Possible Duplicate:
Equal simple field extensions?
Let $\alpha$ be algebraic over the field $F$ and let the minimal polynomial of $\alpha$, denoted by $m_{\alpha}(x)$, have odd order. Then i need to show that $F(\alpha^2)=F(\alpha)$.
Here is my line of thought: We have the extension of fields $F \subset F(\alpha^2) \subset F(\alpha)$. The first observation is that $m_{\alpha}(x)$ divides $m_{\alpha^2}(x^2)$. The second observation is that, if $deg(m_{\alpha})=n_1$ and $deg(m_{\alpha^2})=n_2$, then $n_2 \le n_1 < 2n_2$. The third observation is that $m_{\alpha^2}(x^2)$ contains only monomials of even orders. My idea is to to assume that $d>n$ and arrive at a contradiction (since if $d=n$, then we are done). I am missing though the link between all these above observations and what this contradiction should be. I have been trying to manipulate the division equation $m_{\alpha^2}(x^2)=m_{\alpha}(x)q(x)$ for some quotient $q(x)$ but it feels messy. Any hints?
Thanks.