## Minimal polynomial

I am asking a question about Sheet 2 Problem 9 (solutions are provided for up to 8).

Why $\cos(\frac{4pi}{7})$ and $\cos(\frac{6pi}{7})$ are conjugates of $\alpha=\cos(\frac{2pi}{7})$?

I’ve managed to find minimal polynomial using DeMoivre’s Theorem and I’ve shown that $\cos(\frac{4pi}{7})$, $cos(\frac{6pi}{7})$ satisfy it ($m_{\alpha}=8x^3+4x^2-4x-1$). But the question asks to do the opposite: show that they are conjugates and hence find the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

——————————————-

In the problem, you have $K=Q(\alpha)\subset L=Q(\beta)$, where $\beta=\exp(2\pi i/7).$ Thus, $\alpha= [\beta+\bar{\beta}]/2$, where we use the bar to denote the complex conjugate. Now, the key point is this: every embedding of $K$ into $C$ is the restriction of an embedding of $L$ into $C$, except some of the embeddings will become equal when restricted. Now, $L$ has six embeddings $\sigma_j$ for $j=1,\ldots, 6$ such that $\sigma_j(\beta)=\beta^j$. This can be seen by considering the roots of the minimal polynomial for $\beta$. Since $\bar{\beta}=\beta^6$, we see that for each embedding, $\sigma_j(\bar{\beta})=\sigma_j(\beta^6)=\beta^{6j}=\bar{\beta^j}$. Therefore, the conjugates of $\alpha$ are simply $[\beta^j+\bar{\beta^j}]/2=\cos(2\pi j/7)$. Hence, you get exactly the three numbers indicated.

By the way, the minimal polynomial needs to be monic.