The Galois correspondence

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $$\mathrm{Gal}\left(𝔽\right)={\mathrm{Aut}}_{𝔽}\left(𝔼\right)= {g\in \mathrm{Aut}\left(𝔼\right)|g\alpha =\alpha \; for\; all\; \alpha \in 𝔽} .$$
• If $H$ is a subgroup of $\mathrm{Aut}\left(𝔼\right)$ the fixed field of $H$ is $$\mathrm{Fix}\left(H\right)={𝔼}^H= {\alpha \in 𝔼|h\alpha =\alpha \; for\; all\; h\in H} .$$

Thus $$\begin{array}{rrcl}\mathrm{Gal}:& \left\{subfields\; of\; 𝔼\right\}& \to & \left\{subgroups\; of\; \mathrm{Aut}\right(𝔼\left)\right\}\\ \mathrm{Fix}:& \left\{subfields\; of\; 𝔼\right\}& \leftarrow & \left\{subgroups\; of\; \mathrm{Aut}\right(𝔼\left)\right\}\end{array}$$ and the definitions imply that, for $𝔽,𝕂\subseteq 𝔼$ and $H,G\subseteq \mathrm{Aut}(𝔼),$

1. $\mathrm{Fix}\left(\mathrm{Gal}\right(𝔽\left)\right)\supseteq 𝔽,$
2. $\mathrm{Gal}\left(\mathrm{Fix}\right(H\left)\right)\supseteq H,$
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 $H\subseteq \mathrm{Aut}\left(𝔼\right),$ $$\mathrm{Fix}\mathrm{Gal}\mathrm{Fix}\left(H\right)=\mathrm{Fix}\left(H\right)and\mathrm{Gal}\mathrm{Fix}\mathrm{Gal}\left(𝔽\right)=\mathrm{Gal}\left(𝔽\right).$$ If $\sigma \in \mathrm{Aut}\left(𝔼\right)$ then $$\mathrm{Gal}\left(\sigma 𝔽\right)=\sigma \mathrm{Gal}\left(𝔽\right){\sigma }^{-1}and\mathrm{Fix}\left(\sigma H{\sigma }^{-1}\right)=\sigma \mathrm{Fix}\left(H\right).$$

$$\mathrm{Gal}:\{subfields\; of\; 𝔼 \; such\; that\; 𝔼/𝔽 \; is\; finite\}\to \left\{finite\; subgroups\; of\; \mathrm{Aut}\right(𝔼\left)\right\}$$

		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 $$f\left(x\right)=m_{{\alpha }_1,𝔽}\left(x\right)\cdots m_{{\alpha }_d,𝔽}\left(x\right)\in 𝔽\left[x\right].$$ 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 $$\left\{roots\; of\; f\right(x\left) \; which\; lie\; in\; 𝔼\right\}.$$ So $\mathrm{Gal}\left(𝔽\right)\subseteq S_r,$ where $r=\mathrm{Card}\left\{roots\; of\; f\right(x\left) \; which\; lie\; in\; 𝔼\right\}.$ $\square$

$$\mathrm{Fix}:\{subfields\; of\; 𝔼 \; such\; that\; 𝔼/𝔽 \; is\; Galois\}\leftarrow \left\{finite\; subgroups\; of\; \mathrm{Aut}\right(𝔼\left)\right\}$$

		Proof.
	Let $H\subseteq \mathrm{Aut}\left(𝔼\right)$ be finite and let $𝔽=\mathrm{Fix}\left(H\right).$ Then, for $\alpha \in 𝔼,$ $$m_{\alpha ,𝔽}\left(x\right)=\prod _{\beta \in H\alpha }(x-\beta ),$$ since $p\left(x\right)=\prod _{\beta \in H\alpha }(x-\beta )$ 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 $(x-\alpha )$ to all other $(x-\beta )$. 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 $$[𝔼:\mathrm{Fix}(H\left)\right]=\left|H\right|and\mathrm{Gal}\mathrm{Fix}\left(H\right)=H.$$

		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 ???, $$m_{\alpha ,𝔽}\left(x\right)=\prod _{\beta \in H\alpha }(x-\beta ).$$ Thus each element of of $\mathrm{Gal}(𝔽)$ 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 $$\mathrm{deg}{m}_{\alpha ,𝔽}\left(x\right)=\left|\mathrm{Gal}\mathrm{Fix}\right(H\left)\right|\ge \left|H\right|\ge \left|H\alpha \right|=\mathrm{deg}{m}_{\alpha ,𝔽}\left(x\right),$$ is follows that $$[𝔼:𝔽]=\left[𝔽\right(\alpha ):𝔽]=\mathrm{deg}{m}_{\alpha ,𝔽}\left(x\right)=\left|H\right|=\left|\mathrm{Gal}\right(𝔽\left)\right|.$$ So $\mathrm{Gal}\mathrm{Fix}\left(H\right)=H.$ $\square$

