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1 vote
0 answers
69 views

How exactly to adapt Brown's collapse from monoids to algebras?

In The Geometry of Rewriting Systems (Springerlink behind paywall), Kenneth Brown describes a method to collapse the bar resolution of a monoid. Roughly: Given a simplicial set $X$ equipped with a ...
2 votes
1 answer
229 views

Has the "semidirect monoid of a semiring" been considered anywhere?

Given a semiring $S$, we get a monoid $M(S)$ as follows: The underlying set of $S$ is $S^2$ The identity element is $(0,1)$ The law of composition is given by $$(a,A)(b,B) = (Ba+b,AB),$$ where $a,A,b$...
1 vote
0 answers
116 views

A generalized Cauchy type functional equation

Let $(S,+)$ be an abelian semigroup . Let $f:S \to \mathbb C$ be a function such that for some positive integer $n>1$, $f(x+y)^n=(f(x)+f(y))^n,\forall x,y \in S$. Then is it true that $f(x+y)=f(x)...
10 votes
1 answer
274 views

A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids

In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following. ...
2 votes
0 answers
81 views

A semigroup property related to von Neumann regularity

A very common and useful notion in rings is that of von Neumann regular elements: those elements $a\in R$ such that there exists $b\in R$ satisfying $aba=a$. As this is a property defined solely ...
3 votes
0 answers
79 views

Semigroups containing an ideal with a local identity

I'm looking for some classes of semigroups containing a (proper) ideal with a local identity (i.e., ideal submonoid). Can somebody give some examples or/and theorems for the followings cases: (a) ...
4 votes
1 answer
434 views

Regarding a new algebraic structure

By "left semigroup-joined-semigroup" I mean an algebraic structures $(S,\cdot,*)$ such that both $\cdot,*$ are associative, and the following property holds (see this ) $$ x*(y\cdot z)=x*y*z\;\; ; \;...
21 votes
1 answer
759 views

Extending $\Bbb N$ to a semiring with isomorphic additive and multiplicative structure

Seen $(\Bbb N,+,\cdot)$ as a semiring, is it possible to extend it to a semiring $(R,+,\cdot)$ so that the additive and multiplicative monoids become isomorphic? This means there is some monoid-...
4 votes
1 answer
154 views

Right reversibility of submonoids of nilpotent groups

Let $G$ be a finitely generated group (optionally torsion-free). Let $N$ be a submonoid of $G$ (that is, a subsemigroup with $1$). A (cancellative) monoid/semigroup $S$ is right reversible if for ...
7 votes
1 answer
167 views

For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup?

Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \...
3 votes
0 answers
47 views

Counting the monic atoms $f$ in the semiring $\mathbf N[x]$ with $f(0)=1$, bounded coefficients, and degree $k$ (in the limit as $k \to \infty$)

Let $H$ be the multiplicative monoid of the (usual) semiring of polynomials in one variable $x$ with coefficients in $\mathbf N$. Given $\alpha, k \in \mathbf N$, denote by $\mathcal A_k(\alpha)$ the ...
4 votes
1 answer
173 views

On the factorization of powers of atoms in the ring of integers of a number field

Let $H$ be a multiplicatively written monoid with identity $1_H$. An atom of $H$ is a non-unit element $a \in H$ that doesn't split into the product of two non-unit elements. Given $x \in H$, we ...
2 votes
0 answers
62 views

Extensions of an ideal-theoretic criterion for a monoid to be BF

Let $H$ be a multiplicatively written, commutative monoid. We denote by $H^\times$ the set of units (or invertible elements) of $H$, and by $\mathcal A(H)$ the set of atoms (or irreducible elements) ...
1 vote
1 answer
96 views

If $H$ is commutative and unit-cancellative, then so is the monoid of non-empty ideals of $H$

Let $H$ be a (multiplicatively written) commutative monoid with identity $1_H$. Given $X, Y \subseteq H$, we take $$XY := \{xy: x \in X,\, y \in Y\}.$$ We call a set $I \subseteq H$ an ideal of $H$ ...
4 votes
1 answer
385 views

Which monoids can be realized as the monoid of ideals of a commutative monoid?

Let $H$ be a commutative monoid (written multiplicatively). We say that a set $I \subseteq H$ is an ideal of $H$ if $IH = I$. The set $\mathcal I(H)$ of all ideals of $H$ is made into a (commutative) ...
3 votes
2 answers
165 views

Weak ideal systems $r$ for which the $r$-coheight satisfies a kind of triangle inequality

Let $H$ be a multiplicatively written, commutative monoid with identity $1_H$, and let $\mathcal P(H)$ be the power set of $H$. If $X, Y \subseteq H$, we will set $$XY := \{xy: x \in X,\, y \in Y\}.$$ ...
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/...
6 votes
2 answers
422 views

Monoids in which every prime is an atom

Let $H$ be a multiplicatively written monoid with identity $1_H$. We write $H^\times$ for the set of units (or invertible elements) of $H$. We say that an element $a \in H$ is an atom if $a \notin H^\...
2 votes
1 answer
122 views

If $H$ is essentially equimorphic to $K$, then is $H$ atomic only if so is $K$?

