I am working on a problem in which I need to prove that
$\mathbb{Z}[i]/I_n$ is a field if, and only if, $q = n^2 + 1$ is prime.
Here, $I_n = _{\mathbb{Z}[i]}\langle i -n \rangle$ is a given ideal of the Gaussian integers.
Earlier on, we were introduced to a ring homomorphism $$f: \mathbb{Z}[i] \rightarrow \mathbb{Z}_q, \quad f(r+is) =[r] + [n][s]$$ where the square brackets denote equivalence classes modulo $q = n^2 + 1$.
I have found the kernel of this homomorphism to be $$\{\alpha qr + \beta qsi,\; -ns + q\alpha+si : \alpha, \beta, q,r \in \mathbb{Z}\}$$ after working through various scenarios where the equivalence classes would be $0$.
My thought process is that if I can show that $I_n = \ker f$, then by the First Isomorphism Theorem we know that $$\mathbb{Z}[i]/I_n \cong \mathbb{Z}_q.$$ From here, I would simply be using that for $q$ prime, $\mathbb{Z}_q = \mathbb{F}_q$ is a finite field.
I am really struggling to show the link between $I_n$ and $\ker f$.
Any advice on how to proceed would be great, or if this is the totally wrong tree to be barking up I would really appreciate a nudge in the right direction. :)
Below working (not really leading to any progress) is what I have been playing around with so far.
$$\begin{align*}I_n &= _{\mathbb{Z}[\sqrt{-1}]}\langle \sqrt{-1} -n \rangle\\ &= \{z(i-n) : r \in \mathbb{Z}[i] \\ &= \{(r + si)(i-n) : r,s \in \mathbb{Z}\} \\ &=\{(-nr-s) + (r-ns)i : r,s \in \mathbb{Z}\}\end{align*}$$
It it obvious that $I_n\subseteq\ker f$.
Conversely, if $r+is\in\ker f$ then $n^2+1\mid r+ns$. It follows that $n^2+1\mid nr-s$. Now notice that $r+is=(i-n)z$ for some $z\in \mathbb Z[i]$, so $r+is\in I_n$. (In order to find $z$ start with $\frac{r+is}{i-n}$ and multiply the numerator and denominator by $i+n$.)