Finding the Kernel and Image of $\mathbb Z \to \mathbb Z[i]/(1+3i)$, $x\mapsto x+(1+3i)$ I'm trying to apply the homomorphism theorem to the following function:
$$h:\mathbb Z \to \mathbb Z[i]/(1+3i)$$
$$x\mapsto x+(1+3i)$$
Where $(1+3i)$ is the ideal generated by $1+3i$.
I know that Because $\ker(h)$ is an ideal in $\mathbb Z$ which is an Principal Ideal Domain, then $\ker(h)=m\mathbb Z$ for some $m \in \mathbb Z$, but I'm having some trouble finding that $m$.
And I have no clue how I can find the elements of $\text{im}(h)$.
How can I do this?

Note that $\ker{h}=\{n \in \mathbb{Z} : n \in \langle 1+3i \rangle\}=\langle 1+3i \rangle \cap \mathbb{Z}$. $n \in \ker{h} \implies n=(1+3i)(a+bi)=a+bi+3ai-3b=(a-3b)+i(b+3a)$. Since $n \in \mathbb{Z}$, $b=-3a$. Thus, every $n \in \ker{h}$ is of the form $a(1+3i)(1-3i)=10a$ for some $a \in \mathbb{Z}$. Clearly, the least such element is $10$ because $|10a|=|10||a|=10|a|\geq 10$ and for $a=1$, equality is satisfied. So, $\ker{h}=\langle 10 \rangle$.
