Finding the minimal polynomial of $w+w^{-1}$ with $w$ is a primitive nth root of unity Let $w$ be primitive nth root of unity over $\mathbb{Q}$. Find the minimal polynomial of $w+w^{-1}$ over $\mathbb{Q}$
 A: What happens here is similar to what happens with the polynomial $M_w$ 
of $w$ : there is no general formula, the answers depends on the prime 
factorization of $n$ (for example, its degree is $\phi(n)$). In general, we only know
that $M_w$ divides $X^{n-1}+X^{n-2}+\ldots +1$, and $M_w$ coincides with
this polynomial when $n$ is prime.
Similarly, if $n$ is odd say $n=2m+1$, we may define
$$
T_m=\Bigg(\sum_{j=0}^{\lfloor \frac{m}{2} \rfloor}
\binom{m-j}{j} (-1)^j x^{m-2j}\Bigg) +
\Bigg(\sum_{j=0}^{\lfloor \frac{m-1}{2} \rfloor}
\binom{m-1-j}{j} (-1)^j x^{m-1-2j}\Bigg) \tag{1}
$$
Then, we have the algebraic identity
$$
T_m\bigg(w+\frac{1}{w}\bigg)=\frac{1}{w^m}\sum_{i=0}^{2m} w^i \tag{2}
$$
which shows that $T_m$ annihilates $v=w+\frac{1}{w}$, so the minimal
polynomial $M_v$ of $v$ divides $T_m$. When $n$ is an odd
prime we have $M_v=T_m$.
A: The minimum polynomial satisfied by $w$ is palindromic. Making the substitution $z=w+w^{-1}$ is a standard method of treating such polynomials, since it divides the degree of the polynomial by half. Look up palindromic polynomials.
