TITLE: Square of algebraic element of odd order generates the same field
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: You start very well: Note that $F\subseteq F(\alpha^2)\subseteq F(\alpha)$.
Now, what is $[F(\alpha):F(\alpha^2)]$? 
Hint: Can you come up with a polynomial with coefficients in $F(\alpha^2)$ that is satisfied by $\alpha$?

Alternate, bull-headed proof (just push forward until you get what you need; no finesse whatsoever).
Let $p(x)$ be the monic irreducible polynomial of $\alpha$ with coefficients in $F$. Write:
$$p(x) = x^{2n+1} + a_{2n}x^{2n} + \cdots + a_1x + a_0.$$
Then
$$\alpha^{2n+1} + a_{2n}\alpha^{2n} + \cdots+ a_1\alpha + a_0 = 0.$$
Move all the even powers to one side, and all the odd powers to the other:
$$\begin{align*}
\alpha^{2n+1}+a_{2n-1}\alpha^{2n-1}+\cdots + a_1\alpha &= a_{2n}\alpha^{2n}+\cdots + a_2\alpha^2 + a_0\\
\alpha(\alpha^{2n}+ a_{2n-1}\alpha^{2n-2} + \cdots + a_1) &= a_{2n}\alpha^{2n}+\cdots + a_2\alpha^2 + a_0\\
&\vdots
\end{align*}$$