# Number of unramified quadratic extensions of a number field

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 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?

• "is maybe adjacent": it is in fact completely equivalent. – abx Aug 20 '20 at 8:59
• The unramified abelian extensions of $K$ are in bijection with the subgroups of $\text{Cl}_K$. – GH from MO Aug 20 '20 at 9:07
• @abx In particular, it is "maybe adjacent". – RP_ Aug 20 '20 at 9:17
• @GHfromMO I agree, by HIlbert Class Field I guess. – bean Aug 20 '20 at 9:22
• @bean However, #A equals #Hom(A,Z/2Z) by duality. – RP_ Aug 20 '20 at 9:39

## 1 Answer

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. $$\#\text{Hom}(\text{Cl}_K, \mathbb{Z}/2\mathbb{Z}) = \#\text{Cl}_K$$ 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 $$\#\text{Cl}_K$$ when $$K$$ is not a quadratic extension appears to be under active study and seems to be a challenging problem in general.