Relative consistency of a ZFC counterexample to PGF equals GP

Establish whether the existence of a ring R with PGF(R) properly contained in GP(R) is consistent relative to ZFC, namely whether Con(ZFC) implies Con(ZFC + ∃R (R is a ring and PGF(R) ⊊ GP(R))).

Background

The main theorem constructs a ring witnessing PGF(R) ⊊ GP(R) under the stronger assumption that an uncountable strongly compact cardinal exists. The paper conjectures that this large-cardinal assumption can be removed at the level of relative consistency, so that a counterexample may consistently exist over ZFC alone.

References

The following relative-consistency implication holds:

\operatorname{Con}(\textsf{ZFC}) \implies \operatorname{Con}!\left( \textsf{ZFC}+\exists R\, \bigl(R\text{ is a ring}\mathbin{\wedge} PGF(R)\subsetneqGP(R)\bigr) \right).

The main theorem proves the corresponding implication with the stronger antecedent \begin{align*} & \operatorname{Con}\bigl( \textsf{ZFC}+\text{``there is an uncountable strongly compact cardinal''}\bigr) \ &\qquad\implies \operatorname{Con}!\left( \textsf{ZFC}+\exists R\, \bigl(R\text{ is a ring}\mathbin{\wedge} PGF(R)\subsetneqGP(R)\bigr) \right). \end{align*} The conjecture asserts that the strongly compact cardinal assumption can be removed.

— A strongly compact cardinal yields a left and right coherent ring with $\mathcal{PGF}(R)\subsetneq\mathcal{GP}(R)$  (2608.17748 - Zhang, 18 Aug 2026) in Section 4, Discussion of related consistency questions, Conjecture \ref{conj:negative-in-zfc}