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0 votes
1 answer
61 views

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 ...
1 vote
0 answers
122 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 ...
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 ...
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 $f(\phi(x,y))=f(x)+f(y)$....
1 vote
1 answer
81 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 ...
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 ...
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 ...
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 $M := H \setminus H^\times$,...
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 ...
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 $S_n$. ...
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 ...
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}, ...
5 votes
2 answers
1k views

An example where finitistic dimension does not equal right global dimension?

The (right) big finitistic dimension of a ring is Findim$(R) =$ sup{proj.dim(M) | $M$ a right $R$-module of finite projective dimension}. The (right) little finitistic dimension findim$(R)$ is the sup ...
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, ...
1 vote
1 answer
84 views

Simple-direct-injective modules

A right $R$-module $M$ is called a simple-direct-injective module if it satisfies any of the following equivalent conditions: For any simple submodules $A,B$ of $M$ with $A \cong B \subseteq^{\oplus} ...
2 votes
1 answer
404 views

Reference request: a cousin to the log semiring

Let $f$ be strictly increasing on $\mathbb{R}$. Then $x \oplus y := f^{-1}(f(x)+f(y))$ gives rise to a strict symmetric monoidal ($\Rightarrow$ commutative monoid) structure on $(\mathbb{R},\ge)$ with ...
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 \...
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, ...
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=...
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 ...
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 ...
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 ...
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,...
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 ...
64 votes
4 answers
8k views

What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group. A couple of good ...
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 ...
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 $KG$-...
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 ...
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 ...
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 $b = ab = ba$, ...
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 ...
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 ...
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 ...
12 votes
0 answers
542 views

Does Wedderburn's Little Theorem hold constructively?

Wedderburn's Little Theorem states that every finite division ring is commutative. Perhaps even more surprising, this implies that every finite reduced ring is commutative. The proofs that I am aware ...
4 votes
0 answers
148 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 ...
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 $$A = S \oplus J$$...
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 ...
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 ...
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 ...
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 ...
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?
6 votes
0 answers
632 views

Generating functions in countable commutative monoids

Let $f: \mathbb{N}_0 \rightarrow \mathbb{C}$ be a function. The power series of $f$ can be viewed as the function $\mathscr{P}_f : q \mapsto \sum_{n \in \mathbb{N}_0}^{} f(n)q^n$ where $q \in \mathbb{...
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 ...
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 ...
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 $x=y$.
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 ...
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 ...
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 ...
9 votes
1 answer
889 views

Combinatorics for the action of Virasoro / Kac–Schwarz operators: partition polynomials of free probability theory

In the background sections below, I establish the relations among characterizations of the action of Virasoro / Kac–Schwarz operators of 2D gravity models presented in terms of Laurent series by ...

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