Timeline for Are all free monoids residually finite?
Current License: CC BY-SA 4.0
17 events
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Jul 27, 2023 at 14:32 | history | edited | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
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S Jul 24, 2023 at 12:03 | history | suggested | einpoklum | CC BY-SA 4.0 |
Clarified what A* means, since the question doesn't used that expression + only need a shorter length for the ideal
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Jul 22, 2023 at 21:21 | comment | added | Carl-Fredrik Nyberg Brodda | @einsupportsModeratorStrike I don't need it -- it's sufficient! Of course $\operatorname{max}(|u|, |v|)$ works too. | |
Jul 22, 2023 at 21:07 | review | Suggested edits | |||
S Jul 24, 2023 at 12:03 | |||||
Jul 22, 2023 at 21:05 | comment | added | einpoklum | Why do you need $|u| + |v|$ instead of $\max(|u|,|v|)$? | |
Jul 21, 2023 at 21:19 | history | edited | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
added 7 characters in body
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Jul 21, 2023 at 17:43 | history | edited | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
added 5 characters in body
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Jul 21, 2023 at 17:43 | comment | added | Carl-Fredrik Nyberg Brodda | @cody Yep — thank you! ($F$ for finite, not free :-). | |
Jul 21, 2023 at 15:54 | comment | added | cody | Is the first sentence of your second paragraph supposed to be "consider the ideal $I$ of $A^*$"? | |
Jul 21, 2023 at 11:45 | comment | added | Carl-Fredrik Nyberg Brodda | @SalvoTringali Sure, in this question there is no reference to finite alphabets, so you're quite right to point it out! | |
Jul 21, 2023 at 11:43 | comment | added | Salvo Tringali | @Carl-FredrikNybergBrodda I don't see any reference to finite alphabets in the OP (whence my comment). With that said, I love your proof. | |
Jul 21, 2023 at 11:40 | comment | added | Carl-Fredrik Nyberg Brodda | @SalvoTringali To modify it for infinitely generated free monoids is just as BenjaminSteinberg puts it. Most answers (and the question there) on the free group question assume finite generation, so I did too. | |
Jul 21, 2023 at 0:40 | comment | added | Salvo Tringali | @BenjaminSteinberg I agree. It's just that this would be better mentioned in Carl-Fredrik Nyberg Brodda's answer. | |
Jul 21, 2023 at 0:28 | comment | added | Benjamin Steinberg | @SalvoTringali, If you have two words in an infinitely generated free monoid then you can send all letters not in the words to 1 to reduce to the finitely generated case. Or you can add to your ideal all words containing a letter not appearing in any of the two words. | |
Jul 20, 2023 at 23:38 | comment | added | Salvo Tringali | Doesn't this only work with finitely generated free monoids? If the alphabet is not finite, then the Rees quotient you construct is infinite. Or am I missing something? | |
Jul 20, 2023 at 12:54 | history | edited | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
added 7 characters in body
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Jul 20, 2023 at 12:49 | history | answered | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |