# 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.

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What is the motivation of this question? – Martin Brandenburg Feb 15 '13 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 '13 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 '13 at 14:16
@François: By zero-dimensional I mean every prime ideal is maximal. – user30230 Feb 15 '13 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. – user30230 Feb 15 '13 at 18:46