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Is there a characterization of monoids that distribute over each other?

Let $(M, e_1, \times_1, e_2, \times_2)$ be an algebraic structure such that $(M, e_1, \times_1)$ and $(M, e_2, \times_2)$ are monoids $x \times_1 (y \times_2 z) = (x \times_1 y) \times_2 (x \times_1 ...
Keith's user avatar
  • 621
1 vote
0 answers
88 views

What is a quantum condensed space?

Due to a categorical equivalence involving compact topological spaces and unital commutative $\mathrm{C}^*$-algebras, there is a practise involved in so-called quantum mathematics where a ...
JP McCarthy's user avatar
  • 1,037
8 votes
1 answer
437 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
1 vote
1 answer
80 views

Hilbert symbol of a quaternion algebra given ramified places

I am reading the paper: https://projecteuclid.org/journals/experimental-mathematics/volume-17/issue-3/Derived-Arithmetic-Fuchsian-Groups-of-Genus-Two/em/1227121388.full in order to find an explicit ...
ah--'s user avatar
  • 155
11 votes
0 answers
427 views

Is there a theory of completions of semirings similar to $I$-adic completions of rings?

Let $L = \text{Con } (\mathbb{N}, 0, +) \setminus \Delta$ be the lattice of monoid congruences on the naturals, excluding the trivial congruence. As it happens, every $\theta \in L$ is the meet of ...
Keith's user avatar
  • 621
47 votes
10 answers
6k views

Algebraic theorems with no known algebraic proofs

What are some good examples of algebraic theorems that have no known algebraic proofs? A few I know concern classifications of (not necessarily associative) division algebras over $\mathbb{R}$: the ...
2 votes
0 answers
92 views

Geometric interpretation of flags and the role of the rook monoid and Kazhdan–Lusztig theory in $M_n(\mathbb{C})$

Let $G = GL_n(\mathbb{C})$, $B$ be its Borel subgroup, and $P$ a parabolic subgroup. The space $G/B$ corresponds to complete flags in $ \mathbb{C}^n$, and $G/P$ corresponds to partial flags. The ...
Learner's user avatar
  • 141
2 votes
0 answers
95 views

Free, easy-to-use program for noncommutative algebra over finite fields

I am looking for a computer program that can handle computations in noncommutative algebra over a finite field of prime order $p$. My requirements are: The program should be free, as I do not have ...
gualterio's user avatar
  • 1,013
0 votes
0 answers
61 views

Defining rank of an abelian subgroup using the second centralizer

I recently posted this on MSE, but didn't receive any feedback; so I'm posting it on MO. I recently came across this article which explored the maximal abelian subgroups of the symmetric group $S_n$. ...
dbossaller's user avatar
0 votes
0 answers
32 views

Right maximal ideals in skew-Laurent rings over division Rings

Let $R$ be a Noetherian domain, and let $D$ denote its division ring. Define $S = D_q[x_1^{\pm 1}, x_2^{\pm 1}]$ as an iterated skew-Laurent polynomial ring with the relation $x_1 x_2 = q x_2 x_1$. Is ...
Sky's user avatar
  • 923
0 votes
0 answers
92 views

Finitely generated module over a skew Laurent polynomial ring $\mathbb{K}[x_1^{\pm 1},x_2^{\pm1}]$

Let $A := \mathbb{K}_{\Lambda}[x_1^{\pm 1}, x_2^{\pm 1}, x_3^{\pm 1}, x_4^{\pm 1}]$, $B := \mathbb{K}_{\Lambda'}[x_1^{\pm 1}, x_2^{\pm 1}, x_3^{\pm 1}]$, and $C := \mathbb{K}_{\Lambda''}[x_1^{\pm 1}, ...
Sky's user avatar
  • 923
0 votes
0 answers
104 views

Non-degenerate bilinear pairing of finite dimensional algebras

A finite dimensional algebra (over $\mathbb{C}$, say) is said to be Frobenius if it comes equipped with a nondegenerate bilinear pairing \begin{align*} \langle -, - \rangle : A \times A \rightarrow \...
James Steele's user avatar
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
5 votes
1 answer
883 views

Is this ring isomorphic to a quotient of a group algebra?

Consider the quotient of the free algebra $\mathbb{Q}\langle \alpha, \beta, \gamma, \delta, \varepsilon, \zeta \rangle$ by the two-sided ideal $I$ subject to the relations $$ \alpha\delta=\delta\alpha=...
Bumblebee's user avatar
  • 1,093
8 votes
1 answer
322 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
6 votes
3 answers
551 views

Conjecture about commutative semigroups

Conjecture: given any commutative semigroup $S$ of order $n \ge 4$, there exist $a, b \in S$ with $a \ne b$, an integer $m \ge \lfloor (n-1)/2 \rfloor$, and two $m$-element subsets $X = \{x_1, \ldots, ...
Fabius Wiesner'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
6 votes
0 answers
79 views

Examples of $\mathbb{N}$-graded algebras whose global dimension is strictly less than the GK dimension

