Questions tagged [noncommutative-algebra]
For questions about rings which are not necessarily commutative and modules over such rings.
1,370
questions
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Every order in a finite-dimensional $\mathbb{Q}$-algebra is contained in a maximal order
Definition. A lattice in a finite-dimensional $\mathbb{Q}$-algebra $V$ is a finitely generated $\mathbb{Z}$-submodule $\mathcal{L} \subset V$ such that $\mathcal{L}\mathbb{Q}=V$ (i.e., $\mathcal{L}$ ...
1
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0
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33
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Orders in quaternion algebras
Definition. An algebra $B$ over a field $F$ is a quaternion algebra if there exists $i,j\in B$ such that $1,i,j,ij$ is a basis for $B$ as a vector space over $F$.
Throughout, we fix $F=\mathbb{Q}$.
...
5
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Examples of rings where every left ideal is two-sided but not every right ideal
Is there an example of a ring where every left-ideal is two-sided but not every right ideal?
WHAT FOLLOWS IS A FAILED EXAMPLE
As an argument but not a proof that the example fails, consider:
$ f(a) \;...
3
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1
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Ideals of the Lipschitz quaternions
Consider the subring $\mathcal{O}:=\mathbb{Z}+\mathbb{Z}i+\mathbb{Z}j+\mathbb{Z}k$ of the ring $\mathbb{H}=\mathbb{R}+\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ of real Hamiltonians. Let $I$ be a right ...
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If $R$ is a product of matrix rings over division rings then every $R$ module is completely reducible
In Dummit and Foote's Abstract Algebra, the proof of the Wedderburn-Artin theorem is sketched out in a series of exercises for reader to solve. I'm currently working on the final exercise in the ...
1
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1
answer
42
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Every $R$-module is injective implies $R$ is a product of simple rings
In Dummit and Foote's Abstract Algebra, all the main results about representations of finite groups are derived from the Wedderburn-Artin theorem. The proof of the Wedderburn-Artin theorem itself is ...
2
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1
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How to define the non-commutative ring $\mathbb{F}_{4}+e\mathbb{F}_{4}$, $e^2=1$, $ae=ea^2$ in MAGMA(Computational Algebra System)?
I'm trying to learn to use MAGMA(Computational Algebra System) for research in coding theory over non-commutative rings, but it's been slow going. I feel like it's hard to find anything in the ...
4
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0
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Commuting operators vs. commutative rings, and localization in the quantum mechanical sense vs. commutative algebra sense.
One of the answers in this MO post https://mathoverflow.net/questions/7917/non-commutative-algebraic-geometry draws a connection between localization in commutative algebra (or the failure thereof), ...
1
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0
answers
14
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Power function in non-commutative algebras
Is there a canonical form of the power function, $a^b=x$, that extends to non-commutative algebras like matrices and hypercomplex numbers? It is known that $a^b=e^{b\log{a}}$ for commutative algebras, ...
1
vote
1
answer
64
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Spectral triple on $\mathcal C(M)$ where $M$ is a compact Riemannian manifold, not necessarily spin
I have been reading Alain Connes' Compact metric spaces, Fredholm modules and hyperfiniteness.
In proposition 1, it is mentioned that an unbounded Fredholm module (nowadays: spectral triple) over $C(M)...
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2
answers
74
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Meaning of "free on the basis"
T.Y. Lam in his book "A First Course in Noncommutative Rings" gave an example of "Hurwitz ring of integral quaternions" which is $$R=\lbrace(a+bi+cj+dk)/2 \mid a,b,c,d\in\mathbb{Z} ...
0
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1
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39
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Example of a quasi nilpotent element which is not a nilpotent element
Let $R$ be a ring with unity. An element $a\in R$ is said to be a quasi nilpotent element of $R$ if $1-ax$ is unit for all $x\in comm(a)$ where $comm(a)=\lbrace x \in R | ax=xa\rbrace $. It is obvious ...
