Is there a general formula for the number of unramified quadratic extensions of a number field $K$?


When $K$ is quadratic, this is known (by genus theory) to be $2^{\omega(\Delta\_K)-1}$, where $\omega(n)$ denotes the number of distinct prime factors of $n$ and $\Delta\_K$ is the discriminant of $K$. I'm interested in results for when $K$ is of higher degree.


It seems like this problem might be much harder and is maybe adjacent to understanding the two-torsion of the class group $\text{Cl}\_K$ (which seems [hard](https://arxiv.org/pdf/1701.02458.pdf) when $K$ is not quadratic), but I'm pretty new to the area and could be totally off-base. Is there any hope of a more direct approach?



Number of unramified quadratic extensions of a number field
The answer seems to be no.


1. The number of unramified quadratic extensions of $K$ is equal to the number of index-two subgroups of the ideal class group $\text{Cl}\_K$ by class field theory.
2. The index-two subgroups of $\text{Cl}\_K$ correspond to the non-zero elements of $\text{Hom}(\text{Cl}\_K, \mathbb{Z}/2\mathbb{Z})$.
3. $\#\text{Hom}(\text{Cl}\_K, \mathbb{Z}/2\mathbb{Z}) = \#\text{Cl}\_K[2]$ by Pontryagin duality, as pointed out to me by @RP\_ and @abx in the comments.
4. The problem of computing (or even bounding) the size of $\#\text{Cl}\_K[2]$ when $K$ is not a quadratic extension appears to be under [active study](https://arxiv.org/pdf/1701.02458.pdf) and seems to be a challenging problem in general.


