Integral Rings, The Norm of an Ideal

The Norm of an Ideal

Let R be a pid, and let S be the integral ring of R inside a finite extension E/F. Thus S is dedekind. Let the element p generate the prime ideal P in R. Let Q be a prime ideal in S lying over P. Let c be the residue degree, and let d be the ramification degree. What is the norm of the ideal Q?

First let R and S be dvrs, and let t be any generator of Q. If the valuation of p is 1, then the valuation of t is 1/d. This holds for every conjugate of t. Assume the extension is separable, so that the norm of t is a product of conjugates. Each conjugate has the same valuation. Multiply d conjugates together and the valuation becomes 1. The result is p times some unit in R. This happens c times over, for all possible embeddings of S into the algebraic closure of E. Therefore the norm of t is p^c times some unit in R. This holds for every t, so let |Q| = the associate class of p^c, or if you prefer, P^c.

In general, norm and localization should commute. (They do when taking the norm of an element.) Localize about Q, and both rings become dvrs. So |Q|, localized about P, yields P^c. Localize about any other prime and Q becomes the whole ring. Represent this by 1, or any other unit (having valuation 0), and take the norm, and find a unit in R. Thus |Q|, localized about any other prime, is R. The factorization of |Q|, in the dedekind domain R, has to be P^c.

We can say that norm and product commute by definition, and that's ok, but we would like this to be consistent with the traditional definition of norm. Assume x generates a principal ideal that is a product of prime ideals, that may not themselves be principal. Now we have the norm of x (in the traditional sense), and the norm of the ideal {x} generated by x. The key observation is that these definitions agree locally. Write {x} as a product of prime ideals, localize about any prime ideal (thus giving a dvr), represent ideals with elements, recall that norm and product commute when dealing with elements, and establish equality between the two definitions locally. Since the norm of x commutes with localization, and the norm of {x} commutes with localization by definition, pull back to find the norms must be the same globally. All is well.

This definition extends to fractional ideals in a natural way. The norm of Q^-1 is P^-c, and so on.

There are more general ways to define the norm of an ideal, but I'm not going to get into them here.