Timeline for Is there a finite equational basis for the join of the commutative and associative equations?
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| Jul 15, 2020 at 7:32 | comment | added | YCor | Here's a purely algebraic (rough) formulation of the question: roughly: what are identities satisfied by all associative magmas, and by all commutative magmas? is there a finite number of identities generating all those? | |
| Jul 15, 2020 at 7:29 | history | edited | YCor |
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| Jul 15, 2020 at 1:05 | comment | added | user44143 | @bof, thanks; that’s good enough to make me think the answer to the question is no. | |
| Jul 15, 2020 at 0:35 | history | edited | bof |
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| Jul 15, 2020 at 0:27 | history | edited | bof | CC BY-SA 4.0 |
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| Jul 15, 2020 at 0:25 | comment | added | bof | @MattF. $((ab)c)(a(bc))=(a(bc))((ab)c)$ at least if you don't insist on "interesting". | |
| Jul 14, 2020 at 23:21 | comment | added | user44143 | If I understand this right, one theory in the join is the theory of $a(ba)=(ab)a$, since that follows from either commutativity or associativity. Are there other interesting theories in this join? | |
| Jul 14, 2020 at 18:19 | comment | added | Gerhard Paseman | For this example, I don't know. I recall some work about chains of varieties, probably semi group varieties ,where every other member was not finitely based. I suspect nfb (and thus fb) is not well behaved under join. I don't recall the author names, but I would be unsurprised if one of them was Mark Sapir. Gerhard "Look Up Chains And NFB" Paseman, 2020.07.14. | |
| Jul 14, 2020 at 15:24 | history | asked | user107952 | CC BY-SA 4.0 |