

Let be a number field, and let be the ring of integers of . To employ our geometric intuition, as the Lenstras did on the cover of [LL93], it is helpful to view as a onedimensional scheme
Ideals were originally introduced by Kummer because, as we proved last Tuesday, in rings of integers of number fields ideals factor uniquely as products of primes ideals, which is something that is not true for general algebraic integers. (The failure of unique factorization for algebraic integers was used by Liouville to destroy Lamé's purported 1847 ``proof'' of Fermat's Last Theorem.)
If is a prime number, then the ideal of factors uniquely as a product , where the are maximal ideals of . We may imagine the decomposition of into prime ideals geometrically as the fiber (with multiplicities).
How can we compute in practice?
> R<x> := PolynomialRing(RationalField()); > K<a> := NumberField(x^5 + 7*x^4 + 3*x^2  x + 1); > OK := MaximalOrder(K); > I := 2*OK; > Factorization(I); [ <Principal Prime Ideal of OK Generator: [2, 0, 0, 0, 0], 1> ] > J := 5*OK; > Factorization(J); [ <Prime Ideal of OK Two element generators: [5, 0, 0, 0, 0] [2, 1, 0, 0, 0], 1>, <Prime Ideal of OK Two element generators: [5, 0, 0, 0, 0] [3, 1, 0, 0, 0], 2>, <Prime Ideal of OK Two element generators: [5, 0, 0, 0, 0] [2, 4, 1, 0, 0], 1> ] > [K!OK.i : i in [1..5]]; [ 1, a, a^2, a^3, a^4 ]Thus is already a prime ideal, and
The exponent of in the factorization of above suggests ``ramification'', in the sense that the cover has less points (counting their ``size'', i.e., their residue class degree) in its fiber over than it has generically. Here's a suggestive picture: