Suppose I have a collection of "elements" together with operations that satisfy the axioms for a commutative ring with identity --- except that these elements form not a set, but a proper class.


Must such a thing contain a maximal ideal (where an "ideal" is allowed to be proper-class-size)? Obviously, the usual Zorn's Lemma argument is not available.



Proper-class sized "ring" with no maximal ideals
The [axiom of global choice](http://en.wikipedia.org/wiki/Axiom_of_global_choice) is sufficient. More generally, any proper class ring whose elements can be enumerated along the ordinals will have a maximal ideal. The argument is the usual one...


Suppose $x\_\alpha$, $\alpha \in \mathrm{Ord}$, enumerates the elements of the proper class ring $R$. Without loss of generality $x\_0$ is the zero of the ring $R$. Define $h(0) = 0$ and for each $\alpha \gt 0$, let $h(\alpha)$ be the first ordinal $\eta$ (if any) such that $x\_\eta$ is not in $I\_\alpha = \sum\_{\beta\lt\alpha} Rx\_{h(\beta)}$ and $Rx\_\eta + I\_\alpha \neq R$. These $R$-ideals are all *uniformly* definable from $R$, $\langle x\_\alpha\rangle\_{\alpha\in\mathrm{Ord}}$ and $\langle h(\beta)\rangle\_{\beta\lt\alpha}$, so there is no trouble carrying out this construction in NBG. The elements indexed by the function $h$ will enumerate a (not necessarily proper) class of generators for a maximal $R$-ideal.


Since the issue is choice, a counterexample in the absence of global choice could be done by extending the ideas of Hodges in [Six Impossible Rings](http://dx.doi.org/10.1016/0021-8693%2874%2990065-9) [J. Algebra 31 (1974), 218–244; [MR0347814](http://www.ams.org/mathscinet-getitem?mr=347814)] to proper classes. This seems plausible but I strongly suspect that nobody has ever done it in public...


