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2 votes
0 answers
126 views

Isoartinian and isosimple modules

I'm reading this article by A. Facchini and Z. Nazemian, in wich they discuss modules with chain conditions up to isomorphism. A couple of the main concepts are the following: Definition We say that ...
0 votes
1 answer
46 views

Equivalent condition for distributive property of $\mathcal{S}(M)$ [closed]

$\textbf{Definitions:}$ Let $M$ be an $A$-module. The collection $\mathcal{S}(M)$ of all submodules of $M$ is partially ordered by the inclusion relation. The collection $\mathcal{S}(M)$ is said to ...
5 votes
2 answers
706 views

Zero tensor product over a complex algebra?

Let $A$ be an algebra over $\mathbb{C}$. Let $M$ be a left $A$-module, let $N$ be a right $A$-module and consider the tensor product $N \otimes_A M$, which is a complex vector space. Q1: Can this ...
4 votes
1 answer
616 views

When does the forgetful functor from modules to vector spaces have a right adjoint?

Given any algebra $R,$ when does the forgetful functor $R\text{-}Mod \rightarrow Vec$ have a right adjoint? Does this imply any finiteness conditions on R? Is there a book/paper discussing this? I'...
2 votes
0 answers
91 views

What are all pairs $(R,M)$ of a ring $R$ and a two-sided $R$-module $M$ such that all endomorphisms of $M$ are scalar multiples of $\text{id}_M$?

I was playing with some endomorphism rings and got curious whether there is a classification of all two-sided (not necessarily unitary on any side) modules $M$ over a (not necessarily unital) ring $R$ ...
2 votes
1 answer
310 views

Flatness of submodules of free modules

Are submodules of free $\mathbb{Z}[G]$-modules flat? if not what conditions on $G$ makes it true? $G$ is an infinite group. If $\mathbb{Z}[G]$ is a Prüfer domain then this is true. Can a group ring $...
4 votes
1 answer
158 views

Dimension of a module over a left-Ore domain

If $R$ is a domain, and $M$ a (left) $R$-module, what are the different notions of dimension of $M$ and their respective assets, what do they measure? I found out that if $\dim_RM$ is the cardinal of ...
2 votes
1 answer
143 views

A weak Schur's lemma for non-semisimple finite dimensional algebras

Let $B \subseteq C$ be an inclusion of finite dimensional (associative) algebras over a field $k$. Assume that $C$ is a free $B$-module. Let $\bigoplus_i U_i$ be a decomposition of $B$ into ...
5 votes
0 answers
187 views

Any f.p. faithful simple module over a primitive group ring?

Recall that a ring $R$ is primitive if it has a faithful simple left module. Let $G$ be a countable discrete group and $R=\mathbb{k}G$, where $\mathbb{k}$ denotes some field or $\mathbb{Z}$. There ...
13 votes
1 answer
595 views

Inverse of the Structure Theorem for Finitely Generated Modules over PID

We know that for a PID $R$, any finitely generated module is of the form $\frac{R}{(a_1)} \oplus \dots \oplus \frac{R}{(a_s)} $. I was wondering if the converse of this statement is true, that is, is ...
1 vote
0 answers
151 views

Lifting idempotents and projective coverings --- reference request

Let $R$ be a (non-commutative, unital) ring with Jacobson radical $J$, write $\overline{R}=R/J$ and denote the quotient map $R\to \overline{R}$ by $a\mapsto \overline{a}$. It is easy to see that if ...
7 votes
1 answer
242 views

Injective indecomposable modules over Laurent polynomial rings

What does the injective envelope of $\mathbb C[x,x^{-1}]/(p(x,x^{-1}))$ as a $\mathbb C[x,x^{-1}]$-module look like where $p(x,x^{-1})$ is an irreducible element? I’m sure this is well known, but ...
7 votes
1 answer
262 views

Description of modules over self-injective algebras of finite representation type

Is there any description of indecomposable modules and irreducible morphisms over self-injective algebras of finite representation type? I am interested mainly in such a description for nonstandard ...
6 votes
1 answer
780 views

Equivalence of idempotents and projective modules over nonunital rings

For a nonunital ring $R$ (or "rng") one has to be a little bit careful when considering the category of (left or right) $R$-module and, furthermore, the standard equivalent definitions of projective ...
5 votes
1 answer
192 views

Given a representation-infinite algebra, when is every AR component infinite?

Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. The Auslander-Reiten quiver $\Gamma_A$ of $A$ is a means of presenting the category of finitely generated right $A$-...
3 votes
1 answer
111 views

Extensions of modules of type $FP_n$

Let $R$ be a ring (not necessarily commutative, but with a unit). Recall that an $R$-module $M$ is of type $FP_n$ if $M$ has a partial projective resolution of length $n$ whose terms are all finitely ...
9 votes
1 answer
473 views

Why do we want $p$-permutation modules in splendid equivalences?

