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?


  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.