## § 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.

#### § $H = Gal(M/Fix(H))$

• 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 $|H| = |Gal(M/Fix(H))|$..

#### § $L = Fix(Gal(L/K))$

• 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 $[M:L] = [M:Fix(Gal(L/K))]$.

#### § Proof Strategy

• 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.

#### § Part 1: $H \to Fix(H)$ preserves size

• 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

#### § Part 2: $L$ to $Gal(M/L)$ preserves size

• 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 $|Gal(M/L)| = [M:L]$.