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Feb 24 at 20:52 vote accept Pace Nielsen
Feb 24 at 8:23 answer added Jeremy Rickard timeline score: 14
Feb 23 at 15:43 comment added Jeremy Rickard Or $(\mathbb{N}\setminus\{1\},+)$, as some people prefer to call that monoid.
Feb 23 at 15:32 comment added Pace Nielsen @JeremyRickard Ah, so we take the commutative monoid generated by two elements $a,b$, subject to the single relation $a+a=b+b+b$. Taking $R$ to be the endomorphism ring of the projective module corresponding to $a$, and taking $S$ to be the endomorphism ring of the projective module corresponding to $b$, we are done (because, clearly, no element $c$ exists with $6c=2a=3b$).
Feb 23 at 15:16 comment added Jeremy Rickard I don’t have time just now to write a full answer, but doesn’t this follow from Bergman’s theorem that every monoid is the monoid of finitely generated projective modules for some ring unless it obviously isn’t?
Feb 23 at 13:54 history asked Pace Nielsen CC BY-SA 4.0