# Number Theory Lecture 4

This lecture ( NumTh4 ) was given on 9/8/09

### Excerpts

To do: find a finite extension of number fields that isn’t Galois.

Continuing from last time with the frame work $\mathbb{Z} \to O_F \to O_E$.  Say a $p$ downstairs has $m$ primes $P_i$ upstairs, then if its a Galois extension, then

$|G(E/F)| = m \cdot f \cdot e$

The action of the Galois group is transitive on the $P_i$ so the orbit has size m and basic group theory say $|G(P_i)| = e\cdot f$.  It then become relatively easy to prove that for $\eta \colon G(P_i) \to Gal(F_{P}/F_p)$, $|\ker \eta| = e$; this is the inertia subgroup.  Note $F_p \to F_P$ is a finite extension of finite fields and is generated by the Frobenious morphisms $Frob_{P_i}$, so it can be checked directly that its a Galois extension.

In the case $e = 1$ then $\eta$ is an isomorphism and define $(E/F, P_i) = \eta^{-1}(Frob_{P_i})$.  Now replacing $P_i$ with $P_j$ another prime lying over $p$ conjugates $(E/F, P_i)$ by an element of the Galois group: $(E/F, P_j) = g(E/F, P_i)g^{-1}$.  So associate to $p$ the resulting conjugacy class

$Frob_p = \{ g(E/F, P_i)g^{-1} \} \subset G(E/F)$

its called the Frobenius class at $p$.

There was some stuff on the compositum of fields and as a result I have this question: If $K/E/F$ are all Galois, and if I take any $g \in Gal(K/F)$, this clearly by restricting I get a map $g \colon E \to K$, why does this land in $E$?

Basically because everything is Galois: if $a \in E$ then it satisfies some polynomial $p(a) = 0$ with coefficients in $F$, so the action of $Gal(K/F)$ permutes the roots of $p(x)$ so sends $a$ to another element in $E$; $a, g(a)$ both satisfy the same polynomial equation, so they are in the same extension field where $g \in Gal(K/F)$.

Thm (Tchebotarev Density thm): Let $E/F$ be a Galois ext. of number fields.  Let $C \subset G(E/F)$ be a conjugacy class, then the set of primes $p \subset O_F$ such that $C = Frob_p$ has density $|C|/|Gal(E/F)|$.