Number Theory Lecture 1

This lecture was given on August 27th 2009

Reciporcity

Thm (Gauss): Let $p \ne q$ be odd primes.  Then $x^2 \equiv p \mbox{mod } q$ and $x^2 \equiv q \mbox{mod } p$ are either both soluble or insoluble except when p,q are both congruent to 3 mod 4, in which case exactly one is soluble.

Perspective: fix q and let p vary.  Then to determine ( for $x \in \mathbb{Z}$) if $p|x^2 - q$ it suffices (in most cases) to determine if $q|x^2 - p$

frob p and S(E)

Now consider some finite Galois extensions.  In let $E$ be the splitting field of a monic $f(x) \in \mathbb{Z}$ of degree n.  Let $a_1, ..., a_n$ be its roots (notice at this point we’ve picked a way to order the roots).  The Galois group $G = G(E/\mathbb{Q})$ permutes to roots and this action in fact gives an injection into $S_n$.

Say in $F_p[x]$ there is a factorization $f= f_1 \dotsm f_r$ into distinct irreducible factors (note this is an additional assumption on f(x)) of degree $n_i$.  So for each prime p we get a partition of n $(n_1, ..., n_r)$.  Denote this partition as $t_p$.  Recall in $S_n$ conjugacy classes are determined by cycle type = partition of n.  Its possible that two different conjugacy classes in G end up having the same cycle type when considered in the larger group $S_n$.  However when $p \not | descrim(f)$ (descriminant) its a thm that there is a unique conjugacy class in G that has cycle type $t_p$.  In other words, for such p, we get a unique conjugacy we’ll call $frob_p$.  In fact its also a (hard) thm that every conjugacy class of G is realized as some $frob_p$.

A prime p splits completely in E if it does not divide the descriminant of f(x) and $frob_p = \{1\}$.  This says in particular that $f(x)$ spits into linear factors in $F_p[x]$.  Define

$S(E) = \{ p | \mbox{ splits completely in } E\}$

Thm: The map from Galois number fields to subsets of primes given by $E \mapsto S(E)$ injective ( meaning if $E,E'$ are not isomorphic then $S(E) \ne S(E')$ as sets but could have nonempty intersection).

Langlands Stuff

Consider now faithful representations (injections) $G = G(E/\mathbb{Q}) \xrightarrow{\rho} GL(m, \mathbb{C})$.  For p not dividing the discriminant, $\rho$ gives a well defined semi-simple (meaning diagonalizable) conj. class.  For such p define (s is the complex variable)

$L_p(\rho, s) = \det [I - p^{-s}\cdot rho(Frob_p)]^{-1}$

For example when $f(x) = x$ (so $\rho = 1$ is trivial ) plugging everything in as above yields

$L_p(1, s) = \frac{1}{1 - p^{-s}}$

In particular taking the product over all primes give $\prod_p L_p(1,s) = \zeta(s) = \sum_n \frac{1}{n^s}$.  This motivates the global L function

$L(\rho, s) = \prod_{p \not | descrim} L_p(\rho,s)$

Fact: this converges to a mermorphic function on $\mathbb{C}$.  A theorem of Dirichlet says the $L_p$ can be recovered from the global L function.  Notice when $p \in S(E)$, $L_p(\rho, s)$ is a power of $L_p(1,s)$ so in some sense, the global L function “knows” when $p \in S(E)$.

So from a finite Galois extension we’ve obtained $L(\rho, s)$, a rather analytic object.  Langlands said this meromorphic function should in fact be associated to something else: an automorphic representation.