# “Converse” of dimension counting formula for fields

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 $F=\mathbb Q$, $E=\mathbb Q(\sqrt2)$, $L=\mathbb Q(\sqrt[3]2)$.
Just another query, if we have another field extension of $\mathbb{R}$ of finite degree, say $L$ then is it true that $L$ must contain the complex numbers? – user38268 Mar 16 '12 at 12:47
@BenjaminLim, that is a separate question, but yes, and there is just one field extension of $\mathbb R$ of finite degree, the complex numbers, because $\mathbb C$ is algebraically closed. – lhf Mar 16 '12 at 13:16
Thanks. I think because if we have another field extension $L$ of $\mathbb{R}$ that is of finite degree, $L$ is algebraic over $\mathbb{R}$ and since $\mathbb{C}$ contains $\mathbb{R}$, $L$ would be algebraic over $\mathbb{C}$. But then $L = \mathbb{C}$ for the complex numbers are algebraically closed. I think this does it. – user38268 Mar 16 '12 at 13:26