Number Theory Lecture 5

This lecture ( NumTh5 ) was given on 9/10/09

Excerpts

An aboslute value on a field $F$ is a fucntion $|-|\colon F \to \mathbb{R}_+$ s.t

1. $|x| = 0 \Longleftrightarrow x = 0$
2. $\forall x,y \in F, |xy| = |x|\cdot|y|$
3. $\forall x,y \in F, |x+y| \le |x| + |y|$

With these properties it can be checked the norm of 1 is 1 and $|1/x| = 1/|x|$ and you can define a metric via $d(x,y) = |x - y|$.

If you have a field with a discrete valuation $v$ then you can get a absolute value essentially by turning the valuation on its head: $c \in (0,1)$, then define $|x|_{v,c} = c^{v(x)}$.

An aboslute value is non archimedean if $|x+y| \le \max (|x|,|y|), \forall x,y$.  If this fails then $|-|$ is called archimedean.

Facts on this stuff

1. Any $F$ with an archimedean absolute value is isomorphic to a subfield of $\mathbb{C}$.
2. For a non archimedean absolute values, $\bar B_r(a) = \{|x - a|\le r\}$ is open and all triangles are isoceles.
3. If $F, |-|$ is non archimedean, then the induced topology makes $F$ totally disconnected.
4. With $F, |-|$ as above, $R = \{x| |x| \le 1\}$ is a ring with maximal ideal $M = \{|x|<1\}$.

Back to number fields

The basic story we’ve been following consists at looks at extension of number fields $F \to E$, and its ring of integers $O_F \to O_E$.  In the case of Galois extension there’s a nice story about a prime downstairs $p \subset O_F$ and the primes lying over it $P_i \subset O_E$.  This lead to looking at the Galois group $Gal(E/F)$ which permutes the $P_i$ and by looking at the stabilizer we get a map $\eta \colon Stab(P_i) \to Gal(F_P/F_p)$ which gives us a canonical Frobenius conjugacy class when the prime $p$ doesn’t ramify, and we can test for this with the descriminant.

Now taking $a \in F^*$ we can look at the factional ideal this generates and this factor into a product of prime ideals to various powers $(a) = \prod_i p_i^{n_i}$, so naturally a prime gives a valuation.  Taking $c = 1/|F_p| \in (0,1)$ provides an absolute value (non archimedean) on $F^*$.  Since everything gets turned on its head, its not hard to check that the ring $R,m$ associated to an absolute value in this case is just $O_{F,p}$.

Here is the upshot.  For a number field $F/\mathbb{Q}$, taking a prime in $O_F$ gives a non archimedean absolute value on $F$.  An embedding $\sigma \colon F \to \mathbb{C}$ given an archimedean absolute value.

Thm: All non archimedean and archimedean absolute values on $F$ arise in this way.  Further, two different embedding $F \to \mathbb{C}$ give rise to the same absolute value exactly when they differ by complex conjugation.

A place on a number field $F$ is an equivalence class of a non-trivial absolute value.  For example, $F = \mathbb{Q}[x]/(x^2 + 1)$ has two embeddings and one place.  Archimedean places are called infinite places, non archimedean are finite places.

Thm: Let $F$ a number field and $x \in F^*$, let $v$ index places on $F$.

$\prod_{v \infty} |x|_v^{e(v)} \cdot \prod_{v \mbox{ finite}} |x|_v = 1$

Let’s verify this for $F = \mathbb{Q}$.  For a nonzero rational, $r/s = \prod p_i^n_i \prod q_j^{-n_j}$.  Let $v$ be the place corresponding $p_i$, then $|x|_v = (1/p_i)^{n_i}$, in the case of negative exponent, we get $|x|_v = (1/q_j)^{-n_j} = q_j^{n_j}$.  Of course there is only one infinite place that is just the usual absolute value on $\mathbb{C}$.