Is $\Bbb Q(\sqrt{15}) \subset \Bbb Q(\zeta_{15})$?

Question

Is $\Bbb Q(\sqrt{15}) \subset \Bbb Q(\zeta_{15})$ ?

$Gal(\Bbb Q(\zeta_{15}):\Bbb Q)=\varphi(15)=8$ . Thus $[\Bbb Q(\zeta_{15}) \cap \Bbb R : \Bbb Q]=\frac{\varphi(15)}{2}=4$, so it does not quite help! How to proceed? Thanks in advance for help!

Answer 1

Here's a proof without Gauss sums. The Galois group of $\mathbb Q (\zeta_{15}) |_{\mathbb Q}$ is $\mathbb Z_{15}^* \cong \mathbb Z_4 \times \mathbb Z_2$ which has $3$ subgroups of order $4$

Thus $\mathbb Q(\zeta _{15}) | _{\mathbb Q}$ has exactly $3$ subfields of degree $2$ over $\mathbb Q$. It is easy to see that $\mathbb Q(\zeta_ 3)= \mathbb Q(\sqrt{-3})$ and $\mathbb Q(\zeta_5 +\mathbb \zeta_5^{-1})= \mathbb Q(\sqrt 5) $ are two such. Thus we have a third such field $\mathbb Q(\sqrt {-15}) $ and these are all different from $\mathbb Q(\sqrt {15})$

Answer 2

Let $G$ denote the galois group. Then $|G|=8$ . Using the facts that

(i) $\Bbb Q(\zeta_{15}) \cong \Bbb Q(\zeta_3,\zeta_5)$ (since 3 and 5 are distinct primes)

(ii) If $F\subset K_1,K_2 \subset \bar{F}$ be fields such that $K_1,K_2$ are finite Galois extensions over $F$ and $K_1 \cap K_2=F$ then $Gal(K_1 K_2 | F) \cong Gal (K_1 |F) \times Gal (K_2 |F)$ .

(iii) $G \cong {\Bbb Z_{15}}^*$ is abelian

It follows that $G \cong Z_4 \times Z_2$ and $Z_4 \times Z_2$ has 3 subgroups of order 4 so that we can then argue that $\Bbb Q(\sqrt{-3}),\Bbb Q(\sqrt{5}),\Bbb Q(\sqrt{-15})$ are the all intermediate quadratic extensions over $\Bbb Q$ and thus there is no room for $\Bbb Q(\sqrt{15})$