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Let $K/F$ be a finite field extension.

If $K/F$ is Galois then it is well known that there is a bijection between subgroups of $Gal(K/F)$ and subfields of $K/F$.

Since finding subgroups of a finite group is always easy (at least in the meaning that we can find every subgroup by brute-force or otherwise) this gives a nice way of finding subfields and proving they are the only ones.

What can we do in the case that $K/F$ is not a Galois extension ? that is: How can I find all subfields of a non-Galois field extension ?

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  • $\begingroup$ An easy answer to this would be to compute the Galois closure of $K/F$ and consider the subfields of the Galois closure which are in $K$. Would that work for you? $\endgroup$ Commented Oct 14, 2012 at 18:54
  • $\begingroup$ @PatrickDaSilva - what if the extension is not separable ? this answer seems good for the cases of characteristic $0$ and for finite fields so it's very nice on but doesn't answer all of the problem. this is just a question of intrest, I don't need an application for something so any information is good. Thank you for your answer! $\endgroup$ Commented Oct 14, 2012 at 18:58
  • $\begingroup$ Hm. I must say I assumed the extension was separable because I don't see very often non-separable extensions. =P $\endgroup$ Commented Oct 14, 2012 at 19:09
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    $\begingroup$ @PatrickDaSilva - quickmeme.com/meme/3rc44q :P $\endgroup$ Commented Oct 14, 2012 at 19:19
  • $\begingroup$ Very nice meme, I like it $\endgroup$ Commented Oct 15, 2012 at 2:40

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An easy answer in the separable case to this would be to compute the Galois closure of K/F and consider the subfields of the Galois closure which are in K.

Hope that helps,

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In the inseparable case there is an idea for a substitute Galois correspondence due, I think, to Jacobson: instead of considering subgroups of the Galois group, we consider (restricted) Lie subalgebras of the Lie algebra of derivations. I don't know much about this approach, but "inseparable Galois theory" seems to be a good search term.

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