# Number Theory lecture 3

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

### Excerpts:

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.