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2 votes
0 answers
69 views

Is anything known about the center of the Fomin-Kirillov algebra?

Let $\mathcal{B}_{\mathbb{S}_m}$ be the quotient of the Fomin-Kirillov algebra so that its pairing becomes certainly nondegenerate. This algebra is conjecturally isomorphic to the Fomin-Kirillov ...
Christoph Mark's user avatar
1 vote
1 answer
110 views

Particular example of a quadratic extension of a nonunital ring

I want to construct a concrete non-unital ring $R$ with the following properties: $R$ is a noncommutative non-unital ring with a right unite $r$ i.e $t.r=t$ for any $t\in R$. $S\subset R$ is a ...
GSM's user avatar
  • 223
2 votes
0 answers
80 views

An alternative definition for finitely generated (and principal) ideals in a semigroup

Let $S$ be a semigroup. An ideal (of $S$) is a subset $I$ of $S$ such that $SI$ and $IS$ are both contained in $I$. The non-empty ideals constitute a subsemigroup, $\mathfrak I(S)$, of the power ...
Salvo Tringali's user avatar
1 vote
0 answers
69 views

On a lemma of projective dimension

Let $R$ be a finite-dimensional algebra, and $A=R\oplus A_1\oplus A_2\oplus \dotsb$ be an $\mathbb{N}$-graded algebra which is locally finite (i.e. all $A_i$'s are of finite dimension). Let $\text{...
Noto_Ootori's user avatar
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{...
Tian Vlašić's user avatar
10 votes
1 answer
221 views

Matrix ring isomorphisms of different sizes

Do there exist (unital, associative, noncommutative) rings $R$ and $S$, where $\mathbb{M}_2(R)\cong \mathbb{M}_3(S)$, but these matrix rings are not isomorphic to $\mathbb{M}_6(T)$ for any ring $T$?
Pace Nielsen's user avatar
  • 18.7k
3 votes
0 answers
137 views

Composition of Frobenius $n$-homomorphisms, characteristic-free?

This question is, as so often, a crossbreed of curiosity and laziness. The former has led me to an interesting, but somewhat unsatisfactory paper by Khudaverdian and Voronov (arXiv:2002.02395v2) and, ...
darij grinberg's user avatar
2 votes
0 answers
91 views

A recursive description of the smallest divisor-closed subsemigroup containing a set

Let $S$ be a semigroup and $\widehat{S}$ be its unitization, i.e., the monoid obtained from $S$ by adjoining an identity element if necessary (so that $\widehat{S} = S$ when $S$ is already a monoid). ...
Salvo Tringali's user avatar
3 votes
1 answer
171 views

Every homomorphism between (rational) Puiseux monoids is multiplication by a non-negative rational

Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (...
Salvo Tringali's user avatar
1 vote
1 answer
368 views

Wedderburn theorem for finite-dimensional algebras over the complex numbers

I'm trying to understand how to apply the Wedderburn theorem in the context of unitary algebras over $\mathbb{C}$ that are finite-dimensional and semisimple. Let $\mathcal{A}$ be a $\mathbb{C}$-...
Henrique Assumpção's user avatar
7 votes
1 answer
129 views

Semi-simple algebras over operads

I believe people thought about this questions, however I couldn't find any reference. I appreciate if someone could direct me to some detailed discussions about it. The categories of associative ...
Li Guanyu's user avatar
  • 449
9 votes
1 answer
236 views

Formal smoothness of path algebras and connections

Let $k$ be a field of characteristic zero and $A = kQ$ the path algebra associated with a quiver $Q$. The algebra $A$ is said to be formally smooth over $k$ if $$ \Omega^1_kA = \operatorname{Ker}(\...
Qwert Otto's user avatar
2 votes
2 answers
226 views

Minimal ideals and subalgebras of semisimple algebras

I'm considering an algebra to be a ring which is also a vector space over some field $F$, and the algebra $A$ is said to be semisimple if it is semisimple as a ring, i.e., $A$ can be written as a ...
Henrique Assumpção's user avatar
6 votes
1 answer
173 views

References on semigroup actions

I posted this question on Math Stack Exchange about 10 days ago, but received no answer (https://math.stackexchange.com/q/4843881/1223994). I would like to ask for references on semigroup actions on ...
Marco Farotti's user avatar
7 votes
2 answers
1k views

Is there a notion of point in noncommutative geometry?

It is not clear to me whether there is a general notion of point in NCG. I have heard (more through physics) that the notion of a point becomes meaningless or ill-defined in noncommutative spaces, but ...
Esmond's user avatar
  • 136
4 votes
0 answers
143 views

On the conditions for Artin-Schelter Gorenstein algebras

Let $ k $ be a field and $ A $ a connected graded $ k $-algebra ($ A $ is associative, but not assumed to be commutative). The algebra $ A $ is called Artin-Schelter Gorenstein* of dimension $ d $ if ...
Cranium Clamp's user avatar
3 votes
0 answers
75 views

Are the automorphisms of the power semigroup of a cancellative semigroup cardinality-preserving?

