Which ideal contains which? Or are they the same? If I'm understanding this correctly, $13$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{-23})}$ is irreducible but not prime. From A & W we see that $x^2 \equiv -23 \bmod 13$ has solutions and therefore $\langle 13 \rangle$ factors as $\langle 13, x - \sqrt{-23} \rangle \langle 13, x + \sqrt{-23} \rangle$, where any $x$ that satisfies the congruence will work, e.g., $x = 4, 9, 17, 22,$ etc.
But where do ideals like $$\left\langle 13, \frac{9}{2} + \frac{\sqrt{-23}}{2}\right\rangle$$ fit into all this? Would that ideal be contained in $\langle 13, 4 - \sqrt{-23} \rangle$ or $\langle 13, 4 + \sqrt{-23} \rangle$? Or is it the other way around? Or is the same ideal as one of those? I'm getting confused.
 A: You can tell the ideals apart by simple congruences: Let 
$I = (13,(9+\sqrt{-23})/2)$; then clearly $\sqrt{-23} \equiv -9 \bmod I$
and $-9 \equiv 4 \bmod 13$. Since $13 \in I$, you have 
$\sqrt{-23} \equiv 4 \bmod I$.Therefore $I$ divides (and so contains)
$(13,4 - \sqrt{-23})$. In fact they are equal:
$$  (13,4 - \sqrt{-23}) = (13, -9 - \sqrt{-23}) 
    = (13, 9 + \sqrt{-23}) = (13,2)I = I. $$
A: (Edit:) This is one of the ideals in the factorization of $(13)$. Here's how you can see this by writing the factorization in a different way. The ring of integers is $\mathbb{Z}[\alpha]$ where $\alpha = \frac{1 + \sqrt{-23}}{2}$ satisfies $\alpha^2 - \alpha + 6 = 0$. To factor $(13)$ we look for solutions  $x^2 - x + 6 \equiv 0 \bmod 13$. The two roots $\bmod 13$ are $x = 9, x = -8$, which gives the factorization
$$(13) = (13, 9 - \alpha)(13, 8 + \alpha).$$
The ideal you're confused about is $(13, 4 + \alpha) = (13, 9 - \alpha)$. The ideals in your factorization are (if we pick $x = 4$) 
$$(13) = (13, 5 - 2 \alpha)(13, 3 + 2 \alpha)$$
but $6(5 - 2 \alpha) = 30 - 12 \alpha \equiv 4 + \alpha \bmod 13$ and similarly $7(3 + 2 \alpha) = 21 + 14 \alpha \equiv 8 + \alpha \bmod 13$. 
A: Another approach is to look for signs of mutual containment; this is something that I'm just barely beginning to understand myself.
If a number is in $\left\langle 13, \frac{9 + \sqrt{-23}}{2} \right\rangle$, it can be expressed in the form $13x +  \left(\frac{9 + \sqrt{-23}}{2}\right)y$, where $x, y \in \mathcal{O}_{\textbf{Q}(\sqrt{-23})}$. Obviously $$9 + \sqrt{-23} = 2 \left(\frac{9 + \sqrt{-23}}{2}\right).$$
It's a little tougher to show that $$\frac{9 + \sqrt{-23}}{2} \in \langle 13, 9 + \sqrt{-23} \rangle.$$ It is, in fact. We're looking to show that this number is of the form $13x + (9 + \sqrt{-23})y$. The important thing is that $x$ and $y$ don't have to be ordinary integers, so long as they are algebraic integers in this ring. And so: $$\left(\frac{13 + 13 \sqrt{-23}}{2}\right) + (9 + \sqrt{-23}) \left(\frac{-3 - \sqrt{-23}}{2}\right) = \frac{9 + \sqrt{-23}}{2}.$$
This was not easy for me to find, which is why I'm the third to answer this question after a week, rather than the first after a few minutes.
