TITLE: Why is $(2, 1+\sqrt{-5})$ not principal?
QUESTION [13 upvotes]: Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?

Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.
Or is there a simpler method ? (Maybe not UFD would imply not PID)

REPLY [3 votes]: Yes, there is a simpler method, not that different from what you've done so far, but you have to keep in mind that there are still principal ideals even if the domain is not a principal ideal domain. For example, $\langle 5, \sqrt{-5} \rangle$ is a principal ideal; a little thought should reveal that it's generated by a single element and is in fact $\langle \sqrt{-5} \rangle$.
So if $\langle 2, 1 + \sqrt{-5} \rangle$ is a principal ideal, we should be able to find a single element $x \in \mathbb{Z}[\sqrt{-5}]$ to generate it, an element that is not a unit. There isn't one: $N(2) = 4$ and $N(1 + \sqrt{-5}) = 6$, as you already know, so we'd need for $x$ to satisfy $N(x) = 2$, which as you also already know, has no solutions.
It should be noted that $\sqrt{-5} \not\in \langle 2, 1 + \sqrt{-5} \rangle$. In fact, this ideal does not contain any numbers with odd norm, which means no purely real odd integers. There is no combination of $r, s \in \mathbb{Z}[\sqrt{-5}]$ that will give you $2r + s + s \sqrt{-5} = 3$, for example, because $N(2r + s + s \sqrt{-5})$ must be even. This confirms that $\langle 2, 1 + \sqrt{-5} \rangle$ is not the whole ring.

It might be helpful to compare a similar ideal in a similar domain: $\langle 2, 1 + \sqrt{5} \rangle$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$. As before, $N(2) = 4$. But $N(1 + \sqrt{5}) = 4$ as well. This may or may not be in a PID (spoiler alert: it is), but this particular ideal is a principal ideal, and it is in fact $\langle 2 \rangle$ (verify that the familiar number $\frac{1}{2} + \frac{\sqrt{5}}{2}$ is an algebraic integer).