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I'm trying to show that $x^5-1/2$ is the minimal polynomial of $2^{-1/5}$.
I know I have to show that it's irreducible over Q. First I have to show that it has no rational roots (no problem!) then I have to show that it cannot be written as a product of a second degree and a third degree polynomial (also no problem). The thing is that I would like to have a cleaner solution. For instance, I've already shown that the minimal polynomial of the inverse element $2^{1/5}$ is $x^5-2$ using Eisenstein's criterion. It seems I should be able to use that somehow. Is it for example true that an element and its inverse have the same degree of the minimal polynomial? How would I go about to prove that?

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    $\begingroup$ Consider the two extensions $K(\alpha)$ and $K(\alpha^{-1})$. Well, actually, they are the same extension. $\endgroup$ Dec 2, 2013 at 20:26

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Is it for example true that an element and its inverse have the same degree of the minimal polynomial?

Yes. Let $K$ any field, and $\alpha \neq 0$ algebraic over $K$. Since $\alpha^{-1} \in K(\alpha)$ we have $K(\alpha^{-1}) \subset K(\alpha)$. Symmetrically, we have $K(\alpha) \subset K(\alpha^{-1})$, hence equality.

Now, $[K(\alpha) : K]$ is the degree of the minimal polynomial of $\alpha$, and of course $[K(\alpha^{-1}) : K]$ is the degree of the minimal polynomial of $\alpha^{-1}$. Since the extensions are the same, the degrees are the same.

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