Quotient rings of Gaussian integers I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:

For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism
  $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$

Could someone show me an isomorphism between these rings to prove this?
 A: Hint: Define a map $\phi : \mathbb{Z}[i] \to {\mathbb Z}_{a^2 + b^2}$ by $\phi (x + yi) = x-ab^{-1}y$. Next, show that $\phi$ is surjective homomorphism  and find its kernel. For your hint $\ker(\phi) = \langle a+bi\rangle$.
A: $$\mathbb{Z}[i]/(a+ib)=\mathbb{Z}[i]/(a^2+b^2,a+ib)=\mathbb{Z}_{a^2+b^2}[i]/(a+ib) =\mathbb{Z}_{a^2+b^2}[x]/( x^2+1,a+bx)\\=\mathbb{Z}_{a^2+b^2}[x]/(a^2 x^2+a^2,a+bx) =\mathbb{Z}_{a^2+b^2}[x]/(-b^2 x^2+a^2,a+bx)\\ =\mathbb{Z}_{a^2+b^2}[x]/(a+bx) \simeq \mathbb{Z}_{a^2+b^2}$$
Where I used  $a,b$ is invertible $\bmod a^2+b^2$
A: $\rm\overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ Let $\rm\ w = a\!+\!b\,{\it i}\,.\, $ In $\rm\, \langle w\rangle\,$ is $\,\rm(a\!+\!b\,{\it i}\,)(j\!+\!k\,{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{#d0f}{ e\,\ +\,\ {\it i}}}  $
$\rm  \smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/\langle w\rangle\ $ is $\rm\,\color{#0a0}{onto,\ }$ by $\rm\bmod w\!:\,\ \color{#d0f}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\,{\it i}\:\equiv\, c\!-\!de\in \Bbb Z.\  $ Let $\rm\ n = ww'$
$\!\rm\begin{eqnarray}\rm Note\ \  \color{#c00}{m\in ker\ h} &\iff&\rm w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{ww'}\!=\dfrac{ma\!-\!mb\,{\it i}}n\in \Bbb Z[{\it i}\,]\\ &\iff&\rm n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $
$\rm Thus  \ \ \Bbb Z[{\it i}\,]/\langle w\rangle\! \color{#0a0}{= Im\:h}\,\cong\: \Bbb Z/\color{#c00}{ker\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.
