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Let $F$ be a field, and $K$ a field extension of $F$. Prove that if $[K:F]$ is prime, then there is no field $L$ such that $F\subset L\subset K$ and $F\neq L,L\neq K$.

Well, if $L$ is an extension of $F$ and $K$ is an extension of $L$, then we can say $[K:F]=[K:L][L:F]$, yielding that either $K=L$ or $L=F$.

But here $L$ is any field between $F$ and $K$. What should we do?

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  • $\begingroup$ If $F\subset L\subset K$ then $L$ is an extension of $F$ and $K$ is an extension of $L$. $\endgroup$ – egreg Jan 18 '14 at 1:06
  • $\begingroup$ @egreg I guess I'm understanding that "$L$ is an extension of $F$" means $L=F(a_1,a_2,\ldots,a_n)$ for some $a_1,\ldots,a_n$. If $F\subset L$, then is $L$ necessarily an extension of $F$ in this sense? $\endgroup$ – Kunal Jan 18 '14 at 1:10
  • $\begingroup$ No, usually “$L$ is an extension of $F$” just means that $F$ is a subfield of $L$. $\endgroup$ – egreg Jan 18 '14 at 1:12
  • $\begingroup$ @egreg I see, thank you. $\endgroup$ – Kunal Jan 18 '14 at 1:13
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You're misunderstanding the meaning of

$L$ is an extension of $F$

which just means that $F$ is a subfield of $L$.

However, in this case of $F\subset L\subset K$, you can say that $L$ is of the form $F(a_1,\dots,a_n)$, because, by hypothesis, $K$ is finite dimensional over $F$. Hence also $L$ is finite dimensional over $F$ and so it can be obtained by adjoining to $F$ a finite number of elements (for instance, a basis of $L$ over $F$). The same applies to $K$ over $L$.

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