The Galois correspondence

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 02 February 2012

The Galois correspondence

Let 𝔼 be a field.

Thus Gal: {subfields of   𝔼} {subgroups of   Aut(𝔼)} Fix: {subfields of   𝔼} {subgroups of   Aut(𝔼)} and the definitions imply that, for 𝔽,𝕂𝔼 and H,GAut(𝔼),

  1. Fix(Gal(𝔽))𝔽,
  2. Gal(Fix(H))H,
  3. If 𝕂𝔽 then Gal(𝕂)Gal(𝔽),
  4. If HG then Fix(H)Fix(G).
These facts imply that, for 𝔽𝔼 and HAut(𝔼), FixGalFix(H) =Fix(H) and GalFixGal(𝔽) =Gal(𝔽). If σAut(𝔼) then Gal(σ𝔽) =σGal(𝔽) σ - 1 and Fix(σH σ - 1 ) =σFix(H).

Gal: { subfields of   𝔼   such that   𝔼/𝔽   is finite } { finite subgroups of   Aut(𝔼) }

Let 𝔼 be a finite extension of 𝔽 and let α1...αd be a basis of 𝔼 over 𝔽. Let f(x) = mα1,𝔽(x) mαd,𝔽(x) 𝔽[x]. Since 𝔼 is generated by α1...αd over 𝔽, every element of Gal(𝔽) is determined by its action on { roots of   f(x)   which lie in   𝔼 }. So Gal(𝔽) Sr, where r=Card { roots of   f(x)   which lie in   𝔼 }.

Fix: { subfields of   𝔼   such that   𝔼/𝔽   is Galois } { finite subgroups of   Aut(𝔼) }

Let HAut(𝔼) be finite and let 𝔽=Fix(H). Then, for α𝔼, mα,𝔽 (x) = βHα (x-β), since
  1. p(x)= βHα (x-β) is an element of 𝔽[x] since it is a polynomial in 𝔼[x] fixed by H,
  2. mα,𝔽 (x) divides p(x) since α is a root of p(x),
  3. p(x) divides mα,𝔽 (x) since H fixes mα,𝔽 (x) and takes the factor (x-α) to all other (x-β).
Thus the minimal polynomial mα,𝔽 (x) has roots of multiplicity one and splits in 𝔼[x]. So 𝔼 is normal and separable over 𝔽.

If HAut(𝔼) is finite then [𝔼:Fix(H)] =|H| and GalFix(H)= H.

By Proposition ??? 𝔼 is finite and separable over 𝔽. Thus, by the Theorem of the Primitive Element 𝔼=𝔽(α) for some α𝔼. By Proposition ???, mα,𝔽 (x) = βHα (x-β). Thus each element of of Gal(𝔽) is determined by where it sends α. So the elements of Gal(𝔽) are in correspondence with the roots of mα,𝔽 (x). Since deg mα,𝔽 (x) = |GalFix(H)| |H| |Hα| = deg mα,𝔽 (x), is follows that [𝔼:𝔽] = [𝔽(α):𝔽] = deg mα,𝔽 (x) =|H| =|Gal(𝔽)| . So GalFix(H)=H.

If 𝔼 is a Galois extension of 𝔽 then FixGal(𝔽) =𝔽 and [𝔼:𝔽] = |Gal(𝔽)|.

Let G=Gal(𝔽). Since 𝔼 is Galois over 𝔽, the Theorem of the Primitive Element implies that 𝔼=𝔽(α) for some α𝔼. Then mα,𝔽 = βGα (x-β), since
  1. mα,𝔽 (x) divides p(x) = βGα (x-β) since mα,𝔽 (x) is separable, all roots of mα,𝔽 (x) are in 𝔼, and G takes α to other roots of mα,𝔽 (x),
  2. p(x) divides mα,𝔽 (x) since x-α is a factor of each of these polynomials and mα,𝔽 (x) is fixed by G.
Since 𝔼=𝔽(α) the roots β of mα,𝔽 (x) are in one to one correspondence with the elements of G. So Card(Gα) =|G|. So [𝔼:𝔽] = deg mα,𝔽 (x) =|G| = |GalFixGal(𝔽)| = [𝔼:FixGal(𝔽)], where the last equality follows from Proposition ???.

  1. If 𝔼/𝕂 is Galois and 𝔼𝔽𝕂 then 𝔼/𝔽 is Galois.
  2. If 𝔼/𝕂 is Galois and 𝔼𝔽𝕂 then 𝔽/𝕂   is Galois σ𝔽=𝔽   for all   σ Gal(𝔼/𝕂) Gal(𝔼/𝔽)   is normal in   Gal(𝔼/𝕂) .
  3. If 𝔼/𝕂 is Galois and 𝔼𝔽𝕂 then Gal ( 𝔼/𝕂 ) Gal ( 𝔽/𝕂 ) σ σ |𝔽 is a group homomorphism with kernel Gal (𝔼/𝔽).

  1. Follows from the definitions.
  2. If 𝔽/𝕂 is Galois and α𝔽 then mα,𝕂 (x) splits in 𝔽[x] and so all roots of mα,𝕂 (x) = βGα (x-β) (where G=Gal(𝔼/𝕂)) are in 𝔽. So σ𝔽=𝔽 for all σ Gal(𝔼/𝕂) .

If 𝔽𝕂 is a finite separable extension then there is a finite extension 𝔼𝔽𝕂 with 𝔼/𝕂 Galois.

Let α1 ... αd be a basis of 𝔽 over 𝕂. Let 𝔼 be a splitting field of f(x) = m α1,𝕂 (x) m αr,𝕂 (x) 𝕂[x]. Then m α1,𝕂 (x) is separable and splits in 𝔼[x]. For every root β of m α1,𝕂 (x) the isomorphism θ:𝕂 α1 𝕂β extends to θ:𝔼𝔼 , and so m α1,𝕂 (x) = βGα (x-β), where   G= Gal(𝔼/𝕂) . Let 𝕃=FixGal (𝔼/𝕂). Then 𝔼 is Galois over 𝕃 and m α1,𝕃 (x) = βGα1 (x-β), where   G=Gal (𝔼/𝕃) = GalFixGal(𝕂) = Gal(𝔼/𝕂) . By induction, since 𝔼 is the splitting field of f(x) over 𝕂(α1), 𝕂(α1) = FixGal(𝔼/𝕂 α1) FixGal (𝔼/𝕂) =𝕃. So 𝕂α1 = 𝕃α1 𝕃𝕂 and [𝕂α1:𝕃] = [𝕃α1:𝕃] = Card(Gα1) and [𝕂α1:𝕂] = Card(Gα1). So 𝕃=𝕂.

Notes and References

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