Let $n > 1$ be an integer, and $\zeta$ an $n^{\text{th}}$ root of 1. Consider the field $\mathbb{Q}(\zeta)$ and its group of field automorphisms $G$. It is easy to see that every element of $G$ fixes every element of $\mathbb{Q}$.
I know from the Galois correspondence that the converse is also true: if $x \in \mathbb{Q}(\zeta)$ is fixed by every element of $G$, then $x \in \mathbb{Q}$. Is there a way to prove this without using the Galois correspondence?
I have an argument for $n$ prime.
The field $\mathbb{Q}(\zeta)$ can be seen as a $\mathbb{Q}$-vectorial space of dimension equal to the degree of the extension, and whose basis is formed by the nth primitive roots of unity.
On the other hand, we have that the field automorphisms of $\mathbb{Q}(\zeta)$ are completely described by the images of any primitive root.
We can express an arbitrary element of the field as $x = a_1 \zeta_1 + \dots + a_n \zeta_{n-1}$, where $\zeta_i$ are the primitive roots of unity.
Then let $\phi\in G$, so that $\phi(x) = a_1 \phi (\zeta_1) + \dots + a_{n-1} \phi (\zeta_n)$. Since automorphisms shuffle around the roots, the effect of $\phi$ on $x$ will be equal to $\phi(x) = a_{\sigma{1}} \zeta_{1} + \dots + a_{\sigma{n-1}} \zeta_r$ where $\sigma$ is a permutation.
Thus, the condition of being fixed by any $\phi$ is equivalent to $x = a_0 + a_1 \zeta_1 + \dots + a_r \zeta_r = a_{\sigma{1}} \zeta_{1} + \dots + a_{\sigma{n-1}} \zeta_{n-1}$ for any $\sigma$.
That is, $(a_1-a_{\sigma 1}) \phi (\zeta_1) + \dots + (a_{n-1} - a_{\sigma {n-1}}) \phi (\zeta_{n-1}) = 0$.
But since the elements of the basis are $\mathbb{Q}$-lineally independent, this means that for any $\sigma$ the coefficients $(a_i-a_{\sigma i})$ should be $0$, so all of them should be equal to a constant $k$.
But since $n$ is prime, the nth cyclotomic polynomial is of the form $1+\zeta +\dots + \zeta^{n-1} = 0$, so $x=k\zeta +\dots + k\zeta^{n-1}=-k\in \mathbb{Q}$.