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Let $d\in \mathbb Q$ , how to show that there exist $n \in \mathbb N$ such that $\mathbb Q(\sqrt d) \subseteq \mathbb Q(e^{2i\pi/n})$ ?

NOTE : I want to do this using basic field extension and Galois theory , without invoking Kronecker-Weber theorem . If possible I would also want to avoid Gauss sum or Legendre symbol

Please help . Thanks in advance

1 Answers1

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I will assume that $d \in \Bbb N_{>0}$ (this will cover all the quadratic extensions of $\Bbb Q$). Looking at the decomposition of $d$ into prime factors, we know that $\Bbb Q(\sqrt d)$ is the compositum of some fields $\Bbb Q(\sqrt p)$, where $p$ is either prime or $-1$. It is therefore sufficient to prove the claim for the fields $\Bbb Q(\sqrt p)$ (where $p$ is either prime or $-1$), since a compositum of cyclotomic fields is again a cyclotomic field.

First of all, the case $p=-1, K=\mathbb{Q}(i) = \mathbb{Q}(\zeta_4)$ is trivial.

Secondly, consider $K=\mathbb{Q}(\sqrt{2})$. We have $\zeta_8 = \dfrac{\sqrt{2}}{2}(1+i),$ so that $$\sqrt{2} = \dfrac{2\zeta_8}{1+\zeta_8^2}$$ and $K \subset \mathbb{Q}(\zeta_8)$.

Finally, let $p>2$ be an odd prime and consider $K = \mathbb{Q}(\sqrt{p})$. The discriminant of the $p$-th cyclotomic polynomial is

$$\Delta := \mathrm{disc}(\Phi_p) = \prod_{1 \leq i < j \leq p-1} (\zeta_p^i - \zeta_p^j)^2 = (-1)^{\frac{p-1}{2}} p^{p-2} \in \mathbb{Q}(\zeta_p).$$ This yields $$\sqrt{p} = \dfrac{\sqrt{(-1)^{\frac{p-1}{2}} \Delta}}{ p^{\frac{p-3}{2}} } \in \mathbb{Q}(i,\zeta_p) = \mathbb{Q}(\zeta_{4p}).$$ Hence, we get $K \subset \mathbb{Q}(\zeta_{4p})$.

Watson
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  • In the previous to the last equation, it should have been $\sqrt \Delta \in \mathbb Q(\zeta_p)$ – user Mar 04 '18 at 17:23
  • @users : the discriminant is defined as in the equation above (there is a square because of the condition $i<j$). The fact $\sqrt{\Delta} \in \Bbb Q(\zeta_p)$ is indeed needed for the last equation, but it immediately comes from the description of $\sqrt{\Delta}$ as a product of $p$-roots of unity. Therefore, it is not necessary to change this answer, I guess. – Watson Mar 04 '18 at 17:34
  • Alternatively, one may proceed as follows: the extension $\Bbb Q(\zeta_p) / \Bbb Q$ is cyclic of even order $p-1$, so there is a unique quadratic subfield $K \subset \Bbb Q(\zeta_p)$. But the only ramified prime in $\Bbb Q(\zeta_p)$ is $p$, so the same holds for $K = \Bbb Q(\sqrt d)$. Looking at the discriminant of a quadratic field, we conclude that $d = \pm p$ depends on the class of $p \pmod 4$. – Watson Mar 20 '20 at 07:14