1
vote
0answers
32 views

Tensor product of two simple modules

Let $k$ be a field of arbitrary characteristic. Suppose that $A$ and $B$ are finite dimensional $k$-algebras. Let $S$ be a finite dimensional simple $A$-module and let $T$ be a finite dimensional ...
0
votes
1answer
31 views

How to show Not a Free Module

Let $\mathbb K$ be a field, $A= \mathbb K [x,y]$ and $ M = Ax + Ay$. prove that $M$ is NOT a free module!
1
vote
0answers
26 views

Tensoring and retaining projectiveness

Let $A$be a unital associative ring, If $A$ is not a projective $A$-bimodule, however $A\otimes A$ is a projective $A$-bimodule, can we conclude that $A\otimes A \otimes A$ is also projective?
2
votes
1answer
35 views

Example of a faithful irreducible module

Let $R$ be a non trivial simple ring. I am trying to show that there is a faithful irreducible left $R$-module. Is the ring $R$ considered as a left module over itself such a module? I think ...
1
vote
1answer
41 views

Dimension of dual module

Let $R$ be a finite dimensional semisimple $k$ algebra ($R$ is not necessarily commutative) and $M$ be an $R$-bimodule such that $M$ has finite dimension over $k$. Define ...
3
votes
1answer
54 views

“Vector spaces” over a skew-field are free?

Are modules over a skew field free? That is, if $F$ is a skewfield then can any module $M$ be written as $\underset{i \in I}{\bigoplus} F$ for some indexing set $I$?
1
vote
1answer
27 views

Cotensor and counit?

If M is a C-bicomodule, then considering C as a $C$-bicomodule also, is $M \square_C C \cong C$, where $\square_C$ is the cotensor product in $^C\mathscr{M}^C$.
0
votes
0answers
15 views

Commutativity of comodules

If C is a cocommutative R-coalgebra, R is some commutative semi-simple artinian ring and A and B are C-bicomodules, then is $A\otimes_R B \cong B \otimes_R A$ as $R-modules$. However, this also true ...
0
votes
1answer
78 views

Relation between Jacobson radical and composition series

Let $R$ be a not necessarily commutative ring with 1. Suppose $R$, viewed as a right $R$-module, has a finite composition series with non-isomorphic composition factors. Prove that the Jacobson ...
1
vote
0answers
41 views

Enveloping Algebra equal to algebra

Let $R$ be a unital associative ring, $A$ be an associative $R$-algebra of finite dimension, and $A^e$ its enveloping algebra. What are the requirements on $A$, so that $A^e \cong A$ (as ...
2
votes
1answer
21 views

Smooth algebras and quasi-freeness

Let A be a unital associative algebra over a field k. Then A is smooth if and only if X:=Spec(A) is smooth. That is $\Omega_{X|Spec(k)}$ is locally-free. The later module is isomorphic to ...
1
vote
1answer
66 views

for any ring $A$ the matrix ring $M_n(A)$ is simple if and only if $A$ is simple

Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring ...
0
votes
0answers
47 views

Is the universal enveloping algebra of the free Poisson algebra generated by finite set (left) noetherian?

Let $P$ be the free Poisson algebra over $k$ (a field) generated by finite set $x_1,...,x_n$. Let's consider the universal enveloping algebra $P^e$ of the free Poisson algebra $P$. Hence a Poisson ...
6
votes
0answers
137 views

A few questions about a specific ring

My question is kinda long, so please bear with me... And I only need hints to get me started. So, I'm working on the ring $R =\left( \begin{matrix} \mathbb{Z} & \mathbb{Q} \\ 0 & ...
3
votes
1answer
91 views

Problem with Wedderburn Theorem proof.

This is a Wedderburn Theorem proof in Frank W. Anderson, Kent R. Fuller: Rings and Categories of Modules Please explain that: "Therefore $_RR$ has a composition series of length $n$". Exercise ...
4
votes
1answer
56 views

Weak flat condition?

Let $R$ be a unit ring (not necessarily commutative). Then it is clear that for a right $R$-module $M$ we have: $M$ is flat $R$-module $\Rightarrow$ for any left $R$-module $E$ with $E\otimes_{R}M=0$ ...
1
vote
1answer
73 views

Is there a proof of this that does not use idempotents?

