<s>[INST] Why $x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$? 
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.
 [/INST] 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 $.
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