All Questions
113 questions
4
votes
1
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173
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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 ...
- 10.5k
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 $H$ ...
- 10.5k
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) ...
- 10.5k
4
votes
1
answer
364
views
Values attained by the coheight of $(H \setminus H^\times)^k$ as a function of $H$ and $k$
Edit (Apr 24, 2017). I'm updating this post in the light of the latest developments of a related thread.
Let $H$ be a multiplicatively written, commutative monoid, and set $M := H \setminus H^\times$,...
- 10.5k
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 $$XY := \{xy: x \in X,\, y \in Y\}.$$
...
- 10.5k
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) ...
- 10.5k
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^\...
- 10.5k
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 ...
- 10.5k
9
votes
2
answers
1k
views
Ternary associative multiplication
In this answer Brian M. Scott describes the following generalization of a binary associative multiplication to a ternary one: it is a function $$[\cdot,\cdot,\cdot] : G\times G \times G \to G$$ such ...
- 4,780
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 $a \in H$ ...
- 10.5k
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 ~~~ \...
- 5,482
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 ...
- 10.5k
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 \...
- 10.5k
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 ...
- 10.5k
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 ...
- 501
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 ...
- 143
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?
$$\text{(P) If }\,xy = x\...
- 10.5k
7
votes
1
answer
350
views
Pushouts of injective monoid homomorphisms
Given a pushout square in the category of monoids
$$\begin{array}{ccc}A & \rightarrow & M \\ \downarrow && \downarrow \\ N & \rightarrow & P\end{array}$$such that $A \to M$ and ...
- 5,482
5
votes
1
answer
342
views
Can the trivial module be stably free for a monoid ring?
Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ ...
- 38.6k
5
votes
2
answers
364
views
Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?
For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes:
$$
\forall f,g,h\in G:hg(f)=h(g(f))
$$
Now suppose there is additional axiom, or constraint if you prefer, ...
- 505
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 ...
- 1,445
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 ...
- 6,206
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 ...
- 190
5
votes
3
answers
851
views
What are some examples of non-commutative $\mathbb{Q}$-monoids and/or $\mathbb{R}$-monoids?
Definition 0. Let $R$ denote a commutative semiring with $0$ and $1$. By an $R$-monoid, I mean a monoid $M$ equipped with an action $R \times M \rightarrow M$ denoted $r,m \mapsto m^r$, satisfying the ...
- 3,793
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 ...
- 6,870
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 ...
- 1,445
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$ ...
- 1,445
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 ...
- 1,445
6
votes
2
answers
237
views
Realizing a monoid as $\mathrm{End}(G)$ for some graph $G$
This question relates to Realizing groups as automorphism groups of graphs.
Given a monoid $M$, is there a graph $G$ such that the endomorphism monoid $\textrm{End}(G)$ is isomorphic to $M$?
- 48.9k
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,$...
- 1,445
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 ...
- 2,464
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 $$ then the same generators with the same relations can ...
- 233
16
votes
1
answer
548
views
Does every commutative variety of algebras have a cogenerator?
By a commutative variety $\mathcal{V}$ I mean a classical variety of algebras for some $(\Sigma,E)$, such that each pair of operations in $\Sigma$ commutes.
Equivalently (i) every interpretation of ...
- 1,271
4
votes
2
answers
507
views
Are algebraic structures uniquely identifed by their free objects?
It might be a naive question, as I am not a specialist in this field.
This is a follow-up to this question.
I want to study varieties of objects generalizing ordered monoids, in particular using an ...
- 1,341
3
votes
3
answers
345
views
Examples of cancellative normal semigroups
I've got a couple of things to test against normality in cancellative semigroups. A normal semigroup $S$ is one in which for any $x\in S$ we have $xS=Sx.$ This implies the Ore condition $$x,y\in S\...
- 31
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 ...
- 798
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 ...
- 221
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,\...
- 19.7k
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 ...
- 3,880
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 ...
- 1,445
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 ...
- 2,050
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. ...
- 233
6
votes
2
answers
416
views
For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring?
Let $(M,\times)$ be a monoid with zero. Let $\Sigma(M,\times)$ be the set of binary operations $+$ on $M$ such that $(M,+,\times)$ is a ring. Let $\sim$ be an equivalence relation on $\Sigma(M,\times)$...
- 1,445
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 ...
- 1,445
10
votes
5
answers
1k
views
On the notion of partial semigroup
A partial binary operation on a set $X$ is just a (partial) function $\varphi: X \times X \rightharpoonup X$ (I'm using \rightharpoonup for partial maps), and a partial magma is a pair $\mathbb M = (M,...
- 10.5k
3
votes
2
answers
1k
views
Semiring naturally associated to any monoid?
For any monoid $M$, we can naturally construct a semiring $S$ as follows:
Let the additive monoid of $S$ be the free commutative monoid on $M$
Let the multiplicative monoid of $S$ be $M$
Then, if ...
- 4,897
4
votes
1
answer
215
views
Name for a regular band
Is there a name for regular bands that satisfy $xyx=yx$ for all $x$,$y$?
user22363
1
vote
2
answers
375
views
What are the monoids in which every globally idempotent subsemigroup contains the identity element?
A semigroup is called globally idempotent when for any $x\in S$ there are $y,z\in S$ such that $x=yz$.
Is there a name for monoids whose every globally idempotent subsemigroup contains the identity ...
- 1,445
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 ...
- 1,445
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 ...
- 801