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Assuming Zorn's lemma, we can prove that in every ring with unity satisfying Ascending-Chain-Condition, every ideal is finitely generated. Is this statement equivalent to Zorn's lemma? Can we prove it without assuming Axiom of choice?

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  • $\begingroup$ You've asked two questions which are somewhat similar, and both of them include nothing about your efforts to tackle this problem. At least in searching through standard literature or Google. Have you tries something like that? $\endgroup$ – Asaf Karagila Apr 9 '15 at 15:57
  • $\begingroup$ As far as I know, it requires at least the denumerable axiom of choice. $\endgroup$ – Bernard Apr 9 '15 at 16:00
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    $\begingroup$ I think it suffices to use axiom of dependent choice (abbreviated as DC). So you should not expect it to be as strong as Zorn's Lemma, which is strictly stronger than DC. I think a more suitable question you could ask is that if the statement equivalent to DC. $\endgroup$ – Censi LI Apr 10 '15 at 9:22
  • $\begingroup$ @AsafKaragila : Both of the two questions came up while dealing with Ascending chain condition ( the prime ideal question , while dealing with commutative rings ) in face while proving the equivalence of the very same statements as you have mentioned in your answer ... $\endgroup$ – user228168 Apr 10 '15 at 9:51
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If you look at the fourth section of Hodges paper,

Wilfrid Hodges, Six impossible rings, J. Algebra 31 (1974), 218--244.

He proves (Section 3, Th. 1) that the implication that you are asking for is not provable without the axiom of choice. Namely, the following chain of implications is irreversible without using Zorn's lemma:

$$\text{Every non-empty set of ideals has a maximal element}\implies\text{Every ideal is finitely generated}\implies\text{Every strictly increasing chain of ideals is finite}$$

Of course, one can note by looking at the usual proof that with the principle of dependent choice, we can prove these are indeed equivalent. So it is strictly weaker than the axiom of choice.

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