If $c \in \mathbb{Q}(\zeta)$ ,where $\zeta=\zeta_{14}$ is a constructable number ,then $c \in \mathbb{Q}$?

Question

If $c \in \mathbb{Q}(\zeta)$ ,where $\zeta=\zeta_{14}$, is a constructable number ,then $c \in \mathbb{Q}$ ?
I believe the statement is true .
Let $c=\zeta^d$ , for $d:gcd(d,14)=1$. The number $\zeta^d$ must be constructable and thus be an extension over $\mathbb{Q}$ of degree $2^k$ , but we know that $[\mathbb{Q}(\zeta),\mathbb{Q}]=\phi(14)=6 \neq 2^k$.
Thus we conclude that $c\in \mathbb{Q}$.

I'm really not happy with how i approach the matter but I'm out of ideas too.
If anyone could help it would be appreciated !

Answer 1

If $c$ is a constructible number then the extension $\mathbb{Q}(c)|\mathbb{Q}$ has degree $2^k$ for some $k$ integer. We know $[\mathbb{Q}(\zeta):\mathbb{Q}]=6$, which implies that $\mathbb{Q}(c)|\mathbb{Q}$ has degree $2$ , also that implies that $[\mathbb{Q}(\zeta):\mathbb{Q}(c)]=3$. Also we know thath the Galois group of our cyclotomic extension is isomorphic to $\mathbb{Z}_{14}^{*}$ whose order is 6, so it must be isomorphic to $\mathbb{Z}_{6}$ because the abelian groups classification theorem. $\mathbb{Z}_{6}$ has only one subgroup of order 3, so by the fundamental theorem of Galois theory there only exists one subfield $K\subset\mathbb{Q}(\zeta)$ such as $[\mathbb{Q}(\zeta):K]=3$. It's a litle tedious but we can prove that $K=\mathbb{Q}(\eta)$ where $\eta=\zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}$, with minimal polynomial $x^2-x+2$. So c must be $\eta$ that clearly is not a real number, so it does not make sense to talk about its constructibility. That implies that c must be a rational number.