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 $\sum\limits_{k=0}^{12}\zeta^k=0$.
Because $7$ is prime, we know that, $Aut(\mathbb{Q}[\zeta_7]/\mathbb{Q}) \cong \mathbb{Z}_7^\times \cong \mathbb{Z}_6$.
From here, we wish to find a generator for that automorphism group. Note that $\mathbb{Q}[\zeta_7]$ is the splitting field for the irreducible polynomial $f(x) = x^6 + x^5 + \cdots + x + 1$ whose roots are the $6$ nontrivial $7$th roots of unity. Now, it is a theorem that the automorphism group acts transitively on the roots of the polynomial $\iff$ that polynomial is irreducible. Therefore, we know there must exist an automorphism within this group such that $\zeta \mapsto \zeta^n$ for any $n \in \{1, 2, ..., 5\}$. The easiest way I know how to do this is through experimentation.
Once you've found your generator, $\phi$, then your automorphism group is simply $\{id, \phi, \phi^2, ..., \phi^5\}$.
The motivation behind all this work was that you need to let your entire automorphism group act on $\zeta + \zeta_5$ to generate the minimal polynomial. If we denote $\{\alpha_1, \alpha_2, ..., \alpha_k \}$ as the orbit of $\zeta + \zeta^5$ under this action, then the minimal polynomial will be be $m(x) = \displaystyle \prod_{i=1}^k\Big(x-\alpha_i\Big)$.
The moral of the story is that you had the right idea, but the previous sentence is where you went astray.
Edit: It is not necessary to actually find the generator, but it is necessary to let the entire automorphism group act on $\zeta + \zeta^5$. Note that there will be an automorphism that maps $\zeta \mapsto \zeta^k$ for all $k$ by the reasoning above, and each one must (obviously) be unique. Thus, the $6$ of them are:
- $\zeta \mapsto \zeta$
- $\zeta \mapsto \zeta^2$
- $\zeta \mapsto \zeta^3$
etc.
To conclude, if possible, I would recommend you find a copy of Artin and read Chapter 14, Section 4, Proposition 4.4. I think that would get to the crux of where you went wrong initially. You can find a .pdf version of this text here: http://www.drchristiansalas.org.uk/MathsandPhysics/AbstractAlgebra/ArtinAlgebra.pdf
Best Answer
There's another way. You need a fourth degree divisor of x10 - 1 that doesn't have the six non-primitive roots of 1 as roots. Those are the fifth roots and -1. Thus, you divide x10 - 1 by x5 - 1 to get x5 + 1, and then divide that by x + 1 to get x4 - x3 + x2 - x + 1 .