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In my book (Jantzen, Algebra, 2014), Noetherian rings are defined by three equivalent conditions. I wonder how the first two can be equivalent:

Every ascending chain of ideals $(a_1) \subset (a_2) \subset \ldots$ becomes stationary. Every set of of ideals, ordered by set inclusion, has a maximum element.

The proof is just a sentence, my problem is: Isn't the set in the ascending chain condition countable while the other set is not? The proof says, suppose there is a non empty set without maximum element, then we can construct a chain of ideals that is ascending, but does not become stationary (- I see why).

Is the Axiom of Choice used to construct the chain? How could you build such a chain when your set in the second condition is uncountable?

Thanks.

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    $\begingroup$ Suppose thre is no maximal element (not maximum element...) Pick any element $I_1$. It is not maximal, so it is contained in some other different element $I_2$, which is not maximal so it is contained in some other dfferent element $I_3$, etc. Uncountability of anything is irrelevant to the construction. $\endgroup$ – Mariano Suárez-Álvarez Jul 19 '16 at 8:41
  • $\begingroup$ Such an ascending chain is an element of $R^{\mathbb{N}}$, isn't it? I wonder how the whole chain is constructed at once, not just the first finitely many elements. $\endgroup$ – Greg P. Jul 19 '16 at 8:48
  • $\begingroup$ @MarianoSuárez-Alvarez : '... so it is contained in some other different element' - I was always under the impression that we need some form of Choice now to actually pick one of these to continue by induction. $\endgroup$ – Matthias Klupsch Jul 19 '16 at 8:48
  • $\begingroup$ I was not adressing the use or not of AoC (I am in the I-do-not-really-care-if-AoC-is-used-or-not team...) just observing that the uncountability is irrelevant. $\endgroup$ – Mariano Suárez-Álvarez Jul 19 '16 at 8:50
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You're correct that some choice is needed to construct the countable ascending chain. In your comment above you noted that this countable chain is an element of $P(R)^\omega$ (where $P(R)$ is the power set of $R$), and this might suggest that you need to use the Countable Axiom of Choice, but in fact you need more. You are not just choosing an element of $P(R)^\omega$ (which countable choice guarantees the existence of), you are choosing an element in such a way that $a_{n+1}$ depends on your choice of $a_n$. What you need for this is called the Axiom of Dependent Choice (it says exactly that you can choose such a sequence). Of course, this is still a weakened version of the regular Axiom of Choice.

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