## Galois is finished

Over the past month or so, two issues have arisen that have prevented me from posting regularly. The first is that free time has been scarce. The second is that I have grown slightly bored of Galois Theory. I find that my explanations are much worse than I’d like them to be when I’m uninterested in a subject, so let me put a halt to Galois theory for the time being. What we didn’t cover are normal and separable extensions, a precise statement of the fundamental theorem of Galois theory, and the insolvability of the quintic equation.

For the sake of completeness, below is a link to a collection of short proofs in Galois theory. I hope you find them useful.

http://www.math.ucla.edu/~ryanr/galois_theory.pdf

I haven’t decided what is to come next, or when it is to come. I would like to cover some category theory, but drawing diagrams is time consuming. I’ll take some time to figure out what I want to do, and start posting when I feel confident I can get to a satisfactory point within the topic.

## Galois Connection

So we mentioned two extremely important concepts to Galois theory already.

1. The Galois group of a field extension $E$ of a field $k$. This is the group of automorphisms of $E$ that fix $k$. We’ll denote this with $\Phi(E/k)$
2. The fixed field of a subgroup $H$ of the automorphism group $\mbox{Aut}E$. This is the field of all elements fixed by every automorphism in $H$. We’ll denote this with $\Psi(H)$.

If you had to pick the coolest possible theorem that you could involving $\Phi$ and $\Psi$, what would it be? Without a doubt, it would be that $\Phi(\Psi(H))=H$ and $\Psi(\Phi(E/k))=k$. Is it true? Well, no, but mostly yes. Confused? Good. Regardless, we aren’t going to prove it today.

Take a big field $E$, and a subfield $k$. Look at all the automorphisms of $E$ which fix $k$. Then look at all the elements fixed by these automorphisms. We know immediately that all of $k$ is fixed by all of these automorphisms. After all, we chose the automorphisms to be the things that fix $k$. So this tells us that $\Psi(\Phi(E/k)\ge k$. The theorem would give us equality, meaning that $k$ is exactly what is fixed by the automorphisms of $k$. There is a similar statement about the automorphism groups for the other statement in the “theorem.”

As I said, we won’t prove it today. After all, it isn’t even true. But we will prove a slightly weaker result:

$\Phi(E/\Psi(\Phi(E/k)))=\Phi(E/k)$.

So the first result which isn’t always true says that going there and back gets you back to where you started. The result we’re about to prove says that going there, back, and then there again is the same as just going there. See why? This picture should show you the problem.

Ready for some trickery? Here goes. We just saw an argument that $\Psi(\Phi(E/k))\ge k$. Applying $\Phi$ to this means I’m asking about the Galois group. As mentioned previously, if the subfield I require to be fixed is bigger, then the automorphism group is smaller, so $\Phi(E/\Psi(\Phi(E/k)))\le \Phi(E/k)$. But, I also know that $\Phi(E/\Psi(H))\ge H$. If I let $H$ be $\Phi(E/k)$, I get $\Phi(E/\Psi(\Phi(E/k)))\ge \Phi(E/k)$. Now I have inequalities in both directions, so I’m done. Once again, the result about groups that $\Psi(\Phi(E/\Psi(H)))=\Psi(H)$ is basically the same idea.

It feels like I did basically nothing. This is true. You can get very close to the fundamental theorem of Galois theory while doing virtually no work. There’s more to be done if we want to really have equality (in a there-and-back sense), and we’ll get to it soon enough.

## Galois Group

Last time we started with a subgroup of the automorphism group of a field $k$, and asked about it’s fixed field, the field of elements which are fixed by every automorphism in the subgroup. We noticed that if we took a big group, we would get a small fixed field. If we picked a small group, we would get a big fixed field.

Today I want to do the same thing but in the other direction. Instead of starting with a subgroup and finding it’s fixed field, I want to start with a field and compute the group that fixes it. Of course, we always need to do this relative to some base field, so really what we’re asking about is the $E$ is a field extension of $k$, what are the automorphisms of $E$ which fix $k$. We think of $E$ as being set in stone, and $k$ as being allowed to vary.

