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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
7 votes
3 answers
525 views

Is the class of inverse semigroups globally determined?

This question is a follow-up to this one I asked on math.stackexchange. I've decided to ask here because I believe this is a research-level question. I'm sorry if I'm wrong -- I'm not a researcher ...
Michał Masny's user avatar
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
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
11 votes
3 answers
939 views

What is the smallest variety of algebras containing all fields?

A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative ...
Thomas Klimpel's user avatar
15 votes
2 answers
1k views

Exact sequence of monoids

What is the right definition of an exact sequence of monoid homomorphisms? I can't seem to find a consistent in my searches; indeed Balmer (Remark 2.6, http://www.math.ucla.edu/~balmer/Pubfile/...
John Voight's user avatar
  • 3,009
11 votes
1 answer
948 views

Magma "actions" (or alternatively, "What is the Yoneda lemma for magmas?")

Arguably the most import thing about groups, semigroups and more generally categories, is that they can act on sets (or even collections of sets in the case of a category). This is the basis for all ...
Mikola's user avatar
  • 2,392
5 votes
2 answers
754 views

Do all finitely generated nilpotent semigroups have polynomial growth?

The notion of nilpotency passes nicely from groups to semigroups. Define $q_1(x,y)=xy$ and $$q_{i+1}(x,y,z_1,\cdots,z_i)=q_i(x,y,z_1,\cdots,z_{i-1})z_iq_i(y,x,z_1,\cdots,z_{i-1})$$ inductively for all ...
Victor's user avatar
  • 1,437
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
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
4 votes
2 answers
364 views

General linear inverse monoid

Let $V$ be a finite dimensional vector space over some field (say, $\mathbb C$). Consider the set $\operatorname{GLI}(V)$ of all linear isomorphisms between subspaces of $V$. This is a monoid under ...
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

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