For questions about rings which are not necessarily commutative and modules over such rings.

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4
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0answers
24 views

Crossed products and division algebras

I am currently reading some introductory material on Brauer groups ("Noncommutative Algebra", by Farb and Dennis) and the following two questions came to my mind: 1) Are all crossed products ...
0
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2answers
61 views

Example of a ring for which $rs \neq 0$ but $sr = 0$. [duplicate]

I am looking for an example of an associative noncommutative ring $R$ with the following property: for $r,s \in R$, $$ rs \neq 0, \text{ but } sr = 0. $$ Moreover, do rings for which this cannot ...
1
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0answers
8 views

$F\{x,y\}/(xy-yx-x)$ is primitive

Let $F$ be a field of characteristic $0$ and $F\{x,y\}$ denote the free algebra over $F$ generated by $x,y$. Then show that $R=F\{x,y\}/(xy-yx-x)$ is primitive. I tried to use the usual trick that to ...
0
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0answers
39 views

nil Jacobson radicals

Let $f$ be an idempotent element in a ring $S$ with Jacobson radical $J$ so that both $fJf$ and $(1-f)J(1-f)$ are nil. I guess that $J$ is nil too, but I am not sure. I know that the former is ...
4
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1answer
81 views

Ore extensions of noetherian $k$-algebras are again noetherian (proof explanation)

Let $k$ be a field. All occuring $k$-algebras are required to be associative and unital. By noetherian I always mean left noetherian. In a lecture I’m currently taking the notion of an Ore ...
4
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1answer
59 views

If $e$ is idempotent and both $eRe$ and $(1-e)R(1-e)$ are orthogonally finite, is $R$ orthogonally finite?

Let $R$ be a ring with $1$ possessing a non-zero idempotent $e$. It usually happens that if $eRe$ and $(1-e)R(1-e)$ have a property $P$ then so does $R$. A ring is said to be orthogonally finite if ...
1
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2answers
26 views

A question about Semisimple ring and their Jacobson radical

I'm reading "Noncommutative Rings" by Herstein, and I got stuck in theorem 1.2.5/page 16. It says: "if $A$ is an two -sided ideal of a noncommutive ring $R$ (may be not unity) then $J(A)=A \cap ...
1
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1answer
29 views

Uncountably many left ideals?

Let $R$ be a following subring of $M_2(\mathbb{C}):$ \begin{equation*} R = \left\{ \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} ~:~ a\in \mathbb{Q} ...
3
votes
1answer
84 views

Examples of non-principal free ideals

If $R$ is a commutative ring, and $I\subset R$ is a non-zero free ideal, then it is principal generated by a non-zerodivisor. If $R$ is a non-commutative ring having the IBN property and $I\subset R$ ...
10
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1answer
105 views

Example of a ring such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules)

The usual example of unitary ring without the IBN property is the ring of column finite matrices, and in this case we have $R\simeq R^2$ as (left) $R$-modules. (See also here.) In particular, we have ...
3
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1answer
25 views

Maximal right ideal of $\mathbb{H}[x]$

Hi I'm trying to prove the right ideal $(x-i)\mathbb{H}[x]$ of $\mathbb{H}[x]$ is maximal. I've tried defining a surjective function $f:\mathbb{H}[x] \to \mathbb{H}$ by $g(x) \mapsto g(i)$ and using ...
0
votes
1answer
39 views

Idempotent ideals having only idempotents

I search for sufficient conditions for a ring $R$ so that any idempotent ideal constitutes only of idempotent elements of $R$. Of course, in the commutative case, any finitely generated idempotent ...
0
votes
1answer
28 views

Does the tensor product distribute over a direct sum exactly if all the involved modules are bimodules?

Given a non-commutative ring $R$, a right module $M$, and a left module $N$, we can define the tensor product $M \otimes_R N$. I suspect that $$ \left(\bigoplus_{i=1}^n M_i\right) \otimes_R N \simeq ...
0
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0answers
46 views

Idempotent direct summands of rings

I know that if an ideal $I$ is a direct summand of a ring $R$ then it is an idempotent ideal, i.e. $I^2=I$. My question concerns the rings all of whose idempotent ideals are direct summands. ...
0
votes
1answer
34 views

A probable equality between two ideals

Let $I$ and $J$ be ideals of a ring $R$. I want to know whether the ideal $(I+J)^2$ equals $I^2+IJ+JI+J^2$. By taking elements and using the definition of product of two ideals $I$ and $J$ as the set ...
1
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1answer
23 views

Idempotent subideals of $J(R)$

If $R$ is a unital ring, it is well-known that its Jacobson radical $J(R)$ contains no non-zero idempotent element of $R$. My question: Is there a ring $R$ such that $J(R)$ contains a non-zero ...
3
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1answer
51 views

Compute the Jacobson radical of the group ring $\mathbb{F}_2S_3$.

