${\mathcal}{D}_k(p)$ is *admissible* on ${\underline}{\ell}$ if $p$ is prime, divides $\ell_i$ for all $i\neq k$ and does not divide $\ell_k$.

Classically, two (coprime) weight vectors correspond to isomorphic weighted projective spaces if and only if one can be obtained from the other with an iterated application (in any order) of admissible multiplications and divisions [@BFNR13]. So for example, one has ${\mathbb{P}}(1,2,2)\simeq{\mathbb{P}}(2,3,6)\simeq{\mathbb{C}\mathrm{P}}^2$ while ${\mathbb{P}}(1,1,2)\not\simeq{\mathbb{C}\mathrm{P}}^2$ (in particular, ${\mathcal}{D}_0(2)$ is admissible in the former case, while ${\mathcal}{M}_2(2)$ is not admissible in the latter case).

A weight vector ${\underline}{\ell}={\underline}{m}^{\sharp}$ is coprime if and only if ${\underline}{m}$ is pairwise coprime.

Note that $\gcd(\ell_i,\ell_j)=\gcd(m_i,m_j)\prod_{k\neq i,j}m_k$ for all $i\neq j$.

Let ${\underline}{m}$ be pairwise coprime. We prove by induction that $\gcd(\ell_0,\ell_1,\ldots,\ell_k)=\prod_{i>k}m_i$ for all $1\leq k\leq n$ (with the convention that empty products are $1$). It follows from the equation above when $k=1$. Assume it is true for some $k\in\{1,\ldots,n-1\}$. Then $$\gcd(\ell_0,\ell_1,\ldots,\ell_{k+1})=
\big(\gcd(\ell_0,\ell_1,\ldots,\ell_k),\ell_{k+1}\big)=
\gcd(1,m_0\cdots m_{k})\prod_{i>k+1}m_i=
\prod_{i>k+1}m_i \;.$$

*Proof of “$\,\Rightarrow$”.* Fix $i,j$ with $i\neq j$. Since $m_i$ divides $\ell_k$ for all $k\neq i$ and $m_j$ divides $\ell_i$, $\gcd(m_i,m_j)$ divides $\ell_k$ for all $k$. Hence if ${\underline}{\ell}$ is coprime, $\gcd(m_i,m_j)=1$, for all $i\neq j$.

\[thm:6\] Let ${\underline}{\ell}$ be coprime. Then ${\mathbb{P}}({\underline}{\ell})\simeq{\mathbb{C}\mathrm{P}}^n$ if and only if there is a pairwise 