left brace if and only if $f_{z}-{\operatorname{id}}$ is bijective for all $z\in {\mathbb{Z}}/(n)$.

To prove the necessity of the stated condition it is sufficient (and easy) to check that $J=\{ ((t_1,\dots ,t_n),(s_1,\dots
,s_n))\in T\rtimes_{\circ} S \mid t_z =(u_{1}, \dots , u_{m_z})$ such that $u_{i} \in \text{Im}(f_{z}-{\operatorname{id}})  \}$ is an ideal of $T\rtimes_{\circ} S$.

For the converse, suppose that $f_{z}-{\operatorname{id}}$ is bijective for all $z\in {\mathbb{Z}}/(n)$. Let $I$ be a nonzero ideal of $T\rtimes_{\circ}S$ and let $0\neq ((t_1,\dots ,t_n),(s_1,\dots ,s_n))\in I$. Since the $p_{z}$’s are distinct prime numbers, we may assume, without loss of generality, that for some $k\in {\mathbb{Z}}/(n)$ we have $(t_k,s_k)\neq 0$ and $(t_z,s_z)=0$ for all $z\neq k$. Write $t_{k}=( u_{1}, \dots,
u_{m_{k}})$ and $s_k= \sum_{i=1}^{r_{k}} \mu_{i} e_{i}^{(k)}$. Thus $((t_1,\dots ,t_n),(s_1,\dots ,s_n))\in I\cap A_k$.

For $z\in {\mathbb{Z}}/(n)$, let $a_{z,i}$ be the element in $A_{z}$ defined by $$a_{z,i} = ((0,\dots,0),(s'_1,\dots ,s'_n)),$$ with $s'_z=e_{i}^{(z)}$.

Note that, for all $t,t'\in T$ and all $s,s'\in S$, $$(t,
s)-(t', s')=(t-t', s-s'-b(t-t',t'))$$ by the definition of the sum in the asymmetric product $T\rtimes_{\circ} S$. Throughout the proof we will use this formula several times.

[*Claim 1*]{}: If there exists a positive integer $j$ such that $r_{k} \leq j \leq m_{k}$ and $u_{j}\neq 0$, then $a_{k,r_{k}}\in 