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Let $F$ be a number field (or a function field with constants $\mathbb{F}_q$ where $2 \nmid q$). If $L$ is some extension such that $[L:F]=4$ and $ \operatorname{Gal}(L'/F)=D_4$, the dihedral group with 8 elements, then there is a unique quadratic extension $K$ of $F$ between $F$ and $L$.

Does this at least in part go the other way? That is,

Given a field $F$ as above and a quadratic extension $K/F$, is there at least one quadratic extension $L/K$ such that $ \operatorname{Gal}(L'/F)=D_4$?

Bernard
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user10039910
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    What is $L'$ w.r.t. $L$? – Bernard May 31 '19 at 18:06
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    It has to be the Galois closure. Given degree $4$ polynomials are radical you should be able to find a few polynomials of $F[t,x]$ whose specialization $t=a$ are all the dihedral polynomials of degree $4$, their discriminant will be polynomials of $t$ – reuns May 31 '19 at 21:21
  • You can find a solution here https://math.stackexchange.com/a/2046653/300700 : first consider a biquadratic extension L/F containing your quadratic K, then construct a quadratic extension E/L s.t. E/F is galois of degree 8. There is a precise criterion for E/F to have Galois group $\cong D_4$. Walking backwards, you see how to construct L. – nguyen quang do Jun 01 '19 at 07:54

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