- 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 $|Gal(M/L)| = [M:L]$.