Simplification rules for roots of unity I am trying to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7+\zeta_7^2+\zeta_7^4$ over $\Bbb Q$.
For the first one, I found the following solution on the site (Minimal polynomial of $\omega:=\zeta_7+\overline{\zeta_7}$).
Here is the solution from Ewan Delanoy:
$$
\omega=\zeta+\zeta^6, \ \ \ \omega^2=\zeta^2+\zeta^5+2, \ \ \
\omega^3=\zeta^3+\zeta^4+3\omega
$$
Adding all those three up, you obtain
$$
\omega^3+\omega^2+\omega =\sum_{k=1}^{6} \zeta^k+(3\omega+2)=
-1+(3\omega+2)=3\omega+1
$$
So
$$
\omega^3+\omega^2-2\omega-1=0
$$
I can verify that this is indeed the the minimal polynomial of $\zeta_7+\zeta_7^{-1}$ over $\Bbb Q$.But I don't really understand the intermediate steps involved in finding that polynomial.I suspect that there might be some simplification rules for the root of unity which I don't know. So I am asking whether there are some properties of the root of unity used in the above calculation?
Also, how would I go about to find the minimal polynomial of $\zeta_7+\zeta_7^2+\zeta_7^4$ over $\Bbb Q$?
Could someone give me a hand? Thanks so much. 
 A: Surely all there is to know (algebraically) about $\zeta=\zeta_7$ is that it is a root of the irreducible polynomial $X^6+X^5+X^4+X^3+X^2+X+1$ ?
The other roots are the powers of $\zeta$ (except $1$), and algebraically they are indistinguishable.
So what we can say algebraically about $\zeta+\zeta^{-1}$ we can say about $\zeta^2+\zeta^{-2}$ and $\zeta^4+\zeta^{-4}$. [Any choice of $\zeta^i\not=1$ leads to one of these.] 
So a bold guess for the minimal polynomial of $\zeta+\zeta^{-1}$ is
$$
(X-(\zeta+\zeta^{-1}))(X-(\zeta^2+\zeta^{-2}))(X-(\zeta^4+\zeta^{-4}))
$$
which, using $\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$, is
$$
X^3+X^2-3X-1,$$
which is irreducible and so the answer.
Similarly, $\alpha=\zeta+\zeta^2+\zeta^4$ and $\beta=\zeta^{-1}+\zeta^{-2}+\zeta^{-4}$ are the roots of 
$$
(X-\alpha)(X-\beta)=\dots=X^2+X+2,
$$
again using only $\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$. 
A: For $\zeta=\zeta_7$, here’s how I find the minimal polynomial of $\xi=\zeta+1/\zeta$:
\begin{align}
\zeta^3+\zeta^2+\zeta+1&+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0\\
\xi^3=\zeta^3+3\zeta&+3\zeta^{-1}+\zeta^{-3}\\
=-\zeta^2+2\zeta-1&+2\zeta^{-1}-\zeta^{-2}\\
\xi^3+\xi^2=2\zeta+1&+2\zeta^{-1}\,,
\end{align}
and thus $\xi^3+\xi^2=2\xi+1$.
You can use the same method for any $\zeta_n+\zeta_n^{-1}$.
