Number Theory lecture 3

This lecture (lecture3) was given on September 3rd 2009.


The lecture explain why the following story is true.  Consider an extension of number fields \mathbb{Q} \to F \to E.  There is also a map on the ring of integers \mathbb{Z} \to O_F \to O_E.  The convention is P \subset O_E is prime upstairs and p = P \cap O_F.  Its a standard result that p is prime.  Let q = p \cap \mathbb{Z}.  Imagine appropriate vertical arrows that would make the following diagram commute

\mathbb{Q} \to F \to E

\mathbb{Z} \to O_F \to O_E

q \to p \to P

It follows there is a chain of fields \mathbb{Z}/(q) \to O_F/p \to O_E/P = \mathbb{Z}/(q) \to F_p \to F_P.  Taking any lifts of a \in O_E/P and using that O_E is integral over \mathbb{Z}, it follows that the lift satisfies some polynomial relation with integer coefficients, reducing mod P gives a relation for a with coefficients in \mathbb{Z}/(q).  It follows that the latter residue fields above are finite extensions of a finite field, hence themselves finite fields.

Result: P|p; more generally in a Dedekind domain I \subset J iff J|I.

So the upshot is for any prime upstairs there is a unique prime downstairs, and for any prime p \subset O_F we can consider the ideal pO_E and since we’re in a Dedekind domain so there is some finite set of primes

pO_E = \prod_i P_i^{n_i}

Ramification index: e_i = e(P_i/p) = n_i.  The residue class degree: f_i = f(P_i/p) = [F_{P_i}: F_p].

Thm: fix a prime p downstairs, then [E:F] = \sum_i e_i f_i.

a prime downstairs ramifies if some e_i > 1, splitting completely means all e_i = 1.

Now thinking of E as an F vector spaces and looking at left multiplication as an F linear map, we can define determinant and trace maps.  In particular, for x \in E, let m_x denote the linear operator that is multiplication by x.  Then Tr_{E/F} \colon E \to F is x \mapsto trace(m_x) and N_{E/F}\colon E \to F is \det (m_x).  These things sends O_E to O_F. (prove this)

Proposition: B \colon E \times E \to F given by u,v \mapsto Tr_{E/F}(uv) is a nondegenerate symmetric form.

Now define the relative discriminant as the ideal D_{E/F} generated by \det B as the matrix for B ranges over all bases \subset O_E for E over F.

Thm: p ramifies in E if and only if p|D_{E/F}.

Here is the rest of the story.  A wonderful fact is that when F \subset E is Galois, then G = Gal(E/F) permutes the primes P_i above any given p \subset O_F, in a transitive manner.  In this case e = e(P_i/p), f = f(P_i/p) don’t vary as i varies.  Let G(P_i) = stab(P_i) under this action.  Naturally g \in G(P_i) given an automorphism E \to E preserving O_E and fixing P_i, hence an element of Gal(F_{P_i}/F_p); this is called the Artin map.

Proposition: \eta \colon G(P_i) \to Gal(F_{P_i}/p) is surjective.

A question

Consider F = \mathbb{Q} and E = F[i], this is a degree 2 extension and is the splitting field of x^2 + 1.  The poly has discriminant -4, and the relative discriminant as defined above seems to be -2, in either case they pick out the prime 2, but his seems to split completely 2 = (1 + i)(1 -i), what’s going on?

Actually nothing is going on, (2) = (1+i)(1-i) = i(1 - i)^2, so indeed the prime does ramify.


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