The noncommutative-algebra tag has no wiki summary.
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3answers
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Are Clifford groups very *non-commutative*?
Clifford groups seem to be very non-commutative by the relation \begin{equation}
\gamma_{i}\gamma_{j}=-\gamma_{j}\gamma_{i}.
\end{equation} But is it really so? Can we put this degree of ...
0
votes
2answers
32 views
tensor product and direct product of algebra presentations
Let $R$ be a commutative unital ring and $R\langle x_i\mid f_j\rangle$ denote a unital $R$-algebra presentation.
Q1: What is the presentation of $R\langle x_i\mid f_k\rangle\otimes R\langle y_j\mid ...
7
votes
1answer
99 views
Projective objects in the category of rings
What are the projective objects in the category of rings with identity ?
Remarks:
The only projectives I could find so far are $\{ 0\}$ and $\mathbb{Z}$.
If $R$ is projective and ...
6
votes
1answer
265 views
A graded ring $R$ is graded-local iff $R_0$ is a local ring?
Update: I've copied this question over to mathoverflow.net:
http://mathoverflow.net/questions/100755/a-graded-ring-r-is-graded-local-iff-r-0-is-a-local-ring
to see if I get any answers there.
Let ...
3
votes
1answer
77 views
Universal enveloping algebra of a Poisson algebra
For a Lie algebra, $\mathfrak{g}$, one has an equivalence of categories between Mod($\mathfrak{g}$) and Mod($U(\mathfrak{g})$), where $U(\mathfrak{g})$ is the universal enveloping algebra of ...
3
votes
1answer
70 views
Selfinjectivity and Frobenius algebras
Suppose that $R$ is a finite dimensional $k$-algebra. I say that $R$ is Frobenius if it is locally bounded (see this question for a definition) and indecomposable projectives and injectives coincide. ...
2
votes
1answer
23 views
In the lattice of ideals, what are the lowerbounds of the prime ideals?
Take the lattice of ideals in a non-commutative ring with 1. It is well known that the lowerbounds of all the maximal ideals are the superfluous ideals. Is there a similar characterization for the ...
5
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0answers
113 views
Is there a simple example of a ring that satifies the DCC on two-sided ideals, but doesn't satisfy the ACC on two-sided ideals?
It follows from the Hopkins–Levitzki theorem that if a ring satisfies the DCC on left ideals, then it also satisfies the ACC on left ideals. I've been trying to find a counterexample to the following ...
5
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0answers
157 views
A ring that has exactly 7 left ideals (T. Y. Lam)
Exercise 3.25 in Lam's First Course states:
Let $R$ be a ring that has exactly seven nonzero left ideals. Prove that one of them is an ideal (i.e. left and right) and provide an example of such a ...
5
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0answers
135 views
Examples of Noncommutative Noetherian Rings in which Lasker-Noether Fails?
I'm writing a paper on Emmy Noether for my introductory Abstract Algebra class, and I'm looking for examples of noncommutative Noetherian rings in which the Lasker-Noether theorem fails to hold.
...
4
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0answers
124 views
Problem 24 of section 2 of *Noncommutative Algebra* by Farb & Dennis
Problem 24 of section 2 of Noncommutative Algebra by Farb & Dennis states:
Let $R$ be an artinian ring and let $G$ be a finite group. Show that $R[G]$ is semisimple if and only if $R$ is ...
3
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0answers
59 views
When is $\mathbb{Z}\Gamma$ a left Noetherian ring?
Denote $\Gamma$ to be a countable discrete group, let $\mathbb{Z}\Gamma$ to be its integer group ring.
A ring is left-Noetherian if it satisfies the ascending chain condition on left ideals.(c.f. ...
3
votes
0answers
61 views
Applications of Govorov-Lazard Theorem?
The Govorov-Lazard Theorem states that a (right) module over an unital ring is flat iff it is a direct limit of finitely generated free (right) modules.
I wonder if this theorem has interesting ...
3
votes
0answers
68 views
Computation of determinant of a matrix with elements from an arbitrary commutative ring
The cofactor formula for computing the determinant of a matrix is applicable when elements of the matrix are from a commutative ring. However, the complexity of this method is extremely high and I ...
3
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0answers
142 views
Coproduct in the category of (noncommutative) associative algebras
For the case of commutative algebras, I know that the coproduct is given by the tensor product, but how is the situation in the general case? (for associative, but not necessarily commutative algebras ...
3
votes
0answers
111 views
Integral forms of loop algebras.
The question following is about integral forms for semisimple Lie algebras and loop algebras constructed from them.