0
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2
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55
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Spectral triple for a (real) full matrix algebra
Let $\mathcal A = \mathbb R^{N \times N}$ be the real full matrix algebra, $N \in \mathbb N_{> 1}$, which is represented by the Hilbert space $H := \mathbb R^N$ (that is, $\mathcal A \to B(H)$, $A \...
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2
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Is there are a ring where $\frac{1}{2}$ doesn't commute with everything?
Is there a ring with an element $x$ such that $2x=1$ which is non-central (there is some $y$ such that $xy \neq yx$)?
Suppose there is, then by taking the subring generated by $x$ and $y$, there had ...
2
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1
answer
39
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Noncommutative extension of algebra of smooth functions
Let $A$ denote the algebra of smooth real functions on $\mathbb{R}^n$.
Let $j : \mathbb{R}[x_1,\ldots,x_n] \rightarrow A$ be the injective homomorphism sending a polynomial to its corresponding ...
1
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1
answer
28
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Set of all elements of a ring which semicommute with other elements of a ring
I am wondering that as we define $Z(R)$, center of a ring is a subring of $R$. Can we define a subset which is collection of all those elements which semicommute with other elements. Will this set ...
5
votes
1
answer
63
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Central simple algebras and semisimple subalgebras
Let $B$ be a central simple algebra over an algebraically closed field $k$, and let $A\subset B$ be a semi-simple $k$-subalgebra. By this I also mean $k$ is central in $A$ as well, and the map $k\to A\...
2
votes
1
answer
87
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The C$^*$ algebras of matrices, continuous functions, measures and matrix-valued measures / continuous functions and their state spaces
I have no background in Algebra, but want to understand matrix-valued measures and matrix-valued continuous functions from the C$^*$-algebra perspective to identify what definition of matrix-valued ...
0
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0
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31
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Either $r$ or $1-r$ is left-invertible in a ring $R$ with $1$ $\iff$ either $r$ or $1-r$ is a unit.
Let $R$ be a noncommutative ring with $1$. I need to show that for any $r\in R$ either $r$ or $1-r$ is left-invertible if and only if either $r$ or $1-r$ is a unit.
Proof. Indeed, if either $r$ or $1-...
0
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1
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Every non-unit in a non-commutative ring with unity is contained in a maximal left ideal.
Suppose that $x$ is a non-unit in a non-commutative ring $R$ with unity $1$. Is it true that $x$ is contained in some maximal left ideal of $R$. I tried to prove the statement by Zorn's Lemma by ...
2
votes
2
answers
110
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Fredholm module over $\mathbb R^{N \times N}$ and over matrix-valued measures
The following is taken from Compact metric spaces, Fredholm modules,
and hyperfiniteness by Alain Connes, cf. also his paper Non-commutative differential geometry.
Definition.
Let $A$ be a unital $C^*$...
1
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1
answer
47
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Semi-commutative but not commutative ring.
So, P.P Nielsen created an example of a semicommutative ring which is not McCoy (or say commutative).
He took, $k=\mathbb{F}_2\left\langle a_0,a_1,a_2,a_3,b_0,b_1\right\rangle$ be the free algebra (...
0
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0
answers
50
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(local) Artinian rings with nilpotent Jacobson Radical and Maximal ideals
Problem 1: If $R$ is an Artinian ring, then show that the Jacobson
Radical $J(R)$ is nilpotent.
Problem 2: If $(R, \mathfrak{m})$ is a local Artinian ring, then show
that $\mathfrak{m}$ is nilpotent.
...
1
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1
answer
45
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Let $A$ be a central division algebra (of finite dimension) over a field $k$. Show that $[A,A] \neq A$.
I am looking at the post A central division algebra is not its commutator and I have a few questions regarding the proof that was provided in the answer.
Why is $A$ a simple $k$-algebra?
My first ...
2
votes
1
answer
50
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Are Weyl algebra $A_1$ and it's opposite algebra isomorphic?
Let $A$ be a noncommutative ring and $ab=c$ in $A$. $A'$ is it's opposite ring if $ba=c$ in $A'$. If $A$ is a Weyl algebra $A_1$, are $A$ and $A'$ isomorphic?
