A weird Dedekind domain

Here is a random question that I was wondering about at some point (just out of curiosity):

Question. Does there exist a Dedekind domain R that has infinitely many points whose residue field has characteristic 0 and finitely many points (at least one) whose residue field has positive characteristic?

I don’t fully recall why this question came up, but it had something to do with a similar property that was satisfied by an object involved in the definition of the Fargues–Fontaine curve. However, we don’t need such deep theory to discuss this elementary commutative algebra question.

Lemma. Such a Dedekind domain exists.

Proof. We will construct R as a localisation of \mathbb Z[x]. Recall that prime ideals of \mathbb Z[x] come in four types:

  • The generic point (0), of height 0;
  • Height 1 primes (p) for every prime p \in \mathbb Z;
  • Height 1 primes (f) for every irreducible polynomial f \in \mathbb Z[x];
  • Height 2 closed points (p,f) for p \in \mathbb Z a prime and f \in \mathbb Z[x] a polynomial whose reduction \bar f \in \mathbb F_p[x] is irreducible.

We first localise at (p,x); then the only primes we have left are the ones contained in (p,x). This is the generic point, the height 1 prime (p), the height 1 primes (f) where f \in \mathbb Z[x] is an irreducible polynomial whose constant coefficient is divisible by p (e.g. (x), (x-p), \ldots), and exactly one height 2 prime (p,x).

Next, we invert x; denoting the resulting ring by R. This gets rid of all prime ideals containing x, which are (p,x) and (x). In particular, there are no more height 2 primes, so R is 1-dimensional. It is a normal Noetherian domain because it a localisation of a normal Noetherian domain. Therefore, R is a Dedekind domain.

The primes of R are (0) with residue field \mathbb Q(x); the prime (p) with residue field \mathbb F_p(x); and the prime ideals (f) with f \neq x a polynomial whose constant coefficient is divisible by p, whose residue field is a finite extension of \mathbb Q. \qedsymbol

Remark. The ring R we constructed is essentially of finite type over \mathbb Z (a localisation of a finite type \mathbb Z-algebra). There are no examples of finite type over \mathbb Z, because by Chevalley’s theorem the image of \operatorname{Spec} R \to \operatorname{Spec} \mathbb Z would be constructible. However, no set of the form \{(0)\} \cup S for S \subseteq \operatorname{Spec} \mathbb Z finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type \mathbb Z-algebra has residue characteristic p > 0.)

I was a little surprised that we can make examples when we drop the finite type assumption. I don’t know if this type of ring has ever been used for anything.

1 thought on “A weird Dedekind domain

  1. Your example is basically Z_p((T)) without any completions. We should have thought of Z_p((T)) when we were talking about this before. Also, I think the inclusion of your example in Z_p((T)) induces a bijection on Spec’s, by Weierstrass Preparation.

Leave a Reply

Your e-mail address will not be published. Required fields are marked *