Tagged Questions

1answer
36 views

1answer
63 views

About Rees homomorphism

I am came across the notion Rees Congruence for semigroups. J. Howie defines it as $$\rho_I=(I\times I)\cup {1_S}$$ wherein $I$ is an ideal of semigroup $S$ ...
1answer
35 views

One-sided and two-sided cancellability

Is there a semigroup $S$ with right- and left-cancellable elements but no elements cancellable from both sides? $s\in S$ is left-cancellable if for any $a,b\in S$ we have $$sa=sb\implies a=b$$ and ...
1answer
75 views

An estimate involving gaps in a subsemigroup of $(\mathbb N,+)$

I think this question can be solved by a high school student, maybe there is some trick on it or I'm forgetting something. Before my question, some background is required: Definition: A ...
1answer
116 views

About automorphisms of commutative semigroups

Suppose that $M$ is a commutative monoid and that the product $P$ of $M$ and the nonnegative integers $\mathbb{N}$ with addition has no nontrivial automorphisms. The set $S$ of pairs $(m,n)$ in $P$ ...
1answer
23 views

Finite Linearly Ordered Abelian Monoids

This question concerns the proper definition of the phrase "finite linearly ordered abelian monoids". The sequence A030453 of OEIS counts the number of "finite linearly ordered abelian monoids". The ...
1answer
109 views

Who first studied semilattices?

Historically, who first studied semilattices, as opposed to lattices or Boolean algebras? (With or without identity, I do not mind.)
0answers
26 views

Krull dimension and Hilbert polynomial of graded rings

Let $P \subseteq \mathbb{R}^n$ be a polytope with integral vertices, and $Q := \mathbb{R}_+ (P \times \{1\}) \cap \mathbb{Z}^{n+1}$ be the monoid of all lattice points of the cone generated by $P$ in ...
1answer
143 views

The lattice points of a f.g. rational cone form a f.g. monoid.

In their book "Polytopes, Rings, and K-Theory" Bruns and Gubeladze sketch an alternative approach to Gordan's Lemma, which is stated in the headline (f.g. = finitely generated). I don't understand the ...
1answer
64 views

If $G,H$ are monoids and not groups, prove that $f(e_G)=e_H$ may be wrong.

Problem: If $f:G\longrightarrow H$ is a homomorphism of groups, then $f(e_G)=e_H$ and $f(a^{-1})=f(a)^{-1}, \forall a\in G$. Show by example that the first conclusion may be false if $G, H$ are ...
0answers
63 views

How many associative ternary operations there are on a finite set?

We know that algebraic operation a is function $f:\underbrace{\left ( X\times X\times \cdots \times X\right )}_{t\ \text{times}}\rightarrow{X}$ If $X$ is a set and and cardinality is $|X|=n$ then ...
1answer
50 views

Commutative Cancellative Semigroup: When is an irreducible element prime?

Suppose $(S,\cdot)$ is a semigroup with neutral element $e$ (i.e. $xe=ex=x$ for all $x\in S$) and the following properties: S is commutative: $xy=yx$. S is cancellative: $xy = xz$ implies $y = z$. ...
1answer
34 views

Involution on semigroups with identity

I'm trying to understand the following: Let $S$ be a semigroup. By an involution on $S$ we mean a map $* : S \to S$ satisfying for all $a,b\in S$ $(ab)^*=b^*a^*$ $(a^*)^*=a$ My problem is the ...
1answer
22 views

Does a discrete semilattice of finite width have finite number of points between any pair of elements.

Let $(S,<)$ be a countably infinite semilattice such that: 1) $S$ is no-where dense - i.e. there does not exist a subset $T$ such that for all $a,b\in T$ with $a<b$, there exists $c\in T$ with ...
1answer
34 views

A construction using a semiring

In page number 3 of J.S. Golan's book Semirings and their applications, there is a result which says that if $R$ is a hemiring and $S$ is a subhemiring of $R$ which is a semiring having ...
1answer
29 views

Semigroups — Underlying Set Non Empty or it can be Empty?

