A very basic ring theory question, which I am not able to solve.
Let $\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(2+3i) \text{ where } \phi(z) = z + (2+3i)\mathbb{Z}[i]$.
Because I want to prove that $\mathbb{Z}[i]/(2+3i) \cong \mathbb{Z}/13\mathbb{Z}$
How does one show that $\ker \phi = 13\mathbb{Z}$?
This is my attempt:
$\ker \phi = (2+3i)\mathbb{Z}[i] \cap \mathbb{Z}$. So for any such $z\in\ker\phi$, we have $z = (2+3i)(a+ib)$ for some $a,b \in \mathbb{Z}$ happens if and only if $3a+2b = 0$. So $\ker \phi = (2+3i)\mathbb{Z}[i] \cap \mathbb{Z}=\{(2+3i)(a+ib):3a+2b=0, a,b\in\mathbb{Z}\}.$ How can I continue?
$3a+2b=0$ is equivalent to $a=2k$ and $b=-3k$ for $k\in\mathbb Z$. Now the product $z=(2+3i)(a+bi)$ equals $2a-3b=13k$ and you are done. (Well, this shows that $\ker\phi\subseteq13\mathbb Z$, but I suppose you already noticed that $13\in\ker\phi$.)