Subfield of $\mathbb{Q}[\zeta]$ fixed by conjugation I am reading Milne's Fields and Galois Theory notes (I am reviewing these topics).
On pages 39-40, the extension $\mathbb{Q}[\zeta]/\mathbb{Q}$ is analysed, where $\zeta$ is a primitive $7$th root of $1$.
It is claimed the the subfield of $\mathbb{Q}[\zeta]$ fixed by complex conjugation is $K=\mathbb{Q}[\zeta+\overline{\zeta}]$. Obviously, $K$ is fixed by conjugation.

Why does every element of $\mathbb{Q}[\zeta]$ fixed by complex conjugation belong to $K$? My attempt below turned up to be a solution when I wrote it down, but - is there a simpler way to see it, or at least a shorter way to write it?

Here's what I tried: The minimum polynomial of $\zeta$ is $X^6+X^5+X^4+X^3+X^2+X^1+X^0$. Therefore, $\zeta^0,\dotsc,\zeta^5$ is a basis for $\mathbb{Q}[\zeta]$ over $\mathbb{Q}$. I take an arbitrary element $x=\sum_{i=0}^5a_i\zeta^i$ (with $a_i\in\mathbb{Q}$). Then $\overline{x}=\sum_{i=0}^5a_i\zeta^{-i}=a_0+a_1\zeta^6+a_2\zeta^5+a_3\zeta^4+a_4\zeta^3+a_5\zeta^2=(a_0-a_1)+(-a_1)\zeta^1+(a_2-a_1)\zeta^5+(a_3-a_1)\zeta^4+(a_4-a_1)\zeta^3+(a_5-a_1)\zeta^2$;
Now, assume that $x=\overline{x}$. Then, we get


*

*$a_0=a_0-a_1$

*$a_1=-a_1$

*$a_2=a_5-a_1$

*$a_3=a_4-a_1$

*$a_4=a_3-a_1$

*$a_5=a_5-a_1$


Which is equivalent to $a_1=0,a_2=a_5,a_3=a_4$.
So, $x=a_0+a_2(\zeta^2+\zeta^{-2})+a_3(\zeta^3+\zeta^{-3})$.
So, it remains to show that both $\zeta^2+\zeta^{-2}$ and $\zeta^3+\zeta^{-3}$ belong to $K$. This is done by writing $(\zeta+\zeta^{-1})^2-2=\zeta^2+\zeta^{-2}$ and similarly for $\zeta^3+\zeta^{-3}$.
 A: Let the subfield fixed by conjugation be $L$. We have $K \subseteq L \subsetneq \mathbb Q[\zeta]$.
Since $|\zeta| = 1$, $\overline\zeta = \zeta^{-1}$. I'll use the latter form because it makes things easier.
Now, let's compute $[\mathbb Q[\zeta] : K]$. One good way to do this would be to find a polynomial with coefficients in $K$ with a root $\zeta$. Well, that's not too hard: $\zeta(\zeta + \zeta^{-1}) = \zeta^2 + 1$, so $x^2 - (\zeta + \zeta^{-1})x + 1$ would do the trick.
So $[\mathbb Q[\zeta] : K]$ is at most $2$, and it's certainly not $1$ because $\mathbb Q[\zeta]$ has complex elements, but $K \subset \mathbb R$. So it's $2$.
Now, recall that $[\mathbb Q[\zeta] : K] = [\mathbb Q[\zeta] : L][L : K]$. So $[L : K]$ is either $1$ or $2$, which means that $L$ is equal to either $\mathbb Q[\zeta]$ or $K$: there just isn't room for any extensions in between. But we know it's not the former, so it must be the latter, i.e. $L = K$.
(An earlier version of this answer had me mistakenly calculating the degree to be $3$ instead. Observe that that would still have worked fine, because it's prime, so it still forces one of the RHS degrees to be $1$.)
A: We could note that the complex conjugation generates a proper subgroup (of order two) in $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$, and that $\mathbb{Q}\subsetneq\mathbb{Q}(\zeta+\overline{\zeta})\subset L\subsetneq\mathbb{Q}(\zeta)$ where $L$ is the field fixed by that subgroup. Order considerations would suffice to conclude that there could be no other proper field extensions in between.
