TITLE: Confused about the splitting of 2 in $\mathbb{Q}(i).$
QUESTION [9 upvotes]: How do we split 2 in the cyclotomic ring $S = \mathbb{Z}[\sqrt{-1}]?$ Clearly the field $\mathbb{Q}(i)$ is a degree 2 normal extension. However, $(2)$ in $S$ is equal to $(2) = (1 - i)^2.$ Thus, the ideal cannot split anymore so we can deduce that $(1 - i)$ is in fact a prime ideal. But $S$ is a Dedekind domain and $(1 - i)$ is not a maximal ideal...I say this because $S/(1 - i) \cong \mathbb{Z}.$ Did I mistakenly assume that $\mathbb{Z}[i]/(1 - i) \cong \mathbb{Z}?$

REPLY [7 votes]: That's because it's a ramifying ideal, not a splitting ideal. Any time you have $\langle p \rangle = \mathfrak P^2$ in $\textbf Q(\sqrt d)$, the ideal is ramifying rather than splitting.
This is easier to see when $\gcd(p, d) > 1$. For example, consider $\langle 13 \rangle$ in $\textbf Z[\sqrt{-13}]$. Obviously $(-1)(\sqrt{-13})^2 = 13$ and so $\langle 13 \rangle = \langle \sqrt{-13} \rangle^2$.
It is much less obvious that $\langle 2 \rangle = \langle 2, 1 + \sqrt{-13} \rangle^2$. Since $(1 - \sqrt{-13})(1 + \sqrt{-13}) = 14$, one might think that $\langle 2 \rangle$ splits. However, since $\langle 2, 1 + \sqrt{-13} \rangle$ consists of all numbers of the form $2x + (y + y \sqrt{-13})$, where both $x$ and $y$ are arbitrary numbers in $\textbf Z[\sqrt{-13}]$, it follows that $\langle 1 - \sqrt{-13} \rangle \subset \langle 2, 1 + \sqrt{-13} \rangle$ since $2(-\sqrt{-13}) + (1 + \sqrt{-13}) = 1 - \sqrt{-13}$, thus confirming that $\langle 2 \rangle = \langle 2, 1 + \sqrt{-13} \rangle^2$. We could just as easily express it as $\langle 2 \rangle = \langle 2, 1 - \sqrt{-13} \rangle^2$ if we want.
Of course $\textbf Z[\sqrt{-1}]$ "enjoys" unique factorization (a silly choice of word), vastly simplifying things for us. Since $(1 - i)(1 + i) = 2$ rather than some larger even composite number... 
I guess I better stop there, since I'd just be repeating Robert's answer. Except for this point about ramifying and splitting, I think Robert's answer and David's comment pretty much cover everything. (David's no relation to me). Well, there is one point that's still missing...