proposition 7.5], the Krull dimension of $S/I_{L_\bb}$, which is $r-{\operatorname{height}}I_{L_\bb}$, equals $r-{\operatorname{rank}}{L_\bb}=r-n$, whence the result.

(**iii**) Since intersection of homogeneous ideals is homogeneous, we have $$\displaystyle I(Y)=\bigcap_{[P] \in Y} I([P]).$$ In order to show that this decomposition is minimal we prove that $I([P])$ is a minimal prime of $I(Y)$, for every $[P]\in Y$. This follows from the following observation. $ (\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P)\x^{{\mathfra{{\mathbb K}}{m}}^-})(P_0)=0$ if and only if $\x^{{\mathfra{{\mathbb K}}{m}}}(P)=\x^{{\mathfra{{\mathbb K}}{m}}}(P_0)$ if and only if $ (\x^{{\mathfra{{\mathbb K}}{m}}^+}-\x^{{\mathfra{{\mathbb K}}{m}}}(P_0)\x^{{\mathfra{{\mathbb K}}{m}}^-})(P)=0$. Thus, $I([P])\subset I([P_0])$ if and only if $I([P_0])\subset I([P])$, meaning that $I([P])\subset I([P_0])$ if and only if $I([P])= I([P_0])$.

(**iv**) This follows directly from the previous decomposition and the fact that each $I([P])$ is prime.

(**v**) ${\operatorname{height}}I(Y)=n$, since all the minimal primes are of height $n$ and thus the maximum is $n$.

(**vi**) $\dim I(Y)$ is the maximum of $\dim I([P])$ as $[P]$ vary in $Y$. Since $\dim I([P])=r-{\operatorname{height}}I([P])$, for all points $[P] \in Y$, the result follows from item (**ii**).

Evaluation codes on degenerate tori {#S:codes}
===================================

In this section we apply our results in previous sections to evaluation codes on degenerate tori in a toric variety.

Recall the basic definitions from coding theory. Let ${\mathbb F}_q$ be a finite field of $q$ elements and ${\mathbb F}_q^*={\mathbb F}_q\setminus\{0\}$ its multiplicative group. A subspace ${{\mathcal{C}}}$ of $ {\mathbb F}_q^{N}$ is called a *linear code*, and its elements ${c}=(c_{1},\dotsc,c_{N})$ are called *codewords*. The number $N$ is called the [*block-length*]{} of ${{\mathcal{C}}}$. The [*weight*]{} of $c$ in ${{\mathcal{C}}}$ is the number of non-zero entries in $c$. The [*distance*]{} between two codewords $a$ and $b$ in $\mathcal{C}$ is the weight of $a-b\in{{\mathcal{C}}}$. The minimum distance between distinct codewords in ${{\mathcal{C}}}$ is the same as the minimum weight of non-zero codewords in ${{\mathcal{C}}}$. The block-length $N$, the dimension $k=\dim_{{\mathbb F}_q}({{\mathcal{C}}})$, and the minimum distance $d=d({{\mathcal{C}}})$ are the basic parameters of ${{\mathcal{C}}}$.

Now, let $X$ be a simplicial complete toric variety over ${\mathbb F}_q$ with torsion-free class group and $S$ its homogeneous coordinate ring as in the previous sections.

Now, we recall evaluation codes defined on subsets $Y=\{[P_1],\dots,[P_N]\}$ of the torus $T_X$. Fix a degree $\aa\in{\mathbb N}\beta$ and a monomial $F_0=\x^{\phi({\mathfra{{\mathbb K}}{m}}_0)+\a} \in S_{\aa}$, where ${\mathfra{{\mathbb K}}{m}}_0\in {\mathbb Z}^n$, $\a$ is any element of ${\mathbb Z}^r$ with $\deg(\a)=\aa$, and $\phi$ as in the exact sequence $\mathfrak{P}$. This defines the [*evaluation map*]{} $$\label{e:evalmap}
\text{ev}_{Y}:S_\aa\to {\mathbb F}_q^N,\quad F\mapsto \left(\frac{F(P_1)}{F_0(P_1)},\dots,\frac{F(P_N)}{F_0(P_N)}\right).$$ The image ${{\mathcal{C}}}_{\aa,Y}=\text{ev}_{Y}(S_\aa)$ is a linear code, called the [*generalized toric code*]{}. It can be readily seen that different choices of $F_0\in S_\aa$ yield to equivalent codes. Clearly, the block-length $N$ of ${{\mathcal{C}}}_{\aa,Y}$ equals $|T_X|=(q-1)^n$ in the case of toric codes where $Y=T_X$, which was introduced for the first time by Hansen in [@Ha0; @Ha1]. A way to compute the dimension of a toric code is given in [@Ru]. In the generalized case, multigraded Hilbert function can be used to compute the length and dimension as shown by Şahin and Soprunov in [@sasop]. For general information about algebraic geometry codes we refer the reader to [@TVN] and [@Li].

The next proposition provides a way to calculate the dimension of the code as the value of the Hilbert function $H_Y(\aa):=\dim_{{{\mathbb K}}} S_{\aa}-\dim_{{{\mathbb K}}} I(Y)_{\aa}$.

\[p:dim=HF\] $\dim_{{{\mathbb K}}} {{\mathcal{C}}}_{\aa,Y}$ equals $H_{Y}(\aa)$.

This follows as the kernel of the evaluation map is exactly the the degree $\aa$ part of the vanishing ideal $I(Y)$.

We write $\aa \preceq \aa'$ if $\aa'-\aa$ lies in ${\mathbb N}\beta$, where ${\mathbb N}\beta\subset {{\mathcal{A}}}$ denotes the semigroup generated by the degrees $\bb_i$ of the variables $x_i$.

\[c:dimcode\] If the set $Y=Y_{A,H}$ and $\aa \preceq d_1\bb_1+\cdots +d_r \bb_r$, then $ev_{Y}$ is injective, that is, $\dim_{{{\mathbb K}}} {{\mathcal{C}}}_{\aa,Y}=\dim_{{{\mathbb K}}} S_{\aa}$.

In this case we have $I(Y)=I_L$ for $L=D(L_{\bb D})$ by Theorem \[t:I(Y\_A)\], where $D=diag (d_1,\dots,d_r)$. By Proposition \[p:ci\], the generators of $I(Y)$ are obtained from the generators of the toric ideal $I_{L_{\bb}}$ by substituting $x_i^{d_i}$ in $x_i$. Thus, their degrees are at least $d_i\bb_i$. In particular, the degree $\aa$ part of the kernel $I(Y)_{\aa}=\{0\}$ when $\aa \preceq d_1\bb_1+\cdots +d_r \bb_r$.

\[c:lengthcode\] If the set $Y=Y_{A,H}$ and $d_1,\dots ,d_r$ are pairwise relatively prime, then the length of ${{\mathcal{C}}}_{\aa,Y}$ is $d_1\cdots d_r$.

This follows directly from Proposition \[p:degY\].

We finish with the following example.

\[ex:idealnotcyclicCode\] Take $X={\mathcal{H}}_{2}$ to be the Hirzebruch surface over ${\mathbb F}_{11}$ and consider $(a_1,a_2,a_3,a_4)=(2,5,4,5)$. Then, we have $(d_1,d_2,d_3,d_4)=(5,2,5,2)$ and $\gcd(d_1,d_2,d_3,d_4)=1$. Thus, $I(Y_{A,{{\mathbb K}}^*})={\langle}x_1^5-x_3^5, x_1^{20}x_2^{10}-x_4^{10} {\rangle}$ by Example \[ex:idealnotcyclic\]. The fan ${\Sigma}$ of $X$ determines an important subsemigroup ${\mathcal{K}}$ of the semigroup ${\mathbb N}\bb$. Namely, ${\displaystyle}{\mathcal{K}}=\cap_{{\sigma}\in {\Sigma}} {\mathbb N}\hat{{\sigma}}$, where ${\mathbb N}\hat{{\sigma}}$ is the semigroup generated by the subset $\{\bb_j \; : \; \rho_j \notin {\sigma}\}$. In this example, ${\displaystyle}{\mathcal{K}}={\mathbb N}^2$. Since the degrees $(5,0)$ and $(0,10)$ lie in ${\mathcal{K}}$, it follows that the Hilbert function at $(5,10)$ gives us the length by [@sasop].

    i1 : hilbertFunction({5,10},IYQ)
    o1 = 50

On the other hand, if we take $(a_1,a_2,a_3,a_4)=(5,2,5,4)$, then $(d_1,d_2,d_3,d_4)=(2,5,2,5)$ and $\gcd(d_1,d_2,d_3,d_4)=1$ as before. Similarly, the vanishing ideal is found to be $I(Y_{A,{{\mathbb K}}^*})={\langle}x_1^2-x_3^2, x_1^{10}x_2^{5}-x_4^{5} {\rangle}$. Note that the matrix Gamma above is mixed-dominating and that the ideal $I(Y_{A,{{\mathbb K}}^*})$ is a complete intersection, in both cases.

Dimensions of various codes are encoded in the matrix below:

    i2 : apply(6,j-> apply(18,i-> hilbertFunction({i-5,5-j},IYQ)));
    i3 : oo / print @@ print
    {06,07,08,09,10,10,10,10,10,10,10,10,10,10,10,10,10,10}
    {04,05,06,07,08,09,10,10,10,10,10,10,10,10,10,10,10,10}
    {02,03,04,05,06,07,08,08,08,08,08,08,08,08,08,08,08,08}
    {00,01,02,03,04,05,06,06,06,06,06,06,06,06,06,06,06,06}
    {00,00,00,01,02,03,04,04,04,04,04,04,04,04,04,04,04,04}
    {00,00,00,00,00,01,02,02,02,02,02,02,02,02,02,02,02,02}

Acknowledgements {#acknowledgements .unnumbered}
================

All the examples were computed by using the computer algebra system Macaulay $2$, see [@Mac2].

[10]{} E. Baran and M. Şahin. preprint. Eisenbud, D., Sturmfels, B., 1997. [*Binomial ideals*]{}, Duke Math. J., 84, 1–45.

D. A. Cox, [*The homogeneous coordinate ring of a toric variety*]{}, J. Algebr. Geom. [**4**]{} (1995), 17–50.

Cox, D. A., Little, J., Schenck, H., 2011, [Toric Varieties]{}, Graduate Studies in Mathematics, 124, AMS, Providence, RI

E. Dias and J. Neves, [Codes over a weighted torus]{}, Finite Fields and Their Appl., 2015, 33, 66-79.

Duursma, I., . Rentería, C, Tapia-Recillas, H., 2001. [Reed-Muller codes on complete intersections]{}, Appl. Algebra Engrg. Comm. Comput. 11, 455–462.

Gold, L., Little, J., Schenck, H., 2005. [*Cayley–Bacharach and evaluation codes on complete intersections*]{}, J. Pure Appl. Algebra 196, no. 1, 91–99.

D. Grayson, M. Stillman, Macaulay2–A System for Computation in Algebraic Geometry and Commutative Algebra. [math.uiuc.edu/Macaulay2]{}.

Hansen, J., 2003. [*Linkage and codes on complete intersections*]{}, Appl. Algebra Engrg. Comm. Comput. 14, 175–185.

Hansen, J., 2000. [*Toric Surfaces and Error–correcting Codes*]{} in: Buchmann, J. et al. (Eds.), Coding Theory, Cryptography, and Related Areas, Springer, pp. 132–142.

Hansen, J., 2001. [*Error-Correcting Codes from Higher-Dimensional Varieties*]{}, Finite Fields Appl., [ 7]{} no. 4, 530–552.

Joyner, D., 2004. [*Toric codes over finite fields*]{}, Appl. Algebra Engrg. Comm. Comput., 15, 63–79.

Little, J., 2008. [Algebraic geometry codes from higher dimensional varieties]{}, in Martinez-Moro, E. et al. (Eds.), Advances in algebraic geometry codes, Series on Coding Theory Cryptology, [5]{}, World Sci. Publ., Hackensack, NJ., pp. 257–293.

H. H. Lopez, R. H. Villarreal and L. Zarate, [Complete Intersection Vanishing Ideals on Degenerate Tori over Finite Fields]{}, Arab J. Math, 2013, 189-197.

Miller, E. and Sturmfels, B., Combinatorial commutative algebra. Graduate Texts in Mathematics, 227. Springer-Verlag, New York, (2005).

M. Morales and A. Thoma, [Complete Intersection Lattice Ideals]{}, J. Algebra, 2005, 284, 755-770.

J. Neves, M. V. Pinto, R. H. Villarreal [*Regularity and algebraic properties of certain lattice ideals*]{}, Bull. Brazilian Math. Soc. 45 (4) (2014),777–806.

Renteria, C., Simis, A., Villarreal, R. H., 2011. [*Algebraic methods for parameterized codes and invariants of vanishing over finite fields*]{}, Finite Fields Appl.,[17]{} no. 1, 81–104.

Ruano, Diego, 2007. [ On the parameters of $r$-dimensional toric codes]{}, Finite Fields and Their Applications [13]{}, 962–976.

M. Şahin and I. Soprunov, [*Multigraded Hilbert functions and toric complete intersection codes*]{}, J. Algebra, 459 (2016), 446–467.

I. Soprunov, [*Toric complete intersection codes*]{}, J. Symbolic Comput. 50 (2013), 374–385.

A. Tochimani and R. H. Villarreal, [Binomial Vanishing Ideals]{}, J. Algebra Comb. Discrete Appl., 2015, 151-156.

Tsfasman M., Vlăduţ, S., Nogin, D.,2007. [Algebraic geometric codes: basic notions]{}, Mathematical Surveys and Monographs, 139, AMS, Providence.

[^1]: The author is supported by TÜBİTAK Project No:114F094


xbay wrote:What if I was to swing you a trade? Say Flacco for Ryan Grant and a QB of yours... either Leftwich or McNabb. Would you do it?

Could be. Although we'd have to work it out somehow, because the Ravens schedule is about to get real tough, with games against the Pats, Bengals, Vikings... so Flacco's numbers are probably gonna suffer a bit (I do expect him to still be good, just not as impressive as he's been til now).

Oh, and I certainly doubt you'd want to add Leftwich, considering he's been demoted straight to #3 in the Bucs depth chart. Josh Johnson is the starter now (I probably shouldn't tell you this, but I'd feel like cheating if not... and I know you're fully aware the Leftwich era is over in Tampa).

I've sent you a trade proposal. And, before you freak out, keep in mind that Fred Jackson is gonna lose lots of touches starting next game now that Lynch is back from his suspension. I think it's a pretty fair deal, actually, for both of us.

You're right, it is pretty fair. Having Fred Jackson there is sorta like me having Frank Gore and having Coffee on my squad as well.

xbay wrote:What if I was to swing you a trade? Say Flacco for Ryan Grant and a QB of yours... either Leftwich or McNabb. Would you do it?

Could be. Although we'd have to work it out somehow, because the Ravens schedule is about to get real tough, with games against the Pats, Bengals, Vikings... so Flacco's numbers are probably gonna suffer a bit (I do expect him to still be good, just not as impressive as he's been til now).

Oh, and I certainly doubt you'd want to add Leftwich, considering he's been demoted straight to #3 in the Bucs depth chart. Josh Johnson is the starter now (I probably shouldn't tell you this, but I'd feel like cheating if not... and I know you're fully aware the Leftwich era is over in Tampa).

I've sent you a trade proposal. And, before you freak out, keep in mind that Fred Jackson is gonna lose lots of touches starting next game now that Lynch is back from his suspension. I think it's a pretty fair deal, actually, for both of us.

You're right, it is pretty fair. Having Fred Jackson there is sorta like me having Frank Gore and having Coffee on my squad as well.

Cool. Now I have the full running game of the Bills and three offensive performers of the Ravens. I'll go as far as those two teams take me.

I think it's a good trade, tho. It hurts to lose Ryan Grant, but with Lynch, Jackson and McGahee I think I've got my RB positions well covered, and I still have Wells if the Cards ever find a way to use him. I also get the backup I needed for McNabb, so that side of the deal is pretty sweet for me.

btw, this goes into the other thread, you know. You probably were just jelous I was dealing with Bada in this thread and decided to steal the spotlight... but you gotta understand I'm not that loyal when it comes to business. There's room for more than one dealing partner in my life.

As punishment, I'm calling deal interference on this one. You'll have one week penalty in which you won't be able to strike any more deals. Next time, it will be a two week penalty, Flaccogate.

bada wrote:Kevin Smith is a solid #2 fantasy RB. You dont think he is in the same tier as Vincent Jackson and DeSaun Jackson?

Eh... no. Not even close. Both Jacksons play for contenders and have QBs that are gonna look at them as their first options to score and make big plays (on the passing game).

Kevin Smith is a very solid RB that doesn't have great big play potential and who's playing for a bad team that's gonna trail more often than not, which means relying more on the passing game than the running game. Also, the Lions o-line ain't exactly opening big holes for him.

I think he's, as you say, a solid #2 for fantasy, and could have been better if the Lions weren't so... offensively challenged. But, in their current situation, and with the schedule the Lions will face (Chicago, Pittsburgh, Green Bay, Rams, Seahawks, Vikings, Browns, Packers, Bengals, Ravens, Cards, Niners and Bears), it's not like Smith is gonna be a big time weapon this season.

In fact, with that schedule, they'll be lucky if they can win a couple more games... so Smith's value is reduced a lot by the strength of the defenses he's gonna face for the rest of the season.

I accept and understand you valuing those WR's more than Smith and not wanting to make the deal, but to say they are not even close and I am being greedy in asking for Vincent Jackson or DeSaun Jackson for Kevin Smith is rediculous.

Smith in his rookie year for an 0-16 team put up 1200 combined yards and 8 TD's. (both figures are more than Vincent Jackson put up in his outstanding season last year). The Lions are not worse then they were last year. And even if they do throw, he is their receiving RB so he will be involved there as well.

I am not saying Kevin Smith is a world beater, but there are very few RB's that are 3 down backs anymore. He is one of them and that helps his value considerably.

Again, you may not wish to make the trade, but these players are very much on par with each other.

bada wrote:I accept and understand you valuing those WR's more than Smith and not wanting to make the deal, but to say they are not even close and I am being greedy in asking for Vincent Jackson or DeSaun Jackson for Kevin Smith is rediculous.

Smith in his rookie year for an 0-16 team put up 1200 combined yards and 8 TD's. (both figures are more than Vincent Jackson put up in his outstanding season last year). The Lions are not worse then they were last year. And even if they do throw, he is their receiving RB so he will be involved there as well.

I am not saying Kevin Smith is a world beater, but there are very few RB's that are 3 down backs anymore. He is one of them and that helps his value considerably.

Again, you may not wish to make the trade, but these players are very much on par with each other.

Phillip Rivers did whatever he could against the Steelers giving me about a 24 point lead over Joel, but I doubt I'll win. He still has Greg Jennings and Adrian Peterson left. Looks like I'll be taking my first loss this week.


Oil Drillinghttp://nwpr.org
enSeattle City Council Looks At Shell Drill Rigshttp://nwpr.org/post/seattle-city-council-looks-shell-drill-rigs

Seattle City Council members heard testimony on a resolution today.

