$\ell'_2,\ldots,\ell'_n$ are coprime to $p_0$ and $p_1$, the vector ${\underline}{p}:=(p_0,\ldots,p_n)$ is pairwise coprime. Moreover, $\ell_i=\prod_{j\neq i}p_j$ for all $i\geq 2$. It remains to prove that $p'_2=p_2p_3\ldots p_n$.

For all $i\neq j$ with $i,j\geq 2$, $\ell_{i:j}=p_j$ divides $\ell_0$. Hence $\ell_0=rp_1p_2^{k_2}p_3^{k_3}\ldots p_n^{k_n}$ for integers $r,k_2,\ldots,k_n\geq 1$. Since $\ell_{0:2}=rp_2^{k_2}p_3^{k_3-1}\ldots p_n^{k_n-1}$ divides $\ell_3$, $p_3^{k_3-1}$ divides $\ell_3$, and then $k_3=1$. Similarly one shows that $k_i=1$ for all $i\geq 2$. So, $p'_2=rp_2p_3\ldots p_n$. From the above expression for $\ell_{0:2}$ it follows that $r$ divides $\ell_3$, and similarly $\ell_i$ for all $i\geq 2$. On the other hand, $r$ divides $p'_2$ and then $\ell_0$ and $\ell_1$. Since ${\underline}{\ell}$ is a coprime vector, it must be $r=1$.

\[thm:11\] Let ${\underline}{\ell}$ be a coprime weight vector. Then ${\mathcal{O}}({\mathbb{P}}_q({\underline}{\ell}))$ is generated by the elements $\xi_{i,j}$ if and only if there is a pairwise coprime weight vector ${\underline}{p}$ such that ${\underline}{\ell}={\underline}{p}^{\,\sharp}$.

Since the statement is trivial for $n=1$, we assume $n\geq 2$. We assume the thesis is false, and prove that this contradicts the hypothesis, i.e. we assume there is no ${\underline}{p}$ such that ${\underline}{\ell}={\underline}{p}^{\,\sharp}$, and prove the existence of an irreducible monomial with length $3$.

By Lemma \[lemma:divide\], there exists indices $i,j,k$ with $i\neq j$ and $j\neq k$ such that $\ell_{i:j}$ 