Consider the free module $\mathbb{Z}[\zeta]$ over $\mathbb{Z}$ of rank 2, where $\zeta=e^{\frac{\pi i} 3}$. $I$ is an ideal of $\mathbb{Z}[\zeta]$ satisfying $\frac{\mathbb{Z}[\zeta]}I= \frac{\mathbb{Z}}{N\mathbb{Z}},$ where $N$ is a positive integer.
Why there exists a basis $\lbrace w_1, w_2\rbrace$ of $\mathbb{Z}[\zeta]$ over $\mathbb{Z}$ such that $\lbrace w_1, Nw_2\rbrace$ is basis of $I$ over $\mathbb{Z}?$
Recall the following theorems, which are related.
Thm 1. Let $M$ be a free $\mathbb{Z}$-module of rank $n$, and let $N$ be a submodule of $M$. Then there exists a basis $(e_1,\ldots,e_n)$ of $M$, an integer $r\geq 0$ and positive integers $q_1,\ldots, q_r$ such that $q_1\mid\cdots\mid q_r$ such that $(q_1e_1,\ldots,q_re_r)$ is a basis of $N$. Moreover, $r, q_1,\ldots,q_r$ are unique.
Furthermore, if $q_1=\cdots=q_\ell=1$ and $q_i\geq 2$, for $i>\ell$, we have $M/N\simeq \mathbb{Z}^{n-r}\times \prod_{i>\ell} \mathbb{Z}/q_i\mathbb{Z}$.
Thm 2. Any finitely generated $\mathbb{Z}$-module $P$ is isomorphic to $\mathbb{Z}^k\times \prod_{i=1}^s \mathbb{Z}/a_i\mathbb{Z}$, where $k\geq 0$, $a_i\geq 2$, $a_1\mid a_2\mid \ldots\mid a_s$, and $s,a_1,\ldots,a_s$ are unique.
Since $\mathbb{Z}[\zeta]$ is free of rank $2$, $I$ is free of rank $r\leq 2$. Using notation of thm 1, we have $\mathbb{Z}[\zeta]/I\simeq \mathbb{Z}^{n-r}\times \mathbb{Z}/q_1\mathbb{Z}\times \mathbb{Z}/q_2\mathbb{Z}$.
The uniqueness part of thm 2, and the last part of thm 1 show that $2-r=0$, $s=1$ ,so$q_1=1$ and $q_2=N$.