Number of ring homomorphisms I'm trying to solve the following problem:
Determine how many different ring homomorphism from $\mathbb{Z}[i] \to \mathbb{Z}/(85)$ exist.
For a previous question I had to determine the unique integer factorization of $85=(1-2i)(1+2i)(1-4i)(1+4i)$. I thought of creating the following ring homomorphisms:
$$f_{\pm 2}:\mathbb{Z}[i] \to \mathbb{Z}/(85), i \mapsto \pm 2 $$
and another two mapping $ i \mapsto \pm 4$ which gives me a total of four ring homormoprhism. I was wondering if there are any more beside these ring homomorphisms.
edit: fixed notation to the quotient ring
 A: There's four.
My reasoning is that $\varphi (i)^2=\varphi (i^2)=\varphi (-1)=-1.$
So we need $-1$ to be a quadratic residue in $\Bbb Z_{85}$.
So $(\dfrac{-1}5)=(-1)^{\frac {5-1}2}=1$ and $(\dfrac {-1}{17})=(-1)^{\frac {17-1}2}=1$ are the relevant Legendre symbols.
The solutions are $\pm13$ and $\pm38$.  So those are the possibilities for $\varphi (i)$.

As we know,  $\Bbb Z_5×\Bbb Z_{17}\cong \Bbb Z_{85}$.
One way is with Bezout coefficients $\color {blue}7×5-\color {purple}2×17=1.$
Under the isomorphism $(\color {purple}x,\color {blue}y)\mapsto \color {blue}{7}×5×\color {blue}y-\color {purple}2×17×\color {purple}x$, we have $(-2,-4)\mapsto 13$ and $(-2,4)\mapsto 38.$
A: As in the other answer, we need $\varphi: \Bbb Z[i]\to\Bbb Z/(85)$ to satisfy
$$
-1 = \varphi(-1) = \varphi(i^2) = \varphi(i)^2
$$
We could use the Chinese remainder theorem to see that we need $\varphi(i)$ to be either $2$ or $3$ modulo $5$, and we need it to be either $4$ or $13$ modulo $17$. These two binary choices are independent. That means 4 solutions, even if it takes a bit of work to find exactly what those four are modulo 85 from this (I never could get the hang of the constructive part of the Chinese remainder theorem).
