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Jul 27, 2023 at 14:32 history edited Carl-Fredrik Nyberg Brodda CC BY-SA 4.0
deleted 20 characters in body
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
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
Jul 21, 2023 at 17:43 history edited Carl-Fredrik Nyberg Brodda CC BY-SA 4.0
added 5 characters in body
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
Jul 20, 2023 at 12:49 history answered Carl-Fredrik Nyberg Brodda CC BY-SA 4.0