The resolution urges the Port of Seattle to reconsider its controversial decision to host Shell Oil’s arctic drill rigs.

The resolution warns that allowing Shell to use Terminal 5 in West Seattle will cause: “disruption, division and direct conflict with the Port’s stated values and policies.”

At the hearing, environmentalists made their opposition to the Shell lease known. But others who testified felt differently.
Wed, 06 May 2015 00:17:11 +0000Ashley Ahearn62485 at http://nwpr.orgSeattle City Council Looks At Shell Drill RigsShell Oil Rig Arrives In Puget Soundhttp://nwpr.org/post/shell-oil-rig-arrives-puget-sound

A 400 foot tall oil rig has arrived in Port Angeles. Shell Oil plans to use the Polar Pioneer to drill this summer, off Alaska’s North Slope. Environmentalists oppose the exploration - many were in Port Angeles Friday for what they called an "unwelcoming".

Among them, Michael Bittner. He said the arrival of the Polar Pioneer felt like an invasion.
Sat, 18 Apr 2015 00:38:27 +0000Cathy Duchamp61659 at http://nwpr.orgShell Oil Rig Arrives In Puget SoundShell Arctic Drilling Rig Expected In Port Angeleshttp://nwpr.org/post/shell-arctic-drilling-rig-expected-port-angeles

Shell Oil’s Arctic drilling rig, the Polar Pioneer, is expected to arrive Friday in Port Angeles, Washington.

Shell has just received the necessary federal permits to drill for oil in the Arctic and will be staging its fleet in Seattle, despite a lawsuit filed by environmental groups and an investigation launched by the Seattle City Council.

Activists have warned of a flotilla of kayaks that would extend a less-than-warm welcome to Shell when it arrives at the Port of Seattle.
Fri, 17 Apr 2015 00:29:31 +0000Ashley Ahearn61607 at http://nwpr.orgShell Arctic Drilling Rig Expected In Port AngelesCantwell Asks White House To Stop Alaska Mine Projecthttp://nwpr.org/post/cantwell-asks-white-house-stop-alaska-mine-project-0

Senator Maria Cantwell wants the White House to stop a proposed mine in Alaska that she says will threaten jobs in Washington state. The proposed Pebble Mine, in Western Alaska, sits near Bristol Bay, which is one of the world’s most abundant salmon habitats. About 1,000 Washington residents hold commercial fishing permits there. At a rally at Fisherman’s Terminal in Seattle, Cantwell described the mine as a “giant cauldron of toxic waste.”

Fri, 24 Jan 2014 00:47:37 +0000Ed Ronco40397 at http://nwpr.orgCantwell Asks White House To Stop Alaska Mine ProjectShell Cancels Arctic Oil Drillinghttp://nwpr.org/post/shell-cancels-arctic-oil-drilling

Shell Oil has put Arctic drilling on hold. The company announced Wednesday that it will not attempt to drill for oil in the Arctic Ocean this year. The announcement comes after a year of accidents and setbacks for Shell’s Arctic drilling efforts. KUOW’s John Ryan reports from Seattle.Thu, 28 Feb 2013 14:33:05 +0000John Ryan23486 at http://nwpr.orgShell Cancels Arctic Oil DrillingAnother Setback For Shell’s Arctic Drilling Planshttp://nwpr.org/post/another-setback-shell-s-arctic-drilling-plans

Another costly setback for Shell Oil and its efforts to drill in the Arctic Ocean: Shell needs to send its beleaguered oil rigs to Asia for major repairs, instead of to Seattle as planned. KUOW’s John Ryan reports.Wed, 13 Feb 2013 14:44:05 +0000John Ryan22690 at http://nwpr.orgAnother Setback For Shell’s Arctic Drilling PlansShell’s Beached Kulluk Oil Rig Towed To Safe Harborhttp://nwpr.org/post/shell-s-beached-kulluk-oil-rig-towed-safe-harbor

A shipwrecked oil rig that was bound for Seattle has been floated off the rocks and towed to a safe harbor in the Gulf of Alaska. A fleet of nine ships accompanied Shell Oil’s Kulluk drill rig on the 45-mile tow.Tue, 08 Jan 2013 00:46:32 +0000John Ryan20831 at http://nwpr.orgShell’s Beached Kulluk Oil Rig Towed To Safe HarborGrounded Drilling Rig Remains Stable In Alaskahttp://nwpr.org/post/grounded-drilling-rig-remains-stable-alaska

A grounded oil drilling rig remains in stable shape off the coast of Alaska Wednesday. Shell Oil’s Kulluk ran aground on New Year’s Eve on an island in the Gulf of Alaska. The vessel hasn’t spilled any oil so far, according to the Alaska Department of Environmental Conservation.Thu, 03 Jan 2013 14:46:54 +0000Jay Barrett20624 at http://nwpr.orgGrounded Drilling Rig Remains Stable In Alaska


using System;
using Microsoft.Extensions.Configuration;

namespace BaGet.Core
{
    /// <summary>
    /// Attempts to provide the <typeparamref name="TService"/>.
    /// </summary>
    /// <typeparam name="TService">The service that may be provided.</typeparam>
    public interface IProvider<TService>
    {
        /// <summary>
        /// Attempt to provide the <typeparamref name="TService"/>.
        /// </summary>
        /// <param name="provider">The dependency injection container.</param>
        /// <param name="configuration">The app's configuration.</param>
        /// <returns>
        /// An instance of <typeparamref name="TService"/>, or, null if the
        /// provider is not currently active in the <paramref name="configuration"/>.
        /// </returns>
        TService GetOrNull(IServiceProvider provider, IConfiguration configuration);
    }

    /// <summary>
    /// Implements <see cref="IProvider{TService}"/> as a delegate.
    /// </summary>
    internal class DelegateProvider<TService> : IProvider<TService>
    {
        private readonly Func<IServiceProvider, IConfiguration, TService> _func;

        /// <summary>
        /// Create an <see cref="IProvider{TService}"/> using a delegate.
        /// </summary>
        /// <param name="func">
        /// A delegate that returns an instance of <typeparamref name="TService"/>, or,
        /// null if the provider is not currently active due to the app's configuration.</param>
        public DelegateProvider(Func<IServiceProvider, IConfiguration, TService> func)
        {
            _func = func ?? throw new ArgumentNullException(nameof(func));
        }

        public TService GetOrNull(IServiceProvider provider, IConfiguration configuration)
        {
            return _func(provider, configuration);
        }
    }
}


School of Law to move to North Haven by 2014

Quinnipiac University’s School of Law is projected to move to the North Haven campus by the summer of 2014, which will affect many current law students and those who plan to attend in the next few years.