The relationship between the global and GK dimensions of Artin-Schelter regular algebras remains to be mysterious, yet both dimensions are conjectured to be equal. In a more broad setting, are there ...
A. Kabbaj's user avatar
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
3 votes
0 answers
250 views

Action (of a graded monoid) required

Reference request: Did the construction below appear anywhere before? Any mentions of it or especially any links to something commonly known would be really helpful. I feel that it might be related to ...
Nikita Safonkin's user avatar
0 votes
0 answers
45 views

projections on minimal left ideals of semisimple algebras

Let $KG$ be a semisimple group algebra of a finite group $G$ over $K$. Consider $W=KGe$ as a minimal left ideal of this algebra and $e$ as a primitive idempotent. Here, $W$ is a simple left $KG$-...
khashayar's user avatar
  • 143
3 votes
2 answers
255 views

Is being graded commutative a necessary condition on $A$ such that $H^*(A)$ is commutative?

If we consider any differential graded algebra $A^\bullet$, then its homology is a graded algebra, since the tensor product interacts well with homology. A sufficient condidtion for the homology to be ...
curious math guy's user avatar
1 vote
0 answers
30 views

Star-algebra isomorphism

I have asked this question: When an algebra isomorphism preserves positive involution, but now I want to modify it. Let $A$ and $B$ be $K$-algebras where $K$ is a field with a unique ordering. We say ...
khashayar's user avatar
  • 143
2 votes
1 answer
103 views

When an algebra isomorphism preserves positive involution

Let $A$ be a $K$-algebra where $K$ is a field with a unique ordering. We say a $K$-linear involution $*$ is positive if the map $A \to K$ via $a \mapsto tr(a^*a)$ is positive definite with respect to ...
khashayar's user avatar
  • 143
3 votes
0 answers
89 views

Ordering the elements of a semigroup by $a \le b$ iff $a=b$ or $b=ab=ba$

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, ...
Salvo Tringali's user avatar
8 votes
1 answer
685 views

The state of the art on topological rings - the Jacobson topology

I was recently studying the Jacobson density theorem and I found it quite interesting. Most textbooks I've seen, including Jacobson's own Basic Algebra, only spend a few lines about the reason why it ...
Melanzio's user avatar
  • 183
4 votes
0 answers
131 views

Is there anything like a Čech complex for calculating local cohomology over *noncommutative* rings?

$\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\colim{colim}$Let $R$ be a ring, and consider a two-sided ideal $I = (r_1, r_2, \dots, r_j)$ in $R$. The corresponding $n$th local cohomology functor ...
user509184's user avatar
  • 1,335
0 votes
0 answers
114 views

Clarifications sought on the paper on the semigroup associated with a free polynomial by Ali Abbas and Abdallah Assi

I have three questions regarding the proof of Proposition 4 on page 4 of this paper here. For those interested in addressing these questions, please refer to some definitions in the first two or three ...
Mousa hamieh's user avatar
4 votes
1 answer
223 views

Recent research on polynomial identities

I work in computational complexity, where I work on the problem of polynomial identity testing over arithmetic circuits. One particular case is when the variables over the polynomial ring don't ...
Anagha's user avatar
  • 49
4 votes
0 answers
158 views

Wedderburn-Malcev principal theorem for graded-finite algebras

Let $k$ be a field and $A$ be a noncommutative $k$-algebra with Jacobson radical $J$. If $A$ is finite-dimensional, the Wedderburn-Malcev says that $A$ has a subalgebra $S$ such that $$A = S \oplus J$$...
Alvaro Martinez's user avatar
3 votes
0 answers
161 views

On generators of the multiplicative semigroup $\{r\in\mathbb Q:\ r>1\}$

The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the ...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
0 answers
92 views

Reference for the monoidal category structure $X \otimes Y = X + Y + X \times Y$ on a distributive category

Given a distributive category $\mathscr C$ (more generally a rig category), we can define a (semicocartesian) monoidal category structure on $\mathscr C$ with tensor product given by $X \otimes Y := X ...
varkor's user avatar
  • 10.7k
5 votes
0 answers
288 views

Representation functor on modules

Let $k$ be a field and $A$ a unital associative $k$-algebra. The representation functor associates, to each object in non-commutative geometry, a genuine geometric object on the representation variety ...
Qwert Otto's user avatar
2 votes
0 answers
66 views

Noncommutative transcendence degree of representation algebras

Let $G$ be a reductive group, for example $\text{GL}_n(\mathbb{C})$. Let $V$ denote its defining representation, and let $R$ denote the tensor algebra on the irreducible representations of $G$. It may ...
kindasorta's user avatar
  • 2,907
3 votes
0 answers
83 views

Non-commutative Gorenstein Koszul algebras

I was wondering if there exists a characterization of finite-dimensional Gorenstein Koszul algebras in terms of their Koszul dual?
Paulo Rossi's user avatar
4 votes
0 answers
147 views

Isomorphism between the reduced C*-algebra of a groupoid and the crossed product of inverse semigroups

In Paterson's book "Groupoids, Inverse Semigroups and their Operator Algebras" he proves that for any r-discrete groupoid $G$ with unit space $G^0$, its full $C^* $-algebra $C^* (G)$ is ...
Tomás Pacheco's user avatar
5 votes
0 answers
191 views

Do most semigroups have a zero?

