in generate the algebra ${\mathcal{O}}({\mathbb{P}}_q({\underline}{\ell}))$ if and only if the weight vector ${\underline}{\ell}$ is as in Theorem \[thm:6\], hence classically ${\mathbb{P}}({\underline}{\ell})\simeq{\mathbb{C}\mathrm{P}}^n$. We need some preliminary lemmas. First of all, let us recall:

\[lemma:Bez\] Let $R$ be a principal ideal domain. For any $a,b\in R$ there exists $x,y\in R$ such that $\gcd(a,b)=ax+by$.

When $R={\mathbb{Z}}$ and $a,b\geq 1$, it is an easy exercise to prove that one can always choose $x,y$ different from zero and with opposite sign. So, as a corollary:

\[lemma:coprime\] For any three positive integers $a,b,k$ there exists two non-zero integers $r,s$ with opposite sign such that $ar+bs=k\gcd(a,b)$.

\[lemma:divide\] Let ${\underline}{\ell}$ be a coprime weight vector. There is a pairwise coprime weight vector ${\underline}{p}$ such that ${\underline}{\ell}={\underline}{p}^{\,\sharp}$ if and only if $\ell_{i:j}=\ell_i/{\mathrm}{gcd}(\ell_i,\ell_j)$ divides $\ell_k$ for all $i\neq j$ and $j\neq k$.

If ${\underline}{\ell}={\underline}{p}^{\,\sharp}$ with ${\underline}{p}$ pairwise coprime, one has $\ell_{i:j}=p_j$ for any $i\neq j$. Hence $\ell_{i:j}$ divides $\ell_k$ for all $k\neq j$.

*Proof of “$\,\Leftarrow$”.* We prove it by induction. The statement is trivial if $n=0,1$. Let $n\geq 2$.

Define $p_0=\ell_{1:0}$, $p_1=\ell_{0:1}$ and $p'_2=\gcd(\ell_0,\ell_1)$. Note that $p_0$ and $p_1$ are coprime. By construction $\ell_0=p_1p'_2$ and $\ell_1=p_0p'_2$. By hypothesis $p_0$ and $p_1$ divide $\ell_2,\ldots,\ell_n$. Since they are coprime, for all 