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I want to find a semi-constructive example of a unitary commutative ring without any maximal ideals assuming that axiom of choice is incorrect and/or a model of $\sf ZF$ where we have such a concrete ring.

This question is similar to the questions Vector space bases without axiom of choice and A confusion about Axiom of Choice and existence of maximal ideals..

What I tried is to use $\mathbb{R}$ as a $\mathbb{Q}$ vector space without a basis and try to construct some chains of ideals on a related ring and try to show that a maximal ideal corresponds to a basis but didn't achieve much.

Thank you in advance!

aeyalcinoglu
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  • As far as I know, existence of maximal ideals in rings is properly weaker than choice. – egreg Jul 07 '19 at 13:59
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    @egreg https://math.stackexchange.com/questions/317028/a-confusion-about-axiom-of-choice-and-existence-of-maximal-ideals - they seem equal – aeyalcinoglu Jul 07 '19 at 14:24
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    I stand corrected. – egreg Jul 07 '19 at 14:28
  • @egreg: Sit down, prime ideals are weaker than choice as they are equivalent to BPI. – Asaf Karagila Jul 07 '19 at 14:29
  • See also https://math.stackexchange.com/a/317038/622 – Asaf Karagila Jul 07 '19 at 14:40
  • @AsafKaragila I was looking to read that but thought maybe there is an easier construction, especially using $\mathbb{R}$ as a vector space over $\mathbb{Q}$, or a possibility for a more general answer. – aeyalcinoglu Jul 07 '19 at 14:43
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    Well. You can either construct special models with particular examples, and that normally involves techniques like forcing and stuff; or you can sort of repeat the proof that the existence of maximal ideals implies choice and start with a set that cannot be well-ordered as a counterexample. This set can be $\Bbb R$ in models where there is no Hamel basis, for example. – Asaf Karagila Jul 07 '19 at 14:46
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    Near-duplicate: https://math.stackexchange.com/questions/2182014/proving-specific-unital-rings-have-maximal-ideals-without-zorns-lemma – Eric Wofsey Jul 07 '19 at 18:29

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Let $k$ be a field, let $I\subset k^{\mathbb{N}}$ be the ideal of sequences that are eventually zero, and let $S=k^{\mathbb{N}}/I$. Then maximal ideals in $S$ are in bijection with nonprincipal ultrafilters on $\mathbb{N}$. In particular, in any model of ZF in which there are no nonprincipal ultrafilters on $\mathbb{N}$, there will be no maximal ideals in $S$.

Eric Wofsey
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