Solution: Define a map
\[
f: \mathbb{Z} \rightarrow \frac{\mathbb{Z} [\iota ]}{\left\langle a + \iota b \right\rangle },\quad m \mapsto [m] = m + \left\langle a + \iota b \right\rangle.
\]
It is clear that the map is a homomorphism. We will show that the above map is surjective with kernel being $\left\langle a^2 + b^2 \right\rangle $.
-
The map $f$ is surjective. Since $\gcd (a,b) = 1$, there exists $p,q\in \mathbb{Z} $ such that $ap + bq = 1$. Consider
\begin{align*}
(a + \iota b)(q + \iota p) & = (aq - bp) + \iota (ap + bq) = aq - bp + \iota.
\end{align*}
This implies,
\begin{align*}
[a + \iota b] = [aq - bp + \iota] \implies [\iota] = [bp - aq] \in \frac{\mathbb{Z} [\iota ]}{\left\langle a+\iota b \right\rangle }.
\end{align*}
Let us denote $bp - aq$ by $A$. Since, $A \in \mathbb{Z} $, for any $x + \iota y \in \mathbb{Z} [\iota ]$, we have
\begin{align*}
f(x + Ay) = x + Ay + \left\langle a + \iota b \right\rangle = x + \iota y + \left\langle a + \iota b \right\rangle .
\end{align*}
This shows that $f$ is surjective.
-
We now claim that $\ker f $ is generated by $a^2 + b^2$.
\begin{align*}
m \in \ker f & \iff f(m) = 0 \iff m \in a + \iota b \iff (a + \iota b) \mid m \\
& \iff \frac{m}{a + \iota b} \in \mathbb{Z} \iff \frac{m}{a + \iota b}\times \frac{a - \iota b}{a - \iota b} \in \mathbb{Z} \\
& \iff \frac{m(a - \iota b)}{a^2 + b^2} \in \mathbb{Z} \iff (a^2 + b^2) \mid ma, mb \\
& \iff (a^2 + b^2) \mid \gcd (ma, mb) \\
& \iff (a^2 + b^2) \mid m.
\end{align*}
Thus, $\ker f = \left\langle a^2 + b^2 \right\rangle $.
Hence, from the first isomorphism theorem, we conclude that
\[
\textcolor{teal}{\boxed{
\frac{\mathbb{Z}[i]}{\left\langle a + \iota b \right\rangle} \cong \frac{\mathbb{Z}}{\left\langle a^2 + b^2 \right\rangle} = \frac{\mathbb{Z}}{(a^2 + b^2)\mathbb{Z} }
}}
\]