TITLE: Express the ideal $(6) \subset\mathbb Z\left[\sqrt {-5}\right]$ as a product of prime ideals.
QUESTION [8 upvotes]: Express the ideal $(6) \subset \mathbb Z\left[\sqrt {-5}\right]$ as a product of prime ideals.

I know I can write $(6)=(2)(3)=\left(1+\sqrt {-5}\right)\left(1-\sqrt {-5}\right)$. But I guess these factors might not be prime. 
What's more, how to solve this kind of problem more systematically?

REPLY [3 votes]: In a principal ideal domain, all ideals are principal, whether generated by prime numbers or composite numbers. But you already know that $\mathbb Z[\sqrt{-5}]$ is not a principal ideal domain.
In domains such as these, it helps to remember that all prime ideals are maximal ideals (with the possible exception of $\langle 0 \rangle$, but that's a digression you may or may not care to get into).
So clearly neither $\langle 2 \rangle$ nor $\langle 3 \rangle$ are maximal ideals of $\mathbb Z[\sqrt{-5}]$, since they are properly contained in $\langle 2, 1 \pm \sqrt{-5} \rangle$ and $\langle 3, 1 \pm \sqrt{-5} \rangle$, respectively. Might either of those ideals be principal?
Oh, wait, you said you wanted a more systematic way of solving this sort of problem. In that case, you should memorize these facts:

$\langle 2 \rangle$ is prime in $\mathcal O_{\mathbb Q(\sqrt d)}$ if $d \equiv 5 \pmod 8$. 
$\displaystyle \langle 2 \rangle = \left\langle 2, \frac{1}{2} - \frac{\sqrt d}{2} \right\rangle \left\langle 2, \frac{1}{2} + \frac{\sqrt d}{2} \right\rangle$ if $d \equiv 1 \pmod 8$ 
$\langle 2 \rangle = \langle 2, 1 + \sqrt d \rangle^2$ if $d \equiv 3 \pmod 4$
$\langle 2 \rangle = \langle 2, \sqrt{d} \rangle^2$ if $d$ is even

and if $p$ is an odd prime in $\mathbb Z$,

$\langle p \rangle$ is prime if $\gcd(p, d) = 1$ and $d$ is not a square modulo $p$
but if $x^2 \equiv d \pmod p$ can be solved ($\gcd(p, d) = 1$ still holds), then $\langle p \rangle = \langle p, x - \sqrt d \rangle \langle p, x + \sqrt d \rangle$
or if $\gcd(p, d) > 1$, then $\langle p \rangle = \langle p, \sqrt d \rangle^2$.

These are well known facts, but the only book I've ever seen them neatly summarized like this is the algebraic number theory book by Alaca and Williams.
Then, given non-prime $n$ an algebraic integer of degree 1 (that is, purely real, and rational), then the factorization of $\langle n \rangle$ into ideals of $\mathcal O_{\mathbb Q(\sqrt d)}$ is a simple matter of factorizing the ideals generated by the prime factors of $n$ in $\mathbb Z$.
So in this particular example we have $n = 6$ and $d = -5$. Remember that $-5 \equiv 3 \pmod 8$, not 5, and $-5 \equiv 3 \pmod 4$, not 1. Then $\langle 2 \rangle = \langle 2, 1 + \sqrt{-5} \rangle^2$. So 2 is a square? I know, that seems super weird. The thing is that $\langle 2, 1 - \sqrt{-5} \rangle = \langle 2, 1 + \sqrt{-5} \rangle$. Verify that $1 - \sqrt{-5} = 2 (-\sqrt{-5}) + (1 + \sqrt{-5})1$.
As for 3, we see that $-5 \equiv 1 \pmod 3$ and $1^2 = 1$, so $\langle 3 \rangle = \langle 3, 1 - \sqrt{-5} \rangle \langle 3, 1 + \sqrt{-5} \rangle$.