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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?

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