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Theorem: Every finite extension, normal and separable is a Galois extension.

Is the theorem equivalent to: $\mathbb K:\mathbb F$ is Galois $\iff \mathbb K:\mathbb F$ is normal & $\mathbb K:\mathbb F$ is separable ?

thus, $\mathbb K:\mathbb F$ is not a Galois extension $\iff \mathbb K:\mathbb F$ is not normal or $\mathbb K:\mathbb F$ is not separable, but only one holds, either normal or separable.

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    $\begingroup$ What's your definition of a Galois extension? One definition is that an extension is Galois if and only if it's algebraic, normal, and separable. But maybe you have a different definition and that's a theorem for you. Anyway, you can't leave out some kind of assumption of algebraicity or finiteness. (Normal and separable alone are not enough.) $\endgroup$ – Zach Teitler Jul 11 '17 at 21:02
  • $\begingroup$ @ZachTeitler : An algebraic extension $K:F$ is Galois if $G(K:F)^+=f(F)$, where $f$ is the monomorphism. $\endgroup$ – user441848 Jul 11 '17 at 21:19
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    $\begingroup$ @Annelise The fixed field characterization (for an algebraic but not necessarily finite extension) turns out to be equivalent to algebraic + normal + separable. A first course in Galois theory usually only deals with finite Galois extensions however. $\endgroup$ – Trevor Gunn Jul 11 '17 at 23:33
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For a finite extension $K/k$ the following are equivalent:

  1. $\require{begingroup}\begingroup\DeclareMathOperator{\Aut}{Aut}|\Aut(K/k)| = [K:k]$
  2. $k$ is the fixed field of $\Aut(K/k)$
  3. $K/k$ is normal and separable
  4. $K$ is the splitting field of a finite number of separable polynomials over $k$
  5. $K$ is the splitting field of a single separable polynomial over $k$
  6. There is a one to one correspondence between subgroups of $\Aut(K/k)$ and intermediate extensions of $K/k$

For a possibly infinite algebraic extension $K/k$ the following are equivalent:

  1. $k$ is the fixed field of $\Aut(K/k)$
  2. $K/k$ is normal and separable
  3. $K$ is the splitting field of a possibly infinite family of separable polynomials over $k$
  4. There is a one to one correspondence between closed subgroups of $\Aut(K/k)$ and intermediate extensions of $K/k$

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  • $\begingroup$ Does this theorem: 'Every finite extension, normal and separable is a Galois extension.' has the form $p\Rightarrow q$ or $p\iff q?$ $\endgroup$ – user441848 Jul 18 '17 at 18:06
  • $\begingroup$ @AnneliseToft An extension is finite, normal and separable if and only if it is finite and Galois. This is the most general usage of "Galois" which includes infinite extensions. Sometimes, when dealing only with finite extensions one drops the word "finite" and simply declares at the beginning of the section or chapter that all extensions are finite. $\endgroup$ – Trevor Gunn Jul 18 '17 at 18:20
  • $\begingroup$ So is the theorem missing the ' finite' word ? (Because only says ...Galois extension. ). I am asking you this because I used that theorem in my last exam and my professor told me that I was wrong in my answer because the theorem is not an 'if and only if', the theorem has the form 'if..., then..' . And I lost 4 points in my exam 😭, which is the main reason why I failed $\endgroup$ – user441848 Jul 18 '17 at 18:30
  • $\begingroup$ @AnneliseToft That seems like the most likely explanation. $\endgroup$ – Trevor Gunn Jul 18 '17 at 19:13
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    $\begingroup$ @Annelise The problem is that given any book on field and Galois theory, these equivalences tend to be either scattered throughout different sections or not mentioned at all. One book I have that puts this in one place is Rowan's Graduate Algebra: Commutative View (Theorem 4.55) for 1,2,3,5 of the finite case (4 and 5 is trivial: if $K/k$ is the splitting field of $f_1,\dots,f_n$ then it is the splitting field of the product $f_1\cdots f_n$). Equivalence with 6 is generally a separate theorem called Galois Correspondence. (6 implies 2 is easy and often not mentioned). $\endgroup$ – Trevor Gunn Jul 19 '17 at 20:46

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