All Questions
195 questions
3
votes
2
answers
831
views
Divisible torsion $\mathbb{Z}$-modules
I am trying to prove that for any divisible torsion $\mathbb{Z}$-module $V$,
this map
$$f:\mathbb{Q}/\mathbb{Z}\otimes_E\text{Hom}(\mathbb{Q}/\mathbb{Z},V)\longrightarrow V\mbox{ defined by }
f((q+\...
- 33
12
votes
7
answers
1k
views
Properties of rings that have an elegant description in terms of the associated category of modules
Suppose $A$ is a ring. Then $A$ happens to be a division ring iff every left $A$-module is free. (See here for proofs). I think this is very beautiful; what other properties of rings have an elegant ...
5
votes
1
answer
480
views
Re-interpreting vector spaces in a choice-less model of ZF as modules over a regular ring in ZFC
I am searching a module $M$ over a (von Neumann) regular ring $A$ ($\forall a\in A$, $\exists x\in A$: $axa=a$) with two properties:
(1) every finitely generated submodule of $M$ is projective (...
- 51
5
votes
0
answers
195
views
Sum of projective submodules of a projective over a semihereditary ring
Sorry in advance if this is too silly. Let $R$ be a right semihereditary ring and $P$ a projective right $R$-module. It is well-known that finitely generated (thus projective) submodules of $P$ form a ...
- 409
0
votes
2
answers
345
views
Dual of a module
Let $M$ be a $ \mathbb{Z}_{p}[[T]] $-module and $X=Hom(M,\mathbb{Q}_{p}/\mathbb{Z}_{p})$ be the dual of $M$. Let $X[p^n]$ denotes the $p^n$-torsion points of $X$. Is $X/X[p^n]$ the dual of $M[p^n]$ $?$...
- 303
18
votes
2
answers
1k
views
Does base extension reflect the property of being isomorphic?
Let L/K be a (separable?) field extension, let A be a finite dimensional algebra over K, and let M and N be two A-modules. Let $A' = L \otimes_K A$ be the algebra given by extension of scalars, and ...
- 28.1k
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 ?
- 13
7
votes
1
answer
246
views
Rings in which every module has an injective image
Consider the class of rings $R$ with identity such that any left $R$-module has a non-zero injective homomorphic image. Any such ring is clearly a left V-ring. Is it true that any such ring must be ...
- 157
3
votes
3
answers
1k
views
Is it true that simple projective modules are injective?
It is known that simple modules are either projective or singular. Is it true that simple projective modules over (commutative) rings are injective ?
user39125
3
votes
0
answers
260
views
simple tensor product of modules over algebras
Let $M$, $N$ be simple modules over associative algebras $A$ and $B$ (over $\mathbb{C}$), respectively. When is $M\otimes N$ simple as a $A\otimes B$-module?
It is right if $A$ or $B$ has a ...
- 31
22
votes
4
answers
2k
views
Freeness of a Z[x]-module
Definition: Call a mapping $f: \mathbb{Z} \rightarrow \mathbb{Z}$
a generalized polynomial if for any distinct integers $m$ and $n$
we have $(m - n)|(f(m)-f(n))$.
It is easy to check that polynomial ...
- 19.6k
3
votes
3
answers
483
views
Support of a module over a polynomial algebra
In Atiyah and Bott's paper "The Moment Map and Equivariant Cohomology", they say that for any exact sequence of modules over $\mathbb{C}[u_1,...,u_l]$
$$D \to E \to F,$$
we have that
Supp $E \subset$ ...
- 502
1
vote
0
answers
781
views
How to find the tensor product of modules that we don't know a basis for them?
Hi
I know how to calculate some easy tensor products like $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}\cong_{\mathbb{Z}} \mathbb{Z}/(m,n)\mathbb{Z} $ or $F[X] \otimes_{F} F[Y] \...
