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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 $AC$

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While I don't know the answer to this question off-hand, you should remember that for the past 50 years or so people have been using the axiom of choice in its full power for a lot more than it is needed. People usually don't mind how much choice is needed, at best they would like to know if the axiom of choice is used, and rarely whether or not it has been used to its full extent. In this aspect for most people BPI is practically AC, and DC is practically AC - despite both being pretty far from actually being AC in its full glorious power. –  Asaf Karagila Jun 3 '12 at 19:54
    
@Asaf: Yes, that is a very said state of affairs. It's like the old times when people didn't care whether a function was differentiable, they just differentiated it. –  Andrej Bauer Jun 3 '12 at 21:26
    
Andrej, I was drinking coffee with a friend today, and two guests were in the lounge talking about mathematics. I overheard one say the words "you need the axiom of choice for that" and I inquired further. The topic was discontinuous linear operators from a Hilbert space (to a normed space, I presume). It took me a second to think about it, and then I recalled that you actually need quite less than AC itself for that. One of the guests concurred, and the other simply mentioned it was good knowing you don't need all of AC for that. It is not the first time this sort of thing happen to me, too. –  Asaf Karagila Jun 3 '12 at 21:56
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1 Answer 1

up vote 13 down vote accepted

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.

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Hello Dear sawin. I have enjoyed from your easy and short Argument For my Long discerption.I'd like to say my best thanks to you for this. But I have two Questions for disillusion. 1.Is the free polynomial ring which you defined is the usual polynomials with real coefficients and indeterminants {$A_\alpha$}....2.You assumed that you could choose $x$, $y$ different elements in the same set. But if we have infinitely many sets with one elements without Knowing the indexes of these sets, can we go on As your argument without any problem? –  Ali Reza Jun 3 '12 at 21:21
    
Coefficients in any field, including the reals, should work. I'm taking all pairs of elements in the same set, which could be empty. This does not require choice or anything like that. Infinitely many one-element sets do not require the axiom of choice to choose one element from each. –  Will Sawin Jun 3 '12 at 22:36
    
Will, I believe you need the sets in the set to be disjoint (and nonempty) since the argument doesn't seem to work for $\left\lbrace\lbrace1,2\rbrace,\lbrace1,3\rbrace,\lbrace2,3\rbrace\right\rbrace$‌​. –  François G. Dorais Jun 3 '12 at 23:48
    
Fixed. Disjoint & nonempty are all you need for choice, obviously. –  Will Sawin Jun 4 '12 at 0:07
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My proof is to just take the subset of it that is elements of the union of the original set of sets, then look at the idea generated by that. Since the ideal contains all but one element from each set, the sub-ideal satisfies the same property, which makes it clear that the quotient is a free polynomial algebra, thus an integral domain. –  Will Sawin Jun 4 '12 at 16:18
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