Minimal polynomial of $\zeta+\zeta^{-1}$

Solution 1:

Galois theory provides some machinery for this:

Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then

$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,

$\quad \rm(x-\sigma\alpha)$ all coprime, $\rm \sigma\in G/S$, $\rm S=Stab_G(\alpha)$, implies $\rm f(x):=\prod\limits_{\sigma\in G/S}(x-\sigma\alpha)\mid m$ in $\rm L[x]$,

$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,

$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.

Therefore the zeros of $\rm\alpha$'s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.

As ${\rm Gal}\big({\bf Q}(\zeta_n)/{\bf Q}\big)=({\bf Z}/n{\bf Z})^\times$ it suffices to consider $\sigma:\zeta\mapsto\zeta^k$ for $k=1,\cdots,12$.

By symmetry we need only consider $1,\cdots,6$ for $\alpha=\zeta+\zeta^{-1}$. Thus

$$\begin{array}{ll} {\rm minpoly}_{\zeta+\zeta^{-1},\bf Q}(x) = & ~~~~\left(x-(\zeta+\zeta^{-1})\right)\left(x-(\zeta^2+\zeta^{-2})\right)\left(x-(\zeta^3+\zeta^{-3})\right) \\ & \times\left(x-(\zeta^4+\zeta^{-4})\right)\left(x-(\zeta^5+\zeta^{-5})\right)\left(x-(\zeta^6+\zeta^{-6})\right). \end{array}$$

Simplify the resulting expansion via the rules $\zeta^n=\zeta^{n\,\bmod\,13}$ and $\sum\limits_{k=0}^{12}\zeta^k=0$.

Solution 2:

Zero-theory computational approach:

Since $\zeta^{13}-1=0$, $\sum\limits_{k=0}^{12}\zeta^k=0$ hence $1+\sum\limits_{k=1}^6x_k=0$, where $x_k=\zeta^k+\zeta^{-k}$. Let $x=x_1$, then the system $x^k=\sum\limits_{2i\leqslant k}{k\choose i}x_{k-2i}$ for every $k\geqslant1$ is lower triangular with unknowns $(x_k)_{k\geqslant1}$ and unit diagonal hence each $x_k$ is a linear combination with integer coefficients of $x^i$ for $i\leqslant k$, for example $x_1=x$, $x_2=x^2-2$, $x_3=x^3-3x$, $x_4=x^4-4x^2+2$, and so on.

Thus, $\sum\limits_{k=1}^6x_k=P_6(x)$ where $P_6$ is a monic polynomial of degree $6$ with integer coefficients, and $P_6(x)+1=0$.