Many authors consider Semigroups being Non Empty sets. Others include empty sets as Semigroups. What is the rationale behind both choices ?
0answers
54 views

semidirect product of semigroups

Let $S,T$ be two semigroups. A function $f:S\to T$ is called multiplicative if for any $x,y\in S$ we have $f(xy)=f(x)f(y)$. If $T=S$ then $f:S\to S$ is called automorphism on $S$. A function ...
2answers
35 views

Semigroups and units

I am working my way though Basic Algebra 1. I am currently on chapter 1 , and more specifically I busy with the following exercise: "Let $S$ be a set a define a product in $S$ by $ab=b$. Show that ...
1answer
45 views

How to show $c-b\lt b-a$

The question: Let $G$ be an Arf semigroup and $a\lt b\lt c$ be three consecutive elements in $G$. How to show that $c-b\lt b-a$ and how to show that this is not necessarily the case for every ...
0answers
66 views

What are $\Gamma$-semigroups?

I have some problems with $\Gamma$-Semigroups, the definition that I've found is A $\Gamma$-Semigroup is a pair $(M,\Gamma)$ defined as follow If $x,y$ and $z$ are in $M$ and $\alpha$ and ...
1answer
34 views

Isomorphism of direct product of semigroups

I would appreciate some help with the following problem. Consider four semigroups $A,B,C,D$. I was able to prove that $A\cong C\wedge B\cong D$ implies $A\times B\cong C\times D$. But does also ...
1answer
41 views

positive integer polynomial under the usual polynomial multiplication

consider the set of polynomials with positive integer coefficients together with the operation, usual multiplication of polynomials. now my first question is does this set together with the ...
1answer
36 views

1answer
86 views

What does $\mathcal{J}$ stand for in Green relations?

Following this book The Algebraic Theory of Semigroups, Volume I , we see that: $a\mathcal{L}b$ means $a$ and $b$ generate the same principle left ideal of the semigroup $S$. $a\mathcal{R}b$ means ...
1answer
49 views

Testing for the cancellation laws

When I'm given a presentation of a finitely generated semigroup/monoid, are there any tricks I could use to check if it is cancellative on both sides? I'm not asking for a general algorithm, as I ...
1answer
107 views

Naturally Ordered Semigroup: Why does axioms imply order of group is infinite countable ? Why are every group equal up to isomorphism?

Naturally Ordered Semigroup: http://www.proofwiki.org/wiki/Definition:Naturally_Ordered_Semigroup I know $\mathbb N$ is a naturally ordered semigroup by definition. Also I know that the naturally ...
0answers
50 views

What can we learn about a magma by studying these monoids?

Given a magma $(X,*)$, we get three monoids in the following way. First, define a pair of functions $L,R : X \rightarrow (X \rightarrow X).$ $$(Lx)(y) = x*y,\quad (Rx)(y) = y*x$$ Then each of the ...
1answer
48 views

Does Cartesian Product and Collection of all Sets Perform a Semigroup?

We know that the Cartesian Product is a binary operation. Also it is an associative operation. We know that Cartesian Product of two set is again set, there is even closure axiom. So I need to know ...
0answers
60 views

Principal ideal in a semigroup ring.

Let $S= \langle(0,1),(3,2),(5,2) \rangle \subset \mathbb{N}^2$ be a semigroup and consider the semigroup ring $K[S]$, with $K$ a field. We can consider the principal ideal generated by ...
2answers
56 views

If a semigroup satisfies these identities, is it necessarily commutative?

Suppose a semigroup $X$ satisfies the following identities. $$xya\equiv yxa,\quad axy \equiv ayx$$ Without assuming anything further, can we deduce that $X$ satisfies $xy \equiv yx$? In particular, ...
1answer
28 views

Do cosets of this semigroup partition this ring?

Let $R = \Bbb{Z}^3$ be the usual direct product ring, and let $S = \{ (k^a, k^b, k^c) : k \in \Bbb{Z} - \{0\}\}$ be a subsemigroup of $R$. Then do the cosets $aS$ partition $R$? Let $R^* = S^{-1}R$ ...
1answer
53 views

Unique idempotents in semigroups

In a finite semigroups every element has a unique idempotent power, just take $s^{n!}$ where $n = |S|, s \in S$. In an infinite semigroup there are clearly elements without idempotents, just take ...
2answers
76 views

On the powerset of a monoid, does the second operation play nice with the first? Indeed, is it even useful?

Let $X$ denote a monoid. Then we can make $Y = \mathcal{P}(X)$ into a monoid, too. Define $$AB = \{ab \mid a \in A, b \in B\}$$ for all $A,B \in Y.$ We see immediately that $1$ (shorthand for $\{1\}$) ...
1answer
52 views