An interesting application of ideal norm

Proposition. Let $m,n\in \mathbb{Z}$$m, n \in \mathbb{Z}$ such that $gcd(m,n)=1$$\gcd(m, n) = 1$, then $\mathbb{Z}[i]/(m+ni)\cong \mathbb{Z}/N(m+ni)\mathbb{Z}$$\mathbb{Z}[i]/(m + ni) \cong \mathbb{Z}/N(m + ni)\mathbb{Z}$.

We require the following lemma.

Lemma. If $(m+ni)\mid k\in \mathbb{Z}$$(m + ni) \mid k \in \mathbb{Z}$, then $N(m+ni)\mid k$$N(m + ni) \mid k$.

Proof. Let $(m+ni)(a+bi)=k$$(m + ni)(a + bi) = k$. Then $ma+nbi+nai-mb=k$$ma + nbi + nai - mb = k$, so $ma-mb+i(na+nb)=k$$ma - mb + i(na + nb) = k$. Since $k\in \mathbb{Z}$$k \in \mathbb{Z}$, we must have $na+nb=0$$na + nb = 0$, which gives us $an=-bm$$an = -bm$.

Hence $m\mid an$$m \mid an$. Since $gcd(m,n)=1$$\gcd(m, n) = 1$, we have $m\mid a$$m \mid a$, so $a=m{a}^{\prime }$$a = ma'$ for some $a^{\prime }\in \mathbb{Z}$$a' \in \mathbb{Z}$.

Therefore, we have $a^{\prime }n=-b$$a'n = -b$. Hence $a+bi=m{a}^{\prime }-a^{\prime }ni=a^{\prime }(m-ni)$$a + bi = ma' - a'ni = a'(m - ni)$.

It follows that $k=(m+ni)\cdot a^{\prime }(m-ni)=a^{\prime }(m+ni)(m-ni)=a^{\prime }\cdot N(m+ni)$$k = (m + ni) \cdot a'(m - ni) = a'(m + ni)(m - ni) = a' \cdot N(m + ni)$.

Therefore, if $(m+ni)\mid k\in \mathbb{Z}$$(m + ni) \mid k \in \mathbb{Z}$, then $N(m+ni)\mid k$$N(m + ni) \mid k$. $◻$$\square$

We now prove the proposition.

Proof. By the properties of the absolute ideal norm, we have $|\mathbb{Z}[i]/(m+ni)|=N(m+ni)$$|\mathbb{Z}[i]/(m + ni)| = N(m + ni)$.

Now let us prove that $\mathbb{Z}[i]/(m+ni)$$\mathbb{Z}[i]/(m + ni)$ is a cyclic group. We will show that the additive order of $1\in \mathbb{Z}[i]/(m+ni)$$1 \in \mathbb{Z}[i]/(m + ni)$ is $N(m+ni)$$N(m + ni)$.

Let $k\cdot 1\equiv 0(\mathrm{mod}m+ni)$$k \cdot 1 \equiv 0 \pmod{m + ni}$. Then $(m+ni)\mid k$$(m + ni) \mid k$, which by our lemma implies $N(m+ni)\mid k$$N(m + ni) \mid k$.

Hence the additive order of $1$$1$ is exactly $N(m+ni)$$N(m + ni)$. Since $\mathbb{Z}[i]/(m+ni)$$\mathbb{Z}[i]/(m + ni)$ is generated by $1$$1$ as an additive group, we conclude that $\mathbb{Z}[i]/(m+ni)\cong \mathbb{Z}/N(m+ni)\mathbb{Z}$$\mathbb{Z}[i]/(m + ni) \cong \mathbb{Z}/N(m + ni)\mathbb{Z}$ as additive groups. $◻$$\square$