Dedekind Domains, Finitely Many Prime Ideals

One Prime Ideal

Let R be a dedekind domain with one prime ideal P. All ideals are powers of P, and are linearly ordered by containment. This makes R a valuation ring. The valuation group is isomorphic to the integers, as dictated by the powers of P. This makes R a dvr, and a pid.

Finitely Many Primes

Let R be a dedekind domain with finitely many prime ideals. Since prime ideals do not contain each other, ideals are not linearly ordered, and R is not a valuation ring, or a dvr. However, R is still a pid.

By criteria 2, it is enough to show each prime ideal is principal.

Let H be the sum of any two prime ideals. If this does not span 1 then drive H up to a maximal ideal M, which contains our two prime ideals. Now M contains two different maximal ideals, hence one of them is not maximal after all. This is a contradiction. The prime ideals are pairwise coprime, and if J is the product of these prime ideals, the chinese remainder theorem tells us R is isomorphic to the direct product of fields R/P_i for each prime ideal P_i.

In a generalization of the above, let H be the sum of two prime powers. If this does not span 1 then a maximal ideal M contains both prime powers, and since M is prime, it contains both prime ideals. Once again M contains two maximal ideals, a contradiction. If J is the product of prime powers, R/J is isomorphic to the direct product of the quotient rings R mod these prime powers.

Let P be one of the aforementioned prime ideals and let c lie in P-P². Apply the chinese remainder theorem to P², and all other prime ideals. Let g = c mod P², and 1 mod all other primes. If g lies in some other prime Q then g becomes 0 in the field R/Q, which is a contradiction. Thus g, and the ideal generated by g, does not lie in Q or any power of Q. If Q is part of the unique factorization of g*R, then g would lie in Q. Thus g generates a power of P.

If g were a unit it would be invertible mod P, yet it is 0 mod P. Thus g*R is a positive power of P, at least P¹. If g*R is P² or higher than g becomes 0 mod P². Yet g = c mod P², which is nonzero. Therefore g*R = P, and P is primciple. This holds for every prime ideal P, and R is a pid.