Let $\frak g$ a finite-dimensional Lie algebra over $\mathbb C$ and $L(\frak ...
2
votes
0answers
16 views
Partial cycles in projective resolutions of square-free algebra
Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand?
I suspect not, but have not ...
2
votes
0answers
39 views
Defining the Rank of a Projective Module
I am trying to understand the definition of rank for a projective module over a noncommutative ring. The definition I am using is:
A sufficient condition for the rank of a free module over a ring ...
2
votes
0answers
41 views
Left ideals of central simple algebra generated by symmetric element
Let $F$ be a field of characteristic $\neq 2$. Suppose $(A,\sigma)$ is central simple algebra over a field $F$ with an orthogonal involution. Assume $(A,\sigma)$ is non-split. How to show that every ...
2
votes
0answers
79 views
On the Nakayama functor
Let $A$ be a finite dimensional $k$-algebra with 1. Denote by $_AP$ the category of projective left $A$-modules finite dimensional. And with $_AI$ the category of injective left $A$-modules finite ...
2
votes
0answers
69 views
Non-commutative integral extensions?
In Commutative algebra there is a notion of an integral extension: Let $P$ be a subring of $R$. Then $R$ is the integral extension of $P$ if each element of $R$ is a root of a monic polynomial with ...
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0answers
59 views
Existance of inverse operators for Hermitian adjoint operators
I have one assumption that I can't prove. Maybe there are some other conditions which I didn't take into account.
My assumption:
Suppose $a$ and $a^\dagger$ is Hermitian adjoint operators and ...
1
vote
0answers
26 views
Center of a quantum matrix algebra
Let $p \in k^\times$ be a nonroot of unity. It seems to be a well-known fact that the center of the quantum matrix algebra $\mathcal{O}_p(M_n(k))$ is generated by the quantum determinant $D_p$. It is ...
1
vote
0answers
36 views
Total divisor in a Principal Ideal Domain.
Let $R$ be a right and left principal ideal domain. An element $a\in R$ is said to be a right divisor of $b\in R$ if there exists $x \in R$ such that $xa=b$ . And similarly define left divisor.
$a$ ...
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0answers
109 views
All finite-dimensional simple modules are $1$-dimensional
Let $A$ be a (non-commutative) $k$-algebra, where $k$ is an algebraically closed, characteristic zero field. Let $M$ be a finite-dimensional simple $A$-module. If $A/\operatorname{ann}(M)$ is ...
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0answers
84 views
Unit group of quotient of noncommutative polynomial ring
In this recent post the original question led people to look for rigid, noncommutative rings. (Rigid means that the only endomorphisms are zero and the identity). Several (somewhat complicated) ...
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0answers
44 views
Nil ring version of Burnside's problem
The famous Burnside group of exponent $n$ over a set $G$ of generators is the
group defined by generators $g\in G$ and relations $h^n=1$ where $h$ is any
product of elements of $G$.
Likewise, let ...
1
vote
0answers
168 views
A classic example in noncommutative ring theory
A classic example of a left-not-right Noetherian ring is the quotent of the free algebra $\mathbb{Z}\langle x,y\rangle$ (noncommuting indeterminates) by the ideal generated by $y^2$ and $yx$. (It's ...
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0answers
89 views
PBW Theorem applied to graded Lie algebras
Fix a $\mathbb Z_+^n$-graded Lie algebra ${\frak a}=\oplus_{r \in\mathbb Z_+^n}^{} {\frak a}[r]$ such that ${\frak g}:={\frak a}[0]$ is a finite-dimensional semisimple Lie algebra over the complex ...
0
votes
0answers
21 views
A good description of $M^{\vee\vee}$?
Let $R$ be a f.g. $\mathbb{N}$-graded non-commutative $\mathbb{C}$-algebra. Assume $R$ is connected, i.e. $R_0=\mathbb{C}$. Let $M$ be a f.g. torsion free graded right $R$-module of rank $1$. Is there ...
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0answers
62 views
Left-modules over a bialgebra form a monoidal category
Let $B = (B, \nabla, \eta, \Delta, \epsilon )$ be a bialgebra over a commutative ring $k$. Let $M$ and $N$ be two left $B$-modules. Then the tensor product $M \otimes_k N$ becomes a left $B$-module ...
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Idempotent isomorphic to $1$
Let $R$ be a ring with unit element $1$. An idempotent element $e \in R$ is called isomorphic to $1$ if there are elements $u, v \in R$ such that $vu=1$ and $uv=e$ and this is written $e \cong 1$.
...
