All Questions
Tagged with modules ra.rings-and-algebras
195 questions
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 ...
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 ...
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 ...
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.
...
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 ...
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 ...
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 ...
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) = \...
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 ...
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 = \...
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 ...
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
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 ...
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 ...
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 ...
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 ...
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 ...
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(...
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 $S$ ...
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.
...
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
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 ...
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 $(R,S)$-...
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 ...
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
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[\...
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 ...
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?
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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.
...