If [F(α):F] is odd, then F(α²) = F(α)

Let F be a field, and let \alpha be algebraic over F. Prove that if [F(\alpha):F] is odd, then F(\alpha^2) = F(\alpha).

The inclusion F(\alpha^2) \subseteq F(\alpha) is immediate.

Let p(x) = \sum c_ix^i be the minimal polynomial of \alpha over F. Now write p(x) = \sum c_{2i}x^{2i} + \sum c_{2i+1}x^{2i+1}, separating the even and odd terms. Now p(x) = \sum c_{2i}(x^2)^i + \sum c_{2i+1}x(x^2)^i = r(x^2) + xs(x^2), and we have r(\alpha^2) + \alpha s(\alpha^2) = 0. Since p(x) has odd degree and the powers of \alpha are linearly independent over F, s(\alpha^2) \neq 0. Thus \alpha = -r(\alpha^2)/s(\alpha^2) \in F(\alpha^2), and we have F(\alpha) \subseteq F(\alpha^2).

Note that our proof only needs some nonzero odd-degree term in the minimal polynomial of \alpha, so in some cases the proof holds if \alpha has even degree.

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