Equal simple field extensions? I have a question about simple field extensions.
For a field $F$, if $[F(a):F]$ is odd, then why is $F(a)=F(a^2)$?

Since $[F(a):F]$ is odd the minimal polynomial of $a$ is an odd degree polynomial, say $p(x)=b_{0}x^{2k+1}+b_1x^{2k}+...b_{2k+1}$, now since $a$ satisfies $p(x)$ we have: $b_{0}a^{2k+1}+b_1a^{2k}+...b_{2k+1}=0$ $\implies$ $a(b_0a^{2k}+b_2a^{2k-2}+...b_{2k})+b_1a^{2k}+b_3a^{2k-2}+...b_{2k+1}=0$ $\implies$ $a= -(b_1a^{2k}+b_3a^{2k-2}+...+b_{2k+1})/(b_0a^{2k}+b_2a^{2k-2}+...b_{2k})$ $\implies$ $a$ is in $F(a^2)$,so obviously $F(a^2)=F(a)$  
