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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