I have an idea, but I think it's wrong. ...
3
votes
1
answer
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Uniqueness of a submodule
Suppose I want to show that $\mathbb{C}[X]$ is the unique simple $A_1$-submodule of $\mathbb{C}[X,X^{-1}]$, where $A_1 = A_1(\mathbb{C})$ is the first Weyl Algebra.
It is not difficult to show it is ...
0
votes
1
answer
42
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quintic solution to noncommutative polynomial
So, we know there are generally no solutions to an arbitrary quintic polynomial (EDIT from comments: Of course, I mean you can't write the roots using radicals, not that the roots don't exist). I don'...
0
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0
answers
47
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Computing the nilradical of a ring
Let $R=\begin{pmatrix}
\mathbb{C} & \mathbb{C} & \mathbb{C}\\
0 & \mathbb{C} & \mathbb{C}\\
0 & \mathbb{C} & \mathbb{C}
\end{pmatrix}$. I want to find the nilradical: $$N(R)=\...
2
votes
1
answer
37
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Product of linearly dependent vectors is 0 in an anti-commutative algebra
From the 3rd edition of the book "The Linear Algebra a Beginning Graduate Student Ought to Know" by Jonathan S. Golan, we find the following statement in chapter 4:
Note that if ${v,w}$ is ...
0
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0
answers
45
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How to determine if a non-commutative algebra is semisimple
Let $\cal{Q}$ be the field of rationals and $L=\cal{Q}(\sqrt 2, \sqrt 3)$ be a Galois extension of degree 4 ($[L:\cal{Q}]=4$).
Using theory of Drinfeld twists on the galois group of $L$, I have ...
2
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1
answer
89
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A division quaternion algebra in which the integral elements don't form a ring
I wish to find a division quaternion algebra $B$ over $\mathbb{Q}$ and elements $\alpha, \beta\in B$ such that $\alpha, \beta$ are integral over $\mathbb{Z}$ but both $\alpha + \beta$ and $\alpha \...
3
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1
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Prove: $P(XY=YX) \leq 5/8$ in a finite no commutative group
Question:
Let $(G;.)$ a finite no commutative group. Find that an upper bound that to the probability that two element $X,Y$ of $G$ randomly chosen with an uniforme probability (that means the ...
0
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1
answer
49
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How is invertibility defined for square matrices over non-commutative rings?
In this article Wikipedia defines invertibility for square matrices over commutative rings as follows:
...in the case of the ring being commutative, the condition for a square matrix to be invertible ...
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0
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33
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Reference of automorphism with respect to locally nilpotent derivations
Let $ R $ be a commutative algebra. Let $ \delta $ be a locally nilpotent derivation, that is, there exists some $ n $ such that $ \delta^{n} = 0 $. Then we can define a automorphism, namely $ \exp(...
3
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2
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Why do we say that the tensor product of vector spaces is commutative, but the tensor product of vectors is not?
The Wikipedia article on the tensor product says
The tensor product of two vector spaces $V$ and $W$ is commutative in the sense that there is a canonical isomorphism V ⊗ W ≅ W ⊗ V that maps v ⊗ w to ...
0
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1
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Definition of a basic algebra over a field $K$
Let $K$ be an algebraically closed field and let $A$ be a $K$-algebra with a complete set $\{e_1,…,e_n\}$ of primitive orthogonal idempotents. So, the algebra $A$ is called basic if
$$e_iA \cong e_jA \...
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0
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Seeing that a product of quasi-free algebras is quasi-free in terms of lifting homomorphisms
A $k$-algebra over a field $k$ is quasi-free if for any bimodule $M$ the second Hochschild cohomology $H^2(A,M)$ vanishes or, equivalently, if for any square-zero extension $R/M=A$ there is a lifting ...
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0
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A generalization of the Clifford algebra
A minimal example of the Clifford algebra is the $\mathbb{C}$-algebra (unital, associative) generated by $x,y$ quotient over the relations
\begin{eqnarray}
x^2&=&1,\tag{1}\\
y^2&=&1,\...
