+ \cdots + \alpha_{n-1} u_{\ell-n+1}\theta_d^{n-1}
\in J+(\overline \Theta,\theta_d^n).$$ Let $s = \min \{k : \alpha_k \ne 0\}$. Then there is an $h \in A$ such that $$\theta_d^s \alpha_s u_{\ell-s} + \theta_d^{s+1} h \in J+ (\overline \Theta).$$ Since $\Theta$ is an l.s.o.p. of a Cohen-Macaulay ring $A/J$, $\theta_d$ is a non-zero divisor of $A/(J+(\overline \Theta))$. Thus $$\alpha_s u_{\ell-s}  + \theta_d h \in  J +(\overline \Theta).$$ This fact implies $u_{\ell-s} \in J+(\Theta)$. Since $u_{\ell-s}$ divides $u_\ell$, we have $$x_{v_d}x_{v_{d-1}} \cdots x_{v_{d-\ell+1}}
= u_{\ell}
\in J+(\Theta),$$ as desired.

Next, we study the condition (7). To simplify the notation, we write $R=A/J \cong K[P]$ and let $$C=(L+J)/J=
\big((x_\sigma: \sigma \in P \setminus \partial P)+J \big)/J.$$ Thus $C$ is the ideal of $R$ generated by the interior faces of $P$.

\[2.3.1\] $C$ is a $d$-dimensional Cohen-Macaulay $A$-module such that $H_C(t)={(h_d+h_{d-1}t + \cdots + h_0t^d)} /{(1-t)^d}$.

Consider the exact sequence $$\begin{aligned}
\label{E2.4}
0 \longrightarrow
C \longrightarrow
R \longrightarrow
R/C \longrightarrow
0.\end{aligned}$$ Observe $R \cong K[P]$ and $R/C = A/(L+J) \cong K[\partial P]$. Then $$H_C(t)=H_R(t)-H_{R/C}(t)=H_{K[P]}(t)-H_{K[\partial P]}(t)
=\frac {(h_d+h_{d-1}t + \cdots + h_0t^d)} {(1-t)^d}.$$ Also, $C$ is a $d$-dimensional Cohen-Macaulay $A$-module since $R/C$ is a $(d-1)$-dimensional Cohen-Macaulay $A$-module and $R$ is a $d$-dimensional Cohen-Macaulay $A$-module (e.g., use [@E 