Number Theory Lecture 1

This lecture was given on August 27th 2009


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.



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