Compute the Jacobson radical and the maximal semisimple quotient of the group ring $\mathbb{F}_2S_3$ of the symmetric group on three letters over the field with two elements, and compute the ...
0
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0answers
15 views

central closure of semiprime rings

I have studied ring of quotients with 'Rings with generalized identities'.$R$:semiprime ring.$Q_{mr}$={$[f;J]| J$ is dense and $f:J\longrightarrow R$ right R-module homomorphism}$H$={$(f;J)| J$ is ...
0
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0answers
16 views

Martindale quotient rings

Let R be prime ring.We know ($Q,+,.$) is ring under following operation: $(U,f)+(V,g)=(U+V,f+g)$$(U,f).(V,g)=(VU,fog)$ Iwant to find additive identity of $(Q,+)$ .So, I am looking any $(V,g)$ ...
7
votes
1answer
41 views

Augmentation ideal and the abelianization of $G$

On a qual problem recently, I came across the following fact: If $G$ is a finite group, and $\mathfrak{a}$ is the augmentation ideal of the integral group ring $\mathbb{Z}G$, then ...
0
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0answers
21 views

Units in Semiperfect Skew Group Rings

Let $k$ be a field and $S$ the ring $k[[x_1,\ldots, x_n]]$. Let $G$ be a finite subgroup of $GL_n(k)$ that does not contain any nontrvial pseudo-reflections and such that $|G|$ is invertible in $k$. ...
6
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1answer
48 views

Ring with Unique Simple Module

Let $A$ be a not necessarily commutative unital ring with a unique simple module (up to isomorphism). Let $\mathfrak m$ be the annihilator of this simple module, which is a two-sided ideal. We claim ...
1
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1answer
25 views

Let $R = M_n(k)$, where $k$ is a field. Then any R-module that is finite dimensional over K is a direct sum of

Let $R = M_n(k)$, where $k$ is a field. Then any $R$-module that is finite dimensional over $K$ is a direct sum of isomorphic copies of $V$, where $V = k^n$. I was able to show that $R$ has a ...
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0answers
32 views

$M/IM$ finitely generated $\Rightarrow$ $M$ finitely generated

Let $R$ be a ring, $M$ a left $R$-module and $I$ a two-sided ideal in $R$. $M/IM$ is a left $R$-module via $$ \begin{align} r\cdot (m+IM)=rm+IM \quad \text{ for } r\in R, m\in M. \end{align} ...
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2answers
41 views

How to show that two left ideals over $R$ are not isomorphic $R$-modules?

Let $R=\begin{bmatrix}F&0&0\\ F&F&0\\F&0&F\end{bmatrix}$ with a field $F$. Show that $$\begin{align} ...
1
vote
1answer
37 views

matrix ring Noetherian, Artinian, semisimple?

Let $k$ be a field and $\Lambda=\begin{bmatrix} k & 0 \\ k^2 & k[x]/(x^2) \end{bmatrix}$. This ring is an algebra over $k$. (a) What is $\dim_k \Lambda$? (b) Is $\Lambda$ a left ...
0
votes
0answers
30 views

derivation of non-zero ideal of prime rings

$d_{1},d_{2}$ are derivations of a prime ring $R$ and $I$ is a non-zero ideal of $R$. How can we show that if $d_{1},d_{2}$ are non-zero, then $d_{i}(I^2)\neq 0$ for $i=1,2$? $I^2=\{\sum ...
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0answers
43 views

free associative algebras as tensor product

Let $A,B$ be $k$-algebras with $k$ a commutative ring. Suppose that $A\otimes_k B$ is a free associative $k$-algebra of rank $n$. Does it imply that $A,B$ are both free associative $k$-algebras? ...
0
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0answers
25 views

Derivation of central closure of prime rings

enter link description here My question is about lemma 4. How can we show $d (U^{2}) \subset U$ what is the meaning of $d(f)$ how is it going from $U^{2}$ to $R$.$d(f)$ is not acomposition.Because ...
1
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2answers
38 views

Does a simple $k$-algebra has finite dimension?

Let $k$ be a field, $A$ a non-commutative ring over $k$. Suppose $A$ has no two sided ideals. Is there an example for such an $A$ with infinite dimension over $k$?
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0answers
14 views

derivation on prime ring

Suppose that R is a prime ring and charR $\neq2$.$d_{1},d_{2}$ are derivations of R such that for some non-zero ideal of I of R and some $c\in C$ (C is an extenden centroid of R) $d_{1}d_{2}(x)=cx$ ...
0
votes
0answers
40 views

Center of a ring

Is there a prime ring $R$ such that $Z(R)=0$ ? There, $Z$ is a center. It is ohvious that If R is a commutative then there isn't prime ring R such that $Z(R)=0$,because $Z(R)=R$.How should we think ...
2
votes
1answer
36 views

Inner derivations determined by idempotent elements and nilpotent elements

If R is a prime ring, $e\in R$ is a non-central idempotent and $d$ is an inner derivation determined by $e$, then $0\neq d=d^{3}=d^{5}...$ If $b\in R$ is a non-central nilpotent element of a degree ...
0
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1answer
40 views

Simple modules of an algebra

How can we find the simple modules of this algebra $$ \begin{pmatrix} k & 0 &0 \\ k & k & 0 \\ k&0&k \end{pmatrix} $$ And why this algebra is not semisimple(i,e it is ...
0
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1answer
36 views

Aside from Matrix Multiplication, when else is multiplication not commutative?

