A very basic ring theory question, which I am not able to solve. How does one show that 


$\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.
Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.


My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful. 

Define
$$\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(3-i) \text{ where } \phi(z) = z + (3-i)\mathbb{Z}[i].$$
It follows simply that
$\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z}$.
So for any such $z \in \ker \phi$, we have $z = (3-i)(a+bi)$ for some $a,b \in \mathbb{Z}$. But $(3-i)(a+bi) \in \mathbb{Z}$ happens if and only if $3b-a=0$.
So $$\begin{align*}
\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z} &= \{(3-i)(3b+bi)\mid b \in \mathbb{Z}\}\\
&= \{(9b + b) + i(3b-3b)\mid b \in \mathbb{Z}\}\\
&= \{10b\mid b \in \mathbb{Z}\}\\
&= 10\mathbb{Z}.
\end{align*}$$

To see $\phi$ is surjective, let $(a+bi) + (3-i)\mathbb{Z}[i] \in \mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $\phi(a+3b) = (a+bi) + (3-i)\mathbb{Z}[i]$.

Hence $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.
