Show that the ideal $I := \langle 2, 1+\sqrt{-5} \rangle$ is maximal in $\mathbb{Z}\left[\sqrt{-5}\right]$ by considering the parities of $a,b$ for every $a + b\sqrt{-5} \in I$.
I have already shown that $I = \{ a + b\sqrt{-5} \mid a,b \in \mathbb{Z} : a \equiv b \pmod 2 \}$. However, when I know try to show that $\mathbb{Z}\left[\sqrt{-5}\right] / I$ is a field, I obtain by the parity condition that $\mathbb{Z}\left[\sqrt{-5}\right] / I = \{ I, 1 + I, \sqrt{-5} + I \}$, but I do not think that this is indeed a field.
Am I missing something?
Instead of number theory, you can do it with algebra: $\mathbb{Z}[\sqrt{-5}] \simeq \mathbb{Z}[x]/(x^2+5)$. Now your quotient ring is isomorphic to $$\mathbb{Z}[x]/\langle x^2+5, 2, 1+x\rangle$$
You can factor in a different order and get
$$\mathbb({Z}/2) [x]/\langle x^2+1, x+1\rangle \simeq \mathbb{Z}/2$$
a field indeed.