Let $S$ be a semigroup (written multiplicatively) and $f$ be an automorphism of the power semigroup $\mathcal P(S)$ of $S$, that is, a bijective function on the family of all non-empty subsets of $S$ ...
Salvo Tringali's user avatar
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} ...
Hussein Eid's user avatar
8 votes
2 answers
577 views

Faithful flatness and non-commutative algebras

$\DeclareMathOperator\Spec{Spec}$When dealing with commutative algebras, a usefull criterion for faithful flatness is the following: Let $f:A\rightarrow B$ be a morphism of commutative algebras. Then $...
FPV's user avatar
  • 541
1 vote
1 answer
142 views

Congruences that aren't "finite from above," take 2: semigroups

This is a hopefully less trivial version of this question. Briefly, say that a congruence is parafinite if it is the largest congruence contained in some equivalence relation with finitely many ...
Noah Schweber's user avatar
3 votes
0 answers
103 views

An isomorphism problem for semigroups of ideals

An ideal of a semigroup $S$ (written multiplicatively) is a set $I \subseteq S$ such that $IS$ and $SI$ are both contained in $I$ (here, $XY$ means, for all $X, Y \subseteq S$, the setwise product of $...
Salvo Tringali's user avatar
14 votes
4 answers
742 views

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$. I have verified the statement for $n \leq 4$ with a Mathematica code. I have ...
Geoffrey Critzer's user avatar
12 votes
1 answer
1k views

A problem in commutative algebra whose solution requires algebraic geometry (resp., noncommutative algebra)?

One can argue that commutative algebra is affine algebraic geometry. However, a great deal of commutative algebra generalizes to non-commutative algebra, and in that setting there is little geometry, ...
Jesse Elliott's user avatar
0 votes
0 answers
92 views

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known?

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known? Given that there are $3{,}684{,}030{,}417$ semigroups of order $8$, I guess $n\...
Shaun's user avatar
  • 379
2 votes
1 answer
194 views

Continuity of Moore-Penrose generalized inversion

Any matrix $A\in\mathbb{C}^{m\times n}$ has a unique generalized inverse $A^{\dagger}\in\mathbb{C}^{n\times m}$ with the properties $$AA^{\dagger}A=A,\qquad A^{\dagger}AA^{\dagger}=A^{\dagger},\qquad (...
Bumblebee's user avatar
  • 1,093
5 votes
0 answers
305 views

Arithmetic derivatives and non-commutative generalizations

In the theory of arithmetic derivatives, in the simplest case an arithmetic derivative on $\mathbb{N}$ is defined via the rule $(a \times b)'= a \times b' + a' \times b$, mirroring the product rule ...
Ilk's user avatar
  • 1,347
5 votes
0 answers
135 views

Confusion around a (necklace) cobracket in Ginzburg's article Calabi-Yau Algebras

Something has been puzzling me for quite a while in Ginzburg's article Calabi-Yau Algebras. At some point he considers the free graded algebra $\mathbb{C}\langle x_1, \dots, x_n, \theta_1, \dots \...
Vik S.'s user avatar
  • 437
5 votes
0 answers
160 views

$S$ and $T$ globally isomorphic semigroups, with $S$ (commutative and) cancellative, iff $S$ is isomorphic to $T$?

Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication ...
Salvo Tringali's user avatar
2 votes
1 answer
271 views

Apropos of two groups being globally isomorphic iff they are isomorphic

Denote by $\mathcal P(S)$ the semigroup obtained by endowing the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced ...
Salvo Tringali's user avatar
0 votes
1 answer
152 views

Name for a monoid on the basis of a vector space?

Is there a name for the structure of a vector space with a monoid defined on its basis? Given a vector space V over a field F, we can choose a basis and define a monoid on it. Now we can use each ...
Spencer Woolfson's user avatar
0 votes
1 answer
327 views

Can we generalise groupoids to monoid-oids? [closed]

Groups correspond to one object categories where every morphism is an isomorphism. Monoids correspond to one object categories. Groupoids correspond to small categories where every morphism is an ...
Diego de la Paz's user avatar
7 votes
0 answers
227 views

On the structure of an algebra as a bimodule

$\DeclareMathOperator\End{End}\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Ker{Ker}\newcommand{\bi}{\mathrm{bi}}\newcommand{\op}{\mathrm{op}}$Let $K$ be a field (say of characteristic zero), and $...
FPV's user avatar
  • 541
2 votes
1 answer
265 views

Gluing data for modules over a ring with idempotents

Let $A$ be a ring. If $e$ is an idempotent, then there is an abelian recollement involving the categories $A\text{-}\mathrm{Mod}$ and $eAe\text{-}\mathrm{Mod}$. This is Example 2.7 in Homological ...
Sergey Guminov's user avatar
5 votes
2 answers
396 views

Algebra with three anti-commutator relations

Let $u,v,w \in \mathbb{F}_p^{\times}$. Consider the $\mathbb{F}_p$-algebra $V$ generated by $ a,b,c$ and the relations $$u a^2 = bc + cb$$ $$v b^2 = ac + ca$$ $$w c^2 = ab + ba$$ Is $V$ generated by ...
Martin Brandenburg's user avatar
5 votes
0 answers
293 views

