Let $K\subseteq K(\beta)$ be a finite field extension of odd degree. Does this imply $K(\beta)=K(\beta^2)$?
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Since $[K(\beta):K]$ is odd we have $[K(\beta):K]=2k+1$ for some $k\in\mathbb N^*.$ Or $\beta^2\in K(\beta)$ implies that $K(\beta^2)\subset K(\beta)\Longrightarrow$ $[K(\beta):K(\beta^2)]|2k+1$ so $[K(\beta):K(\beta^2)]=1$ thus $K(\beta)=K(\beta^2).$ – Med Feb 06 '14 at 00:21
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The degree $[K(\beta):K(\beta^2)]$ divides $[K(\beta):K]$, but it can be only $1$ or $2$. Now conclude. :-)
Martin Brandenburg
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