is $\Bbb Q(2^{1/4}) \subseteq \Bbb Q(\zeta_{16})$

Question

Is $\Bbb Q(2^{1/4}) \subseteq \Bbb Q(\zeta_{16})?$

$\Bbb Q(2^{1/4}) \subseteq \Bbb Q(\zeta_{16})\implies x^4 - 2 $ splits in $\Bbb Q(\zeta_{16})\implies \Bbb Q(2^{1/4},i)\subseteq \Bbb Q(\zeta_{16})\implies Q(2^{1/4},i)= Q(\zeta_{16}) $ since both have degree of extension 8 over $\Bbb Q$. I don't know how to proceed. Thanks in advance for help!

Answer 1

No. The Galois group for the normal closure of $\Bbb Q(2^{1/4})$ (which is $\Bbb Q(2^{1/4},i)$) is non-Abelian. But every cyclotomic field has Abelian Galois group, so that all its subfields are normal also with Abelian Galois group.