This collection of automorphisms is called the Galois Group of the extension $E/k$ (this is notation for saying that $E$ is an extension of $k$). We denote it by $\mbox{Gal}(E/k)$. More techncially,

$\mbox{Gal}(E/k)=\{\sigma\in\mbox{Aut}E\mid\sigma|_k=\mbox{id}_k\}$

Of course, we need to really check that this is a group, but it’s not so bad. If two automorphisms both fix $k$, then so does their composition (first the first one fixes $k$, then the second fixes $k$). It’s similarly easy to check for inverses and identity. This proof is neither difficult nor enlightening. I’d rather get to the math that is both.

Let’s work out a simple example. Let $k$ be a field. What is $\mbox{Gal}(k/k)$? These are the automorphism of $k$ which fix $k$. There’s exactly one of those. The identity (or lame) automorphism.

There’s something else you may notice. $k/k$ is a particularly small extension. However, I’m thinking of it as $E/k$ where $E=k$. Furthermore, I’m thinking of $E$ being set in stone. So then the subfield $k$ is particularly big inside $E$. This is the correct notion, because, I really do want to think of the bigger field as unchanging. So a very big subfield (as big as they come) gives us a very small Galois group (as small as they come). This is true in general too. Small subfields give big Galois groups and big subfields give small Galois groups.

This idea of matching groups with fields in such a way that big groups get matched with small fields and vice versa is incredibly important and incredibly powerful. We can now start exploring just how deep the rabbit hole goes.

## Fixed fields

If I have a field $k$ and a field automorphism $\sigma$, I can ask about the fixed points of $\sigma$. That is, all of the $x\in k$ such that $\sigma(x)=x$. Notice that if $\sigma$ fixes $x$ and $y$, then $\sigma(x+y)=\sigma(x)+\sigma(y)=x+y$ and $\sigma(xy)=\sigma(x)\sigma(y)=xy$, so $\sigma$ fixes $x+y$ and $xy$. In fact, the set of fixed points of $\sigma$ is itself a field! I just checked two of the rules. You can check the rest.

This field is called the fixed field of $\sigma$, and it’s super super important. If we have two automorphisms, $\sigma$ and $\tau$, we can take the intersection of their fixed fields. The intersection of fields is always a field. What field is it? It’s the field consisting of everything fixed by both $\sigma$ and $\tau$.

But why stop at two automorphisms? What if I take a group $G$ of automorphisms of $k$. That has a fixed field. It’s called the fixed field of $G$. I could take the entire automorphism group of $k$, or I could just take a subgroup of it. The bigger the group I take, the more fields I intersect, so the smaller my fixed field is.

That’s an important idea. Big group means small fixed field. Small group means big fixed field. We’ll make this precise very soon.

## Automorphisms of Q

We’ve talked a bit about automorphisms, but we haven’t seen very much of them, so I wanted to do an example. What is $\mbox{Aut}(\mathbb Q)$ (the group of automorphisms of the field of rational numbers)?

It definitely contains the identity (a.k.a. lame) automorphism, since every field has that one. Is there anything else? No, and here is why. Let $\phi$ be an automorphism of $\mathbb Q$. That is, let $\phi\in\mbox{Aut}(\mathbb Q)$. Let $p/q$ be a rational number. Then we can write $p/q$ as

$p/q=\frac{\overbrace{1+\cdots+1}^{p\ times}}{\underbrace{1+\cdots+1}_{q\ times}}$

If we apply $\phi$,, we get that

$\phi(p/q)=\frac{\overbrace{\phi(1)+\cdots+\phi(1)}^{p\ times}}{\underbrace{\phi(1)+\cdots+\phi(1)}_{q\ times}}=\frac{p\cdot\phi(1)}{q\cdot\phi(1)}=p/q.$

This is to say, any field automorphism of a field extending $\mathbb Q$ fixes $\mathbb Q$. Another way to say this is that we can reach any number in $\mathbb Q$ just from 1 and operations which any field automorphism respects (namely addition and division).

This is to say, the automorphism group of the field of rationals is trivial!

## The group of field automorphisms

Recall that a field automorphism is a function that “respects all of the properties of fields.” For instance, if $f$ is a field automorphism, then $\phi(0)=0$, $\phi(1)=1$, $\phi(x+y)=\phi(x)+\phi(y)$, and $\phi (x y)=\phi(x)\phi(y)$.

I did leave out one rule of note $\phi(x^{-1})=\phi(x)^{-1}$. Why did I leave this one out? Because I can prove it from the other ones…

$1=\phi(1)=\phi(x\cdot x^{-1})=\phi(x)\phi(x^{-1})$,

so $\phi(x)^{-1}=\phi(x^{-1})$. So basically, anything remotely reasonable I can say about fields passes through the automorphism.

If I were to write down all of the automorphisms of a given field, what could I say about them? As I mentioned last time, there is always the trivial (a.k.a. lame) automorphism that doesn’t do anything. You plug in $x$, it gives you back $x$ exactly as is. We call this the identity automorphism. In general in mathematics, “identity” can be loosely translated as “lame.”

If $\phi$ is an automorphism on some field $F$, we required it to be a bijection (one-to-one and onto). So it has an inverse, which we will write as $\phi^{-1}$. This is a field automorphism as well. This one simply undoes whatever $\phi$ did.

If I have two field automorphisms $\phi$ and $\psi$, then I can compose them. You can verify that this is also an automorphism. I’ll show that it respects addition. The rest is just as easy.

$\psi(\phi(x+y))=\psi(\phi(x)+\phi(y))=\psi(\phi(x))+\psi(\phi(y))$

So we can compose automorphisms and take inverses. Oh, and there is an identity automorphism. Sound familiar? Such objects are called groups. It turns out, whatever structure we are talking about, the collection of automorphisms form a group. This will be an extremely important tool, so we should give it a name. We’ll denote the group of automorphisms of a field $F$ by $\mbox{Aut}(F)$.

## Field automorphisms

It’s worth, I suppose, talking about automorphisms in general. An automorphism is a “renaming” of stuff so that it keeps the same structure. The “auto” means self. The “morph” means shape. The “same” is suppressed, because it really comes from the word “isomorphism,” which means same shape. So this is like a self-similar shape… but not in a fractally sense.

What is required for an automorphism? Well, any special constants need to stay the same. And the operations need to be “preserved,” whatever that means. For a field, it means that 0 and 1 are fixed, and that you can move $+$ and $\cdot$ outsied the function. More specifically, $\phi$ is an automorphism of a field $F$ if it is a function $\phi:F\to F$ with

• $\phi(0)=0$
• $\phi(1)=1$
• $\phi(x+y)=\phi(x)+\phi(y)$
• $\phi(xy)=\phi(x)\phi(y)$
• $\phi$ is a bijection. That is, $\phi$ doesn’t send two different points to the same point, and it sends something to every point. (Bijection is a fancy word meaning “1-1 and onto.”)
Can you think of any field automorphisms? In any field, you always have the trivial automorphism which sends a point $x$ to $x$. It’s lame, but it counts.
What about something a bit more complicated? How about complex conjugation? What if $\phi:\mathbb C\to \mathbb C$ by $\phi(a+bi)=a-bi$ (for $a,b\in \mathbb C$)? Yes, this works. It’s not hard to check any of these rules, but I’m too lazy to do so.
How about something really crazy? Take the field $\mathbb Q(\sqrt2)$, and the automorphism $\psi$ defined by $\psi(a+b\sqrt 2)=a-b\sqrt 2$ (where $a,b\in \mathbb Q$). What? But that sends negative things to positive things? It doesn’t preserve the shape at all!
I assure you, it is a field automorphism. If you don’t believe me, check the rules yourself. The issue you’re potentially having, is that you’re thinking of $\mathbb Q(\sqrt 2)$ as sitting inside $\mathbb R$, and thinking of the ordering on $\mathbb R$. But alas, I said nothing about order. It’s damn near impossible, but try to forget everything you know about the structure of $\mathbb R$, because it will only misguide you.