# Tagged Questions

The noncommutative-algebra tag has no wiki summary.

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### Deformations of Ext rings

Let $k$ be a base ring and $k[x]$ the ring of polynomials in an indeterminate $x$ over $k$. Consider a (not necessarily commutative) algebra $A$ over $k[x]$ and two $A$-modules $M$ and $N$. Then for ...

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### Do representations of the universal enveloping algebra $\mathrm{U}\mathfrak{su}_2$ retain the Hopf algebra structure?

A Lie algebra $\mathfrak{g}$ generates its universal enveloping algebra $\mathrm{U}\mathfrak{g}$, which has the structure of a Hopf algebra. Modules of $\mathrm{U}\mathfrak{g}$ are exactly the of ...

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### Why Jacobson, but not the left (right) maximals individually?

I firstly asked the following question on MathStackExchange a couple of months ago. I did not receive any answers, but a short comment. So, I decided to post it here, hoping to receive answers from ...

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46 views

### Semisimple Coquasitriangular Hopf Algebras

Let $(G,R)$ be a coquasitriangular Hopf algebra, which we furthermore assume to be cosemisimple. Generalising classical Peter-Weyl, cosemisimplicity of $G$ is well-known to imply an isomorphism
$$
G ...

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754 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 ...

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142 views

### How to prove that a projective module is not free?

Let $A$ be a noncommutative (perhaps $\ast$-) algebra (over $\mathbb{C}$) and let $M$ be a projective module defined via a projector $P\in M_n(A)$; i.e. $M=P(A^n)$. Furthermore, assume that all ...

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378 views

### $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?

Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by finite group.
Denote the ...

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### Locally nilpotent operators of the Weyl algebra

$\newcommand{\ad}{\operatorname{ad}}$As my recent post (here) did not receive any answers yet, I thought I would ask a similar question in which I'm also interested.
Let $A=$ $^{k \langle x,y\rangle ...

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80 views

### Nilpotent operator of the Weyl algebra

For a research project I'm currently working on, I came across the following problem:
Let $A=$ $^{k <x,y> }\Big/_{(yx-xy-1)}$ be the Weyl Algebra over a field $k$ of characteristic $p$, where ...

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### How would you solve this tantalizing Halmos problem?

1-ab invertible => 1-ba invertible has a slick power series "proof" as below, where Halmos asks for an explanation of why this tantalizing derivation succeeds. Do you know one?
Geometric series. In ...

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### 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 ...

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575 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 ...

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182 views

### NCG with all noncommutativity in a nilpotent ideal

While in general non-commutative geometry behaves rather differently from commutative geometry when it comes to local-to-global properties (descent), there are versions of "mild" noncommutative ...

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### If $R$ is generated by idempotents, then $\text{Ann}(R)=0$?

Let $R$ be a ring (not necessarily commutative or unital) that is generated by idempotents. I'd like to know if $\text{Ann}(R)=0$ must hold. Here I use $\text{Ann}(R)$ to denote the set of all ...

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855 views

### What do you do if you believe a problem is undecidable?

While the title of this question is subjective, I hope to make what I'm looking for quite concrete. The first, and main question is this: If you believe that a problem you are working on is formally ...

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166 views

### The Jordan Plane and Enveloping Algebras

Let $k$ denote a field of characteristic $0$ (assume algebraically closed for convenience). Define $J=k\langle x,y|[x,y]=y^{2}\rangle$. This noncommutative algebra (which can be viewed as a derivation ...

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### Motivation for the Preprojective Algebra

Let $Q=(Q_0,Q_1)$ be a quiver and $k$ a field. We construct a new quiver $\bar{Q}$ in the following way. Let the vertices of $\bar{Q}$ be the same as the vertices of $Q$, and let the arrows of ...

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### Making an algebra the uniqe maximal one-sided ideal in another 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 ...

