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If $k[S]$ is noetherian, is S finitely generated?

Let $S$ be a semigroup. If $S$ is abelian, then it follows that the semigroup algebra $k[S]$ is finitely generated if and only if $S$ is. What if we relax the condition on $k[S]$, so that $k[S]$ is ...
J.C. Ottem's user avatar
  • 11.6k
5 votes
1 answer
334 views

Short proof a monoid is a group iff every splitting is right homogeneous

In the paper "Schreier split epimorphisms between monoids" by Bourn, Nelson, Martins-Ferreira, Montoli and Sobral, Semigroup Forum June 2014, the authors prove a characterization of groups among ...
Arrow's user avatar
  • 10.5k
5 votes
3 answers
718 views

Subsets of $\mathbb{R}^+$ closed under addition

No one's answered the question cumulant problem so here's a simpler question: Has anyone described or catalogued all sets of non-negative real numbers that are closed under addition? In particular, ...
Michael Hardy's user avatar
5 votes
1 answer
208 views

Conjugacy classes of monoids II: Abelianising a monoid, wrongly

$\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is $$ G/\left(ab\sim ba\ \middle|\ a,b\in G\right)? $$ Answer: not $G^{\mathrm{ab}}$, but the set of conjugacy classes of $G$. When ...
Emily's user avatar
  • 11.8k
5 votes
2 answers
137 views

The smallest number of vertices for a graph with the same endomorphism monoid

Let $G$ be a directed graph without loop and suppose that $M$ is its endomorphism monoid. First how we can create a simple graph $\Gamma$ (using $G$), with the same endomorphism monoid? and then how ...
khers's user avatar
  • 237
5 votes
1 answer
304 views

Sets of natural numbers such that sums of a bounded number of its elements form a semigroup

This is a naive question and I'm afraid it might be better placed on math.se. I would like to leave it to your judgement. I would like to know what is known about sets $A$ of natural numbers such ...
Michał Masny's user avatar
5 votes
1 answer
655 views

an example of a semigroup with solvable word problem but unsolvable power problem

We say that a semigroup $S$ has solvable power problem if there is an algorithm that takes as input an element $s \in S$ and decides whether or not there exist $m,n \in \mathbb{N}$ with $m \neq n$ and ...
dan's user avatar
  • 549
5 votes
2 answers
478 views

Generalization of the concept of a measure

Consider the following generalization of the concept of a measure: Let $L = (X, \lor, \land, \bot)$ be a semi-bounded lattice. Let $M = (Y, \bullet, e)$ be a commutative monoid. An $(L, M)$-measure is ...
user76284's user avatar
  • 2,203
5 votes
1 answer
152 views

Cartan matrix of the full transformation monoid ring

Let $T_n$ be the full transformation monoid of an $n$-set and $A_n=KT_n$ its monoid algebra over the complex numbers. Question 1: Is the Cartan matrix of $A_n$ known? Im especially interested to see ...
Mare's user avatar
  • 26.5k
5 votes
1 answer
359 views

Computations of divisor class monoids

Let me first recall some definitions from the very first pages of Bourbaki, Commutative Algebra, Chapter 7, "Divisors". Let $A$ be a (commutative) domain, $K$ its field of fractions. A ...
Joël's user avatar
  • 26k
5 votes
1 answer
597 views

Can every cancellative invertible-free monoid be embedded in a group?

A monoid is invertible-free if $xy=1$ implies $x=y=1$ for all $x,y$. Question: Can every cancellative invertible-free monoid be embedded in a group? I'm fairly sure that a quotient of the free product ...
David Pokorny's user avatar
5 votes
1 answer
170 views

Which homotopy types can be realized as the classifying space of a right-cancellative discrete monoid?

