Fractional Ideals, the Class Group, and the Class Number

Fractional ideals make the ideals into a group

Ordinary ideals multiply, but they have no inverses — you can't divide. The fix is to enlarge: a [[fractional-ideal|fractional ideal]] of O_K is a nonzero finitely generated O_K-submodule of K, equivalently (1/d)·I for an ordinary ideal I and some d in O_K. In a Dedekind domain every nonzero fractional ideal is invertible: its inverse is { x in K : x·a ⊆ O_K }. So the nonzero fractional ideals form an abelian group J_K under multiplication, with identity O_K.

By unique factorization of ideals, J_K is in fact the free abelian group on the prime ideals: every fractional ideal is a finite product ∏ p^(n_p) with integer exponents n_p (negative exponents allowed). This makes computation in J_K bookkeeping with exponent vectors.

Quotient by the principal ideals: the class group

Inside J_K sit the principal fractional ideals P_K = { (x) : x in K* }, which form a subgroup (since (x)(y) = (xy)). The quotient Cl(K) = J_K / P_K is the [[ideal-class-group|ideal class group]]. Two ideals are in the same class iff they differ by multiplication by a principal ideal — i.e. iff a = (x)·b for some x in K. This is a quotient-style construction, but of groups.

Finiteness and a worked class-number computation

The deep theorem is that Cl(K) is finite — proved via the Minkowski bound, a geometry-of-numbers estimate guaranteeing every ideal class contains an integral ideal of norm at most M_K. So to find Cl(K) you only factor the finitely many primes p ≤ M_K and track the classes of the prime ideals above them.

1. Compute the Minkowski bound M_K from the discriminant and the number of real/complex embeddings.
2. List all rational primes p ≤ M_K; factor each (p) into prime ideals of O_K.
3. Cl(K) is generated by the classes of these prime ideals; find relations among them by spotting principal ideals (small-norm elements).

K = Q(sqrt(-5)),  O_K = Z[sqrt(-5)],  disc = -20.
Minkowski bound  M_K = (4/pi) * sqrt(20) / 4 ≈ 2.84.
==> only need primes p <= 2,  i.e.  p = 2.

Factor (2):   x^2+5 ≡ x^2+1 ≡ (x+1)^2 (mod 2)
   ==>  (2) = p^2  with  p = (2, 1+sqrt(-5)).

Is p principal?  If p = (a+b sqrt(-5)) then N(p) = 2 = a^2+5b^2,
which has no integer solution.  So p is NON-principal: [p] != 1.
But  p^2 = (2)  is principal, so  [p]^2 = 1  in Cl(K).

Cl(K) is generated by [p] of order 2.
==>  Cl(Q(sqrt(-5))) ≅ Z/2Z,   class number  h_K = 2.

Reading: h_K = 2 != 1 confirms Z[sqrt(-5)] is NOT a UFD,
exactly matching the 6 = 2*3 = (1+s)(1-s) failure.