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10 votes
4 answers
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Strongly Noetherian property. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian rings $A$ and $B$?

Let $k$ be a field. It is well-known that $A\otimes_{k}B$ is not necessarily Noetherian even if $k$-algebras $A$ and $B$ are Noetherian. For example $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$. When ...
user2013's user avatar
  • 1,663
2 votes
0 answers
137 views

Noncommutative Castelnuovo-Mumford regularity

I am looking for noncommutative version of Castelnuovo-Mumford regularity. To be more precise, let $A=\oplus_{i=0}^{\infty}A_{i}$ be a $good$ (finite global dimension, connected etc) noncommutative ...
Eric D's user avatar
  • 21
5 votes
2 answers
639 views

Moduli space of modules over non-commutative rings

Let $X=Proj(A)$ be a projective scheme, one can the moduli space of coherent sheaves on $X$ with fixed Hilbert polynomial and stability. Since coherent sheaves on $X$ are all obtained as the ...
user2013's user avatar
  • 1,663
4 votes
0 answers
290 views

Deformation of modules over noncommutaitve rings

Let $M$ be a finitely generated module over a commutative ring $R$. The first order deformation of module $M$ is parametrized by $Ext^{1}(M,M)$ and the obstruction is parametrized by $Ext^{2}(M,M)$. ...
user2013's user avatar
  • 1,663
1 vote
0 answers
86 views

Finite dimensional consistently graded Lie superalgebras of depth greater than 2

Victor Kac, in the paper "Classification of infinite-dimensional simple linearly compact Lie superalgebras", http://www.mat.univie.ac.at/~esiprpr/esi605.pdf writes at the beginning of section 5 (p....
AndreA's user avatar
  • 971
2 votes
2 answers
475 views

On the notion of torsion-freeness in semigroup theory

The following seems to be the "official" notion of torsion-freeness in the context of semigroups: TF1. A (multiplicatively written) semigroup $\mathfrak A$ is torsion-free if there do not ...
Salvo Tringali's user avatar
4 votes
1 answer
518 views

Strictly totally ordered semigroups - Looking for references

Let $\mathfrak A = (A, \cdot)$ be a semigroup (written multiplicatively). We say that $\mathfrak A$ is linearly orderable if there exists a total order $\le$ on $A$ such that $ac < bc$ and $ca < ...
Salvo Tringali's user avatar
17 votes
3 answers
1k views

Isomorphisms of quantum planes

Let $k$ be a field and $q\in k^{*}$. The quantum plane $k_{q}[x,y]$ is the algebra $k\langle x,y\rangle/\langle xy=qyx \rangle$ (i.e. the quotient of the free non-commutative $k$-algebra on two ...
Jan Grabowski's user avatar
3 votes
1 answer
212 views

Liftability of a submodule from an associated graded module

Let $k$ be a field, $A$ a $k$-algebra (probably noncommutative), and $M$ an $A$-module that's finite-dimensional as a vector space over $k$. Let $Gr(M;k)$ denote the set of all $k$-subspaces of $M$, ...
Allen Knutson's user avatar
5 votes
0 answers
442 views

A reference on semisimple linear algebra

Is there any literature where the tools familiar from (multi)linear algebra are systematically transferred to the setting of semisimple modules over noncommutative rings? In fact this question is a ...
Alexander Shamov's user avatar
5 votes
2 answers
2k views

Tensor product of simple modules

Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$. So my questions are: Can ...
Alexander Shamov's user avatar
12 votes
0 answers
443 views

Nullstellensatz for quaternionic plane curves?

By a quaternionic plane curve I mean the zero locus of a noncommutative polynomial in two variables, $x$ and $y$ say, over ${\Bbb H}$, Hamilton's quaternions. It is evidently well-known that, after ...
David Feldman's user avatar
13 votes
1 answer
694 views

Classification of long exact sequences

Let $\mathcal C$ be the category of long exact sequences of finitely generated abelian groups almost all of whose entries vanish. The category $\mathcal C$ is naturally additive as a subcategory of ...
Rasmus's user avatar
  • 3,184
1 vote
1 answer
130 views

Projective dimension over hypersurface

Let $R$ be (not necessarily commutative) ring and $S$ a simple right $R$-module. Let $f\in Ann(S)$ be normalizng and a non-zero divisor. Is it always true that $$ pdim_{R}(S)=pdim_{R/(f)}(S)+1? $$
M Simon's user avatar
  • 63
3 votes
1 answer
307 views

Global dimension of quantum $\mathbb{P}^{n}$

Let $k$ be a field. Given a (not necessarily commutative) $k$ graded ring $A$, M. Artin and J.J. Zhang introduced a notion of "noncommutative projective scheme" $Proj(A)$ in this paper. It is defined ...
Michel's user avatar
  • 65
4 votes
1 answer
297 views

