Minimal prime ideals and Axiom of Choice(revised version)

2012-06-03T19:24:48Z

From the page:

http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice,

I have found that The existence of prime ideals in commutative rings is equivalent to the Boolean Prime Ideal theorem. But $BPI$ is weaker than Axiom of choice. this means that The existence of prime ideal in commutative rings with unity is weaker than $AC$. Know Another Question came in my mind that I think It is a bit different from that one. Let me recall the following theorem:

Theorem:For any commutative unitary ring $R$ there exists a minimal prime ideal.

To proving this result One can pickup a prime ideal, and throw it in a maximal chain of prime ideals(Zorn's lemma) and then the intersection of this chain gives a minimal prime ideal at hand.

You Know that the existence of minimal prime ideal needs to apply one of the equivalences of $AC$ (i.e.Zorn's Lemma) But I didn't see anything about the converse of Above theorem.

STATEMENT:Is it true that The existence of minimal prime ideals in commutative unitary rings is equivalent to $AC$.

I am interested in To Know if the situation changes When we give minimality Condition on Prime ideals.

I think its better to recall the difference of two following situations in topology:

The statement "product of compact Hausdorff spaces is compact", does not implies $AC$

But

The statement "product of compact spaces is compact" is equivalent to $AC$

Answer by Will Sawin for Minimal prime ideals and Axiom of Choice(revised version)

2012-06-03T20:03:36Z

Suppose I have a set of disjoint, nonempty sets, and I want to choose one element from each. Consider the free polynomial ring generated by all the elements of all the sets, then take the quotient by the ideal generated by $xy$ for each pair $x$ and $y$ different elements in the same set. Any prime ideal must contain all but one element from each set.

We need to show that a minimal prime ideal does not contain all the elements from any set. Then a minimal prime ideal will give us a choice function. We can take the minimal prime to be generated by the elements, since every prime ideal contains a prime ideal generated by elements. Then remove one element from a set entirely contained in the prime ideal. The ideal will still be prime, and smaller, this is a contradiction.

So minimal primes give a choice function.

Minimal prime ideals and Axiom of Choice(revised version) - MathOverflow

Minimal prime ideals and Axiom of Choice(revised version)

Answer by Will Sawin for Minimal prime ideals and Axiom of Choice(revised version)