Galois correspondence

After meeting the term of an automorphism, and, of more importantly, the Galois group of an extension, we would like to learn a bit more about the connection between the group and the extension by definig a correspondece betwee sub-group and sub-fields. Let’s see how we can do such a thing:

Sub-groups to sub-fields

Suppose that $H\leq \text{Gal}(K/F)$ is a subgroup. I’ll define:

K^H=\{a\in K|\forall \sigma\in H:\sigma(a)=a\}

This is the set of all the elements of $K$ , that are fixed under all the automorphisms in the subgroup $H$ .

I state that it is indeed a sub-field, let’s verify it:

Closed under addition and multiplication

a,b\in K^H\Rightarrow \forall{\sigma\in H}:\sigma(a)=a,\sigma(b)=b

\Rightarrow\forall{\sigma\in H}:\sigma(a+b)=\sigma(a)+\sigma(b)=a+b\ \land\sigma(ab)=\sigma(a)\sigma(b)=ab

\Rightarrow a+b,ab\in K^H

Inverse elements

a\in K^H\Rightarrow \forall\sigma\in H:\sigma(a)=a\Rightarrow\forall\sigma\in H:\sigma(-a)=-\sigma(a)=-a\Rightarrow -a\in K^H

a\in K^H\Rightarrow \forall\sigma\in H:\sigma(a)=a\Rightarrow\forall\sigma\in H:\sigma(a^{-1})=\sigma(a)^{-1}=a^{-1}\Rightarrow a^{-1}\in K^H

0 and 1

\forall\sigma\in \text{Gal}(K/F):\sigma(1)=1,\sigma(0)=0\Rightarrow 0,1\in H

The Fixed Field

So after we proved all the properties – we are now ready to give it an appropriate name. The field $F\subseteq K^H\subseteq K$ is called The Fixed Field of $H$ . That’s a good name for this field since it’s elements are the one that the automorphisms in $H$ fix!

Sub-Fields to Sub-Groups

The first two ways you would think about in order to map a sub-field to a sub-group are probably:

$L\mapsto \text{Gal}(L/F)$
$L\mapsto \text{Gal}(K/L)$

However, the first one is not the right way to do so:

Automorphisms in the group $\text{Gal}(L/F)$ don’t even ‘aware’ of the bigger field $K$ – so that can’t be right.

On the other hand, the group $\text{Gal}(K/L)$ is indeed a sub-group of $K$ . This is just the set of all the automorphisms that fix the elements of $F$ (check that it is indeed a subgroup!).

The correspondence

So we have found a correspondence:

\{\text{sub-groups of}\text{ Gal}(K/F)\}\leftrightarrows\{\text{sub-fields of }K\}

Defined as:

L^{\circ}:=\text{Gal}(K/L)\longleftarrow L

H\longmapsto K^{H}=:H^{*}

This correspondence is called the Galois correspondence. Let’s see some properties of it:

First, let’s check what is $K^\circ$ . By definition, $K^\circ=\text{Gal}(K/K)$ . This is the set of all the automorphisms that fix $K$ . However, the only automorphism with that property is the identity, thus:

K^\circ=\text{Gal}(K/K)=\{id_K\}

That’s pretty unexpected – the biggest field is being mapped to the smallest subgroup.

What about $F$ ? After we’ve seen this inversion, we would probably expect it to be mapped to the whole group. Let’s verify it:

By definition $F^\circ=\text{Gal}(K/F)$ – as we thought.

Now, what about the other direction? Is the biggest /smallest sub-group is being mapped to the smallest / biggest subfield? Let’s find out:

I’ll start with the trivial subgroup: $\{id_K\}$ . By definition:

\{id_K\}^*=K^{\{id_K\}}=\{a\in K|\forall \sigma\in\{id_K\}:\sigma(a)=a\}

Those are all the elements of $K$ that the identity fixes. But the identity fixes all of the elements, thus: $\{id_K\}^*=K^{\{id_K\}}=K$ .

On the other hand, the ‘biggest’ sub-group is the group itself: $\text{Gal}(K/F)$ . It is being mapped to it’s fixed field:

\text{Gal}(K/F)^*=K^{\text{Gal}(K/F)}=\{a\in K|\forall \sigma\in\text{Gal}(K/F):\sigma(a)=a\}

Ok, by definition of automorphism of an extension we know that every $a\in F$ has to be mapped to itself. So $F\subseteq \text{Gal}(K/F)^*$ .

However, maybe some elements outside of $F$ are mapped to themselves in all the automorphisms?

So we cannot say that $F = \text{Gal}(K/F)^*$ (and there are cases where the equality is false). Now the question is, when can we? Well, that’s actually one of the biggest questions in this theory! We’ll need to work a bit more in order to answer it.

I think this diagram concludes what we’ve discussed so far pretty well – It kind of explains the whole thing in a really easy way.

There is some sort of an asymmetry here – when mapping fields to groups, everything works fine. However, when mapping groups to fields – that’s not always the case!

