# 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.