Reference for subsemigroups of $\mathbb{N}^n$

A well known result about the natural numbers $\mathbb{N}$ says that for any finite subset $A \subset \mathbb{N}$ there exists $R \ge 0$ such that if $n$ is in the subgroup of $\mathbb{Z}$ generated ...
Lee Mosher's user avatar
  • 15.4k
9 votes
2 answers
2k views

Global dimensions of non-commutative rings

This is related to my previous question: When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? I now would like to know the global dimension of the ring $R=\mathbb{C}\langle x_1,\dots,x_n\rangle/...
user2013's user avatar
  • 1,663
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 ...
zacarias's user avatar
  • 801
6 votes
0 answers
618 views

Duality between conjugacy classes and irreducible characters for finite monoids?

Qiaochu's answer to this question suggests that the proper way to view the bijection between conjugacy classes and irreducible complex representations of a finite group is via a duality. My question ...
Benjamin Steinberg's user avatar
30 votes
4 answers
3k views

A mysterious Heisenberg algebra identity from Sylvester, 1867

I am trying to understand two papers by James Joseph Sylvester: P92: "Note on the properties of the test operators which occur in the calculus of invariants, their derivatives, analogues, and laws of ...
darij grinberg's user avatar
8 votes
2 answers
1k views

Explicit description of a quaternion algebra with a prescribed set of ramified places

Let $k$ be an algebraic number field. I understand that given a finite set of non-complex places $S\subset V(k)$ of even cardinality, there exists a unique quaternion algebra $Q$ over $k$ such that $Q$...
anonymous's user avatar
4 votes
2 answers
357 views

Minimal right ideals in finite semigroup

Let $E$ be a finite semigroup. According to N. Bourbaki (Algèbre I p. 121 exerc. 14 c), if $M$ and $M'$ are minimal right ideals in $E$, then they are isomorphic. I spent some time browsing through ...
Pnine's user avatar
  • 43
3 votes
1 answer
297 views

Is the universal inverse semigroup of a commutative semigroup an embedding?

The question of existence of a universal inverse semigroup of an arbitrary semigroup has been answered before (this is a construction similar to the Grothendieck group). Let's refer to the universal ...
Thomas Klimpel's user avatar
5 votes
1 answer
329 views

Free left = free right ?

Let $R \subseteq S$ be an extension of rings with unit. Suppose that $S$ is free as left $R$-module. I wonder what can said about the freeness of $S$ as right $R$-module. To be a little more precise ...
Ralph's user avatar
  • 16.2k
16 votes
4 answers
2k views

What's the name for the analogue of divided power algebras for x^i/i?

I recently came across divided power algebras here: http://amathew.wordpress.com/2012/05/27/lazards-theorem-ii/ It interests me because the free divided power algebra on one variable $x$, where $x^{(i)...
Vipul Naik's user avatar
  • 7,320
2 votes
2 answers
286 views

Idempotent semigroups: Are they all residually finite?

As pointed out by Mark Sapir in his answer to a related question, every residually finite divisible semigroup is idempotent (hence uniquely divisible). On another hand, it is not difficult to prove ...
Salvo Tringali's user avatar
2 votes
1 answer
360 views

On a property of subsemigroups

Let $H$ denote a subsemigroup of a semigroup $G$. I'm interested in the following property: $$\forall g\in G\exists h\in H:gh\in H.$$ This property is weaker than the property that $H$ is an ideal ...
John's user avatar
  • 21
4 votes
1 answer
314 views

A semigroup with the property that $x^n = a$ has at least one solution

Is there a standard name for a (multiplicatively-written) semigroup $(A, \cdot)$ such that, given an arbitrary $a \in A$, the equation $x^n = a$ has at least one solution $x \in A$ for each $n \in \...
Salvo Tringali's user avatar
1 vote
1 answer
434 views

A Version of Nullstellensatz for Rings of Dİfferential Operators

Here is one of the classical versions of the nullstellensatz: Let $K$ be a field and let $\mathfrak{m}$ be a maximal ideal of the polynomial ring $K[T_1,\ldots,T_n]$. Then $K[T_1,\ldots,T_n]/\mathfrak{...
Sonat Suer's user avatar
13 votes
5 answers
3k views

Noncommutative localization of a ring: complete construction

I've been looking for the following construction in the literature, but I've only been able to find (very) partial proofs or proofs of special cases. Let $R$ be a non-commutative ring and $S$ a ...
Steve's user avatar
  • 465
4 votes
2 answers
1k views

strong nilpotent elements

An element x in a noncommutative ring R is strongly nilpotent if any chain $x_1=x, x_2, ... $, with $x_{n+1}\in x_n R x_n$ terminates at zero. It becomes clear why this is a good definition once one ...
Roman 's user avatar
  • 43
43 votes
8 answers
3k views

How to quantify noncommutativity?

