Timeline for Structure theorem for a class of idempotent monoids (where $xy = x$ or $xy = y$ for all $x, y$)
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 5, 2023 at 19:23 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed an incorrect statement
|
| Oct 18, 2022 at 12:39 | vote | accept | Salvo Tringali | ||
| Oct 18, 2022 at 12:35 | answer | added | Benjamin Steinberg | timeline score: 7 | |
| Oct 18, 2022 at 12:18 | comment | added | Benjamin Steinberg | @YCor, sorry I was just defining it among bands. There are lots of tricks to show identities among bands imply other ones. | |
| Oct 18, 2022 at 11:04 | comment | added | YCor | @BenjaminSteinberg great! (It should be generated by $zxyz=zxzyz$ among bands, hence, by this and the idempotent identity $z^2=z$ among semigroups, since the latter doesn't follow from the former.) I didn't realize that $xzxyz=xzyz$ follows from these identities but this is indeed very easy. | |
| Oct 18, 2022 at 10:56 | comment | added | Benjamin Steinberg | @YCor, it should be the variety of regular bands. This is the variety generated by a left zero semigroup with adjoined identity and a right zero semigroup with adjoined zero and it is defined by $zxyz=zxzyz$. en.m.wikipedia.org/wiki/Band_(algebra)#Varieties_of_bands | |
| Oct 18, 2022 at 10:55 | comment | added | Salvo Tringali | @BenjaminSteinberg Maybe not necessary, but you provided the reference that nails the problem and the original question includes the "reference request" tag. | |
| Oct 18, 2022 at 10:48 | comment | added | Benjamin Steinberg | @SalvoTringali, is it necessary? Two people have already given the complete structural description so it's more a historical remark | |
| Oct 18, 2022 at 10:46 | comment | added | Benjamin Steinberg | @YCor, every variety of bands of finitely based. There is a complete description of the lattice of band varieties | |
| Oct 18, 2022 at 10:03 | comment | added | YCor | (My initial comments are quite obsolete now, but anyway in the terminology I suggested, the classification result says that the simple breakable semigroups are precisely the nonempty sets endowed with either the law $xy=x$, or the law $xy=y$ — which are nonisomorphic unless the set is a singleton.) | |
| Oct 18, 2022 at 9:57 | comment | added | YCor | I'd be curious whether the variety generated by breakable semigroups is finitely based. It is locally finite (easy exercise, given the structural result). Identities in $\le 3$ variables are generated by $x^2=x$, $xyxzx=xyzx$, $xzyxy=xzxy$, $xyxzy=xyzy$. Two semigroup words in a free breakable semigroup over a set $K$ represent the same element iff they have the same letters (not counting multiplicities) and for every subset $J$ of $K$, whenever removing all letters of $J$ from these words, the resulting words have the same first and last letter. | |
| Oct 18, 2022 at 7:04 | comment | added | Salvo Tringali | @BenjaminSteinberg Why not move your comments on Rédei's work to an answer? | |
| Oct 18, 2022 at 6:50 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
deleted 2 characters in body
|
| Oct 18, 2022 at 6:43 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed a phrase.
|
| Oct 18, 2022 at 1:01 | answer | added | Joseph Van Name | timeline score: 5 | |
| Oct 18, 2022 at 0:48 | comment | added | Benjamin Steinberg | These seemed to be first studied by Redei in Rédei, L. Algebra I; Pergamon Press: Oxford, UK, 1967. See Theorem 50 of section 27 where it gives Yemon's result. I've seen them called breakable semigroups and Redei semigroups in old papers. See link.springer.com/article/10.1007/BF02276097 for example | |
| Oct 18, 2022 at 0:41 | comment | added | Benjamin Steinberg | I had not heard this name before. But then again this is such a rigid notion I wouldn't be surprised if it has been rediscovered again and again. But I.think Yemon's answer says pretty much what the paper says in a more efficient way modulo counting | |
| Oct 18, 2022 at 0:24 | comment | added | Joseph Van Name | For problems like this, it may be a good idea to count the number of such algebras and then put the sequence into the online encyclopedia of integer sequences. I did a brute force search (backtracking is more efficient but harder to code), and I got 1,4,20,138 which is oeis.org/A292932. These algebras are known as quasi-trivial semigroups. And I found a paper on this arxiv.org/pdf/1709.09162.pdf. | |
| Oct 18, 2022 at 0:01 | comment | added | Benjamin Steinberg | @YemonChoi, i saw that paper | |
| Oct 17, 2022 at 23:49 | comment | added | Yemon Choi | @BenjaminSteinberg Seeing as no one in BanAlg world took an interest in maths.lancs.ac.uk/~choiy1/pubmath/papers/HHbands.html I feel compelled to try and get maximum mileage out of the hours I spent reading Howie's book :) | |
| Oct 17, 2022 at 23:47 | comment | added | Benjamin Steinberg | @YemonChoi, your answer is great. | |
| Oct 17, 2022 at 23:45 | comment | added | Yemon Choi | @BenjaminSteinberg I was just finishing typing up a long-winded answer which I think amounts to what you just said | |
| Oct 17, 2022 at 23:45 | answer | added | Yemon Choi | timeline score: 4 | |
| Oct 17, 2022 at 23:44 | comment | added | Benjamin Steinberg | My first impression is all you can do is build a chain of J-classes where each J-class acts as the identity on those below it and each J-class is a left or right zero semigroup | |
| Oct 17, 2022 at 23:08 | comment | added | YCor | A little further: any finite strongly idempotent semigroup $S$ has a unique partition $S=S_1\sqcup S_2\sqcup\dots\sqcup S_n$, such that $S_i\neq\emptyset$ and $S_iS_j\subset S_i$ for all $i\le j$, and $S_i$ is simple ($\neq\emptyset$, no 2-sided ideal other that $S_i$ and $\emptyset$). The law on each $S_i$ determines the whole law (clear). So the classification reduces to the simple case. The smallest simple strongly idempotent semigroups are 2-elements sets with law $xy=x$, and its opposite; if I'm correct there's no other on $\le 3$ elements. | |
| Oct 17, 2022 at 22:44 | comment | added | YCor | More precisely, call this "strongly idempotent semigroup". For a semigroup $S$ write $mS$ for the unitization $S\sqcup\{1\}$. Then every finite semigroup as you consider can be written canonically as $m^nS$ for some strongly idempotent semigroup $S$ that is not a monoid (namely, remove units step by step). This decomposition is canonical. Thus (in the finite case) one is reduced to classify those strongly idempotent semigroups without unit. For instance, $m^n\emptyset$ is an $n$-element chain with union as law. | |
| Oct 17, 2022 at 22:35 | comment | added | YCor | For a semigroup this is the same as a semigroup in which every subset is a subsemigroup. [In particular in a monoid, the complement of $\{1\}$ will be a subsemigroup, so restricting to monoids looks a bit artificial.] So one might get a lot of information by classifying small finite such semigroups. | |
| Oct 17, 2022 at 22:21 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed a typo and the definition of the restricted power monoid
|
| Oct 17, 2022 at 22:15 | history | asked | Salvo Tringali | CC BY-SA 4.0 |