- 11
1
vote
0
answers
263
views
Average weighted value of a linear functional over increasing bounded subsets of Z^n
Say you're working within the finite-dimensional free Z-module $\mathbb{Z}^n$, and you want to impose a "norm" on this module. By a "norm" I mean a function $\|·\|: \mathbb{Z}^n \to \mathbb{R}$ which ...
- 4,897
7
votes
1
answer
262
views
Description of modules over self-injective algebras of finite representation type
Is there any description of indecomposable modules and irreducible morphisms over self-injective algebras of finite representation type? I am interested mainly in such a description for nonstandard ...
- 71
14
votes
0
answers
1k
views
Kaplansky's theorem and Axiom of choice
Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow:
Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...
- 1,788
5
votes
2
answers
2k
views
Tensor product of simple modules
Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$. So my questions are:
Can ...
- 4,000
5
votes
1
answer
446
views
Projective dimension of simple module
Let $R$ be a (not necessarily commutative) ring and $M$ a simple right $R$-module. Then $\mathfrak{m}=Ann(M)$ is a maximal ideal of $R$. It is seems known that
$$
pdim_{R}(M)=pdim_{R_{\mathfrak{m}}}(...
- 63
0
votes
1
answer
333
views
Free Module with a Projective Sub- Module, and Tensor Products
Let us consider a unital algebra $A$, with a subalgebra $B \subseteq A$, along with an $A$-$A$-bimodule $M$ which is free as a right module, and a subspace $N$ (with respect to the action of the field ...
1
vote
1
answer
434
views
A Version of Nullstellensatz for Rings of Dİfferential Operators
Here is one of the classical versions of the nullstellensatz: Let $K$ be a field and let $\mathfrak{m}$ be a maximal ideal of the polynomial ring $K[T_1,\ldots,T_n]$. Then $K[T_1,\ldots,T_n]/\mathfrak{...
- 355
3
votes
1
answer
252
views
Bimodule version of IBN
Hello all,
Does anyone have an example in mind of a ring $R$ for which $R^n\cong R^m$ as $R,R$ bimodules for some positive integers $n\neq m$?
I would be a little surprised if someone showed no such ...
- 1,507
1
vote
0
answers
427
views
Ring such that any submodule of an injective module is flat?
Does anyone know examples of rings $R$ with the property that any submodule of an injective (right) $R$-module is flat? If I'm not missing something, this class of rings includes the (Von Neumann) ...
- 21
9
votes
3
answers
2k
views
Module category equivalent to graded module category?
Main Question
Let $R$ be a graded ring, graded by the nonnegative integers. Denote by $\mathrm{gr}R-\mathrm{Mod}$ the category of $\mathbb{Z}$-graded left $R$-modules with morphisms that preserve ...
- 8,559
3
votes
3
answers
3k
views
Generalization of eigenvalues/vectors to modules?
What is the generalization of eigenvalues/vectors to modules?
To be specific, given a "vector" v in a module over some ring, and a linear "operator" O from the module to itself (please feel free to ...
3
votes
2
answers
927
views
Modules of finite support
I'm reading Dwyer and Fried's paper "Homology of free abelian covers, I". In it, they make the following claim, which I'm having trouble verifying.
Let $F$ be a field and $A = F[x_1^{\pm 1},\ldots,...
- 33
4
votes
1
answer
433
views
SBN and IBN rings
Hello, I can not figure out why a ring that is not IBN (invariant basis number) must be SBN (single basis number). More precisely: Let $R$ be a ring (with unit, generally non-commutative) such that ...
16
votes
1
answer
1k
views
Are there non-reflexive modules isomorphic to their bi-dual?
Let $M$ be an $R$-module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism.
I'd like to know if there exists a module isomorphic to its bi-dual but not ...
- 279
1
vote
1
answer
570
views
Rank of a module
I have seen the definition of a module,not neccessary free, the alternatin sum of free modules in a free resolution of that module. it's clear that when the module is free our definition Coincide the ...
