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We recently learned in class that if we have a tower of fields

$F \subseteq E \subseteq L$

then $[L:F]$ is finite iff both $[E:F]$ and $[L:E]$ are. In this case we have $[L:F]= [L:E][E :F]$. So then I thought that perhaps it is natural to ask the following question:

If we have a field $F$ and two fields $E$ and $L$ that contain $F$ such that $[E:F] = m$ and $[L : F] = n$, with $n > m$, is it the case that $L$ must contain $E$? Obviously if $L$ contains $E$ then $n > m$ because then $E$ would be an $F$ - vector subspace of $L$.

Thanks.

Edit: I have observed that if we have a field extension that is finite over the reals (Say $L$) then $L$ is algebraic over $\mathbb{R}$ so that it is over $\mathbb{C}$. But then this is only possible when $L = \mathbb{C}$ because $\mathbb{C}$ is algebraically closed.

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Certainly not. Take $F=\mathbb Q$, $E=\mathbb Q(\sqrt2)$, $L=\mathbb Q(\sqrt[3]2)$.

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    I don't want to ask a new question for I think it would be closed due to the similarity with this one. If $L$ does not contain $\mathbb{C}$ then how do I make an argument that in fact $L = \mathbb{C}$ or something like that?2012-03-16