Timeline for Invertible elements in monoid rings of unital monoids without non-trivial invertible elements
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2011 at 11:45 | comment | added | Andreas Thom | Benjamin, your second comment is precisely the difficult part. Why should there be an invertible element in the support? | |
| Jun 27, 2011 at 20:55 | comment | added | Benjamin Steinberg | I checked in the literature and it seems that every unit in the algebra of the monoid defined by the presentation $\langle a,b\mid ab=1\rangle$ does have $1$ in its support. | |
| Jun 25, 2011 at 14:00 | comment | added | Benjamin Steinberg | Sorry, I had the impression from the previous answers that you were looking at finite monoids. More generally, the set of elements $L$ of the monoid $M$ that are not left invertible is a proper left ideal. Therefore, the span $kL$ is a proper left ideal of $kM$. It follows that any invertible element must contain a left invertible element in its support and dually a right invertible element. I will try to think whether this situation can come up without having an invertible element. | |
| Jun 24, 2011 at 6:08 | comment | added | Andreas Thom | Yes, the same holds if left-invertible is equivalent to right-invertible; which if holds in finite monoids. The question is only interesting if these notions are not equivalent. | |
| Jun 23, 2011 at 21:19 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |