A Universe of Sorts

§ Fundamental theorem of galois theory

Let $K \subseteq M$ is a finite galois extension (normal + separable), then there a 1:1 correspondence between intermediate fields $L$ and subgroups of the galois group $G = Gal(M/K)$ .
Recall that a finite extension has finitely many subfields iff it can be written as an extension $K(\theta)/K$ . This is the primitive element theorem.
We send $L \mapsto Gal(M/L)$ , the subgroup of $Gal(M/K)$ that fixes $L$ pointwise.
We send $H$ to $fix(H)$ , the subfield of $K$ that is fixed pointwise.

It is clear that $H \subseteq Gal(M/Fix(H))$ , by definition, since every element of $H$ fixes $Fix(H)$ pointwise.
To show equality, we simply need to show that they are the same size, in terms of cardinality.
So we will show that $|H| = |Gal(M/Fix(H))|$ ..

It is clear that $L \subseteq Fix(Gal(M/L)))$ , by definition, since every element of $Gal(M/L)$ fixes $L$ pointwise.
To show equality, we simply need to show that they are the same size.
Here, we measure size using $[M:L]$ . This means that as $L$ becomes larger, the "size" actually becomes smaller!
However, this is the "correct" notion of size, since we will have the size of $L$ to be equal to $Gal(L/K)$ .
As $L$ grows larger, it has fewer automorphisms.
So, we shall show that $[M:L] = [M:Fix(Gal(L/K))]$ .

Rather than show that the "round trip" equalities are correct, we will show that the intermediates match in terms of size.
We will show that the map $H \to Fix(H)$ is such that $|H| = [M:H]$ .
Similarly, we will show that the map $L \mapsto Gal(L/K)$ is such that $[M:K] = |L|$ .
This on composing $Gal$ and $Fix$ show both sides shows equality.

Consider the map which sends $H \mapsto Fix(H)$ . We need to show that $|H| = [M:Fix(H)]$ .
Consider the extension $M/Fix(H)$ . Since $M/K$ is separable, so in $M/Fix(H)$ [polynomials separable over $K$ remain separable over super-field $Fix(H)$ ]
Since the extension is separable, we have a $\theta \in M$ such that $M = Fix(H)(\theta)$ by the primitive element theorem.
The galois group of $M/Fix(H) = Fix(H)(\theta)/Fix(H)$ must fix $Fix(H)$ entirely. Thus we are trying to extend the function $id: Fix(H) \to Fix(H)$ to field automorphisms $\sigma: M \to M$ .
Since $M/K$ is normal, so is $M/Fix(H)$ , since $M/K$ asserts that automorphisms $\sigma: M \to \overline K$ that fix $K$ stay within $M$ . This implies that automorphisms $\tau: M \to \overline K$ that fix $Fix(H)$ stay within $M$ .
Thus, the number of field automorphisms $\sigma: M \to \overline M$ that fix $Fix(H)$ is equal to the number of field automorphisms $M \to M$ that fix $Fix(H)$ .
The latter is equal to the field of the separable extension $[M:Fix(H)]$ , since the only choice we have is where we choose to send $\theta$ , and there are $[M:Fix(H)]$ choices.
The latter is also equal to the size of the galois group

We wish to show that $[M:L] = |Gal(M/L)|$
Key idea: Start by writing $M = L(\alpha)$ since $M$ is separable by primitive element theorem. Let $\alpha$ have minimal polynomial $p(x)$ . Then $deg(p(x))$ equals $[M:L]$ equals number of roots of $p(x)$ since the field is separable.
Next, any automorphism $\sigma: M \to M$ which fixes $L$ is uniquely determined by where it sends $\alpha$ . Further, such an automorphism $\sigma$ must send $\alpha$ to some other root of $p(x)$ [by virtue of being a field map that fixes $L$ , $0 = \sigma(0) = \sigma(p(\alpha)) = p(\sigma(\alpha))$ ].
There are exactly number of roots of $p$ (= $[M:L]$ ) many choices. Each gives us one automorphism. Thus $|Gal(M/L)| = [M:L]$ .