Nearly all of my experience with math is in the "applied math" realm, so I haven't had any formal study of rings, or other fundamental algebraic concepts that help to prove all the relevant applied ...
2
votes
1answer
25 views

Understanding dual (noncommutative) modules / vector spaces functorially

For vector spaces over a fixed field $\Bbbk$, there's a natural transformation from the identity functor to the double dual functor. I think here's a way to construct it. Start from the identity arrow ...
1
vote
1answer
76 views

Show $R$ is right-Artinian but not left-Artinian

Let $R$ be the subring of $M_2(\mathbb{R})$ defined by $$R=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}\mid a\in\mathbb{Q},b,d\in\mathbb{R}\right\}$$ Show that $R$ is right-Artinian but not ...
0
votes
1answer
54 views

Proof for a ring being right Artinian but not left Artinian

I am currently looking at the following example (and other similar examples) and I can follow the proof that it is a right Artinian ring and I also follow the example given as to why it is not a left ...
1
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1answer
45 views

Version of Wedderburn's theorem on central simple algebras

Suppose that $A$ be a central simple algebra over a field $k$. Then by Wedderburn's theorem $A\cong M_n(D)$ for some division $k$-algbera $D$. But to define the 'Brauer equivalence' I need that $D$ is ...
2
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1answer
61 views

Non-abelian groups of order less than or equal to 150

Please is there anywhere one could see a classification of nonabelian groups of orders less than or equal to 150?
0
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1answer
28 views

Showing a $k$-algebra isomorphism

Suppose $A$ is a finite dimensional $k$-algebra. I want to show that $A\otimes_{k}M_n(k)\cong M_n(A)$ for any positive integr $n$. Can you show how to proceed?
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0answers
15 views

Solving non-commutative “quadratic” equation with inhomogenously typed coefficients

Is there a general method to solve for $z\in\mathbb{R}^d$ in the non-commutative $z^\intercal\alpha z + \beta z + \gamma = 0$ where $\alpha\in\mathcal{M}_d(\mathbb{R})$ (real $d\times d$-matrix), ...
-1
votes
1answer
30 views

Tensor Product of irreducible modules

Let $A$ be a $\mathbb C$ algebra. Let $S$ be an irreducible $A$ module? Then what $ S \otimes_A Hom_A(S,S)$? Is it equal to S? I know that $S \otimes_{\mathbb C} Hom_A(S,S)$ is isomorphic to $S$ as a ...
0
votes
0answers
25 views

Noncommutative free resolution

This is probably something very simple but I got stucked with it. Consider the polynomial ring $k[x_1,\dots,x_n]$. How can one construct a free resolution of it in the category of $k$-algebras(not ...
0
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0answers
21 views

Tensor product of non-commutative algebras

Define the affine Hecke algebra, $H_n=H_n(q)$, as the algebra with generators $T_1, \dots, T_{n-1}, X_1^{\pm 1}, \dots, X_n^{\pm 1}$ that satisfy $$ (T_i + 1)(T_i − q) = 0$$ $$T_i T_j = T_j T_i \text{ ...
1
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1answer
34 views

Indecomposable representations of an algebra

Let $\rho$ be an indecomposable representation of Algebra A on a finite dimensional vector space V. Let $a \in Z(A)$, how do I show $\rho(a)$ has exactly one eigenvalue. $Z(A)$ represents the centre ...
6
votes
2answers
86 views

Is it true that $(R\times S)[G]\cong R[G]\times S[G]$?

I know for two groups $G, H$ (not necessarily finite) we have $R[G\times H]\cong (R[G])[H]$, but I was wondering if we had a similar statement for rings $R,\,S$. In other words, if $R,\,S$ are two ...
1
vote
0answers
17 views

Determinants of mixed matrices involving complex and non-commuting Grassmann variables.

I was working on a problem involving a mixed matrix of Grassmann variables and complex numbers. Essentially I found two expressions for a so called "superdeterminant" of the below matrix $G$. At the ...
1
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1answer
37 views

An algebra with no two-sided ideal has only one irreducible representation [closed]

Suppose an algebra has no sided ideal. Then why does it have only one irreducible representation upto isomorphism?
1
vote
2answers
67 views

The Jacobson radical of an Ideal

I have edited my question according to your suggestions and comments. Suppose now that we have a ring $R$ and a subring $S$ of $R,$ so we can define the Jacobson radicals, for example as the ideal ...