## The Galois correspondence

Last update: 02 February 2012

## The Galois correspondence

Let $𝔼$ be a field.

• If $𝔽$ is a subfield of $𝔼$ the Galois group of $𝔼$ over $𝔽$ is
• If $H$ is a subgroup of $\mathrm{Aut}\left(𝔼\right)$ the fixed field of $H$ is

Thus and the definitions imply that, for $𝔽,𝕂\subseteq 𝔼$ and $H,G\subseteq \mathrm{Aut}\left(𝔼\right),$

1. $\mathrm{Fix}\left(\mathrm{Gal}\left(𝔽\right)\right)\supseteq 𝔽,$
2. $\mathrm{Gal}\left(\mathrm{Fix}\left(H\right)\right)\supseteq H,$
3. If $𝕂\subseteq 𝔽$ then $\mathrm{Gal}\left(𝕂\right)\supseteq \mathrm{Gal}\left(𝔽\right),$
4. If $H\subseteq G$ then $\mathrm{Fix}\left(H\right)\supseteq \mathrm{Fix}\left(G\right).$
These facts imply that, for $𝔽\subseteq 𝔼$ and $H\subseteq \mathrm{Aut}\left(𝔼\right),$ $FixGalFix(H) =Fix(H) and GalFixGal(𝔽) =Gal(𝔽).$ If $\sigma \in \mathrm{Aut}\left(𝔼\right)$ then $Gal(σ𝔽) =σGal(𝔽) σ - 1 and Fix(σH σ - 1 ) =σFix(H).$

 Proof. Let $𝔼$ be a finite extension of $𝔽$ and let $\left({\alpha }_{1},...,{\alpha }_{d}\right)$ be a basis of $𝔼$ over $𝔽$. Let $f(x) = mα1,𝔽(x) ⋯ mαd,𝔽(x) ∈𝔽[x].$ Since $𝔼$ is generated by $\left({\alpha }_{1},...,{\alpha }_{d}\right)$ over $𝔽$, every element of $\mathrm{Gal}\left(𝔽\right)$ is determined by its action on So $\mathrm{Gal}\left(𝔽\right)\subseteq {S}_{r},$ where $\square$

 Proof. Let $H\subseteq \mathrm{Aut}\left(𝔼\right)$ be finite and let $𝔽=\mathrm{Fix}\left(H\right).$ Then, for $\alpha \in 𝔼,$ $mα,𝔽 (x) = ∏ β∈Hα (x-β),$ since $p\left(x\right)=\prod _{\beta \in H\alpha }\left(x-\beta \right)$ is an element of $𝔽\left[x\right]$ since it is a polynomial in $𝔼\left[x\right]$ fixed by $H$, ${m}_{\alpha ,𝔽}\left(x\right)$ divides $p\left(x\right)$ since $\alpha$ is a root of $p\left(x\right)$, $p\left(x\right)$ divides ${m}_{\alpha ,𝔽}\left(x\right)$ since $H$ fixes ${m}_{\alpha ,𝔽}\left(x\right)$ and takes the factor $\left(x-\alpha \right)$ to all other $\left(x-\beta \right)$. Thus the minimal polynomial ${m}_{\alpha ,𝔽}\left(x\right)$ has roots of multiplicity one and splits in $𝔼\left[x\right].$ So $𝔼$ is normal and separable over $𝔽$. $\square$

If $H\subseteq \mathrm{Aut}\left(𝔼\right)$ is finite then $[𝔼:Fix(H)] =|H| and GalFix(H)= H.$

 Proof. By Proposition ??? $𝔼$ is finite and separable over $𝔽$. Thus, by the Theorem of the Primitive Element $𝔼=𝔽\left(\alpha \right)$ for some $\alpha \in 𝔼.$ By Proposition ???, $mα,𝔽 (x) = ∏ β∈Hα (x-β).$ Thus each element of of $\mathrm{Gal}\left(𝔽\right)$ is determined by where it sends $\alpha$. So the elements of $\mathrm{Gal}\left(𝔽\right)$ are in correspondence with the roots of ${m}_{\alpha ,𝔽}\left(x\right).$ 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 $\mathrm{Gal}\mathrm{Fix}\left(H\right)=H.$ $\square$

