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I've been trying to come up with a simple algebraic extension $F(\alpha)$ over a field $F$ that has $[F(\alpha):F]$ not divisible by 3, but has $F(\alpha^3)$ properly contained in $F(\alpha)$. I haven't had any luck - maybe I'm thinking incorrectly, but all I can think of is cube roots, fourth roots and the like.

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    @peoplepower: I think it'd be great if you put that comment as answer now that the question is cleared up.2012-12-22

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Consider the extension $\mathbb Q(\omega)/\mathbb Q$ with $\omega=e^{2\pi i/3}$, and note that $\omega$ is a root of the polynomial $x^3-1=(x-1)(x^2+x+1)$. Since $\omega$ is distinct from $1$, it must satisfy $p(x)=x^2+x+1$. Finally, $p(x+1)=x^2+3x+3$ is irreducible over $\mathbb{Q}$ by Eisenstein's Criterion.

Therefore the extension is a degree 2, simple extension generated by a cube root of an element of the base field.

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    Looks OK to me, +12012-12-22