Minimal prime ideals and Axiom of Choice(revised version) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:55:39Z http://mathoverflow.net/feeds/question/98731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98731/minimal-prime-ideals-and-axiom-of-choicerevised-version Minimal prime ideals and Axiom of Choice(revised version) AliReza Olfati 2012-06-03T19:24:48Z 2012-06-04T00:07:05Z <p>From the page: </p> <p><a href="http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice" rel="nofollow">http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice</a>,</p> <p>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:</p> <p><em><strong>Theorem:For any commutative unitary ring \$R\$ there exists a minimal prime ideal.</em></strong></p> <p>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.</p> <p>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.</p> <p><strong>STATEMENT:Is it true that The existence of minimal prime ideals in commutative unitary rings is equivalent to \$AC\$</strong>.</p> <p>I am interested in To Know if the situation changes When we give minimality Condition on Prime ideals.</p> <hr> <p>I think its better to recall the difference of two following situations in topology:</p> <p>The statement "product of compact Hausdorff spaces is compact", does not implies \$AC\$</p> <p>But</p> <p>The statement "product of compact spaces is compact" is equivalent to \$AC\$ </p> http://mathoverflow.net/questions/98731/minimal-prime-ideals-and-axiom-of-choicerevised-version/98734#98734 Answer by Will Sawin for Minimal prime ideals and Axiom of Choice(revised version) Will Sawin 2012-06-03T20:03:36Z 2012-06-04T00:07:05Z <p>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.</p> <p>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.</p> <p>So minimal primes give a choice function.</p>