Proving $\mathbb{Z}[i] / (7+5i) \cong \mathbb{Z}_{74}$. Here's my attempt as I think I vaguely remember something similar:
We have $\phi : \mathbb{Z} \to \mathbb{Z}[i]/(7+5i)$ by $\phi(n) = n + (7+5i)$. I would like now to prove that $\ker\phi = \langle 74 \rangle$ and so by the first isomorphism theorem $$\mathbb{Z}[i]/(7+5i) \cong \mathbb{Z} / \langle 74 \rangle= \mathbb{Z}_{74}.$$
Questions:


*

*How do we prove that $\ker \phi = \langle 74 \rangle $

*Is the last equals sign correct?

 A: Hint $\rm\: n\in ker\ \phi\iff 7\!+\!5{\it i}\:|\:n \iff (7\!-\!5{\it i})(7\!+\!5{\it i})\:|\:(7\!-\!5{\it i})\,n\iff 74\:|\:7n\!-\!5n{\it i}\iff 74\:|\:n$
Written in fraction form it amounts to simply rationalizing denominators, viz. for $\rm\:w\ne 0$
$$\rm\ w\:|\:n\ in\ R \iff \frac{n}w \in R\iff \frac{nw'}{ww'}\in R$$
so reducing the problem of divisibility by the complex number $\rm\:w\:$ to the much simpler problem of divisibility by the integer $\rm\:ww'\in \Bbb Z.$
See also my post here on analogous ways of determining units, i.e. divisors of $1$ (vs. $\rm\:n\in \mathbb Z).$
A: One can even prove a more general result (and it's a fun exercise, too). Given any $n\in\Bbb Z$ not a perfect square, and any $a,b\in\Bbb Z$, we have that $\Bbb Z[\sqrt{n}]/\langle a+b\sqrt{n}\rangle\cong\Bbb Z/\langle a^2-b^2n\rangle$ iff $a,b$ are coprime in $\Bbb Z$. This problem, then, is just the special case of this more general result with $n=-1$. A key observation, here, is that $a,b$ are coprime in $\Bbb Z$ iff there exist $k,l\in\Bbb Z$ such that $ak+bl=-1$.
If you decide to play with this and get stuck, just let me know, and I can give you some more hints.