McDuff showed that every connected homotopy type can be realized as the classifying space of a discrete monoid, but the monoid she constructs has lots of idempotents. Question: Which homotopy types ...
Tim Campion's user avatar
  • 63.9k
5 votes
2 answers
236 views

Descent of flatness from algebras to monoids

Consider a morphism of commutative monoids $u\colon M\rightarrow N$. We say that $u$ is flat, if the tensor product functor $\bullet\otimes_MN$ from the category of $M$-modules to the category of $N$-...
Fred Rohrer's user avatar
  • 6,700
5 votes
1 answer
383 views

Projective resolutions for commutative monoids

What is the right notion of a projective resolution of a commutative monoid? The category Mon of commutative monoids has plenty of projective (and even free) objects. Indeed, for every set $X$ one ...
Hannes Thiel's user avatar
  • 3,497
5 votes
1 answer
284 views

Directed homotopy in the Cayley graph of a monoid

There is a the notion of the Cayley graph $C(G)$ of a group $G$ (which depends on a given presentation $G \cong \mathcal F(S) / \sigma$ where $\mathcal F$ is the free group functor and $\sigma$ some ...
JustAskin's user avatar
  • 190
5 votes
1 answer
378 views

Representations of products of groups (and monoids)

I have very little knowledge of representation theory, but the following has come up in my summer undergrad research project (relates to conformal field theory and geometric function theory). Suppose ...
ismythe's user avatar
  • 51
5 votes
1 answer
142 views

On the width of the Catalan monoid and the rank of K-groups of the Furstenberg transformation group

The semigroup algebra of the Catalan monoid is isomorphic to the incidence algebra of $P_n$, where $P_n$ is the poset consisting of subsets of { 1,...,n } where for two subsets $X \leq Y$ if and only ...
Mare's user avatar
  • 26.5k
5 votes
1 answer
303 views

A characterisation of faces of rational polyhedral cones

This is about a (seemingly) basic lemma about rational polyhedral cones that is sometimes used when working with toric varieties and is usually "left to the reader". Unfortunately, I could ...
Fred Rohrer's user avatar
  • 6,700
5 votes
1 answer
205 views

Topological category of topological monoids / operads

The category of topological monoids can be made into a topological category in a naive way. Namely, the space of all continuous homomorphisms between two topological monoids is a closed subspace of ...
Keke Zhang's user avatar
5 votes
1 answer
342 views

Can the trivial module be stably free for a monoid ring?

Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ ...
Benjamin Steinberg's user avatar
5 votes
1 answer
227 views

"Tietze-like transformations" for defining interesting bijections between algebraic structures

Consider the following two definitions of the natural numbers: The natural numbers are the algebraic structure $\mathbb{N}_1$ generated by one constant, $0$ and one unary function, $S$ (and no ...
Sophie Swett's user avatar
  • 1,173
5 votes
2 answers
252 views

Monoid of continuous self-maps of (real) surfaces

Let $S$ be a closed surface of genus $g > 0$ and $[S,S] = Hom(\pi_{1}(S),\pi_{1}(S))$ be the monoid of (homotopy classes of) continuous maps from $S$ to itself. Consider the semi-group $A$ of ...
Nick L's user avatar
  • 6,995
5 votes
2 answers
317 views

Proving that a semigroup is regular

In a number of diverse situations of interest to me (mostly associated with something called the abelian sandpile model), one can define a nonabelian semigroup generated by commuting elements $a_1,\...
James Propp's user avatar
  • 19.7k
5 votes
2 answers
341 views

Existence of a possible counterexample in automaton semigroups

In an attempt to resolve a question posed by Cain in his paper on Automaton Semigroups (open problem 6.12), I would like to know if there exists a finite semigroup $S$ satisfying the following ...
Alex McLeman's user avatar
5 votes
2 answers
332 views

Questions on weakly symmetric algebras

A finite dimensional algebra $A$ over a field $K$ is called weakly symmetric in case $soc(P)=top(P)$ for every indecomposable projective module $P$ and it is called symmetric in case $D(A) \cong A$ as ...
Mare's user avatar
  • 26.5k
5 votes
2 answers
402 views

