condition becomes $$\sum_{i=0}^nk_i\prod_{j\neq i}p_j\in (p_0\ldots p_n) {\mathbb{Z}}.$$ Every summand in the equation above besides the $0$-th is divisible by $p_0$, hence $k_0\ell_0$ must be divisible by $p_0$, i.e. $k_0$ is divisible by $p_0$. Similarly $k_i\in p_i{\mathbb{Z}}$ for all $i$. Thus $z^{{\underline}{k}}=\zeta_0^{h_0}\zeta_1^{h_1}\ldots\zeta_n^{h_n}$ where $h_i=k_i/p_i$. Since each $\zeta_i$ is ${\mathbb{Z}}_p$-invariant, such a monomial clearly belongs to ${\mathcal{O}}(L_q(p;{\underline}{\ell}))$, thus concluding the proof.

We close this section by computing the relations among the generating elements .

\[prop:relLens\] The elements $x_i$, $\zeta_i$, $\zeta_i^*$ satisfy the commutation relations

\[eq:relLens\] $$\begin{aligned}
x_ix_j &= x_jx_i && \textup{for all} \quad i,j \;, \label{eq:relLensA} \\
x_i\zeta_j &=\zeta_jx_i && \textup{for all} \quad 0\leq i<j\leq n \;, \label{eq:relLensB} \\
x_j\zeta_i &=q^{2p_i}\zeta_ix_j && \textup{for all} \quad 0\leq i<j\leq n \;, \label{eq:relLensC} \\
\zeta_i\zeta_j &=q^{-p_ip_j}\zeta_j\zeta_i && \textup{for all} \quad 0\leq i<j\leq n \;, \label{eq:relLensD} \\
\zeta_i^*\zeta_j &=q^{p_ip_j}\zeta_j\zeta_i^* && \textup{for all} \quad i\neq j \;, \label{eq:relLensE} \\
[x_i,\zeta_i] &=(1-q^{2p_i})\,\zeta_i\sum\nolimits_{j>i}x_j
    \hspace{-1cm} && \textup{for all} \quad i=0,\ldots,n \;, \label{eq:relLensF} \\
\intertext{together with the relations:}
x_0 & +x_1+\ldots +x_n =1 \;, \label{eq:relLensG} \\
\zeta_i\zeta_i^* &=\prod_{k=0}^{p_i-1}\Big\{x_i+(1-q^{-2k})\sum\nolimits_{j>i}x_j\Big\}
    && \textup{for all} \quad i=0,\ldots,n \;, \label{eq:relLensH} \\
\zeta_i^*\zeta_i &=\prod_{k=1}^{p_i}\Big\{x_i+(1-q^{2k})\sum\nolimits_{j>i}x_j\Big\}
    && \textup{for all} \quad i=0,\ldots,n \;. \label{eq:relLensL} \end{aligned}$$

It is understood that an empty sum is $0$.

The relations from to are easy 