Timeline for Monoids (or semigroups) with a "finite decomposition" property
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 2, 2017 at 7:22 | comment | added | Salvo Tringali | @AlešBizjak I think that, if you are only interested in monoids with many idempotents, this should be made clear in the OP: Some of your examples and non-examples give a completely different impression. I see your edits of the original post, and you didn't address this issue (raised by Mark Sapir's comments to my answer). So, I do actually presume that you're interested in any sort of monoids that satisfy the condition stated in the OP, though your motivation comes from applications where (infinitely) many non-trivial idempotents exist. Is my interpretation of your request correct? | |
| May 1, 2017 at 9:10 | comment | added | Aleš Bizjak | I've edited the question to address some of the points raised in the comments. | |
| May 1, 2017 at 9:09 | history | edited | Aleš Bizjak | CC BY-SA 3.0 |
Terminology as suggested in the comments, fixed a bug in the example.
|
| May 1, 2017 at 0:00 | comment | added | user6976 | Probably there is a way to describe this class of commutative monoids. Take an example $S$, and decompose it as a semilattice of archimedian commutative semigroups $S_\alpha$, $\alpha\in I$. Then each $S_\alpha$ may contain at most one idempotent, $e_i$. So I think the archimedian parts can be characterized. Then look at the semilattice structure. There should be some sort of Noetherian property satisfied by the semilattice, and the structure homomorphisms should be possible to characterize. | |
| Apr 30, 2017 at 20:54 | comment | added | Salvo Tringali | @AlešBizjak I think you don't mean "A non-example is [something] and things like positive rationals or reals under multiplication", since the latter are groups, so they are already included in [something]. I think you mean the positive rationals (respectively, reals) under addition: More in general, any non-trivial, divisible, cancellative monoid is a non-example. | |
| Apr 30, 2017 at 19:57 | comment | added | Benjamin Steinberg | Every element of finite order must be an idempotent. If x is finite order and not an idempotent it has arbitrarily long powers that are the same element | |
| Apr 30, 2017 at 16:45 | comment | added | Benjamin Steinberg | This will fail in any semigroup containing generalized inverses a,b with aba=a and bab=b with a,b not idempotents. This is quite rare for finite semigroups. | |
| Apr 30, 2017 at 16:36 | comment | added | user6976 | > A non-example is any monoid with an element which has an inverse, but is not its own inverse, You probably meant "a unit which is not the identity element". | |
| Apr 30, 2017 at 15:26 | comment | added | Salvo Tringali | @BenjaminSteinberg It does really depend. What I find strange about this request, is that the OP doesn't allow for the "free use" of units in the factorizations: It may reflect the kind of applications he has in mind, but it is unusual from the point of view of factorization theory in the way it has been developed so far, where units are just absorbed by the atoms and treated as a kind of "negligible factors". | |
| Apr 30, 2017 at 14:53 | comment | added | Benjamin Steinberg | A more usual thing to ask is that any sufficiently long factorization of x, some subword has product an idempotent. This weaker statement is true for finite monoids and is essentially the pumping lemma from automata theory. | |
| Apr 30, 2017 at 14:47 | answer | added | Salvo Tringali | timeline score: 1 | |
| Apr 30, 2017 at 13:42 | history | asked | Aleš Bizjak | CC BY-SA 3.0 |