Maximal commuting subsets of $\text{End}(X)$

Let $X$ be a set and let $\text{End}(X)$ be the set of all functions $f:X\to X$. We say that $f, g\in \text{End}(X)$ commute if $g\circ f = f\circ g$, and $S\subseteq \text{End}(X)$ is a commuting ...
Dominic van der Zypen's user avatar
5 votes
1 answer
304 views

flat maps of monoids which are not localizations

It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module. Rather, I am looking for extensions of rings which share certain properties of localizations, like ...
Ricardo Andrade's user avatar
5 votes
1 answer
498 views

Percolation in Cayley graphs of semigroups.

Percolation in Cayley graphs of groups are studied by many researchers. There are also the concept Cayley graphs for semigroups. Are there any research about percolation in Cayley graphs for ...
Jianrong Li's user avatar
  • 6,201
5 votes
1 answer
251 views

Monoid associated to $>2$-player Hackenbush

There is some literature on multiplayer combinatorial game theory, but as far as I can tell none of it follows the line of attack below. I'd love a pointer to a similar approach taken in the ...
Noah Schweber's user avatar
5 votes
1 answer
254 views

Examples of Yang-Baxter monoids

Then we say that an algebra $(X,f,g,\circ,1)$ is a Yang-Baxter monoid if it satisfies the following identities: $(X,\circ,1)$ is a monoid, $f(x,1)=1,f(1,x)=x,g(x,1)=x,g(1,x)=1$ $x\circ y=f(x,y)\circ ...
Joseph Van Name's user avatar
5 votes
2 answers
364 views

Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?

For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes: $$ \forall f,g,h\in G:hg(f)=h(g(f)) $$ Now suppose there is additional axiom, or constraint if you prefer, ...
James Smith's user avatar
5 votes
1 answer
293 views

semigroups acting as continuous functions on regular rooted trees

Let $T$ be a regular rooted tree. Make $T$ into a metric space by making each edge isometric to the unit interval. What is known about what semigroups can act as continuous functions on $T$ such ...
user12232's user avatar
5 votes
0 answers
191 views

Do most semigroups have a zero?

It is widely believed in finite semigroup theory that asymptotically almost all finite semigroups $S$, up to isomorphism, are 3-nilpotent, i.e., they satisfy $\#\{abc\,:\,a,b,c\in S\} = 1$. My ...
user513093's user avatar
5 votes
0 answers
187 views

Isbell duality for monoids and groups

Isbell Duality $\newcommand{\IsbellSpec}{\mathsf{Spec}}\newcommand{\IsbellO}{\mathsf{O}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\rmL}{\mathrm{L}}\newcommand{\rmR}{\mathrm{R}}\newcommand{\B}{\...
Emily's user avatar
  • 11.8k
5 votes
0 answers
107 views

Structure of well-ordered commutative monoids

Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where $\forall a\in M,\ 0\leq a$ $\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$ The first condition means $M$ will be ...
Pace Nielsen's user avatar
  • 18.7k
5 votes
0 answers
160 views

$S$ and $T$ globally isomorphic semigroups, with $S$ (commutative and) cancellative, iff $S$ is isomorphic to $T$?

Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication ...
Salvo Tringali's user avatar
5 votes
0 answers
138 views

Can we define partial group actions on (finite) sets via generators and relators?

Let $G = \langle Y | R \rangle$ be a finitely presented group. A partial group action on a set $X$ is a premorphism into the inverse semigroup $$ \mathcal I (X) = \{ f: A \to B : A, B \subseteq X, f\...
jpmacmanus's user avatar
5 votes
0 answers
225 views

The forgetful functor from Groups to Semigroups

While teaching this term I found myself reminded of the fact that the "usual" definition of a group homomorphism is really the definition of a semigroup homomorphism, applied to semigroups ...
Yemon Choi's user avatar
  • 25.8k
5 votes
0 answers
107 views

