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Let $f(x) \in {\mathbb Q}[x]$ be a polynomial that is irreducible over ${\mathbb Q}$ with $D_{n}$ (the dihedral group of order $2n$) as its Galois group. Let $\alpha$ be a root of $f(x)$ and put $K={\mathbb Q}(\alpha)$. The splitting field of $f(x)$, denote it by $L$, will be a quadratic extension of $K$.

Is there an ''nice'' element, $\beta \in K$ such that $L=K(\sqrt{\beta})$? Ideally ''nice'' here means that $\beta$ has a simple expression in terms of $\alpha$.

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    $\begingroup$ The discriminant of $f$ works, at least if $n$ is odd, and I think in general. $\endgroup$
    – Ari Shnidman
    Commented Jan 11, 2020 at 21:58
  • $\begingroup$ @Ari Shnidman thanks for that. Would you have a reference, please? And do we know what happens for $n$ even too? $\endgroup$
    – user150983
    Commented Jan 12, 2020 at 4:42
  • $\begingroup$ Sorry, I was being too confident. Even when $n$ is odd the discriminant does not always work. The square root of disc($f$) lives in $L$, since it is the product of the difference of the roots of $f$. So if disc($f$) is not a square, then its squareroot gives a quadratic extension, which is not contained in $K$ (since $n$ is odd). The problem is that disc($f$) may be a square. If $n$ is an odd prime, then disc($f$) is a square if and only if $n$ is 1 mod 4 (see page 2 here: arxiv.org/pdf/1609.09153.pdf). So when $n$ is a prime that is 3 mod 4, disc(f) does what you want, but not for all $n$. $\endgroup$
    – Ari Shnidman
    Commented Jan 12, 2020 at 12:16
  • $\begingroup$ @Ari Shnidman thanks for your follow-up comment, Ari. And for the reference too. That is very helpful to me. I do appreciate you taking your time to help. $\endgroup$
    – user150983
    Commented Jan 13, 2020 at 2:44
  • $\begingroup$ The formula in Cohen-Thorne only says that the ideal of the discriminant is a square, not that the discriminant is a square. I am not convinced that your first remark is wrong. $\endgroup$
    – Franz Lemmermeyer
    Commented Jan 15, 2020 at 18:59

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