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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 ...
Adi Ostrov's user avatar
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 ...
Uriya First's user avatar
  • 2,928
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 ...
Benjamin Steinberg's user avatar
6 votes
1 answer
779 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 ...
Joakim Arnlind's user avatar
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 ...
Drike's user avatar
  • 1,555
14 votes
1 answer
581 views

Group rings such that every (countably generated) module has a maximal submodule

Every (non-zero) finitely generated module over a ring has a maximal proper submodule by a simple application of Zorn's lemma. I am interested in the following question, with two variants. ...
Benjamin Steinberg's user avatar
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 $A$-...
Iteraf's user avatar
  • 482
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 ...
Sarah's user avatar
  • 39
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 ...
Fred Rohrer's user avatar
  • 6,700
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 ...
Symmetric's user avatar
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) = \...
user76284's user avatar
  • 2,203
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 ...
Ali Taghavi's user avatar
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 ...
karparvar's user avatar
  • 355
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 ...
Ioannis Zolas's user avatar
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 ...
Ioannis Zolas's user avatar
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 ...
Jian's user avatar
  • 496
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 ...
Lennart Meier's user avatar
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 ...
Mare's user avatar
  • 26.5k
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 ...
Sergei Ivanov's user avatar
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 ...
Mare's user avatar
  • 26.5k
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 = \...
Manny Reyes's user avatar
  • 5,407
6 votes
2 answers
2k views

Direct sum of injective modules is injective

By the Bass-Papp Theorem, for a unital ring $R$, any direct sum of injective left $R$-modules is injective if and only if $R$ is left Noetherian. I would like to restrict my consideration to an ...
Batominovski's user avatar
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 $M$?
Marco Farinati's user avatar
4 votes
1 answer
382 views

Non-existence of projective covers

I didn't get the argument of Example 7.3.11, page 123, from the representation theory book of Peter Webb, available also online at: http://www-users.math.umn.edu/~webb/RepBook/RepBookLatex.pdf In ...
user103474's user avatar
3 votes
2 answers
227 views

Operations on semi-hereditary rings

I am learning about left (right) semi-hereditary rings, and got stuck with the following questions. Just to recall that being left semi-hereditary means that every finitely generated submodule of a ...
SKuzin's user avatar
  • 31
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 ...
e.r's user avatar
  • 41
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 ...
truebaran's user avatar
  • 9,330
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 ...
Jakob W's user avatar
  • 349
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 ...
Iteraf's user avatar
  • 482
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 ...
user114539's user avatar
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 ...
Najmeh Dehghani's user avatar
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(...
eddie's user avatar
  • 255
8 votes
2 answers
1k views

Polynomial roots in the ring extension

Let $R$ be a ring with identity (not necessarily commutative) and $R[x]$ be a ring of polynomials over $R$. We say that a ring $S$ is an extension of $R$ if there is a subring $\tilde{R}$ in $S$ ...
Mikhail Goltvanitsa's user avatar
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. ...
eddie's user avatar
  • 255
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 ...
eddie's user avatar
  • 255
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 ...
eddie's user avatar
  • 255
7 votes
1 answer
455 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 ...
Mark Wildon's user avatar
  • 11.2k
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 $(R,S)$-...
tzelin1016's user avatar
9 votes
1 answer
472 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 ...
AnalysisStudent0414's user avatar
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 $M$...
Najmeh Dehghani's user avatar
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[\...
Dr. Evil's user avatar
  • 2,751
8 votes
1 answer
387 views

Jordan-Hölder-like statements for modules with $\Delta$-filtrations over a quasihereditary algebra

Definitions Let $A$ be an Artin algebra (for instance, take $A$ to be a finite dimensional algebra over some field) and label the isomorphism classes of simple $A$-modules by the elements of a ...
3 A's's user avatar
  • 425
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?
user78678's user avatar
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 ...
azimut's user avatar
  • 253
6 votes
1 answer
491 views

Homomorphisms from projective modules

Let $B$ be a $A$-algebra which is free of finite rank as $A$-module. Let $X$ be a finitely generated projective left $B$ module. (So $X$ is also a f.g. projective $A$ module.) Are these homomorphism ...
Mostafa - Free Palestine's user avatar
4 votes
1 answer
133 views

Existence of small projective dimensioned modules

Suppose $A$ is a (if necessary unital) associative ring and $I$ is a left ideal in $A$. Let $\operatorname{pd}(M)$ denote the projective dimension of a left $A$-module $M$. Then do either of the ...
Homologizer's user avatar
2 votes
1 answer
404 views

Example of a Frobenius algebra that is not projective over a Frobenius subalgebra

I'd like to know an example of a Frobenius algebra $A$, with a subalgebra $B$ that is itself a Frobenius algebra, such that $A$ is not projective as a left $B$-module. I don't require any ...
Alistair Savage's user avatar
0 votes
1 answer
139 views

Reference request for stably free modules

I am looking for some references on the theory of stably free modules. I will call (F) the following property for a ring $R$: every f.g. stably free module over $R$ is free. 1) Is there a standard ...
Ferra's user avatar
  • 509
2 votes
1 answer
458 views

General criterion to find a Z-basis in a fixed generating subset

Let $V=\mathbf{Z}^N$ be a free $\mathbf{Z}$-module of rank $N$. Let $S\subseteq V$ be a fixed finite subset. Consider the submodule $M:=\langle S\rangle\leq V$ generated by $S$. We know form the ...
Hugo Chapdelaine's user avatar
7 votes
1 answer
749 views

Injective flat module

Let $R$ be a (right noetherian) ring. Is there always a right $R$-module which is both flat and injective? If $R$ is an integral domain, then the answer is indeed yes, as the quotient field is such. ...
Fred.Fred's user avatar
  • 409