Sarah Wagner

In the State of the QUnion address on Jan. 30, President John Lahey detailed the university’s plans to move the law school to the North Haven campus.

By the fall 2014 semester, both Lahey and Senior Vice President for Academic & Student Affairs Mark Thompson are confident that the current law school will be vacated and ready to use for undergraduate classes.

“There are a couple of reasons that we decided to make the move,” Thompson said. “We’re developing the North Haven campus to be primarily a graduate campus, and to leave the Law School on Mount Carmel wouldn’t be as effective.”

Construction will most likely begin this summer once the medical school construction is completed, according to Thompson. The focus will then shift to renovating an existing building on the North Haven campus, Building 3, for the law school.

“The current law school building is as large as three Echlin Centers,” Thompson said. “We need that space for so many other undergraduate programs right now, and also for faculty office space. There is no shortage of needs. It’s just figuring out how the space will be utilized.”

In addition to creating more space for undergraduate classes, the move may also relieve the parking situation on the Mount Carmel campus.

“Parking on the Mount Carmel campus will definitely become easier. There’s no question about that,” Thompson said. “Most of the current law school classes are day classes. There are just under 400 law students, so the move will open up a lot of spaces for undergraduates.”

Thompson said that the reception of the plans for the law school has been positive so far, and that many of the Quinnipiac University members look forward to the move.

“Faculty and staff members have been part of the planning throughout the whole process,” Thompson said. “They have also involved students to find out their opinions.”

Erica Rocha, who is currently a first-year law student, was a member of the student panel working with architects to create the ideal building for law students.

“Coming from undergrad, we all had ideas about what we liked and would like to see in the new building,” Rocha said. “The architects were extremely open to hearing about the elements of the current law school that we wanted to keep.”

Rocha said that the panel consisted of about 10 Quinnipiac students in all years of law school. The students and architects discussed not only the layout and private study rooms, but also even smaller details such as cubbies and the ideal amount of desk space.

“From working with the architects, we know that the campus will be very practical for our needs,” Rocha said. “But it will also be great professionally. To eventually have all of the graduate programs on one campus will create a great networking opportunity to forge strong professional relationships.”

However, not all students agree with the decision.

“I really enjoyed where the law school was,” said Chris Klepps, who graduated from Quinnipiac’s School of Law in May. “I thought it was great to have such a beautiful building on main campus, and I think the building itself is what attracts some students to attend.”

Although the location is a concern for past students, some prospective students are concerned about the undergraduate to graduate ratio.

“I don’t think it’s a good idea to move the law school to North Haven,” said junior JP Orlando, who intends to go to law school after completing his undergraduate degree in legal studies. “The legal studies majors will not have the same atmosphere that they have now while preparing themselves for law school.”

Orlando said that having the law school on the Mount Carmel campus helps undergraduates interact with law students, which creates a more welcoming and familiar environment for transitioning after completing their undergraduate degrees.

Despite the reservations that some students have about the move, Rocha said she is sure that students will warm up to the new building.

“I know students are a little nervous because they are attached to the current situation and building,” Rocha said. “I can definitely understand that, but I visited the new building and I thought it was beautiful. Students may be unsure at first, but once they see it I’m sure they will fall in love with it.”


Chaumet: Discover new 'Laurier' collection

December 21, 2018

From the moment it was founded, the jewellery house Chaumet has placed nature at the heart of its designs, creating a herbarium of jewellery drenched in symbolism. For over two centuries, one of its symbols, laurel, has been the French brand’s constant companion — and recently it has taken a more central role in the new ‘Laurier’ collection.

Why laurel, you ask? The answer can be found in the 2016 high jewellery collection La Naturede Chaumet. 64 pieces, dedicated to ears of corn, oak, lily and laurel were included in the collection, which contained unique items for each theme. Each plant told its own story, with the laurel telling the tale of the Apollo’s passionate love for the beautiful nymph Daphne.

The masculine element of the line was embodied in a blue colour scheme, in the 'Firmament Apollinien' line. The pieces were embellished with carved moonstones, and Burmese and Ceylon sapphires sang the ancient god’s glory and beauty. The feminine principle was represented in the 'Métamorphose de Daphné' pieces — a necklace, earrings, a ring and an elegant brooch with diamonds and a luxurious pink spinel from Tajikistan and Tanzania.

The 2018 collection Laurier returns to this theme again, but this time in more delicate jewellery, designed for everyday wear. The earrings, rings and pendants of this collection are draped in the reincarnation of the laurel theme, an amazing plant which symbolises immortality - as well as the victory and power of Napoléon, the lover of Josephine. Just like in the high art jewellery collection 'La Nature de Chaumet' (especially in the 'Firmament Apollinien') the laurel comes to life in an explosion of movement — with wild and lush curves of a particularly sophisticated nature. However, this time the shape and architecture of the plant is given an even more prominent role, as all the pieces in the collection are monochrome.

The 'Laurier' collections contains six pieces: a flexible necklace whose clasp resembles laurel branches swaying gently in them wind, hidden amongst diamond leaves, and on it, a medallion pendant with a delicate relief forms its own ring as the branch softly curves around it — an embodiment of the plant which bends but does not break when beaten. Two rings are designed to cover whole fingers, but their foliage boldly extends to the neighbouring fingers too. The earrings resemble whole laurel bouquets, gathered together with ribbons (wreaths traditionally bequeathed to competition winners,) and the watch secretly hides its dial under a lid scattered with diamonds, whose treasures are revealed only to the wearer.

Each piece demonstrates acute mastery in jewellery techniques, as well as the design virtuosity of the Chaumet house: exceptional materials, asymmetrical compositions, bold leaves, fashioned from gold and diamonds. In the 'Laurier 'collection, every detail is an exquisite example of the commendable high jewellery tradition of the house.


Lucas warns against Anfield slip-ups

CAPTAIN’S COMMENTS： Bolton travel to Anfield today having climbed into the top 10 of the Premier League and with skipper Kevin Nolan praising boss Gary Megson

AFP , LIVERPOOL, ENGLAND

Fri, Dec 26, 2008 - Page 20

Liverpool midfielder Lucas Leiva has warned the Premier League leaders they can’t afford any more slip-ups at Anfield if they are to end their 19-year wait for the English title.

Despite going into the Christmas fixture list as Premier League leaders for the first time since 1996, Liverpool face Gary Megson’s Bolton at Anfiled today having drawn their last three home games, against Fulham, West Ham and Hull.