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

 Proof. Let $G=\mathrm{Gal}\left(𝔽\right)$. Since $𝔼$ is Galois over $𝔽$, the Theorem of the Primitive Element implies that $𝔼=𝔽\left(\alpha \right)$ for some $\alpha \in 𝔼$. Then $mα,𝔽 = ∏ β∈Gα (x-β),$ since ${m}_{\alpha ,𝔽}\left(x\right)$ divides $p\left(x\right)=\prod _{\beta \in G\alpha }\left(x-\beta \right)$ since ${m}_{\alpha ,𝔽}\left(x\right)$ is separable, all roots of ${m}_{\alpha ,𝔽}\left(x\right)$ are in $𝔼$, and $G$ takes $\alpha$ to other roots of ${m}_{\alpha ,𝔽}\left(x\right),$ $p\left(x\right)$ divides ${m}_{\alpha ,𝔽}\left(x\right)$ since $x-\alpha$ is a factor of each of these polynomials and ${m}_{\alpha ,𝔽}\left(x\right)$ is fixed by $G$. Since $𝔼=𝔽\left(\alpha \right)$ the roots $\beta$ of ${m}_{\alpha ,𝔽}\left(x\right)$ are in one to one correspondence with the elements of $G$. So $\mathrm{Card}\left(G\alpha \right)=|G|.$ So $[𝔼:𝔽] = deg mα,𝔽 (x) =|G| = |GalFixGal(𝔽)| = [𝔼:FixGal(𝔽)],$ where the last equality follows from Proposition ???. $\square$

1. If $𝔼/𝕂$ is Galois and $𝔼\supseteq 𝔽\supseteq 𝕂$ then $𝔼/𝔽$ is Galois.
2. If $𝔼/𝕂$ is Galois and $𝔼\supseteq 𝔽\supseteq 𝕂$ then
3. If $𝔼/𝕂$ is Galois and $𝔼\supseteq 𝔽\supseteq 𝕂$ then $Gal ( 𝔼/𝕂 ) → Gal ( 𝔽/𝕂 ) σ ↦ σ |𝔽$ is a group homomorphism with kernel $\mathrm{Gal}\left(𝔼/𝔽\right).$

 Proof. Follows from the definitions. If $𝔽/𝕂$ is Galois and $\alpha \in 𝔽$ then ${m}_{\alpha ,𝕂}\left(x\right)$ splits in $𝔽\left[x\right]$ and so all roots of ${m}_{\alpha ,𝕂}\left(x\right)=\prod _{\beta \in G\alpha }\left(x-\beta \right)$ (where $G=\mathrm{Gal}\left(𝔼/𝕂\right)$) are in $𝔽$. So $\sigma 𝔽=𝔽$ for all $\sigma \in \mathrm{Gal}\left(𝔼/𝕂\right).$ $\square$

If $𝔽\subseteq 𝕂$ is a finite separable extension then there is a finite extension $𝔼\supseteq 𝔽\supseteq 𝕂$ with $𝔼/𝕂$ Galois.

 Proof. Let ${\alpha }_{1},...,{\alpha }_{d}$ 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 ${m}_{{\alpha }_{1},𝕂}\left(x\right)$ is separable and splits in $𝔼\left[x\right]$. For every root $\beta$ of ${m}_{{\alpha }_{1},𝕂}\left(x\right)$ the isomorphism $\theta :𝕂\left({\alpha }_{1}\right)\to 𝕂\left(\beta \right)$ extends to $\theta :𝔼\to 𝔼,$ and so Let $𝕃=\mathrm{Fix}\mathrm{Gal}\left(𝔼/𝕂\right).$ Then $𝔼$ is Galois over $𝕃$ and By induction, since $𝔼$ is the splitting field of $f\left(x\right)$ over $𝕂\left({\alpha }_{1}\right),$ $𝕂(α1) = FixGal(𝔼/𝕂 α1) ⊇FixGal (𝔼/𝕂) =𝕃.$ So $𝕂\left({\alpha }_{1}\right)=𝕃\left({\alpha }_{1}\right)\supseteq 𝕃\supseteq 𝕂$ and $[𝕂α1:𝕃] = [𝕃α1:𝕃] = Card(Gα1) and [𝕂α1:𝕂] = Card(Gα1).$ So $𝕃=𝕂.$ $\square$

## Notes and References

Where are these from?

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