Catenarity of monoid algebras

Let R be a commutative ring, let M be a commutative monoid, and let R[M] denote the corresponding monoid algebra. Suppose further that R is universally catenary. One may ask for conditions on M such that the ring R[M] is catenary. I know of the following results: If M is of finite type then R[M] … Read more

Do you know rings without involutions, auto-anti-isomorphics? In that case, what is the minimal example?

Do you know rings without involutions, but auto-anti-isomorphic (isomorphic to their opposite)? In that case, what is the minimal example? If a ring has an involution f, then f is an anti-automorphism; but there can be auto-anti-isomorphic ring without involutions. Answer AttributionSource : Link , Question Author : José María Grau Ribas , Answer Author … Read more

Is the group ring of an amenable group, viewed as multiplicative monoid, amenable?

Motivated by this question, it seems natural to ask the following: Question 1: Is there a [finitely generated discrete] torsion-free virtually Abelian (but not Abelian) group G so that the multiplicative monoid R∖{0} of R=F2[G] its group ring over F2 is [right-]amenable? Note that for virtually Abelian groups the zero-divisor conjecture holds, so it makes … Read more

What is a “cusp” (“кусок”) in relation to Guba’s embedding theorem?

I’m confused by the definition of a “cusp” as found in V.S. Guba, Conditions for the embeddability of semigroups into groups, Math. Notes 56 (1994), Nos. 1-2, 763-769 (link). In the words of Mark Sapir (from an answer that has meanwhile been removed from this thread), “cusp” is a «weird translation» into English of the … Read more

Is every invertible-free cancellative monoid action represented by “shifting” certain maps?

[Note: This question is closed. It’s current content reflects a draft of a potential new question, modified from the original by adding conditions to the premises; see comments] Let W,X be cancellative invertible-free [1] monoids. A map e:W→X is a homography [2] if it is non-decreasing in the prefix order and e(1)=1. The shift of … Read more

Strictly totally ordered semigroups – Looking for references

Let A=(A,⋅) be a semigroup (written multiplicatively). We say that A is linearly orderable if there exists a total order ≤ on A such that ac<bc and ca<cb for all a,b,c∈A with a<b (note strict inequalities). Some examples of linearly orderable semigroups are: the real numbers with the usual addition; the positive integers divisible only … Read more

Embedding a linearly ordered free monoid into a linearly ordered group

A linearly ordered (shortly, l.o.) monoid is a triple $\mathbb M = (M, \cdot, \le)$ for which $(M, \cdot)$ is a (multiplicatively written) monoid and $\le$ is a total order on $M$ such that $xy < xz$ and $yx < zx$ for all $x,y,z \in M$ with $y < z$. In particular, $\mathbb M$ is … Read more

Cancellable elements of a power semigroup

For a semigroup $S,$ its power semigroup $P(S)$ is the semigroup of all non-empty subsets of $S$ with the operation given by $AB=\{ab\,|\,a\in A,b\in B\}.$ I would like to know about the cancellable elements of $P(S)$ given some knowledge of cancellability in $S.$ If $s\in S$ is left-cancellable in $S,$ then $\{s\}$ is also left-cancellable … Read more

Endomorphism of Brandt Semigroup Bn(G)B_n(G), where GG is a finite group

I want to show that End0(Bn(G))=∪ϕσ,g∪CI(Bn(G)), where ϕσ,g:Bn(G)→Bn(G) is an endomorphism is defined by (i,a,j)ϕσ,g=(iσ,ag,jσ) and σ∈Sn and g∈End(G), CX is the set of all constant map on X and I(Bn(G)) is the set of all idempotents in Bn(G). I have proved that ϕσ,g and constant maps are endomorphism, so ∪ϕσ,g∪CI(Bn(G))⊆End0(Bn(G)) we have proved the … Read more