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I just saw someone prove the equivalence for a commutative ring $R$:

(1) $R$ is Noetherian;

(2) Every ideal of $R$ is finitely generated;

(3) Every non-empty set of ideals of $R$ has a maximal element.

For $(2) \Rightarrow (3)$, he used the axiom of choice (although Zorn's Lemma would have been more direct). Suppose I would only want to prove $(1) \iff (2)$, could there exists a proof without AC, or is that impossible since (1), (2), (3) are in fact equivalent?

(I have already proven $(2) \Rightarrow (1)$, so in fact my question pertains to $(1) \Rightarrow (2)$.)

Jos van Nieuwman
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  • Are you sure this is in fact a duplicate? The post you refer to seems to deal only with (1) and (3). It makes sense that there AC is needed, because it is about maximal elements which immediately sends you to Zorn, i.e. to AC, whereas I'm only interested in finite generation. – Jos van Nieuwman Oct 06 '19 at 23:58
  • I think I've found the answer. In a ring with the ascending chain condition, a maximal element of my arbitrary ideal presents itself immediately. Quite obvious actually, if I see it correctly. – Jos van Nieuwman Oct 07 '19 at 00:21

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