I am going to present a statement and a proof. The proof makes use of idempotents which makes it a little cumbersome. Is there a proof that does not use idempotents? (using well-known theorems ...
2
votes
2answers
118 views

Multiplicity of the simple $R$-module $M$ in the semisimple ring $R$

I'm confused about the conclusion of Wedderburn's structure theorem for semisimple rings. Let's consider the special case where $R=M^n$ as modules for some simple module $M$. Wedderburn's theorem says ...
3
votes
1answer
89 views

Question about Noncommutative Rings by I. N. Herstein

Here's an example that I do not quite understand it fully, it's on page 6 of the book. And here's what it says: Let $F$ be a field, and $F_n$ be a ring of all $n\times n$ matrices over $F$. We ...
2
votes
2answers
38 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
68 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
1answer
111 views

An ideal generated by a nontrivial idempotent is not a free module

Let $R = \Bbb QG$ with $G$ isomorphic to a cyclic group of order $2$, and $x$ its generator. I'm trying to show that $$\frac{1}{2}(1+x)R$$ is not a free $R$-module. I found out that ...
2
votes
1answer
72 views

When do finiteness conditions coincide on modules?

Let $R$ be a (possibly noncommutative, but unital) ring. There are several finiteness conditions on a left $R$-module that we can consider. The ones I have come across include finitely generated ...
1
vote
1answer
99 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. $a$ ...
0
votes
1answer
56 views

How to prove finite dimensionality of Hom-spaces between modules of finite length?

Let $R$ be a ring and $k$ be a field such that $k\hookrightarrow R$. Thus, given two $R$-modules $M$ and $N$, we can regard $\operatorname{Hom}(M,N)=\operatorname{Hom}_R(M,N)$ as a vector space over ...
3
votes
1answer
90 views

Natural homomorphism $\mathfrak{a} \prod E_\lambda \to \prod \mathfrak{a} E_\lambda$

Let $A$ be a ring (we don't assume that $A$ is commutative), and $\frak a$ a left ideal of $A$. Let $(E_\lambda)$ be a system of left $A$-modules. There is a natural homomorphism of modules $$ \phi ...
3
votes
0answers
190 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 ...
3
votes
1answer
229 views

Trivial extension of an algebra

Suppose that $A$ is a finite dimensional $k$-algebra. Call $Q=\mathrm{Hom}_k(A,k)$. $Q$ admits an $A$-$A$-bimodule structire in the obvious way. The trivial extension of $A$ is defined as follows: ...
3
votes
1answer
112 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
66 views

On locally bounded algebras

Let $k$ be a field and $R$ an associative $k$-algebra suppose $R^2=R$. We say that $R$ is locally bounded if there exists a complete set of pairwise orthogonal primitive idempotents $\{e_x:x\in I\}$ ...
5
votes
1answer
295 views

Why are these all the indecomposable projective modules?

Let $A$ be a finite dimensional associative algebra with unit over a commutative field $k$. Suppose that $M$ is a finitely generated left module. Denote by $I(M)$ the injective hull of $M$ and by ...
1
vote
2answers
83 views

Minimal Right Ideals in $\Bbb{C}[G]$

Let $G$ be a finite group. Consider the group algebra $\Bbb{C}[G]$ as a right module over itself. By Maschke's Theorem, $\Bbb{C}[G]$ is semisimple. How can we identify all the minimal right ideals of ...
8
votes
1answer
237 views

Tensor product of simple modules

Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. My questions are: How can we describe $M \otimes_R N$ explicitly? Well, I guess that it is a quotient of $R$ by a sum of ...
1
vote
1answer
269 views

Annihilator of a simple module

Let $R$ be a finitely generated commutative ring and $C$ an $R$-algebra ($C$ is not necessarily commutative). Assume that $C$ is a finitely generated $R$-module. If $S$ is a simple $C$-module, then ...
3
votes
0answers
85 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 ...
1
vote
1answer
222 views

Finding all simple $R$ modules of a ring.

I was hoping someone had an idea on how to go about solving the following; Find (up to isomorphism) all simple R-modules where i) $R = \begin{pmatrix} \mathbb{Z}/15 \mathbb{Z} & \mathbb{Z}/15 ...
2
votes
1answer
116 views

On the length of a ring

Suppose that $R$ is a ring, and suppose that $\lambda_R(_RR)<\infty$ and $\lambda_R(R_R)<\infty$ (where $\lambda_R$ is the length of an $R$-module). Is it true that then ...
3
votes
3answers
204 views

Why is this ring semisimple?

Let $R$ be a simple ring (i.e. a ring with no nontrivial two-sided ideals) which contains a left ideal which is simple as a left $R$-module. How can I prove that $R$ is semisimple?
0
votes
1answer
45 views

Why is this module simple?

Define $A:=F\{x,y\}/\langle xy-yx-1\rangle$ where $F\{x,y\}$ denotes the free algebra generated by $x,y$ and $F$ is a field. Let $I$ be a maximal left ideal of $A$, then $E=A/I$ is a simple ...
0
votes
2answers
239 views

A simple module versus a simple ring

In the book I've been reading recently (Algebra by Thomas W. Hungerford) the following definition is given: A left module $A$ over a ring $R$ is simple provided that $RA\neq 0$ and $A$ has no ...