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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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What about $\mathbb{R}(i)$? – Michael Albanese Dec 21 '12 at 23:48
@MichaelAlbanese, I'm looking for proper containment. $\mathbb{R}(i^3)=\mathbb{R}(i)$. – Frank White Dec 21 '12 at 23:50
Sorry, I missed the word 'properly'. – Michael Albanese Dec 21 '12 at 23:52
Although this does not help if $E$ is not supposed to be $F$ in one spot and $F(\alpha)$ in another, an example might be $\mathbb Q(\omega)$ where $\omega=\mathrm{exp}(2\pi i/3)$, since $\omega^2+\omega+1=0$. – peoplepower Dec 21 '12 at 23:52
@Frank: Sorry, not to nitpick, but the standard notation for the degree of an extension $L/K$ is $[L:K]$, i.e. the larger field is on the left. Now that I think I understand what you're asking, I'll edit. – Zev Chonoles Dec 22 '12 at 0:15

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, +1 – Belgi Dec 22 '12 at 14:04

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