Timeline for Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2016 at 7:31 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Removed some text that was no longer meaningful (after Benjamin Steinberg's remark)
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| Mar 13, 2016 at 20:23 | vote | accept | Salvo Tringali | ||
| Mar 13, 2016 at 20:15 | vote | accept | Salvo Tringali | ||
| Mar 13, 2016 at 20:15 | |||||
| Mar 13, 2016 at 17:15 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Mar 13, 2016 at 16:51 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Embedded BS's remark
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| Mar 13, 2016 at 16:49 | comment | added | Benjamin Steinberg | It seems to me the answer should be no. Take a fg cancellative monoid $M$ not embeddable in a group. Enumerate the elements and invert them one by one. Then the original monoid would embed in the unit group of the direct limit. | |
| Mar 13, 2016 at 16:46 | comment | added | Salvo Tringali | To my shame, I hadn't thought of it. Let me edit and include your remark. | |
| Mar 13, 2016 at 16:44 | comment | added | Benjamin Steinberg | Why do you say left invertible? In a cancellative monoid left invertible elements are right invertible. If yx =1 then xyx=x and so xy =1 by cancellation | |
| Mar 13, 2016 at 16:23 | history | asked | Salvo Tringali | CC BY-SA 3.0 |