Timeline for Monoids where every two non-unit elements have a common power
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2017 at 20:10 | vote | accept | Salvo Tringali | ||
| Apr 8, 2017 at 19:22 | comment | added | Benjamin Steinberg | It is also satisfied in semigroups in which all elements generate the same two sided ideal. | |
| Apr 8, 2017 at 19:20 | comment | added | Benjamin Steinberg | I don't know interesting examples. For a finite monoids this means the complement of the group of units I'd a nilpotent extension of a completely simple semigroup. | |
| Apr 8, 2017 at 19:01 | comment | added | Salvo Tringali | Got it, thanks! I will wait to see if anyone can at least provide some interesting examples of monoids that satisfy the condition in the OP. If not, I will eventually accept your answer. | |
| Apr 8, 2017 at 18:58 | comment | added | Benjamin Steinberg | People allow noncommutative archimedean semigroups and it is the same as what you write of you ignore the restriction about non units. Otherwise there is no name. | |
| Apr 8, 2017 at 18:54 | comment | added | Salvo Tringali | (...) an archimedean monoid $H$ in the sense of this definition would have the property that, for all $x \in H$, there is $n \in \mathbf N^+$ with $x^n = 1_H$, and hence would be a group. But groups are not really interesting from the point of view of (classical) factorization theory. So I guess the restriction to non-units in the OP is what makes the difference here. | |
| Apr 8, 2017 at 18:54 | comment | added | Salvo Tringali | At least in (classical) factorization theory, a (multiplicatively written) commutative monoid $H$ is called archimedean if $\bigcap_{n \ge 0} a^n H = \emptyset$ for all $a \in H \setminus H^\times$, cf., e.g., Halter-Koch's book Ideal Systems: An Introduction to Multiplicative Ideal Theory (Ch. 3, Exercise 6). This, however, seems quite different from the property stated in the OP. On the other hand, if I get your answer correctly, you take a semigroup $H$ to be archimedean if, for all $x, y \in H$, there exist $m,n \in \mathbf N^+$ with $x^m = y^n$: In particular, (...) | |
| Apr 8, 2017 at 18:19 | review | Low quality posts | |||
| Apr 8, 2017 at 18:23 | |||||
| Apr 8, 2017 at 17:59 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |