Why $f(x) = x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$ (where $\zeta = \zeta_7$ is a primitive root of the unit) over $\mathbb{Q}$?
Of course it's irreducible by the Eisenstein criterion, however it apparently does not satisfies $1 + 2(\zeta + \zeta^2 + \zeta^4)$ as a root, I tried to calculate several times however I couldn't get $f(1 + 2(\zeta + \zeta^2 + \zeta^4))$ = 0$.
Thanks in advance.
Compute: $$\begin{align*} (1+2(\zeta+\zeta^2+\zeta^4))^2+7&=\bigg[1^2+4(\zeta+\zeta^2+\zeta^4)+4(\zeta+\zeta^2+\zeta^4)^2\bigg]+7\\[0.1in] &=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta^8+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in] (\mathsf{\text{because }}\zeta^8=\zeta)\quad&=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in] &=\bigg[1+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6\bigg]+7\\[0.1in] &=(-7)+8+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6+7\\[0.2in] &=(-7)+8\Phi_7(\zeta)+7\\[0.1in] \left(\begin{array}{c}\mathsf{\text{because }}\Phi_7\mathsf{\text{ is the}}\\ \mathsf{\text{minimal poly of }}\zeta\end{array}\right)\quad&=(-7)+7\\[0.1in] &=0 \end{align*}$$
Slightly easier:
$(x-1)^2 = 4 ( \zeta^2 + \zeta^4 + \zeta^8 + 2 \zeta^3 + 2 \zeta^5 + 2 \zeta^6) = 4 ( - 2 - \zeta -\zeta^2 - \zeta^4) = 4 ( -2 - \frac{x-1}{2})$
Hence $ x^2 - 2x + 1 = -8 - 2x + 2$ or $x^2 + 7 = 0 $.
If you don't already know the primitive polynomial, you can find it with Galois theory. The element given is an element of the cyclotomic field, and so it's conjugates are all the roots of the primitive polynomial. In fact, there is only one different conjugate, obtained for example by cubing each primitive root in the original expression. So $1+2(\zeta^{3}+\zeta^{5}+\zeta^{6})$ is also a root, and there are no others. Call these $r_1$ and $r_2$. The minimum polynomial must be $(x-r_1)(x-r_2)$. The sum of the roots is zero, so we only need to compute the product, which is easily found to equal 7.