If $𝔼$ is a Galois extension of $𝔽$ then $$\mathrm{Fix}\mathrm{Gal}\left(𝔽\right)=𝔽and[𝔼:𝔽]=\left|\mathrm{Gal}\right(𝔽\left)\right|.$$

		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 $$m_{\alpha ,𝔽}=\prod _{\beta \in G\alpha }(x-\beta ),$$ since $m_{\alpha ,𝔽}\left(x\right)$ divides $p\left(x\right)=\prod _{\beta \in G\alpha }(x-\beta )$ 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 $\mathrm{Card}\left(G\alpha \right)=\left|G\right|.$ So $$[𝔼:𝔽]=\mathrm{deg}{m}_{\alpha ,𝔽}\left(x\right)=\left|G\right|=\left|\mathrm{Gal}\mathrm{Fix}\mathrm{Gal}\right(𝔽\left)\right|=[𝔼:\mathrm{Fix}\mathrm{Gal}(𝔽\left)\right],$$ 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 $$𝔽/𝕂 \; is\; Galois\iff \sigma 𝔽=𝔽 \; for\; all\; \sigma \in \mathrm{Gal}(𝔼/𝕂)\iff \mathrm{Gal}(𝔼/𝔽) \; is\; normal\; in\; \mathrm{Gal}(𝔼/𝕂).$$
3. If $𝔼/𝕂$ is Galois and $𝔼\supseteq 𝔽\supseteq 𝕂$ then $$\begin{array}{ccc}\mathrm{Gal}(𝔼/𝕂)& \to & \mathrm{Gal}(𝔽/𝕂)\\ \sigma & \mapsto & \sigma {|}_{𝔽}\end{array}$$ is a group homomorphism with kernel $\mathrm{Gal}(𝔼/𝔽).$

		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 $m_{\alpha ,𝕂}\left(x\right)=\prod _{\beta \in G\alpha }(x-\beta )$ (where $G=\mathrm{Gal}(𝔼/𝕂)$) are in $𝔽$. So $\sigma 𝔽=𝔽$ for all $\sigma \in \mathrm{Gal}(𝔼/𝕂).$ $\square$

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 $$f\left(x\right)=m_{{\alpha }_1,𝕂}\left(x\right)\cdots m_{{\alpha }_r,𝕂}\left(x\right)\in 𝕂\left[x\right].$$ 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 $$m_{{\alpha }_1,𝕂}\left(x\right)=\prod _{\beta \in G\alpha }(x-\beta ),where\; G=\mathrm{Gal}(𝔼/𝕂).$$ Let $𝕃=\mathrm{Fix}\mathrm{Gal}(𝔼/𝕂).$ Then $𝔼$ is Galois over $𝕃$ and $$m_{{\alpha }_1,𝕃}\left(x\right)=\prod _{\beta \in G{\alpha }_1}(x-\beta ),where\; G=\mathrm{Gal}(𝔼/𝕃)=\mathrm{Gal}\mathrm{Fix}\mathrm{Gal}\left(𝕂\right)=\mathrm{Gal}(𝔼/𝕂).$$ By induction, since $𝔼$ is the splitting field of $f\left(x\right)$ over $𝕂\left({\alpha }_1\right),$ $$𝕂\left({\alpha }_1\right)=\mathrm{Fix}\mathrm{Gal}(𝔼/𝕂\left({\alpha }_1\right))\supseteq \mathrm{Fix}\mathrm{Gal}(𝔼/𝕂)=𝕃.$$ So $𝕂\left({\alpha }_1\right)=𝕃\left({\alpha }_1\right)\supseteq 𝕃\supseteq 𝕂$ and $$[𝕂\left({\alpha }_1\right):𝕃]=[𝕃\left({\alpha }_1\right):𝕃]=\mathrm{Card}\left(G{\alpha }_1\right)and[𝕂\left({\alpha }_1\right):𝕂]=\mathrm{Card}\left(G{\alpha }_1\right).$$ So $𝕃=𝕂.$ $\square$

Notes and References

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