Timeline for non-associative but commutative algebra [closed]
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2020 at 10:46 | review | Reopen votes | |||
| Aug 30, 2020 at 12:32 | |||||
| Aug 29, 2020 at 10:31 | comment | added | YCor | Answers are useful, I eventually voted to reopen. | |
| Jul 13, 2020 at 2:00 | history | closed |
Jeremy Rickard abx YCor Andreas Blass Konstantinos Kanakoglou |
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| Jul 12, 2020 at 17:07 | comment | added | Richard Stanley | @YCor I shouldn't have called it a monoid. It is the free commutative magma with one generator. There is some information at en.wikipedia.org/wiki/Wedderburn-Etherington_number. | |
| Jul 12, 2020 at 15:23 | answer | added | Andreas Blass | timeline score: 7 | |
| Jul 12, 2020 at 15:22 | comment | added | YCor | @RichardStanley I can't find any occurrence of this monoid, but a monoid is associative and so is its semigroup algebra. You probably have some non-associative magma in mind rather than a monoid? | |
| Jul 12, 2020 at 15:10 | answer | added | Phil Tosteson | timeline score: 3 | |
| Jul 12, 2020 at 14:20 | comment | added | jg1896 | The polynomial algebra is the free object in the category of commutative associative algebras. If you begin with the free associative algebra and mod out the commutators, it is a matter of checking that the resulting algebra will also be free in the category of commutative associative algebras. Since such categories are, in fact, what is known as a variety of algebras (universal algebra parlance), the result follows. | |
| Jul 12, 2020 at 14:10 | history | edited | Sunny | CC BY-SA 4.0 |
deleted 6 characters in body
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| Jul 12, 2020 at 14:10 | comment | added | Sunny | But how do you know that it's polynomial algebra. | |
| Jul 12, 2020 at 14:04 | review | Close votes | |||
| Jul 13, 2020 at 2:03 | |||||
| Jul 12, 2020 at 14:03 | comment | added | jg1896 | In your question, it should be quotient by the ideal generated by the commutators. The result would just be an polynomial algebra, so it does not work. | |
| Jul 12, 2020 at 13:57 | answer | added | jg1896 | timeline score: 8 | |
| Jul 12, 2020 at 13:57 | comment | added | Sunny | Isn't abelian lie algebra is associative? | |
| Jul 12, 2020 at 13:49 | comment | added | Richard Stanley | One could take the semigroup algebra (over a field, say) of the Wedderburn-Etherington monoid. See H. W. Becker, Solution to Advanced Problem 4277, Amer. Math. Monthly 56 (1949), 697-699. | |
| Jul 12, 2020 at 13:40 | history | edited | Qfwfq | CC BY-SA 4.0 |
edited title
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| Jul 12, 2020 at 13:35 | history | asked | Sunny | CC BY-SA 4.0 |