Question: Let $R=\mathbb{Z}[\sqrt{-5}]$ and let
$$ \phi: \mathbb{Z}[\sqrt{-5}] \rightarrow \mathbb{Z} / 3 \mathbb{Z}, \quad a+b \sqrt{-5} \mapsto \overline{a+b} \quad(a, b \in \mathbb{Z}) . $$ (a) Prove $\phi$ is a surjective ring homomorphism.
So we have to prove that $\phi(a+b) = \phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$.
$\phi(a+b)=\phi((a+b\sqrt{-5})+(c+d\sqrt{-5})) = \overline{a+b+c+d}$
And, $\phi(a)+\phi(b)=\overline{a+b}+\overline{c+d} = \overline{a+b+c+d}$
So we have $\phi(a+b)=\phi(a)+\phi(b)$.
I struggle with $\phi(ab)=\phi(a)\phi(b)$.
$\phi(ab)=\phi((a+b\sqrt{-5})(c+d\sqrt{-5}) = \phi(ac+ad\sqrt{-5}+bc\sqrt{-5}-5bd) = \overline{ac-5bd+ad+bc}$.
$\phi(a)\phi(b)=(a+b\sqrt{-5})(c+d\sqrt{-5})= (\overline{a+b})(\overline{c+d})$
What do I do wrong?
Let me go line by line.
The problem here is that $a+b\neq a+ b\sqrt{-5} + c + d\sqrt{-5}$ in general and what you want to write is something like $r = a + b\sqrt{-5}$ and $s = c + d\sqrt{-5}$, so $\phi(r+s) = \phi((a+b\sqrt{-5})+(c+d\sqrt{-5})).$ Furthermore, it's not clear what exactly you are using in the next equality, to clarify you could write something like $$ \phi((a+b\sqrt{-5})+(c+d\sqrt{-5})) = \phi ((a+c)+(b+d)\sqrt{-5}) = \overline{(a+c)+(b+d)}$$ where we used how addition is defined in $\mathbb Z[\sqrt{-5}]$ (or simply commutativity and distributivity in $\mathbb C$, however you want to look at it) and then definition of $\phi$.
Similarly, instead of
you want $$\phi(r) + \phi(s) = \overline{a+b}+\overline{c+d} = \overline{(a+b)+(c+d)}.$$
Again, you have the same problem with checking multiplicativity, so I won't repeat it.
The ingredient that you are missing to finish the proof, as mentioned in comments is that $$-5bd \equiv bd \pmod 3.$$
Also, you did not write explicitly how you'd prove surjectivity, but this is really easy since $\phi(n) = \overline n$, for any integer $n$.