Let $[K(\alpha):K]$ be odd, $\alpha$ algebraic over field $K$, proof that $K(\alpha) = K(\alpha^2)$.

Question

Let $K$ be a field. Suppose $\alpha$ is algebraic over $K$ and let $[K(\alpha):K]$ be an odd number. Proof: $K(\alpha) = K(\alpha^2)$.

Obviously $\alpha^2 \in K(\alpha)$ implying $K(\alpha^2)\subset K(\alpha)$. However $\alpha\in K(\alpha^2)$ is more difficult.

As an example $\alpha = \sqrt[3]{2}$ and $\alpha^2 = \sqrt[3]{2^2}$ over $\mathbb{Q}$. We see $ \sqrt[3]{2^2}^2 = 2\sqrt[3]{2}$ so $\sqrt[3]{2}\in \mathbb{Q}(\alpha^2)$.

In general, let $[K(\alpha):K] = 2n-1$, we see $K(\alpha^2)\ni(\alpha^2)^n = \alpha^{2n} = \alpha \alpha^{2n-1}$. It seems to be that $\alpha^{2n-1}\in K(\alpha^2)$ such that $\alpha\in K(\alpha^2)$, just like $\sqrt[3]{2}^3 = 2\in\mathbb{Q}(\sqrt[3]{2^2})$. However I am not sure how to proof this statement formally. Does anybody has suggestions on solving this question?

Answer 1

We have

$K \subset K(\alpha^2) \subset K(\alpha), \tag 1$

and thus

$[K(\alpha):K] = [K(\alpha):K(\alpha^2)][K(\alpha^2):K]; \tag 2$

note that

$p(x) = x^2 - \alpha^2 \in K(\alpha^2)[x], \tag 3$

and

$p(\alpha) = \alpha^2 - \alpha^2 = 0; \tag 4$

thus

$[K(\alpha):K(\alpha^2)] \le \deg p(x) = 2$ $\Longrightarrow [K(\alpha):K(\alpha^2)] = 1 \; \text{or} \; [K(\alpha):K(\alpha^2)] = 2; \tag 5$

the hypothesis that

$[K(\alpha):K] \; \text{is odd} \tag 6$

allows us to rule out, via (2),

$[K(\alpha):K(\alpha^2)] = 2, \tag 7$

since

$2 \not \mid [K(\alpha):K]; \tag 8$

thus

$[K(\alpha):K(\alpha^2)] = 1, \tag 8$

and we conclude that

$K(\alpha^2) = K(\alpha). \tag 9$

$OE\Delta$.