If I have two operators or finite-dimensional matrices $A$ and $B$, how can I quantify the amount to which they commute or don't commute? (I would consider it a big plus if it is computable easily for ...
Jiahao Chen's user avatar
  • 1,890
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
2 answers
284 views

Making a non-unital algebra the unique maximal one-sided ideal in a unital algebra

If $R$ is an algebra without a unit, then the standard unitisation $R^\sharp$ can have maximal one-sided ideals other than $R$. Thus, it is natural to ask about the following. Let $R$ be an algebra ...
princehall's user avatar
10 votes
2 answers
1k views

Connective spectra versus simplicial abelian groups - very basic question

Hello, I have very general , "introductory" questions (It is quite hard for me to seek for specific things in the algebraic topology literature). I guess that connective spectra have a model ...
Sasha's user avatar
  • 5,562
4 votes
2 answers
427 views

The octonion equations

A good treatment have been given to the quaternion equations. Indeed, Ivan Niven in his paper Equations in Quaternion given in this link http://jones.math.unibas.ch/~massierer/algebra-hs11/niven(...
Motaz Hammouda's user avatar
8 votes
2 answers
427 views

Is there a general result that theorems about finite structures proved in ZFC can be proved in ZF?

The title question is too vague so let me be specific. Much of modern finite semigroup theory uses profinite semigroups and properties of profinite semigroups that depend on the existence of prime ...
Benjamin Steinberg's user avatar
3 votes
1 answer
706 views

Center of universal enveloping algebra of nilpotent lie algebra

Let g be a finite dimensional nilpotent lie algebra over a field k of characteristic zero. Let U(g) be the universal enveloping algebra and Z(g) be its center. Denote by Z_1(g) the augmentation ideal ...
Daniel Pomerleano's user avatar
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
8 votes
2 answers
826 views

Which semigroups can be linearly ordered?

As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the ...
boumol's user avatar
  • 768
1 vote
2 answers
442 views

submonoids of Z_n

Anyone knows how to describe explicitly the submonoids of Z_n, regarded as a multiplicative monoid?
Camilo Salazar's user avatar
11 votes
3 answers
942 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
0 votes
1 answer
117 views

Local algebras with small maximal left ideals

Is there an infinite-dimensional, non-commutative complex local algebra $A$ (which is not a field) with the (unique) maximal left-ideal finitely generated as a left ideal? Or as a right ideal?
Alex Ortega's user avatar
5 votes
0 answers
918 views

Commutator Baker-Campbell-Hausdorff formula

Consider the Baker-Campbell-Hausdorff formula $\Phi(X,Y)\in\mathbb{Q}\langle\!\langle X,Y\rangle\!\rangle$ in non-commutative variables. Define $X*Y:=\Phi(X,Y)$ and $[X,Y]=(-X)*(-Y)*X*Y$, and then (as ...
Diego Sulca's user avatar
4 votes
0 answers
396 views

Is there a homological way to compute quiver presentations?

I have recently been studying with colleagues the representation theory of certain finite monoids that come up in probability theory and combinatorics, see Ken Brown's beautiful survey here. These ...
Benjamin Steinberg's user avatar
22 votes
2 answers
977 views

Mapping from a finite index subgroup onto the whole group

Dear All, here is the question: Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$? My guess ...
Victor's user avatar
  • 1,437
13 votes
4 answers
1k views

Injective dimension of graded-injective modules

In "Existence theorems..." Van den Bergh proposes the following "pleasant excercise in homological algebra": Let $A$ be a connected graded noetherian $k$-algebra (that is, $\mathbb N$-graded with $...
Pablo Zadunaisky's user avatar
4 votes
0 answers
694 views

Noncommutative analog of Koszul complex

Let $R=R_0 \oplus R_1 \oplus R_2 ...$ be a graded not necessarily commutative algebra over field $k$ and $R$ is generated by $R_1$, $R_0=k$. In commutative situation if one wants free resolution of $R/...
Sasha Pavlov's user avatar
  • 1,545
15 votes
1 answer
588 views

Is the ring of quaternionic polynomials factorial?

Denote by $\mathbb{H}[x_1,\dots,x_n]$ the ring of polynomials in $n$ variables with quaternionic coefficients, where the variables commute with each other and with the coefficients. Two polynomials $P,...
mikhail skopenkov's user avatar
5 votes
1 answer
304 views

flat maps of monoids which are not localizations

It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module. Rather, I am looking for extensions of rings which share certain properties of localizations, like ...
Ricardo Andrade's user avatar

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