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For $K$ a field and $|K(\alpha):K| = 5$, why do we have $K(\alpha^2) = K(\alpha)$?

middlethird_cantor
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  • Well, $K(\alpha^2)\subset K(\alpha)$, so we know that $|K(\alpha^2):K|\leq5$. We know, then, that $|K(\alpha^2):K|=1,5$. If it is $1$, that implies that $\alpha^2\in K$, which would suggest that $\alpha$ is a root of $x^2-\alpha^2$, which would imply that $|K(\alpha):K|\leq2$, a contradiction. So, $|K(\alpha^2):K|=5$. However, that means that $K(\alpha^2)=K(\alpha)$. – Rushabh Mehta Oct 17 '19 at 05:56

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You have the extensions $ K \subseteq K(\alpha^2) \subseteq K(\alpha)$ and therefore regarding the degrees:

$$[K(\alpha):K(\alpha^2)][K(\alpha^2):K]=[K(\alpha):K]=5.$$

Therefore $[K(\alpha^2):K]\in \{1,5\}$. $[K(\alpha^2):K]=1$ can’t be as it means that $\alpha^2 \in K$ in contradiction with the degree of $\alpha$ over $K$. Hence $[K(\alpha^2):K]=5$ and the conclusion.

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