------------------------------------------------------------
581b -- finiteness of the class group

  * Make sure to mention that 581d will be on a number theory topic
    this week -- elliptic curve computation in Sage.

  * Announce ECC is next week: http://2010.eccworkshop.org/

  * Plan for course:

     * Prove the main theorem of the course (starts this week, may
       continue to next week): finiteness of class group; Dirichlet's
       unit theorem (both for ring of integers of number fields)

     * Key theorems and structure that make it possible to compute (next week): 
        computing O_K, factoring p*O_K, computing class group
         (and examples of how to use Sage to compute these things...)

     * Local structure (and Galois representations): 
          Theorems about decomposition and inertia groups,
          Definition of Frobenius elements, zeta functions, L-series

     * Adeles, ideles, and finiteness of the class group: a language
       that you must know to understand a lot of number theory
       literature.

     * Class field theory: statements using both ideal and idelic language 

     * If time permits -- Automorphic forms and representations, the
       Langlands program, what did that new Fields Medalist do? (prove
       the "Fundamental Lemma") (Adeles are required to talk about
       this stuff...)
          

The class group of a Dedekind domain
------------------------------------

Recall:

  R = Dedekind domain (noetherian, krull.dim(R)<=1,
                       R integrally closed in K=Frac(R))

  Div(R) = group of fractional ideals

Using "Div" since like divisors on a curve; notation only good because
of following theorem, we proved completely last week:

Theorem: Div(R) is a free abelian group (free on the nonzero prime ideals of R).

Defn: A *principal fractional ideal* is one of the form:
               I = alpha*R for 0 != alpha in K.

Defn: Prin(R) = group of principal fractional ideals

Defn: Cl(R) = Div(R) / Prin(R)   <-------  so it's an abelian group

Prop: Prin(R) isom K^* / R^*. 
Proof: K^* --> Prin(R), by definition.  
     kernel = {u in K^* : u*R has no prime factors} = {u in K^* : u*R = R } = R^*
   Note: if u in K^* with u*R = R, then u in R, since 1 in R, so u=u*1 in R.


Thus exact sequence:

   1 --> R^* --> K^* --> Div(R) --> Cl(R) --> 1


Our main goal is to prove the following *deep theorem* (the deepest in
this class?):

Theorem: If R = O_K is ring of integers of number field, then Cl(R) is *finite*.

Strategy of proof:
  
  *  (easy) Use maps K \--> C and log to embed O_K into some Euclidean
    space R^n.

  * (hard) Use a geometric argument ("geometry of numbers") to show that each
    ideal class in Cl(R) contains an ideal I with
           Norm(I) <= (4/pi)^s*(n!/n^n)*sqrt(|d_K|)
    Here, Norm(I) = #(R/I) and d_K = "discriminant" of K.

  * (trivial) Observe that there are finitely many ideals of bounded norm.

Remark: Above theorem not true in general!  Even "Norm(I)" doesn't
make sense in general, since R/I need not be finite, e.g., if R=Q[x]
and I=(x), then R/I = Q is infinite.  Also, whatever d_K is, it
wouldn't make sense in general either.

Example in which Cl(R) is not finite. 

  R = C[x,y]/(y^2 - (x^3 + 1)),    C = complex numbers

The nonzero prime ideals of R are the ideals

        P_{a,b} =  (x-a, y-b)

where (a,b) is a complex point on the affine curve y^2 = x^3+1.

A principal fractional ideal is got by a taking any rational function
alpha(x,y) = f(x,y)/g(x,y), with f,g polys, and considering the
fractional ideal it generates.  We think about this fractional ideal
in terms of its prime factorization (divisor!), so

    alpha*R = prod P_{a_i,b_i} / prod Q_{c_j, d_j}

where the (a_i,b_i) are the zeros of f(x,y) and (c_j,d_j) the poles,
with appropriate multiplicities.

Claim:
    P_{a,b} is not in Prin(R)

Proof: If alpha=f/g and alpha*R = P_{a,b}, then alpha is a rational
function on y^2=x^3+1 which has no poles and one zero.  It thus
extends to a rational function of degree 1 on the projective closure C
of y^2=x^3+1, which would extend to an isomorphism to P^1 (see ch 1 of
Hartshorne), a contradiction since C has genus 1 and P^1 has genus 0.

NOTE: Totally false if we instead use a genus 0 curve, e.g., C[X].

----

To see that Cl(R) is infinite, take any nonzero point z = (a,b) 
and note that P_{a,b} defines a nonzero element of Cl(R).

  * The group law is compatible with the the group operation on Cl(R).
    (explain this)

  * For n=1,2,3,..., get P_{n*z} distinct primes that are all nonzero
    elements of Cl(R), so Cl(R) is infinity.  

In fact, Cl(R) is *uncountable*.
   
---

So there is something very special with R = O_K that we haven't seen
so far, which makes the classgroup small.

DISCRIMINANTS:

A key step in our argument is to introduce a notion of discriminant D of
O_K, and note that there are only finitely many ideals with norm at
most |D|.

Definition: Let a1, ..., an be a Q-basis for K.  Then 
   Disc(a1,...,an) = det(Tr(ai*aj))

Let R = ring of integers O_K of K.  

Definition:

    Disc(R) = Disc(a1,...,an) 

where a1,...,an any basis for R as a ZZ-module.  Often one
writes Disc(K) := Disc(R).

Remark: Disc(R) is nonzero and well defined.  (Exercise)

More generally, if S is any finite index subring of R, let Disc(S) be
the discriminant of any ZZ-basis for S.

Proposition: Disc(S) = Disc(R) * [R:S]^2

NORMS OF IDEALS:

Definition (Lattice Index): 

   L, M -- "lattices" in vector space V over Q
           so L, M are Z-module of rank dim(V) st Q*L=Q*M = V.

   [L:M] =defn= |det(A)| where A any linear automorphism st A(L)=M.

If M contained in L, then [L:M]=#(L/M) is usual index
In general, for any M,L,N:

    [L:N] = [L:M]*[M:N]

by basic properties of linear transformations and determinants.

Defn: 

   I - fractional ideal of R
   Norm(I) = [R : I] 

which is a nonzero rational number. 

Prop: 
  B = positive integer
  Then set of integral ideals I in R with norm(I) <= B is finite.

Proof:

An integral ideal I is a subgroup of R of index equal to the norm of
I. If G is any finitely generated abelian group, then there are only
finitely many subgroups of G of index at most B, since the subgroups
of index dividing an integer n are all subgroups of G that contain nG,
and the group G/nG is finite.