Inversion of the order

Suppose that $F\subseteq L\subseteq L^\prime\subseteq K$ . What can we say about the groups:

L^\circ=\text{Gal}(K/L)\ \ \ ,\ \ \ {L^{\prime}}^\circ=\text{Gal}(K/L^\prime)

Notice that if $\sigma \in {L^{\prime}}^\circ$ , then $\sigma(a)=a$ for every $a\in L^\prime$ . However, since $L$ is a sub-field of $L^\prime$ then $\sigma$ is in particular, fixes the elements of $L$ as well. Thus, $\sigma\in L^\circ$ , which shows that ${L^\prime}^\circ\subseteq L^\circ$

So, as it turns out, we found out that this map is reversing the order of the inclusion:

L\subseteq L^\prime\Rightarrow {L^\prime}^\circ\subseteq L^\circ

Is the opposite map reverses the order as well? Let’s find out:

Suppose that $\{id_K\}\subseteq H\subseteq H^\prime\subseteq \text{Gal}(K/F)$ .

Let $a\in {H^\prime}^*$ be some element of the fixed field. Thus, every automorphism in $H^\prime$ maps $a$ to itself. However, every automorphism of $H$ is in particular, an automorphism of $H^\prime$ ( $H$ is a subset of $H^\prime$ ). Therefore, $a\in H^*$ . And by that we proved that this map is reversing the order as well:

H\subseteq H^\prime\Rightarrow {H^\prime}^*\subseteq H^*

Back and forth

What happens if we pick a subgroup $H$ , map it to a field $H^*$ and then, map it back to a group $H^{*\circ}$ .

My first thought was that it would be nice that we will return to the original subgroup again. i.e. $H=H^{*\circ}$ . However, that’s a pretty strong condition that means the maps are inverting each other – is that true? We need to verify it:

By definition, $H^*=K^H$ is the field of all the elements that are mapped to themselves under automorphisms in $H$ .

Now what are the automorphism in the sub-group $H^{*\circ}$ ? Those are all the automorphism that fix the elements of $H^*$ .

If $\sigma\in H$ , then for every $a\in H^*$ , $\sigma(a)=a$ . Therefore, $\sigma\in H^{*\circ}$ . This gives us:

H\subseteq H^{*\circ}

Which is the same as:

H\subseteq \text{Gal}(K/K^H)

On the other hand, If $\sigma\in H^{*\circ}$ , then it fixes all the elements of $H^*$ . However, this does not necessarily mean that $\sigma \in H$ . There could be other automorphisms outside of $H$ that fix the elements that the automorphisms in $H$ fix.

So again, we were only able to prove inclusion… It tells us that when we go ‘back and forth’, we can only make the subgroup bigger.

Try it yourself, to get in a similar way that:

L\subseteq L^{\circ*}

Which is the same as:

L\subseteq K^{\text{Gal}(K/L)}

We can now use the information in the above and the fact that those maps are order-inversing to get:

H^*\subseteq (H^{*})^{\circ*}=H^{*\circ*}=(H^{*\circ})^*\underset{H^{*\circ}\supseteq H}{\subseteq} H^*

We got that $H^*=H^{*\circ*}$ and that’s a little comforting, Note that this is the same as $K^H=K^{\text{Gal}(K/K^H)}$ . Moreover:

L^\circ\subseteq(L^\circ)^{*\circ}=L^{\circ*\circ}=(L^{\circ*})^{\circ}\underset{L^{\circ*}\supseteq L}{\subseteq} L^\circ

Great, we also know that $L^\circ=L^{\circ*\circ}$ , which is equivalent to $\text{Gal}(K/L)=\text{Gal}(K/K^{\text{Gal}(K/L)})$ .

(I think it’s pretty clear now why I prefer working with starts and circles – it’s a lot less confusing…)

Those facts show us that if you apply a map once, and then, you ‘go back and forth’ you will return to where you were after you applied the first map. In other words – the maps are inverting each other on the images.

So our question is – “Is any sub-field / sub-group an image of some sub-group / sub-field?” If the answer is yes, then the maps are inverting each other, and we will also get that $\text{Gal}(K/F)^*=K^{\text{Gal}(K/F)}$ and that solves all of our problems!

Summary

So we’ve met the correspondence, and found exactly what we want from an extension – we would that every subfield of it is an image of some subgroup in it’s galois group, and vice versa.

There are indeed such extensions – and they have a special name? Can you guess what can it be? Well, it’s not a big suprise but such extensions are called Galois extensions. And those extensions have a lot of amazing properties, as we’ll see in the next post.

I know this correspondence may seem very confusing at first, and the fact that it’s inversing the order makes it even more confusing! I found that the only way to actually understand it, is just go over it again and again, maybe try playing with concrete field extensions (I’ll do such examples myself in the future), you have to put some work in it – and that’s no wonder, those correspondences came straight from the mind of the brilliant Galois!