- 11
6
votes
2
answers
1k
views
Lemma on infinitely generated projective modules
Is it true that every finitely generated submodule of a non-finitely generated projective over a (not necessarily commutative!) ring is contained in a proper summand?
N.B.: I asked this already on ...
- 47.3k
9
votes
2
answers
1k
views
Modules over Laurent series rings
Let $k[x]$ be the ring of polynomials over a field k in one variable x. A $k[x]$-module is a k-vector space together with a linear endomorphism (the action of x).
The field $k(x)$ of rational ...
- 66.7k
7
votes
1
answer
2k
views
structure theorem for modules
Can structure theorem for modules be extended to modules over UFDS , to modules over Neotherian rings ? if yes then can one get the statement and reference?
Since operations on matrices with ...
- 201
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 ...
11
votes
1
answer
4k
views
Are there any finitely generated artinian modules that are not Noetherian?
It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example ...
- 2,027
32
votes
7
answers
4k
views
"Sums-compact" objects = f.g. objects in categories of modules?
Hello,
Let us call an object of an additive category sumpact (contraction of "sums" and "compact") if taking $Hom$ from it (considered as functor from the category to $Ab$) commutes with coproducts. ...
- 5,562
1
vote
1
answer
600
views
Unimodular column property
Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property.
I would like to know if there is a ring $R$ that doesn't ...
- 113
8
votes
1
answer
1k
views
Direct sum of injective modules over non-Noetherian rings
By the Bass-Papp theorem, if every direct sum of injective $R$-modules is injective then $R$ is Noetherian. I would like to know if there exists an injective module over $R$ non-Noetherian, that ...
- 113
2
votes
2
answers
389
views
Related to fractional ideals
$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define
$$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$
Then it is easy to see that
$$M\subset A\Longleftrightarrow A\subset (A:_{...
- 2,857
3
votes
2
answers
755
views
What is the characteristic of the module over Jacobson semisimple ring？
We know a ring R is semisimple ring iff every module over R is semisimple，a ring R is von-Neumann regular ring iff every module over R is flat，What about the Jacobson semisimple ring？
- 391
3
votes
3
answers
1k
views
Structure theorem for finitely generated Z[G] modules
For a finite abelian group $G$ is there an analogue of structure theorem for finitely generated modules like for P.I.D. rings but with $Z[G]$ group ring over integers instead ?
- 303
2
votes
0
answers
237
views
Classification of finitely generated bimodules over "skew PID"s?
Let R be a noncommutative ring with a 1 and no zero divisors, such that all (two-sided) ideals of R are principally generated. Is there a classification theorem for finitely generated bimodules over ...
user5810
5
votes
1
answer
410
views
Behavior of the projective dimension of modules in a continuous chain of extensions
Let $R$ be an arbitrary ring. Let $D$ be the class of $R$-modules of projective dimension less than or equal to a natural number $n$. If $L$ is the direct union of a continuous chain of submodules ${...
- 111
3
votes
2
answers
2k
views
Extension problem
As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what ...
- 2,857
28
votes
4
answers
5k
views
When are modules and representations not the same thing?
I've been trying for a while to get a real concrete handle on the relationship between representations and modules. To frame the question, I'll put here the standard situation I have in mind:
A ring $...
- 1,872
1
vote
2
answers
652
views
Understanding the modules of semiprimitive rings
As far as I understand, a semiprimitive ring can be fully 'explored' by its simple modules, in the sense that a semiprimitive ring is the subdirect product of its simple modules (for brevity, I'll use ...
- 2,513
4
votes
0
answers
325
views
Localization of power series and module structure
Let $R=\mathbb{Q}[X,Y]$ be the polynomial ring of two commuting variable.
Let $S$ be the multiplicative subset of $R$ generated by homogeneous linear polynomials.
Let also $\widehat{R}$ be the ring of ...
- 41