#Hom$(\mathbb{Z}[i], \mathbb{Z}/85\mathbb{Z})$

Question

I'm trying to determine the number of ring homomorphisms from $\mathbb{Z}[i]$ to $\mathbb{Z}/85\mathbb{Z}$. My reasoning is as follows: We know that for all $f \in \text{Hom}(\mathbb{Z}[i], \mathbb{Z}/85\mathbb{Z})$ we must have $f(1) = \bar{1} = 1 + 85\mathbb{Z}$. We also know that for any $z = a+bi \in \mathbb{Z}[i]$, $f(z) = f(a) + bf(i)$, so we only need to find out what the possibilities for $f(i)$ are. As $\bar{84} = -\bar{1} = f(-1) = f(i^2) = f(i)^2$, we need to find solutions to the equation $f(i)^2 + 1 = 85n$ for some $n \in \mathbb{Z}$. I've found that $f(i) = \bar{13}$ and $f(i) = \bar{16}$ satisfy this relation (for $n = 2, n = 3$, respectively). Now I could continue on trying $\bar{n^2}$ for all $n \in \mathbb{N}_{\leq 85}$, and have done so for $n \leq 32$, but it's getting more and more cumbersome to do this. Is there a way to know if I'm done already?

Answer 1

As you say, any such (unital) homomorphism verifies

$$ f(a+ib) = a + f(i)b. $$

Now, note that defining $f(a+ib) := a+f(i)b$ we already get a group morphism for $+$ which sends $0$ to $0$ and $1$ to $1$. Hence what we should study is the presevation of multiplication, that is when does it hold that

$$ ac-bd + f(i)(ad+bc) = f((ac-bd) + i(ad+bc)) = f((a+ib)(c+id)) = f(a+ib)f(c+id) = (a+f(i)b)(c+f(i)d) = ac + f(i)(ad+bc) + f(i)^2bd $$

Consicely, we need $f(i)$ such that for all $b,d$ integers,

$$ f(i)^2bd \equiv -bd \pmod{85} $$

In particular when $b = d = 1$, a necessary (and sufficient!) condition is

$$ f(i)^2 \equiv -1 \pmod{85} $$

Since $85 = 5 * 17$, by the Chinese Remainder Theorem this is equivalent to

$$ \cases{ f(i)^2 \equiv -1 \pmod{5} \\ f(i)^2 \equiv -1 \pmod{17} }$$

By a case by case inspection, the first equation gives $f(i) \equiv 2,3 \pmod{5}$ as possible solutions and the second $f(i) \equiv 4,13 \pmod{17}$. This gives four possible cases. Note that for each one, $f(i)^2$ will be $1$ modulo $85$ and that is all we need. Some verification is needed in case some solutions are actually equal $\pmod{85}$. A computer assisted (or manual but rather long) confirms that these cases are all different, corresponding to the solutions $$f(i) \in \{13,38,47,72\} =: S \subset \mathbb{Z}_{85}$$ Hence, the maps

$$ f(a+ib) := a+(s + 85k)b $$

with $s \in S$ are ring homomorphisms.

Edit: Since $s+85k \equiv s \pmod{85}$ for all $k$, as you correctly said, this only gives $4$ distinct maps.