1

Let $K\subseteq K(\beta)$ be a finite field extension of odd degree. Does this imply $K(\beta)=K(\beta^2)$?

Martin Brandenburg
  • 163,620
noah
  • 21
  • 1
    Yes $;;;;;;;;;;;;$ – Old John Feb 05 '14 at 23:24
  • 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

1 Answers1

2

The degree $[K(\beta):K(\beta^2)]$ divides $[K(\beta):K]$, but it can be only $1$ or $2$. Now conclude. :-)

Martin Brandenburg
  • 163,620