on the Axiom of Choice and the Spectrum of Rings

consider the following theorem, when $R$ is a commutative ring with a non-zero identity:

A ring $R$ is zero-dimensional if and only if $\mbox{Spec(R)}$ is Hausdorff.

The proof uses the Axiom of Choice. So, I am wondering if this theorem is equivalent to the Axiom of choice or not.

edited Feb 15 at 13:18

Joel David Hamkins
76.7k●8●192●355

asked Feb 15 at 11:05

chatish
2,203●4●20

What is the motivation of this question? – Martin Brandenburg Feb 15 at 13:31

Choice is often used in such situation for the sole purpose of giving the spectrum points. It can very often be avoided. What is "the proof" you are referring to? – Andrej Bauer Feb 15 at 13:39

This depends on precise definitions. How do you define zero-dimensional ring? Every prime ideal is maximal? In terms of lengths of chains of prime ideals? – François G. Dorais♦ Feb 15 at 14:16

@François: By zero-dimensional I mean every prime ideal is maximal. – chatish Feb 15 at 18:42

@Martin: I am interested in algebraic equivalents of AC. So as an old habit whenever I see an application of AC I ask myself is that necessary to use AC ? In this particular problem, I don't know the answer. – chatish Feb 15 at 18:46

show 1 more comment