\cdots x_{v_n} \in L+(\Theta)$ by Lemma \[int1\](i). Since $n+ \ell \leq d$, by the same argument as in the proof of Theorem \[2-3\], it is enough to prove that $$x_{v_d}x_{v_{d-1}}\cdots x_{v_{d-\ell+1}} \in J+(\Theta).$$

Consider the exact sequence $$0 \longrightarrow
N\longrightarrow
A/\big(J+(\overline \Theta,\theta_d^n)\big)
\longrightarrow
A/\big(L+J+(\overline \Theta,\theta_d^n)\big)
\longrightarrow 
0,$$ where $N=(L+J+(\overline \Theta,\theta_d^n))
/(J+(\overline \Theta,\theta_d^n))$. Since $A/(L+J)\cong K[\partial P]$ and since $\overline \Theta$ is an l.s.o.p. of $A/(L+J)$, we have $\dim_K (A/(L+J+(\overline \Theta)))_n=\partial h_n=0$ and $$A_n=\big(L+J+(\overline \Theta)\big)_n.$$ In particular, we have $L+J+(\overline \Theta)=L+J+(\overline \Theta,\theta_d^n)$. Then, by Lemma \[trivial\] we have $$\begin{aligned}
\label{qqq}
\hspace{16pt} \dim_K N_\ell&=&\dim_K \big(A/\big(J+(\overline \Theta,\theta_d^n)\big)
\big)_\ell
-
\dim_K \big(A/\big(L+J+(\overline \Theta,\theta_d^n)\big)
\big)_\ell\\
\nonumber & =&
(h_\ell+ h_{\ell-1}+\cdots + h_{\ell-n+1}) - \partial h_\ell \\
\nonumber &\leq& n-1,\end{aligned}$$ where the last inequality follows from the assumption.

Let $u_k=x_{v_d} x_{v_{d-1}} \cdots x_{v_{d-k+1}}$ for $k=1,2,\dots,\ell$. Since $A_n=(L+J+(\overline \Theta))_n$, any element in $A$ of degree $n$ is contained in $L+J+(\overline \Theta)$. Since $\ell \geq n$, this fact says that the elements $$u_\ell,u_{\ell-1} \theta_d,\dots,u_{\ell-n+1}\theta_d^{n-1}$$ are contained in $L+J+(\overline \Theta)$ since they are products of elements of degree $1$. Then says that the above $n$ elements are $K$-linearly dependent in $N$. Thus there are $\alpha_0,\alpha_1,\dots,\alpha_{n-1} \in K$ with $(\alpha_0,\alpha_1,\dots,\alpha_{n-1}) \ne (0,0,\dots,0)$ such that $$\alpha_0 u_\ell + \alpha_1 u_{\ell-1} \theta_d 