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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?


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marked as duplicate by Arturo Magidin, Asaf Karagila, t.b., Henning Makholm, J. M. Dec 5 '11 at 13:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I think this is similar to the question here:… since $m_\alpha(x)$ having odd degree is the same as $[F(\alpha):F]$ being odd. – Vika Dec 2 '11 at 19:50
$[F(\alpha):F(\alpha^2)]$ is either 1 or 2 (since $\alpha$ satisfies $x^2-\alpha^2=0$). if it is 2, then $2|[F(\alpha):F]$, a contradiction since $[F(\alpha):F]$ is the degree of $m_{\alpha}$ which is odd. – yoyo Dec 2 '11 at 19:56
up vote 6 down vote accepted

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: $$\begin{align*} \alpha^{2n+1}+a_{2n-1}\alpha^{2n-1}+\cdots + a_1\alpha &= a_{2n}\alpha^{2n}+\cdots + a_2\alpha^2 + a_0\\ \alpha(\alpha^{2n}+ a_{2n-1}\alpha^{2n-2} + \cdots + a_1) &= a_{2n}\alpha^{2n}+\cdots + a_2\alpha^2 + a_0\\ &\vdots \end{align*}$$

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