I'm working in this setting: $K = \mathbb{Q}[\sqrt{2},\sqrt{3}]$, and let $K_1= \mathbb{Q}[\sqrt{2}],K_2= \mathbb{Q}[\sqrt{3}],K_3= \mathbb{Q}[\sqrt{6}]$ the three subfields. I know that $2 \in \mathbb{Z}$ is totally ramified in each of the $K_i$ ($2$ divides $2,6$ and $3 \equiv 3 \bmod 4$) and I know how to compute generators of $2$ in each of the $K_i$. Moreover, I know that $2$ is totally ramified in $K$ namely, $2R=Q^4$ where $R$ is the number ring of $K$ and $Q$ is the only prime of $R$ lying over $2$.
I want to compute $Q$ finding its generators. I did it using SageMath, but I want to achieve the result without software. Can you help me? (I have no idea about the starting point)
I'm using Marcus'book "Number fields" if you want a reference.
The $\Bbb{Q}$-minimal polynomial of $\sqrt{3}-1$ is $(x+1)^2-3=x^2+2x-2$
The $\Bbb{Q}(\sqrt{2})$-minimal polynomial of $\frac{\sqrt{3}-1}{\sqrt{2}}$ is $x^2+\sqrt{2}x-1$
The $\Bbb{Q}(\sqrt{2})$-minimal polynomial of $\frac{\sqrt{3}-1}{\sqrt{2}}+1$ is $$(x-1)^2+\sqrt{2}(x-1)-1= x^2+(\sqrt{2}-2)x-\sqrt{2}$$
Thus $K/\Bbb{Q}$ is totally ramified of degree $4$ and its uniformizer is $\frac{\sqrt{3}-1}{\sqrt{2}}+1$
ie. $$2O_K=(2,\frac{\sqrt{3}-1}{\sqrt{2}}+1)^4$$
The same algorithm works for arbitrary totally ramified extensions.