Computing prime factorization of ideals? I want to compute the prime factorizations of the ideals $\langle 4\sqrt{-14}\rangle$, $\langle 6\sqrt{-6} \rangle$ and $\langle 4\sqrt{-5} \rangle$ in the ring of algebraic integers of $\mathbb{Q}(\sqrt{-14})$,$\mathbb{Q}(\sqrt{-6})$ and $\mathbb{Q}(\sqrt{-5})$. How can I do that? 
Also, is there any way to prove every prime ideal in prime factorization of the ideals above is stable under complex conjugation?
 A: As an example, take $I = \langle 4\sqrt{-14}\rangle\subset\mathbb Q(\sqrt{-14})$. It's clear that $I = \langle 2\rangle^2\cdot\langle \sqrt{-14}\rangle$. To factorise $I$, it is sufficient to factorise $\langle2\rangle$ and $\langle \sqrt{-14}\rangle$ separately.
To factorise $\langle 2\rangle$ we can use the Kummer-Dedekind theorem: $\mathcal O_{\mathbb Q(\sqrt{-14})}=\mathbb Z[\sqrt{-14}]$, and 
$$X^2+14\equiv X^2\pmod 2$$
so $\langle 2\rangle = \langle2,\sqrt{-14}\rangle^2$.
To factorise $\langle \sqrt{-14}\rangle$, observe that $\langle \sqrt{-14}\rangle\cap\mathbb Z = 14\mathbb Z$, so $\langle \sqrt{-14}\rangle$ is not prime, and it has norm $14$, so it should factorise as a product of a prime above $2$ and a prime above $7$. Using the Kummer-Dedekind theorem, we see that $7\mathcal O_{\mathbb Q(\sqrt{-14})} = \langle 7,\sqrt{-14}\rangle^2$, and it is simple to check that $\langle\sqrt{-14}\rangle=\langle 2,\sqrt{-14}\rangle\cdot\langle 7,\sqrt{-14}\rangle$.
It is straightforward to observe that each of these prime ideals is stable under complex conjugation. However, there are ways of seeing this without fully computing the factorisation.
To illustrate this, consider $J=\langle 4\sqrt{-5}\rangle\subset \mathbb Q(\sqrt{-5})$. Suppose that $\mathfrak p$ is a prime ideal dividing $J$, which is not stable under complex conjugation. Since $J$ is stable under complex conjugation, it follows that $\overline{\mathfrak p}$ also divides $J$, and therefore $\mathfrak p\overline{\mathfrak p} = p\mathcal O_{\mathbb Q(\sqrt{-5})}$ divides $J$, where $p = \mathfrak p\cap\mathbb Z$. In this case $p$ must therefore be $2$. It is simple to check that $2$ is totally ramified in $\mathbb Q(\sqrt{-5})$, which rules out this case.
More generally, if $I$ is an ideal in an imaginary quadratic field that is stable under conjugation, and for every rational prime $p$ such that $p\mathcal O\mid I$, there is a unique prime ideal of $\mathcal O$ lying above $p$ (i.e. $p$ is totally ramified or inert), then the prime ideals dividing $I$ are stable under complex conjugation.