I will first state my question, and then give all the relevant definitions. Q. Let $H$ and $K$ be monoids, and assume $H$ is essentially equimorphic to $K$. Is it true that $H$ is atomic only if so ...
2 votes
1 answer
211 views

Terminology for a monoid $(H, \cdot)$ s.t. $ax=a$ or $xa =a$ only if $x$ is a unit

Let $(H, \cdot)$ be a (multiplicative) monoid. Is there any consolidated name for the following Property $\text{(P)}$, or for the class of monoids for which it is satisfied? $$\text{(P) If }\,xy = x\...
3 votes
2 answers
101 views

A non-reduced, commutative BF-monoid s.t. $au = u$ for all $a \in \mathcal A(H)$ and $u \in H^\times$

Let $H$ be a monoid, and denote by $H^\times$ and $\mathcal A(H)$, respectively, the set of units (or invertible elements) and the set of atoms (or irreducible elements) of $H$ (an element $a \in H$ ...
5 votes
1 answer
243 views

Terminology for a monoid $H$ s.t. $xy \in H^\times$ only if $x, y \in H^\times$

The title has it all. Is there any consolidated terminology for referring to a (multiplicative) monoid $H$ such that $xy \in H^\times$ (if and) only if $x, y \in H^\times$? Here is a short list of ...
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 ~~~ \...
3 votes
1 answer
122 views

A BF-monoid $H$ s.t. $H^\times$ is not divisor-closed

Let $H$ be a (multiplicative) monoid, and denote by $H^\times$ the set of units of $H$ and by $\mathcal A(H)$ the set of atoms of $H$ (let me recall that an element $a \in H$ is an atom if (i) $a \...
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 ...
0 votes
0 answers
101 views

Spherical Rings

My question is concerned with filtered rings. It is a classical result that if $R$ is a finitely generated commutative ring graded by a semigroup $S$ then $S$ is also finitely generated. The reverse ...
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, ...
4 votes
1 answer
91 views

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

I want to show that $End_0 (B_n(G)) = \cup\phi_{\sigma,g} \cup C_{I(B_n(G))}$, where $\phi_{\sigma,g} : B_n(G) \rightarrow B_n(G) $ is an endomorphism is defined by $(i,a,j)\phi_{\sigma,g} = (i\sigma ...
4 votes
1 answer
215 views

Name for a regular band

Is there a name for regular bands that satisfy $xyx=yx$ for all $x$,$y$?
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 $mz=z=zm$ ...
2 votes
0 answers
139 views

Goldie's Theorem for Semigroups

Goldie's theorem is a theorem in noncommutative ring theory that gives a clear picture of semiprime Noetherian rings (actually a slightly broader class). Let $R$ be a semiprime Noetherian ring. The ...
10 votes
1 answer
243 views

Can a semigroup with zero be globally isomorphic to a semigroup without zero?

This is not a great question for sure and it may even be trivial for all I know, but a couple of years ago, when I still thought I'd be a mathematician, I spent quite a lot of time thinking about it ...
3 votes
1 answer
191 views

Is every $n$-ary semigroup a subalgebra of an algebra derived from a binary semigroup?

Let $(A,f)$ be an $n$-ary semigroup ($n \ge 2$). Then there exists a ($2$-ary) semigroup $(\overline A,*)$ with an inclusion homomorphism $A \hookrightarrow \overline A$ such that that the restriction ...
8 votes
2 answers
1k views

Sets of matrices which are irreducible but not strongly irreducible

A set of $d \times d$ real or complex matrices is commonly called irreducible if those matrices do not jointly preserve a linear subspace with dimension strictly between zero and $d$. A stronger ...
6 votes
1 answer
1k views

Who coined "mob" and "clan" and why these words?

A mob is a word used for a topological semigroup which is a Hausdorff space. A clan is a compact connected mob with a two-sided identity element. Who used these words with these meanings first and ...
3 votes
0 answers
314 views

Certain conditions on cancellative semigroups

This is extracted from this question following Benjamin Steinberg's suggestion. For a semigroup $S,$ let $P(S)$ denote the power semigroup of $S,$ which is made up of all non-empty subsets of $S$ ...
4 votes
1 answer
428 views

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 ...
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,\...
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 $X,$...
4 votes
3 answers
1k views

Is there a general notion of semigroup action?

The analogous result to Cayley’s theorem or Yoneda lemma in semigroup theory represents semigroups as semigroups of functions from a set to itself. This suggests that a semigroup action consists of a ...
0 votes
2 answers
240 views

Using group presentation for its corresponding semigroup?

Somewhere Colin M. Campbell noted: If $A$ is a semigroup defined as $$A=Sg(\pi)=\langle a_1,\cdots, a_d\mid u_1=v_1,\cdots,u_e=v_e\rangle $$ then the same generators with the same relations can ...
4 votes
2 answers
507 views

Are algebraic structures uniquely identifed by their free objects?

It might be a naive question, as I am not a specialist in this field. This is a follow-up to this question. I want to study varieties of objects generalizing ordered monoids, in particular using an ...
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 ...
2 votes
2 answers
278 views

How much information does the multiplicative semigroup of an algebra contain?

How much do we know about an given algebra when we only know its semigroup strucure under the product law? How far can two algebras be distinguished by knowing only their semigroup strucure? The ...
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 ...
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. ...
4 votes
1 answer
423 views

What is the formula for the commutative multiplication on CP(infinity)?

There is a classic formula for maps $\mathrm{CP}(r) \times \mathrm{CP}(s) \to \mathrm{CP}(r+s)$ or maybe $r+s+1$ using Plücker coordinates - IF memory serves. In the limit we get the abelian ...
15 votes
1 answer
2k views

Automorphisms of $P(\Bbb N)$

I believe I've proved that the power semigroup of non-negative integers with addition has a trivial automorphism group. The proof is a bit long, completely elementary and rather unremarkable (as the ...
6 votes
1 answer
903 views

Monoids and groups of fractions

Let $G$ be a group containing a monoid $M$ that spans $G$ as a group. Is it possible to have a proper quotient $\varphi \colon G \to Q$ of $G$ such that the restriction of $\varphi$ to $M$ is ...
11 votes
1 answer
949 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 ...