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### When is the essential extension commutes with colimits(or push forward)

Let $M$ be an $R$-module,where $R$ is a hereditary (or cohomological dimension less or equal to 1).Take $E(R)$ to be injective hull of $R$, then we have the essential extension
$i:R^I\rightarrowtail ...

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### Anticommuting operators with positive properties

Which classes of $M\in \mathsf M_k(\Bbb R)^{n\times n}$ admit solutions $N\in \mathsf M_k(\Bbb R)^{n\times n}$ such that
$$(M\otimes N+N\otimes M)(u\otimes u)=0$$ forall $u\in \mathsf D_k(\Bbb ...

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**1**answer

475 views

### How to make the Capelli's identity less mysterious?

The formulation of the Capelli's identity is very elementary; it has important applications in invariant theory and representation theory, see http://en.wikipedia.org/wiki/Capelli%27s_identity
To ...

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### Turning left modules into right modules over a homotopy Gerstenhaber algebra

For simplicity's sake, let $A$ be a dg-algebra over $\mathbb{Z}/2\mathbb{Z}$.
In the case when $A$ is a commutative algebra, we can turn a left $A$ module into a right $A$ module trivially. Of course ...

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728 views

### Semiprime (but not prime) ring whose center is a domain

The center of a prime ring is a domain and the center of a semiprime ring is reduced.
Now I have no evidence to believe that if the center of a semiprime ring R is a domain,
then R has to be a ...

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### A Non-Commutative Nullstellensatz

In studying presentations of pro-$p$-groups via generators and relations, one is led (via the so-called Magnus embedding) to questions involving power series in non-commuting variables. Results from ...

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### What properties “should” spectrum of noncommutative ring have?

There are already a lot of discussion about the motivation for prime spectrum of commutative ring. In my perspective(highly non original), there are following reasons for the importance of prime ...

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### What are the main open problems in the theory of quasigroups and loops?

What are the main open problems in the theory of quasigroups and loops?
A short survey would be welcome.
Thanks

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### Does this kind of non-noetherian bimodule exist?

Question: Do there exist simple rings $R$ and $S$ (i.e., rings with no proper nonzero ideals) and an $(R,S)$-bimodule $M$ such that
$M$ is finitely generated both as a left $R$-module and a right
...

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68 views

### Projective dimension of ring over its center

If $A$ is a ring and $Z(A)$ is its center then what is a sufficient condition for the projective dimension of $A$ over $Z(A)$ (ie: $pd_{Z(A)}(A)$) to be finite?
(Assuming that $A\neq Z(A)$).

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### Existence of small projective dimensioned modules

Suppose $A$ is a (if necessary unital) associative ring and $I$ is a left ideal in $A$. Let $\operatorname{pd}(M)$ denote the projective dimension of a left $A$-module $M$.
Then do either of the ...

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**1**answer

192 views

### Hochschild cohomology of commutative quotients

Notation:
Let $k$ be a commutative local ring and let $HH^{i}(A,N)$ denote the $i^{th}$ Hochschild cohomology $k$-module of a $k$-algebra A with coefficients in an $(A,A)$-bi-module $N$.
If ...

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230 views

### Non-commutative normalization

Let $A$ be a (non-commutative) associative algebra with 1. Assume that $A$ contains a cental subalgebra $Z$ such that
a) $Z$ is a noetherian domain
b) $A$ is a finitely generated module over $Z$.
...

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### Is the “algebraic closure” of the quaternions, finite dimensional? [closed]

This post is a sequel of: What's the algebraic closure of the quaternions?
$\mathbb{H}$ is algebraically closed for the polynomials of the form $\sum a_r x^r$, but it is not for the polynomials ...

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478 views

### Noncommutative arithmetic mean geometric mean inequality and symmetric polynomials

While analyzing convergence speed of stochastic-gradient methods for convex optimization problems, Recht et al (2011) posed a tantalizing conjecture. It seems quite tricky, so after having struggled a ...

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### Is the square diagram of index and exponential maps in $K$-theory of $C^*$-algebras anti-commutative?

