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.


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