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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 $\mathbb{R}(i)$?2012-12-21
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    @MichaelAlbanese, I'm looking for proper containment. $\mathbb{R}(i^3)=\mathbb{R}(i)$.2012-12-21
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    Sorry, I missed the word 'properly'.2012-12-21
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    Your question doesn't seem to make sense to me. Perhaps you mean to say $F(\alpha)$ is a simple extension over $F$? If that's the case, what is $E$ doing?2012-12-21
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    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 $\omega^2+\omega+1=0$.2012-12-21
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    I apologize there, I fixed it.2012-12-22
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    @Frank: Sorry, not to nitpick, but [the standard notation for the degree of an extension](http://en.wikipedia.org/wiki/Degree_of_a_field_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.2012-12-22
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    @hwhm: $[F(\alpha) : F(\alpha^3)]$ is not necessarily $3$.2012-12-22
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    I attempted to answer the question, but given the hour its best if someone reviews it2012-12-22
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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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