§ Galois extension

• Let $M$ be a finite extension of $K$. Let $G = Gal(M/K)$. Then $M$ is said to be Galois iff:

1. $M$ is normal and separable (over $K$).
2. $deg(M/K) = |G|$. We will show that $|G| \leq deg(M/K)$. So $M$ is "symmetric as possible" --- have the largest possible galois group
3. $K = M^G$ [The fixed poits of $M$ under $G$]. This is useful for examples.
4. $M$ is the splitting field of a separable polynomial over $K$. Recall that a polynomial is separable over $K$ if it has distinct roots in the algebraic closure of $K$. Thus, the number of roots is equal to the degree.
5. $K \subseteq L \subseteq M$ and $1 \subseteq H \subseteq G$: There is a 1-1 correspondece between $L \mapsto Gal(M/L)$ [NOT $L/K$! ], and the other way round, to go from $H$ to $M^H$. This is a 1-1 correspondence. $L$ is in the denominator because we want to fix $L$ when we go back.

• We'll show (1) implies (2) implies (3) implies (4) implies (1)

§ (4) implies (1)

• We've shown that splitting fields of sets of polynomials are normal, so this case is trivial.
• Just to recall the argument, let $M$ be the splitting field of some separable polynomial $p \in K[x]$ over $K$. We need to show that $M$ is normal and separable.
• It's separable because it only adds elements to new elements to $K$ which are the roots of $p$, a separable polynomial. Thus, the minimal polynomial of new elements will also be separable, and the base field is trivially separable.
• We must now show that $M$ is normal. We proceed by induction on degree. Normality is trivial for linear polynomials, if $M$ contains one root it contains all of the roots (the only one).
• Let $q \in K[x]$ have a root $\alpha \in M$. If $\alpha \in K$, then divide by $(x - \alpha)$ and use induction. So suppose $\alpha \not \in K$.
• Then $\alpha$ is some element that is generated by the roots

• Borcherds lecture