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Let $F\subset E$ be a finite degree field extension. Then $E=F[\alpha]$ for some element $\alpha \in E$ iff there are only finitely many fields $K$ with $F \subset K \subset E$. In other words, a finite degree field extension is a simple extension if and only if there are only finitely many intermediate fields.

I don't know how should begin the proof. Is there any book you can suggest to read this proof of theorem?

Thanks a lot.

Jyrki Lahtonen
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bozcan
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    Serge Lang's Algebra. It's the Primitive element theorem. – Bernard Mar 26 '17 at 19:23
  • I want to see a motivation example. For instance, let $F[\alpha]=\mathbb{Q}[\sqrt 6]$ and $F=\mathbb{Q}$. Are the intermediate fields $\mathbb{Q}[\sqrt 2]$ and $\mathbb{Q}[\sqrt 3]$? – bozcan Mar 26 '17 at 19:58
  • No. None of them is contained in $\mathbf Q[\sqrt 6]$. Actually, the latter has only two subfields, itself and $\mathbf Q$, since it is a degree $2$ extension of $\mathbf Q$. – Bernard Mar 26 '17 at 20:04
  • Can you give a valid example? – bozcan Mar 26 '17 at 20:11
  • A valid example of what? – Bernard Mar 26 '17 at 20:12
  • I mean that I want to see an example like: The intermediate fields of $\mathbb{Q}[\sqrt 2,\sqrt 3]$ are $\mathbb{Q}[\sqrt 2]$, $\mathbb{Q}[\sqrt 3]$ and $\mathbb{Q}[\sqrt 6]$. – bozcan Mar 26 '17 at 20:22
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    I don't see anything about finite fields so I removed the tag. $\Bbb{Q}$ and its extensions all have infinitely many elements. – Jyrki Lahtonen Mar 26 '17 at 21:00

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