First Rickard (in Splendid Equivalences: Derived Categories and Permutation Modules ) and then Rouquier (Block theory via stable and Rickard equivalences, Appendix A.1) define splendid equivalences ...
8 votes
1 answer
2k views

Adjoints of scalar extension and scalar coextension

Let $h\colon R\rightarrow S$ be a morphism of commutative rings. We consider the following functors (I am aware that the notations may be different in other contexts): $h^*$: Scalar extension by ...
14 votes
0 answers
1k views

Kaplansky's theorem and Axiom of choice

Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow: Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...
6 votes
3 answers
446 views

Is the category of symmetric bimodules over a commutative ring closed under extensions?

Let $A$ be a commutative ring, and consider the category of bimodules over $A$. An $A$-bimodule $M$ is called symmetric if $a\cdot m = m \cdot a$ for all $a \in A$, $m \in M$. Is the category of ...
4 votes
2 answers
664 views

Modules over infinite rings which can not be a finite union of their proper submodules

It is well known that a vector space over an infinite field cannot be a finite union of its proper subspaces. Does this fact have an immediate and obvious generalization to modules over infinite ...
14 votes
2 answers
618 views

"Inner product" between prime factorizations

Let $x \in \mathbb{Q}^+$. Then $x$ can be expressed uniquely as a product over primes: $$x = \prod_{p \text{ prime}} p^{\nu_p(x)}$$ where $\nu_p(x)$ is the p-adic valuation of $x$. Let $\nu(x) = \...
1 vote
1 answer
124 views

A nonreduced quotient ring

I am searching for a commutative ring $R$ and a semisimple $R$-module $M$ such that the quotient ring $\frac {R}{soc(R)\cap ann_R(M)}$ has a nonzero nilpotent element. Here, $soc(R)$ means the socle ...
0 votes
0 answers
110 views

Hochster-Roberts theorem

I am trying to understand this theorem, and as an example I try this case: Let $R=\mathbb{C}[z_1^{\pm},\ldots,z_n^{\pm}]^{S_n}$ be the algebra of Laurent polynomials. Now $R$ should be somehow ...
2 votes
1 answer
243 views

Isomorphism of irreducible R-modules

Let $R$ be a $k-$algebra and $M,N$ two irreducible $R-$modules, isomorphic as vector spaces. If we know that for every $r\in R$ we have the same eigenvalues on $M$ and $N$ (with multiplicities) is it ...
7 votes
1 answer
1k views

indecomposable module over a local ring

I ask this in mathematics for some days.it doesn't have an answer up to now. https://math.stackexchange.com/questions/2565828/indecomposable-module-over-a-local-ring As we all know, for an arbitrary ...
13 votes
1 answer
670 views

When is $A\otimes R$ a free $R$-module?

Let $R$ be a commutative ring. If I am not mistaken, there is the following fact: For a finitely generated abelian group $A$, the $R$-module $A\otimes R$ is free if and only if we can write the ...
1 vote
1 answer
149 views

Finding modules to check for finite global dimension

Given a finite dimensional algebra $A$ and a generator-cogenerator $M$ and let $B:=End_A(M)$. $B$ has finite global dimension iff every $A$-module has finite $add(M)$-resolution dimension, which is ...
1 vote
0 answers
294 views

Is it true that the functor of completion of a module over a local ring is injective on isomorphism classes?

Let $A$ be a commutative Noetherian local ring and $\hat A$ be its completion. Then we have the functor of completion from the category of finitely generated $A$-modules to the category of finitely ...
6 votes
1 answer
302 views

Endomorphism ring of bimodules

Given an algebra $A$ with a right $A$-module $M$ with $End_A(M) \cong A$. Then we can view $M$ as a natural $A$-bimodule. When is $M$ as a bimodule indecomposable and what is its endomorphism ring as ...
8 votes
1 answer
335 views

Graded and projective (but not bounded below) module that is not graded-projective?

Let $A = \bigoplus_{n = 0}^\infty A_n$ be a graded algebra over a field $k$ that is locally finite: each $A_n$ is a finite-dimensional $k$-vector space. We say that a graded left $A$-module $P = \...
8 votes
1 answer
556 views

In modules over finite rings $Rv = Rw \iff R^\times v = R^\times w$

Let $R$ be a finite ring (with unit, possibly non-commutative), and $M$ a left module over $R$. Let $v,w\in M$. Then $$Rv = Rw \iff R^\times v = R^\times w.$$ This follows from Lemma 6.4 in Hyman ...
2 votes
0 answers
214 views

"Strongly finite rings" or similar term for a condition implying IBN

While (once again) perusing T.Y. Lam's excellent GTM 189 "Lectures on Modules and Rings" I compared the various conditions given that typically imply IBN, e.g. the (left) strong rank condition, stably ...
5 votes
2 answers
512 views

