Number Theory Lecture 4

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


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)|.


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