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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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    $\begingroup$ What is the motivation of this question? $\endgroup$ Feb 15, 2013 at 13:31
  • $\begingroup$ 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? $\endgroup$ Feb 15, 2013 at 13:39
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    $\begingroup$ 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? $\endgroup$ Feb 15, 2013 at 14:16
  • $\begingroup$ @François: By zero-dimensional I mean every prime ideal is maximal. $\endgroup$
    – user30230
    Feb 15, 2013 at 18:42
  • $\begingroup$ @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. $\endgroup$
    – user30230
    Feb 15, 2013 at 18:46

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