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If $K$ is an extension of field $F$ such that $[K:F]$ is finite and for two subfields $K_1$ and $K_2$ which contains $F$, either $K_2\subset K_1$ or $K_1\subset K_2$, then $K$ has a primitive element over $F$.

I am trying to prove this by an idea of induction. Since is $[K:F]$ finite we can select a basis that is $B=\{e_1,\cdots, e_n\}$ for $K$ over $F$. Since $K_1$ and $K_2$ are subfields of $K$, we can take $B$ as their generating set. I eventually want to say that for $K=F(e_1, e_2)(e_3,\cdots, e_n)$, I can find $x$ such that $x=c_1e_1+c_2e_2$ and using this I can find my primitive element by substituting as $F(x,e_3)(e_4, \cdots, e_n)$. However I am stuck on how to use my second given condition (the subset one).

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  • $\begingroup$ There is a result of Artin to the effect that a finite extension of fields has a primitive element if and only if the number of intermediate fields is finite. It is in the book Algebra by Lang. $\endgroup$
    – orangeskid
    Jan 29 '15 at 5:26
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A starting point point for a more direct approach could be this one: take any intermediate field of $K/F$ the form $F(x)$. If $K=K(x)$ we are happy. Otherwise for an element $y\in K\setminus K(x)$ we must have $K(x)\subset K(y)$. Proceed ...

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