The splitting of an ideal Let $K = \mathbb{Q}(\sqrt{-5})$. Now the ring of integers $\mathcal{O}_{K}$ is $\mathbb{Z}[i\sqrt{5}]$. 
I want to describe the ideal $(2)$ in $\mathbb{Z}[i\sqrt{5}]$ using the prime factorization.
My idea so far is to assume that we can write $2 = (a + i\sqrt{5}b)(\alpha + i\sqrt{5}\beta)$ for some integers $a,b,\alpha,\beta\in\mathbb{Z}$.
I then take the norm of this equation and obtain:
$4 = (a^{2}+b^{2}5)(\alpha^{2}+\beta^{2}5)$
I then try all the combinations of possible solutions and conclude that $2$ must be a prime ideal. 
I've tried to verify this (i.e. $(2)$ is a prime ideal) by computing the quotient $\mathcal{O}_{K}/(2)$, but this doesn't seem to work.
Is my solution correct?
 A: Your solution is wrong because it's based on a false premise. The number 2 is irreducible but not prime. Given any odd $a, b \in \mathbb{Z}$, we see that $2 \mid (a - b \sqrt{-5})(a + b \sqrt{-5})$, but $2 \nmid (a - b \sqrt{-5})$, likewise $2 \nmid (a + b \sqrt{-5})$.
"An ideal generated by an irreducible element is not necessarily irreducible as an ideal," writes Laura Lynch in a thesis that you should really take the time to read: http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1015&context=mathstudent
Indeed, $\langle 2 \rangle$ is a "composite ideal" (not standard terminology, as the comments will point out), and one that ramifies at that. According to Lynch, $\langle 2 \rangle = \langle 2, 1 + \sqrt{-5} \rangle^2$ (p. 37, or p. 45 in Adobe Reader, Case 1 of the proof of Theorem 5.16; unfortunately Lynch uses parentheses rather than angle brackets to denote ideals, something that will certainly cause occasional confusion).
It might be helpful for you to review the concept of containment of ideals. A prime ideal is a maximal ideal, and a maximal ideal is one that is not properly contained within any other proper ideal. Since $\langle 2 \rangle \subset \langle 2, a + b \sqrt{-5} \rangle \subset \mathbb{Z}[\sqrt{-5}]$, it follows that $\langle 2 \rangle$ is not maximal and therefore not prime.
For the sake of comparison, try $\mathcal{O}_K / \langle 13 \rangle$.