0
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1
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Proof of (left) Ore condition implies there exists Q(R) a left quotient ring of R.
I am trying to understand the proof of Theorem 7.1.1 of `Noncommutative Rings' I. N. Herstein.
Definition: An element in a ring is said to be regular if it is neither a left nor a right zero divisor ...
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0
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Request for reference books, articles, papers on the topic topic of Non Abelian Kummer Extensions.
I am a final year undergraduate student. I am doing my research on Non-Abelian Kummer Extensions. Can someone please introduce me to formal definition of Non-abelian Kummer Extensions. I know it is ...
1
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0
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Help me Understanding Non Abelian Kummer Extensions
[![I am doing my undergraduate research work on the Non-Abelian Kummer Extensions. I am following the book "Algebra" by Serge Lange. I have understood the difference between the abelian ...
1
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1
answer
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Does Mashke's theorem hold for semisimple rings?
I saw a proof of Mashke's theorem using the theory of modules. However, it seems like it works in much more generality than it was stated in the text; can anyone confirm if it does, or if there's ...
1
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0
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24
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Number of solutions to simultaneous equations on lie groups
Say I have $n$ variables, special unitary operators from $SU(k)$, and write a set of $m$ equations that they must satisfy. These have the form $UVW...Z=I$, i.e. each one specifies that some product of ...
1
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1
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If E is a splitting field for A then it is clear that there exists a finitely generated subfield E'/F that is also a splitting field. [closed]
I saw in Jacobson's book Finite Dimensional Division Algebras (1996) page 158 the sentence "If E is a splitting field for A then it is clear that there exists a finitely
generated subfield E'/F ...
2
votes
1
answer
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Necessary and sufficient conditions for the function $f(t) := \mbox{trace}(A^{-1}e^{-tB} A e^{-tB})$ to be monotone decreasing.
Let $A$ and $B$ be a positive-definite $n \times n$ matrices. For any $t \ge 0$, define $f(t) := \mbox{trace}(A^{-1}e^{-tB} A e^{-tB}) = \|A^{-1/2}e^{-tB}A^{1/2}\|_F^2$.
Question. What are necessary ...
3
votes
1
answer
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division rings $D,D'$ both of finite dimension over their center $F$ are isomorphic as rings iff isomorphic as $F$-algebras?
Say I have two division rings, $D,D'$, both with center $F$ and both are of finite dimension over $F$, for some field $F$. Now suppose that $D\cong D'$ as rings, does if follow that $D\cong D'$ as $F$ ...
3
votes
1
answer
90
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Nakayama’s Lemma gives $\mathrm{rad}(S) \subset \mathrm{rad}(R)$ for $S \subset R$ under certain finiteness conditions
I'm preparing for the qualify exam and coming up with the following exercise:
Question: Let $S \subset R$ be a subring contained in the center of $R$. Suppose
that $R$ is finitely generated as a left $...
0
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0
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31
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Subalgebras which are semisimple
Let $k$ be a field.
Let $A$ be a $k$-algebra and $B \subset A$ be a subalgebra of $A$ which is a semisimple algebra.
Note that we do not assume that $A$ is commutative.
We also assume that $A$ is a ...
2
votes
0
answers
67
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Jacobson density theorem, and its relations to Artin-Wedderburn, and double centralizer theorems
$\newcommand{\End}{\operatorname{End}}\newcommand{\Hom}{\operatorname{Hom}}$
On pg. 647 of Lang's Algebra, Lang proves the Jacobson density theorem by doing some stuff with $\End_{\End_R(V^n)}(V^n)$ ...
1
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0
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71
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I've proven Krull-Schmidt for arbitrary decomposition into indecomposables. What's wrong in my proof?
My question is probably stupid and I'm likely committing a very trivial mistake. It's well known that the uniqueness of decomposition of modules into indecomposable submodules ${}_A M = \bigoplus_{I \...