A question with simple and indecomposable modules

Assume $M$ is both noetherian and artinian and fix $S_0\subseteq M$ a simple submodule. How to prove that $S_0$ is contained in some indecomposable direct summand of $M$?
4 votes
0 answers
74 views

self-cogenerator rings

Let $\mathbb{U}$ be a non-empty set (class) of objects of a category $C$. An object $B$ in $C$ is said to be cogenerated by $\mathbb{U}$ or $\mathbb{U}$-cogenerated if, for every pair of distinct ...
2 votes
1 answer
172 views

Morphisms in K-theory: comparison of two pictures

Algebraic $K_0$ group for an algebra $A$ may be defined in terms of stable isomorphism classes of idempotents in $M_n(A)$ or equivalently in terms of isomorphism clasess of finitely generated ...
2 votes
1 answer
263 views

Nonnegatively graded algebra $A$ finitely generated as $k$-algebra iff $A_0$ finitely generated, $A_{>0}$ finitely generated as $A$-module?

This is related to my question here. My question is as follows. How do I see that a nonnegatively graded algebra $A$ is finitely generated as a $k$-algebra if and only if $A_0$ is finitely generated ...
1 vote
1 answer
68 views

On finite Uniform (Goldie) dimensions

1) Is there any characterization of $\Bbb{Z}$-modules with finite uniform dimensions? 2) Find two $\Bbb{Z}$-modules $N, M$ such that $N\leq M$ and $M\hookrightarrow N$ but $N$ is not isomorphic to $M$...
3 votes
0 answers
378 views

$I=mI$, when I is not finitely generated

Let $(R,m)$ be a commutative local ring with unit. Suppose $I$ is an ideal (not finitely generated). If $I=mI$, what can we say about $I$? Can we conclude something interesting? If $I$ were finitely ...
5 votes
1 answer
394 views

Classification of indecomposable modules in tame hereditary algebras

An algebra $A$ is said to be tame if the isomorphism classes of indecomposable $A$-modules in each dimension occur in a finite number of 1-parameter families. $A$ is said to be of finite ...
1 vote
1 answer
357 views

the relation between projective and quasi-projective modules

An $R$-module $M$ is called quasi-projective if $\text{Hom}_R(M,M)\to\text{Hom}_R(M,N)$ is surjective for every surjective homomorphism $M\twoheadrightarrow N$. What are the rings $R$ for which every ...
6 votes
1 answer
302 views

Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?

Let $A$ be a finite dimensional semisimple $F$-algebra and $K_1(A)$ the Whitehead group of $A$. By splitting $A$ into its Wedderburn components, the reduced norm map $\operatorname{nr}_A:K_1(A)\to Z(...
1 vote
1 answer
271 views

Rank of a locally free $\mathbb Z[G]$-module

This is a basic question. Let $G$ be a finite group, $M$ a finitely generated $\mathbb Z[G]$-module so that the $\mathbb Z_p[G]$-module $M_p$ is free for all prime numbers $p$, i.e. is locally free. ...
6 votes
1 answer
2k views

A Hom-Tensor identity - $\text{Hom}_{R}(P,B)\otimes _SC \cong \text{Hom}_{R}(P,B \otimes_S C) $

let $R,S$ be associative algebras over $\mathbb{C}$. Let $\mathcal{C} \subseteq$ $R$-Mod be a full abelain subcategory of $R$-Mod which is the category of $R$-modules. Let $B$ and $C$ be, a $(R,S)$-...
4 votes
0 answers
85 views

An order in $\mathbb Q[G]$ which is a maximal $\mathbb Z_p$-order in $\mathbb Q_p[G]$ for finitely many primes $p$

Let $G$ be a finite group and $S$ a finite set of prime numbers. I know that every separable $\mathbb Q$-algebra $A$ contains a maximal $\mathbb Z$-order but I wonder if the following is true. Is ...
3 votes
1 answer
303 views

An invariant submodule of a projective module

This is a basic question (not research level) which has already been asked on SE by someone else but doesn't yet have an answer so I'd like to repost it on MO. Let $R$ be a commutative ring with ...
7 votes
1 answer
456 views

Hopfian modules

My question is slightly motivated by basic results in linear algebra, for example, that if $F$ is a field then a surjective linear map $F^n \rightarrow F^n$ is injective. More generally, any ...
1 vote
3 answers
365 views

On modules with finite uniform dimension

Is it true that if a module $M$ has finite uniform dimension then the same is true for its homomorphic images ?
3 votes
2 answers
475 views

Rings all of whose torsion modules are cyclic

Let us call a (possibly non-commutative) ring $R$ "very good" if every finitely generated torsion left $R$-module is cyclic. Here is an example of such a ring: Let $k=\mathbb{C}((t))$ and let $R=k[\...
2 votes
1 answer
241 views

Cyclic faithfully flat modules

I am looking for an example of a cyclic faithfully flat $R$-module that is not projective. Could someone help me?