Skip to main content

All Questions

Filter by
Sorted by
Tagged with
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-...
M. Winter's user avatar
  • 13.6k
18 votes
1 answer
783 views

Are there any "simple" monoids with intermediate growth?

The discovery of the Grigorchuk group which has intermediate growth caused a number of other such groups to be found, but they are all fairly complicated, and as far as I know none of them are ...
saolof's user avatar
  • 1,947
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 ...
Rob Myers's user avatar
  • 1,271
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
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 ...
Michał Masny'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
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
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,...
Salvo Tringali's user avatar
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. ...
Benjamin Steinberg's user avatar
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 ...
Michał Masny's user avatar
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 ...
Anton Fetisov's user avatar
8 votes
2 answers
596 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups,...
Salvo Tringali's user avatar
8 votes
1 answer
321 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively ...
Salvo Tringali's user avatar
8 votes
1 answer
436 views

Function $\phi$ such that $f(\phi(x,y)) = f(x) + f(y)$

I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$, and I am looking for a continuous (or at least measurable) function $\phi:\mathbb{R}^{2n}\to\mathbb{R}^n$ such that $f(\phi(x,y))=f(x)+f(y)$....
gmvh's user avatar
  • 3,065
8 votes
1 answer
238 views

Functions over monoids which factor in two different ways

This is a follow-up question to this MO question, which was asked by Richard Stanley in a comment to my answer there. Let $S$ be a commutative monoid and $f(x_1, \dots, x_n)$ be a function from $S^n$ ...
Tony Huynh's user avatar
  • 32.1k
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 ...
Ian Morris's user avatar
  • 6,206
8 votes
0 answers
285 views

Matrix decompositions as monoid isomorphisms. Ever considered before?

I've noticed some correspondences between some matrix decompositions and monoid isomorphisms (always to some free commutative monoid), in addition to the one I asked about in a previous question: ...
wlad's user avatar
  • 4,943
7 votes
2 answers
488 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ...
Salvo Tringali's user avatar
7 votes
1 answer
271 views

Algebraic proof that the monoid ring of a torsion-free monoid is reduced

In what follows, I say that a monoid $M$ is torsion-free if the $n$-th power map is injective for all $n \geq 1$. I have a proof of the following result: Claim: if $M$ is a torsion-free commutative ...
Moinsdeuxcat's user avatar
7 votes
1 answer
444 views

Which monoids have a faithful irreducible representation?

Let $*$ be a binary operation on a set $M$, with an identity element $e\in M$. A monoid representation of $(M,*,e)$ is a map $\delta:M\to (S\to S)$ for some set $S$, such that $\delta(e)=\mathrm{id}_S$...
Bjørn Kjos-Hanssen's user avatar
7 votes
1 answer
349 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 ...
HeinrichD's user avatar
  • 5,482
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 \...
Salvo Tringali'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
7 votes
0 answers
294 views

A minimal semigroup generating subset of the additive reals

I asked this on MSE, but I was told to ask it here because it is a difficult question. Consider the additive magma of the real numbers, $(\mathbb{R};+)$. Does there exist a subset $S$ of the reals ...
user107952's user avatar
  • 2,013
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 $M$?
Dominic van der Zypen's user avatar
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 $\Sigma(M,\times)$...
Michał Masny's user avatar
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^\...
Salvo Tringali's user avatar
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 ...
Michał Masny's user avatar
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 ...
Stefan Witzel's user avatar
6 votes
1 answer
185 views

A name for semigroups in which left and right principal ideals coincide

Is there any standard name for semigroups $S$ in which $xS=Sx$ for all $x\in S$? Examples of such semigroups are commutative semigroups and Clifford inverse semigroups.
Taras Banakh's user avatar
  • 41.8k
6 votes
1 answer
135 views

Automorphisms of special egg-box diagrams

By a egg-box diagram I will simply mean a (possibly infinite) rectangular array of holes, with some of the holes containing an egg (denoted by a filled-in circle) and the rest of the holes are empty (...
Pace Nielsen's user avatar
  • 18.7k
6 votes
0 answers
177 views

Is the monoid of all cancellative finitely generated commutative monoids cancellative?

$\DeclareMathOperator\Mon{Mon}\DeclareMathOperator\Grp{Grp}$Let $\Mon'$ be the set of isomorphism classes of (small) commutative, unital, cancellative ($a + t = b + t$ implies $a = b$) monoids. It is ...
Leo Herr's user avatar
  • 1,084
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 ...
goblin GONE's user avatar
  • 3,793
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 ...
Salvo Tringali's user avatar
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
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
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 $mz=z=zm$ ...
Benjamin Steinberg's user avatar
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,\...
James Propp's user avatar
  • 19.7k
5 votes
2 answers
332 views

Questions on weakly symmetric algebras

A finite dimensional algebra $A$ over a field $K$ is called weakly symmetric in case $soc(P)=top(P)$ for every indecomposable projective module $P$ and it is called symmetric in case $D(A) \cong A$ as ...
Mare's user avatar
  • 26.5k
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, ...
James Smith's user avatar
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 ~~~ \...
HeinrichD's user avatar
  • 5,482
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 ...
Thomas Klimpel's user avatar
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 ...
Jim Stasheff's user avatar
  • 3,880
4 votes
1 answer
239 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, ...
Salvo Tringali's user avatar
4 votes
1 answer
146 views

When is semigroup algebra local?

Let $G$ be a finite semigroup (or monoid if that helps) and $K$ a field. Question: When is the semigroup algebra $KG$ local? Here local means that there is a unique maximal right (or left) ideal. ...
Mare's user avatar
  • 26.5k
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\;\; ; \;...
M.H.Hooshmand's user avatar
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 ...
Denis's user avatar
  • 1,341
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
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 ...
David Handelman's user avatar