More specifically, I just want to prove that in a Noetherian (even Dedekind) ring, every ideal in contained in a maximal ideal. Is the axiom oh choice needed here? The usual proofs (for general rings use Zorn's lemma; for Noetherian rings one can build an infinite ascending chain if there is no maximal element) seem to rely on it.

If the axiom of dependent choice can be avoided as well, even better.

  • $\begingroup$ According to an answer of Joel David Hamkins on MO, the axiom of dependent choice is sufficient, and we do not need the full axiom of choice. But I don't know if some form of AC is needed. $\endgroup$ – Zhen Lin Aug 22 '11 at 14:54
  • $\begingroup$ Here is a similar question math.stackexchange.com/questions/41590/… $\endgroup$ – Asaf Karagila Aug 22 '11 at 17:10

It actually depends on how you define Noetherian rings. The two most common definitions are:

  1. A ring in which every non-empty set of ideals has a maximal element.

  2. A ring which satisfies the ascending chain condition on ideals — every nondecreasing sequence of ideals is eventually constant.

The fact that (1) implies (2) is clear, but the implication from (2) to (1) is not provable without some weak form of the Axiom of Choice (specifically, the Axiom of Dependent Choice).

Now, using the first definition, it is trivial that every proper ideal $I$ in a Noetherian ring $R$ is contained in a maximal one. Take the collection $\mathcal{A}$ of proper ideals of $R$ that contain $I$. Note that $\mathcal{A}$ is nonempty since $I \in \mathcal{A}$. Therefore, $\mathcal{A}$ has a maximal element $M$, which must be a maximal ideal of $R$ by definition of $\mathcal{A}$.

Using the second definition, it is not provable without any form of choice that every proper ideal $I$ in a Noetherian ring $R$ is contained in a maximal one. It is tempting to proceed by contradiction. Assume there is no maximal ideal that contains $I$. Since $I$ is not maximal, we can find a proper ideal $I' \supsetneq I$. Since $I'$ is not maximal, we can find an ideal $I'' \supsetneq I'$. And so on, thereby obtaining a strictly ascending chain $$I \subsetneq I' \subsetneq I'' \subsetneq \cdots$$ However, each step of this construction requires choosing an ideal extending the previous one, infinitely many choices in total. In the general case, one needs the Axiom of Dependent Choice to justify this.

Note that the Axiom of Dependent Choice is only needed for the general case. For example, if the ring $R$ is countable then the process outlined above can be made effective. Let $x_0,x_1,\dots$ be a fixed enumeration of $R$. To pick $I'$, scan through the enumeration of $R$ until you find an element $x_i$ such that $x_i \notin I$ and $I + Rx_i \neq R$, then let $I' = I + Rx_i$ and continue in the same manner to find $I'', I''', \dots$ If $I$ is not contained in a maximal ideal, then such an $x_i$ can always be found and thus we contradict the ascending chain condition.

  • $\begingroup$ You asserted that ‘it is not provable without any form of choice ...’ — does that mean there is a model of ZF containing a ring satisfying the ascending chain condition for which there is an ideal not contained in any maximal ideal? $\endgroup$ – Zhen Lin Aug 22 '11 at 15:36
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    $\begingroup$ Yes, see W. Hodges, Six impossible rings, J. Algebra 31 (1974), 218-244. $\endgroup$ – François G. Dorais Aug 22 '11 at 15:51
  • $\begingroup$ Thanks for the reference! I think that conclusively answers the OP's question. $\endgroup$ – Zhen Lin Aug 22 '11 at 15:58
  • $\begingroup$ Another common definition is "Every ideal is finitely generated". This implies ACC without the need for choice, but the implication going the other way needs dependent choice, I believe. $\endgroup$ – Arturo Magidin Aug 22 '11 at 16:21
  • $\begingroup$ Good point, Arturo. Note that Hodges separates the three definitions in the paper cited above. $\endgroup$ – François G. Dorais Aug 23 '11 at 3:33

Fred Richman's webpage has some papers on mathematics without countable choice (and without the principle of excluded middle), including a choiceless development of the theory of Noetherian rings.



Defining Noetherian by the ascending chain condition causes a reliance on countable dependent choice. To build a theory without nonconstructive principles of choice or excluded middle, one has to formulate an alternative definition of Noetherian that (in the presence of those principles) is equivalent to the usual chain condition, is inequivalent without the constructive principles, and is strong enough to allow the important parts of the nonconstructive theory to be constructivized.

I did not see whether Richman's paper discusses maximal ideals (in their classical formulation), since they are less useful constructively, or whether he proves the result you are looking for. But I think his work goes some way toward showing that in principle the whole theory could be rebuilt without any use of AC, DC or CC. This is maybe similar to the gradual elimination of Noetherian hypotheses in (some) theorems of algebraic geometry where it is habitual to assume finite dimensionality but the results could be stated in greater generality, after rephrasing in a form that is equivalent in the Noetherian case but also true in the general case.


No you don't need AC for noetherian rings. Indeed, fix an ideal $I$ of $A$. If $I$ is maximal, you're done. Else, there exists $I_1$ containing $I$ which is a proper ideal. If $I_1$ is not maximal, you continue the procedure, so that you obtain an increasing sequence $(I_n)$ of proper ideals of $A$. By the noetherianity of $A$, it is stationary, that is for $n \geqslant N$, $I_n=I_N$, and thus $I_N$ is maximal and contains $I$.

Moreover, you can see that the ascending chain condition (ACC) used in the proof is equivalent to the finite generation of every ideal.

  • 2
    $\begingroup$ You have used the axiom of dependent choice here. $\endgroup$ – Zhen Lin Aug 22 '11 at 14:56
  • $\begingroup$ Indeed, I was misleading - DC is not good as well. $\endgroup$ – Gadi A Aug 22 '11 at 15:11
  • $\begingroup$ @Gadi: Ok, my bad! $\endgroup$ – Henri Aug 22 '11 at 15:13
  • $\begingroup$ Yours is a perfectly good answer to my original question. This is my bad. $\endgroup$ – Gadi A Aug 22 '11 at 15:24

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