It is widely believed in finite semigroup theory that asymptotically almost all finite semigroups $S$, up to isomorphism, are 3-nilpotent, i.e., they satisfy $\#\{abc\,:\,a,b,c\in S\} = 1$. My ...
user513093's user avatar
1 vote
1 answer
215 views

Reference about cancellation property for semigroups

Have the semigroups with the following cancellation property been studied? Property: Let $S$ be a semigroup and $x,y\in S$ such that $xz=yz,$ for all $z\in S,$ then $x=y$.
Hector Pinedo's user avatar
2 votes
1 answer
57 views

Are simplicial commutative inverse semigroups fibrant?

Let $X$ be a simplicial object in the category of commutative inverse semigroups (or monoids, if needed). Is the underlying simplicial set of $X$ always a Kan complex? If so, are there some nice ...
Aurélien Djament's user avatar
7 votes
1 answer
653 views

Which CAS can do basic non-commutative differential algebra?

This is a repost of my question at MSE from 7 months ago, to which I haven't been able to find an answer yet. I am looking for a CAS (possibly incl. additional packages/libraries) that can compute ...
M.G.'s user avatar
  • 7,127
3 votes
0 answers
130 views

Trace map on Ext group

Let $R$ be a (possibly non-commutative) unital ring and $M$ be a perfect left $R$-module. Then, we have the trace map $$ \operatorname{Tr}\colon \mathrm{Hom}_R(M,M)\to R/[R,R]\,. $$ According to the ...
Qwert Otto's user avatar
1 vote
0 answers
90 views

Multiplicative bases, path algebras, and Ext algebras

I am interested in understanding when a multiplicative basis exists for finite dimensional algebras over an algebraically closed field, and, in particular, Ext-algebras that are finite dimensional. It ...
James Steele's user avatar
5 votes
2 answers
199 views

Determining the multiplication via addition and some unary operation

It is known that the addition operation in a skew-field $F$ (more generally, in a quasifield) is uniquely determined by the multiplication operation and the unary involutive operation $1_{-}:F\to F$, ...
Taras Banakh's user avatar
  • 41.9k
4 votes
0 answers
174 views

Centers and conjugacy classes of groups relative to a pair of group homomorphisms

$\newcommand{\defeq}{\mathbin{\overset{\mathrm{def}}{=}}}$Given a group $G$, its center $\mathrm{Z}(G)$ and set of conjugacy classes $\mathrm{Cl}(G)$ are defined by \begin{align*} \mathrm{Z}(G) &\...
Emily's user avatar
  • 11.8k
6 votes
0 answers
151 views

On dual notions of morphisms of algebraic structures obtained by replacing equaliser with coequalisers

This question is based on this discussion from the Category Theory Zulip. See also the earlier question Natural cotransformations and "dual" co/limits. Let $G$ and $H$ be groups. We define ...
Emily's user avatar
  • 11.8k
2 votes
1 answer
423 views

Conjecture about semigroups

Let $G$ be a finite semigroup with order $n$ odd. Let $S_i \in G, i=1,\ldots,\binom{n}{(n+1)/2}$ be all the subsets of $G$ of size $(n+1)/2$. Let $E(S_i)$ be the set obtained "expanding" $...
Fabius Wiesner's user avatar
0 votes
0 answers
95 views

Which algebraic structure characterizes the set of non-trivial qudratic residues in a finite field?

I understand this question may be too naive to ask, but I am unable to figure it out. Suppose, $\mathbb{QR^*}$ denotes the set of all quadratic residues in a finite field except the identity element $...
Somudro Gupto's user avatar
5 votes
0 answers
187 views

Isbell duality for monoids and groups

Isbell Duality $\newcommand{\IsbellSpec}{\mathsf{Spec}}\newcommand{\IsbellO}{\mathsf{O}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\rmL}{\mathrm{L}}\newcommand{\rmR}{\mathrm{R}}\newcommand{\B}{\...
Emily's user avatar
  • 11.8k
8 votes
1 answer
236 views

Quiver and relations for a monoid related to Catalan numbers

Let $C_n$ be the monoid consisting of monotone maps $\{1,...,n\} \rightarrow \{1,...,n\}$ with $f(i) \leq i$ for all $i$. The cardinality of $C_n$ is given by the Catalan numbers. Consider $A_n= \...
Mare's user avatar
  • 26.5k
5 votes
0 answers
107 views

Structure of well-ordered commutative monoids

Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where $\forall a\in M,\ 0\leq a$ $\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$ The first condition means $M$ will be ...
Pace Nielsen's user avatar
  • 18.7k

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