On each occasion Liverpool had been expected to comfortably claim all three points having already beaten champions Manchester United and title rivals Chelsea in the league this term.

But on top of those costly dropped points Rafael Benitez’s players also failed to claim three points against ten-man Arsenal on Sunday.

With newly-crowned world club champions United now just seven points adrift of Liverpool with two games in hand — at home to Wigan and Fulham next month — Brazilian midfielder Lucas admits that Benitez’s men must quickly rediscover their best form at Anfield.

“We should have beaten Arsenal last Sunday, but we must now focus on our next game and try to get the three points at Anfield against Bolton,” Lucas said. “We have had three draws in a row at home, so it’s important we win this game. We need to win our home games.”

“We are at the top of the league and we have to keep going, be confident and just do our job,” he said.

Liverpool manager Benitez is expected to return to the dug-out today after being laid low with kidney stones over the past week, a problem which forced him to miss the 1-1 draw at Arsenal.

Spain striker Fernando Torres is also in with a chance of a return to action following a month on the sidelines with a hamstring injury.

Bolton travel to Anfield having climbed into the top ten in recent weeks, however, and the club are beginning to display the same qualities that made them a feared Premier League force under former manager Sam Allardyce.

And captain Kevin Nolan, a boyhood Liverpool supporter, insists his team’s recent revival is down to the efforts of manager Megson.

“Sam Allardyce had a plan when he was here and Gary has got a plan too,” Nolan said. “It was about stabilizing at first, getting back into the top 10, then maybe next year top eight and then Europe after that.”

“I thought it was completely unfair the criticism he got when he first came in because you have to give someone a chance to show what they can do,” he said. “He stood tall, took it all on his shoulders and never put the blame on the players. He wants to prove everyone wrong and I’m sure he’s doing that.”


