Commutative unitary ring without maximal ideal without axiom of choice

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!

Answer

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 $S$.