In a quadratic extension $\mathbb{Q}(\sqrt{d})$of $\mathbb{Q}$ it is clear that 2 ramifies if and only if $d\equiv 2,3\mod 4$ (easy to see if you compute the discriminant). But if I take a relative quadratic extension, to make it simple let's say $L=K(\sqrt{\pi})$ where $K$ is an arbitrary number field and $\pi$ is an irreducible element of the ring of integers of $K$, when does a dyadic prime ramify from $K$ to $L$?

Certainly $\pi$ itself ramifies in this case, that's easy to see without knowing the discriminant or even ring of integers. But how do I know when primes above $2$ are ramifying, i.e. being contributed to the discriminant?

In general, I know that computing the discriminant (or ring of integers) of such a field, is computationally complex. But is there a known congruence condition on $\pi$ or some algebraic way to decide when primes above 2 ramify?

irreducibleelement of the ring of integers of $K$, and you're looking at the ring of integers of $K(\sqrt{\pi})$. But the set of irreducible elements is not so nice in general. So I think that your attempted simplification is actually confusing the issue. Why not just ask: Is there a congruence condition on $\delta\in O_K$ that determines whether primes above 2 are ramified in $O_{K(\sqrt{\delta})}$? $\endgroup$ – Joe Silverman Mar 16 '14 at 13:09