Algebraic Extension about 7th primitive roots I find an interesting question about 7th primitive roots.
Suppose that $\psi$ is a 7th primitive root of $1$.
Try to find the number of elements in $$\{a=a_1\psi+a_2\psi^2+a_3\psi^3+a_4\psi^4+a_5\psi^5+a_6\psi^6: a_i \in (0,1)\}$$ satisfy that $\mathbb{Q}(a)=\mathbb{Q}(\psi)$.
Anyone got some ideals?
 A: Denote an automorphism on $\sigma:\mathbb{Q}(\psi) \to \mathbb{Q}(\psi)$ by $\sigma(\psi) = \psi^3$. It is a generator for the Galois group $G$. Let $X=\{a_1\psi+a_2\psi^2+a_3\psi^3+a_4\psi^4+a_5\psi^5+a_6\psi^6| a_i \in \{0,1\}\}$. Then $G$ acts on $X$. We wish to know the number of elements in $X$ with trivial stabilizer.
Let $x=a_1\psi+a_2\psi^2+a_3\psi^3+a_4\psi^4+a_5\psi^5+a_6\psi^6$.
There are only two elements of $X$ with stabilizer $G$: if $x$ is to be fixed by $\sigma$, then all $a_i$ are equal. 
If $x$ is to be fixed by $\sigma^2$, then $a_1=a_2=a_4$, $a_3=a_5=a_6$, so there are $2^2=4$ such $x$ which will be fixed $\sigma^2$, but among them, 2 are also fixed by $\sigma$. Therefore there are 2 elements of $X$ which have stabilizer $\{1,\sigma^2,\sigma^4\}$.
If $x$ is to be fixed by $\sigma^3$, then $a_1=a_6$, $a_3=a_4$ and $a_2=a_5$, so there are $2^3=8$ such $x$ which will be fixed $\sigma^3$, but among them, 2 are also fixed by $\sigma$. Therefore there are 6 elements of $X$ which have stabilizer $\{1,\sigma^3\}$.
Hence there are $10$ elements with nontrivial stabilizer, your desired number is $2^6-10=54$.
