Odd subfield of cyclotomic field with 2-ramification

Question

I have a tower of fields $\mathbb{Q} \subset L \subset \mathbb{Q}(\zeta_{m})$ with the following hypothesis. $[L:\mathbb{Q}]$ is odd, $2$ is the only rational prime that ramifies in $L$. I want to prove that $L=\mathbb{Q}$. I tried to compute the discriminant of $L$ in relation to that of $\mathbb{Q}(\zeta_{m})$, but didn't manage to do that. What can I do?

Answer 1

$m=2^k \prod_j p_j^{d_j}$.

$\Bbb{Q}(\zeta_m)$ is totally ramified at the primes of $\Bbb{Q}(\zeta_{m/{p_j}^{d_j}})$ above $p_j$ so $L\subset \Bbb{Q}(\zeta_m)$ is unramified at $p_j$ implies that $L\subset \Bbb{Q}(\zeta_{m/p_j^{d_j}})$.

Whence $$L\subset \bigcap_j \Bbb{Q}(\zeta_{m/p_j^{d_j}})=\Bbb{Q}(\zeta_{2^k})$$

Since $[\Bbb{Q}(\zeta_{2^k}):\Bbb{Q}]= 2^{k-1}$ then $[L:\Bbb{Q}]$ is odd means that $L=\Bbb{Q}$.