Prove $\mathbb{Z[i]}/(1+4i) \cong \mathbb{Z_{17}}$ Prove that $\dfrac{\mathbb{Z[i]}}{(1+4i)} \cong \mathbb{Z_{17}}$.
My attempt:
$\phi:\mathbb{Z} \rightarrow \mathbb{Z_{17}}$ be the natural (onto) ring homomorphism, sending $a \mapsto \bar{a}$. Hence by universal property of polynomial rings there exists a (onto) ring homomorphism $\phi':\mathbb{Z[X]} \rightarrow \mathbb{Z_{17}}$ such that $\phi'(x)=\bar{4}$  and $\phi'(n)=\phi(n)$ for all $n \in \mathbb{Z}$.
Clearly, $(1+x^2) \subset \operatorname{Ker}\phi' \implies$ we can obtain a (onto) ring homomorphism $\phi'':\mathbb{Z}[X]/(1+x^2) \rightarrow \mathbb{Z_{17}}$ where $\bar{x} \mapsto \bar{4}$
Since $\mathbb{Z}[X]/(1+x^2) \cong \mathbb{Z[i]}$ with $i \mapsto \bar{x}$, one can conclude there is an onto ring homomorphism $\Phi:\mathbb{Z[i]} \rightarrow \mathbb{Z_{17}}$ where $i \mapsto \bar{4}$.
$1+4i \mapsto 0$ one has $(1+4i) \subset \operatorname{Ker}\Phi$.
How can I conclude the reverse inclusion?
 A: With these problems, the third isomorphism theorem is usually your friend (divide out by $17$ and $x-4$ before you divide out by $x^2 +1$). Here is a proof of what you want to show without doing any explicit quotienting, although it works just as well if you rephrase it to be about the map $\Bbb Z[x]/(17, x-4, x^2+1)\to \Bbb Z_{17}$.
The kernel of $\phi'$ clearly contains $(17, x^2+1, x-4)$, as you have shown. We want to show that this is the entire kernel. Note that $x^2+1\in (17, x-4)$, because $(x-4)(x+4) + 17 = x^2 + 1$. So that's a superfluous generator. Now we just have $(17, x-4)$, and things are much simpler.
Take an element $f$ in the kernel of $\phi'$. We want to study the coset $f + (17, x-4)$. It has a representative among the elements $\{0, 1, 2, \ldots, 16\}$ (we can use the $x-4$ generator to get rid of any $x$'s, and then the $17$ generator to bring the remaining constant term into this range). And since any element in this coset is in the kernel of $\phi'$, the chosen representative must be in the kernel. But among these, only $0$ is in the kernel of $\phi'$. So $$f + (17, x-4) = 0 + (17, x-4)$$and we are done.
