Intuitive motivation to try to factor an ideal In $\mathbb{Z}[\sqrt{- 5}]$,  $2$ is irreducible, but the ideal $(2)$ factors into non-units:
$$(2) = (2, 1 + \sqrt{- 5})(2, 1 - \sqrt{- 5}).$$
In general, what gives one the intuitive motivation (or maybe a priori confidence) to attempt to factor an ideal?
I could say in the specific case above that $(1 + \sqrt{- 5})$ and $(1 - \sqrt{- 5})$ are easy combinations of the elements of the basis of $\mathbb{Z}[\sqrt{- 5}]$, $\{1, \sqrt{- 5}\}$ and their product is divisible by $2$, so it's pretty easy to give them a try along with $2$.
But I was wondering if there is something that gives a clue that an ideal can be further factored. 
Thanks
 A: The key insight into all of this (at least for quadratic fields) is something called the absolute ideal norm. Let $K$ be a quadratic extension of $\Bbb{Q}$ and let $\mathcal{O}_K$ denote the integral closure of $\Bbb{Z}$ in $K$. For any ideal $\mathfrak{a} \in \mathcal{O}_K$, we define $||\mathfrak{a}||$ to be the number of elements in the finite group $\mathcal{O}_K/\mathfrak{a}$. To see why $\mathcal{O}_K/\mathfrak{a}$ is finite consider the ses
$$0 \to \mathfrak{a} \to \mathcal{O}_K \to \mathcal{O}_K/\mathfrak{a} \to 0$$
and apply the exact functor $-\otimes_{\Bbb{Z}} \Bbb{Q}$ to the conclude that the free part of $\mathcal{O}_K/\mathfrak{a}$ is zero. Right back to our problem. Suppose now that $\mathfrak{a}$ is a principal ideal. For a concrete example let's take $K = \Bbb{Q}(\sqrt{-21})$ and $\mathfrak{a} = (3 + \sqrt{-21})$. 
Now factor $\mathfrak{a}$ into primes $p_1\ldots p_n$ (including multiplicities). Then using the fact that the absolute norm is multiplicative and that the absolute norm of a principal ideal is the field norm of the generator we get that 
$$30 = N_{K/\Bbb{Q}}(3 + \sqrt{-21})) = ||p_1|| \ldots ||p_n||.$$
Here is now the killer blow: The right hand side is just a plain old product of integers!!! 
Thus it must be the case that upto rearrangement, $||p_1|| = 2$, $||p_2||= 3$ and $||p_3||| =5$. Thus we now ask ourselves: What are the possibilities for a prime $p_1$ of norm $2$? Well it has to lie over $2$ of course! Why? Because we recall that for $P$ a prime in $\mathcal{O}_K$ lying over a prime $p \in \Bbb{Z}$, the number of elements in $\mathcal{O}_K$ is just
$$|\Bbb{Z}/p\Bbb{Z}|^{f(P|p)}$$
namely a power of a prime. Thus we see that it will suffice to factor $2\mathcal{O}_K$,$3\mathcal{O}_K$ and $5\mathcal{O}_K$. I get that
$$\begin{eqnarray*} 2\mathcal{O}_K &=& (2, 1 + \sqrt{-21})^2\\
3\mathcal{O}_K &=& (3,\sqrt{-21})^2 \\
5\mathcal{O}_K &=& (5,2+\sqrt{-21})(5,2-\sqrt{-21}). \end{eqnarray*}$$
and so $(3+\sqrt{-21})$ must necessarily factor as
$$(3+\sqrt{-21}) = (2, 1 + \sqrt{-21})(3,\sqrt{-21})\mathfrak{p}_5$$
where $\mathfrak{p}_5$ is one of the primes lying over $5$ of which there are two possibilities. To determine exactly which one it is takes some bashing, but at least you get the general idea of how we factor principal ideals. Lastly, I should say that it is not so hard to determine how a prime splits in a quadratic extension. Let me give an example. Consider $\Bbb{Q}(\sqrt{-21})$ as before and the prime $p = 5$ in $\Bbb{Z}$. We know that $\mathcal{O}_K = \Bbb{Z}[\sqrt{-21}]$ and so 
$$\begin{eqnarray*}\Bbb{Z}[\sqrt{-21}]/(5) &\cong& \Bbb{Z}[x]/(x^2 + 21)/(5,x^2+ 21)/(x^2 + 21) \\
&\cong& \Bbb{Z}[x]/(5,x^2+21)\\
&\cong& \left(\Bbb{Z}/5\Bbb{Z}\right)[x]/(x^2 + 1) \\
&\cong& \left(\Bbb{Z}/5\Bbb{Z}\right)[x]/\left((x+2)(x+3)\right) \\
&\cong& \Bbb{Z}/5\Bbb{Z} \times \Bbb{Z}/5\Bbb{Z} \end{eqnarray*}$$
by the Chinese remainder theorem. This tells you that $5$ must split into two prime ideals, namely $(5,2+\sqrt{-21})$ and $(5,3+\sqrt{-21})$. The latter is of course the same as $(5,2-\sqrt{-21})$, which is exactly like what I wrote down above.
