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.


The 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 [J. Algebra 31 (1974), 218–244; MR0347814] to proper classes. This seems plausible but I strongly suspect that nobody has ever done it in public...


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