On the deformation theory of associative algebras

Let us start by recalling the notion of a formal deformation: Let $K$ be a field of characteristic zero and $A$ be an associative $K$-algebra. Consider a commutative augmented $K$-algebra $R$, with ...
FPV's user avatar
  • 541
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 ...
Martin Brandenburg's user avatar
5 votes
2 answers
478 views

Generalization of the concept of a measure

Consider the following generalization of the concept of a measure: Let $L = (X, \lor, \land, \bot)$ be a semi-bounded lattice. Let $M = (Y, \bullet, e)$ be a commutative monoid. An $(L, M)$-measure is ...
user76284's user avatar
  • 2,203
4 votes
1 answer
270 views

Kaplansky inverse element theorem on group C-star algebra

In a class talking about $C^*$ algebra and (higher) index theory, I heard a theorem (related to Kaplansky, proved?), that is Suppose $\Gamma$ is a group (admitting Haar measure if necessary) while $\...
YOTAL's user avatar
  • 193
2 votes
0 answers
144 views

Zero divisors in the extra-special group algebra $\mathbb{R}[2^{1+6}_+]$

Can you characterize the unit-group of the real group-algebra of the extraspecial plus-type 2-group of order 128? (That is $\mathbb{R}[2_+^{1+6}]$ using Conway's notation.) (Please choose any irrep ...
Eric Downes's user avatar
3 votes
0 answers
176 views

The monoid of stably-free modules over integral group rings

Fix a torsion-free group G, let $M_G$ be the monoid of stably-free $\mathbb{Z}G$-modules under operation $\oplus$, the direct sum of modules. In studying objects related to Wall’s D2 problem on CW-...
William Thomas's user avatar
2 votes
0 answers
71 views

Classification of polynomials leading to finite dimensional admissible algebras

Let $K \langle x , y \rangle $ ($K$ a field, we can assume it has only two elements if it helps) be the non-commutative polynomial ring in 2 variables. Question 1: For which non-commutative ...
Mare's user avatar
  • 26.5k
7 votes
1 answer
183 views

Is lambda calculus polymorphism a type of generalized monad?

Let $\mathbf{C}$ be a Cartesian closed category. Then simply typed lambda calculus in $\mathbf{C}$ in one type variable can be interpreted as a category $\mathbf{STLC}_{\mathbf C}$ where the objects ...
Johan Thiborg-Ericson's user avatar
1 vote
0 answers
86 views

Inner product on Standard form of von Neumann algebra

Let $(M, H, H_+,J)$ be a standard form of a von Neumann algebra $M$ acting on a complex Hilbert space $H$ endowed with a self-dual cone $H_+$. Is it true that $$\langle x,yz\rangle=\langle zx,y\rangle$...
Guest's user avatar
  • 131
2 votes
0 answers
176 views

On the origin of power semigroups

Let $S$ be a (multiplicatively written) semigroup. Equipped with the (binary) operation of setwise multiplication $(X, Y) \mapsto \{xy \colon x \in X, \, y \in Y\}$, the family of all non-empty ...
Salvo Tringali's user avatar
8 votes
3 answers
1k views

Are all free monoids residually finite?

I cannot manage to prove that a free monoid with operation concatenation, and with at least two generators is residually finite. If there is just one generator, the free monoid $\{a\}^*$ is isomorphic ...
iguessarian's user avatar
1 vote
1 answer
87 views

Semigroup algebras with one dimensional center

Let $S$ be a finite semigroup and $K$ a field of characteristic 0 (we can assume the complex numbers for simplicity). Question: Is there a characterization when the center of the semigroup algebra $...
Mare's user avatar
  • 26.5k
1 vote
1 answer
129 views

Cycles in almost breakable semigroups

Last October, I learned from Benjamin Steinberg's answer to another question of mine that a semigroup $S$ is called breakable if $xy \in \{x, y\}$ for all $x, y \in S$. Let's now say that $S$ is an ...
Salvo Tringali's user avatar
4 votes
0 answers
80 views

Indecomposable injectives over Weyl algebras

Let $A=A_n(\mathbb{C})$ be the $n$-th Weyl algebra over the complex field. Then $A$ is a left Noetherian noncommutative ring. Is there a complete classification of indecomposable injective $A$-modules?...
Doug Liu's user avatar
  • 615
10 votes
0 answers
248 views

What is the tiling semigroup for an einstein "hat" tiling?

My undergraduate dissertation was on inverse semigroups and the key text I used for it was Lawson's, "Inverse Semigroups: The Theory of Partial Symmetries". In said book, Lawson describes ...
Shaun's user avatar
  • 379
0 votes
0 answers
63 views

A construction that sort of merges two semigroups to build a new one

Suppose $H$ and $K$ are semigroups and assume without loss of generality that (the underlying sets of) $H$ and $K$ are disjoint. We can then extend the operations of both $H$ and $K$ to a binary ...
Salvo Tringali's user avatar

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