Assume we have a $3\times 3$ grid with rows and columns being short exact sequences of $C^*$-algebras.
This gives a grid of 6-term exact sequences: 3 "horizontal" sequences and 3 "vertical" ...

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### Criterion for global dimension of subring

All rings are assumed to be associative and unital.
If $B$ is a commutative sub-ring of $A$ (which itself needs not be commutative) then what properties of $B$ are both necessary and sufficient for ...

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### Orders of Clifford algebra

Let $C_n$ be the Clifford algebra over $\mathbb{Q}$ associated to negative definite quadratic form $-I_n$ (i.e. $-x_1^2-\dots-x_n^2$). Let $\mathcal{O}$ be a $\mathbb{Z}$-order of $C_n$.
Q1) Is it ...

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### Compute Lie algebra cohomology

Is there a computer algebra system that is able to compute the Lie algebra cohomology in a given representation? What if the Lie algebra is finite dimensional?
In my case I would like to be able to ...

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**1**answer

220 views

### Endomorphism Ring of Indecomposable MCM Modules

Let $R = k[[x, y]]/(f)$, where $k$ is algebraically closed of characteristic zero. I'm particularly interested in studying the endomorphism ring of indecomposable MCM (maximal Cohen-Macaulay) modules ...

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### 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 ...

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**1**answer

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### Isomorphism of matrix ring over ore domain

Let $R_1,R_2$ be (left and right) ore domains. Does $ Mat_n(R_1)\cong Mat_m(R_2)$ implie m=n and $q.f.(R_1)\cong q.f.(R_2)$?
An counter example, a proof or a reference is welcomed.
Thanks

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### Are there any finitely generated artinian modules that are not Noetherian?

It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example ...

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### Noncommutative HKR theorem

What is the analog of HKR theorem in the noncommutative world?
Recall that the well-known theorem by Hochschild-Kostant-Rosenberg says that for a smooth commutative algebra $A$ of finite type ...

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### Interpretation of Hochschild Homology groups

In all the literature I've come across there are many concrete interpretations of the first few Hochschild Cohomology groups. For example $HH^1(A,M)\cong Derivation/Inner Derivations$ etc....
In ...

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### Tensor product of commutators vs. commutator in a tensor product

Let $R$ be a (noetherian) commutative ring, and let $V$ and $W$ be finitely generated free $R$-modules. Let $X \subseteq \mathrm{End}_R(V)$ and $Y \subseteq \mathrm{End}_R(W)$ be finite subsets, and ...

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### nth term in the Baker-Campbell-Hausdorff formula

I am trying to prove a result for which I need the nth term of the Baker-Campbell-Hausdorff formula. I came at this particular result (which is not of significance for the question, but mentioning for ...

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### Balanced dualizing complex vs rigid dualizing complex?

In noncommutative projective geometry, there is a counterpart of dualizing complex in commutative world. It seems to me that they are called either a balanced dualizing complex or rigid dualizing ...

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### Smooth Affine algebras are Calabi-Yau

Are all smooth affine algebras over a field Calabi-Yau?
I'm thinking yes since they satisfy Van den Bergh duality with dualizing module themselves (have I made a mistake in this reasoning)/

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### Intersection of Maximal Left Ideals with Finite Dimensional Quotient

Let $\Gamma$ be a finitely generated group and let $A=\mathbb{C}[\Gamma]$ be the corresponding group algebra over $\mathbb{C}$. Let $X$ be the set of all maximal left ideals of $A$ and let $X_0=\{I ...

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### Finitely generated projective = finitely presented flat over a noncommutative Noetherian ring

Let $R$ be a possibly noncommutative left Noetherian ring and $M$ an $R$-module. I am looking for a reference or a proof for the following fact: $M$ is finitely generated and projective if and only if ...

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### Global dimension of a subalgebra with all units

(All rings here are always assumed to be unital and associative).
Setup
Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:
If $u$ is a unit ...