/(-2))/2). Suppose m*j + 2*j = 55. Is j a multiple of 11?
True
Suppose -2*h + 2*p = -23 - 23, 0 = 5*p. Let b = 4 - 12. Let s = h + b. Does 14 divide s?
False
Let j(y) = 4*y - 5 - 6 + 4 + 3. Is j(4) a multiple of 12?
True
Suppose -5 = 5*s - 15. Suppose -3 - 11 = -k. Suppose -2*l = -k - s. Is l a multiple of 4?
True
Let u(w) = w**2 - 9*w - 1. Let a be u(9). Is (-10)/a*(-70)/(-25) a multiple of 17?
False
Let n be (-2)/(-8) + 1673/28. Suppose 4*o - n = -4*x, 4*x - x - 85 = 5*o. Is 6 a factor of x?
False
Let w(d) = -d**3 + 5*d**2 + d. Let v be w(7). Suppose 2*m + 210 = -3*m. Let l = m - v. Is 20 a factor of l?
False
Let h(x) = -3*x**2 + 2*x**2 - x + 2*x**2. Let a = -2 + -1. Is 6 a factor of h(a)?
True
Let a(b) = b**3 - 5*b**2 + b + 2. Is 18 a factor of a(6)?
False
Suppose 0 = -4*i - 5*f - 7, -3*i - f + 7 = 2*i. Does 15 divide i/(-8) - (-242)/8?
True
Does 30 divide (-1)/(-1 - (-87)/90)?
True
Suppose 16 + 60 = 2*r. Is 38 a factor of r?
True
Let c(u) = -44*u - 51. Is c(-5) a multiple of 13?
True
Let s(f) = 3*f - 4. Is s(14) a multiple of 29?
False
Suppose 6*x = 2*x. Suppose -2*a = 4 - x. Is 15 a factor of (28 - a) + 1 + -1?
True
Let p(z) = 4*z + 12. Is p(4) a multiple of 28?
True
Let m = -1 + -1. Let g = m + 5. Is 3 a factor of g?
True
Let r(d) = -d**3 + d**2 + d + 4. Let s be r(0). Suppose -i + 30 = -5*j, -i + j - 42 = -s*i. Is i a multiple of 15?
True
Let q(u) = 6*u**2 + 3*u - 5. Let x(c) = 5*c**2 + 4*c - 4. Let y(b) = -6*q(b) + 7*x(b). Is 11 a factor of y(9)?
True
Suppose 0 = 6*l - 34 + 10. Suppose l*u - 6 - 30 = 0. Is 9 a factor of u?
True
Suppose -45*b + 84 = -41*b. Is 3 a factor of b?
True
Suppose -g + 0*k = -3*k - 106, 0 = -k - 4. Does 14 divide g?
False
Let h(l) = l**3 - 3*l**2 + 4*l - 2. Let y be h(2). Suppose 82 = 4*i + y*x, 119 = 2*i + 3*i - 3*x. Is i a multiple of 10?
False
Suppose -2*u = 3*u - 395. Does 18 divide u?
False
Let h be 4*(3 + 161/4). Let g = h - 111. Let b = 92 - g. Does 18 divide b?
False
Suppose g - 6*a + 3*a = -15, 5*a = 25. Suppose m + 6 = -g*m. Does 18 divide (-213)/m + (-1)/(-2)?
True
Suppose 0 = 5*i + 4*t - 158, 4*i - 86 = 4*t + 26. Is 15 a factor of i?
True
Let f(m) = -m + 1. Let q be f(-5). Let p(l) = 3*l**2 + 14*l**2 - 1 - q*l**2. Is 10 a factor of p(-1)?
True
Suppose h = -2*y + 2, -y + 1 = h - 2*h. Suppose h*z + 111 = 3*z. Is z a multiple of 15?
False
Let b be (90 + 3 - 3) + 1. Let w = b - 48. Is w a multiple of 21?
False
Let s be 8*((-4 - -3) + -3). Let i be -4*4/(s/306). Suppose 5*b - i = -18. Does 9 divide b?
True
Let d(m) = 2*m**2 + 4*m - 1. Suppose -8 = -3*i + 5*f + 1, 0 = -i + 4*f + 3. Let v be d(i). Let g = -19 + v. Is 5 a factor of g?
True
Suppose -4*x - 5*w = -x - 6, 2*w - 12 = 2*x. Let v(a) = 2*a + 2. Let h be v(x). Is 42/h*(-14)/21 a multiple of 2?
False
Let i(k) = 2*k - 5. Let b be i(5). Suppose 4*a - 4 + 1 = -y, 0 = b*a - 5*y - 35. Is 11/((a - 2) + 1) a multiple of 6?
False
Suppose -2*d + 4 = -5*o, -3*d - 2 = 4*o - 8. Let u be (d/(-12))/((-2)/12). Is 7 a factor of (-4 - (0 - u)) + 11?
False
Let p(r) = 128*r**2 + 2*r + 1. Does 10 divide p(-1)?
False
Suppose -3*c = 5*h - 0*h, -c = -4*h. Suppose c = 3*w - 46 + 7. Is 9 a factor of w?
False
Let r(t) = 5*t - 5 - 7 - 2 + 0. Is r(8) a multiple of 26?
True
Let y = -15 + 32. Is 7 a factor of y?
False
Suppose -5*p = -p. Suppose -330 = -4*j - 2*t, -j - 5*t + 78 = -p*t. Is j a multiple of 23?
False
Let p = 140 + -79. Does 24 divide p?
False
Let n(m) = 4*m - 4 + 0 + 0 + 0. Does 2 divide n(2)?
True
Let l(x) = x - 11. Let t be l(8). Let h = t - -20. Is h a multiple of 17?
True
Suppose 323 = 2*n - 353. Is n a multiple of 13?
True
Let g = 3 - 1. Suppose -g*y + 0*y = -48. Does 12 divide y?
True
Let r be (-48)/(-2 + -2 - -3). Suppose -c = -2*u - r, -c + 157 = 3*c - u. Is c a multiple of 19?
True
Suppose 32 - 7 = 5*p. Let w(n) = n**3 - 4*n**2 - 4*n + 2. Is w(p) even?
False
Let q(f) = 9*f**2 - 3*f + 1. Is q(2) a multiple of 11?
False
Suppose 4 + 0 = f. Is 15 a factor of 128/24*30/f?
False
Let x(d) = -d**2 - 13*d - 3. Let a be x(-12). Suppose a = 2*n + 95. Let w = n - -70. Is w a multiple of 9?
True
Let i = 17 - -13. Does 15 divide i?
True
Suppose 3*b - 127 = -2*u, 8*u - 221 = 4*u + 5*b. Let a = u - 11. Is 24 a factor of a?
True
Let f be (-10)/((-21)/(-9) - 2). Let b = f + 20. Is (-4)/10 - 244/b a multiple of 12?
True
Let j = 226 - 103. Let b = 135 + -74. Suppose 5*c - b - j = -p, 56 = c + 5*p. Is 12 a factor of c?
True
Let o(y) = y**3 + 5*y**2 + 4*y + 5. Let v be o(-4). Suppose 3*r + 136 = v*s - 0*r, r + 132 = 5*s. Is s a multiple of 15?
False
Suppose p - 6*p + 25 = 0. Suppose -25 - 40 = p*o. Let w = o - -24. Is 7 a factor of w?
False
Let d = 487 - 326. Let a = -113 + d. Let g = -32 + a. Is g a multiple of 15?
False
Let c be -1 + -1 + 1 - -3. Suppose 3*p = -c*x - 3 + 4, -11 = -4*x + 3*p. Suppose -a - 40 = -3*a - x*f, 4*a + 5*f - 83 = 0. Is 6 a factor of a?
False
Suppose 488 - 63 = 5*a. Suppose 10 = s + s. Suppose 4*c - 16 = 0, -21 + a = s*z + c. Is 12 a factor of z?
True
Suppose 0 = 3*a + 2*c - 30 - 7, -5*c = -2*a + 31. Let p = a + -23. Does 4 divide (-84)/p + (-4)/10?
True
Suppose 7 = -5*l + 2, -2*u - l = 13. Let i = u + 13. Is 7 a factor of i?
True
Suppose -3*j - 2*b = -6*j + 25, -14 = -j - 5*b. Suppose -j = -z + m, -m = m - 8. Does 9 divide z?
False
Let n = 1056 + -741. Does 31 divide n?
False
Let z(d) = 3*d**2 - 3*d - 7. Does 15 divide z(5)?
False
Suppose -193 = -i + 236. Is i a multiple of 33?
True
Let x = -17 + 21. Is 17 a factor of -4*2/x + 19?
True
Let n = 6 + -4. Suppose -3*k - 24 = -3*z, n*z - 5*k - 2 = 23. Is z even?
False
Suppose 0 = 2*r - 5*b, -3*r + 4*b = 2*r + 17. Let a(u) = 10*u + 1. Let p(c) = -9*c - 1. Let x(i) = 5*a(i) + 6*p(i). Is x(r) a multiple of 9?
False
Suppose 207 = 4*p - 105. Does 16 divide p?
False
Let l(f) = -225*f**3 + 2*f**2 + f - 1. Is l(-1) a multiple of 13?
False
Let y = -71 - -110. Suppose 0 = -3*c + g + 58, 5*g - 4*g = 2*c - y. Does 19 divide c?
True
Let q(g) = 159*g + 1. Let y be q(1). Suppose t - y = -4*t. Is t a multiple of 16?
True
Let b = -10 - -14. Let k(p) = 2 - b + p + 3. Is k(8) a multiple of 4?
False
Let t(z) = -40*z**3 + z**2 + 4*z + 1. Is t(-1) a multiple of 3?
False
Let i(f) be the first derivative of f**2 - f - 2. Let x be i(-6). Let a = 29 + x. Is a a multiple of 9?
False
Is (-6)/(-33) + 361/11 a multiple of 11?
True
Let n = 9 - 40. Let y = n + 67. Does 17 divide y?
False
Let o(c) = c**2 + 2*c + 3. Let q be o(-3). Let v(p) = p**2 - 7*p - 7. Let z be v(q). Let x = z + 32. Is x a multiple of 12?
False
Let z = 1 + 4. Suppose -z*i = -2*t - 249 + 53, t - 122 = -3*i. Does 15 divide i?
False
Let b = 71 + -48. Is b a multiple of 7?
False
Let m(g) = 4*g + 15. Let c(l) = -2*l - 8. Let f(i) = 7*c(i) + 4*m(i). Let x be f(3). Suppose -o - 3 = -x. Is o a multiple of 7?
True
Suppose 0 = -3*v + 2*v + 61. Does 18 divide v?
False
Let u(r) = -r + 49. Does 28 divide u(0)?
False
Let h = 6 + -2. Is h/(-3)*585/(-26) a multiple of 17?
False
Let x(l) = -l**3 - l**2 + 112. Let o be x(0). Let i = o + -56. Does 28 divide i?
True
Let j(x) = 4*x - 6. Let p be j(6). Suppose -2*z + 2 + p = 4*n, -5*n = -2*z - 25. Suppose 3*c - 30 = -5*b, -c + 4*b - 8*b + 17 = z. Is 5 a factor of c?
True
Let y(w) = 2*w**2 - 6*w**2 + 4*w + 5 + 5*w**2. Let j be y(-4). Suppose 0 = -0*h + h + 2*r - 9, -j*r - 23 = -4*h. Is h a multiple of 7?
True
Suppose 0*t = -5*t + 110. Is 11 a factor of t?
True
Suppose -8 - 10 = 3*t. Let u = t + 6. Suppose b + u*b = 21. Is 7 a factor of b?
True
Let w = -12 - -16. Suppose -5*f - w*g + 170 = 0, -f + 4*g = 5*g - 35. Does 17 divide f?
False
Let y = 52 - 32. Is y a multiple of 20?
True
Let i be (-149)/(-6) + (-2)/(-12). Suppose -4*m = 5 - i. Is 5 a factor of m/(3/18*2)?
True
Let t be 3/((-4)/(-4)*1). Let u(g) = 4*g - 2. Is u(t) a multiple of 10?
True
Let v(c) = c + 5. Suppose -u + 35 = 4*u. Let j be v(u). Does 15 divide (66/(-4))/(j/(-16))?
False
Is 0 + (-2 - 4)*10/(-4) a multiple of 15?
True
Let y(o) = 55*o. Does 19 divide y(3)?
False
Let t = -93 - -185. Is ((-9)/6)/((-3)/t) a multiple of 16?
False
Suppose 5*f + 5*n = 5, 5 - 3 = -2*f - 4*n. Let y(p) = 14*p - 2. Let i be y(f). Let u = 66 - i. Is u a multiple of 7?
False
Let y(m) = 2*m - 7. Let o be y(5). Suppose -o - 9 = -4*p + 3*i, -5*p = -4*i - 16. Let t = p - -8. Is t a multiple of 5?
False


