How can we determine the factorisation of $2\mathcal{O}_L$? Let $L =\Bbb{Q}(\sqrt{2},\sqrt{-3})$ and $ K = \Bbb{Q}(\sqrt{2})$. Now the prime $2$ is totally ramified in two of the three subfields of $L$ and from ramification theory we know that $2$ splits as $P^2$ in $\mathcal{O}_L$. How can I determine the exact prime decomposition of $2\mathcal{O}_L$? I do this because I want to prove that every prime factor of $2\mathcal{O}_L$ is principal, thus showing that $\mathcal{O}_L$ is a PID (The minkowski bound is less than 3).
However my problem now is that the usual way to do this via factoring a certain polynomial mod $p$ does not work. To simplify this, I tried to see if I could factor the ideal $(\sqrt{2})$ of $\mathcal{O}_K$ in the bigger ring $\mathcal{O}_L$. However this is not so simple: we have 
$$\operatorname{disc} \mathcal{O}_L = 16 \cdot 36$$
and $\operatorname{disc} \mathcal{O}_K[\sqrt{-3}] = (-96)^2 $ (this latter calculation was done here) and thus the index $$\left[\mathcal{O}_L : \mathcal{O}_K[\sqrt{-3}] \right] = 4.$$
Since $2|4$ I can't apply the usual procedure even to factoring $\sqrt{2}$ in $\mathcal{O}_L$. 


My question is: How can I find the exact factorisation of $2\mathcal{O}_L$?  More generally how can I find the factorisation of prime ideals when the usual process does not work?


Thanks.
 A: Let $e$, $f$, and $g$ be the usual prime splitting parameters for 2 in the extension $L$. Looking at how 2 splits in ${\mathbf Q}(\sqrt{2})$ shows $2|e$. Looking at how 2 splits in ${\mathbf Q}(\sqrt{-3})$ shows $2|f$. Since $efg = 4$, we must have $e = 2, f = 2$, and $g = 1$. Thus $2{\mathcal O}_L = {\mathfrak p}_4^2$, where ${\mathfrak p}_4$ is a prime ideal in ${\mathcal O}_L$ of norm 4. But also $2{\mathcal O}_L = (\sqrt{2}{\mathcal O}_L)^2$, so $(\sqrt{2}{\mathcal O}_L)^2 = {\mathfrak p}_4^2$. By unique factorization, ${\mathfrak p}_4 = \sqrt{2}{\mathcal O}_L$ is principal. 
This is simply saying that the prime over 2 in ${\mathbf Q}(\sqrt{2})$ stays prime in ${\mathbf Q}(\sqrt{2},\sqrt{-3}) = {\mathbf Q}(\sqrt{2},\zeta_3)$, which in retrospect is clear since the residue field at $(\sqrt{2})$ in ${\mathbf Q}(\sqrt{2})$ is ${\mathbf F}_2$ and $x^2 + x + 1$ - the minimal polynomial of $\zeta_3$ over ${\mathbf Q}(\sqrt{2})$ - is irreducible over ${\mathbf F}_2$. That is, if you'd think of $L$ as ${\mathbf Q}(\sqrt{2})(\zeta_3) = K(\zeta_3)$, a quadratic extension of $K$, and ${\mathcal O}_L$ as the integral closure of ${\mathcal O}_K$ in $L$, then you can use the same methods to factor $\sqrt{2}{\mathcal O}_K$ in ${\mathcal O}_L$ as you'd use to factor a rational prime in a quadratic field.  After all, $K$ has class number 1, just like ${\mathbf Q}$. When you write that the "usual" methods don't work, it's not true! You need to think of number fields in a relative way, one number field as an extension of another, instead of everything as an extension of ${\mathbf Q}$. 
A: As in my question I would like to determine how $\sqrt{2}$ splits in the $\mathcal{O}_L = \Bbb{Z}[\sqrt{2},\alpha]$ where $\alpha = \frac{1 + \sqrt{-3}}{2}$. Now consider the composite
$$ \Bbb{Z}[\alpha] \to \Bbb{Z}[\sqrt{2},\alpha] \to \Bbb{Z}[\sqrt{2},\alpha]/(\sqrt{2}).$$
The kernel of this composite contains $(2)$ and now I claim it is exactly that: For the quotient $$\Bbb{Z}[\alpha]/(2) \cong \Bbb{Z}[x]/(x^2 - x + 1, 2) \cong \Bbb{F}_4$$
which is the field with four elements. Since the composite I wrote above is not the zero map, we get that the kernel is exactly $(2)$. Now $\Bbb{F}_4$ is a field and so $\sqrt{2}\mathcal{O}_L$ is maximal. In particular this means that $\sqrt{2}\mathcal{O}_K$ is inert in $\mathcal{O}_L$.
