3
$\begingroup$

If $K$ is a number field, does there always exist an unramified extension of $K$, i.e. an extension $L/K$ so that the extension is unramified at all $\mathfrak{p} \subseteq \mathcal{O}_K$? For example, given $K=\mathbb{Q}(\sqrt{-2})$, we have $\mathcal{O}_K=\mathbb{Z}[\sqrt{-2}]$. How would one construct a unramified extension?

I know ramification indexes and inertia degrees are multiplicative and a prime ramifies in $K$ if and only if it divides the discriminant of $K$ but I do not see how this helps me.

EDIT: As LordShark helpfully pointed out, these do not always exist. What I am really interested in is how does one construct them assuming they exist. Looking up the narrow class group, I think it should exist in the case I have above. So let's just say $K=\mathbb{Q}(\sqrt{d})$ for some square free $d \in \mathbb{Z}$, i.e. $d=-2,-5,7$, etc. for which such an extension exists.

$\endgroup$
  • 3
    $\begingroup$ There is not always a nontrivial unramified extension of $K$; e.g. $K=\Bbb Q$. By class field theory, there is a nontrivial Abelian unramified extension iff the narrow class group is non-trivial. $\endgroup$ – Lord Shark the Unknown Mar 26 '18 at 3:43
  • 1
    $\begingroup$ @LordSharktheUnknown Thanks! I had assumed one could always exist. I had not heard of the narrow class group so I will have to read. I suppose I mean how does one construct them generally so I will have to edit the question $\endgroup$ – TinyTim Mar 26 '18 at 3:49
  • 2
    $\begingroup$ Construction of unramified extensions is hard! Like. Really hard! One topic you could read about is the Hilbert Class Field over a number field. $\endgroup$ – User0112358 Mar 26 '18 at 4:01
  • $\begingroup$ @User0112358 Is it still hard even in the simple case of a quadratic extension like the example? $\endgroup$ – TinyTim Mar 26 '18 at 4:07
  • $\begingroup$ Imaginary quadratic fields are one of the few cases where explicit class field theory can actually be done. Let $E$ be the elliptic curve $\mathbf{C}/\mathcal{O}_K$. Then the Hilbert class field of $K$ is given by adjoining the $j$-invariant of $E$, and the ray class fields are essentially given by adjoining the coordinates of the torsion points on $E$. There are lots of references for this, including Silverman's Advanced Topics in the Arithmetic of Elliptic Curves. $\endgroup$ – Brandon Carter Mar 26 '18 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.