Find a maximal ideal in $R = \mathbb{Z}[\sqrt{−5}]$ containing the principal ideal (3) I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.
 A: The ideals $I\subset R$ containing the principal ideal $(3)\subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $\Bbb{F}_3\times\Bbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
A: In general, given a positive number $p$ that is prime in $\mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$\left(\frac{d}{p}\right) = 1,$$ find the smallest positive integer $n$ such that $n^2 \equiv d \pmod p$, then $\langle p \rangle = \langle p, n - \sqrt d \rangle \langle p, n + \sqrt d \rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$\left(\frac{-5}{3}\right) = 1,$$ $-5 \equiv 1 \pmod 3$, and obviously $1^2 \equiv 1 \pmod 3$. Then $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $\mathbb Z[\sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $\langle 3, 1 - \sqrt{-5} \rangle$ or $\langle 3, 1 + \sqrt{-5} \rangle$. For example, $4 + \sqrt{-5} = 3 + (1 + \sqrt{-5})$ and hence $(4 + \sqrt{-5}) \in \langle 3, 1 + \sqrt{-5} \rangle$. Likewise $(4 - \sqrt{-5}) \in \langle 3, 1 - \sqrt{-5} \rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + \sqrt{-5}) \not\in \langle 3, 1 + \sqrt{-5} \rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $\langle 3 \rangle$ have norms divisible by 9 and all nonzero numbers in $\langle 1 - \sqrt{-5} \rangle$ have norms divisible by 6. Same goes for $\langle 1 + \sqrt{-5} \rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 \pm \sqrt{-5})$, with both $x$ and $y$ being any numbers from $\mathbb Z[\sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $\langle 3, 1 - \sqrt{-5} \rangle$ or $\langle 3, 1 + \sqrt{-5} \rangle$ is the whole ring.