Heuristics for the word problem for monoids

The question is about a purely practical problem: Given is a list of identities, as in http://www.findstat.org/MapsDatabase/Mp00069: ...
Martin Rubey's user avatar
  • 5,822
5 votes
0 answers
64 views

Characters on monotone functions

Characters on the semigroup $(C_{+}^{b}(\mathbb{R}^{d}),+)$, i.e. on bounded positive continuous functions with the ususal pointwise addition, are known to be of the form $C_{+}^{b}(\mathbb{R}^{d})\ni ...
Tobsn's user avatar
  • 289
5 votes
0 answers
395 views

Derived tensor products and Tor of commutative monoids

Two commutative monoids $M,N$ have a tensor product $M\otimes N$ satisfying the universal property that there is a tensor-Hom adjunction for any other commutative monoid $L$: $$\text{Hom}(M\otimes N,L)...
John Berman's user avatar
5 votes
0 answers
99 views

Zappa-Szép products of the monoid of integers with itself

Question What are all the functions $\alpha , \beta : \mathbb{N} \to \mathbb{N}$ satisfying the following functional equations? $\bullet ~~~~ \alpha(0)=0, \quad \beta(0)=0\\ \bullet ~~~ \...
HeinrichD's user avatar
  • 5,482
5 votes
0 answers
637 views

Unique product groups (and semigroups)

A group $G$ is called a u.p.-group (short for unique product group) if for all nonempty finite subsets $A,B\subseteq G$, there exists an element $g\in A \cdot B$ which can be uniquely written as a ...
Pace Nielsen's user avatar
  • 18.7k
5 votes
0 answers
137 views

Pseudovarieties of monoids

All (pseudo)varieties considered here are (pseudo)varieties of monoids. It is known that any (finite or infinite) monoid that satisfies the identities \begin{equation} xhxyty = xhyxty, \quad xhytxy=...
E W H Lee's user avatar
  • 563
5 votes
0 answers
295 views

Orbit-Stabilizer theorem for continuous groups

The orbit-stabilizer relationship (also known as the orbit-stabilizer theorem) is very clear for finite groups. Is there an equivalent relation for continuous groups? Also, is there a similar notion ...
Arnab's user avatar
  • 615
5 votes
0 answers
2k views

Is the radical of a homogeneous ideal homogeneous?

Let $S$ be an $M$-graded $R$-algebra, where $M$ is some monoid, and $I\subset S$ an homogeneous ideal. The original, naïve, question, was: is it true that $\sqrt{I}$ is homogeneous? In this generality,...
quim's user avatar
  • 1,811
5 votes
0 answers
350 views

Chain/Hierarchy of Monoids

Let's assume that we have the following collection of structures: Some space $P$. Monoids $(M_{i+1},\circ_{i+1})$, and Actions $\bullet_{i+1}:M_{i+1}\times M_i\to M_i$, for $i\ge 0$ And $\bullet_{0}:...
supercooldave's user avatar
4 votes
4 answers
1k views

Why do we choose the standard total order on the integers?

I understand why the set of natural numbers $\mathbb N = \{ 0, 1, 2, \cdots \}$ is equipped with a total order. Indeed, every monoid has a pre-order, where $$n' \succeq n \quad \mathrm{if~and~only~if} ...
Tom LaGatta's user avatar
  • 8,512
4 votes
1 answer
417 views

Is a solvable group satisfying a semigroup law?

Let $S$ be the free semigroup on the set $\{x_1,\ldots ,x_n\}$, where $n$ is a positive integer. Suppose that $\mu=\mu (x_1,\ldots ,x_n)$ and $\nu = \nu (x_1,\ldots ,x_n)$ are two elements in $S$. We ...
mahdi meisami's user avatar
4 votes
2 answers
544 views

Membership problem in monoids

What is the simplest example of a monoid with undecidable membership problem? In other words, I'm looking for a concrete monoid $S$ such that there is no algorithm which takes elements $s_1,...,s_n$ ...
dan's user avatar
  • 41

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