Hypothalamic network for thermoregulatory vasomotor control.
Warming one side of a rat's preoptic area and anterior hypothalamus (POAH) causes skin vasodilation on both sides of the body, and the present study evaluated the extent to which signals mediating this vasodilation cross the midline within and below the POAH. Hind paw vasomotion during unilateral POAH thermal or electrical stimulation was measured for rats in which the POAH had been midsagittally transected and for rats in which one side of the hypothalamus had been coronally transected just below the POAH. In midsagittally transected rats, unilateral POAH thermal or electrical stimulation produced bilateral paw vasodilation, but the ipsilateral dilation occurred at a hypothalamic temperature lower than that at which contralateral dilation occurred. In the unilaterally transected rats, unilateral POAH warming produced bilateral vasodilation and, regardless of which side of the POAH was warmed, the threshold stimulus temperature was always lower for vasodilation on the intact side. Unilateral transection of a part of the hypothalamus that included the medial forebrain bundle had the same effect as did unilateral transection of the whole hypothalamus. Information controlling thermoregulatory vasomotion thus crosses the midline both within and below the POAH. Although efferent signals descending through the medial forebrain bundle innervate skin blood vessels on both sides of the body, this innervation is stronger on the ipsilateral side.


A multifunctional docking site mediates signaling and transformation by the hepatocyte growth factor/scatter factor receptor family.
Signaling by tyrosine kinase receptors is mediated by selective interactions between individual Src homology 2 (SH2) domains of cytoplasmic effectors and specific phosphotyrosine residues in the activated receptor. Here, we report the existence in the hepatocyte growth factor/scatter factor (HGF/SF) receptor of a multifunctional docking site made of the tandemly arranged degenerate sequence YVH/NV. Phosphorylation of this site mediates intermediate- to high-affinity interactions with multiple SH2-containing signal transducers, including phosphatidylinositol 3-kinase, phospholipase C gamma, pp60c-src, and the GRB-2-Sos complex. Mutation of the two tyrosines results in loss of biological function, as shown by abrogation of the transforming activity in the oncogenic counterpart of the receptor. The same bidentate motif is conserved in the evolutionarily related receptors Sea and Ron, suggesting that in all members of the HGF/SF receptor family, signal transduction is channeled through a multifunctional binding site.


Parents of former Thai princess jailed for insulting monarchy

The elderly parents of a former Thai princess were both jailed for two and a half years today after pleading guilty to defaming the monarchy, an institution protected by one of the world's toughest lese majeste laws.

Bangkok: The elderly parents of a former Thai princess were both jailed for two and a half years today after pleading guilty to defaming the monarchy, an institution protected by one of the world's toughest lese majeste laws.

Apiruj Suwadee, 72, and his 66-year-old wife Wanthanee are the latest close relatives of fallen former princess Srirasmi to face legal proceedings after a corruption scandal involving her family erupted late last year.

Both had pleaded guilty to insulting the royal family, lodging a malicious claim and asking officials to file false charges against a woman in 2003.

"The judge convicted two defendants on insulting the monarchy and sentenced them to five years in prison," a verdict published by the court stated today.

"Due to their confession the judge halved the sentence to two years and six months without probation," it added.

At least eight of Srirasmi's family have now been convicted and jailed for lese majeste -- her elder sister and brother-in-law, two brothers, a nephew and her parents.

Another relative, Pongpat Chayapun, the former head of Thailand's elite Central Investigation Bureau, was handed a 31-year jail term for a series of convictions linked to an alleged criminal empire that spanned illegal gambling, extortion and kidnapping. He was initially sentenced to six years for lese majeste.

Srirasmi's siblings were arrested in the wake of that scandal alongside a slew of senior police officers.

But the case against her parents came later and centred on a former neighbour's complaint that she had been jailed for 18 months on a fraud charge brought maliciously by the Suwadee couple more than a decade ago.

According to the court verdict, the Suwadees made false claims about Crown Prince Maha Vajiralongkorn -- who was married to their daughter at the time -- in a bid to threaten their neighbour.

Vajiralongkorn divorced Srirasmi and she renounced her royal title following the scandal, which has gripped a public unused to seeing palace intrigue play out in the open.

She and a younger sister have not been arrested but have retreated from public view.

The spectacular demise of the former princess's family comes at a time of heightened anxiety over the health of the country's revered but ailing monarch.

Experts say the last decade of political turmoil in Thailand is intertwined with concerns among competing elites over the direction of the kingdom once the reign of 87-year-old King Bhumibol Adulaydej comes to an end.

Bhumibol has largely been confined to a Bangkok hospital in recent months, but he made a rare public appearance on Monday.