This is kind of a two-parter, I am reading this proof that has some steps that suggest that this true:
Let $K=\mathbb{Q}(\theta)$ with $\mathbb{Q} \subset K$ Galois, and $\mathbb{Q(\theta_1)}, \mathbb{Q(\theta_2)},...... \mathbb{Q(\theta_n)}$ be all the subfields of $K$.
Is $K$ $\mathbb{Q}$-linearly generated by $\theta_1,..\theta_n$? (i.e for every $x \in K$, $x$ can be expressed as $x=a_1\theta_1+...+a_n\theta_n$ for $a_i \in \mathbb{Q}$)
Is there a relation between the linear expression above given and $\theta$'s conjugates? (i.e if $x=a_1\theta_1+...+a_n\theta_n$, can I find $x'$ ($x$ first conjugate) using that equation?)
Now I am guessing the first can be explained by absurd reasoning and generating another subfield. The second seems logical, since each $\mathbb{Q}(\theta_i)$ is directly associated to a conjugate of $\theta$ by Galois. But I can't seem to figure it out.
If $K/\mathbb{Q}$ is cyclic, with Galois group $\mathbb{Z}/n\mathbb{Z}$, then there are as many subfields of $K$ as there are divisors of $n$. Of course unless $n\leqslant 2$ this number is strictly less than $n$, so generators of subfields cannot form a linear basis of $K$, since the dimension of $K$ is $n$.