Let $\mathbb{Q}_3$ be the $3$-adic field and $\mathbb{Q}_3(\zeta_3)$ be the cyclotomic extension and $\mathbb{Z}_3[\zeta_3]$ and $M=\pi \mathbb{Z}_3[\zeta_3]$ be its maximal ideal. Let $(\mathbb{Z}_3[\zeta_3])^{*}$ be the units in the ring of integers $\mathbb{Z}_3[\zeta_3]$.
I am trying to show $l^2a+\pi l b \in M^2$, where $|a|_3 \leq \frac{1}{3}$, $|b|_3 \leq \frac{1}{3}$ and $l \in (\mathbb{Z}_3[\zeta_3])^{*}$ and $\pi=\sqrt{-3}$.
For simplicity, assume $|a|_3=|b|_3=\frac{1}{3}$ i.e., $a=b=3u^n$, $u \in (\mathbb{Z}_3[\zeta_3])^{*}$. Then,
\begin{align*} (l^2a+\pi lb)&= l^2a(1+\pi \cdot U), \ \text{where} \ U=1/l \in (\mathbb{Z}_3[\zeta_3])^{*}, \\ &=(3) \cdot \left\langle l^2 \right\rangle \cdot (1+\pi U)\\ &= (\pi^2) \cdot \left\langle l^2 \right\rangle \cdot (1+\pi U) \\&=(\pi \mathbb{Z}_3[\zeta_3])^2 \\&=m^2. \end{align*}
So in cases $|a|_3=|b|_3=\frac{1}{3}$, we see $$l^2a+\pi lb \in m^2. $$ If we assume $|a|_3=|b|_3=\frac{1}{3^n}<\frac{1}{3}$,then also we get similar result. (please check it)
But how to deal the case when $|a|_3 \neq |b|_3$ provided $|a|_3 \leq \frac{1}{3}$, $|b|_3 \leq \frac{1}{3}$ ?
Thanks for help
If $|a|_3,|b|_3\leq\tfrac13$ then $a,b\in3\Bbb{Z}_3[\zeta_3]=M^2$, hence $l^2a+\pi lb\in M^2$.