minimal polynomial of $7^{th}$ root of unity

Question

Let $\alpha=\zeta_7+\zeta_7^2+\zeta_7^4$. What's the minimal polynomial of $\zeta_7$ over $\mathbb{Q}(\alpha).$

I have managed to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and $[\mathbb{Q}(\zeta):\mathbb{Q}]=6$. By tower rule, the minimal polynomial has degree 3. As long as we find a polynomial with degree 3 such that $\zeta$ is a root, then we are done. But I got stuck in finding such polynomial. Any hint is appreciated.

Answer 1

Following you're analysis of the indices, the Galois group of the extension $\mathbb{Q}(\zeta_7)\supseteq \mathbb{Q}(\alpha)$ is a (the) subgroup of order $3$ of $\text{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong (\mathbb{Z}/7\mathbb{Z})^\times\cong\mathbb{Z}/6\mathbb{Z}$. This subgroup is the cyclic subgroup generated by the element $$ \sigma\in\text{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q}), \,\,\,\sigma(\zeta_7)=\zeta_7^2 $$ A great trick which often helps in finding polynomials (or really any type of object in the context of galois/representation theory) which are invariant under the action of some group of transformations is taking the 'average'. More explicitly, if we're interested in finding a polynomial in $\mathbb{Q}(\alpha)[x]=(\mathbb{Q}(\zeta_7))^{\langle \sigma \rangle}[x]$ which has $\zeta_7$ as a root, we can construct it by taking the product of the orbit of $(x-\zeta_7)$ under $\langle \sigma \rangle$: $$ p(x)=(x-\zeta_7)(x-\sigma(\zeta_7))(x-\sigma^2(\zeta_7))=(x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4) $$ Because of the construction, clearly $p(x)$ is invariant under the action of $\langle \sigma \rangle$ (thought of as an automorphism of the ring $\mathbb{Q}(\zeta_7)[x]$ by acting on coefficients) and hence the coefficients of the polynomial $p(x)$ must all lie in $(\mathbb{Q}(\zeta_7))^{\langle \sigma\rangle}=\mathbb{Q}(\alpha)$.

At this point, I think it's quite satisfying to verify this last statement directly:$$ p(x)=(x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4)= $$ $$=x^3-(\zeta_7+\zeta_7^2+\zeta_7^4)x^2+(\zeta_7^3+\zeta_7^5+\zeta_7^6)x-1=x^3-\alpha x^2 + \frac{\alpha^2-\alpha}{2}x-1. $$ And low and behold the coefficients in fact are in $\mathbb{Q}(\alpha)$.

Hope this helps :)

POST: bare in mind that this is all possible because $\mathbb{Q}(\zeta_7)\supseteq \mathbb{Q}$ is Galois.

Answer 2

Put $\beta=\zeta^{-1}+\zeta^{-2}+\zeta^{-4}$ and note $\beta=-1-\alpha\in\mathbb{Q}(\alpha)$. Now form the cubic with roots $\zeta, \zeta^2,\zeta^4$. The sum of the roots is $\alpha$, the sum of the roots two at a time is $\beta$, the product of the roots is 1.