$\Bbb{Z}/10\Bbb{Z}$ isomorphic to $\Bbb{Z}[i]/\langle 1+3i\rangle$. I need to prove that 
$\Bbb{Z}/10\Bbb{Z}$ is isomorphic to $\Bbb{Z}[i]/\langle 1+3i\rangle$. 
I know I can use the third isomorphism, but I would like to use the first one. I consider a homomorphism $\Bbb{Z}$ to $\Bbb{Z}[i]/(1+3i)$ : 
$$\phi : \Bbb{Z}\rightarrow \Bbb{Z}[i]/\langle 1+3i\rangle,\quad x\mapsto \overline{x}.$$ The kernel is a subgroup of $\Bbb{Z}$, so it looks like $p\Bbb{Z}$ for a unique $p>0$.  
Plus $\overline{x}$ means (here) that it's equal to $(a+ib)+\langle 1+3i\rangle =(a+ib)+(1+3i)(a+ib)$, isn't it ? I am not sure, I used the fact that $\Bbb{Z}[i]$ is euclidean so principal and then all ideal are is that form. On the other hand $\Bbb{Z}\cap\langle 1+3i\rangle=10\Bbb{Z}$.

So the question is how can I write, correctly, the morphism $\phi$  ?

 A: As you said, we define a homomorphism $\phi:\mathbb{Z}\to\mathbb{Z}[i]/\langle 1+3i\rangle$ by
$$\phi(n)=n+\langle 1+3i\rangle$$
Then obviously $\phi(n)=0 \iff n\in\langle 1+3i\rangle$, but for any integer $n$, we have
$$n\in\langle a+bi\rangle\iff n\in\langle a-bi\rangle$$
because $n=\overline{n}$. Moreover, if $n\in\langle \alpha\rangle$ and $n\in\langle \beta\rangle$ for any two $\alpha,\beta\in\mathbb{Z}[i]$, then $n\in\langle \mathrm{lcm}(\alpha,\beta)\rangle$, which is true (and makes sense) because $\mathbb{Z}[i]$ is a PID. Thus
$$\begin{align*}
\phi(n)=0 &\iff n\in\langle 1+3i\rangle\;\text{ and }\;n\in\langle 1-3i\rangle\\
&\iff n\in\langle (1+3i)(1-3i)\rangle=\langle 10\rangle
\end{align*}$$
Therefore, the kernel of $\phi$ is the ideal $10\mathbb{Z}\subset\mathbb{Z}$.
Check that $\phi$ is surjective because $\phi(1)=1+\langle 1+3i\rangle$ and $$\begin{align*}
\phi(3)&=3+\langle 1+3i\rangle\\
&=3+i(1+3i)+\langle 1+3i\rangle\\
&=i+\langle 1+3i\rangle
\end{align*}$$
so that $\phi(a+3b)=a+bi+\langle 1+3i\rangle$ for any $a,b\in\mathbb{Z}$.
By the first isomorphism theorem, we get an induced isomorphism $\widetilde{\phi}:\mathbb{Z}/10\mathbb{Z}\to\mathbb{Z}[i]/\langle 1+3i\rangle$.
A: Write $\, w=1\!+\!3i.\ $  The natural map $\, \phi\color{#0a0}{ \ {\rm is\ surjective\  (onto)}}\,$ since $\,{\rm mod}\ w\!:\ \,w\bar w=10\equiv 0,\,\ {-}3i\equiv 1\,\overset{\times\,3}\Rightarrow\,i\equiv 3\,\Rightarrow\, a+bi\equiv a+3b\in\Bbb Z$   
$\color{#c00}{I = \ker\phi = 10\,\Bbb Z}\ $ follows by rationalizing a denominator
$$ n\in I\iff 1\!+\!3i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \dfrac{n}{1\!+\!3i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!3i)}{10}\in\Bbb Z[i]\iff \color{#c00}{10\mid n}$$
Therefore, applying the First Isomorphism Theorem, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{10\,\Bbb Z}.$
