When does a dyadic prime ramify in a relative quadratic extension? 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?  
 A: There's an appendix on Kummer theory to Milne's class field theory notes (available here) that gives an answer to this question (EDIT: in some cases). Assuming that $K$ contains a primitive $p$th root of unity $\zeta$, $\pi = (\zeta - 1)$ and
$a \in K^{\times}$ is such that
$\bullet$ $a$ is not a $p$th power in $K$,
$\bullet$ $a$ is relatively prime to $p$, and
$\bullet$ $X^{p} \equiv a \pmod{p \pi}$ has a nonzero solution $X \in \mathcal{O}_{K}$,
then none of the primes $\mathfrak{p}$ above $p$ ramify in $K[a^{1/p}]/K$.(This is Remark A.6 on page 223.)
Milne doesn't give a proof, instead referring to exercises in Washington's Cyclotomic Fields text (pages 182 and 183 there), and Cassels and Frohlich.
A: While I was away I realized that I didn’t have a complete answer to this question. But the key to finding whether $K(\sqrt\delta)\supset K$ is ramified over $2$ is to take a prime $\mathfrak q$ of $K$ above $2$, and consider $V=v_{\mathfrak q}$, the (additive) valuation at $\mathfrak q$ and $V(\delta-1)$. I’ll normalize $V$ so that $V(2)=1$.
Now, in view of the fact that the binomial expansion $(1+4t)^{1/2}$ has its coefficients in $\mathbb Z$, you see that if $V(\delta-1)>2$, the prime $\mathfrak q$ splits in the extension. Similarly, if $\delta=1+4u$ for a $\mathfrak q$-unit $u$, you see directly that the extension is unramified: may or may not split.
The same kind of computation shows that if $V(\delta-1)<2$, $V(\sqrt\delta-1)=\frac12V(\delta-1)$. In case $V(\delta)=m/e$ with $m$ odd, and where the local ramification is $e$, you’re home free: the extension is definitely ramified. But I don’t see offhand how you can exclude the possibility that $m$ is even, and there I’m not so sure.
