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5 votes
2 answers
537 views

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
0 votes
1 answer
403 views

When does a power semigroup have a zero, and what can the zero be?

Let $S$ be a semigroup. The power semigroup of $S$ is the set $P(S)=2^S\setminus\lbrace\varnothing\rbrace $ with the operation $$AB=\lbrace ab\ |\ a\in A,\ b\in B\rbrace.$$ This operation is ...
Michał Masny's user avatar
1 vote
2 answers
442 views

submonoids of Z_n

Anyone knows how to describe explicitly the submonoids of Z_n, regarded as a multiplicative monoid?
Camilo Salazar's user avatar
2 votes
0 answers
51 views

Existence of a canonical embedding from a quotient of $\mathscr{F}^\ast(\mathcal A(H))$ to another

Let $H$ be a multiplicative, commutative monoid, and denote by $H^\times$ the group of units of $H$, by $\mathcal A(H)$ the set of atoms (or irreducible elements) of $H$, by $\mathscr{F}^\ast(\mathcal ...
Salvo Tringali's user avatar
3 votes
1 answer
297 views

Is the universal inverse semigroup of a commutative semigroup an embedding?

The question of existence of a universal inverse semigroup of an arbitrary semigroup has been answered before (this is a construction similar to the Grothendieck group). Let's refer to the universal ...
Thomas Klimpel's user avatar
3 votes
1 answer
243 views

Embedding Semigroups in Rings

Let $S$ be a finite commutative semigroup with identity. Under what conditions (on the semigroup $S$) it is possible to find a ring $R$ such that the multiplicative structure of $R - \{0\}$ is ...
zacarias's user avatar
  • 801
2 votes
1 answer
290 views

Idempotents in Green J classes

I recently read this article Syntactic semigroups. In page $8$, he speaks about a J class having an idempotent is called regular: A $\mathcal J$-class containing an idempotent is called regular. ...
Mikasa's user avatar
  • 233
4 votes
2 answers
607 views

Invertible elements in monoid rings of unital monoids without non-trivial invertible elements

This question is somewhat related to Tilmans notorious problem in this post. Let $(M,\cdot)$ be a monoid with unit $1$ and set $$(M,\cdot)^{\times} := \lbrace x \in M \mid \exists y \in M : xy=yx=1 \...
Andreas Thom's user avatar
  • 25.5k
3 votes
0 answers
126 views

dual composition of binary relations

I'm not sure if this is of any interest at all, but I spent some time looking at it a couple of years ago so I'd like to ask for input on this. Given two binary relations $\rho,\,\sigma$ on a set $X,$...
Michał Masny's user avatar
0 votes
1 answer
128 views

What are the semigroups in which congruence classes can be multplied like sets?

For a semigroup $S$ and a congruence $\rho$ on $S$, let's say that $\rho$ is good when for all $a,b\in S$ we have that $[ab]=[a][b],$ where $[x]$ denotes the congruence class of $x$ modulo $\rho$ and ...
Michał Masny's user avatar
2 votes
1 answer
1k views

monoid ring and some structure within it - how is it called?

I am amateur - mathematics is my hobby, and I find some strange structure working with toy matrices structure so I try to ask some questions regarding it. Let me allow to introduce some structure ...
kakaz's user avatar
  • 1,626
1 vote
2 answers
394 views

Relations in matrix semigroups

Suppose that $A_1, \dots, A_k \in M_n(\mathbb{Q})$ and $S$ is the semigroup generated by them. Two questions: are there always a finite set of relations $\{R_i\}$ among the $A_j$ such that $S$ is ...
Victor Miller's user avatar
0 votes
0 answers
72 views

Decomposition results for locally commutative semigroups

Every finite abelian group is the direct product of its cyclic groups of prime order, and every commutative monoid divides a product of its cyclic submonoids. Could these results generalized to ...
StefanH's user avatar
  • 798

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