An Ideal is Maximal

Question

In $\mathbb Z[\sqrt{-5}]$, define the ideals $=\mathbf p \langle 2, 1+\sqrt{-5}\rangle $, $\mathbf q=\langle3, 1+\sqrt{-5}\rangle$, and $\mathbf r=\langle3, 1-\sqrt{-5}\rangle$. Show that each ideals are maximal.

I have solved for p.

$Z[\sqrt{-5}]/\langle2\rangle=2^2=4$

Since $\langle2\rangle\subsetneq\langle2, 1+ \sqrt{-5} \rangle$, it has order dividing $4$.

Hence $\mathbb Z[\sqrt{-5}]/\mathbf p$ is congruent to $\mathbb Z_2,$ Where $\mathbb Z_2$ is a field.

Therefore, p is maximal.

I have a good understanding, however I have noticed that qr$=<3>$. Should I show this form of an ideal is maximal, or should I show that q and r are maximal separately?

Answer 1

Hint: modulo $1+ \sqrt{-5},$ $a+b \sqrt{-5}$ is congruent to $a-b\in\mathbb Z.$ Modulo $3,$ $a-b$ is congruent to one of $0,1,2.$

One step you skipped with $\mathbf p$ is proving the ideal is not all of $\mathbb Z[\sqrt{-5}].$ You’d need to do this in all cases.