does not divide $\ell_k$. It must be $k\neq i$, since $\ell_{i:j}$ always divides $\ell_i$. To simplify the notations, we may assume $i=0$, $j=1$, $k=2$, the general case being the same.

Using Corollary \[lemma:coprime\] for $a=\ell_0$, $b=\ell_2$ and $k=\ell_1$ one can find non-zero integers $r,s$ with opposite sign such that $\ell_0 r+\ell_2 s=\ell_1\gcd(\ell_0,\ell_2)$. If $r$ is positive, the length $3$ monomial $$\label{eq:notgen}
(z_0^*)^{|r|}z_1^{\gcd(\ell_0,\ell_2)}z_2^{|s|}$$ belongs to ${\mathcal{O}}({\mathbb{P}}_q({\underline}{\ell}))$. By hypothesis $\ell_{0:1}$ does not divide $\ell_2$, thus it does not divide $\gcd(\ell_0,\ell_2)$. On the other hand, if an invariant element of the form $(z_0^*)^{k_0}z_1^{k_1}z_2^{k_2}$ (with positive $k_0,k_1,k_2$) is generated by element $\xi_{i,j}$ as in , then $\ell_{0:1}$ divides $k_1$ (since $\xi_{0,1}^\alpha\xi_{0,2}^\beta=(z_0^*)^{\alpha\ell_{1:0}+\beta\ell_{2:0}}
z_1^{\alpha\ell_{0:1}}z_2^{\beta\ell_{0:2}}$ for all $\alpha,\beta\geq 1$). This proves that is not a product of elements in . For $r$ negative, we repeat the proof with the monomial $z_0^{|r|}z_1^{\gcd(\ell_0,\ell_2)}(z_2^*)^{|s|}$.

*Proof of “$\,\Leftarrow$”.* Let ${\underline}{\ell}={\underline}{p}^{\,\sharp}$ with ${\underline}{p}$ pairwise coprime. If $z^{{\underline}{k}}$ is ${\mathrm{U}}(1)$-invariant, one has that $$-k_0\ell_0=k_1\ell_1+\ldots+k_n\ell_n \;.$$ Since $p_0$ divides every weight in the right hand side, it divides $k_0\ell_0$; since it does not divide $\ell_0$, it has to divide $k_0$. Similarly $p_i$ divides $k_i$ for all $i=1,\ldots,n$. After reparametrization, any invariant monomial $z^{{\underline}{k}}$ in Lemma \[lemma:2\] is of the form $\zeta^{{\underline}{k}}:=\zeta_0^{k_0}\zeta_1^{k_1}\ldots \zeta_n^{k_n}$ with $\zeta_i:=z_i^{p_i}$. 